Automated MNLP evaluation report (2026-06-11)

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:** 2026-06-11T06:23:10+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:** [`37abb85`](https://huggingface.co/cs-552-2026-vibe-trainers/math_model/commit/37abb85a14cf832bc2b6ecb7ff5edd20213c56b3)
+- **Message:** Update chat_template.jinja
+- **Committed:** 2026-06-04T20:08:52+00:00
+
+## Summary
+
+| Benchmark | Accuracy | Status |
+|---|---:|---|
+| Math | 0.4300 | 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**: `3`
+- **overall** (8/8 completions correct)
+- **extracted** (✓): `3`
+- **completion**:
+
+    ```text
+    We are given the equation:
+    
+    $$
+    (x-1)g(x) = 3x^4 + x^3 - 25x^2 + 38x - 17
+    $$
+    
+    We are to find the **degree** of the polynomial $ g(x) $.
+    
+    ---
+    
+    ### Step 1: Understand the structure
+    
+    We are told that $ (x - 1)g(x) $ is a polynomial of degree 4. That is:
+    
+    $$
+    \deg((x - 1)g(x)) = 4
+    $$
+    
+    But the degree of a product of two polynomials is the **sum of their degrees**. So:
+    
+    $$
+    \deg(x - 1) + \deg(g(x)) = 4
+    $$
+    
+    We know that $ x - 1 $ is a **linear polynomial**, so:
+    
+    $$
+    \deg(x - 1) = 1
+    $$
+    
+    So:
+    
+    $$
+    1 + \deg(g(x)) = 4 \Rightarrow \deg(g(x)) = 3
+    $$
+    
+    ---
+    
+    ### ✅ Final Answer:
+    
+    $$
+    \boxed{3}
+    $$
+    ```
+
+**Incorrect** (1 shown)
+
+- **reference**: `$g(x)=c, g(x)=\lceil x \rceil, g(x)=\lfloor x \rfloor$`
+- **overall** (0/8 completions correct)
+- **extracted** (✗): `h(x) = 0 \text{ or } h(x) = x`
+- **completion**:
+
+    ```text
+    We are given a functional equation involving a mapping $ h : \mathbb{Q} \to \mathbb{Z} $, and the condition:
+    
+    $$
+    h(x) = h\left( \frac{h(bx - a) + a}{b} \right)
+    $$
+    
+    for all $ x \in \mathbb{Q} $, $ a \in \mathbb{Z} $, and $ b \in \mathbb{N} $.
+    
+    ---
+    
+    ### Step 1: Understand the structure of the equation
+    
+    Let’s rewrite the equation for clarity:
+    
+    $$
+    h(x) = h\left( \frac{h(bx - a) + a}{b} \right)
+    $$
+    
+    Thi
+    
+    ... [truncated 823 chars] ...
+    
+    ht) = c
+    $$
+    
+    So this works.
+    
+    Try $ n = 1 $:
+    
+    Then:
+    
+    $$
+    h(x) = x + c
+    $$
+    
+    Check the original equation:
+    
+    $$
+    h(x) = x + c
+    $$
+    
+    Compute RHS:
+    
+    $$
+    h\left( \frac{h(bx - a) + a}{b} \right) = h\left( \frac{(bx - a) + c + a}{b} \right) = h\left( \frac{bx + c}{b} \right) = \frac{bx + c}{b} + c = x + \frac{c}{b} + c
+    $$
+    
+    Now compare with LHS:
+    
+    $$
+    h(x) = x + c
+    $$
+    
+    So the RHS is $ x + c + \frac{c}{b} $, which is **not equal** to $ x + c $ unless $ \frac{c}{b} = 0 $, i.e., $ c = 0 $.
+    
+    So only when $ c = 0 $, $ h(x) = x $ works.
+    
+    ---
+    
+    ### Step 5: Conclude possible forms of $ h $
+    
+    From the above, the only possible forms that satisfy the equation are:
+    
+    - $ h(x) = 0 $, for all $ x \in \mathbb{Q} $
+    - $ h(x) = x $, for all $ x \in \mathbb{Q} $
+    
+    Let’s check both.
+    
+    ---
+    
+    ### Step 6: Verify both possibilities
+    
+    #### Case 1: $ h(x) = 0 $
+    
+    Then:
+    
+    $$
+    h(x) = h\left( \frac{h(bx - a) + a}{b} \right) = h\left( \frac{0 + a}{b} \right) = h\left( \frac{a}{b} \right) = 0
+    $$
+    
+    So this works.
+    
+    #### Case 2: $ h(x) = x $
+    
+    Then:
+    
+    $$
+    h(x) = x = h\left( \frac{h(bx - a) + a}{b} \right) = \frac{h(bx - a) + a}{b} = \frac{(bx - a) + a}{b} = x
+    $$
+    
+    So this also works.
+    
+    ---
+    
+    ### Final Answer:
+    
+    $$
+    \boxed{h(x) = 0 \text{ or } h(x) = x}
+    $$
+    ```