Update Automated MNLP evaluation report (2026-05-17)

EVAL_REPORT.md

@@ -2,7 +2,7 @@
 
 - **Model repo:** [`cs-552-2026-bilko/math_model`](https://huggingface.co/cs-552-2026-bilko/math_model)
 - **Owner(s):** group **bilko**
-- **Generated at:** 2026-05-16T04:57:46+00:00 (UTC)
+- **Generated at:** 2026-05-17T04:52:19+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._
@@ -24,105 +24,82 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `0`
-- **overall** (1/8 completions correct)
-- **extracted** (✓): `0`
+- **reference**: `194`
+- **overall** (3/8 completions correct)
+- **extracted** (✓): `194`
 - **completion**:
 
     ```text
     <think>
-    1. **Identify the polynomial form:**
-       Given polynomial \( p(x) \) is expressed as:
-       \[
-       p(x) = \frac{(x-b)(x-c)}{(a-b)(a-c)} + \frac{(x-a)(x-c)}{(b-a)(b-c)} + \frac{(x-a)(x-b)}{(c-a)(c-b)}
-       \]
+    1. **Understanding the problem**: We need to find the smallest positive integer \( k \) such that we can select \( k \) cells on a \( 194 \times 194 \) board, ensuring that the board can be tiled completely by \( 1 \times 2 \) and \( 2 \times 1 \) dominoes, with each domino containing at most one marked cell.
     
-    2. **Substitute \( x = a \):**
-       When \( x = a \):
-       \[
-       p(a) = \frac{(a-b)(a-c)}{(a-b)(a-c)} + \frac{(a-a)(a-c)}{(b-a)(b-c)} + \frac{(a-a)(a-b)}{(c-a)(c-b)}
-       \]
-       Simplifying each te
+    2. **Observation**: The problem resembles a \( 2 \times 2 \) chessboard where ea
     
-    ... [truncated 239 chars] ...
+    ... [truncated 17 chars] ...
     
-    [
-       p(b) = \frac{(b-c)(b-a)}{(a-b)(a-c)} + 1 + \frac{(b-a)^2}{(c-a)(c-b)}
-       \]
-       This expression simplifies to:
-       \[
-       p(b) = \frac{(b-a)(b-c)}{(a-b)(a-c)} + 1 + \frac{(b-a)^2}{(c-a)(c-b)}
-       \]
-       \[
-       p(b) = \frac{(b-a)(b-c)(c-a) + (b-a)(b-c) + (b-a)^2}{(a-b)(a-c)(c-a)}
-       \]
-       \[
-       p(b) = \frac{(b-a)[(b-c)(c-a) + (b-c) + (b-a)]}{(a-b)(a-c)(c-a)}
-       \]
-       This simplifies to:
-       \[
-       p(b) = \frac{(b-a)(b-c)(c-a) + (b-a)(b-c) + (b-a)^2}{(a-b)(a-c)(c-a)}
-       \]
-       \[
-       p(b) = 1
-       \]
+    er two cells. If we mark one of the four cells, the other three cells must be covered by dominoes without any marked cell in the same domino. This can be visualized as a chessboard where the diagonals are marked.
     
-    4. **Substitute \( x = c \):**
-       When \( x = c \):
-       \[
-       p(c) = \frac{(c-b)(c-a)}{(a-b)(a-c)} + \frac{(c-a)(c-b)}{(b-a)(b-c)} + \frac{(c-a)(c-b)}{(c-a)(c-b)}
-       \]
-       Simplifying each term:
-       \[
-       p(c) = \frac{(c-b)(c-a)}{(a-b)(a-c)} + 1 + \frac{(c-a)(c-b)}{(b-a)(b-c)}
-       \]
-       This expression simplifies to:
-       \[
-       p(c) = \frac{(c-a)(c-b)(b-a) + (c-a)(c-b) + (c-a)(c-b)}{(a-b)(a-c)(b-a)}
-       \]
-       \[
-       p(c) = \frac{(c-a)(c-b)(b-a) + (c-a)(c-b) + (c-a)(c-b)}{(a-b)(a-c)(b-a)}
-       \]
-       \[
-       p(c) = 1
-       \]
+    3. **Pattern for smaller boards**: For a \( 2 \times 2 \) board, marking one of the four cells allows the other three to be covered by dominoes without any marked cell in the same domino. This pattern extends to larger boards.
     
-    5. **Conclusion:**
-       Since \( p(a) = p(b) = p(c) = 1 \), the polynomial \( p(x) \) is a constant polynomial. Therefore, the degree of the polynomial \( p(x) \) is zero.
+    4. **Generalizing to \( 194 \times 194 \) board**: We need to extend this pattern to a \( 194 \times 194 \) board. If we mark cells in a diagonal pattern, each domino will cover cells from different diagonals, ensuring that no two marked cells are in the same domino.
+    
+    5. **Counting the marked cells**: The diagonal pattern on a \( 194 \times 194 \) board consists of 194 cells. Each domino covers exactly two cells, and since the board is tiled completely, the number of dominoes is \( \frac{194 \times 194}{2} \).
+    
+    6. **Conclusion**: Therefore, marking 194 cells in a diagonal pattern ensures that the board can be tiled completely with \( 1 \times 2 \) and \( 2 \times 1 \) dominoes, with each domino containing at most one marked cell.
     
     \[
-    \boxed{0}
-    \
+    \boxed{194}
+    \]
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `$(n-2)2^n +1$`
+- **reference**: `$g(x)=c, g(x)=\lceil x \rceil, g(x)=\lfloor x \rfloor$`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `2^n - 3`
+- **extracted** (✗): `h(x) = 0 \text{ or } h(x) = x + 1`
 - **completion**:
 
     ```text
     <think>
-    Okay, let's try to solve this problem step by step. So, we have a set C_n defined as {2^n - 2^i : i is an integer and 0 ≤ i < n}. We need to find the greatest positive integer that cannot be represented as a sum of elements from C_n, allowing repeated use of any element.
+    Okay, let's try to solve this problem step by step. The goal is to determine all mappings \( h: \mathbb{Q} \to \mathbb{Z} \) that satisfy the given condition for every \( x \in \mathbb{Q} \), every \( a \in \mathbb{Z} \), and every positive integer \( b \):
+    
+    \[
+    h(x) = h\left(\frac{h(bx - a) + a}{b}\right).
+    \]
     
-    First, let's understand the set C_n better. For example, if n = 2, then C_2 = {2^2 - 2^0, 2^2 - 2^1} = {4 - 1, 4 - 2} =
+    First, let's understand the condition better. We need to find a function \( h \)
     
     ... [truncated 823 chars] ...
     
-    eneral formula for this.
+    tion.
     
-    Notice that 1 = 2^2 - 3, 5 = 2^3 - 3, and 11 = 2^4 - 3. It seems like the largest number that cannot be expressed as a sum of elements from C_n is 2^n - 3.
+    5. **Checking for non-zero solutions:**
     
-    To verify this, let's assume that for some positive integer k, there exists a representation of k as a sum of elements from C_n. We need to show that k can be represented as a sum of elements from C_{n-1}.
+       Suppose \( h(x) = x + 1 \) for some \( x \). Then:
+       \[
+       h\left(\frac{h(bx - a) + a}{b}\right) = \frac{h(bx - a) + a}{b} + 1.
+       \]
+       Substituting \( h \):
+       \[
+       \frac{(bx - a) + a}{b} + 1 = x + 1.
+       \]
+       Simplifying:
+       \[
+       x + 1 = x + 1.
+       \]
+       This is true, so \( h(x) = x + 1 \) is a valid solution.
+    
+    6. **Conclusion:**
     
-    If k is not representable as a sum of elements from C_n, then k must be less than 2^n - 3. This is because 2^n - 3 is the largest number that cannot be expressed as a sum of elements from C_n.
+       We have found that \( h(x) = 0 \) and \( h(x) = x + 1 \) are solutions. To ensure these are the only solutions, we need to check if there can be other functions that satisfy the condition.
     
-    Now, let's consider the representation of k - 1 as a sum of elements from C_n. If k - 1 is representable as a sum of elements from C_n, then k is representable as a sum of elements from C_n. If k - 1 is not representable as a sum of elements from C_n, then k - 1 must be less than 2^n - 3.
+       Suppose there exists another function \( h \) that satisfies the condition. Then, for any \( x \in \mathbb{Q} \), \( h(x) \) must be an integer. Moreover, for any \( a \in \mathbb{Z} \), \( h(bx - a) \) must also be an integer. This implies that \( h \) is a linear function, as previously shown.
     
-    Since k - 1 is less than 2^n - 3, it follows that k - 1 is representable as a sum of elements from C_{n-1}. Therefore, k is representable as a sum of elements from C_n.
+       Therefore, \( h(x) \) must be of the form \( h(x) = kx + c \). We have already shown that \( k = 0 \) or \( k = 1 \) are the only possibilities, leading to \( h(x) = 0 \) or \( h(x) = x + 1 \).
     
-    This completes the proof by induction.
+       Hence, the only solutions are \( h(x) = 0 \) and \( h(x) = x + 1 \).
     
-    Therefore, the greatest positive integer that cannot be expressed as a sum of elements of C_n is $\boxed{2^n - 3}$.
+    \[
+    \boxed{h(x) = 0 \text{ or } h(x) = x + 1}
+    \
     ```