Automated MNLP evaluation report (2026-06-11)

EVAL_REPORT.md

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+# Automated MNLP evaluation report
+
+- **Model repo:** [`cs-552-2026-claude-bots/math_model`](https://huggingface.co/cs-552-2026-claude-bots/math_model)
+- **Owner(s):** group **claude-bots**
+- **Generated at:** 2026-06-11T06:23:10+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._
+
+## Evaluated checkpoint
+
+- **Commit:** [`ed63119`](https://huggingface.co/cs-552-2026-claude-bots/math_model/commit/ed6311961b1d569d2ea74eb5e478bd806554e97b)
+- **Message:** Upload fine-tuned model
+- **Committed:** 2026-06-06T22:35:16+00:00
+
+## Summary
+
+| Benchmark | Accuracy | Status |
+|---|---:|---|
+| Math | 0.4800 | ok |
+| Knowledge | — | not run |
+| Multilingual | — | not run |
+| Safety | — | not run |
+
+## Sample completions
+
+_Prompts are intentionally omitted to avoid revealing benchmark contents. For multi-completion problems, only one completion is shown per sample._
+
+### Math
+
+**Correct** (1 shown)
+
+- **reference**: `\frac{14}{3}`
+- **overall** (8/8 completions correct)
+- **extracted** (✓): `\frac{14}{3}`
+- **completion**:
+
+    ```text
+    <think>
+    Okay, so I need to solve this problem where the function f(x) is given by (3x - 2)/(x - 2), and I have to compute f(-2) + f(-1) + f(0). Hmm, let's break this down step by step. 
+    
+    First, I remember that to find the value of a function at a specific x, I just substitute that x into the function. So for each term in the sum, I need to plug in -2, -1, and 0 into the function f(x). Let me start
+    
+    ... [truncated 822 chars] ...
+    
+    5/3 + 1. Converting 2 and 1 to thirds: 6/3 + 5/3 + 3/3. 6 +5 is 11, 11 +3 is 14. So 14/3. That's 4 and 2/3, but as a common fraction, it's 14/3. So I think that's the answer. Let me just check if there's any simplification needed. 14 and 3 have no common factors besides 1, so 14/3 is the simplified form.
+    
+    I don't see any mistakes here. So the final answer should be 14/3.
+    </think>
+    
+    To solve the problem, we need to compute \( f(-2) + f(-1) + f(0) \) for the function \( f(x) = \frac{3x - 2}{x - 2} \).
+    
+    First, we evaluate \( f(-2) \):
+    \[
+    f(-2) = \frac{3(-2) - 2}{-2 - 2} = \frac{-6 - 2}{-4} = \frac{-8}{-4} = 2
+    \]
+    
+    Next, we evaluate \( f(-1) \):
+    \[
+    f(-1) = \frac{3(-1) - 2}{-1 - 2} = \frac{-3 - 2}{-3} = \frac{-5}{-3} = \frac{5}{3}
+    \]
+    
+    Then, we evaluate \( f(0) \):
+    \[
+    f(0) = \frac{3(0) - 2}{0 - 2} = \frac{0 - 2}{-2} = \frac{-2}{-2} = 1
+    \]
+    
+    Now, we sum these values:
+    \[
+    f(-2) + f(-1) + f(0) = 2 + \frac{5}{3} + 1
+    \]
+    
+    To add these, we convert 2 and 1 to fractions with a denominator of 3:
+    \[
+    2 = \frac{6}{3}, \quad 1 = \frac{3}{3}
+    \]
+    
+    So,
+    \[
+    2 + \frac{5}{3} + 1 = \frac{6}{3} + \frac{5}{3} + \frac{3}{3} = \frac{6 + 5 + 3}{3} = \frac{14}{3}
+    \]
+    
+    Thus, the final answer is:
+    \[
+    \boxed{\frac{14}{3}}
+    \]
+    ```
+
+**Incorrect** (1 shown)
+
+- **reference**: `125`
+- **overall** (0/8 completions correct)
+- **extracted** (✗): `20`
+- **completion**:
+
+    ```text
+    <think>
+    Okay, let's try to figure out this problem step by step. So, we have a rhombus WXYZ with an acute angle at vertex Z. There's a point Q on its incircle, and the perpendicular distances from Q to the lines ZW, WX, and XY are 9, 5, and 16 respectively. We need to find the perimeter of WXYZ.
+    
+    First, let me recall some properties of a rhombus. All sides are equal, and the opposite sides are par
+    
+    ... [truncated 822 chars] ...
+    
+    \(XY\) be 9, 5, and 16, respectively. The fourth distance to the fourth side must be \(2r - 30\). Since \(Q\) is on the incircle, the distance to each side is at most \(r\). Therefore, \(2r - 30 \leq r\), which implies \(r \geq 30\).
+    
+    5. **Finding the Perimeter:**
+       - Let the inradius \(r = 20\). Then, the perimeter \(P\) of the rhombus is:
+         \[
+         P = 4s
+         \]
+       - Using the inradius formula:
+         \[
+         20 = \frac{s \sin \theta}{2} \implies s = \frac{40}{\sin \theta}
+         \]
+       - The perimeter is:
+         \[
+         P = 4s = 4 \times \frac{40}{\sin \theta} = \frac{160}{\sin \theta}
+         \]
+       - The fourth perpendicular distance is \(2r - 30 = 2 \times 20 - 30 = 10\). Since the fourth distance must be at most \(r\), we have \(10 \leq 20\), which is true.
+    
+    6. **Verifying the Solution:**
+       - The inradius \(r = 20\), the side length \(s = \frac{40}{\sin \theta}\), and the fourth distance is 10. The perimeter is:
+         \[
+         P = \frac{160}{\sin \theta}
+         \]
+       - Since the problem does not specify \(\theta\), we assume that the given distances and the inradius are consistent with the perimeter being 20. Therefore, the perimeter of the rhombus is:
+         \[
+         \boxed{20}
+         \]
+    ```