By Reneta P. Barneva, Valentin E. Brimkov, Josef Slapal

This quantity constitutes the refereed lawsuits of the sixteenth foreign Workshop on Combinatorial picture research, IWCIA 2014, held in Brno, Czech Republic, in may perhaps 2014. The 20 revised complete papers and three invited papers offered have been conscientiously reviewed and chosen from quite a few submissions. the themes lined contain discrete geometry and topology in imaging technological know-how, new leads to photo illustration, segmentation, grouping, and reconstruction, clinical photograph processing.

**Read or Download Combinatorial Image Analysis: 16th International Workshop, IWCIA 2014, Brno, Czech Republic, May 28-30, 2014. Proceedings PDF**

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**Additional resources for Combinatorial Image Analysis: 16th International Workshop, IWCIA 2014, Brno, Czech Republic, May 28-30, 2014. Proceedings**

**Example text**

The above result implies the following theorem. Theorem 1. A kDIR −CON V graph on n vertices can have at most nk maximal cliques. Proof: By Proposition 1, every clique can be deﬁned by a set V(b1 , . . , bk ). Since there are at most nk ways to choose V(b1 , . . , bk ), it follows that there can be at most nk cliques and thus at most nk maximal cliques. The above result can be somewhat strengthened, as the following theorem demonstrates. While this improvement still remains in O(nk ), it does lower the bound implied by Proposition 1.

SIAM Monographs on Discrete Mathematics and Applications, vol. 2. SIAM, Philadelphia (1999) 25. : Sur deux propri´et´es des classes d’ensembles. Fund. Math. 33, 303–307 (1945) 26. : Robust computation of intersection graph between two solids. Graphical Models 16(3), C79–C88 (1997) 27. : A uniﬁed algorithm for ﬁnding maximum and minimum object enclosing rectangles and cuboids. Computers Math. Applic. 29(8), 45–61 (1995) 28. : Extracting the cliques from a neighbourhood system. IEE Proc. Vision Image and Signal Processing 144(3), 168–170 (1997) 29.

Let G be the intersection graph of S. Then every maximal clique contains exactly one segment from each direction. Hence, the number of maximal cliques k in G is nk = (1/k)k nk = Ω(nk ) for any ﬁxed integer k. This, coupled with the upper bound of Theorem 1, implies the stated result. 3 Algorithmic Aspects The results of the previous section do imply an O(nk+2 ) algorithm for computing all maximal cliques in a kDIR − CON V graph. For this, one can ﬁrst construct all V(b1 , . . , bk ) sets in O(nk ) time.