A Deterministic Algorithm for Summarizing Asynchronous Streams ...

assigned to each of the subsequent nodes of p ′ , we get:|X 1 | ≤ 2 i · (|p ′ | − 1) + 2 i+1 = 2 i · |p| ≤ 2 i · (1 + log W ). (2)With a similar analysis (the details are omitted due to space constraints) incan be shown that:|X 2 ∪ X 3 | ≤ 2 i · (1 + log W ). (3)Combining Equations 1, 2, and 3 we can bound s t − e i t :−2 i · (1 + log W ) ≤ −|X 1 | ≤ s t − e i t ≤ |X 2 ∪ X 3 | ≤ 2 i · (1 + log W ).Therefore, |s t − e i t | ≤ 2i · (1 + log W ). In case S i ′ = ∅, ei t = s t − |X 3 | = 0, and thesame bound follows immediately.⊓⊔Lemma 2. When asked for an estimate of the number of timestamps in [c−w, c](1)There exists a level i ∈ [0, M] such that T i < c − w, and(2)Algorithm 3 returns e l c−w where l ∈ [0, M] is the smallest level such thatT l < c − w.The proof of Lemma 2 is omitted due to space constraints, and can be foundin the full version [3]. Let l denote the level used by Algorithm 3 to answer aquery for the number of timestamps in [c − w, c]. From Lemma 2 we know lalways exists.Lemma 3. If l > 0, then s c−w ≥(1+log W )·2lɛ.Proof. If l > 0, it must be true that T l−1 ≥ c − w, since otherwise level l − 1would have been chosen. Let t = T l−1 + 1. Then, t > c − w, and thus s c−w ≥ s t .From Lemma 1, we know s t ≥ e l−1t − (1 + log W ) · 2 l−1 . Thus we have:s c−w ≥ e l−1t − (1 + log W ) · 2 l−1 (4)We know that for each bucket b ∈ S l−1 , l(b) ≥ t. Further we know that eachbucket in S l−1 has a weight of at least 2 l−1 (only the initial bucket in S l−1may have a smaller weight, but this bucket must have split, since otherwise T l−1would still be −1). Since there are α buckets in S l−1 , we have:e l−1t ≥ α2 l−1 ≥ (1 + log W ) · 2 + ɛ · 2 l−1ɛ(5)The lemma follows from Equations 4 and 5.⊓⊔Theorem 1. The answer returned by Algorithm 3 is within an ɛ relative errorof s c−w .Proof. Let X denote the value returned by Algorithm 3. If l = 0, it can beverified that Algorithm 3 returns exactly s c−w (proof omitted due to space constraints).If l > 0, from Lemmas 1 and 2, we have |X − s c−w | ≤ (1 + log W ) · 2 l .Using Lemma 3, we get |X − s c−w | ≤ ɛ · s c−w as needed.⊓⊔