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【正文】 例如我們會看到遞歸表達式（ ）通過等式兩邊加 1 來簡化： T0 + 1= 1。 2 找到并證明關(guān)系量的一個數(shù)學表達式。假設(shè)當 n 被 n?1 取代后（ ）成立且 n0： Tn= 2Tn1+1= 2(2n?1?1)+1=2n?1. 因此對于（ ），問題規(guī)模為 n 也成立。因此，我們依次計算T3=23+1=7； T4=27+1=15； T5=215+1=31； T6=231+1=63。但當 n 很大時，沒有人真的喜歡使用遞推 。但在最后一次移動最大的磁盤后，我們必須要移動 n1 個較小磁盤（必須在一根柱上）到最大磁盤上面，這也需要 Tn1步數(shù)。=39。我們還可以輕易的得到另一個值 ，考慮到最小的問題規(guī)模，顯然是 T0= 0，因為轉(zhuǎn)移 n＝ 0 的塔不需要移動步數(shù)。 解這樣一個問題的最好辦法是要一般 化一下，梵天塔有 64 個磁盤而漢諾塔有 8 個 。盧卡斯在 1883 年提出來的。ll frequently skip stages I and 2 entirely, because a mathematical expression will be given to 武漢科技大學本科畢業(yè)論文外文翻譯 5 us as a starting point. But even then, we39。 T5 = 215+1= 31。=39。ll see repeatedly in this book that it39。s not immediately obvious that the puzzle has a solution, but a little thought (or having seen the problem before) convinces us that it does. Now the question arises： What39。s change our perspective and try to think big； how can we transfer a large tower? Experiments with three disks show that the winning idea is to transfer the top two disks to the middle peg, then move the third, then bring the other two onto it. This gives us a clue for transferring n disks in general： We first transfer the n?1 smallest to a different peg (requiring Tn1 moves), then move the largest (requiring one move), and finally transfer the n?1 smallest back onto the largest (requiring another Tn1 moves). Thus we can transfer n disks (for n 0)in at most 2Tn1+1 moves： Tn ≤2Tn—1+1， for n 0. This formula uses 39。t made a foolish error. Such checks will be especially valuable when we get into more plicated maneuvers in later chapters.) A set of equalities like () is called a recurrence (a. k. a. recurrence relation or recursion relation). It gives a boundary value and an equation for the general value in terms of earlier ones. Sometimes we refer to the general equation alone as a recurrence, although technically it needs a boundary value to be plete. The recurrence allows us to pute Tn for any n we like. But nobody really like to pute from a recurrence， when n is large； it takes too long. The recurrence only gives indirect, local information. A solution to the recurrence would make us much happier. That is, we39。t ended； they39。他們有兩個共同特點：一是他們都被數(shù)學家反復研究 。據(jù)說，牧師們不分晝夜的忙碌這項工作，到他們完成那一天，塔將碎裂世界也將到盡頭。 武漢科技大學本科畢業(yè)論文外文翻譯 3 解決這個問題的下一步是引入適當?shù)姆枺喝∶⑶蠼?。因此，我們最多可以使?2Tn1+1 步來移動 n 個磁盤（ n0）： Tn ≤ 2 Tn1+1， (n＞ 0)． 此公式使用 39。在某些時候，我們必須移動最大的磁盤。它給出了一個邊界值和根據(jù)叫造紙表達一般只的一個方程。依據(jù)一個封閉的形式，我們可以了解到真正的 Tn。這樣的證明提供無限多的結(jié)果，只用數(shù)量有限的工作。 漢諾塔的遞推思想在解決各種應(yīng)用提出的問題中是很典型的。但即使這樣，子問題的解仍然將通過所有的三個階段
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