Update Automated MNLP evaluation report (2026-05-24)

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:** 2026-05-23T14:37:35+00:00 (UTC)
+- **Generated at:** 2026-05-24T16:13:44+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._
@@ -24,124 +24,45 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `\frac{448}{15625}`
-- **overall** (1/8 completions correct)
-- **extracted** (✓): `\frac{448}{15625}`
+- **reference**: `\frac{7}{4}`
+- **overall** (6/8 completions correct)
+- **extracted** (✓): `\frac{7}{4}`
 - **completion**:
 
     ```text
-    <think>
-    
-    </think>
-    
-    The problem involves finding the probability of exactly 4 out of 7 islands containing treasure, where each island has a probability of \(\frac{1}{5}\) of containing treasure. This is a binomial probability problem.
-    
-    1. **Identify the parameters**:
-       - Number of trials \( n = 7 \)
-       - Probability of success \( p = \frac{1}{5} \)
-       - Desired number of successes \( k = 4 \)
-    
-    2. **Use the binomial probability formula**:
-       \[
-       P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
-       \]
-       Substituting the values, we get:
-       \[
-       P(X = 4) = \binom{7}{4} \left(\frac{1}{5}\right)^4 \left(\frac{4}{5}\right)^{3}
-       \]
-    
-    3. **Calculate the binomial coefficient**:
-       \[
-       \binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7 \times 6 \times 5 \times 4!}{4! \times 3 \times 2 \times 1} = 35
-       \]
-    
-    4. **Calculate the probabilities**:
-       \[
-       \left(\frac{1}{5}\right)^4 = \frac{1}{625}
-       \]
-       \[
-       \left(\frac{4}{5}\right)^3 = \frac{64}{125}
-       \]
-    
-    5. **Combine the results**:
-       \[
-       P(X = 4) = 35 \times \frac{1}{625} \times \frac{64}{125} = 35 \times \frac{64}{78125} = \frac{2240}{78125}
-       \]
-    
-    6. **Simplify the fraction**:
-       \[
-       \frac{2240}{78125} = \frac{224}{7812.5} = \frac{448}{15625}
-       \]
-    
-    7. **Final probability**:
-       \[
-       \boxed{\frac{448}{15625}}
-       \]
+    The slope of the line segment from $A$ to $B$ is $\frac{3-4}{-1-(-5)} = \frac{-1}{4}$. The direction vector of the line is $\begin{pmatrix} -7 \\ b \end{pmatrix}$. The slope of the line can also be calculated as $\frac{b}{-7}$. Setting these slopes equal gives:
+    \[
+    \frac{b}{-7} = \frac{-1}{4}
+    \]
+    Solving for $b$, we multiply both sides by $-7$:
+    \[
+    b = \frac{-1}{4} \cdot -7 = \frac{7}{4}
+    \]
+    Thus, $b = \boxed{\frac{7}{4}}$.
     ```
 
 **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) = c \text{ for all } x \in \mathbb{Q} \text{ and some } c \in \mathbb{Z}`
+- **extracted** (✗): `0`
 - **completion**:
 
     ```text
-    <think>
-    
-    </think>
-    
-    To solve the problem, we need to find all functions \( h: \mathbb{Q} \to \mathbb{Z} \) such that for every \( x \in \mathbb{Q} \), every \( a \in \mathbb{Z} \), and every positive integer \( b \), the following equation holds:
-    \[
-    h(x) = h\left( \frac{h(bx - a) + a}{b} \right).
-    \]
+    Let the centers of the balloons be \(A\), \(B\), and \(C\) with radii \(11\), \(13\), and \(19\) respectively. The distances between centers are \(AB = 11 + 13 = 24\), \(BC = 13 + 19 = 32\), and \(CA = 11 + 19 = 30\). 
     
-    #### Step 1: Analyzing the functional equation
+    Since the cross-sections are congruent and centered at \(X\), \(Y\), and \(Z\), and given \(XY^2 = 560\), we can use the Law of Cosines in triangle \(XYZ\) to find \(XZ^2\). We need to find \(YZ^2\) first using the Law of Cosines in triangle \(ABC\).
     
-    Given the equation:
+    Using the Law of Cosines in triangle \(ABC\):
     \[
-    h(x) = h\left( \frac{h(bx -
-    
-    ... [truncated 629 chars] ...
-    
-    t).
+    \cos \angle ABC = \frac{AB^2 + BC^2 - CA^2}{2 \cdot AB \cdot BC} = \frac{24^2 + 32^2 - 30^2}{2 \cdot 24 \cdot 32} = \frac{576 + 1024 - 900}{1536} = \frac{1300}{1536} = \frac{1625}{2048}
     \]
-    Since \( a \) and \( b \) are arbitrary, we can choose \( a = b \):
     \[
-    h(1) = h(0).
+    YZ^2 = 560 + 32^2 - 2 \cdot 560 \cdot \frac{1625}{2048} = 560 + 1024 - \frac{1840000}{2048} = 1584 - 900 = 684
     \]
-    Thus, \( h(1) = h(0) \).
-    
-    #### Step 4: Assume \( h \) is a constant function
-    
-    Assume \( h(x) = c \) for all \( x \in \mathbb{Q} \). Then:
-    \[
-    c = h\left( \frac{c + a}{b} \right) = c,
-    \]
-    which is always true. Therefore, \( h(x) = c \) is a solution for any constant \( c \in \mathbb{Z} \).
-    
-    #### Step 5: Verify if \( h(x) = c \) is the only solution
-    
-    Let's suppose \( h(x) \neq c \) for some \( x \). Consider \( x = 1 \):
-    \[
-    h(1) = c.
-    \]
-    Now, let's set \( x = \frac{1}{b} \):
-    \[
-    h\left(\frac{1}{b}\right) = h\left( \frac{h(1) - 1}{b} \right) = h\left( \frac{c - 1}{b} \right).
-    \]
-    If \( c \neq b \), then \( \frac{c-1}{b} \neq 0 \), and we can choose \( a = c - 1 \) and \( b = c \):
-    \[
-    h\left(\frac{c-1}{c}\right) = h(0).
-    \]
-    This implies \( h\left(\frac{c-1}{c}\right) = c \), which contradicts \( h(x) \neq c \).
-    
-    Therefore, \( h(x) = c \) is the only constant solution.
-    
-    #### Conclusion:
-    
-    The only function \( h: \mathbb{Q} \to \mathbb{Z} \) that satisfies the given functional equation is the constant function:
+    Now, applying the Law of Cosines again in triangle \(XYZ\):
     \[
-    \boxed{h(x) = c \text{ for all } x \in \mathbb{Q} \text{ and some } c \in \mathbb{Z}}.
+    XZ^2 = 560 + 684 - 2 \sqrt{560 \cdot 684} = 1244 - 2 \sqrt{386640} = 1244 - 2 \cdot 622 = 1244 - 1244 = 0
     \]
+    Thus, \(XZ^2 = \boxed{0}\).
     ```