Power calcualation once more :(

[mathon]

hi,

i already posted an entry because of this problem, unfortunately i
havent solved it so far..

I have created a recursion for the calculation of the power like this:

[code]
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 ist that i should use this formula x^2n = x^n x^n in
any way in order to have a runtime of log(n).
I read in the previous posting something like i should use this:

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

Maybe someone could help me here once more in order to reach my final
aim...?

matti

 

[Steve Pope]

hi,

i already posted an entry because of this problem, unfortunately i
havent solved it so far..

I have created a recursion for the calculation of the power like this:

[code]
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 ist that i should use this formula x^2n = x^n x^n in
any way in order to have a runtime of log(n).
I read in the previous posting something like i should use this:

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

Maybe someone could help me here once more in order to reach my final
aim...?

You're pretty close. Try:

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

You may have been confused by the shorthand use of ^ which
shouldn't appear in your code.

Steve

 

[mathon]

hi,

ah you mean like that:

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

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

Have you meant it in that way...?

matti

 

[Steve Pope]

hi,

ah you mean like that:

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

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

Have you meant it in that way...?

Yes, that looks okay to me, but I haven't tried to run it.
Good luck.

Steve

 

[mathon]

yes it runs.

one last question, shouldnt i modify this line:

return 1/power(x,-n);

as well, to reach the runtime of log(n)?

matti

hi,

ah you mean like that:

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

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

Have you meant it in that way...?

Yes, that looks okay to me, but I haven't tried to run it.
Good luck.

Steve

matti

 

[Steve Pope]

yes it runs.

one last question, shouldnt i modify this line:

return 1/power(x,-n);

as well, to reach the runtime of log(n)?

No, because that line gets run exactly once (and then only
if the input is negative). It will not affect the order
of your runtime.

S.

 

[Stuart Redmann]

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

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

>
> yes it runs.
>
> one last question, shouldnt i modify this line:
>
> return 1/power(x,-n);
>
> as well, to reach the runtime of log(n)?

1. Please don't top-post.
2. Remove any greetings from the posts you're replying to (they have no
informational content, so they just lengthen the post).
3. If you put an exponent of the form 2^n - 1 into your power function,
you still end up with a complexity of O(n), since the step
return x*power(x,n-1) will be called 1/2 * 2^n - 1 times.
I will give you the advice not to try to tackle this problem
recursively, as you will need to have all potences of x available (you
still can do this recursively, but the resulting code would be rather
messy). I won't spoil your fun of finding the right solution, so all I'm
saying is that the solution you have found so far is good enough.

Regards,
Stuart

 

[Steve Pope]

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

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

3. If you put an exponent of the form 2^n - 1 into your power function,
you still end up with a complexity of O(n), since the step
return x*power(x,n-1) will be called 1/2 * 2^n - 1 times.

That doesn't seem right. Any pair of consecutive calls
beginning with power(n) must reduce its argument by at least a factor
of 2 since the only possibilities are:

n = 2*k (n even) which results in a call to power(k)

n = 2*k+1 (n odd) which results in a call to power(2*k), then a
call to power(k)

Therefore the runtime is bounded by 2 * log2(n). The runtime
will be equal to this bound if n, (n-1)/2, ((n-1)/2)/2 etc. are
all odd. Otherwise it will be lower than this bound.

Or am I missing something?

Steve

 

[Steve Pope]

I wrote,

That doesn't seem right. Any pair of consecutive calls
beginning with power(n) must reduce its argument by at least a factor
of 2 since the only possibilities are:

n = 2*k (n even) which results in a call to power(k)

n = 2*k+1 (n odd) which results in a call to power(2*k), then a
call to power(k)

Therefore the runtime is bounded by 2 * log2(n). The runtime
will be equal to this bound if n, (n-1)/2, ((n-1)/2)/2 etc. are
all odd. Otherwise it will be lower than this bound.

Oops, that should be: n, (n-1)/2, ((n-1)/2 - 1)/2 etc.

S.

 

[Stuart Redmann]

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

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

3. If you put an exponent of the form 2^n - 1 into your power function,
you still end up with a complexity of O(n), since the step
return x*power(x,n-1) will be called 1/2 * 2^n - 1 times.

That doesn't seem right. Any pair of consecutive calls
beginning with power(n) must reduce its argument by at least a factor
of 2 since the only possibilities are:

n = 2*k (n even) which results in a call to power(k)

n = 2*k+1 (n odd) which results in a call to power(2*k), then a
call to power(k)

Therefore the runtime is bounded by 2 * log2(n). The runtime
will be equal to this bound if n, (n-1)/2, ((n-1)/2)/2 etc. are
all odd. Otherwise it will be lower than this bound.

Or am I missing something?

Nope. Obviously I did (*shame*). I didn't read your code properly (no
comments, BTW). Of course, the computational complexity you have given
is right. Using this algorithm is certainly even the most elegant
solution since recursion is the shortest way to implement it.

Sorry for giving bad advice (at least point 3

Stuart