Automated MNLP evaluation report (2026-05-23)

EVAL_REPORT.md

@@ -0,0 +1,87 @@
+# Automated MNLP evaluation report
+
+- **Model repo:** [`cs-552-2026-the-transformers/math_model`](https://huggingface.co/cs-552-2026-the-transformers/math_model)
+- **Owner(s):** group **the-transformers**
+- **Generated at:** 2026-05-23T14:37:35+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 | 0.3600 | 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** (3/8 completions correct)
+- **extracted** (✓): `194`
+- **completion**:
+
+    ```text
+    <think>
+    To solve this problem, we need to understand the conditions under which a \(194 \times 194\) board can be tiled with \(1 \times 2\) and \(2 \times 1\) dominoes, and how marking cells affects this tiling.
+    
+    First, we note that a \(194 \times 194\) board has an area of \(194^2 = 37636\) cells. Each domino covers exactly 2 cells, so the number of dominoes required to tile the board is \(37636
+    
+    ... [truncated 147 chars] ...
+    
+    ontains at most one marked cell. This means that the marked cells can be placed in such a way that no two marked cells are in the same domino.
+    
+    To find the minimum number of marked cells, we need to consider the maximum number of dominoes that can be placed such that no two marked cells are in the same domino. This is equivalent to placing the marked cells in such a way that they are not in the same row or column, as dominoes span two rows and two columns.
+    
+    If we place a marked cell in each row and each column, we can ensure that no two marked cells are in the same row or column. This is similar to placing non-attacking rooks on a chessboard, where no two rooks attack each other.
+    
+    For a \(194 \times 194\) board, the maximum number of non-attacking rooks we can place is 194 (one per row and one per column). Therefore, the minimum number of marked cells required to ensure that no two marked cells are in the same row or column is 194.
+    
+    Thus, the least positive integer \(k\) for which one can choose \(k\) cells to mark so that 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 is \(\boxed{194}\).
+    ```
+
+**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) = c \text{ for all } x \in \mathbb{Q} \text{ and some } c \in \mathbb{Z}`
+- **completion**:
+
+    ```text
+    <think>
+    Given the problem, we need to find all mappings $h:\mathbb{Q}\to\mathbb{Z}$ such that for every $x\in\mathbb{Q}$, every $a\in\mathbb{Z}$, and every positive integer $b$, the following equation holds:
+    \[
+    h(x)=h\!\left(\frac{h(bx-a)+a}{b}\right).
+    \]
+    
+    First, we can simplify the given equation by setting $x=0$. Substituting $x=0$ into the equation, we get:
+    \[
+    h(0) = h\!\left(\frac{h(0-a)+a}{b}
+    
+    ... [truncated 371 chars] ...
+    
+    r all $a$ and $b$. Therefore, $h$ must be constant on the set of all rational numbers. Let's denote this constant value by $c$. So, we have:
+    \[
+    h(x) = c \quad \text{for all} \quad x \in \mathbb{Q}.
+    \]
+    Now, we need to check if this constant function satisfies the given equation. Substituting $h(x) = c$ into the original equation, we get:
+    \[
+    c = c \quad \text{which is true for all} \quad c \in \mathbb{Z}.
+    \]
+    Thus, the constant function $h(x) = c$ for all $x \in \mathbb{Q}$ is a solution. To confirm that this is the only solution, we need to ensure that no other functions can satisfy the given equation. Suppose there exists a non-constant function $h$ that satisfies the equation. Then, there must exist some $x$ for which $h(x) \neq c$. However, the condition $h(x) = h\!\left(\frac{h(bx-a)+a}{b}\right)$ must hold for all $x$, $a$, and $b$, and since $h$ is constant on all rational numbers, this condition can only be satisfied if $h$ is constant. Therefore, the only possible solution is the constant function $h(x) = c$ for all $x \in \mathbb{Q}$.
+    
+    Hence, the solution to the problem is:
+    \[
+    </think>
+    
+    \boxed{h(x) = c \text{ for all } x \in \mathbb{Q} \text{ and some } c \in \mathbb{Z}}.
+    \]
+    ```