Magic Graphs - download pdf or read online

By Alison M. Marr

ISBN-10: 0817683909

ISBN-13: 9780817683900

ISBN-10: 0817683917

ISBN-13: 9780817683917

Magic squares are one of the extra renowned mathematical recreations. over the past 50 years, many generalizations of “magic” principles were utilized to graphs. lately there was a resurgence of curiosity in “magic labelings” as a result of a couple of effects that experience purposes to the matter of decomposing graphs into bushes.

Key gains of this moment version include:

· a brand new bankruptcy on magic labeling of directed graphs

· functions of theorems from graph conception and engaging counting arguments

· new examine difficulties and workouts protecting a number difficulties

· a completely up to date bibliography and index

This concise, self-contained exposition is exclusive in its specialise in the speculation of magic graphs/labelings. it could possibly function a graduate or complex undergraduate textual content for classes in arithmetic or laptop technology, and as reference for the researcher.

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Extra resources for Magic Graphs

Sample text

Partition the integers from 1 to 2n into two classes A and B so that exactly one member of {i, 2n + 1 − i} is in each set, for every i. 2. Order the elements of A in a cyclic sequence so that the n sums of consecutive pairs comprise the integers from 2n + 2 to 3n + 1 in some order (or, what is equivalent, the elements of B must be ordered so that the consecutive sums are a permutation of {n + 1, . . , 2n}). 6 Wheels b1 an + a1 bn an b2 a1+a2 a1 43 a 2 + a3 a2 a3 b 3 Fig. 6. Labeling a wheel To visualize this process, it is perhaps easiest to consider a graph with vertices 1, 2, .

5 Complete Bipartite Graphs 41 For K3,3 one has 18 ≤ k ≤ 30, and k must be even. Cases k = 26, 28, 30 are dual to cases k = 22, 20, 18. The solutions are k = 18, no solutions k = 20, S1 = {1, 2, 3}, S2 S1 = {1, 2, 9}, S2 k = 22, S1 = {1, 2, 3}, S2 S1 = {1, 3, 5}, S2 S1 = {1, 5, 12}, S2 k = 24, no solutions. 26 In any edge-magic total labeling of a star, the center receives label 1, n + 1 or 2n + 1. Proof. Suppose the center receives label x. Then kn = 2n + 2 + (n − 1)x. 20) modulo n we find x ≡ (n + 1)(2n + 1) ≡ 1 and the result follows.

If T = di , define τ and Q by T = 2τ ·Q, τ integral, Q odd. If G has an edge-magic total labeling λ, then τ = 0 ⇒ v = 2δ V for some V ≡ 1 mod 2 e = 2δ E for some E ≡ 1 mod 2 τ = 1 ⇒ v = 2δ+1 V for some V ≡ 1 mod 2 e = 2δ+1 E for some E ≡ 1 mod 2 τ ≥ 2 ⇒ 2δ+2 divides v and 2δ+2 divides e Proof. Since e is even, let us write e = 2ν E for some odd integer E. The familiar equation 2e = d(xi ) yields 2e = v + 2δ T, so v + e = 3e − 2δ+τ ·Q = 2ν ·3E − 2δ+τ ·Q. 3) is (2ν ·3E − 2δ+τ ·Q)(2ν ·3E − 2δ+τ ·Q + 1)/2 = 2ν ·Ek − 2δ · di λ(xi ), which is of the form 2δ ·(2R − 2τ ·X) = 2ν ·Y where R = di λ(xi ), X = (2δ+τ Q−3 · 2ν+1 −1)Q and Y = 2Ek−2ν · 9E 2 − 3E.

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Magic Graphs by Alison M. Marr


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