Automated MNLP evaluation report (2026-06-11)

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:** 2026-05-19T03:22:52+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:** [`b808807`](https://huggingface.co/cs-552-2026-thinking-tokens/math_model/commit/b80880717f0a37b231da02f24ae9fa34b70747b9)
+- **Message:** Upload folder using huggingface_hub
+- **Committed:** 2026-06-06T17:57:58+00:00
+
 ## Summary
 
 | Benchmark | Accuracy | Status |
 |---|---:|---|
-| Math | 0.3900 | ok |
+| Math | 0.2200 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -24,184 +30,31 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `194`
-- **overall** (2/8 completions correct)
-- **extracted** (✓): `194`
+- **reference**: `.35625`
+- **overall** (4/8 completions correct)
+- **extracted** (✓): `0.35625`
 - **completion**:
 
     ```text
-    We are given a $194 \times 194$ board, and we are to find the **least positive integer $k$** such that we can choose $k$ cells to mark, and there is **exactly one tiling** of the board by $1 \times 2$ and $2 \times 1$ dominoes in which **every domino contains at most one marked cell**.
-    
-    ---
-    
-    ### Step 1: Understand the problem
-    
-    We are to find the **smallest number of marked cells** such that:
-    
-    - Th
-    
-    ... [truncated 823 chars] ...
-    
-    t can be **involving marked cells** is $k$.
-    - Since there are 18818 dominoes, the number of dominoes that **do not involve marked cells** is $18818 - k$.
-    
-    We want **exactly one** tiling where **every domino has at most one marked cell**.
-    
-    This is equivalent to saying that the **marked cells** must be placed in such a way that **each domino is uniquely determined** by the marked cells.
-    
-    This is a very strong condition.
-    
-    ---
-    
-    ### Step 4: What is the minimum number of marked cells?
-    
-    This is a classic problem in **combinatorics and tiling theory**.
-    
-    We are to find the **minimum number of marked cells** such that:
-    
-    - There is **exactly one tiling** of the board with dominoes.
-    - In this tiling, **every domino contains at most one marked cell**.
+    <think>
+    Solution: $57: 160=0.35625$
+    The solution can be verified by multiplying the result by 160: $0.35625 \cdot 160=57$.
+    </think>
     
-    This is equivalent to saying that the **marked cells** are **independent** in the tiling, and the tiling is **unique**.
-    
-    This is a well-known result in combinatorics and tiling theory.
-    
-    ---
-    
-    ### Step 5: Known result
-    
-    It is known that the **minimum number of marked cells** needed to ensure **exactly one tiling** with dominoes, where **every domino contains at most one marked cell**, is:
-    
-    $$
-    \boxed{194}
-    $$
-    
-    ---
-    
-    ### Final Answer:
-    
-    $$
-    \boxed{194}
-    $$
+    Therefore, the final answer is \boxed{0.35625}.
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `$g(x)=c, g(x)=\lceil x \rceil, g(x)=\lfloor x \rfloor$`
+- **reference**: `5,-3`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `h(x) = x`
+- **extracted** (✗): `11`
 - **completion**:
 
     ```text
-    We are given a functional equation:
-    
-    $$
-    h(x) = h\left( \frac{h(bx - a) + a}{b} \right)
-    $$
-    
-    for all $ x \in \mathbb{Q} $, $ a \in \mathbb{Z} $, and $ b \in \mathbb{N} $.
-    
-    We are to find all functions $ h : \mathbb{Q} \to \mathbb{Z} $ satisfying this equation.
-    
-    ---
-    
-    ### Step 1: Try simple candidate functions
-    
-    Let’s try a function of the form:
-    
-    $$
-    h(x) = kx + c
-    $$
-    
-    for some constants $ k, c \in \math
-    
-    ... [truncated 822 chars] ...
-    
-    n, only possible for specific $ a $ and $ b $, not all.
-    
-    ---
-    
-    ### Step 7: Try the function $ h(x) = \frac{a}{b} $
-    
-    Let’s try:
-    
-    $$
-    h(x) = \frac{a}{b}
-    $$
-    
-    But this is not a function from $ \mathbb{Q} \to \mathbb{Z} $, because $ \frac{a}{b} \in \mathbb{Q} $, not necessarily an integer.
-    
-    ---
-    
-    ### Step 8: Try $ h(x) = \lfloor x \rfloor $ again
-    
-    We tried this, and it failed.
-    
-    ---
-    
-    ### Step 9: Try a function that maps rational numbers to integers in a specific way
-    
-    Let’s consider the **identity function**:
-    
-    $$
-    h(x) = x
-    $$
-    
-    Then:
-    
-    $$
-    h(x) = x
-    $$
-    $$
-    h(bx - a) = bx - a
-    $$
-    $$
-    \frac{h(bx - a) + a}{b} = \frac{bx - a + a}{b} = x
-    $$
-    
-    So the equation becomes:
-    
-    $$
-    x = x
-    $$
-    
-    Which is **true**.
-    
-    So $ h(x) = x $ is a solution.
-    
-    ---
-    
-    ### Step 10: Check if there are other solutions
-    
-    Suppose $ h(x) = x $ is a solution.
-    
-    Now suppose $ h(x) = x + c $ for some constant $ c \in \mathbb{Z} $
-    
-    Then:
-    
-    $$
-    h(x) = x + c
-    $$
-    $$
-    h(bx - a) = bx - a + c
-    $$
-    $$
-    \frac{h(bx - a) + a}{b} = \frac{bx - a + c + a}{b} = x + \frac{c}{b}
-    $$
-    
-    So:
-    
-    $$
-    x + c = x + \frac{c}{b} \Rightarrow c = \frac{c}{b}
-    $$
-    
-    This is only possible if $ c = 0 $, so $ h(x) = x $ is the **only** solution.
-    
-    ---
-    
-    ### ✅ Final Answer:
+    <think>
+    If \(n\) is an integer such that \(8n-41\) and \(n+3\) are non-zero, then \(8n-41=(n+3)^{2} k^{3}\) for some integer \(k\). This is equivalent to \((n+3)^{2} k^{3}+41=8n\). Now, \(k^{3} \leq 8n-41 \leq 8n-41+41=8n\), so \(k^{3} \leq 8n \leq 8k^{3}\), which implies \(k^{3} \leq k^{3}\) and \(k^{2} \leq 8k\) which is always true for \(k \geq 1\) and false for \(k=-1, -2\). Thus, \(k=1\) or \(k=0\). If \(k=0\), then \(n=41/8\) which is not an integer. Therefore, \(k=1\) and \(n=11\). Finally, \(n+3=14\) and \(8n-41=57\).
+    </think>
     
-    $$
-    \boxed{h(x) = x}
-    $$
+    Therefore, the final answer is \boxed{11}.
     ```