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Unformatted text preview: Fall 2009 CS131 – Combinatorial Structures Homework 5 Homework 5, due Oct 20 You must prove your answer to every question. Do not rely only on the homework for exercise: there are several selfcheck ex ercises of the easier kind in the book, try to solve them, too! The following two problems are designed to train you in careful, systematic thinking and work. Problem 1. In class, we looked at necklaces containing 8 blue and 2 red beads, say at positions 0 9 around a circle. Two necklaces were considered equivalent, when one could be rotated into the other one. Now, let us allow a necklace to be turned over: allow reflection around any axis as in a mirror that results in a permutation of the beads. (Exchanging bead 0 with 9, 1 with 8, and so on, is one such reflec tion. Exchanging bead 2 with 9, 3 with 8, and so on, leaving 0 and 5 in place, is another.) We call two necklaces equivalent, if one can be turned into the other by a combination of rotations and (possibly) reflections. (a) (10pt) How many equivalence classes are there now for 2 red beads and 8 blue beads? Solution. In class we have seen that when rotations are allowed, there were 4 equivalence classes of necklaces each containing 10 elements, and one equiva lence class containing 5 elements: the one in which the red beads are opposite to each other. Each class was determined by the distance between the two red beads. Reflection does not change the distance of the red beads, therefore it does not unite any different classes: we still have the same classes as before. (b) (20pt) How many equivalence classes are there for 3 red beads and 7 blue beads, if reflections are allowed? Solution. You did not have to count the size of each equivalence class, but I will do it anyway, for selfchecking. The total number of necklaces is ( 10 3 ) = 120....
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 Fall '11
 FredPhelps
 Math, Equivalence relation, combinatorial structures, red beads

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