complexity of recersive function nestedin a loop

Question

Im trying to calculate the complexity of a recursive call inside a loop:

Calculate(n,m){
for(i=1; i<n; i++){
    if(n==1) return 1;
    Calculate(n-1,m);
 }
for(j=1; i<m; i++){
    Calculate(n,m-1);
    if(m==1) return 1;
} 
}

here is what i did to calculate it :

C(m,n) = 1 + C(m-1,n) + C(m-2,n) + .... + C(1,n) 
           + C(m,n-1) + C(m,n-2) + .... + C(m,1)

is the complexity = 2^(m+n) ??

Ken Wei · Accepted Answer · 2017-10-26 07:34:44Z

1

You are correct: to reduce either m or n by 1, you need to make 2 calls; since you need to reduce both of them to zero, you need to go through O(m+n) 'levels'.

Thus the number of calls is O(2^(m+n)).

answered Oct 26, 2017 at 7:34

Ken Wei

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10 Comments

Achraf Over a year ago

i'm not sure but to reduce m or n by 1 we need to make m-1 + n-1 call .aren't we ??

Ken Wei Over a year ago

Yes, but when we talk about complexity, we are only concerned about the highest-order terms. O(m+n-2) is the same as O(m+n).

Achraf Over a year ago

Ok thanks but i still didn't really understand why it's 2^??

Ken Wei Over a year ago

Because every recursion makes 2 calls, so you're going to have 1+2+4+8+...+2^(m+n-2) calls, which is equal to 2^(m+n-1).

Achraf Over a year ago

There is two calls each one in loop then it will not be only two calls

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