Finding log2() using sqrt()

Question

This is an interview question I saw on some site.

It was mentioned that the answer involves forming a recurrence of log2() as follows:

double log2(double x )
{
if ( x<=2 ) return 1;
 if ( IsSqureNum(x) )
   return log2(sqrt(x) ) * 2;
 return log2( sqrt(x) ) * 2 + 1; // Why the plus one here.
}

as for the recurrence, clearly the +1 is wrong. Also, the base case is also erroneous. Does anyone know a better answer? How is log() and log10() actually implemented in C.

Answer 1

Perhaps I have found the exact answers the interviewers were looking for. From my part, I would say it's little bit difficult to derive this under interview pressure. The idea is, say you want to find log2(13), you can know that it lies between 3 to 4. Also 3 = log2(8) and 4 = log2(16),

from properties of logarithm, we know that log( sqrt( (8*16) ) = (log(8) + log(16))/2 = (3+4)/2 = 3.5

Now, sqrt(8*16) = 11.3137 and log2(11.3137) = 3.5. Since 11.3137<13, we know that our desired log2(13) would lie between 3.5 and 4 and we proceed to locate that. It is easy to notice that this has a Binary Search solution and we iterate up to a point when our value converges to the value whose log2() we wish to find. Code is given below:

double Log2(double val)
{
    int lox,hix;
    double rval, lval;
    hix = 0;
    while((1<<hix)<val)
        hix++;
    lox =hix-1;
    lval = (1<<lox) ;
    rval = (1<<hix);
    double lo=lox,hi=hix;
   // cout<<lox<<" "<<hix<<endl;
    //cout<<lval<<" "<<rval;
    while( fabs(lval-val)>1e-7)
    {
        double mid = (lo+hi)/2;
        double midValue = sqrt(lval*rval);

        if ( midValue > val)
        {
             hi = mid;
             rval = midValue;
        }
        else{
            lo=mid;
            lval = midValue;
        }
    }
    return lo;

}