We consider cycle decomposition and path decomposition on undirected graph. The following are some results:
Theorem 1.1. [10] (1) For all odd integers n and all non-negative integer r satisfying 3r = n(n−1)2 there is a decomposition of Kn into r 3-cycles which partitions the edge set of Kn. (2) For all even integers n and all non-negative integers r satisfying 3r = n(n−2)2 there is a decomposition of Kn− F into r 3-cycles which partitions the edges set of Kn− F .
We can establish the existence of cycle systems not only the 3-cycle system but also the m-cycle system for any m. There are some results below:
Theorem 1.2. [16] (1) For all odd integers n and all non-negative integer r and m satisfying mr = n(n−1)2 there is a decomposition of Kn into r m-cycles which partitions the edge set of Kn. (2) For all even integers n and all non-negative integers r and m satisfying mr = n(n−2)2 there is a decomposition of Kn− F into r m-cycles which partitions the edges set of Kn− F .
Theorem 1.3. [1] (1) For all odd integers n and all non-negative integer r and s satisfying 3r + 5s = n(n−1)2 there is a decomposition of Kn into r 3-cycles and s 5-cycles which partitions the edge set of Kn. (2) For all even integers n and all non-negative integers r and s satisfying 3r + 5s = n(n−2)2 there is a decomposition of Kn− F into r 3-cycles and s 5-cycles which partitions the edges set of Kn− F .
Theorem 1.4. [7] (1) For all odd integers n and all non-negative integer r, s and t satisfying 3r + 4s + 6t = n(n−1)2 there is a decomposition of Kn into r 3-cycles, s 4-cycles, and t 6-cycles which partitions the edge set of Kn. (2) For all even integers n and all non-negative integers r, s and t satisfying 3r + 4s + 6t = n(n−2)2 there is a decomposition of Kn− F into r 3-cycles, s 4-cycles, and t 6-cycles which partitions the edges set of Kn− F . Theorem 1.5. [4] (1) For all odd integers n and all non-negative integer r and s satisfying 4r + 5s = n(n−1)2 there is a decomposition of Kn into r 4-cycles and s 5-cycles which partitions the edge set of Kn. (2) For all even integers n and all non-negative integers r
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and s satisfying 4r + 5s = n(n−2)2 there is a decomposition of Kn− F into r 4-cycles and s 5-cycles which partitions the edges set of Kn− F .
The following useful contains three different lengths which are n, n − 1, n − 2.
Theorem 1.6. [7] Let S = {n−2, n−1, n}. If n is odd and a(n−2)+b(n−1)+cn = n(n−1)2 , then Kn = aCn−2+ bCn−1+ cCn. If n is even and a(n − 2) + b(n − 1) + cn = n(n−2)2 , then Kn− F = aCn−2+ bCn−1+ cCn.
Alspach Conjecture is also true if the cycles lengths mi are bounded by some linear function of n and n is sufficiently large.
Theorem 1.7. [2] Assume n must be larger than N2 which is very large absolute constants.
If m1, . . . , mt are integers with 3 ≤ mi ≤ ⌊n−112120 ⌋ andPt
i=1mi = (n2) (n odd) or (n2) −n2 (n even), then one can pick Kn (n odd) or Kn− I(n even) with cycles of length m1, . . . , mt. Theorem 1.8. [6] Let n be a n even positive integer. Then Kn can be decomposed into
n
2hamiltonian paths.
Theorem 1.9. [15] If n is odd and {ai : 1 ≤ i ≤ r} is a multiset of r positive integers satisfying 1 ≤ ai ≤ n − 2 and Pr
i=1ai = (n2). Then Kn can be decomposed into {Pai|1 ≤ i ≤ r}.
Theorem 1.10. [21] Let m|λ(n2), and m ≤ n − 1. Then λKn caon be decomposed into isomorphic paths of length m.
Theorem 1.11. [5] If v is odd. Let m1, m2, . . . , mt be t positive integers such that 1 ≤ mi ≤ n − 2, Pt
i=1mi + k(n − 1) = (n2), and k ∈ {1, 2,n−12 }, then Kv can be decomposed into t + k paths P1, P2, . . . , Pt+k such that the length of Pi is mi for i = 1, 2, . . . , t and the length of Pi is n − 1 for i > t.
Theorem 1.12. [5] If v is odd. Let n − 1 ≥ m1 ≥ m2 ≥ · · · ≥ mt ≥ 1 and h ≤ mt ≤ n − h − 1 such that Pt
i−1mi = (n2), m1 = m2 = · · · = mh = n − 1. Then Kv can be decomposed into t paths P1, P2, . . . , Pt such that the length of Pi is mi for i = 1, 2, . . . , t. Moreover, if there exists a h < t′ ≤ t such that h ≤ mt′ ≤ n − h − 1 or
h ≤Pt
i=t′mi ≤ n − h − 1, then Kv can be decomposed into t paths P1, P2, . . . , Pt such that the length of Pi is mi for i = 1, 2, . . . , t.
Theorem 1.13. [5] If v is odd. Let n − 1 ≥ m1 ≥ m2 ≥ · · · ≥ mt ≥ 1, mt < h, and mt−1mt ≤ n − h − 1 such that P
i=1tmi = (n2), m1 = m2 = . . . = mh = n − 1. Then Kv can be decomposed into t paths P1, P2, . . . , Pt such that the length of Pi is mi for i = 1, 2, . . . , t.
Theorem 1.14. [5] If v is odd. Let n−1 ≥ m1 ≥ m2 ≥ · · · ≥ mt≥ 1 and n+h−2 ≤ mt+ mt−1 ≤ 2n−h−3 such thatPt
i−1mi = (n2), m1 = m2 = . . . = mh = n−1. Then Kv can be decomposed into t paths P1, P2, . . . , Pt such that the length of Pi is mi for i = 1, 2, . . . , t.
Moreover, if there exists a h < t′ ≤ t such that n + h − 2 ≤ Pt
i=t′mi ≤ 2n − h − 3, then Kv can be decomposed into t paths P1, P2, . . . , Pt such that the length of Pi is mi
for i = 1, 2, . . . , t.
Next, we consider some results of RPPDC and EPPDC.
Proposition 1.15. [19] Suppose that G is a graph with an eulerian perfect path double cover. Then for 1 ≤ d(x) ≤ 3, G + x has an eulerian perfect double cover.
Proposition 1.16. [19] For any n ≥ 1, Kn,n has an RPPDC. Moreover, if n is odd, then Kn,n has an ERPPDC.
Proposition 1.17. [19] For any m, n ≥ 1, Km,n has an EPPDC.
Proposition 1.18. [19] If G is a k-regular graph, k ≥ 1, then L(G) has an RPPDC.
Proposition 1.19. [19] Let G be a graph with m edges. Suppose 2G has an Euler circuit e1, e2, . . . , e2m such that S1 = {e1, e3, . . . , e2m−1} and S2 = {e2, e4, . . . , e2m} are both the set E(G) of all edges of G. Furthermore, suppose that for each v ∈ V (G) there is ordering, C(v), of the edges incident to v such that every pair of consecutive edges inC(v) occurs exactly once as a pair of consecutive edges in the Euler circuit. Then L(G) has an EPPDC.
Proposition 1.20. [19] For all m ≥ 2, L(Km) has an ERPPDC.
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Proposition 1.21. [19] For all m, n ≥ 1, L(Km,n) has an RPPDC. Furthermore, if gcd(n, m) = 1 or gcd(n, n − m + 2) = 1, then L(Km,n) has an ERPPDC.
Proposition 1.22. [19] For every positive odd integer n, L(Kn,n) has an ERPPDC.
Proposition 1.23. [19]
• If G and H have RPPDCs, then GH has an RPPDC.
• If G and H have EPPDCs and (|G|, |H|) = 1, then GH has an EPPDC.
Proposition 1.24. [19] If G has an EPPDC, then the Cartesian product GK2 has an EPPDC.
Proposition 1.25. [19] For all n ≥ 0, the n-cube, Qn, has an EPPDC.
In our main result we concentrate on oriented version. Now, we consider some results below:
Lemma 1.26. [13] K2n has an OPPDC.
Proof. Let V (K2n) = {v0, v1, . . . , v2n−1}. For 0 ≤ i ≤ 2n−1 set Pi = (vi, vi+1, vi−1, vi+2, vi−2, . . . , vi+n) where all subscripts are read modulo 2n. It is easy to verity that P = {Pi|0 ≤ i ≤ 2n − 1}
is an OPPDC of K2n.
Lemma 1.26 gives an easy construction of OPPDC for all K2n. Tillson proved in [22]
that all K2n+1 have an OPPDC for n ≥ 3.
Example 1.27. [13] K7 has an OPPDC as follow:
P1 = 1263547 P2 = 2731465 P3 = 3742516 P4 = 4536721 P5 = 5764132 P6 = 6175243 P7 = 7156234
We can check that the collection P = {P1, . . . , P7} is an OPPDC of K7.
Next, we consider the minimal (i.e. , with minimal number of edges) connected graph G such that G 6= K3, G 6= K5 and G has no OPPDC. J. Maxov´a had show that G has no vertices of degree 1, 2.
Lemma 1.28. [13] Let G1, G2 be two graphs which have an OPPDC. Suppose that G1∩ G2 = {v}.Then the union G1∪ G2has an OPPDC.
Proof. Denote by Pian OPPDC of Gi, i = 1, 2. Let P1 ∈ P1be the path that starts in v and P2 ∈ P2be the path that ends at v. Then the collection P1∪ P2∪ {P1∪ P2}\{P1, P2} is an OPPDC of G1∪ G2.
Corollary 1.29. [13] Let G be a simple graph; G 6= K3, and v ∈ V {G} a vertex of degree 1. If G \ v has an OPPDC then G has an OPPDC.
By applying this corollary, we get that if we add a new vertex of degree 1 to a graph with an OPPDC then the resulting graph also has an OPPDC. Hence every tree has an OPPDC.
Theorem 1.30. [13] Let G be a simple graph; G 6= K3,and v ∈ V {G} a vertex of degree 2. If G \ v has an OPPDC then G has an OPPDC.
By applying Corollary 1.29 and Theorem 1.30, we get that every 2-degenerate graph has an OPPDC, except K3. The following are some results
Corollary 1.31. [13] If G is a union of two arbitrary trees; G 6= K3; then G has an OPPDC.
Another construct which preserves the property of having an OPPDC is the so-called arrow construction.
Definition 1.32. [14] A graph I with two distinguished vertices a, b, a, b /∈ E(I), is called an indicator. For a given directed graph D = (V, E) and an indicator (I, a, b) we define an (undirected) graph D ∗ (I, a, b) = (W, F ) sa follows:
W = (E × V (I))/ ∼,
where the equivalence is generated by the following pairs:
((x, y), a), ((x, y′), a), ((x, y), b), ((x′, y), b), ((x, y), b), ((y, z), a).
For a pair (e, x) ∈ E × V (I) its equivalence class is denoted by [e, x].
We put {[e, x], [e′, x′]} ∈ F ⇐⇒ e = e′ and {x, x′} ∈ E(I).
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Figure 1: Arrow construction
This arrow construction is schematically indicated in Fig. 1 (One can check that the indicator I in Fig. 1 satisfies the assumptions of Theorem1.33 below.)
Theorem 1.33. [14] Suppose an indicator (I, a, b) has an OPPDC Π containing two paths P1, P2 ∈ Π such that P1 begins in a and ends in b, and P2 begins in b and ends in a. Further suppose G has an OPPDC. Then for any orientation D of G the graph D ∗ (I, a, b) has an OPPDC.
Proposition 1.34. [14] If G is a 2-connected graph with |E(G)| ≤ 2n − 1; G 6= K3; then G has an OPPDC.
Conjecture 1.35. [14] K3 and K5 are the only connected graphs which do not have an OPPDC.
2 Main Results
In this section, we focus on the minimal degree of a graph G first. We show that if we add a new vertex of degree 3 to a graph with an OPPDC then the resulting graph also has an OPPDC. And we use this theorem to prove that if G is a 3-degenerate graph and G has no components which isomorphism to K3 then G has an OPPDC. Next, we show that the complete graph Kn,n and the multipartite graph Km(n) has an OPPDC by a special construction.
2.1 3-degenerate graph
Theorem 2.1. Let G be a simple graph; G 6= K3,and v ∈ V {G} a vertex of degree 3. If G \ v has an OPPDC then G has an OPPDC.
Proof. Let N(v) = {a, b, c} be the neighbors of the vertex v. Denote by P an OPPDC of the graph G \ v. For u ∈ V (G \ v), letPu (resp. Pu) denote the path of P beginning (resp. ending) with u. We call Pu (resp. Pu) is the outer (resp. inner) path of u in G \ v.
Case 1. There exists an outer path Pu,u ∈ N(v),Pu pass through N(v).
Without loss of generality , we assume Pa pass through b and then c.
Subcase 1-1: Pc 6= Pb
Separate Pa into two paths, P1 and P2,where P1 is the path that beginning at a and ending at b along Pa and P2 is the path that beginning at b and along Pa. Let Pc∗= (c, v) ∪ (v, a) ∪ P1
Pa∗ = (a, v) ∪ (v, b) ∪ P2
Pv∗ = Pb∪ (b, v) Pv∗ = (v, c) ∪ Pc
Then the collection P \ {Pa, Pb, Pc} ∪ {Pv∗, Pv∗, Pc∗, Pa∗}is an OPPDC of G.
Subcase 1-2: Pc = Pb
Subcase 1-2-1: Pb 6= Pa
Separate Pa into three paths , P1, P2, and P3, where P1 is the path that beginning at a and ending at b along Pa , P2 is the path that beginning at b
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and ending at c along Pa , P3 is the path that beginning at c and along Pa. Let Pa∗ = P1 ∪ (b, v) ∪ (v, c) ∪ P3
Pb∗ = P2∪ (c, v) ∪ (v, a) Pv∗= (v, b) ∪ Pb
Pv∗ = Pa∪ (a, v)
Then the collection P \ {Pa, Pa, Pb, } ∪ {Pa∗, Pb∗, Pv∗, Pv∗} is an OPPDC of G.
Subcase 1-2-2: Pb = Pa
Let Pa∗ = (a, v) ∪ (v, c) ∪ Pc Pc∗ = (c, v) ∪ (v, b) ∪ Pb Pv∗= (v, a) ∪ Pa
Pv∗ = (b, v)
Then the collection P \ {Pa, Pb, Pc} ∪ {Pa∗, Pc∗, Pv∗, Pv∗} is an OPPDC of G.
Case 2. There is no outer path Pu,u ∈ N(v),Pu pass through N(v).
Without loss of generality , we assume Pa doesn’t pass through c.
Subcase 2-1: Pc doesn’t pass through a.
Let Pc∗= (c, v) ∪ (v, a) ∪ Pa Pa∗ = (a, v) ∪ (v, c) ∪ Pc Pv∗ = (v, b) ∪ Pb
Pv∗ = (b, v)
Then the collection P \ {Pa, Pb, Pc} ∪ {Pa∗, Pc∗, Pv∗, Pv∗} is an OPPDC of G.
Subcase 2-2: Pc pass through a.
Since Pc passes through a, it can’t pass through b. If Pb doesn’t pass through c, it will return to case 2-1. So Pb passes through c. If Pb passes through a, it will return to case 1. So Pb doesn’t pass through a.
Let Pc∗= (c, v) ∪ (v, a) ∪ Pa Pa∗ = (a, v) ∪ (v, b) ∪ Pb Pv∗ = (v, c) ∪ Pc
Pv∗ = (b, v)
Then the collection P \ {Pa, Pb, Pc} ∪ {Pa∗, Pc∗, Pv∗, Pv∗} is an OPPDC of G.
Thus, we have the prove.
Lemma 2.2. G1 and G2 has an OPPDC.
Proof. Since degG1(V1) = 3 degG1(V2) = 2 and G1\V1, G1\V1 are paths. By Theorem 1.30 and Theorem 2.1 we know that G1 and G2 has an OPPDC.
Theorem 2.3. If G has no components which isomorphism to K3 and G is a 3-degenerate graph, then G has an OPPDC.
Proof. We proceed by induction on n = V |(G)|. Since G is a 3-degenerate graph, there is a vertex v ∈ V (G) of degree at most 3. We denote that G′ = G \ v. If G′ is isomorphic to K3 then G is isomorphic to K4 or one of the graphs G1, G2 in Lemma 2.2, that all have an OPPDC. If G′ is a disconnected graph with some components which are isomorphic to K3. Then we choose another vertex v′ which in K3 and let G′ = G \ v′. Since degG(v′) ≤ 3 we know that G′ applies to the induction hypothesis. If degG(v) = 1 by induction hypothesis the graph G′ has an OPPDC. Then by applying Corollary 1.29 the graph G has an OPPDC. If degG(v) = 2 by induction hypothesis the graph G′ has an OPPDC. Then by applying Theorem 1.30 the graph G has an OPPDC. If degG(v) = 3 by induction hypothesis the graph G′ has an OPPDC. Then by applying Theorem 2.1 the graph G has an OPPDC.
Thus, we have the prove.
Corollary 2.4. Every cubic graph has an OPPDC.
Proof. We know that every cubic graph is 3 − degenerate. By Theorem 2.3 we have the prove.
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