图灵机上欧几里得最大公约数算法的复杂性 [英] The complexity of Euclid's greatest common divisor algorithm on a Turing Machine
问题描述
考虑Euclid算法的这种实现:
Consider this implementation of Euclid's algorithm:
function gcd(a, b)
while b ≠ 0
t := b
b := a mod b
a := t
return a
关于维基百科的一个很好的证据表明,该算法总是需要少于O(h)除法,其中h是较小数字b中的位数.
A nice proof on wikipedia shows that the algorithm "always needs less than O(h) divisions, where h is the number of digits in the smaller number b".
但是,在图灵机上,计算mod b的过程的时间复杂度为O(a + b).我的直觉和一些大型测试告诉我,在图灵机上,欧几里得算法的复杂度仍然是O(a + b).
However, on a Turing Machine a procedure that computes a mod b has a time complexity of O(a+b). My intuition and some large tests tell me that the complexity for Euclid's algorithm is still O(a+b) on a Turing Machine.
关于如何证明这一点的任何想法?
Any thoughts on how to prove that?
推荐答案
如果我要在Turing机器上的二进制数字上实现GCD,我将实现以下算法.我看不到它会比O((ln a + ln b)^ 2)小.我认为,最有效的表示方法是在步骤2之后按位交织两个值.
If I were implementing GCD on binary numbers on a Turing machine, I'd implement the following algorithm. I can't see how it would be smaller than O((ln a + ln b)^2) though. The most efficient representation I think would be bitwise interleaving both values after step 2.
- 让s1 = a的最低有效位中的零个数.删除这些最低的s1零位.
- 让s2 = b的最低有效位中的零个数.删除这些最低的s2零位.
- 让s = min(s1,s2)
- 现在a和b都是奇数.如果b < a然后交换a和b.
- b> = a.设置b = b-a,然后从b中删除所有最低有效零位.
- 如果b!= 0,则转到4.
- 在a的末尾添加s个零位.这就是结果.
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