208 lines
5.0 KiB
Markdown
208 lines
5.0 KiB
Markdown
# Automated MNLP evaluation report
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- **Model repo:** [`cs-552-2026-thinking-tokens/math_model`](https://huggingface.co/cs-552-2026-thinking-tokens/math_model)
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- **Owner(s):** group **thinking-tokens**
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- **Generated at:** 2026-05-19T03:22:52+00:00 (UTC)
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- **Pipeline:** [mnlp-project-ci](https://github.com/eric11eca/mnlp-project-ci)
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_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._
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## Summary
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| Benchmark | Accuracy | Status |
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|---|---:|---|
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| Math | 0.3900 | ok |
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| Knowledge | — | not run |
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| Multilingual | — | not run |
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| Safety | — | not run |
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## Sample completions
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_Prompts are intentionally omitted to avoid revealing benchmark contents. For multi-completion problems, only one completion is shown per sample._
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### Math
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**Correct** (1 shown)
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- **reference**: `194`
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- **overall** (2/8 completions correct)
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- **extracted** (✓): `194`
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- **completion**:
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```text
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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**.
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---
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### Step 1: Understand the problem
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We are to find the **smallest number of marked cells** such that:
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- Th
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... [truncated 823 chars] ...
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t can be **involving marked cells** is $k$.
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- Since there are 18818 dominoes, the number of dominoes that **do not involve marked cells** is $18818 - k$.
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We want **exactly one** tiling where **every domino has at most one marked cell**.
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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.
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This is a very strong condition.
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---
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### Step 4: What is the minimum number of marked cells?
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This is a classic problem in **combinatorics and tiling theory**.
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We are to find the **minimum number of marked cells** such that:
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- There is **exactly one tiling** of the board with dominoes.
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- In this tiling, **every domino contains at most one marked cell**.
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This is equivalent to saying that the **marked cells** are **independent** in the tiling, and the tiling is **unique**.
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This is a well-known result in combinatorics and tiling theory.
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---
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### Step 5: Known result
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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:
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$$
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\boxed{194}
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$$
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---
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### Final Answer:
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$$
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\boxed{194}
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$$
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```
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**Incorrect** (1 shown)
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- **reference**: `$g(x)=c, g(x)=\lceil x \rceil, g(x)=\lfloor x \rfloor$`
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- **overall** (0/8 completions correct)
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- **extracted** (✗): `h(x) = x`
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- **completion**:
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```text
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We are given a functional equation:
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$$
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h(x) = h\left( \frac{h(bx - a) + a}{b} \right)
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$$
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for all $ x \in \mathbb{Q} $, $ a \in \mathbb{Z} $, and $ b \in \mathbb{N} $.
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We are to find all functions $ h : \mathbb{Q} \to \mathbb{Z} $ satisfying this equation.
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---
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### Step 1: Try simple candidate functions
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Let’s try a function of the form:
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$$
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h(x) = kx + c
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$$
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for some constants $ k, c \in \math
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... [truncated 822 chars] ...
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n, only possible for specific $ a $ and $ b $, not all.
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---
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### Step 7: Try the function $ h(x) = \frac{a}{b} $
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Let’s try:
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$$
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h(x) = \frac{a}{b}
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$$
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But this is not a function from $ \mathbb{Q} \to \mathbb{Z} $, because $ \frac{a}{b} \in \mathbb{Q} $, not necessarily an integer.
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---
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### Step 8: Try $ h(x) = \lfloor x \rfloor $ again
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We tried this, and it failed.
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---
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### Step 9: Try a function that maps rational numbers to integers in a specific way
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Let’s consider the **identity function**:
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$$
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h(x) = x
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$$
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Then:
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$$
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h(x) = x
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$$
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$$
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h(bx - a) = bx - a
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$$
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$$
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\frac{h(bx - a) + a}{b} = \frac{bx - a + a}{b} = x
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$$
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So the equation becomes:
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$$
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x = x
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$$
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Which is **true**.
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So $ h(x) = x $ is a solution.
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---
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### Step 10: Check if there are other solutions
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Suppose $ h(x) = x $ is a solution.
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Now suppose $ h(x) = x + c $ for some constant $ c \in \mathbb{Z} $
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Then:
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$$
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h(x) = x + c
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$$
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$$
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h(bx - a) = bx - a + c
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$$
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$$
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\frac{h(bx - a) + a}{b} = \frac{bx - a + c + a}{b} = x + \frac{c}{b}
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$$
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So:
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$$
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x + c = x + \frac{c}{b} \Rightarrow c = \frac{c}{b}
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$$
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This is only possible if $ c = 0 $, so $ h(x) = x $ is the **only** solution.
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---
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### ✅ Final Answer:
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$$
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\boxed{h(x) = x}
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$$
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```
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