Calucalting the power

[mathon]

hi,

i develeoped a recursive function for the power and it works fine

double power(double x, int n) {
if (n < 0)
return 1/power(x, -n);
else if (n == 0)
return 1.0;
else
return x * power(x, n-1);
}
 

But the problem is this function should have a runtime of log(n) and
therefore i should use this formula x^2n = x^n x^n in any way. does
maybe someone knows here how i can modifiy it to reach my aim...?

matti

[Victor Bazarov]

ma****@gmx.at wrote:

i develeoped a recursive function for the power and it works fine

double power(double x, int n) {
 if (n < 0)
   return 1/power(x, -n);
 else if (n == 0)
   return 1.0;
 else
   return x * power(x, n-1);
}
 

But the problem is this function should have a runtime of log(n) and
therefore i should use this formula x^2n = x^n x^n in any way. does
maybe someone knows here how i can modifiy it to reach my aim...?

If 'n' is even, calculate the power as (x^(n/2))*(x^(n/2))
If 'n' is odd, calculate the power as x*(x^(n/2))*(x^(n/2))

V
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[mathon]

so you mean in that way:

double power(double x, int n)
{
if (n==0)
return 1.0;
 
if (0 == n%2)
{
return (x^((n-1)/2))*(x^((n-1)/2))
}
else
{
return x*(x^((n-1)/2))*(x^((n-1)/2))
}
}
 

or did i interpret something wrong...?

matti

[Victor Bazarov]

ma****@gmx.at wrote:

so you mean in that way:

double power(double x, int n)
{
   if (n==0)
       return 1.0;

 
You could also create a few other predefined return values,
for example for 'n==1' or 'n==2', just to optimize a bit.
 
        
                >
  if (0 == n%2)
  {
       return (x^((n-1)/2))*(x^((n-1)/2))
  }
   else
  {
      return x*(x^((n-1)/2))*(x^((n-1)/2))
  }
}
 
 
 

>
or did i interpret something wrong...?

Did I use "-1" anywhere in my reply (which you didn't quote)?
First of all, don't use ^, instead use the recursion which you
already have. Second, use -1 only where needed.

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

[Mark P]

Victor Bazarov wrote:

ma****@gmx.at wrote:

>so you mean in that way:

double power(double x, int n)
{
   if (n==0)
       return 1.0;

You could also create a few other predefined return values,
for example for 'n==1' or 'n==2', just to optimize a bit.

        
                >  if (0 == n%2)
  {
       return (x^((n-1)/2))*(x^((n-1)/2))
  }
   else
  {
      return x*(x^((n-1)/2))*(x^((n-1)/2))
  }
}