问题
Why is the recurrence relation of recursive factorial algorithm this?
T(n)=1 for n=0
T(n)=1+T(n-1) for n>0
Why is it not this?
T(n)=1 for n=0
T(n)=n*T(n-1) for n>0
Putting values of n i.e 1,2,3,4...... the second recurrence relation holds(The factorials are correctly calculated) not the first one.
回答1:
This question is very confusing... You first formula is not factorial. It is simply T(n) = n + 1, for all n. Factorial of n is the product of the first n positive integers: factorial(1) = 1. factorial(n) = n * factorial(n-1). Your second formula is essentially correct.
回答2:
Looks like T(n) is the recurrence relation of the time complexity of the recursive factorial algorithm, assuming constant time multiplication. Perhaps you misread your source?
回答3:
we generally use recurrence relation to find the time complexity of algorithm.
Here, the function T(n) is not actually for calculating the value of an factorial but it is telling you about the time complexity of the factorial algorithm.
It means for finding a factorial of n it will take 1 more operation than factorial of n-1 (i.e. T(n)=T(n-1)+1) and so on.
so correct recurrence relation for a recursive factorial algorithm is T(n)=1 for n=0 T(n)=1+T(n-1) for n>0 not that you mentioned later.
like recurrence for tower of hanoi is T(n)=2T(n-1)+1 for n>0;
回答4:
I assume that you have bad information. The second recurrence relation you cite is the correct one, as you have observed. The first one just generates the natural numbers.
回答5:
Where did you find the first one ? It's completely wrong.
It's only going to add 1 each time whatever the value is .
回答6:
What he put was not the factorial recursion, but the time complexity of it.
Assuming this is the pseudocode for such a recurrence:
1. func factorial(n)
2. if (n == 0)
3. return 1
4. return n * (factorial - 1)
- I am assuming that tail-recursion elimination is not involved.
Line 2 and 3 costs a constant time, c1 and c2.
Line 4 costs a constant time as well. However, it calls factorial(n-1) which will take some time T(n-1). Also, the time it takes to multiply factorial(n-1) by n can be ignored once T(n-1) is used.
Time for the whole function is just the sum: T(n) = c1 + c2 + T(n-1).
This, in big-o notation, is reduced to T(n) = 1 + T(n-1).
This is, as Diam has pointed out, is a flat recursion, therefore its running time should be O(n). Its space complexity will be enormous though.
回答7:
T(n) = T(n-1) + 1 is correct recurrence equation for factorial of n. This equation gives you the time to compute factorial of n NOT value of the factorial of n.
回答8:
First you have to find a basic operation and for this example it is multiplication. Multiplication happens once in every recursion. So T(n) = T(n-1) +1 this +1 is basic operation (mutliplication for this example) T(n-1) is next recursion call.
回答9:
TL;DR: The answer to your question actually depends on what sequence your recurrence relation is defining. That is, whether the sequence Tn in your question represents the factorial function or the running-time cost of computing the factorial functionX.
The factorial function
The recursive defintion of the factorial of n, fn, is:
fn = n • fn-1 for n > 0 with f0 = 1.
As you can see, the equation above is actually a recurrence relation, since it is an equation that, together with the initial term (i.e., f0 = 1), recursively defines a sequence (i.e., the factorial function, fn).
Modelling the running-time cost of computing the factorial
Now, we are going to find a model for representing the running-time cost of computing the factorial of n. Let's call Tn the running-time cost of computing fn.
Looking at the definition above of the factorial function fn, its running-time cost Tn will consist of the running-time cost of computing fn-1 (i.e., this cost is Tn-1) plus the running-time cost of performing the multiplication between n and fn-1. The multiplication is achieved in constant time. Therefore we could say that Tn = Tn-1 + 1.
However, what is the value of T0? T0 represents the running-time cost of computing f0. Since the value of f0 is initially known by definition, the running-time cost for computing f0 is actually constant. Therefore, we could say that T0 = 1.
Finally, what we obtain is:
Tn = Tn-1 + 1 for n > 0 with T0 = 1.
This equation above is also a recurrence relation. However, what it defines (together with the initial term), is a sequence that models the running-time cost of computing the factorial function.
XTaking into account how the sequence in your recurrence relation is called (i.e., Tn), I think it very likely represents the latter, i.e., the running-time cost of computing the factorial function.
来源：https://stackoverflow.com/questions/1379929/recurrence-relation