4 DECOMPOSITION OF COMPLETE GRAPHS INTO CIRCULANT...
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International Journal of Computer Engineering and Technology (IJCET), ISSN 0976-6367(Print),
ISSN 0976 - 6375(Online), Volume 4, Issue 6, November - December (2013), © IAEME
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DECOMPOSITION OF COMPLETE GRAPHS INTO CIRCULANT GRAPHS
AND ITS APPLICATION
1Jayanta Kr. Choudhury,
2Anupam Dutta,
3Bichitra Kalita
1Swadeshi Academy Junior College, Guwahati-781005
2Patidarrang College, Muktapur, Loch, Kamrup, Assam
3Department of Computer Application, Assam Engineering College, Guwahati-781013
ABSTRACT
In this paper, the decomposition of complete graphs 12 +mK for 2≥m into circulant graphs has
been discussed. Two theorems have been established for different values of 2≥m relating to 12 +m
is prime, n
m 3312 ⋅=+ for 1≥n and sm 312 =+ for 5≥s is also prime. Finally, an algorithm under
different situation for the traveling salesman problem have been discussed when the weights of edges
are non-repeated of the complete graph 12 +mK for 2≥m .
Keywords: Algorithm, Circulant Graphs, Complete Graphs, Hamiltonian Graphs,
Traveling Salesman Problem (TSP).
1. INTRODUCTION
Circulant graph have many applications in areas like telecommunication network, VLSI
design and distributed computing. Circulant graph is a natural extension of a ring, with increased
connectivity. Zbgniew R. Bogdanowicz [1] give the necessary and sufficient conditions under which
a directed circulant graph G of order n and with k jumps can be decomposed into k pair wise arc-
disjoint anti-directed Hamiltonian cycles, each induced by two jumps. In addition, the necessary
condition for complete decomposition of G into arbitrary anti-directed Hamiltonian cycles has been
discussed. Matthew Dean [2] shown that there is a Hamiltonian cycle decomposition of every 6-
regular circulant graph n
S in which S has an element of order n , { }0\nZS ⊆ has vertex set nZ
and edge set [ ]{ }SsZxsxx n ∈∈+ ,|, . S. El-Zanati, Kyle King and Jeff Mudrock [3] discussed the
cyclic decomposition of circulant graphs into almost bipartite graphs. Dalibor Froncek [4] used
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labeling to show the cyclic decomposition of complete graphs into eK nm +, : the missing case. J.C.
Bermond, O. Favaron and M. Maheo [7] prove that any 4-regular connected Cayley graph on a finite
abelian group can be decomposed into two Hamiltonian cycles. Daniel K. Biss [5] shown that the
circulant graph ( )dcdGm , is Hamiltonian decomposable for all positive integers dc, and m with
dc < where the graph ( )dNG , has vertex set { }1.,,.........2,1,0 −= NV , with { }wv, an edge if
( )Ndwv mod±≡− for some 1log0 −≤≤ Ni d . This extends work of Michenean [6]. Marsha F.
Foregger [8] settles Kotzig’s [9] conjecture that the Cartesian product of any three cycles can be
decomposed into three Hamiltonian cycles. He also show that the Cartesian product of ba
32 graphs,
each decomposed into mba32 Hamiltonian cycles. Dave Morris [10] discussed the status of the
search for Hamiltonian cycles in circulant graphs and circulant digraphs. R. Anitha and R. S. Lekshmi
[13] have proved that the complete graph nK2 can be decomposed into 2−n n -suns, a Hamiltonian
cycle and a perfect matching, when n is even and for odd cases, the decomposition is 1−n n -suns
and a perfect matching. They also discussed that a spanning tree decomposition of even order
complete graph using the labeling scheme of n -suns decomposition. A complete bipartite graph nnK ,
can be decomposed into 2
n n -suns when
2
n is even. When
2
n is odd, nnK , can be decomposed into
( )2
2−n n -suns and a Hamiltonian circuit [13]. Tay-Woei Shyu [14] investigated the decomposition of
complete graph nK cycles tC ’ s and stars kS ’ s and studied necessary or sufficient conditions for
such a decomposition to exist. Dalibor Froncek [15] show that every bipartite graph H which
decomposes kK and nK also decomposes knK . Darryn E. Bryant and Peter Adams [16] discussed
that for all 3≥v , the complete graph vK on v vertices can be decomposed into 2−v edge disjoint
cycles whose lengths are 1,,5,4,3,3 −vKK and for all 7≥v , vK can be decomposed into 3−v edge
disjoint cycles whose lengths are vvvv ,1,2,4,,4,3 −−−KK . Fu, Hwang, Jimbo, Mutoh and Shiue
[17] considered the problem of decomposing a complete graph into the Cartesian product of two
complete graphs rK and cK and they found a general method of constructing such decomposition
using various sorts of combinatorial designs. The traveling salesman problem (TSP) in the circulant
weighted undirected graph case have been discussed by I. Gerace and F. Greco [11]. They discussed
an upper bound and a lower bound for the Hamiltonian case and analyzed the two stripe case.
Moreover, I. Gerace and F. Greco [12] studied short SCTSP (symmetric circulant traveling salesman
problem) and they presented an upper bound, a lower bound and a polynomial time 2-approximation
algorithm for the general case of SCTSP (symmetric circulant traveling salesman problem). Further,
different types of algorithms and results have also been focused by B.Kalita [19-22], J.K. choudhury
[23-24] and A. Dutta[25-26].
In this paper, we present a theorem for decomposition of complete graphs 12 +mK into m edge
disjoint circulant graphs ( )jC m 12 + for 2≥m where j is the jump, mj ≤≤1 . Thereafter, we present
a theorem for the decomposition of 12 +mK into edge disjoint circulant graphs whenever 12 +m is a
prime, 1,3312 ≥⋅=+ nmn
and sm 312 =+ , 5≥s is a prime and this circulant graphs lead to
determine edge disjoint Hamiltonian cycle. Finally, an algorithm has been developed to determine the
least cost route of a traveler.
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The paper is organized as follows The section 1 includes the introduction part containing works of other researcher. Section 2
includes notations and terminologies. In section 3, two theorems are stated and proved. Section 4
includes an algorithm. Section 5 explains experimental result and the conclusion is included in
section 6.
2. NOTATION AND TERMINOLOGY
The notation and terminologies have been considered from the standard references [1-29]
Definition 2.1[18] (circulant matrix): Every nn × matrix C of the form
=
−
−
−−
−−
0121
12
01
2101
1210
cccc
cc
cc
cccc
cccc
C
n
n
nn
nn
L
OO
MOM
L
is called a circulant matrix. The matrix C is completely determined by its first row because other
rows are rotations of the first row. C is symmetric if iin cc =− for 1,.......,2,1 −= ni . Further, C is
an adjacency matrix if 00 =c and { }1,01 ∈=− in cc .
Definition 2.2[18] (circulant graph): A circulant graph is a graph which has a circulant adjacency
matrix.
Examples of circulant graphs are the cycle nC , the complete graph nK , and the complete
bipartite graph nnK , .
Circulant Graph 2.3 [27] : For a given positive integer, let knnn ,,, 21 KK be a sequence of integers
where
( )2
10 21
+<<<<<
pnnn kKK .
Then the circulant graph ( )kp nnnC ,,, 21 KK is the graph on p nodes pvvv ,,, 21 KK with vertex
iv adjacent to each vertex ( )pni jv mod± . The values in are called jump sizes.
Definition 2.4[28]: A graph G is said to be decomposable into the sub
graphs iHHHH ,,,, 321 KK , ki ≤≤1 , having no isolated vertices and
( ) ( ) ( ) ( ){ }kHEHEHEHE ,,,, 321 KK is a partition of ( )GE .
3. THEOREMS
3.1 Theorem: The complete graph 12 +mK for 2≥m can be decomposed to m edge disjoint circulant
graph ( )jC m 12 + where j is the jump and mj ≤≤1 .
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Proof: Consider the complete graph 1225 +×= KK . Then vertex set of 5K is { }54321 ,,,, vvvvvV = and
it has ( )
102
45
2
1=
×=
−nn edges. Let us partitioned the edges of 5K
into 2 distinct sets containing
5 edges each as { }15544332211 ,,,, vvvvvvvvvvE = and { }25145342312 ,,,, vvvvvvvvvvE = respectively.
Adjoining edges of 1E and 2E , we can construct 2 edge disjoint circulant graphs ( )15C and ( )25C
for the jumps 1=j and 2=j respectively and their respective permutations can be represented as 1f
and 2f where
=
15432
54321
1vvvvv
vvvvvf and
=
21543
54321
2vvvvv
vvvvvf
i.e., ( )543211 vvvvvf = and ( )425312 vvvvvf = . Figure-1 shows two circulant graphs ( )15C and ( )25C
for the complete graph 5K .
( )15C ( )25C
Figure-1
Now, we append the edges to the circulant graphs as
( )
=
≤≤=
−
+
5
41
4 iifv
iifvvf
i
ji
ij , 1=j
and ( )
≤≤
≤≤=
−
+
54
31
3 iifv
iifvvf
i
ji
ij , 2=j
Similarly, consider the complete graph 1327 +×= KK whose vertex set is { }7654321 ,,,,,, vvvvvvvV =
and have ( )
212
67
2
1=
×=
−nn edges. Let us partitioned the edges of 7K into 3 distinct sets
consisting of 7 edges each, as { }177665544332211 ,,,,,, vvvvvvvvvvvvvvE = ,
{ }271675645342312 ,,,,,, vvvvvvvvvvvvvvE = and { }372615746352413 ,,,,,, vvvvvvvvvvvvvvE = . Adjoining
7 edges each of 1E , 2E and 3E , we can construct 3 edge disjoint circulant graphs ( )17C , ( )27C and
( )37C for the jumps 1=j , 2=j and 3=j respectively and their respective permutations are 1f , 2f
and 3f where
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=
1765432
7654321
1vvvvvvv
vvvvvvvf ,
=
2176543
7654321
2vvvvvvv
vvvvvvvf and
=
3217654
7654321
3vvvvvvv
vvvvvvvf
i.e., ( )76543211 vvvvvvvf = , ( )64275312 vvvvvvvf = and ( )52637413 vvvvvvvf = . Figure-2 shows circulant
graphs ( )17C , ( )27C and ( )37C .
v
1
v2
v3
v4
v5
v6
v7
v1
v2
v3
v4 v
5
v6
v7
v1
v2
v3
v4
v5
v6
v7
( )17C ( )27C ( )37C
Figure-2
The edges to the circulant graphs can be appended as
( )
=
≤≤=
−
+
7
61
6 iifv
iifvvf
i
ji
ij , 1=j
( )
≤≤
≤≤=
−
+
76
51
5 iifv
iifvvf
i
ji
ij , 2=j
and ( )
≤≤
≤≤=
−
+
75
41
4 iifv
iifvvf
i
ji
ij , 3=j
Similarly, partitioning the edges of 9K into 4 distinct sets containing 9 edges each as
{ }1998877665544332211 ,,,,,,,, vvvvvvvvvvvvvvvvvvE = ,
{ }2918978675645342312 ,,,,,,,, vvvvvvvvvvvvvvvvvvE = ,
{ }3928179685746352413 ,,,,,,,, vvvvvvvvvvvvvvvvvvE =
and { }4938271695847362514 ,,,,,,,, vvvvvvvvvvvvvvvvvvE =
respectively, we can construct 4 edge disjoint circulant graphs ( )19C , ( )29C , ( )39C and ( )49C for the
jumps 1=j , 2=j , 3=j and 4=j respectively and their corresponding permutations are given by
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( )987654321
198765432
987654321
1 vvvvvvvvvvvvvvvvvv
vvvvvvvvvf =
= ,
( )864297531
219876543
987654321
2 vvvvvvvvvvvvvvvvvv
vvvvvvvvvf =
= ,
( )( )( )963852741
321987654
987654321
3 vvvvvvvvvvvvvvvvvv
vvvvvvvvvf =
=
and ( )627384951
432198765
987654321
4 vvvvvvvvvvvvvvvvvv
vvvvvvvvvf =
=
respectively, whose structures are shown in Figure-3.
V
1
V2
V3
V4
V5 V
6
V7
V8
V9
V1
V2
V3
V4
V5 V
6
V7
V8
V9
V1
V2
V3
V4
V5 V6
V7
V8
V9
( )19C ( )29C ( )39C ( )49C
Figure-3
Append the edges to the circulant graphs as
( )
=
≤≤=
−
+
9
81
8 iifv
iifvvf
i
ji
ij , 1=j
( )
≤≤
≤≤=
−
+
98
71
7 iifv
iifvvf
i
ji
ij , 2=j
( )
≤≤
≤≤=
−
+
97
61
6 iifv
iifvvf
i
ji
ij , 3=j
and ( )
≤≤
≤≤=
−
+
96
51
5 iifv
iifvvf
i
ji
ij , 4=j
V1
V2
V3
V4
V5 V6
V7
V8
V9
V1
V2
V3
V4
V5 V
6
V7
V8
V9
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From the above discussion, it is seen that the complete graph 12 +mK for 2≥m can be
decomposed into m edge disjoint circulant graph ( )jC m 12 + . Here j represents the jump of the
circulant also depends upon m . If 2=m , then 2,1=j and we have 2 circulant graphs ( )15C and
( )25C .
Again, if 3=m , then 3,2,1=j and then 7K decomposes into 3 edge disjoint circulant
graphs ( )17C , ( )27C and ( )37C and so on. Moreover, in their respective permutations, the images of
first element and last element in the first row are respectively 1+jv and jv . Here we excluded the
identity permutations as well as permutations which produce the same circulant graphs,
i.e.
15432
54321
vvvvv
vvvvv and
43215
54321
vvvvv
vvvvv represents the same circulant graph ( )15C .
Therefore, as discussed above, we get 2 permutations if 2=m , 3 permutations if 3=m etc, i.e.,
number of permutations are depends upon the value of 2≥m .
Finally, we can comment that the complete graph 12 +mK for 2≥m can be decomposed into
m edge disjoint circulant graph ( )jC m 12 + for the jump mj ≤≤1 and their corresponding m
permutations are defined as
( )( )
( ) ( )
+≤≤−−
−−≤≤=
−−+
+
1222
121
21 mijmifv
jmiifvvf
mji
ji
ij
for mj ≤≤1 and which complete the theorem.
Note: Here, complete graphs 12 +mK for 2≥m has odd number of vertices and therefore they play
different role for different values of 2≥m . In the next theorem we discuss only three cases.
3.2 Theorem: Let 12 +mK be a complete graph for 2≥m . Then following conditions are satisfied
(a) If 12 += mp is a prime number, then pK can be decomposed into m edge disjoint circulant
graphs ( )jC p where j is the jump and mj ≤≤1 .
(b) If nmq 3312 ⋅=+= where ( )13
2
1 1 −= +nm for 1≥n , then qK can be decomposed into
(i) n3 edge disjoint circulant graphs ( )jCq for the jump mj ≤≤1 but lj 3≠ for
( )132
11 −≤≤ n
l and
(ii) a circulant graph ( )lCq 3,,9,6,3 KK for the jump l3 where ( )132
11 −≤≤ n
l .
(c) If smr 312 =+= where 5≥s is a prime number, then rK can be decomposed into
(i) 1−s edge disjoint circulant graphs ( )jCr for the jump mj ≤≤1 where sj ≠ and ej 3≠ ,
( )13
11 −≤≤ me
(ii) a circulant graph ( )eCr 3,,9,6,3 KK for the jump ( )13
11 −≤≤ me and
(iii) a circulant graph ( )sCr for the jump s .
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Proof: (a) When =+= 12mp prime, then it follows that the complete graph 5K , 7K , 11K ,
13K ,……….. for KKK,6,5,3,2=m exist and the proof is same as discussed in Theorem 3.1. Hence
the complete graph pK where 12 += mp is a prime for 2≥m can be decomposed into ( )12
1−= pm
edge disjoint circulant graphs ( )jC p for the jump mj ≤≤1 .
In the above discussion, we observed that each edge disjoint circulant graph ( )jC p for the
jump ( )12
11 −≤≤ pj is regular of degree two. In other words, they are cycle pC of length p and
therefore they represent edge disjoint Hamiltonian circuits. So in this discussion, number of edge
disjoint Hamiltonian circuits in the complete graph pK is m if 12 += mp is a prime for 2≥m .
Proof: (b) Let 12 +mK be a complete graph for 2≥m . Suppose that n
mq 3312 ⋅=+= such that
( )132
1 1 −= +nm for 1≥n . Then { }qvvvvV ..,,.........,, 321= is the vertex set of qK and it has ( )
2
1−qq edges.
Let us define ( ) ( )qqj KVKVf →: as follows:
(i) when ( )132
11 1 −≤≤ +n
j but lj 3≠ for ( )132
11 −≤≤ n
l 1≥∀n
( )
≤≤+−
−≤≤=
−+
+
qijqifv
jqiifvvf
qji
ji
ij1
1
and (ii) when ( )132
11 −≤≤ n
l 1≥∀n , ( )
≤≤+−
−≤≤=
−+
+
qilqifv
lqiifvvf
qli
li
il13
31
3
3
3
For 1=n , 4=m and 1339142 ⋅==+×=q . Then { }987654321 ,,,,,.,, vvvvvvvvv is the vertex set of
9K have ( )
36942
89
2
1=×=
×=
−qq edges. Since 4=m , we can partition the edges of 9K into 4 edge
disjoint sets, each set containing 9 edges, by the above rule and their permutations are given as
follows.
For 1=j , ( )987654321
198765432
987654321
1 vvvvvvvvvvvvvvvvvv
vvvvvvvvvf =
=
For 2=j , ( )864297531
219876543
987654321
2 vvvvvvvvvvvvvvvvvv
vvvvvvvvvf =
=
For 4=j , ( )627384951
432198765
987654321
4 vvvvvvvvvvvvvvvvvv
vvvvvvvvvf =
=
Again, for 1=l , ( )( )( )963852741
321987654
987654321
3 vvvvvvvvvvvvvvvvvv
vvvvvvvvvf =
=
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From these permutations, we get 4 edge disjoint sets
{ }1998877665544332211 ,,,,,,,, vvvvvvvvvvvvvvvvvvE = ,
{ }2918978675645342312 ,,,,,,,, vvvvvvvvvvvvvvvvvvE = ,
{ }3928179685746352413 ,,,,,,,, vvvvvvvvvvvvvvvvvvE = and
{ }4938271695847362514 ,,,,,,,, vvvvvvvvvvvvvvvvvvE = .
Adjoining edges of 1E , 2E and 4E , we get 331 = edge disjoint circulant graphs ( )19C ,
( )29C and ( )49C for the jumps 1=j , 2=j and 4=j respectively [Figure-3]. Also, adjoining the
edges of 3E , we get a circulant graph ( )39C for the jump l3 where 1=l . Thus 9K decomposed into
13 edge disjoint circulant graphs ( )jC9 for 41 ≤≤ j , 3≠j and a circulant graph ( )lC 39 for 1=l
[Figure-3].
For 2=n , 13=m and 233271132 ⋅==+×=q . Then { }27321 ,,, vvvv KKK is the vertex set of the
complete graph 27K and have ( )3512713
2
2627
2
1=×=
×=
−qq edges. Since 13=m , we can
partition the edges of 27K into 13 edge disjoint sets, each set containing 27 edges, by the rule
defined above.
For 1=j , ( )2726252423222120191817161514131211109876543211 vvvvvvvvvvvvvvvvvvvvvvvvvvvf =
For 2=j , ( )2624222018161412108642272523211917151311975312 vvvvvvvvvvvvvvvvvvvvvvvvvvvf =
For 4=j , ( )2420161284272319151173262218141062252117139514 vvvvvvvvvvvvvvvvvvvvvvvvvvvf =
For 5=j , ( )2318138325201510527221712722419149426211611615 vvvvvvvvvvvvvvvvvvvvvvvvvvvf =
For 7=j , ( )2114727201362619125251811424171032316922215817 vvvvvvvvvvvvvvvvvvvvvvvvvvvf =
For 8=j , ( )2012423157261810221135241682719113221462517918 vvvvvvvvvvvvvvvvvvvvvvvvvvvf =
For 10=j , ( )18825155221221992616623133201027177241442111110 vvvvvvvvvvvvvvvvvvvvvvvvvvvf =
For 11=j , ( )17622112716521102615420925143198241321872312111 vvvvvvvvvvvvvvvvvvvvvvvvvvvf =
For 13=j , ( )15216317418519620721822923102411251226132714113 vvvvvvvvvvvvvvvvvvvvvvvvvvvf =
Again,
For 1=l , ( )( )( )2724211815129632623201714118522522191613107413 vvvvvvvvvvvvvvvvvvvvvvvvvvvf =
For 2=l , ( )( )( )2418126272115932317115262014822216104251913716 vvvvvvvvvvvvvvvvvvvvvvvvvvvf =
For 3=l , ( )( )( )( )( )( )( )( )( )2718926178251672415623145221342112320112191019 vvvvvvvvvvvvvvvvvvvvvvvvvvvf =
For 4=l , ( )( )( )18621924122715317520823112614216419722102513112 vvvvvvvvvvvvvvvvvvvvvvvvvvvf =
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From these permutations we can define corresponding edge disjoint sets 1E , 2E , 3E , 4E , 5E ,
6E , 7E , 8E , 9E , 10E , 11E , 12E and 13E respectively.
Adjoining edges of 1E , 2E , 4E , 5E , 7E , 8E , 10E , 11E and 13E , we can construct 932 = edge
disjoint circulant graphs ( )127C , ( )227C , ( )427C , ( )527C , ( )727C , ( )827C . ( )1027C , ( )1127C and
( )1327C for the jumps 1=j , 2=j , 4=j , 5=j , 7=j , 8=j , 10=j , 11=j and 13=j
respectively. Also, adjoining edges of 4E , 6E , 9E and 12E , we get a circulant graph ( )12,9,6,327C
for the jump l3 where 41 ≤≤ l . Thus the complete graph 27K can be decomposed into 932 =
edge disjoint circulant graph ( )jC27 where j is the jump such that 131 ≤≤ j , 12,9,6,3≠j and a
circulant graph ( )12,9,6,327C .
Let kn = so that ( )132
1 1 −= +km and
133312 +=⋅=+= kkmq . Then { }qvvvv ,,,, 321 KKK is
the vertex set the complete graph 13 += kKK q and it has ( ) ( )2
133
2
1 11 −=
− ++ kkqq edges.
Since ( )132
1 1 −= +km , so we can partition edges of 13 += kKK q into ( )132
1 1−
+k distinct sets,
each set containing 13 +k edges, given by
(i) when ( )132
11 1 −≤≤ +kj but lj 3≠ for ( )13
2
11 −≤≤
kl
( )
≤≤+−
−≤≤=
−+
+
qijqifv
jqiifvvf
qji
ji
ij1
1
and (ii) when ( )132
11 −≤≤ kl
( )
≤≤+−
−≤≤=
−+
+
qilqifv
lqiifvvf
qli
li
il13
31
3
3
3
Now, we can have edge disjoint sets 1E , 2E , 3E , 4E ,……………,( )13
2
1 1−+kE and their corresponding
permutations are 1f , 2f , 3f , 4f ,…………….,( )13
2
1 1−+kf .
Adjoining edges of 1E , 2E , 4E , 5E , 7E ,……, ( )53
2
1 1 −+kE ,
( )132
1 1 −+kE , we can construct
k3 edge
disjoint circulant graphs ( )jC k 13 + where j is the jump and ( )132
11 1 −≤≤ +k
j , lj 3≠ for
( )132
11 −≤≤ k
l . Also, adjoining edges of 3E , 6E , 9E ,……,( )73
2
1 1 −+kE ,
( )332
1 1 −+kE , we get a circulant
graph ( )lC k 3,,9,6,313KK+ for the jump ( )13
2
11 −≤≤ kl .
Now, adding 132 +⋅ k additional vertices to the vertex set and ( )11 394 ++ −⋅ kk
additional edges to the
edge set of the complete graph 13 +kK . Then { }2221 313133321 ,,,,,,,, ++++ −+ kkkk vvvvvvv KK will be the vertex
set for the complete graph 12 333 ++ ⋅= kk KK i.e., 23 += kKK q and
( ) ( ){ }132
113
2
13123 11222 −=−=⇒=+⇒= +++++ kkkk mmq .
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The complete graph 23 +kK has ( ) ( ) ( ) ( ){ }
2
133
2
133
2
1111122
−=
−=
−++++++ kkkk
qq edges. These edges can be
partition into ( ) ( ){ }132
113
2
1 112 −=− +++ kk disjoint sets and their respective permutations are given by
(i) when ( )132
11 2 −≤≤ +kj but lj 3≠ for ( )13
2
11 1 −≤≤ +kl
( )
≤≤+−
−≤≤=
−+
+
qijqifv
jqiifvvf
qji
ji
ij1
1
and (ii) when ( )132
11
1−≤≤
+kl , ( )
≤≤+−
−≤≤=
−+
+
qilqifv
lqiifvvf
qli
li
il13
31
3
3
3
Now, we can have ( ){ }13
2
1 11 −++k edge disjoint sets 1E , 2E , 3E , 4E , …,
( )132
1 2 −+kE and their
respective permutations are 1f , 2f , 3f , 4f ,………,( )13
2
1 2 −+kf . Adjoining edges of 1E , 2E , 4E , 5E ,
7E ,……..,( )73
2
1 2 −+kE ,
( )532
1 2 −+kE ,
( )132
1 2 −+kE , we can construct
13 +k edge disjoint circulant graph
( )jC k 23 + i.e., ( ) ( )jC k 113 ++ where j is the jump and ( )( )132
11 11 −≤≤ ++k
j , lj 3≠ for ( )132
11 1 −≤≤ +kl .
Also, adjoining edges of 3E , 6E , 9E ,….., ( )93
2
1 2 −+kE ,
( )332
1 2 −+kE , we have a circulant graph
( ) ( )lC k 3,,9,6,3113KK++ for the jump ( )13
2
11 1 −≤≤ +k
l . Thus the result is true for 1+= kn whenever
it is true for kn = . Hence by principle of mathematical induction, the result is true for all Nn ∈ .
In this discussion, we observed that each edge disjoint circulant graph ( )jCq
for the jump
( )132
11 1 −≤≤ +n
j , lj 3≠ for ( )132
11 −≤≤ n
l are regular of degree two. In other words they are
cycle qC of length q and therefore they represent n3 edge disjoint Hamiltonian circuits 1≥n .
Further the circulant graph ( )lC q 3,,9,6,3 KK for the jump ( )132
11 −≤≤
nl , as we have observed,
can be decomposed into ( )132
1−n edge disjoint circulant graphs ( )tCq 3 for the jump t3 ,
( )132
11 −≤≤ n
t for all 1≥n . If the jump is
nt 33 = , then
n3 3K -decomposition
133 −= n
t , then n3 9C -decomposition
233 −= nt , then
n3 27C -decomposition
………………………………………..
………………………………………..
333 =t , then
n3 23 −nC -decomposition
233 =t , then
n3 13 −nC -decomposition
133 =t , then
n3 nC3
-decomposition
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Clearly above cycles does not represent Hamiltonian circuits. So in this discussion, number of edge
disjoint Hamiltonian circuits in the complete graph qK is n if
nq 33⋅= for 1≥n .
(c) Let 12 +mK be a complete graph for all 2≥m . Suppose that smr 312 =+= where 5≥s is a
prime number and 2
13 −=
sm . Then { }rvvvvV ..,,.........,, 321= is the vertex set of the complete graph
rK i.e., sK3 and have ( )2
1−rr edges.
Let us define ( ) ( )ssj KVKVf →: as follows:
(i) when ( )132
11 −≤≤ sj but sj ≠ and ej 3≠ for ( )1
2
11 −≤≤ se
( )
≤≤+−
−≤≤=
−+
+
rijrifv
jriifvvf
rji
ji
ij1
1
(ii) when ( )12
11 −≤≤ se , ( )
≤≤+−
−≤≤=
−+
+
rierifv
eriifvvf
rei
ei
ie13
31
3
3
3
and (iii) ( )
≤≤+−
−≤≤=
−+
+
risrifv
sriifvvf
rsi
si
is1
1
For 5=s , a prime number, we have 1553 =×=r and 72
153=
−×=m .
Then { }1514321 ,..,,.........,, vvvvv is the vertex set of the complete graph 15K and have
( )105157
2
1415
2
1=×=
×=
−rredges. Since 7=m , we can partition the edges of 15K into 7 disjoint
sets, each set containing 15 edges and their corresponding permutations by the above rule are as
follows:
For 1=j , ( )1514131211109876543211 vvvvvvvvvvvvvvvf = and
{ }1151514141313121211111010998877665544332211 ,,,,,,,,,,,,,, vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvE =
For 2=j , ( )1412108642151311975312 vvvvvvvvvvvvvvvf = and
{ }2151141513141213111210119108978675645342312 ,,,,,,,,,,,,,, vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvE =
For 4=j , ( )1284151173141062139514 vvvvvvvvvvvvvvvf = and
{ }4153142131121511141013912811710695847362514 ,,,,,,,,,,,,,, vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvE =
For 7=j , ( )9210311412513614715817 vvvvvvvvvvvvvvvf = and
{ }7156145134123112101915814713612511410392817 ,,,,,,,,,,,,,, vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvE =
For 1=e , ( )( )( )1512963141185213107413 vvvvvvvvvvvvvvvf = and
{ }3152141131512141113101291181079685746352413 ,,,,,,,,,,,,,, vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvE =
For 2=e , ( )( )( )1261593115148210413716 vvvvvvvvvvvvvvvf = and
{ }6155144133122111101591481371261151049382716 ,,,,,,,,,,,,,, vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvE =
For 5=s , ( )( )( )( )( )1510514941383127211615 vvvvvvvvvvvvvvvf = and
{ }5154143132121111510149138127116105948372615 ,,,,,,,,,,,,,, vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvE =
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Adjoining edges of 1E , 2E , 4E and 7E , we can construct 4151 =−=−s edge disjoint circulant
graph ( )115C , ( )215C , ( )415C and ( )715C respectively for jumps 1=j , 2=j , 4=j
and 7=j [Figure-5]. Also, adjoining edges of 3E and 6E , we can construct a circulant graph
( )6,315C [Figure-6]. Again, adjoining the edges of 5E , we can construct a circulant graph ( )515C for
the jump 5=s [Figure-7]. V
1
V2
V3
V5
V6
V7
V8V
9
V11
V10
V12
V13
V14
V15
V4
V1
V2
V3
V5
V6
V7
V8
V9
V11
V10
V12
V13
V14
V15
V4
( )115C ( )215C ( )415C ( )715C
Figure-5
V1
V2
V3
V5
V6
V7
V8 V
9
V11
V10
V12
V13
V14
V15
V4
V1
V2
V3
V5
V6
V7
V8V
9
V11
V10
V12
V13
V14
V15
V4
( )6,315C ( )515C
Figure-6 Figure-7
Thus the complete graph 15K can be decomposed into 4151 =−=−s edge disjoint circulant
graphs ( )jC15 for 71 ≤≤ j , 6,5,3≠j and a circulant graph ( )6,315C and a circulant graph ( )515C
for 5=s . Similarly, we can show for KK,13,11,7=s etc.
Let ks = , a prime number. Then kr 3= , 2
13 −=
km and the complete graph kK3 has
( ) ( )2
133
2
1 −=
− kkrr edges. Since 2
13 −=
km , so we can partition the edges of kK3 into ( )13
2
1−k
edge disjoint sets, each set containing k3 edges and their respective permutations are given by
(i) when ( )132
11 −≤≤ kj but kj ≠ and ej 3≠ for ( )1
2
11 −≤≤ ke
( )
≤≤+−
−≤≤=
−+
+
rijrifv
jriifvvf
rji
ji
ij1
1
(ii) when ( )12
11 −≤≤ ke , ( )
≤≤+−
−≤≤=
−+
+
rierifv
eriifvvf
rei
ei
ie13
31
3
3
3
and (iii) ( )
≤≤+−
−≤≤=
−+
+
rikrifv
kriifvvf
rki
ki
ik1
1
V1
V2
V3
V5
V6
V7
V8V
9
V11
V10
V12
V13
V14
V15
V4
V1V
2
V3
V5
V6
V7
V8
V9
V11
V10
V12
V13
V14
V15
V4
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Now, we can have edge disjoint sets 1E , 2E , 3E , 4E ,……,( )13
2
1−k
E and their respective permutations
are 1f , 2f , 3f , 4f ,………,( )13
2
1−k
f . Adjoining edges of jE ’s where ( )132
11 −≤≤ kj ; kj ≠ and
ej 3≠ for ( )12
11 −≤≤ ke , we can construct 11 −=− ks edge disjoint circulant graphs ( )jC k3
where j is the jump. Again, adjoining the edges of eE3 ’s, ( )12
11 −≤≤ ke we can construct a
circulant graph ( )eC k 3,,9,6,33 KK for the jump e3 . Also, adjoining the edges of kE , we can
construct a circulant graph ( )kC k3 for the jump k .
In this discussion we observed that each edge disjoint circulant graphs ( )jCr for the jump
( )132
11 −≤≤ sj , sj ≠ and ej 3≠ for ( )1
2
11 −≤≤ se where 5≥s is a prime, are regular of degree
two. In other words they are cycles rC of length r and therefore they represent edge disjoint
Hamiltonian circuits. Again the circulant graph ( )eCr 3,,9,6,3 KK for the jump e3 where
( )12
11 −≤≤ se can be decomposed into ( )sm − edge disjoint circulant graphs t3 where
( )12
11 −≤≤ st and these circulants are regular of degree two each and they can be further
decomposed into ( )1−m cycles sC of length rs < and so they do not represent Hamiltonian
circuits. Also, the circulant graph ( )sCr for the jump s is regular of degree two and from which we
can have s 3K -decompositions and they do not represent a Hamiltonian circuit. So in this discussion,
number of edge disjoint Hamiltonian circuits in the complete graph rK is 1−s if sr 3= , 5≥s is a
prime.
4. ALGORITHM
Input: Let 12 +mK for 2≥m be a complete graph.
Output: Find least cost route.
Case 1: When 12 += mp is a prime number and ( )12 +m consecutive greatest weights are incident
at different vertices of the complete graph 12 +mK for 2≥m along edges of a cycle 12 +mC , next
( )12 +m greatest weights incident at different vertices along edges of a cycle 12 +mC other than the
previous one, next ( )12 +m greatest weights incident at different vertices along edges of a cycle
12 +mC other than the previous two cycles and so on.
Step 1: Study the graph 12 +mK for 2≥m .
Step 2: Delete ( )12 +m number of consecutive greatest weights incident at different
vertices along the edges of a cycle 12 +mC of the complete graph 12 +mK for 2≥m .
Step 3: Observed that the graph obtained in Step 2 is a regular sub graph of the complete graph
12 +mK of degree 2 . Otherwise go to Step 6.
Step 4: Observed that the graph obtained in Step 3 is a cycle 12 +mC for 2≥m attached with
minimum weights and this is the required Hamiltonian circuit i.e., least cost route.
Step 5: Stop
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Step 6: From the graph obtained in Step 2, delete another ( )12 +m number of edges attached with
consecutive greatest weights other than ( )12 +m consecutive greatest weights already deleted in
Step 2.
Step 7: Go to Step 3 to Step 5. Otherwise go to Step 8.
Step 8: Repeat Step 6 till reaching the stage of Step 3 to Step 4 and go to Step 5.
Case 2: When nmq 3312 ⋅=+= for 1≥n and
13 +n number of consecutive greatest weights incident
at different vertices along edges of n3 edge disjoint 3K , next
23 +n number of consecutive greatest
weights attached with edges of n3 edge disjoint cycles 23
C , next 33 +n number of consecutive greatest
weights attached with edges of n3 edge disjoint cycles 33
C and so on the remaining 123 +n number of
consecutive greatest weights are attached with edges of n3 circuits 13 +nC .
Step 9: Delete 13 +n number of consecutive greatest weights incident at different vertices along edges
of n3 edge disjoint 3K .
Step 10: Observed that the graph obtained in Step 9 is a regular sub graph of the complete graph
nK33⋅
of degree n32 ⋅ for 1≥n .
Step 11: From the graph obtained in Step 11, delete another 13 +n number of edges forming the
circuit 13 +nC attached with consecutive greatest weights other than 13 +n consecutive greatest weights
already deleted in Step 9, successively ( )13 −n times.
Step 12: Observed that the graph obtained in Step 11 is a regular sub graph of the complete graph
nK33⋅
of degree 2 for 1≥n . Otherwise go to Step 15.
Step 13: Observed that graph obtained in Step 12 is a circuit 13 +nC attached with minimum weights,
which is the required least cost route.
Step 14: Go to Step 5.
Step 15: Observed that the graph obtained in Step 10 is a regular sub graph of the complete graph
nK33⋅
of degree ( )133 1 −+n for 1≥n .
Step 16: From the graph obtained in Step 15, delete another 23 +n number of consecutive greatest
weights other than the greatest weights already deleted in Step 11, incident at different vertices
forming n3 edge disjoint cycles 23
C for 1≥n .
Step 17: Go to Step 10 to Step 14. Otherwise go to Step 18.
Step 18: From the graph obtained in Step 17, delete another 33 +n number of consecutive greatest
weights other than the greatest weights already deleted in Step 16, incident at different vertices
forming n3 edge disjoint cycles 33
C for 1≥n .
Step 19: Go to Step 10 to Step 14. Otherwise go to Step 20.
Step 20: From the graph obtained in Step 18, delete another 43 +n number of consecutive greatest
weights other than the greatest weights already deleted in Step 18, incident at different vertices
forming n3 edge disjoint cycles 43
C for 1≥n and continue this process till the graph becomes
regular of degree n32 ⋅ for 1≥n .
Step 21: Go to Step 10 to Step 14.
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Case 3: When smr 312 =+= , 5≥s is a prime and s3 number of consecutive greatest weights
incident at different vertices forming s edge disjoint 3K , next ( )sms −3 number of consecutive
greatest weights incident at different vertices forming ( )sm −3 edge disjoint cycles sC and the
remaining ( )13 −ss number of consecutive greatest weights incident at different vertices forming
( )sm − edge disjoint cycles sC3 .
Step 22: Delete s3 number of consecutive greatest weights incident at different vertices forming s
edge disjoint 3K for 5≥s .
Step 23: Observed that the graph obtained in Step 22 is a regular sub graph of the complete graph
sK3 of degree ( )13 −s for 5≥s .
Step 24: From the graph obtained in Step 23, delete another ( )sms −3 number of edges forming
( )sm −3 edge disjoint circuits sC attached with greatest consecutive weights, other than s3 number
of consecutive greatest weights already deleted in Step 22.
Step 25: Observed that the graph obtained in Step 24 is a regular sub graph of the complete graph
sK3 of degree ( )12 −s having ( )13 −ss number of remaining edges for 5≥s .
Step 26: From the graph obtained in Step 25, delete another s3 number of edges forming the cycle
sC3 attached with consecutive greatest weights which are different from greatest weights already
deleted in Step 24.
Step 27: Observed that the graph obtained in Step 26 is a regular sub graph of the complete graph
sK3 of degree ( )22 −s for 5≥s .
Step 28: Repeat Step 26 on the graph obtained in Step 23 for ( )3−s times where 5≥s .
Step 29: Observed that the graph obtained in Step 28 is a cycle sC3 attached with minimum weights
and this is the required Hamiltonian circuit i.e. , least cost route.
Step 30: Go to Step 5
5. EXPERIMENTAL RESULTS FOR ALGORITHM
Case 1: The following cost matrix (Table-1) is considered for seven cities (i.e., for 3=m of 12 +mK
and 712 =+m , a prime)
A B C D E F G
A ∞ 20 3 35 51 18 34
B 20 ∞ 23 13 47 43 10
C 3 23 ∞ 27 6 41 40
D 35 13 27 ∞ 30 16 37
E 51 47 6 30 ∞ 31 7
F 18 43 41 16 31 ∞ 33
G 34 10 40 37 7 33 ∞
Table-1
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From the Table-1, we have a complete graph of seven vertices, which is shown in Figure-8.
Now, applying the Step 2 of the algorithm, we have a graph as shown in Figure-9.
Now, applying the Step 6 to Step 8 of the algorithm, we have a graph as shown in Figure-10 and this
is the least cost route as A→C→E→G→B→D→F→A with weights equal to 63.
Figure 8 Figure 9 Figure 10
Case 2: The following cost matrix (Table-2) is considered for nine cities (i.e., for 4=m of 12 +mK
and 133912 ⋅==+m )
A B C D E F G H I
A ∞ 31 2 81 59 80 86 29 57
B 31 ∞ 34 21 88 78 76 91 20
C 2 34 ∞ 39 7 93 73 70 97
D 81 21 39 ∞ 41 24 84 69 64
E 59 88 7 41 ∞ 47 13 90 61
F 80 78 93 24 47 ∞ 49 26 95
G 86 76 73 84 13 49 ∞ 52 18
H 29 91 70 69 90 26 52 ∞ 55
I 57 20 97 64 61 95 18 55 ∞
Table-2
From the Table-2, we have a complete graph of nine vertices, which is shown in Figure-11.
Now, applying the Step 9 of the algorithm, we delete 9 greatest weights attached with
edges, 81=AD , 84=DG , 86=AG , 88=BE , 90=EH , 91=BH , 93=CF , 95=FI , 97=CI
forming 3 edge disjoint 3K . Then we have a graph as shown in Figure-12.
Now, applying the Step10 to Step14 of the algorithm, we delete a cycle 9C whose edges are attached
with greatest weights 59=AE , 61=EI , 64=ID , 69=DH , 70=HC , 73=CG , 76=GB ,
78=BF , 80=FA , and then we delete another cycle 9C whose edges are attached with next greatest
weights 31=AB , 34=BC , 39=CD , 41=DE , 47=EF , 49=FG , 52=GH , 55=HI , 57=IA .
Then we have a graph as shown in Figure-13 and this is the least cost route of the traveler as
A→C→E→G→I→B→D→F→H→A with weights equal to 160.
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Figure 11 Figure 12 Figure 13
Case 3: The following cost matrix (Table-3) is considered for fifteen cities (i.e., for 7=m of 12 +mK
and 531512 ×==+m )
A B C D E F G H I J K L M N O
A ∞ 94 69 122 44 166 143 3 41 149 169 66 128 93 121
B 94 ∞ 96 81 129 50 170 150 39 37 156 175 48 135 80
C 69 96 ∞ 97 70 136 55 176 158 34 31 163 180 54 141
D 122 81 97 ∞ 99 83 123 61 181 147 29 26 146 184 60
E 44 129 70 99 ∞ 101 71 130 45 185 155 23 21 153 190
F 166 50 136 83 101 ∞ 104 86 138 51 167 162 18 17 161
G 143 170 55 123 71 104 ∞ 105 73 125 57 173 144 14 10
H 3 150 176 61 130 86 105 ∞ 109 87 132 63 178 152 6
I 41 39 158 181 45 138 73 109 ∞ 110 76 139 46 183 159
J 149 37 34 147 185 51 125 87 110 ∞ 112 89 127 53 187
K 169 156 31 29 155 167 57 132 76 112 ∞ 115 77 133 59
L 66 175 163 26 23 162 173 63 139 89 115 ∞ 116 90 140
M 128 48 180 146 21 18 144 178 46 127 77 116 ∞ 118 79
N 93 135 54 184 153 17 14 152 183 53 133 90 118 ∞ 119
O 121 80 141 60 190 161 10 6 159 187 59 140 79 119 ∞
Table: 3
From the Table-3, we have a complete graph of nine vertices, which is shown in Figure-14.
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Figure: 14
Now, applying the Step 22 of the algorithm, we delete 15 edges forming 5 edge disjoint 3K
attached with greatest weights AE=166, FK=167, AK=169, BG=170, GL=173, BL=175, CH=176,
HM=178, CM=180, DI=181, IN=183, DN=184, EJ=185, JO=187, EO=190. Then we have a regular
sub graph of degree 12 as shown in Figure-15.
Figure-15
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Now, applying the Step 24 of the algorithm, we delete 30 edges forming 6 edge disjoint cycles
5C attached with next greatest weights AG=143, GM=144, MD=146, DJ=147, AJ=149, BH=150,
HN=146, DJ=147, AJ=149, BH=150, HN=152, NE=153, EK=155, BK=156, CI=158, IO=159,
OF=161, FL=162, CL=163, AD=122, DG=123, GJ=125, JM=127, MA=128, BE=129, EH=130,
HK=132, KN=133, NB=135, CF=136, FI=138, IL=139, LO=140, OC=141. Then we have a regular
sub graph of 15K of degree 8 as shown in Figure-16.
Figure-16
Now, applying the Step 26 of the algorithm, we delete 15 edges forming the cycle 15C
attached with next greatest weights AB=94, BC=96, CD=97, DE=99, EF=101, FG=104, GH=105,
HI=109, IJ=110, JK=112, KL=115, LM=116, MN=118, NO=119, OA=121. Then we have a regular
sub graph of 15K of degree 6 as shown in Figure-17.
Figure-17
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Applying the Step 28 of the algorithm, we delete 30 edges forming 2 edge disjoint cycles 15C
attached with next greatest weights AC=69, CE=70, EG=71, GI=73, IK=76, KM=77, MO=79,
OB=80, BD=81, DF=83, FH=86, HJ=87, JL=89, LN=90, NA=93, AE=44, EI=45, IM=46, MB=48,
BF=50, FJ=51, JN=53, NC=54, CG=55, GK=57, KO=59, OD=60, DH=61, HL=63, LA=66. Then
we have a graph as shown in Figure-18 and this is the least cost route as
A→H→O→G→N→F→M→E→L→D→K→C→J→B→I→A with weights equal to 349.
Figure-18
6. CONCLUSION
Here we have discussed decomposition of complete graphs 12 +mK for 2≥m into circulant
graphs if 12 +m = a prime number, 1,3312 ≥⋅=+ nmn
and sm 312 =+ , 5≥s is a prime. However,
one can study the other situation like 1,5512 ≥⋅=+ nm n or sm 512 =+ , 7≥s and find the circulant
graph which may lead to study the Hamiltonian circuit related with the traveling salesman problems
in near future.
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