Update Automated MNLP evaluation report (2026-05-27)

EVAL_REPORT.md

@@ -2,7 +2,7 @@
 
 - **Model repo:** [`cs-552-2026-vibe-trainers/math_model`](https://huggingface.co/cs-552-2026-vibe-trainers/math_model)
 - **Owner(s):** group **vibe-trainers**
-- **Generated at:** 2026-05-26T12:29:10+00:00 (UTC)
+- **Generated at:** 2026-05-27T13:02:22+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._
@@ -11,16 +11,173 @@ _This PR is opened automatically by the course CI. It is **non-blocking** — yo
 
 | Benchmark | Accuracy | Status |
 |---|---:|---|
-| Math | — | error (inference) |
+| Math | 0.4000 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
 
-## Errors
+## Sample completions
 
-### math — `inference`
+_Prompts are intentionally omitted to avoid revealing benchmark contents. For multi-completion problems, only one completion is shown per sample._
 
-```
-expected token 'end of print statement', got 'name'
-```
-_Full traceback recorded at `results/errors/cs-552-2026-vibe-trainers_math_model_math.log` in the CI run._
+### Math
+
+**Correct** (1 shown)
+
+- **reference**: `194`
+- **overall** (3/8 completions correct)
+- **extracted** (✓): `194`
+- **completion**:
+
+    ```text
+    We are given a $194 \times 194$ board and asked 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** where **every domino contains at most one marked cell**.
+    
+    ---
+    
+    ### Step 1: Understanding the Problem
+    
+    - The board is $194 \times 194$, so it has $194^2 = 37636$ cells.
+    - A dom
+    
+    ... [truncated 823 chars] ...
+    
+    ondition that each domino contains at most one marked cell means that **no two marked vertices are connected by an edge**.
+    
+    This is equivalent to saying that the set of marked cells forms an **independent set** in the graph.
+    
+    ---
+    
+    ### Step 4: Tiling and Independent Sets
+    
+    We are told that there is **exactly one** tiling of the board by dominoes with **at most one marked cell per domino**.
+    
+    This means that the **set of marked cells** must be such that:
+    - It is an **independent set** (no two marked cells are adjacent).
+    - The domino tiling is **unique** (i.e., the tiling is uniquely determined by the set of marked cells).
+    
+    ---
+    
+    ### Step 5: Minimum Size of Independent Set
+    
+    We are to **minimize** the number of marked cells $k$, such that:
+    - The set of marked cells is an **independent set**.
+    - The tiling of the board with dominoes is **unique**.
+    
+    This is a **combinatorial optimization** problem.
+    
+    ---
+    
+    ### Step 6: Known Result
+    
+    This is a classic result in **graph theory** and **domino tiling**. Specifically, the **minimum number of marked cells** needed to ensure a **unique tiling** with **at most one marked cell per domino** 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) = c \text{ for some constant } c \in \mathbb{Z}`
+- **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: Analyze the structure of the equation
+    
+    Let’s denote:
+    
+    $$
+    y = \frac{h(bx - a) + a}{b}
+    $$
+    
+    Then the equation becomes:
+    
+    $$
+    h(x)
+    
+    ... [truncated 414 chars] ...
+    
+    $, a constant function, for some constant $ c \in \mathbb{Z} $.
+    
+    Then:
+    
+    $$
+    h(x) = c, \quad h(bx - a) = c, \quad h\left(\frac{c + a}{b}\right) = c
+    $$
+    
+    So the equation becomes:
+    
+    $$
+    c = c
+    $$
+    
+    This is true for all $ x, a, b $. So **constant functions** are solutions.
+    
+    ---
+    
+    ### Step 3: Are there non-constant solutions?
+    
+    Suppose $ h $ is not constant. Then there exist $ x_1, x_2 \in \mathbb{Q} $ such that $ h(x_1) \ne h(x_2) $.
+    
+    Let’s test with specific values:
+    
+    #### Try $ x = 1 $, $ a = 1 $, $ b = 1 $
+    
+    Then:
+    
+    $$
+    h(1) = h\left(\frac{h(1 - 1) + 1}{1}\right) = h(h(0) + 1)
+    $$
+    
+    So:
+    
+    $$
+    h(1) = h(h(0) + 1)
+    $$
+    
+    If $ h $ is constant, then this is fine. But if $ h $ is not constant, then $ h(0) + 1 $ must be mapped to the same value as $ h(1) $, which may not be possible unless $ h $ is constant.
+    
+    So, **constant functions** are the only possible solutions.
+    
+    ---
+    
+    ### Step 4: Conclusion
+    
+    The only function $ h : \mathbb{Q} \to \mathbb{Z} $ satisfying the given equation for all $ x, a, b \in \mathbb{Q}, \mathbb{Z}, \mathbb{N} $ is the **constant function**:
+    
+    $$
+    \boxed{h(x) = c}
+    $$
+    
+    where $ c \in \mathbb{Z} $.
+    
+    ---
+    
+    ### Final Answer:
+    
+    $$
+    \boxed{h(x) = c \text{ for some constant } c \in \mathbb{Z}}
+    $$
+    ```