A Tight Bound for EMAC - CWI

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Lemma 4. Let n, q ≥ 1 and 1 ≤ m1,...,mq ≤ ℓ and M = {M1,...,Mq} with Mi ∈ B mi n be distinct messages, then |G 1 col(M)| ≤q(q + ℓ + ℓ 2 )/2 Now combining (6)-(8) and the above Lemma we get: Lemma 5. CPq,n,ℓ ≤ q(q+ℓ+ℓ2 ) 2(2n−ℓq) + 4q4ℓ 4 22n This already gives CPq,n,ℓ ∈ O(q 2 /2 n )forq 2 ℓ 4 ∈ O(2 n )andq ∈ Ω(ℓ 2 ). But we can do better. The reason why this bound is not so great is that the term which bounds the “two or more” accident case is of rather large order q 4 ℓ 4 /2 n as we consider a graph (i.e. a total message length) of size qℓ. Weachievethe bound claimed by Lemma 1 by splitting the messages in chunks of size ℓ 2 (with foresight) and then applying the following lemma which is a generalisation of (1). Lemma 6. If r divides q then CPq,n,ℓ ≤ CP2r,n,ℓ · q(q−r) 2·r 2 Proof. Consider q messages M1,...,Mq where CPq,n,ℓ = CPn(M1,...,Mq). We split the q messages into q/r sets S1,...,Sq/r, each containing r messages. If two messages collide, then there are two sets containing this two messages, so using the union bound CPn(M1,...,Mq) ≤ i,j,1≤i 1 as otherwise the above is ≥ 1 which is a trivial upper bound for CPq,n,ℓ. Wealsohaveq
We will prove two claims, which then will imply the statement of the lemma. The first claim — which is basically Lemma 19 from [BPR05] — states that if i ∈ Pj, then there’s at most one structure graph with exactly one accident where Mi and Mj collide. The second claim bounds the number of structure graphs having one accident and a collision between Mj and any other message Mi where i ∈Pj by ℓ(ℓ+1)/2 (note that this bound only depends on the length, but not on the number of messages considered). To prove this claim we use the simple observation that if there’s a collision between Mj and any Mi where i ∈Pj, then it must be the case that the j-path makes a loop. So we can upper bound the number of structure graphs having such a collision by the number of structure graphs having where the j-path loops. P Claim 1 For each i ∈ Pj, |Gi,j| ≤1. S Fig. 3. Figure for proof of Claim 1 Proof (of Claim). Let P denote the common prefix and S the common suffix of Mi and Mj. SoMi = P M ′ i S and Mj = P M ′ j S where the M ′ i and M ′ j are nonempty as i ∈ Pj. Letp = |P |/n, s = |S|/n. By definition G ∈Gi,j means V mi mj i = Vj , this implies that also V mi−s i = (as for the last s steps the i and j path must go in parallel). Now as M mi−s−1 i = M mj−s−1 j (otherwise we could extend the suffix) we have V mi−s−1 i = (because in a structure graph two edges with distinct labels cannot be parallel). So there’s a true collision in G [i,j] which hits the vertex V mi−s i . V mj−s j V mj−s−1 j As by fact (i) 9 there can be only one true collision in G [i,j] this means that the “suffix path” V mi−s i = V mj−s j same reason the “prefix path” V 1 i → ... → V mi i = V 1 j mj = Vj has no loops. For the p p → ... → V = V makes no loop and also the prefix and suffix paths must be disjoint. So the subgraph of G [i,j] built by the first p +1 and the last s +1edges of the i and j path looks like shown on the left in Figure 3. There’s only way to extend this subgraph to the full G [i,j] without introducing more true collisions, this is the second graph in Figure 3. So there’s only one possible G [i,j], and by fact (ii) it uniquely determines the whole structure graph, thus there’s just one G ∈Gi,j. △ Claim 2 ≤ ℓ(ℓ +1)/2. i∈P Gi,j 9 We refer to the facts stated at the end of Section 3. i j
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Page 9: e an accident. And if we only consi

A Tight Bound for EMAC - CWI

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