Update Automated MNLP evaluation report (2026-06-03)

EVAL_REPORT.md

@@ -2,7 +2,7 @@
 
 - **Model repo:** [`cs-552-2026-bilko/math_model`](https://huggingface.co/cs-552-2026-bilko/math_model)
 - **Owner(s):** group **bilko**
-- **Generated at:** 2026-06-01T06:13:57+00:00 (UTC)
+- **Generated at:** 2026-06-03T08:04:32+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._
@@ -11,7 +11,7 @@ _This PR is opened automatically by the course CI. It is **non-blocking** — yo
 
 | Benchmark | Accuracy | Status |
 |---|---:|---|
-| Math | 0.3400 | ok |
+| Math | 0.3200 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -24,97 +24,121 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `\frac{7}{4}`
-- **overall** (6/8 completions correct)
-- **extracted** (✓): `\frac{7}{4}`
+- **reference**: `194`
+- **overall** (2/8 completions correct)
+- **extracted** (✓): `194`
 - **completion**:
 
     ```text
-    To find the value of $b$ in the direction vector of the red line given as $\begin{pmatrix} -7 \\ b \end{pmatrix}$, we need to determine the slope of the red line and compare it with the slope of the line segment $AB$ on the graph.
+    1. **Establishing Domino Tiling**:
+       - Consider a \(194 \times 194\) chessboard.
+       - Since the board has an even number of squares, it can be completely tiled with \(1 \times 2\) and \(2 \times 1\) dominoes.
     
-    First, we calculate the slope of line segment $AB$. The coordinates of points $A$ and $B$ are given as $(-5,4)$ and $(-1,3)$, respectively. The slope $m_{AB}$ is calculated using the formula for the slope between two points $(x_1, y_1)$ and $(x_2, y_2)$, which is $\frac{y_2 - y_1}{x_2 - x_1}$. Applying this formula to points $A$ and $B$ gives:
+    2. **Tiling with Marked Squares**:
+       - Let \( k \) be the number of squares we choose to mark.
+       - Each domino that covers exactly one marked square must cover two adjacent squares.
     
-    \[
-    m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 4}{-1 - (-5)} = \frac{-1}{4}.
-    \]
+    3. *
     
-    Next, we compare this slope with the slope of the direction vector $\begin{pmatrix} -7 \\ b \end{pmatrix}$, which can be calculated similarly. The slope $m$ is given by $\frac{b - 0}{-7 - 0} = \frac{b}{-7}$. To find the value of $b$ that makes the slope of the direction vector equal to the slope of line segment $AB$, we set the two slopes equal to each other and solve for $b$:
+    ... [truncated 263 chars] ...
     
-    \[
-    \frac{b}{-7} = \frac{-1}{4}.
-    \]
+    by dominoes, the number of marked squares \( k \) is at most:
+         \[
+         k \leq 18818
+         \]
     
-    Solving for $b$, we multiply both sides of the equation by $-7$:
+    4. **Ensuring Unique Tiling**:
+       - To ensure that there is exactly one tiling with at most one marked square per domino, we need to consider the placement of marked squares.
+       - If we choose \( k \geq 194 \), we can arrange the marked squares such that no two marked squares are adjacent, ensuring each domino covering a marked square is unique.
     
-    \[
-    b = \frac{-1}{4} \cdot -7 = \frac{7}{4}.
-    \]
+    5. **Verification for Larger \( k \)**:
+       - Suppose \( k = 194 \), and we divide the board into \( 194 \times 194 / 194 = 194 \) regions, each of size \( 194 \times 1 \).
+       - In each \( 194 \times 1 \) region, mark one square. The number of marked squares is \( 194 \), and each domino covers exactly one marked square.
+    
+    6. **Ensuring a Unique Tiling**:
+       - Each \( 194 \times 1 \) region can be tiled with dominoes without overlap, ensuring a unique tiling configuration. Since each marked square is isolated and only one per domino, the tiling remains unique.
     
-    Therefore, the value of $b$ that makes the slope of the direction vector equal to the slope of line segment $AB$ is
+    7. **Conclusion**:
+       - Therefore, the least positive integer \( k \) ensuring a unique tiling with each domino covering at most one marked square is:
+         \[
     </think>
     
-    $\boxed{\frac{7}{4}}$.
+    \boxed{194}
+         \]
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `756`
+- **reference**: `104`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `0.1064349083`
+- **extracted** (✗): `390`
 - **completion**:
 
     ```text
-    Given three spherical balloons with radii \(11\), \(13\), and \(19\), and each pair externally tangent, we need to find \(XZ^2\) where \(XY^2 = 560\).
+    Given the tetrahedron \( WXYZ \) with the following edge lengths:
+    \[
+    WX = YZ = \sqrt{41}, \quad WY = XZ = \sqrt{80}, \quad WZ = XY = \sqrt{89}
+    \]
     
-    1. **Identify the distances between centers**:
-       - Since the balloons are externally tangent, the distance between the centers of any two balloons is the sum of their radii:
-         \[
-         XY = 11 + 13 = 24 \quad \text{and} \quad XZ = 11 + 19 = 30
+    We are to find a point \( P \) inside the tetrahedron such that \( P \) is equidistant from all four faces of the tetrahedron, and then calculate the value of \( a + b + c \) where the distance \( d \) from \( P \) to the faces is expressed as \( \frac{
     
-    ... [truncated 655 chars] ...
+    ... [truncated 823 chars] ...
     
-    XY^2 = 560 \Rightarrow \cos \angle XYZ = \frac{560 + XZ^2 - 576}{2 \cdot 4\sqrt{35} \cdot XZ}
-         \]
-         Substituting \(XZ^2 = 290 + 304\cos \angle XYZ\):
-         \[
-         \cos \angle XYZ = \frac{560 + 290 + 304\cos \angle XYZ - 576}{2 \cdot 4\sqrt{35} \cdot XZ}
-         \]
-         \[
-         \cos \angle XYZ = \frac{294 + 304\cos \angle XYZ - 576}{80\sqrt{35} \cdot XZ}
-         \]
-         Simplifying:
-         \[
-         \cos \angle XYZ = \frac{-282 + 304\cos \angle XYZ}{80\sqrt{35} \cdot XZ}
-         \]
-         \[
-         80\sqrt{35} \cdot XZ (\cos \angle XYZ) = -282 + 304\cos \angle XYZ
-         \]
-         Solving for \(XZ\):
-         \[
-         80\sqrt{35} XZ + 304XZ = 282
-         \]
-         \[
-         XZ (80\sqrt{35} + 304) = 282
-         \]
-         \[
-         XZ = \frac{282}{80\sqrt{35} + 304}
-         \]
-         Approximating:
-         \[
-         XZ^2 \approx \frac{282^2}{(80\sqrt{35} + 304)^2}
-         \]
+    ): Area of triangle opposite to \( Y \)
+       - \( A_4 \): Area of triangle opposite to \( Z \)
     
-    3. **Calculation**:
-       - Calculate \(XZ^2\):
-         \[
-         XZ^2 = \left( \frac{282}{80\sqrt{35} + 304} \right)^2 \approx \frac{79524}{(80 \cdot 5.916 + 304)^2} \approx \frac{79524}{(473.28 + 304)^2} \approx \frac{79524}{777.28^2} \approx 0.1064349083
-         \]
-       - Therefore,
-         \[
-         XZ^2 = \boxed{0.1064349083}
-         \]
+       Using Heron's formula:
+    
+       \[
+       A_1 = \frac{\sqrt{41}}{2} \sqrt{81} = \frac{9 \sqrt{41}}{2}
+       \]
+       \[
+       A_2 = \frac{\sqrt{41}}{2} \sqrt{89} = \frac{\sqrt{41}}{2} \sqrt{89}
+       \]
+       \[
+       A_3 = \frac{\sqrt{41}}{2} \sqrt{89} = \frac{\sqrt{41}}{2} \sqrt{89}
+       \]
+       \[
+       A_4 = \frac{\sqrt{41}}{2} \sqrt{81} = \frac{9 \sqrt{41}}{2}
+       \]
     
+    4. **Solve for \( d \)**:
+       Substituting the areas back into the volume equation:
+    
+       \[
+       \left( \frac{1}{12} \sqrt{12342} \right)^2 = d^2 \left( \frac{9 \sqrt{41}}{6} + \frac{\sqrt{80}}{6} + \frac{\sqrt{89}}{6} + \frac{9 \sqrt{41}}{6} \right)
+       \]
+    
+       Simplify the equation:
+    
+       \[
+       \frac{12342}{144} = d^2 \left( \frac{18 \sqrt{41} + 2 \sqrt{80} + \sqrt{89}}{6} \right)
+       \]
+    
+       \[
+       \frac{12342}{144} = d^2 \left( \frac{18 \sqrt{41} + 2 \sqrt{80} + \sqrt{89}}{6} \right)
+       \]
+    
+       Solving for \( d \):
+    
+       \[
+       d = \sqrt{\frac{12342}{144 \times \frac{18 \sqrt{41} + 2 \sqrt{80} + \sqrt{89}}{6}}}
+       \]
+    
+       \[
+       d = \frac{a \sqrt{b}}{c}
+       \]
+    
+       After simplifying, we get:
+    
+       \[
+       a + b + c = 390
+       \]
+    
+    ### Conclusion:
+    \[
     </think>
     
-    $\boxed{0.1064349083}$
+    \boxed{390}
+    \]
     ```