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

EVAL_REPORT.md

@@ -2,7 +2,7 @@
 
 - **Model repo:** [`cs-552-2026-thinking-tokens/math_model`](https://huggingface.co/cs-552-2026-thinking-tokens/math_model)
 - **Owner(s):** group **thinking-tokens**
-- **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 | 0.3900 | ok |
+| Math | 0.4000 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -40,7 +40,9 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
     
     ... [truncated 823 chars] ...
     
-    uch that **only one tiling** satisfies the condition.
+    t **every domino contains at most one marked cell**.
+    
+    This is a **combinatorial problem** where we want to **minimize the number of marked cells** such that **only one tiling** satisfies the condition.
     
     ---
     
@@ -58,15 +60,17 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
     
     ### Step 6: Final Answer
     
-    This is a known result in combinatorics. For a $n \times n$ board, the **minimum number of marked cells** required to ensure that **only one tiling** exists where **every domino contains at most one marked cell** is:
+    This is a known result in combinatorics. For a $n \times n$ board, the **minimum number of marked cells** required to ensure that **only one tiling** of the board with $1 \times 2$ dominoes has **at most one marked cell per domino** is:
     
     $$
     \boxed{194}
     $$
     
-    This is because in a $194 \times 194$ board, there are 194 rows and 194 columns. To ensure that **only one tiling** satisfies the condition, we need to mark **exactly one cell per row and one cell per column**, i.e., a **permutation** of the rows and columns.
+    This is the **minimum** number of marked cells required to ensure that **only one tiling** satisfies the condition.
+    
+    ---
     
-    Thus, the **minimum number of marked cells** is:
+    ### ✅ Final Answer:
     
     $$
     \boxed{194}
@@ -77,7 +81,7 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 - **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 some } c \in \mathbb{Z}, \text{ and } h(x) = x`
+- **extracted** (✗): `h(x) = c \text{ or } h(x) = x \text{ for some } c \in \mathbb{Z}`
 - **completion**:
 
     ```text
@@ -103,18 +107,7 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
     
     ... [truncated 822 chars] ...
     
-    = k^2x + \frac{-k^2a + km + ka}{b} + m
-    $$
-    
-    Matching coefficients:
-    
-    - Coefficient of $ x $: $ k = k^2 \Rightarrow k = 0 $ or $ k = 1 $
-    - Constant term: $ m = \frac{-k^2a + km + ka}{b} + m $
-    
-    Try $ k = 1 $:
-    
-    - Coefficient of $ x $: OK
-    - Constant term: $ m = \frac{-1^2a + 1\cdot m + 1\cdot a}{b} + m = \frac{-a + m + a}{b} + m = \frac{m}{b} + m $
+    - a)}{b} + m = - \frac{-m}{b} + m = \frac{m}{b} + m $
     
     So:
     
@@ -124,16 +117,17 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
     
     So $ h(x) = x $ is a solution.
     
-    Check:
+    Let’s verify:
     
     $$
-    h(x) = x, \quad h(bx - a) = bx - a, \quad \frac{h(bx - a) + a}{b} = \frac{bx - a + a}{b} = x
+    h(x) = x \Rightarrow h(bx - a) = bx - a
     $$
     
-    So:
+    Then:
     
     $$
-    h(x) = h(x)
+    \frac{h(bx - a) + a}{b} = \frac{bx - a + a}{b} = x
+    \Rightarrow h(x) = x
     $$
     
     So $ h(x) = x $ is a solution.
@@ -142,22 +136,29 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
     
     ### Step 5: Are there other solutions?
     
-    We have found:
+    We have found two types of solutions:
+    
+    1. **Constant functions**: $ h(x) = c $, for some $ c \in \mathbb{Z} $
+    2. **Linear functions**: $ h(x) = x $, which is also a solution
     
-    - Constant functions: $ h(x) = c \in \mathbb{Z} $
-    - Linear function: $ h(x) = x $
+    Are there any other solutions?
     
-    Are there other solutions?
+    Suppose $ h $ is not constant or linear. Let’s try $ h(x) = c $, a constant function, and $ h(x) = x $, and see if they are the **only** solutions.
     
-    Suppose $ h(x) = c $ for all $ x $, and $ h(x) = x $ for some $ x $, then we must have $ c = x $ for all $ x $, which is only possible if $ c = x $ for all $ x \in \mathbb{Q} $, which is not possible unless $ c $ is not a constant.
+    Suppose $ h $ is not constant, and not linear. Then we must show that it must be constant or linear.
     
-    So **the only solutions are constant functions and $ h(x) = x $**.
+    But from the earlier analysis, **no non-constant linear functions** work, and **constant functions** and $ h(x) = x $ both work.
+    
+    So the only possible solutions are:
+    
+    - Constant functions: $ h(x) = c $, for some $ c \in \mathbb{Z} $
+    - $ h(x) = x $
     
     ---
     
-    ### ✅ Final Answer:
+    ### Final Answer:
     
     $$
-    \boxed{h(x) = c \text{ for some } c \in \mathbb{Z}, \text{ and } h(x) = x}
+    \boxed{h(x) = c \text{ or } h(x) = x \text{ for some } c \in \mathbb{Z}}
     $$
     ```