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A small note on the additive composition of variant functions.

This is only a small note, for its observation is nearly trivial. The note is made, however, because it describes a method of some generality, by which results can be derived that —as I have observed— tend to strike the reader as “a trick” if he has never seen the like and is unaware of the method behind it.

It deals with finding a variant function for a repetitive construct with more than one alternative:

            do B1 → S1 ▯ ... ▯ Bn → Sn od .
The simple idea is to contruct the variant function t as the sum
            t = t1 + ... + tn ,
where ti is decreased by at least one by the i-th guarded command and left constant by the others.

We shall illustrate it with an example of matching simplicity. On a shunting yard are a finite number of trains; a train is either atomic (iff it consists of a single car) or composite (if? it consists of a finite number of cars greater than one). The yard is empty, when it contains no trains. Repeatedly we select a train: if it is atomic, it is removed from the yard, if it is composite, it is split into two (shorter) trains that, in this move, remain on the yard. The process terminates, when the yard is empty. The point is to construct the decreasing function t that is bounded from below, so that termination is proved. We sketch the process by

            do atomic train selected → remove it
  ▯ composite train selected → split it
od
With CARS = the number of cars on the yard
            t1 = CARS
satisfies the requirement that the first guarded command decreases it and the other one leaves it constant. with
            TRAINS = the number of trains on the yard ,
        t2 = - TRAINS
would satisfy the requirement that it is decreased by the second guarded commend (because that increases + TRAINS). This remains true when we add a function left constant by it, e.g.
                  t2 = f(CARS) - TRAINS .
(Note that we have already verified that CARS , and therefore f(CARS) for any function f , is unaffected by the second guarded command.) Because t2 has to remain constant when the first guarded command is selected, we must choose for f the identity function:
                  t2 = CARS - TRAINS
and thus we find t = 2 * CARS - TRAINS .
(With CLOSCONNS = the number of closed car-to-car connections,
                  t2 = CLOSCONNS
also does the job: it is left unchanged by the first and decreased by the second guarded command. This is not a new solution, for CLOSCONNS = CARS - TRAINS . )

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For the asymmetric computation for x = gcd(X, Y) with X > 0 and Y > 0 :
x, y := X, Y; do X > y → x:=x - y ▯ y > x → x, y == y, x od
we find quite analogously t1 = x + y and t2 = y .

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Note. Obviously the condition that ti is left unchanged by the j-th (j ≠ i) guarded commands is stronger than necessary: it suffices to establish that ti is not increased by the other guarded commands. (End of note.)

Note. Instead of choosing t = t1 + ... + tn , we could also take —after having chosen the ti — t = a1*t1 + ... + an*tn with all ai’s positive; in other words, the variant function is by no means unique, and we can choose the most convenient one. (End of note.)

The above has been prompted by my impression that some people seem to think that that there is something mysterious about the termination of the type of programs discussed. Its moral is that there is no mystery.

Plataanstraat 5prof.dr.Edsger W.Dijkstra
NUENEN NL-4565Burroughs Research Follow
The Netherlands

Transcribed by Martin P.M. van der Burgt
Last revision 2014-12-08 .