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Xiaofan Yang , David J. Evans ,Hongjian Lai , GrahamM. Megson
Information Processing Letters 92
(2004) 3137
As presented by Ying-Jhih Chen
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A (di)graph is called hamiltonian if itcontains Hamilton cycle.
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Let n be a positive even integer, mbe a positive integer, and dbe anonnegative integerthat is less than
n and is of the same parity as m. An(m,n, d) generalized honeycombtorus, denotedby GHT(m,n, d), is a
graph with vertex set V( GHT(m, n, d) )={ : i { 0, 1,
, m1 }, j { 0, 1, , n1} }.
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Note: Here and in what follows, all arithmetic operationscarried out on the first and second components aremodulo m and n, respectively.
Two vertices and with i k are adjacentif
and only if one of the following three conditions issatisfied:
(a) = or = ;
(b) 0 i m2, i+j is odd, and = ;
(c) i = 0, j is even, and = .
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Let p and q be two positive integers.Let g(p, q) denote the smallestpositive integer s satisfying p s = 0
(mod q). Then g(p, q) = q / gcd(p,q) .
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p ( q / gcd(p, q) ) = ( p / gcd(p, q) ) q 0 (modq)
Let s be an integer with 1 s ( q / gcd(p,q) ) 1.
Ifp s = 0 (mod q)
Since gcd( p / gcd(p,q) , q / gcd(p,q) ) = 1
g(p, q) q /gcd(p,q)
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g(p, q) q /gcd(p,q)
r N s.t.p s =r q.( p/gcd(p,q) ) s = r ( q /
gcd(p,q) )
q / gcd(p,q) | s ( 1 s ( q / gcd(p,q) )
1 )
g(p, q) = q / gcd(p,q)
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Given two positive integers a and b,we need to consider a graph G(a, b)that has {0, 1, . . .,a 1} as the
vertex set and { < i, i + b >: 0 i a 1} as the edge set, where thearithmetic is modulo a.
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If gcd(a, b) = 1, then G(a, b) is acycle (loop and multiple edgesinclusive).
e. g. G( 3, 2)
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2
1
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Consider the infinite sequence (0, b,2b, 3b, . . .) of neighboring vertices.
By Lemma 1
(0, b, 2b, 3b, . . ., (a 1)b, 0)forms a cycle.
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.
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If m is even, then set {P(k gcd(n,d), gcd(n, d)): 0 k ( n / gcd(n,d) ) 1} constitutes a path
decomposition of GHT(m,n, d). (Wecall this path decomposition asstandard path decomposition.)
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GHT(4, 12, 4):
P(0, 4)
P(4, 4)
P(8, 4)
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If m is even. Then GHT(m,n, d) isHamiltonian.
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We construct a graph G( n /gcd(n,d) , d/gcd(n,d) )
By Lemma 2, the sequence of neighboring
vertices ( 0, d/gcd(n, d), 2 d/gcd(n, d), . . . ,((n /
gcd(n, d)) 1 ) d/gcd(n, d), 0 )
forms a cycle.
This cycle can be extended to aHamiltonian cycle of GHT(m,n, d) accordingto the following steps:
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Step 1: For each i with 0 i (n /gcd(n,d)) 1, let
V (i)= { : 0 p m1, i gcd(n, d) q (i + 1) gcd(n, d) 1.
Step 2: Replace each vertex i of G( n
/gcd(n,d) , d/gcd(n,d) ) with V (i).
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Step 3: Replace each edge (i, i + d)of G( n /gcd(n,d) , d/gcd(n,d) ) witha path of GHT(m,n, d) obtained from
path P(igcd(n, d), gcd(n, d)) byadding the following edge:
(, ).
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P(0, 4)
P(4, 4)
P(8, 4)
GHT(4, 12 ,4)
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If m is odd, then GHT(m,n, d) is Hamiltonian. Poof: We construct a graph G( n /gcd(n,d+1) , d+1
/gcd(n,d+1) ) in the way given in Lemma 2.
By Lemma 2, the sequence
( 0, (d + 1)/gcd(n, d + 1), 2 ((d +1)/gcd(n, d + 1)),. . . , (n/gcd(n, d + 1) 1) ((d + 1)/gcd(n, d + 1)),0 )
Forms a cycle.This cycle can be extended to a
Hamiltonian cycle of GHT(m,n, d) according to thesome steps.
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GHT (5, 12, 5 )
P(6, 2), P(8, 4)
P(0, 2), P(2, 4)
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Every generalized honeycomb torusis Hamiltonian.
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