# Automated MNLP evaluation report

- **Model repo:** [`cs-552-2026-thinking-tokens/math_model`](https://huggingface.co/cs-552-2026-thinking-tokens/math_model)
- **Owner(s):** group **thinking-tokens**
- **Generated at:** 2026-05-19T03:22:52+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._

## Summary

| Benchmark | Accuracy | Status |
|---|---:|---|
| Math | 0.3900 | 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**: `194`
- **overall** (2/8 completions correct)
- **extracted** (✓): `194`
- **completion**:

    ```text
    We are given a $194 \times 194$ board, and we are to find the **least positive integer $k$** such that we can choose $k$ cells to mark, and there is **exactly one tiling** of the board by $1 \times 2$ and $2 \times 1$ dominoes in which **every domino contains at most one marked cell**.
    
    ---
    
    ### Step 1: Understand the problem
    
    We are to find the **smallest number of marked cells** such that:
    
    - Th
    
    ... [truncated 823 chars] ...
    
    t can be **involving marked cells** is $k$.
    - Since there are 18818 dominoes, the number of dominoes that **do not involve marked cells** is $18818 - k$.
    
    We want **exactly one** tiling where **every domino has at most one marked cell**.
    
    This is equivalent to saying that the **marked cells** must be placed in such a way that **each domino is uniquely determined** by the marked cells.
    
    This is a very strong condition.
    
    ---
    
    ### Step 4: What is the minimum number of marked cells?
    
    This is a classic problem in **combinatorics and tiling theory**.
    
    We are to find the **minimum number of marked cells** such that:
    
    - There is **exactly one tiling** of the board with dominoes.
    - In this tiling, **every domino contains at most one marked cell**.
    
    This is equivalent to saying that the **marked cells** are **independent** in the tiling, and the tiling is **unique**.
    
    This is a well-known result in combinatorics and tiling theory.
    
    ---
    
    ### Step 5: Known result
    
    It is known that the **minimum number of marked cells** needed to ensure **exactly one tiling** with dominoes, where **every domino contains at most one marked cell**, is:
    
    $$
    \boxed{194}
    $$
    
    ---
    
    ### Final Answer:
    
    $$
    \boxed{194}
    $$
    ```

**Incorrect** (1 shown)

- **reference**: `$g(x)=c, g(x)=\lceil x \rceil, g(x)=\lfloor x \rfloor$`
- **overall** (0/8 completions correct)
- **extracted** (✗): `h(x) = x`
- **completion**:

    ```text
    We are given a functional equation:
    
    $$
    h(x) = h\left( \frac{h(bx - a) + a}{b} \right)
    $$
    
    for all $ x \in \mathbb{Q} $, $ a \in \mathbb{Z} $, and $ b \in \mathbb{N} $.
    
    We are to find all functions $ h : \mathbb{Q} \to \mathbb{Z} $ satisfying this equation.
    
    ---
    
    ### Step 1: Try simple candidate functions
    
    Let’s try a function of the form:
    
    $$
    h(x) = kx + c
    $$
    
    for some constants $ k, c \in \math
    
    ... [truncated 822 chars] ...
    
    n, only possible for specific $ a $ and $ b $, not all.
    
    ---
    
    ### Step 7: Try the function $ h(x) = \frac{a}{b} $
    
    Let’s try:
    
    $$
    h(x) = \frac{a}{b}
    $$
    
    But this is not a function from $ \mathbb{Q} \to \mathbb{Z} $, because $ \frac{a}{b} \in \mathbb{Q} $, not necessarily an integer.
    
    ---
    
    ### Step 8: Try $ h(x) = \lfloor x \rfloor $ again
    
    We tried this, and it failed.
    
    ---
    
    ### Step 9: Try a function that maps rational numbers to integers in a specific way
    
    Let’s consider the **identity function**:
    
    $$
    h(x) = x
    $$
    
    Then:
    
    $$
    h(x) = x
    $$
    $$
    h(bx - a) = bx - a
    $$
    $$
    \frac{h(bx - a) + a}{b} = \frac{bx - a + a}{b} = x
    $$
    
    So the equation becomes:
    
    $$
    x = x
    $$
    
    Which is **true**.
    
    So $ h(x) = x $ is a solution.
    
    ---
    
    ### Step 10: Check if there are other solutions
    
    Suppose $ h(x) = x $ is a solution.
    
    Now suppose $ h(x) = x + c $ for some constant $ c \in \mathbb{Z} $
    
    Then:
    
    $$
    h(x) = x + c
    $$
    $$
    h(bx - a) = bx - a + c
    $$
    $$
    \frac{h(bx - a) + a}{b} = \frac{bx - a + c + a}{b} = x + \frac{c}{b}
    $$
    
    So:
    
    $$
    x + c = x + \frac{c}{b} \Rightarrow c = \frac{c}{b}
    $$
    
    This is only possible if $ c = 0 $, so $ h(x) = x $ is the **only** solution.
    
    ---
    
    ### ✅ Final Answer:
    
    $$
    \boxed{h(x) = x}
    $$
    ```