. Remarquons-que-sm-i, ) sm i [x][y].sn est le nombre d'écritures effectuées par p y dans mem[y] simulées jusque là par q x et (2) sm i [x]

?. Ensuite, Pour cela, il extrait de sm i [1..t + 1][y] la valeur écrite par le simulateur q s le plus avancé dans sa simulation de p y . Ceci est exprimé dans les lignes 02-03. operation sim_snapshot i,j () : (01) smi ? MEM .snapshot() : (02) for each y : 1 ? y ? n : do inputi[y] = smi[s][y].value (03) where ?x : 1 ? x ? t + 1 : smi[s][y].sn ? smi[x][y].sn end for ; (04) snap_sni[j] ? snap_sni[j] + 1 ; let snapsn = snap_sni[j] ; (05) enter_mutex ; SAFE _AG[j, snapsn].propose i (inputi) ; exit_mutex, ) res ? SAFE _AG[j, snapsn].decidei() (07) return(res)

L. and A. Er, j, snapsn] est utilisé par un simulateur q i quand il simule son snapsn-ième invocation mem.snapshot() pour le processus simulé p j pour 1 ? j ? t + 1. (Comme nous le verrons, quand t + 1 < j ? n, la simulation de mem.snapshot(

A. Le-but-de and . _val, snapsn][1] est de contenir la valeur qui doit être renvoyée pour la snapsnième invocation mem.snapshot() par le processus simulé p j si le propriétaire q j est désigné comme gagnant par l'objet ARBIT ER[j, snapsn] correspondant. Si le propriétaire q j n'est pas le gagnant

G. Représentant-canonique-d-'une-tâche and . Étant-donnée-une-tâche-n, son représentant canonique est la tâche n, m, ?, u ? -GSB telle que la tâche n, m, ?, u ? ? 1-GSB n'est pas ?-ancrée. Une définition similaire s'applique aux tâches u-ancrées. Une tâche qui n'est ni u-ancrée ni ?-ancrée, ou qui est (?, u)-ancrée, est son propre représentant

. La-tâche-gsb, La tâche 6, 3, 1, 4-GSB, qui est ?-ancrée, est la représentante des trois tâches associées à l'ensemble noyau {-GSB, qui n'est pas ancrée, est sa propre représentante : c'est la seule tâche associée à l'ensemble noyau {, Quand on considère la table 6.1, il y a sept représentants canoniques. Ces tâches canoniques sont représentées dans la figure 6.1 où " A ? B " signifie " A inclut strictement B " . Remarquons que la tâche représentante 6 n'est pas ancrée, p.3, 2003.

. Démonstration, Comme n ? ?(m ? 1) ? u + 1, il existe un vecteur (avec m entrées) dont la première entrée est égale à u + 1 qui est un vecteur noyau de la tâche n, m, ?, u + 1-GSB. Mais comme u + 1 > u, ce vecteur ne peut pas être un vecteur noyau de la tâche n, m, ?, u-GSB. La tâche n, m, ?, u-GSB ne peut donc pas être ?-ancrée

. Démonstration, Le raisonnement est similaire à celui du théorème 6

. Démonstration, T. La-seule-différence-entre, and T. , borne supérieure sur le nombre de processus qui peuvent décider chaque valeur Si au plus u processus décident chaque valeur, alors nécessairement moins de u ? processus décident chaque valeur, et donc chaque vecteur de sortie de la tâche n, m, ?, u-GSB T est aussi un vecteur de sortie de la tâche n, m, ?

. Démonstration, Le raisonnement est similaire à celui du lemme 6

. De-la-tâche, (n + 1)-renommage (c'est-à-dire la tâche n, n + 1, 0, 1-GSB) de manière sans attente en utilisant des registres et une solution à la tâche du (n ? 1)-slot (c'est-à-dire la tâche n, n ? 1, 1, 2-GSB). L'objet qui résout la tâche n, n ? 1, 1, 2-GSB utilisé dans l'algorithme est noté KS . Il fournit aux processus une seule opération notée slot_request n?1 (

?. Le-processus-p-i, Si il n'observe aucun autre processus avec le même numéro de slot, ce numéro de slot devient son nouvel identifiant. (lignes 03-04) Dans le cas contraire, les propriétés de l'objet KS impliquent qu'il y a exactement un autre processus p j qui a le même numéro de slot (ligne 05) Les processus p i et p j sont alors en compétition pour un nouvel identifiant. De plus, il est possible que p j considère déjà son numéro de slot comme son nouvel identifiant. Le processus p i résout ce conflit selon l'ordre entre son identité et celle de p j : si l

. Démonstration and {. La-terminaison-sans-attente-et-le-fait-que-les-nouveaux-identifiants-appartiennent-À-l-'ensemble, n+1} sont des conséquences directes du texte de l'algorithme. Nous nous concentrons donc sur la démonstration du fait que deux processus ne peuvent pas obtenir le même nouvel identifiant. À cause des propriétés de l'objet KS , n ? 2 obtiennent des slots uniques parmi [1..n ? 1]. Soient p x et p y les processus qui reçoivent le même slot s. La démonstration se base sur le fait que les invocations snapshot(

R. Castañeda, . Ont, and . Dans, Weak Symmetry Breaking) peut être résolue de manière sans-attente pour certaines valeurs de n, mais n'ont pas donné d'algorithme permettant de le faire. Il serait intéressant de découvrir un tel algorithme ; cela permettrait de résoudre le (2n ? 2)-renommage non adaptatif (tâche n, 2n ? 2, 0, 1-GSB) [61] pour ces valeurs particulières de n. Les tâches GSB définies dans ce document ne sont pas adaptatives, p.1

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