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Unformatted text preview: Math 3124 Tuesday, September 6 September 6 Ungraded Homework The first position is 8 10 7 9 6 12 5 11 4 14 3 13 2 1 15 . We first move the space to the bottom right hand corner; there will be several ways of doing this, and it doesnt matter which we choose. One way will give the position 8 10 7 9 6 12 5 11 4 14 3 13 2 1 15 . The corresponding permutation is 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 8 10 7 9 6 12 5 11 4 14 3 13 2 1 15 Writing this as a product of disjoint cycles, we obtain ( 1 8 11 3 7 5 6 12 13 2 10 14 )( 4 9 ) The 12cycle is a product of 11 ( = 12 1) transpositions, so the whole permutation is a product of 11 + 1 transpositions. 12 is even, so we conclude that the given position can be obtained from the starting position. The second position is 1 2 3 4 8 7 6 5 9 10 11 12 15 14 13 . We first move the space to the bottom right hand corner; there will be several ways of doing this, and it doesnt matter which we choose. One way will give the position 1 2 3 4 8 7 6 5 9 10...
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 Fall '08
 PARRY
 Math, Algebra

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