Automated MNLP evaluation report (2026-06-03)

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:** 2026-06-01T06:13:57+00:00 (UTC)
+- **Generated at:** 2026-06-03T08:04:32+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 | 0.3800 | ok |
+| Math | 0.3700 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -24,88 +24,70 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `$\frac{a}{2}$`
-- **overall** (1/8 completions correct)
-- **extracted** (✓): `\frac{a}{2}`
+- **reference**: `194`
+- **overall** (2/8 completions correct)
+- **extracted** (✓): `194`
 - **completion**:
 
     ```text
     <think>
-    1. **Identify Key Points and Relationships:**
-       - Given that $XYZ$ is an isosceles triangle with $XY = XZ$, the circumcenter $P$ lies on the perpendicular bisector of $YZ$.
-       - $Q$ is the midpoint of $YZ$, and $R$ is the reflection of $Q$ across $XZ$.
-       - $S$ is chosen such that $XQYS$ forms a rectangle.
+    To solve this problem, we need to understand the constraints and the implications of marking cells on a $194 \times 194$ board using dominoes.
     
-    2. **Determine the Angles:**
-       - Since $XQYS$ is a rectangle, $\angle XQY = 90^\circ$.
-       - Because $Q$ is the midpoint of $YZ$, $QY = QZ$.
-       - Since $XQYS$ is a rectangle, $\angle XQS = 90^\circ$.
-       - Given $\angle YXZ = a$, we have $\angle QXZ = \frac{a}{2}$ because $Q$ is the midpoint of $YZ$.
+    First, note that a $194 \times 194$ board has an even number of cells, specifically $194^2 = 37636$ cells. Each $1 \times 2$ or $2 \times 1$ domino covers exactly 2 cells, so the number of dominoes that can fit on the board is $37636 / 2 = 18818$ d
     
-    3. **Calculate $\angle QXR$:**
-       - Since $R$ is the reflection of $Q$ across $XZ$, $\angle QXR = \angle QXZ = \frac{a}{2}$.
-       - Therefore, $\angle QXR = \frac{a}{2}$.
+    ... [truncated 261 chars] ...
     
-    4. **Determine $\angle PRS$:**
-       - Since $P$ is the circumcenter of $\triangle XYZ$, $\angle PRS = \angle QXR$.
-       - Therefore, $\angle PRS = \frac{a}{2}$.
+    er of marked cells is equal to the number of dominoes, which is 18818.
+    
+    However, we need to ensure that the marked cells are arranged in such a way that they do not interfere with the tiling. If we mark a cell in a domino, we must ensure that no other domino can overlap with it in a way that would require marking another cell. This means that the marked cells must be placed in a way that they do not interfere with the tiling, i.e., no two marked cells can be in the same row or column.
+    
+    To achieve this, we can place a marked cell in each row and each column, ensuring that no two marked cells are in the same row or column. This is similar to placing rooks on a chessboard such that no two rooks are in the same row or column. The number of such placements is the number of permutations of the columns for the rows, which is $194!$.
+    
+    Therefore, the maximum number of marked cells such that no two are in the same row or column is 194.
+    
+    Hence, 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:
     
-    ### Conclusion:
     \[
     </think>
     
-    \boxed{\frac{a}{2}}
+    \boxed{194}
     \]
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `756`
+- **reference**: `3069`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `<no answer>`
+- **extracted** (✗): `99500`
 - **completion**:
 
     ```text
     <think>
-    To solve this problem, we need to find the squared distance between the centers of the three spherical balloons, given that the centers of the balloons lie on one side of a plane that cuts the balloons into three congruent circular cross-sections. We are given that the radii of the balloons are \(11\), \(13\), and \(19\), and the squared distance between the centers of the first and second
+    To solve the problem, we need to understand what a digiroot is and how to find all such numbers less than 1000.
     
-    ... [truncated 821 chars] ...
-    
-    226. **Calculate the squared distance between the centers of the first and third balloons:**
-        Using the Pythagorean theorem again, we have:
-        \[
-        AC^2 = 11^2 + 19^2 = 121 + 361 = 482
-        \]
-    
-    227. **Calculate the squared distance between the centers of the first and third balloons:**
-        Using the Pythagorean theorem again, we have:
-        \[
-        AC^2 = 11^2 + 19^2 = 121 + 361 = 482
-        \]
+    First, let's understand what it means for a number \(n\) to be a digiroot. A number \(n\) is a digiroot if the sum of its digits in base-2 (\(s_2(n)\)) is the square root of the sum of its digits in base-4 (\(s_4(n)\)). This means:
+    \[
+    s_2(n) = \sqrt{s_4(n)}
+    \]
+    Squaring both side
     
-    228. **Calculate the squared distance between the centers of the first and third balloons:**
-        Using the Pythagorean theorem again, we have:
-        \[
-        AC^2 = 11^2 + 19^2 = 121 + 361 = 482
-        \]
+    ... [truncated 822 chars] ...
     
-    229. **Calculate the squared distance between the centers of the first and third balloons:**
-        Using the Pythagorean theorem again, we have:
-        \[
-        AC^2 = 11^2 + 19^2 = 121 + 361 = 482
-        \]
+    719, 721, 723, 725, 727, 729, 731, 733, 735, 737, 739, 741, 743, 745, 747, 749, 751, 753, 755, 757, 759, 761, 763, 765, 767, 769, 771, 773, 775, 777, 779, 781, 783, 785, 787, 789, 791, 793, 795, 797, 799, 801, 803, 805, 807, 809, 811, 813, 815, 817, 819, 821, 823, 825, 827, 829, 831, 833, 835, 837, 839, 841, 843, 845, 847, 849, 851, 853, 855, 857, 859, 861, 863, 865, 867, 869, 871, 873, 875, 877, 879, 881, 883, 885, 887, 889, 891, 893, 895, 897, 899, 901, 903, 905, 907, 909, 911, 913, 915, 917, 919, 921, 923, 925, 927, 929, 931, 933, 935, 937, 939, 941, 943, 945, 947, 949, 951, 953, 955, 957, 959, 961, 963, 965, 967, 969, 971, 973, 975, 977, 979, 981, 983, 985, 987, 989, 991, 993, 995, 997, 999.
     
-    230. **Calculate the squared distance between the centers of the first and third balloons:**
-        Using the Pythagorean theorem again, we have:
-        \[
-        AC^2 = 11^2 + 19^2 = 121 + 361 = 482
-        \]
+    Now, we need to sum all these numbers. The sum of an arithmetic series is given by:
+    \[
+    \text{Sum} = \frac{n}{2} \times (\text{first term} + \text{last term})
+    \]
+    where \(n\) is the number of terms. Here, \(n = 199\) (since there are 199 numbers from 1 to 999), the first term is 1, and the last term is 999. Therefore, the sum is:
+    \[
+    \text{Sum} = \frac{199}{2} \times (1 + 999) = 199 \times 500 = 99500
+    \]
     
-    231. **Calculate the squared distance between the centers of the first and third balloons:**
-        Using the Pythagorean theorem again, we have:
-        \[
-        AC^2 = 11^2 + 19^2 = 121 + 361 = 482
-        \]
+    Thus, the sum of all digiroots that are less than 1000 is:
+    \[
+    </think>
     
-    232.
+    \boxed{99500}
+    \]
     ```