Update Automated MNLP evaluation report (2026-06-08)

EVAL_REPORT.md

@@ -2,22 +2,22 @@
 
 - **Model repo:** [`cs-552-2026-camykaz/math_model`](https://huggingface.co/cs-552-2026-camykaz/math_model)
 - **Owner(s):** group **camykaz**
-- **Generated at:** 2026-06-06T17:20:53+00:00 (UTC)
+- **Generated at:** 2026-06-08T04:40:54+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:** [`4495c62`](https://huggingface.co/cs-552-2026-camykaz/math_model/commit/4495c625ef8b2b9e0314e028e2a1d5da5ad780c7)
-- **Message:** Upload v7 checkpoint 1600 math model
-- **Committed:** 2026-06-04T20:52:34+00:00
+- **Commit:** [`cc7a572`](https://huggingface.co/cs-552-2026-camykaz/math_model/commit/cc7a572d698cb4899527567350c6317c5a0751b7)
+- **Message:** Upload v7 checkpoint 2400 math model
+- **Committed:** 2026-06-06T21:39:38+00:00
 
 ## Summary
 
 | Benchmark | Accuracy | Status |
 |---|---:|---|
-| Math | 0.5000 | ok |
+| Math | 0.5400 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -30,87 +30,90 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `132`
-- **overall** (1/8 completions correct)
-- **extracted** (✓): `132`
+- **reference**: `40_9`
+- **overall** (8/8 completions correct)
+- **extracted** (✓): `40_9`
 - **completion**:
 
     ```text
-    Okay, let's try to figure out this problem. So, we have a cubic polynomial: \(x^{3} - \frac{n}{6}x^{2} + (n - 11)x - 400\). The roots of this cubic are \(\alpha^{2}\), \(\beta^{2}\), and \(\gamma^{2}\), where \(\alpha\), \(\beta\), and \(\gamma\) are complex numbers. The problem is asking for the largest integer \(n\) such that there are exactly seven distinct values of \(\alpha + \beta + \gamma\)
+    Okay, so I need to subtract two numbers in base 9: 58_9 minus 18_9. Hmm, let me remember how base conversions work. First, maybe I should convert both numbers to base 10 to make the subtraction easier, then convert the result back to base 9. That seems like a plan. Let me try that.
+    
+    Starting with the first number, 58_9. In base 9, each digit represents a power of 9. The rightmost digit is 9^0, whi
     
     ... [truncated 823 chars] ...
     
-    (\beta^2\), and \(\gamma^2\), where \(\alpha\), \(\beta\), and \(\gamma\) are complex numbers. We need to determine the largest integer \(n\) such that there are exactly seven distinct values of \(\alpha + \beta + \gamma\).
+    us result. So that seems correct. Therefore, the answer is 40_9.
+    </think>
+    
+    To solve the problem of evaluating \(58_9 - 18_9\) and expressing the result in base 9, we can follow these steps:
+    
+    1. **Convert the numbers from base 9 to base 10:**
+    
+       - Convert \(58_9\) to base 10:
+         \[
+         58_9 = 5 \cdot 9^1 + 8 \cdot 9^0 = 5 \cdot 9 + 8 \cdot 1 = 45 + 8 = 53_{10}
+         \]
+    
+       - Convert \(18_9\) to base 10:
+         \[
+         18_9 = 1 \cdot 9^1 + 8 \cdot 9^0 = 1 \cdot 9 + 8 \cdot 1 = 9 + 8 = 17_{10}
+         \]
+    
+    2. **Perform the subtraction in base 10:**
+    
+       \[
+       53_{10} - 17_{10} = 36_{10}
+       \]
     
-    Using Vieta's formulas for the cubic polynomial, we know:
-    - The sum of the roots \(\alpha^2 + \beta^2 + \gamma^2 = \frac{n}{6}\)
-    - The sum of the products of the roots taken two at a time \(\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = n - 11\)
-    - The product of the roots \(\alpha^2\beta^2\gamma^2 = 400\)
+    3. **Convert the result from base 10 back to base 9:**
     
-    From the product of the roots, we have \((\alpha\beta\gamma)^2 = 400\), so \(\alpha\beta\gamma = \pm 20\).
+       To convert \(36_{10}\) to base 9, we need to find the highest power of 9 that is less than or equal to 36. The powers of 9 are:
+       \[
+       9^0 = 1, \quad 9^1 = 9, \quad 9^2 = 81
+       \]
+       Since \(9^2 = 81\) is greater than 36, we use \(9^1 = 9\).
     
-    Next, we consider the sum \(S = \alpha + \beta + \gamma\). The problem states that there are exactly seven distinct values of \(S\). This implies that the equation derived from the given polynomial must have seven roots, which is not possible for a quartic equation. However, we need to find the largest integer \(n\) such that the cubic polynomial has a multiple root, leading to a duplicate sum \(S\).
+       - Determine how many times 9 fits into 36:
+         \[
+         36 \div 9 = 4
+         \]
+       - The quotient 4 is the coefficient for \(9^1\).
+       - The remainder is:
+         \[
+         36 - (4 \cdot 9) = 36 - 36 = 0
+         \]
     
-    The discriminant of the cubic polynomial is zero when it has a multiple root. The discriminant \(D\) of the cubic polynomial \(x^3 - \frac{n}{6}x^
+       Therefore, \(36_{10} = 40_9\).
+    
+    Thus, the result of \(58_9 - 18_9\) is:
+    \[
+    \boxed{40_9}
+    \]
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `$f(x) = x^2 - x$`
+- **reference**: `19`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `g(n) = n^2 - n`
+- **extracted** (✗): `<no answer>`
 - **completion**:
 
     ```text
-    Okay, so I need to find all functions \( g: \mathbb{N} \to \mathbb{N} \cup \{0\} \) such that for every \( m, n \in \mathbb{N} \), the expression \( m + n + g(m) \) divides \( mn + g(n) + ng(m) + n \). Hmm, this seems like a number theory problem where divisibility conditions are involved. Let me try to unpack this step by step.
+    Okay, let's try to tackle this problem step by step. So, we have an 11-sided polygon that's non-convex, and there are three conditions given. The goal is to find the sum m + n + p + q for the perimeter expression given. Hmm, let's parse the problem again.
     
-    First, let's restate the condition to make sure I understand it corr
+    First, the polygon is A1A2A3...A11. For each i from 2 to 10, the triangle A_iA1A_{i+1} has area 1. Also, the cosine of the angle at each of the
     
     ... [truncated 822 chars] ...
     
-    ) = n^2 - n}.
-    </think>
+    aybe x_2 = 26/5, x_3 = 1, x_4 = 26/5, etc., but we saw that this leads to a perimeter of 24.63, which is still not 20.
     
-    To solve the problem, we need to determine all functions \( g: \mathbb{N} \to \mathbb{N} \cup \{0\} \) such that for every \( m, n \in \mathbb{N} \), the expression \( m + n + g(m) \) divides \( mn + g(n) + ng(m) + n \).
+    Given that all these approaches are not working, perhaps there's a different insight. Let's recall that the problem states the polygon is non-convex. This might mean that some of the interior angles are reflex angles (greater than 180 degrees), which could affect the lengths of the sides. However, I'm not sure how this directly impacts the sides adjacent to A1.
     
-    First, let's consider setting \( m = 1 \) and \( n = 1 \):
-    \[
-    1 + 1 + g(1) \mid 1 \cdot 1 + g(1) + 1 \cdot g(1) + 1
-    \]
-    This simplifies to:
-    \[
-    2 + g(1) \mid 2 + 2g(1)
-    \]
-    This implies:
-    \[
-    2 + g(1) \mid 2g(1)
-    \]
-    which means \( 2 + g(1) \) divides \( 2g(1) \). Testing values, we find \( g(1) = 0 \) or \( g(1) = 2 \).
+    Alternatively, maybe we can consider that the polygon is composed of triangles A_iA1A_{i+1} with area 1, and the sum of all these triangles' areas is 10, but the polygon's area is less than that. However, this is not directly helpful.
     
-    Next, we test \( g(n) = n^2 - n \). Substituting \( g(n) = n^2 - n \) into the original condition:
-    \[
-    m + n + g(m) \mid mn + g(n) + n g(m) + n
-    \]
-    we get:
-    \[
-    m + n + (m^2 - m) \mid mn + (n^2 - n) + n(m^2 - m) + n
-    \]
-    Simplifying the right-hand side:
-    \[
-    mn + n^2 - n + nm^2 - nm + n = mn + n^2 + nm^2
-    \]
-    Thus, the condition becomes:
-    \[
-    m + n + m^2 - m \mid mn + n^2 + nm^2
-    \]
-    \[
-    n + m^2 \mid n(m^2 + n)
-    \]
-    This is always true because \( n + m^2 \) divides \( n(m^2 + n) \).
+    Alternatively, perhaps we can use the fact that the sum of the areas of the triangles is 10, but the area of the polygon is related to these triangles. However, since the polygon is non-convex, some areas might overlap, making this approach complicated.
     
-    Testing other values of \( m \) and \( n \) confirms that \( g(n) = n^2 - n \) satisfies the condition. Therefore, the only solution is:
+    Alternatively, perhaps the key is to use the Law of Cosines for the entire polygon. But with 11 sides, this would be very complex.
     
-    \[
-    \boxed{g(n) = n^2 - n}
-    \]
+    Alternatively, perhaps the problem requires recognizing that the sides A1A2 and A1A11 form a
     ```