CC BY
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2019 Прикладная теория графов №45
UDC 519.1 DOI 10.17223/20710410/45/8
ON THE ONE EDGE ALGORITHM FOR THE ORTHOGONAL DOUBLE COVERS
R. El-Shanawany, A. El-Mesady
Menoufia University, Menouf, Egypt
E-mail: [email protected], [email protected]
The existing problem of the orthogonal double covers of the graphs is well-known in the theory of combinatorial designs. In this paper, a new technique called the one edge algorithm for constructing the orthogonal double covers of the complete bipartite graphs by copies of a graph is introduced. The advantage of this algorithm is that it is accessible to discrete mathematicians not intimately familiar with the theory of the orthogonal double covers.
Keywords: graph decomposition, symmetric starter, orthogonal double covers.
Nomenclature:
Km complete graph on m vertices;
Kmn complete bipartite graph with independent sets of sizes m and n;
mG m disjoint copies of G ;
Ck(ni, n2,..., nk) caterpillar (tree) obtained from the path Pk = xix2 ... xk by joining vertex Xi to ni new vertices where k ^ 1, ni,n2,... ,nk are positive integers, ni,nk ^ 1 and ni ^ 0 for i E {2, 3,... ,k — 1}.
Introduction
Graphs serve as a mathematical model to solve many real-world problems successfully. Some problems in chemistry, physics, computer technology, communication science, psychology, genetics, linguistics, and sociology can be formulated as problems in graph theory. Also, many branches of mathematics, such as topology, probability, group theory, and matrix theory, have close connections with the graph theory. Some puzzles of a practical nature have been instrumental in the development of various topics in graph theory. The cyclic graphs theory was developed for solving many problems of electrical networks, and the study of "trees" is considered a helping tool for enumerating isomers of organic compounds.
An orthogonal double cover (ODC) of H by G is a collection G = {n(x) : x E V(H)} of isomorphic subgraphs (to G) of H (called pages) such that (i) every edge of H is contained in exactly two pages of G and (ii) n(a) and n(b)share an edge if and only if a and b are adjacent in H.
The existence problem of ODC has attracted much attention during the last few years. The problem is known to be hard in general as it includes some long-standing open problems like the existing problems of biplanes. The ODC problem originally stems from problems in database optimization, statistical design of experiments, and design theory. In [1, 2], J. Demetrovics et al. have inspected the key of the Armstrong databases of minimum size. The ODC of Kn whose elements consist of distinct cliques is equivalent to Armstrong database of size n. Also, ODCs are related to several graph decomposition problems [3, 4]. The initial interest was concerned with the complete graphs, but the specialists have solved
the ODC problem for several graphs such as the Cayley graphs [5] and the complete bipartite graphs (e.g., [6, 7]).
The labeling of the vertices of the complete bipartite graph Knn is defined by the bijective mapping 0 : V(Kn,n) ^ Zn x Z2. The product (v, j) E Zn x Z2 will be written as Vj referring to the corresponding vertex and the edge (cY, d«s) E E(Kn,n) if y = 5 for all c, d E Zn and y, 5 E Z2. We shall denote by (c, d) the edge between the vertices c0 and di. If G is a subgraph of Knn and i E Zn, then G + i is called i-translate of G. The edges of G + i are obtained from G by rotating the edges of G, i.e., by mapping the edge (c,d) in G to the edge (c + i, d + i) in G + i (with calculations modulo n). If e = (p, q) E E(G), then it has a length defined by d(e) = q — p (with calculations modulo n). For the graph G, if |E(G)| = n and the lengths of all edges in G are mutually distinct and equal to Zn, then G is said to be a half starter w.r.t. Zn. R. El-Shanawany et al. [8] have proved the following three results.
A. The union of all translates of G forms an edge decomposition of Kn,n, i.e., (J E(G+z) =
zGZn
= E(Kn,n), if and only if G is a half starter.
In what follows, a half starter G will be represented by the vector v(G) = (v0, v1,..., vn-1) E Zn where vj, i E Zn and (vj)0 is the unique vertex ((vj, 0) E Zn x {0}) that belongs to the unique edge of length i in G. The two half starter vectors v(G0) and v(G1) are said to be orthogonal if {vj(G0) — vi(G1) : i E Zn} = Zn.
B. If v(G0) and v(G1) are orthogonal two half starter vectors, then G = {G^ : (z,l) E E Zn x Z2} with Gz,i = G^ + z is considered the ODC of Kn,n.
If the graph Gs is a subgraph of Knn with E(Gs) = {(p0,q1) : (q0,p1) E E(G)}, then Gs is the symmetric graph of G. It is easy to prove that if G is a half starter, then Gs is also a half starter. Also, if v(G) and v(Gs) are orthogonal, then the half starter G is called a symmetric starter w.r.t. Zn.
C. If n is any positive integer and G is a half starter with v(G) = (v0, v1,..., vn-1), then G is symmetric starter iff {vj — v-j + i : i E Zn} = Zn.
For illustrative purposes, Fig. 1 exhibits the graph K44 and its ODC by K1 4.
Fig. 1. K4 4 and its ODC by KM
The aim of this paper is to present a new algorithm which is called the one edge algorithm. This algorithm is a helping tool for constructing the ODCs of the complete
bipartite graphs. In addition, we present a number of new results as a direct application of this algorithm. The difference between our paper and [9] is as follows. In [9], El-Serafi et al. constructed the orthogonal double cover of complete bipartite graphs by a complete bipartite graph and the disjoint union of complete bipartite graphs with the cartesian product of symmetric starter vectors. For several results of ODCs by different graph classes, see [10-12].
1. Main results
In this section, we construct the orthogonal double covers of the complete bipartite graphs by copies of stars, copies of a caterpillar, and by copies of a complete bipartite graph.
Definition 1. For k ^ 2, a caterpillar graph Ck(mi,m2,... ,mk) is obtained from a path Pk = v1v2 .. .vk by attaching n, ^ 0 pendant vertices ui,j (1 ^ j ^ m,) to each v,.
Definition 2. A complete bipartite graph of the form K1n is called a star.
In what follows, we prove the theorems on existence of ODCs for particular classes of complete bipartite graphs Kmn,mn and K2mn2mn with n > 0 and m = 1, 5 (mod 6). Really, our effort was concentrated on these classes and in the future we hope to solve the problem for the other classes to be used in the design theory. The proofs of these theorems base on the direct constructions of ODCs. The main strengths of represented algorithms is that they are clear for discrete mathematicians not intimately familiar with the theory of the orthogonal double covers.
Theorem 1. Let m and n be positive integers with m = 1, 5 (mod 6). Then, there is an ODC of Kmn,mn by mKi,n.
Proof. Algorithm 1 proves the Theorem 1. ■
Algorithm 1. ODC of Kmn,mn by mK1n
1: [Inauguration.] Choose the values m,n E N, m = 1, 5 (mod 6).
2: [Initial edge.] Construct the edge (0, 0).
3: [Complementary edges.] Add (0, —3) to (0,0) where 3 E Zn \ {0}.
4: [Initial graph The union of the edge set in steps 2, 3 gives
= G0.] the edge set of Ki,„.
5: [Complementary graphs Add (an, 2an) to the edge set of step 4 where a E Zm \ {0}.
m— 1
^ u Ga.]
a=1
6: [Symmetric starter The union of the edge set in steps 4, 5 gives the edge set
m— 1
^ u Ga.] of G = mK 1^.
a=0
7: [Symmetric graph.] If the edge (a, b) belongs to the edge set of step 6, then
(b, a) is an edge of the graph Gs = mK1,n.
8: [Translation.] Add (7,7) to the edge set in steps 6, 7 where 7 E Zmn.
9: [Done.] Output the edge set of step 8 which represents the ODC _of Kmn,mn by mK^._
For more illustration to Theorem 1, let m = 5 and n = 3, then there is an ODC of K1515 by 5K1,3, see Fig. 2.
Fig. 2. Symmetric starter of an ODC of K15,15 by 5K1,3
Theorem 2. Let m and n ^ 5 be positive integers with m = 1, 5 (mod 6). Then, :e is an ODC of Kmn,mn by 2mK2 U mK1n-2. Proof. Algorithm 2 proves the Theorem 2. ■
Algorithm 2.
ODC of K
by 2mK2 U mK1,n-2
1: [Inauguration.]
2: [Initial edge.] 3: [Complementary edges.]
4: [Initial graph = G0.]
5: [Complementary graphs
m— 1
= u
a=1
6: [Symmetric starter
m— 1
of G = U Ga.]
a=0
7: [Symmetric graph.]
8: [Translation.] 9: [Done.]
Choose the values of m and n where m and n ^ 5 are positive integers with m = 1, 5 (mod 6). Construct the edge (0, 0).
Add (1 + n, 2n — 1), (2n — 1, 3n — 2), and (0, —2^) to (0, 0) where ^ E Zn \ Z3.
The union of the edge set in steps 2, 3 gives the edge set of 2K2 U K1,n-2.
Add (an, 2an) to the edge set of step 4 where a E Zm \ {0}.
The union of the edge set in steps 4, 5 gives the edge set = 2mK2 U mK1n-2.
If the edge (a, b) belongs to the edge set of step 6, then (b, a) is an edge of the graph Gs = 2mK2 U mK1,n-2. Add (y, Y) to the edge set in steps 6, 7 where y E Zmn. Output the edge set of step 8 which represents the ODC of Kmn,mn by 2mK2 U mK1,n-2.
For more illustration to Theorem 2, let m = 5 and n = 5, then there is an ODC of K25,25 by 5Ki,3 U IOK2, see Fig. 3.
Ol 21, 22i I«, 6, 7, 20, I«, 17, S, I, 2, IS, II, 12,
G" G' G2 G' G"
Fig. 3. Symmetric starter of an ODC of K25,25 by 5K1,3 U 10K2
Theorem 3. Let m and n ^ 4 be positive integers with m = 1, 5 (mod 6). Then, there is an ODC of Kmn,mn by mC2(1, n — 2).
Proof. Algorithm 3 proves the Theorem 3.
Algorithm 3.
ODC of Kmn,mn by mC2(1, n - 2)
1: [Inauguration.]
[Initial edge.]
[Complementary edges.] [Initial graph = G0.]
5: [Complementary graphs
m-1
= U Ga.]
a=1
6: [Symmetric starter
m-1
^ u Ga.]
a=0
7: [Symmetric graph.]
8: [Translation.] 9: [Done.]
Choose the values of m and n where m and n ^ 4 are
positive integers with m = 1, 5 (mod 6).
Construct the edge (0, 0).
Add (1,1 + 3 — n) to (0, 0) where 3 E Zn \ {0}.
The union of the edge set in steps 2, 3 gives the edge set
of C2(1, n — 2).
Add (an, 2an) to the edge set of step 4 where a E Zm \{0}.
The union of the edge set in steps 4, 5 gives the edge set of G = mC2(1,n — 2).
If the edge (a, b) belongs to the edge set of step 6, then (b, a) is an edge of the graph Gs = mC2(1, n — 2). Add (7,7) to the edge set in steps 6, 7 where 7 E Zmn. Output the edge set of step 8 which represents the ODC of Kmn,mn by mC2(1, n — 2).
For more illustration to K20,20 by 5C2(1, 2), see Fig.
Theorem 3, let m 4.
5 and n = 4, then there is an ODC of
Fig. 4. Symmetric starter of an ODC of K20,20 by 5C2(1,2)
Theorem 4. Let m and n ^ 2 be positive integers with m = 1, 5 (mod 6). Then, there is an ODC of K2mn,2mn by mK2,n.
Proof. Algorithm 4 proves the Theorem 4. ■
Algorithm 4.
ODC of K2mn,2mn by mK2,i
1: [Inauguration.]
[Initial edge.]
[Complementary edges.] [Initial graph = G0.]
5: [Complementary graphs
m-1
= u Ga.
a=1
6: [Symmetric starter
m—1
^ u Ga.]
a=0
7: [Symmetric graph.]
Choose the values of m and n where m and n ^ 2 are positive integers with m = 1, 5 (mod 6). Construct the edge (0, 0).
Add (0, —23) and (1, —23) to (0,0) where 3 E Z,n \ {0}. The union of the edge set in steps 2, 3 gives the edge set of K2,n.
Add (2an, 4an) to the edge set of step 4 where a E Zm\{0}.
The union of the edge set in steps 4, 5 gives the edge set of G = mK2,n.
If the edge (a, b) belongs to the edge set of step 6, then (b, a) is an edge of the graph Gs = mK2,n.
8: [Translation. 9: [Done.]
Add (y, Y) to the edge set in steps 6, 7 where y E Z2mn. Output the edge set of step 8 which represents the ODC of K2mn,2mn by mK2,„.
For more illustration to Theorem 4, let m K30,30 by 5K2,3, see Fig. 5.
5 and n = 3, then there is an ODC of
Fig. 5. Symmetric starter of an ODC of K30,30 by 5K2,3
Theorem 5. Let m and n ^ 5 be positive integers with m = 1, 5 (mod 6). Then, there is an ODC of Kmn,mn by mC5(1, 0, 0, 0, n — 5).
Proof. Algorithm 5 proves the Theorem 5. ■
Algorithm 5.
ODC of Kmn,mn by mC5(1, 0, 0, 0, n — 5).
1: [Inauguration.]
2: [Initial edge.] 3: [Complementary edges.]
4: [Initial graph = G0.]
5: [Complementary graphs
m— 1
^ u Ga.]
a=1
6: [Symmetric starter
m— 1
^ u Ga.]
a=0
7: [Symmetric graph.]
8: [Translation.] 9: [Done.]
Choose the values of m and n where m and n ^ 5 are positive integers with m = 1, 5 (mod 6). Construct the edge (0, 0).
Add (0, 2 — n), (2, 2 — n + ft), ((m + 7)n/2,Y — n) to (0, 0) where ft E Zn \ {0, 2, n — 2, n — 1} and y E {2, 3}. The union of the edge set in steps 2, 3 gives the edge set of C5(1, 0, 0, 0, n — 5).
Add (an, 2an) to the edge set of step 4 where a E Zm\{0}.
The union of the edge set in steps 4, 5 gives the edge set of G = mC5(1,0, 0, 0, n — 5).
If the edge (a, b) belongs to the edge set of step 6, then (b, a) is an edge of the graph Gs = mC5(1, 0,0, 0, n — 5). Add (y, Y) to the edge set in steps 6, 7 where y E Zmn. Output the edge set of step 8 which represents the ODC of Kmn,mn by mC5(1, 0, 0, 0, n — 5).
For more illustration to Theorem 5, let m K35 35 by 5C5(1, 0,0, 0, 2), see Fig. 6.
5 and n = 7, then there is an ODC of
Fig. 6. Symmetric starter of an ODC of K3535 by 565(1, 0, 0,0,2)
Conclusion
In this paper, we study the orthogonal double covers of the complete bipartite graphs
by algorithms for generating the orthogonal double covers by one edge.
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