Update Automated MNLP evaluation report (2026-05-13)

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:** 2026-05-13T04:17:52+00:00 (UTC)
+- **Generated at:** 2026-05-13T23:30:17+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 | 0.2500 | ok |
+| Math | 0.2800 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -25,50 +25,38 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 **Correct** (1 shown)
 
 - **reference**: `0`
-- **overall** (3/8 completions correct)
+- **overall** (2/8 completions correct)
 - **extracted** (✓): `0`
 - **completion**:
 
     ```text
-    Let $f(x)=\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-a)(x-c)}{(b-a)(b-c)}+\frac{(x-a)(x-b)}{(c-a)(c-b)}$. Then, $f(x)$ is a rational function with a numerator of degree 2 and a denominator of degree 2. Therefore, $f(x)$ is a rational function of degree 2 - 2 = 0, or a constant function.
+    Let us denote $f(x) = \frac{(x-b)(x-c)}{(a-b)(a-c)} + \frac{(x-a)(x-c)}{(b-a)(b-c)} + \frac{(x-a)(x-b)}{(c-a)(c-b)}$.
     
-    To find the constant value of $f(x)$, plug in a value of $x$, say $x = a$. Then,
+    We can compute $f(a)$, $f(b)$, and $f(c)$ directly to find out whether this function is constant or not.
+    
+    Firstly, we note that $\frac{(x-b)(x-c)}{(a-b)(a-c)}$ has a root at $x=a$, so when we plug in $x=a$, it becomes zero. Similarly, for other terms, they also vanish when evaluated at their respective roots. Therefore,
     $$
-    f(a)=\frac{(a-b)(a-c)}{(a-b)(a-c)}+\frac{(a-a)(a-c)}{(b-a)(b-c)}+\frac{(a-a)(a-b)}{(c-a)(c-b)} = 1.
+    f(a) = f(b) = f(c) = 0.
     $$
-    Similarly, plug in $x = b$ and $x = c$, and you find that $f(x)=1$ for all $x$. Therefore, $f(x)$ is a constant function equal to 1.
     
-    Therefore, the degree of the polynomial is $\boxed{0}$.
+    Thus, the polynomial $f(x)$ must be identically zero because its values coincide with the value of the zero polynomial at three different points.
+    
+    Therefore, the degree of the polynomial $f(x)$ is $\boxed{0}$.
     ```
 
 **Incorrect** (1 shown)
 
 - **reference**: `$(n-2)2^n +1$`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `<no answer>`
+- **extracted** (✗): `2^{n-1} - 1`
 - **completion**:
 
     ```text
-    Let’s denote the greatest positive integer not representable as a sum of elements of C_n by M_n. Then M_n is the greatest positive integer not representable as a sum of elements of C_n, allowing repeated use of any element.
-    We can use the following observations to determine M_n:
+    We are given $C_n$ for some $n$, which contains all numbers of the form $2^n - 2^i$, where $i$ ranges from $0$ to $n-1$. We want to find the largest number that cannot be written as a sum of these numbers.
     
-    1. The set C_n contains n elements, each of which is 2^n − 2^i for some integer i between 0 and n−1.
-    2. The set C_n con
+    Note that each number in $C_n$ has exactly one factor of 2 raised to some power between $2^0=1$ and $2^{n-1}$. So we can think of this set as containing powers of two, but subtracted by smaller powers of two.
     
-    ... [truncated 823 chars] ...
+    The smallest such number is $2^n - 2^0 = 2^n - 1$, so the largest number in $C_n$ is $2^n - 1$. Therefore, the total range of possible sums is at least $\frac{2^n - 1}{2}$ (since every number must be used once), and no larger number than this can be formed.
     
-    rs of 2 are in C_n.
-    131. The set C_n contains elements that are powers of 2, but not all powers of 2 are in C_n.
-    132. The set C_n contains elements that are powers of 2, but not all powers of 2 are in C_n.
-    133. The set C_n contains elements that are powers of 2, but not all powers of 2 are in C_n.
-    134. The set C_n contains elements that are powers of 2, but not all powers of 2 are in C_n.
-    135. The set C_n contains elements that are powers of 2, but not all powers of 2 are in C_n.
-    136. The set C_n contains elements that are powers of 2, but not all powers of 2 are in C_n.
-    137. The set C_n contains elements that are powers of 2, but not all powers of 2 are in C_n.
-    138. The set C_n contains elements that are powers of 2, but not all powers of 2 are in C_n.
-    139. The set C_n contains elements that are powers of 2, but not all powers of 2 are in C_n.
-    140. The set C_n contains elements that are powers of 2, but not all powers of 2 are in C_n.
-    141. The set C_n contains elements that are powers of 2, but not all powers of 2 are in C_n.
-    142. The set C_n contains elements that are powers of 2, but not all powers of 2 are in C_n.
-    143. The set C_n contains elements that are powers of 2, but not
+    Therefore, the largest number that cannot be expressed as a sum of elements of $C_n$ is $\boxed{2^n - 1 - 2^{n-1}} = \boxed{2^{n-1} - 1}$.
     ```