今天在课堂上,我的老师在黑板上写下了这种递归阶乘算法:
Today in class my teacher wrote on the blackboard this recursive factorial algorithm:
int factorial(int n) { if (n == 1) return 1; else return n * factorial(n-1); }她说费用为T(n-1) + 1.
然后她用迭代方法说T(n-1) = T(n-2) + 2 = T(n-3) + 3 ... T(n-j) + j,所以算法在n - j = 1时停止,所以j = n - 1.
Then with iteration method she said that T(n-1) = T(n-2) + 2 = T(n-3) + 3 ... T(n-j) + j, so the algorithm stops when n - j = 1, so j = n - 1.
此后,她用T(n-j) + j替换为T(n-j) + j,并获得了T(1) + n-1.
After that, she substituted j in T(n-j) + j, and obtained T(1) + n-1.
她直接说,对于n-1 = 2 (log 2 n-1),因此算法的成本是指数级的.
She directly said that for that n-1 = 2(log2n-1), so the cost of the algorithm is exponential.
我真的失去了最后两个步骤.有人可以给我解释一下吗?
I really lost the last two steps. Can someone please explain them to me?
推荐答案让我们从分析该算法开始.我们可以为已完成的工作量编写一个递归关系.作为基本情况,当算法在大小为1的输入上运行时,您需要执行一个工作单元,因此
Let's start off with the analysis of this algorithm. We can write a recurrence relation for the total amount of work done. As a base case, you do one unit of work when the algorithm is run on an input of size 1, so
T(1)= 1
T(1) = 1
对于大小为n + 1的输入,您的算法在函数本身内执行一个工作单元,然后在大小为n的输入上调用同一函数.因此
For an input of size n + 1, your algorithm does one unit of work within the function itself, then makes a call to the same function on an input of size n. Therefore
T(n + 1)= T(n)+1
T(n + 1) = T(n) + 1
如果您扩展重复条款,您将得到
If you expand out the terms of the recurrence, you get that
- T(1)= 1
- T(2)= T(1)+1 = 2
- T(3)= T(2)+1 = 3
- T(4)= T(3)+1 = 4
- ...
- T(n)= n
因此,一般而言,此算法需要完成n个工作单元(即T(n)= n).
So in general this algorithm will require n units of work to complete (i.e. T(n) = n).
您的老师接下来要说的是
The next thing your teacher said was that
T(n)= n = 2 log 2 n
T(n) = n = 2log2 n
此陈述是正确的,因为对于任何非零x,2 log 2 x = x,因为幂和对数是彼此的逆运算.
This statement is true, because 2log2x = x for any nonzero x, since exponentiation and logarithms are inverse operations of one another.
所以问题是:这是多项式时间还是指数时间?我将其归为伪多项式时间.如果将输入x视为数字,则运行时确实是x中的多项式.但是,多项式时间是正式定义的,因此算法的运行时间必须是关于用于指定问题输入的位数的多项式.此处,只能在θ(log x)位中指定数字x,因此从技术上讲2 log x 的运行时间是指数时间.我在上写过有关长度的信息,我建议您对其进行详细了解.
So the question is: is this polynomial time or exponential time? I'd classify this as pseudopolynomial time. If you treat the input x as a number, then the runtime is indeed a polynomial in x. However, polynomial time is formally defined such that the runtime of the algorithm must be a polynomial with respect to the number of bits used to specify the input to the problem. Here, the number x can be specified in only Θ(log x) bits, so the runtime of 2log x is technically considered exponential time. I wrote about this as length in this earlier answer, and I'd recommend looking at it for a more thorough explanation.
希望这会有所帮助!
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