Update Automated MNLP evaluation report (2026-06-01)

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:** 2026-05-29T06:54:15+00:00 (UTC)
+- **Generated at:** 2026-06-01T06:08:54+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._
@@ -11,7 +11,7 @@ _This PR is opened automatically by the course CI. It is **non-blocking** — yo
 
 | Benchmark | Accuracy | Status |
 |---|---:|---|
-| Math | 0.3700 | ok |
+| Math | 0.3500 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -24,169 +24,97 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `\frac{639}{40}`
-- **overall** (6/8 completions correct)
-- **extracted** (✓): `\frac{639}{40}`
+- **reference**: `194`
+- **overall** (1/8 completions correct)
+- **extracted** (✓): `194`
 - **completion**:
 
     ```text
-    We are given:
-    
-    - A **length of 3 inches** on the map represents **10 miles** in real life.
-    - The **actual distance** between two buildings is **53.25 miles**.
-    - We are to find the **distance on the map** (in inches) between the two buildings.
-    
-    ---
-    
-    ### Step 1: Find the scale factor
-    
-    The scale is:
-    
-    $$
-    \text{Scale} = \frac{3 \text{ inches}}{10 \text{ miles}} = \frac{3}{10} \text{ inches per mile}
-    $$
-    
-    So, **1 mile = 3/10 inches**.
-    
-    ---
-    
-    ### Step 2: Convert actual distance to map distance
-    
-    We are given:
-    
-    $$
-    \text{Actual distance} = 53.25 \text{ miles}
-    $$
-    
-    Use the scale to find the map distance:
-    
-    $$
-    \text{Map distance} = 53.25 \times \frac{3}{10}
-    $$
-    
-    First, multiply:
-    
-    $$
-    53.25 \times 3 = 159.75
-    $$
-    
-    Then divide by 10:
-    
-    $$
-    \frac{159.75}{10} = 15.975 \text{ inches}
-    $$
-    
-    ---
-    
-    ### Step 3: Express the result as a fraction
-    
-    $$
-    15.975 = \frac{15975}{1000}
-    $$
-    
-    Simplify the fraction:
-    
-    $$
-    \frac{15975}{1000} = \frac{639}{40} \quad \text{(divided numerator and denominator by 25)}
-    $$
-    
-    So, the distance between the buildings on the map is:
-    
-    $$
-    \boxed{\frac{639}{40}} \text{ inches}
-    $$
-    
-    ---
-    
-    ### ✅ Final Answer:
-    
-    $$
-    \boxed{\frac{639}{40}} \text{ inches}
-    $$
+    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.
+    
+    2. **Initial Observation:**
+       It's evident that \(k = 193\) is not suffi
+    
+    ... [truncated 823 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.
+    
+    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.
+    
+    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.
+    
+    8. **Conclusion:**
+       Since \(194\) marked cells allow for a valid tiling and \(193\) marked cells do not, the minimal \(k\) required is:
+       \[
+       \boxed{194}
+       \]
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `279`
+- **reference**: `$g(x)=c, g(x)=\lceil x \rceil, g(x)=\lfloor x \rfloor$`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `12`
+- **extracted** (✗): `h(x) = kx + c \quad \text{for some } k \in \mathbb{Z}, \ c \in \mathbb{Z}`
 - **completion**:
 
     ```text
-    We are given a function $ f(n) $ defined as follows:
-    
-    - For a positive integer $ n < 1000 $, take its base-ten digits (e.g., $ n = 72 $ has digits $ 7 $ and $ 2 $).
-    - Let $ b $ be the **smallest integer strictly larger than every digit** in $ n $.
-    - Interpret the digit string of $ n $ as a number in base $ b $, and call that value $ f(n) $.
-    
-    We are to count how many such numbers $ n < 1000 $ satis
-    
-    ... [truncated 467 chars] ...
-    
-    o the maximum digit is $ 7 $, so $ b = 8 $.
-    
-    Then, interpret the digits $ d_1 d_2 \dots d_k $ as a base-$ b $ number:
-    
-    $$
-    f(n) = d_1 \cdot b^{k-1} + d_2 \cdot b^{k-2} + \dots + d_k \cdot b^0
-    $$
-    
-    We are to find all $ n < 1000 $ such that:
-    
-    $$
-    f(n) = n
-    $$
-    
-    ---
-    
-    ### Step 2: Consider all $ n < 1000 $
-    
-    We are to count all $ n < 1000 $ such that $ f(n) = n $.
-    
-    Let’s consider all numbers $ n < 1000 $, i.e., all numbers from 1 to 999 (inclusive).
-    
-    We'll use a **brute-force approach** to compute $ f(n) $ for each $ n $, and count how many satisfy $ f(n) = n $.
-    
-    ---
-    
-    ### Step 3: Implement the function $ f(n) $ for $ n < 1000 $
-    
-    We can define a helper function:
-    
-    ```python
-    def f(n):
-        digits = list(map(int, str(n)))
-        max_digit = max(digits)
-        b = max_digit + 1
-        value = 0
-        for d in digits:
-            value = value * b + d
-        return value
-    ```
-    
-    Now, we'll iterate over all $ n < 1000 $ and check if $ f(n) = n $.
-    
-    ---
-    
-    ### Step 4: Count the valid numbers
-    
-    Let’s compute this:
-    
-    ```python
-    count = 0
-    for n in range(1, 1000):
-        if f(n) == n:
-            count += 1
-    print(count)
-    ```
-    
-    ---
-    
-    ### ✅ Final Answer:
-    
-    $$
-    \boxed{12}
-    $$
-    
-    There are **12** positive integers $ n < 1000 $ such that $ f(n) = n $.
+    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 \( b \in \mathbb{N} \).
+    
+    2. **Substitution and Simplification**:
+       Substitute \( x = \frac{a}{b} \) into the functional equation:
+       \[
+       h\left(\frac{a}{b
+    
+    ... [truncated 823 chars] ...
+    
+    kx + c \quad \text{for some } k \in \mathbb{Z}, \ c \in \mathbb{Z}
+       \]
+    
+    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:
+       \[
+       kx + c = k\left(\frac{kbx - ka + c + a}{b}\right) + c
+       \]
+       Simplify the right-hand side:
+       \[
+       kx + c = \frac{k^2bx - k^2a + kc + ka}{b} + c
+       \]
+       Multiply through by \( b \) to clear the fraction:
+       \[
+       b(kx + c) = k^2bx - k^2a + kc + ka + bc
+       \]
+       Simplify and equate coefficients:
+       \[
+       bkx + bc = k^2bx - k^2a + kc + ka + bc
+       \]
+       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:
+       \[
+       \boxed{h(x) = kx + c \quad \text{for some } k \in \mathbb{Z}, \ c \in \mathbb{Z}}
+       \]
     ```