[{"data":1,"prerenderedAt":161},["ShallowReactive",2],{"topic-page:\u002Fhigh-school\u002Fmathematics-a\u002Fmathematics-and-human-activities\u002Ftower-of-hanoi-and-recursion":3},{"topic":4,"category":142,"seo":146,"breadcrumbs":151,"prerequisites":158,"nextTopics":159,"relatedTopics":160},{"id":5,"slug":6,"title":7,"summary":8,"canonicalPath":9,"categoryPath":10,"difficulty":11,"targetAudience":12,"estimatedMinutes":13,"generatedAt":14,"publishedAt":14,"updatedAt":14,"reviewStatus":15,"tags":16,"expectedKnowledge":22,"learningGoal":26,"blocks":27},"high-school-mathematics-a-mathematics-and-human-activities-tower-of-hanoi-and-recursion","tower-of-hanoi-and-recursion","ハノイの塔と再帰","ハノイの塔を一段小さい同じ問題へ分け、最小手数の再帰関係と、その式が成り立つ手順上の理由を理解します。","\u002Fhigh-school\u002Fmathematics-a\u002Fmathematics-and-human-activities\u002Ftower-of-hanoi-and-recursion","\u002Fhigh-school\u002Fmathematics-a\u002Fmathematics-and-human-activities","beginner","student_high",6,"2026-07-27","ai_generated",[17,18,19,20,21],"数学A","数学と人間の活動","ハノイの塔","再帰","最小手数",[23,24,25],"2の累乗","規則性","簡単な式の代入","n枚の移動をn−1枚の同じ問題へ分解し、最小手数の再帰関係と値を説明できる。",[28,34,44,67,73,82,88,110,116,132],{"id":29,"type":30,"title":31,"subtitle":32,"shortDefinition":33},"hero","HeroBlock","大きい問題を同じ小問題へ分ける","ハノイの塔で再帰を体験する","一見複雑なn枚の円盤移動も、最大の円盤を動かす前後にn−1枚の移動が必要だと見抜けば、同じ形の小さな問題二つと一手へ分解できます。手順と手数の理由を同時に説明できる分解です。",{"id":35,"type":36,"term":37,"definition":38,"plainExplanation":39,"keywords":40},"definition","DefinitionBlock","再帰的な分解","ある大きさの問題を、同じ種類で一段小さい問題を使って表し、最も小さい場合までたどる考え方です。","最大円盤を出発棒から目的棒へ動かすには、上のn−1枚を補助棒へよける必要があります。最大円盤を一手で動かした後、同じn−1枚の問題をもう一度解いて目的棒へ重ねます。n=1なら一手で終わるという基準があるので、分解をいつ止めるかも決まります。",[41,42,43],"同じ小問題","基準となる場合","前後に二回",{"id":45,"type":46,"title":47,"lead":48,"formulas":49,"strategy":66},"formula","FormulaBlock","最小手数の関係","n枚の最小手数をT(n)とすると、n−1枚の移動が前後に一回ずつ必要です。",[50,62],{"label":51,"expression":52,"meaning":53,"symbols":54,"tip":61},"再帰関係","T(n)=2T(n−1)+1","n−1枚を二回動かす手数に、最大円盤を動かす中央の一手を加えます。二回とも最小である必要があります。",[55,58],{"symbol":56,"meaning":57},"T(n)","n枚を移す最小手数",{"symbol":59,"meaning":60},"n","円盤の枚数","最初の値はT(1)=1です。ここから順にT(2)、T(3)を作れます。",{"label":21,"expression":63,"meaning":64,"tip":65},"T(n)=2^n−1","枚数が1増えるたびに、手数は前の2倍より1多くなります。指数のnは円盤枚数です。","n=1、2、3で1、3、7になるか確かめます。","式だけを覚えず、「上をよける・最大を動かす・上を戻す」の三段階へ対応させます。小問題を一手と数えないことも重要です。",{"id":68,"type":69,"src":70,"alt":71,"caption":72},"diagram","DiagramBlock","\u002Fassets\u002Fhigh-school\u002Fmathematics-a\u002Fmathematics-and-human-activities\u002Ftower-of-hanoi-and-recursion\u002Ftower-of-hanoi-and-recursion.svg","n枚のハノイの塔をnマイナス1枚、最大円盤1枚、nマイナス1枚の三段階へ分ける図","最大円盤を動かす一手を中央に置くと、その前後がどちらもn−1枚の同じ移動問題になります。塊の下のT(n−1)は一手ではなく、小問題全体の最小手数です。",{"id":74,"type":75,"steps":76,"title":81},"steps","StepBlock",[77,78,79,80],"上のn−1枚を、目的棒を一時利用して補助棒へ移す","空いた最大円盤を出発棒から目的棒へ1回動かす","n−1枚を、出発棒を一時利用して目的棒へ移す","n=1なら円盤を直接1回動かして分解を終える","n枚を移す三段階",{"id":83,"type":84,"scenario":85,"explanation":86,"result":87},"example","ExampleBlock","3枚の円盤を最小何手で移せるか。","まず上の2枚を補助棒へ移すのにT(2)=3手、最大円盤を目的棒へ1手、2枚を目的棒へ重ねるのに再び3手です。合計3+1+3=7手です。式でもT(3)=2×T(2)+1=2×3+1=7となります。前半か後半を3手未満にできないため、全体も7手未満にはできません。","3枚の最小手数は7手です。三段階は実行方法だけでなく、それより短くできない理由にもなっています。",{"id":89,"type":90,"columns":91,"rows":94,"summary":109},"compare","CompareBlock",[92,93,21],"枚数n","分解",[95,98,102,105],[96,97,96],"1","直接1手",[99,100,101],"2","1手分を二回+1","3",[101,103,104],"3手分を二回+1","7",[106,107,108],"4","7手分を二回+1","15","1、3、7、15と、前の値を2倍して1を加える規則が続きます。差ではなく、問題の分解から規則が導かれます。",{"id":111,"type":112,"warningTitle":113,"message":114,"severity":111,"action":115},"warning","WarningBlock","円盤をまとめて一手にしない","n−1枚を一つの塊として図に描いても、実際には一度に1枚しか動かせません。T(n−1)手を一手と数えてはいけません。","塊の下にT(n−1)手と書き、再帰的に必要な手数だと区別します。最大円盤の前後で二回必要なことも確認しましょう。",{"id":117,"type":118,"title":119,"questions":120},"quiz","QuizBlock","確認テスト",[121],{"question":122,"choices":123,"answerIndex":127,"choiceExplanations":128},"4枚の最小手数T(4)をT(3)=7から求める式はどれですか。",[124,125,126],"2×7+1=15","7+1=8","2×7=14",0,[129,130,131],"正解です。3枚の移動を最大円盤の前後に二回行い、中央の最大円盤の一手を加えます。","n−1枚の移動は最大円盤を空ける前半だけでなく、目的棒へ重ね直す後半にも必要です。","3枚の移動は二回ありますが、最大円盤を出発棒から目的棒へ動かす中央の一手が抜けています。",{"id":133,"type":134,"title":135,"points":136},"summary","SummaryBlock","まとめ",[137,138,139,140,141],"n枚の問題はn−1枚の同じ問題二つへ分かれる","最大円盤の一手を二つの小問題の間に置く","最小手数はT(n)=2T(n−1)+1で表せる","T(n−1)は一手でなく小問題全体の手数である","基準T(1)=1から2ⁿ−1へつながる",{"path":10,"title":18,"description":143,"parentPath":144,"accentToken":145},"数量や図形と人間の活動、整数の約数・倍数、ユークリッドの互除法、二進法、座標、数学史、ゲームやパズルを整理していくカテゴリです。","\u002Fhigh-school\u002Fmathematics-a","signal-blue",{"title":147,"description":148,"canonicalUrl":149,"ogImage":150},"ハノイの塔と再帰 | 数学A","ハノイの塔を小さい同じ問題へ分け、最小手数が2倍して1増える理由を学びます。","https:\u002F\u002Fzeqnilo.com\u002Fhigh-school\u002Fmathematics-a\u002Fmathematics-and-human-activities\u002Ftower-of-hanoi-and-recursion","https:\u002F\u002Fzeqnilo.com\u002Fassets\u002Fsite\u002Fdefault-og.svg",[152,155,156,157],{"title":153,"path":154},"高校","\u002Fhigh-school",{"title":17,"path":144},{"title":18,"path":10},{"title":7,"path":9},[],[],[],1785565320313]