Update Automated MNLP evaluation report (2026-05-19)

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:** 2026-05-18T02:30:25+00:00 (UTC)
+- **Generated at:** 2026-05-19T03:22:52+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 | 0.3200 | ok |
+| Math | 0.3300 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -24,64 +24,109 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `\frac{639}{40}`
-- **overall** (4/8 completions correct)
-- **extracted** (✓): `\frac{639}{40}`
+- **reference**: `\frac{448}{15625}`
+- **overall** (1/8 completions correct)
+- **extracted** (✓): `\frac{448}{15625}`
 - **completion**:
 
     ```text
-    <think>
+    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**:
+       \[ \binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]
     
-    1. **Set up the proportion**: We start by setting up a proportion to find the distance on the map for $53.25$ miles in real life. Let $x$ be the distance on the map in inches.
-       \[
-       \frac{3 \text{ inches}}{10 \text{ miles}} = \frac{x \text{ inches}}{53.25 \text{ miles}}
-       \]
+    2. **Calculate the probability of success raised to the power of 4**:
+       \[ \left(\frac{1}{5}\right)^4 = \frac{1}{625} \]
     
-    2. **Cross-multiply to solve for $x$**: Cross-multiplying gives us:
-       \[
-       3 \times 53.25 = 10 \times x
-       \]
-       Simplifying the left side:
-       \[
-       159.75 = 10x
-       \]
+    3. **Calculate the probability of failure raised to the power of 3**:
+       \[ \left(1 - \frac{1}{5}\right)^3 = \left(\frac{4}{5}\right)^3 = \frac{64}{125} \]
     
-    3. **Divide both sides by 10**: Solving for $x$:
-       \[
-       x = \frac{159.75}{10} = 15.975
-       \]
+    4. **Combine these results to find the total probability**:
+       \[ P(X = 4) = 35 \times \frac{1}{625} \times \frac{64}{125} = \frac{35 \times 64}{625 \times 125} = \frac{2240}{78125} = \frac{1792}{62500} = \frac{448}{15625} \]
     
-    4. **Express as a fraction**: Converting $15.975$ into a fraction, we note that $0.975 = \frac{39}{40}$, thus:
-       \[
-       x = 15 + \frac{39}{40} = \frac{600}{40} + \frac{39}{40} = \frac{639}{40}
-       \]
-    
-       Therefore, the distance between the two buildings on the map is $\boxed{\frac{639}{40}}$ inches.
+    Conclusion:
+    The probability that exactly 4 out of the 7 islands contain treasure is:
+    \[ \boxed{\frac{448}{15625}} \]
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `2 - \frac{\pi}{2}`
+- **reference**: `$g(x)=c, g(x)=\lceil x \rceil, g(x)=\lfloor x \rfloor$`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `\frac{1}{2}`
+- **extracted** (✗): `h(x) = -x^2`
 - **completion**:
 
     ```text
-    <think>
+    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:
+       \[
+       h\left(\frac{q}{b}\right) = h\left(\frac{h(q - 1) + b}{b}\right) = 0
+       \]
+       Therefore:
+       \[
+       h(q - 1) = -b
+       \]
     
-    </think>
+    5. **Conclusion for \(h(q)\):**
+       By induction, we can generalize that for any integer \(n\):
+       \[
+       h(q - n) = -bn
+       \]
+       This implies that for any integer \(n\):
+       \[
+       h(nq) = -bn^2
+       \]
     
-    Let $p$ be the probability that Derek becomes debt-free starting from a debt of 1 MIT. On each toss:
-    - With probability $\frac{1}{2}$, he reduces his debt by 1 MIT.
-    - With probability $\frac{1}{2}$, he reduces his debt by 2 MITs.
+    6. **Final Verification:**
+       We need to verify that \(h(x) = -x^2\) satisfies the original equation. Let:
+       \[
+       h(x) = -x^2
+       \]
+       Substitute \(h(x) = -x^2\) into the original equation:
+       \[
+       h(x) = h\left(\frac{h(bx - a) + a}{b}\right)
+       \]
+       Substituting \(h(x) = -x^2\):
+       \[
+       -x^2 = h\left(\frac{-\left(bx - a\right)^2 + a}{b}\right)
+       \]
+       Simplify the argument of \(h\):
+       \[
+       -x^2 = h\left(\frac{-b^2x^2 + 2abx - a^2 + a}{b}\right)
+       \]
+       Since \(h\) is defined as \(-y^2\):
+       \[
+       -x^2 = -\left(\frac{-b^2x^2 + 2abx - a^2 + a}{b}\right)^2
+       \]
+       This simplifies correctly, confirming that \(h(x) = -x^2\) is indeed a solution.
     
-    Thus, the probability of becoming debt-free starting from a debt of 1 MIT is:
-    $$p = \frac{1}{2} \cdot 1 + \frac{1}{2} \cdot 2p.$$
-    Solving for $p$:
-    $$p = \frac{1}{2} + p.$$
-    This implies:
-    $$p = \frac{1}{2}.$$
-    Therefore, Derek has a $\boxed{\frac{1}{2}}$ probability of eventually becoming debt-free.
+    Therefore, the solution is:
+    \[
+    \boxed{h(x) = -x^2}
+    \]
     ```