Automated MNLP evaluation report (2026-05-29)

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:** 2026-05-27T13:02:22+00:00 (UTC)
+- **Generated at:** 2026-05-29T06:54:15+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 | 0.3600 | ok |
+| Math | 0.3800 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -24,84 +24,95 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `194`
-- **overall** (4/8 completions correct)
-- **extracted** (✓): `194`
+- **reference**: `\frac{639}{40}`
+- **overall** (5/8 completions correct)
+- **extracted** (✓): `\frac{639}{40}`
 - **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 the marking of cells affects this tiling.
+    To solve this problem, we need to find the distance between two buildings on the map given the actual distance and the scale of the map.
     
-    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 481 chars] ...
-    
-    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.
-    
-    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.
+    First, let's understand the scale of the map. According to the problem, a length of 3 inches on the map represents 10 miles in the real world. This means that for every 3 inches on the map, the actual distance is 10 miles.
     
-    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.
+    We can set up a proportion to find the distance on the map that corresponds to 53.25 miles in the real world. Let's denote the distance on the map as \(x\) inches.
     
-    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.
-    
-    After analyzing the problem, we find that the least positive integer \(k\) for which this is possible is:
+    The proportion can be written as:
     \[
-    \boxed{194}
+    \frac{3 \text{ inches}}{10 \text{ miles}} = \frac{x \text{ inches}}{53.25 \text{ miles}}
     \]
-    ```
-
-**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) = 0 \text{ for all } x \in \mathbb{Q}`
-- **completion**:
-
-    ```text
-    <think>
-    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 \(b \in \mathbb{N}\):
     
+    To solve for \(x\), we can cross-multiply:
     \[
-    h(x) = h\left(\frac{h(bx - a) + a}{b}\right).
+    3 \times 53.25 = 10 \times x
     \]
     
-    First, let's consider the case when \(a = 0\). The equation becomes:
-    
+    Calculating the left side:
     \[
-    h(x) = h\left(\frac{h(bx)}{b}\right) = h\left(\frac{h(x)}{b}\ri
-    
-    ... [truncated 823 chars] ...
-    
-    efore, the solution is:
+    3 \times 53.25 = 159.75
+    \]
     
+    So, we have:
     \[
-    h(x) = c \text{ for all } x \in \mathbb{Q},
+    159.75 = 10x
     \]
     
-    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:
-    
+    To find \(x\), we divide both sides by 10:
     \[
-    \boxed{h(x) = c \text{ for all } x \in \mathbb{Q}},
+    x = \frac{159.75}{10} = 15.975
     \]
     
-    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:
-    
+    Now, we need to express 15.975 as a fraction. We can write 15.975 as:
     \[
-    \boxed{h(x) = c \text{ for all } x \in \mathbb{Q}},
+    15.975 = \frac{15975}{1000}
     \]
     
-    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:
+    We can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 25:
+    \[
+    \frac{15975 \div 25}{1000 \div 25} = \frac{639}{40}
+    \]
     
+    Therefore, the distance between the two buildings on the map, written as a fraction, is:
     \[
-    \boxed{h(x) = 0 \text{ for all } x \in \mathbb{Q}}.
+    \boxed{\frac{639}{40}}
     \]
     ```
+
+**Incorrect** (1 shown)
+
+- **reference**: `756`
+- **overall** (0/8 completions correct)
+- **extracted** (✗): `<no answer>`
+- **completion**:
+
+    ```text
+    <think>
+    To solve this problem, we need to use the properties of spheres and the distances between their centers. Let's denote the centers of the three balloons as points $X$, $Y$, and $Z$, and their radii as $r_1 = 11$, $r_2 = 13$, and $r_3 = 19$, respectively.
+    
+    Since the balloons are externally tangent to each other, the distance between any two centers is the sum of their radii. Therefore, we ha
+    
+    ... [truncated 822 chars] ...
+    
+    XZ^2 = 1600 - 2 \cdot 24 \cdot \frac{10912}{480}. \]
+    \[ XZ^2 = 1600 - 2 \cdot 24 \cdot \frac{10912}{480}. \]
+    \[ XZ^2 = 1600 - 2 \cdot 24 \cdot \frac{10912}{480}. \]
+    \[ XZ^2 = 1600 - 2 \cdot 24 \cdot \frac{10912}{480}. \]
+    \[ XZ^2 = 1600 - 2 \cdot 24 \cdot \frac{10912}{480}. \]
+    \[ XZ^2 = 1600 - 2 \cdot 24 \cdot \frac{10912}{480}. \]
+    \[ XZ^2 = 1600 - 2 \cdot 24 \cdot \frac{10912}{480}. \]
+    \[ XZ^2 = 1600 - 2 \cdot 24 \cdot \frac{10912}{480}. \]
+    \[ XZ^2 = 1600 - 2 \cdot 24 \cdot \frac{10912}{480}. \]
+    \[ XZ^2 = 1600 - 2 \cdot 24 \cdot \frac{10912}{480}. \]
+    \[ XZ^2 = 1600 - 2 \cdot 24 \cdot \frac{10912}{480}. \]
+    \[ XZ^2 = 1600 - 2 \cdot 24 \cdot \frac{10912}{480}. \]
+    \[ XZ^2 = 1600 - 2 \cdot 24 \cdot \frac{10912}{480}. \]
+    \[ XZ^2 = 1600 - 2 \cdot 24 \cdot \frac{10912}{480}. \]
+    \[ XZ^2 = 1600 - 2 \cdot 24 \cdot \frac{10912}{480}. \]
+    \[ XZ^2 = 1600 - 2 \cdot 24 \cdot \frac{10912}{480}. \]
+    \[ XZ^2 = 1600 - 2 \cdot 24 \cdot \frac{10912}{480}. \]
+    \[ XZ^2 = 1600 - 2 \cdot 24 \cdot \frac{10912}{480}. \]
+    \[ XZ^2 = 1600 - 2 \cdot 24 \cdot \frac{10912}{480}. \]
+    \[ XZ^2 = 1600 - 2 \cdot 24 \cdot \frac{10912}{480}. \]
+    \[ XZ^2 = 1600 - 2 \cdot 24 \cdot \frac{10912}{480}. \]
+    \[ XZ^2 = 1600 - 2 \cdot 2
+    ```