Update Automated MNLP evaluation report (2026-05-23)

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:** 2026-05-22T03:04:12+00:00 (UTC)
+- **Generated at:** 2026-05-23T14:37:35+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._
@@ -30,103 +30,118 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 - **completion**:
 
     ```text
-    To solve this problem, we use the binomial probability formula, which is given by:
-    \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
-    where:
-    - \( n \) is the total number of trials (in this case, 7 islands),
-    - \( k \) is the number of successes (in this case, 4 islands with treasure),
-    - \( p \) is the probability of success on a single trial (in this case, \(\frac{1}{5}\)).
+    <think>
     
-    1. **Calculate the binomial coefficient**:
-       \[ \binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]
+    </think>
     
-    2. **Calculate the probability of success raised to the power of 4**:
-       \[ \left(\frac{1}{5}\right)^4 = \frac{1}{625} \]
+    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.
     
-    3. **Calculate the probability of failure raised to the power of 3**:
-       \[ \left(1 - \frac{1}{5}\right)^3 = \left(\frac{4}{5}\right)^3 = \frac{64}{125} \]
+    1. **Identify the parameters**:
+       - Number of trials \( n = 7 \)
+       - Probability of success \( p = \frac{1}{5} \)
+       - Desired number of successes \( k = 4 \)
     
-    4. **Combine these results to find the total probability**:
-       \[ P(X = 4) = 35 \times \frac{1}{625} \times \frac{64}{125} = \frac{35 \times 64}{625 \times 125} = \frac{2240}{78125} = \frac{1792}{62500} = \frac{448}{15625} \]
-    
-    Conclusion:
-    The probability that exactly 4 out of the 7 islands contain treasure is:
-    \[ \boxed{\frac{448}{15625}} \]
-    ```
-
-**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) = -x^2`
-- **completion**:
-
-    ```text
-    1. **Initial Equation Setup:**
-       We start with the given functional equation:
-       \[
-       h(x) = h\left(\frac{h(bx - a) + a}{b}\right)
-       \]
-       for all \(x \in \mathbb{Q}\), \(a \in \mathbb{Z}\), and \(b \in \mathbb{N}\).
-    
-    2. **Assumption of a Solution:**
-       Suppose there exists a solution \(h\) such that \(h(q) = 0\) for some \(q \in \mathbb{Q}\).
-    
-    3. **Substitution and Simplification:**
-       Let \(x
-    
-    ... [truncated 578 chars] ...
-    
-    1) + b}{b}\right)
-       \]
-       This simplifies to:
-       \[
-       h\left(\frac{q}{b}\right) = h\left(\frac{h(q - 1) + b}{b}\right)
-       \]
-       Given \(h(q) = 0\), we have:
+    2. **Use the binomial probability formula**:
        \[
-       h\left(\frac{q}{b}\right) = h\left(\frac{h(q - 1) + b}{b}\right) = 0
+       P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
        \]
-       Therefore:
+       Substituting the values, we get:
        \[
-       h(q - 1) = -b
+       P(X = 4) = \binom{7}{4} \left(\frac{1}{5}\right)^4 \left(\frac{4}{5}\right)^{3}
        \]
     
-    5. **Conclusion for \(h(q)\):**
-       By induction, we can generalize that for any integer \(n\):
+    3. **Calculate the binomial coefficient**:
        \[
-       h(q - n) = -bn
-       \]
-       This implies that for any integer \(n\):
-       \[
-       h(nq) = -bn^2
+       \binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7 \times 6 \times 5 \times 4!}{4! \times 3 \times 2 \times 1} = 35
        \]
     
-    6. **Final Verification:**
-       We need to verify that \(h(x) = -x^2\) satisfies the original equation. Let:
+    4. **Calculate the probabilities**:
        \[
-       h(x) = -x^2
+       \left(\frac{1}{5}\right)^4 = \frac{1}{625}
        \]
-       Substitute \(h(x) = -x^2\) into the original equation:
        \[
-       h(x) = h\left(\frac{h(bx - a) + a}{b}\right)
+       \left(\frac{4}{5}\right)^3 = \frac{64}{125}
        \]
-       Substituting \(h(x) = -x^2\):
+    
+    5. **Combine the results**:
        \[
-       -x^2 = h\left(\frac{-\left(bx - a\right)^2 + a}{b}\right)
+       P(X = 4) = 35 \times \frac{1}{625} \times \frac{64}{125} = 35 \times \frac{64}{78125} = \frac{2240}{78125}
        \]
-       Simplify the argument of \(h\):
+    
+    6. **Simplify the fraction**:
        \[
-       -x^2 = h\left(\frac{-b^2x^2 + 2abx - a^2 + a}{b}\right)
+       \frac{2240}{78125} = \frac{224}{7812.5} = \frac{448}{15625}
        \]
-       Since \(h\) is defined as \(-y^2\):
+    
+    7. **Final probability**:
        \[
-       -x^2 = -\left(\frac{-b^2x^2 + 2abx - a^2 + a}{b}\right)^2
+       \boxed{\frac{448}{15625}}
        \]
-       This simplifies correctly, confirming that \(h(x) = -x^2\) is indeed a solution.
+    ```
+
+**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 all } x \in \mathbb{Q} \text{ and some } c \in \mathbb{Z}`
+- **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:
+    \[
+    h(x) = h\left( \frac{h(bx - a) + a}{b} \right).
+    \]
+    
+    #### Step 1: Analyzing the functional equation
+    
+    Given the equation:
+    \[
+    h(x) = h\left( \frac{h(bx -
+    
+    ... [truncated 629 chars] ...
+    
+    t).
+    \]
+    Since \( a \) and \( b \) are arbitrary, we can choose \( a = b \):
+    \[
+    h(1) = h(0).
+    \]
+    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:
     
-    Therefore, the solution is:
+    The only function \( h: \mathbb{Q} \to \mathbb{Z} \) that satisfies the given functional equation is the constant function:
     \[
-    \boxed{h(x) = -x^2}
+    \boxed{h(x) = c \text{ for all } x \in \mathbb{Q} \text{ and some } c \in \mathbb{Z}}.
     \]
     ```