PPCL is Sequentially Consistent
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Memory Model Background Definitions
Earlier Version With Additional Assumptions
This is a new version of the proof, updated 03/06/08. It does not make any extra assumptions, and defines Load Completeness as k <k l implies k <l l, and k <k l, l <l m implies k <k m (for any three loads k,l,m).
Proof outline
Let Z = (E,<w, <1,..., <n) be an execution.
We label each element in E with a phase. The phase of a store s is its rank in the total order of S(a(s)) given by <w. The phase of a load l is 0 if it reads the initial value and the rank of the store s s.t. l in load(s) otherwise. For each event e we let phi(e) denote its phase.
We now define a graph, the execution graph G(Z) on E. There are two kinds of edges in G(Z):
- cross-edges
- for all stores s, t and loads l:
- s -> t if a(s)=a(t) and phi(t)=phi(s)+1,
- s -> l if a(s)=a(l) and phi(t)=phi(s),
- l -> s if a(s) = a(l)+1 and phi(s)=phi(l)+1
- for all stores s, t and loads l:
- down-edges
- for all processors p and events d, e in E(p): d -> e if d <p e (and there is no other event in E(p) in between).
Say that a total ordering << of E is consistent if d << e implies d ->* e.
Claim: Any consistent total ordering of E is a sequentially consistent execution of E generating the same value for each load.
Therefore to show that PPCL is sequentially consistent we merely have to show that such a total ordering exists for every execution of a program P. (Assume that every two load/store instructions in the program for a processor are separated by a fence.)
Now a necessary and sufficient condition for a consistent total ordering << of E to exist is that G(Z) has no cycles.
We show that G(Z) has no cycles.
- Notation
- We call a path consisting only of stores a store sequence. We will let Y,Z range over store sequences. We let s,t,u,v range over stores and k,l,m over loads. We use the notation d <all e as shorthand for forall p in P: d <p e.
- Lemma 1
- s <all t whenever sZt is a path in G(Z).
- Proof
- By induction. Details are routine.
End of Proof
Let Q be a path in G(Z) with a segment skt. We say k is skipped (in Q) if s <all t. For instance, k is skipped if the edges s->k and k->t are cross-edges.
- Lemma 2
- Let Q be a path all of whose loads (if any) are skipped.
- (a) If sQt is a path in G(Z), then s <all t.
- (b) If kQl is a path in G(Z), then k <k l.
- Proof
- By structural induction on the number of skipped loads in Q, Lemma 1 and case analysis.
- Lemma 3
- Let Q be a path in G(Z), containing kRlu, with all loads in R (if any) skipped. Then k <k l whenever l has an incoming down-edge. Otherwise (l has an incoming cross-edge), l is skipped.
- Proof
- By use of Lemmas 1 and 2, and analysis of the ways in which a load is connected to its preceding and succeeding stores.
Suppose R is empty. Then the edge k->l must be a down-edge, hence k <l l.
Otherwise, let R=Ut be non-empty. Then k <k t by Lemma 2 and analysis of the out-edge from k. If t->l is a down-edge then t <all l, and hence k <k l. Therefore by Load Completeness, k <l l.
Let t->l be a cross-edge. If l->u is a cross-edge the t <all u (since t and u are writes to the same variable, in phase order). If l->u is a down-edge, then t <all u (since t <l l <l u and there is an hwsync between l and u). Thus l is skipped in both cases.
End of Proof
- Lemma 4
- Let Q be a path in G(Z). Let k1,..., kn be the only non-skipped loads in Q enumerated in the order in which they occur in Q. Then k1 <k1 kn.
- Proof
Consider the subsequence ki R ki+1 of Q (for i less than n). By Lemma 3, ki <ki ki+1. Hence, by application of transitivty of loads, (T), k1 <k1 kn.
End of Proof
- Main Lemma
- Let dQe be a path in G(Z) with d,e both loads or both stores. If p(d)=p(e) then d <d e.
- Proof
If Q is empty, the conclusion follows by a simple case analysis.
Suppose all loads in Q are skipped. Now d <all e if d,e in S (Lemma 2) or d <d e in L (Lemma 3). Thus in both cases, d <d e.
Now suppose Q has at least one non-skipped load. Let k be the first such load and l the last. By Lemma 4, k <k l.
If d in L, then d <d k by Lemma 3. If d in S, d <all s by Lemma 1, where s is is the store leading through a down-edge into k. Since there is an hwsync between s and k, d <all k. Thus in both cases d <k k.
Now either l=e or there is a path lRe with all loads in R skipped and l <l e (Lemma 2). Hence l =<l e.
From k <k l we get k <l l (by Load Completeness). Adding d <all k we get d <l l. Together with l =<l e we get d <l e.
Now since p(d)=p(e) either d <all e or e <all d (since there must be an hwsync between d and e). Since d <l e it follows that d<all e. Hence d <d e.
End of Proof
- Theorem
- All PPCL execution sequences of a program with fences between each pair of store/load operations are sequentially consistent.
- Proof
Let Z be a PPCL execution sequence. For no path dQe in G(Z) is it the case that d=e. Clearly d cannot be equal to e if one is a load and the other a store. But if they are both loads or both stores then by the Main Lemma d <d e. Hence G(Z) has no cycles. Therefore any total order << consistent with G(Z) gives an SC execution with the same values for each load as Z.
End of Proof

