By Mario Szegedy (auth.), Andrei A. Bulatov, Arseny M. Shur (eds.)

This e-book constitutes the lawsuits of the eighth foreign desktop technological know-how Symposium in Russia, CSR 2013, held in Ekaterinburg, Russia, in June 2013. The 29 complete papers provided during this quantity have been conscientiously reviewed and chosen from fifty two submissions. furthermore the e-book comprises eight invited lectures. The papers are equipped in topical sections on: algorithms; automata; common sense and facts complexity; complexity; phrases and languages; and common sense and automata.

**Read or Download Computer Science – Theory and Applications: 8th International Computer Science Symposium in Russia, CSR 2013, Ekaterinburg, Russia, June 25-29, 2013. Proceedings PDF**

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**Extra resources for Computer Science – Theory and Applications: 8th International Computer Science Symposium in Russia, CSR 2013, Ekaterinburg, Russia, June 25-29, 2013. Proceedings**

**Example text**

T. n. Let us deﬁne a continuous function F : R>0 → R by F (x) = x · lg x + 2 {x} . It is piecewise diﬀerentiable with right derivative lg x + 2. Furthermore, for x ≥ y > δ ≥ 0 we have the inequalities: F (x) + F (y) ≤ F (x + δ) + F (y − δ) and F (x) + F (y) ≤ F (x + y). Applying them we can prove the following (for details see [6]): 1. Let 1 ≤ ν ∈ R. For all sequences x1 , x2 , . . t. ν, we have i=1 2. lg n i=1 F n 2i F (xi ) ≤ lg ν F i=1 ≤ F (n) − 2n + O(lg n). ν 2i . 30 V. Diekert and A. Weiß Combining these facts with (2) yields the proof of Thm.

Heap extraction (sorting phase) with at most Text (n) comparisons (worst case). We analyze the three parts separately and put them together at the end. The partitioning is the only randomized part of our algorithm. The expected number of comparisons depends on the selection method for the pivot. For the expected number of comparisons by QuickHeapsort on the input array we obtain E[ T (n) ] ≤ Tcon (n) + Text (n) + E[Tpart (n)]. Theorem 1. The expected number E[ T (n) ] of comparisons by basic resp.

Wn )), and C consists of all the remaining indices. At each step we select the median of {frac(wi ) : i ∈ C} and compare it with frac(M (w1 , w2 , . . , wn )) using Lemma 1 and Lemma 3. Depending on the outcome of this comparison we move half of the elements of C into L or R. Observe that at each step of the computation values of wi with i ∈ L ∪ R are permanently round up or down. This suggest that if a whole segment i, i + 1, i + 2, . . , j belongs to L ∪ R already, we could try to somehow preprocess the function gwi ◦ gwi+1 ◦ gwi+2 ◦ .