Automated MNLP evaluation report (2026-06-11)

EVAL_REPORT.md

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+# Automated MNLP evaluation report
+
+- **Model repo:** [`cs-552-2026-OAAA/math_model`](https://huggingface.co/cs-552-2026-OAAA/math_model)
+- **Owner(s):** group **OAAA**
+- **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:** [`a7fc550`](https://huggingface.co/cs-552-2026-OAAA/math_model/commit/a7fc55016982451246691a3530624ef17acea9a2)
+- **Message:** Upload generation_config.json with huggingface_hub
+- **Committed:** 2026-06-07T14:19:42+00:00
+
+## Summary
+
+| Benchmark | Accuracy | Status |
+|---|---:|---|
+| Math | 0.4800 | 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**: `1012`
+- **overall** (6/8 completions correct)
+- **extracted** (✓): `1012`
+- **completion**:
+
+    ```text
+    <think>
+    Okay, so I need to find the maximum possible value of this sum:
+    
+    $$
+    \sum_{i=1}^{2023} \frac{a_i - a_{i-1}}{a_{i+1} - a_i}
+    $$
+    
+    given that $a_1, \ldots, a_{2024}$ are positive real numbers with $a_{i+1} \ge 2a_i$ for each $i = 1, \ldots, 2023$, and $a_0 = 0$. 
+    
+    First, let me try to understand the problem. We have a sequence starting at 0, and each term is at least twice the previous one. So,
+    
+    ... [truncated 822 chars] ...
+    
+    4a, \quad \ldots, \quad a_{2024} = 2^{2023}a
+    $$
+    
+    This is a geometric progression with ratio 2.
+    
+    ---
+    
+    ### Step 3: Computing the Sum in the Optimal Case
+    
+    Let’s compute the sum when $ a_{i+1} = 2a_i $ for all $ i $, which gives:
+    
+    $$
+    \frac{a_i - a_{i-1}}{a_{i+1} - a_i} = \frac{a_i - a_{i-1}}{a_i} = 1 - \frac{a_{i-1}}{a_i}
+    $$
+    
+    So the entire sum becomes:
+    
+    $$
+    \sum_{i=1}^{2023} \left(1 - \frac{a_{i-1}}{a_i} \right) = 2023 - \sum_{i=1}^{2023} \frac{a_{i-1}}{a_i}
+    $$
+    
+    Now, compute the second term:
+    
+    $$
+    \sum_{i=1}^{2023} \frac{a_{i-1}}{a_i} = \frac{a_0}{a_1} + \frac{a_1}{a_2} + \frac{a_2}{a_3} + \cdots + \frac{a_{2023}}{a_{2024}}
+    $$
+    
+    Since $ a_{i} = 2^{i-1}a_1 $, we have:
+    
+    $$
+    \frac{a_{i-1}}{a_i} = \frac{2^{i-2}a_1}{2^{i-1}a_1} = \frac{1}{2}
+    $$
+    
+    This holds for all $ i \ge 2 $, and the first term is $ \frac{a_0}{a_1} = 0 $.
+    
+    So the sum becomes:
+    
+    $$
+    \sum_{i=1}^{2023} \frac{a_{i-1}}{a_i} = 0 + 2022 \cdot \frac{1}{2} = 1011
+    $$
+    
+    Thus, the total sum is:
+    
+    $$
+    2023 - 1011 = \boxed{1012}
+    $$
+    
+    ---
+    
+    ### Final Conclusion
+    
+    The maximum possible value of the given sum is achieved when the sequence grows exponentially with a ratio of 2 at each step. In this optimal case, the sum evaluates to:
+    
+    $$
+    \boxed{1012}
+    $$
+    ```
+
+**Incorrect** (1 shown)
+
+- **reference**: `237`
+- **overall** (0/8 completions correct)
+- **extracted** (✗): `60671`
+- **completion**:
+
+    ```text
+    <think>
+    Okay, so I need to solve this problem where I have to find the probability that a randomly chosen non-empty subset B of the positive divisors of 2025 has a least common multiple (LCM) equal to 2025. Then express that probability as a reduced fraction m/n and find m + n. Alright, let's start by understanding the problem step by step.
+    
+    First, let me recall that the divisors of a number can b
+    
+    ... [truncated 822 chars] ...
+    
+    least one** element from the set of elements with $ a = 4 $ **and** at least one from the set with $ b = 2 $.
+    
+    Let’s denote:
+    
+    - $ N $: the number of subsets of $ A $ that contain **at least one** element with $ a = 4 $ and **at least one** element with $ b = 2 $
+    - The number of such subsets is given by the inclusion-exclusion principle:
+    
+    $$
+    N = 2^{15} - 2^{12} - 2^{10} + 2^8
+    $$
+    
+    $$
+    N = 32768 - 4096 - 1024 + 256 = 27904
+    $$
+    
+    This counts **all subsets** (including the empty set) that contain at least one element with $ a = 4 $ and at least one with $ b = 2 $. Since the **empty set** is not in this set, the number of **non-empty** subsets is:
+    
+    $$
+    N = 27904
+    $$
+    
+    ---
+    
+    ### Step 4: Compute the Probability
+    
+    The probability is:
+    
+    $$
+    \frac{27904}{32767}
+    $$
+    
+    We now check whether this fraction is in its lowest terms.
+    
+    - $ 32767 = 7 \cdot 31 \cdot 151 $
+    - $ 27904 = 2^8 \cdot 109 $
+    
+    Since 109 is a prime number and does **not** divide 32767, and 2 is not a factor of 32767, the two numbers are **coprime**.
+    
+    Thus, the fraction is in its simplest form:
+    
+    $$
+    \frac{27904}{32767}
+    $$
+    
+    ---
+    
+    ### Final Step: Compute $ m + n $
+    
+    Here, $ m = 27904 $, $ n = 32767 $, so:
+    
+    $$
+    m + n = 27904 + 32767 = \boxed{60671}
+    $$
+    ```