整数除法算法分析

Question

For an assignment we are required to write a division algorithm in order to complete a certain question using just addition and recursion. I found that, without using tail recursion, the naive repeated subtraction implementation can easily result in a stack overflow. So doing a quick analysis of this method, and correct me if I'm wrong, shows that if you divide A by B, with n and m binary digits respectively, it should be exponential in nm. I actually get

O( (n-m)*2^(n-m) ) 

since you need to subtract an m binary digit number from an n binary digit number 2^(nm) times in order to drop the n digit number to an n-1 digit number, and you need to do this nm times to get a number with at most m digits in the repeated subtraction division, so the runtime should be as mentioned. Again, I very well may be wrong so someone please correct me if I am. This is assuming O(1) addition since I'm working with fixed size integers. I suppose with fixed size integers one could argue the algorithm is O(1).

Back to my main question. I developed a different method to perform integer division which works much better, even when using it recursively, based on the idea that for

P = 2^(k_i) + ... 2^(K_0)

we have

A/B = (A - B*P)/B + P

The algorithm goes as follows to caclulate A/B :

input:
    A, B

    i) Set Q = 0

   ii) Find the largest K such that B * 2^K <= A < B * 2(K + 1)

  iii) Q -> Q + 2^K

   iv) A -> A - B * 2^k

    v) Repeat steps ii) through iv) until A <= B

   vi) Return Q  (and A if you want the remainder)

with the restrictions of using only addition, I simply add B to itself on each recursive call, however here is my code without recursion and with the use of shifts instead of addition.

int div( unsigned int m, unsigned int n )
{
    // q is a temporary n, sum is the quotient
    unsigned int q, sum = 0;
    int i;

    while( m > n )
    {
         i = 0;
         q = n;

         // double q until it's larger than m and record the exponent
         while( q <= m )
         {
              q <<= 1;
              ++i;
         }

         i--;
         q >>= 1;         // q is one factor of 2 too large
         sum += (1<<i);   // add one bit of the quotient
         m -= q;          // new numerator
    }

    return sum;
}

I feel that sum |= (1<<i) may be more appropriate in order to emphasize I'm dealing with a binary representation, but it didn't seem to give any performance boost and may make it harder to understand. So, if M and N are the number of bits in m and n respectively, an analysis suggests the inner loop is performed M - N times and each time the outer loop is completed that m looses one bit, and it must also be completed M - N times in order for the condition m <= n so I get that it's O( (M - N)^2 ).

So after all of that, I am asking if I am correct about the runtime of the algorithm and whether it can be improved upon?

Answer 1

Your algorithm is pretty good and your analysis of the running time is correct, but you don't need to do the inner loop every time:

unsigned div(unsigned num, unsigned den)
{
    //TODO check for divide by zero
    unsigned place=1;
    unsigned ret=0;
    while((num>>1) >= den) //overflow-safe check
    {
        place<<=1;
        den<<=1;
    }
    for( ;place>0; place>>=1,den>>=1)
    {
       if (num>=den)
       {
           num-=den;
           ret+=place;
       }
    }
    return ret;
}

That makes it O(MN)