By Kurt Mehlhorn (auth.), Tiziana Calamoneri, Irene Finocchi, Giuseppe F. Italiano (eds.)
This booklet constitutes the refereed court cases of the sixth Italian convention on Algorithms and Computation, CIAC 2006, held in Rome, Italy, in may well 2006.
The 33 revised complete papers offered 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 video game concept, computational biology, computational complexity, conversation networks, computational geometry, cryptography, discrete optimization, graph drawing, mathematical programming, and quantum algorithms.
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Extra resources for Algorithms and Complexity: 6th Italian Conference, CIAC 2006, Rome, Italy, May 29-31, 2006. Proceedings
There is a source s that feeds all the lines. The capacity of each arc (s, S) emanating from the source equals x(S) · c(S). There is a sink t that is fed by all the rectangles. The capacity of every arc (u, t) entering the sink equals 1. Observation 1. There is a one-to-one correspondence between vectors y such that (x, y) is a partial cover and ﬂows f in Nx . The correspondence y ↔ fy satisﬁes fy (u, t) = S|u∈S y(S, u), for every rectangle u ∈ U, and fy (s, S) = u∈S y(S, u), for every line S ∈ S.
6: for each rank ρ in DR ∩ [c, . . , c + ci1 ] do 7: Let be the line stored at A2 [ρ − c]. 8: Update DI to record the pair ( 1 , ) as the intersection with rank ρ. } 10: Let c := c + ci1 . 11: i1 := i1 + 1. } 13: i2 := i2 + 1. 14: end if 15: end for algorithm CountAndRecord is an iterative algorithm that computes, during the i-th iteration the element with rank i in the ﬁnal sorted order (Line 3 of Algorithm 2). Thus we only need to be able to ﬁnd out whether the line that will be the i-th element in sorted order is an element involved in an intersection.
We denote the LP-relaxation by lp-soft. The integrality gap of both lp-hard and lp-soft is at least 2 − o(1) even in the one-dimensional case. Consider an instance that contains k + 1 rectangles and two lines of capacity k that intersect all the rectangles. A fractional optimal solution is x∗ (S) = (k + 1)/(2k) for each line S and y ∗ (S, u) = 1/2 for every line S and rectangle u. This means that the value of the fractional minimum is 1 + k1 , while the integral optimum is 2. The following deﬁnitions apply to both lp-hard and lp-soft.
Algorithms and Complexity: 6th Italian Conference, CIAC 2006, Rome, Italy, May 29-31, 2006. Proceedings by Kurt Mehlhorn (auth.), Tiziana Calamoneri, Irene Finocchi, Giuseppe F. Italiano (eds.)