Addendum to A Distributed Implementation of a Task

The algorithm given on page 6 may deadlock because communications do not follow the directedness of the tree but may occur in either direction. The algorithm, and its proof, are correct if we restrict ourselves to trees, rather than RDAG`s. RDAG's ((1]) do not contain directed cycles, but may contain cycles. In those cases the algorithm, and the 'proof' of the absence of deadlock on p.7 are incorrect. We now introduce a version of the algorithm which will operate correctly on any RDAG. First we note that whenever deadlock occurs, at least one process is dedicated to a communication with a predecessor in the RDAG. We therefore allow the main process to commit to a communication with a successor only, and require only that the condition under which a communication with a predecessor was originally initiated, lead to a communication at some point in the future. A more formal speciication of this problem, and an elegant solution are given in a note at the end of this section. We also modiied the guards according to the remark on page 7. The program can be found in Figure 1. In some cases communications are used for synchronization only. In those cases the variable communicated is irrelevant and has been replaced by a dot. The channels are (OUT ij ; I N ji) for all pairs i; j that satisfy p-edge ij _p-edge ji , and (TO ij ; F ROM ji) for all pairs i; j that satisfy p-edge ij. For those last i and j we have introduced variables op ij in process P i to keep track of the value of p i last communicated from process P i to process P j. The variables q ij hold the values of the p j last communicated by process P j. We need the following invariant: (8i; j : p-edge ij : q ji = op ij). The invariant holds initially and holds on every repetition of the rst outermost do construct because the innermost do loop establishes p i = q ji and therefore the following assignment op ij := p i reestablishes the invariant. Finally, the other statements do not change q ij or op ij. We check the three requirements, but in a diierent order : (0)". There is no deadlock. We assume that all communications between the taskpool and its corresponding C i terminate. Since these …