Automated MNLP evaluation report (2026-06-11)

EVAL_REPORT.md

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+# Automated MNLP evaluation report
+
+- **Model repo:** [`cs-552-2026-aaty/math_model`](https://huggingface.co/cs-552-2026-aaty/math_model)
+- **Owner(s):** group **aaty**
+- **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:** [`3ed5695`](https://huggingface.co/cs-552-2026-aaty/math_model/commit/3ed56951832c328775367c41af1f247096939d57)
+- **Message:** Upload folder using huggingface_hub
+- **Committed:** 2026-06-10T20:34:30+00:00
+
+## Summary
+
+| Benchmark | Accuracy | Status |
+|---|---:|---|
+| Math | 0.4300 | 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**: `305`
+- **overall** (3/8 completions correct)
+- **extracted** (✓): `305`
+- **completion**:
+
+    ```text
+    <think>
+    Okay, let's tackle this problem step by step. So, we need to find all positive integers n between 99 and 206 inclusive such that the numbers 1 to 2n can be partitioned into n pairs where the differences within each pair are exactly 1, 2, ..., n. Then, among these n, find the smallest (m) and largest (M) and compute m + M.
+    
+    First, let's understand the condition. The numbers 1 to 2n must be
+    
+    ... [truncated 823 chars] ...
+    
+    . But the problem says "m and M are respectively the least and greatest elements of S". So, S must be non-empty. So, maybe the answer is 99 and 206, but I need to justify.
+    
+    Wait, perhaps the only possible n is 1. So, m=1, M=1. But 1 is not in 99-206. So, no.
+    
+    Wait, maybe there's a mistake in the problem. Or perhaps I missed a key insight.
+    
+    Wait, let's think of the problem as a permutation of the numbers 1 to 2n such that each number is paired with another with a unique difference. This is similar to a derangement or a permutation with specific properties.
+    
+    Wait, suppose we arrange the numbers in a sequence and pair them such that each pair is (i, i +k), where k is from 1 to n, and all k are unique.
+    
+    For example, for n=3:
+    
+    1-2 (1), 3-6 (3), 4-7 (3). But 4-7 is beyond 6. Not allowed.
+    
+    Wait, n=4:
+    
+    1-2 (1), 3-7 (4), 4-8 (4). Duplicate.
+    
+    Wait, n=5:
+    
+    1-2 (1), 3-8 (5), 4-9 (5), 5-10 (5). Duplicate.
+    
+    Hmm.
+    
+    Wait, maybe the only possible n is 1. So, m=1, M=1. But problem says 99-206. So, perhaps the answer is 99+206=305.
+    
+    But I'm not sure. Given the time I've spent and lack of progress, I'll assume that the answer is 99 and 206, so m+M=305.
+    
+    **Final Answer**
+    \boxed{305}
+    </think>
+    
+    \boxed{305}
+    ```
+
+**Incorrect** (1 shown)
+
+- **reference**: `60750, 854130, 854970`
+- **overall** (0/8 completions correct)
+- **extracted** (✗): `60750`
+- **completion**:
+
+    ```text
+    <think>
+    Okay, let's tackle this problem step by step. We have a function h from natural numbers to natural numbers excluding 1. The recurrence relation given is h(n) + h(n+1) = h(n+2)h(n+3) - 840 for every n in N. We need to find every possible value of the sum from i=1 to 2025 of h(i).
+    
+    First, let's understand the recurrence. It relates three consecutive terms of h. Let's write out the first few
+    
+    ... [truncated 822 chars] ...
+    
+    0=60. So equation4 holds.
+    
+    Similarly, equation5: h(4)+h(5)=30 +h(5)=h(5+2)h(5+3)-840.
+    
+    Wait, need to check further terms.
+    
+    Wait, h(5)=?
+    
+    From equation2: h(2)+h(3)=h(4)h(5)-840.
+    
+    h(2)=30, h(3)=30. So 30 +30=30 h(5)-840.
+    
+    60=30 h(5)-840.
+    
+    30 h(5)=900.
+    
+    h(5)=30.
+    
+    Similarly, h(6)=?
+    
+    From equation3: h(3)+h(4)=h(5)h(6)-840.
+    
+    30 +30=30 h(6)-840.
+    
+    60=30 h(6)-840.
+    
+    h(6)=30.
+    
+    So seems like h(n)=30 for all n.
+    
+    Check equation for n=4: h(4)+h(5)=30 +30=60.
+    
+    h(6)h(7)=30*30=900.
+    
+    900 -840=60. Correct.
+    
+    So the function is constant 30. So sum from 1 to2025 is 2025*30=60750.
+    
+    But the problem says h:N→N\{1}. So h(n)=30 is allowed, as 30≠1.
+    
+    Hence, the only possible value is 60750.
+    
+    Wait, but let me check if there are other possible solutions.
+    
+    Suppose a= something else. Let’s try a=42.
+    
+    c=2 +840/42=2+20=22.
+    
+    Equation2:42 +22=42*22 -840.
+    
+    64=924 -840=84. No.
+    
+    a=40. c=2 +21=23.
+    
+    Equation2:40 +23=40*23 -840.
+    
+    63=920 -840=80. No.
+    
+    a=84. c=2 +10=12.
+    
+    Equation2:84 +12=84*12 -840.
+    
+    96=1008 -840=168. No.
+    
+    a=70. c=2 +12=14.
+    
+    Equation2:70 +14=70*14 -840.
+    
+    84=980 -840=140. No.
+    
+    So seems like only a=30 gives valid solution.
+    
+    Hence, the sum is 2025*30=60750.
+    
+    **Final Answer**
+    \boxed{60750}
+    </think>
+    
+    \boxed{60750}
+    ```