Automated MNLP evaluation report (2026-05-27)

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:** 2026-05-26T12:29:10+00:00 (UTC)
+- **Generated at:** 2026-05-27T13:02:22+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 | 0.2200 | ok |
+| Math | 0.2000 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -24,60 +24,58 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `1600`
+- **reference**: `28`
 - **overall** (8/8 completions correct)
-- **extracted** (✓): `1600`
+- **extracted** (✓): `28`
 - **completion**:
 
     ```text
-    Okay, so I need to figure out what this expression equals: 441 plus 2 times 21 times 19 plus 361. Let me write that down again to make sure I have it right:
+    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. 
     
-    $441 + 2 \times 21 \times 19 + 361 = x$
+    First, let me make sure I understand the problem correctly. It says there are at least six of each type available. So, for exa
     
-    Hmm, let me break this down step by step. First, maybe I should calculate each part separately and then add them up. Let me start with the multiplication part because there's a 2 times 21 times 19 insi
+    ... [truncated 823 chars] ...
     
-    ... [truncated 822 chars] ...
+    e previous result. 
     
-    o recognize this expression without calculating each term separately? Let me think. The original expression is 441 + 2×21×19 + 361. Wait, 441 and 361 are both squares. 21 squared is 441, right? 21×21 is 441. And 19 squared is 361. Oh! Wait, that's a key insight. So the expression is 21² + 2×21×19 + 19². Which is a perfect square. Specifically, it's (a + b)² where a = 21 and b = 19. Let me check that.
+    Therefore, the answer is 28. 
     
-    (a + b)² = a² + 2ab + b². Exactly! So that's exactly the expression given. Therefore, x = (21 + 19)². Let me compute that. 21 + 19 is 40. Then 40 squared is 1600. There we go! So that's a much quicker way. Instead of computing each term separately, recognizing the pattern as a perfect square. That makes sense. So x is 1600.
+    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 just to be thorough, let me confirm once more. If I expand (21 + 19)^2, it's 21^2 + 2*21*19 + 19^2. Which is exactly the original expression. Therefore, substituting 21 + 19 = 40, so 40 squared is 1600. Yep, that's correct.
+    Therefore, I think the answer is 28. 
     
-    Therefore, the answer is 1600. I think that's solid. I don't see any mistakes in my reasoning. Both methods gave me the same result, so I feel confident about this.
+    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. 
+    
+    Therefore, I think the answer is 28, so \boxed{28}.
     
     **Final Answer**
-    The value of $x$ is \boxed{1600}.
+    \boxed{28}
     </think>
     
-    \boxed{1600}
+    \boxed{28}
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `$\frac{1}{3}$`
+- **reference**: `31`
 - **overall** (0/8 completions correct)
 - **extracted** (✗): `<no answer>`
 - **completion**:
 
     ```text
-    Okay, so I need to solve this geometry problem about a tetrahedron and some points related to it. Let me try to parse the problem step by step.
-    
-    First, we have a tetrahedron $ T_1T_2T_3T_4 $ in 3D space. There's a point $ C $ such that its distance to all four vertices is equal. So, $ CT_1 = CT_2 = CT_3 = CT_4 $. That means $ C $ is equidistant from all four vertices of the tetrahedron. 
+    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.
     
-    Then, th
+    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.
     
-    ... [truncated 823 chars] ...
-    
-    ,-1,1), C(-1,1,-1), D(1,-1,-1). Let me find the equation of the plane through these three points.
+    ... [truncated 822 chars] ...
     
-    First, compute two vectors in the plane: vector BC = (0,2,-2) as before, and vector BD = (2,0,-2). Then, the normal vector is (-4, -4, -4) as before. Therefore, the equation of the plane is -4(x +1) -4(y +1) -4(z -1) = 0? Wait, no. Wait, the plane passes through point B(-1,-1,1). So the equation is -4(x - (-1)) -4(y - (-1)) -4(z - 1) = 0? Wait, no. Wait, the normal vector is (-4, -4, -4), so the plane equation is -4(x - x0) -4(y - y0) -4(z - z0) = 0. Using point B(-1,-1,1):
+    (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.
     
-    -4(x +1) -4(y +1) -4(z -1) = 0 => -4x -4 -4y -4 -4z +4 = 0 => -4x -4y -4z -4 = 0 => divide by -4: x + y + z + 1 = 0. So the plane equation is x + y + z = -1.
+    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.
     
-    Now, the line $ T_1K $ is the line from A(1,1,1) to K(0,0,0). Parametric equations: x = 1 - t, y = 1 - t, z = 1 - t, where t ranges from 0 to 1. We need to find where this line intersects the plane x + y + z = -1. Substitute x = 1 - t, y = 1 - t, z = 1 - t into the plane equation:
+    If Q1 has one root (discriminant zero) and Q2 has two roots (discriminant positive). Then, similar logic applies. Let me try.
     
-    (1 - t) + (1 - t) + (1 - t) = -1 => 3 - 3t = -1 => 3t = 4 => t = 4/3.
+    In this case, Q1: x² + a x + (b - c) = 0 has discriminant zero ⇒ a² = 4(b - c).
     
-    Therefore, the intersection point X is at t = 4/3: x = 1 - 4/3 = -1/3, y = -1/3, z = -1/3. So X is (-1/3, -1
+    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
     ```