Update Automated MNLP evaluation report (2026-06-08)

EVAL_REPORT.md

@@ -2,22 +2,22 @@
 
 - **Model repo:** [`cs-552-2026-thinking-tokens/math_model`](https://huggingface.co/cs-552-2026-thinking-tokens/math_model)
 - **Owner(s):** group **thinking-tokens**
-- **Generated at:** 2026-06-06T17:20:53+00:00 (UTC)
+- **Generated at:** 2026-06-08T04:40:54+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:** [`49dec59`](https://huggingface.co/cs-552-2026-thinking-tokens/math_model/commit/49dec590cc507292c1b49514d50cffe3835ae4ac)
+- **Commit:** [`b808807`](https://huggingface.co/cs-552-2026-thinking-tokens/math_model/commit/b80880717f0a37b231da02f24ae9fa34b70747b9)
 - **Message:** Upload folder using huggingface_hub
-- **Committed:** 2026-05-20T12:47:18+00:00
+- **Committed:** 2026-06-06T17:57:58+00:00
 
 ## Summary
 
 | Benchmark | Accuracy | Status |
 |---|---:|---|
-| Math | 0.3700 | ok |
+| Math | 0.2900 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -30,179 +30,42 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `\frac{639}{40}`
+- **reference**: `4`
 - **overall** (8/8 completions correct)
-- **extracted** (✓): `\frac{639}{40}`
+- **extracted** (✓): `4`
 - **completion**:
 
     ```text
-    We are given:
-    
-    - A map scale: $ 3 $ inches on the map represent $ 10 $ miles in real life.
-    - Actual distance between two buildings: $ 53.25 $ miles.
-    - We are to find the map distance in inches.
-    
-    ---
-    
-    ### Step 1: Determine the scale factor
-    
-    We know that:
-    
-    $$
-    3 \text{ inches} = 10 \text{ miles}
-    $$
-    
-    We want to find how many inches represent $ 1 $ mile.
-    
-    $$
-    \text{Scale factor} = \frac{3}{10} \text{ inches per mile}
-    $$
-    
-    ---
-    
-    ### Step 2: Use the scale factor to find the map distance
-    
-    We are given:
-    
-    $$
-    53.25 \text{ miles} \times \frac{3}{10} \text{ inches/mile} = ?
-    $$
-    
-    $$
-    \text{Map distance} = 53.25 \times \frac{3}{10}
-    $$
-    
+    <think>
+    Solution 1
+    We rewrite the equation as
     $$
-    \text{Map distance} = \frac{53.25 \times 3}{10}
+    10^x=10+9990=10000
     $$
-    
-    $$
-    \text{Map distance} = \frac{159.75}{10}
-    $$
-    
+    Since $10^4=10000$, we have $x=4$.
+    Solution 2
+    We rewrite the equation as
     $$
-    \text{Map distance} = 15.975 \text{ inches}
+    10^x=10+9990=10+999 \cdot 10=10(1+999)=10000
     $$
+    Since $10^4=10000$, we have $x=4$.
+    </think>
     
-    ---
-    
-    ### Step 3: Convert to a fraction
-    
-    We now convert $ 15.975 $ into a fraction.
-    
-    $$
-    15.975 = 15 + 0.975
-    $$
-    
-    $$
-    0.975 = \frac{975}{1000}
-    $$
-    
-    $$
-    \frac{975}{1000} = \frac{39}{40} \quad \text{(simplify by dividing numerator and denominator by 25)}
-    $$
-    
-    So:
-    
-    $$
-    15.975 = 15 + \frac{39}{40} = \frac{600}{40} + \frac{39}{40} = \frac{639}{40}
-    $$
-    
-    ---
-    
-    ### Final Answer:
-    
-    $$
-    \boxed{\frac{639}{40}} \text{ inches}
-    $$
+    Therefore, the final answer is \boxed{4}.
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `279`
+- **reference**: `1305`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `27`
+- **extracted** (✗): `18`
 - **completion**:
 
     ```text
-    We are given a function $ f(n) $ defined as follows:
-    
-    - For a positive integer $ n < 1000 $, write $ n $ in base 10 as a string of digits.
-    - Let $ b $ be the smallest integer strictly larger than **every digit** in $ n $.
-    - Interpret the digit string of $ n $ as a number in base $ b $, and compute its value as $ f(n) $.
-    
-    We are to find how many such $ n < 1000 $ satisfy $ f(n) = n $.
-    
-    ---
-    
-    ### Ste
-    
-    ... [truncated 823 chars] ...
-    
-    1 \cdot 2^0 = 4 + 2 + 1 = 7 \neq 111 $: ❌
-    
-    Try $ n = 199 $:
-    
-    - $ b = 10 $
-    - $ f(199) = 1 \cdot 10^2 + 9 \cdot 10^1 + 9 \cdot 10^0 = 100 + 90 + 9 = 199 $: ✅
-    
-    So $ n = 199 $ works.
-    
-    Try $ n = 299 $:
-    
-    - $ f(299) = 2 \cdot 100 + 9 \cdot 10 + 9 = 299 $: ✅
-    
-    Try $ n = 399 $:
-    
-    - $ f(399) = 3 \cdot 100 + 9 \cdot 10 + 9 = 399 $: ✅
-    
-    Try $ n = 499 $:
-    
-    - $ f(499) = 4 \cdot 100 + 9 \cdot 10 + 9 = 499 $: ✅
-    
-    Try $ n = 599 $:
-    
-    - $ f(599) = 5 \cdot 100 + 9 \cdot 10 + 9 = 599 $: ✅
-    
-    Try $ n = 699 $:
+    <think>
+    The number of divisors of a positive integer $n$ is smallest when it is a product of the first few primes. We will try small prime factors first.
+    The smallest imbalanced multiple of $2000$ is $2000 \times 2 = 4000$, which has divisors $1, 2, 4, 5, 8, 10, 16, 20, 25, 40, 50, 80, 100, 200, 250, 400, 500, 1000, 2000$. This number has 18 divisors.
+    </think>
     
-    - $ f(699) = 6 \cdot 100 + 9 \cdot 10 + 9 = 699 $: ✅
-    
-    Try $ n = 799 $:
-    
-    - $ f(799) = 7 \cdot 100 + 9 \cdot 10 + 9 = 799 $: ✅
-    
-    Try $ n = 899 $:
-    
-    - $ f(899) = 8 \cdot 100 + 9 \cdot 10 + 9 = 899 $: ✅
-    
-    Try $ n = 999 $:
-    
-    - $ f(999) = 9 \cdot 100 + 9 \cdot 10 + 9 = 999 $: ✅
-    
-    So all three-digit numbers where the last two digits are 99 (i.e., 199, 299, ..., 999) satisfy $ f(n) = n $.
-    
-    That's 9 numbers.
-    
-    ---
-    
-    ### Step 7: Count all valid numbers
-    
-    - **Single-digit numbers**: 1 to 9 → 9 numbers
-    - **Two-digit numbers**: 19, 29, ..., 99 → 9 numbers
-    - **Three-digit numbers**: 199, 299, ..., 999 → 9 numbers
-    
-    So total number of such $ n < 1000 $ is:
-    
-    $$
-    \boxed{9 + 9 + 9 = 27}
-    $$
-    
-    ---
-    
-    ### ✅ Final Answer:
-    
-    $$
-    \boxed{27}
-    $$
+    Therefore, the final answer is \boxed{18}.
     ```