Update Automated MNLP evaluation report (2026-06-08)

EVAL_REPORT.md

@@ -2,7 +2,7 @@
 
 - **Model repo:** [`cs-552-2026-kth/math_model`](https://huggingface.co/cs-552-2026-kth/math_model)
 - **Owner(s):** group **kth**
-- **Generated at:** 2026-06-06T17:20:53+00:00 (UTC)
+- **Generated at:** 2026-06-08T04:40:54+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._
@@ -17,7 +17,7 @@ _This PR is opened automatically by the course CI. It is **non-blocking** — yo
 
 | Benchmark | Accuracy | Status |
 |---|---:|---|
-| Math | 0.2800 | ok |
+| Math | 0.3500 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -30,84 +30,80 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `\frac{14}{3}`
-- **overall** (8/8 completions correct)
-- **extracted** (✓): `\frac{14}{3}`
+- **reference**: `194`
+- **overall** (1/8 completions correct)
+- **extracted** (✓): `194`
 - **completion**:
 
     ```text
     <think>
-    We can compute each value of \(f(x)\) separately and then add them together.
+    ## Step 1: Understand the problem
+    We need to find the least positive integer $k$ for which we can choose $k$ cells to mark on a $194 \times 194$ board so that there is exactly one tiling of the board by $1 \times 2$ and $2 \times 1$ dominoes in which every domino contains at most one marked cell.
     
-    1. Compute \(f(-2)\):
-    \[ f(-2) = \frac{3(-2) - 2}{-2 - 2} = \frac{-6 - 2}{-4} = \frac{-8}{-4} = 2 \]
+    ## Step 2: Analyze the tiling condition
+    For a $194 \times 194$ board to be tiled by $1 \times
     
-    2. Compute \(f(-1)\):
-    \[ f(-1) = \frac{3(-1) - 2}{-1 - 2} = \frac{-3 - 2}{-3} = \frac{-5}{-3} = \frac{5}{3} \]
+    ... [truncated 823 chars] ...
     
-    3. Compute \(f(0)\):
-    \[ f(0) = \frac{3(0) - 2}{0 - 2} = \frac{0 - 2}{-2} = \frac{-2}{-2} = 1 \]
+    Since the board is $194 \times 194$, which is an even number, the minimum $k$ would be related to the number of cells that need to be marked to restrict the possible arrangements of dominoes.
     
-    Now add these values together:
-    \[ f(-2) + f(-1) + f(0) = 2 + \frac{5}{3} + 1 = \frac{6}{3} + \frac{5}{3} + \frac{3}{3} = \frac{14}{3} \]
+    ## Step 13: Realize the mistake in the previous step
+    The previous step was incorrect because it assumed a specific pattern for marking cells. In reality, the minimum $k$ depends on the specific constraints of the tiling problem and the number of cells that need to be marked to achieve the desired outcome.
     
-    The answer is:
-    \[
+    ## Step 14: Correct the approach
+    To solve this problem, we need to consider the constraints imposed by the tiling condition and the requirement for exactly one tiling. This involves finding the minimum number of marked cells required to restrict the possible arrangements of dominoes in a way that meets the given conditions.
+    
+    ## Step 15: Consider the solution
+    After reevaluating the problem, we realize that the minimum $k$ is related to the size of the board and the constraints imposed by the tiling condition. The correct solution involves finding the minimum number of marked cells required to achieve the desired outcome.
+    
+    The final answer is: $
     </think>
     
-    Therefore, the answer is \boxed{\frac{14}{3}}.
+    Therefore, the answer is \boxed{194}.
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `(3,2,5)`
+- **reference**: `$g(x)=c, g(x)=\lceil x \rceil, g(x)=\lfloor x \rfloor$`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `\text{None}`
+- **extracted** (✗): `0`
 - **completion**:
 
     ```text
     <think>
-    Let's start by examining the given conditions:
-    
-    1. \(p\) is a prime greater than 2.
-    2. \(m\) and \(n\) are positive integers with \(m \le n\).
-    3. \(m^{p-1}+n\) and \(n^{p-1}+m\) are both powers of \(p\).
-    
-    We can start by considering the possible values of \(p\). Since \(p\) is a prime greater than 2, the smallest possible value of \(p\) is 3.
-    
-    Let's consider the case where \(p = 3\). We ne
+    Let's use the given functional equation to derive conditions for $h$.
     
-    ... [truncated 526 chars] ...
+    Given:
+    \[ h(x) = h\left(\frac{h(bx - a) + a}{b}\right) \]
     
-    p > 3\).
+    For any $a \in \mathbb{Z}$ and any positive integer $b$, we want to find $h$ that satisfies this equation for all $x \in \mathbb{Q}$.
     
-    We can rewrite the given conditions as:
-    \[ m^{p-1}+n = a^{p} \]
-    \[ n^{p-1}+m = b^{p} \]
+    Substitute $x = 0$ into the given equation:
+    \[ h(0) = h\left(\frac{h(-a) + a}{b}\right) \]
     
-    where \(a\) and \(b\) are positive integers.
+    Since $h$ maps to $\mathbb{Z}$, $h(-a)$ must also be an integer.
     
-    Subtracting the two equations, we get:
-    \[ m^{p-1}-n^{p-1} = a^{p}-b^{p} \]
+    Now, let's consider what happens when we set $a = 0$:
+    \[ h(0) = h\left(\frac{h(0)}{b}\right) \]
     
-    Using the fact that \(p\) is a prime greater than 2, we can factor the left-hand side of the equation as:
-    \[ (m-n)(m^{p-2}+m^{p-3}n+\cdots+n^{p-2}) = (a-b)(a^{p-1}+a^{p-2}b+\cdots+ab^{p-2}) \]
+    For this to hold for any $b$, $h(0)$ must be $0$, because otherwise, the right side would not be an integer for all $b$.
     
-    Since \(m \le n\), we have \(m-n \le 0\). Therefore, the left-hand side of the equation is non-positive.
+    So, $h(0) = 0$.
     
-    On the other hand, the right-hand side of the equation is non-negative, since \(a \ge b\) and \(a^{p-1}+a^{p-2}b+\cdots+ab^{p-2} \ge 0\).
+    Next, let's examine the case when $b = 1$:
+    \[ h(x) = h\left(h(x - a) + a\right) \]
     
-    This is a contradiction, since the left-hand side of the equation is non-positive and the right-hand side of the equation is non-negative.
+    Since $h$ maps to integers and $a$ is an integer, $h(x - a)$ must also be an integer.
     
-    Therefore, our assumption that there exists an ordered triple \((m, n)\) such that \(m^{p-1}+n\) and \(n^{p-1}+m\) are both powers of \(p\), and \(p > 3\), is false.
+    However, for the equation to hold for any $x$ and any integer $a$, we must have $h(x) = 0$ for all $x \in \mathbb{Q}$.
     
-    Hence, there are no solutions for \(p > 3\).
+    This is because if $h(x) \neq 0$ for some $x$, then for any integer $a$, $h(x - a)$ would not be an integer, contradicting the fact that $h$ maps to integers.
     
-    In conclusion, the only possible value of \(p\) is 3, and there are no solutions for \(p = 3\).
+    Therefore, the only solution is $h(x) = 0$ for all $x \in \mathbb{Q}$.
     
-    Therefore, the final answer is:
+    So, the answer is:
     \[
     </think>
     
-    Therefore, the answer is \boxed{\text{None}}.
+    Therefore, the answer is \boxed{0}.
     ```