Automated MNLP evaluation report (2026-05-19)

EVAL_REPORT.md

@@ -2,7 +2,7 @@
 
 - **Model repo:** [`cs-552-2026-emainelpe/math_model`](https://huggingface.co/cs-552-2026-emainelpe/math_model)
 - **Owner(s):** group **emainelpe**
-- **Generated at:** 2026-05-16T04:57:46+00:00 (UTC)
+- **Generated at:** 2026-05-19T03:22:52+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 | 0.3400 | ok |
+| Math | 0.3100 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -24,64 +24,104 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `\frac65`
-- **overall** (8/8 completions correct)
-- **extracted** (✓): `\frac{6}{5}`
+- **reference**: `4`
+- **overall** (1/8 completions correct)
+- **extracted** (✓): `4`
 - **completion**:
 
     ```text
     <think>
-    To find the sum of the solutions \(a + b + c\), we first need to rewrite the given equation in the standard quadratic form \(ax^2 + bx + c = 0\). We do this by moving all terms to one side of the equation:
+    The given equation can be rewritten as
+    \[ x^3 - 3x^2 + 3x + y^3 - 3y^2 + 3y + z^3 - 3z^2 + 3z = 4 \]
     
-    \[ \frac{1}{x} + 5x^2 = 6x - 24 \Rightarrow 5x^2 - 6x + 24 + \frac{1}{x} = 0 \]
+    Factor the expression:
+    \[ (x + 1)^3 - 1 + (y + 1)^3 - 1 + (z + 1)^3 - 1 = 4 \]
     
-    To eliminate the fraction, multiply every term by \(x\):
+    \[ (x + 1)^3 + (y + 1)^3 + (z + 1)^3 = 7 \]
     
-    \[ 5x^3 - 6x^2 + 24x + 1 = 0 \]
+    Now, we want to find the largest possible value of
+    \[ x^2 + y^2 + z^2 - x - y - z \]
     
-    Now, we have a cubic equation. To find the sum of its roots, we can use Vieta's formulas, which state that for a polynomial of the form \(ax^3 + bx^2 + cx + d = 0\), the sum of the roots is given by \(-\frac{b}{a}\).
+    We can rewrite this expression as
+    \[ (x^2 - x) + (y^2 - y) + (z^2 - z) \]
     
-    In our case, \(a = 5\) and \(b = -6\), so the sum of the roots \(a + b + c\) is:
+    Comp
     
-    \[ a + b + c = -\frac{b}{a} = -\frac{-6}{5} = \boxed{\frac{6}{5}} \]
+    ... [truncated 223 chars] ...
+    
+    \right) - 3 \]
+    
+    Since the sum of squares is always nonnegative, the largest possible value occurs when $(x - 1)^2 + (y - 1)^2 + (z - 1)^2$ is maximized.
+    
+    Notice that $(x + 1)^3 + (y + 1)^3 + (z + 1)^3 = 7$ can be rewritten as
+    \[ (x + 1)^3 + (y + 1)^3 + (z + 1)^3 - 3(x + 1)(y + 1)(z + 1) = 4 \]
+    
+    \[ (x + 1)^3 + (y + 1)^3 + (z + 1)^3 - 3(x + 1)(y + 1)(z + 1) - 3 = 1 \]
+    
+    \[ (x + 1 - y - 1)(y + 1 - z - 1)(z + 1 - x - 1) = 1 \]
+    
+    \[ (x - y)(y - z)(z - x) = 1 \]
+    
+    This implies that $x, y, z$ cannot be equal, and the differences between them are not zero.
+    
+    Now, we can use the fact that the sum of squares is maximized when the variables are as far apart as possible.
+    
+    Since $(x + 1)^3 + (y + 1)^3 + (z + 1)^3 = 7$, the largest possible value of $(x - 1)^2 + (y - 1)^2 + (z - 1)^2$ occurs when $x, y, z$ are as far apart as possible.
+    
+    This happens when $x = 1 + \sqrt[3]{2}, y = 1 - \sqrt[3]{2}, z = 1$.
+    
+    Substituting these values into the expression, we get
+    \[ x^2 + y^2 + z^2 - x - y - z = (1 + \sqrt[3]{2})^2 + (1 - \sqrt[3]{2})^2 + 1^2 - (1 + \sqrt[3]{2}) - (1 - \sqrt[3]{2}) - 1 \]
+    
+    \[ = 2 + 2\sqrt[3]{2} + 2 - 2\sqrt[3]{2} = 4 \]
+    
+    So, the largest possible value is $\boxed{4}$.
     </think>
     
-    \boxed{\frac{6}{5}}
+    \boxed{4}
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `$(n-2)2^n +1$`
+- **reference**: `504`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `2^{n-1} - 1`
+- **extracted** (✗): `120`
 - **completion**:
 
     ```text
     <think>
-    ## Step 1: Understand the problem
-    We are given a set C_n, which consists of numbers of the form 2^n - 2^i, where i is an integer and 0 ≤ i < n. We need to find the greatest positive integer that cannot be represented as a sum of elements from C_n, allowing repeated use of any element.
+    Let the height of the trapezoid be \(h\). Since the circle is inscribed in the trapezoid, the height of the trapezoid is equal to the diameter of the circle, which is \(2r\). Therefore, \(h = 2r\).
+    
+    The area of the trapezoid is given by:
+    \[ \text{Area} = \frac{1}{2} (r + s) h \]
+    
+    Substitute the given area and the expression for \(h\):
+    \[ 72 = \frac{1}{2} (r + s) (2r) \Rightarrow 72 = r(r +
+    
+    ... [truncated 631 chars] ...
     
-    ## Step 2: Analyze the set C_n
-    Let's analyze the set C_n. We can see that the elements of C_n are of the
+    rrow r = 12 \]
     
-    ... [truncated 207 chars] ...
+    Now, we can find \(s\):
+    \[ s = 6 - r = 6 - 12 = -6 \]
     
-    cause when i = n-1, the expression 2^n - 2^i becomes 2^n - 2^{n-1} = 2^{n-1}(2 - 1) = 2^{n-1}.
+    However, since \(r\) and \(s\) are the lengths of the parallel sides of the trapezoid, they must be positive. Therefore, we made a mistake in our previous steps.
     
-    ## Step 4: Consider the binary representation of the largest element
-    The largest element in C_n, which is 2^{n-1}, can be represented in binary as a 1 followed by n-1 zeros. This means that when we add up elements of C_n, we can only get numbers that have a binary representation with at most n-1 zeros.
+    Let's re-examine the problem. We know that the area of the trapezoid is \(72\), and the height is \(2r\). Therefore:
+    \[ 72 = \frac{1}{2} (r + s) (2r) \Rightarrow 72 = r(r + s) \Rightarrow 72 = r^2 + rs \]
     
-    ## Step 5: Determine the largest positive integer that cannot be represented as a sum of elements of C_n
-    Based on the analysis above, we can see that the largest positive integer that cannot be represented as a sum of elements of C_n is 2^n - 1 - 2^{n-1} = 2^{n-1} - 1. This is because we can always subtract 2^{n-1} from 2^n - 1 to get a number that cannot be represented as a sum of elements of C_n.
+    We also know that the circle is inscribed in the trapezoid, so the diameter of the circle is equal to the height of the trapezoid, which is \(2r\). Therefore:
+    \[ 2r = h \Rightarrow 2r = \sqrt{(r - s)^2 + h^2} \Rightarrow 4r^2 = (r - s)^2 + h^2 \Rightarrow 4r^2 = (r - s)^2 + (2r)^2 \Rightarrow 4r^2 = r^2 - 2rs + s^2 + 4r^2 \Rightarrow 2rs = s^2 \Rightarrow 2r = s \]
     
-    ## Step 6: Find the greatest positive integer that is not representable as a sum of elements of C_n
-    The greatest positive integer that is not representable as a sum of elements of C_n is 2^n - 1 - 2^{n-1} = 2^{n-1} - 1.
+    Now, we can substitute \(s = 2r\) into the equation \(72 = r(r + s)\):
+    \[ 72 = r(r + 2r) \Rightarrow 72 = r(3r) \Rightarrow 72 = 3r^2 \Rightarrow r^2 = 24 \Rightarrow r = \sqrt{24} = 2\sqrt{6} \]
     
-    ## Step 7: Simplify the expression
-    We can simplify the expression 2^{n-1} - 1 to get the final answer.
+    Now, we can find \(s\):
+    \[ s = 2r = 2(2\sqrt{6}) = 4\sqrt{6} \]
     
-    The final answer is: $\boxed{2^{n-1} - 1}
+    Finally, we can find \(r^2 + s^2\):
+    \[ r^2 + s^2 = (2\sqrt{6})^2 + (4\sqrt{6})^2 = 24 + 96 = \boxed{120} \]
     </think>
     
-    \boxed{2^{n-1} - 1}
+    \boxed{120}
     ```