Automated MNLP evaluation report (2026-05-13)

EVAL_REPORT.md

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+# Automated MNLP evaluation report
+
+- **Model repo:** [`cs-552-2026-MandMP/math_model`](https://huggingface.co/cs-552-2026-MandMP/math_model)
+- **Owner(s):** group **MandMP**
+- **Generated at:** 2026-05-13T04:17:52+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._
+
+## Summary
+
+| Benchmark | Accuracy | Status |
+|---|---:|---|
+| Math | 0.2000 | 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**: `0`
+- **overall** (3/8 completions correct)
+- **extracted** (✓): `0`
+- **completion**:
+
+    ```text
+    <think>
+    
+    </think>
+    
+    The polynomial $p(x)$ is equal to $\frac{1}{2}$ when $x=a$, $x=b$, and $x=c$.  Thus, $p(x)$ is a constant polynomial and the degree is $\boxed{0}$.
+    ```
+
+**Incorrect** (1 shown)
+
+- **reference**: `$(n-2)2^n +1$`
+- **overall** (0/8 completions correct)
+- **extracted** (✗): `2^{2^m} - 1`
+- **completion**:
+
+    ```text
+    <think>
+    
+    </think>
+    
+    Let $C_n$ be the set of numbers we are trying to sum.  Then $2^n - 2^i = 2^i (2^{n - i} - 1)$, so each element of $C_n$ is a multiple of $2^i$ for some $i$ (and in particular, for $i = 0$).
+    
+    If $n$ is a power of 2, say $n = 2^m$, then $2^{2^m} - 2^{2^m - 1} = 2^{2^m - 1} (2 - 1) = 2^{2^m - 1}$, and all elements of $C_n$ are multiples of $2^{2^m - 1}$.  Thus the largest number not in $C_n$ is $2^{2^m - 1} - 1$.
+    
+    If $n$ is not a power of 2, then $n = 2^m + 1$ for some $m$.  Then $2^{2^m + 1} - 2^{2^m} = 2^{2^m} (2 - 1) = 2^{2^m}$, and all elements of $C_n$ are multiples of $2^{2^m}$.  Thus the largest number not in $C_n$ is $2^{2^m} - 1$.
+    
+    Hence the answer is $\boxed{2^{2^m} - 1}$, where $n = 2^m + 1$.
+    ```