Adv Alegbra HW Solutions 80

Adv Alegbra HW Solutions 80 - 80 Solution. True. 2.138 How...

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Unformatted text preview: 80 Solution. True. 2.138 How many flags are there with n stripes each of which can be colored any one of q given colors? Solution. We use Proposition 2.164. Here, G = τ , where τ = (1 n )(2 n − 1) · · · (k k + 1) if n = 2k , and τ = (1 n )(2 n − 1) · · · (k − 1 k + 1)(k ) if n = 2k + 1. Therefore, PG (x1 , . . . , xn ) = 1n 2 (x1 1n 2 (x1 k + x2 ) k + x1 x2 ) if n = 2k if n = 2k + 1. Thus, PG (q , . . . , q ) = 1 (q n + q k ) or 1 (q n + q k +1 ), depending on the 2 2 parity of n . One can give a ”master” formula: PG (q , . . . , q ) = 1 (q n + q [n /2] ). 2 2.139 Let X be the squares in an n × n grid, and let ρ be a rotation by 90◦ . Define a chessboard to be a (q , G )-coloring, where the cyclic group G = ρ of order 4 is acting. Show that the number of chessboards is 1 4 2 qn + q (n 2 +1)/2 + 2q (n 2 +3)/4 , where x is the greatest integer in the number x . Solution. The group G that acts here is a cyclic group ρ of order 4, where ρ is (clockwise) rotation by 90◦ . As n ≡ 0, 1, 2, or 3 mod 4, we have n 2 ≡ 0 or 1 mod 4. Now G is acting on the n 2 squares, and one sees that ρ is a product of n 2 /4 disjoint 4-cycles in the first case, whereas it is a similar product with a 1-cycle otherwise. Now ρ 3 = ρ −1 has the same cycle structure as ρ , while each 4-cycle splits into two disjoint 2-cycles in ρ 2 . Thus, when n 2 ≡ 0 mod 4, n 2 /4 n PG (x1 , . . . , xn ) = 1 (x1 n + 2x4 4 and PG (q , . . . , q ) = 1 (q n + 2q n 4 2 /4 n 2 /2 + x2 + qn 2 /2 ) ). There is a similar formula when n 2 ≡ 1 mod 4, and both can be combined into the formula using greatest integer notation. ...
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

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