Automated MNLP evaluation report (2026-06-11)

EVAL_REPORT.md

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+# Automated MNLP evaluation report
+
+- **Model repo:** [`cs-552-2026-mvte/math_model`](https://huggingface.co/cs-552-2026-mvte/math_model)
+- **Owner(s):** group **mvte**
+- **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:** [`1399ce3`](https://huggingface.co/cs-552-2026-mvte/math_model/commit/1399ce371a8748c92bedd84a0cd4400a855ad1d4)
+- **Message:** Delete .ipynb_checkpoints
+- **Committed:** 2026-06-10T06:05:07+00:00
+
+## Summary
+
+| Benchmark | Accuracy | Status |
+|---|---:|---|
+| Math | 0.3600 | 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**: `1600`
+- **overall** (4/8 completions correct)
+- **extracted** (✓): `1600`
+- **completion**:
+
+    ```text
+    To solve the given expression, we follow the order of operations (PEMDAS/BODMAS). We start by calculating the multiplication and division, then proceed to addition and subtraction.
+    
+    First, calculate the multiplication:
+    \[2 \times 21 \times 19 = 798\]
+    
+    Next, we add this result to the given numbers:
+    \[441 + 798 + 361\]
+    
+    Adding the numbers together:
+    \[441 + 798 = 1239\]
+    \[1239 + 361 = 1600\]
+    
+    Therefore, the value of \(x\) is \(\boxed{1600}\).
+    ```
+
+**Incorrect** (1 shown)
+
+- **reference**: `60750, 854130, 854970`
+- **overall** (0/8 completions correct)
+- **extracted** (✗): `56700`
+- **completion**:
+
+    ```text
+    1. **Initial Observations and Recurrence Relation:**
+       We are given a function \( h : \mathbb{N} \to \mathbb{N} \setminus \{1\} \) that satisfies the recurrence relation:
+       \[
+       h(n) + h(n+1) = h(n+2)h(n+3) - 840 \quad \text{for all } n \in \mathbb{N}.
+       \]
+       We need to determine the possible values of the sum \( \sum_{i=1}^{2025} h(i) \).
+    
+    2. **Assume a Form for \( h(n) \):**
+       Let's assume
+    
+    ... [truncated 820 chars] ...
+    
+    ) + b) - 840.
+       \]
+       Simplifying, we get:
+       \[
+       2a \cdot n + 2b = (a \cdot (n+2) + b)(a \cdot (n+3) + b) - 840.
+       \]
+    
+    5. **Matching Coefficients:**
+       By expanding the right-hand side and equating coefficients, we find:
+       \[
+       a = 0, \quad b = -\frac{840}{2} = -420.
+       \]
+       This form does not satisfy the original equation either.
+    
+    6. **Conclusion:**
+       After exploring various forms of \( h(n) \), we find that the function \( h(n) \) must be a constant function. This means:
+       \[
+       h(n) = C \quad \text{for all } n \in \mathbb{N}.
+       \]
+       Substituting this into the recurrence relation, we get:
+       \[
+       C + C = C^2 - 840.
+       \]
+       Solving this, we find:
+       \[
+       2C = C^2 - 840 \implies C^2 - 2C - 840 = 0.
+       \]
+       Solving the quadratic equation, we get:
+       \[
+       C = 28 \quad \text{or} \quad C = -28.
+       \]
+       Since \( h(n) \in \mathbb{N} \setminus \{1\} \), we discard \( C = -28 \) and accept \( C = 28 \).
+    
+    7. **Summing the Series:**
+       The sum \( \sum_{i=1}^{2025} h(i) \) is:
+       \[
+       \sum_{i=1}^{2025} 28 = 28 \cdot 2025 = 56700.
+       \]
+       Therefore, the only possible value for \( \sum_{i=1}^{2025} h(i) \) is:
+       \[
+       \boxed{56700}
+       \]
+    The final answer is $\boxed{56700}$.
+    ```