[離散] 離散(3)

看板Math作者 (批踢踢基金只進不出)時間20年前 (2005/07/24 16:51), 編輯推噓1(1推 0噓 5→)
留言6則, 2人參與, 最新討論串1/1
The number f(n) of steps required to solve the "chinese rings puzzle" with n ring satisfies f(1) = 1 and f(n+1) = 2f(n) if n is odd, 2f(n) + 1 if n is even. Prove that f(n+2) = f(n+1) + 2f(n) + 1. Hence or otherwise find a formula for f(n) in term of n. -- 我好窮啊，我好缺批幣啊 ，你有摳摳ㄋㄟ 可憐可憐我吧，施捨一點吧 請到(P)LAY-->(P)AY-->(0)GIVE-->PttFund-->吧 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.112.218.142
推
61.229.112.108 07/25, , 1F
2^(n+1)/3 + (-1)^(n+1)/6 - 1/2
61.229.112.108 07/25, 1F
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04/29 23:10, , 2F
我也做出來了，謝謝JGU的答案讓我能把答案更化簡
04/29 23:10, 2F
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04/29 23:10, , 3F
本來我做出來的是
04/29 23:10, 3F
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04/29 23:11, , 4F
當 n 是奇數時，f(n) = 2^(n+1)/3 - 1/3
04/29 23:11, 4F
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04/29 23:11, , 5F
當 n 是偶數時，f(n) = 2^(n+1)/3 - 2/3
04/29 23:11, 5F
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04/29 23:13, , 6F
加強了對振盪數列的體會，感恩啊！
04/29 23:13, 6F
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文章代碼(AID): #12urQOCu (Math)