MAT 101 Lecture 12: MAT 101 lecture 12
9 Mar 2018
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How does anyone solve one of these puzzles. We"re looking at one way that a solution (cid:272)ould (cid:271)e (cid:271)uilt (cid:271)y (cid:272)onsidering properties of permutations and the cycle structure of permutations: today we"ll pra(cid:272)ti(cid:272)e those ideas. Easy example: f: cycle structure (ufl, ufr, dfr, dfl) & (uf, fr, df, fl) cycle structure of the inverse? (dfl, dfr, ufr, ufl) & (fl, df, fr, uf) it"s just the first (cid:272)y(cid:272)le stru(cid:272)ture (cid:271)a(cid:272)kwards. Easy example: f: cycle structure (ufl, ufr, dfr, dfl) & (uf, fr, df, fl) order, the order is the least common multiple of the cycle lengths. Fi or r r fi or ri ri fi) how many pieces were affected by. Cycle lengths of 6, 5, and 2 the smallest number they all go into is 30. Example: fr: how many pieces were affected by fr, there are 13 pieces affected, six corners and 7 edges.
