Automated MNLP evaluation report (2026-05-20) (#4)

- Automated MNLP evaluation report (2026-05-20) (241883a733d7ead36c20f71e469deb1f3e2de67d)


Co-authored-by: Zeming Chen - Eric <zechen-nlp@users.noreply.huggingface.co>

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:** 2026-05-19T03:22:52+00:00 (UTC)
+- **Generated at:** 2026-05-20T05:49:46+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._
@@ -30,42 +30,23 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 - **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. 
+    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. 
     
-    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:
-    
-    10^x - 10 + 10 = 9990 + 10
-    
-    Simplifying b
+    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
     
     ... [truncated 823 chars] ...
     
-    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)
+    ation: 10^x - 10 = 9990
     
-    Which becomes:
+    Add 10 to both sides: 10^x = 10,000
     
-    x * ln(10) = ln(10,000)
+    Recognize that 10,000 is 10^4, so x = 4
     
-    Therefore, x = ln(10,000) / ln(10)
+    Check by substitution: 10^4 - 10 = 10,000 - 10 = 9,990, which matches the right side. 
     
-    But ln(10,000) is ln(10^4) = 4*ln(10). Therefore:
+    Therefore, all steps are consistent and correct. 
     
-    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. 
+    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}.
@@ -77,23 +58,31 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
     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:
+    ---
+    
+    ### Step 1: Isolate the Exponential Term
+    
+    We begin by isolating the exponential term $ 10^x $ by adding 10 to both sides of the equation:
     
     $$
     10^x - 10 + 10 = 9990 + 10
     $$
     
     $$
-    10^x = 10000
+    10^x = 10,000
     $$
     
-    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:
+    ---
+    
+    ### Step 2: Recognize the Exponential Form
+    
+    Now that we have:
     
     $$
-    10^4 = 10000
+    10^x = 10,000
     $$
     
-    we immediately conclude that:
+    we recognize that $ 10,000 = 10^4 $. Therefore, we can equate the exponents:
     
     $$
     x = 4
@@ -101,17 +90,19 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
     
     ---
     
-    To verify, we substitute $x = 4$ back into the original equation:
+    ### Step 3: Verify the Solution
+    
+    To confirm, substitute $ x = 4 $ back into the original equation:
     
     $$
-    10^4 - 10 = 10000 - 10 = 9990
+    10^4 - 10 = 10,000 - 10 = 9,990
     $$
     
     This matches the right-hand side of the original equation, confirming that the solution is correct.
     
     ---
     
-    **Final Answer:**
+    ### Final Answer
     
     $$
     \boxed{4}
@@ -120,22 +111,31 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Incorrect** (1 shown)
 
-- **reference**: `\frac{270}7\text{ degrees}`
+- **reference**: `1012`
 - **overall** (0/8 completions correct)
 - **extracted** (✗): `<no answer>`
 - **completion**:
 
     ```text
-    <think>
-    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
+    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
     
     ... [truncated 823 chars] ...
     
-    oordinates.
+    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. 
     
-    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.
+    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. 
     
-    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.
+    Therefore, the maximum is achieved when all b_i's are equal. 
     
-    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
+    Therefore, the maximum value of the sum is 2023
     ```