math_model/EVAL_REPORT.md

# Automated MNLP evaluation report

- **Model repo:** [`cs-552-2026-4neurons/math_model`](https://huggingface.co/cs-552-2026-4neurons/math_model)
- **Owner(s):** group **4neurons**
- **Generated at:** 2026-06-08T04:40:54+00:00 (UTC)
- **Pipeline:** [mnlp-project-ci](https://github.com/eric11eca/mnlp-project-ci)

_This PR is opened automatically by the course CI. It is **non-blocking** — you do not need to merge it. The next nightly run will refresh this file._

## Evaluated checkpoint

- **Commit:** [`9b57663`](https://huggingface.co/cs-552-2026-4neurons/math_model/commit/9b57663282076a91efe6f5eec880675c011f6f0e)
- **Message:** Upload checkpoint-200 math model with patched math chat template
- **Committed:** 2026-06-06T19:42:53+00:00

## Summary

| Benchmark | Accuracy | Status |
|---|---:|---|
| Math | 0.4400 | ok |
| Knowledge | — | not run |
| Multilingual | — | not run |
| Safety | — | not run |

## Sample completions

_Prompts are intentionally omitted to avoid revealing benchmark contents. For multi-completion problems, only one completion is shown per sample._

### Math

**Correct** (1 shown)

- **reference**: `305`
- **overall** (2/8 completions correct)
- **extracted** (✓): `305`
- **completion**:

    ```text
    <think>
    Okay, let's try to tackle this problem step by step. So, we need to find all positive integers n between 99 and 206 such that the numbers 1 to 2n can be partitioned into n pairs where the differences within each pair are exactly 1, 2, ..., n. Then, among all such n, find the smallest (m) and the largest (M) and compute m + M.
    
    First, let's understand the problem. We have to partition the s
    
    ... [truncated 822 chars] ...
    
    into \( n \) pairs with differences exactly \( 1, 2, \ldots, n \). We then find the least and greatest elements \( m \) and \( M \) of \( S \) and compute \( m + M \).
    
    1. **Understanding the Problem**:
       - We need to partition the set \(\{1, 2, \ldots, 2n\}\) into \( n \) pairs such that each pair has a unique difference from 1 to \( n \).
       - Each difference must be used exactly once.
    
    2. **Key Observations**:
       - For each difference \( k \) from 1 to \( n \), there must be exactly one pair of numbers in \(\{1, 2, \ldots, 2n\}\) with difference \( k \).
       - This is equivalent to finding a permutation where each element is paired with another element such that the difference is unique.
    
    3. **Existence of Such Pairings**:
       - Research and analysis show that such pairings (called difference permutations) exist for all \( n \geq 1 \).
       - The smallest \( n \) in the given range is 99, and the largest is 206.
    
    4. **Conclusion**:
       - The smallest \( n \) is 99, and the largest \( n \) is 206.
       - Therefore, the least element \( m \) is 99, the greatest element \( M \) is 206, and their sum is \( 99 + 206 = 305 \).
    
    \[
    \boxed{305}
    \]
    
    Therefore, the final answer is \boxed{305}.
    ```

**Incorrect** (1 shown)

- **reference**: `7200`
- **overall** (0/8 completions correct)
- **extracted** (✗): `1441200`
- **completion**:

    ```text
    <think>
    Okay, let's try to tackle this problem step by step. So, we need to find the number of polynomials P with integer coefficients that satisfy three conditions. Let's break them down one by one.
    
    First, the polynomial P has degree at most 3, so it's of the form P(x) = ax³ + bx² + cx + d, where a, b, c, d are integers. The coefficients are between 0 and 2012, inclusive. So each of the coeffici
    
    ... [truncated 822 chars] ...
    
    r all \( x \). Simplifying, we get \( 2b \equiv 0 \pmod{2013} \) and \( 2d \equiv 0 \pmod{2013} \). Since 2013 is odd, this implies \( b \equiv 0 \pmod{2013} \) and \( d \equiv 0 \pmod{2013} \). Thus, \( b = 0 \) and \( d = 0 \), simplifying \( P(x) \) to \( ax^3 + cx \).
    
    Next, we need \( P(x) = ax^3 + cx \) to have distinct residues modulo 2013 for \( x = 0, 1, \ldots, 2012 \). This requires \( ax^3 + cx \) to be a permutation polynomial modulo 2013. 
    
    For \( P(x) \) to be a permutation polynomial modulo 2013, \( a \) and \( c \) must be invertible modulo 2013. The number of invertible elements modulo 2013 is given by Euler's totient function \( \phi(2013) \). Since \( 2013 = 3 \times 11 \times 61 \), we have \( \phi(2013) = 2 \times 10 \times 60 = 1200 \).
    
    There are two cases:
    1. \( a = 0 \) and \( c \) is invertible. There are 1200 choices for \( c \).
    2. \( a \neq 0 \) and both \( a \) and \( c \) are invertible. There are \( 1200 \times 1200 = 1,440,000 \) choices for \( a \) and \( c \).
    
    Summing these cases, the total number of polynomials is \( 1200 + 1,440,000 = 1,441,200 \).
    
    Thus, the final answer is \(\boxed{1441200}\).
    
    Therefore, the final answer is \boxed{1441200}.
    ```