Automated MNLP evaluation report (2026-05-13)

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:** 2026-05-13T04:17:52+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 | 0.2000 | ok |
+| Math | 0.1800 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -24,9 +24,9 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `0`
-- **overall** (3/8 completions correct)
-- **extracted** (✓): `0`
+- **reference**: `\frac65`
+- **overall** (4/8 completions correct)
+- **extracted** (✓): `\frac{6}{5}`
 - **completion**:
 
     ```text
@@ -34,14 +34,14 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
     
     </think>
     
-    The polynomial $p(x)$ is equal to $\frac{1}{2}$ when $x=a$, $x=b$, and $x=c$.  Thus, $p(x)$ is a constant polynomial and the degree is $\boxed{0}$.
+    Multiplying both sides of the equation by $x$ yields \[1+5x^3=6x^2-24x\]and we can move all the terms to one side to obtain \[5x^3-6x^2+24x+1=0\]By Vieta's formulas, we have \[a+b+c=\boxed{\frac{6}{5}}.\]
     ```
 
 **Incorrect** (1 shown)
 
 - **reference**: `$(n-2)2^n +1$`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `2^{2^m} - 1`
+- **extracted** (✗): `2^{n-1} - 1`
 - **completion**:
 
     ```text
@@ -49,11 +49,11 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
     
     </think>
     
-    Let $C_n$ be the set of numbers we are trying to sum.  Then $2^n - 2^i = 2^i (2^{n - i} - 1)$, so each element of $C_n$ is a multiple of $2^i$ for some $i$ (and in particular, for $i = 0$).
+    Let $n$ be a positive integer with $n \geq 2$, and define $C_n = \{2^n - 2^i : i \text{ is an integer and } 0 \le i < n\}$. Note that $2^{n} - 2^i$ is always even for all $i$. Therefore, we can divide the problem into two parts: finding the greatest positive integer that is not representable as a sum of even integers, and finding the greatest positive integer that is not representable as a sum of odd integers.
     
-    If $n$ is a power of 2, say $n = 2^m$, then $2^{2^m} - 2^{2^m - 1} = 2^{2^m - 1} (2 - 1) = 2^{2^m - 1}$, and all elements of $C_n$ are multiples of $2^{2^m - 1}$.  Thus the largest number not in $C_n$ is $2^{2^m - 1} - 1$.
+    To find the greatest positive integer that is not representable as a sum of even integers, we note that $2^{n} - 2^i$ is divisible by $2^i$, for $0 \le i < n$. Therefore, we can find the greatest positive integer that is not representable as a sum of even integers by subtracting $2^{n-1}$ from $2^n - 2^0 = 2^n - 1$. This gives us $2^n - 2^{n-1} - 1 = 2^{n-1} - 1$, which is the greatest positive integer that is not representable as a sum of even integers.
     
-    If $n$ is not a power of 2, then $n = 2^m + 1$ for some $m$.  Then $2^{2^m + 1} - 2^{2^m} = 2^{2^m} (2 - 1) = 2^{2^m}$, and all elements of $C_n$ are multiples of $2^{2^m}$.  Thus the largest number not in $C_n$ is $2^{2^m} - 1$.
+    To find the greatest positive integer that is not representable as a sum of odd integers, we note that $2^{n} - 2^i$ is divisible by $2^i$, for $0 \le i < n$. Therefore, we can find the greatest positive integer that is not representable as a sum of odd integers by subtracting $2^{n-1} - 1$ from $2^n - 2^1 = 2^n - 2$. This gives us $2^n - 2^{n-1} - 1 = 2^{n-1} - 1$, which is the greatest positive integer that is not representable as a sum of odd integers.
     
-    Hence the answer is $\boxed{2^{2^m} - 1}$, where $n = 2^m + 1$.
+    Therefore, the greatest positive integer that is not representable as a sum of elements of $C_n$ is $\boxed{2^{n-1} - 1}$.
     ```