Suppose you have a rule for addition of some objects (most generally, elements of a semigroup), and you wish to compute the sum of n copies of the same object. How could you achieve this?
The primary school answer is to do just that: given g, compute 2g=g+g; then compute 3g=2g+g and so on, for n steps. But this rapidly becomes tedious, and isn’t (I hope) how we’d perform such a calculation in our heads: instead, we’d try to take bigger jumps. For instance, to compute the number 16g, we can find 2g, add that to itself to reach 4g, and again for 8g, hitting 16g after just 4 steps. Had we wanted 17, we’d then just add g again.
The sequence of intermediate values correspond to an addition chain for n: this is a sequence starting at a0=1 and ending at al=n with the property that for any term ak there exist terms ai,aj with ak=ai+aj. That is, every term is the sum of two earlier terms (not necessarily distinct), so we can reach ng by computing each akg in turn without needing anything fancier than our addition law.
As a side effect, an addition chain also tells us how to exponentiate if we know how to multiply, since xa+b=xaxb so we can build up to xn by finding the powers xak in turn.
Thus our tedious sequence for 16 is 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16, with the more sophisticated one being 1,2,4,8,16. If I asked you to compute 16x for a number x, you’d probably do something different: compute 10x (easy in base 10!) and 6x, then add those. With access to a multiplication procedure, this is definitely more efficient: but without that luxury it can still be used to give an addition chain for 16, by finding chains for 10 and 6. These are 1,2,4,5,10 and 1,2,3,6 respectively, so we can merge them into the chain 1,2,3,4,5,6,10,16.
We describe the length of the chain as l: that is, there are l terms after the 1. Notice that at worst, therefore, we need n-1 terms. If we move as quickly as possible by taking ak+1=ak+ak=2ak then the greatest value we can reach after m terms is 2m, as demonstrated with the chain for 24=16. However, if we can compute n and m in ln,lm terms, it needn’t take as many as ln+lm+1 terms to compute n+m, since there may be common terms in the chains. Finally, as defined an addition chain needn’t make use of all it’s terms: for example, 1,2,4,6,8,16 is still a chain for 16, although the 6 isn’t needed. We will obviously be interested in chains without any such wasted terms.
The observation for powers of two motivates a first attempt at an efficient algorithm for building addition chains. if n is 1, then we have nothing to do: the chain 1 suffices. Otherwise, we can work recursively: for an even number n=2k, we ask for the chain for k and then add 2k to the end; whereas for an odd number n=2k+1, we ask for the chain for k and add 2k then 2k+1 to the end. This will require at most log2n calls, and each step adds 1 or 2 entries to the chain.
This can also be captured by considering the binary representation of n, which will also indicate whether we hit an even or odd number at each stage. Suppose n is of the form
_2)
where each di is either 0 or 1, and dk=1 so there are no leading zeros. Then we start the chain with 1, and read from left to right, ignoring the leading 1: if we see a 0, then we double the last term of the chain; if we see a 1, add 1 to this new term as well. More formally:
Input: the binary expansion _2)
of n.
Output: an addition chain for n.
Set a0=1
Set t=1
for i from k-1 down to 0:
Set at=2at-1 and t=t+1.
If d_i=1:
Set at=at-1+1 and t=t+1
return a0…at
For instance, with n a power of 2 we will see only a run of zeros, and thus perform the appropriate number of doublings, for a chain of length log2n. For n one less than a power of two, the expansion consists entirely of 1s, costing 2log2n. For an ‘average’ number where each digit has around a 50% chance of being a 1 or a 0, the chain will therefore be of approximate length (3/2)log2n. More formally, we can define the Hamming Weight v(n) to be the number of 1s in the binary expansion of n; then the chain will have length floor(log2n) + v(n) -1.
Is this optimal? A counterexample suffices to show it is not, the first being 15, which is 1111 in binary. Using the binary expansion we get the sequence 1,2,3,6,7,14,15 of length 6 (=3+4-1 as claimed). But the sequence would be 1,2,3,5,10,15 of length 5 and hence 1 shorter.
Nonetheless, our binary chain gives a much better upper bound than the naive chain of (n-1) steps, and is very simple to implement both in logic and memory. The idea also generalises to arbitrary bases, as follows.
Suppose n is of the form
_m)
where each di is in the range 0…m-1, dk non-zero. Then we can compute an m-ary addition chain for n:
Input: the m-ary expansion _m)
of n.
Output: an addition chain for n.
Set a=0
Start the chain with 1,2,3,…,m-1.
for i from k down to 0:
Extend the chain to m*a (if not already present) and set a=m*a.
Add a+di to the chain (if not already present), and set a=a+di.
return chain.
Notice that a tracks the last term computed, and that extending the chain to m*a is itself an addition chain problem: this makes some 2k a good choice for m, since this extension can then be accomplished in k additions.
The shorter chain for 15 arises from its ternary expansion: considered as (120)3 1*9+2*3+0*1, we have an initial chain of 1,2 then apply the loop. With a=0,d2=1 we set a=3a=0 then a=a+1=1; neither of these need to be added to the chain; for a=1, d1=2 we set a=3a=3 which requires the addition of 3 to the chain; then a=a+2=5 which extends the chain to 1,2,3,5; finally for a=5,d2=0 we set a=3a=15, adding first 10 then 15 to the chain; then a=a+0=15 so we terminate with chain 1,2,3,5,10,15.
The optimal choice of m depends on n, since although there will be less steps (logmn) for greater m, each step will require more additions in the computation of m*a. Nor are m-ary expansions the end of the story- there are various other tricks available to finesse addition chains. However, there is no known algorithm for finding shortest addition chains- and a generalisation of the problem to finding shortest addition sequences (chains that contain a desired list of values n1,n2,…,nk) is NP-complete. Searching for short chains is therefore an interesting challenge, and is constrained by the fact that unless you are precomputing them for future use, any shorter chain must offer a speedup in computation greater than the time taken to find it.
There are, however, bounds for the length of the chain:
 \leq l(n) \leq log_2 n + v(n) -1)
Where v(n) is the Hamming weight as before.
However, if you have access to basic operations other than addition, all bets are off. In some contexts (such as elliptic curves), finding the inverse of an element is easy and so we can achieve subtraction, giving rise to addition-subtraction chains. For the pairing computations I’ve been interested in, I’ve never actually described a block addition. It is possible, following an algorithm of Shipsey, to do so, although it is much more expensive than the block double. But even if it cost no more, and we had an oracle for optimal chains, such an approach would only rarely beat the current algorithm. This is because access to the double-add procedure makes the Hamming weight irrelevant: for instance, with 15 we could form the chain 1,3,7,15. This means that we can compute in log2n block operations, and comparing to the lower bound for addition chains above shows that this is almost always better.
Still, ideas from addition chains can motivate construction of other operations: access to triple, tripleadd and tripleadd2 procedures would allow for chain of log3n length, although it is likely that these block operations would be more expensive than the double and doubleadds. Even if they don’t turn out to be useful for my work, the subtleties of this innocent-looking problem still makes for interesting mathematics!
References/Further Reading
Knuth- The Art of Computer Programming Vol. 2: Seminumerical Algorithms.
Bos and Coster- Addition Chain Heuristics in Advances in Cryptology- Crypto’89 (Lecture Notes in Computer Science) (Available via SpringerLink.)
Gordon- A Survey of Fast Exponentiation Methods. (Available via citeseer.)
