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Thread: rewriting a recursive function
 04192012, 07:10 AM #1Senior Member
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rewriting a recursive function
Hello, I have to rewrite a recursive function that take a number and raises it to a power. It's sounds simple, but we have to do it in Logn time using X^2n = X^n times X^n. I am having a hard time implementing this. Here is the code for the other power function that I have to modify.
Java Code:public static double pow(double x, int n ) { if(x == 0 && n <= 0) throw new IllegalArgumentException("Wrong inputs"); else if (x == 0) return 0; else if (n == 0) return 1; else if (n > 0) return x * pow(x, n1); else return 1/pow(x,n); }
 04192012, 07:34 AM #2Moderator
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Re: rewriting a recursive function
Java Code:else if (n > 0) return x * pow(x, n1);
I think what you are asked to do is divide n by 2 instead of subtracting one. There is a small complication to deal with if n is odd. But even that shouldn't be too difficult since, in that case, n1 is even.
 04192012, 03:49 PM #3Senior Member
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Re: rewriting a recursive function
Here's the deal. The original algorithm will work just fine for solving x^cn where c is some constant such as 2, 3, 21, etc. It will just take longer than if you take advantage of the algebraic rule that x^cn = (x^c)^n. Note that x^cn does NOT equal x^c * x^n
The original algorithm requires (c*n)1 passes to achieve the answer. Taking advantage of factoring the exponents will take (n1)+(c1) passes.
I think I have that correct upon rereading it before posting.
 04192012, 05:27 PM #4Senior Member
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 04192012, 05:33 PM #5
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Re: rewriting a recursive function
Write down the recurrence relations for the pow operator:
1) x^0 == 1
2) x^1 == x
3) x^n == x^(n/2)*x^(n/2) if n is even
4) x^n == x*x^(n/2)*x^(n/2) if n is odd
cases 1) and 2) are the sentinel cases while case 3) and 4) describe the recursive step.
kind regards,
JosBuild a wall around Donald Trump; I'll pay for it.
 04192012, 05:47 PM #6Senior Member
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 04192012, 06:48 PM #7
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Re: rewriting a recursive function
Build a wall around Donald Trump; I'll pay for it.
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