Update Automated MNLP evaluation report (2026-05-21)

EVAL_REPORT.md

@@ -2,7 +2,7 @@
 
 - **Model repo:** [`cs-552-2026-clankers-builder/math_model`](https://huggingface.co/cs-552-2026-clankers-builder/math_model)
 - **Owner(s):** group **clankers-builder**
-- **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._
@@ -24,47 +24,34 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `\frac{639}{40}`
+- **reference**: `106`
 - **overall** (1/8 completions correct)
-- **extracted** (✓): `\frac{639}{40}`
+- **extracted** (✓): `106`
 - **completion**:
 
     ```text
     <think>
-    Okay, so I need to figure out the distance between two buildings on a map. The problem says that 3 inches on the map represent 10 miles in real life. And the actual distance between the buildings is 53.25 miles. I need to find out how far that is on the map, expressed as a fraction. Alright, let's break this down step by step.
+    Okay, let me try to figure out this problem. So, we have this product from k=4 to k=63 of [log base k of (5^{k² -1}) divided by log base (k+1) of (5^{k² -4})]. And we need to simplify this product and express it as m/n where m and n are coprime, then find m + n. Hmm, that seems complicated, but maybe there's a pattern or some telescoping product here.
     
-    First, let me understand the scale. The map scale is 3 inches
+    First, let me recall that log_b(a) ca
     
     ... [truncated 823 chars] ...
     
-    ays "written as a fraction", so improper fraction is acceptable. So, 639/40. Let me check if that reduces. 639 and 40. Let's see, 639 divided by 4 is 159.75, not integer. 639 divided by 5 is 127.8, not integer. 639 divided by 3 is 213, which is 213/40. Wait, 639 divided by 3 is 213. 213 divided by 3 is 71. So, 639 is 3*3*71. 40 is 2^3*5. So, no common factors. So, 639/40 is the simplest form. Therefore, the answer is 639/40.
+    )/ln(4) = 3 ln4 / ln4 = 3. Correct.
     
-    So, after going through all these steps and checking my calculations, I believe that 639/40 inches is the correct answer. Let me just recap to make sure I didn't skip any steps or make a mistake.
+    First product: Product_{k=4}^{63} [(k² -1)/(k² -4)] = [Product (k -1)/(k -2)] * [Product (k +1)/(k +2)] = [Product (k -1)/(k -2)] = 62/2 = 31. Wait, but 62/2 is 31, yes. And [Product (k +1)/(k +2)] = 5/65 = 1/13. So 31 * 1/13 = 31/13. Then multiply by 3: 31/13 * 3 = 93/13. Then 93 and 13: 13*7=91, 93-91=2, so 93/13 is reduced. So m=93, n=13, m + n=106.
     
-    Starting with the scale: 3 inches =10 miles. So, to find the map distance for 53.25 miles, set up the proportion:
+    But let me check the first product again. The product of (k -1)/(k -2) from k=4 to 63. Let's see: when k=4, (4-1)/(4-2)=3/2; k=5:4/3; k=6:5/4; ... up to k=63:62/61. So the product is (3/2)*(4/3)*(5/4)*...*(62/61). All the terms cancel except the first denominator and the last numerator: 62/2 = 31. Correct.
     
-    3/10 = x/53.25
+    Similarly, the product of (k +1)/(k +2) from k=4 to 63. When k=4, (5/6); k=5:6/7; ... up to k=63:64/65. So the product is (5/6)*(6/7)*...*(64/65). All terms cancel except the first numerator and the last denominator: 5/65 = 1/13. Correct. So 31 * 1/13 = 31/13. Then multiplied by 3 gives 93/13. Correct.
     
-    Cross-multiplying gives 3*53.25 =10x
-    
-    Calculating 3*53.25: 53*3=159, 0.25*3=0.75, so 159+0.75=159.75
-    
-    Then 159.75=10x => x=159.75/10=15.975
-    
-    Convert 15.975 to a fraction: 15.975=15 +0.975
-    
-    0.975=975/1000=39/40
-    
-    Therefore, 15.975=15 39/40=639/40
-    
-    Yes, that all checks out. So, the answer is 639/40 inches.
+    So 93/13 is in lowest terms because 93 divided by 13 is 7 with a remainder of 2. So yes, m=93, n=13, m + n=106. Therefore, the answer is 106.
     
     **Final Answer**
-    The distance between the two buildings on the map is \boxed{\dfrac{639}{40}} inches.
-     
+    \boxed{106}
     </think>
     
-    \boxed{\frac{639}{40}}
+    \boxed{106}
     ```
 
 **Incorrect** (1 shown)
@@ -76,19 +63,13 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
     ```text
     <think>
-    Okay, so Derek starts with a debt of π MIT, which is approximately 3.1416 MIT. Each day he flips a fair coin. If it's heads, he reduces his debt by 1 MIT, and if it's tails, he reduces it by 2 MITs. If his debt ever becomes nonpositive, he's debt-free. Otherwise, after the payment, his debt doubles. The question is asking for the probability that Derek eventually becomes debt-free. Since M
-    
-    ... [truncated 823 chars] ...
-    
-    : 2^1 -1 =1. P(1)=1/2. Hmm, not quite.
-    
-    Wait, maybe the numerator is (2^{n} - 1)/something. Let's see:
+    Okay, so Derek starts with a debt of π MIT, which is approximately 3.1416 MIT. Each day he flips a fair coin. If it's heads, he reduces his debt by 1 MIT, and if it's tails, he reduces it by 2 MITs. If his debt ever becomes nonpositive, he's debt-free. Otherwise, after the payment, his debt is doubled. We need to find the probability that Derek eventually becomes debt-free. 
     
-    For n=1: numerator 1, denominator 2. (2^1 -1)/2 = 1/2. That works.
+    Hmm, this see
     
-    For n=2: numerator 3, denominator 4. (2^2 -1)/2 = 3/2. Not matching.
+    ... [truncated 822 chars] ...
     
-    Wait, maybe another approach. Let's see the recursion. The recurrence is P(n) = (1/2) P(n-1) + (1/2) P(n-2). With P(0) =1, P(-1)=0. Wait, but if we have P(n) = (1/2) P(n-1) + (1/2) P(n-2), then the general solution would be P(n) = A (1/2)^n + B (1/2)^{n} * something. Wait, but with boundary conditions P(0) =1 and P(-1)=0. Wait, but actually, for n ≥1, P(n) = (1/2) P(n-1) + (1/2) P(n-2). So this is a linear recurrence with constant coefficients. The characteristic equation is r^2 - (1/2) r - (1/2) =0. Solving:
+    Similarly, for x = 2, P(2) = 0.5*P(2*(2 - 1)) + 0.5*P(2*(2 - 2)) = 0.5*P(2) + 0.5*P(0). Therefore, P(2) = 0.5*P(2) + 0.5*1. Solving for P(2), we get 0.5*P(2) = 0.5*1 => P(2) = 1. Similarly, for x = 3, P(3) = 0.5*P(2*(3 - 1)) + 0.5*P(2*(3 - 2)) = 0.5*P(4) + 0.5*P(2). But P(4) = 1, P(2) = 1, so P(3) = 0.5*1 + 0.5*1 = 1. So again, P(x) = 1. 
     
-    r = [1/2 ± sqrt(1/4 + 2)] / 2 = [1/2 ± sqrt(9/4)] / 2 = [1/2 ± 3/2]/2. So the roots are (1/2 + 3/2)/2 = (2)/2 =1, and (1/2 - 3/2)/2 = (-1)/2 = -1/2. Therefore, the general solution is P(n) = A(1)^n + B(-1/2)^n = A + B(-1/2)^n. Applying the boundary conditions: P(0) =1 = A + B. And for n ≥1, P(n) = (1/2) P(n-1) + (1/2) P(n-2). Let's check for n=1: P(1) = (1/2) P(0) + (1/2) P(-1). But P(-1)=0, so P(1) = (1/2)(1) + 0 = 1/2. Similarly, for n=
+    Therefore, it seems that P(x) = 1 for all x >= 0. Therefore, the probability that Derek eventually becomes debt-free is 1. But that seems surprising. Let me test with a smaller example. Suppose Derek starts with a debt of 1. Then, regardless of whether he pays 1 or 2, he becomes debt-free. So P(1) = 1. If he starts with a debt of 2, then he can pay 1, leading to 1, which is debt-free, or pay 2, leading to 0, which is debt-free. So again, P(2) = 1. If he starts with a debt of 3, he can pay 1, leading to 2, which is debt-free (since after paying 1, the debt is 2, which is positive, so it's doubled to 4. Then, he pays 1 again, leading to 3, which is debt-free (after paying 1, it's 3, which is positive, so doubled to 6. Then, he pays 1 again, leading to 5, doubled to 10. This seems like an infinite loop? Wait, but according to our previous reasoning
     ```