sqrt()

[tomcant]

Yay! This works nicely:
[syntax="cpp"]#include <iostream>

double sqr(double num,double tol) {
double hi=num,lo=0.0,m;
while(lo<hi) {
m=(double)(lo+hi)/2.0;
if(m*m>=num-tol&&m*m<=num+tol)
return m;
else if(m*m<num) lo=m;
else hi=m;
}
return 0.0;
}

int main() {
std::cout<<sqr(25,0.000005);
std::cin.get();
}[/syntax]

[Darryl]

I'm still getting 6 for 25


















just kiddin

[Wizard]

Code: Select all

if (fabs(m*m - num) < eps) return m;

for some really small value of eps. try giving it a value of 0.00001 or something.

edit: ok, nevermind. Solve your own problems, see if I care

[hmily]

maybe hmily thought that was a solution for getting sqrt.

I tried it out and it gives 6 for sqrt of 25, so I guess it needs more work

D

the code have something wrong.I have modiflied it:

Code: Select all

nt i;
long sqrt(long num)
long i;
  for(i=0;num>0;++i){
      num=num-(2*i+1);
  }
return i;
}

[tomcant]

hmily, your version only works for the square numbers.

I might keep my version and stick it in some library I havn't decided to make yet.

[hmily]

hmily, your version only works for the square numbers.

I might keep my version and stick it in some library I havn't decided to make yet.

hehe,but Doe only want a long return sqrt(),so I think mine is better

[tomcant]

hehe,but Doe only want a long return sqrt(),so I think mine is better

Sorry to dig this up, but hmily, this is so blatantly better:
[syntax="cpp"]double sqr(double num,double tol=.000001) {
double hi=num,lo=0,m;
while(lo<hi) {
m=(lo+hi)/2;
if(m*m>num+tol) hi=m;
else if(m*m<num-tol) lo=m;
else return m;
} return 0;
}[/syntax]

I just had to post that


[edit]: Thats my `fast as I can make it' version. Thats the reason I'm posting it.