Integer Division Algorithm Analysis - c

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 n-m. 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^(n-m) times in order to drop the n digit number to an n-1 digit number, and you need to do this n-m 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(M-N)