Better ways to implement a modulo operation (algorithm question)

Question

I've been trying to implement a modular exponentiator recently. I'm writing the code in VHDL, but I'm looking for advice of a more algorithmic nature. The main component of the modular exponentiator is a modular multiplier which I also have to implement myself. I haven't had any problems with the multiplication algorithm- it's just adding and shifting and I've done a good job of figuring out what all of my variables mean so that I can multiply in a pretty reasonable amount of time.

The problem that I'm having is with implementing the modulus operation in the multiplier. I know that performing repeated subtractions will work, but it will also be slow. I found out that I could shift the modulus to effectively subtract large multiples of the modulus but I think there might still be better ways to do this. The algorithm that I'm using works something like this (weird pseudocode follows):

result,modulus : integer (n bits) (previously defined)
shiftcount : integer (initialized to zero)
while( (modulus<result) and  (modulus(n-1) != 1) ){
     modulus = modulus << 1
     shiftcount++
}
for(i=shiftcount;i>=0;i--){
     if(modulus<result){result = result-modulus}
     if(i!=0){modulus = modulus >> 1}
}

So...is this a good algorithm, or at least a good place to start? Wikipedia doesn't really discuss algorithms for implementing the modulo operation, and whenever I try to search elsewhere I find really interesting but incredibly complicated (and often unrelated) research papers and publications. If there's an obvious way to implement this that I'm not seeing, I'd really appreciate some feedback.

Answer 1

For modulo itself, I'm not sure. For modulo as part of the larger modular exponential operation, did you look up Montgomery multiplication as mentioned in the wikipedia page on modular exponentiation? It's been a while since I've looked into this type of algorithm, but from what I recall, it's commonly used in fast modular exponentiation.

edit: for what it's worth, your modulo algorithm seems ok at first glance. You're basically doing division which is a repeated subtraction algorithm.

Answer 2

I'm not sure what you're calculating there to be honest. You talk about modulo operation, but usually a modulo operation is between two numbers a and b, and its result is the remainder of dividing a by b. Where is the a and b in your pseudocode...?

Anyway, maybe this'll help: a mod b = a - floor(a / b) * b.

I don't know if this is faster or not, it depends on whether or not you can do division and multiplication faster than a lot of subtractions.

Another way to speed up the subtraction approach is to use binary search. If you want a mod b, you need to subtract b from a until a is smaller than b. So basically you need to find k such that:

a - k*b < b, k is min

One way to find this k is a linear search:

k = 0;
while ( a - k*b >= b )
    ++k;

return a - k*b;

But you can also binary search it (only ran a few tests but it worked on all of them):

k = 0;
left = 0, right = a
while ( left < right )
{
    m = (left + right) / 2;
    if ( a - m*b >= b )
       left = m + 1;
    else
       right = m;
}

return a - left*b;

I'm guessing the binary search solution will be the fastest when dealing with big numbers.

If you want to calculate a mod b and only a is a big number (you can store b on a primitive data type), you can do it even faster:

for each digit p of a do
    mod = (mod * 10 + p) % b
return mod

This works because we can write a as a_n*10^n + a_(n-1)*10^(n-1) + ... + a_1*10^0 = (((a_n * 10 + a_(n-1)) * 10 + a_(n-2)) * 10 + ...

I think the binary search is what you're looking for though.

Answer 3

That test (modulus(n-1) != 1) //a bit test?

-seems redundant combined with (modulus<result).

Designing for hardware implementation i would be conscious of the smaller/greater than tests implying more logic (subtraction) than bitwise operations and branching on zero.

If we can do bitwise tests easily, this could be quick:

m=msb_of(modulus)

while( result>0 ) 
{
  r=msb_of(result) //countdown from prev msb onto result
  shift=r-m        //countdown from r onto modulus or 
                   //unroll the small subtraction 

  takeoff=(modulus<<(shift))  //or integrate this into count of shift

  result=result-takeoff;  //necessary subtraction

  if(shift!=0 && result<0)
  { result=result+(takeoff>>1); }

  } //endwhile

if(result==0) { return result }
else          { return result+takeoff }

(code untested may contain gotchas)

result is repetively decremented by modulus shifted to match at most significant bits.

After each subtraction: result has a ~50/50 chance of loosing more than 1 msb. It also has ~50/50 chance of going negative, addition of half what was subtracted will always put it into positive again. > it should be put back in positive if shift was not=0

The working loop exits when result is underrun and 'shift' was 0.