2
$\begingroup$

Originally posted on PHYS, however, obviously it has more to do with CS

I am currently trying to implement a circuit for computing the greatest common divisor in the Quantum Computing Language. In my work, I partially proceed from the paper Quantum Circuits for GCD Computation with O(nlogn) Depth and O(n) Ancillae published by I. Markov and M. Saeedi in 2013, which may be found at arXiv http://arxiv.org/abs/1304.7516.

The paper introduces a binary GCD algorithm analogue and works out a design of the quantum circuit which is clarified in least details, however, in the same time it is somewhat confusing due to several statements that actually do not correspond with the rest of the work. First of all, let me enclose an image denoting design of the circuit. enter image description here (Credit goes to the paper)

As you may see, the circuit entirely corresponds to the classical binary GCD algorithm. It takes advantage of quantum circuits for comparison, bit shift, addition and multiplication. Nevertheless, in our case those functions are controlled by state of ancillary qubits and thus, they may not regarded reversible on their own. It means, we have to account for a separate register of auxiliary qubits which cannot be erased till the end of our computation as it would simply violate desired reversibility otherwise.

But in contrast to that, in the paper we may find at least three statements which do actually deny those simple facts. Let me cite: (About preceding works) The authors did not clear all zero-initialized ancillae which limits the applicability of their techniques. (About the circuit) The zero-initialized ancillae may be modified in-side a given circuit, but should be returned to zero at the end of computation to be reused. (About implementation) To count ancillae, note that all computational ancillae are cleared inside each block. After the final multiplication block for R·B, one can copy (in log depth) the final GCD result to another n-qubit zero-initialized register and apply the whole circuit (except for copying the result) in reverse order to recover A,B, and zero-initialized ancillae. Given that all components use O(n) ancillae, the total number of ancillae remains linear.

I ask myself: How can be all ancillaes cleared at the end of each computational block while the binary GCD seems to be reversible only with regard to an auxiliary register of control qubits? It is apparent this question reduces to the problem of binary GCD reversibility which is actually something I am doubtful about. Maybe I lack of insight and the usage of other control bits is so obvious so as to omit it, however, in every case I would like to ask you about your view on the problem.

$\endgroup$
  • 3
    $\begingroup$ This question is unanswerable unless you tell us what they mean by "computational block" (unless we look at the original paper, which you should avoid making us do). $\endgroup$ – Peter Shor Feb 9 '16 at 13:56
  • $\begingroup$ Thanks for your notice - your comment actually helped me to solve the problem as it all rather lied in my misunderstanding. $\endgroup$ – JBI Feb 11 '16 at 17:31
1
$\begingroup$

It looks like they just directly translated the algorithm into a circuit, so it's not too hard to reproduce.

  1. Write out the binary gcd algorithm as a loop:

    def gcd(a, b):
        let shift = 0
        while a != 0 or b != 0:
            let pa = a & 1
            let pb = b & 1
            let pc = a >= b
    
            if pa == 0 and pb == 0:
                a >>= 1
                b >>= 1
                shift++
            elif pa == 0:
                a >>= 1
            elif pb == 0:
                b >>= 1
            elif pc:
                a -= b
                a >>= 1
            else:
                b -= a
                b >>= 1
    
        return (a | b) << shift
    
  2. Keep track of any intermediate values so you can uncompute them instead of leaking them. Also switch to a fixed number of iterations to avoid leaking the size of the inputs.

    def gcd(a, b, len_a, len_b):
        let shift = 0
        # Note: >> and << are now bit rotations instead of bit shifts
        for i in range(len_a + len_b):
            let g = a > b
            let pa = (a & 1) == 0
            let pb = (b & 1) == 0
    
            if pa and pb and a != 0 and b != 0: shift++
            if pa: a >>= 1
            if pb: b >>= 1
            if not pa and not pb:
                if g:
                    a -= b
                    a >>= 1
                else:
                    b -= a
                    b >>= 1
    
            # We can't clear these yet, so just set them aside until later
            push g, pa, pb
    
        let result = (a | b) << shift
    
        # Uncompute intermediate values
        for i in range(n + m):
            pop pb, pa, g
    
            if not pa and not pb:
                if g:
                    a <<= 1
                    a += b
                else:
                    b <<= 1
                    b += a
            if pb: b <<= 1
            if pa: a <<= 1
            if pa and pb and a != 0 and b != 0: shift--
    
            unlet pb = (b & 1) == 0
            unlet pa = (a & 1) == 0
            unlet g = a > b
    
        unlet shift = 0
        return result
    
  3. Convert to a circuit.

    (Edit: oops, I forgot the shift part. You'll have to add that part in yourself.)

    gcd circuit

It takes $O(n)$ iterations to guarantee the inputs have been reduced all the way down, and each iteration requires $\Omega(\lg n)$ depth and $O(n)$ gates so the overall bound is $O(n^2)$ gates and $O(n \lg n)$ depth.

Now, all that being said, you need to be wary of the fact that the result is not totally coherent. Before doing interference-y stuff with it, I'd want to be a lot more careful about ensuring the stashed ancilla are independent of the computed gcd. Also you'd want to do any computation in the middle so that the result doesn't have to co-exist with the inputs.

But I hope this gives the general idea.

$\endgroup$
  • $\begingroup$ You are completely right. As I have already written in a reply to P. Shor´s comment, my problem - the reason why I asked this question - turned out to arise from a foolish misunderstanding of the cited paper. Yes, their algorithm is an 1:1 copy of the common binary GCD alg., and it can be even optimized in the quantum case (it is unnecessary to store such many ancillae as we do not need to replicate whole conditions checking process). However, I am surprised someone noticed this old question and I feel really grateful for your comprehensive answer! $\endgroup$ – JBI Jul 26 '16 at 0:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.