Update Automated MNLP evaluation report (2026-06-11)

EVAL_REPORT.md

@@ -2,7 +2,7 @@
 
 - **Model repo:** [`cs-552-2026-camykaz/math_model`](https://huggingface.co/cs-552-2026-camykaz/math_model)
 - **Owner(s):** group **camykaz**
-- **Generated at:** 2026-06-10T21:20:55+00:00 (UTC)
+- **Generated at:** 2026-06-11T04:19:46+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._
@@ -17,7 +17,7 @@ _This PR is opened automatically by the course CI. It is **non-blocking** — yo
 
 | Benchmark | Accuracy | Status |
 |---|---:|---|
-| Math | 0.4700 | ok |
+| Math | 0.5400 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -30,85 +30,90 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `\frac{14}{3}`
+- **reference**: `40_9`
 - **overall** (8/8 completions correct)
-- **extracted** (✓): `\frac{14}{3}`
+- **extracted** (✓): `40_9`
 - **completion**:
 
     ```text
-    Okay, let's see. I need to compute f(-2) + f(-1) + f(0) where the function f is given by (3x - 2)/(x - 2). Hmm, right. So, first, I should probably figure out what each of these individual values are by plugging in -2, -1, and 0 into the function. Then add them up. Let me take it step by step.
+    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 f(-2). Let's substitute x with -2. So, f(-2) would be (3*(-2) - 2)/( (-2) - 2 ). Let me ca
+    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] ...
     
-    14/3. Wait, no, that's not correct. Wait, 2 is 6/3, 5/3 is 5/3, 1 is 3/3. So, 6/3 + 5/3 + 3/3 = (6 + 5 + 3)/3 = 14/3. Wait, but 2 + 5/3 + 1 is the same as (2 + 1) + 5/3 = 3 + 5/3 = 14/3. Yes, that's correct. So, 3 is 9/3, 9/3 + 5/3 = 14/3. Yep, that's right. So, 14/3 is the answer. I think that's correct. No mistakes here. So, the final answer is 14/3.
+    us result. So that seems correct. Therefore, the answer is 40_9.
     </think>
     
-    To solve the problem, we need to compute \( f(-2) + f(-1) + f(0) \) for the function \( f(x) = \frac{3x - 2}{x - 2} \).
+    To solve the problem of evaluating \(58_9 - 18_9\) and expressing the result in base 9, we can follow these steps:
     
-    First, we calculate \( f(-2) \):
-    \[
-    f(-2) = \frac{3(-2) - 2}{-2 - 2} = \frac{-6 - 2}{-4} = \frac{-8}{-4} = 2
-    \]
+    1. **Convert the numbers from base 9 to base 10:**
     
-    Next, we calculate \( f(-1) \):
-    \[
-    f(-1) = \frac{3(-1) - 2}{-1 - 2} = \frac{-3 - 2}{-3} = \frac{-5}{-3} = \frac{5}{3}
-    \]
+       - Convert \(58_9\) to base 10:
+         \[
+         58_9 = 5 \cdot 9^1 + 8 \cdot 9^0 = 5 \cdot 9 + 8 \cdot 1 = 45 + 8 = 53_{10}
+         \]
     
-    Finally, we calculate \( f(0) \):
-    \[
-    f(0) = \frac{3(0) - 2}{0 - 2} = \frac{0 - 2}{-2} = \frac{-2}{-2} = 1
-    \]
+       - Convert \(18_9\) to base 10:
+         \[
+         18_9 = 1 \cdot 9^1 + 8 \cdot 9^0 = 1 \cdot 9 + 8 \cdot 1 = 9 + 8 = 17_{10}
+         \]
     
-    Now, we sum these results:
-    \[
-    f(-2) + f(-1) + f(0) = 2 + \frac{5}{3} + 1
-    \]
+    2. **Perform the subtraction in base 10:**
     
-    To add these fractions, we convert 2 and 1 to fractions with a common denominator:
-    \[
-    2 = \frac{6}{3}, \quad 1 = \frac{3}{3}
-    \]
+       \[
+       53_{10} - 17_{10} = 36_{10}
+       \]
     
-    So,
-    \[
-    2 + \frac{5}{3} + 1 = \frac{6}{3} + \frac{5}{3} + \frac{3}{3} = \frac{6 + 5 + 3}{3} = \frac{14}{3}
-    \]
+    3. **Convert the result from base 10 back to base 9:**
+    
+       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 \(9^1 = 9\).
+    
+       - Determine how many times 9 fits into 36:
+         \[
+         36 \div 9 = 4
+         \]
+       - The quotient 4 is the coefficient for \(9^1\).
+       - The remainder is:
+         \[
+         36 - (4 \cdot 9) = 36 - 36 = 0
+         \]
+    
+       Therefore, \(36_{10} = 40_9\).
     
-    Thus, the final answer is:
+    Thus, the result of \(58_9 - 18_9\) is:
     \[
-    \boxed{\frac{14}{3}}
+    \boxed{40_9}
     \]
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `125`
+- **reference**: `19`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `30 + 2\sqrt{590}`
+- **extracted** (✗): `<no answer>`
 - **completion**:
 
     ```text
-    Okay, let's see. I need to find the perimeter of rhombus WXYZ with an acute angle at ZWX. There's a point Q on its incircle, and the perpendicular distances from Q to the lines ZW, WX, and XY are 9, 5, and 16 respectively. Hmm, rhombus properties... All sides equal, opposite angles equal, diagonals bisect each other at right angles. The incircle of a rhombus touches all four sides, so the inradius
+    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.
     
-    ... [truncated 823 chars] ...
+    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
     
-    rcle, the distances from \(Q\) to the sides should be equal to the inradius \(r\). However, the problem states different distances (9, 5, 16), which suggests a misinterpretation or a trick in the problem setup.
+    ... [truncated 822 chars] ...
     
-    3. **Reinterpreting the Problem**:
-       - The problem might be referring to the distances from \(Q\) to the three sides \(ZW\), \(WX\), and \(XY\) as 9, 5, and 16, and the fourth side \(YZ\) is not considered. However, this is not logically consistent because the incircle is tangent to all four sides.
-       - The given distances might be related to the inradius and the perimeter. The perimeter of the rhombus can be calculated using the inradius and the distances from \(Q\) to the sides.
+    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.
     
-    4. **Calculation**:
-       - The inradius \(r\) can be approximated by the average of the given distances: \(\frac{9 + 5 + 16}{3} = 10\).
-       - The perimeter of the rhombus is then \(4 \times 2r = 80\). However, this is a guess and not based on the problem's specific setup.
+    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.
     
-    5. **Final Answer**:
-       - Given the problem's setup and the need for a logical answer, the perimeter is calculated using the distances from \(Q\) to the sides and the inradius. The final answer is:
+    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.
     
-    \[
-    \boxed{30 + 2\sqrt{590}}
-    \]
+    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.
+    
+    Alternatively, perhaps the key is to use the Law of Cosines for the entire polygon. But with 11 sides, this would be very complex.
+    
+    Alternatively, perhaps the problem requires recognizing that the sides A1A2 and A1A11 form a
     ```