HYUKI5/simplescaling__s1K

simplescaling/s1K（複製）

データの内容

難しい問題 1,000 件に推論の過程と解答を付けたデータ。1 行が 1 問で、question に問い、solution に解答、thinking_trajectories に推論の過程、cot_type に分野が入る。カードは推論を Gemini から蒸留したと書くが、調査メモでは Gemini と R1 の 2 種類の出力が混ざっていると記録されている。

データの概要

項目	値
行数	1,000（1 構成・1 分割の合計）
列数	8
容量	6.91 MB（ファイル 3 本）
ファイルの形式	parquet 1
公開	public

構成と分割

構成（config）	分割（split）ごとの行数	列数
default	train 1,000	8

列（構成 default・分割 train）

列	型	先頭の行の値
solution	string	1. **Rewrite the function using trigonometric id…
question	string	Let $a,b,A,B$ be given reals. We consider the …
cot_type	string	math
source_type	string	AI-MO/NuminaMath-CoT/aops_forum
metadata	string	{'source': 'aops_forum', 'messages': [{'content'…
cot	null	null
thinking_trajectories	list<string>	[1 件]
attempt	string	Let $P(x) = 1 - a \cos x - b \sin x - A \cos 2x …

行数と列は HF のデータセットビューアの集計（元のデータセットの現在の版で数えたもの）。全データセットの一覧は HYUKI5/dataset-catalog にある。

複製について

simplescaling/s1K をリビジョン 278d72baaa2b887a7e76a70a0ae254a5a45536e4 の時点でそのまま複製したもの。中身は変えていない。判断モデル Laya（convaiinnovations/laya）の微調整に使う候補として、調査メモ（Laya 学習用データセット調査）に載せたデータを集めている。

出典

• 元のデータセット: https://huggingface.co/datasets/simplescaling/s1K
• 複製したリビジョン: 278d72baaa2b887a7e76a70a0ae254a5a45536e4（2025-02-11 01:14:45+00:00）
• 元のライセンス表記: apache-2.0

論文

• https://arxiv.org/abs/2501.19393

アノテーションの方法

（元のカードに該当の見出しが無いので、ラベルの付け方に触れた段落を抜き出した）

目次

• 目次
• データセットの説明
  • データセットの概要
• データセットの構造
  • データのインスタンス
  • データフィールド
  • データ分割
• データセットの作成
  • 選定の理由
  • 元データ
  • アノテーション
• 追加情報
  • ライセンス情報
  • 引用情報
  • 貢献者

各フィールドは次のとおりである。

• solution: 問題の正解の解答（ground truth）
• question: 問題
• cot_type: 問題が属する大分類の 1 つ
• source_type: 問題の出典となったデータセットとそのカテゴリ
• metadata 元のデータセットにあるその他のフィールド
• cot: None
• thinking_trajectories: Gemini が生成した推論トレース。常に長さ 1 のリストである
• attempt: Gemini が生成した応答

微調整に使うときの注意

• 調査メモの台帳での区分: B（継承条件・gated・モデル由来の条件・上流の規約などが付く）。区分は記載の文言で機械的に付けたもので、法的な判定ではない。学習に入れる前に元のカードとライセンスの原文で確かめる。
• Laya の入力は 2,048 token まで（ModernBERT-large / mmBERT）。長い本文は先頭から黙って切られるので、切れる行は事前に除くか、要約・切り出しをしてから入れる。
• 分割は記事・文書単位で行い、同じ本文から作った別の問いを学習側と較正側に分けない。温度と閾値は学習に使っていない分割で別々に当てる。

調査メモの記述

メモ	節	ライセンス（記載）	確認	注記
frontier_closed_reasoning_logs.md	Q4. 正誤・採点ラベルが付いたクローズドモデルの推論ログ・回答（金融・判断に近いものを含む）は何があるか (L112)	s1K: apache-2.0、s1K-1.1: mit	確認済	複数/混在/生成元モデルの出力規約が別に及ぶ(Q4) ／ 原文注記: 思考は Gemini が中央値 11,154 文字（8,000 文字超 87.7%）、R1 が中央値 26,228 文字（94.2%、最大 80,254）。8,000 文字を超える行が 87.7% と 94.2% あり、

元のカード（日本語訳）

以下は元のデータセットのカードの本文を日本語に訳したもの（機械翻訳）。訳に疑問があるときは、末尾の原文を確かめる。

---

データセットカード: s1K

目次

• 目次
• データセットの説明
  • データセットの概要
• データセットの構造
  • データのインスタンス
  • データフィールド
  • データ分割
• データセットの作成
  • 選定の理由
  • 元データ
  • アノテーション
• 追加情報
  • ライセンス情報
  • 引用情報
  • 貢献者

データセットの説明

• リポジトリ: https://github.com/simplescaling/s1
• 論文: s1: Simple test-time scaling

データセットの概要

s1K は、Gemini Thining から蒸留した推論トレースと解答を付けた、多様で高品質かつ難しい問題 1,000 件のデータセットである。詳細は s1 論文 を参照。

使い方

# pip install -q datasets
from datasets import load_dataset
ds = load_dataset("simplescaling/s1K")["train"]
ds[0]

データセットの構造

データのインスタンス

例は次のとおりである。

{
    'solution': '1. **Rewrite the function using trigonometric identities:**...(shortened)',
    'question': 'Let  $a,b,A,B$  be given reals. We consider the function defined by \\[ f(x) = 1 - a \\cdot \\cos(x) - b \\cdot \\sin(x) - A \\cdot \\cos(2x) - B \\cdot \\sin(2x). \\] Prove that if for any real number  $x$  we have  $f(x) \\geq 0$  then  $a^2 + b^2 \\leq 2$  and  $A^2 + B^2 \\leq 1.$ ',
    'cot_type': 'math',
    'source_type': 'AI-MO/NuminaMath-CoT/aops_forum',
    'metadata': "{'source': 'aops_forum', 'messages': [{'content': 'Let  $a,b,A,B$  be given reals. We consider the function defined by \\\\[ f(x) = 1 - a \\\\cdot \\\\cos(x) - b \\\\cdot \\\\sin(x) - A \\\\cdot \\\\cos(2x) - B \\\\cdot \\\\sin(2x). \\\\] Prove that if for any real number  $x$  we have  $f(x) \\\\geq 0$  then  $a^2 + b^2 \\\\leq 2$  and  $A^2 + B^2 \\\\leq 1.$ ', 'role': 'user'}, {'content': '1. **Rewrite the function using trigonometric identities:**\\n   \\\\[\\n   f(x) = 1 - a \\\\cos(x) - b \\\\sin(x) - A \\\\cos(2x) - B \\\\sin(2x)\\n   \\\\]\\n   We can use the angle addition formulas for sine and cosine:\\n   \\\\[\\n   \\\\cos(x + \\\\theta) = \\\\cos(x)\\\\cos(\\\\theta) - \\\\sin(x)\\\\sin(\\\\theta)\\n   \\\\]\\n   \\\\[\\n   \\\\sin(x + \\\\theta) = \\\\sin(x)\\\\cos(\\\\theta) + \\\\cos(x)\\\\sin(\\\\theta)\\n   \\\\]\\n   Let:\\n   \\\\[\\n   R_1 = \\\\sqrt{a^2 + b^2}, \\\\quad \\\\theta_1 = \\\\arctan\\\\left(\\\\frac{b}{a}\\\\right)\\n   \\\\]\\n   \\\\[\\n   R_2 = \\\\sqrt{A^2 + B^2}, \\\\quad \\\\theta_2 = \\\\arctan\\\\left(\\\\frac{B}{A}\\\\right)\\n   \\\\]\\n   Then:\\n   \\\\[\\n   a \\\\cos(x) + b \\\\sin(x) = R_1 \\\\sin(x + \\\\theta_1)\\n   \\\\]\\n   \\\\[\\n   A \\\\cos(2x) + B \\\\sin(2x) = R_2 \\\\sin(2x + \\\\theta_2)\\n   \\\\]\\n   Therefore, the function becomes:\\n   \\\\[\\n   f(x) = 1 - R_1 \\\\sin(x + \\\\theta_1) - R_2 \\\\sin(2x + \\\\theta_2)\\n   \\\\]\\n\\n2. **Analyze the function for non-negativity:**\\n   Given \\\\( f(x) \\\\geq 0 \\\\) for all \\\\( x \\\\), we need to ensure that the maximum values of the sine terms do not exceed 1:\\n   \\\\[\\n   1 - R_1 \\\\sin(x + \\\\theta_1) - R_2 \\\\sin(2x + \\\\theta_2) \\\\geq 0\\n   \\\\]\\n   The maximum value of \\\\( \\\\sin(x + \\\\theta_1) \\\\) and \\\\( \\\\sin(2x + \\\\theta_2) \\\\) is 1. Therefore:\\n   \\\\[\\n   1 - R_1 - R_2 \\\\geq 0\\n   \\\\]\\n   This implies:\\n   \\\\[\\n   R_1 + R_2 \\\\leq 1\\n   \\\\]\\n\\n3. **Consider the individual terms:**\\n   Since \\\\( R_1 = \\\\sqrt{a^2 + b^2} \\\\) and \\\\( R_2 = \\\\sqrt{A^2 + B^2} \\\\), we need to show:\\n   \\\\[\\n   \\\\sqrt{a^2 + b^2} \\\\leq 1 \\\\quad \\\\text{and} \\\\quad \\\\sqrt{A^2 + B^2} \\\\leq 1\\n   \\\\]\\n\\n4. **Prove the inequalities:**\\n   - For \\\\( \\\\sqrt{a^2 + b^2} \\\\leq 1 \\\\):\\n     \\\\[\\n     a^2 + b^2 \\\\leq 1^2 = 1\\n     \\\\]\\n   - For \\\\( \\\\sqrt{A^2 + B^2} \\\\leq 1 \\\\):\\n     \\\\[\\n     A^2 + B^2 \\\\leq 1^2 = 1\\n     \\\\]\\n\\n5. **Conclusion:**\\n   Therefore, we have shown that if \\\\( f(x) \\\\geq 0 \\\\) for all \\\\( x \\\\), then:\\n   \\\\[\\n   a^2 + b^2 \\\\leq 2 \\\\quad \\\\text{and} \\\\quad A^2 + B^2 \\\\leq 1\\n   \\\\]\\n\\n\\\\(\\\\blacksquare\\\\)\\n\\nThe final answer is \\\\( \\\\boxed{ a^2 + b^2 \\\\leq 2 } \\\\) and \\\\( A^2 + B^2 \\\\leq 1 \\\\)', 'role': 'assistant'}]}",
    'cot': None,
    'thinking_trajectories': ["Let $f(x) = 1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x \\geq 0$ for all $x \\in \\mathbb{R}$.\nWe want to show that $a^2 + b^2 \\leq 2$ and $A^2 + B^2 \\leq 1$.\n\nConsider specific values of $x$.\nFor $x=0$, $f(0) = 1 - a - A \\geq 0$, so $a+A \\leq 1$.\nFor $x=\\pi$, $f(\\pi) = 1 - a \\cos \\pi - b \\sin \\pi - A \\cos 2\\pi - B \\sin 2\\pi = 1 - a(-1) - b(0) - A(1) - B(0) = 1+a-A \\geq 0$, so $A-a \\leq 1$.\nFor $x=\\pi/2$, $f(\\pi/2) = 1 - a \\cos(\\pi/2) - b \\sin(\\pi/2) - A \\cos(\\pi) - B \\sin(\\pi) = 1 - a(0) - b(1) - A(-1) - B(0) = 1 - b + A \\geq 0$, so $b-A \\leq 1$.\nFor $x=3\\pi/2$, $f(3\\pi/2) = 1 - a \\cos(3\\pi/2) - b \\sin(3\\pi/2) - A \\cos(3\\pi) - B \\sin(3\\pi) = 1 - a(0) - b(-1) - A(-1) - B(0) = 1 + b + A \\geq 0$, so $-b-A \\leq 1$.\n\nFrom $a+A \\leq 1$ and $A-a \\leq 1$, we have $-1 \\leq a+A \\leq 1$ and $-1 \\leq a-A \\leq 1$.\nFrom $b-A \\leq 1$ and $-b-A \\leq 1$, we have $-1 \\leq b-A \\leq 1$ and $-1 \\leq b+A \\leq 1$.\n\nConsider the integral of $f(x)$ over a period.\n$\\int_0^{2\\pi} f(x) dx = \\int_0^{2\\pi} (1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x) dx \\geq 0$.\n$\\int_0^{2\\pi} 1 dx = 2\\pi$.\n$\\int_0^{2\\pi} \\cos x dx = [\\sin x]_0^{2\\pi} = 0$.\n$\\int_0^{2\\pi} \\sin x dx = [-\\cos x]_0^{2\\pi} = -1 - (-1) = 0$.\n$\\int_0^{2\\pi} \\cos 2x dx = [\\frac{1}{2} \\sin 2x]_0^{2\\pi} = 0$.\n$\\int_0^{2\\pi} \\sin 2x dx = [-\\frac{1}{2} \\cos 2x]_0^{2\\pi} = -\\frac{1}{2} - (-\\frac{1}{2}) = 0$.\nSo $\\int_0^{2\\pi} f(x) dx = 2\\pi \\geq 0$. This doesn't give any information.\n\nMultiply $f(x)$ by $\\cos x$ and integrate.\n$\\int_0^{2\\pi} f(x) \\cos x dx = \\int_0^{2\\pi} (1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x) \\cos x dx \\geq 0$.\n$\\int_0^{2\\pi} \\cos x dx = 0$.\n$\\int_0^{2\\pi} \\cos^2 x dx = \\int_0^{2\\pi} \\frac{1+\\cos 2x}{2} dx = [\\frac{x}{2} + \\frac{\\sin 2x}{4}]_0^{2\\pi} = \\pi$.\n$\\int_0^{2\\pi} \\sin x \\cos x dx = \\int_0^{2\\pi} \\frac{1}{2} \\sin 2x dx = [-\\frac{1}{4} \\cos 2x]_0^{2\\pi} = -\\frac{1}{4} - (-\\frac{1}{4}) = 0$.\n$\\int_0^{2\\pi} \\cos 2x \\cos x dx = \\int_0^{2\\pi} \\frac{1}{2} (\\cos 3x + \\cos x) dx = [\\frac{1}{6} \\sin 3x + \\frac{1}{2} \\sin x]_0^{2\\pi} = 0$.\n$\\int_0^{2\\pi} \\sin 2x \\cos x dx = \\int_0^{2\\pi} \\frac{1}{2} (\\sin 3x + \\sin x) dx = [-\\frac{1}{6} \\cos 3x - \\frac{1}{2} \\cos x]_0^{2\\pi} = (-\\frac{1}{6} - \\frac{1}{2}) - (-\\frac{1}{6} - \\frac{1}{2}) = 0$.\nSo $\\int_0^{2\\pi} f(x) \\cos x dx = 0 - a \\pi - 0 - 0 - 0 = -a\\pi \\geq 0$, so $a \\leq 0$. This is not necessarily true.\n\nConsider $f(x) \\geq 0$.\n$1 - (a \\cos x + b \\sin x) - (A \\cos 2x + B \\sin 2x) \\geq 0$.\nLet $a = r \\cos \\alpha, b = r \\sin \\alpha$. Then $a \\cos x + b \\sin x = r (\\cos \\alpha \\cos x + \\sin \\alpha \\sin x) = r \\cos(x-\\alpha)$.\nLet $A = R \\cos \\beta, B = R \\sin \\beta$. Then $A \\cos 2x + B \\sin 2x = R (\\cos \\beta \\cos 2x + \\sin \\beta \\sin 2x) = R \\cos(2x-\\beta)$.\nSo $1 - r \\cos(x-\\alpha) - R \\cos(2x-\\beta) \\geq 0$.\nLet $y = x-\\alpha$. Then $2x-\\beta = 2(y+\\alpha) - \\beta = 2y + 2\\alpha - \\beta$. Let $\\gamma = 2\\alpha - \\beta$.\n$1 - r \\cos y - R \\cos(2y+\\gamma) \\geq 0$.\n$1 - r \\cos y - R (\\cos 2y \\cos \\gamma - \\sin 2y \\sin \\gamma) \\geq 0$.\n$1 - r \\cos y - R ((2\\cos^2 y - 1) \\cos \\gamma - 2 \\sin y \\cos y \\sin \\gamma) \\geq 0$.\n$1 - r \\cos y - 2R \\cos^2 y \\cos \\gamma + R \\cos \\gamma + 2R \\sin y \\cos y \\sin \\gamma \\geq 0$.\n\nConsider $f(x) \\geq 0$.\nMultiply by $\\cos(x-\\phi)$ and integrate from $0$ to $2\\pi$.\n$\\int_0^{2\\pi} f(x) \\cos(x-\\phi) dx \\geq 0$.\n$\\int_0^{2\\pi} (1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x) (\\cos x \\cos \\phi + \\sin x \\sin \\phi) dx \\geq 0$.\n$\\cos \\phi \\int_0^{2\\pi} (\\cos x - a \\cos^2 x - b \\sin x \\cos x - A \\cos 2x \\cos x - B \\sin 2x \\cos x) dx + \\sin \\phi \\int_0^{2\\pi} (\\sin x - a \\cos x \\sin x - b \\sin^2 x - A \\cos 2x \\sin x - B \\sin 2x \\sin x) dx \\geq 0$.\n$\\cos \\phi (0 - a\\pi - 0 - 0 - 0) + \\sin \\phi (0 - 0 - b\\pi - 0 - 0) \\geq 0$.\n$-a\\pi \\cos \\phi - b\\pi \\sin \\phi \\geq 0$.\n$a \\cos \\phi + b \\sin \\phi \\leq 0$.\nThis must be true for all $\\phi$. Let $\\phi$ such that $\\tan \\phi = b/a$. Then $a \\frac{a}{\\sqrt{a^2+b^2}} + b \\frac{b}{\\sqrt{a^2+b^2}} = \\sqrt{a^2+b^2} \\leq 0$, so $a=b=0$. This is wrong.\n\nConsider $f(x) \\geq 0$.\n$1 - (a \\cos x + b \\sin x) - (A \\cos 2x + B \\sin 2x) \\geq 0$.\nThe minimum value of $a \\cos x + b \\sin x$ is $-\\sqrt{a^2+b^2}$.\nThe minimum value of $A \\cos 2x + B \\sin 2x$ is $-\\sqrt{A^2+B^2}$.\n$1 - (a \\cos x + b \\sin x) \\geq A \\cos 2x + B \\sin 2x$.\nMinimum of LHS is $1 - \\sqrt{a^2+b^2}$. Maximum of RHS is $\\sqrt{A^2+B^2}$.\n$1 - \\sqrt{a^2+b^2} \\geq \\sqrt{A^2+B^2}$.\n\nConsider $g(x) = 1 - A \\cos 2x - B \\sin 2x \\geq a \\cos x + b \\sin x$.\nMinimum of LHS is $1 - \\sqrt{A^2+B^2}$. Maximum of RHS is $\\sqrt{a^2+b^2}$.\n$1 - \\sqrt{A^2+B^2} \\geq \\sqrt{a^2+b^2}$.\n\nLet $x=0$, $1-a-A \\geq 0$.\nLet $x=\\pi$, $1+a-A \\geq 0$.\nLet $x=\\pi/2$, $1-b+A \\geq 0$.\nLet $x=3\\pi/2$, $1+b+A \\geq 0$.\n\nConsider the case where $b=B=0$.\n$f(x) = 1 - a \\cos x - A \\cos 2x \\geq 0$.\n$1 - a \\cos x - A (2 \\cos^2 x - 1) \\geq 0$.\n$1 - a \\cos x - 2A \\cos^2 x + A \\geq 0$.\n$A+1 - a \\cos x - 2A \\cos^2 x \\geq 0$.\nLet $y = \\cos x$, $-1 \\leq y \\leq 1$.\n$g(y) = -2Ay^2 - ay + A+1 \\geq 0$ for $y \\in [-1, 1]$.\nIf $A=0$, $-ay+1 \\geq 0$ for $y \\in [-1, 1]$. If $a>0$, $-a+1 \\geq 0$, $a \\leq 1$. If $a<0$, $a+1 \\geq 0$, $a \\geq -1$. So $|a| \\leq 1$. $a^2 \\leq 1 \\leq 2$.\nIf $A>0$, the parabola opens downwards. We need $g(-1) = -2A+a+A+1 = -A+a+1 \\geq 0$, $a \\geq A-1$. And $g(1) = -2A-a+A+1 = -A-a+1 \\geq 0$, $a \\leq 1-A$. Also the vertex must be outside $[-1, 1]$ or the vertex is in $[-1, 1]$ and the value is non-negative. Vertex at $y = -(-a)/(2(-2A)) = -a/4A$. If $|-a/4A| \\leq 1$, then $a^2 \\leq 16A^2$. $g(-a/4A) = -2A (-a/4A)^2 - a(-a/4A) + A+1 = -2A a^2/16A^2 + a^2/4A + A+1 = -a^2/8A + a^2/4A + A+1 = a^2/8A + A+1 \\geq 0$.\nIf $A<0$, the parabola opens upwards. We need $g(-1) \\geq 0$ and $g(1) \\geq 0$.\n$-A+a+1 \\geq 0$, $a \\geq A-1$.\n$-A-a+1 \\geq 0$, $a \\leq 1-A$.\n\nConsider integrals again.\n$\\int_0^{2\\pi} f(x) dx \\geq 0 \\implies 2\\pi \\geq 0$.\n$\\int_0^{2\\pi} f(x) \\cos x dx \\geq 0 \\implies -a \\pi \\geq 0 \\implies a \\leq 0$.\n$\\int_0^{2\\pi} f(x) \\sin x dx \\geq 0 \\implies -b \\pi \\geq 0 \\implies b \\leq 0$.\n$\\int_0^{2\\pi} f(x) \\cos 2x dx \\geq 0 \\implies -A \\pi \\geq 0 \\implies A \\leq 0$.\n$\\int_0^{2\\pi} f(x) \\sin 2x dx \\geq 0 \\implies -B \\pi \\geq 0 \\implies B \\leq 0$.\nThis is not true.\n\nConsider the vector $v = (1, \\cos x, \\sin x, \\cos 2x, \\sin 2x)$.\nThe condition is $v \\cdot c \\geq 0$, where $c = (1, -a, -b, -A, -B)$.\n\nConsider the case where $f(x) = (1 - \\alpha \\cos x - \\beta \\sin x)^2 \\geq 0$.\n$1 + \\alpha^2 \\cos^2 x + \\beta^2 \\sin^2 x - 2 \\alpha \\cos x - 2 \\beta \\sin x + 2 \\alpha \\beta \\cos x \\sin x \\geq 0$.\n$1 + \\frac{\\alpha^2+\\beta^2}{2} + \\frac{\\alpha^2-\\beta^2}{2} \\cos 2x - 2 \\alpha \\cos x - 2 \\beta \\sin x + \\alpha \\beta \\sin 2x \\geq 0$.\nCompare with $1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x \\geq 0$.\n$-a = -2\\alpha$, $-b = -2\\beta$, $-A = \\frac{\\alpha^2-\\beta^2}{2}$, $-B = \\alpha \\beta$.\n$a = 2\\alpha$, $b = 2\\beta$.\n$A = \\frac{\\beta^2-\\alpha^2}{2}$, $B = -\\alpha \\beta$.\n$a^2+b^2 = 4\\alpha^2+4\\beta^2$.\n$A^2+B^2 = \\frac{(\\beta^2-\\alpha^2)^2}{4} + \\alpha^2 \\beta^2 = \\frac{\\beta^4 - 2\\alpha^2 \\beta^2 + \\alpha^4 + 4\\alpha^2 \\beta^2}{4} = \\frac{(\\alpha^2+\\beta^2)^2}{4}$.\n\nConsider $f(x) \\geq 0$.\nConsider the Fourier series of $|f(x)| = f(x)$.\nThe coefficients must match.\n\nLet $z = e^{ix}$.\n$f(x) = 1 - \\frac{a}{2}(z+z^{-1}) - \\frac{b}{2i}(z-z^{-1}) - \\frac{A}{2}(z^2+z^{-2}) - \\frac{B}{2i}(z^2-z^{-2}) \\geq 0$.\n$2if(x) = 2i - ia(z+z^{-1}) + b(z-z^{-1}) - iA(z^2+z^{-2}) + B(z^2-z^{-2}) \\geq 0$.\n$2if(x) = 2i - i a z - i a z^{-1} + b z - b z^{-1} - i A z^2 - i A z^{-2} + B z^2 - B z^{-2} \\geq 0$.\n$2if(x) = 2i + (b-ia)z + (-b-ia)z^{-1} + (B-iA)z^2 + (-B-iA)z^{-2} \\geq 0$.\n\nConsider the integral $\\frac{1}{2\\pi} \\int_0^{2\\pi} f(x) e^{inx} dx$.\nFor $n=0$, $\\frac{1}{2\\pi} \\int_0^{2\\pi} f(x) dx = 1 \\geq 0$.\nFor $n=1$, $\\frac{1}{2\\pi} \\int_0^{2\\pi} f(x) (\\cos x + i \\sin x) dx = \\frac{1}{2\\pi} \\int_0^{2\\pi} (1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x) (\\cos x + i \\sin x) dx = -a/2 - ib/2$.\nSince $f(x) \\geq 0$, the coefficients of the Fourier series must satisfy certain conditions.\n\nConsider trigonometric polynomials that are non-negative.\nIf $P(\\theta) = \\sum_{k=-n}^n c_k e^{ik\\theta} \\geq 0$, then $c_{-k} = \\overline{c_k}$ and $c_0 \\geq 0$.\nIn our case, the coefficients are:\n$c_0 = 1$.\n$c_1 = -a/2 - ib/2$.\n$c_{-1} = -a/2 + ib/2$.\n$c_2 = -A/2 - iB/2$.\n$c_{-2} = -A/2 + iB/2$.\nWe need a condition on the coefficients such that the polynomial is non-negative.\n\nIf $1 - a \\cos x - b \\sin x \\geq 0$, then $1 \\geq \\sqrt{a^2+b^2}$.\nIf $1 - A \\cos 2x - B \\sin 2x \\geq 0$, then $1 \\geq \\sqrt{A^2+B^2}$.\n\nConsider $f(x) = |1 - \\alpha e^{ix} - \\beta e^{2ix}|^2 \\geq 0$.\n$(1 - \\alpha e^{ix} - \\beta e^{2ix})(1 - \\bar{\\alpha} e^{-ix} - \\bar{\\beta} e^{-2ix})$.\n$1 - \\bar{\\alpha} e^{-ix} - \\bar{\\beta} e^{-2ix} - \\alpha e^{ix} + |\\alpha|^2 - \\alpha \\bar{\\beta} e^{-ix} - \\beta e^{2ix} - \\beta \\bar{\\alpha} e^{ix} + |\\beta|^2$.\n$1 + |\\alpha|^2 + |\\beta|^2 - (\\bar{\\alpha} + \\alpha + \\alpha \\bar{\\beta}) e^{-ix} - (\\bar{\\beta} + \\beta + \\beta \\bar{\\alpha}) e^{-2ix}$.\n$1 + |\\alpha|^2 + |\\beta|^2 - (2 \\text{Re}(\\alpha) + \\alpha \\bar{\\beta}) e^{-ix} - (2 \\text{Re}(\\beta) + \\beta \\bar{\\alpha}) e^{-2ix}$.\nCompare coefficients.\n\nIf $P(\\theta) = |\\sum_{k=0}^n z_k e^{ik\\theta}|^2$, then $P(\\theta) \\geq 0$.\nConsider $P(x) = |1 - \\alpha e^{ix}|^2 = (1-\\alpha e^{ix})(1-\\bar{\\alpha} e^{-ix}) = 1 - \\bar{\\alpha} e^{-ix} - \\alpha e^{ix} + |\\alpha|^2 = 1+|\\alpha|^2 - 2 \\text{Re}(\\alpha) \\cos x + 2 \\text{Im}(\\alpha) \\sin x$.\nCompare with $1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x$.\n\nConsider $f(x) \\geq 0$.\n$1 - (a \\cos x + b \\sin x) - (A \\cos 2x + B \\sin 2x) \\geq 0$.\nLet $a = r_1 \\cos \\phi_1, b = r_1 \\sin \\phi_1$.\nLet $A = r_2 \\cos \\phi_2, B = r_2 \\sin \\phi_2$.\n$1 - r_1 \\cos(x-\\phi_1) - r_2 \\cos(2x-\\phi_2) \\geq 0$.\nLet $x - \\phi_1 = y$. Then $2x - \\phi_2 = 2(y+\\phi_1) - \\phi_2 = 2y + 2\\phi_1 - \\phi_2$. Let $\\delta = 2\\phi_1 - \\phi_2$.\n$1 - r_1 \\cos y - r_2 \\cos(2y+\\delta) \\geq 0$.\n$1 - r_1 \\cos y - r_2 (\\cos 2y \\cos \\delta - \\sin 2y \\sin \\delta) \\geq 0$.\n$1 - r_1 \\cos y - r_2 ( (2\\cos^2 y - 1) \\cos \\delta - 2 \\sin y \\cos y \\sin \\delta ) \\geq 0$.\n$1 - r_1 \\cos y - 2 r_2 \\cos^2 y \\cos \\delta + r_2 \\cos \\delta + 2 r_2 \\sin y \\cos y \\sin \\delta \\geq 0$.\n$1 + r_2 \\cos \\delta - r_1 \\cos y - 2 r_2 \\cos \\delta \\cos^2 y + r_2 \\sin \\delta \\sin 2y \\geq 0$.\n\nIf $f(x) \\geq 0$, then the average value is non-negative.\nConsider $1 - a \\cos x - b \\sin x \\geq 0$. Square both sides? No.\nConsider $1 - A \\cos 2x - B \\sin 2x \\geq |a \\cos x + b \\sin x|$.\n$1 - \\sqrt{A^2+B^2} \\leq 1 - A \\cos 2x - B \\sin 2x \\leq 1 + \\sqrt{A^2+B^2}$.\n$|a \\cos x + b \\sin x| \\leq \\sqrt{a^2+b^2}$.\nSo $1 - \\sqrt{A^2+B^2} \\geq \\sqrt{a^2+b^2}$. This is not necessarily true.\n\nConsider $f(x) \\geq 0$.\n$\\int_0^{2\\pi} f(x) d x = 2\\pi$.\n$\\int_0^{2\\pi} f(x) \\cos x d x = -a\\pi$.\n$\\int_0^{2\\pi} f(x) \\sin x d x = -b\\pi$.\n$\\int_0^{2\\pi} f(x) \\cos 2x d x = -A\\pi$.\n$\\int_0^{2\\pi} f(x) \\sin 2x d x = -B\\pi$.\nSince $f(x) \\geq 0$, consider the Fejer-Riesz theorem. A non-negative trigonometric polynomial can be written as the squared modulus of another trigonometric polynomial.\n$1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x = |c_0 + c_1 e^{ix} + c_2 e^{2ix}|^2$.\n$= (c_0 + c_1 e^{ix} + c_2 e^{2ix})(\\bar{c_0} + \\bar{c_1} e^{-ix} + \\bar{c_2} e^{-2ix})$.\n$= |c_0|^2 + |c_1|^2 + |c_2|^2 + (c_0 \\bar{c_1} + c_1 \\bar{c_2}) e^{ix} + (\\bar{c_0} c_1 + \\bar{c_1} c_2) e^{-ix} + c_0 \\bar{c_2} e^{2ix} + \\bar{c_0} c_2 e^{-2ix}$.\n$= |c_0|^2 + |c_1|^2 + |c_2|^2 + 2 \\text{Re}(c_0 \\bar{c_1} + c_1 \\bar{c_2}) \\cos x - 2 \\text{Im}(c_0 \\bar{c_1} + c_1 \\bar{c_2}) \\sin x + 2 \\text{Re}(c_0 \\bar{c_2}) \\cos 2x - 2 \\text{Im}(c_0 \\bar{c_2}) \\sin 2x$.\nCompare coefficients:\n$1 = |c_0|^2 + |c_1|^2 + |c_2|^2$.\n$-a = 2 \\text{Re}(c_0 \\bar{c_1} + c_1 \\bar{c_2})$.\n$-b = -2 \\text{Im}(c_0 \\bar{c_1} + c_1 \\bar{c_2})$.\n$-A = 2 \\text{Re}(c_0 \\bar{c_2})$.\n$-B = -2 \\text{Im}(c_0 \\bar{c_2})$.\n$A^2+B^2 = 4 (\\text{Re}(c_0 \\bar{c_2})^2 + \\text{Im}(c_0 \\bar{c_2})^2) = 4 |c_0 \\bar{c_2}|^2 = 4 |c_0|^2 |c_2|^2$.\nLet $c_0 = 1$.\n$1 = 1 + |c_1|^2 + |c_2|^2 \\implies |c_1|^2 + |c_2|^2 = 0 \\implies c_1 = c_2 = 0$. Then $f(x) = 1$, $a=b=A=B=0$.\n"],
    'attempt': 'Let $P(x) = 1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x$.\nSince $P(x) \\geq 0$ for all real $x$, by Fejer-Riesz theorem, there exists a polynomial $Q(z) = c_0 + c_1 z + c_2 z^2$ such that $P(x) = |Q(e^{ix})|^2$.\nWe have $|Q(e^{ix})|^2 = (c_0 + c_1 e^{ix} + c_2 e^{2ix})(\\bar{c_0} + \\bar{c_1} e^{-ix} + \\bar{c_2} e^{-2ix})$.\nExpanding this, we get\n\\begin{align*} |Q(e^{ix})|^2 &= |c_0|^2 + |c_1|^2 + |c_2|^2 \\\\ &+ (c_0 \\bar{c_1} + c_1 \\bar{c_2}) e^{ix} + (\\bar{c_0} c_1 + \\bar{c_1} c_2) e^{-ix} \\\\ &+ c_0 \\bar{c_2} e^{2ix} + \\bar{c_0} c_2 e^{-2ix} \\\\ &= |c_0|^2 + |c_1|^2 + |c_2|^2 \\\\ &+ (c_0 \\bar{c_1} + \\bar{\\bar{c_1}} c_2 + \\overline{c_0 \\bar{c_1} + \\bar{\\bar{c_1}} c_2}) \\cos x + i (c_0 \\bar{c_1} + \\bar{\\bar{c_1}} c_2 - \\overline{c_0 \\bar{c_1} + \\bar{\\bar{c_1}} c_2}) \\sin x \\\\ &+ (c_0 \\bar{c_2} + \\bar{c_0} c_2) \\cos 2x + i (c_0 \\bar{c_2} - \\bar{c_0} c_2) \\sin 2x \\\\ &= |c_0|^2 + |c_1|^2 + |c_2|^2 + 2 \\text{Re}(c_0 \\bar{c_1} + c_1 \\bar{c_2}) \\cos x - 2 \\text{Im}(c_0 \\bar{c_1} + c_1 \\bar{c_2}) \\sin x + 2 \\text{Re}(c_0 \\bar{c_2}) \\cos 2x - 2 \\text{Im}(c_0 \\bar{c_2}) \\sin 2x\\end{align*}\nComparing the coefficients of $P(x)$ with $|Q(e^{ix})|^2$, we have:\n$1 = |c_0|^2 + |c_1|^2 + |c_2|^2$\n$-a = 2 \\text{Re}(c_0 \\bar{c_1} + c_1 \\bar{c_2})$\n$-b = -2 \\text{Im}(c_0 \\bar{c_1} + c_1 \\bar{c_2})$\n$-A = 2 \\text{Re}(c_0 \\bar{c_2})$\n$-B = -2 \\text{Im}(c_0 \\bar{c_2})$\nFrom the last two equations, $A^2 + B^2 = 4 (\\text{Re}(c_0 \\bar{c_2})^2 + \\text{Im}(c_0 \\bar{c_2})^2) = 4 |c_0 \\bar{c_2}|^2 = 4 |c_0|^2 |c_2|^2$.\nSince $1 = |c_0|^2 + |c_1|^2 + |c_2|^2$, we have $|c_0|^2 \\leq 1$ and $|c_2|^2 \\leq 1$.\nThus $A^2 + B^2 = 4 |c_0|^2 |c_2|^2 \\leq 4 \\cdot 1 \\cdot 1 = 4$. This is not the desired result.\n\nConsider $Q(z) = 1 - \\alpha z - \\beta z^2$. Then $|Q(e^{ix})|^2 = |1 - \\alpha e^{ix} - \\beta e^{2ix}|^2$.\n$|1 - \\alpha e^{ix} - \\beta e^{2ix}|^2 = (1 - \\alpha e^{ix} - \\beta e^{2ix})(1 - \\bar{\\alpha} e^{-ix} - \\bar{\\beta} e^{-2ix})$\n$= 1 - \\bar{\\alpha} e^{-ix} - \\bar{\\beta} e^{-2ix} - \\alpha e^{ix} + |\\alpha|^2 - \\alpha \\bar{\\beta} e^{-ix} - \\beta e^{2ix} - \\beta \\bar{\\alpha} e^{ix} + |\\beta|^2$\n$= 1 + |\\alpha|^2 + |\\beta|^2 - (\\alpha + \\alpha \\bar{\\beta} + \\bar{\\alpha}) \\cos x + i (\\alpha + \\alpha \\bar{\\beta} - \\bar{\\alpha}) \\sin x - (\\beta + \\beta \\bar{\\alpha} + \\bar{\\beta}) \\cos 2x + i (\\beta + \\beta \\bar{\\alpha} - \\bar{\\beta}) \\sin 2x$.\nCompare with $1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x$.\n$a = \\alpha + \\bar{\\alpha} + \\alpha \\bar{\\beta} + \\bar{\\alpha} \\beta = 2 \\text{Re}(\\alpha) + 2 \\text{Re}(\\alpha \\bar{\\beta})$.\n$b = -i (\\alpha - \\bar{\\alpha} + \\alpha \\bar{\\beta} - \\bar{\\alpha} \\beta) = 2 \\text{Im}(\\alpha) + 2 \\text{Im}(\\alpha \\bar{\\beta})$.\n$A = \\beta + \\bar{\\beta} = 2 \\text{Re}(\\beta)$.\n$B = -i (\\beta - \\bar{\\beta}) = 2 \\text{Im}(\\beta)$.\n$A^2+B^2 = 4 (\\text{Re}(\\beta)^2 + \\text{Im}(\\beta)^2) = 4 |\\beta|^2$.\n$a^2+b^2 = |2\\alpha + 2\\alpha \\bar{\\beta}|^2 = 4 |\\alpha|^2 |1+\\bar{\\beta}|^2$.\n$1 = 1 + |\\alpha|^2 + |\\beta|^2$. So $|\\alpha|^2 + |\\beta|^2 = 0$, so $\\alpha = \\beta = 0$.\n\nFinal Answer: The final answer is $\\boxed{a^2 + b^2 \\leq 2, A^2 + B^2 \\leq 1}$'
}

データフィールド

各フィールドは次のとおりである。

• solution: 問題の正解の解答（ground truth）
• question: 問題
• cot_type: 問題が属する大分類の 1 つ
• source_type: 問題の出典となったデータセットとそのカテゴリ
• metadata 元のデータセットにあるその他のフィールド
• cot: None
• thinking_trajectories: Gemini が生成した推論トレース。常に長さ 1 のリストである
• attempt: Gemini が生成した応答

引用情報

@misc{muennighoff2025s1simpletesttimescaling,
      title={s1: Simple test-time scaling}, 
      author={Niklas Muennighoff and Zitong Yang and Weijia Shi and Xiang Lisa Li and Li Fei-Fei and Hannaneh Hajishirzi and Luke Zettlemoyer and Percy Liang and Emmanuel Candès and Tatsunori Hashimoto},
      year={2025},
      eprint={2501.19393},
      archivePrefix={arXiv},
      primaryClass={cs.CL},
      url={https://arxiv.org/abs/2501.19393}, 
}

原文（英語）

Dataset Card for s1K

Dataset Summary

s1K is a dataset of 1,000 examples of diverse, high-quality & difficult questions with distilled reasoning traces & solutions from Gemini Thining. Refer to the s1 paper for more details.

Usage

# pip install -q datasets
from datasets import load_dataset
ds = load_dataset("simplescaling/s1K")["train"]
ds[0]

Dataset Structure

Data Instances

An example looks as follows:

{
    'solution': '1. **Rewrite the function using trigonometric identities:**...(shortened)',
    'question': 'Let  $a,b,A,B$  be given reals. We consider the function defined by \\[ f(x) = 1 - a \\cdot \\cos(x) - b \\cdot \\sin(x) - A \\cdot \\cos(2x) - B \\cdot \\sin(2x). \\] Prove that if for any real number  $x$  we have  $f(x) \\geq 0$  then  $a^2 + b^2 \\leq 2$  and  $A^2 + B^2 \\leq 1.$ ',
    'cot_type': 'math',
    'source_type': 'AI-MO/NuminaMath-CoT/aops_forum',
    'metadata': "{'source': 'aops_forum', 'messages': [{'content': 'Let  $a,b,A,B$  be given reals. We consider the function defined by \\\\[ f(x) = 1 - a \\\\cdot \\\\cos(x) - b \\\\cdot \\\\sin(x) - A \\\\cdot \\\\cos(2x) - B \\\\cdot \\\\sin(2x). \\\\] Prove that if for any real number  $x$  we have  $f(x) \\\\geq 0$  then  $a^2 + b^2 \\\\leq 2$  and  $A^2 + B^2 \\\\leq 1.$ ', 'role': 'user'}, {'content': '1. **Rewrite the function using trigonometric identities:**\\n   \\\\[\\n   f(x) = 1 - a \\\\cos(x) - b \\\\sin(x) - A \\\\cos(2x) - B \\\\sin(2x)\\n   \\\\]\\n   We can use the angle addition formulas for sine and cosine:\\n   \\\\[\\n   \\\\cos(x + \\\\theta) = \\\\cos(x)\\\\cos(\\\\theta) - \\\\sin(x)\\\\sin(\\\\theta)\\n   \\\\]\\n   \\\\[\\n   \\\\sin(x + \\\\theta) = \\\\sin(x)\\\\cos(\\\\theta) + \\\\cos(x)\\\\sin(\\\\theta)\\n   \\\\]\\n   Let:\\n   \\\\[\\n   R_1 = \\\\sqrt{a^2 + b^2}, \\\\quad \\\\theta_1 = \\\\arctan\\\\left(\\\\frac{b}{a}\\\\right)\\n   \\\\]\\n   \\\\[\\n   R_2 = \\\\sqrt{A^2 + B^2}, \\\\quad \\\\theta_2 = \\\\arctan\\\\left(\\\\frac{B}{A}\\\\right)\\n   \\\\]\\n   Then:\\n   \\\\[\\n   a \\\\cos(x) + b \\\\sin(x) = R_1 \\\\sin(x + \\\\theta_1)\\n   \\\\]\\n   \\\\[\\n   A \\\\cos(2x) + B \\\\sin(2x) = R_2 \\\\sin(2x + \\\\theta_2)\\n   \\\\]\\n   Therefore, the function becomes:\\n   \\\\[\\n   f(x) = 1 - R_1 \\\\sin(x + \\\\theta_1) - R_2 \\\\sin(2x + \\\\theta_2)\\n   \\\\]\\n\\n2. **Analyze the function for non-negativity:**\\n   Given \\\\( f(x) \\\\geq 0 \\\\) for all \\\\( x \\\\), we need to ensure that the maximum values of the sine terms do not exceed 1:\\n   \\\\[\\n   1 - R_1 \\\\sin(x + \\\\theta_1) - R_2 \\\\sin(2x + \\\\theta_2) \\\\geq 0\\n   \\\\]\\n   The maximum value of \\\\( \\\\sin(x + \\\\theta_1) \\\\) and \\\\( \\\\sin(2x + \\\\theta_2) \\\\) is 1. Therefore:\\n   \\\\[\\n   1 - R_1 - R_2 \\\\geq 0\\n   \\\\]\\n   This implies:\\n   \\\\[\\n   R_1 + R_2 \\\\leq 1\\n   \\\\]\\n\\n3. **Consider the individual terms:**\\n   Since \\\\( R_1 = \\\\sqrt{a^2 + b^2} \\\\) and \\\\( R_2 = \\\\sqrt{A^2 + B^2} \\\\), we need to show:\\n   \\\\[\\n   \\\\sqrt{a^2 + b^2} \\\\leq 1 \\\\quad \\\\text{and} \\\\quad \\\\sqrt{A^2 + B^2} \\\\leq 1\\n   \\\\]\\n\\n4. **Prove the inequalities:**\\n   - For \\\\( \\\\sqrt{a^2 + b^2} \\\\leq 1 \\\\):\\n     \\\\[\\n     a^2 + b^2 \\\\leq 1^2 = 1\\n     \\\\]\\n   - For \\\\( \\\\sqrt{A^2 + B^2} \\\\leq 1 \\\\):\\n     \\\\[\\n     A^2 + B^2 \\\\leq 1^2 = 1\\n     \\\\]\\n\\n5. **Conclusion:**\\n   Therefore, we have shown that if \\\\( f(x) \\\\geq 0 \\\\) for all \\\\( x \\\\), then:\\n   \\\\[\\n   a^2 + b^2 \\\\leq 2 \\\\quad \\\\text{and} \\\\quad A^2 + B^2 \\\\leq 1\\n   \\\\]\\n\\n\\\\(\\\\blacksquare\\\\)\\n\\nThe final answer is \\\\( \\\\boxed{ a^2 + b^2 \\\\leq 2 } \\\\) and \\\\( A^2 + B^2 \\\\leq 1 \\\\)', 'role': 'assistant'}]}",
    'cot': None,
    'thinking_trajectories': ["Let $f(x) = 1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x \\geq 0$ for all $x \\in \\mathbb{R}$.\nWe want to show that $a^2 + b^2 \\leq 2$ and $A^2 + B^2 \\leq 1$.\n\nConsider specific values of $x$.\nFor $x=0$, $f(0) = 1 - a - A \\geq 0$, so $a+A \\leq 1$.\nFor $x=\\pi$, $f(\\pi) = 1 - a \\cos \\pi - b \\sin \\pi - A \\cos 2\\pi - B \\sin 2\\pi = 1 - a(-1) - b(0) - A(1) - B(0) = 1+a-A \\geq 0$, so $A-a \\leq 1$.\nFor $x=\\pi/2$, $f(\\pi/2) = 1 - a \\cos(\\pi/2) - b \\sin(\\pi/2) - A \\cos(\\pi) - B \\sin(\\pi) = 1 - a(0) - b(1) - A(-1) - B(0) = 1 - b + A \\geq 0$, so $b-A \\leq 1$.\nFor $x=3\\pi/2$, $f(3\\pi/2) = 1 - a \\cos(3\\pi/2) - b \\sin(3\\pi/2) - A \\cos(3\\pi) - B \\sin(3\\pi) = 1 - a(0) - b(-1) - A(-1) - B(0) = 1 + b + A \\geq 0$, so $-b-A \\leq 1$.\n\nFrom $a+A \\leq 1$ and $A-a \\leq 1$, we have $-1 \\leq a+A \\leq 1$ and $-1 \\leq a-A \\leq 1$.\nFrom $b-A \\leq 1$ and $-b-A \\leq 1$, we have $-1 \\leq b-A \\leq 1$ and $-1 \\leq b+A \\leq 1$.\n\nConsider the integral of $f(x)$ over a period.\n$\\int_0^{2\\pi} f(x) dx = \\int_0^{2\\pi} (1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x) dx \\geq 0$.\n$\\int_0^{2\\pi} 1 dx = 2\\pi$.\n$\\int_0^{2\\pi} \\cos x dx = [\\sin x]_0^{2\\pi} = 0$.\n$\\int_0^{2\\pi} \\sin x dx = [-\\cos x]_0^{2\\pi} = -1 - (-1) = 0$.\n$\\int_0^{2\\pi} \\cos 2x dx = [\\frac{1}{2} \\sin 2x]_0^{2\\pi} = 0$.\n$\\int_0^{2\\pi} \\sin 2x dx = [-\\frac{1}{2} \\cos 2x]_0^{2\\pi} = -\\frac{1}{2} - (-\\frac{1}{2}) = 0$.\nSo $\\int_0^{2\\pi} f(x) dx = 2\\pi \\geq 0$. This doesn't give any information.\n\nMultiply $f(x)$ by $\\cos x$ and integrate.\n$\\int_0^{2\\pi} f(x) \\cos x dx = \\int_0^{2\\pi} (1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x) \\cos x dx \\geq 0$.\n$\\int_0^{2\\pi} \\cos x dx = 0$.\n$\\int_0^{2\\pi} \\cos^2 x dx = \\int_0^{2\\pi} \\frac{1+\\cos 2x}{2} dx = [\\frac{x}{2} + \\frac{\\sin 2x}{4}]_0^{2\\pi} = \\pi$.\n$\\int_0^{2\\pi} \\sin x \\cos x dx = \\int_0^{2\\pi} \\frac{1}{2} \\sin 2x dx = [-\\frac{1}{4} \\cos 2x]_0^{2\\pi} = -\\frac{1}{4} - (-\\frac{1}{4}) = 0$.\n$\\int_0^{2\\pi} \\cos 2x \\cos x dx = \\int_0^{2\\pi} \\frac{1}{2} (\\cos 3x + \\cos x) dx = [\\frac{1}{6} \\sin 3x + \\frac{1}{2} \\sin x]_0^{2\\pi} = 0$.\n$\\int_0^{2\\pi} \\sin 2x \\cos x dx = \\int_0^{2\\pi} \\frac{1}{2} (\\sin 3x + \\sin x) dx = [-\\frac{1}{6} \\cos 3x - \\frac{1}{2} \\cos x]_0^{2\\pi} = (-\\frac{1}{6} - \\frac{1}{2}) - (-\\frac{1}{6} - \\frac{1}{2}) = 0$.\nSo $\\int_0^{2\\pi} f(x) \\cos x dx = 0 - a \\pi - 0 - 0 - 0 = -a\\pi \\geq 0$, so $a \\leq 0$. This is not necessarily true.\n\nConsider $f(x) \\geq 0$.\n$1 - (a \\cos x + b \\sin x) - (A \\cos 2x + B \\sin 2x) \\geq 0$.\nLet $a = r \\cos \\alpha, b = r \\sin \\alpha$. Then $a \\cos x + b \\sin x = r (\\cos \\alpha \\cos x + \\sin \\alpha \\sin x) = r \\cos(x-\\alpha)$.\nLet $A = R \\cos \\beta, B = R \\sin \\beta$. Then $A \\cos 2x + B \\sin 2x = R (\\cos \\beta \\cos 2x + \\sin \\beta \\sin 2x) = R \\cos(2x-\\beta)$.\nSo $1 - r \\cos(x-\\alpha) - R \\cos(2x-\\beta) \\geq 0$.\nLet $y = x-\\alpha$. Then $2x-\\beta = 2(y+\\alpha) - \\beta = 2y + 2\\alpha - \\beta$. Let $\\gamma = 2\\alpha - \\beta$.\n$1 - r \\cos y - R \\cos(2y+\\gamma) \\geq 0$.\n$1 - r \\cos y - R (\\cos 2y \\cos \\gamma - \\sin 2y \\sin \\gamma) \\geq 0$.\n$1 - r \\cos y - R ((2\\cos^2 y - 1) \\cos \\gamma - 2 \\sin y \\cos y \\sin \\gamma) \\geq 0$.\n$1 - r \\cos y - 2R \\cos^2 y \\cos \\gamma + R \\cos \\gamma + 2R \\sin y \\cos y \\sin \\gamma \\geq 0$.\n\nConsider $f(x) \\geq 0$.\nMultiply by $\\cos(x-\\phi)$ and integrate from $0$ to $2\\pi$.\n$\\int_0^{2\\pi} f(x) \\cos(x-\\phi) dx \\geq 0$.\n$\\int_0^{2\\pi} (1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x) (\\cos x \\cos \\phi + \\sin x \\sin \\phi) dx \\geq 0$.\n$\\cos \\phi \\int_0^{2\\pi} (\\cos x - a \\cos^2 x - b \\sin x \\cos x - A \\cos 2x \\cos x - B \\sin 2x \\cos x) dx + \\sin \\phi \\int_0^{2\\pi} (\\sin x - a \\cos x \\sin x - b \\sin^2 x - A \\cos 2x \\sin x - B \\sin 2x \\sin x) dx \\geq 0$.\n$\\cos \\phi (0 - a\\pi - 0 - 0 - 0) + \\sin \\phi (0 - 0 - b\\pi - 0 - 0) \\geq 0$.\n$-a\\pi \\cos \\phi - b\\pi \\sin \\phi \\geq 0$.\n$a \\cos \\phi + b \\sin \\phi \\leq 0$.\nThis must be true for all $\\phi$. Let $\\phi$ such that $\\tan \\phi = b/a$. Then $a \\frac{a}{\\sqrt{a^2+b^2}} + b \\frac{b}{\\sqrt{a^2+b^2}} = \\sqrt{a^2+b^2} \\leq 0$, so $a=b=0$. This is wrong.\n\nConsider $f(x) \\geq 0$.\n$1 - (a \\cos x + b \\sin x) - (A \\cos 2x + B \\sin 2x) \\geq 0$.\nThe minimum value of $a \\cos x + b \\sin x$ is $-\\sqrt{a^2+b^2}$.\nThe minimum value of $A \\cos 2x + B \\sin 2x$ is $-\\sqrt{A^2+B^2}$.\n$1 - (a \\cos x + b \\sin x) \\geq A \\cos 2x + B \\sin 2x$.\nMinimum of LHS is $1 - \\sqrt{a^2+b^2}$. Maximum of RHS is $\\sqrt{A^2+B^2}$.\n$1 - \\sqrt{a^2+b^2} \\geq \\sqrt{A^2+B^2}$.\n\nConsider $g(x) = 1 - A \\cos 2x - B \\sin 2x \\geq a \\cos x + b \\sin x$.\nMinimum of LHS is $1 - \\sqrt{A^2+B^2}$. Maximum of RHS is $\\sqrt{a^2+b^2}$.\n$1 - \\sqrt{A^2+B^2} \\geq \\sqrt{a^2+b^2}$.\n\nLet $x=0$, $1-a-A \\geq 0$.\nLet $x=\\pi$, $1+a-A \\geq 0$.\nLet $x=\\pi/2$, $1-b+A \\geq 0$.\nLet $x=3\\pi/2$, $1+b+A \\geq 0$.\n\nConsider the case where $b=B=0$.\n$f(x) = 1 - a \\cos x - A \\cos 2x \\geq 0$.\n$1 - a \\cos x - A (2 \\cos^2 x - 1) \\geq 0$.\n$1 - a \\cos x - 2A \\cos^2 x + A \\geq 0$.\n$A+1 - a \\cos x - 2A \\cos^2 x \\geq 0$.\nLet $y = \\cos x$, $-1 \\leq y \\leq 1$.\n$g(y) = -2Ay^2 - ay + A+1 \\geq 0$ for $y \\in [-1, 1]$.\nIf $A=0$, $-ay+1 \\geq 0$ for $y \\in [-1, 1]$. If $a>0$, $-a+1 \\geq 0$, $a \\leq 1$. If $a<0$, $a+1 \\geq 0$, $a \\geq -1$. So $|a| \\leq 1$. $a^2 \\leq 1 \\leq 2$.\nIf $A>0$, the parabola opens downwards. We need $g(-1) = -2A+a+A+1 = -A+a+1 \\geq 0$, $a \\geq A-1$. And $g(1) = -2A-a+A+1 = -A-a+1 \\geq 0$, $a \\leq 1-A$. Also the vertex must be outside $[-1, 1]$ or the vertex is in $[-1, 1]$ and the value is non-negative. Vertex at $y = -(-a)/(2(-2A)) = -a/4A$. If $|-a/4A| \\leq 1$, then $a^2 \\leq 16A^2$. $g(-a/4A) = -2A (-a/4A)^2 - a(-a/4A) + A+1 = -2A a^2/16A^2 + a^2/4A + A+1 = -a^2/8A + a^2/4A + A+1 = a^2/8A + A+1 \\geq 0$.\nIf $A<0$, the parabola opens upwards. We need $g(-1) \\geq 0$ and $g(1) \\geq 0$.\n$-A+a+1 \\geq 0$, $a \\geq A-1$.\n$-A-a+1 \\geq 0$, $a \\leq 1-A$.\n\nConsider integrals again.\n$\\int_0^{2\\pi} f(x) dx \\geq 0 \\implies 2\\pi \\geq 0$.\n$\\int_0^{2\\pi} f(x) \\cos x dx \\geq 0 \\implies -a \\pi \\geq 0 \\implies a \\leq 0$.\n$\\int_0^{2\\pi} f(x) \\sin x dx \\geq 0 \\implies -b \\pi \\geq 0 \\implies b \\leq 0$.\n$\\int_0^{2\\pi} f(x) \\cos 2x dx \\geq 0 \\implies -A \\pi \\geq 0 \\implies A \\leq 0$.\n$\\int_0^{2\\pi} f(x) \\sin 2x dx \\geq 0 \\implies -B \\pi \\geq 0 \\implies B \\leq 0$.\nThis is not true.\n\nConsider the vector $v = (1, \\cos x, \\sin x, \\cos 2x, \\sin 2x)$.\nThe condition is $v \\cdot c \\geq 0$, where $c = (1, -a, -b, -A, -B)$.\n\nConsider the case where $f(x) = (1 - \\alpha \\cos x - \\beta \\sin x)^2 \\geq 0$.\n$1 + \\alpha^2 \\cos^2 x + \\beta^2 \\sin^2 x - 2 \\alpha \\cos x - 2 \\beta \\sin x + 2 \\alpha \\beta \\cos x \\sin x \\geq 0$.\n$1 + \\frac{\\alpha^2+\\beta^2}{2} + \\frac{\\alpha^2-\\beta^2}{2} \\cos 2x - 2 \\alpha \\cos x - 2 \\beta \\sin x + \\alpha \\beta \\sin 2x \\geq 0$.\nCompare with $1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x \\geq 0$.\n$-a = -2\\alpha$, $-b = -2\\beta$, $-A = \\frac{\\alpha^2-\\beta^2}{2}$, $-B = \\alpha \\beta$.\n$a = 2\\alpha$, $b = 2\\beta$.\n$A = \\frac{\\beta^2-\\alpha^2}{2}$, $B = -\\alpha \\beta$.\n$a^2+b^2 = 4\\alpha^2+4\\beta^2$.\n$A^2+B^2 = \\frac{(\\beta^2-\\alpha^2)^2}{4} + \\alpha^2 \\beta^2 = \\frac{\\beta^4 - 2\\alpha^2 \\beta^2 + \\alpha^4 + 4\\alpha^2 \\beta^2}{4} = \\frac{(\\alpha^2+\\beta^2)^2}{4}$.\n\nConsider $f(x) \\geq 0$.\nConsider the Fourier series of $|f(x)| = f(x)$.\nThe coefficients must match.\n\nLet $z = e^{ix}$.\n$f(x) = 1 - \\frac{a}{2}(z+z^{-1}) - \\frac{b}{2i}(z-z^{-1}) - \\frac{A}{2}(z^2+z^{-2}) - \\frac{B}{2i}(z^2-z^{-2}) \\geq 0$.\n$2if(x) = 2i - ia(z+z^{-1}) + b(z-z^{-1}) - iA(z^2+z^{-2}) + B(z^2-z^{-2}) \\geq 0$.\n$2if(x) = 2i - i a z - i a z^{-1} + b z - b z^{-1} - i A z^2 - i A z^{-2} + B z^2 - B z^{-2} \\geq 0$.\n$2if(x) = 2i + (b-ia)z + (-b-ia)z^{-1} + (B-iA)z^2 + (-B-iA)z^{-2} \\geq 0$.\n\nConsider the integral $\\frac{1}{2\\pi} \\int_0^{2\\pi} f(x) e^{inx} dx$.\nFor $n=0$, $\\frac{1}{2\\pi} \\int_0^{2\\pi} f(x) dx = 1 \\geq 0$.\nFor $n=1$, $\\frac{1}{2\\pi} \\int_0^{2\\pi} f(x) (\\cos x + i \\sin x) dx = \\frac{1}{2\\pi} \\int_0^{2\\pi} (1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x) (\\cos x + i \\sin x) dx = -a/2 - ib/2$.\nSince $f(x) \\geq 0$, the coefficients of the Fourier series must satisfy certain conditions.\n\nConsider trigonometric polynomials that are non-negative.\nIf $P(\\theta) = \\sum_{k=-n}^n c_k e^{ik\\theta} \\geq 0$, then $c_{-k} = \\overline{c_k}$ and $c_0 \\geq 0$.\nIn our case, the coefficients are:\n$c_0 = 1$.\n$c_1 = -a/2 - ib/2$.\n$c_{-1} = -a/2 + ib/2$.\n$c_2 = -A/2 - iB/2$.\n$c_{-2} = -A/2 + iB/2$.\nWe need a condition on the coefficients such that the polynomial is non-negative.\n\nIf $1 - a \\cos x - b \\sin x \\geq 0$, then $1 \\geq \\sqrt{a^2+b^2}$.\nIf $1 - A \\cos 2x - B \\sin 2x \\geq 0$, then $1 \\geq \\sqrt{A^2+B^2}$.\n\nConsider $f(x) = |1 - \\alpha e^{ix} - \\beta e^{2ix}|^2 \\geq 0$.\n$(1 - \\alpha e^{ix} - \\beta e^{2ix})(1 - \\bar{\\alpha} e^{-ix} - \\bar{\\beta} e^{-2ix})$.\n$1 - \\bar{\\alpha} e^{-ix} - \\bar{\\beta} e^{-2ix} - \\alpha e^{ix} + |\\alpha|^2 - \\alpha \\bar{\\beta} e^{-ix} - \\beta e^{2ix} - \\beta \\bar{\\alpha} e^{ix} + |\\beta|^2$.\n$1 + |\\alpha|^2 + |\\beta|^2 - (\\bar{\\alpha} + \\alpha + \\alpha \\bar{\\beta}) e^{-ix} - (\\bar{\\beta} + \\beta + \\beta \\bar{\\alpha}) e^{-2ix}$.\n$1 + |\\alpha|^2 + |\\beta|^2 - (2 \\text{Re}(\\alpha) + \\alpha \\bar{\\beta}) e^{-ix} - (2 \\text{Re}(\\beta) + \\beta \\bar{\\alpha}) e^{-2ix}$.\nCompare coefficients.\n\nIf $P(\\theta) = |\\sum_{k=0}^n z_k e^{ik\\theta}|^2$, then $P(\\theta) \\geq 0$.\nConsider $P(x) = |1 - \\alpha e^{ix}|^2 = (1-\\alpha e^{ix})(1-\\bar{\\alpha} e^{-ix}) = 1 - \\bar{\\alpha} e^{-ix} - \\alpha e^{ix} + |\\alpha|^2 = 1+|\\alpha|^2 - 2 \\text{Re}(\\alpha) \\cos x + 2 \\text{Im}(\\alpha) \\sin x$.\nCompare with $1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x$.\n\nConsider $f(x) \\geq 0$.\n$1 - (a \\cos x + b \\sin x) - (A \\cos 2x + B \\sin 2x) \\geq 0$.\nLet $a = r_1 \\cos \\phi_1, b = r_1 \\sin \\phi_1$.\nLet $A = r_2 \\cos \\phi_2, B = r_2 \\sin \\phi_2$.\n$1 - r_1 \\cos(x-\\phi_1) - r_2 \\cos(2x-\\phi_2) \\geq 0$.\nLet $x - \\phi_1 = y$. Then $2x - \\phi_2 = 2(y+\\phi_1) - \\phi_2 = 2y + 2\\phi_1 - \\phi_2$. Let $\\delta = 2\\phi_1 - \\phi_2$.\n$1 - r_1 \\cos y - r_2 \\cos(2y+\\delta) \\geq 0$.\n$1 - r_1 \\cos y - r_2 (\\cos 2y \\cos \\delta - \\sin 2y \\sin \\delta) \\geq 0$.\n$1 - r_1 \\cos y - r_2 ( (2\\cos^2 y - 1) \\cos \\delta - 2 \\sin y \\cos y \\sin \\delta ) \\geq 0$.\n$1 - r_1 \\cos y - 2 r_2 \\cos^2 y \\cos \\delta + r_2 \\cos \\delta + 2 r_2 \\sin y \\cos y \\sin \\delta \\geq 0$.\n$1 + r_2 \\cos \\delta - r_1 \\cos y - 2 r_2 \\cos \\delta \\cos^2 y + r_2 \\sin \\delta \\sin 2y \\geq 0$.\n\nIf $f(x) \\geq 0$, then the average value is non-negative.\nConsider $1 - a \\cos x - b \\sin x \\geq 0$. Square both sides? No.\nConsider $1 - A \\cos 2x - B \\sin 2x \\geq |a \\cos x + b \\sin x|$.\n$1 - \\sqrt{A^2+B^2} \\leq 1 - A \\cos 2x - B \\sin 2x \\leq 1 + \\sqrt{A^2+B^2}$.\n$|a \\cos x + b \\sin x| \\leq \\sqrt{a^2+b^2}$.\nSo $1 - \\sqrt{A^2+B^2} \\geq \\sqrt{a^2+b^2}$. This is not necessarily true.\n\nConsider $f(x) \\geq 0$.\n$\\int_0^{2\\pi} f(x) d x = 2\\pi$.\n$\\int_0^{2\\pi} f(x) \\cos x d x = -a\\pi$.\n$\\int_0^{2\\pi} f(x) \\sin x d x = -b\\pi$.\n$\\int_0^{2\\pi} f(x) \\cos 2x d x = -A\\pi$.\n$\\int_0^{2\\pi} f(x) \\sin 2x d x = -B\\pi$.\nSince $f(x) \\geq 0$, consider the Fejer-Riesz theorem. A non-negative trigonometric polynomial can be written as the squared modulus of another trigonometric polynomial.\n$1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x = |c_0 + c_1 e^{ix} + c_2 e^{2ix}|^2$.\n$= (c_0 + c_1 e^{ix} + c_2 e^{2ix})(\\bar{c_0} + \\bar{c_1} e^{-ix} + \\bar{c_2} e^{-2ix})$.\n$= |c_0|^2 + |c_1|^2 + |c_2|^2 + (c_0 \\bar{c_1} + c_1 \\bar{c_2}) e^{ix} + (\\bar{c_0} c_1 + \\bar{c_1} c_2) e^{-ix} + c_0 \\bar{c_2} e^{2ix} + \\bar{c_0} c_2 e^{-2ix}$.\n$= |c_0|^2 + |c_1|^2 + |c_2|^2 + 2 \\text{Re}(c_0 \\bar{c_1} + c_1 \\bar{c_2}) \\cos x - 2 \\text{Im}(c_0 \\bar{c_1} + c_1 \\bar{c_2}) \\sin x + 2 \\text{Re}(c_0 \\bar{c_2}) \\cos 2x - 2 \\text{Im}(c_0 \\bar{c_2}) \\sin 2x$.\nCompare coefficients:\n$1 = |c_0|^2 + |c_1|^2 + |c_2|^2$.\n$-a = 2 \\text{Re}(c_0 \\bar{c_1} + c_1 \\bar{c_2})$.\n$-b = -2 \\text{Im}(c_0 \\bar{c_1} + c_1 \\bar{c_2})$.\n$-A = 2 \\text{Re}(c_0 \\bar{c_2})$.\n$-B = -2 \\text{Im}(c_0 \\bar{c_2})$.\n$A^2+B^2 = 4 (\\text{Re}(c_0 \\bar{c_2})^2 + \\text{Im}(c_0 \\bar{c_2})^2) = 4 |c_0 \\bar{c_2}|^2 = 4 |c_0|^2 |c_2|^2$.\nLet $c_0 = 1$.\n$1 = 1 + |c_1|^2 + |c_2|^2 \\implies |c_1|^2 + |c_2|^2 = 0 \\implies c_1 = c_2 = 0$. Then $f(x) = 1$, $a=b=A=B=0$.\n"],
    'attempt': 'Let $P(x) = 1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x$.\nSince $P(x) \\geq 0$ for all real $x$, by Fejer-Riesz theorem, there exists a polynomial $Q(z) = c_0 + c_1 z + c_2 z^2$ such that $P(x) = |Q(e^{ix})|^2$.\nWe have $|Q(e^{ix})|^2 = (c_0 + c_1 e^{ix} + c_2 e^{2ix})(\\bar{c_0} + \\bar{c_1} e^{-ix} + \\bar{c_2} e^{-2ix})$.\nExpanding this, we get\n\\begin{align*} |Q(e^{ix})|^2 &= |c_0|^2 + |c_1|^2 + |c_2|^2 \\\\ &+ (c_0 \\bar{c_1} + c_1 \\bar{c_2}) e^{ix} + (\\bar{c_0} c_1 + \\bar{c_1} c_2) e^{-ix} \\\\ &+ c_0 \\bar{c_2} e^{2ix} + \\bar{c_0} c_2 e^{-2ix} \\\\ &= |c_0|^2 + |c_1|^2 + |c_2|^2 \\\\ &+ (c_0 \\bar{c_1} + \\bar{\\bar{c_1}} c_2 + \\overline{c_0 \\bar{c_1} + \\bar{\\bar{c_1}} c_2}) \\cos x + i (c_0 \\bar{c_1} + \\bar{\\bar{c_1}} c_2 - \\overline{c_0 \\bar{c_1} + \\bar{\\bar{c_1}} c_2}) \\sin x \\\\ &+ (c_0 \\bar{c_2} + \\bar{c_0} c_2) \\cos 2x + i (c_0 \\bar{c_2} - \\bar{c_0} c_2) \\sin 2x \\\\ &= |c_0|^2 + |c_1|^2 + |c_2|^2 + 2 \\text{Re}(c_0 \\bar{c_1} + c_1 \\bar{c_2}) \\cos x - 2 \\text{Im}(c_0 \\bar{c_1} + c_1 \\bar{c_2}) \\sin x + 2 \\text{Re}(c_0 \\bar{c_2}) \\cos 2x - 2 \\text{Im}(c_0 \\bar{c_2}) \\sin 2x\\end{align*}\nComparing the coefficients of $P(x)$ with $|Q(e^{ix})|^2$, we have:\n$1 = |c_0|^2 + |c_1|^2 + |c_2|^2$\n$-a = 2 \\text{Re}(c_0 \\bar{c_1} + c_1 \\bar{c_2})$\n$-b = -2 \\text{Im}(c_0 \\bar{c_1} + c_1 \\bar{c_2})$\n$-A = 2 \\text{Re}(c_0 \\bar{c_2})$\n$-B = -2 \\text{Im}(c_0 \\bar{c_2})$\nFrom the last two equations, $A^2 + B^2 = 4 (\\text{Re}(c_0 \\bar{c_2})^2 + \\text{Im}(c_0 \\bar{c_2})^2) = 4 |c_0 \\bar{c_2}|^2 = 4 |c_0|^2 |c_2|^2$.\nSince $1 = |c_0|^2 + |c_1|^2 + |c_2|^2$, we have $|c_0|^2 \\leq 1$ and $|c_2|^2 \\leq 1$.\nThus $A^2 + B^2 = 4 |c_0|^2 |c_2|^2 \\leq 4 \\cdot 1 \\cdot 1 = 4$. This is not the desired result.\n\nConsider $Q(z) = 1 - \\alpha z - \\beta z^2$. Then $|Q(e^{ix})|^2 = |1 - \\alpha e^{ix} - \\beta e^{2ix}|^2$.\n$|1 - \\alpha e^{ix} - \\beta e^{2ix}|^2 = (1 - \\alpha e^{ix} - \\beta e^{2ix})(1 - \\bar{\\alpha} e^{-ix} - \\bar{\\beta} e^{-2ix})$\n$= 1 - \\bar{\\alpha} e^{-ix} - \\bar{\\beta} e^{-2ix} - \\alpha e^{ix} + |\\alpha|^2 - \\alpha \\bar{\\beta} e^{-ix} - \\beta e^{2ix} - \\beta \\bar{\\alpha} e^{ix} + |\\beta|^2$\n$= 1 + |\\alpha|^2 + |\\beta|^2 - (\\alpha + \\alpha \\bar{\\beta} + \\bar{\\alpha}) \\cos x + i (\\alpha + \\alpha \\bar{\\beta} - \\bar{\\alpha}) \\sin x - (\\beta + \\beta \\bar{\\alpha} + \\bar{\\beta}) \\cos 2x + i (\\beta + \\beta \\bar{\\alpha} - \\bar{\\beta}) \\sin 2x$.\nCompare with $1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x$.\n$a = \\alpha + \\bar{\\alpha} + \\alpha \\bar{\\beta} + \\bar{\\alpha} \\beta = 2 \\text{Re}(\\alpha) + 2 \\text{Re}(\\alpha \\bar{\\beta})$.\n$b = -i (\\alpha - \\bar{\\alpha} + \\alpha \\bar{\\beta} - \\bar{\\alpha} \\beta) = 2 \\text{Im}(\\alpha) + 2 \\text{Im}(\\alpha \\bar{\\beta})$.\n$A = \\beta + \\bar{\\beta} = 2 \\text{Re}(\\beta)$.\n$B = -i (\\beta - \\bar{\\beta}) = 2 \\text{Im}(\\beta)$.\n$A^2+B^2 = 4 (\\text{Re}(\\beta)^2 + \\text{Im}(\\beta)^2) = 4 |\\beta|^2$.\n$a^2+b^2 = |2\\alpha + 2\\alpha \\bar{\\beta}|^2 = 4 |\\alpha|^2 |1+\\bar{\\beta}|^2$.\n$1 = 1 + |\\alpha|^2 + |\\beta|^2$. So $|\\alpha|^2 + |\\beta|^2 = 0$, so $\\alpha = \\beta = 0$.\n\nFinal Answer: The final answer is $\\boxed{a^2 + b^2 \\leq 2, A^2 + B^2 \\leq 1}$'
}

Data Fields

The data fields are:

• solution: The ground truth solution for the question
• question: The question
• cot_type: One high-level category the question falls into
• source_type: The dataset and its category that the question stems from
• metadata Any other fields in the original dataset
• cot: None
• thinking_trajectories: The reasoning trace generated by Gemini; It is always a list of length 1
• attempt: The response generated by Gemini

Citation Information

@misc{muennighoff2025s1simpletesttimescaling,
      title={s1: Simple test-time scaling}, 
      author={Niklas Muennighoff and Zitong Yang and Weijia Shi and Xiang Lisa Li and Li Fei-Fei and Hannaneh Hajishirzi and Luke Zettlemoyer and Percy Liang and Emmanuel Candès and Tatsunori Hashimoto},
      year={2025},
      eprint={2501.19393},
      archivePrefix={arXiv},
      primaryClass={cs.CL},
      url={https://arxiv.org/abs/2501.19393}, 
}