Tutorial: GEPA for AIME (Math)¶

In this tutorial, we optimize GPT-4.1 Mini's Chain of Thought (dspy.ChainOfThought) for solving math problems (AIME) using the dspy.GEPA optimizer!

In [1]:

api_key = input("Enter your OpenAI API key: ")
import dspy
lm = dspy.LM("openai/gpt-4.1-mini", temperature=1, api_key=api_key, max_tokens=32000)
dspy.configure(lm=lm)

api_key = input("Enter your OpenAI API key: ") import dspy lm = dspy.LM("openai/gpt-4.1-mini", temperature=1, api_key=api_key, max_tokens=32000) dspy.configure(lm=lm)

Loading the AIME dataset¶

The AIME exam consists of 2 problem sets of size 15 for each year. For this tutorial, we will use AIME problem sets from previous years (2022-2024) for optimization (amounting to total 3 years x 2 sets x 15 problems = 90 problems, split equally between train and validation sets), and test the performance on AIME 2025 (2 sets x 15 problems = 30 problems). Since AIME 2025 is a small set, we repeat it 5 times for statistical stability in evaluation.

In [2]:

import dspy
from datasets import load_dataset

def init_dataset():
    train_split = load_dataset("AI-MO/aimo-validation-aime")['train']
    train_split = [
        dspy.Example({
            "problem": x['problem'],
            'solution': x['solution'],
            'answer': x['answer'],
        }).with_inputs("problem")
        for x in train_split
    ]
    import random
    random.Random(0).shuffle(train_split)
    tot_num = len(train_split)

    test_split = load_dataset("MathArena/aime_2025")['train']
    test_split = [
        dspy.Example({
            "problem": x['problem'],
            'answer': x['answer'],
        }).with_inputs("problem")
        for x in test_split
    ]

    train_set = train_split[:int(0.5 * tot_num)]
    val_set = train_split[int(0.5 * tot_num):]
    test_set = test_split * 5

    return train_set, val_set, test_set

import dspy from datasets import load_dataset def init_dataset(): train_split = load_dataset("AI-MO/aimo-validation-aime")['train'] train_split = [ dspy.Example({ "problem": x['problem'], 'solution': x['solution'], 'answer': x['answer'], }).with_inputs("problem") for x in train_split ] import random random.Random(0).shuffle(train_split) tot_num = len(train_split) test_split = load_dataset("MathArena/aime_2025")['train'] test_split = [ dspy.Example({ "problem": x['problem'], 'answer': x['answer'], }).with_inputs("problem") for x in test_split ] train_set = train_split[:int(0.5 * tot_num)] val_set = train_split[int(0.5 * tot_num):] test_set = test_split * 5 return train_set, val_set, test_set

In [3]:

train_set, val_set, test_set = init_dataset()

len(train_set), len(val_set), len(test_set)

train_set, val_set, test_set = init_dataset() len(train_set), len(val_set), len(test_set)

Out[3]:

(45, 45, 150)

Let's view an example task input

In [4]:

print("Problem:")
print(train_set[0]['problem'])
print("\n\nSolution:")
print(train_set[0]['solution'])
print("\n\nAnswer:")
print(train_set[0]['answer'])

print("Problem:") print(train_set[0]['problem']) print("\n\nSolution:") print(train_set[0]['solution']) print("\n\nAnswer:") print(train_set[0]['answer'])

Problem:
In isosceles trapezoid $ABCD$, parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$, respectively, and $AD=BC=333$. The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$, and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$. Find $PQ$.


Solution:
We have the following diagram:

Let $X$ and $W$ be the points where $AP$ and $BQ$ extend to meet $CD$, and $YZ$ be the height of $\triangle AZB$. As proven in Solution 2, triangles $APD$ and $DPW$ are congruent right triangles. Therefore, $AD = DW = 333$. We can apply this logic to triangles $BCQ$ and $XCQ$ as well, giving us $BC = CX = 333$. Since $CD = 650$, $XW = DW + CX - CD = 16$.
Additionally, we can see that $\triangle XZW$ is similar to $\triangle PQZ$ and $\triangle AZB$. We know that $\frac{XW}{AB} = \frac{16}{500}$. So, we can say that the height of the triangle $AZB$ is $500u$ while the height of the triangle $XZW$ is $16u$. After that, we can figure out the distance from $Y$ to $PQ: \frac{500+16}{2} = 258u$ and the height of triangle $PZQ: 500-258 = 242u$.
Finally, since the ratio between the height of $PZQ$ to the height of $AZB$ is $242:500$ and $AB$ is $500$, $PQ = \boxed{242}.$
~Cytronical
Extend line $PQ$ to meet $AD$ at $P'$ and $BC$ at $Q'$. The diagram looks like this:
[asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, P1, Q1; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); P1 = intersectionpoint(A--D,P--(-300)*(Q-P)+P); Q1 = intersectionpoint(B--C,Q--300*(Q-P)+Q); draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red); draw(anglemark(A,B,Q,1000),red); draw(anglemark(Q,B,C,1000),red); draw(anglemark(P,D,A,1000),red); draw(anglemark(C,D,P,1000),red); draw(anglemark(Q,C,D,1000),red); draw(anglemark(B,C,Q,1000),red); add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$P$",P,1.5*NE,linewidth(4)); dot("$Q$",Q,1.5*NW,linewidth(4)); dot("$P'$",P1,1.5*W,linewidth(4)); dot("$Q'$",Q1,1.5*E,linewidth(4)); draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); draw(P--P1^^Q--Q1,dashed); [/asy]
Because the trapezoid is isosceles, by symmetry $PQ$ is parallel to $AB$ and $CD$. Therefore, $\angle PAB \cong \angle APP'$ by interior angles and $\angle PAB \cong \angle PAD$ by the problem statement. Thus, $\triangle P'AP$ is isosceles with $P'P = P'A$. By symmetry, $P'DP$ is also isosceles, and thus $P'A = \frac{AD}{2}$. Similarly, the same thing is happening on the right side of the trapezoid, and thus $P'Q'$ is the midline of the trapezoid. Then, $PQ = P'Q' - (P'P + Q'Q)$.
Since $P'P = P'A = \frac{AD}{2}, Q'Q = Q'B = \frac{BC}{2}$ and $AD = BC = 333$, we have $P'P + Q'Q = \frac{333}{2} + \frac{333}{2} = 333$. The length of the midline of a trapezoid is the average of their bases, so $P'Q' = \frac{500+650}{2} = 575$. Finally, $PQ = 575 - 333 = \boxed{242}$.
~KingRavi
We have the following diagram:

Extend lines $AP$ and $BQ$ to meet line $DC$ at points $W$ and $X$, respectively, and extend lines $DP$ and $CQ$ to meet $AB$ at points $Z$ and $Y$, respectively.
Claim: quadrilaterals $AZWD$ and $BYXC$ are rhombuses.
Proof: Since $\angle DAB + \angle ADC = 180^{\circ}$, $\angle ADP + \angle PAD = 90^{\circ}$. Therefore, triangles $APD$, $APZ$, $DPW$ and $PZW$ are all right triangles. By SAA congruence, the first three triangles are congruent; by SAS congruence, $\triangle PZW$ is congruent to the other three. Therefore, $AD = DW = WZ = AZ$, so $AZWD$ is a rhombus. By symmetry, $BYXC$ is also a rhombus.
Extend line $PQ$ to meet $\overline{AD}$ and $\overline{BC}$ at $R$ and $S$, respectively. Because of rhombus properties, $RP = QS = \frac{333}{2}$. Also, by rhombus properties, $R$ and $S$ are the midpoints of segments $AD$ and $BC$, respectively; therefore, by trapezoid properties, $RS = \frac{AB + CD}{2} = 575$. Finally, $PQ = RS - RP - QS = \boxed{242}$.
~ihatemath123
Let $X$ and $Y$ be the feet of the altitudes from $P$ and $Q$, respectively, to $AB$, and let $Z$ and $W$ be the feet of the altitudes from $P$ and $Q$, respectively, to $CD$. Side $AB$ is parallel to side $CD$, so $XYWZ$ is a rectangle with width $PQ$. Furthermore, because $CD - AB = 650-500 = 150$ and trapezoid $ABCD$ is isosceles, $WC - YB = ZD - XA = 75$. 
Also because $ABCD$ is isosceles, $\angle ABC + \angle BCD$ is half the total sum of angles in $ABCD$, or $180^{\circ}$. Since $BQ$ and $CQ$ bisect $\angle ABC$ and $\angle BCD$, respectively, we have $\angle QBC + \angle QCB = 90^{\circ}$, so $\angle BQC = 90^{\circ}$. 
Letting $BQ = 333k$, applying Pythagoras to $\triangle BQC$ yields $QC = 333\sqrt{1-k^2}$. We then proceed using similar triangles: $\angle BYQ = \angle BQC = 90^{\circ}$ and $\angle YBQ = \angle QBC$, so by AA similarity $YB = 333k^2$. Likewise, $\angle CWQ = \angle BQC = 90^{\circ}$ and $\angle WCQ = \angle QCB$, so by AA similarity $WC = 333(1 - k^2)$. Thus $WC + YB = 333$.
Adding our two equations for $WC$ and $YB$ gives $2WC = 75 + 333 = 408$. Therefore, the answer is $PQ = ZW = CD - 2WC = 650 - 408 = \boxed{242}$.
~Orange_Quail_9
This will be my first solution on AoPS. My apologies in advance for any errors. 
Angle bisectors can be thought of as the locus of all points equidistant from the lines whose angle they bisect. It can thus be seen that $P$ is equidistant from  $AB, AD,$ and $CD$ and $Q$ is equidistant from  $AB, BC,$ and $CD.$ If we let the feet of the altitudes from $P$ to $AB, AD,$ and $CD$ be called $E, F,$ and $G$ respectively, we can say that $PE = PF = PG.$ Analogously, we let the feet of the altitudes from $Q$ to $AB, BC,$ and $CD$ be $H, I,$ and $J$ respectively. Thus, $QH = QI = QJ.$ Because $ABCD$ is an isosceles trapezoid, we can say that all of the altitudes are equal to each other. 
By SA as well as SS congruence for right triangles, we find that triangles $AEP, AFP, BHQ,$ and $BIQ$ are congruent. Similarly, $DFP, DGP, CJQ,$ and $CIQ$ by the same reasoning. Additionally, $EH = GJ = PQ$ since $EHQP$ and $GJQP$ are congruent rectangles. 
If we then let $x = AE = AF = BH = BI,$ let $y = CI = CJ = DG = DF,$ and let $z = EH = GJ = PQ,$ we can create the following system of equations with the given side length information:
\begin{align*} 2x + z &= 500, \\ 2y + z &= 650, \\ x + y &= 333. \end{align*}
Adding the first two equations, subtracting by twice the second, and dividing by $2$ yields $z = PQ = \boxed{242}.$
~regular
Extend line $PQ$ to meet $AD$ at $P'$ and $BC$ at $Q'$. The diagram looks like this: 
[asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, P1, Q1; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); P1 = intersectionpoint(A--D,P--(-300)*(Q-P)+P); Q1 = intersectionpoint(B--C,Q--300*(Q-P)+Q); draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red); draw(anglemark(A,B,Q,1000),red); draw(anglemark(Q,B,C,1000),red); draw(anglemark(P,D,A,1000),red); draw(anglemark(C,D,P,1000),red); draw(anglemark(Q,C,D,1000),red); draw(anglemark(B,C,Q,1000),red); add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$P$",P,1.5*NE,linewidth(4)); dot("$Q$",Q,1.5*NW,linewidth(4)); dot("$P'$",P1,1.5*W,linewidth(4)); dot("$Q'$",Q1,1.5*E,linewidth(4)); draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); draw(P--P1^^Q--Q1,dashed); [/asy]
Since $\angle A + \angle D=\angle B + \angle C = 180^{\circ}$, it follows that $\angle P'AP+\angle P'DP = \angle Q'BQ + \angle Q'CQ = 90^{\circ}$. Thus, $\angle APD = \angle BQC = 90^{\circ}$, implying that $\triangle APD$ and $\triangle BQC$ are right triangles. Since $P'P$ and $Q'Q$ are medians, $P'P+Q'Q=\frac{333\times2}{2}=333$. Since $P'Q'=\frac{500+650}{2}=575$, we have $PQ+P'P+Q'Q=575$, or $PQ=575-333=\boxed{242}$.
~sigma
Let $PQ = x$. Note that since $AP$ bisects $\angle{A}$ and $DP$ bisects $\angle{D}$, we have \[\angle{APD} = 180^{\circ}-\tfrac12 \angle{A}-\tfrac12 \angle{D}=90^{\circ}.\] Let $\angle{ADP}=\theta$. We have that $\angle{ADC} = 2\theta.$ Now, drop an altitude from $A$ to $CD$ at $E$. Notice that $DE=\tfrac{650-500}{2}=75$. By the definition of cosine, we have \[\cos{2\theta}=1-2\cos^2{\theta}=\tfrac{75}{333}=\tfrac{25}{111} \implies \cos{\theta}=\tfrac{2\sqrt{1887}}{111}.\] Notice, however, that we can also apply this to $\triangle{APD}$; we have \[\cos{\theta}=\tfrac{DP}{333} \implies DP=6\sqrt{1887}.\] By the Pythagorean Theorem, we get \[AP=\sqrt{333^2-(6\sqrt{1887})^2}=3\sqrt{4773}.\] Then, drop an altitude from $P$ to $AB$ at $F$; if $AF=y$, then $PQ=x=500-2y$. Because $AP$ is an angle bisector, we see that $\angle{BAP}=\angle{DAP}=90^{\circ}-\theta$. Again, by the definition of cosine, we have \[\cos{(90^{\circ}-\theta)}=\sin{\theta}=\tfrac{\sqrt{4773}}{111}=\tfrac{y}{3\sqrt{4773}} \implies y=129.\] Finally, $PQ=500-2y=\boxed{242}$.
~pqr.
As in solution 4, $\angle APD = 90^{\circ}$. Set $k = AX$ and $x = DP$.
We know that $DZ = AX + \frac{DC-AB}{2}$, so $DZ = k + \frac{650-500}{2} = k + 75$.
$\triangle DPZ \sim \triangle APD$ by AA, so we have $\frac{PD}{AD} = \frac{ZD}{PD}$, resulting in
\[\frac{x}{333} = \frac{k+75}{x} \text{ (1)}\]
$\triangle APX \sim \triangle ADP$ by AA, so we have $\frac{AP}{AD} = \frac{AX}{AP}$, resulting in
\[\frac{\sqrt{333^2-x^2}}{333} = \frac{k}{\sqrt{333^2-k^2}} \text{ (2)}\]
From $\text{(1)}$, we have $x^2 = 333k + 333(75) = 333k + 24975$. From $\text{(2)}$, we have $333^2 - x^2 = 333k$, or $x^2 = 333^2 - 333k$. Thus, $333k + 24975 = 333^2 - 333k$. Solving for $k$ yields $k = 129$.
By symmetry, $YB = AX = 129$. Thus, $PQ = XY = AB - 2AX = 500 - 2(129) = \boxed{242}$.
~ adam_zheng


Answer:
242

Let's define the program: A simple dspy.ChainOfThought¶

In [5]:

class GenerateResponse(dspy.Signature):
    """Solve the problem and provide the answer in the correct format."""
    problem = dspy.InputField()
    answer = dspy.OutputField()

program = dspy.ChainOfThought(GenerateResponse)

class GenerateResponse(dspy.Signature): """Solve the problem and provide the answer in the correct format.""" problem = dspy.InputField() answer = dspy.OutputField() program = dspy.ChainOfThought(GenerateResponse)

Defining the evaluation metric¶

We simply check exact match between the predicted answer and the correct answer.

In [6]:

def metric(example, prediction, trace=None, pred_name=None, pred_trace=None):
    correct_answer = int(example['answer'])
    try:
        llm_answer = int(prediction.answer)
    except ValueError as e:
        return 0
    return int(correct_answer == llm_answer)

def metric(example, prediction, trace=None, pred_name=None, pred_trace=None): correct_answer = int(example['answer']) try: llm_answer = int(prediction.answer) except ValueError as e: return 0 return int(correct_answer == llm_answer)

Evaluating unoptimized Chain Of Thought¶

In [7]:

import dspy
evaluate = dspy.Evaluate(
    devset=test_set,
    metric=metric,
    num_threads=32,
    display_table=True,
    display_progress=True
)

evaluate(program)

import dspy evaluate = dspy.Evaluate( devset=test_set, metric=metric, num_threads=32, display_table=True, display_progress=True ) evaluate(program)

Average Metric: 70.00 / 150 (46.7%): 100%|██████████████████████████████████████████████████████████████████████████████████████████████████| 150/150 [00:01<00:00, 119.75it/s]
2025/08/12 21:49:36 INFO dspy.evaluate.evaluate: Average Metric: 70 / 150 (46.7%)

2025/08/12 21:49:36 INFO dspy.evaluate.evaluate: Average Metric: 70 / 150 (46.7%)

	problem	example_answer	reasoning	pred_answer	metric
0	Find the sum of all integer bases $b>9$ for which $17_b$ is a divi...	70	We are looking for integer bases \( b > 9 \) such that \( 17_b \) ...	70	✔️ [1]
1	On $\triangle ABC$ points $A, D, E$, and $B$ lie in that order on ...	588	Let's analyze the problem step-by-step. We have triangle \( ABC \)...	588	✔️ [1]
2	The 9 members of a baseball team went to an ice-cream parlor after...	16	We have 9 players, each choosing one of three flavors: chocolate (...	16	✔️ [1]
3	Find the number of ordered pairs $(x,y)$, where both $x$ and $y$ a...	117	We start with the given equation: \[12x^2 - xy - 6y^2 = 0.\] Our g...	117	✔️ [1]
4	There are $8!= 40320$ eight-digit positive integers that use each ...	279	We are given that there are \(8! = 40320\) eight-digit numbers tha...	279	✔️ [1]
...	...	...	...	...	...
145	Let $S$ be the set of vertices of a regular $24$-gon. Find the num...	113	We have a regular 24-gon with vertex set \( S \) of size 24. We wa...	1666
146	Let $A_1 A_2 A_3 \ldots A_{11}$ be an $11$-sided non-convex simple...	19	Let's analyze the problem step-by-step. We are given an 11-sided p...	19	✔️ [1]
147	Let $x_1, x_2, x_3, \ldots$ be a sequence of rational numbers defi...	248	We have the sequence: \[ x_1 = \frac{25}{11}, \quad x_{k+1} = \fra...	589
148	Let $\triangle ABC$ be a right triangle with $\angle A = 90^\circ$...	104	Given a right triangle \(\triangle ABC\) with \(\angle A = 90^\cir...	98
149	There are exactly three positive real numbers $k$ such that the fu...	240	We are given the function \[ f(x) = \frac{(x - 18)(x - 72)(x - 98)...	240	✔️ [1]

150 rows × 5 columns

Out[7]:

EvaluationResult(score=46.67, results=<list of 150 results>)

Optimize the program with dspy.GEPA¶

GEPA is a reflective prompt optimizer, and it's strength lies in being able to leverage additional sources of information, like the DSPy program's execution and evaluation pipelines, which provides GEPA more visibility into why the system got the score that it did, and then GEPA can introspect to identify how to improve the score. GEPA can also leverage additional supervision provided in this manner. For example, during optimization, we can return the correct solution's to the problems the program failed to solve.

We note that while such explicit supervision is not available in all scenarios, GEPA can work very flexibly with different forms of feedback (for example, using LLM-as-a-judge feedback shown in the PAPILLON tutorial, or just using answer labels, as shown in the facility-support tutorial).

Let's quickly modify the evaluation metric to become an optimization metric for GEPA, that can provide this additional supervision!

In [8]:

def metric_with_feedback(example, prediction, trace=None, pred_name=None, pred_trace=None):
    correct_answer = int(example['answer'])
    written_solution = example.get('solution', '')
    try:
        llm_answer = int(prediction.answer)
    except ValueError as e:
        feedback_text = f"The final answer must be a valid integer and nothing else. You responded with '{prediction.answer}', which couldn't be parsed as a python integer. Please ensure your answer is a valid integer without any additional text or formatting."
        feedback_text += f" The correct answer is '{correct_answer}'."
        if written_solution:
            feedback_text += f" Here's the full step-by-step solution:\n{written_solution}\n\nThink about what takeaways you can learn from this solution to improve your future answers and approach to similar problems and ensure your final answer is a valid integer."
        return dspy.Prediction(score=0, feedback=feedback_text)

    score = int(correct_answer == llm_answer)

    feedback_text = ""
    if score == 1:
        feedback_text = f"Your answer is correct. The correct answer is '{correct_answer}'."
    else:
        feedback_text = f"Your answer is incorrect. The correct answer is '{correct_answer}'."
    
    if written_solution:
        feedback_text += f" Here's the full step-by-step solution:\n{written_solution}\n\nThink about what takeaways you can learn from this solution to improve your future answers and approach to similar problems."

    return dspy.Prediction(score=score, feedback=feedback_text)

def metric_with_feedback(example, prediction, trace=None, pred_name=None, pred_trace=None): correct_answer = int(example['answer']) written_solution = example.get('solution', '') try: llm_answer = int(prediction.answer) except ValueError as e: feedback_text = f"The final answer must be a valid integer and nothing else. You responded with '{prediction.answer}', which couldn't be parsed as a python integer. Please ensure your answer is a valid integer without any additional text or formatting." feedback_text += f" The correct answer is '{correct_answer}'." if written_solution: feedback_text += f" Here's the full step-by-step solution:\n{written_solution}\n\nThink about what takeaways you can learn from this solution to improve your future answers and approach to similar problems and ensure your final answer is a valid integer." return dspy.Prediction(score=0, feedback=feedback_text) score = int(correct_answer == llm_answer) feedback_text = "" if score == 1: feedback_text = f"Your answer is correct. The correct answer is '{correct_answer}'." else: feedback_text = f"Your answer is incorrect. The correct answer is '{correct_answer}'." if written_solution: feedback_text += f" Here's the full step-by-step solution:\n{written_solution}\n\nThink about what takeaways you can learn from this solution to improve your future answers and approach to similar problems." return dspy.Prediction(score=score, feedback=feedback_text)

In [9]:

from dspy import GEPA

optimizer = GEPA(
    metric=metric_with_feedback,
    auto="light",
    num_threads=32,
    track_stats=True,
    reflection_minibatch_size=3,
    reflection_lm=dspy.LM(model="gpt-5", temperature=1.0, max_tokens=32000, api_key=api_key)
)

optimized_program = optimizer.compile(
    program,
    trainset=train_set,
    valset=val_set,
)

from dspy import GEPA optimizer = GEPA( metric=metric_with_feedback, auto="light", num_threads=32, track_stats=True, reflection_minibatch_size=3, reflection_lm=dspy.LM(model="gpt-5", temperature=1.0, max_tokens=32000, api_key=api_key) ) optimized_program = optimizer.compile( program, trainset=train_set, valset=val_set, )

2025/08/12 21:49:36 INFO dspy.teleprompt.gepa.gepa: Running GEPA for approx 560 metric calls of the program. This amounts to 6.22 full evals on the train+val set.
3
2025/08/12 21:49:36 INFO dspy.teleprompt.gepa.gepa: Using 45 examples for tracking Pareto scores. You can consider using a smaller sample of the valset to allow GEPA to explore more diverse solutions within the same budget.
2025/08/12 21:52:15 INFO dspy.evaluate.evaluate: Average Metric: 17.0 / 45 (37.8%)
2025/08/12 21:52:15 INFO dspy.teleprompt.gepa.gepa: Iteration 0: Base program full valset score: 0.37777777777777777
2025/08/12 21:52:15 INFO dspy.teleprompt.gepa.gepa: Iteration 1: Selected program 0 score: 0.37777777777777777
Average Metric: 2.00 / 3 (66.7%): 100%|██████████████████████████████████████████████████████████████████████████████████████████████████████████| 3/3 [03:10<00:00, 63.51s/it]
2025/08/12 21:55:26 INFO dspy.evaluate.evaluate: Average Metric: 2.0 / 3 (66.7%)

2025/08/12 21:56:50 INFO dspy.teleprompt.gepa.gepa: Iteration 1: Proposed new text for predict: You will be given a single math problem as plain text under an input key like “problem.” Your task is to solve it correctly and return:

- reasoning: a concise, logically ordered solution that uses appropriate identities/constraints to minimize brute force and ends with a quick verification.
- answer: the final result only (just the number or expression requested, no extra words).

Formatting:
- Use two top-level fields exactly named “reasoning” and “answer.”
- Keep the reasoning succinct but complete. Avoid heavy markup. Bullet points are fine.
- The answer field must contain only the final value requested (e.g., 227, 585, 601).

General problem-solving guidance:
- Parse the problem type (e.g., base representation, palindrome constraints across bases, symmetric sums with constraints, counting ordered triples).
- Always enforce domain constraints (e.g., base-b digits are 0..b-1; leading digit in base-10 three-digit numbers is 1..9; ordered triples count permutations unless otherwise specified).
- Use algebraic identities and modular arithmetic to reduce the search space and ensure correctness.

Domain-specific strategies (from common contest problems):

1) Base-conversion/digit rearrangement:
- Translate positional notation correctly:
  For base 10: abc = 100a + 10b + c.
  For base 9 (or b): (b c a)_9 = b·9^2 + c·9 + a.
- Enforce digit ranges: in base 9, digits a, b, c ∈ {0,…,8}; if the base-10 number is three-digit, a ∈ {1,…,9} and also must satisfy the base-9 digit limit (so a ∈ {1,…,8}).
- Set up equality and simplify. Example pattern: 100a + 10b + c = 81b + 9c + a ⇒ 99a = 71b + 8c.
- Use modular constraints to prune:
  • Mod 9: 99a ≡ 0 and 71 ≡ 8 ⇒ 0 ≡ 8(b + c) ⇒ b + c ≡ 0 (mod 9), so b + c ∈ {0, 9} within digit bounds.
  • Mod 8: 99 ≡ 3, 71 ≡ 7 ⇒ 3a ≡ 7b (mod 8) ⇒ b ≡ −3a (mod 8).
- Solve within digit bounds and verify.

2) Palindromes across bases (e.g., base 10 and base 8):
- Determine the base-8 length given the magnitude (n < 1000 ⇒ base-8 has 3 or 4 digits).
- Characterize base-8 palindromes:
  • 3-digit: (A B A)_8 = 64A + 8B + A = 65A + 8B.
  • 4-digit: (A B B A)_8 = 512A + 64B + 8B + A = 513A + 72B.
- For n < 1000 and 4-digit octal palindrome, A must be 1. Enumerate B ∈ {0,…,7} to get candidates 513 + 72B and test which are decimal palindromes. Choose the greatest valid n.

3) Symmetric sums with a + b + c fixed (ordered triples of nonnegative integers):
- Convert sums like a^2b + a^2c + b^2a + b^2c + c^2a + c^2b using:
  S = ab(a + b) + bc(b + c) + ca(c + a) = (a + b + c)(ab + bc + ca) − 3abc.
- With a + b + c given (e.g., 300), plug into S = given constant to derive:
  100(ab + bc + ca) − abc = constant.
- Use the shift a = 100 + x, b = 100 + y, c = 100 + z, so x + y + z = 0. Then:
  (a − 100)(b − 100)(c − 100) = abc − 100(ab + bc + ca) + 2,000,000.
  Setting S correctly yields (a − 100)(b − 100)(c − 100) = 0.
- Count solutions:
  • If exactly one equals 100, the other two sum to 200 with both ≠ 100. Count all integer splits respecting nonnegativity; multiply by 3 for which variable is 100.
  • Include the case a = b = c = 100 once.
  • Treat (a, b, c) as ordered unless the problem explicitly asks for unordered.

Quality checks:
- Verify digit/base constraints and final equality numerically.
- For “greatest/least” questions, justify why your candidate is optimal (structural argument or bounded enumeration).
- For counting, avoid over/undercounting; be explicit about ordered vs unordered.

Finally:
- Place the clean final numeric result in the “answer” field.
2025/08/12 21:57:19 INFO dspy.evaluate.evaluate: Average Metric: 3.0 / 3 (100.0%)
2025/08/12 22:00:37 INFO dspy.evaluate.evaluate: Average Metric: 19.0 / 45 (42.2%)
2025/08/12 22:00:37 INFO dspy.teleprompt.gepa.gepa: Iteration 1: New program is on the linear pareto front
2025/08/12 22:00:37 INFO dspy.teleprompt.gepa.gepa: Iteration 1: Full valset score for new program: 0.4222222222222222
2025/08/12 22:00:37 INFO dspy.teleprompt.gepa.gepa: Iteration 1: Full train_val score for new program: 0.4222222222222222
2025/08/12 22:00:37 INFO dspy.teleprompt.gepa.gepa: Iteration 1: Individual valset scores for new program: [0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0]
2025/08/12 22:00:37 INFO dspy.teleprompt.gepa.gepa: Iteration 1: New valset pareto front scores: [0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0]
2025/08/12 22:00:37 INFO dspy.teleprompt.gepa.gepa: Iteration 1: Full valset pareto front score: 0.5111111111111111
2025/08/12 22:00:37 INFO dspy.teleprompt.gepa.gepa: Iteration 1: Updated valset pareto front programs: [{0, 1}, {0, 1}, {0, 1}, {0, 1}, {0, 1}, {1}, {0, 1}, {0}, {0, 1}, {0, 1}, {0}, {0, 1}, {0, 1}, {0, 1}, {0, 1}, {0}, {0, 1}, {1}, {0, 1}, {0, 1}, {1}, {0, 1}, {0, 1}, {0, 1}, {0, 1}, {0, 1}, {0, 1}, {0, 1}, {0, 1}, {0, 1}, {0, 1}, {1}, {1}, {0, 1}, {1}, {0, 1}, {0, 1}, {0, 1}, {0, 1}, {0, 1}, {0, 1}, {0, 1}, {0, 1}, {0}, {0, 1}]
2025/08/12 22:00:37 INFO dspy.teleprompt.gepa.gepa: Iteration 1: Best valset aggregate score so far: 0.4222222222222222
2025/08/12 22:00:37 INFO dspy.teleprompt.gepa.gepa: Iteration 1: Best program as per aggregate score on train_val: 1
2025/08/12 22:00:37 INFO dspy.teleprompt.gepa.gepa: Iteration 1: Best program as per aggregate score on valset: 1
2025/08/12 22:00:37 INFO dspy.teleprompt.gepa.gepa: Iteration 1: Best score on valset: 0.4222222222222222
2025/08/12 22:00:37 INFO dspy.teleprompt.gepa.gepa: Iteration 1: Best score on train_val: 0.4222222222222222
2025/08/12 22:00:37 INFO dspy.teleprompt.gepa.gepa: Iteration 1: Linear pareto front program index: 1
2025/08/12 22:00:37 INFO dspy.teleprompt.gepa.gepa: Iteration 1: New program candidate index: 1
2025/08/12 22:00:37 INFO dspy.teleprompt.gepa.gepa: Iteration 2: Selected program 0 score: 0.37777777777777777
Average Metric: 2.00 / 3 (66.7%): 100%|██████████████████████████████████████████████████████████████████████████████████████████████████████████| 3/3 [02:57<00:00, 59.20s/it]
2025/08/12 22:03:34 INFO dspy.evaluate.evaluate: Average Metric: 2.0 / 3 (66.7%)

2025/08/12 22:04:41 INFO dspy.teleprompt.gepa.gepa: Iteration 2: Proposed new text for predict: You will be given a single field:
- problem: a math contest-style problem (algebra/number theory/geometry/combinatorics).

Your task:
1) Solve the problem correctly with clear, exact reasoning (avoid decimal approximations unless explicitly required).
2) Output in this exact format:
   ### reasoning
   [succinct, rigorous solution steps]
   ### answer
   [final answer only, no extra words]

General guidance and domain-specific strategies:
- Keep the reasoning concise but complete enough to verify correctness. Prefer clean theoretical arguments over coordinate bashing or numeric approximations.
- For geometry (especially symmetric figures like isosceles trapezoids):
  - Exploit symmetry and parallelism.
  - Angle-bisector facts: the bisector at a vertex is the locus of points equidistant from the adjacent sides; in a cyclic or trapezoidal setup adjacent-angle bisectors may form right angles because A + D = 180° implies half-angles sum to 90°.
  - In right triangles, the midpoint of the hypotenuse is equidistant from the endpoints; medians and midlines are powerful.
  - In trapezoids, the midline length equals the average of the bases. A useful identity for an isosceles trapezoid with bases AB, CD and legs AD = BC is:
    PQ = (AB + CD)/2 − (AD + BC)/2; in the isosceles case this simplifies to PQ = (AB + CD)/2 − AD.
  - Angle-bisector constructions often yield congruent right triangles and rhombuses by equal-distance properties—use these to get exact lengths.
- For products over roots of unity:
  - If ω is a primitive n-th root (or any ω with ω^n = 1 and ω ≠ 1), then {ω^k}_{k=0}^{n−1} runs over all n-th roots; products of the form ∏ f(ω^k) are independent of the chosen ω.
  - Factor polynomials in terms of linear factors at convenient complex numbers and use
    ∏_{k=0}^{n−1} (a − ω^k) = a^n − 1.
    Example: for f(x) = x^2 − 2x + 2 = (x − (1+i))(x − (1−i)),
    ∏_{k=0}^{n−1} f(ω^k) = ((1+i)^n − 1)((1−i)^n − 1).
  - Use polar form for fast powers: (1 ± i) = √2 e^{±iπ/4}, so (1 ± i)^n = (√2)^n e^{±inπ/4}. Magnitude-squared yields clean integers like 65^2 + 64^2.
  - Reduce modulo as requested at the end.
- For counting rectangles in a regular polygon:
  - Rectangles correspond to choosing two perpendicular direction classes of chords (there are 6 distinct directions in a regular dodecagon).
  - Do careful casework by slope/direction classes. Typical split for a regular 12-gon: directions at 0°, 30°, 60° (and their perpendiculars), and directions at 15°, 45°, 75°. Count pairs of parallel chords in each case using combinatorial counts (often via binomial coefficients) and inclusion–exclusion to avoid double counting.
  - Do not assume only axis-aligned (or diameter-based) rectangles; many rectangles have sides along diagonals.

Quality checks:
- Avoid overcounting; use inclusion–exclusion where overlaps occur.
- Prefer exact algebraic expressions; avoid rounding mid-solution.
- If a cleaner identity or symmetry shortcut exists, use it.
- The final line under ### answer must be only the final value/expression (e.g., 242 or 321), with no extra text.
2025/08/12 22:08:01 INFO dspy.evaluate.evaluate: Average Metric: 2.0 / 3 (66.7%)
2025/08/12 22:08:01 INFO dspy.teleprompt.gepa.gepa: Iteration 2: New subsample score is not better, skipping
2025/08/12 22:08:01 INFO dspy.teleprompt.gepa.gepa: Iteration 3: Selected program 1 score: 0.4222222222222222
Average Metric: 2.00 / 3 (66.7%): 100%|██████████████████████████████████████████████████████████████████████████████████████████████████████████| 3/3 [03:16<00:00, 65.66s/it]
2025/08/12 22:11:18 INFO dspy.evaluate.evaluate: Average Metric: 2.0 / 3 (66.7%)

2025/08/12 22:12:39 INFO dspy.teleprompt.gepa.gepa: Iteration 3: Proposed new text for predict: You will be given a single math problem as plain text under an input key like “problem.” Your task is to solve it correctly and return:

- reasoning: a concise, logically ordered solution that uses appropriate identities/constraints to minimize brute force and ends with a quick verification.
- answer: the final result only (just the number or expression requested, no extra words).

Formatting:
- Use two top-level fields exactly named “reasoning” and “answer.”
- Keep the reasoning succinct but complete. Avoid heavy markup. Bullet points are fine.
- The answer field must contain only the final value requested (e.g., 227, 585, 601).

General problem-solving guidance:
- Parse the problem type (e.g., base representation, palindromes across bases, symmetric sums under constraints, combinatorics with “exactly k,” geometry with slices/planes, 3D vector setups).
- Always enforce domain constraints (e.g., base-b digits ∈ {0,…,b−1}, no leading zero for a base-10 three-digit number, sphere radii > 0, lengths/areas/volumes ≥ 0, ordered vs unordered if counting).
- Use algebraic identities and modular arithmetic to reduce the search space and ensure correctness.
- Prefer structure/symmetry and key constraints over coordinate bash; introduce coordinates/vectors only when they simplify.

Domain-specific strategies and identities:

1) Base-conversion/digit rearrangement:
- Translate positional notation correctly:
  • Base 10: abc = 100a + 10b + c.
  • Base b: (x y z)_b = x·b^2 + y·b + z.
- Enforce digit bounds: in base 9, digits are 0..8; if also a base-10 three-digit integer, leading digit 1..9 and ≤ base-9 max digit ⇒ a ∈ {1,…,8}.
- Set up equality and simplify. Use modular constraints to prune:
  • Mod 9: reduce coefficients to constrain sums like b + c ≡ 0 (mod 9).
  • Mod 8 (or others): reduce to get congruences between digits.

2) Palindromes across bases:
- Bound the number of digits from magnitude (e.g., n < 1000 ⇒ octal has 3–4 digits).
- Characterize palindromes:
  • 3-digit base-8: (A B A)_8 = 65A + 8B.
  • 4-digit base-8: (A B B A)_8 = 513A + 72B.
- For 4-digit octal palindrome < 1000, A must be 1. Enumerate B ∈ {0,…,7}, test decimal-palindrome property, and pick the greatest/least as required.

3) Symmetric sums with a + b + c fixed (ordered triples of nonnegative integers):
- Use identities to collapse symmetric expressions:
  • S = Σ a^2b + a^2c + … = (a + b + c)(ab + bc + ca) − 3abc.
- With a + b + c given (e.g., 300), convert to a linear relation in (ab + bc + ca) and abc.
- Apply the shift a = m + x, b = m + y, c = m + z with x + y + z = 0 to factor:
  • (a − m)(b − m)(c − m) = abc − m(ab + bc + ca) + 2m^3 − m^3 = abc − m(ab + bc + ca) + m^3.
- Use the given constants to force one factor zero and count ordered solutions carefully:
  • Count cases where exactly one equals m and the others sum appropriately (excluding double-counts), plus the all-equal case once.
  • Treat triples as ordered unless the problem states otherwise.

4) Sets with “exactly k” owners (inclusion-exclusion, baseline item owned by all):
- Let x1, x2, x3, x4 be counts of residents owning exactly 1, 2, 3, 4 of the given sets.
- Everyone owning a baseline item (e.g., candy hearts) still counts toward the “k of four” totals; do not drop it.
- Use:
  • x1 + x2 + x3 + x4 = N (total people),
  • Sum of set sizes = 1·x1 + 2·x2 + 3·x3 + 4·x4.
- Plug given x2, x3 (if provided) and total set counts to solve a small linear system for x4 (owners of all four).

5) Spheres intersected by a plane (congruent circle sections, mutually tangent spheres):
- Mutually externally tangent spheres with radii R_i have center distances R_i + R_j.
- If a plane intersects spheres in congruent circles of radius r, and the perpendicular distances from centers to the plane are h_i, then:
  • r^2 = R_i^2 − h_i^2 for each sphere i (constant r across all).
  • For projections A, B, C of the centers onto the plane, the in-plane squared distances satisfy:
    |A_iA_j|^2 = |O_iO_j|^2 − (h_i − h_j)^2,
    hence |O_iO_j|^2 = |A_iA_j|^2 + (h_i − h_j)^2.
- Use AB^2 (in-plane) and known |O_iO_j| = R_i + R_j to get (h_i − h_j)^2, then solve for r^2 via R_i^2 − h_i^2 = constant. Verify by computing the target in-plane distance with |O_iO_k|^2 − (h_i − h_k)^2.

6) Tilted cube, vertex heights, and a horizontal water plane:
- If the problem specifies a set of vertices with given heights above a fixed horizontal plane, and a specific rectangle/plane (e.g., ABDC) is perpendicular to the horizontal plane, exploit this geometry:
  • The height function over the cube is linear along cube edges.
  • First, find the side length s using robust constraints:
    - Face diagonals have length s√2; space diagonal s√3; edges length s.
    - Use similarity or slope along lines within the stated perpendicular face/rectangle (e.g., compare vertical rises along an edge vs along a face diagonal) to solve for s before proceeding.
  • Water volume at height H equals the volume below the horizontal plane z = H within the cube. Two reliable approaches:
    A) Frustum method (when two parallel cross-sections are similar):
       - If the complement above/below forms a prismatic frustum between two parallel planes cutting similar cross-sections (e.g., squares cut along a face direction), use:
         V_frustum = (h/3)·(S1 + √(S1 S2) + S2),
         where h is the distance between the planes (measured along the line perpendicular to the planes within the prism), and S1, S2 are the areas of the parallel cross-sections. Linear dimensions along a given straight direction scale linearly with height; areas scale with the square of that ratio.
       - Cube water volume = total cube volume − frustum volume (or vice versa), whichever is simpler.
    B) Linear-inequality integration over a unit cube:
       - Represent any point as A + u·e1 + v·e2 + w·e3 with u,v,w ∈ [0,1], where e1,e2,e3 are orthogonal edges of length s. The height is z = z0 + a u + b v + c w (linear).
       - To compute volume where z ≤ H, integrate over {u,v,w ∈ [0,1]: a u + b v + c w ≤ t} and multiply by s^3. If some coefficients are negative, perform a variable flip (e.g., u' = 1 − u converts coefficient sign and shifts t) to reduce to all-positive coefficients or use the complement region.
       - Partition the (u,v) domain by the line a u + b v = thresholds to integrate piecewise. Ensure the final volume is between 0 and s^3.
  • Sanity checks:
    - s must be determined consistently from the geometry (e.g., in a classic configuration, s = 6).
    - Cross-section areas and volumes must be nonnegative and not exceed bounding values.
    - If an attempted coordinate labeling yields contradictions (e.g., heights inconsistent with linearity), reconsider the assignment order or use a different method (similar triangles/frustum).

Quality checks:
- Verify digit/base constraints and final equalities numerically.
- For “greatest/least” questions, justify optimality by bounding or short enumeration.
- For counting, avoid over/undercounting; be explicit about ordered vs unordered.
- For geometry, check intermediate values (e.g., differences of heights, r^2, s) and plausibility (lengths/areas/volumes ≥ 0, within expected ranges).
- End with a quick numeric verification plugging back into the original relation.

Finally:
- Place the clean final numeric result in the “answer” field.
2025/08/12 22:14:20 INFO dspy.evaluate.evaluate: Average Metric: 2.0 / 3 (66.7%)
2025/08/12 22:14:20 INFO dspy.teleprompt.gepa.gepa: Iteration 3: New subsample score is not better, skipping
2025/08/12 22:14:20 INFO dspy.teleprompt.gepa.gepa: Iteration 4: Selected program 0 score: 0.37777777777777777
Average Metric: 2.00 / 3 (66.7%): 100%|██████████████████████████████████████████████████████████████████████████████████████████████████████████| 3/3 [02:36<00:00, 52.14s/it]
2025/08/12 22:16:56 INFO dspy.evaluate.evaluate: Average Metric: 2.0 / 3 (66.7%)

2025/08/12 22:18:15 INFO dspy.teleprompt.gepa.gepa: Iteration 4: Proposed new text for predict: Instructions for solving and formatting contest-style math problems:

1) Read the prompt carefully and determine exactly what to compute and how the answer must be reported.
   - If the problem asks for r^2 = p/q with p, q relatively prime and asks for p+q, return the integer p+q.
   - If it asks for a probability m/n in lowest terms and then m+n, return that integer.
   - If it asks for a single count, return that integer.
   - Always reduce fractions to lowest terms before extracting p+q or m+n.
   - Provide the final answer exactly in the requested format (usually a single integer). Keep any reasoning concise if shown; ensure the final line is the answer only.

2) General problem-solving practices:
   - Translate given conditions into equations/constraints and identify what quantity should be maximized or minimized. Be precise about “for each” vs “exists” and “contains every” vs “contains a specific.”
   - Use symmetry to reduce variables when optimizing (extrema for symmetric constraints often occur when two variables are equal).
   - For boxes/spheres: for a rectangular box with edges x, y, z, the sphere that contains it must have radius at least half the space diagonal: r = sqrt(x^2 + y^2 + z^2)/2. Surface area SA = 2(xy + yz + zx). Volume V = xyz.
     • If asked for the smallest sphere that can contain every box in a set defined by constraints, you must maximize the required radius over that feasible set (worst case), not minimize it.
     • With fixed S2 = xy + yz + zx and S3 = xyz, note x^2 + y^2 + z^2 = (x + y + z)^2 − 2S2, so maximizing the diagonal corresponds to maximizing S1 = x + y + z. At extrema under symmetric constraints, set two variables equal (e.g., y = z) to reduce to one variable and solve (Lagrange multipliers or symmetry argument). Use Rational Root Theorem when a cubic arises. Select the root that actually maximizes the target.
   - For parity-based arrangement problems with pairs of identical items:
     • “Even number of items between identical colors” means the two copies occupy positions of opposite parity (one odd, one even).
     • If there are k colors with two copies each and positions 1..2k, count valid arrangements by: (i) pairing each odd position with an even position (k! ways), and (ii) assigning colors to the k pairs (k! ways). Total valid = (k!)^2. Total permutations with duplicates = (2k)! / (2!)^k. Simplify the resulting probability fully before computing m+n.
   - For grid/monochromatic row-column constraints with maximality:
     • “All chips in the same row and all chips in the same column have the same color” implies the grid is determined by sets of white rows W_r and white columns W_c: all cells in W_r × W_c are white; all cells in the complementary set of rows and columns B_r × B_c are black; the cross rectangles W_r × B_c and B_r × W_c must be empty to keep row/column uniformity.
     • “Any additional chip would violate” (maximality) rules out having both some empty rows and some empty columns simultaneously where their intersection would admit a valid placement. A clean equivalent counting for a 5×5 grid is:
       - Choose rows and columns occupied by white chips as any nonempty, not-all subsets: (2^5 − 2) choices for rows and (2^5 − 2) for columns. This yields (2^5 − 2)^2 configurations with both colors present.
       - Add 2 for the all-white and all-black full boards. The empty board is disallowed by “some chips.”
       - For 5×5, the total is (32 − 2)^2 + 2 = 900 + 2 = 902.
     • Chips are indistinguishable; once W_r and W_c are fixed, the configuration is uniquely determined.

3) Algebraic care and verification:
   - Keep track of whether you are maximizing or minimizing; re-check interpretations like “smallest sphere that can contain each box” (maximize over the set) vs “for a given box” (minimize for that box).
   - When radicals appear but the problem expects a rational r^2, express r^2 directly from symmetric sums if possible to avoid unnecessary radicals.
   - Rationalize and reduce fractions; ensure p and q are coprime before summing.
   - Sanity-check candidate extrema numerically to ensure you selected the correct root (e.g., larger S1 for maximizing diagonal).

4) Output formatting:
   - End with only the requested final answer (e.g., a single integer), with no extra text.
2025/08/12 22:18:35 INFO dspy.evaluate.evaluate: Average Metric: 2.0 / 3 (66.7%)
2025/08/12 22:18:35 INFO dspy.teleprompt.gepa.gepa: Iteration 4: New subsample score is not better, skipping
2025/08/12 22:18:35 INFO dspy.teleprompt.gepa.gepa: Iteration 5: Selected program 1 score: 0.4222222222222222
Average Metric: 0.00 / 3 (0.0%): 100%|███████████████████████████████████████████████████████████████████████████████████████████████████████████| 3/3 [00:27<00:00,  9.33s/it]
2025/08/12 22:19:03 INFO dspy.evaluate.evaluate: Average Metric: 0.0 / 3 (0.0%)

2025/08/12 22:20:52 INFO dspy.teleprompt.gepa.gepa: Iteration 5: Proposed new text for predict: You will be given one math problem as plain text under a key like “problem.” Your job is to solve it correctly and return:

- reasoning: a concise, logically ordered solution that uses identities/structure to avoid brute force, ends with a quick verification.
- answer: the final requested number/expression only (no extra words).

Formatting:
- Use exactly two top-level fields named “reasoning” and “answer.”
- Keep reasoning succinct but complete. Bullet points are fine.
- The answer field must contain only the final value requested (e.g., 227, 585, 601).

General problem-solving guidance:
- Parse the problem type (e.g., base representation, intersecting families of subsets, avoiding arithmetic progressions, symmetric sums with constraints, ordered tuples counting).
- Always enforce domain constraints (e.g., base-b digits in 0..b−1; no leading zero for base-10 “three-digit”; ordered vs unordered families; strict increase conditions in sequences).
- Use algebraic identities and modular arithmetic to reduce the search space; prefer structural arguments over naive enumeration.
- For “greatest/least” questions, derive tight bounds and give a construction that attains them.

Domain-specific strategies and pitfalls (learned from typical contest problems and prior feedback):

1) Base-conversion/digit rearrangement:
- Translate positional notation correctly: in base b, (a b c)_b = a·b^2 + b·b + c; in base 10: abc = 100a + 10b + c.
- Enforce digit ranges strictly (e.g., in base 9, digits ∈ {0,…,8}; if also a is a base-10 leading digit, then a ∈ {1,…,8}).
- Set up equality and simplify. Use modular constraints to prune:
  • Mod 9 often collapses coefficients; e.g., 99a = 71b + 8c ⇒ mod 9 gives b + c ≡ 0 (mod 9).
  • Mod 8: 99 ≡ 3, 71 ≡ 7 ⇒ 3a ≡ 7b (mod 8) ⇒ b ≡ −3a (mod 8).
- Solve within digit bounds and verify numerically.

2) Palindromes across bases:
- Bound the base length by magnitude (e.g., n < 1000 ⇒ octal has 3–4 digits).
- Characterize palindromes:
  • 3-digit octal: (A B A)_8 = 65A + 8B.
  • 4-digit octal: (A B B A)_8 = 513A + 72B (with A ≥ 1).
- Enumerate small parameter ranges and test the other-base palindrome constraint. For “greatest”, check candidates in descending order with justification.

3) Symmetric sums with a + b + c fixed (ordered triples of nonnegative integers):
- Use identities to compress expressions:
  S = ab(a + b) + bc(b + c) + ca(c + a) = (a + b + c)(ab + bc + ca) − 3abc.
- With a + b + c known (e.g., 300), convert the given sum into a relation among ab + bc + ca and abc.
- Use the shift a = A + x etc. to isolate a product like (a−A)(b−A)(c−A) and deduce factorization constraints, enabling clean counting.
- Count ordered solutions carefully; include/exclude symmetric/degenerate cases precisely.

4) Intersecting families of subsets (collections from the power set):
- Intersecting means every pair has nonempty intersection. The empty set cannot be included.
- Complement pairs: S and S^c cannot both be present. Use this to structure counts.
- Use size-based pigeonhole facts: In [n], any two subsets of size > n/2 must intersect. For n = 5, any two subsets of size ≥ 3 intersect; thus “all subsets of size ≥ 3” is an intersecting family (size 16).
- Do not assume that “stars” (all subsets containing a fixed element) are the only intersecting families of maximum size. For odd n, both the star and “all subsets of size > n/2” have size 2^{n−1}.
- When counting collections of a fixed size:
  • Consider the minimum set size N in the family and do casework on how many 2-element sets are included (for n=5), as these control which 3-sets must be excluded (complements).
  • Ensure completeness of cases and avoid double counting by parameterizing canonical patterns (e.g., how many 2-sets, how they overlap, whether they share a common element).
  • Remember order of subsets in a collection does not matter; count distinct families.

5) Avoiding 4-term arithmetic progressions in a strictly increasing sequence with fixed anchors:
- First bound the variable terms by strict increase (e.g., if fixed terms are 3,4,5,...,30,40,50 then 6 ≤ a < b ≤ 29).
- Pre-eliminate values that cause a 4-term AP with three fixed terms:
  • 3,4,5,a forbids a = 6.
  • b,30,40,50 forbids b = 20.
  • Similarly, a,30,40,50 forbids a = 20.
- Start with the count of pairs from allowed values and then subtract specific pairs that complete APs with two fixed endpoints:
  • 3,5,a,b ⇒ (a,b) = (7,9).
  • 3,a,b,30 ⇒ (a,b) = (12,21).
  • 4,a,b,40 ⇒ (a,b) = (16,28).
  • 5,a,b,50 ⇒ (a,b) = (20,35) but may be outside bounds or pre-excluded (e.g., 20 banned).
- Systematically check all endpoint combinations; use the fact that if endpoints differ by Δ, then Δ must be divisible by 3 for a 4-term AP, and solve for integer a,b within bounds.
- Avoid double subtraction; ensure monotonicity and domain constraints are respected.

6) Order statistics with sum and absolute-sum constraints (e.g., x_1 ≤ ... ≤ x_n, sum |x_i| = 1, sum x_i = 0):
- Total positive mass equals total negative mass: both = 1/2.
- For maximizing x_k (k near the top): if there are T largest terms from k to n (T = n − k + 1), then sum of these T terms ≥ T·x_k. Since the total positive mass ≤ 1/2, we get x_k ≤ (1/2)/T.
- For minimizing x_l (l near the bottom): if there are l smallest terms, sum of these l terms ≤ l·x_l. Since the total negative mass is −1/2, we get x_l ≥ (−1/2)/l.
- To attain these bounds, concentrate masses evenly on exactly those positions: set the smallest l terms equal to −1/(2l), the largest T terms equal to 1/(2T), and the middle to 0 (respecting monotonicity). Verify sums and absolute sums.
- Example: For n=100, maximize x_76 − x_16: T = 25 ⇒ x_76 ≤ 1/50; l = 16 ⇒ x_16 ≥ −1/32; construction with 16 negatives at −1/32, 59 zeros, 25 positives at 1/50 attains 1/50 − (−1/32) = 41/800.

Quality checks:
- Verify digit/base constraints and final equalities numerically if applicable.
- For extremal problems, provide both a tight bound and an explicit construction achieving it.
- For counting, explicitly handle ordered vs unordered, exclude impossible/duplicate cases, and check complements/forbidden pairs.
- For AP-avoidance, confirm integrality and bounds; ensure no missed endpoint combinations.
- For “greatest/least” questions, justify optimality structurally (e.g., convexity/majorization/pigeonhole).

Finally:
- Put the clean final numeric result in the “answer” field only.
2025/08/12 22:21:18 INFO dspy.evaluate.evaluate: Average Metric: 1.0 / 3 (33.3%)
2025/08/12 22:23:54 INFO dspy.evaluate.evaluate: Average Metric: 24.0 / 45 (53.3%)
2025/08/12 22:23:54 INFO dspy.teleprompt.gepa.gepa: Iteration 5: New program is on the linear pareto front
2025/08/12 22:23:54 INFO dspy.teleprompt.gepa.gepa: Iteration 5: Full valset score for new program: 0.5333333333333333
2025/08/12 22:23:54 INFO dspy.teleprompt.gepa.gepa: Iteration 5: Full train_val score for new program: 0.5333333333333333
2025/08/12 22:23:54 INFO dspy.teleprompt.gepa.gepa: Iteration 5: Individual valset scores for new program: [0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0]
2025/08/12 22:23:54 INFO dspy.teleprompt.gepa.gepa: Iteration 5: New valset pareto front scores: [0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0]
2025/08/12 22:23:54 INFO dspy.teleprompt.gepa.gepa: Iteration 5: Full valset pareto front score: 0.6222222222222222
2025/08/12 22:23:54 INFO dspy.teleprompt.gepa.gepa: Iteration 5: Updated valset pareto front programs: [{0, 1, 2}, {0, 1}, {0, 1, 2}, {0, 1, 2}, {0, 1, 2}, {1, 2}, {0, 1}, {0, 2}, {0, 1, 2}, {0, 1, 2}, {0, 2}, {0, 1, 2}, {0, 1, 2}, {2}, {2}, {0, 2}, {0, 1, 2}, {1, 2}, {0, 1, 2}, {0, 1, 2}, {1, 2}, {0, 1, 2}, {0, 1, 2}, {0, 1, 2}, {0, 1, 2}, {2}, {0, 1, 2}, {0, 1, 2}, {0, 1, 2}, {2}, {0, 1, 2}, {1}, {1, 2}, {2}, {1, 2}, {0, 1, 2}, {0, 1, 2}, {0, 1, 2}, {0, 1, 2}, {0, 1, 2}, {0, 1, 2}, {0, 1, 2}, {0, 1, 2}, {0}, {0, 1, 2}]
2025/08/12 22:23:54 INFO dspy.teleprompt.gepa.gepa: Iteration 5: Best valset aggregate score so far: 0.5333333333333333
2025/08/12 22:23:54 INFO dspy.teleprompt.gepa.gepa: Iteration 5: Best program as per aggregate score on train_val: 2
2025/08/12 22:23:54 INFO dspy.teleprompt.gepa.gepa: Iteration 5: Best program as per aggregate score on valset: 2
2025/08/12 22:23:54 INFO dspy.teleprompt.gepa.gepa: Iteration 5: Best score on valset: 0.5333333333333333
2025/08/12 22:23:54 INFO dspy.teleprompt.gepa.gepa: Iteration 5: Best score on train_val: 0.5333333333333333
2025/08/12 22:23:54 INFO dspy.teleprompt.gepa.gepa: Iteration 5: Linear pareto front program index: 2
2025/08/12 22:23:54 INFO dspy.teleprompt.gepa.gepa: Iteration 5: New program candidate index: 2
2025/08/12 22:23:54 INFO dspy.teleprompt.gepa.gepa: Iteration 6: Selected program 1 score: 0.4222222222222222
Average Metric: 3.00 / 3 (100.0%): 100%|█████████████████████████████████████████████████████████████████████████████████████████████████████████| 3/3 [00:20<00:00,  6.84s/it]
2025/08/12 22:24:14 INFO dspy.evaluate.evaluate: Average Metric: 3.0 / 3 (100.0%)
2025/08/12 22:24:14 INFO dspy.teleprompt.gepa.gepa: Iteration 6: All subsample scores perfect. Skipping.
2025/08/12 22:24:14 INFO dspy.teleprompt.gepa.gepa: Iteration 6: Reflective mutation did not propose a new candidate
2025/08/12 22:24:14 INFO dspy.teleprompt.gepa.gepa: Iteration 7: Selected program 2 score: 0.5333333333333333

Average Metric: 1.00 / 3 (33.3%): 100%|██████████████████████████████████████████████████████████████████████████████████████████████████████████| 3/3 [01:10<00:00, 23.53s/it]
2025/08/12 22:25:25 INFO dspy.evaluate.evaluate: Average Metric: 1.0 / 3 (33.3%)

2025/08/12 22:27:47 INFO dspy.teleprompt.gepa.gepa: Iteration 7: Proposed new text for predict: You will be given one math problem as plain text under a key like “problem.” Your job is to solve it correctly and return:

- reasoning: a concise, logically ordered solution that uses identities/structure to avoid brute force, ends with a quick verification.
- answer: the final requested number/expression only (no extra words).

Formatting:
- Use exactly two top-level fields named “reasoning” and “answer.”
- Keep reasoning succinct but complete. Bullet points are fine.
- The answer field must contain only the final value requested (e.g., 227, 585, 601).

General problem-solving guidance:
- Parse the problem type (e.g., base representation, intersecting families of subsets, avoiding arithmetic progressions, symmetric sums with constraints, ordered tuples counting, classical Euclidean geometry with circles, secants/tangents).
- Always enforce domain constraints (digits in range, no leading zeros, ordered vs unordered, strict increases, segment/angle/parallel/perpendicular statements).
- Use algebraic identities and modular arithmetic to reduce the search space; prefer structural arguments over naive enumeration.
- For “greatest/least” questions, derive tight bounds and give a construction that attains them.
- Prefer classic geometry tools (power of a point, radical axis, homothety, parallelogram/rectangle characterizations, Apollonius/Stewart, cyclic angle/arc relations) over coordinate bashes unless necessary.

Domain-specific strategies and pitfalls:

1) Base-conversion/digit rearrangement:
- Translate positional notation correctly: in base b, (a b c)_b = a·b^2 + b·b + c; in base 10: abc = 100a + 10b + c.
- Enforce digit ranges strictly (e.g., in base 9, digits ∈ {0,…,8}; base-10 three-digit means a ∈ {1,…,9}).
- Use modular constraints to prune:
  • Example: 99a = 71b + 8c ⇒ mod 9 gives b + c ≡ 0 (mod 9).
  • Mod 8: 99 ≡ 3, 71 ≡ 7 ⇒ 3a ≡ 7b (mod 8) ⇒ b ≡ −3a (mod 8).
- Solve within digit bounds and verify numerically.

2) Palindromes across bases:
- Bound the base length by magnitude (e.g., n < 1000 ⇒ octal has 3–4 digits).
- Characterize palindromes:
  • 3-digit octal: (A B A)_8 = 65A + 8B.
  • 4-digit octal: (A B B A)_8 = 513A + 72B (with A ≥ 1).
- Enumerate small parameter ranges and test the other-base palindrome constraint. For “greatest”, check candidates in descending order with justification.

3) Symmetric sums with a + b + c fixed:
- Compress via identities:
  S = ab(a + b) + bc(b + c) + ca(c + a) = (a + b + c)(ab + bc + ca) − 3abc.
- With a + b + c known, convert into a relation among ab + bc + ca and abc.
- Use shifts (a = A + x, etc.) to isolate products like (a−A)(b−A)(c−A), enabling clean counting.
- Count ordered solutions carefully; handle symmetric/degenerate cases precisely.

4) Intersecting families of subsets:
- Intersecting ⇒ every pair has nonempty intersection; empty set excluded.
- Complement pairs: S and S^c cannot both be present.
- Size-based facts: In [n], any two subsets of size > n/2 intersect. For n = 5, all subsets of size ≥ 3 is intersecting (size 16).
- When counting fixed-size families:
  • Casework on minimal set size; 2-sets often control exclusions of complementary 3-sets (for n=5).
  • Avoid double counting by parameterizing canonical overlap patterns.
  • Collections are sets of sets (order doesn’t matter).

5) Avoiding 4-term arithmetic progressions with fixed anchors:
- Bound variable terms by strict increase.
- Pre-eliminate values completing APs with three fixed terms (e.g., 3,4,5,a forbids a=6).
- Count allowed pairs then subtract those completing APs with two fixed endpoints; use Δ divisible by 3 for 4-term APs.
- Ensure integrality, bounds, and no double subtraction.

6) Order statistics with sum and absolute-sum constraints:
- Total positive mass = total negative mass = 1/2.
- Bound x_k via T = n−k+1 largest terms: x_k ≤ (1/2)/T.
- Bound x_l via l smallest terms: x_l ≥ (−1/2)/l.
- Attain bounds by concentrating masses on extremes; verify sums.

7) Two intersecting circles, common tangent, parallel through intersection (key lemmas):
- Setup: Circles ω1, ω2 intersect at P, Q. Common external tangent closer to P touches ω1 at A and ω2 at B. Line through P parallel to AB meets ω1 (again) at X and ω2 (again) at Y.
- Perpendicular-bisector fact: AO1 ⟂ AB and AO1 ⟂ XY; since AO1 is perpendicular to chord XY in ω1, it bisects chord PX. Thus CP = PX/2, where C = AO1 ∩ XY. Similarly, PD = PY/2 for D = BO2 ∩ XY.
- Rectangle: Because AO1 and BO2 are both perpendicular to AB and XY, ABCD is a rectangle with C on XY and D on XY. Along XY, CD = CP + PD = PX/2 + PY/2. Hence AB = CD = (PX + PY)/2.
- Radical axis: PQ is the radical axis of ω1 and ω2, so the locus of equal tangent lengths. Therefore PQ passes through the midpoint M of AB, and AM = MB = AB/2.
- Tangent–secant at M on ω1: MA^2 = MP · MQ with MQ = MP + PQ. Solve for MP.
- Height of trapezoid XABY: The distance between lines AB and XY equals AC (since AO1 ⟂ both). In right triangle with diagonal MP across the rectangle, AC^2 = MP^2 − (AM − CP)^2. Then area[XABY] = 1/2 · (AB + XY) · AC with XY = AB + (PX + PY)/2.
- Quick numeric template (from PX, PY, PQ): AB = (PX + PY)/2, M on PQ gives MP from MA^2 = MP(MP+PQ), AC from MP and AM, area as above.

8) Triangle with midpoint on a chord of circumcircle; “unique Q” with equal angles (key lemmas):
- Apollonius/Stewart for median: In any triangle, with M midpoint of BC,
  AM^2 = (AB^2 + AC^2)/2 − (BC^2)/4. For 13–14–15, AM = √148 = 2√37.
- Power of a point at M wrt circumcircle (of ABC): MB·MC = power(M). If line through M meets the circumcircle at A and P (i.e., M lies on AP), then MA·MP = MB·MC ⇒ PM = (MB·MC)/AM. For 13–14–15: MB = MC = 7 ⇒ PM = 49/√148.
- Unique-angle point implies parallelogram: If M is the midpoint of BC and lies on AP (A,P concyclic with ABC), the unique point Q on AM satisfying ∠PBQ = ∠PCQ is the reflection of P across M. Hence M is the midpoint of PQ (QM = PM), and BQCP is a parallelogram (diagonals BC and PQ bisect each other). Then, with PM < AM, Q lies on segment AM and AQ = AM − PM.
- Alternative ratio route (if needed): In cyclic ABCP, ∠BPA = ∠C and ∠CPA = ∠B. From law of sines in triangles BPQ and CPQ with ∠PBQ = ∠PCQ, deduce BQ/sin C = CQ/sin B ⇒ BQ/CQ = AB/AC. Combine with law of cosines for BQ^2 and CQ^2 in terms of AQ and known sides/AM to solve for AQ.
- 13–14–15 quick facts (if needed): Altitude from A to BC is 12, with foot L splitting BC into BL = 5, LC = 9.

9) Rational “sum of numerator+denominator equal after scaling” (e.g., r and 55r):
- Let r = a/b in lowest terms, d = gcd(55a, b). Then 55r reduces to (55a/d)/(b/d).
- Sum equality: a + b = (55a + b)/d ⇒ a(d − 55) = b(1 − d).
- Since gcd(a,b)=1 and d | 55 and d | b, d ∈ {1,5,11,55}. Test each d, write b = d·k, and solve for a integral and positive; enforce gcd(a,b)=1 to restrict k. Typical resulting solutions: r = 2/25 and r = 5/22; sum is 169/550 (already reduced since 169=13^2 and 550=2·5^2·11).

Quality checks:
- Verify all geometric constructions/length relations (power of a point, radical axis midpoint, perpendicular bisectors) and that derived points lie on specified segments (e.g., Q ∈ AM).
- Keep radicals exact; avoid rounding. Ensure n in m√n is squarefree; for m/√n, present as stated (do not rationalize unless asked).
- For extremal problems, provide both bound and attaining construction.
- For counting, handle ordered vs unordered carefully; include/exclude degenerate cases.
- For AP-avoidance, confirm integrality and bounds; ensure no missed endpoint combinations.

Common pitfalls to avoid (seen in prior attempts):
- Unnecessary coordinate bashes in clean geometric configurations; prefer power of a point, radical axis, and symmetry/reflective arguments.
- Guessing parameters numerically; instead, derive exact values via structure.
- Mishandling expression forms (e.g., incorrectly “forcing” m/√n by ad hoc rationalization). Output the exact requested form.
2025/08/12 22:28:03 INFO dspy.evaluate.evaluate: Average Metric: 2.0 / 3 (66.7%)
2025/08/12 22:30:10 INFO dspy.evaluate.evaluate: Average Metric: 21.0 / 45 (46.7%)
2025/08/12 22:30:10 INFO dspy.teleprompt.gepa.gepa: Iteration 7: Full valset score for new program: 0.4666666666666667
2025/08/12 22:30:10 INFO dspy.teleprompt.gepa.gepa: Iteration 7: Full train_val score for new program: 0.4666666666666667
2025/08/12 22:30:10 INFO dspy.teleprompt.gepa.gepa: Iteration 7: Individual valset scores for new program: [0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1]
2025/08/12 22:30:10 INFO dspy.teleprompt.gepa.gepa: Iteration 7: New valset pareto front scores: [0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1]
2025/08/12 22:30:10 INFO dspy.teleprompt.gepa.gepa: Iteration 7: Full valset pareto front score: 0.7333333333333333
2025/08/12 22:30:10 INFO dspy.teleprompt.gepa.gepa: Iteration 7: Updated valset pareto front programs: [{0, 1, 2, 3}, {0, 1, 3}, {0, 1, 2, 3}, {0, 1, 2, 3}, {0, 1, 2, 3}, {1, 2}, {0, 1, 3}, {0, 2, 3}, {0, 1, 2}, {0, 1, 2, 3}, {0, 2}, {0, 1, 2, 3}, {0, 1, 2, 3}, {2}, {2}, {0, 2, 3}, {0, 1, 2, 3}, {1, 2}, {3}, {0, 1, 2, 3}, {1, 2, 3}, {0, 1, 2}, {0, 1, 2, 3}, {0, 1, 2, 3}, {0, 1, 2, 3}, {2}, {0, 1, 2, 3}, {0, 1, 2}, {0, 1, 2, 3}, {2, 3}, {3}, {1}, {1, 2, 3}, {2}, {1, 2, 3}, {0, 1, 2, 3}, {0, 1, 2, 3}, {0, 1, 2, 3}, {0, 1, 2, 3}, {0, 1, 2, 3}, {3}, {3}, {0, 1, 2, 3}, {0}, {3}]
2025/08/12 22:30:10 INFO dspy.teleprompt.gepa.gepa: Iteration 7: Best valset aggregate score so far: 0.5333333333333333
2025/08/12 22:30:10 INFO dspy.teleprompt.gepa.gepa: Iteration 7: Best program as per aggregate score on train_val: 2
2025/08/12 22:30:10 INFO dspy.teleprompt.gepa.gepa: Iteration 7: Best program as per aggregate score on valset: 2
2025/08/12 22:30:10 INFO dspy.teleprompt.gepa.gepa: Iteration 7: Best score on valset: 0.5333333333333333
2025/08/12 22:30:10 INFO dspy.teleprompt.gepa.gepa: Iteration 7: Best score on train_val: 0.5333333333333333
2025/08/12 22:30:10 INFO dspy.teleprompt.gepa.gepa: Iteration 7: Linear pareto front program index: 2
2025/08/12 22:30:10 INFO dspy.teleprompt.gepa.gepa: Iteration 7: New program candidate index: 3
2025/08/12 22:30:10 INFO dspy.teleprompt.gepa.gepa: Iteration 8: Selected program 2 score: 0.5333333333333333
Average Metric: 1.00 / 3 (33.3%): 100%|██████████████████████████████████████████████████████████████████████████████████████████████████████████| 3/3 [01:38<00:00, 32.75s/it]
2025/08/12 22:31:48 INFO dspy.evaluate.evaluate: Average Metric: 1.0 / 3 (33.3%)

2025/08/12 22:33:09 INFO dspy.teleprompt.gepa.gepa: Iteration 8: Proposed new text for predict: You will be given one math problem as plain text under a key like “problem.” Your job is to solve it correctly and return:

- reasoning: a concise, logically ordered solution that uses structure/identities to avoid brute force and ends with a quick verification.
- answer: the final requested number/expression only (no extra words).

Formatting:
- Use exactly two top-level fields named “reasoning” and “answer.”
- Keep reasoning succinct but complete. Bullet points are fine.
- The answer field must contain only the final value requested (e.g., 227, 585, 601).

General problem-solving guidance:
- Parse the problem type (e.g., base representation, combinatorial configurations with symmetry, coefficient extraction via generating functions, motion with relative velocities).
- Enforce domain constraints (e.g., digit bounds in base problems; “three-digit” means no leading zero; ordered vs unordered; strict increase; geometric constraints).
- Use algebraic identities, modular arithmetic, and symmetry to reduce the search space; prefer structural arguments over naive enumeration.
- For extremal/counting problems, derive tight bounds or clean case structures and justify completeness; for “greatest/least,” give both a bound and a construction.

Quality checks:
- Verify digit/base constraints and final equalities numerically if applicable.
- For counting with cases, ensure cases are disjoint and exhaustive; check complements/forbidden patterns.
- Avoid unjustified heuristics; keep arithmetic exact (avoid floating approximations when exact values are available).
- Finish with a brief verification (e.g., plug back, check constraints).

Domain-specific strategies and pitfalls (including lessons from prior feedback):

A) Rectangles in regular polygons (2-colorings on a regular 2n-gon; e.g., dodecagon):
- Geometry fact: In a regular 2n-gon, a quadrilateral formed by vertices is a rectangle if and only if its two diagonals pass through the center, i.e., it consists of two pairs of opposite vertices.
- For the regular dodecagon (n = 6), partition vertices into 6 opposite pairs. A monochromatic rectangle occurs exactly when there exist two distinct opposite pairs both colored the same color.
- Counting colorings with no monochromatic rectangle reduces to casework on opposite pairs:
  • Case 0: No same-colored opposite pair (each pair has one red and one blue): 2^6.
  • Case 1: Exactly one same-colored opposite pair (choose the pair 6 ways, choose its color 2 ways; remaining 5 pairs opposite-colored): 6·2·2^5.
  • Case 2: Exactly two same-colored opposite pairs, but of different colors (choose the red pair 6 ways and the blue pair 5 ways; remaining 4 pairs opposite-colored): 6·5·2^4.
  • Any additional same-colored opposite pair forces two of the same color, creating a monochromatic rectangle; hence disallowed.
- Do not incorrectly reduce this to classical “monochromatic rectangle in an m×m grid” Ramsey facts; the structure here is specific to opposite pairs in the 12-gon.

B) Coefficient extraction for polynomials of the form (x^N − 1)^k / ∏(x^{m_i} − 1) with 0 < x < 1:
- Use x^n − 1 = −(1 − x^n). For 0 < x < 1, expand 1/(1 − x^{m}) as a geometric series; equivalently,
  P(x) = (1 − x^N)^k / ∏(1 − x^{m_i}).
- Expand the numerator via binomial theorem:
  (1 − x^N)^k = ∑_{r=0}^k (−1)^r C(k, r) x^{Nr}.
- The coefficient of x^t in P(x) equals:
  ∑_{r=0}^{⌊t/N⌋} (−1)^r C(k, r) · a_{t − Nr}, where a_s is the number of nonnegative integer solutions to ∑ m_i n_i = s.
  • If t < N, only r = 0 contributes.
- Count a_s by:
  • Modular pruning to constrain variables (e.g., reduce mod primes dividing some m_i but not others to fix residues).
  • LCM structure: if L = lcm(m_i), then s modulo L often isolates a small adjustment term; decompose s = L·q + r and solve the small r-part explicitly.
  • Stars-and-bars after reducing by gcds (e.g., forcing one variable’s congruence class mod a prime).
- Example (as in t = 2022, N = 2310): since 2022 < 2310, only r = 0 contributes. Use modular constraints (e.g., mod 7) to fix one variable’s residue, reduce to a linear equation with equal coefficients, then apply stars-and-bars to get a binomial coefficient (e.g., C(12, 3) = 220).
- Keep all arithmetic exact; avoid unnecessary cyclotomic factorization unless it simplifies the coefficient logic.

C) Motion in a current (relative velocities; meeting at a point equidistant from two starts):
- Set coordinates with the river as x-axis (east positive), south bank y = 0, north bank y = width W.
- If they aim to a point on the north bank equidistant from their starts, its x-coordinate is the midpoint of the starts: x* = (x1 + x2)/2. If starts are (0,0) and (D,0), target is at (D/2, W).
- Let current be (v_c, 0). Let swimmers’ speeds relative to water be s1, s2. They choose heading vectors so that ground velocity equals path direction.
- Two standard exact methods:
  1) Static-water reduction: If they swim for the same time t and arrive simultaneously, in still water they would both aim at a common point B shifted upstream by v_c·t from the actual target. Distances in still water are s1·t and s2·t. Use Pythagorean relations with horizontal shifts ±v_c·t to form two equations; subtract to solve for D in terms of t, then use vertical distance W to get t and hence D.
     • For example, with speeds 80 and 60, current 14, width 264:
       264^2 + (D/2 − 14t)^2 = (60t)^2,
       264^2 + (D/2 + 14t)^2 = (80t)^2.
       Subtract to get D = 100t; then from the first, solve t (here t = 11/2), giving D exactly (here 550).
  2) Vector components: Let net (ground) velocity components be (±x, y) toward the target. Relative-to-water velocity = ground − current; impose speeds s1, s2 via squared norms to get two equations:
     (x − 14)^2 + y^2 = s2^2 and (x + 14)^2 + y^2 = s1^2; solve for x, y, then t = W / y and D = 2x·t.
- Avoid decimal approximations; keep all values exact.

D) Other included standard tactics:
1) Base-conversion/digit rearrangement:
   - In base b, (a b c)_b = a·b^2 + b·b + c; digits in 0..b−1; no leading zero for “three-digit” base-10.
   - Use modular constraints to prune digit choices and verify within bounds.

2) Palindromes across bases:
   - Bound digit lengths; parameterize palindromic forms; test constraints from the other base; for “greatest,” justify by descending search with structure.

3) Symmetric sums with a + b + c fixed:
   - Use identities like S = (a + b + c)(ab + bc + ca) − 3abc; shift to isolate products; count ordered solutions carefully.

4) Intersecting families of subsets:
   - Intersecting means every pair intersects; empty set excluded.
   - Use complement pairs structure; size > n/2 pigeonhole facts; count families with careful casework on small-set inclusions.

5) Avoiding 4-term APs in increasing sequences with anchors:
   - Pre-eliminate values that form APs with fixed terms; count allowed pairs, subtract those completing APs with endpoints; ensure bounds/integrality and no double-subtraction.

6) Order statistics with sum and absolute-sum constraints:
   - Total positive mass = total negative mass; bound extremes by averaging over slots; construct distributions attaining bounds; verify sums.

Finally:
- Put the clean final numeric result in the “answer” field only.
2025/08/12 22:33:58 INFO dspy.evaluate.evaluate: Average Metric: 3.0 / 3 (100.0%)
2025/08/12 22:36:28 INFO dspy.evaluate.evaluate: Average Metric: 19.0 / 45 (42.2%)
2025/08/12 22:36:28 INFO dspy.teleprompt.gepa.gepa: Iteration 8: Full valset score for new program: 0.4222222222222222
2025/08/12 22:36:28 INFO dspy.teleprompt.gepa.gepa: Iteration 8: Full train_val score for new program: 0.4222222222222222
2025/08/12 22:36:28 INFO dspy.teleprompt.gepa.gepa: Iteration 8: Individual valset scores for new program: [0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0]
2025/08/12 22:36:28 INFO dspy.teleprompt.gepa.gepa: Iteration 8: New valset pareto front scores: [0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1]
2025/08/12 22:36:28 INFO dspy.teleprompt.gepa.gepa: Iteration 8: Full valset pareto front score: 0.7333333333333333
2025/08/12 22:36:28 INFO dspy.teleprompt.gepa.gepa: Iteration 8: Updated valset pareto front programs: [{0, 1, 2, 3, 4}, {0, 1, 3}, {0, 1, 2, 3, 4}, {0, 1, 2, 3, 4}, {0, 1, 2, 3, 4}, {1, 2}, {0, 1, 3, 4}, {0, 2, 3, 4}, {0, 1, 2, 4}, {0, 1, 2, 3, 4}, {0, 2, 4}, {0, 1, 2, 3, 4}, {0, 1, 2, 3, 4}, {2}, {2, 4}, {0, 2, 3, 4}, {0, 1, 2, 3, 4}, {1, 2, 4}, {3, 4}, {0, 1, 2, 3, 4}, {1, 2, 3, 4}, {0, 1, 2, 4}, {0, 1, 2, 3, 4}, {0, 1, 2, 3, 4}, {0, 1, 2, 3, 4}, {2}, {0, 1, 2, 3, 4}, {0, 1, 2}, {0, 1, 2, 3, 4}, {2, 3}, {3}, {1, 4}, {1, 2, 3}, {2}, {1, 2, 3}, {0, 1, 2, 3, 4}, {0, 1, 2, 3, 4}, {0, 1, 2, 3, 4}, {0, 1, 2, 3, 4}, {0, 1, 2, 3, 4}, {3}, {3}, {0, 1, 2, 3, 4}, {0}, {3}]
2025/08/12 22:36:28 INFO dspy.teleprompt.gepa.gepa: Iteration 8: Best valset aggregate score so far: 0.5333333333333333
2025/08/12 22:36:28 INFO dspy.teleprompt.gepa.gepa: Iteration 8: Best program as per aggregate score on train_val: 2
2025/08/12 22:36:28 INFO dspy.teleprompt.gepa.gepa: Iteration 8: Best program as per aggregate score on valset: 2
2025/08/12 22:36:28 INFO dspy.teleprompt.gepa.gepa: Iteration 8: Best score on valset: 0.5333333333333333
2025/08/12 22:36:28 INFO dspy.teleprompt.gepa.gepa: Iteration 8: Best score on train_val: 0.5333333333333333
2025/08/12 22:36:28 INFO dspy.teleprompt.gepa.gepa: Iteration 8: Linear pareto front program index: 2
2025/08/12 22:36:28 INFO dspy.teleprompt.gepa.gepa: Iteration 8: New program candidate index: 4
2025/08/12 22:36:28 INFO dspy.teleprompt.gepa.gepa: Iteration 9: Selected program 2 score: 0.5333333333333333
Average Metric: 1.00 / 3 (33.3%): 100%|██████████████████████████████████████████████████████████████████████████████████████████████████████████| 3/3 [02:28<00:00, 49.56s/it]
2025/08/12 22:38:57 INFO dspy.evaluate.evaluate: Average Metric: 1.0 / 3 (33.3%)

2025/08/12 22:40:06 INFO dspy.teleprompt.gepa.gepa: Iteration 9: Proposed new text for predict: Your job: Solve one math problem and return exactly two top-level fields:
- reasoning: a concise, logically ordered solution that leverages structure (identities, modular arithmetic, symmetry) and ends with a quick verification.
- answer: the final requested number/expression only (no extra words or formatting).

Formatting:
- Use exactly the two fields “reasoning” and “answer.”
- Keep the reasoning succinct but complete; bullet points are fine.
- The answer field must contain only the final value requested (e.g., 227, 585, 601).

General problem-solving guidance:
- Identify the problem type (e.g., base/digit relations, repeating decimals, palindromes across bases, symmetric sums with fixed totals, intersecting subset families, avoiding arithmetic progressions, order statistics with absolute-sum constraints, floor-sum optimization).
- Enforce domain constraints strictly (digit ranges in base b; no leading zero where prohibited; ordered vs unordered; strictly increasing where specified).
- Prefer structural arguments over brute force: factor, complete squares, use modular arithmetic, apply symmetry/identities, bound and construct extremals.
- For greatest/least questions, derive a tight bound and exhibit a construction attaining it.
- Verify at the end (e.g., plug back into conditions, check mod constraints, count consistency).

Domain-specific strategies and pitfalls:

A) Base conversion/digit equations:
- Translate positional notation correctly: (a b c)_b = a·b^2 + b·b + c; in base 10: abc = 100a + 10b + c.
- Enforce digit ranges: in base 9, digits ∈ {0,…,8}; leading decimal digits ∈ {1,…,b−1} if specified.
- Use modular constraints to prune search (e.g., mod 9, mod 8).
- Solve within digit bounds and verify numerically.

B) Repeating decimals 0.\overline{abcd} and counting reduced numerators:
- 0.\overline{abcd} = m/9999 with m ∈ {1,…,9999}, and 9999 = 3^2·11·101.
- Reduced numerator is x = m / gcd(m, 9999). The mapping m → x is not injective across m, so do NOT apply ∑_{d|n} φ(d) = n to count distinct numerators; that identity counts reduced fractions by denominator divisors, not distinct reduced numerators arising from a bounded numerator range.
- Characterize x by existence of a divisor y | 9999 such that y ≥ x and gcd(x, y) = 1 (since m = x·(9999/y) with gcd(x, y) = 1 and m ≤ 9999).
- Count x via casework on divisibility by the primes of 9999 with correct cancellation:
  • Track whether x is divisible by 3, 9, 11, 101 and what remains in the denominator after canceling to keep m ≤ 9999.
  • “3 vs 9” matters: if x is divisible by 3 but not 9, you can cancel one factor of 3; if divisible by 9, you may cancel two.
  • Use inclusion–exclusion over the allowable ranges after dividing 9999 by the canceled factors.
- Example outline (for abcd length 4): Totals break into mutually exclusive cases:
  • gcd(x, 9999) = 1 contributes φ(9999) = 6000 (note: mod-1000 suffices if only remainder asked).
  • x divisible by 3 only (not 9,11,101): count using floor(1111/3) minus multiples of 33 and 303, etc.
  • x divisible by 11 only (not 3,101): analogously with 9999/11 = 909.
  • x divisible by 3 and 11 (not 101): work with 9999/99 = 101.
  • x divisible by 101: often yields 0 due to range constraints.
- Sum cases carefully and reduce modulo as requested. Verify small endpoint counts and complementary exclusions.

C) Palindromes across bases:
- Bound number of digits by magnitude; characterize k-digit palindromes in base b (e.g., octal 3-digit ABA: 65A + 8B; 4-digit ABBA: 513A + 72B with A ≥ 1).
- Test other-base palindrome constraints for candidates in justified order.

D) Symmetric sums with a + b + c fixed:
- Use identities to compress expressions, e.g.,
  S = ab(a + b) + bc(b + c) + ca(c + a) = (a + b + c)(ab + bc + ca) − 3abc.
- For counting, use shifts like a = A + x to isolate products and deduce factorization constraints.

E) Intersecting families of subsets:
- Intersecting means every pair intersects; empty set excluded.
- Complement pairs cannot both be present. For n odd, both a star and “all subsets of size > n/2” have size 2^{n−1}.
- When counting fixed-size collections, case on smallest set sizes and their overlaps; avoid double-counting via canonical patterns.

F) Avoiding 4-term arithmetic progressions with fixed anchors:
- Restrict variable terms by monotonicity.
- Pre-eliminate values that create 4-term APs with fixed endpoints.
- Count allowed pairs then subtract those forming APs with two endpoints; if endpoints differ by Δ, then Δ must be divisible by 3 for a 4-term AP; solve for integer interior terms within bounds. Avoid double subtraction.

G) Order statistics with sum and absolute-sum constraints:
- Sum |x_i| = 1, sum x_i = 0 ⇒ total positive mass = total negative mass = 1/2.
- For maximizing x_k, with T = n−k+1 largest positions, x_k ≤ (1/2)/T; for minimizing x_l, x_l ≥ −(1/2)/l.
- Achieve bounds by concentrating masses evenly on those positions; set the rest to 0. Verify sums.

H) Systems with square-root bilinear forms (e.g., √(2x−xy)+√(2y−xy)=const):
- Factor under roots: √(x(2−y)) + √(y(2−x)).
- Two robust approaches:
  • Algebraic: Let A=1−x, B=1−y, C=1−z; after squaring, obtain identities like
    (AB−1/2)^2 = (1−A^2)(1−B^2), (BC)^2 = (1−B^2)(1−C^2), (AC+1/2)^2 = (1−A^2)(1−C^2),
    which yield relations such as B^2 + C^2 = 1 and lead to a clean evaluation of [(1−x)(1−y)(1−z)]^2 = 1/32.
  • Trig: Set x=2cos^2 α, etc., convert sums to sin(α+β)=k/2, solve angles, and compute the desired product, often reducing to standard trig constants (e.g., 1/32).
- Conclude with exact value and minimal arithmetic.

I) Sums of floors of quadratic forms with a parameter (e.g., U = ∑⌊(n^2 − na)/5⌋):
- Separate continuous part U' = (1/5)(∑n^2 − a∑n) and fractional correction:
  U = U' − ∑{(n^2 − na)/5}.
- Choose a to minimize |U| by canceling the main term: with N terms, a ≈ (2N+1)/3; for N=2023, a = 1349 exactly makes U' = 0.
- Argue uniqueness: the range constraint (e.g., −1000 < U < 1000) plus the step size |ΔU| = (∑n)/5 when a changes by 1 ensures only that a works.
- Compute the correction via residue cycles:
  • Let r_n be (n^2 − na) mod 5; fractional part is r_n/5.
  • Count residues modulo 5 over 1..N: for N=2023, counts are 405 each for residues 1,2,3 and 404 each for residues 0,4.
  • With a ≡ 4 (mod 5), the pattern r(n) for n ≡ 0,1,2,3,4 is (0,2,1,2,0); sum of fractional parts equals (405·(2+1+2))/5 = 405, hence U = −405.
- Return the requested combination (e.g., a+U) and verify bounds.

Quality checks:
- Verify digit/base constraints and equalities numerically if applicable.
- For extremal problems, provide both bound and construction achieving it.
- For counting, be explicit about ordered vs unordered and avoid double counting.
- For AP-avoidance, check integrality and bounds for all endpoint combinations.
- For floor sums, ensure residue counts over partial cycles (non-multiples of modulus) are correct.

Final step:
- Put the clean final numeric result in the “answer” field only. No extra text or formatting.
2025/08/12 22:40:42 INFO dspy.evaluate.evaluate: Average Metric: 2.0 / 3 (66.7%)
2025/08/12 22:45:58 INFO dspy.evaluate.evaluate: Average Metric: 16.0 / 45 (35.6%)
2025/08/12 22:45:58 INFO dspy.teleprompt.gepa.gepa: Iteration 9: Full valset score for new program: 0.35555555555555557
2025/08/12 22:45:58 INFO dspy.teleprompt.gepa.gepa: Iteration 9: Full train_val score for new program: 0.35555555555555557
2025/08/12 22:45:58 INFO dspy.teleprompt.gepa.gepa: Iteration 9: Individual valset scores for new program: [0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0]
2025/08/12 22:45:58 INFO dspy.teleprompt.gepa.gepa: Iteration 9: New valset pareto front scores: [0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1]
2025/08/12 22:45:58 INFO dspy.teleprompt.gepa.gepa: Iteration 9: Full valset pareto front score: 0.7555555555555555
2025/08/12 22:45:58 INFO dspy.teleprompt.gepa.gepa: Iteration 9: Updated valset pareto front programs: [{0, 1, 2, 3, 4, 5}, {0, 1, 3}, {0, 1, 2, 3, 4, 5}, {0, 1, 2, 3, 4, 5}, {0, 1, 2, 3, 4, 5}, {1, 2}, {0, 1, 3, 4}, {0, 2, 3, 4, 5}, {0, 1, 2, 4, 5}, {0, 1, 2, 3, 4, 5}, {0, 2, 4}, {0, 1, 2, 3, 4, 5}, {0, 1, 2, 3, 4, 5}, {2, 5}, {2, 4}, {0, 2, 3, 4, 5}, {0, 1, 2, 3, 4, 5}, {1, 2, 4}, {3, 4}, {0, 1, 2, 3, 4, 5}, {1, 2, 3, 4, 5}, {0, 1, 2, 4, 5}, {0, 1, 2, 3, 4, 5}, {0, 1, 2, 3, 4, 5}, {0, 1, 2, 3, 4, 5}, {2}, {0, 1, 2, 3, 4, 5}, {0, 1, 2}, {0, 1, 2, 3, 4, 5}, {2, 3}, {3}, {1, 4}, {1, 2, 3}, {2}, {1, 2, 3, 5}, {0, 1, 2, 3, 4, 5}, {5}, {0, 1, 2, 3, 4, 5}, {0, 1, 2, 3, 4, 5}, {0, 1, 2, 3, 4, 5}, {3}, {3}, {0, 1, 2, 3, 4, 5}, {0}, {3}]
2025/08/12 22:45:58 INFO dspy.teleprompt.gepa.gepa: Iteration 9: Best valset aggregate score so far: 0.5333333333333333
2025/08/12 22:45:58 INFO dspy.teleprompt.gepa.gepa: Iteration 9: Best program as per aggregate score on train_val: 2
2025/08/12 22:45:58 INFO dspy.teleprompt.gepa.gepa: Iteration 9: Best program as per aggregate score on valset: 2
2025/08/12 22:45:58 INFO dspy.teleprompt.gepa.gepa: Iteration 9: Best score on valset: 0.5333333333333333
2025/08/12 22:45:58 INFO dspy.teleprompt.gepa.gepa: Iteration 9: Best score on train_val: 0.5333333333333333
2025/08/12 22:45:58 INFO dspy.teleprompt.gepa.gepa: Iteration 9: Linear pareto front program index: 2
2025/08/12 22:45:58 INFO dspy.teleprompt.gepa.gepa: Iteration 9: New program candidate index: 5
2025/08/12 22:45:58 INFO dspy.teleprompt.gepa.gepa: Iteration 10: Selected program 3 score: 0.4666666666666667
Average Metric: 1.00 / 3 (33.3%): 100%|██████████████████████████████████████████████████████████████████████████████████████████████████████████| 3/3 [01:09<00:00, 23.23s/it]
2025/08/12 22:47:07 INFO dspy.evaluate.evaluate: Average Metric: 1.0 / 3 (33.3%)

2025/08/12 22:49:24 INFO dspy.teleprompt.gepa.gepa: Iteration 10: Proposed new text for predict: You will receive one math problem as plain text under the key “problem.” Solve it and return:

- reasoning: a concise, logically ordered solution that leverages structure/identities (not brute force), ending with a quick verification.
- answer: the final requested number/expression only (no extra words).

Formatting:
- Use exactly two top-level fields named “reasoning” and “answer.”
- Keep reasoning succinct but complete; bullet points are fine.
- The answer field must contain only the final value requested (e.g., 227, 585, 601).

General approach:
- Parse the problem type (e.g., base representation, subset families, AP-avoidance, symmetric sums, ordered tuples, classical Euclidean geometry, polynomial/root structure).
- Enforce domain constraints strictly (digit ranges, no leading zeros, ordered vs unordered, strict increases, integer bounds, segment/angle conditions).
- Use modular arithmetic, algebraic identities, and structure to reduce cases; avoid naive enumeration.
- For extremal problems, prove bounds tight and provide a construction attaining them.
- Prefer classic geometry tools (power of a point, radical axis, homothety, cyclic angle relations, Ptolemy, Stewart/Apollonius, right-triangle/chord/arc formulas) over coordinates unless unavoidable.
- End with a quick verification (substitution/check of conditions).

Quality checks and common pitfalls:
- Respect integer/divisibility constraints; ensure parameters that represent counts are integers and within bounds.
- For polynomials with integer coefficients: if a monic cubic has two integer roots, the third is also integer (Vieta).
- For “unique solution” conditions, consider multiplicities and all structural cases that yield uniqueness (not just one pattern).
- Keep radicals exact and simplify; ensure squarefree radicands if appropriate.
- When counting, avoid double-counting and handle degenerate/symmetric cases precisely.

Domain-specific strategies and templates:

1) Base conversion/digit rearrangements:
- Positional notation: (a b c)_b = a·b^2 + b·b + c; in base 10: abc = 100a + 10b + c.
- Enforce digit ranges (e.g., base 9 digits ∈ {0,…,8}; leading digit ≥ 1 if indicated).
- Use modular constraints to prune and relate digits.

2) Palindromes across bases:
- Bound number of digits by magnitude in each base.
- Characterize palindromes (e.g., octal 3-digit (A B A)_8 = 65A + 8B; 4-digit (A B B A)_8 = 513A + 72B).
- For “greatest,” test candidates in decreasing order with structural justification.

3) Symmetric sums with fixed a + b + c:
- Use identities such as S = (a + b + c)(ab + bc + ca) − 3abc.
- Shift variables to isolate symmetric products, e.g., (a−A)(b−A)(c−A), and count/solve cleanly.

4) Intersecting families of subsets:
- Intersecting ⇒ every pair intersects; empty set excluded.
- Complement pairs cannot both be present.
- Use size-based facts (e.g., in [5], all subsets of size ≥ 3 is intersecting).
- Parameterize by minimal set sizes; avoid double counting.

5) Avoiding 4-term arithmetic progressions with fixed anchors:
- Maintain strict increase.
- Pre-eliminate values that complete APs with anchored terms.
- Count allowed pairs, then subtract those completing 4-term APs (Δ divisible by 3 consideration).
- Ensure integrality and avoid double subtraction.

6) Order statistics with sum and absolute-sum constraints:
- Total positive mass = total negative mass = 1/2.
- Bound kth values via concentration on extremes; verify with constructions that attain bounds.

7) Two intersecting circles with common tangent and parallel through intersection (key lemmas):
- Setup: ω1, ω2 intersect at P,Q; common external tangent touches at A,B. Line through P parallel to AB meets ω1 at X and ω2 at Y.
- Facts: AO1 ⟂ AB and XY, so it bisects chord PX in ω1; similarly for BO2 and PY.
- Rectangle ABCD with C,D on XY; AB = (PX + PY)/2.
- Radical axis PQ passes through midpoint M of AB; AM = MB. Tangent–secant at M on ω1: MA^2 = MP·MQ = MP(MP+PQ). Solve MP, then trapezoid height and area as needed.

8) Triangle with midpoint on a chord of circumcircle; “unique Q” with equal angles:
- Median length via Apollonius/Stewart.
- Power of a point from the midpoint on the chord through A gives PM = (MB·MC)/AM.
- Unique-angle point Q is reflection of P across M (M is midpoint of PQ); BQCP is a parallelogram; deduce AQ = AM − PM.

9) Rational “sum equal after scaling” (e.g., r and 55r):
- Let r = a/b in lowest terms; reduce 55r accordingly via d = gcd(55a, b).
- Solve a(d − 55) = b(1 − d) over d ∈ {1,5,11,55}, enforcing coprimality; sum or count as required.

10) Ratio/crowd problems with added people (bus arrives):
- Let initial total N and adult count A with A/N a given fraction ⇒ N is multiple of denominator, A of numerator.
- After adding T people (e.g., 50), new total N+T must be a multiple of the new denominator.
- If adults on bus = a (0 ≤ a ≤ T), then A + a = new fraction × (N + T). Parameterize N by LCM of denominators; deduce a must be integer in range; minimize/maximize adults as requested. Verify both initial and final ratios.

11) Cubic polynomials with unique integer m ≠ 2 such that p(m) = p(2) (critical details):
- Let p(x) = x^3 + ax^2 + bx + c with a, b, c in a bounded integer range (e.g., −20…20).
- Consider q(x) = p(x) − p(2), a monic cubic with integer coefficients and q(2) = 0.
- Because q has at least two integer roots (2 and m), the third root is an integer (Vieta).
- “Unique integer m ≠ 2 with p(m)=p(2)” arises in exactly two multiplicity patterns:
  • Case A: q(x) = (x − 2)^2 (x − m), m ≠ 2.
    - Expand: q(x) = x^3 − (m + 4)x^2 + (4 + 4m)x − 4m.
    - Thus p(x) = x^3 + ax^2 + bx + c with:
      a = −(m + 4), b = 4 + 4m, c = −4m + p(2).
    - Coefficient bounds impose |b| ≤ B and |a| ≤ A; typically |4 + 4m| ≤ 20 ⇒ −6 ≤ m ≤ 4; exclude m = 2. For each valid m, c can vary freely over its 41 allowed integer values (since c and p(2) shift together), giving 41 polynomials per m.
  • Case B: q(x) = (x − 2)(x − m)^2, m ≠ 2.
    - Expand: q(x) = x^3 − (2m + 2)x^2 + (m^2 + 4m)x − 2m^2.
    - Thus a = −2m − 2, b = m^2 + 4m, c = p(2) − 2m^2.
    - Coefficient bounds typically yield m ∈ {−6, −5, −4, −3, −2, −1, 0, 1}. Again 41 choices for c per m.
- Alternatively (algebraic route): From p(m) = p(2) and m ≠ 2, divide by (m − 2) to get the quadratic in m:
  m^2 + (a + 2)m + (4 + 2a + b) = 0.
  • Case 1 (one m): discriminant zero ⇒ (a − 2)^2 = 4(4 + b); ensure m ≠ 2; count valid (a, b).
  • Case 2 (two m’s with one equal to 2): enforce m = 2 is a root ⇒ 4a + b + 12 = 0, and exclude the double-root outlier (a, b) = (−6, 12).
  • In both cases, c is free (41 values).
- Whichever route, count both cases and sum; do not forget c contributes an independent factor of 41 when unconstrained by the equality.

12) Geometry with two externally tangent circles and a third circle Ω through the centers (AIME 2022 II #15 pattern):
- Setup: ω1 ⟂ ω2 externally tangent at T with centers O1, O2 and O1O2 = r1 + r2. Ω passes through O1 and O2 and meets ω2 at A, D and ω1 at B, C. Hexagon ABO1CDO2 is convex with given chord lengths AB and CD.
- Symmetry via reflection: Reflect A, B across the perpendicular bisector of O1O2 to A′, B′ on Ω. Then:
  • Quadrilaterals ABO1O2 and B′A′O2O1 are congruent; hexagons ABO1CDO2 and A′B′O1CDO2 have equal area.
  • A′B′CD is an isosceles trapezoid; also B′D = O1O2 and A′C = O1O2 (often equal to the center distance).
  • Use Ptolemy on cyclic A′B′CD to solve unknown diagonals (e.g., A′D = B′C = √193 when AB = 2, CD = 16, O1O2 = 15), and compute height (often 12) ⇒ area of trapezoid = 1/2 · height · (AB + CD).
  • Radii: let O1C = O2A′ = r1 and O2D = O1B′ = r2, with r1 + r2 = O1O2. From triangle A′O2D and the angle from the trapezoid (e.g., cos = −3/5, sin = 4/5), apply Law of Cosines to get r1r2 (e.g., r1r2 = 40), so area of each triangle A′O1C and B′O2D is (1/2) r1r2 sin(angle) (e.g., 16 each). Sum areas: trapezoid + two triangles (e.g., 108 + 16 + 16 = 140).
- Alternative trig on Ω: Let 2θ = ∠O1OO2, α = ∠OOB, β = ∠OOA. Then O1O2 = 2R sin θ, AB = 2R sin((2θ − α − β)/2), CD = 2R sin((2π − 2θ − α − β)/2). From AB, CD, and O1O2, solve for cos((α+β)/2), sin((α+β)/2), 2R cos θ; then deduce height and areas.

Verification:
- Substitute/check final values against all given constraints (lengths/ratios/root-counts/uniqueness).
- Ensure coefficient bounds and integrality are satisfied; confirm exclusions (e.g., m ≠ 2).

Output exactly two fields: “reasoning” and “answer”.
2025/08/12 22:50:00 INFO dspy.evaluate.evaluate: Average Metric: 3.0 / 3 (100.0%)
2025/08/12 22:52:15 INFO dspy.evaluate.evaluate: Average Metric: 18.0 / 45 (40.0%)
2025/08/12 22:52:15 INFO dspy.teleprompt.gepa.gepa: Iteration 10: Full valset score for new program: 0.4
2025/08/12 22:52:15 INFO dspy.teleprompt.gepa.gepa: Iteration 10: Full train_val score for new program: 0.4
2025/08/12 22:52:15 INFO dspy.teleprompt.gepa.gepa: Iteration 10: Individual valset scores for new program: [0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0]
2025/08/12 22:52:15 INFO dspy.teleprompt.gepa.gepa: Iteration 10: New valset pareto front scores: [0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1]
2025/08/12 22:52:15 INFO dspy.teleprompt.gepa.gepa: Iteration 10: Full valset pareto front score: 0.7777777777777778
2025/08/12 22:52:15 INFO dspy.teleprompt.gepa.gepa: Iteration 10: Updated valset pareto front programs: [{0, 1, 2, 3, 4, 5, 6}, {0, 1, 3}, {0, 1, 2, 3, 4, 5, 6}, {0, 1, 2, 3, 4, 5, 6}, {0, 1, 2, 3, 4, 5, 6}, {1, 2}, {0, 1, 3, 4}, {0, 2, 3, 4, 5, 6}, {0, 1, 2, 4, 5, 6}, {6}, {0, 2, 4}, {0, 1, 2, 3, 4, 5, 6}, {0, 1, 2, 3, 4, 5, 6}, {2, 5, 6}, {2, 4, 6}, {0, 2, 3, 4, 5, 6}, {0, 1, 2, 3, 4, 5, 6}, {1, 2, 4, 6}, {3, 4}, {0, 1, 2, 3, 4, 5, 6}, {1, 2, 3, 4, 5, 6}, {0, 1, 2, 4, 5}, {0, 1, 2, 3, 4, 5, 6}, {0, 1, 2, 3, 4, 5, 6}, {0, 1, 2, 3, 4, 5, 6}, {2}, {0, 1, 2, 3, 4, 5, 6}, {0, 1, 2}, {0, 1, 2, 3, 4, 5, 6}, {2, 3}, {3}, {1, 4}, {1, 2, 3}, {2}, {1, 2, 3, 5, 6}, {0, 1, 2, 3, 4, 5, 6}, {5}, {0, 1, 2, 3, 4, 5, 6}, {0, 1, 2, 3, 4, 5, 6}, {0, 1, 2, 3, 4, 5, 6}, {3}, {3, 6}, {0, 1, 2, 3, 4, 5, 6}, {0}, {3}]
2025/08/12 22:52:15 INFO dspy.teleprompt.gepa.gepa: Iteration 10: Best valset aggregate score so far: 0.5333333333333333
2025/08/12 22:52:15 INFO dspy.teleprompt.gepa.gepa: Iteration 10: Best program as per aggregate score on train_val: 2
2025/08/12 22:52:15 INFO dspy.teleprompt.gepa.gepa: Iteration 10: Best program as per aggregate score on valset: 2
2025/08/12 22:52:15 INFO dspy.teleprompt.gepa.gepa: Iteration 10: Best score on valset: 0.5333333333333333
2025/08/12 22:52:15 INFO dspy.teleprompt.gepa.gepa: Iteration 10: Best score on train_val: 0.5333333333333333
2025/08/12 22:52:15 INFO dspy.teleprompt.gepa.gepa: Iteration 10: Linear pareto front program index: 2
2025/08/12 22:52:15 INFO dspy.teleprompt.gepa.gepa: Iteration 10: New program candidate index: 6
2025/08/12 22:52:15 INFO dspy.teleprompt.gepa.gepa: Iteration 11: Selected program 0 score: 0.37777777777777777
Average Metric: 3.00 / 3 (100.0%): 100%|█████████████████████████████████████████████████████████████████████████████████████████████████████████| 3/3 [00:24<00:00,  8.24s/it]
2025/08/12 22:52:40 INFO dspy.evaluate.evaluate: Average Metric: 3.0 / 3 (100.0%)
2025/08/12 22:52:40 INFO dspy.teleprompt.gepa.gepa: Iteration 11: All subsample scores perfect. Skipping.
2025/08/12 22:52:40 INFO dspy.teleprompt.gepa.gepa: Iteration 11: Reflective mutation did not propose a new candidate
2025/08/12 22:52:40 INFO dspy.teleprompt.gepa.gepa: Iteration 12: Selected program 2 score: 0.5333333333333333

Average Metric: 1.00 / 3 (33.3%): 100%|██████████████████████████████████████████████████████████████████████████████████████████████████████████| 3/3 [01:20<00:00, 26.70s/it]
2025/08/12 22:54:00 INFO dspy.evaluate.evaluate: Average Metric: 1.0 / 3 (33.3%)

2025/08/12 22:56:06 INFO dspy.teleprompt.gepa.gepa: Iteration 12: Proposed new text for predict: You will be given one math problem as plain text under a key like “problem.” Your job is to solve it correctly and return:

- reasoning: a concise, logically ordered solution that uses identities/structure to avoid brute force, ends with a quick verification.
- answer: the final requested number/expression only (no extra words).

Formatting:
- Use exactly two top-level fields named “reasoning” and “answer.”
- Keep reasoning succinct but complete. Bullet points are fine.
- The answer field must contain only the final value requested (e.g., 227, 585, 601). If the problem implies the answer is an integer (e.g., asks for m+n, or a contest-style “Find n”), ensure you simplify to the integer; if you get a non-integer, recheck your work.

General problem-solving guidance:
- Parse the problem type (e.g., base representation, intersecting families of subsets, avoiding arithmetic progressions, symmetric sums with constraints, ordered tuples counting, expected value with optimal strategy, planar geometry with parallels/similarity).
- Always enforce domain constraints (e.g., digit ranges for bases; reduce rational numbers to lowest terms before summing m+n; ordered vs unordered; strict increase; no leading zero).
- Use algebraic identities and modular arithmetic to reduce the search space; prefer structural arguments over naive enumeration.
- For “greatest/least” questions, derive tight bounds and give a construction that attains them.
- For “m+n” style answers: reduce the fraction m/n to lowest terms first, then compute m+n.

Domain-specific strategies and pitfalls:

1) Base-conversion/digit rearrangement:
- In base b, (a b c)_b = a·b^2 + b·b + c; in base 10: abc = 100a + 10b + c.
- Enforce digit ranges (0..b−1) and leading-digit constraints.
- Use modular constraints to prune (e.g., mod 9, mod 8) and then solve within bounds; verify numerically.

2) Palindromes across bases:
- Bound the number of digits by magnitude.
- Characterize palindromes:
  • 3-digit octal: (A B A)_8 = 65A + 8B.
  • 4-digit octal: (A B B A)_8 = 513A + 72B (A ≥ 1).
- Enumerate small parameter ranges and test constraints; for “greatest,” check candidates in descending order with justification.

3) Symmetric sums with a + b + c fixed:
- Compress expressions using identities:
  S = ab(a + b) + bc(b + c) + ca(c + a) = (a + b + c)(ab + bc + ca) − 3abc.
- With a + b + c known, convert to relations among ab + bc + ca and abc.
- Use shifts to isolate products like (a−A)(b−A)(c−A) and deduce factorization constraints; count ordered solutions carefully.

4) Intersecting families of subsets:
- Intersecting means every pair has nonempty intersection; the empty set cannot be included.
- Complement pairs: S and S^c cannot both be present.
- Use size-based facts: in [n], any two subsets of size > n/2 intersect. For n = 5, any two subsets of size ≥ 3 intersect.
- When counting families of fixed size, parameterize by small-set structure (e.g., 2-sets) and avoid double counting; collections are unordered.

5) Avoiding 4-term arithmetic progressions with fixed anchors:
- Use strict increase bounds for variables.
- Pre-eliminate values that create a 4-term AP with fixed terms.
- Count allowed pairs then subtract those that complete APs with two fixed endpoints. If endpoints differ by Δ, then Δ must be divisible by 3; solve for integer interior terms within bounds.
- Avoid double subtraction; respect bounds and monotonicity.

6) Order statistics with sum and absolute-sum constraints:
- Total positive mass equals total negative mass (both = 1/2).
- For maximizing x_k (k near the top): if T = n − k + 1 largest terms, sum ≥ T·x_k ⇒ x_k ≤ (1/2)/T.
- For minimizing x_l: if l smallest terms, sum ≤ l·x_l ⇒ x_l ≥ (−1/2)/l.
- To attain bounds, concentrate masses evenly on exactly those positions; verify sums and |sums|.

7) Optimal guessing with finite multiset of colors (e.g., 3 red and 3 black revealed in random order):
- Optimal policy: at each step, guess the color with more remaining cards (equivalently, the color that has occurred less so far). If tied, either color gives probability 1/2.
- Two efficient solution methods:
  a) Linearity of expectation by stage:
     • At each stage, determine the split (a vs b) of remaining colors under the optimal policy and compute the probability of a correct guess as a/(a+b); account for branching where needed by conditioning on prior correctness. Sum stage probabilities over all stages.
  b) Dynamic programming N(a,b):
     • N(a,b) = max over guessing red/black of the appropriate expectation:
       N(a,b) = (a/(a+b))·(1 + N(a−1,b)) + (b/(a+b))·N(a,b−1) if guessing red is optimal (a ≥ b); similarly for black if b ≥ a.
     • Base cases: N(0,b) = b, N(a,0) = a; symmetry: N(a,b) = N(b,a).
- Sanity check: expected total lies between min and max possible correct guesses; for (3,3), the optimal EV is 41/10.

8) Equilateral hexagon with opposite sides parallel; triangle from extensions of AB, CD, EF:
- Let ABCDEF be convex, equilateral with side length x, and opposite sides parallel (AB ∥ DE, BC ∥ EF, CD ∥ FA).
- Let P = AB ∩ CD, Q = CD ∩ EF, R = EF ∩ AB. The lines AB, CD, EF form triangle PQR with given side lengths (e.g., QR, RP, PQ).
- Key similarity facts (due to parallelism):
  • Small “corner” triangles like BCP are similar to RQP (BC ∥ RQ, BP ∥ RP), giving x/BP = QR/RP ⇒ BP in terms of x and given sides.
  • Similarly, AFR ∼ PQR gives x/AR = PQ/RP ⇒ AR in terms of x.
- Use a linear relation along one side of triangle PQR composed of contiguous segments from the hexagon construction (e.g., RA + AB + BP = RP). Substitute BP, AR, AB = x to solve for x.
- Example: if QR = 200, RP = 300, PQ = 240, then:
  • From BCP ∼ RQP: x/BP = 200/300 ⇒ BP = 3x/2.
  • From AFR ∼ PQR: x/AR = 240/300 ⇒ AR = 5x/4.
  • RA + AB + BP = 300 ⇒ (5x/4) + x + (3x/2) = 300 ⇒ x = 80.
- Prefer this similarity/intercept approach over ad hoc vector algebra; verify the found x fits all proportions.

9) Linear systems from logarithms:
- Let a = log_b x, etc., to convert multiplicative equations into linear systems in a,b,c.
- Sum or combine equations to solve quickly; compute the target linear form (e.g., 4a+3b+2c) and take absolute value if required.
- Reduce rational results to lowest terms before computing m+n.

Quality checks:
- Verify digit/base constraints and final equalities numerically when applicable.
- For extremal problems, provide both a tight bound and an explicit construction achieving it.
- For expected values, ensure the policy is optimal (majority-remaining rule) and the computed EV is in a plausible range.
- For “m+n” answers, reduce to lowest terms before summing; ensure the final answer is an integer if the problem implies it.

Finally:
- Put the clean final result in the “answer” field only (no units/symbols/words).
2025/08/12 22:56:30 INFO dspy.evaluate.evaluate: Average Metric: 3.0 / 3 (100.0%)
2025/08/12 23:00:10 INFO dspy.evaluate.evaluate: Average Metric: 19.0 / 45 (42.2%)
2025/08/12 23:00:10 INFO dspy.teleprompt.gepa.gepa: Iteration 12: Full valset score for new program: 0.4222222222222222
2025/08/12 23:00:10 INFO dspy.teleprompt.gepa.gepa: Iteration 12: Full train_val score for new program: 0.4222222222222222
2025/08/12 23:00:10 INFO dspy.teleprompt.gepa.gepa: Iteration 12: Individual valset scores for new program: [0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0]
2025/08/12 23:00:10 INFO dspy.teleprompt.gepa.gepa: Iteration 12: New valset pareto front scores: [0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1]
2025/08/12 23:00:10 INFO dspy.teleprompt.gepa.gepa: Iteration 12: Full valset pareto front score: 0.8
2025/08/12 23:00:10 INFO dspy.teleprompt.gepa.gepa: Iteration 12: Updated valset pareto front programs: [{0, 1, 2, 3, 4, 5, 6, 7}, {0, 1, 3, 7}, {0, 1, 2, 3, 4, 5, 6, 7}, {0, 1, 2, 3, 4, 5, 6, 7}, {0, 1, 2, 3, 4, 5, 6, 7}, {1, 2}, {0, 1, 3, 4, 7}, {0, 2, 3, 4, 5, 6, 7}, {0, 1, 2, 4, 5, 6, 7}, {6}, {0, 2, 4}, {0, 1, 2, 3, 4, 5, 6, 7}, {0, 1, 2, 3, 4, 5, 6, 7}, {2, 5, 6}, {2, 4, 6}, {0, 2, 3, 4, 5, 6}, {0, 1, 2, 3, 4, 5, 6, 7}, {1, 2, 4, 6, 7}, {3, 4}, {0, 1, 2, 3, 4, 5, 6, 7}, {1, 2, 3, 4, 5, 6, 7}, {0, 1, 2, 4, 5}, {0, 1, 2, 3, 4, 5, 6, 7}, {0, 1, 2, 3, 4, 5, 6, 7}, {0, 1, 2, 3, 4, 5, 6, 7}, {2, 7}, {0, 1, 2, 3, 4, 5, 6, 7}, {0, 1, 2, 7}, {0, 1, 2, 3, 4, 5, 6, 7}, {2, 3}, {3, 7}, {1, 4}, {1, 2, 3}, {2}, {1, 2, 3, 5, 6, 7}, {0, 1, 2, 3, 4, 5, 6, 7}, {5}, {0, 1, 2, 3, 4, 5, 6, 7}, {7}, {0, 1, 2, 3, 4, 5, 6, 7}, {3}, {3, 6}, {0, 1, 2, 3, 4, 5, 6, 7}, {0}, {3}]
2025/08/12 23:00:10 INFO dspy.teleprompt.gepa.gepa: Iteration 12: Best valset aggregate score so far: 0.5333333333333333
2025/08/12 23:00:10 INFO dspy.teleprompt.gepa.gepa: Iteration 12: Best program as per aggregate score on train_val: 2
2025/08/12 23:00:10 INFO dspy.teleprompt.gepa.gepa: Iteration 12: Best program as per aggregate score on valset: 2
2025/08/12 23:00:10 INFO dspy.teleprompt.gepa.gepa: Iteration 12: Best score on valset: 0.5333333333333333
2025/08/12 23:00:10 INFO dspy.teleprompt.gepa.gepa: Iteration 12: Best score on train_val: 0.5333333333333333
2025/08/12 23:00:10 INFO dspy.teleprompt.gepa.gepa: Iteration 12: Linear pareto front program index: 2
2025/08/12 23:00:10 INFO dspy.teleprompt.gepa.gepa: Iteration 12: New program candidate index: 7
2025/08/12 23:00:10 INFO dspy.teleprompt.gepa.gepa: Iteration 13: Selected program 6 score: 0.4
Average Metric: 2.00 / 3 (66.7%): 100%|██████████████████████████████████████████████████████████████████████████████████████████████████████████| 3/3 [01:52<00:00, 37.66s/it]
2025/08/12 23:02:03 INFO dspy.evaluate.evaluate: Average Metric: 2.0 / 3 (66.7%)

2025/08/12 23:03:48 INFO dspy.teleprompt.gepa.gepa: Iteration 13: Proposed new text for predict: You will receive one math problem as plain text under the key “problem.” Solve it and return:

- reasoning: a concise, logically ordered solution that leverages structure/identities (not brute force), ending with a quick verification.
- answer: the final requested number/expression only (no extra words).

Formatting:
- Use exactly two top-level fields named “reasoning” and “answer.”
- Keep reasoning succinct but complete; bullet points are fine.
- The answer field must contain only the final value requested (e.g., 227, 585, 601).

General approach:
- Identify the problem type (e.g., logarithms/functional equations, symmetric sums, combinatorics on sets, ordered tuples, classical Euclidean geometry, polynomial/root structure).
- Enforce all domain constraints strictly (digit ranges, no leading zeros, positive/acute/parallel conditions, ordered vs unordered, strict increases, integer bounds).
- Exploit algebraic structure, classic identities, and projective/metric geometry tools; avoid unnecessary coordinates, decimal approximations, or brute force.
- End with a quick verification (substitution or geometric consistency check).

Essential algebra/logarithm tactics (from common patterns):
- If log expressions are equal, set their common value t and convert to exponential form to eliminate variables cleanly. Example: if log_{a}(b) = log_{c}(d) = t ⇒ a^t = b and c^t = d; dividing gives (a/c)^t = b/d ⇒ t = log_{a/c}(b/d).
- Change of base only when needed; prefer exact symbolic relations over numeric approximation.
- Ratio trick: if a/b = c/d = k (with b ≠ d and denominators nonzero), then (a − c)/(b − d) = k; useful for simplifying equal-fraction equations.
- For monic cubics with integer coefficients: if two integer roots are known (counting multiplicities), the third is also integer (Vieta).

Core Euclidean geometry tools (prioritize these over coordinates):
- Power of a Point: for a circle intersecting a line at P and Q, and tangents from a point X, XP·XQ = (tangent from X)^2.
- Tangent properties: lengths from the same external point to a circle are equal; radius to point of tangency is perpendicular to the tangent line.
- Parallel lines and distances: the distance between two parallel lines is constant; in a tangential figure, this often equals an easy multiple of the inradius.
- Similar triangles, cyclic angle relations, homothety, right-triangle/chord relations, Ptolemy, Stewart/Apollonius as appropriate.

Geometric patterns frequently useful (capture and reuse when they appear):
- Parallelogram with a circle tangent to AB, AD, and BC:
  • AB ∥ DC and AD ∥ BC.
  • If a circle is tangent to AD and BC (a pair of parallels), the distance between these lines equals 2r (r = circle radius). This is the altitude for base BC, so Area = BC · 2r.
  • If the circle intersects diagonal AC at P and Q (with A–P–Q–C), then:
    - From A: tangent length AT satisfies AT^2 = AP·AQ.
    - From C: tangent length CT satisfies CT^2 = CQ·CP (with CP = PQ + QC and AQ = AP + PQ).
  • Equal-tangent facts from vertices let you parameterize side lengths (e.g., BX = BE = x when tangency points on AB and BC are X, E).
  • Combine base = (known tangent segment) + x and height = 2r to get area.
- Rhombus with an incircle (all sides equal, diagonals are perpendicular and meet at the incenter):
  • Distances from any point P on the incircle to two opposite parallel sides sum to the diameter: if distances to AD and BC are u and v (with AD ∥ BC), then u + v = 2r.
  • Inradius r relates side s and acute angle θ by r = (s sin θ)/2 ⇒ s = 2r / sin θ and perimeter = 4s.
  • With P on the incircle and perpendicular foot constructions, Power of a Point and right triangles can recover r, chord lengths (e.g., PQ), and trigonometric relations to solve for s.

Quality checks and common pitfalls:
- Respect all integer/divisibility constraints; ensure parameters that represent counts are integers and within bounds.
- Keep radicals exact and simplified; use squarefree radicands where appropriate.
- For “unique solution” conditions, consider multiplicities and all structural cases that yield uniqueness.
- For base/digit problems, enforce digit sets and leading-digit rules; for lengths/angles, ensure geometric configurations are consistent (e.g., acute angle constraints).
- Avoid numeric approximations unless the problem explicitly asks for a decimal; prefer exact algebraic/log expressions.
- Verify by substitution or a short geometric consistency check (e.g., recompute area as base·height; check sums like AP + PQ + QC = AC).

Verification:
- Substitute back or re-derive a key relation (e.g., recomputed t from derived expression; recomputed area/height/segment sums) to confirm all given constraints.

Output exactly two fields: “reasoning” and “answer”.
2025/08/12 23:04:39 INFO dspy.evaluate.evaluate: Average Metric: 1.0 / 3 (33.3%)
2025/08/12 23:04:39 INFO dspy.teleprompt.gepa.gepa: Iteration 13: New subsample score is not better, skipping
2025/08/12 23:04:39 INFO dspy.teleprompt.gepa.gepa: Iteration 14: Selected program 2 score: 0.5333333333333333
Average Metric: 3.00 / 3 (100.0%): 100%|█████████████████████████████████████████████████████████████████████████████████████████████████████████| 3/3 [00:26<00:00,  8.86s/it]
2025/08/12 23:05:05 INFO dspy.evaluate.evaluate: Average Metric: 3.0 / 3 (100.0%)
2025/08/12 23:05:05 INFO dspy.teleprompt.gepa.gepa: Iteration 14: All subsample scores perfect. Skipping.
2025/08/12 23:05:05 INFO dspy.teleprompt.gepa.gepa: Iteration 14: Reflective mutation did not propose a new candidate
2025/08/12 23:05:05 INFO dspy.teleprompt.gepa.gepa: Iteration 15: Selected program 3 score: 0.4666666666666667

Average Metric: 1.00 / 3 (33.3%): 100%|██████████████████████████████████████████████████████████████████████████████████████████████████████████| 3/3 [01:33<00:00, 31.19s/it]
2025/08/12 23:06:39 INFO dspy.evaluate.evaluate: Average Metric: 1.0 / 3 (33.3%)

2025/08/12 23:08:04 INFO dspy.teleprompt.gepa.gepa: Iteration 15: Proposed new text for predict: You will be given one math problem as plain text under a key like “problem.” Your job is to solve it correctly and return:

- reasoning: a concise, logically ordered solution that uses identities/structure to avoid brute force, ends with a quick verification.
- answer: the final requested number/expression only (no extra words).

Formatting:
- Use exactly two top-level fields named “reasoning” and “answer.”
- Keep reasoning succinct but complete. Bullet points are fine.
- The answer field must contain only the final value requested (e.g., 227, 585, 601).

General problem-solving guidance:
- Parse the problem type (e.g., base representation, intersecting families of subsets, avoiding arithmetic progressions, symmetric sums with constraints, ordered tuples counting, classical Euclidean geometry with circles/secants/tangents, arrangements/planar graphs).
- Always enforce domain constraints (digits in range, no leading zeros, ordered vs unordered, strict increases, “general position” assumptions like “no three concurrent” when stated).
- Use algebraic identities and modular arithmetic to reduce the search space; prefer structural arguments over naive enumeration.
- For “greatest/least” questions, derive tight bounds and give a construction that attains them.
- Prefer classic geometry tools (power of a point, radical axis, homothety, similar triangles in symmetric cross-sections, Apollonius/Stewart, cyclic angle/arc relations) over coordinate bashes unless necessary.
- When the problem statement provides a small example (e.g., m=3, n=2 -> 8 regions), use it to sanity-check your derived formula.
- If the wording appears ambiguous (e.g., repeated phrases like “outside”/“inside”), infer the intended meaning from context and validate by consistency checks.

Domain-specific strategies, templates, and pitfalls:

A) Random pairings in 4-player single-elimination tournaments:
- There are exactly 3 equally likely semifinal matchings: {(A,C),(J,S)}, {(A,J),(C,S)}, {(A,S),(C,J)}; each with probability 1/3.
- Condition on the pairing, multiply by match win probabilities; outcomes of different matches are independent if stated.
- If two “other” players (like J vs S) have no specified advantage, treat their match as 1/2–1/2.
- Sum over cases. If the answer is requested as p+q for a reduced fraction p/q, reduce before summing.

B) Segments between two parallel lines (complete bipartite geometric graph K_{m,n}):
- Setup: m distinct points on line ℓ_A and n distinct points on parallel line ℓ_B. All mn segments A_iB_j are drawn. The intended “general position” condition for AIME 2022 II #9 is: no three segments are concurrent in the open strip (i.e., any two segments intersect in at most one interior point, and no interior point lies on more than two segments).
- Key counts in general position:
  • Number of interior intersection points: C(m,2)·C(n,2) (choose two A’s and two B’s; the corresponding pair of segments cross once).
  • Total bounded regions:
    f(m,n) = C(m,2)·C(n,2) + m·n − 1.
- Derivations/checks:
  • Euler route: Promote every interior intersection to a vertex, include edges along ℓ_A between consecutive A_i and along ℓ_B between consecutive B_j; apply V−E+F=2 and subtract 1 for the unbounded outside face to get bounded regions.
  • Sanity check with the given example: for (m,n)=(3,2), C(3,2)·C(2,2)+3·2−1 = 3·1+6−1=8.
- Beware the common pitfall of assuming “no intersections.” If the example contradicts that assumption, reinterpret the condition as “no three concurrent” and proceed with the formula above.

C) Torus–sphere tangency along circles (surface of revolution contact):
- Torus parameters: major radius R (distance from axis to center of generating circle) and minor radius r (radius of generating circle). Sphere radius s.
- Reduce to a meridian (axis-containing) 2D cross-section. Use similar triangles stemming from the homothety at the sphere center:
  • Let OE = distance from sphere center O to the center E of the generating circle in the cross-section when tangent occurs. For external tangency (torus outside sphere) and internal tangency (torus “inside”/hugging sphere), OE equals s ± r respectively.
  • The “lever arm” from the axis to E is constant and equals R.
  • The circle of tangency on the sphere has radius GH, with OG = s. From similarity ΔOEF ~ ΔOGH, EF/OE = GH/OG ⇒ GH = s·R / (s ± r).
- Therefore:
  • Tangency circle radius for the “inner” position: r_i = s·R / (s − r).
  • Tangency circle radius for the “outer” position: r_o = s·R / (s + r).
  • Difference: r_i − r_o = s·R·(2r) / (s^2 − r^2).
- For R=6, r=3, s=11: r_i = 33/4, r_o = 33/7, so r_i − r_o = 99/28, and m+n = 99+28 = 127.
- Verify units and that the radii are positive and less than s.

D) Planar graphs and Euler:
- When counting faces in drawings with intersections, convert each interior crossing into a vertex, update the edge count accordingly (each crossing increases E by 2 relative to the two original edges), then apply Euler’s formula V−E+F=2 (F includes the unbounded outer face). Bounded faces = F − 1.
- Use symmetry to reduce overcounting; check small cases.

Quality checks:
- Always test derived formulas on provided small examples.
- Keep radicals exact; reduce fractions fully; ensure squarefree radicands in m√n if relevant.
- For combinatorics/probability, confirm independence assumptions, case partition completeness, and that probabilities sum to 1 when expected.
- For geometry, confirm that constructed points/lengths satisfy all given constraints and that the locus (circle/line) is correct.

Output discipline:
- Only two top-level JSON-like fields: “reasoning” and “answer.”
- The “answer” field must be only the final requested value (e.g., 244 or 99/28 or 127), with no extra text.
2025/08/12 23:08:25 INFO dspy.evaluate.evaluate: Average Metric: 3.0 / 3 (100.0%)
2025/08/12 23:11:05 INFO dspy.evaluate.evaluate: Average Metric: 21.0 / 45 (46.7%)
2025/08/12 23:11:05 INFO dspy.teleprompt.gepa.gepa: Iteration 15: Full valset score for new program: 0.4666666666666667
2025/08/12 23:11:05 INFO dspy.teleprompt.gepa.gepa: Iteration 15: Full train_val score for new program: 0.4666666666666667
2025/08/12 23:11:05 INFO dspy.teleprompt.gepa.gepa: Iteration 15: Individual valset scores for new program: [0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0]
2025/08/12 23:11:05 INFO dspy.teleprompt.gepa.gepa: Iteration 15: New valset pareto front scores: [0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1]
2025/08/12 23:11:05 INFO dspy.teleprompt.gepa.gepa: Iteration 15: Full valset pareto front score: 0.8
2025/08/12 23:11:05 INFO dspy.teleprompt.gepa.gepa: Iteration 15: Updated valset pareto front programs: [{0, 1, 2, 3, 4, 5, 6, 7, 8}, {0, 1, 3, 7, 8}, {0, 1, 2, 3, 4, 5, 6, 7, 8}, {0, 1, 2, 3, 4, 5, 6, 7, 8}, {0, 1, 2, 3, 4, 5, 6, 7, 8}, {8, 1, 2}, {0, 1, 3, 4, 7, 8}, {0, 2, 3, 4, 5, 6, 7, 8}, {0, 1, 2, 4, 5, 6, 7, 8}, {8, 6}, {0, 2, 4}, {0, 1, 2, 3, 4, 5, 6, 7, 8}, {0, 1, 2, 3, 4, 5, 6, 7, 8}, {8, 2, 5, 6}, {8, 2, 4, 6}, {0, 2, 3, 4, 5, 6, 8}, {0, 1, 2, 3, 4, 5, 6, 7, 8}, {1, 2, 4, 6, 7, 8}, {3, 4}, {0, 1, 2, 3, 4, 5, 6, 7, 8}, {1, 2, 3, 4, 5, 6, 7, 8}, {0, 1, 2, 4, 5}, {0, 1, 2, 3, 4, 5, 6, 7}, {0, 1, 2, 3, 4, 5, 6, 7, 8}, {0, 1, 2, 3, 4, 5, 6, 7, 8}, {8, 2, 7}, {0, 1, 2, 3, 4, 5, 6, 7, 8}, {0, 1, 2, 7, 8}, {0, 1, 2, 3, 4, 5, 6, 7, 8}, {2, 3}, {3, 7}, {1, 4}, {1, 2, 3}, {2}, {1, 2, 3, 5, 6, 7, 8}, {0, 1, 2, 3, 4, 5, 6, 7, 8}, {5}, {0, 1, 2, 3, 4, 5, 6, 7, 8}, {7}, {0, 1, 2, 3, 4, 5, 6, 7, 8}, {3}, {3, 6}, {0, 1, 2, 3, 4, 5, 6, 7, 8}, {0}, {3}]
2025/08/12 23:11:05 INFO dspy.teleprompt.gepa.gepa: Iteration 15: Best valset aggregate score so far: 0.5333333333333333
2025/08/12 23:11:05 INFO dspy.teleprompt.gepa.gepa: Iteration 15: Best program as per aggregate score on train_val: 2
2025/08/12 23:11:05 INFO dspy.teleprompt.gepa.gepa: Iteration 15: Best program as per aggregate score on valset: 2
2025/08/12 23:11:05 INFO dspy.teleprompt.gepa.gepa: Iteration 15: Best score on valset: 0.5333333333333333
2025/08/12 23:11:05 INFO dspy.teleprompt.gepa.gepa: Iteration 15: Best score on train_val: 0.5333333333333333
2025/08/12 23:11:05 INFO dspy.teleprompt.gepa.gepa: Iteration 15: Linear pareto front program index: 2
2025/08/12 23:11:05 INFO dspy.teleprompt.gepa.gepa: Iteration 15: New program candidate index: 8
2025/08/12 23:11:05 INFO dspy.teleprompt.gepa.gepa: Iteration 16: Selected program 2 score: 0.5333333333333333
Average Metric: 2.00 / 3 (66.7%): 100%|██████████████████████████████████████████████████████████████████████████████████████████████████████████| 3/3 [00:55<00:00, 18.48s/it]
2025/08/12 23:12:01 INFO dspy.evaluate.evaluate: Average Metric: 2.0 / 3 (66.7%)

2025/08/12 23:13:17 INFO dspy.teleprompt.gepa.gepa: Iteration 16: Proposed new text for predict: You will be given one math problem as plain text under a key like “problem.” Your job is to solve it correctly and return:

- reasoning: a concise, logically ordered solution that uses identities/structure to avoid brute force, ends with a quick verification.
- answer: the final requested number/expression only (no extra words).

Formatting:
- Use exactly two top-level fields named “reasoning” and “answer.”
- Keep reasoning succinct but complete. Bullet points are fine.
- The answer field must contain only the final value requested (e.g., 227, 585, 601).

General problem-solving guidance:
- Parse the problem type (e.g., base representation, intersecting families of subsets, avoiding arithmetic progressions, symmetric sums with constraints, ordered tuples counting, polynomial equal-value/root-structure, complex-argument maximization, rhombic parallelepipeds).
- Always enforce domain constraints (e.g., base-b digits in 0..b−1; no leading zero for “three-digit”; coefficient bounds; |z| fixed radius; ordered vs unordered; strict increase).
- Use algebraic identities and modular arithmetic to reduce the search space; prefer structural arguments over naive enumeration.
- For “greatest/least” questions, derive tight bounds and give a construction that attains them.
- Verify final candidates numerically and confirm they satisfy all constraints; check edge/degenerate cases explicitly.

Domain-specific strategies and pitfalls (from common contests and prior feedback):

1) Base-conversion/digit rearrangement:
- Positional notation: in base b, (a b c)_b = a·b^2 + b·b + c; in base 10: abc = 100a + 10b + c.
- Enforce digit ranges strictly (e.g., in base 9, digits ∈ {0,…,8}; leading digit nonzero when specified).
- Set up equalities and simplify. Use modular constraints to prune:
  • Mod 9 often collapses coefficients.
  • Mod 8/11 helpful for parity and alternating sums.
- Solve within digit bounds and verify numerically.

2) Polynomials with equal values at two integers (e.g., find count of cubics with a unique m ≠ k such that p(m)=p(k)):
- Let q(x) = p(x) − p(k). Then q(k) = 0, so q(x) = (x−k)·r(x) where r is degree one less.
- Integer-coefficient, monic case: any rational root is an integer. For quadratics with integer coefficients:
  • If discriminant is a nonzero perfect square, both roots are integers.
  • Exactly one integer root occurs only for a double root (D=0), and that root is integer if parity allows (for monic: −p must be even).
- Crucial split to count “exactly one integer m ≠ k”:
  Case A (double non-k root): r(x) = (x−m)^2 with m ≠ k. Match coefficients (e.g., via Vieta) to get explicit a,b,… in bounds.
  Case B (k is repeated): r(k)=0, so q(x) has (x−k)^2(x−m). Impose “k is a root of r”: r(k)=0 gives linear relations among coefficients (Vieta). Solve within bounds, exclude the degenerate m=k, and respect any additional constraints.
- Don’t forget coefficients that do not affect q (like constant term c in p(x)=x^3+…): count all possibilities for them at the end.
- Verify uniqueness: ensure no extra integer roots besides k and the intended m.

3) Palindromes across bases:
- Bound number of digits by size (e.g., n<1000 ⇒ octal has 3–4 digits).
- Palindrome forms:
  • 3-digit octal: (A B A)_8 = 65A + 8B.
  • 4-digit octal: (A B B A)_8 = 513A + 72B (A ≥ 1).
- Enumerate small parameter ranges; test other-base palindrome constraints. For “greatest,” check candidates in descending order with a proof of optimality.

4) Symmetric sums with a + b + c fixed (ordered triples of nonnegative integers):
- Use identities to compress expressions:
  S = ab(a + b) + bc(b + c) + ca(c + a) = (a + b + c)(ab + bc + ca) − 3abc.
- With a + b + c known, convert into relations among ab + bc + ca and abc.
- Use shifts like a = A + x to isolate factors such as (a−A)(b−A)(c−A), enabling clean counting.
- Count ordered solutions carefully; include/exclude symmetric/degenerate cases precisely.

5) Intersecting families of subsets:
- Intersecting: every pair has nonempty intersection; ∅ cannot be included.
- Complement pairs: S and S^c cannot both be present; leverage to structure counts.
- Size pigeonhole: In [n], any two subsets of size > n/2 intersect. For n=5, all subsets of size ≥3 form an intersecting family (size 16).
- When counting families of fixed size:
  • Casework by minimum set size and how many 2-sets are included (they force exclusions via complements).
  • Avoid double counting by canonical patterns; order of subsets in a collection does not matter.

6) Avoiding 4-term arithmetic progressions in a strictly increasing sequence with fixed anchors:
- Bound variable terms by monotonicity.
- Pre-eliminate values completing APs with three fixed terms (anchor-triplet checks).
- Count allowed pairs, then subtract specific pairs that close 4-term APs with two fixed endpoints.
- Use that endpoints difference Δ must be divisible by 3 for a 4-term AP; solve for integer interior terms within bounds.
- Avoid double subtraction; re-check integrality and bounds.

7) Order statistics with sum and absolute-sum constraints (x_1 ≤ ... ≤ x_n, sum |x_i| = 1, sum x_i = 0):
- Total positive mass = total negative mass = 1/2.
- For maximizing x_k: with T = n − k + 1 largest terms, sum ≥ T·x_k ⇒ x_k ≤ (1/2)/T.
- For minimizing x_l: with l smallest terms, sum ≤ l·x_l ⇒ x_l ≥ (−1/2)/l.
- Attain bounds by concentrating masses evenly on those positions and setting the middle to 0. Verify sums and order.

8) Complex maximization for expressions like A z + B/z with |z|=r:
- Let z = r e^{iθ}, so 1/z = e^{−iθ}/r.
- Expression reduces to α e^{iθ} + β e^{−iθ}; its real part is M cos θ + N sin θ.
- Maximum over θ is √(M^2 + N^2). Compute M,N explicitly; verify arithmetic.

9) Parallelepipeds whose faces are rhombi with given diagonals:
- For a rhombus generated by equal-length vectors of length s with angle θ:
  • d_long^2 = 2s^2(1 + cos θ), d_short^2 = 2s^2(1 − cos θ).
  • Sum: d_long^2 + d_short^2 = 4s^2 ⇒ s^2 = (d_long^2 + d_short^2)/4.
  • cos θ = (d_long^2 − d_short^2)/(d_long^2 + d_short^2) =: ρ (and each face angle cosine can be ±ρ).
- For a parallelepiped with edge length s and pairwise angle cosines c12, c23, c31, squared volume factor:
  D = 1 + 2 c12 c23 c31 − c12^2 − c23^2 − c31^2, so V = s^3 √D.
- With all faces the same rhombus, c_ij ∈ {+ρ, −ρ}. The two noncongruent realizations correspond to different sign patterns:
  • “All acute” near: (+ρ, +ρ, +ρ) gives D_plus = 1 + 2ρ^3 − 3ρ^2.
  • “One obtuse” pattern (two +, one − up to symmetry) gives D_minus = 1 − 2ρ^3 − 3ρ^2.
  • Ratio of volumes: √(D_plus/D_minus). For d_long^2=31, d_short^2=21 ⇒ s^2=13, ρ=5/26, yielding ratio 63/62.
- Alternative routes: Gram determinant, inscribed tetrahedra with edges equal to face diagonals, or centroid-height computation all lead to the same ratio.

Quality checks:
- Verify digit/base constraints, coefficient bounds, and final equalities numerically.
- For polynomial-equality problems, handle both “double non-k root” and “k repeated” cases; use Vieta and discriminant parity; exclude m=k and out-of-range coefficients; include free parameters like c.
- For extremal problems, provide both a tight bound and an explicit construction achieving it (or the θ that attains equality).
- For counting, explicitly handle ordered vs unordered, exclude impossible/duplicate cases, and check complements/forbidden pairs.
- For AP-avoidance, confirm integrality and bounds; ensure all endpoint combinations checked and no double subtraction.
- Put only the clean final numeric result in the “answer” field.

Finally:
- Return exactly two top-level fields: “reasoning” and “answer”. Keep reasoning concise but complete, and end with a brief verification.
2025/08/12 23:14:14 INFO dspy.evaluate.evaluate: Average Metric: 2.0 / 3 (66.7%)
2025/08/12 23:14:14 INFO dspy.teleprompt.gepa.gepa: Iteration 16: New subsample score is not better, skipping
2025/08/12 23:14:14 INFO dspy.teleprompt.gepa.gepa: Iteration 17: Selected program 8 score: 0.4666666666666667
Average Metric: 2.00 / 3 (66.7%): 100%|██████████████████████████████████████████████████████████████████████████████████████████████████████████| 3/3 [00:38<00:00, 12.98s/it]
2025/08/12 23:14:53 INFO dspy.evaluate.evaluate: Average Metric: 2.0 / 3 (66.7%)

2025/08/12 23:16:01 INFO dspy.teleprompt.gepa.gepa: Iteration 17: Proposed new text for predict: You will be given one math problem as plain text under a key like “problem.” Your job is to solve it correctly and return:

- reasoning: a concise, logically ordered solution that uses identities/structure to avoid brute force, ends with a quick verification.
- answer: the final requested number/expression only (no extra words).

Formatting:
- Use exactly two top-level fields named “reasoning” and “answer.”
- Keep reasoning succinct but complete. Bullet points are fine.
- The answer field must contain only the final value requested (e.g., 227, 585, 601, 29/96). Do not include units or words.
- Reduce and simplify completely. If the result is an integer, output the integer (not an unsimplified expression or radical).
- If the prompt asks for m+n or p+q, first reduce the fraction to lowest terms, then output the integer sum.

General problem-solving guidance:
- Parse problem type (e.g., base representation, subset families, avoidance problems, symmetric sums with constraints, ordered tuple counts, classical Euclidean geometry, arrangements/planar graphs).
- Enforce domain constraints (digit ranges, no leading zeros unless explicitly required, ordered vs unordered, strict inequalities, “general position” assumptions like “no three concurrent” when stated).
- Prefer structural arguments over brute force: algebraic identities, invariants, parity/modular arithmetic, symmetry, bijections, recursion/D.P., and geometric power/similarity.
- For “greatest/least” questions, derive tight bounds and give/describe a construction that attains them.
- Quick verification: check boundary conditions, recompute via a second route if short, or plug back to confirm constraints. If the problem style strongly suggests an integer answer (e.g., AIME/AMC-style), and you’ve obtained a non-integer, re-examine your steps.

Output discipline and simplification:
- Integers: output plain digits only, no signs unless negative, no leading zeros unless explicitly requested.
- Fractions: output as a/b in lowest terms with positive denominator.
- Radicals: use exact simplified radicals with squarefree radicands only if the problem explicitly asks for an expression; otherwise evaluate to an integer if it simplifies to one.
- For prompts like “find m+n” or “find p+q,” reduce first (e.g., p/q in lowest terms) before summing.

Domain-specific strategies, templates, and pitfalls:

A) 4-player single-elimination tournaments with random semifinal pairings:
- Exactly 3 equally likely semifinal matchings (each with probability 1/3):
  {(A,C),(J,S)}, {(A,J),(C,S)}, {(A,S),(C,J)}.
- Condition on the pairing, multiply by match win probabilities; assume independence across matches if stated.
- If two “other” players (like J vs S) have no specified advantage, treat their match as 1/2–1/2.
- Sum over cases. If the final asks for p+q for a reduced fraction p/q, reduce before summing.

B) Segments between two parallel lines (complete bipartite geometric graph K_{m,n}):
- Setup: m distinct points on line ℓ_A and n distinct points on parallel line ℓ_B. All mn segments A_iB_j are drawn. The intended “general position” condition (AIME 2022 II #9 style) is: no three segments are concurrent in the open strip (i.e., any two segments intersect in at most one interior point, and no interior point lies on more than two segments).
- Counts:
  • Number of interior intersections: C(m,2)·C(n,2).
  • Total bounded regions: f(m,n) = C(m,2)·C(n,2) + m·n − 1.
- Derivation: Promote interior intersections to vertices; add edges along ℓ_A, ℓ_B between consecutive points; apply Euler V−E+F=2; bounded faces = F−1.
- Sanity check example: (m,n)=(3,2) → 8 bounded regions.

C) Torus–sphere tangency along circles (surface of revolution contact):
- Torus parameters: major radius R, minor radius r; sphere radius s.
- In meridian cross-section, use similarity from the sphere center:
  • Tangency circle radius for “inner” position: r_i = s·R / (s − r).
  • Tangency circle radius for “outer” position: r_o = s·R / (s + r).
  • Difference: r_i − r_o = s·R·(2r) / (s^2 − r^2).
- Check positivity and that radii < s.

D) Planar graphs/Euler with crossings:
- Convert each interior crossing to a vertex; update edges accordingly; apply V−E+F=2 (F includes the unbounded outer face). Bounded faces = F−1.
- Use symmetry and small-case checks to avoid over/undercounting.

E) Optimal guessing with known counts revealed sequentially (e.g., 3 red and 3 black, random order):
- Optimal policy at each step: guess the color with more remaining cards; if equal, either guess (symmetry).
- Dynamic programming recurrence for expected correct guesses E(x,y):
  • If x>y: E(x,y) = (x/(x+y))·(1 + E(x−1,y)) + (y/(x+y))·E(x,y−1).
  • If y>x: swap roles.
  • If x=y: E(x,x) = 0.5·(1 + E(x−1,x)) + 0.5·E(x,x−1).
- Linearity-of-expectation approach can also sum stage-wise correctness probabilities.
- Reduce any resulting fraction before extracting m+n if asked.

F) Classical Euclidean geometry preferences:
- Favor Power of a Point, radical axis, perpendicular bisectors to locate circle centers, cyclic angle/arc relations, homothety, and similar triangles over coordinate bashes unless coordinates simplify dramatically.
- In rectangle/circle configurations, exploit equal radii from the center to chord endpoints, midpoints/perpendicular bisectors, and right-angle structures. If points are collinear across rectangles, consider Power of a Point on intersection lines.
- Quick check: lengths must be positive and satisfy triangle inequalities; if an expected integer length comes out non-integer, revisit.

Quality checks:
- Reduce fractions fully; ensure squarefree radicands if radicals are required; simplify expressions.
- For combinatorics/probability, confirm independence assumptions, that case partitions are complete/disjoint, and that intermediate probabilities are in [0,1]; if summing probabilities over exhaustive cases, totals should be plausible.
- For geometry, confirm all constraints (parallelism, cyclicity, tangency) are met by the computed values.
- Use any provided small example to sanity-check derived formulas.

Common pitfalls to avoid:
- Returning an unsimplified expression when an integer is expected.
- Failing to reduce p/q before computing p+q.
- Adding extraneous text/formatting in the “answer” field.
- Ignoring constraints like no leading zeros, digit ranges, or general position.

Deliverable:
- Exactly two fields: “reasoning” and “answer”.
- “answer” is a single simplified numeric value in the requested form (integer, reduced fraction, or the specified sum like m+n). No other text.
2025/08/12 23:16:30 INFO dspy.evaluate.evaluate: Average Metric: 2.0 / 3 (66.7%)
2025/08/12 23:16:30 INFO dspy.teleprompt.gepa.gepa: Iteration 17: New subsample score is not better, skipping
2025/08/12 23:16:30 INFO dspy.teleprompt.gepa.gepa: Iteration 18: Selected program 5 score: 0.35555555555555557
Average Metric: 2.00 / 3 (66.7%): 100%|██████████████████████████████████████████████████████████████████████████████████████████████████████████| 3/3 [02:12<00:00, 44.22s/it]
2025/08/12 23:18:43 INFO dspy.evaluate.evaluate: Average Metric: 2.0 / 3 (66.7%)

2025/08/12 23:20:08 INFO dspy.teleprompt.gepa.gepa: Iteration 18: Proposed new text for predict: Your job
- Solve exactly one math problem and return exactly two top-level JSON-like fields:
  - reasoning: a concise, logically ordered solution that leverages structure (identities, modular arithmetic, symmetry) and ends with a quick verification.
  - answer: the final requested number/expression only (no extra words or formatting).

Formatting
- Use exactly the two fields “reasoning” and “answer.”
- Keep the reasoning succinct but complete; bullet points are fine.
- The answer field must contain only the final value requested (e.g., 227, 585, 601).
- Avoid heavy formatting (no LaTeX blocks, tables). Bullet lists are acceptable.

General problem-solving guidance
- Identify the problem type early (e.g., base/digit relations, repeating decimals, palindromes across bases, symmetric sums with fixed totals, intersecting subset families, avoiding arithmetic progressions, order statistics with absolute-sum constraints, floor-sum optimization, circle/geometry with radical axes or cyclic quadrilaterals, logarithm equations with shared unknowns).
- Enforce domain constraints strictly:
  • Base-b digits are in 0..b−1; leading digit nonzero if a positional numeral.
  • Logarithms: argument > 0, base > 0 and ≠ 1; respect any excluded x-values given.
  • Ordered vs unordered; strictly increasing where specified.
- Prefer structural arguments over brute force:
  • Factorization, completing the square, modular arithmetic, symmetry/reflection, similar triangles, cyclic identities, Ptolemy’s theorem, Law of Cosines/Sines, power of a point, bounding plus construction for extremals.
- For greatest/least questions, derive a tight bound and exhibit a construction attaining it.
- Always verify at the end (e.g., plug back into conditions, check modular constraints, recompute a key equality numerically).

Domain-specific strategies and pitfalls

A) Base conversion/digit equations
- Translate positional notation correctly: abc (base 10) = 100a + 10b + c; (a b c)_b = a·b^2 + b·b + c.
- Enforce digit ranges: in base 9, digits ∈ {0,…,8}. Leading digit in base-b must be 1..b−1 if a k-digit numeral.
- Use modular constraints to prune search (e.g., mod 9 or mod 71 from coefficient comparison).
- When solving 100a + 10b + c = 81b + 9c + a, reduce to 99a = 71b + 8c; check moduli (e.g., mod 71 or mod 9) and small ranges.
- Verify numerically by converting the base-b representation back to decimal.

B) Repeating decimals 0.\overline{abcd} and counting reduced numerators
- 0.\overline{abcd} = m/9999 with m ∈ {1,…,9999}, and 9999 = 3^2·11·101.
- Reduced numerator is x = m / gcd(m, 9999). Do NOT use ∑φ(d) = n here to count distinct reduced numerators; that counts reduced fractions by denominator divisors, not distinct numerators from a bounded m.
- Characterize x via existence of y | 9999 with y ≥ x and gcd(x, y) = 1 (since m = x·(9999/y) ≤ 9999).
- Use inclusion–exclusion over divisibility by 3, 9, 11, 101 carefully; treat “3 vs 9” distinctly.

C) Palindromes across bases
- Bound number of digits by magnitude; write k-digit palindrome formulas (e.g., base-8 ABA: 65A + 8B; ABBA: 513A + 72B, A ≥ 1).
- Test candidate bases/lengths using congruences and magnitude bounds.

D) Symmetric sums with a + b + c fixed
- Compress using identities, e.g., S = (a + b + c)(ab + bc + ca) − 3abc.

E) Intersecting families of subsets
- Intersecting means every pair intersects; empty set excluded.
- Complement pairs cannot both be present; for n odd, star and “all subsets of size > n/2” both have size 2^{n−1}.
- For fixed-size counting, case on minimum sizes and overlaps; use canonical patterns to avoid double-counting.

F) Avoiding 4-term arithmetic progressions
- Pre-eliminate values that force a 4-term AP with anchors.
- If endpoints differ by Δ, then interior step is Δ/3; require integrality and bounds.

G) Order statistics with sum and absolute-sum constraints
- Sum |x_i| = 1 and sum x_i = 0 ⇒ positive mass = negative mass = 1/2.
- For maximizing x_k, with T = n−k+1 largest positions, x_k ≤ (1/2)/T; minimize x_l similarly.
- Achieve bounds by concentrating masses evenly; rest zero.

H) Systems with square-root bilinear forms
- Factor under roots: √(x(2−y)) style.
- Algebraic approach (A=1−x etc.) or trigonometric substitution (x=2cos^2 α); common elegant results like products equal constants (e.g., 1/32).

I) Floor-sum with a parameter
- Separate continuous main term and fractional correction.
- Pick parameter to cancel main term; argue uniqueness via step sizes.
- Count residues modulo m precisely, including partial cycles.

J) Circle geometry with two tangent circles and a circumcircle through centers (key patterns from AoPS AIME 2022 II #15)
- Setup:
  • Two externally tangent circles ω1, ω2 with centers O1, O2 and radii r1, r2, with r1 + r2 = O1O2.
  • A third circle Ω through O1, O2 intersects ω1 at B, C and ω2 at A, D; A, B, C, D cyclic on Ω.
- Symmetry/reflection:
  • Reflect A, B across the perpendicular bisector of O1O2 to A′, B′ on Ω. Then ABO1O2 ≅ B′A′O2O1 and hexagons ABO1CDO2 and A′B′O1CDO2 have equal area.
  • Quadrilateral A′B′CD is an isosceles trapezoid; also B′O1DO2 is an isosceles trapezoid.
  • From these, deduce chord equalities: e.g., B′D = O1O2 and A′C = O1O2; apply Ptolemy on cyclic A′B′CD.
- Ptolemy on A′B′CD:
  • If AB and CD are known chords on Ω and O1O2 is known, Ptolemy yields A′D·B′C + AB·CD = (O1O2)^2.
  • Often forces A′D = B′C = √((O1O2)^2 − AB·CD_sum_correction), e.g., √193 when AB=2, CD=16, O1O2=15.
- Heights/areas:
  • Compute angle in triangle A′B′D via Law of Cosines; typical values yield cos α = 3/5 ⇒ sin α = 4/5 (a 3–4–5 relation), giving height 12 and trapezoid area = 1/2·height·(sum of bases).
  • For triangles involving radii r1, r2 spanning angle (π − α), Law of Cosines gives r1r2 from chord length, e.g., (r1 + r2)^2 − (4/5)r1r2 = known ⇒ r1r2 = 40.
  • Then area of each triangle with sides r1, r2 and included angle α is (1/2) r1 r2 sin α = 16. Sum areas: trapezoid plus two triangles.
- Alternative trigonometric approach:
  • Let ∠O1OO2 = 2θ for Ω’s center O; derive AB = 2R sin(θ − (α+β)/2), CD = 2R sin(θ + (α+β)/2).
  • Adding/subtracting AB and CD with O1O2 = 2R sin θ yields cos((α+β)/2) and sin((α+β)/2) (e.g., 3/5 and 4/5), then compute R, cos θ, etc., and total area via r^2-weighted sine sums.
- Avoid coordinate bloat; prefer radical axis, cyclic/quadrilateral relations, Ptolemy, symmetry, and concise trigonometry.

K) Logarithm equations with same unknown on bases/arguments
- Let the common value be y; rewrite as exponential equations and divide to eliminate x:
  • If (A x)^y = B x and (C x)^y = D x, then (A/C)^y = B/D ⇒ y = log_{A/C}(B/D) = log_{10}(B/D) when A/C = 10, etc.
- Check domain: x > 0, bases > 0 and ≠ 1, arguments > 0; obey any excluded x. If problem asserts existence of such x, you can proceed to compute y; a quick verification by back-substitution suffices.
- If the final asks for log_{10}(m/n), reduce the fraction and return m + n.

Quality checks
- Verify digit/base constraints and equalities numerically if applicable.
- In geometry, confirm derived equal lengths/angles meet cyclic and tangency conditions and that area decomposition pieces sum correctly.
- For logarithms, verify the deduced y makes both given logs equal; ensure m,n are coprime before summing.
- For extremal problems, present both a bound and a construction achieving it.

Final step
- Put the clean final numeric result in the “answer” field only. No extra text or formatting.
2025/08/12 23:20:17 INFO dspy.evaluate.evaluate: Average Metric: 3.0 / 3 (100.0%)
2025/08/12 23:22:13 INFO dspy.evaluate.evaluate: Average Metric: 23.0 / 45 (51.1%)
2025/08/12 23:22:13 INFO dspy.teleprompt.gepa.gepa: Iteration 18: Full valset score for new program: 0.5111111111111111
2025/08/12 23:22:13 INFO dspy.teleprompt.gepa.gepa: Iteration 18: Full train_val score for new program: 0.5111111111111111
2025/08/12 23:22:13 INFO dspy.teleprompt.gepa.gepa: Iteration 18: Individual valset scores for new program: [0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0]
2025/08/12 23:22:13 INFO dspy.teleprompt.gepa.gepa: Iteration 18: New valset pareto front scores: [0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1]
2025/08/12 23:22:13 INFO dspy.teleprompt.gepa.gepa: Iteration 18: Full valset pareto front score: 0.8
2025/08/12 23:22:13 INFO dspy.teleprompt.gepa.gepa: Iteration 18: Updated valset pareto front programs: [{0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, {0, 1, 3, 7, 8, 9}, {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, {8, 1, 2}, {0, 1, 3, 4, 7, 8, 9}, {0, 2, 3, 4, 5, 6, 7, 8, 9}, {0, 1, 2, 4, 5, 6, 7, 8, 9}, {8, 6}, {0, 9, 2, 4}, {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, {2, 5, 6, 8, 9}, {2, 4, 6, 8, 9}, {0, 2, 3, 4, 5, 6, 8, 9}, {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, {1, 2, 4, 6, 7, 8, 9}, {9, 3, 4}, {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, {1, 2, 3, 4, 5, 6, 7, 8, 9}, {0, 1, 2, 4, 5, 9}, {0, 1, 2, 3, 4, 5, 6, 7, 9}, {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, {8, 2, 7}, {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, {0, 1, 2, 7, 8}, {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, {2, 3}, {3, 7}, {1, 4}, {1, 2, 3, 9}, {2}, {1, 2, 3, 5, 6, 7, 8, 9}, {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, {9, 5}, {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, {7}, {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, {3}, {3, 6}, {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, {0}, {3}]
2025/08/12 23:22:13 INFO dspy.teleprompt.gepa.gepa: Iteration 18: Best valset aggregate score so far: 0.5333333333333333
2025/08/12 23:22:13 INFO dspy.teleprompt.gepa.gepa: Iteration 18: Best program as per aggregate score on train_val: 2
2025/08/12 23:22:13 INFO dspy.teleprompt.gepa.gepa: Iteration 18: Best program as per aggregate score on valset: 2
2025/08/12 23:22:13 INFO dspy.teleprompt.gepa.gepa: Iteration 18: Best score on valset: 0.5333333333333333
2025/08/12 23:22:13 INFO dspy.teleprompt.gepa.gepa: Iteration 18: Best score on train_val: 0.5333333333333333
2025/08/12 23:22:13 INFO dspy.teleprompt.gepa.gepa: Iteration 18: Linear pareto front program index: 2
2025/08/12 23:22:13 INFO dspy.teleprompt.gepa.gepa: Iteration 18: New program candidate index: 9
2025/08/12 23:22:13 INFO dspy.teleprompt.gepa.gepa: Iteration 19: Selected program 8 score: 0.4666666666666667
Average Metric: 3.00 / 3 (100.0%): 100%|█████████████████████████████████████████████████████████████████████████████████████████████████████████| 3/3 [01:03<00:00, 21.21s/it]
2025/08/12 23:23:16 INFO dspy.evaluate.evaluate: Average Metric: 3.0 / 3 (100.0%)
2025/08/12 23:23:16 INFO dspy.teleprompt.gepa.gepa: Iteration 19: All subsample scores perfect. Skipping.
2025/08/12 23:23:16 INFO dspy.teleprompt.gepa.gepa: Iteration 19: Reflective mutation did not propose a new candidate
2025/08/12 23:23:16 INFO dspy.teleprompt.gepa.gepa: Iteration 20: Selected program 4 score: 0.4222222222222222

Average Metric: 2.00 / 3 (66.7%): 100%|██████████████████████████████████████████████████████████████████████████████████████████████████████████| 3/3 [00:46<00:00, 15.62s/it]
2025/08/12 23:24:03 INFO dspy.evaluate.evaluate: Average Metric: 2.0 / 3 (66.7%)

2025/08/12 23:25:38 INFO dspy.teleprompt.gepa.gepa: Iteration 20: Proposed new text for predict: You will be given one math problem as plain text under a key like “problem.” Your job is to solve it correctly and return:

- reasoning: a concise, logically ordered solution that uses structure/identities to avoid brute force and ends with a quick verification.
- answer: the final requested number/expression only (no extra words).

Formatting:
- Use exactly two top-level fields named “reasoning” and “answer.”
- Keep reasoning succinct but complete. Bullet points are fine.
- The answer field must contain only the final value requested (e.g., 227, 585, 601).

General problem-solving guidance:
- Parse the problem type (e.g., base representation, combinatorial configurations with symmetry, coefficient extraction via generating functions, motion with relative velocities, products over roots of unity, modular divisibility on ratios).
- Enforce domain constraints (e.g., digit bounds in base problems; “three-digit” means no leading zero; counts are integers; ordered vs unordered; strict increase; geometric constraints).
- Prefer structural identities, symmetry, and modular arithmetic over brute force; derive clean case structures and justify completeness.
- For extremal/“least/greatest” and counting problems, give both a bound and a construction/argument attaining it.

Quality checks:
- Verify base/digit constraints and final equalities numerically if applicable.
- For counting with cases, ensure cases are disjoint and exhaustive; check complements/forbidden patterns.
- Keep arithmetic exact (no unnecessary decimal approximations; reduce modulo only at the end if asked).
- Finish with a brief verification (e.g., plug back, check constraints, or a quick numeric check).

Domain-specific strategies and pitfalls (including lessons from prior feedback):

A) Rectangles in regular polygons (2-colorings on a regular 2n-gon; e.g., dodecagon):
- Geometry fact: A vertex quadrilateral is a rectangle iff its diagonals pass through the center; i.e., it is formed by two opposite vertex pairs.
- Partition the 2n-gon into n opposite pairs. A monochromatic rectangle occurs iff there exist two distinct opposite pairs of the same color.
- Counting colorings with no monochromatic rectangle reduces to casework on opposite pairs:
  • Case 0: No same-colored opposite pair (each pair has one of each color): 2^n.
  • Case 1: Exactly one same-colored opposite pair: n·2·2^{n−1}.
  • Case 2: Exactly two same-colored opposite pairs of different colors: n·(n−1)·2^{n−2}.
  • Any additional same-colored opposite pair forces a monochromatic rectangle.
- Do not confuse with m×m grid Ramsey rectangles; here the structure is via opposite pairs.

B) Products over roots of unity (key identities and sign handling):
- For ω primitive n-th root, the set {ω^k}_{k=0}^{n−1} are the roots of z^n − 1.
- Core identity: ∏_{k=0}^{n−1} (a − ω^k) = a^n − 1.
- If f(x) factors as ∏ (x − r_i), then ∏_{k} f(ω^k) = ∏_{i} ∏_{k} (r_i − ω^k) = ∏_{i} (r_i^n − 1).
  • Example: For f(x) = x^2 − 2x + 2 = (x − (1+i))(x − (1−i)), we get
    ∏_{k=0}^{n−1} f(ω^k) = ((1+i)^n − 1)((1−i)^n − 1).
- Shifted-root trick: If z_k = ω^k − 1, then H(t) = ∏_{k} (t − z_k) = (t+1)^n − 1.
  • Then ∏_{k} (z_k^2 + 1) = ∏ (z_k − i) ∏ (z_k + i) = H(i)·H(−i). Sign caution:
    With H(t) = ∏ (t − z_k), we have ∏ (z_k − a) = (−1)^n H(a). Using both (a = i and a = −i)
    makes the signs cancel: ∏ (z_k − i) ∏ (z_k + i) = (−1)^n H(i) · (−1)^n H(−i) = H(i)H(−i).
  • This avoids the common sign error.
- For (1±i)^n, use 1±i = √2 e^{± iπ/4} to compute exactly via De Moivre; reduce mod at the end.

C) Coefficient extraction for (x^N − 1)^k / ∏(x^{m_i} − 1), 0 < x < 1:
- Rewrite with 1 − x^N and geometric series: P(x) = (1 − x^N)^k / ∏(1 − x^{m_i}).
- Binomial-expand numerator: (1 − x^N)^k = ∑_{r=0}^k (−1)^r C(k,r) x^{Nr}.
- Coefficient of x^t equals ∑_{r=0}^{⌊t/N⌋} (−1)^r C(k,r) · a_{t−Nr}, where a_s counts nonnegative integer solutions to ∑ m_i n_i = s.
- Use modular constraints, gcd/lcm structure, and stars-and-bars to compute a_s exactly.

D) Motion in a current (relative velocities; equidistant target on bank):
- Coordinates: river along x-axis; banks y = 0 and y = W.
- If target is midpoint horizontally, x* = (x1 + x2)/2. Speeds relative to water s1, s2; current v_c.
- Two methods:
  1) Still-water reduction: shift the common aim point by v_c t; write right-triangle equations, subtract to get D in terms of t, then solve using W.
  2) Vector components: For net ground velocities (±x, y), impose (x ∓ v_c)^2 + y^2 = s^2; solve for x,y, then t = W/y and D = 2xt.
- Keep values exact; avoid decimals.

E) Ratios that change after adding/removing a fixed number (divisibility/minimization):
- Let initial total N and initial ratio p/q imply N ≡ 0 (mod q).
- After adding/removing K people, if the new ratio is r/s, then N+K ≡ 0 (mod s).
- To minimize totals (or counts like number of adults), take the least N satisfying both congruences (CRT) and then compute the implied count (e.g., adults = (r/s)(N+K)).
- If composition of the added group is unknown, you generally don’t need to split it (x adults among the K) when only the final ratio is given; divisibility on totals suffices.

F) Other included standard tactics:
1) Base-conversion/digit rearrangement:
   - In base b, (abc)_b = a b^2 + b b + c; digits in 0..b−1; no leading zero for “three-digit”.
2) Palindromes across bases: bound lengths, parametrize palindromes, use constraints from both bases; justify “greatest” by structured descending search.
3) Symmetric sums with fixed sum: use identities like S = (a+b+c)(ab+bc+ca) − 3abc; count ordered vs unordered carefully.
4) Intersecting families: empty set excluded; use complement pairs and size > n/2 pigeonhole facts; stratify by small-set inclusions.
5) Avoiding 4-term APs in increasing sequences with anchors: pre-eliminate values forming APs; count allowed pairs; subtract completions; avoid double counts.
6) Order statistics with sum/absolute-sum constraints: positive mass equals negative mass; bound extremes by averaging slots; construct extremizers and verify.

Common pitfalls to avoid:
- Roots-of-unity products: incorrect overall sign; forgetting to include k=0 term; mixing up (a − ω^k) vs (ω^k − a).
- Using grid-rectangle facts on polygon vertex problems (use opposite-pair structure instead).
- Ignoring integer/divisibility constraints in ratio problems; assuming composition of added group when unnecessary.
- Sloppy modular arithmetic or premature modular reduction before exact evaluation.

Finally:
- Put the clean final numeric result in the “answer” field only.
2025/08/12 23:26:30 INFO dspy.evaluate.evaluate: Average Metric: 1.0 / 3 (33.3%)
2025/08/12 23:26:30 INFO dspy.teleprompt.gepa.gepa: Iteration 20: New subsample score is not better, skipping
2025/08/12 23:26:30 INFO dspy.teleprompt.gepa.gepa: Iteration 21: Selected program 9 score: 0.5111111111111111
Average Metric: 1.00 / 3 (33.3%): 100%|██████████████████████████████████████████████████████████████████████████████████████████████████████████| 3/3 [00:43<00:00, 14.35s/it]
2025/08/12 23:27:13 INFO dspy.evaluate.evaluate: Average Metric: 1.0 / 3 (33.3%)

2025/08/12 23:29:37 INFO dspy.teleprompt.gepa.gepa: Iteration 21: Proposed new text for predict: Your job
- Solve exactly one math problem and return exactly two top-level JSON-like fields:
  - reasoning: a concise, logically ordered solution that leverages structure (identities, modular arithmetic, symmetry) and ends with a quick verification.
  - answer: the final requested number/expression only (no extra words or formatting).

Formatting
- Use exactly the two fields “reasoning” and “answer.”
- Keep the reasoning succinct but complete; bullet lists are fine.
- The answer field must contain only the final value requested (e.g., 227, 585, 601).
- Avoid heavy formatting (no LaTeX blocks, tables). Bullet lists are acceptable.

General problem-solving guidance
- Identify the problem type early (e.g., base/digit relations, repeating decimals, palindromes across bases, symmetric sums with fixed totals, intersecting subset families, avoiding arithmetic progressions, order statistics with absolute-sum constraints, floor-sum optimization, circle/geometry with radical axes or cyclic quadrilaterals, logarithm equations with shared unknowns, counting rectangles via chord-direction classes, equilateral hexagon with opposite sides parallel via support-triangle similarity).
- Enforce domain constraints strictly:
  • Base-b digits are in 0..b−1; leading digit nonzero if a positional numeral.
  • Logarithms: argument > 0, base > 0 and ≠ 1; respect any excluded x-values given.
  • Ordered vs unordered; strictly increasing where specified.
- Prefer structural arguments over brute force:
  • Factorization, completing the square, modular arithmetic, symmetry/reflection, similar triangles, cyclic identities, Ptolemy’s theorem, Law of Cosines/Sines, power of a point, bounding plus construction for extremals.
- For greatest/least questions, derive a tight bound and exhibit a construction attaining it.
- Always verify at the end (e.g., plug back into conditions, check modular constraints, recompute a key equality numerically).

Domain-specific strategies and pitfalls

A) Base conversion/digit equations
- Translate positional notation correctly: abc (base 10) = 100a + 10b + c; (a b c)_b = a·b^2 + b·b + c.
- Enforce digit ranges: in base 9, digits ∈ {0,…,8}. Leading digit in base-b must be 1..b−1 if a k-digit numeral.
- Use modular constraints to prune search (e.g., mod 9 or mod 71 from coefficient comparison).
- Verify numerically by converting the base-b representation back to decimal.

B) Repeating decimals 0.\overline{abcd} and counting reduced numerators
- 0.\overline{abcd} = m/9999 with 9999 = 3^2·11·101.
- Reduced numerator is x = m / gcd(m, 9999).
- Characterize x via existence of y | 9999 with y ≥ x and gcd(x, y) = 1 (since m = x·(9999/y) ≤ 9999).
- Use inclusion–exclusion over divisibility by 3, 9, 11, 101 carefully; treat “3 vs 9” distinctly.

C) Palindromes across bases
- Bound number of digits by magnitude; write k-digit palindrome formulas (e.g., base-8 ABA: 65A + 8B; ABBA: 513A + 72B, A ≥ 1).
- Test candidate bases/lengths using congruences and magnitude bounds.

D) Symmetric sums with a + b + c fixed
- Compress using identities, e.g., S = (a + b + c)(ab + bc + ca) − 3abc.

E) Intersecting families of subsets
- Intersecting means every pair intersects; empty set excluded.
- Complement pairs cannot both be present; for n odd, star and “all subsets of size > n/2” both have size 2^{n−1}.
- For fixed-size counting, case on minimum sizes and overlaps; use canonical patterns to avoid double-counting.

F) Avoiding 4-term arithmetic progressions
- Pre-eliminate values that force a 4-term AP with anchors.
- If endpoints differ by Δ, then interior step is Δ/3; require integrality and bounds.

G) Order statistics with sum and absolute-sum constraints
- Sum |x_i| = 1 and sum x_i = 0 ⇒ positive mass = negative mass = 1/2.
- For maximizing x_k, with T = n−k+1 largest positions, x_k ≤ (1/2)/T; minimize x_l similarly.
- Achieve bounds by concentrating masses evenly; rest zero.

H) Systems with square-root bilinear forms
- Factor under roots: √(x(2−y)) style.
- Algebraic approach (A=1−x etc.) or trigonometric substitution (x=2cos^2 α); typical products/constants often emerge; verify by back-substitution.

I) Floor-sum with a parameter
- Separate continuous main term and fractional correction.
- Pick parameter to cancel main term; argue uniqueness via step sizes.
- Count residues modulo m precisely, including partial cycles.

J) Circle geometry with two tangent circles and a circumcircle through centers
- Prefer radical axis, cyclic/quadrilateral relations, Ptolemy, symmetry, and concise trigonometry over coordinates.
- Use reflections across the perpendicular bisector of O1O2; exploit isosceles trapezoids; Ptolemy on A′B′CD; infer 3–4–5 angle patterns frequently.

K) Logarithm equations with same unknown on bases/arguments
- Let the common value be y; rewrite as exponential equations and divide to eliminate x:
  • If (A x)^y = B x and (C x)^y = D x, then (A/C)^y = B/D ⇒ y = log_{A/C}(B/D).
- Check domain: x > 0, bases > 0 and ≠ 1; ensure arguments > 0. Verify numerically.

L) Counting rectangles formed by chords in a regular polygon (crucial for regular 12-gon)
- “Each side lies on a side or a diagonal” means each rectangle side is a chord line of the polygon; vertices of the rectangle need not be polygon vertices.
- Classify by chord directions (slopes). For a regular dodecagon:
  • All chord directions occur at multiples of 15° (not just edges at 30°). Perpendicular pairs differ by 90°.
  • The six relevant direction pairs are:
    - {0°, 90°}, {30°, 120°}, {60°, 150°} (call these the “even” families),
    - {15°, 105°}, {45°, 135°}, {75°, 165°} (call these the “odd” families).
- Allowed chord “distances” (minor-arc skipped vertices) by family:
  • Even families: even distances 0, 2, 4.
  • Odd families: odd distances 1, 3, 5 (note distance 5 has only one chord, so cannot be the shorter in a pair of parallel chords).
- Counting template for each family:
  • Fix the shorter distance d_short and count pairs of parallel chords in that family: count = 1 + 2·(D_max − d_short), where D_max = 4 for even families and 3 for odd families.
  • For the perpendicular family, the number of ways to choose the two parallel lines is C(d_short + 2, 2) for even families with d_short = 0,2,4 interpreted as 2i (i=0..2), and C(d_short + 2, 2) for odd families with d_short = 1,3 interpreted as 2i+1 (i=0,1).
  • Sum over admissible d_short in the family, then multiply by 3 for the three rotations of that family type.
- Inclusion–exclusion and overcounting:
  • The above classification by perpendicular direction pairs avoids double-counting across families.
  • Don’t restrict to rectangles aligned with diameters or vertex-anchored; many valid rectangles use non-vertex chords.

M) Equilateral hexagon with opposite sides parallel and the “support triangle” (lines AB, CD, EF)
- Let ABCDEF be equilateral with AB ∥ DE, BC ∥ EF, CD ∥ FA. Extend lines AB, CD, EF to form triangle PQR where:
  • P = AB ∩ CD, Q = CD ∩ EF, R = EF ∩ AB.
- Powerful similarity relations:
  • Small corner triangles cut off by the hexagon are similar to ΔPQR. Typical equalities (with hex side length x):
    - From ΔBCP ~ ΔRQP: x / BP = (side on ΔPQR parallel to BC) / (adjacent side). If the sides of ΔPQR opposite AB, CD, EF are known (say lengths along RQ, PR, PQ), plug the corresponding numbers to get BP in terms of x.
    - From ΔAFR ~ ΔPQR: x / AR = (corresponding side ratio), yielding AR in terms of x.
  • Use a side of ΔPQR as a sum of consecutive segments cut by the hexagon: e.g., RA + AB + BP equals the side of ΔPQR that AB is parallel to.
- Linear solve:
  • Set up 2–3 linear relations like BP = αx, AR = βx, then add along a side of ΔPQR (e.g., RA + AB + BP = given side length L) to solve for x.
- Avoid pitfalls:
  • Do not assume the support triangle is isosceles or has sides in simple fixed ratios like s, s√3, 2s; directions need not be 120° apart affinely.
  • The inradius/semiperimeter shortcut (e.g., s = 2r of the support triangle) is not generally valid here.

N) GCD-sum conditions for scaled rationals (e.g., r and 55r)
- Let r = a/b in lowest terms; write 55r = (55a)/b and reduce by d = gcd(55a, b).
- Equate sums of numerator and denominator: a + b = (55a + b)/d ⇒ (d − 55)a = (1 − d)b.
- Since gcd(a, b) = 1 and d | b and d | 55a, conclude d | 55, so d ∈ divisors of 55.
- Test each feasible d, enforce integrality and gcd=1, and verify both reduced fractions share the same sum.

Quality checks
- Verify digit/base constraints and equalities numerically if applicable.
- In geometry, confirm derived lengths/angles meet the stated parallel/perpendicular and cyclic conditions; ensure area/length decompositions sum correctly.
- For logarithms, verify the deduced y makes both given logs equal; ensure m,n are coprime before summing if requested.
- For extremal problems, present both a bound and a construction achieving it.

Final step
- Put the clean final numeric result in the “answer” field only. No extra text or formatting.
2025/08/12 23:29:51 INFO dspy.evaluate.evaluate: Average Metric: 1.0 / 3 (33.3%)
2025/08/12 23:29:51 INFO dspy.teleprompt.gepa.gepa: Iteration 21: New subsample score is not better, skipping

Let's see the prompt generated¶

In [10]:

print(optimized_program.predict.signature.instructions)

print(optimized_program.predict.signature.instructions)

You will be given one math problem as plain text under a key like “problem.” Your job is to solve it correctly and return:

- reasoning: a concise, logically ordered solution that uses identities/structure to avoid brute force, ends with a quick verification.
- answer: the final requested number/expression only (no extra words).

Formatting:
- Use exactly two top-level fields named “reasoning” and “answer.”
- Keep reasoning succinct but complete. Bullet points are fine.
- The answer field must contain only the final value requested (e.g., 227, 585, 601).

General problem-solving guidance:
- Parse the problem type (e.g., base representation, intersecting families of subsets, avoiding arithmetic progressions, symmetric sums with constraints, ordered tuples counting).
- Always enforce domain constraints (e.g., base-b digits in 0..b−1; no leading zero for base-10 “three-digit”; ordered vs unordered families; strict increase conditions in sequences).
- Use algebraic identities and modular arithmetic to reduce the search space; prefer structural arguments over naive enumeration.
- For “greatest/least” questions, derive tight bounds and give a construction that attains them.

Domain-specific strategies and pitfalls (learned from typical contest problems and prior feedback):

1) Base-conversion/digit rearrangement:
- Translate positional notation correctly: in base b, (a b c)_b = a·b^2 + b·b + c; in base 10: abc = 100a + 10b + c.
- Enforce digit ranges strictly (e.g., in base 9, digits ∈ {0,…,8}; if also a is a base-10 leading digit, then a ∈ {1,…,8}).
- Set up equality and simplify. Use modular constraints to prune:
  • Mod 9 often collapses coefficients; e.g., 99a = 71b + 8c ⇒ mod 9 gives b + c ≡ 0 (mod 9).
  • Mod 8: 99 ≡ 3, 71 ≡ 7 ⇒ 3a ≡ 7b (mod 8) ⇒ b ≡ −3a (mod 8).
- Solve within digit bounds and verify numerically.

2) Palindromes across bases:
- Bound the base length by magnitude (e.g., n < 1000 ⇒ octal has 3–4 digits).
- Characterize palindromes:
  • 3-digit octal: (A B A)_8 = 65A + 8B.
  • 4-digit octal: (A B B A)_8 = 513A + 72B (with A ≥ 1).
- Enumerate small parameter ranges and test the other-base palindrome constraint. For “greatest”, check candidates in descending order with justification.

3) Symmetric sums with a + b + c fixed (ordered triples of nonnegative integers):
- Use identities to compress expressions:
  S = ab(a + b) + bc(b + c) + ca(c + a) = (a + b + c)(ab + bc + ca) − 3abc.
- With a + b + c known (e.g., 300), convert the given sum into a relation among ab + bc + ca and abc.
- Use the shift a = A + x etc. to isolate a product like (a−A)(b−A)(c−A) and deduce factorization constraints, enabling clean counting.
- Count ordered solutions carefully; include/exclude symmetric/degenerate cases precisely.

4) Intersecting families of subsets (collections from the power set):
- Intersecting means every pair has nonempty intersection. The empty set cannot be included.
- Complement pairs: S and S^c cannot both be present. Use this to structure counts.
- Use size-based pigeonhole facts: In [n], any two subsets of size > n/2 must intersect. For n = 5, any two subsets of size ≥ 3 intersect; thus “all subsets of size ≥ 3” is an intersecting family (size 16).
- Do not assume that “stars” (all subsets containing a fixed element) are the only intersecting families of maximum size. For odd n, both the star and “all subsets of size > n/2” have size 2^{n−1}.
- When counting collections of a fixed size:
  • Consider the minimum set size N in the family and do casework on how many 2-element sets are included (for n=5), as these control which 3-sets must be excluded (complements).
  • Ensure completeness of cases and avoid double counting by parameterizing canonical patterns (e.g., how many 2-sets, how they overlap, whether they share a common element).
  • Remember order of subsets in a collection does not matter; count distinct families.

5) Avoiding 4-term arithmetic progressions in a strictly increasing sequence with fixed anchors:
- First bound the variable terms by strict increase (e.g., if fixed terms are 3,4,5,...,30,40,50 then 6 ≤ a < b ≤ 29).
- Pre-eliminate values that cause a 4-term AP with three fixed terms:
  • 3,4,5,a forbids a = 6.
  • b,30,40,50 forbids b = 20.
  • Similarly, a,30,40,50 forbids a = 20.
- Start with the count of pairs from allowed values and then subtract specific pairs that complete APs with two fixed endpoints:
  • 3,5,a,b ⇒ (a,b) = (7,9).
  • 3,a,b,30 ⇒ (a,b) = (12,21).
  • 4,a,b,40 ⇒ (a,b) = (16,28).
  • 5,a,b,50 ⇒ (a,b) = (20,35) but may be outside bounds or pre-excluded (e.g., 20 banned).
- Systematically check all endpoint combinations; use the fact that if endpoints differ by Δ, then Δ must be divisible by 3 for a 4-term AP, and solve for integer a,b within bounds.
- Avoid double subtraction; ensure monotonicity and domain constraints are respected.

6) Order statistics with sum and absolute-sum constraints (e.g., x_1 ≤ ... ≤ x_n, sum |x_i| = 1, sum x_i = 0):
- Total positive mass equals total negative mass: both = 1/2.
- For maximizing x_k (k near the top): if there are T largest terms from k to n (T = n − k + 1), then sum of these T terms ≥ T·x_k. Since the total positive mass ≤ 1/2, we get x_k ≤ (1/2)/T.
- For minimizing x_l (l near the bottom): if there are l smallest terms, sum of these l terms ≤ l·x_l. Since the total negative mass is −1/2, we get x_l ≥ (−1/2)/l.
- To attain these bounds, concentrate masses evenly on exactly those positions: set the smallest l terms equal to −1/(2l), the largest T terms equal to 1/(2T), and the middle to 0 (respecting monotonicity). Verify sums and absolute sums.
- Example: For n=100, maximize x_76 − x_16: T = 25 ⇒ x_76 ≤ 1/50; l = 16 ⇒ x_16 ≥ −1/32; construction with 16 negatives at −1/32, 59 zeros, 25 positives at 1/50 attains 1/50 − (−1/32) = 41/800.

Quality checks:
- Verify digit/base constraints and final equalities numerically if applicable.
- For extremal problems, provide both a tight bound and an explicit construction achieving it.
- For counting, explicitly handle ordered vs unordered, exclude impossible/duplicate cases, and check complements/forbidden pairs.
- For AP-avoidance, confirm integrality and bounds; ensure no missed endpoint combinations.
- For “greatest/least” questions, justify optimality structurally (e.g., convexity/majorization/pigeonhole).

Finally:
- Put the clean final numeric result in the “answer” field only.

It can be seen that what GEPA is doing here, is precomputing some reasoning to come up with a good plan for future task instances. Due to the improved performance in unseen validation set, we expect this prompt to generalize!

Evaluating the Chain Of Thought optimized with GEPA¶

In [11]:

evaluate(optimized_program)

evaluate(optimized_program)

Average Metric: 85.00 / 150 (56.7%): 100%|██████████████████████████████████████████████████████████████████████████████████████████████████| 150/150 [00:00<00:00, 476.89it/s]

2025/08/12 23:53:14 INFO dspy.evaluate.evaluate: Average Metric: 85 / 150 (56.7%)

	problem	example_answer	reasoning	pred_answer	metric
0	Find the sum of all integer bases $b>9$ for which $17_b$ is a divi...	70	- Interpret the numbers in base \( b \): \[ 17_b = 1 \cdot b + 7 =...	70	✔️ [1]
1	On $\triangle ABC$ points $A, D, E$, and $B$ lie in that order on ...	588	1. Set up coordinate system: - Place \( A \) at the origin \((0,0)...	588	✔️ [1]
2	The 9 members of a baseball team went to an ice-cream parlor after...	16	- We have 9 players and 3 flavors: Chocolate (C), Vanilla (V), Str...	16	✔️ [1]
3	Find the number of ordered pairs $(x,y)$, where both $x$ and $y$ a...	117	We need to find all integer pairs \((x,y)\) with \(x,y \in [-100, ...	117	✔️ [1]
4	There are $8!= 40320$ eight-digit positive integers that use each ...	279	- We consider all 8-digit numbers using each of the digits 1 throu...	567
...	...	...	...	...	...
145	Let $S$ be the set of vertices of a regular $24$-gon. Find the num...	113	- The problem is to find the number of ways to pair up the 24 vert...	11
146	Let $A_1 A_2 A_3 \ldots A_{11}$ be an $11$-sided non-convex simple...	19	We are given a simple polygon \(A_1 A_2 \dots A_{11}\) with vertic...	19	✔️ [1]
147	Let $x_1, x_2, x_3, \ldots$ be a sequence of rational numbers defi...	248	Given the recurrence: \[ x_1 = \frac{25}{11}, \quad x_{k+1} = \fra...	728
148	Let $\triangle ABC$ be a right triangle with $\angle A = 90^\circ$...	104	- Given the right triangle \( \triangle ABC \) with right angle at...	104	✔️ [1]
149	There are exactly three positive real numbers $k$ such that the fu...	240	We are given \[ f(x) = \frac{(x-18)(x-72)(x-98)(x-k)}{x}, \quad x>...	252

150 rows × 5 columns

Out[11]:

EvaluationResult(score=56.67, results=<list of 150 results>)

GEPA was able to optimize the GPT-4.1 Mini's performance on AIME 2025 from 46.6% score to 56.6%, a 10% improvement, with just a budget of auto="light"!