Update Automated MNLP evaluation report (2026-05-26)

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:** 2026-05-25T02:20:01+00:00 (UTC)
+- **Generated at:** 2026-05-26T12:29:10+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._
@@ -25,20 +25,20 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 **Correct** (1 shown)
 
 - **reference**: `\frac{7}{4}`
-- **overall** (4/8 completions correct)
+- **overall** (2/8 completions correct)
 - **extracted** (✓): `\frac{7}{4}`
 - **completion**:
 
     ```text
     <think>
-    We know that the red line is parametrized with a direction vector of the form $\begin{pmatrix} -7 \\ b \end{pmatrix}$.
+    The slope of the red line is $\frac{4-3}{-5-(-1)} = \frac{1}{-4} = -\frac{1}{4}$.
     
-    The slope of the line is $\frac{b}{-7}$.
+    Since the direction vector of the red line is $\begin{pmatrix} -7 \\ b \end{pmatrix}$, we have $\frac{b}{-7} = -\frac{1}{4}$.
     
-    The slope of the line passing through the points $(-5,4)$ and $(-1,3)$ is $\frac{3-4}{-1-(-5)} = \frac{-1}{4} = -\frac{1}{4}$.
+    Solving for $b$, we get $b = \frac{7}{4}$.
     
-    Since the line is the same, the slopes are equal:
-    \[ \frac{b}{-7} = -\frac{1}{4} \Rightarrow b = -\frac{1}{4} \times -7 \Rightarrow b =
+    So, the answer is:
+    \[
     </think>
     
     Therefore, the answer is \boxed{\frac{7}{4}}.
@@ -46,23 +46,55 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Incorrect** (1 shown)
 
-- **reference**: `$g(x)=c, g(x)=\lceil x \rceil, g(x)=\lfloor x \rfloor$`
+- **reference**: `756`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `h(x) = 0`
+- **extracted** (✗): `102000000000`
 - **completion**:
 
     ```text
     <think>
-    Let's start by analyzing the given functional equation.
+    Let's denote the radii of the balloons as \(r_1 = 11\), \(r_2 = 13\), and \(r_3 = 19\). The centers of the balloons are denoted as \(X\), \(Y\), and \(Z\), respectively.
+    
+    Since the balloons are externally tangent, the distance between the centers of any two balloons is equal to the sum of their radii. Let's denote this distance as \(d_{ij}\), where \(i\) and \(j\) are the indices of the tw
+    
+    ... [truncated 173 chars] ...
+    
+    Given that \(XY^2 = 560\), we can substitute this value into the equation:
+    \[d_{12}^2 = (11 + 13)^2 - (560 - (11 + 13)^2)^2\]
+    \[d_{12}^2 = 24^2 - (560 - 24^2)^2\]
+    \[d_{12}^2 = 576 - (560 - 576)^2\]
+    \[d_{12}^2 = 576 - (-16)^2\]
+    \[d_{12}^2 = 576 - 256\]
+    \[d_{12}^2 = 320\]
+    
+    Similarly, we can find \(d_{13}^2\) and \(d_{23}^2\):
+    \[d_{13}^2 = (11 + 19)^2 - (560 - (11 + 19)^2)^2\]
+    \[d_{13}^2 = 30^2 - (560 - 30^2)^2\]
+    \[d_{13}^2 = 900 - (560 - 900)^2\]
+    \[d_{13}^2 = 900 - (-340)^2\]
+    \[d_{13}^2 = 900 - 115600\]
+    \[d_{13}^2 = -114700\]
+    
+    \[d_{23}^2 = (13 + 19)^2 - (560 - (13 + 19)^2)^2\]
+    \[d_{23}^2 = 32^2 - (560 - 32^2)^2\]
+    \[d_{23}^2 = 1024 - (560 - 1024)^2\]
+    \[d_{23}^2 = 1024 - (-464)^2\]
+    \[d_{23}^2 = 1024 - 215296\]
+    \[d_{23}^2 = -214272\]
+    
+    Now, we can use the fact that the centers of the balloons lie on one side of the plane to find \(XZ^2\).
     
-    1.  For $x = 0$, we have $h(0) = h\left(\frac{h(0) - a + a}{1}\right) = h(h(0))$. This implies that $h(0)$ is a fixed point of $h$.
-    2.  For $a = 0$, we have $h(x) = h\left(\frac{h(0) + 0}{b}\right) = h\left(\frac{h(0)}{b}\right)$. This means that $h(x)$ is a constant function for all $x \in \mathbb{Q}$, and the constant value is $h(0)$.
+    Using the formula for the distance between two points in 3D space, we can express \(XZ^2\) as:
+    \[XZ^2 = (d_{12}^2 + d_{13}^2 - d_{23}^2)^2\]
     
-    Therefore, the only possible solutions are $h(x) = 0$ for all $x \in \mathbb{Q}$ or $h(x) = c$ for some constant $c \in \mathbb{Z}$ for all $x \in \mathbb{Q}$.
+    Substituting the values we found earlier, we get:
+    \[XZ^2 = (320 - 114700 + 214272)^2\]
+    \[XZ^2 = (100872)^2\]
+    \[XZ^2 = 102000000000\]
     
-    The answer is:
+    So, the answer is:
     \[
     </think>
     
-    Therefore, the answer is \boxed{h(x) = 0}.
+    Therefore, the answer is \boxed{102000000000}.
     ```