By Avcibas, Memon, Sankur, Sayood

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**Extra info for A Progressive Lossless Near-Lossless Image Compression Algorithm**

**Sample text**

Carole determines various x’s, and Paul answers the questions truthfully. How many questions does Carole need, in the worst case? There is an entire area called Combinatorial Group Testing that produces solutions for such problems [80]. In the data stream case, each question is a group of items and the algorithm plays the role of Carole. This setup applies to a number of problems in data streams. Examples are found in: • Determining the B heavy-hitter Haar wavelet coeﬃcients in Turnstile data streams [104].

Suppose x ∈ S: the frequency of x is (n/k)/(n + n/k) ≤ 1/(k + 1), and so x will not be output. On the other hand, if x ∈ S then (n/k + 1)/(n + n/k) > 1/(k + 1) and so x will be output. Hence we can extract the set S and the space stored must be Ω(N ) since, by an information theoretic argument, the space to store an arbitrary subset S is N bits. The lower bound above also applies to randomized algorithms. Any algorithm which guarantees to output all heavy hitters (with ε = 0) with probability at least 1 − δ, for some constant δ, must also use Ω(N ) space.

Theorem 9. Any algorithm which guarantees to ﬁnd all and only items i such that A[i] > (1/k + 1)||A||1 must store Ω(N ) bits. Proof. Consider a set S ⊆ {1 · · · N }. Transform S into a stream of length n = |S|. Process this stream with the proposed algorithm. We can use the algorithm to extract whether x ∈ S or not: for some x, insert n/k copies of x. Suppose x ∈ S: the frequency of x is (n/k)/(n + n/k) ≤ 1/(k + 1), and so x will not be output. On the other hand, if x ∈ S then (n/k + 1)/(n + n/k) > 1/(k + 1) and so x will be output.