Automated MNLP evaluation report (2026-05-27)

EVAL_REPORT.md

@@ -2,7 +2,7 @@
 
 - **Model repo:** [`cs-552-2026-4neurons/math_model`](https://huggingface.co/cs-552-2026-4neurons/math_model)
 - **Owner(s):** group **4neurons**
-- **Generated at:** 2026-05-18T02:30:25+00:00 (UTC)
+- **Generated at:** 2026-05-27T13:02:22+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 | 0.3100 | ok |
+| Math | 0.2700 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -24,72 +24,44 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `4`
-- **overall** (8/8 completions correct)
-- **extracted** (✓): `4`
+- **reference**: `305`
+- **overall** (1/8 completions correct)
+- **extracted** (✓): `305`
 - **completion**:
 
     ```text
     <think>
-    To solve for x in the equation 10^x − 10 = 9990, we need to isolate the term 10^x.
     
-    Add 10 to both sides of the equation:
-    \[ 10^x = 9990 + 10 \Rightarrow 10^x = 10000 \]
-    
-    Now, we can take the logarithm of both sides to solve for x. Since the base of the exponential term is 10, we can use the logarithm base 10.
+    </think>
     
-    Take the logarithm base 10 of both sides:
-    \[ \log_{10} 10^x = \log_{10} 10000 \Rightarrow x \log_{10} 10 = \log_{10} 10000 \]
+    Suppose $n$ is odd.  Then the pairs are $(1,2), (3,4), \ldots, (2n-1,2n)$.  The difference between the first two numbers in each pair is $1$, and the difference between the last two numbers in each pair is $1$.  The difference between the first and last number in the list is $2n-1$, which is odd.  The sum of the first $n-1$ numbers in the list is $(n-1)(n+1)/2$, which is odd.  T
     
-    Since log base 10 of 10 is 1, we have:
-    \[ x = \log_{10} 10000 \]
+    ... [truncated 304 chars] ...
     
-    Now, calculate the value of log base 10 of 10000:
-    \[ x = \log_{10} 10000 \approx 4 \]
+    rs in each pair is $1$, and the difference between the first and last number in the list is $1$, so this works.
     
-    So, the answer is:
-    \[ \boxed{4} \]
-    </think>
+    If $n$ is even, then the pairs are $(1,3), (2,4), \ldots, (n-1,n+1), (n,n+2)$.  The difference between the first two numbers in each pair is $2$, and the difference between the last two numbers in each pair is $2$.  The difference between the first and last number in the list is $2n-1$, which is odd.  The sum of the first $n-1$ numbers in the list is $(n-1)(n+1)/2$, which is odd.  Therefore, the difference between the first and last number in the list is odd.  This means that the difference between the first and last number in the list is $1$, and we can pair up the numbers as $(1,3), (2,4), \ldots, (n-1,n+1), (n,n+2)$, where the last number is $(n-1,n+2)$.  The difference between the numbers in each pair is $1$, and the difference between the first and last number in the list is $1$, so this works.
     
-    Therefore, the final answer is \boxed{4}.
+    Therefore, we can partition the numbers $1,2,\ldots,2n$ into $n$ pairs so that the differences within the pairs are precisely $1,2,\ldots,n$ for every $n$.  This means that $S = \{99,100, \ldots,206\}$, so $m = 99$ and $M = 206$.  Therefore, $m + M = 99 + 206 = \boxed{305}$.
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `2 - \frac{\pi}{2}`
+- **reference**: `91/6`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `1`
+- **extracted** (✗): `<no answer>`
 - **completion**:
 
     ```text
     <think>
-    The problem is asking for the probability that Derek eventually becomes debt-free, which means he never has a nonnegative debt after any number of days.
-    
-    Let's analyze the problem step-by-step:
-    
-    1. Derek starts with a debt of $\pi$ MIT.
-    2. Each day, he tosses a fair coin:
-       - If the coin lands on heads, his debt decreases by 1 MIT.
-       - If the coin lands on tails, his debt decreases by 2 MITs.
-    3. If his debt becomes nonpositive at any point, he is considered debt-free.
-    4. If his debt is still positive after the payment, it is doubled.
     
-    To find the probability that Derek eventually becomes debt-free, we can use the concept of expected value and recurrence relations.
-    
-    Let $P(x)$ be the probability that Derek becomes debt-free starting with a debt of $x$ MIT.
-    
-    We can write the recurrence relation as:
-    \[ P(x) = \frac{1}{2} P(x - 1) + \frac{1}{2} P(x - 2) \]
+    </think>
     
-    The base cases are:
-    \[ P(0) = 1 \quad \text{(if his debt is 0, he is debt-free)} \]
-    \[ P(x) = 0 \quad \text{if } x < 0 \quad \text{(if his debt is negative, he is debt-free)} \]
+    Let $E_r$ be the expected number of turns it takes to complete the ritual when Triton has $r$ marbles of each color. We will solve for $E_3$ by first solving for $E_2$.
     
-    We can use the recurrence relation and base cases to compute the values of $P(x)$ for different values of $x$.
+    Note that in order to complete the ritual, Triton must get one of his marbles to the bag, and then have exactly 4 marbles of one color in the bag. So, suppose he has $r$ marbles of each color. He must put one of
     
-    After computing the values, we find that:
-    \[ P(\pi) = \boxed{1} \]
-    </think>
+    ... [truncated 823 chars] ...
     
-    Therefore, the final answer is \boxed{1}.
+    frac{2}{1594323} + \frac{2}{4782969} + \frac{2}{4782969}E_{-14} \\ &= \frac{1321}{729} + \frac{3}{6561} + \frac{2}{19683} + \frac{2}{19683} + \frac{2}{59049} + \frac{2}{59049} + \frac{2}{177147} + \frac{2}{177147} + \frac{2}{531441} + \frac{2}{531441} + \frac{2}{1594323} + \frac{2}{1594323} + \frac{2}{4782969} + \frac{2}{4782969} + \frac{2}{14348907} + \frac{2}{14348907}E_{-15} \\ &= \frac{1321}{729} + \frac{3}{6561} + \frac{2}{19683} + \frac{2}{19683} + \frac{2}{59049} + \frac{2}{59049} + \frac{2}{177147} + \frac{2}{177147} + \frac{2}{531441} + \frac{2}{531441} + \frac{2}{1594323} + \frac{2}{1594323} + \frac{2}{4782969} + \frac{2}{4782969} + \frac{2}{14348907} + \frac{2}{14348907} + \frac{2}{428758907} + \frac{2}{428758907}E_{-16} \\ &= \frac{1321}{729} + \frac{3}{6561} + \frac{2}{19683} + \frac{2}{19683} + \frac{2}{59049} + \frac{2}{59049} + \frac{2}{177147} + \frac{2}{177147} + \frac{2}{531441} + \frac{2}{531441} + \frac{2}{1594323} + \frac{2}{1594323} + \frac{2}{4782969} + \frac{2}{4782969} + \frac{2}{14348907} + \frac{2}{14348907} + \frac{2}{428758907} + \frac{2}{428758907} + \frac{2}{12884901987} + \frac{2}{12884901987}E_{-17} \\ &= \frac{1321}{729} + \frac{3}{6561} + \frac{2
     ```