notes-091105 - MATH 681 Notes Combinatorics and Graph...

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Unformatted text preview: MATH 681 Notes Combinatorics and Graph Theory I 1 Restricted symmetry-enumerations So far we’ve built what is essentially a magic bullet to answer the question: “if we freely paint an object with k colors, how many distinct colorings are there subject to symmetry group G ?” Recall that our answer to that question was: Cyc( G ) x i = k | G | However, if the painting is restricted in some way, then we’ll need a different formula; recall that our central conceit Inv( π ) = Cyc( π ) x i = k was based on the idea that each cycle can be freely painted any of k colors, which might not be true if we restrict our color choices. By analogy to our old friend, the twelvefold way, we can look at two particular coloring restrictions. 1.1 Using each color no more than once If we restrict ourselves to using each color no more than once, we start by reducing the size of the non-symmetry-reduced S : instead of | S | = k n , we now have | S | = k ! ( n- k )! (and if the number of features n of the object being colored exceeds the number of colors, no such colorings exist, and | S | = 0). Otherwise, however, this case will be easy to analyze, because most permutations have no invariants in S . Proposition 1. If S consists of the colorings of an object not using the same color twice, and π is a nontrivial permutation, then Inv( π ) = 0 . Proof. Since π is nontrivial, there must be some i 6 = j such that π ( i ) = j . Suppose x is an invariant of π . Then it must be the case that x i = x j , since i is mapped onto j by the permutation π . But since i 6 = j , it would thus follow that two distinct elements of x are the same color, which would preclude x being in S ; thus our premise that x is an invariant of π is contradictory, and there can be no suce x . Therefore, Burnside’s Lemma is easy to apply in this circumstance, since all but one element of the sum of the invariants will be zero: 1 | G | X π ∈ G Inv( π ) = 1 | G | Inv( e ) = | S | | G | Note that, in this special case, Burnside’s Lemma reduces down to the simple symmetry-reduction techniques we learned very early in this course! Examples of these cases were given as the bridge- table and first beaded-necklace problem from October 22nd’s notes. 1.2 Using each color no less than once Here we have a case which is in a number of ways unfamiliar, and not quite as simplifiable as either the free selection or the unique-color case. We shall start with an example to work our way towards a general theory: Question 1: We have 3 colors which we shall use to color the faces of an octahedron. If we must use each color at least once, how many ways (up to rotation-equivalence) are there to do this? What if we had n colors?...
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notes-091105 - MATH 681 Notes Combinatorics and Graph...

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