Automated MNLP evaluation report (2026-05-19)

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:** 2026-05-18T02:30:25+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 | 0.1700 | ok |
+| Math | 0.1300 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -24,63 +24,118 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `$\frac{1}{3}$`
-- **overall** (1/8 completions correct)
-- **extracted** (✓): `\frac{1}{3}`
+- **reference**: `4`
+- **overall** (8/8 completions correct)
+- **extracted** (✓): `4`
 - **completion**:
 
     ```text
     <think>
+    Okay, so I need to solve the equation 10^x minus 10 equals 9990. Let me write that down first: 10^x - 10 = 9990. Hmm, let me think about how to approach this. 
     
-    </think>
+    First, maybe I should isolate the term with the exponent. That means I need to get rid of the -10 on the left side. To do that, I can add 10 to both sides of the equation. Let me try that:
     
-    </think>
+    10^x - 10 + 10 = 9990 + 10
     
-    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
+    Simplifying b
     
     ... [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 $\boxed{\frac{1}{3}}$.
+    answer. 
+    
+    But just to be absolutely sure, let me check with another method. Let's say I use natural logarithm. 
+    
+    Starting with 10^x = 10,000. Taking natural logarithm on both sides:
+    
+    ln(10^x) = ln(10,000)
+    
+    Which becomes:
+    
+    x * ln(10) = ln(10,000)
+    
+    Therefore, x = ln(10,000) / ln(10)
+    
+    But ln(10,000) is ln(10^4) = 4*ln(10). Therefore:
+    
+    x = (4*ln(10)) / ln(10) = 4. 
+    
+    Same result. 
+    
+    Therefore, regardless of the method, x is 4. 
+    
+    So, I think that's solid. 
+    
+    Therefore, the solution is x = 4. 
+    
+    **Final Answer**
+    The solution is \boxed{4}.
+    </think>
+    
+    To solve the equation:
+    
+    $$
+    10^x - 10 = 9990
+    $$
+    
+    we begin by isolating the exponential term $10^x$. This is done by adding 10 to both sides of the equation:
+    
+    $$
+    10^x - 10 + 10 = 9990 + 10
+    $$
+    
+    $$
+    10^x = 10000
+    $$
+    
+    Now, we need to solve for $x$ in the equation $10^x = 10000$. Since 10 is the base of the exponential function, and we know that:
+    
+    $$
+    10^4 = 10000
+    $$
+    
+    we immediately conclude that:
+    
+    $$
+    x = 4
+    $$
+    
+    ---
+    
+    To verify, we substitute $x = 4$ back into the original equation:
+    
+    $$
+    10^4 - 10 = 10000 - 10 = 9990
+    $$
+    
+    This matches the right-hand side of the original equation, confirming that the solution is correct.
+    
+    ---
+    
+    **Final Answer:**
+    
+    $$
+    \boxed{4}
+    $$
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `82`
+- **reference**: `\frac{270}7\text{ degrees}`
 - **overall** (0/8 completions correct)
 - **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.
-    
-    First, the grid is 2x2, which means there are four squares. Each square has four sid
+    Okay, so I need to find the measure of angle BAC in this figure where a square and a regular heptagon share the side AD. Let me try to visualize the Asymptote figure they provided. There's a square and a regular heptagon overlapping with side AD. The square has vertices labeled A, C, D, and F? Wait, the Asymptote code draws the square and heptagon with shared side AD. Let me parse the Asym
     
-    ... [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.
+    ... [truncated 823 chars] ...
     
-    But maybe I need to think of the entire grid's edges. Let me think of the horizontal edges:
+    oordinates.
     
-    There are three horizontal edges: top, middle, bottom. Each horizontal edge is between two squares. Similarly, three vertical edges: left, middle, right.
+    First, let me recall that in a regular heptagon, the central angles between adjacent vertices are 360/7 ≈ 51.4286 degrees. So the angle between any two adjacent vertices from the center is 360/7 degrees.
     
-    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.
+    But since the Asymptote code defines the heptagon with vertices at angles 90°, 90 + 360/7°, etc., as mentioned earlier. So point A is at angle 90° + 3*360/7° ≈ 244.2857°, as computed earlier. Point F is at angle 90° + 4*360/7° ≈ 295.7143°. So the angle between A and F as viewed from the center is 295.7143° - 244.2857° ≈ 51.4286°, which is exactly 360/7 degrees, as expected for a regular heptagon.
     
-    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
+    Now, the square shares side AD. So points A and D are adjacent vertices of the square. Wait, the square has vertices A, C, D, F? Wait, the Asymptote code draws A--C--D--F. So the square has sides AC, CD, DF, FA? Wait, but the problem says the square and heptagon share side AD. Wait, the Asymptote code draws A--C--D--F. So maybe the square is A, C, D, F? So that would make AD a diagonal of the square? Wait, but the problem says they share side AD. So maybe AD is a side of both the square and the heptagon? Wait, but in the Asymptote code, they have a square and a heptagon. The
     ```