By Kurt Mehlhorn (auth.), Tiziana Calamoneri, Irene Finocchi, Giuseppe F. Italiano (eds.)

This publication constitutes the refereed lawsuits of the sixth Italian convention on Algorithms and Computation, CIAC 2006, held in Rome, Italy, in may perhaps 2006.

The 33 revised complete papers awarded including three invited papers have been rigorously reviewed and chosen from eighty submissions. one of the subject matters addressed are sequential, parallel and disbursed algorithms, facts buildings, approximation algorithms, randomized algorithms, online algorithms, graph algorithms, research of algorithms, set of rules engineering, algorithmic online game thought, computational biology, computational complexity, conversation networks, computational geometry, cryptography, discrete optimization, graph drawing, mathematical programming, and quantum algorithms.

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Claim 3. The coverage of every rectangle u is greater than (1 − 4dε) in the partial cover (x, y). Proof. Consider a rectangle u. We show that, in each dimension, the coverage of u decreases by less than 4ε due to the transition from y ∗ to y. By definition, coverage by lines in H is preserved. In addition, if a rectangle u intersects all the ∗ lines in a block Lhj , then the coverage of u by lines in Lhj is now covered by Sh,j . ∗ ∗ Namely, S∈Lh y (S, u) = y(Sh,j , u). It follows that u may lose coverage only j in the “leftmost” and “rightmost” blocks that u intersects.

Lhb(h) and the (possibly empty) leftover block by L construction, ε ≤ S∈Lh x∗ (S) < 2ε for every j ≤ b(h) and S∈L˜ h x∗ (S) < ε. j The same type of partitioning is applied to the vertical lines in Lv to obtain the ˜v. blocks Lv1 , . . , Lvb(v) and the leftover block L Observation 5. The number of blocks (not including the leftover block) in each dimension satisfies b(h) ≤ 1ε · S∈Lh x∗ (S) and b(v) ≤ 1ε · S∈Lv x∗ (S). ∗ ∗ Let Sh,j and Sv,j denote lines of maximum capacity in Lhj and Lvj , respectively.

Fy (S) < x(S) · c(S)), then obviously S is not thirsty, so S is a dam. , fy (S) = c(S)) and yet not thirsty. Such a case is easily described using the network flow formalism: the arc (s, S) belongs to a min-cut in Nx but not to every min-cut. Lemma 1. Let (x, y) be a partial cover such that x is integral and y is maximum with respect to x. , fy (u) = 1), and (2) if u ∈ S and y(S , u) > 0, then S is also a dam. Proof. Proof of (1). If u is not covered, then an increase in c(S) can be used to increase y(S, u), contradicting the assumption that S is a dam.

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