abs function itself has a condition??
I think it is not that good solution.
On Wed, Aug 24, 2011 at 6:33 PM, priyanka raju priyark...@gmail.com wrote:
int a,b,max,min;
max=(a+b+abs(a-b))/2;
min=(a+b-abs(a-b))/2;
--
cheers
priyanka
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@Rajesh: abs() can be done without conditional operations. There
probably are many ways. The first two that come to mind are:
abs(x) = (x 31) x | ~(x 31) ~x
abs(x) = (x 31) ^ x + (x 31) 1
Dave
On Aug 26, 2:56 am, rajesh singarapu rajesh0...@gmail.com wrote:
abs function itself has a
int a,b,max,min;
max=(a+b+abs(a-b))/2;
min=(a+b-abs(a-b))/2;
--
cheers
priyanka
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nice solution priyanka :)
Sanju
:)
On Wed, Aug 24, 2011 at 6:03 AM, priyanka raju priyark...@gmail.com wrote:
int a,b,max,min;
max=(a+b+abs(a-b))/2;
min=(a+b-abs(a-b))/2;
--
cheers
priyanka
--
You received this message because you are subscribed to the Google Groups
Algorithm
@Abhishek:
int c = (a - b) 31;
max = c b | ~c a;
Explanation: c = 0 if a = b, else c = all one bits.
Then c b = 0 if c = 0, but c b = b if c = all ones, i.e. if the
max is b,
and ~c a = a if c = 0, i.e., if the max is a, but ~c a = 0 if c =
all ones.
Dave
On Aug 23, 8:07 am, Abhishek
