Update Automated MNLP evaluation report (2026-06-09)

EVAL_REPORT.md

@@ -2,22 +2,22 @@
 
 - **Model repo:** [`cs-552-2026-flab/math_model`](https://huggingface.co/cs-552-2026-flab/math_model)
 - **Owner(s):** group **flab**
-- **Generated at:** 2026-06-08T04:40:54+00:00 (UTC)
+- **Generated at:** 2026-06-09T08:56:40+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._
 
 ## Evaluated checkpoint
 
-- **Commit:** [`e38dc63`](https://huggingface.co/cs-552-2026-flab/math_model/commit/e38dc632b04087730c509fa3a27efe6e93f3daa5)
-- **Message:** grpo-dsr-sft-8g-ckpt50-budgetRL
-- **Committed:** 2026-06-07T18:47:01+00:00
+- **Commit:** [`d509c36`](https://huggingface.co/cs-552-2026-flab/math_model/commit/d509c3676174e188181cd577372c8125d27020ab)
+- **Message:** g2rpo-ds4-ckpt120-tuned-sampling-T0.6-topp1.0-topk-1
+- **Committed:** 2026-06-08T15:31:44+00:00
 
 ## Summary
 
 | Benchmark | Accuracy | Status |
 |---|---:|---|
-| Math | 0.3800 | ok |
+| Math | 0.4000 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -30,68 +30,59 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `279`
-- **overall** (2/8 completions correct)
-- **extracted** (✓): `279`
+- **reference**: `194`
+- **overall** (4/8 completions correct)
+- **extracted** (✓): `194`
 - **completion**:
 
     ```text
     <think>
-    Okay, so I need to figure out how many positive integers n less than 1000 satisfy f(n) = n. The function f(n) is defined by taking the usual base-10 digits of n, finding the smallest integer b that's strictly larger than every digit in n, then interpreting the same digits as a base-b number. For example, if n is 72, the digits are 7 and 2, so the smallest b larger than both is 8. Then 72 i
+    Okay, so I need to figure out the least positive integer k for a 194x194 board where if I mark k cells, then there's exactly one tiling of the board with dominoes (1x2 and 2x1) such that every domino in this tiling has at most one marked cell. Hmm, this seems related to tiling problems and maybe some linear algebra concepts. Let me start by understanding the problem better.
     
-    ... [truncated 822 chars] ...
+    First, a domin
     
-    single-digit numbers satisfy \( f(n) = n \).
-       - There are 9 such numbers.
+    ... [truncated 822 chars] ...
     
-    2. **Two-digit numbers (10-99):**
-       - For a two-digit number \( n = 10a + b \) where \( a \) and \( b \) are digits, the maximum digit is \( m \). Then \( b = m + 1 \).
-       - For \( f(n) = n \), we need \( 10a + b = a(m + 1) + b \). This simplifies to \( 10a = a(m + 1) \) which implies \( m = 9 \).
-       - Two-digit numbers with maximum digit 9 are \( 19, 29, \ldots, 99 \), totaling 18 numbers.
+    e tiling to be unique, the marked cells must be arranged such that they are covered by dominoes in the unique tiling, and any other tiling must cover at least two marked cells.
     
-    3. **Three-digit numbers (100-999):**
-       - For a three-digit number \( n = 100a + 10b + c \) where \( a, b, c \) are digits, the maximum digit is \( m \). Then \( b = m + 1 \).
-       - For \( f(n) = n \), we need \( 100a + 10b + c = a(m + 1)^2 + b(m + 1) + c \). This simplifies to \( 100a + 10b = a(m + 1)^2 + b(m + 1) \).
-       - For \( m = 9 \), \( b = 10 \), and the equation holds for all three-digit numbers with at least one digit 9. The number of such three-digit numbers is 252.
+    3. **Pattern Recognition**: Observing smaller grids (e.g., 2x2, 4x4) shows that marking cells in a diagonal (or corner) pattern forces a unique tiling. For a 2x2 grid, marking two cells (the corners) forces a unique tiling. For a 4x4 grid, marking four cells (the corners) forces a unique tiling. This pattern suggests that marking \( n \) cells (where \( n \) is the side length of the grid) forces a unique tiling.
     
-    ### Total Count:
-    - Single-digit numbers: 9
-    - Two-digit numbers: 18
-    - Three-digit numbers: 252
+    4. **Generalizing to \(194 \times 194\)**: Extending this pattern, marking 194 cells (the corners) in a \(194 \times 194\) grid forces a unique tiling. This is because the remaining cells form a grid where the dominoes must cover the marked cells in a unique way, and any other tiling would cover at least two marked cells.
     
-    Summing these, we get \( 9 + 18 + 252 = 279 \).
+    5. **Verification**: Marking fewer than 194 cells results in an odd number of remaining cells, which is impossible to tile with dominoes. Therefore, 194 is the minimal number of cells needed to ensure a unique tiling.
     
-    Thus, the number of positive integers \( n < 1000 \) that satisfy \( f(n) = n \) is \(\boxed{279}\).
+    Thus, the least positive integer \( k \) is \(\boxed{194}\).
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `75`
+- **reference**: `$g(x)=c, g(x)=\lceil x \rceil, g(x)=\lfloor x \rfloor$`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `<no answer>`
+- **extracted** (✗): `h(x) = x`
 - **completion**:
 
     ```text
     <think>
-    Okay, so I need to solve this geometry problem about a parallelogram. Let me start by understanding the given information and what's being asked.
+    Okay, let's try to tackle this problem. So, we need to find all functions \( h: \mathbb{Q} \to \mathbb{Z} \) such that for every \( x \in \mathbb{Q} \), every integer \( a \), and every positive integer \( b \), the equation \( h(x) = h\left( \frac{h(bx - a) + a}{b} \right) \) holds. Hmm, that's a mouthful. Let me break it down.
     
-    First, the problem states that in parallelogram PQRS, angle S is 60 degrees. PS is 2 units long, and PQ is sqrt(3) + 1 units. N is the midpoint of PS, so that means PN = NS = 1. Then, there's a point E on segment NQ such that RE bisects angle R.
+    First, let's parse the equation. For any rational number \(
     
     ... [truncated 823 chars] ...
     
-    t the problem involves sqrt(3) and 60 degrees, the answer might be 30 degrees, 45 degrees, or 60 degrees. However, our calculation shows approximately 51 degrees, which is close to 45 degrees but not exact.
-    
-    Alternatively, maybe there's a geometric insight I'm missing. Let me think again.
+    y \( x \in \mathbb{Q} \), every \( a \in \mathbb{Z} \), and every positive integer \( b \):
     
-    Given that angle R is 120 degrees, and RE bisects it into two 60-degree angles. If we can show that triangle REQ is a 30-60-90 triangle, then angle REQ would be 30 degrees. Alternatively, if triangle EQN is equilateral, but that seems unlikely.
+    \[
+    h(x) = h\left( \frac{h(bx - a) + a}{b} \right),
+    \]
     
-    Alternatively, since N is the midpoint of PS, and PS is 2, so PN = NS =1. Maybe there's a symmetry or using properties of midpoints and angle bisectors.
+    we start by testing linear functions. We hypothesize that \( h(x) = kx + c \) might be a solution. For \( k = 0 \), we get \( h(x) = 0 \), which satisfies the equation. For \( k = 1 \), we get \( h(x) = x \), which also satisfies the equation. 
     
-    Alternatively, consider triangle PNS. Since N is the midpoint of PS, and PN =1, PS=2. If we can relate this to other parts of the parallelogram.
+    Next, we consider other possibilities. We test the zero function \( h(x) = 0 \) and find it satisfies the equation. For the identity function \( h(x) = x \), we verify it works by substituting back into the equation, showing both sides equal \( x \).
     
-    Alternatively, since PQ = sqrt(3)+1, and PS=2, maybe using the Law of Cosines in triangle PQS or something.
+    We also explore the possibility of non-linear functions, but through detailed analysis and testing various functions, we find that any deviation from the identity function leads to contradictions. For example, assuming \( h(q) \neq q \) for some \( q \) results in inconsistencies when substituting specific values of \( x \), \( a \), and \( b \).
     
-    Alternatively, since the problem involves angle bisectors and midpoints, maybe using coordinate geometry to find the required angle.
+    Thus, after thorough testing and reasoning, we conclude that the only solutions are the zero function and the identity function. 
     
-    But given the time I've spent and the calculations leading to approximately 51 degrees, which is close to 45 degrees but
+    ### Final Answer
+    The solutions are \boxed{h(x) = 0} and \boxed{h(x) = x}.
     ```