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

EVAL_REPORT.md

@@ -2,16 +2,22 @@
 
 - **Model repo:** [`cs-552-2026-bilko/math_model`](https://huggingface.co/cs-552-2026-bilko/math_model)
 - **Owner(s):** group **bilko**
-- **Generated at:** 2026-06-05T03:01:58+00:00 (UTC)
+- **Generated at:** 2026-06-06T17:20:53+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:** [`b63a9ed`](https://huggingface.co/cs-552-2026-bilko/math_model/commit/b63a9edf27f2abbede2deed75d3ebd6ae87b17b4)
+- **Message:** GRPO ontop of SFT model name: grpo_openr1__qwen3_maxtokens_lr2e-6_checkpoint-450_merged
+- **Committed:** 2026-06-04T14:39:53+00:00
+
 ## Summary
 
 | Benchmark | Accuracy | Status |
 |---|---:|---|
-| Math | 0.4000 | ok |
+| Math | 0.4400 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -24,58 +30,69 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `194`
+- **reference**: `305`
 - **overall** (2/8 completions correct)
-- **extracted** (✓): `194`
+- **extracted** (✓): `305`
 - **completion**:
 
     ```text
-    Okay, so I need to figure out the smallest number k of cells to mark on a 194x194 board such that there's exactly one way to tile the board with dominoes, and in that tiling, every domino has at most one marked cell. Hmm, domino tiling problems often involve pairing adjacent cells, so maybe this has something to do with pairing or matching?
+    Okay, so I have this problem here where I need to find the set S of positive integers n between 99 and 206, inclusive, such that the numbers 1, 2, ..., 2n can be partitioned into n pairs where the differences within each pair are exactly 1, 2, ..., n. Then, I need to find the smallest and largest elements in S and add them together. Hmm, let me try to wrap my head around this.
     
-    First, let me recall that a domino tiling covers the ent
+    First, let's parse
     
-    ... [truncated 823 chars] ...
+    ... [truncated 822 chars] ...
     
-    , meaning they cannot be adjacent in the tiling.
+    that the only possible n are 99 and 206, and their sum is 305. This might be a trick question where the answer is simply 99 + 206 = 305. But why?
     
-    3. **Graph Theory Insight**: The problem can be modeled using graph theory, where each cell is a vertex and edges represent adjacency. A domino tiling corresponds to a perfect matching in this graph. The marked cells must form an independent set in this graph.
+    Wait, perhaps the problem is related to the fact that 2n -1 must be divisible by n +1. For n=99, 2*99 -1=197, and 99 +1=100. 197 divided by 100 is 1.97, not an integer. For n=206, 2*206 -1=411, and 206 +1=207. 411 divided by 207 is 2. So, 207*2=414, which is more than 411. So, 411 is not divisible by 207. Hence, no.
     
-    4. **Key Insight**: To ensure a unique tiling, the marked cells must be structured in a way that they are isolated in the tiling. This can be achieved by marking cells in a diagonal pattern, ensuring that each marked cell is isolated from others in the tiling.
+    Alternatively, perhaps the problem is related to the fact that 2n must be one more than a multiple of n +1. For n=99, 2*99=198, and 198 divided by 99 +1=100 is 1.98. Not an integer. For n=206, 412 divided by 207≈1.98. Not an integer. So, no.
     
-    5. **Pattern Recognition**: For smaller boards, such as \( 2 \times 2 \), marking two diagonal cells forces a unique tiling. Extending this to a \( 194 \times 194 \) board, marking one cell in each row and column (diagonally) ensures that the remaining cells can be tiled with dominoes, and the marked cells are isolated in the tiling.
+    Given that I've spent a significant amount of time on this and haven't found a mathematical condition that fits the given range, I think the answer must be 99 + 206 = 305. Perhaps the problem is designed such that only n=99 and n=206 satisfy the condition, and their sum is 305. Therefore, the final answer is 305.
     
-    6. **Conclusion**: By marking one cell in each row and column (diagonally), we ensure that the remaining cells form a unique tiling. This approach minimizes the number of marked cells while satisfying the problem's constraints.
+    **Final Answer**
+    \boxed{305}
+    </think>
     
-    Thus, the least positive integer \( k \) is \(\boxed{194}\).
+    To solve the problem, we need to determine the set \(S\) of positive integers \(n\) such that \(99 \leq n \leq 206\) and the numbers \(
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `$g(x)=c, g(x)=\lceil x \rceil, g(x)=\lfloor x \rfloor$`
+- **reference**: `60750, 854130, 854970`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `h(x) = x`
+- **extracted** (✗): `60750`
 - **completion**:
 
     ```text
-    Okay, so I have this problem here where I need to find all functions \( h: \mathbb{Q} \to \mathbb{Z} \) that satisfy a certain condition. The condition is that for every rational number \( x \), every integer \( a \), and every positive integer \( b \), the function \( h \) evaluated at \( x \) must equal \( h \) of \( \frac{h(bx - a) + a}{b} \). Hmm, that seems a bit abstract, but maybe if I brea
-    
-    ... [truncated 823 chars] ...
+    Okay, so I need to solve this problem where there's a function h from the natural numbers to the natural numbers excluding 1. The condition given is that for every n in natural numbers, h(n) + h(n+1) = h(n+2)h(n+3) - 840. And I have to find all possible values that the sum from i=1 to 2025 of h(i) can take. Hmm, interesting. Let me try to break this down step by step.
     
-    Assume \( h(q) = kq + c \). Substituting this into the equation leads to the conclusion that \( k = 1 \) and \( c = 0 \), resulting in \( h(q) = q \). This function satisfies the condition.
+    First, let me understand the
     
-    2. **Testing Constant Functions**:
-       - Consider \( h(q) = c \). For this to satisfy the condition, \( \frac{c + a}{b} \) must be an integer for all \( a \in \mathbb{Z} \) and \( b \in \mathbb{N} \). This is only possible if \( c = 0 \), leading to the constant function \( h(q) = 0 \).
+    ... [truncated 822 chars] ...
     
-    3. **Verification**:
-       - For \( h(q) = q \), substituting into the condition confirms the equality holds for all \( x \), \( a \), and \( b \).
-       - For \( h(q) = 0 \), substituting into the condition confirms the equality holds for all \( x \), \( a \), and \( b \).
+    that can occur for the sum \( \sum_{i=1}^{2025} h(i) \).
     
-    4. **Uniqueness**:
-       - If \( h(q) \) were not constant or identity, contradictions arise when considering specific values of \( x \), \( a \), and \( b \). For example, fixing \( x = 0 \) and \( a = 1 \) leads to the conclusion that \( h(0) = 0 \), which must hold for all integers \( a \).
+    First, we test if the function \( h(n) \) is constant. Suppose \( h(n) = c \) for all \( n \). Substituting into the given equation:
+    \[
+    c + c = c \cdot c - 840 \implies 2c = c^2 - 840 \implies c^2 - 2c - 840 = 0.
+    \]
+    Solving this quadratic equation, we find:
+    \[
+    c = \frac{2 \pm \sqrt{4 + 3360}}{2} = \frac{2 \pm 58}{2}.
+    \]
+    The positive solution is \( c = 30 \). Checking \( h(n) = 30 \), we verify:
+    \[
+    30 + 30 = 30 \cdot 30 - 840 \implies 60 = 900 - 840 \implies 60 = 60.
+    \]
+    Thus, \( h(n) = 30 \) is a valid constant solution.
     
-    Thus, the only solutions are the zero function and the identity function. The final answer is:
+    Next, we check if there could be non-constant solutions. Assume \( h(n) \) is not constant. Using the functional equation and induction, we find that any deviation from 30 leads to inconsistencies. For example, if \( h(1) \neq 30 \), the subsequent terms would not satisfy the equation unless they are also 30. This leads to a contradiction unless all terms are 30.
     
+    Thus, the only solution is the constant function \( h(n) = 30 \). The sum from \( i = 1 \) to \( 2025 \) is:
     \[
-    \boxed{h(x) = 0} \quad \text{and} \quad \boxed{h(x) = x}
+    \sum_{i=1}^{2025} 30 = 2025 \times 30 = 60750.
     \]
+    
+    The only value that can occur for \( \sum_{i=1}^{2025} h(i) \) is \(\boxed{60750}\).
     ```