This task is almost duplicate of another one: Very fast 3D distance check?
And there was link to great article: http://www.azillionmonkeys.com/qed/sqroot.html
In the article you can find different aproaches for approximation of root. For example maybe this one is suitable for you:
int isqrt (long r) {
float tempf, x, y, rr;
int is;
rr = (long) r;
y = rr*0.5;
*(unsigned long *) &tempf = (0xbe6f0000 - *(unsigned long *) &rr) >> 1;
x = tempf;
x = (1.5*x) - (x*x)*(x*y);
if (r > 101123) x = (1.5*x) - (x*x)*(x*y);
is = (int) (x*rr + 0.5);
return is + ((signed int) (r - is*is)) >> 31;
}
If you can calculate root operation fast, then you can calculate distance in regular way:
return isqrt(a*a+b*b)
And one more link: http://www.flipcode.com/archives/Fast_Approximate_Distance_Functions.shtml
u32 approx_distance( s32 dx, s32 dy )
{
u32 min, max;
if ( dx < 0 ) dx = -dx;
if ( dy < 0 ) dy = -dy;
if ( dx < dy )
{
min = dx;
max = dy;
} else {
min = dy;
max = dx;
}
// coefficients equivalent to ( 123/128 * max ) and ( 51/128 * min )
return ((( max << 8 ) + ( max << 3 ) - ( max << 4 ) - ( max << 1 ) +
( min << 7 ) - ( min << 5 ) + ( min << 3 ) - ( min << 1 )) >> 8 );
}