Automated MNLP evaluation report (2026-05-17)

EVAL_REPORT.md

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+# Automated MNLP evaluation report
+
+- **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:** 2026-05-17T04:52:19+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 | 0.0900 | 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**: `\frac65`
+- **overall** (4/8 completions correct)
+- **extracted** (✓): `\frac{6}{5}`
+- **completion**:
+
+    ```text
+    <think>
+    The equation is $5x^2 -6x + 24 + \frac{1}{x} = 0$ or $5x^3 -6x^2 + 24x + 1 = 0$. Let $f(x) = 5x^3 -6x^2 + 24x + 1$. The sum of the roots of this cubic is $\frac{6}{5}$. Thus $a+b+c = \boxed{\frac{6}{5}}$.
+    </think>
+    
+    #### $\frac{6}{5}$
+    ```
+
+**Incorrect** (1 shown)
+
+- **reference**: `\frac{7}{2}`
+- **overall** (0/8 completions correct)
+- **extracted** (✗): `3`
+- **completion**:
+
+    ```text
+    <think>
+    From the first two equations, we can subtract them and obtain
+    $$
+    a^2+ab+b^2 - (a^2+ac+c^2) = 3(a+b) - 3(a+c) \Rightarrow ab - ac + b^2 - c^2 = 3b - 3c$$
+    $$
+    \Rightarrow (b - c)(a + b + c) = 3(b - c)$$
+    If $b \ne c$, we can divide both sides by $b - c$, obtaining
+    $$
+    a + b + c = 3$$
+    Now, we use the third equation, and the fact that $a + b + c = 3$, to obtain
+    $$
+    b^2 + bc + c^2 = 5(b + c) + 1 \Rightarrow (b + c)^2 - bc = 5(b + c) + 1$$
+    Let $s = b + c$, so we obtain
+    $$
+    s^2 - bc = 5s + 1 \Rightarrow bc = s^2 - 5s - 1$$
+    We also know from the first two equations, that $a = 3 - s$, and from the first equation, we have
+    $$
+    (3 - s)^2 + (3 - s)s + s^2 = 3(3 - s) + 3s$$
+    $$
+    \Rightarrow 9 - 6s + s^2 + 3s - s^2 + s^2 = 9 - 3s + 3s$$
+    $$
+    \Rightarrow s^2 - 3s + 9 = 9 \Rightarrow s^2 - 3s = 0 \Rightarrow s(s - 3) = 0$$
+    Therefore, $s = 0$ or $s = 3$
+    If $s = 0$, then $a = 3 - 0 = 3$, and $bc = 0^2 - 5(0) - 1 = -1$
+    If $s = 3$, then $a = 3 - 3 = 0$, and $bc = 3^2 - 5(3) - 1 = 9 - 15 - 1 = -7$
+    Therefore, we have two possibilities for $a$: either $a = 3$ or $a = 0$
+    We now check each of these possibilities in the third equation
+    If $a = 3$, then $b + c = 0$, and $bc = -1$, so $b$ and $c$ are the roots of $x^2 + 1 = 0$
+    If $a = 0$, then $b + c = 3$, and $bc = -7$, so $b$ and $c$ are the roots of $x^2 - 3x - 7 = 0$
+    Therefore, the possible values for $a$ are $\boxed{0}$ and $\boxed{3}$
+    ```