The Linear 4arboricity of Balanced Complete Bipartite Graphs
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The Linear 4arboricity of Balanced Complete Bipartite Graphs ∗
Liancui Zuo,† Shengjie He, and Ranqun Wang
Abstract
A linear kforest is a graph whose components are paths of length at most k. The linear karboricity of a graph G, denoted by lak(G), is the least number of linear kforests needed to decompose G. In this paper, it is obtained that la4(Kn,n) = ⌈5n/8⌉ for n ≡ 0( mod 5).
Keywords: Linear kforest; linear karboricity; complete bipartite graph; perfect matching; bipartite diﬀerence
a matching, and la1(G) is the edge chromatic number, or chromatic index, χ′(G) of a graph G. Moreover, the linear k−arboricity lak(G) is also a reﬁnement of the ordinary linear arboricity la(G)(or la∞(G)) [14] of a graph G, which is the case when every component of each forest is a path with no length constraint. In 1982, Habib and Peroche [12] proposed the following conjecture for an upper bound on lak(G).
Conjecture 1.1. If G is a graph with maximum degree ∆(G) and k ≥ 2, then
1 Introduction
In this paper, all graphs considered are ﬁnite, undirected, and simple (i. e., loopless and without multiple edges). We refer to [20] for terminology in graph theory. In recent years, many parameters and classes of graphs were studied. For example, in [11], diﬀerent properties of the intrinsic order graph were obtained, namely those dealing with its edges, chains, shadows, neighbors and degrees of its vertices, and some relevant subgraphs, as well as the natural isomorphisms between them. In [18], the ndimensional cubeconnected complete graph was studied. In [24, 25], the hamiltonicity, path tcoloring, and the shortest paths of Sierpin´skilike graphs were researched. In [26], the vertex arboricity of integer distance graph G(Dm,k) was obtained.
A decomposition of a graph is a list of subgraphs such that each edge appears in exactly one subgraph in the list. If a graph G has a decomposition G1, G2, · · · , Gd, then we say that G1, G2, · · · , Gd decompose G, or G can be decomposed into G1, G2, · · · , Gd. A linear k−forest is a forest whose components are paths of length at most k. The linear k−arboricity of a graph G, denoted by lak(G), is the least number of linear k−forests needed to decompose G. Let x be a real number, denoted by ⌊x⌋ the maximum integer no more than x, and denoted by ⌈x⌉ the minimum integer no less than x. For any integers a < b, let [a, b] denote the set of integers {a, a + 1, · · · , b} for simplicity. For any positive integer λ, let Pλ be a path on λ vertices which has length λ − 1.
The notion of linear k−arboricity was ﬁrst introduced by Habib and Peroche [13], which is a natural generalization of edge coloring. Clearly, a linear 1forest is induced by
∗Manuscript received August 4,2014. This work was supported by NSFC for youth with code 61103073.
†Liancui Zuo, Shengjie He and Ranqun Wang are with College of Mathematical Science, Tianjin Normal University, Tianjin, 300387, China. Email: lczuo@mail.tjnu.edu.cn;lczuo@163.com
⌈
∆(G)·V (G)
2⌊
k·V (G) k+1
⌋
⌉,
if
∆(G)
=
V
(G)

−1,
lak(G) ≤
⌈
∆(G)·V (G)+1
2⌊
k·V (G) k+1
⌋
⌉,
if
∆(G)
<
V
(G)

−1.
For k = V (G)  −1, it is the Akiyama’s conjecture [1].
Conjecture
1.2.
[1]
la(G)
≤
⌈
∆(G)+1 2
⌉.
So far, quite a few results on the veriﬁcation of Conjecture 1.1 have obtained in the literature, especially for graphs with particular structures, such as trees [5, 6, 13], cubic graphs [4, 16, 19], regular graphs [2, 3], planar graphs [17], balanced complete bipartite graphs [8, 9, 10], balanced complete multipartite graphs [22] and complete graphs [5, 7, 8, 9, 21]. It is obtained that the linear 2arboricity, the linear 3arboricity and the low bound of linear karboricity of balanced complete bipartite graph in [8, 9, 10], respectively. In [23], Xue and Zuo obtained the linear (n − 1)arboricity of complete multipartite graph Kn(m). In [15], the linear 6arboricity of the graph Km,n was obtained. All the results are coherent with the corresponding cases of Conjecture 1.1. But for the general graph, this conjecture has not been proved yet.
As for a lower bound on lak(G), it is obvious that the following result holds.
Lemma 1.3. For any graph G with maximum degree △(G), then
lak (G)
≥
max{⌈
∆(G) 2
⌉,
⌈
E(G)
⌊
kV (G) k+1
⌋
⌉}.
2 Main results
Note that in the following the index of each vertex is modulo n. Our main result is the following theorem.
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Let Kn,n = G(A, B) be a balanced complete bipartite graph with partite sets A and B, where A = {a0, a1, · · · , an−1} and B = {b0, b1, · · · , bn−1}. It is deﬁned that the bipartite dif f erence of an edge apbq in [9] as the value (q − p)( mod n). The edge set of Kn,n can be partitioned into n pairwise disjoint perfect matchings M0, M1, · · · , Mn−1, where Mj is exactly the set of edges of bipartite diﬀerence j in Kn,n for j ∈ [0, n − 1].
Theorem 2.1. la4(Kn,n) = ⌈5n/8⌉ for n ≡ 0( mod 5).
Proof. Clearly, by Lemma 1.3,
la4(Kn,n) ≥ ⌈n2/⌊4 · 2n/5⌋⌉ = ⌈5n/8⌉,
so we need only to prove the upper bound.
It is easy to see that the following Claim 1 holds.
Claim 1. For each t ∈ [0, ⌊n/8⌋−1], the edges of M8t+2q∪ M8t+2q+1 other than that in ct,q can form one linear 4forest, where
ct,q = {a2+2t+5(i−1)b2+2t+5(i−1)+8t+2q , a4+2t+5(i−1)b4+2t+5(i−1)+8t+2q+1i ∈ [1, n/5]}
for q ∈ {0, 2}, and
ct,q = {a3+2t+5(i−1)b3+2t+5(i−1)+8t+2q , a5+2t+5(i−1)b5+2t+5(i−1)+8t+2q+1i ∈ [1, n/5]}
for q ∈ {1, 3}.
For example, if n = 40, the edges of M0 ∪ M1 other than that in
c0,0 = {a2+5(i−1)b2+5(i−1), a4+5(i−1)b4+5(i−1)+1i ∈ [1, 8]}
can form one linear 4forest (please see Fig. 1). Similarly, the edges of M2 ∪ M3 other than that in
c0,1 = {a3+5(i−1)b3+5(i−1)+2, a5+5(i−1)b5+5(i−1)+3i ∈ [1, 8]}
can form one linear 4forest. The edges of M4 ∪ M5 other than edges in
c0,2 = {a2+5(i−1)b2+5(i−1)+4, a4+5(i−1)b4+5(i−1)+5i ∈ [1, 8]}
can form one linear 4forest. The edges of M6 ∪ M7 other than edges in
c0,3 = {a3+5(i−1)b3+5(i−1)+6, a5+5(i−1)b5+5(i−1)+7i ∈ [1, 8]}
can form another linear 4forest, and so on.
Claim 2. la4(Kn,n) ≤ 5n/8 for n ≡ 0( mod 8) and n ≡ 0( mod 5).
By Claim 1, we have obtained n/2 linear 4forests in total. Thus we have to estimate the number of linear 4forests induced by the union of ct,q.
It is easy to verify that all edges of ∪q3=0ct,q can form n/5 pairwise disjoint paths P5
{b5+2t+5(i−1)+8t+3a5+2t+5(i−1)b5+2t+5(i−1)+8t+7 a5+2t+5(i−1)+7b2+2t+5(i−1)+8t+4+10i ∈ [1, n/5]}
and n/5 pairwise disjoint 4cycles
{a3+2t+5(i−1)b3+2t+5(i−1)+8t+2a4+2t+5(i−1) b4+2t+5(i−1)+8t+5a3+2t+5(i−1)i ∈ [1, n/5]},
for each t ∈ [0, n/8 − 1]. Moreover, for each t ∈ [0, n/8 − 1], we obtain one linear 4forest by deleting edges in
{a4+2t+5(i−1)b4+2t+5(i−1)+8t+1i ∈ [1, n/5]}
of 4cycles of ∪3q=0ct,q, adding edges of
{aubu+8tl+9u = 2tl + 5i + 1, i ∈ [1, n/5]}
to the remained edges of ∪q3=0ctl,q, and adding edges of
{aubu+8(5l−5)+1u = 2(5l − 5) + 5i − 1, i ∈ [1, n/5]}
to the remained edges of ∪3q=0c(5l−1),q, where tl ∈ [5(l − 1), 5(l − 1) + 3] and l ∈ [1, n/40].
Hence, la4(Kn,n) ≤ n/2 + n/8 = 5n/8 in the case of n ≡ 0( mod 40). For example, if n = 40, for t = 0, the edges of ∪q3=0c0,q can form 8 pairwise disjoint paths P5
{b5i+3a5ib5i+7a5i+7b5i+11i ∈ [1, 8]}
and 8 pairwise disjoint 4cycles
{a5i−2b5ia5i−1b5i+4a5i−2i ∈ [1, 8]}.
(Please see Figure 2. Note that the paths P5 of ∪q3=0c0,q have not been displayed in the ﬁgure.) Moreover, for t = 0, we obtain one linear 4forest by deleting edges in
{a4+5(i−1)b4+5(i−1)+1i ∈ [1, 8]}
of 4cycles of ∪3q=0c0,q and adding edges of
{a4+2+5(i−1)b4+2+5(i−1)+8+1i ∈ [1, 8]}
to the remained edges of ∪q3=0c0,q. Similarly, for t = 1, we obtain one linear 4forest by deleting edges in
{a4+2+5(i−1)b4+2+5(i−1)+8+1i ∈ [1, 8]}
of 4cycles of ∪3q=0c1,q and adding edges of
{a4+2·2+5(i−1)b4+2·2+5(i−1)+8·2+1i ∈ [1, 8]}
to the remained edges of ∪3q=0c1,q. For t = 2, we obtain one linear 4forest by deleting edges in
{a4+2·2+5(i−1)b4+2·2+5(i−1)+8·2+1i ∈ [1, 8]}
of 4cycles of ∪3q=0c2,q and adding edges of
{a4+2·3+5(i−1)b4+2·3+5(i−1)+8·3+1i ∈ [1, 8]}
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b0 q b1 q b2 q b3 q b4 q b5 q b6 q b7 q b8 q b9 q b10 q b11 q b12 q b13 q b14 q b15 q b16 q b17 q b18 q b19 q b20 q b21 q b22 q b23 q b24 q b25 q b26 q b27 q b28 q b29 q b30 q b31 q b32 q b33 q b34 q b35 q b36 q b37 q b38 q b39 q
Figure 1. M0 ∪ M1 with c0,0 broken and others normal
q a0 b0 q q a1 b1 q q a2 b2 q q a3 b3 q q a4 b4 q q a5 b5 q q a6 b6 q q a7 b7 q q a8 b8 q q a9 b9 q q a10 b10q q a11 b11q q a12 b12q q a13 b13q q a14 b14q q a15 b15q q a16 b16q q a17 b17q q a18 b18q q a19 b19q q a20 b20q q a21 b21q q a22 b22q q a23 b23q q a24 b24q q a25 b25q q a26 b26q q a27 b27q q a28 b28q q a29 b29q q a30 b30q q a31 b31q q a32 b32q q a33 b33q q a34 b34q q a35 b35q q a36 b36q q a37 b37q q a38 b38q q a39 b39q
Figure 2. broken edges: the edges deleted from 4cycles of ∪q3=0c0,q
heavy edges: the edges adding to the remained edges of ∪3q=0c0,q
q a0 q a1 q a2 q a3 q a4 q a5 q a6 q a7 q a8 q a9 q a10 q a11 q a12 q a13 q a14 q a15 q a16 q a17 q a18 q a19 q a20 q a21 q a22 q a23 q a24 q a25 q a26 q a27 q a28 q a29 q a30 q a31 q a32 q a33 q a34 q a35 q a36 q a37 q a38 q a39
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to the remained edges of ∪q3=0c2,q. For t = 3, we obtain one linear 4forest by deleting edges in
{a4+2·3+5(i−1)b4+2·3+5(i−1)+8·3+1i ∈ [1, 8]}
of 4cycles of ∪q3=0c3,q and adding edges of
{a4+2·4+5(i−1)b4+2·4+5(i−1)+8·4+1i ∈ [1, 8]}
to the remained edges of ∪q3=0c3,q. For t = 4, we obtain one linear 4forest by deleting edges in
{a4+2·4+5(i−1)b4+2·4+5(i−1)+8·4+1i ∈ [1, 8]}
of 4cycles of ∪q3=0c4,q and adding edges of
{a4+5(i−1)b4+5(i−1)+1i ∈ [1, 8]}
to the remained edges of ∪3q=0c4,q.
Hence la4(K40,40) ≤ 25.
Claim 3. la4(Kn,n) ≤ ⌈5n/8⌉ for n ≡ 2( mod 8) and n ≡ 0( mod 5).
The edges of Mn−2 ∪ Mn−1 other than that in
{a4+5(i−1)b2+5(i−1), a1+5(i−1)b5(i−1)i ∈ [1, n/5]}
can form one linear 4forest. Now we have obtained ⌊n/8⌋ · 4 + 1 = n/2 linear 4forests in total by Claim 1. Next we will estimate the number of linear 4forests induced by the edges that are not used in Kn,n.
It is not diﬃcult to verify that all edges of ∪3q=0ct,q can form n/5 pairwise disjoint paths P5
{b5+2t+5(i−1)+8t+3a5+2t+5(i−1)b5+2t+5(i−1)+8t+7 a5+2t+5(i−1)+7b2+2t+5(i−1)+8t+4+10i ∈ [1, n/5]}
and n/5 pairwise disjoint 4cycles
{a3+2t+5(i−1)b3+2t+5(i−1)+8t+2a4+2t+5(i−1) b4+2t+5(i−1)+8t+5a3+2t+5(i−1)i ∈ [1, n/5]},
for each t ∈ [0, ⌊n/8⌋ − 1]. Moreover, for each t ∈ [0, ⌊n/8⌋ − 2], we obtain one linear 4forest by deleting edges in
{a4+2t+5(i−1)b4+2t+5(i−1)+8t+1i ∈ [1, n/5]}
of 4cycles of ∪3q=0ct,q, adding edges of
{aubu+8tl+9u = 2tl + 5i + 1, i ∈ [1, n/5]}
to the remained edges of ∪3q=0ctl,q and adding edges of
{aubu+8(5l−5)+1u = 2(5l − 5) + 5i − 1, i ∈ [1, n/5]}
to the remained edges of ∪q3=0c(5l−1),q, where tl ∈ [5(l − 1), 5(l − 1) + 3] and l ∈ [1, ⌊⌊n/8⌋/5⌋]. For t = ⌊n/8⌋ − 1, we obtain one linear 4forest by deleting edges in
{al+5(i−1)bl+5(i−1)+8(⌊n/8⌋−1)+1i ∈ [1, n/5]}
of 4cycles of ∪3q=0c(⌊n/8⌋−1),q for l = 4 + 2(⌊n/8⌋ − 1). It is easy to verify that the remained edges, i.e., all edges of
{al+5(i−1)bl+5(i−1)+8(⌊n/8⌋−1)+1, a4+5(i−1)b2+5(i−1), a1+5(i−1)b5(i−1)i ∈ [1, n/5]}
with l = 4 + 2(⌊n/8⌋ − 1) can form one linear 4forest.
Hence, la4(Kn,n) ≤ n/2 + (⌊n/8⌋ − 1) + 1 + 1 = n/2 + ⌈n/8⌉ = ⌈5n/8⌉.
For example, if n = 10, then the edges of M0 ∪ M1 other than that in
c0,0 = {albl, a2+lb2+l+1l = 5i − 3, i ∈ [1, 2]}
can form one linear 4forest. The edges of M2 ∪ M3 other than that in
c0,1 = {albl+2, a2+lb2+l+3l = 5i − 2, i ∈ [1, 2]}
can form one linear 4forest. The edges of M4 ∪ M5 other than that in
c0,2 = {albl+4, a2+lb2+l+5l = 5i − 3, i ∈ [1, 2]}
can form one linear 4forest. The edges of M6 ∪ M7 other than that in
c0,3 = {albl+6, a2+lb2+l+7l = 5i − 2, i ∈ [1, 2]}
can form one linear 4forest. Clearly, the edges of ∪q3=0c0,q produce two disjoint paths P5
{bl+3albl+7al+7bl+11l = 5i, i ∈ [1, 2]}
and two disjoint 4cycles
{atbt+2a1+tb1+t+5att = 5i − 2, i ∈ [1, 2]}.
We obtain one linear 4forest by deleting edges
{a4+5(i−1)b4+5(i−1)+1i ∈ [1, 2]}
of 4cycles of ∪q3=0c0,q. Moreover, the edges of M8 ∪ M9 other than that in
{a4+5(i−1)b2+5(i−1), a1+5(i−1)b5(i−1)i ∈ [1, 2]}
can form one linear 4forest. It is not diﬃcult to verify that the remained edges, i.e., all edges of
{a4+lb4+l+1, a4+lb2+l, a1+lbll = 5(i − 1), i ∈ [1, 2]}
can form one linear 4forest. Hence, la4(K10,10) ≤ 7.
Claim 4. la4(Kn,n) ≤ ⌈5n/8⌉ for n ≡ 4( mod 8) and n ≡ 0( mod 5).
The edges of Mn−4 ∪ Mn−3 other than that in
{a6+5(i−1)b2+5(i−1), a3+5(i−1)b5(i−1)i ∈ [1, n/5]}
can form one linear 4forest, and the edges of Mn−2 ∪ Mn−1 other than that in
{a4+5(i−1)b2+5(i−1), a1+5(i−1)b5(i−1)i ∈ [1, n/5]}
can form another one. Now we have obtained ⌊n/8⌋ · 4 + 2 = n/2 linear 4forests in total by Claim 1. In the following we will estimate the number of linear 4forests induced by the edges that are not used in Kn,n.
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It is not diﬃcult to verify that all edges of ∪q3=0ct,q can form n/5 pairwise disjoint paths P5
{bl+3al−8tbl+7al−8t+7bl+11l = 10t + 5i, i ∈ [1, n/5]}
and n/5 pairwise disjoint 4cycles
{albl+8t+2a1+lbl+8t+6all = 2t + 5i − 2, i ∈ [1, n/5]}
for each t ∈ [0, ⌊n/8⌋ − 1]. Moreover, for each t ∈ [0, ⌊n/8⌋ − 3], we obtain one linear 4forest by deleting edges in
{a4+2t+5(i−1)b4+2t+5(i−1)+8t+1i ∈ [1, n/5]} of 4cycles of ∪q3=0ct,q, adding edges of
{aubu+8(tl+1)+1u = 2tl + 5i + 1, i ∈ [1, n/5]} to the remained edges of ∪3q=0ctl,q and adding edges of
{aubu+8(5l−5)+1u = 2(5l − 5) + 5i − 1, i ∈ [1, n/5]} to the remained edges of ∪3q=0c(5l−1),q, where tl ∈ [5(l − 1), 5(l − 1) + 3] and l ∈ [1, ⌊⌊n/8⌋/5⌋]. For t = ⌊n/8⌋ − 2, we obtain one linear 4forest by deleting edges in
{aubu+8⌊n/8⌋−15u = 2⌊n/8⌋ + 5(i − 1), i ∈ [1, n/5]} of 4cycles of ∪q3=0c(⌊n/8⌋−2),q and adding edges of
{a1+5(i−1)b5(i−1)i ∈ [1, n/5]}
of Mn−1 that have not been used. For t = ⌊n/8⌋ − 1, we also obtain one linear 4forest by deleting edges in
{aubu+8⌊n/8⌋−6u = 2⌊n/8⌋ + 5i − 4, i ∈ [1, n/5]} of 4cycles of ∪3q=0c(⌊n/8⌋−1),q and adding edges of
{a3+5(i−1)b5(i−1)i ∈ [1, n/5]}
of Mn−3 that have not been used. It is obvious that the remained edges, i.e., all edges of
{aubu+8(⌊n/8⌋−2)+1, a4+5(i−1)b2+5(i−1), a6+5(i−1)b2+5(i−1), au−1bu+8(⌊n/8⌋−2)+1 u = 2⌊n/8⌋ + 5(i − 1), i ∈ [1, n/5]},
can form one linear 4forest.
Hence, la4(Kn,n) ≤ n/2 + (⌊n/8⌋ − 2) + 1 + 1 + 1 = n/2 + ⌈n/8⌉ = ⌈5n/8⌉.
Claim 5. la4(Kn,n) ≤ ⌈5n/8⌉ for n ≡ 6( mod 8) and n ≡ 0( mod 5).
The edges of Mn−6 ∪ Mn−5 other than that in
{a8+5(i−1)b2+5(i−1), a5+5(i−1)b5(i−1)i ∈ [1, n/5]}
can form one linear 4forest, the edges of Mn−4 ∪ Mn−3 other than that in
{a7+5(i−1)b4+5(i−1), a5+5(i−1)b1+5(i−1)i ∈ [1, n/5]}
can form one linear 4forest, and the edges of Mn−2 ∪ Mn−1 other than edges in
{a4+5(i−1)b2+5(i−1), a1+5(i−1)b5(i−1)i ∈ [1, n/5]}
can form another one. Now we have obtained ⌊n/8⌋ · 4 + 3 = n/2 linear 4forests in total by Claim 1. Thus we have to estimate the number of linear 4forests induced by the edges that are not used in Kn,n.
It is not diﬃcult to see that all edges of ∪3q=0ct,q can form n/5 pairwise disjoint paths P5
{bu+8t+3aubu+8t+7au+7bu+8t+11
u = 2t + 5i i ∈ [1, n/5]
}
and n/5 pairwise disjoint 4cycles
{aubu+8t+2au+1bu+8t+6au
u = 2t + 5i − 2 i ∈ [1, n/5]
}
for each t ∈ [0, ⌊n/8⌋ − 1]. Moreover, for each t ∈ [0, ⌊n/8⌋ − 4], we obtain one linear 4forest by deleting edges in
{a4+2t+5(i−1)b4+2t+5(i−1)+8t+1i ∈ [1, n/5]}
of 4cycles of ∪3q=0ct,q, adding edges of
{aubu+8(tl+1)+1u = 2tl + 5i + 1, i ∈ [1, n/5]}
to the remained edges of ∪3q=0ctl,q and adding edges of
{aubu+8(5l−5)+1u = 10l + 5i − 11, i ∈ [1, n/5]}
to the remained edges of ∪3q=0c(5l−1),q, where tl ∈ [5(l − 1), 5(l − 1) + 3] and l ∈ [1, ⌊⌊n/8⌋/5⌋]. For t = ⌊n/8⌋ − 1, we obtain one linear 4forest by deleting edges in
{aubu+8⌊n/8⌋−7u = 2⌊n/8⌋ + 5i − 3, i ∈ [1, n/5]}
of 4cycles of ∪3q=0c(⌊n/8⌋−1),q and adding edges of {a5+5(i−1)b5(i−1)i ∈ [1, n/5]} of Mn−5 that have not been used. For each t ∈ {⌊n/8⌋ − 2, ⌊n/8⌋ − 3}, we obtain one linear 4forest by deleting edges in
{aubu+8t+1u = 2t + 5i − 1, i ∈ [1, n/5]}
of 4cycles of ∪q3=0ct,q and adding edges of
{aubu+8(t+1)+1u = 2t + 5i + 1, i ∈ [1, n/5]}
that have been deleted from ∪q3=0c(t+1),q. It is easy to verify that the remained edges, i.e., all edges of
{aubu+8(⌊n/8⌋−3)+1, a7+5(i−1)b4+5(i−1), a8+5(i−1)b2+5(i−1), a5+5(i−1)b1+5(i−1), a1+5(i−1)b5(i−1), a4+5(i−1)b2+5(i−1) u = 2⌊n/8⌋ + 5i − 7, i ∈ [1, n/5]},
can form one linear 4forest.
Hence, la4(Kn,n) ≤ n/2 + (⌊n/8⌋ − 3) + 1 + 1 + 1 + 1 = n/2 + ⌈n/8⌉ = ⌈5n/8⌉.
Claim 6. la4(Kn,n) ≤ ⌈5n/8⌉ for n ≡ 1( mod 8) and n ≡ 0( mod 5).
Now we have obtained ⌊n/8⌋·4 = ⌊n/2⌋ linear 4forests in
total by Claim 1. Note that the edges of Mn−1 also have not been used. Thus we have to estimate the number of
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linear 4forests induced by the edges that are not used in Kn,n.
It is obvious that all edges of ∪q3=0ct,q can form n/5 pairwise disjoint paths P5
{bu+3au−8tbu+7au−8t+7bu+11
u = 10t + 5i i ∈ [1, n/5]
}
and n/5 pairwise disjoint 4cycles
It is not diﬃcult to verify that all edges of ∪3q=0ct,q can form n/5 pairwise disjoint paths P5
{bu+8t+3aubu+8t+7au+7bu+8t+11
u = 2t + 5i i ∈ [1, n/5]
}
and n/5 pairwise disjoint 4cycles
{aubu+8t+2au+1bu+8t+6au
u = 2t + 5i − 2 i ∈ [1, n/5]
}
{aubu+8t+2au+1bu+8t+6au
u = 2t + 5i − 2 i ∈ [1, n/5]
}
for each t ∈ [0, ⌊n/8⌋ − 1]. Moreover, for each t ∈ [0, ⌊n/8⌋ − 4], we obtain one linear 4forest by deleting edges in
{a4+2t+5(i−1)b4+2t+5(i−1)+8t+1i ∈ [1, n/5]}
of 4cycles of ∪q3=0ct,q, adding edges of
{aubu+8(tl+1)+1u = 2tl + 5i + 1, i ∈ [1, n/5]}
to the remained edges of ∪q3=0ctl,q, and adding edges of
{aubu+8(5l−5)+1u = 2(5l − 5) + 5i − 1, i ∈ [1, n/5]}
to the remained edges of ∪3q=0c(5l−1),q, where tl ∈ [5(l − 1), 5(l − 1) + 3] and l ∈ [1, ⌊⌊n/8⌋/5⌋]. For t = ⌊n/8⌋ − 1, we obtain one linear 4forest by deleting edges in
{aubu+8⌊n/8⌋−7u = 2⌊n/8⌋ + 5i − 3, i ∈ [1, n/5]}
of 4cycles of ∪3q=0c(⌊n/8⌋−1),q. For each t ∈ {⌊n/8⌋ − 2, ⌊n/8⌋ − 3}, we obtain one linear 4forest by deleting edges in
{a4+2t+5(i−1)b4+2t+5(i−1)+8t+1i ∈ [1, n/5]}
of 4cycles of ∪3q=0ct,q and adding edges of
{aubu+8(t+1)+1u = 2t + 5i + 1, i ∈ [1, n/5]}
that have been deleted from ∪q3=0c(t+1),q. It is easy to verify that the remained edges, i.e., all edges of
{aubu+8⌊n/8⌋−23
u = 2⌊n/8⌋ + 5i − 7 i ∈ [1, n/5]
} ∪ Mn−1,
can form another one.
Hence, la4(Kn,n) ≤ ⌊n/2⌋ + (⌊n/8⌋ − 3) + 1 + 1 + 1 + 1 = ⌈n/2⌉ + ⌊n/8⌋ = ⌈5n/8⌉.
Claim 7. la4(Kn,n) ≤ ⌈5n/8⌉ for n ≡ 3( mod 8) and n ≡ 0( mod 5).
The edges of Mn−3 ∪ Mn−2 other than that in
{a5+5(i−1)b2+5(i−1), a2+5(i−1)b5(i−1)i ∈ [1, n/5]}
can form one linear 4forest. Now we have obtained ⌊n/8⌋ · 4 + 1 = ⌊n/2⌋ linear 4forests in total by Claim 1. Note that the edges of Mn−1 also have not been used. Next we will estimate the number of linear 4forests induced by the edges that are not used in Kn,n.
for each t ∈ [0, ⌊n/8⌋ − 1]. Moreover, for each t ∈ [0, ⌊n/8⌋ − 5], we obtain one linear 4forest by deleting edges in
{aubu+8t+1u = 2t + 5i − 1, i ∈ [1, n/5]}
of 4cycles of ∪3q=0ct,q, adding edges of
{aubu+8(tl+1)+1u = 2tl + 5i + 1, i ∈ [1, n/5]}
to the remained edges of ∪3q=0ctl,q and adding edges of
{aubu+8(5l−5)+1u = 2(5l − 5) + 5i − 1, i ∈ [1, n/5]}
to the remained edges of ∪3q=0c(5l−1),q, where tl ∈ [5(l − 1), 5(l − 1) + 3] and l ∈ [1, ⌊⌊n/8⌋/5⌋]. For t = ⌊n/8⌋ − 1, we obtain one linear 4forest by deleting edges in
{aubu+8⌊n/8⌋−7u = 2⌊n/8⌋ + 5i − 3, i ∈ [1, n/5]}
of 4cycles of ∪3q=0c(⌊n/8⌋−1),q and adding edges of {a2+5(i−1)b5(i−1)i = 1, 2, · · · , n/5} of Mn−2 that have not been used. For each t ∈ [⌊n/8⌋ − 2, ⌊n/8⌋ − 4], we obtain one linear 4forest by deleting edges in
{a4+2t+5(i−1)b4+2t+5(i−1)+8t+1i ∈ [1, n/5]}
of 4cycles of ∪q3=0ct,q and adding edges of
{aubu+8(t+1)+1u = 2t + 5i + 1, i ∈ [1, n/5]}
that have been deleted from ∪q3=0c(t+1),q. It is easy to verify that the remained edges, i.e., all edges of
{aubu+8⌊n/8⌋−31, a5ib5i−3 u = 2⌊n/8⌋ + 5i − 9, i ∈ [1, n/5]}
∪
Mn−1,
can form another linear 4forest.
Hence, la4(Kn,n) ≤ ⌊n/2⌋ + (⌊n/8⌋ − 4) + 1 + 3 + 1 = ⌈n/2⌉ + ⌊n/8⌋ = ⌈5n/8⌉.
Claim 8. la4(Kn,n) ≤ ⌈5n/8⌉ for n ≡ 5( mod 8) and n ≡ 0( mod 5).
The edges of Mn−5 ∪ Mn−4 other than that in
{aubu+n−5, au+2bu+n−2u = 5i − 3, i ∈ [1, n/5]}
can form one linear 4forest, the edges of Mn−3 ∪ Mn−2 other than that in
{aubu+n−3, au+2bu+nu = 5i − 2, i ∈ [1, n/5]}
can form one and the edges of Mn−1 can form another one. Now we have obtained ⌊n/8⌋ · 4 + 3 = ⌈n/2⌉ linear 4forests in total by Claim 1. Thus we have to estimate
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the number of linear 4forests induced by the edges that are not used in Kn,n.
It is not diﬃcult to verify that all edges of ∪3q=0ct,q can form n/5 pairwise disjoint paths P5
{bu+3au−8tbu+7au−8t+7bu+11
u = 10t + 5i i ∈ [1, n/5]
}
and n/5 pairwise disjoint 4cycles
{aubu+8t+2au+1bu+8t+6au
u = 2t + 5i − 2 i ∈ [1, n/5]
}
for each t ∈ [0, ⌊n/8⌋ − 1]. Moreover, for each t ∈ [0, ⌊n/8⌋ − 1], we obtain one linear 4forest by deleting edges in
{a4+2t+5(i−1)b4+2t+5(i−1)+8t+1i ∈ [1, n/5]}
of 4cycles of ∪q3=0ct,q, adding edges of
{aubu+8tl+9u = 2tl + 5i + 1, i ∈ [1, n/5]}
to the remained edges of ∪q3=0ctl,q, and adding edges of
{aubu+8(5l−5)+1u = 2(5l − 5) + 5i − 1, i ∈ [1, n/5]}
to the remained edges of ∪3q=0c(5l−1),q, where tl ∈ [5(l − 1), 5(l − 1) + 3] and l ∈ [1, ⌊n/8⌋/5] (note that ⌊n/8⌋ ≡ 0( mod 5) in this case). It is easy to verify that the remained edges, i.e., all edges of
{aubu+n−5, au+2bu+n−2, au+1bu+n−2, au+3bu+n+1 u = 5i − 3, i ∈ [1, n/5]},
can form another linear 4forest.
Hence, la4(Kn,n) ≤ ⌈n/2⌉ + ⌊n/8⌋ + 1 = ⌈5n/8⌉.
Claim 9. la4(Kn,n) ≤ ⌈5n/8⌉ for n ≡ 7( mod 8) and n ≡ 0( mod 5).
The edges of Mn−7 ∪ Mn−6 other than that in
{a9+5(i−1)b2+5(i−1), a6+5(i−1)b5(i−1)i ∈ [1, n/5]}
can form one linear 4forest, the edges of Mn−5 ∪ Mn−4 other than that in
and n/5 pairwise disjoint 4cycles
{aubu+8t+2au+1bu+8t+6au
u = 2t + 5i − 2 i ∈ [1, n/5]
}
for each t ∈ [0, ⌊n/8⌋ − 1]. Moreover, for each t ∈ [0, ⌊n/8⌋ − 2], we obtain one linear 4forest by deleting edges in
{a4+2t+5(i−1)b4+2t+5(i−1)+8t+1i ∈ [1, n/5]}
of 4cycles of ∪q3=0ct,q, adding edges of
{aubu+8(tl+1)+1u = 2tl + 5i + 1, i ∈ [1, n/5]}
to the remained edges of ∪3q=0ctl,q, and adding edges of
{aubu+8(5l−5)+1u = 2(5l − 5) + 5i − 1, i ∈ [1, n/5]}
to the remained edges of ∪3q=0c(5l−1),q, where tl ∈ [5(l − 1), 5(l − 1) + 3] and l ∈ [1, ⌊⌊n/8⌋/5⌋]. For t = ⌊n/8⌋ − 1, we obtain one linear 4forest by deleting edges in
{aubu+8⌊n/8⌋−7u = 2⌊n/8⌋ + 5i − 3, i ∈ [1, n/5]} of 4cycles of ∪q3=0c(⌊n/8⌋−1),q, and adding edges of
{a6+5(i−1)b5(i−1)i ∈ [1, n/5]}
of Mn−6 that have not been used. It is obvious that all edges of
{aubu+8⌊n/8⌋−7
u = 2⌊n/8⌋ + 5i − 3 i ∈ [1, n/5]
} ∪ Mn−1
can form one linear 4forest, and the remained edges, i.e., all edges of
{a5i+4b5i−3, a5i+3b5i−2, a5(i+1)b5i+1, a5i−2b5(i−1), a5ib5i−2i ∈ [1, n/5]},
can form another one.
Hence, la4(Kn,n) ≤ ⌊n/2⌋ + (⌊n/8⌋ − 1) + 1 + 1 + 1 = ⌈n/2⌉ + ⌈n/8⌉ = ⌈5n/8⌉.
Combining Claims 2 − 9, the theorem is proved completely.
{a5i+3b5i−2, a5(i+1)b5i+1i ∈ [1, n/5]}
can form one, and the edges of Mn−3 ∪ Mn−2 other than that in
{a3+5(i−1)b5(i−1), a5+5(i−1)b3+5(i−1)i ∈ [1, n/5]}
can form another one. Note that the edges of Mn−1 have not been used. Now we have obtained ⌊n/8⌋ · 4 + 3 = ⌊n/2⌋ linear 4forests in total by Claim 1. In the following we will estimate the number of linear 4forests induced by the edges that are not used in Kn,n.
It is not diﬃcult to obtain that all edges of ∪3q=0ct,q can form n/5 pairwise disjoint paths P5
{bu+8t+3aubu+8t+7au+7bu+8t+11
u = 2t + 5i i ∈ [1, n/5]
}
References
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[10] H. L. Fu, K. C. Huang, The linear 2arboricity of complete bipartite graphs, Ars Combin. 38(1994), 309318.
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[12] M. Habib, B. Peroche, Some problems about linear arboricity, Discrete Math. 41(1982), 219220.
[13] M. Habib, B. Peroche, La karboricite´ line´aire des arbres, Ann. Discrete Math.17(1983), 307317.
[14] F. Harary, Covering and packing in graphs I, Ann. New York Acad. Sci. 175(1970),198205.
[15] S. He and L. Zuo, The linear 6arboricity of the complete bipartite graph Km,n, Discrete Mathematics, Algorithms and Applications,Vol 5,No 4,(2013) 1350029.
[16] B. Jackson, N. C. Wormald, On the linear karboricity of cubic graphs, Discrete Math. 162(1996), 293297.
[17] K. W. Lih, L. D. Tong, W. F. Wang, The linear 2arboricity of planar graphs, Graphs Combin. 19(2003), 241248.
[18] J. Liu and X. Zhang, Cubeconnected complete graphs, IAENG International Journal of Applied Mathematics, 44:3, pp.134136, Aug. 2014.
[19] C. Thomassen, Twocoloring the edges of a cubic graph such that each monochromatic component is a path of length at most 5, J. Combin. Theory, Ser B. 75(1999), 100109.
[20] D. B. West, Introduction to Graph Theory, second ed., PrenticeHall, Upper Saddle River,NJ,2001.
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The Linear 4arboricity of Balanced Complete Bipartite Graphs ∗
Liancui Zuo,† Shengjie He, and Ranqun Wang
Abstract
A linear kforest is a graph whose components are paths of length at most k. The linear karboricity of a graph G, denoted by lak(G), is the least number of linear kforests needed to decompose G. In this paper, it is obtained that la4(Kn,n) = ⌈5n/8⌉ for n ≡ 0( mod 5).
Keywords: Linear kforest; linear karboricity; complete bipartite graph; perfect matching; bipartite diﬀerence
a matching, and la1(G) is the edge chromatic number, or chromatic index, χ′(G) of a graph G. Moreover, the linear k−arboricity lak(G) is also a reﬁnement of the ordinary linear arboricity la(G)(or la∞(G)) [14] of a graph G, which is the case when every component of each forest is a path with no length constraint. In 1982, Habib and Peroche [12] proposed the following conjecture for an upper bound on lak(G).
Conjecture 1.1. If G is a graph with maximum degree ∆(G) and k ≥ 2, then
1 Introduction
In this paper, all graphs considered are ﬁnite, undirected, and simple (i. e., loopless and without multiple edges). We refer to [20] for terminology in graph theory. In recent years, many parameters and classes of graphs were studied. For example, in [11], diﬀerent properties of the intrinsic order graph were obtained, namely those dealing with its edges, chains, shadows, neighbors and degrees of its vertices, and some relevant subgraphs, as well as the natural isomorphisms between them. In [18], the ndimensional cubeconnected complete graph was studied. In [24, 25], the hamiltonicity, path tcoloring, and the shortest paths of Sierpin´skilike graphs were researched. In [26], the vertex arboricity of integer distance graph G(Dm,k) was obtained.
A decomposition of a graph is a list of subgraphs such that each edge appears in exactly one subgraph in the list. If a graph G has a decomposition G1, G2, · · · , Gd, then we say that G1, G2, · · · , Gd decompose G, or G can be decomposed into G1, G2, · · · , Gd. A linear k−forest is a forest whose components are paths of length at most k. The linear k−arboricity of a graph G, denoted by lak(G), is the least number of linear k−forests needed to decompose G. Let x be a real number, denoted by ⌊x⌋ the maximum integer no more than x, and denoted by ⌈x⌉ the minimum integer no less than x. For any integers a < b, let [a, b] denote the set of integers {a, a + 1, · · · , b} for simplicity. For any positive integer λ, let Pλ be a path on λ vertices which has length λ − 1.
The notion of linear k−arboricity was ﬁrst introduced by Habib and Peroche [13], which is a natural generalization of edge coloring. Clearly, a linear 1forest is induced by
∗Manuscript received August 4,2014. This work was supported by NSFC for youth with code 61103073.
†Liancui Zuo, Shengjie He and Ranqun Wang are with College of Mathematical Science, Tianjin Normal University, Tianjin, 300387, China. Email: lczuo@mail.tjnu.edu.cn;lczuo@163.com
⌈
∆(G)·V (G)
2⌊
k·V (G) k+1
⌋
⌉,
if
∆(G)
=
V
(G)

−1,
lak(G) ≤
⌈
∆(G)·V (G)+1
2⌊
k·V (G) k+1
⌋
⌉,
if
∆(G)
<
V
(G)

−1.
For k = V (G)  −1, it is the Akiyama’s conjecture [1].
Conjecture
1.2.
[1]
la(G)
≤
⌈
∆(G)+1 2
⌉.
So far, quite a few results on the veriﬁcation of Conjecture 1.1 have obtained in the literature, especially for graphs with particular structures, such as trees [5, 6, 13], cubic graphs [4, 16, 19], regular graphs [2, 3], planar graphs [17], balanced complete bipartite graphs [8, 9, 10], balanced complete multipartite graphs [22] and complete graphs [5, 7, 8, 9, 21]. It is obtained that the linear 2arboricity, the linear 3arboricity and the low bound of linear karboricity of balanced complete bipartite graph in [8, 9, 10], respectively. In [23], Xue and Zuo obtained the linear (n − 1)arboricity of complete multipartite graph Kn(m). In [15], the linear 6arboricity of the graph Km,n was obtained. All the results are coherent with the corresponding cases of Conjecture 1.1. But for the general graph, this conjecture has not been proved yet.
As for a lower bound on lak(G), it is obvious that the following result holds.
Lemma 1.3. For any graph G with maximum degree △(G), then
lak (G)
≥
max{⌈
∆(G) 2
⌉,
⌈
E(G)
⌊
kV (G) k+1
⌋
⌉}.
2 Main results
Note that in the following the index of each vertex is modulo n. Our main result is the following theorem.
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Let Kn,n = G(A, B) be a balanced complete bipartite graph with partite sets A and B, where A = {a0, a1, · · · , an−1} and B = {b0, b1, · · · , bn−1}. It is deﬁned that the bipartite dif f erence of an edge apbq in [9] as the value (q − p)( mod n). The edge set of Kn,n can be partitioned into n pairwise disjoint perfect matchings M0, M1, · · · , Mn−1, where Mj is exactly the set of edges of bipartite diﬀerence j in Kn,n for j ∈ [0, n − 1].
Theorem 2.1. la4(Kn,n) = ⌈5n/8⌉ for n ≡ 0( mod 5).
Proof. Clearly, by Lemma 1.3,
la4(Kn,n) ≥ ⌈n2/⌊4 · 2n/5⌋⌉ = ⌈5n/8⌉,
so we need only to prove the upper bound.
It is easy to see that the following Claim 1 holds.
Claim 1. For each t ∈ [0, ⌊n/8⌋−1], the edges of M8t+2q∪ M8t+2q+1 other than that in ct,q can form one linear 4forest, where
ct,q = {a2+2t+5(i−1)b2+2t+5(i−1)+8t+2q , a4+2t+5(i−1)b4+2t+5(i−1)+8t+2q+1i ∈ [1, n/5]}
for q ∈ {0, 2}, and
ct,q = {a3+2t+5(i−1)b3+2t+5(i−1)+8t+2q , a5+2t+5(i−1)b5+2t+5(i−1)+8t+2q+1i ∈ [1, n/5]}
for q ∈ {1, 3}.
For example, if n = 40, the edges of M0 ∪ M1 other than that in
c0,0 = {a2+5(i−1)b2+5(i−1), a4+5(i−1)b4+5(i−1)+1i ∈ [1, 8]}
can form one linear 4forest (please see Fig. 1). Similarly, the edges of M2 ∪ M3 other than that in
c0,1 = {a3+5(i−1)b3+5(i−1)+2, a5+5(i−1)b5+5(i−1)+3i ∈ [1, 8]}
can form one linear 4forest. The edges of M4 ∪ M5 other than edges in
c0,2 = {a2+5(i−1)b2+5(i−1)+4, a4+5(i−1)b4+5(i−1)+5i ∈ [1, 8]}
can form one linear 4forest. The edges of M6 ∪ M7 other than edges in
c0,3 = {a3+5(i−1)b3+5(i−1)+6, a5+5(i−1)b5+5(i−1)+7i ∈ [1, 8]}
can form another linear 4forest, and so on.
Claim 2. la4(Kn,n) ≤ 5n/8 for n ≡ 0( mod 8) and n ≡ 0( mod 5).
By Claim 1, we have obtained n/2 linear 4forests in total. Thus we have to estimate the number of linear 4forests induced by the union of ct,q.
It is easy to verify that all edges of ∪q3=0ct,q can form n/5 pairwise disjoint paths P5
{b5+2t+5(i−1)+8t+3a5+2t+5(i−1)b5+2t+5(i−1)+8t+7 a5+2t+5(i−1)+7b2+2t+5(i−1)+8t+4+10i ∈ [1, n/5]}
and n/5 pairwise disjoint 4cycles
{a3+2t+5(i−1)b3+2t+5(i−1)+8t+2a4+2t+5(i−1) b4+2t+5(i−1)+8t+5a3+2t+5(i−1)i ∈ [1, n/5]},
for each t ∈ [0, n/8 − 1]. Moreover, for each t ∈ [0, n/8 − 1], we obtain one linear 4forest by deleting edges in
{a4+2t+5(i−1)b4+2t+5(i−1)+8t+1i ∈ [1, n/5]}
of 4cycles of ∪3q=0ct,q, adding edges of
{aubu+8tl+9u = 2tl + 5i + 1, i ∈ [1, n/5]}
to the remained edges of ∪q3=0ctl,q, and adding edges of
{aubu+8(5l−5)+1u = 2(5l − 5) + 5i − 1, i ∈ [1, n/5]}
to the remained edges of ∪3q=0c(5l−1),q, where tl ∈ [5(l − 1), 5(l − 1) + 3] and l ∈ [1, n/40].
Hence, la4(Kn,n) ≤ n/2 + n/8 = 5n/8 in the case of n ≡ 0( mod 40). For example, if n = 40, for t = 0, the edges of ∪q3=0c0,q can form 8 pairwise disjoint paths P5
{b5i+3a5ib5i+7a5i+7b5i+11i ∈ [1, 8]}
and 8 pairwise disjoint 4cycles
{a5i−2b5ia5i−1b5i+4a5i−2i ∈ [1, 8]}.
(Please see Figure 2. Note that the paths P5 of ∪q3=0c0,q have not been displayed in the ﬁgure.) Moreover, for t = 0, we obtain one linear 4forest by deleting edges in
{a4+5(i−1)b4+5(i−1)+1i ∈ [1, 8]}
of 4cycles of ∪3q=0c0,q and adding edges of
{a4+2+5(i−1)b4+2+5(i−1)+8+1i ∈ [1, 8]}
to the remained edges of ∪q3=0c0,q. Similarly, for t = 1, we obtain one linear 4forest by deleting edges in
{a4+2+5(i−1)b4+2+5(i−1)+8+1i ∈ [1, 8]}
of 4cycles of ∪3q=0c1,q and adding edges of
{a4+2·2+5(i−1)b4+2·2+5(i−1)+8·2+1i ∈ [1, 8]}
to the remained edges of ∪3q=0c1,q. For t = 2, we obtain one linear 4forest by deleting edges in
{a4+2·2+5(i−1)b4+2·2+5(i−1)+8·2+1i ∈ [1, 8]}
of 4cycles of ∪3q=0c2,q and adding edges of
{a4+2·3+5(i−1)b4+2·3+5(i−1)+8·3+1i ∈ [1, 8]}
(Advance online publication: 17 February 2015)
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b0 q b1 q b2 q b3 q b4 q b5 q b6 q b7 q b8 q b9 q b10 q b11 q b12 q b13 q b14 q b15 q b16 q b17 q b18 q b19 q b20 q b21 q b22 q b23 q b24 q b25 q b26 q b27 q b28 q b29 q b30 q b31 q b32 q b33 q b34 q b35 q b36 q b37 q b38 q b39 q
Figure 1. M0 ∪ M1 with c0,0 broken and others normal
q a0 b0 q q a1 b1 q q a2 b2 q q a3 b3 q q a4 b4 q q a5 b5 q q a6 b6 q q a7 b7 q q a8 b8 q q a9 b9 q q a10 b10q q a11 b11q q a12 b12q q a13 b13q q a14 b14q q a15 b15q q a16 b16q q a17 b17q q a18 b18q q a19 b19q q a20 b20q q a21 b21q q a22 b22q q a23 b23q q a24 b24q q a25 b25q q a26 b26q q a27 b27q q a28 b28q q a29 b29q q a30 b30q q a31 b31q q a32 b32q q a33 b33q q a34 b34q q a35 b35q q a36 b36q q a37 b37q q a38 b38q q a39 b39q
Figure 2. broken edges: the edges deleted from 4cycles of ∪q3=0c0,q
heavy edges: the edges adding to the remained edges of ∪3q=0c0,q
q a0 q a1 q a2 q a3 q a4 q a5 q a6 q a7 q a8 q a9 q a10 q a11 q a12 q a13 q a14 q a15 q a16 q a17 q a18 q a19 q a20 q a21 q a22 q a23 q a24 q a25 q a26 q a27 q a28 q a29 q a30 q a31 q a32 q a33 q a34 q a35 q a36 q a37 q a38 q a39
(Advance online publication: 17 February 2015)
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to the remained edges of ∪q3=0c2,q. For t = 3, we obtain one linear 4forest by deleting edges in
{a4+2·3+5(i−1)b4+2·3+5(i−1)+8·3+1i ∈ [1, 8]}
of 4cycles of ∪q3=0c3,q and adding edges of
{a4+2·4+5(i−1)b4+2·4+5(i−1)+8·4+1i ∈ [1, 8]}
to the remained edges of ∪q3=0c3,q. For t = 4, we obtain one linear 4forest by deleting edges in
{a4+2·4+5(i−1)b4+2·4+5(i−1)+8·4+1i ∈ [1, 8]}
of 4cycles of ∪q3=0c4,q and adding edges of
{a4+5(i−1)b4+5(i−1)+1i ∈ [1, 8]}
to the remained edges of ∪3q=0c4,q.
Hence la4(K40,40) ≤ 25.
Claim 3. la4(Kn,n) ≤ ⌈5n/8⌉ for n ≡ 2( mod 8) and n ≡ 0( mod 5).
The edges of Mn−2 ∪ Mn−1 other than that in
{a4+5(i−1)b2+5(i−1), a1+5(i−1)b5(i−1)i ∈ [1, n/5]}
can form one linear 4forest. Now we have obtained ⌊n/8⌋ · 4 + 1 = n/2 linear 4forests in total by Claim 1. Next we will estimate the number of linear 4forests induced by the edges that are not used in Kn,n.
It is not diﬃcult to verify that all edges of ∪3q=0ct,q can form n/5 pairwise disjoint paths P5
{b5+2t+5(i−1)+8t+3a5+2t+5(i−1)b5+2t+5(i−1)+8t+7 a5+2t+5(i−1)+7b2+2t+5(i−1)+8t+4+10i ∈ [1, n/5]}
and n/5 pairwise disjoint 4cycles
{a3+2t+5(i−1)b3+2t+5(i−1)+8t+2a4+2t+5(i−1) b4+2t+5(i−1)+8t+5a3+2t+5(i−1)i ∈ [1, n/5]},
for each t ∈ [0, ⌊n/8⌋ − 1]. Moreover, for each t ∈ [0, ⌊n/8⌋ − 2], we obtain one linear 4forest by deleting edges in
{a4+2t+5(i−1)b4+2t+5(i−1)+8t+1i ∈ [1, n/5]}
of 4cycles of ∪3q=0ct,q, adding edges of
{aubu+8tl+9u = 2tl + 5i + 1, i ∈ [1, n/5]}
to the remained edges of ∪3q=0ctl,q and adding edges of
{aubu+8(5l−5)+1u = 2(5l − 5) + 5i − 1, i ∈ [1, n/5]}
to the remained edges of ∪q3=0c(5l−1),q, where tl ∈ [5(l − 1), 5(l − 1) + 3] and l ∈ [1, ⌊⌊n/8⌋/5⌋]. For t = ⌊n/8⌋ − 1, we obtain one linear 4forest by deleting edges in
{al+5(i−1)bl+5(i−1)+8(⌊n/8⌋−1)+1i ∈ [1, n/5]}
of 4cycles of ∪3q=0c(⌊n/8⌋−1),q for l = 4 + 2(⌊n/8⌋ − 1). It is easy to verify that the remained edges, i.e., all edges of
{al+5(i−1)bl+5(i−1)+8(⌊n/8⌋−1)+1, a4+5(i−1)b2+5(i−1), a1+5(i−1)b5(i−1)i ∈ [1, n/5]}
with l = 4 + 2(⌊n/8⌋ − 1) can form one linear 4forest.
Hence, la4(Kn,n) ≤ n/2 + (⌊n/8⌋ − 1) + 1 + 1 = n/2 + ⌈n/8⌉ = ⌈5n/8⌉.
For example, if n = 10, then the edges of M0 ∪ M1 other than that in
c0,0 = {albl, a2+lb2+l+1l = 5i − 3, i ∈ [1, 2]}
can form one linear 4forest. The edges of M2 ∪ M3 other than that in
c0,1 = {albl+2, a2+lb2+l+3l = 5i − 2, i ∈ [1, 2]}
can form one linear 4forest. The edges of M4 ∪ M5 other than that in
c0,2 = {albl+4, a2+lb2+l+5l = 5i − 3, i ∈ [1, 2]}
can form one linear 4forest. The edges of M6 ∪ M7 other than that in
c0,3 = {albl+6, a2+lb2+l+7l = 5i − 2, i ∈ [1, 2]}
can form one linear 4forest. Clearly, the edges of ∪q3=0c0,q produce two disjoint paths P5
{bl+3albl+7al+7bl+11l = 5i, i ∈ [1, 2]}
and two disjoint 4cycles
{atbt+2a1+tb1+t+5att = 5i − 2, i ∈ [1, 2]}.
We obtain one linear 4forest by deleting edges
{a4+5(i−1)b4+5(i−1)+1i ∈ [1, 2]}
of 4cycles of ∪q3=0c0,q. Moreover, the edges of M8 ∪ M9 other than that in
{a4+5(i−1)b2+5(i−1), a1+5(i−1)b5(i−1)i ∈ [1, 2]}
can form one linear 4forest. It is not diﬃcult to verify that the remained edges, i.e., all edges of
{a4+lb4+l+1, a4+lb2+l, a1+lbll = 5(i − 1), i ∈ [1, 2]}
can form one linear 4forest. Hence, la4(K10,10) ≤ 7.
Claim 4. la4(Kn,n) ≤ ⌈5n/8⌉ for n ≡ 4( mod 8) and n ≡ 0( mod 5).
The edges of Mn−4 ∪ Mn−3 other than that in
{a6+5(i−1)b2+5(i−1), a3+5(i−1)b5(i−1)i ∈ [1, n/5]}
can form one linear 4forest, and the edges of Mn−2 ∪ Mn−1 other than that in
{a4+5(i−1)b2+5(i−1), a1+5(i−1)b5(i−1)i ∈ [1, n/5]}
can form another one. Now we have obtained ⌊n/8⌋ · 4 + 2 = n/2 linear 4forests in total by Claim 1. In the following we will estimate the number of linear 4forests induced by the edges that are not used in Kn,n.
(Advance online publication: 17 February 2015)
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It is not diﬃcult to verify that all edges of ∪q3=0ct,q can form n/5 pairwise disjoint paths P5
{bl+3al−8tbl+7al−8t+7bl+11l = 10t + 5i, i ∈ [1, n/5]}
and n/5 pairwise disjoint 4cycles
{albl+8t+2a1+lbl+8t+6all = 2t + 5i − 2, i ∈ [1, n/5]}
for each t ∈ [0, ⌊n/8⌋ − 1]. Moreover, for each t ∈ [0, ⌊n/8⌋ − 3], we obtain one linear 4forest by deleting edges in
{a4+2t+5(i−1)b4+2t+5(i−1)+8t+1i ∈ [1, n/5]} of 4cycles of ∪q3=0ct,q, adding edges of
{aubu+8(tl+1)+1u = 2tl + 5i + 1, i ∈ [1, n/5]} to the remained edges of ∪3q=0ctl,q and adding edges of
{aubu+8(5l−5)+1u = 2(5l − 5) + 5i − 1, i ∈ [1, n/5]} to the remained edges of ∪3q=0c(5l−1),q, where tl ∈ [5(l − 1), 5(l − 1) + 3] and l ∈ [1, ⌊⌊n/8⌋/5⌋]. For t = ⌊n/8⌋ − 2, we obtain one linear 4forest by deleting edges in
{aubu+8⌊n/8⌋−15u = 2⌊n/8⌋ + 5(i − 1), i ∈ [1, n/5]} of 4cycles of ∪q3=0c(⌊n/8⌋−2),q and adding edges of
{a1+5(i−1)b5(i−1)i ∈ [1, n/5]}
of Mn−1 that have not been used. For t = ⌊n/8⌋ − 1, we also obtain one linear 4forest by deleting edges in
{aubu+8⌊n/8⌋−6u = 2⌊n/8⌋ + 5i − 4, i ∈ [1, n/5]} of 4cycles of ∪3q=0c(⌊n/8⌋−1),q and adding edges of
{a3+5(i−1)b5(i−1)i ∈ [1, n/5]}
of Mn−3 that have not been used. It is obvious that the remained edges, i.e., all edges of
{aubu+8(⌊n/8⌋−2)+1, a4+5(i−1)b2+5(i−1), a6+5(i−1)b2+5(i−1), au−1bu+8(⌊n/8⌋−2)+1 u = 2⌊n/8⌋ + 5(i − 1), i ∈ [1, n/5]},
can form one linear 4forest.
Hence, la4(Kn,n) ≤ n/2 + (⌊n/8⌋ − 2) + 1 + 1 + 1 = n/2 + ⌈n/8⌉ = ⌈5n/8⌉.
Claim 5. la4(Kn,n) ≤ ⌈5n/8⌉ for n ≡ 6( mod 8) and n ≡ 0( mod 5).
The edges of Mn−6 ∪ Mn−5 other than that in
{a8+5(i−1)b2+5(i−1), a5+5(i−1)b5(i−1)i ∈ [1, n/5]}
can form one linear 4forest, the edges of Mn−4 ∪ Mn−3 other than that in
{a7+5(i−1)b4+5(i−1), a5+5(i−1)b1+5(i−1)i ∈ [1, n/5]}
can form one linear 4forest, and the edges of Mn−2 ∪ Mn−1 other than edges in
{a4+5(i−1)b2+5(i−1), a1+5(i−1)b5(i−1)i ∈ [1, n/5]}
can form another one. Now we have obtained ⌊n/8⌋ · 4 + 3 = n/2 linear 4forests in total by Claim 1. Thus we have to estimate the number of linear 4forests induced by the edges that are not used in Kn,n.
It is not diﬃcult to see that all edges of ∪3q=0ct,q can form n/5 pairwise disjoint paths P5
{bu+8t+3aubu+8t+7au+7bu+8t+11
u = 2t + 5i i ∈ [1, n/5]
}
and n/5 pairwise disjoint 4cycles
{aubu+8t+2au+1bu+8t+6au
u = 2t + 5i − 2 i ∈ [1, n/5]
}
for each t ∈ [0, ⌊n/8⌋ − 1]. Moreover, for each t ∈ [0, ⌊n/8⌋ − 4], we obtain one linear 4forest by deleting edges in
{a4+2t+5(i−1)b4+2t+5(i−1)+8t+1i ∈ [1, n/5]}
of 4cycles of ∪3q=0ct,q, adding edges of
{aubu+8(tl+1)+1u = 2tl + 5i + 1, i ∈ [1, n/5]}
to the remained edges of ∪3q=0ctl,q and adding edges of
{aubu+8(5l−5)+1u = 10l + 5i − 11, i ∈ [1, n/5]}
to the remained edges of ∪3q=0c(5l−1),q, where tl ∈ [5(l − 1), 5(l − 1) + 3] and l ∈ [1, ⌊⌊n/8⌋/5⌋]. For t = ⌊n/8⌋ − 1, we obtain one linear 4forest by deleting edges in
{aubu+8⌊n/8⌋−7u = 2⌊n/8⌋ + 5i − 3, i ∈ [1, n/5]}
of 4cycles of ∪3q=0c(⌊n/8⌋−1),q and adding edges of {a5+5(i−1)b5(i−1)i ∈ [1, n/5]} of Mn−5 that have not been used. For each t ∈ {⌊n/8⌋ − 2, ⌊n/8⌋ − 3}, we obtain one linear 4forest by deleting edges in
{aubu+8t+1u = 2t + 5i − 1, i ∈ [1, n/5]}
of 4cycles of ∪q3=0ct,q and adding edges of
{aubu+8(t+1)+1u = 2t + 5i + 1, i ∈ [1, n/5]}
that have been deleted from ∪q3=0c(t+1),q. It is easy to verify that the remained edges, i.e., all edges of
{aubu+8(⌊n/8⌋−3)+1, a7+5(i−1)b4+5(i−1), a8+5(i−1)b2+5(i−1), a5+5(i−1)b1+5(i−1), a1+5(i−1)b5(i−1), a4+5(i−1)b2+5(i−1) u = 2⌊n/8⌋ + 5i − 7, i ∈ [1, n/5]},
can form one linear 4forest.
Hence, la4(Kn,n) ≤ n/2 + (⌊n/8⌋ − 3) + 1 + 1 + 1 + 1 = n/2 + ⌈n/8⌉ = ⌈5n/8⌉.
Claim 6. la4(Kn,n) ≤ ⌈5n/8⌉ for n ≡ 1( mod 8) and n ≡ 0( mod 5).
Now we have obtained ⌊n/8⌋·4 = ⌊n/2⌋ linear 4forests in
total by Claim 1. Note that the edges of Mn−1 also have not been used. Thus we have to estimate the number of
(Advance online publication: 17 February 2015)
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linear 4forests induced by the edges that are not used in Kn,n.
It is obvious that all edges of ∪q3=0ct,q can form n/5 pairwise disjoint paths P5
{bu+3au−8tbu+7au−8t+7bu+11
u = 10t + 5i i ∈ [1, n/5]
}
and n/5 pairwise disjoint 4cycles
It is not diﬃcult to verify that all edges of ∪3q=0ct,q can form n/5 pairwise disjoint paths P5
{bu+8t+3aubu+8t+7au+7bu+8t+11
u = 2t + 5i i ∈ [1, n/5]
}
and n/5 pairwise disjoint 4cycles
{aubu+8t+2au+1bu+8t+6au
u = 2t + 5i − 2 i ∈ [1, n/5]
}
{aubu+8t+2au+1bu+8t+6au
u = 2t + 5i − 2 i ∈ [1, n/5]
}
for each t ∈ [0, ⌊n/8⌋ − 1]. Moreover, for each t ∈ [0, ⌊n/8⌋ − 4], we obtain one linear 4forest by deleting edges in
{a4+2t+5(i−1)b4+2t+5(i−1)+8t+1i ∈ [1, n/5]}
of 4cycles of ∪q3=0ct,q, adding edges of
{aubu+8(tl+1)+1u = 2tl + 5i + 1, i ∈ [1, n/5]}
to the remained edges of ∪q3=0ctl,q, and adding edges of
{aubu+8(5l−5)+1u = 2(5l − 5) + 5i − 1, i ∈ [1, n/5]}
to the remained edges of ∪3q=0c(5l−1),q, where tl ∈ [5(l − 1), 5(l − 1) + 3] and l ∈ [1, ⌊⌊n/8⌋/5⌋]. For t = ⌊n/8⌋ − 1, we obtain one linear 4forest by deleting edges in
{aubu+8⌊n/8⌋−7u = 2⌊n/8⌋ + 5i − 3, i ∈ [1, n/5]}
of 4cycles of ∪3q=0c(⌊n/8⌋−1),q. For each t ∈ {⌊n/8⌋ − 2, ⌊n/8⌋ − 3}, we obtain one linear 4forest by deleting edges in
{a4+2t+5(i−1)b4+2t+5(i−1)+8t+1i ∈ [1, n/5]}
of 4cycles of ∪3q=0ct,q and adding edges of
{aubu+8(t+1)+1u = 2t + 5i + 1, i ∈ [1, n/5]}
that have been deleted from ∪q3=0c(t+1),q. It is easy to verify that the remained edges, i.e., all edges of
{aubu+8⌊n/8⌋−23
u = 2⌊n/8⌋ + 5i − 7 i ∈ [1, n/5]
} ∪ Mn−1,
can form another one.
Hence, la4(Kn,n) ≤ ⌊n/2⌋ + (⌊n/8⌋ − 3) + 1 + 1 + 1 + 1 = ⌈n/2⌉ + ⌊n/8⌋ = ⌈5n/8⌉.
Claim 7. la4(Kn,n) ≤ ⌈5n/8⌉ for n ≡ 3( mod 8) and n ≡ 0( mod 5).
The edges of Mn−3 ∪ Mn−2 other than that in
{a5+5(i−1)b2+5(i−1), a2+5(i−1)b5(i−1)i ∈ [1, n/5]}
can form one linear 4forest. Now we have obtained ⌊n/8⌋ · 4 + 1 = ⌊n/2⌋ linear 4forests in total by Claim 1. Note that the edges of Mn−1 also have not been used. Next we will estimate the number of linear 4forests induced by the edges that are not used in Kn,n.
for each t ∈ [0, ⌊n/8⌋ − 1]. Moreover, for each t ∈ [0, ⌊n/8⌋ − 5], we obtain one linear 4forest by deleting edges in
{aubu+8t+1u = 2t + 5i − 1, i ∈ [1, n/5]}
of 4cycles of ∪3q=0ct,q, adding edges of
{aubu+8(tl+1)+1u = 2tl + 5i + 1, i ∈ [1, n/5]}
to the remained edges of ∪3q=0ctl,q and adding edges of
{aubu+8(5l−5)+1u = 2(5l − 5) + 5i − 1, i ∈ [1, n/5]}
to the remained edges of ∪3q=0c(5l−1),q, where tl ∈ [5(l − 1), 5(l − 1) + 3] and l ∈ [1, ⌊⌊n/8⌋/5⌋]. For t = ⌊n/8⌋ − 1, we obtain one linear 4forest by deleting edges in
{aubu+8⌊n/8⌋−7u = 2⌊n/8⌋ + 5i − 3, i ∈ [1, n/5]}
of 4cycles of ∪3q=0c(⌊n/8⌋−1),q and adding edges of {a2+5(i−1)b5(i−1)i = 1, 2, · · · , n/5} of Mn−2 that have not been used. For each t ∈ [⌊n/8⌋ − 2, ⌊n/8⌋ − 4], we obtain one linear 4forest by deleting edges in
{a4+2t+5(i−1)b4+2t+5(i−1)+8t+1i ∈ [1, n/5]}
of 4cycles of ∪q3=0ct,q and adding edges of
{aubu+8(t+1)+1u = 2t + 5i + 1, i ∈ [1, n/5]}
that have been deleted from ∪q3=0c(t+1),q. It is easy to verify that the remained edges, i.e., all edges of
{aubu+8⌊n/8⌋−31, a5ib5i−3 u = 2⌊n/8⌋ + 5i − 9, i ∈ [1, n/5]}
∪
Mn−1,
can form another linear 4forest.
Hence, la4(Kn,n) ≤ ⌊n/2⌋ + (⌊n/8⌋ − 4) + 1 + 3 + 1 = ⌈n/2⌉ + ⌊n/8⌋ = ⌈5n/8⌉.
Claim 8. la4(Kn,n) ≤ ⌈5n/8⌉ for n ≡ 5( mod 8) and n ≡ 0( mod 5).
The edges of Mn−5 ∪ Mn−4 other than that in
{aubu+n−5, au+2bu+n−2u = 5i − 3, i ∈ [1, n/5]}
can form one linear 4forest, the edges of Mn−3 ∪ Mn−2 other than that in
{aubu+n−3, au+2bu+nu = 5i − 2, i ∈ [1, n/5]}
can form one and the edges of Mn−1 can form another one. Now we have obtained ⌊n/8⌋ · 4 + 3 = ⌈n/2⌉ linear 4forests in total by Claim 1. Thus we have to estimate
(Advance online publication: 17 February 2015)
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the number of linear 4forests induced by the edges that are not used in Kn,n.
It is not diﬃcult to verify that all edges of ∪3q=0ct,q can form n/5 pairwise disjoint paths P5
{bu+3au−8tbu+7au−8t+7bu+11
u = 10t + 5i i ∈ [1, n/5]
}
and n/5 pairwise disjoint 4cycles
{aubu+8t+2au+1bu+8t+6au
u = 2t + 5i − 2 i ∈ [1, n/5]
}
for each t ∈ [0, ⌊n/8⌋ − 1]. Moreover, for each t ∈ [0, ⌊n/8⌋ − 1], we obtain one linear 4forest by deleting edges in
{a4+2t+5(i−1)b4+2t+5(i−1)+8t+1i ∈ [1, n/5]}
of 4cycles of ∪q3=0ct,q, adding edges of
{aubu+8tl+9u = 2tl + 5i + 1, i ∈ [1, n/5]}
to the remained edges of ∪q3=0ctl,q, and adding edges of
{aubu+8(5l−5)+1u = 2(5l − 5) + 5i − 1, i ∈ [1, n/5]}
to the remained edges of ∪3q=0c(5l−1),q, where tl ∈ [5(l − 1), 5(l − 1) + 3] and l ∈ [1, ⌊n/8⌋/5] (note that ⌊n/8⌋ ≡ 0( mod 5) in this case). It is easy to verify that the remained edges, i.e., all edges of
{aubu+n−5, au+2bu+n−2, au+1bu+n−2, au+3bu+n+1 u = 5i − 3, i ∈ [1, n/5]},
can form another linear 4forest.
Hence, la4(Kn,n) ≤ ⌈n/2⌉ + ⌊n/8⌋ + 1 = ⌈5n/8⌉.
Claim 9. la4(Kn,n) ≤ ⌈5n/8⌉ for n ≡ 7( mod 8) and n ≡ 0( mod 5).
The edges of Mn−7 ∪ Mn−6 other than that in
{a9+5(i−1)b2+5(i−1), a6+5(i−1)b5(i−1)i ∈ [1, n/5]}
can form one linear 4forest, the edges of Mn−5 ∪ Mn−4 other than that in
and n/5 pairwise disjoint 4cycles
{aubu+8t+2au+1bu+8t+6au
u = 2t + 5i − 2 i ∈ [1, n/5]
}
for each t ∈ [0, ⌊n/8⌋ − 1]. Moreover, for each t ∈ [0, ⌊n/8⌋ − 2], we obtain one linear 4forest by deleting edges in
{a4+2t+5(i−1)b4+2t+5(i−1)+8t+1i ∈ [1, n/5]}
of 4cycles of ∪q3=0ct,q, adding edges of
{aubu+8(tl+1)+1u = 2tl + 5i + 1, i ∈ [1, n/5]}
to the remained edges of ∪3q=0ctl,q, and adding edges of
{aubu+8(5l−5)+1u = 2(5l − 5) + 5i − 1, i ∈ [1, n/5]}
to the remained edges of ∪3q=0c(5l−1),q, where tl ∈ [5(l − 1), 5(l − 1) + 3] and l ∈ [1, ⌊⌊n/8⌋/5⌋]. For t = ⌊n/8⌋ − 1, we obtain one linear 4forest by deleting edges in
{aubu+8⌊n/8⌋−7u = 2⌊n/8⌋ + 5i − 3, i ∈ [1, n/5]} of 4cycles of ∪q3=0c(⌊n/8⌋−1),q, and adding edges of
{a6+5(i−1)b5(i−1)i ∈ [1, n/5]}
of Mn−6 that have not been used. It is obvious that all edges of
{aubu+8⌊n/8⌋−7
u = 2⌊n/8⌋ + 5i − 3 i ∈ [1, n/5]
} ∪ Mn−1
can form one linear 4forest, and the remained edges, i.e., all edges of
{a5i+4b5i−3, a5i+3b5i−2, a5(i+1)b5i+1, a5i−2b5(i−1), a5ib5i−2i ∈ [1, n/5]},
can form another one.
Hence, la4(Kn,n) ≤ ⌊n/2⌋ + (⌊n/8⌋ − 1) + 1 + 1 + 1 = ⌈n/2⌉ + ⌈n/8⌉ = ⌈5n/8⌉.
Combining Claims 2 − 9, the theorem is proved completely.
{a5i+3b5i−2, a5(i+1)b5i+1i ∈ [1, n/5]}
can form one, and the edges of Mn−3 ∪ Mn−2 other than that in
{a3+5(i−1)b5(i−1), a5+5(i−1)b3+5(i−1)i ∈ [1, n/5]}
can form another one. Note that the edges of Mn−1 have not been used. Now we have obtained ⌊n/8⌋ · 4 + 3 = ⌊n/2⌋ linear 4forests in total by Claim 1. In the following we will estimate the number of linear 4forests induced by the edges that are not used in Kn,n.
It is not diﬃcult to obtain that all edges of ∪3q=0ct,q can form n/5 pairwise disjoint paths P5
{bu+8t+3aubu+8t+7au+7bu+8t+11
u = 2t + 5i i ∈ [1, n/5]
}
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