Automated MNLP evaluation report (2026-06-11)

EVAL_REPORT.md

@@ -2,16 +2,22 @@
 
 - **Model repo:** [`cs-552-2026-centralesupechec/math_model`](https://huggingface.co/cs-552-2026-centralesupechec/math_model)
 - **Owner(s):** group **centralesupechec**
-- **Generated at:** 2026-05-13T04:17:52+00:00 (UTC)
+- **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:** [`0e6c355`](https://huggingface.co/cs-552-2026-centralesupechec/math_model/commit/0e6c355e928e4fa6ac64c544927563c5608f0dca)
+- **Message:** SFT-1600
+- **Committed:** 2026-06-09T10:55:18+00:00
+
 ## Summary
 
 | Benchmark | Accuracy | Status |
 |---|---:|---|
-| Math | 0.1700 | ok |
+| Math | 0.4800 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -24,107 +30,157 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `2`
-- **overall** (3/8 completions correct)
-- **extracted** (✓): `2`
+- **reference**: `1012`
+- **overall** (4/8 completions correct)
+- **extracted** (✓): `1012`
 - **completion**:
 
     ```text
     <think>
-    Okay, so I need to solve the equation $3^{2x} + 19 = 10^x$. Let me think about how to approach this. 
+    Okay, so I need to find the maximum possible value of this sum:
     
-    First, let me rewrite the equation to see if I can simplify it or recognize any patterns. The left side has $3^{2x}$, which is the same as $(3^2)^x = 9^x$. So, the equation becomes $9^x + 19 = 10^x$. 
+    $$
+    \sum_{i=1}^{2023} \frac{a_i - a_{i-1}}{a_{i+1} - a_i}
+    $$
     
-    Hmm, so now it's $9^x + 19 = 10^x$. I don't see an obvious way to combine these terms,
+    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 $a_0 = 0$.
     
-    ... [truncated 821 chars] ...
+    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,
     
-    9^x + 19 - 10^x
-    $$
+    ... [truncated 823 chars] ...
     
-    We seek the value(s) of $x$ for which $f(x) = 0$.
+    possible value satisfying the constraint.
     
-    ---
+    So, to maximize each term $ \frac{d_i}{d_{i+1}} $, we set $ d_{i+1} = a_i $, which gives:
+    $$
+    \frac{d_i}{d_{i+1}} = \frac{d_i}{a_i}
+    $$
     
-    ### Step 2: Test Integer Values
+    But since $ d_i = a_i - a_{i-1} $, and $ a_i = \sum_{j=1}^i d_j $, we can write:
+    $$
+    d_i = a_i - a_{i-1} = a_i - a_{i-1}
+    $$
+    So, we can write:
+    $$
+    \frac{d_i}{a_i} = \frac{a_i - a_{i-1}}{a_i}
+    $$
     
-    Let’s test some integer values of $x$:
+    But we can also express this in terms of powers of 2:
+    - If we set $ a_{i+1} = 2a_i $, then $ d_{i+1} = a_i $
+    - Then, $ a_i = 2^{i-1} a_1 $, and $ a_{i-1} = 2^{i-2} a_1 $
+    - So, $ d_i = a_i - a_{i-1} = 2^{i-1}a_1 - 2^{i-2}a_1 = 2^{i-2}a_1 $
     
-    - **At $x = 2$:**
+    Therefore, each term becomes:
+    $$
+    \frac{d_i}{d_{i+1}} = \frac{2^{i-2}a_1}{a_i} = \frac{2^{i-2}a_1}{2^{i-1}a_1} = \frac{1}{2}
+    $$
     
+    But the first term is:
     $$
-    f(2) = 9^2 + 19 - 10^2 = 81 + 19 - 100 = 100 - 100 = 0
+    \frac{d_1}{d_2} = \frac{a_1}{a_1} = 1
     $$
     
-    So, $x = 2$ is a solution.
+    So the sum becomes:
+    $$
+    \sum_{i=1}^{2023} \frac{d_i}{d_{i+1}} = 1 + \sum_{i=2}^{2023} \frac{1}{2} = 1 + \frac{2022}{2} = 1 + 1011 = 1012
+    $$
     
     ---
     
-    ### Step 3: Analyze the Behavior of the Function
+    ### Step 3: General Pattern
+    
+    From the above, we observe that the maximum sum is:
+    $$
+    \text{Max sum} = 1 + \frac{n - 2}{2} = \frac{n}{2}
+    $$
+    where $ n = 2024 $, so:
+    $$
+    \text{Max sum} = \frac{2024}{2} = 1012
+    $$
+    
+    ---
     
-    Define the function:
+    ### Final Answer
     
     $$
-    f(x) = 9^x + 19 - 10^x
+    \boxed{1012}
     $$
+    ```
+
+**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
     
-    As $x \to \infty$, the exponential term $10^x$ dominates, so $f(x) \to -\infty$.
+    ... [truncated 823 chars] ...
     
-    As $x \to -\infty$, both $9^x$ and $10^x$ approach 0, so $f(x) \to 19$.
+    contain any element with exponent 2 for 5: $ 2^{10} $ (5 choices for 3, 2 for 5 ≤ 1).
+    - Subsets that **do not** contain any element with exponent 4 for 3 **and** any element with exponent 2 for 5: $ 2^8 $ (4 choices for 3 ≤ 3, 2 for 5 ≤ 1).
     
-    Now, compute the derivative:
+    Using the inclusion-exclusion principle:
     
     $$
-    f'(x) = \ln(9) \cdot 9^x - \ln(10) \cdot 10^x
+    N = 2^{15} - 2^{12} - 2^{10} + 2^8
     $$
     
-    This derivative is always negative for all $x$, since:
+    Now compute the values:
+    
+    - $ 2^{15} = 32768 $
+    - $ 2^{12} = 4096 $
+    - $ 2^{10} = 1024 $
+    - $ 2^8 = 256 $
     
-    - $\ln(9) \approx 2.1972$
-    - $\ln(10) \approx 2.3026$
-    - $9^x$ grows slower than $10^x$, and the coefficient of $9^x$ is smaller than that of $10^x$
+    So:
     
-    Thus, $f'(x) < 0$ for all $x$, meaning the function is strictly decreasing.
+    $$
+    N = 32768 - 4096 - 1024 + 256 = 27904
+    $$
     
     ---
     
-    ### Step 4: Conclude the Number of Solutions
+    ### Step 3: Total Number of Non-Empty Subsets
     
-    Since $f(x)$ is strictly decreasing and continuous, and it goes from $19$ at $-\infty$ to $-\infty$ at $+\infty$, it can cross the x-axis **exactly once**.
+    The total number of non-empty subsets of $ A $ is:
     
-    We already found that $f(2) = 0$, so this is the **only** solution.
+    $$
+    2^{15} - 1 = 32768 - 1 = 32767
+    $$
     
     ---
     
-    ### Final Answer
+    ### Step 4: Compute the Probability
+    
+    The probability that a randomly chosen non-empty subset $ B $ of $ A $ has LCM 2025 is:
     
     $$
-    \boxed{2}
+    \frac{27904}{32767}
     $$
-    ```
-
-**Incorrect** (1 shown)
-
-- **reference**: `$(n-2)2^n +1$`
-- **overall** (0/8 completions correct)
-- **extracted** (✗): `<no answer>`
-- **completion**:
-
-    ```text
-    <think>
-    Okay, so I need to figure out the greatest positive integer that can't be expressed as a sum of elements from the set C_n, where C_n is defined as {2^n - 2^i | 0 ≤ i < n}. And n is a positive integer greater than or equal to 2. Hmm, let's start by understanding what exactly the set C_n looks like.
     
-    First, let's take a specific example. Let's say n = 2. Then C_2 would be {2^2 - 2^0, 2^2 - 2
+    Now, we simplify the fraction. To do this, we compute the GCD of 27904 and 32767.
     
-    ... [truncated 822 chars] ...
+    Using the Euclidean algorithm:
     
-    3 = 24, okay. So 21 is the largest non-representable number for n=4? But that seems high. Maybe I made a mistake here.
+    $$
+    \gcd(27904, 32767) = 1
+    $$
     
-    Wait, let's check 21 again. How can we get 21? The elements are 8, 12, 14, 15. Let me think. 15 + 8 = 23. 14 + 8 = 22. 12 + 12 = 24. 12 + 14 = 26. 15 + 12 = 27. 15 + 14 = 29. 8 + 14 = 22. 8 + 15 = 23. 12 + 15 = 27. So 21 is not in the list. So 21 can't be formed. 22 is 8 + 14. 23 is 8 + 15. 24 is 8 + 8 + 8. 25: 12 + 13? No. 14 + 11? No. 15 + 10? No. 8 + 12 + 5? No. 8 + 14 + 3? No. So 25 can't be formed? Wait, but 25 is 8 + 8 + 9? No. 12 + 12 + 1? No. 14 + 11? No. 15 + 10? No. So 25 can't be formed. So 25 is not representable.
+    Thus, the fraction is already in its simplest form.
     
-    Wait, so 21, 25, 26? Wait, 26 is 12 + 14. So 26 is okay. So 25 is not representable. Then 26 is okay. So 25 is not representable. Then 27 is 15 + 12. So 27 is okay. So 25 is not representable. So 25 is larger than 21. So maybe the largest non-representable number is 25?
+    ---
     
-    Wait, but how do we get 25? Let me think again. 8, 12, 14, 15. Let me try combinations:
+    ### Step 5: Final Answer
     
-    25 - 15 = 10. 10 can't be formed. 25 - 14 = 11. 11 can't be formed. 25 - 12 = 13. 13 can't be formed. 25 - 8 = 17. 17 can't be formed. So none of these combinations work. So 25 can't be
+    We are asked to find $ m + n $, where the fraction $ \frac{m}{n} = \frac{27904}{32767} $ is in lowest terms.
+    
+    $$
+    m = 27904, \quad n = 32767 \Rightarrow m + n = 27904 + 32767 = \boxed{60671}
+    $$
     ```