Update Automated MNLP evaluation report (2026-06-05)

EVAL_REPORT.md

@@ -2,7 +2,7 @@
 
 - **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:** 2026-06-03T08:04:32+00:00 (UTC)
+- **Generated at:** 2026-06-05T03:00:32+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._
@@ -11,7 +11,7 @@ _This PR is opened automatically by the course CI. It is **non-blocking** — yo
 
 | Benchmark | Accuracy | Status |
 |---|---:|---|
-| Math | 0.3400 | ok |
+| Math | 0.3900 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -24,132 +24,166 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `2`
-- **overall** (1/8 completions correct)
-- **extracted** (✓): `2`
+- **reference**: `194`
+- **overall** (2/8 completions correct)
+- **extracted** (✓): `194`
 - **completion**:
 
     ```text
-    1. **Given Conditions:**
-       - $\angle X = 60^\circ$.
-       - $P$ and $Q$ lie on the sides $XY$ and $XZ$ respectively.
-       - The circumcenter of triangle $XYZ$ is the midpoint of the segment $PQ$.
-    
-    2. **Identifying Triangle Type:**
-       - Since $\angle X = 60^\circ$ and triangle $XYZ$ is inscribed in a circle, let us denote the circumcenter of $\triangle XYZ$ as $O$.
-       - Given $O$ is the midpoint of $PQ
-    
-    ... [truncated 434 chars] ...
-    
-    to the side opposite the vertex through which the line passes.
-       - In $\triangle XYZ$, $HO$ is perpendicular to $PQ$.
-    
-    5. **Angles and Parallelism:**
-       - Since $H$ is the orthocenter, $\angle XHZ = 60^\circ$ because it is an angle formed by the altitudes intersecting at $H$.
-       - This implies that $\triangle XHZ$ is an equilateral triangle.
-       - Therefore, $HZ = XZ$.
-    
-    6. **Using Perpendicular Bisectors:**
-       - Let $K$ be the midpoint of $PQ$ (which is the circumcenter of $\triangle XYZ$).
-       - Since $K$ is the midpoint of $PQ$, $PK = KQ$ and $PK \perp PQ$.
-       - Thus, $\angle POK = 90^\circ$ because $PK$ is perpendicular to $PQ$.
-    
-    7. **Similarity of Triangles:**
-       - Since $\angle POK = 90^\circ$ and $\triangle XHZ$ is equilateral, we have:
-         \[
-         \triangle XHZ \sim \triangle OHP
-         \]
-       - Given that $HZ = XZ$, it follows that $O$ is the midpoint of $HZ$.
-    
-    8. **Calculating Ratios:**
-       - Since $HZ = XZ$ and $HZ = 2HZ$, the ratio of $XZ$ to $HZ$ is:
-         \[
-         \frac{XZ}{HZ} = 2
-         \]
-       - Knowing that $HZ = 2PK$ and $HZ = 2OZ$, the ratio $XQ$ to $PY$ is determined by the fact that:
-         \[
-         \frac{XQ}{PY} = \frac{XZ}{HZ} = 2
-         \]
-    
-    ### Conclusion:
-    \[
-    \boxed{2}
-    \]
+    We are given a $194 \times 194$ board, and we are to find the **least positive integer $k$** such that we can choose $k$ cells to mark, and there exists **exactly one tiling** of the board using $1 \times 2$ and $2 \times 1$ dominoes, where **every domino contains at most one marked cell**.
+    
+    ---
+    
+    ### Step 1: Understand the problem
+    
+    - The board is $194 \times 194$, so it has $194^2 = 37636$ cells.
+    
+    ... [truncated 822 chars] ...
+    
+    ose we mark $k$ cells. We want **exactly one perfect matching** (tiling) of the board such that **each domino contains at most one marked cell**.
+    
+    This is equivalent to: **Find the smallest $k$ such that there is exactly one perfect matching in the graph where each edge is allowed to contain at most one marked cell**.
+    
+    ---
+    
+    ### Step 4: Minimum number of marked cells
+    
+    We want **only one perfect matching** in the graph, and **each domino (edge in the matching) contains at most one marked cell**.
+    
+    This is equivalent to: **Find the smallest $k$ such that the graph has only one perfect matching, and the matching is such that each edge contains at most one marked cell**.
+    
+    This is a known result in combinatorics.
+    
+    ---
+    
+    ### Step 5: Use of known results
+    
+    It is a known result in combinatorics that the **minimum number of marked cells** needed to ensure **exactly one tiling** of the board with **each domino containing at most one marked cell** is:
+    
+    $$
+    \boxed{194}
+    $$
+    
+    This is because the board is $194 \times 194$, and to ensure that **only one tiling** is possible, we need to **mark at least one cell per row and column** to prevent multiple tilings.
+    
+    ---
+    
+    ### ✅ Final Answer:
+    
+    $$
+    \boxed{194}
+    $$
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `60750, 854130, 854970`
+- **reference**: `$g(x)=c, g(x)=\lceil x \rceil, g(x)=\lfloor x \rfloor$`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `26358`
+- **extracted** (✗): `h(x) = x`
 - **completion**:
 
     ```text
-    To solve the problem, we need to determine the function \( h: \mathbb{N} \rightarrow \mathbb{N} \setminus \{1\} \) satisfying the recurrence relation:
+    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 821 chars] ...
+    
+    ### Step 5: Try $ n = 0 $
+    
+    If $ n = 0 $, then $ h(x) = c $, a constant function.
+    
+    Plug into the original equation:
+    
+    $$
+    h(x) = h\left( \frac{h(bx - a) + a}{b} \right)
+    $$
+    
+    If $ h(x) = c $, then:
+    
+    $$
+    c = h\left( \frac{c + a}{b} \right) = c
+    $$
+    
+    So this is valid for any $ c \in \mathbb{Z} $.
+    
+    ---
+    
+    ### Step 6: Try $ n = 1 $
+    
+    If $ n = 1 $, then $ h(x) = x + c $.
+    
+    Let’s check if this works.
+    
+    Then:
+    
+    $$
+    h(bx - a) = bx - a + c
+    $$
+    
+    Then:
     
-    \[
-    h(n) + h(n+1) = h(n+2) h(n+3) - 840 \quad \text{for all } n \in \mathbb{N}.
-    \]
+    $$
+    \frac{h(bx - a) + a}{b} = \frac{bx - a + c + a}{b} = \frac{bx + c}{b} = x + \frac{c}{b}
+    $$
     
-    Let's follow the provided solution and expand it for better clarity.
+    Then:
     
-    1. **Rewriting the Recurrence Relation:**
+    $$
+    h\left( \frac{h(bx - a) + a}{b} \right) = h\left( x + \frac{c}{b} \right) = x + \frac{c}{b} + c
+    $$
     
-       The given recurrence relation can be rearranged t
+    Now compare with $ h(x) = x + c $:
     
-    ... [truncated 525 chars] ...
+    $$
+    h(x) = x + c \quad \text{and} \quad h\left( \frac{h(bx - a) + a}{b} \right) = x + \frac{c}{b} + c
+    $$
     
-    \[
-       \begin{aligned}
-       h(3) + h(4) &= h(5) h(6) - 840, \\
-       h(4) + h(5) &= h(6) h(7) - 840.
-       \end{aligned}
-       \]
+    So unless $ \frac{c}{b} = 0 $, these are not equal.
     
-    4. **Solving for \( h(3) \) and \( h(4) \):**
+    Thus, this only works if $ c = 0 $, and then $ h(x) = x $.
     
-       We will iterate values to find possible sequences that satisfy the conditions. Let's begin with \( h(1) = 1 \) and \( h(2) = 2 \):
+    ---
     
-       \[
-       \begin{aligned}
-       h(3) &= 5, \\
-       h(4) &= 21, \\
-       h(5) &= 50, \\
-       h(6) &= 21, \\
-       h(7) &= 5, \\
-       h(8) &= 21.
-       \end{aligned}
-       \]
+    ### Step 7: Final check
     
-    5. **General Pattern Recognition:**
+    Let’s test $ h(x) = x $ in the original equation:
     
-       Notice that the sequence \( \{h(n)\} \) alternates between \( 5, 21, 50, 21, 5, 21, \ldots \).
+    $$
+    h(x) = h\left( \frac{h(bx - a) + a}{b} \right)
+    $$
     
-    6. **Summation Analysis:**
+    If $ h(x) = x $, then:
     
-       For the summation \( \sum_{i=1}^{2025} h(i) \):
+    $$
+    x = \frac{h(bx - a) + a}{b} = \frac{bx - a + a}{b} = x
+    $$
     
-       \[
-       \begin{aligned}
-       &\sum_{i=1}^{2025} h(i) = h(1) + h(2) + h(3) + \cdots + h(2025) \\
-       &= 1 + 2 + 5 + 21 + 50 + 21 + 5 + 21 + \cdots + 21.
-       \end{aligned}
-       \]
+    So this works.
     
-       Since the sequence alternates between 5 and 21, we can group terms:
+    ---
     
-       \[
-       \begin{aligned}
-       &\sum_{i=1}^{2025} h(i) = 1 + 2 + (5 + 21) \times \frac{2025}{2} \\
-       &= 3 + 26 \times 1012 \\
-       &= 26332 + 26 \\
-       &= 26358.
-       \end{aligned}
-       \]
+    ### ✅ Final Answer:
     
-       Thus, the sum of the first 2025 terms of the sequence \( h \) is:
+    $$
+    \boxed{h(x) = x}
+    $$
     
-       \[
-       \boxed{26358}.
-       \]
+    is the only solution.
     ```