Automated MNLP evaluation report (2026-05-21)

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:** 2026-05-20T05:49:46+00:00 (UTC)
+- **Generated at:** 2026-05-21T03:28:21+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 | 0.1300 | ok |
+| Math | 0.1400 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -30,112 +30,58 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 - **completion**:
 
     ```text
-    Okay, so I need to solve the equation 10^x minus 10 equals 9990. Let me write that down first to make sure I have it right: 10^x - 10 = 9990. Alright, so my goal is to find the value of x that satisfies this equation. 
+    Okay, so I need to solve the equation 10^x minus 10 equals 9990. Let me write that down first to make sure I have it right: 10^x - 10 = 9990. Alright, so my goal is to find the value of x that makes this equation true. 
     
-    Hmm, let me think. I know that 10^x is a common exponential expression, and since the base is 10, maybe I can manipulate this equation to solve for x. Let me start by isolating the
+    Hmm, let me think. I remember that 10^x is a power of ten, right? So if I can figure out what power of ten, when subtracted by 10, gives me 9990. Let me try rearranging the equati
     
-    ... [truncated 823 chars] ...
-    
-    ation: 10^x - 10 = 9990
-    
-    Add 10 to both sides: 10^x = 10,000
-    
-    Recognize that 10,000 is 10^4, so x = 4
-    
-    Check by substitution: 10^4 - 10 = 10,000 - 10 = 9,990, which matches the right side. 
-    
-    Therefore, all steps are consistent and correct. 
-    
-    I think that's thorough enough. I don't see any mistakes in my reasoning, and the answer checks out when substituted back into the original equation. So, I think x = 4 is the correct solution.
-    
-    **Final Answer**
-    The solution is \boxed{4}.
-    </think>
-    
-    To solve the equation:
-    
-    $$
-    10^x - 10 = 9990
-    $$
-    
-    ---
+    ... [truncated 821 chars] ...
     
-    ### Step 1: Isolate the Exponential Term
+    3 and 5. But we already found that x is 4, which is exactly in the middle. 
     
-    We begin by isolating the exponential term $ 10^x $ by adding 10 to both sides of the equation:
+    Wait, but since 10^x is 10,000, which is exactly 10^4, so x is 4. So even though 9990 is close to 10^4, the exact value is there. 
     
-    $$
-    10^x - 10 + 10 = 9990 + 10
-    $$
+    So, I think my answer is correct. 
     
-    $$
-    10^x = 10,000
-    $$
+    But just to make sure, let me plug x = 4 back into the original equation. 
     
-    ---
+    Left side: 10^4 - 10. 
     
-    ### Step 2: Recognize the Exponential Form
+    10^4 is 10,000. 10,000 - 10 is 9,990. Which matches the right side of the equation. 
     
-    Now that we have:
+    Therefore, x = 4 is the correct solution. 
     
-    $$
-    10^x = 10,000
-    $$
+    I don't think I made any mistakes here. The key was recognizing that 10,000 is a power of ten, specifically 10^4, which makes the equation straightforward. 
     
-    we recognize that $ 10,000 = 10^4 $. Therefore, we can equate the exponents:
+    Alternatively, if I didn't remember the powers of ten, I could have used logarithms, but since I already confirmed it with two different methods, I think it's safe. 
     
-    $$
-    x = 4
-    $$
+    Another check: if I didn't remember 10^4 is 10,000, I could have written 10^x = 10,000. Then, since 10^1 = 10, 10^2 = 100, 10^3 = 1,000, 10^4 = 10,000. So x must be 4. 
     
-    ---
+    So yeah, that's solid. 
     
-    ### Step 3: Verify the Solution
+    I think that's all. I can't see any errors in my reasoning, and the answer checks out when I plug it back in. So I'm confident that x is 4.
     
-    To confirm, substitute $ x = 4 $ back into the original equation:
-    
-    $$
-    10^4 - 10 = 10,000 - 10 = 9,990
-    $$
-    
-    This matches the right-hand side of the original equation, confirming that the solution is correct.
-    
-    ---
-    
-    ### Final Answer
+    **Final Answer**
+    The solution is \boxed{4}.
+    </think>
     
-    $$
     \boxed{4}
-    $$
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `1012`
+- **reference**: `\frac{270}7\text{ degrees}`
 - **overall** (0/8 completions correct)
 - **extracted** (✗): `<no answer>`
 - **completion**:
 
     ```text
-    Okay, so I need to find the maximum possible value of the sum $\sum_{i=1}^{2023} \frac{a_i - a_{i-1}}{a_{i+1} - a_i}$, given that $a_0 = 0$ and each $a_{i+1} \geq 2a_i$ for $i = 1, 2, \ldots, 2023$. All the $a_i$ are positive real numbers. 
-    
-    First, let me parse the problem. The variables are $a_1$ through $a_{2024}$, with $a_0 = 0$. The condition is that each subsequent term is at least twice the
+    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. Hmm, let me try to visualize this. The Asymptote code draws a regular heptagon and a square sharing side AD. The square is drawn from point A to C to D to F and back to A, and then there's a square with points labeled A, C, D, F. Wait, maybe I need to think about the positions o
     
     ... [truncated 823 chars] ...
     
-    i=1}^{2023} \frac{a_i - a_{i-1}}{a_{i+1} - a_i}$." So, the sum has 2023 terms, each involving $a_i - a_{i-1}$ and $a_{i+1} - a_i$. 
-    
-    So, for n=2024, there are 2023 terms. So, the problem is for n=2024, with 2023 terms. 
-    
-    But in my earlier examples with n=2 and n=3, the maximum was achieved when the sequence was constant. 
-    
-    But let me think again. Suppose that for some i, we make b_{i+1} larger than b_i, then we have to make b_{i+2} larger than b_{i+1}, etc., which would make the subsequent terms smaller. However, if we make some b_i's larger and some smaller? But since the sequence must be non-decreasing, we can't make some b_i's smaller. 
-    
-    Therefore, the only way to maximize the sum is to have all the b_i's equal. 
-    
-    But let me think again. Suppose that for some i, we have b_{i+1} = b_i, and for other terms, b_{i+1} is larger. But since the sequence is non-decreasing, once we have b_{i+1} larger than b_i, we can't make it smaller. Therefore, the sequence is non-decreasing, so if we have some b_i's equal, and others larger, but the sum would be less than if all are equal. 
+    ernatively, maybe it's a combination of translation and rotation.
     
-    Therefore, the maximum is achieved when all b_i's are equal. 
+    But maybe I don't need to get bogged down in the Asymptote code. Let me try to think geometrically. The square and heptagon share side AD. So, AD is a side of both. The square has four sides, so AD is one side, and the square has another side connected to D. The heptagon has AD as a side, so the heptagon has another side connected to D. The Asymptote code defines point C as A + dir(90)*(F - A), which might be constructing point C such that AC is a side of the square? Wait, if AD is a side of the square, then maybe AC is also a side? But in the Asymptote code, the square is drawn as A--C--D--F. So, if AD is a side of the square, then maybe A is connected to D? But in the square A, C, D, F, A, that would require that A is connected to D, but in the Asymptote code, A is connected to C, then C to D, D to F, F to A. So, if AD is a side of the square, then the square must be A, D, something, something. But the Asymptote code draws A connected to C, so maybe the square is A, C, D, something else?
     
-    Therefore, the maximum value of the sum is 2023
+    Alternatively, maybe the square is A, D, F, and another point. Let me think. If AD is a side of the square, then the square has
     ```