Automated MNLP evaluation report (2026-05-18)

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:** 2026-05-18T02:30:25+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 | 0.1700 | 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**: `$\frac{1}{3}$`
+- **overall** (1/8 completions correct)
+- **extracted** (✓): `\frac{1}{3}`
 - **completion**:
 
     ```text
@@ -34,26 +34,53 @@ _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 $\boxed{0}$.
+    </think>
+    
+    Let $a$, $b$, $c$, and $d$ be the position vectors of the points $T_1$, $T_2$, $T_3$, and $T_4$, respectively, with respect to some coordinate system. Let $K$ be the point with position vector $\vec{k}$, and let $C$ be the point with position vector $\vec{c}$. We are told that $|\vec{k}-\vec{c}|=|\vec{a}-\vec{c}|=|\vec{b}-\vec{c}|=|\vec{c}-\vec{c}|=0$, so $C = K$. That
+    
+    ... [truncated 823 chars] ...
+    
+    $ be the fraction of the way from $T_1$ to $Y$ that $K$ lies. Then:
+    $$K = T_1 + t(Y - T_1)$$
+    $$K = T_1 + t\left(\frac{1}{3}(\vec{b}+\vec{c}+\vec{d}) - T_1\right)$$
+    $$K = T_1 - tT_1 + \frac{1}{3}t(\vec{b}+\vec{c}+\vec{d})$$
+    $$K = (1-t)T_1 + \frac{1}{3}t(\vec{b}+\vec{c}+\vec{d})$$
+    We know that $K = \frac{1}{6}(\vec{b}+\vec{c}+\vec{d})(T_1 - \frac{1}{2})$, so:
+    $$\frac{1}{6}(\vec{b}+\vec{c}+\vec{d})(T_1 - \frac{1}{2}) = (1-t)T_1 + \frac{1}{3}t(\vec{b}+\vec{c}+\vec{d})$$
+    $$\frac{1}{6}(\vec{b}+\vec{c}+\vec{d})(T_1 - \frac{1}{2}) = (1-t)T_1 + \frac{1}{3}t(\vec{b}+\vec{c}+\vec{d})$$
+    $$\frac{1}{6}(\vec{b}+\vec{c}+\vec{d})T_1 - \frac{1}{12}(\vec{b}+\vec{c}+\vec{d}) = (1-t)T_1 + \frac{1}{3}t(\vec{b}+\vec{c}+\vec{d})$$
+    Equating coefficients of $T_1$ and the constant term:
+    $$\frac{1}{6} = 1-t$$
+    $$-\frac{1}{12} = \frac{1}{3}t$$
+    Solving these equations, we find that $t = \frac{1}{4}$ and $t = \frac{1}{4}$. Therefore, $K$ lies $\frac{1}{4}$ of the way from $T_1$ to $Y$. Therefore, the ratio $\frac{KX}{KY}$ is $\frac{1}{3}$, since $K$ lies $\frac{1}{4}$ of the way from $T_1$ to $Y$, and $X$ is $\frac{1}{4}$ of the way from $T_1$ to $Y$. Therefore, the ratio $\frac{KX}{KY}$ is $\boxed{\frac{1}{3}}$.
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `$(n-2)2^n +1$`
+- **reference**: `82`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `2^{2^m} - 1`
+- **extracted** (✗): `<no answer>`
 - **completion**:
 
     ```text
     <think>
+    Okay, so I need to figure out how many colorings there are for a 2x2 grid where each square has exactly two red sides and two blue sides. The example given uses some red and blue lines, but I need to count all possible colorings that satisfy the condition. Let me start by understanding the problem better.
     
-    </think>
+    First, the grid is 2x2, which means there are four squares. Each square has four sid
+    
+    ... [truncated 822 chars] ...
+    
+    ares. Therefore, the colorings must be such that the red sides form a certain pattern.
+    
+    Alternatively, think of the grid as a graph where each square is connected to its neighbors. However, not sure.
+    
+    Alternatively, think of the grid as a graph with horizontal and vertical edges. Each horizontal edge is between two squares, and each vertical edge is between two squares. Each horizontal edge is part of the top and bottom of the squares, and each vertical edge is part of the left and right of the squares.
     
-    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 $i = 0$).
+    But maybe I need to think of the entire grid's edges. Let me think of the horizontal edges:
     
-    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 $2^{2^m - 1} - 1$.
+    There are three horizontal edges: top, middle, bottom. Each horizontal edge is between two squares. Similarly, three vertical edges: left, middle, right.
     
-    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 $2^{2^m} - 1$.
+    Each horizontal edge is part of two squares. For example, the top horizontal edge is part of the top-left and top-right squares. Similarly, the middle horizontal edge is part of the middle-left and middle-right squares. The bottom horizontal edge is part of the bottom-left and bottom-right squares.
     
-    Hence the answer is $\boxed{2^{2^m} - 1}$, where $n = 2^m + 1$.
+    Similarly, vertical edges: left vertical edge is part of the left-top and left-bottom squares, middle vertical edge is part of the middle-top
     ```