Update Automated MNLP evaluation report (2026-06-10)

EVAL_REPORT.md

@@ -2,7 +2,7 @@
 
 - **Model repo:** [`cs-552-2026-kth/math_model`](https://huggingface.co/cs-552-2026-kth/math_model)
 - **Owner(s):** group **kth**
-- **Generated at:** 2026-06-09T08:56:40+00:00 (UTC)
+- **Generated at:** 2026-06-10T21:20:55+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._
@@ -17,7 +17,7 @@ _This PR is opened automatically by the course CI. It is **non-blocking** — yo
 
 | Benchmark | Accuracy | Status |
 |---|---:|---|
-| Math | 0.3500 | ok |
+| Math | 0.3400 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -30,36 +30,53 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `194`
-- **overall** (1/8 completions correct)
-- **extracted** (✓): `194`
+- **reference**: `\frac{7}{4}`
+- **overall** (6/8 completions correct)
+- **extracted** (✓): `\frac{7}{4}`
 - **completion**:
 
     ```text
     <think>
-    ## Step 1: Understand the problem
-    We need to find the least positive integer $k$ for which we can choose $k$ cells to mark on a $194 \times 194$ board 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.
+    To find the value of $b$, we need to use the fact that the red line is parametrized with a direction vector of the form $\begin{pmatrix} -7 \\ b \end{pmatrix}$.
     
-    ## Step 2: Analyze the tiling condition
-    For a $194 \times 194$ board to be tiled by $1 \times
+    The direction vector of the red line is $\begin{pmatrix} -7 \\ b \end{pmatrix}$.
     
-    ... [truncated 823 chars] ...
+    Since the red line is parametrized with a direction vector of the form $\begin{pmatrix} -7 \\ b \end{pmatrix}$, we can write the equation of the red
     
-    Since the board is $194 \times 194$, which is an even number, the minimum $k$ would be related to the number of cells that need to be marked to restrict the possible arrangements of dominoes.
+    ... [truncated 391 chars] ...
     
-    ## Step 13: Realize the mistake in the previous step
-    The previous step was incorrect because it assumed a specific pattern for marking cells. In reality, the minimum $k$ depends on the specific constraints of the tiling problem and the number of cells that need to be marked to achieve the desired outcome.
+    = \frac{\begin{pmatrix} -5 \\ 4 \end{pmatrix}}{\begin{pmatrix} -7 \\ b \end{pmatrix}} \]
     
-    ## Step 14: Correct the approach
-    To solve this problem, we need to consider the constraints imposed by the tiling condition and the requirement for exactly one tiling. This involves finding the minimum number of marked cells required to restrict the possible arrangements of dominoes in a way that meets the given conditions.
+    Since $t$ is a parameter, it must be a constant value.
     
-    ## Step 15: Consider the solution
-    After reevaluating the problem, we realize that the minimum $k$ is related to the size of the board and the constraints imposed by the tiling condition. The correct solution involves finding the minimum number of marked cells required to achieve the desired outcome.
+    Equating the two expressions for $t$, we get:
+    \[ \frac{\begin{pmatrix} -5 \\ 4 \end{pmatrix}}{\begin{pmatrix} -7 \\ b \end{pmatrix}} = \frac{\begin{pmatrix} -1 \\ 3 \end{pmatrix}}{\begin{pmatrix} -7 \\ b \end{pmatrix}} \]
     
-    The final answer is: $
+    Comparing the two expressions, we get:
+    \[ \frac{-5}{-7} = \frac{-1}{-7} \quad \text{and} \quad \frac{4}{b} = \frac{3}{b} \]
+    
+    Simplifying, we get:
+    \[ \frac{5}{7} = \frac{1}{7} \quad \text{and} \quad 4 = 3 \]
+    
+    The first equation is not true, so there is no value of $b$ that satisfies the equation.
+    
+    However, the second equation is not true either, so there is no value of $b$ that satisfies the equation.
+    
+    But, we can see that the red line passes through the points $(-5,4)$ and $(-1,3)$.
+    
+    The slope of the red line is:
+    \[ \frac{4 - 3}{-5 - (-1)} = \frac{1}{-4} = -\frac{1}{4} \]
+    
+    The slope of the red line is also equal to $\frac{b}{-7}$.
+    
+    Equating the two expressions for the slope, we get:
+    \[ \frac{b}{-7} = -\frac{1}{4} \]
+    
+    Solving for $b$, we get:
+    \[ b = \frac{-7}{-4} =
     </think>
     
-    Therefore, the answer is \boxed{194}.
+    Therefore, the answer is \boxed{\frac{7}{4}}.
     ```
 
 **Incorrect** (1 shown)