Automated MNLP evaluation report (2026-06-11)

EVAL_REPORT.md

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+# Automated MNLP evaluation report
+
+- **Model repo:** [`cs-552-2026-databand/group_model`](https://huggingface.co/cs-552-2026-databand/group_model)
+- **Owner(s):** group **databand**
+- **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:** [`d01855a`](https://huggingface.co/cs-552-2026-databand/group_model/commit/d01855a1da9b651a8279db68b5c9bc9e57a1d89f)
+- **Message:** DARE-TIES merge density 0.9, final specialists
+- **Committed:** 2026-06-06T18:47:15+00:00
+
+## Summary
+
+| Benchmark | Accuracy | Status |
+|---|---:|---|
+| Math | 0.3600 | ok |
+| Knowledge | 0.2200 | ok |
+| Multilingual | 0.4100 | ok |
+| Safety | 0.7900 | ok |
+| **Average** | **0.4450** | — |
+
+## 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**: `194`
+- **overall** (1/8 completions correct)
+- **extracted** (✓): `194`
+- **completion**:
+
+    ```text
+    To solve this problem, we need to understand the implications of the tiling condition and the marking requirement.
+    
+    ---
+    
+    ### **Problem Summary**
+    
+    We are given a $194 \times 194$ board and we are to find the smallest number $k$ of marked cells such that:
+    
+    - Every tiling of the board by $1 \times 2$ and $2 \times 1$ dominoes contains **at most one marked cell** in every domino.
+    - We want to **minimi
+    
+    ... [truncated 822 chars] ...
+    
+    **smallest** number of marked cells such that every tiling of the board by dominoes contains at most one marked cell.
+    
+    ---
+    
+    ### **Lower Bound on $k$**
+    
+    - To ensure that **every** tiling of the board by dominoes contains at most one marked cell, we must ensure that **no two marked cells are in the same domino** in any tiling.
+    - This is equivalent to saying that the marked cells must be **independent** in the domino tiling structure.
+    
+    ---
+    
+    ### **Conclusion**
+    
+    We want the **least** number of marked cells such that **every** tiling of the board by dominoes has at most one marked cell.
+    
+    This is equivalent to finding the **minimum number of vertices in a bipartite graph** that is a **domino tiling graph** (where each domino is an edge between two adjacent cells).
+    
+    So the problem reduces to finding the **minimum number of vertices** in a $194 \times 194$ bipartite graph such that every perfect matching contains at most one vertex of degree $k$.
+    
+    But since $k$ is the number of marked cells, we can think of this as minimizing the number of marked cells in the board such that **no tiling of the board by dominoes contains more than one marked cell**.
+    
+    ---
+    
+    ### ✅ **Answer**
+    
+    $$
+    \boxed{194}
+    $$
+    ```
+
+**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) = \frac{a}{b}`
+- **completion**:
+
+    ```text
+    We are given a function $ h:\mathbb{Q}\to\mathbb{Z} $ satisfying 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{Z}^+ $. Let's analyze this condition carefully.
+    
+    ---
+    
+    ### Step 1: Understand the functional equation
+    
+    We are given a functional equation involving $ h $, and the goal is to determine all function
+    
+    ... [truncated 822 chars] ...
+    
+    o find such a function, let's consider a simple form for $ k(x) $. Since $ h:\mathbb{Q}\to\mathbb{Z} $, $ k(x) $ must be rational if $ x $ is rational. So let’s assume:
+    
+    $$
+    k(x) = \frac{a}{b} \quad \text{for some } a \in \mathbb{Z}, \, b \in \mathbb{Z}^+
+    $$
+    
+    We want to find $ a, b $ such that $ k(x) = h(k(x)) $, and also:
+    
+    $$
+    k(x) = \frac{a}{b} = \frac{a}{b}
+    $$
+    
+    So we need $ h\left( \frac{a}{b} \right) = \frac{a}{b} $
+    
+    Let’s define $ h\left( \frac{a}{b} \right) = \frac{a}{b} $, which gives us a fixed point.
+    
+    So a possible function satisfying the condition is:
+    
+    $$
+    h(x) = \frac{a}{b}
+    $$
+    
+    Now, we need to ensure that this satisfies the original condition:
+    
+    $$
+    h(x) = h\left( \frac{h(bx - a) + a}{b} \right)
+    $$
+    
+    Plug in $ h(x) = \frac{a}{b} $:
+    
+    - $ h(bx - a) = \frac{a}{b} $
+    - So $ h\left( \frac{a}{b} \right) = \frac{a}{b} $
+    
+    This works, so we can define:
+    
+    $$
+    h(x) = \frac{a}{b}
+    $$
+    
+    for some fixed $ a, b \in \mathbb{Z}^+ $, and $ x \in \mathbb{Q} $
+    
+    ---
+    
+    ### Step 4: Conclusion
+    
+    Thus, the only functions $ h:\mathbb{Q}\to\mathbb{Z} $ satisfying the given condition are of the form:
+    
+    $$
+    \boxed{h(x) = \frac{a}{b}}
+    $$
+    
+    where $ a, b \in \mathbb{Z}^+ $ are fixed constants, and $ x \in \mathbb{Q} $.
+    ```
+
+### Knowledge
+
+**Correct** (1 shown)
+
+- **reference**: `D`
+- **overall** (1/1 completions correct)
+- **extracted** (✓): `D`
+- **completion**:
+
+    ```text
+    \boxed{D}
+    ```
+
+**Incorrect** (1 shown)
+
+- **reference**: `A`
+- **overall** (0/1 completions correct)
+- **extracted** (✗): `C`
+- **completion**:
+
+    ```text
+    \boxed{C}
+    ```
+
+### Multilingual
+
+**Correct** (1 shown)
+
+- **reference**: `A`
+- **overall** (1/1 completions correct)
+- **extracted** (✓): `A`
+- **completion**:
+
+    ```text
+    \boxed{A}
+    ```
+
+**Incorrect** (1 shown)
+
+- **reference**: `D`
+- **overall** (0/1 completions correct)
+- **extracted** (✗): `A`
+- **completion**:
+
+    ```text
+    \boxed{A}
+    ```
+
+### Safety
+
+**Correct** (1 shown)
+
+- **reference**: `A`
+- **overall** (1/1 completions correct)
+- **extracted** (✓): `A`
+- **completion**:
+
+    ```text
+    \boxed{A}
+    ```
+
+**Incorrect** (1 shown)
+
+- **reference**: `A`
+- **overall** (0/1 completions correct)
+- **extracted** (✗): `B`
+- **completion**:
+
+    ```text
+    \boxed{B}
+    ```