Automated MNLP evaluation report (2026-06-11)

EVAL_REPORT.md

@@ -0,0 +1,133 @@
+# Automated MNLP evaluation report
+
+- **Model repo:** [`cs-552-2026-mnlplus/math_model`](https://huggingface.co/cs-552-2026-mnlplus/math_model)
+- **Owner(s):** group **mnlplus**
+- **Generated at:** 2026-06-11T06:23:10+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._
+
+## Evaluated checkpoint
+
+- **Commit:** [`addcaa1`](https://huggingface.co/cs-552-2026-mnlplus/math_model/commit/addcaa1f93098b72139069143e727bbdb8c02a80)
+- **Message:** Upload Qwen3ForCausalLM
+- **Committed:** 2026-06-09T10:28:08+00:00
+
+## Summary
+
+| Benchmark | Accuracy | Status |
+|---|---:|---|
+| Math | 0.4600 | ok |
+| Knowledge | — | not run |
+| Multilingual | — | not run |
+| Safety | — | not run |
+
+## Sample completions
+
+_Prompts are intentionally omitted to avoid revealing benchmark contents. For multi-completion problems, only one completion is shown per sample._
+
+### Math
+
+**Correct** (1 shown)
+
+- **reference**: `4`
+- **overall** (2/8 completions correct)
+- **extracted** (✓): `4`
+- **completion**:
+
+    ```text
+    <think>
+    Okay, so I need to figure out the largest positive integer \( n \) such that there exist distinct real numbers \( a_1, a_2, \ldots, a_n \) each between 0 and 1, and for every pair \( i \neq j \), the equation \( a_i^{[1/a_j]} = a_j^{[1/a_i]} \) holds. The brackets here represent the floor function, which is the greatest integer less than or equal to the number inside. 
+    
+    First, let me make
+    
+    ... [truncated 823 chars] ...
+    
+    d,e): e = d^k
+    
+    Therefore, from (c,e) and (d,e):
+    
+    2^{-k/2} = d^k ⇒ d = 2^{-k/2k} = 2^{-1/2} = 1/√2
+    
+    Therefore, d=1/√2. But this is already c. Not good. Therefore, no solution for d. Hence, n=5 is impossible. 
+    
+    Therefore, the maximum n is 4. 
+    
+    But let's check if there's another set of numbers for n=4. Suppose we have a=1/2, b=1/4, c=1/√2, and d= something else. Let's define d such that all pairs satisfy the equation. For example:
+    
+    For pair (a,d): (1/2)^k = d^2
+    
+    For pair (b,d): (1/4)^k = d^4
+    
+    For pair (c,d): (1/(2√2))^k = d^1
+    
+    For pair (d,d): Not applicable since d must be distinct.
+    
+    From (a,d) and (b,d):
+    
+    From (a,d): d = 2^{-k/2}
+    
+    From (b,d): d = 4^{-k/4} = 2^{-k/2}
+    
+    From (c,d): d = (1/(2√2))^k = 2^{-3k/2}
+    
+    Therefore, 2^{-k/2} = 2^{-3k/2} ⇒ -k/2 = -3k/2 ⇒ k=0, which is invalid. Therefore, no solution for d. Hence, n=4 is impossible with this approach. 
+    
+    Therefore, the only way to have n=4 is through the previous example where d=1/√2, which is part of the existing set. Hence, n=4 is possible with a=1/2, b=1/4, c=1/(2√2), d=1/√2. 
+    
+    Therefore, after this lengthy exploration, the largest n is 4. 
+    
+    **Final Answer**
+    The largest positive integer \( n \) is \(\boxed{4}\).
+    </think>
+    
+    \boxed{4}
+    ```
+
+**Incorrect** (1 shown)
+
+- **reference**: `237`
+- **overall** (0/8 completions correct)
+- **extracted** (✗): `60671`
+- **completion**:
+
+    ```text
+    <think>
+    Alright, let's tackle this problem step by step. So, we need to find the probability that a randomly chosen non-empty subset \( B \) of the positive divisors of 2025 has a least common multiple (LCM) of 2025. Then, express this probability as a reduced fraction \( m/n \) and find \( m + n \). 
+    
+    First, let's recall that the number of positive divisors of a number can be determined from its
+    
+    ... [truncated 823 chars] ...
+    
+    l to the total number of subsets minus the number of subsets with no b_i=2. However, the intersection is not straightforward. 
+    
+    Wait, but in our previous calculation, we used inclusion-exclusion to subtract the subsets that don't have a_i=4 or don't have b_i=2. That gives us the correct count. Let's verify this with another method.
+    
+    Another way to compute |T ∩ U| is:
+    
+    |T ∩ U| = Total subsets - subsets without T - subsets without U + subsets without T and without U
+    
+    Which is exactly what we did earlier. So, substituting the numbers:
+    
+    Total subsets: 2^15 = 32768
+    
+    Subsets without T: subsets without any a_i=4: 2^12 = 4096
+    
+    Subsets without U: subsets without any b_i=2: 2^10 = 1024
+    
+    Subsets without T and U: subsets without any a_i=4 and without any b_i=2: 2^8 = 256
+    
+    Therefore, |T ∩ U| = 32768 - 4096 - 1024 + 256 = 27904
+    
+    Thus, confirming our previous result. Therefore, N = 27904.
+    
+    Therefore, the probability is 27904 / (2^15 - 1) = 27904 / 32767. Since 27904 and 32767 are coprime (as shown by the GCD calculation), the fraction is already reduced. Hence, m = 27904 and n = 32767, so m + n = 60671.
+    
+    Therefore, the final answer is 60671.
+    
+    **Final Answer**
+    \boxed{60671}
+    </think>
+    
+    \boxed{60671}
+    ```