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

EVAL_REPORT.md

@@ -2,16 +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:** 2026-06-05T03:01:58+00:00 (UTC)
+- **Generated at:** 2026-06-06T17:20:53+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)
+- **Message:** Upload folder using huggingface_hub
+- **Committed:** 2026-05-20T12:47:18+00:00
+
 ## Summary
 
 | Benchmark | Accuracy | Status |
 |---|---:|---|
-| Math | 0.4000 | ok |
+| Math | 0.3700 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -24,141 +30,179 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `194`
-- **overall** (1/8 completions correct)
-- **extracted** (✓): `194`
+- **reference**: `\frac{639}{40}`
+- **overall** (8/8 completions correct)
+- **extracted** (✓): `\frac{639}{40}`
 - **completion**:
 
     ```text
-    We are given a $194 \times 194$ board and asked to find the **least positive integer $k$** such that we can choose $k$ cells to mark, and there exists **exactly one tiling** of the board with $1 \times 2$ and $2 \times 1$ dominoes, such that **every domino contains at most one marked cell**.
+    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: Understanding the Problem
+    ### Step 1: Determine the scale factor
     
-    We are working with a **perfect tiling** of the board using **
+    We know that:
     
-    ... [truncated 823 chars] ...
+    $$
+    3 \text{ inches} = 10 \text{ miles}
+    $$
     
-    t **every domino contains at most one marked cell**.
+    We want to find how many inches represent $ 1 $ mile.
     
-    This is a **combinatorial problem** where we want to **minimize the number of marked cells** such that **only one tiling** satisfies the condition.
+    $$
+    \text{Scale factor} = \frac{3}{10} \text{ inches per mile}
+    $$
     
     ---
     
-    ### Step 5: Use of Linear Algebra and Graph Theory
+    ### Step 2: Use the scale factor to find the map distance
+    
+    We are given:
     
-    This is a **graph problem** where the board is a graph, and dominoes are edges that connect two adjacent cells.
+    $$
+    53.25 \text{ miles} \times \frac{3}{10} \text{ inches/mile} = ?
+    $$
+    
+    $$
+    \text{Map distance} = 53.25 \times \frac{3}{10}
+    $$
     
-    We want to choose a **subset of vertices** (cells) such that **only one perfect matching** (tiling) exists in which **every edge (domino)** contains at most one vertex from the subset.
+    $$
+    \text{Map distance} = \frac{53.25 \times 3}{10}
+    $$
     
-    This is a **matching problem** with a **constraint** on the subset of vertices.
+    $$
+    \text{Map distance} = \frac{159.75}{10}
+    $$
     
-    We are to **minimize the size of the subset** such that **only one perfect matching** satisfies the condition.
+    $$
+    \text{Map distance} = 15.975 \text{ inches}
+    $$
     
     ---
     
-    ### Step 6: Final Answer
+    ### Step 3: Convert to a fraction
+    
+    We now convert $ 15.975 $ into a fraction.
     
-    This is a known result in combinatorics. For a $n \times n$ board, the **minimum number of marked cells** required to ensure that **only one tiling** of the board with $1 \times 2$ dominoes has **at most one marked cell per domino** is:
+    $$
+    15.975 = 15 + 0.975
+    $$
     
     $$
-    \boxed{194}
+    0.975 = \frac{975}{1000}
     $$
     
-    This is the **minimum** number of marked cells required to ensure that **only one tiling** satisfies the condition.
+    $$
+    \frac{975}{1000} = \frac{39}{40} \quad \text{(simplify by dividing numerator and denominator by 25)}
+    $$
+    
+    So:
+    
+    $$
+    15.975 = 15 + \frac{39}{40} = \frac{600}{40} + \frac{39}{40} = \frac{639}{40}
+    $$
     
     ---
     
-    ### ✅ Final Answer:
+    ### Final Answer:
     
     $$
-    \boxed{194}
+    \boxed{\frac{639}{40}} \text{ inches}
     $$
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `$g(x)=c, g(x)=\lceil x \rceil, g(x)=\lfloor x \rfloor$`
+- **reference**: `279`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `h(x) = c \text{ or } h(x) = x \text{ for some } c \in \mathbb{Z}`
+- **extracted** (✗): `27`
 - **completion**:
 
     ```text
-    We are given a functional equation involving a mapping $ h : \mathbb{Q} \to \mathbb{Z} $, and the condition:
+    We are given a function $ f(n) $ defined as follows:
     
-    $$
-    h(x) = h\left( \frac{h(bx - a) + a}{b} \right)
-    $$
+    - 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) $.
     
-    for all $ x \in \mathbb{Q} $, $ a \in \mathbb{Z} $, and $ b \in \mathbb{N} $.
+    We are to find how many such $ n < 1000 $ satisfy $ f(n) = n $.
     
     ---
     
-    ### Step 1: Understand the structure of the equation
+    ### Ste
     
-    Let’s rewrite the equation for clarity:
+    ... [truncated 823 chars] ...
     
-    $$
-    h(x) = h\left( \frac{h(bx - a) + a}{b} \right)
-    $$
+    1 \cdot 2^0 = 4 + 2 + 1 = 7 \neq 111 $: ❌
     
-    Thi
+    Try $ n = 199 $:
     
-    ... [truncated 822 chars] ...
+    - $ b = 10 $
+    - $ f(199) = 1 \cdot 10^2 + 9 \cdot 10^1 + 9 \cdot 10^0 = 100 + 90 + 9 = 199 $: ✅
     
-    - a)}{b} + m = - \frac{-m}{b} + m = \frac{m}{b} + m $
+    So $ n = 199 $ works.
     
-    So:
+    Try $ n = 299 $:
     
-    $$
-    m = \frac{m}{b} + m \Rightarrow 0 = \frac{m}{b} \Rightarrow m = 0
-    $$
+    - $ f(299) = 2 \cdot 100 + 9 \cdot 10 + 9 = 299 $: ✅
     
-    So $ h(x) = x $ is a solution.
+    Try $ n = 399 $:
     
-    Let’s verify:
+    - $ f(399) = 3 \cdot 100 + 9 \cdot 10 + 9 = 399 $: ✅
     
-    $$
-    h(x) = x \Rightarrow h(bx - a) = bx - a
-    $$
+    Try $ n = 499 $:
     
-    Then:
+    - $ f(499) = 4 \cdot 100 + 9 \cdot 10 + 9 = 499 $: ✅
     
-    $$
-    \frac{h(bx - a) + a}{b} = \frac{bx - a + a}{b} = x
-    \Rightarrow h(x) = x
-    $$
+    Try $ n = 599 $:
     
-    So $ h(x) = x $ is a solution.
+    - $ f(599) = 5 \cdot 100 + 9 \cdot 10 + 9 = 599 $: ✅
     
-    ---
+    Try $ n = 699 $:
     
-    ### Step 5: Are there other solutions?
+    - $ f(699) = 6 \cdot 100 + 9 \cdot 10 + 9 = 699 $: ✅
     
-    We have found two types of solutions:
+    Try $ n = 799 $:
     
-    1. **Constant functions**: $ h(x) = c $, for some $ c \in \mathbb{Z} $
-    2. **Linear functions**: $ h(x) = x $, which is also a solution
+    - $ f(799) = 7 \cdot 100 + 9 \cdot 10 + 9 = 799 $: ✅
     
-    Are there any other solutions?
+    Try $ n = 899 $:
     
-    Suppose $ h $ is not constant or linear. Let’s try $ h(x) = c $, a constant function, and $ h(x) = x $, and see if they are the **only** solutions.
+    - $ f(899) = 8 \cdot 100 + 9 \cdot 10 + 9 = 899 $: ✅
     
-    Suppose $ h $ is not constant, and not linear. Then we must show that it must be constant or linear.
+    Try $ n = 999 $:
     
-    But from the earlier analysis, **no non-constant linear functions** work, and **constant functions** and $ h(x) = x $ both work.
+    - $ f(999) = 9 \cdot 100 + 9 \cdot 10 + 9 = 999 $: ✅
     
-    So the only possible solutions are:
+    So all three-digit numbers where the last two digits are 99 (i.e., 199, 299, ..., 999) satisfy $ f(n) = n $.
     
-    - Constant functions: $ h(x) = c $, for some $ c \in \mathbb{Z} $
-    - $ h(x) = x $
+    That's 9 numbers.
     
     ---
     
-    ### Final Answer:
+    ### 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:
     
     $$
-    \boxed{h(x) = c \text{ or } h(x) = x \text{ for some } c \in \mathbb{Z}}
+    \boxed{27}
     $$
     ```