Update Automated MNLP evaluation report (2026-06-03)

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:** 2026-06-01T06:13:57+00:00 (UTC)
+- **Generated at:** 2026-06-03T08:04:32+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 | 0.3500 | ok |
+| Math | 0.3400 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -24,97 +24,132 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `194`
+- **reference**: `2`
 - **overall** (1/8 completions correct)
-- **extracted** (✓): `194`
+- **extracted** (✓): `2`
 - **completion**:
 
     ```text
-    1. **Understanding the Problem:**
-       We are given a \(194 \times 194\) board and need to find the smallest number \(k\) such that we can mark \(k\) cells on the board in a way that allows for exactly one tiling of the board with \(1 \times 2\) and \(2 \times 1\) dominoes, with each domino containing at most one marked cell.
+    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. **Initial Observation:**
-       It's evident that \(k = 193\) is not suffi
+    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 823 chars] ...
+    ... [truncated 434 chars] ...
     
-    n can be divided into two parts:
-       - One part where the removed cell was on a black square.
-       - Another part where the removed cell was on a white square.
-       
-       Each of these parts can be independently tiled using dominoes such that no domino contains more than one marked cell. Therefore, removing any one marked cell does not disrupt the tiling.
+    to the side opposite the vertex through which the line passes.
+       - In $\triangle XYZ$, $HO$ is perpendicular to $PQ$.
     
-    6. **Ensuring Minimality:**
-       To prove that \(194\) is the minimal number, consider the following:
-       - Suppose \(k = 193\). Then, there exists a tiling with \(193\) marked cells.
-       - If we remove one marked cell from this tiling, the remaining \(192\) marked cells must still allow for a tiling.
-       - This implies that \(192\) marked cells would suffice for the tiling.
+    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, $HZ = XZ$.
     
-    7. **Contradiction:**
-       However, if \(k = 192\), it is impossible to tile the board with \(192\) marked cells because marking \(192\) cells leaves \(194\) cells unmarked, which must be part of the tiling. This results in a contradiction, as \(192\) marked cells are insufficient to allow for a valid tiling.
+    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 $PQ$.
     
-    8. **Conclusion:**
-       Since \(194\) marked cells allow for a valid tiling and \(193\) marked cells do not, the minimal \(k\) required is:
-       \[
-       \boxed{194}
-       \]
+    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:
+    \[
+    \boxed{2}
+    \]
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `$g(x)=c, g(x)=\lceil x \rceil, g(x)=\lfloor x \rfloor$`
+- **reference**: `60750, 854130, 854970`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `h(x) = kx + c \quad \text{for some } k \in \mathbb{Z}, \ c \in \mathbb{Z}`
+- **extracted** (✗): `26358`
 - **completion**:
 
     ```text
-    1. **Initial Setup and Substitution**:
-       Let \( h(x) = y \) for some \( y \in \mathbb{Z} \). We need to find the general form of \( h \) that satisfies the given functional equation for all \( x \in \mathbb{Q} \), \( a \in \mathbb{Z} \), and \( b \in \mathbb{N} \).
+    To solve the problem, we need to determine the function \( h: \mathbb{N} \rightarrow \mathbb{N} \setminus \{1\} \) satisfying the recurrence relation:
     
-    2. **Substitution and Simplification**:
-       Substitute \( x = \frac{a}{b} \) into the functional equation:
-       \[
-       h\left(\frac{a}{b
+    \[
+    h(n) + h(n+1) = h(n+2) h(n+3) - 840 \quad \text{for all } n \in \mathbb{N}.
+    \]
     
-    ... [truncated 823 chars] ...
+    Let's follow the provided solution and expand it for better clarity.
     
-    kx + c \quad \text{for some } k \in \mathbb{Z}, \ c \in \mathbb{Z}
-       \]
+    1. **Rewriting the Recurrence Relation:**
     
-    5. **Verification**:
-       To verify that \( h(x) = kx + c \) satisfies the original functional equation, substitute \( h(x) = kx + c \) into the equation:
-       \[
-       h(x) = h\left(\frac{h(bx - a) + a}{b}\right)
-       \]
-       Substituting \( h(x) = kx + c \):
-       \[
-       kx + c = h\left(\frac{k(bx - a) + c + a}{b}\right)
-       \]
-       Simplify the argument inside the function:
-       \[
-       kx + c = h\left(\frac{kbx - ka + c + a}{b}\right)
-       \]
-       Since \( h \) is linear:
-       \[
-       kx + c = k\left(\frac{kbx - ka + c + a}{b}\right) + c
+       The given recurrence relation can be rearranged t
+    
+    ... [truncated 525 chars] ...
+    
+    \[
+       \begin{aligned}
+       h(3) + h(4) &= h(5) h(6) - 840, \\
+       h(4) + h(5) &= h(6) h(7) - 840.
+       \end{aligned}
        \]
-       Simplify the right-hand side:
+    
+    4. **Solving for \( h(3) \) and \( h(4) \):**
+    
+       We will iterate values to find possible sequences that satisfy the conditions. Let's begin with \( h(1) = 1 \) and \( h(2) = 2 \):
+    
        \[
-       kx + c = \frac{k^2bx - k^2a + kc + ka}{b} + c
+       \begin{aligned}
+       h(3) &= 5, \\
+       h(4) &= 21, \\
+       h(5) &= 50, \\
+       h(6) &= 21, \\
+       h(7) &= 5, \\
+       h(8) &= 21.
+       \end{aligned}
        \]
-       Multiply through by \( b \) to clear the fraction:
+    
+    5. **General Pattern Recognition:**
+    
+       Notice that the sequence \( \{h(n)\} \) alternates between \( 5, 21, 50, 21, 5, 21, \ldots \).
+    
+    6. **Summation Analysis:**
+    
+       For the summation \( \sum_{i=1}^{2025} h(i) \):
+    
        \[
-       b(kx + c) = k^2bx - k^2a + kc + ka + bc
+       \begin{aligned}
+       &\sum_{i=1}^{2025} h(i) = h(1) + h(2) + h(3) + \cdots + h(2025) \\
+       &= 1 + 2 + 5 + 21 + 50 + 21 + 5 + 21 + \cdots + 21.
+       \end{aligned}
        \]
-       Simplify and equate coefficients:
+    
+       Since the sequence alternates between 5 and 21, we can group terms:
+    
        \[
-       bkx + bc = k^2bx - k^2a + kc + ka + bc
+       \begin{aligned}
+       &\sum_{i=1}^{2025} h(i) = 1 + 2 + (5 + 21) \times \frac{2025}{2} \\
+       &= 3 + 26 \times 1012 \\
+       &= 26332 + 26 \\
+       &= 26358.
+       \end{aligned}
        \]
-       Since \( k, c \in \mathbb{Z} \), both sides match, confirming that \( h(x) = kx + c \) is indeed a solution.
     
-    6. **Conclusion**:
-       The general form of \( h \) that satisfies the given functional equation is:
+       Thus, the sum of the first 2025 terms of the sequence \( h \) is:
+    
        \[
-       \boxed{h(x) = kx + c \quad \text{for some } k \in \mathbb{Z}, \ c \in \mathbb{Z}}
+       \boxed{26358}.
        \]
     ```