Automated MNLP evaluation report (2026-05-29)

EVAL_REPORT.md

@@ -2,7 +2,7 @@
 
 - **Model repo:** [`cs-552-2026-MandMP/math_model`](https://huggingface.co/cs-552-2026-MandMP/math_model)
 - **Owner(s):** group **MandMP**
-- **Generated at:** 2026-05-28T08:03:33+00:00 (UTC)
+- **Generated at:** 2026-05-29T06:54:15+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._
@@ -11,7 +11,7 @@ _This PR is opened automatically by the course CI. It is **non-blocking** — yo
 
 | Benchmark | Accuracy | Status |
 |---|---:|---|
-| Math | 0.2000 | ok |
+| Math | 0.1900 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -24,35 +24,59 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `28`
-- **overall** (8/8 completions correct)
-- **extracted** (✓): `28`
+- **reference**: `3`
+- **overall** (7/8 completions correct)
+- **extracted** (✓): `3`
 - **completion**:
 
     ```text
-    Okay, so Pat is choosing six cookies from a tray that has chocolate chip, oatmeal, and peanut butter cookies. There's at least six of each kind available. And since cookies of the same kind are indistinct, I need to figure out how many different ways she can choose them. 
+    Okay, so I need to solve this problem about treeks, squigs, and goolees. Let me read it again:
     
-    First, let me make sure I understand the problem correctly. It says there are at least six of each type available. So, for exa
+    "Suppose 10 treeks have the same weight as 3 squigs plus 1 goolee, and 2 treeks together with 1 goolee weigh as much as 1 squig. Determine how many treeks have a combined weight equal to that of 1 squig."
+    
+    Alright, so there are three different units: treeks, squigs, and goolees. The problem gives me two
     
     ... [truncated 823 chars] ...
     
-    e previous result. 
+    herefore, x = 3T / T = 3. So, 3 treeks equal 1 squig.
+    
+    Wait, so the answer is 3?
+    
+    Let me check again.
+    
+    From equation (1): 10T = 3S + G
+    
+    From equation (2): 2T + G = S
+    
+    We can substitute S from equation (2) into equation (1):
+    
+    10T = 3*(2T + G) + G => 10T = 6T + 3G + G => 10T = 6T + 4G => 4T = 4G => T = G.
+    
+    Then S = 2T + G = 2T + T = 3T. Therefore, 1 squig is 3 treeks. So yes, 3 treeks equal 1 squig.
     
-    Therefore, the answer is 28. 
+    So the answer is 3. Therefore, boxed answer is \boxed{3}.
     
-    But just to make sure, let me think of another way. Let me consider that since there are at least six of each kind available, but Pat is only selecting six cookies. So, the number of chocolate chip cookies she can take is between 0 and 6, same with the others. But since the total is 6, the maximum number of any one kind is 6. So, the problem is equivalent to the number of non-negative integer solutions to c + o + p = 6. Which is C(6 + 3 -1, 3 -1) = C(8,2) = 28. 
+    But let me think again if I did everything correctly. Maybe I made an assumption?
     
-    Therefore, I think the answer is 28. 
+    Wait, the problem says "how many treeks have a combined weight equal to that of 1 squig". So, if 1 squig is 3 treeks, then 3 treeks = 1 squig. So yes, answer is 3.
     
-    But maybe I made a mistake in assuming that the upper bounds are redundant? Let me think again. Suppose that there were, for example, only five cookies of each kind available. Then, Pat couldn't take more than five of any kind. But since she is only taking six, that would be impossible. However, in our problem, since there are at least six cookies of each kind available, Pat can take up to six of any kind, which is exactly the number she is taking. Therefore, the upper bounds are not restrictive. Therefore, the answer is indeed 28. 
+    Alternatively, maybe I should check with actual numbers. Let me assign some numbers.
     
-    Therefore, I think the answer is 28, so \boxed{28}.
+    Let me assume that T = G = 1 unit. Then S = 3T = 3 units.
+    
+    Check equation (1): 10T = 10*1 = 10. On the other side, 3S + G = 3*3 + 1 = 9 + 1 = 10. Correct.
+    
+    Equation (2): 2T + G = 2*1 + 1 = 3, which equals S = 3. Correct.
+    
+    So yes, that works. Therefore, 3 treeks equal 1 squig. So answer is 3.
+    
+    Therefore, I think the answer is 3. So \boxed{3} is the answer.
     
     **Final Answer**
-    \boxed{28}
+    \boxed{3}
     </think>
     
-    \boxed{28}
+    \boxed{3}
     ```
 
 **Incorrect** (1 shown)
@@ -63,19 +87,51 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 - **completion**:
 
     ```text
-    Okay, so I need to solve this problem where we have integers a, b, c with c ≤ 2025, and the equation |x² + a x + b| = c has exactly 3 distinct integer solutions for x. Then I have to find how many possible values of c there are.
+    Okay, so I need to solve this problem where we have an equation |x² + a x + b| = c, and there are three distinct integer solutions for x. The variables a, b, and c are integers, and c is less than or equal to 2025. We need to find how many possible values of c there are.
+    
+    First, let me recall that absolute value equations like |f(x)| = c can be split into two cases: f(x) = c and f(x) = -c. So, in
+    
+    ... [truncated 823 chars] ...
+    
+    uting c = b - r1 r2 into a² = 4(b + c):
+    
+    a² = 4(b + b - r1 r2) = 4(2b - r1 r2)
+    
+    But also, from Q1, a = -(r1 + r2). Therefore, a² = (r1 + r2)^2.
+    
+    Therefore:
+    
+    (r1 + r2)^2 = 8b - 4 r1 r2
+    
+    Let me write that equation:
+    
+    (r1 + r2)^2 = 8b - 4 r1 r2
+    
+    Expanding left side:
+    
+    r1² + 2 r1 r2 + r2² = 8b - 4 r1 r2
+    
+    Bring all terms to left side:
+    
+    r1² + 2 r1 r2 + r2² + 4 r1 r2 - 8b = 0 ⇒ r1² + 6 r1 r2 + r2² - 8b = 0
+    
+    Therefore:
+    
+    8b = r1² + 6 r1 r2 + r2²
+    
+    Therefore:
     
-    First, let me parse the equation. The absolute value of a quadratic expression equals c. So, |x² + a x + b| = c implies that either x² + a x + b = c or x² + a x + b = -c.
+    b = [r1² + 6 r1 r2 + r2²]/8
     
-    ... [truncated 822 chars] ...
+    But since b must be integer, the numerator must be divisible by 8.
     
-    (discriminant zero) and Q2 has two roots. Wait, but earlier we thought that Case 2 was impossible because m = c and m < -c. But if we consider that, then m = c and m < -c ⇒ c < -c ⇒ c < 0. But c is non-negative, so this is impossible. Therefore, Case 2 is invalid. Therefore, only Case 1 is possible: Q1 has two roots, Q2 has one root. Therefore, only the c's from the case where Q2 discriminant is zero and Q1 discriminant is positive.
+    Similarly, since c = b - r1 r2, then c = [r1² + 6 r1 r2 + r2²]/8 - r1 r2 = [r1² + 6 r1 r2 + r2² - 8 r1 r2]/8 = [r1² - 2 r1 r2 + r2²]/8 = (r1 - r2)^2 / 8
     
-    But wait, there's another possibility. Suppose that Q2 has two roots, but Q1 has one root. Wait, but that would mean that the equation |quadratic| = c has two roots from Q2 and one root from Q1, totaling three. However, in that case, Q1 would have one root (discriminant zero) and Q2 would have two roots (discriminant positive). Therefore, similar to before, but with different signs. Let me check.
+    Therefore, c = (r1 - r2)^2 / 8
     
-    If Q1 has one root (discriminant zero) and Q2 has two roots (discriminant positive). Then, similar logic applies. Let me try.
+    Since c must be integer, (r1 - r2)^2 must be divisible by 8. Therefore, (r1 - r2)^2 ≡ 0 mod 8 ⇒ r1 - r2 is divisible by 2√2? Wait, no. Wait, if (r1 - r2)^2 is divisible by 8, then (r1 - r2)^2 must be divisible by 8. Let me think.
     
-    In this case, Q1: x² + a x + (b - c) = 0 has discriminant zero ⇒ a² = 4(b - c).
+    If (r1 - r2)^2 is divisible by 8, then (r1 - r2)^2 has factors of 2^3. Therefore, (r1 - r2)^2 must be divisible by 8 ⇒ (r1 - r2)^2 is divisible by 8 ⇒ (r1 - r2) is divisible by 2√2? Wait, no. Let me think again.
     
-    Q2: x² + a x + (b + c) = 0 has discriminant positive ⇒ a² - 4(b + c) > 0 ⇒ 4(b - c) - 4(b + c) = -8c > 0 ⇒ -8c > 0 ⇒ c < 0. But again, c is non-negative
+    Suppose that (r1 - r2
     ```