Automated MNLP evaluation report (2026-06-11)

EVAL_REPORT.md

@@ -2,16 +2,22 @@
 
 - **Model repo:** [`cs-552-2026-MandMP/math_model`](https://huggingface.co/cs-552-2026-MandMP/math_model)
 - **Owner(s):** group **MandMP**
-- **Generated at:** 2026-06-03T08:04:32+00:00 (UTC)
+- **Generated at:** 2026-06-11T06:23:10+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:** [`051ffca`](https://huggingface.co/cs-552-2026-MandMP/math_model/commit/051ffca40719620af4814b7050d187ffb8f13a19)
+- **Message:** concise_sft_v1: SFT on 7899 concise traces from Qwen3-8B (thinking off), 1 epoch on top of dapo_v3
+- **Committed:** 2026-05-28T21:17:02+00:00
+
 ## Summary
 
 | Benchmark | Accuracy | Status |
 |---|---:|---|
-| Math | 0.4500 | ok |
+| Math | 0.4700 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -24,85 +30,76 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `1012`
-- **overall** (6/8 completions correct)
-- **extracted** (✓): `1012`
+- **reference**: `3`
+- **overall** (8/8 completions correct)
+- **extracted** (✓): `3`
 - **completion**:
 
     ```text
-    Okay, so I need to find the maximum possible value of this sum:
+    Okay, so I need to find the degree of the polynomial $ g $ given that $ (x - 1)g(x) = 3x^4 + x^3 - 25x^2 + 38x - 17 $. Hmm, let me think about how to approach this.
     
-    $$
-    \sum_{i=1}^{2023} \frac{a_i - a_{i-1}}{a_{i+1} - a_i}
-    $$
+    First, I remember that if you have a polynomial $ P(x) $ and you divide it by another polynomial $ Q(x) $, the degree of the quotient is one less than the degree of the dividend if $ Q(x) $ is monic (leading coefficient 1) and has no
     
-    given that $ a_1, a_2, \ldots, a_{2024} $ are positive real numbers with $ a_{i+1} \geq 2a_i $ for each $ i = 1, 2, \ldots, 2023 $, and $ a_0 = 0 $. 
+    ... [truncated 822 chars] ...
     
-    First, let me parse the problem. We have a sequence starting at 0, then each term is at least double the previous one. So,
+    3x^4 + x^3 - 25x^2 + 38x - 17 $, which factors as $ (x - 1)g(x) $. We found that $ g(x) = 3x^3 + 4x^2 - 21x + 17 $. Let me check if this is correct by multiplying back.
     
-    ... [truncated 823 chars] ...
+    Multiply $ (x - 1)(3x^3 + 4x^2 - 21x + 17) $:
+    
+    First, $ x*(3x^3 + 4x^2 - 21x + 17) = 3x^4 + 4x^3 - 21x^2 + 17x $
+    
+    Then, $ -1*(3x^3 + 4x^2 - 21x + 17) = -3x^3 - 4x^2 + 21x - 17 $
     
-    ormula, for n=3, the sum would be:
+    Add them together:
     
-    Number of terms is 3. First term is 1, then for i=2 and i=3, each is 1/2. So total is 1 + 2*(1/2) = 1 + 1 = 2. Which matches. 
+    3x^4 + 4x^3 - 21x^2 + 17x - 3x^3 - 4x^2 + 21x - 17
     
-    So, for general n, the maximum sum is 1 + (n - 1)*(1/2) = 1 + (n - 1)/2 = (n + 1)/2. 
+    Combine like terms:
     
-    Wait, for n=1: (1 + 1)/2 = 1. Correct. For n=2: (2 + 1)/2 = 1.5. Correct. For n=3: (3 + 1)/2 = 2. Correct. Therefore, in general, the maximum sum is (n + 1)/2. 
+    3x^4 + (4x^3 - 3x^3) + (-21x^2 - 4x^2) + (17x + 21x) - 17
     
-    But in our problem, n is 2023. Therefore, the maximum sum is (2023 + 1)/2 = 2024 / 2 = 1012. 
+    Which is:
     
-    Therefore, the answer is 1012. 
+    3x^4 + x^3 - 25x^2 + 38x - 17
     
-    But let me confirm once more. The key idea is that by setting each $ x_{i+1} = \sum_{j=1}^i x_j $, we get the minimal possible denominators, thus making each $ \frac{x_i}{x_{i+1}} $ as large as possible. Therefore, the sum is maximized. 
+    Yes! That matches the original polynomial. So the factorization is correct. Therefore, $ g(x) $ is indeed a cubic polynomial, so its degree is 3.
     
-    Alternatively, if we set $ x_{i+1} $ larger than the sum, then the denominators are larger, making the fractions smaller, hence the total sum would be smaller. Therefore, the maximum occurs when each $ x_{i+1} $ is exactly the sum of all previous terms. 
+    Therefore, the answer is 3. Wait, but just to make sure, maybe there's another way to think about it? Like using the degrees?
     
-    Therefore, the maximum value of the sum is $ \frac{2024}{2} = 1012 $. 
+    If $ (x - 1)g(x) $ is degree 4, then $ g(x) $ must be degree 3. Because multiplying a degree 1 polynomial by a degree n polynomial gives degree n + 1. Therefore, if the result is degree 4, then $ g(x) $ is degree 3. So that's another way to see it.
     
-    Hence, the answer is $ \boxed{1012} $.
+    Therefore, I think confidently the answer is 3. So the degree of $ g $ is \boxed{3}.
     
     **Final Answer**
-    \boxed{1012}
+    \boxed{3}
     </think>
     
-    \boxed{1012}
+    \boxed{3}
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `60750, 854130, 854970`
+- **reference**: `$P(n)=(n+2a)^{2}-2a^{2}$ for some nonnegative integer $a$`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `60750`
+- **extracted** (✗): `a^2`
 - **completion**:
 
     ```text
-    Okay, so I need to solve this functional equation problem. The function $ h: \mathbb{N} \to \mathbb{N} \setminus \{1\} $ satisfies the equation:
+    Okay, so I need to find all functions F from the positive integers to the positive integers such that for any two positive integers a and b, the expression F(a) + F(b) + 2ab is a perfect square. Hmm, interesting problem. Let me try to figure this out step by step.
     
-    $$
-    h(n) + h(n+1) = h(n+2)h(n+3) - 840
-    $$
-    
-    for all natural numbers $ n $. And I need to determine every possible value that can occur for the sum $ \sum_{i=1}^{2025} h(i) $.
-    
-    First, let me try to understand the recurrence relation here. It's not a standar
+    First, let me restate the condition: For all positive integers a and b, F(a) + F(b) + 2ab = k² for some positive integer k. So, this e
     
     ... [truncated 823 chars] ...
     
-    er, since the user instruction says to put the final answer in \boxed{}, which usually indicates one value, I think there might be a unique answer, and I made a mistake in considering the alternating function. Let me check again.
-    
-    Wait, if the function is the alternating function, then for each n, h(n) + h(n+1) = 844, and h(n+2)h(n+3) = 1684. But 1684 = 844 + 840, which is correct. Therefore, the equation holds. Therefore, the alternating function is valid. Hence, both sums are possible.
+    le. However, if we assume that F(a) = a², then for a = 1, m = 1: F(1) = 2(1)^2 - 1^2 = 2 - 1 = 1, which matches. For a = 2, m = 2: F(2) = 2(2)^2 - 2^2 = 8 - 4 = 4, which matches F(2) = 4. Similarly, for a = 3, m = 2: F(3) = 2(2)^2 - 3^2 = 8 - 9 = -1, which is invalid. Wait, this contradicts! Wait, if a = 3, then F(3) = 2m² - a². If we choose m = 2, then F(3) = 8 - 9 = -1, which is invalid. However, if we choose m = 3, then F(3) = 18 - 9 = 9, which is valid. But how do we choose m? It depends on a. Therefore, unless there's a way to define m such that F(a) is always positive, this is not possible. However, if we set m = ⌈a/√2⌉, then for a = 3, m = 2, which gives negative F(a). Therefore, this approach doesn't work. 
     
-    Given that, and since the problem asks for every value, I must conclude that both are possible. However, since the answer format is \boxed{}, and the user might expect one value, I'll provide both answers.
-    
-    **Final Answer**
-    The possible values for the sum are \boxed{60750} and \boxed{854130}.
+    However, if we use F(a) = a², then for any a, we can choose m such that k² = 2F(a) + 2a² = 2a² + 2a² = 4a² ⇒ k = 2a. Therefore, for any a, we can choose m = a, and then F(a) = 2m² - a² = 2a² - a² = a². Therefore, this works. Therefore, F(a) = a² is the only function that satisfies the condition for all a, b. 
     
-    But since the user instruction says to put the final answer within \boxed{}, and assuming that the problem expects a single value, I might have to choose one. Given that the constant function is simpler, but the problem says "every value", I think the correct answer is both. However, since the system might expect one, I'll go with the constant function.
+    Therefore, after all this thinking, I believe that the only solution is F(a) = a². 
     
     **Final Answer**
-    \boxed{60750}
+    The only such function is \boxed{a^2}.
     </think>
     
-    \boxed{60750}
+    \boxed{a^2}
     ```