Fastest raise to power.
This tip submitted by Ezzetabi on 2005-03-18 16:31:04. It has been viewed 41403 times.
Rating of 6.0 with 175 votes
If you need raising to power some non standard object, like matrices or something overloaded with a integer exponent you can use this algorith:
a any type that know operator*, b int
(a*a)^(b/2) if b is even
a^b =
a*(a*a)*(int)^(b/2) if b is odd
using a recursive function your calculation will be faster (overall with big exponents) than the usual way.
A little example code:
double mpow (double, int);
double mpow2(double, int);
double mpow(double b, int exp)
{
if (exp < 0)
{
b = 1/b;
exp *= -1;
}
return mpow2(b,exp);
}
double mpow2(double b, int exp)
{
if (exp > 2)
{
if (exp%2)
return b*mpow(b*b,(int)exp/2);
else
return mpow(b*b,exp/2);
}
else if (2 == exp)
return b*b;
else if (1 == exp)
return b;
return 1.0; // exp == 0
}
This example is about floats, so it is nothing more than the usual pow(), yet if you need raising to power try to avoid a*a*a*a*...*a as that is really slow.
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