Download Fun with Algorithms: 5th International Conference, FUN 2010, by Roberto Grossi, Alessio Orlandi, Giuseppe Ottaviano (auth.), PDF

By Roberto Grossi, Alessio Orlandi, Giuseppe Ottaviano (auth.), Paolo Boldi, Luisa Gargano (eds.)

This publication constitutes the lawsuits of the fifth overseas convention, enjoyable 2010, held in June 2010 in Ischia, Italy. enjoyable with algorithms is a three-yearly convention that goals at atractings works which, in addition to a deep and fascinating algorithmic content material, additionally current a laugh and enjoyable features. The 32 complete papers and three invited talks are rigorously chosen from fifty four submissions and concentrate on themes equivalent to distibuted algorithms, graph computations, parallelism, zero-knowledge evidence, iphone, development matching and process video games.

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Additional info for Fun with Algorithms: 5th International Conference, FUN 2010, Ischia, Italy, June 2-4, 2010. Proceedings

Example text

Let x > 1 satisfy nr = n/2 + x, so that nb = n/2 − x. First assume that x ≤ n1/2 . In that case, every red player guesses red with probability x − 1 / n1/2 = ( x − 1)/ n1/2 , and every blue player guesses blue with probability 1 − x / n1/2 . Therefore, the expected number of correct guesses is A Hat Trick 39 (n/2 + x)( x − 1)/ n1/2 + (n/2 − x)(1 − x / n1/2 ) = (n/2 + x) x / n1/2 − (n/2 + x)/ n1/2 + (n/2 − x)(1 − x / n1/2 ) ≥ (n/2 − x) x / n1/2 − (n/2 + x)/ n1/2 + (n/2 − x)(1 − x / n1/2 ) ≥ (n/2 − x) − (n/2 + x)/ n1/2 ≥ n/2 − 4n1/2 , which is at least max{nr , nb } − O(n1/2 ), since max{nr , nb } ≤ n/2 + O(n1/2 ).

For a strip of the cards, a folded state is a flat folding of unit length (where the unit is the width of a card) such that each crease is consistent with its label. ) The main problem in this paper is the following: Input: A strip of n + 1 Kaboozle cards, each with a label of length m. Question: Determine whether the strip has a folded state that is consistent with the labels, and exactly one connected path is drawn on a surface of the folded state. , “MVMV· · ·MV” or “MVMV· · ·MVM”. Proof. Suppose that a mountain-valley pattern has a unique folded state.

Otherwise, |χr (i) − χb (i)| ≥ 2 and so we have either χr (i) = n/2 + c for some c > 0 or χb (i) = n/2 + c for some c > 0 (but not both). In the former case Paula takes a(i) = min{ n1/2 , c } and b(i) = n1/2 , so that p(i) = min{1, c / n1/2 } and in the latter case she takes a(i) = n1/2 − min{ n1/2 , c } and b(i) = n1/2 , so that p(i) = 1 − min{1, c / n1/2 }. Note that a(i), b(i) and p(i) can be computed in polynomial time. Having p(i) at hand, Paula draws a uniformly random real p in the unit interval, guesses red if p ≤ p(i) and blue otherwise.

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