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- 1. International Journal of Computer Engineering COMPUTER ENGINEERING &
INTERNATIONAL JOURNAL OFand Technology (IJCET), ISSN 0976-6367(Print),
ISSN 0976 - 6375(Online), Volume 4, Issue 6, November - December (2013), © IAEME
TECHNOLOGY (IJCET)
ISSN 0976 – 6367(Print)
ISSN 0976 – 6375(Online)
Volume 4, Issue 6, November - December (2013), pp. 25-47
© IAEME: www.iaeme.com/ijcet.asp
Journal Impact Factor (2013): 6.1302 (Calculated by GISI)
www.jifactor.com
IJCET
©IAEME
DECOMPOSITION OF COMPLETE GRAPHS INTO CIRCULANT GRAPHS
AND ITS APPLICATION
1
Jayanta Kr. Choudhury,
2
Anupam Dutta,
3
Bichitra Kalita
1
Swadeshi Academy Junior College, Guwahati-781005
Patidarrang College, Muktapur, Loch, Kamrup, Assam
3
Department of Computer Application, Assam Engineering College, Guwahati-781013
2
ABSTRACT
In this paper, the decomposition of complete graphs K 2m+1 for m ≥ 2 into circulant graphs
has been discussed. Two theorems have been established for different values of m ≥ 2 relating to
2m + 1 is prime, 2m + 1 = 3 ⋅ 3n for n ≥ 1 and 2 m + 1 = 3s for s ≥ 5 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 K 2m+1 for m ≥ 2 .
Keywords: Algorithm, Circulant
Traveling Salesman Problem (TSP).
Graphs,
Complete
Graphs,
Hamiltonian
Graphs,
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 S n in which S has an element of order n , S ⊆ Z n {0} has
vertex set Z n and edge set {[x, x + s ] | x ∈ Z n , s ∈ S }. S. El-Zanati, Kyle King and Jeff Mudrock [3]
discussed the cyclic decomposition of circulant graphs into almost bipartite graphs. Dalibor Froncek
25
- 2. International Journal of Computer Engineering and Technology (IJCET), ISSN 0976-6367(Print),
ISSN 0976 - 6375(Online), Volume 4, Issue 6, November - December (2013), © IAEME
[4] used labeling to show the cyclic decomposition of complete graphs into K m ,n + e : 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
(
m
)
that the circulant graph G cd , d is Hamiltonian decomposable for all positive integers c, d and
m with c < d where the graph G( N , d ) has vertex set V = {0,1,2,.........., N − 1} , with {v, w} an
edge if v − w ≡ ± d (mod N ) for some 0 ≤ i ≤ log d N − 1 . 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 2 a 3b
graphs, each decomposed into 2 a 3b m 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 K 2 n can be decomposed into n − 2 n -suns, a
Hamiltonian cycle and a perfect matching, when n is even and for odd cases, the decomposition is
n − 1 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
n
n
n
graph K n , n can be decomposed into
n -suns when
is even. When
is odd, K n , n can be
2
2
2
decomposed into (n − 2 ) n -suns and a Hamiltonian circuit [13]. Tay-Woei Shyu [14] investigated
2
the decomposition of complete graph K n cycles Ct ’ s and stars S k ’ s and studied necessary or
sufficient conditions for such a decomposition to exist. Dalibor Froncek [15] show that every
bipartite graph H which decomposes K k and K n also decomposes K kn . Darryn E. Bryant and Peter
Adams [16] discussed that for all v ≥ 3 , the complete graph K v on v vertices can be decomposed
into v − 2 edge disjoint cycles whose lengths are 3,3,4,5,KK, v − 1 and for all v ≥ 7 , K v can be
decomposed into v − 3 edge disjoint cycles whose lengths are 3,4,KK, v − 4, v − 2, v − 1, v . Fu,
Hwang, Jimbo, Mutoh and Shiue [17] considered the problem of decomposing a complete graph into
the Cartesian product of two complete graphs K r and K c 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 K 2m+1 into m edge
disjoint circulant graphs C2 m+1 ( j ) for m ≥ 2 where j is the jump, 1 ≤ j ≤ m . Thereafter, we present
a theorem for the decomposition of K 2m+1 into edge disjoint circulant graphs whenever 2m + 1 is a
prime, 2m + 1 = 3 ⋅ 3n , n ≥ 1 and 2 m + 1 = 3s , s ≥ 5 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.
26
- 3. International Journal of Computer Engineering and Technology (IJCET), ISSN 0976-6367(Print),
ISSN 0976 - 6375(Online), Volume 4, Issue 6, November - December (2013), © IAEME
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 n × n matrix C of the form
c0
cn −1
C = M
c2
c
1
c1
L cn−2
c0
c1
c n−1
c0
O
c2
L c n−1
O
O
c n−1
cn−2
M
c1
c0
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 c n −i = ci for i = 1,2,......., n − 1 . Further, C
is an adjacency matrix if c0 = 0 and cn −1 = ci ∈ {0, 1}.
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 C n , the complete graph K n , and the complete
bipartite graph K n , n .
Circulant Graph 2.3 [27] : For a given positive integer, let n1 , n2 ,KK, nk be a sequence of integers
where
( p + 1) .
0 < n1 < n2 < KK < nk <
2
Then the circulant graph C p (n1 , n 2 , KK , n k ) is the graph on p nodes v1 , v 2 , KK , v p with vertex
vi adjacent to each vertex v i ± n j ( mod p ) . The values ni are called jump sizes.
Definition 2.4[28]: A graph G
graphs H 1 , H 2 , H 3 ,KK, H i , 1 ≤ i ≤ k ,
is
said
having
to
be
no
decomposable into the
isolated
vertices
sub
and
{E(H1 ), E(H 2 ), E (H 3 ),KK, E (H k )} is a partition of E (G ) .
3. THEOREMS
3.1 Theorem: The complete graph K 2m+1 for m ≥ 2 can be decomposed to m edge disjoint
circulant graph C 2 m+1 ( j ) where j is the jump and 1 ≤ j ≤ m .
27
- 4. International Journal of Computer Engineering and Technology (IJCET), ISSN 0976-6367(Print),
ISSN 0976 - 6375(Online), Volume 4, Issue 6, November - December (2013), © IAEME
Proof: Consider the complete graph K 5 = K 2×2+1 . Then vertex set of K 5 is V = {v1 , v2 , v3 , v4 , v5 } and
n(n − 1) 5 × 4
=
= 10 edges. Let us partitioned the edges of K 5 into 2 distinct sets containing
it has
2
2
5 edges each as E1 = {v1v2 , v 2 v3 , v3 v4 , v4 v5 , v5 v1 } and E2 = {v1v3 , v2 v4 , v3 v5 , v 4 v1 , v5 v2 } respectively.
Adjoining edges of E1 and E 2 , we can construct 2 edge disjoint circulant graphs C5 (1) and C5 (2)
for the jumps j = 1 and j = 2 respectively and their respective permutations can be represented as
f1 and f 2 where
v
f1 = 1
v
2
v2
v3
v3
v4
v4
v5
v5
v
and f 2 = 1
v
v1
3
v2
v4
v3
v5
v4
v1
v5
v2
i.e., f1 = (v1v2 v3 v4 v5 ) and f 2 = (v1v3 v5 v 2 v 4 ) . Figure-1 shows two circulant graphs C5 (1) and C5 (2)
for the complete graph K 5 .
C5 (1)
C5 (2)
Figure-1
Now, we append the edges to the circulant graphs as
vi + j if 1 ≤ i ≤ 4
f j (vi ) =
vi −4 if i = 5
vi + j if 1 ≤ i ≤ 3
and f j (vi ) =
vi −3 if 4 ≤ i ≤ 5
, j =1
, j=2
Similarly, consider the complete graph K 7 = K 2×3+1 whose vertex set is V = {v1 , v 2 , v3 , v4 , v5 , v6 , v7 }
n(n − 1) 7 × 6
=
= 21 edges. Let us partitioned the edges of K 7 into 3 distinct sets
and have
2
2
E1 = {v1v2 , v2 v3 , v3 v4 , v4 v5 , v5 v6 , v6 v7 , v7 v1 } ,
consisting
of
edges
each,
as
7
E2 = {v1v3 , v2 v4 , v3 v5 , v4 v6 , v5 v7 , v6 v1 , v7 v2 } and E3 = {v1v4 , v2 v5 , v3 v6 , v4 v7 , v5 v1 , v6 v2 , v7 v3 }. Adjoining
7 edges each of E1 , E 2 and E3 , we can construct 3 edge disjoint circulant graphs C 7 (1) , C7 (2) and
C7 (3) for the jumps j = 1 , j = 2 and j = 3 respectively and their respective permutations are f1 ,
f 2 and f 3 where
28
- 5. International Journal of Computer Engineering and Technology (IJCET), ISSN 0976-6367(Print),
ISSN 0976 - 6375(Online), Volume 4, Issue 6, November - December (2013), © IAEME
v v 2 v3 v 4 v5 v 6 v 7
v1
f1 = 1
v v v v v v v , f2 = v
3
4
5
6
7
1
2
3
v v 2 v3 v 4 v5 v6 v7
f3 = 1
v v v v v v v
5
6
7
1
2
3
4
v2
v4
v3
v5
v4
v6
f 2 = (v1v3 v5 v7 v2 v4 v6 ) and
circulant graphs C 7 (1) , C7 (2) and C7 (3) .
i.e.,
f1 = (v1v2 v3 v4 v5 v6 v7 ) ,
v1
v6
v1
v7
and
v2
f 3 = (v1v4 v7 v3v6 v2 v5 ) . Figure-2 shows
v1
v2
v3
v1
v2
v7
v6
v4
v5
v7
v5
C 7 (1)
v2
v7
v3
v6
v4
v5
C7 (2)
v7
v6
v3
v4
v5
C7 (3)
Figure-2
The edges to the circulant graphs can be appended as
vi + j if 1 ≤ i ≤ 6
f j (v i ) =
vi −6 if i = 7
vi+ j if 1 ≤ i ≤ 5
f j (vi ) =
vi−5 if 6 ≤ i ≤ 7
and
, j =1
, j=2
vi+ j if 1 ≤ i ≤ 4
f j (vi ) =
vi−4 if 5 ≤ i ≤ 7
, j =3
Similarly, partitioning the edges of K 9 into 4 distinct sets containing 9 edges each as
E1 = {v1v2 , v2 v3 , v3 v4 , v4 v5 , v5 v6 , v6 v7 , v7 v8 , v8 v9 , v9 v1 } ,
E2 = {v1v3 , v2 v4 , v3 v5 , v4 v6 , v5 v7 , v6 v8 , v7 v9 , v8 v1 , v9 v2 } ,
E3 = {v1v4 , v2 v5 , v3 v6 , v4 v7 , v5 v8 , v6 v9 , v7 v1 , v8 v2 , v9 v3 }
and E4 = {v1v5 , v 2 v6 , v3 v7 , v4 v8 , v5 v9 , v6 v1 , v7 v2 , v8 v3 , v9 v 4 }
respectively, we can construct 4 edge disjoint circulant graphs C9 (1) , C9 (2) , C9 (3) and C9 (4) for
the jumps j = 1 , j = 2 , j = 3 and j = 4 respectively and their corresponding permutations are
given by
29
- 6. International Journal of Computer Engineering and Technology (IJCET), ISSN 0976-6367(Print),
ISSN 0976 - 6375(Online), Volume 4, Issue 6, November - December (2013), © IAEME
v
f1 = 1
v
2
v3
v4
v4
v5
v5
v6
v6
v7
v7
v8
v8
v9
v9
= (v1v 2 v3 v 4 v5 v6 v 7 v8 v9 ) ,
v1
v
f2 = 1
v
3
v2
v4
v3
v5
v4
v6
v5
v7
v6
v8
v7
v9
v8
v1
v9
= (v1v3 v5 v 7 v9 v 2 v 4 v6 v8 ) ,
v2
v
f3 = 1
v
4
v2
v3
v4
v5
v6
v7
v8
v5
v6
v7
v8
v9
v1
v2
v9
= (v1v 4 v 7 )(v 2 v5 v8 )(v3 v6 v9 )
v3
v
f4 = 1
v
5
and
v2
v3
v2
v3
v4
v5
v6
v7
v8
v6
v7
v8
v9
v1
v2
v3
v9
= (v1v5 v9 v 4 v8 v3 v7 v 2 v6 )
v4
respectively, whose structures are shown in Figure-3.
V1
V9
V3
V4
C9 (1)
V7
V7
V5
V6
V8
V8
V4
V7
V5
V6
C9 (2)
Append the edges to the circulant graphs as
vi + j if 1 ≤ i ≤ 8
f j (vi ) =
vi −8 if i = 9
, j =1
vi+ j if 1 ≤ i ≤ 7
f j (vi ) =
vi−7 if 8 ≤ i ≤ 9
, j=2
vi+ j if 1 ≤ i ≤ 6
f j (vi ) =
vi−6 if 7 ≤ i ≤ 9
, j =3
vi+ j if 1 ≤ i ≤ 5
vi−5 if 6 ≤ i ≤ 9
V6
C9 (3)
Figure-3
and f j (vi ) =
V2
V9
V9
V3
V8
V4
V2
V2
V3
V5
V1
V1
V1
V2
, j=4
30
V9
V3
V8
V4
V7
V5
V6
C9 (4)
- 7. International Journal of Computer Engineering and Technology (IJCET), ISSN 0976-6367(Print),
ISSN 0976 - 6375(Online), Volume 4, Issue 6, November - December (2013), © IAEME
From the above discussion, it is seen that the complete graph K 2m+1 for m ≥ 2 can be
decomposed into m edge disjoint circulant graph C 2 m+1 ( j ) . Here j represents the jump of the
circulant also depends upon m . If m = 2 , then j = 1, 2 and we have 2 circulant graphs C5 (1) and
C5 (2) .
Again, if m = 3 , then j = 1, 2, 3 and then K 7 decomposes into 3 edge disjoint circulant
graphs C 7 (1) , C7 (2) and C7 (3) and so on. Moreover, in their respective permutations, the images of
first element and last element in the first row are respectively v j +1 and v j . Here we excluded the
identity permutations as well as permutations which produce the same circulant graphs,
v1 v 2 v3 v 4 v5
v1 v 2 v3 v 4 v5
i.e.
v v v v v and v v v v v represents the same circulant graph C5 (1) .
3
4
5
1
1
2
3
4
2
5
Therefore, as discussed above, we get 2 permutations if m = 2 , 3 permutations if m = 3 etc, i.e.,
number of permutations are depends upon the value of m ≥ 2 .
Finally, we can comment that the complete graph K 2m+1 for m ≥ 2 can be decomposed into m edge
disjoint circulant graph C 2 m+1 ( j ) for the jump 1 ≤ j ≤ m and their corresponding m permutations
are defined as
vi+ j if 1 ≤ i ≤ 2m − ( j −1)
f j (vi ) =
vi+( j−1)−2m if 2m − ( j − 2) ≤ i ≤ 2m + 1
for 1 ≤ j ≤ m and which complete the theorem.
Note: Here, complete graphs K 2m+1 for m ≥ 2 has odd number of vertices and therefore they play
different role for different values of m ≥ 2 . In the next theorem we discuss only three cases.
3.2 Theorem: Let K 2m+1 be a complete graph for m ≥ 2 . Then following conditions are satisfied
(a) If p = 2m + 1 is a prime number, then K p can be decomposed into m edge disjoint circulant
graphs C p ( j ) where j is the jump and 1 ≤ j ≤ m .
1 n +1
3 − 1 for n ≥ 1 , then K q can be decomposed into
2
1
(i) 3n edge disjoint circulant graphs C q ( j ) for the jump 1 ≤ j ≤ m but j ≠ 3l for 1 ≤ l ≤ 3 n − 1
2
and
1
(ii) a circulant graph C q (3,6,9, KK ,3l ) for the jump 3l where 1 ≤ l ≤ 3 n − 1 .
2
(c) If r = 2m + 1 = 3s where s ≥ 5 is a prime number, then K r can be decomposed into
(i) s − 1 edge disjoint circulant graphs C r ( j ) for the jump 1 ≤ j ≤ m where j ≠ s and
1
j ≠ 3e , 1 ≤ e ≤ (m − 1)
3
1
(ii) a circulant graph C r (3,6,9,KK,3e) for the jump 1 ≤ e ≤ (m − 1) and
3
(iii) a circulant graph C r (s ) for the jump s .
(b)
If q = 2m + 1 = 3 ⋅ 3n where m =
(
)
(
(
31
)
)
- 8. International Journal of Computer Engineering and Technology (IJCET), ISSN 0976-6367(Print),
ISSN 0976 - 6375(Online), Volume 4, Issue 6, November - December (2013), © IAEME
Proof: (a) When p = 2m + 1 = prime, then it follows that the complete graph K 5 , K 7 , K11 ,
K13 ,……….. for m = 2,3,5,6,KKK exist and the proof is same as discussed in Theorem 3.1. Hence
the complete graph K p where p = 2m + 1 is a prime for m ≥ 2 can be decomposed into
1
m = ( p − 1) edge disjoint circulant graphs C p ( j ) for the jump 1 ≤ j ≤ m .
2
In the above discussion, we observed that each edge disjoint circulant graph C p ( j ) for the
1
( p − 1) is regular of degree two. In other words, they are cycle C p of length p and
2
therefore they represent edge disjoint Hamiltonian circuits. So in this discussion, number of edge
disjoint Hamiltonian circuits in the complete graph K p is m if p = 2m + 1 is a prime for m ≥ 2 .
jump 1 ≤ j ≤
Proof: (b) Let K 2m+1 be a complete graph for m ≥ 2 . Suppose that q = 2m + 1 = 3 ⋅ 3n such that
1
v
m = (3n +1 − 1) for n ≥ 1 . Then V = { , v , v ,......... .., v } is the vertex set of K and it has q(q − 1)
1
2
2
3
q
q
2
edges.
Let us define f j :V (K q )→ V (K q ) as follows:
(i) when 1 ≤ j ≤ 1 (3 n + 1 − 1 ) but j ≠ 3l for 1 ≤ l ≤ 1 (3n − 1)
2
2
∀n ≥1
vi + j if 1 ≤ i ≤ q − j
f j (v i ) =
vi + j − q if q − j + 1 ≤ i ≤ q
vi +3l if 1 ≤ i ≤ q − 3l
f 3l (vi ) =
vi +3l − q if q − 3l + 1 ≤ i ≤ q
and (ii) when 1 ≤ l ≤ 1 (3 n − 1) ∀n ≥ 1 ,
2
For n = 1 , m = 4 and q = 2 × 4 + 1 = 9 = 3 ⋅ 31 . Then {v1 , v2 , v3 .v4 , v5 , v6 , v7 , v8 , v9 } is the vertex set of
K 9 have q(q − 1) = 9 × 8 = 4 × 9 = 36 edges. Since m = 4 , we can partition the edges of K 9 into 4 edge
2
2
disjoint sets, each set containing 9 edges, by the above rule and their permutations are given as
follows.
v
For j = 1 , f1 = 1
v
2
v
For j = 2 , f 2 = 1
v
3
v
v2
v3
v4
v5
v6
v7
v3
v4
v5
v6
v7
v8
v9
= (v1v 2 v3 v4 v5 v6 v7 v8 v9 )
v1
v8
v9
v2
v3
v4
v5
v6
v7
v4
v5
v6
v7
v8
v9
v1
v
v
v4
v5
v6
v7
v8
v8
v9
v1
v2
v3
v9
= (v1v3 v5 v7 v9 v 2 v 4 v6 v8 )
v2
v8
For j = 4 , f 4 = 1 2 3
v v v
6
7
5
v9
= (v1v5 v9 v 4 v8 v3 v7 v2 v6 )
v4
Again, for l = 1 , f 3 = v1 v2 v3 v4 v5 v6 v7 v8 v9 = (v1v4 v7 )(v2 v5 v8 )(v3 v6 v9 )
v4
v5
v6
v7
v8
v9
v1
v2
32
v3
- 9. International Journal of Computer Engineering and Technology (IJCET), ISSN 0976-6367(Print),
ISSN 0976 - 6375(Online), Volume 4, Issue 6, November - December (2013), © IAEME
From these permutations, we get 4 edge disjoint sets
E1 = {v1v2 , v2 v3 , v3 v4 , v4 v5 , v5 v6 , v6 v7 , v7 v8 , v8 v9 , v9 v1 } ,
E2 = {v1v3 , v2 v4 , v3 v5 , v4 v6 , v5 v7 , v6 v8 , v7 v9 , v8 v1 , v9 v2 } ,
E3 = {v1v4 , v2 v5 , v3 v6 , v4 v7 , v5 v8 , v6 v9 , v7 v1 , v8 v2 , v9 v3 } and
E4 = {v1v5 , v2 v6 , v3 v7 , v4 v8 , v5 v9 , v6 v1 , v7 v2 , v8 v3 , v9 v4 } .
Adjoining edges of E1 , E 2 and E 4 , we get 31 = 3 edge disjoint circulant graphs C9 (1) ,
C9 (2) and C9 (4) for the jumps j = 1 , j = 2 and j = 4 respectively [Figure-3]. Also, adjoining the
edges of E3 , we get a circulant graph C9 (3) for the jump 3l where l = 1 . Thus K 9 decomposed
into 31 edge disjoint circulant graphs C9 ( j ) for 1 ≤ j ≤ 4 , j ≠ 3 and a circulant graph C9 (3l ) for
l = 1 [Figure-3].
For n = 2 , m = 13 and q = 2 × 13 + 1 = 27 = 3 ⋅ 32 . Then {v1 , v2 , v3 ,KKKv 27 } is the vertex set of the
complete graph K 27 and have q (q − 1) = 27 × 26 = 13 × 27 = 351 edges. Since m = 13 , we can
2
2
partition the edges of K 27 into 13 edge disjoint sets, each set containing 27 edges, by the rule
defined above.
For j = 1 , f1 = (v1v2v3v4v5v6v7v8v9v10v11v12v13v14v15v16v17v18v19v20v21v22v23v24v25v26v27 )
For j = 2 , f 2 = (v1v3v5v7v9v11v13v15v17v19v21v23v25v27v2v4v6v8v10v12v14v16v18v20v22v24v26 )
For j = 4 , f4 = (v1v5v9v13v17v21v25v2v6v10v14v18v22v26v3v7v11v15v19v23v27v4v8v12v16v20v24 )
For j = 5 , f 5 = (v1v6 v11v16 v 21v 26 v 4 v9 v14 v19 v 24 v 2 v7 v12 v17 v 22 v 27 v5 v10 v15 v 20 v 25 v3 v8 v13 v18 v 23 )
For j = 7 , f 7 = (v1v8 v15 v22 v2 v9 v16 v 23v3 v10 v17 v24 v4 v11v18v 25v5 v12 v19 v26 v6 v13v20 v27 v7 v14 v21 )
For j = 8 , f 8 = (v1v9 v17 v25v6 v14 v22 v3 v11v19 v27 v8 v16v 24v5 v13v21v2 v10 v18v26 v7 v15v23v4 v12 v20 )
For j = 10 , f10 = (v1v11v21v4 v14v24v7 v17v27v10v20v3v13v23v6 v16v26v9 v19v2 v12v22v5 v15v25v8 v18 )
For j = 11 , f11 = (v1v12v23v7 v18v2 v13v24v8 v19v3v14v25v9 v20v4 v15v26v10v21v5 v16v27v11v22v6 v17 )
For j = 13 , f13 = (v1v14 v 27 v13v26 v12 v25 v11v 24 v10 v 23v9 v22 v8 v 21v7 v20 v6 v19 v5 v18 v4 v17 v3 v16 v 2 v15 )
Again,
For l = 1 , f 3 = (v1v 4 v 7 v10 v13 v16 v19 v 22 v 25 )(v 2 v5 v8 v11v14 v17 v 20 v 23 v 26 )(v3 v 6 v9 v12 v15 v18 v 21v 24 v 27 )
For l = 2 , f 6 = (v1v7 v13v19v25v4 v10v16v22 )(v2 v8 v14 v20v26v5 v11v17 v23 )(v3v9 v15v21v27 v6 v12v18v24 )
For l = 3 , f 9 = (v1v10v19 )(v2v11v20 )(v3v12v21 )(v4v13v22 )(v5v14v23 )(v6v15v24 )(v7v16v25 )(v8v17v26 )(v9v18v27 )
For l = 4 , f12 = (v1v13v25v10v22v7 v19v4v16 )(v2v14v26v11v23v8v20v5v17 )(v3v15v27v12v24v9v21v6v18 )
33
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From these permutations we can define corresponding edge disjoint sets E1 , E 2 , E 3 , E 4 , E5 ,
E 6 , E 7 , E8 , E 9 , E10 , E11 , E12 and E13 respectively. Adjoining edges of
E1 , E 2 , E 4 , E 5 , E 7 , E8 , E10 , E11 and E13 , we can construct 32 = 9 edge disjoint circulant graphs
C27 (1) , C27 (2) , C27 (4) , C27 (5) , C27 (7 ) , C27 (8) . C27 (10) , C27 (11) and C27 (13) for the
jumps j = 1 , j = 2 , j = 4 , j = 5 , j = 7 , j = 8 , j = 10 , j = 11 and j = 13 respectively. Also,
adjoining edges of E 4 , E 6 , E 9 and E12 , we get a circulant graph C27 (3,6,9,12) for the jump 3l where
1 ≤ l ≤ 4 . Thus the complete graph K 27 can be decomposed into 32 = 9 edge disjoint circulant graph
C27 ( j ) where j is the jump such that 1 ≤ j ≤ 13 , j ≠ 3,6,9,12 and a circulant graph C27 (3,6,9,12) .
Let n = k so that m = 1 (3 k +1 − 1) and q = 2m + 1 = 3 ⋅ 3k = 3k +1 . Then {v1 , v 2 , v 3 , KKK , v q } is
2
k +1 k +1
the vertex set the complete graph K q = K 3k +1 and it has q(q − 1) = 3 (3 − 1) edges.
2
2
1 k +1
Since m = 3 − 1 , so we can partition edges of K q = K 3k +1 into 1 (3 k +1 − 1) distinct sets,
2
2
(
)
each set containing 3 k +1 edges, given by
(i) when 1 ≤ j ≤ 1 (3 k + 1 − 1 ) but j ≠ 3l for 1 ≤ l ≤ 1 (3 k − 1)
2
2
vi + j if 1 ≤ i ≤ q − j
f j (vi ) =
vi + j − q if q − j + 1 ≤ i ≤ q
and (ii) when 1 ≤ l ≤ 1 (3k − 1)
2
v i + 3l if 1 ≤ i ≤ q − 3l
f 3l (v i ) =
v i + 3l − q if q − 3l + 1 ≤ i ≤ q
Now, we can have edge disjoint sets E1 , E 2 , E3 , E 4 ,……………, E 1
2
f1 , f 2 , f 3 , f 4 ,……………., f 1
corresponding permutations are
2
E1 , E 2 , E 4 , E 5 , E 7 ,……, E 1
(3
2
k +1
−5
)
, E1
(3
2
k +1
−1
)
(3
k +1
−1
)
(3
k +1
−1
)
and their
. Adjoining edges of
, we can construct 3 k edge disjoint circulant graphs
1
C 3k +1 ( j ) where j is the jump and 1 ≤ j ≤ 1 (3 k + 1 − 1) , j ≠ 3l for 1 ≤ l ≤ (3 k − 1) . Also, adjoining
2
2
edges of E 3 , E 6 , E 9 ,……, E 1
2
jump 1 ≤ l ≤
(3
k +1
−7
)
, E1
2
(3
k +1
−3
)
, we get a circulant graph C 3k +1 (3,6,9, KK ,3l ) for the
1 k
3 −1 .
2
(
)
(
)
Now, adding 2 ⋅ 3k +1 additional vertices to the vertex set and 4 ⋅ 9 k +1 − 3k +1 additional edges to the edge
set of the complete graph K 3k +1 . Then {v1 , v 2 , v3 ,K, v3k +1 , v3k + 2 +1 ,K, v3k + 2 −1 , v3k + 2 } will be the vertex set for
K 3k + 2 = K 3⋅3k +1 i.e., K q = K 3k + 2 and
the complete graph
q = 3 k + 2 ⇒ 2m + 1 = 3 k + 2
1
1
⇒ m = 3k + 2 − 1 = 3(k +1)+1 − 1
2
2
(
)
{
}. The complete graph
34
K 3k + 2 has
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(
)
{
}
q(q − 1) 3k +2 3k +2 − 1 3(k +1)+1 3(k +1)+1 − 1
edges. These edges can be partition into
=
=
2
2
2
1 k +2
(3 − 1) = 1 {3(k +1)+1 − 1} disjoint sets and their respective permutations are given by
2
2
(i) when 1 ≤ j ≤ 1 (3 k +2 − 1) but j ≠ 3l for 1 ≤ l ≤ 1 (3 k +1 − 1)
2
2
vi + j if 1 ≤ i ≤ q − j
f j (vi ) =
vi + j − q if q − j + 1 ≤ i ≤ q
v if 1 ≤ i ≤ q − 3l
1 k +1
and (ii) when 1 ≤ l ≤ 3 − 1 , f 3l (vi ) = i +3l
2
vi +3l − q if q − 3l + 1 ≤ i ≤ q
(
Now, we can have
)
1 (k +1)+1
3
− 1 edge disjoint sets E1 , E 2 , E 3 , E 4 , …, E 1 k + 2 and their respective
(3 −1)
2
2
{
}
permutations are f1 , f 2 , f 3 , f 4 ,………, f 1
2
E 7 ,…….., E 1
(3
2
k +2
−7
)
, E1
(3
2
k +2
−5
)
, E1
(3
2
k+2
−1
)
(3
k +2
−1
)
. Adjoining edges of E1 , E 2 , E 4 , E5 ,
, we can construct 3 k +1 edge disjoint circulant graph C 3k + 2 ( j ) i.e.,
1 k +1
3 − 1 . Also, adjoining
2
, we have a circulant graph C 3( k +1 )+1 (3,6,9, KK ,3l ) for the
C 3( k +1 )+1 ( j ) where j is the jump and 1 ≤ j ≤ 1 (3 (k +1)+1 − 1) ,
2
edges of E3 , E 6 , E 9 ,….., E 1
2
(3
k +2
−9
)
, E1
2
(3
k +2
−3
)
(
j ≠ 3l for 1 ≤ l ≤
)
jump 1 ≤ l ≤ 1 (3 k + 1 − 1 ) . Thus the result is true for n = k + 1 whenever it is true for n = k . Hence by
2
principle of mathematical induction, the result is true for all n ∈ N .
In this discussion, we observed that each edge disjoint circulant graph C q ( j ) for the jump
1≤ j ≤
1 n +1
3
−1
2
(
),
j ≠ 3l for
1≤ l ≤
1 n
3 −1
2
(
) are regular of degree two. In other words they are cycle
Cq
of length q and therefore they represent 3n edge disjoint Hamiltonian circuits n ≥ 1 . Further the circulant
graph
C q (3 , 6 , 9 , K K , 3 l )
1 n
3 −1
2
(
)
for the jump 1 ≤ l ≤ 1 (3 n − 1) , as we have observed, can be decomposed into
2
edge disjoint circulant graphs C q (3t ) for the jump 3t , 1 ≤ t ≤
1 n
3 − 1 for all n ≥ 1 . If the jump
2
(
)
is
3t = 3 n , then 3n K 3 -decomposition
3t = 3n−1 , then 3n C9 -decomposition
3t = 3n−2 , then 3n C27 -decomposition
………………………………………..
………………………………………..
3t = 33 , then 3n C 3 -decomposition
n−2
3t = 32 , then 3n C 3 -decomposition
n −1
1
n
3t = 3 , then 3 C 3 -decomposition
Clearly above cycles does not represent Hamiltonian circuits. So in this discussion, number of edge
disjoint Hamiltonian circuits in the complete graph K q is n if q = 3⋅ 3n for n ≥ 1 .
n
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(c) Let K 2 m+1 be a complete graph for all m ≥ 2 . Suppose that r = 2m + 1 = 3s where s ≥ 5 is a
prime number and m = 3 s − 1 . Then V = {v1 , v2 , v3 ,..........., vr } is the vertex set of the complete graph
2
K r i.e., K 3s and have
r (r − 1 )
2
edges.
Let us define f j : V (K s ) → V (K s ) as follows:
(i) when 1 ≤ j ≤
1
(3s − 1) but j ≠ s and j ≠ 3e for 1 ≤ e ≤ 1 (s − 1)
2
2
v i + j if 1 ≤ i ≤ r − j
f j (v i ) =
v i + j − r if r − j + 1 ≤ i ≤ r
(ii) when 1 ≤ e ≤
1
(s − 1) , f 3 e (v i ) =
2
v i + 3 e if 1 ≤ i ≤ r − 3 e
v i + 3 e − r if r − 3 e + 1 ≤ i ≤ r
vi + s if 1 ≤ i ≤ r − s
and (iii) f s (v i ) =
vi + s − r if r − s + 1 ≤ i ≤ r
For s = 5 , a prime number, we have r = 3 × 5 = 15 and m = 3 × 5 − 1 = 7 .
2
Then {v1 , v2 , v3 ,..........., v14 , v15 } is the vertex set of the complete graph K15 and have
r (r − 1) 15 × 14
=
= 7 × 15 = 105 edges. Since m = 7 , we can partition the edges of K15 into 7
2
2
disjoint sets, each set containing 15 edges and their corresponding permutations by the above rule
are as follows:
For j = 1 , f1 = (v1v2 v3 v4 v5 v6 v7 v8 v9 v10 v11v12 v13v14 v15 ) and
E1 = {v1v2 , v2 v3 , v3v4 , v4 v5 , v5 v6 , v6 v7 , v7 v8 , v8 v9 , v9 v10 , v10v11 , v11v12 , v12 v13 , v13v14 , v14v15 , v15v1 }
For j = 2 , f 2 = (v1v3 v5 v7 v9 v11v13v15v2 v4 v6 v8 v10 v12 v14 ) and
E2 = {v1v3 , v2v4 , v3v5 , v4v6 , v5v7 , v6v8 , v7v9 , v8v10 , v9v11, v10v12 , v11v13 , v12v14 , v13v15 , v14v1 , v15v2 }
For j = 4 , f 4 = (v1v5 v9 v13v2 v6 v10 v14 v3 v7 v11v15v 4 v8 v12 ) and
E4 = {v1v5 , v2 v6 , v3v7 , v4 v8 , v5 v9 , v6 v10 , v7 v11 , v8 v12 , v9 v13 , v10v14 , v11v15 , v12v1 , v13v2 , v14 v3 , v15v4 }
For j = 7 , f 7 = (v1v8 v15 v7 v14 v6 v13v5 v12 v 4 v11v3 v10 v2 v9 ) and
E7 = {v1v8 , v2 v9 , v3 v10 , v4 v11 , v5 v12 , v6 v13 , v7 v14 , v8 v15 , v9 v1 , v10 v2 , v11v3 , v12 v4 , v13v5 , v14 v6 , v15v7 }
For e = 1 , f 3 = (v1v4 v7 v10 v13 )(v2 v5 v8 v11v14 )(v3 v6 v9 v12 v15 ) and
E3 = {v1v4 , v2 v5 , v3 v6 , v4 v7 , v5 v8 , v6 v9 , v7 v10 , v8 v11 , v9 v12 , v10 v13 , v11v14 , v12v15 , v13v1 , v14 v2 , v15v3 }
For e = 2 , f 6 = (v1v7 v13v4 v10 )(v2 v8 v14 v5 v11 )(v3 v9 v15 v6 v12 ) and
E6 = {v1v7 , v2 v8 , v3v9 , v4 v10 , v5 v11 , v6 v12 , v7 v13 , v8 v14 , v9 v15 , v10 v1 , v11v2 , v12 v3 , v13v4 , v14 v5 , v15v6 }
For s = 5 , f 5 = (v1v6 v11 )(v2 v7 v12 )(v3 v8 v13 )(v 4 v9 v14 )(v5 v10 v15 ) and
E5 = {v1v6 , v2 v7 , v3v8 , v4 v9 , v5 v10 , v6 v11 , v7 v12 , v8 v13 , v9 v14 , v10 v15 , v11v1 , v12 v2 , v13v3 , v14 v4 , v15v5 }
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Adjoining edges of E1 , E 2 , E 4 and E 7 , we can construct s − 1 = 5 − 1 = 4 edge disjoint
circulant graph C15 (1) , C15 (2) , C15 (4) and C15 (7 ) respectively for jumps j = 1 , j = 2 , j = 4
and j = 7 [Figure-5]. Also, adjoining edges of E3 and E 6 , we can construct a circulant graph
C15 (3,6) [Figure-6]. Again, adjoining the edges of E 5 , we can construct a circulant graph C15 (5)
for the jump s = 5 [Figure-7].
V1
V1
V2
V2
V15
V3
V4
V13
V12
V5
V1
V15
V2
V3
V14
V14
V4
V13
V4
V14
V13
V13
V4
V12
V5
V12
V5
V15
V3
V14
V12
V5
V1
V2
V15
V3
V6
V6
V6
V11
V11
V6
V11
V11
V7
V10
V7
V9
V10
V7
V8
C15 (1)
V10
V7
V9
V8
V10
V8
V9
V
C15 (2)
C15 (4)
C15 (7 )
Figure-5
V1
V2
V2
V15
V1
V3
V14
V4
V13
V13
V12
V5
V 11
V6
V1 0
V7
V8
V14
V4
V12
V5
V6
V15
V3
V11
V10
V7
V9
V8
C15 (3,6)
Figure-6
V9
C15 (5)
Figure-7
Thus the complete graph K15 can be decomposed into s − 1 = 5 − 1 = 4 edge disjoint circulant
graphs C15 ( j ) for 1 ≤ j ≤ 7 ,
j ≠ 3,5,6 and a circulant graph C15 (3,6) and a circulant graph
C15 (5) for s = 5 . Similarly, we can show for s = 7,11,13,KK etc.
Let s = k , a prime number. Then r = 3k , m = 3 k − 1
2
r (r − 1 ) 3 k (3 k − 1 )
=
2
2
and the complete graph K 3k
edges. Since m = 3 k − 1 , so we can partition the edges of K 3k into 1 (3 k − 1 )
2
2
edge disjoint sets, each set containing 3k edges and their respective permutations are given by
(i) when 1 ≤ j ≤
f
(ii) when
j
has
1
(3k − 1) but j ≠ k and j ≠ 3e for 1 ≤ e ≤ 1 (k − 1)
2
2
(v i ) =
1≤ e ≤
v i + j if 1 ≤ i ≤ r − j
v i + j − r if r − j + 1 ≤ i ≤ r
1
(k − 1 ) ,
2
v i + 3 e if 1 ≤ i ≤ r − 3 e
f 3 e (v i ) =
v i + 3 e − r if r − 3 e + 1 ≤ i ≤ r
and (iii) f k (v i ) = v i + k if 1 ≤ i ≤ r − k
v i + k − r if r − k + 1 ≤ i ≤ r
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Now, we can have edge disjoint sets E1 , E 2 , E3 , E 4 ,……, E 1
2
are f1 , f 2 , f 3 , f 4 ,………, f 1
2
( 3 k −1 )
and their respective permutations
(3 k − 1 )
. Adjoining edges of E j ’s where
1≤ j ≤
1
(3 k − 1 ) ;
2
j ≠ k and
j ≠ 3e for 1 ≤ e ≤ 1 (k − 1 ) , we can construct s − 1 = k − 1 edge disjoint circulant graphs C3k ( j )
2
where j is the jump. Again, adjoining the edges of E 3e ’s,
1≤ e ≤
1
(k − 1 )
2
we can construct a
circulant graph C3k (3,6,9,KK,3e ) for the jump 3e . Also, adjoining the edges of E k , we can
construct a circulant graph C3k (k ) for the jump k .
In this discussion we observed that each edge disjoint circulant graphs C r ( j ) for the jump
1≤ j ≤
1
(3 s − 1 ) , j ≠ s and
2
j ≠ 3e for 1 ≤ e ≤ 1 (s − 1) where s ≥ 5 is a prime, are regular of degree
2
two. In other words they are cycles C r of length r and therefore they represent edge disjoint
Hamiltonian circuits. Again the circulant graph C r (3,6,9, KK,3e ) for the jump 3e where
1
1 ≤ e ≤ (s − 1 ) can be decomposed into (m − s ) edge disjoint circulant graphs 3t where
2
1
1 ≤ t ≤ (s − 1 )
2
and these circulants are regular of degree two each and they can be further
decomposed into
(m − 1)
cycles C s of length s < r and so they do not represent Hamiltonian
circuits. Also, the circulant graph C r (s ) for the jump s is regular of degree two and from which we
can have s K 3 -decompositions and they do not represent a Hamiltonian circuit. So in this
discussion, number of edge disjoint Hamiltonian circuits in the complete graph K r is s − 1 if
r = 3s , s ≥ 5 is a prime.
4. ALGORITHM
Input: Let K 2 m+1 for m ≥ 2 be a complete graph.
Output: Find least cost route.
Case 1: When p = 2m + 1 is a prime number and (2m + 1) consecutive greatest weights are incident
at different vertices of the complete graph K 2 m+1 for m ≥ 2 along edges of a cycle C 2 m+1 , next
(2m + 1)
greatest weights incident at different vertices along edges of a cycle C 2 m+1 other than the
previous one, next (2m + 1) greatest weights incident at different vertices along edges of a cycle
C 2 m+1 other than the previous two cycles and so on.
Step 1: Study the graph K 2m+1 for m ≥ 2 .
Step 2: Delete (2m + 1) number of consecutive greatest weights incident at different
vertices along the edges of a cycle C 2 m+1 of the complete graph K 2m+1 for m ≥ 2 .
Step 3: Observed that the graph obtained in Step 2 is a regular sub graph of the
complete graph
K 2m+1 of degree 2 . Otherwise go to Step 6.
Step 4: Observed that the graph obtained in Step 3 is a cycle C 2 m+1 for m ≥ 2 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 (2m + 1) number of edges attached with
consecutive greatest weights other than (2m + 1) 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 q = 2m + 1 = 3 ⋅ 3n for
n ≥ 1 and 3 n +1 number of consecutive greatest weights
incident at different vertices along edges of 3n edge disjoint K 3 , next 3 n + 2 number of consecutive
greatest weights attached with edges of 3n edge disjoint cycles C 32 , next 3 n +3 number of
consecutive greatest weights attached with edges of 3n edge disjoint cycles C 33 and so on the
remaining 32 n +1 number of consecutive greatest weights are attached with edges of 3n circuits C 3n +1 .
Step 9: Delete 3n +1 number of consecutive greatest weights incident at different vertices along edges
of 3n edge disjoint K 3 .
Step 10: Observed that the graph obtained in Step 9 is a regular sub graph of the complete graph
K 3⋅3n of degree 2 ⋅ 3n for n ≥ 1 .
Step 11: From the graph obtained in Step 11, delete another 3 n +1 number of edges forming the
circuit C 3n +1 attached with consecutive greatest weights other than 3n +1 consecutive greatest weights
(
)
already deleted in Step 9, successively 3n − 1 times.
Step 12: Observed that the graph obtained in Step 11 is a regular sub graph of the complete graph
K 3⋅3n of degree 2 for n ≥ 1 . Otherwise go to Step 15.
Step 13: Observed that graph obtained in Step 12 is a circuit C 3n +1 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
K 3⋅3n of degree 3 3n+1 − 1 for n ≥ 1 .
(
)
Step 16: From the graph obtained in Step 15, delete another 3 n + 2 number of consecutive greatest
weights other than the greatest weights already deleted in Step 11, incident at different vertices
forming 3n edge disjoint cycles C 32 for n ≥ 1 .
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 3 n +3 number of consecutive greatest
weights other than the greatest weights already deleted in Step 16, incident at different vertices
forming 3n edge disjoint cycles C 33 for n ≥ 1 .
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 3 n + 4 number of consecutive greatest
weights other than the greatest weights already deleted in Step 18, incident at different vertices
forming 3n edge disjoint cycles C 34 for n ≥ 1 and continue this process till the graph becomes
regular of degree 2 ⋅ 3n for n ≥ 1 .
Step 21: Go to Step 10 to Step 14.
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Case 3: When r = 2m + 1 = 3s , s ≥ 5 is a prime and 3s number of consecutive greatest weights
incident at different vertices forming s edge disjoint K 3 , next 3s(m − s ) number of consecutive
greatest weights incident at different vertices forming 3(m − s ) edge disjoint cycles C s and the
remaining 3s (s − 1) number of consecutive greatest weights incident at different vertices forming
(m − s ) edge disjoint cycles C3s .
Step 22: Delete 3s number of consecutive greatest weights incident at different vertices forming s
edge disjoint K 3 for s ≥ 5 .
Step 23: Observed that the graph obtained in Step 22 is a regular sub graph of the
complete
graph K 3s of degree 3(s − 1) for s ≥ 5 .
Step 24: From the graph obtained in Step 23, delete another 3s(m − s ) number of edges forming
3(m − s ) edge disjoint circuits C s attached with greatest consecutive weights, other than 3s 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 K 3s of degree 2(s − 1) having 3s (s − 1) number of remaining edges for s ≥ 5 .
Step 26: From the graph obtained in Step 25, delete another 3s number of edges forming the cycle
C3s 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
K 3s of degree 2(s − 2) for s ≥ 5 .
Step 28: Repeat Step 26 on the graph obtained in Step 23 for (s − 3) times where s ≥ 5 .
Step 29: Observed that the graph obtained in Step 28 is a cycle C3s 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 m = 3 of K 2m+1
and 2 m + 1 = 7 , 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
40
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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 m = 4 of K 2m+1
and 2m + 1 = 9 = 3 ⋅ 31 )
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
Table-2
95
18
55
∞
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, AD = 81 , DG = 84 , AG = 86 , BE = 88 , EH = 90 , BH = 91 , CF = 93 , FI = 95 , CI = 97
forming 3 edge disjoint K 3 . Then we have a graph as shown in Figure-12.
Now, applying the Step10 to Step14 of the algorithm, we delete a cycle C9 whose edges are attached
with greatest weights AE = 59 , EI = 61 , ID = 64 , DH = 69 , HC = 70 , CG = 73 , GB = 76 ,
BF = 78 , FA = 80 , and then we delete another cycle C9 whose edges are attached with next greatest
weights AB = 31 , BC = 34 , CD = 39 , DE = 41 , EF = 47 , FG = 49 , GH = 52 , HI = 55 , IA = 57 .
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 m = 7 of K 2m+1
and 2m + 1 = 15 = 3 × 5 )
A
B
C
D
E
F
G
H
I
A
∞
94
69
122
44
166 143
B
94
∞
96
81
129
50
170 150
C
69
96
∞
97
70
136
D
122
81
97
∞
99
E
44
129
70
99
F
166
50
136
G
143 170
L
M
N
O
3
41
149 169
66
128
93
121
39
37
156 175
48
135
80
55
176 158
34
31
163 180
54
141
83
123
61
181 147
29
26
146 184
60
∞
101
71
130
45
185 155
23
21
153 190
83
101
∞
104
86
138
51
167 162
18
17
161
55
123
71
104
∞
105
73
125
57
173 144
14
10
61
130
86
105
∞
109
87
132
63
178 152
6
45
138
73
109
∞
110
76
139
46
183 159
51
125
87
110
∞
112
89
127
53
187
57
132
76
112
∞
115
77
133
59
63
139
89
115
∞
116
90
140
46
127
77
116
∞
118
79
152 183
53
133
90
118
∞
119
159 187
59
140
79
119
∞
H
3
150 176
I
41
39
158 181
J
149
37
34
147 185
K
169 156
31
29
155 167
L
66
175 163
26
23
162 173
21
18
144 178
17
14
190 161
10
M 128
48
180 146
N
93
135
54
184 153
O
121
80
141
60
6
J
K
Table: 3
From the Table-3, we have a complete graph of nine vertices, which is shown in Figure-14.
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ISSN 0976 - 6375(Online), Volume 4, Issue 6, November - December (2013), © IAEME
Figure: 14
Now, applying the Step 22 of the algorithm, we delete 15 edges forming 5 edge disjoint K 3
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 C5 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 K15 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 C15
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 K15 of degree 6 as shown in Figure-17.
Figure-17
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ISSN 0976 - 6375(Online), Volume 4, Issue 6, November - December (2013), © IAEME
Applying the Step 28 of the algorithm, we delete 30 edges forming 2 edge disjoint cycles C15
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 K 2m+1 for m ≥ 2 into circulant
graphs if 2m + 1 = a prime number, 2m + 1 = 3 ⋅ 3n , n ≥ 1 and 2 m + 1 = 3s , s ≥ 5 is a prime. However,
one can study the other situation like 2m + 1 = 5 ⋅ 5 n , n ≥ 1 or 2 m + 1 = 5 s , s ≥ 7 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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