factorial

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); } She said that it has a cost of T(n-1) + 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 . After that, she substituted j in T(n-j) + j , and obtained T(1) + n-1 . She directly said that for that n-1 = 2 (log 2 n-1) , so the cost of the algorithm is exponential. I really lost the last two steps. Can someone please explain them to me?

What's the complexity of a recursive program to find factorial of a number n ? My hunch is that it might be O(n) . Alex Martelli If you take multiplication as O(1) , then yes, O(N) is correct. However, note that multiplying two numbers of arbitrary length x is not O(1) on finite hardware -- as x tends to infinity, the time needed for multiplication grows (e.g. if you use Karatsuba multiplication , it's O(x ** 1.585) ). You can theoretically do better for sufficiently huge numbers with Schönhage-Strassen , but I confess I have no real world experience with that one. x , the "length" or "number

Hi everyone I edited my post because I have another problem with my code.I had a problem with the factorial function yesterday but I managed to solve it thanks to your answers , it was a ridiculous mistake. The problem now is that for some values that are higher than 15, the final results(not the factorials of individual numbers) are always 0 or -1 for the lesser values it works fine.Can someone tell me whats wrong with this code : #include <iostream> #include<time.h> using namespace std; int factorial(int a){ if(a==1) return 1; else if(a==0) return 1; else return factorial(a-1)*a; } int main(

Can someone helping me to find a way to get the inverse factorial in Prolog... For example inverse_factorial(6,X) ===> X = 3 . I have been working on it a lot of time. I currently have the factorial, but i have to make it reversible. Please help me. Prolog's predicates are relations, so once you have defined factorial, you have implicitly defined the inverse too. However, regular arithmetics is moded in Prolog, that is, the entire expression in (is)/2 or (>)/2 has to be known at runtime, and if it is not, an error occurs. Constraints overcome this shortcoming: :- use_module( library(clpfd) ).

how to get the value of an integer x , indicated by x! , it is the product of the numbers 1 to x. Example: 5! 1x2x3x4x5 = 120. int a , b = 1, c = 1, d = 1; printf("geheel getal x = "); scanf("%d", &a); printf("%d! = ", a); for(b = 1; b <= a; b++) { printf("%d x ", c); c++; d = d*a; } printf(" = %d", d); how to get the som of an integer x, indicated by x!, is the product of the numbers 1 to x. Did you mean factorial of x ? Change d = d*a; to d = d*b inside the loop You can simply do: for(b = 1; b <= a; b++) { d *= b; } // d now has a! This is the optimal implementation in size and speed: int

How would you write a non-recursive algorithm to compute n! ? Since an Int32 is going to overflow on anything bigger than 12! anyway, just do: public int factorial(int n) { int[] fact = {1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800, 479001600}; return fact[n]; } in pseudocode ans = 1 for i = n down to 2 ans = ans * i next In the interests of science I ran some profiling on various implementations of algorithms to compute factorials. I created iterative, look up table, and recursive implementations of each in C# and C++. I limited the maximum input value to 12 or less, since