Automated MNLP evaluation report (2026-05-27)

EVAL_REPORT.md

@@ -2,7 +2,7 @@
 
 - **Model repo:** [`cs-552-2026-the-transformers/math_model`](https://huggingface.co/cs-552-2026-the-transformers/math_model)
 - **Owner(s):** group **the-transformers**
-- **Generated at:** 2026-05-23T14:37:35+00:00 (UTC)
+- **Generated at:** 2026-05-27T13:02:22+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._
@@ -25,63 +25,83 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 **Correct** (1 shown)
 
 - **reference**: `194`
-- **overall** (3/8 completions correct)
+- **overall** (4/8 completions correct)
 - **extracted** (✓): `194`
 - **completion**:
 
     ```text
     <think>
-    To solve this problem, we need to understand the conditions under which a \(194 \times 194\) board can be tiled with \(1 \times 2\) and \(2 \times 1\) dominoes, and how marking cells affects this tiling.
+    To solve this problem, we need to understand the conditions under which a \(194 \times 194\) board can be tiled with \(1 \times 2\) and \(2 \times 1\) dominoes, and how the marking of cells affects this tiling.
     
-    First, we note that a \(194 \times 194\) board has an area of \(194^2 = 37636\) cells. Each domino covers exactly 2 cells, so the number of dominoes required to tile the board is \(37636
+    First, we calculate the total number of cells on the board:
+    \[
+    194 \times 194 = 37636
+    \]
+    Since each domino covers exactly 2 cells, the maximum number of dominoes that can tile the
     
-    ... [truncated 147 chars] ...
+    ... [truncated 481 chars] ...
     
-    ontains at most one marked cell. This means that the marked cells can be placed in such a way that no two marked cells are in the same domino.
+    e least positive integer \(k\) for which one can choose \(k\) cells to mark so that there is exactly one tiling of the board by dominoes in which every domino contains at most one marked cell, we need to consider the implications of marking cells on the tiling.
     
-    To find the minimum number of marked cells, we need to consider the maximum number of dominoes that can be placed such that no two marked cells are in the same domino. This is equivalent to placing the marked cells in such a way that they are not in the same row or column, as dominoes span two rows and two columns.
+    If we mark \(k\) cells, the number of possible tilings decreases. However, we need to ensure that there is exactly one tiling in which every domino contains at most one marked cell.
     
-    If we place a marked cell in each row and each column, we can ensure that no two marked cells are in the same row or column. This is similar to placing non-attacking rooks on a chessboard, where no two rooks attack each other.
+    This problem can be approached by considering the implications of marking cells on the tiling. If we mark a cell, it can be part of a domino in either orientation, but it cannot be part of a domino in the other orientation if it is marked. Therefore, marking a cell reduces the number of possible tilings.
     
-    For a \(194 \times 194\) board, the maximum number of non-attacking rooks we can place is 194 (one per row and one per column). Therefore, the minimum number of marked cells required to ensure that no two marked cells are in the same row or column is 194.
+    To find the least positive integer \(k\), we need to consider the worst-case scenario where marking cells reduces the number of tilings to exactly one. This happens when the number of possible tilings is reduced to exactly one, which requires that the number of marked cells is such that it uniquely determines the tiling.
     
-    Thus, the least positive integer \(k\) for which one can choose \(k\) cells to mark 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 is \(\boxed{194}\).
+    After analyzing the problem, we find that the least positive integer \(k\) for which this is possible is:
+    \[
+    \boxed{194}
+    \]
     ```
 
 **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) = c \text{ for all } x \in \mathbb{Q} \text{ and some } c \in \mathbb{Z}`
+- **extracted** (✗): `h(x) = 0 \text{ for all } x \in \mathbb{Q}`
 - **completion**:
 
     ```text
     <think>
-    Given the problem, we need to find all mappings $h:\mathbb{Q}\to\mathbb{Z}$ such that for every $x\in\mathbb{Q}$, every $a\in\mathbb{Z}$, and every positive integer $b$, the following equation holds:
+    To solve the problem, we need to find all mappings \(h:\mathbb{Q}\to\mathbb{Z}\) that satisfy the given functional equation for all \(x \in \mathbb{Q}\), \(a \in \mathbb{Z}\), and \(b \in \mathbb{N}\):
+    
     \[
-    h(x)=h\!\left(\frac{h(bx-a)+a}{b}\right).
+    h(x) = h\left(\frac{h(bx - a) + a}{b}\right).
     \]
     
-    First, we can simplify the given equation by setting $x=0$. Substituting $x=0$ into the equation, we get:
+    First, let's consider the case when \(a = 0\). The equation becomes:
+    
     \[
-    h(0) = h\!\left(\frac{h(0-a)+a}{b}
+    h(x) = h\left(\frac{h(bx)}{b}\right) = h\left(\frac{h(x)}{b}\ri
     
-    ... [truncated 371 chars] ...
+    ... [truncated 823 chars] ...
+    
+    efore, the solution is:
     
-    r all $a$ and $b$. Therefore, $h$ must be constant on the set of all rational numbers. Let's denote this constant value by $c$. So, we have:
     \[
-    h(x) = c \quad \text{for all} \quad x \in \mathbb{Q}.
+    h(x) = c \text{ for all } x \in \mathbb{Q},
     \]
-    Now, we need to check if this constant function satisfies the given equation. Substituting $h(x) = c$ into the original equation, we get:
+    
+    where \(c\) is an integer. Since the problem asks for all such mappings, we conclude that the only possible values for \(c\) are integers. Therefore, the solution is:
+    
     \[
-    c = c \quad \text{which is true for all} \quad c \in \mathbb{Z}.
+    \boxed{h(x) = c \text{ for all } x \in \mathbb{Q}},
     \]
-    Thus, the constant function $h(x) = c$ for all $x \in \mathbb{Q}$ is a solution. To confirm that this is the only solution, we need to ensure that no other functions can satisfy the given equation. Suppose there exists a non-constant function $h$ that satisfies the equation. Then, there must exist some $x$ for which $h(x) \neq c$. However, the condition $h(x) = h\!\left(\frac{h(bx-a)+a}{b}\right)$ must hold for all $x$, $a$, and $b$, and since $h$ is constant on all rational numbers, this condition can only be satisfied if $h$ is constant. Therefore, the only possible solution is the constant function $h(x) = c$ for all $x \in \mathbb{Q}$.
     
-    Hence, the solution to the problem is:
+    where \(c \in \mathbb{Z}\). To specify the integer \(c\), we need to consider the functional equation again. For any \(x \in \mathbb{Q}\), \(h(x) = h\left(\frac{h(bx - a) + a}{b}\right)\). If \(h(x) = c\), then \(c = h\left(\frac{c + a}{b}\right)\), which implies \(c = c + a\), so \(a = 0\) for any \(a \in \mathbb{Z}\). This is always true, so \(c\) can be any integer.
+    
+    Finally, to ensure that the mapping is well-defined, we need to check that \(h(x)\) is an integer for all \(x \in \mathbb{Q}\). Since \(h(x)\) is constant, it is always an integer, and thus the solution is valid.
+    
+    Therefore, the final answer is:
+    
     \[
-    </think>
+    \boxed{h(x) = c \text{ for all } x \in \mathbb{Q}},
+    \]
+    
+    where \(c \in \mathbb{Z}\). To specify the integer \(c\), we can choose any integer, but typically, the simplest solution is \(c = 0\), so the final answer is:
     
-    \boxed{h(x) = c \text{ for all } x \in \mathbb{Q} \text{ and some } c \in \mathbb{Z}}.
+    \[
+    \boxed{h(x) = 0 \text{ for all } x \in \mathbb{Q}}.
     \]
     ```