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Generalized honeycomb torus

Generalized honeycomb torus. Hsun-Jung Cho , Li-Yen Hsu Information Processing Letters 86(2003) 185-190. Abstract.

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Generalized honeycomb torus

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  1. Generalized honeycomb torus Hsun-Jung Cho , Li-Yen Hsu Information Processing Letters 86(2003) 185-190

  2. Abstract • Stojmenovic intorduced three different honeycomb tori by adding wraparound edges on honeycomb meshes, namely honeycomb rectangular torus, honeycomb rhombic torus, and honeycomb hexagonal torus. • Discuss the Hamiltonian property of some generalized honeycomb tori.

  3. Honeycomb Network Ivan Stojmenovic Honeycomb Networks: Topological Properties and Communication Algorithms IEEE Trans. Parallel and Distributed Systems vol. 8, pp. 1036--1042, 1997.

  4. Definition – HRet (m,n)honeycomb rectangular torus m,n are both positive even integers. node set such that and are adjacent if they satisfy one of the conditions:

  5. Example – HRet (6,6)honeycomb rectangular torus j=5 Given the vertices j=4 j=3 j=2 j=1 j=0 i=0 i=1 i=2 i=3 i=4 i=5

  6. Definition – HRot (m,n)honeycomb rhombic torus m,n are positive integer where n is even. node set such that and are adjacent if they satisfy one of the conditions:

  7. Example – HRot (4,6)honeycomb rhombic torus j=8 Given the vertices j=7 j=6 j=5 j=4 j=3 j=2 j=1 j=0 i=0 i=1 i=2 i=3

  8. Definition – HM(n)honeycomb hexagonal mesh The honeycomb hexagonal mesh HM(n) is the graph with the node set : Two nodes and are adjacent if and only

  9. Honeycomb hexagonal mesh

  10. Definition – HT(n)honeycomb hexagonal torus The honeycomb hexagonal torus HT(n) is the graph with the same node set as HM(n). The edge set is the union of E(HM(n)) and the wraparound edge set { ((i,n-i+1, 1-n), (i-n, 1-i, n)) | 1  i  n}  {((1-n, i, n-i+1), (n, i-n, 1-i)) | 1  i  n } {((i, 1-n, n-i+1), (i-n, n, 1-i)) | 1  i  n }

  11. Example – HT(n)honeycomb hexagonal torus

  12. Definition – GHT(m,n,d)generalized honeycomb torus m,n are both positive even integers. d be any integer such that (m-d) is even. node set such that and are adjacent if they satisfy one of the conditions:

  13. Example – GHT(4,6,2)generalized honeycomb torus j=5 Given the vertices j=4 j=3 j=2 j=1 j=0 i=0 i=1 i=2 i=3

  14. j=17 Example – GHT(3,18,9)generalized honeycomb torus j=16 j=15 j=14 j=13 j=12 j=11 j=10 j=9 j=8 j=7 j=6 j=5 j=4 j=3 GHT(m,n,d) is a 3-regular Bipartite graph j=2 j=1 j=0 i=0 i=1 i=2

  15. Isomorphisms –HReT(m,n) GHT(m,n,0) j=5 j=5 j=4 j=4 j=3 j=3 j=2 j=2 j=1 j=1 j=0 j=0 i=0 i=1 i=2 i=3 i=0 i=1 i=2 i=3 HReT(4,6) GHT (4,6, ) 0

  16. Isomorphisms –HRoT(m,n) GHT(m,n,m(mod n)) j=8 j=5 j=7 j=4 j=6 j=3 j=5 j=2 j=4 j=1 j=3 j=2 j=0 i=0 i=1 i=2 i=3 j=1 j=0 i=0 i=1 i=2 i=3 GHT(4,6, ) 4 HRot(4,6)

  17. Isomorphisms –HT(n) GHT(n,6n,3n) • Let h be the function from the node set of HT(n) into the node set of GHT(n, 6n, 3n) by setting

  18. j=17 h(0,-2,3)=(0,0-(-2)+15)=(0,17) j=16 h(3,-2,1)=(1,3-(-2)+6)=(1,11) h(0,3,-2)=(-2+3,0-3+15)=(1,12) j=15 j=14 j=13 j=12 j=11 j=10 j=9 j=8 j=7 j=6 j=5 j=4 j=3 j=2 j=1 j=0 i=0 i=1 i=2

  19. Proof of isomorphisms about HT(n) and GHT(n,6n,3n) for any 1-n  c  n, we use Xc to denote the set of those nodes (x1,x2, x3) in HT(n) with x3=c. we use Yc to denote the set of nodes (i,j) in GHT(n, 6n, 3n) where (1) i = c+n and j  {k | 4n-c-3 < k < 6n}  {k | 0  k < n+c} if c < 0, (2) i =0 and j  { 1  j < 4n} if c = 0, (3) i = c and { j | c  j  4n-c} if 0 < c < n, and (4) i=0 and j  { k | 4n  k < 6n}  {0} if c =n.

  20. Proof of isomorphisms about HT(n) and GHT(n,6n,3n) Let hc denote the function of h induced by Xc. It is easy to check that hc is a one-to-one function from Xc onto Yc. Thus, h is one-to-one and onto. To prove h is an isomorphism, we need to check that h preserves the adjacency.

  21. Proof of isomorphisms about HT(n) and GHT(n,6n,3n) Suppose that e= ((x1,x2,x3),(x1’, x2’, x3’)) be an edge of HT(n). Without loss of generality, we assume that x1+x2+x3 = 2 and x1’+x2’+x3’=1. Suppose that e is an edge of HM(n). Then either x3 = x3’ or x3-x3’ =  1.

  22. Proof of isomorphisms about HT(n) and GHT(n,6n,3n) Case 1: x3 = x3’. Obviously, either (x1’, x2’,x3’) = (x1-1, x2,x3) or (x1’, x2’,x3’) = (x1, x2-1,x3). Suppose that 0  x3 < n. Then h(x1,x2,x3) = (x3,x1-x2+2n). Moreover, h(x1’,x2',x3')= (x3,x1-x2-1+2n) if (x1',x2',x3') = (x1-1, x2,x3) and h(x1', x2',x3')= (x3,x1-x2+1+2n) if (x1', x2',x3') = (x1, x2-1,x3).

  23. Proof of isomorphisms about HT(n) and GHT(n,6n,3n) Suppose that x3 = n. Then h(x1,x2,x3) = (0, x1-x2+5n (mod 6n)). Moreover, h(x1',x2',x3') = (x3,x1-x2-1+5n (mod 6n)) if (x1', x2',x3') = (x1-1,x2,x3) and h(x1', x2',x3')= (x3,x1-x2+1+5n (mod 6n)) if (x1',x2',x3') = (x1, x2-1,x3).

  24. Proof of isomorphisms about HT(n) and GHT(n,6n,3n) Suppose that x3< 0. Then h(x1,x2,x3) = (x3+n, x1-x2+5n (mod6n)). Moreover, h(x1', x2',x3')=(x3,x1-x2-1+5n (mod 6n)) if (x1',x2',x3') = (x1-1,x2,x3) and h(x1', x2',x3')= (x3,x1-x2+1+5n (mod 6n)) if (x1', x2',x3')= (x1, x2-1,x3). Hence, h(x1,x2,x3) and h(x1',x2',x3') are adjacent.

  25. Proof of isomorphisms about HT(n) and GHT(n,6n,3n) Case 2: x3-x3' =  1 Since x1+x2+x3 =2 and x1'+ x2'+x3'=1, (x1', x2',x3')=(x1,x2,x3-1). Suppose that 1  x3 < n. Then h(x1,x2,x3) = (x3,x1-x2+2n) and h(x1',x2',x3')=(x3-1, x1-x2+2n). Suppose that x3=0. Then h(x1,x2,x3) = (0, x1-x2+2n) and h(x1', x2',x3')=(n-1, x1-x2+5n (mod 6n)).

  26. Proof of isomorphisms about HT(n) and GHT(n,6n,3n) Suppose that x3=n. Then h(x1,x2,x3) =(0, x1-x2+5n (mod 6n)) and h(x1',x2',x3')=(n-1, x1-x2+2n). Suppose that 2-n  x3 -1. Then h(x1,x2,x3) = (x3+n, x1-x2+5n (mod 6n)) and h(x1',x2',x3')=(x3+n-1, x1-x2+5n (mod 6n)). Hence, h(x1,x2,x3) and h(x1',x2',x3') are adjacent.

  27. Proof of isomorphisms about HT(n) and GHT(n,6n,3n) Case 3: e  { ((i,n-i+1, 1-n), (i-n, 1-i, n)) | 1  i  n}. e is an wraparound edge of HM(n). Then (x1, x2, x3) = (i, n-i+1,1-n) and (x1', x2',x3') = (i-n, 1-i,n). Obviously, h(x1,x2,x3) = (1, 4n+2i-1) and h(x1', x2',x3') = (0, 4n+2i-1)). Hence, h(x1,x2,x3) and h(x1',x2',x3') are adjacent.

  28. Proof of isomorphisms about HT(n) and GHT(n,6n,3n) Case 4: e  { ((1-n, i, n-i+1), (n, i-n, 1-i)) | 1  i  n}. Hence (x1, x2, x3) = (1-n, i,n-i+1) and (x1', x2',x3')=(n, i-n, 1-i). Obviously, h(x1,x2,x3) is (0,4n) if i =1 and (n-i+1, n-i+1) if 1 < i  n. Similarly, h(x1',x2',x3') is (0, 4n-1)) if i=1 and (n-i+1,n-i) if 1 < i  n. Thus, h(x1,x2,x3) and h(x1', x2',x3') are adjacent.

  29. Proof of isomorphisms about HT(n) and GHT(n,6n,3n) Case 5: e  { ((i, 1-n, n-i+1), (i-n, n,1-i)) | 1  i  n}. Thus (x1, x2, x3) = (i, 1-n,n-i+1) and (x1', x2',x3')=(i-n, n,1-i). Obviously, h(x1,x2,x3) is (0,0) if i =1 and (n-i+1, 3n+i-1) if 1 < i  n. Similarly, h(x1',x2',x3') is (0,1)) if i=1 and (n-i+1,3n+i) if 1 < i  n. Again, h(x1,x2,x3) and h(x1', x2',x3') are adjacent.

  30. Hamiltonian – HRet(m,n) j=5 j=4 j=3 j=2 j=1 j=0 i=0 i=1 i=2 i=3 HReT(4,6)

  31. Hamiltonian – HRot(m,n) j=8 j=7 j=6 j=5 j=4 j=3 j=2 j=1 j=0 i=0 i=1 i=2 i=3

  32. Hamiltonian – GHT(m,2k,k) Case 1 : m is even. k is even

  33. Hamiltonian – GHT(m,2k,k) Case 2 : m is odd. k is odd

  34. Discussion • Generalized honeycomb tori include honeycomb rectangular tori, honeycomb rhombic tori, honeycomb hexagonal tori. • All generalized honeycomb tori are Hamiltonian.

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