AbstractIt is proved in [1] that if a tree T of order n is not a star, then there exists an edge-disjoint placement of two copies of this tree into its fourth power.In this paper, we prove the packing of some trees into their third power

Kheddouci, H.

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A note on the packing of two copies of some trees into their third power 

PERGAMON Applied Mathematics Letters 16 (2003) 1115-1121 
Applied 
Mathematics 
Letters 
A Note on .&he Packing of Two Copies 
of Some Trees into Their Third Power 
H. KHEDDOUCI 
Laboratoire LE21 UMR-CNRS 5158, Universite de Bourgogne 
B.P. 47870, 21078 Dijon Cedex, France 
(Received and accepted July 200.2) 
Communicated by A. Tucker 
Abstract-It is proved in [l] that if a tree T of order n is not a star, then there exists an edge- 
disjoint placement of two copies of this tree into its fourth power. 
In this paper, we prove the packing of some trees into their third power. @  2003 Eisevier Ltd. 
Ali rights reserved. 
Keywords-Embedding, Packing, Permutation, Placement, Power tree. 
1. INTRODUCTION 
Suppose Gi, . . . , Gk are graphs of order n. We say that there is a packing of Gi, . . . , Gk (into 
the complete graph &) if there exist injections cri : V(Gi) -+ V(K,), i = 1,. . . , k, such that 
a;(E(Gi)) fl c$(E(Gj)) = 0 for i # j, where the map of : E(Gi) -+ E(K,) is the one induced 
by CYi. 
A packing of k copies of a graph G will be called a k-placement of G. A packing of two copies 
of G  (i.e., a two-placement) is an embedding of G  (in its complement G). So, an embedding of 
a graph G is a permutation a on V(G) such that if an edge xy belongs to E(G) then act 
does not belong to E(G). If there exists an embedding of G, we say that G  is embeddable. 
Let G  be a graph. The simple graph GP, with p 2 1, is constructed from G by adding edge 
between any pair of vertices at distance at most p on G. Obviously, G1 = G. We denote the 
distance between two vertices x, y on G  by di&(z, y). 
The following theorem was proved, independently, in [2-41. 
THEOREM 1. Let G  = (V, E) be a graph of order n. If [E(G)1 _< n - 2, then G WI be embedded 
in its complement. I 
This result has been improved in many ways. The main references of this paper and of other 
packing problems are to be found in the last chapter of Bollobb’s book [2], the fourth chapter 
of Yap’s book [5], and the survey paper [6]. 
In this paper, we shall consider the case in which G  is a tree on n vertices. The example of 
the star S, shows that Theorem 1 cannot be improved by raising the size of G  even in the case 
when G is a tree. However, in that case, we have the following result. 
08939659/03/$ - see front matter @  2003 Eisevier Ltd. Ail rights reserved. 
doi: 10.1016/S0893-9659(03)00152-6 
‘-M-t by &-G-W 
1116 H. KHEDDOUCI 
THEOREM 2. Let T be a tree of order n, T # S,. Then T is contained in its own complement. a 
Theorem 2 was first proved by Straight (unpublished, cf. [7]). Besides, this result has been 
improved in many ways. For instance, the packing of two trees was considered in [7] and the 
three-placement of a tree in [8]. 
Another possibility of improving Theorem 2 is to consider some additional information about 
embeddings, i.e., packing permutations. 
An example of such a result is the following theorem contained as a lemma in [9] (cf. also [lo]). 
THEOREM 3. Let T be a nonstar tree of order n, with n > 3. Then there exists a two-placement a 
of T such that for every z E V(T), distT(rc, g(z)) 5 3. I 
This theorem immediately implies the following corollary. 
COROLLARY 4. Let T be a nonstar tree of order n, with n > 3. Then there exists an embedding u 
of T such that CT(T) c T7. 
Since T7 is, in general, a proper subgraph of K,, the last corollary can be considered as an 
improvement of Theorem 2. 
Kheddouci et al. considered the problem of the two-placement of a tree T into TP such that p 
is as small as possible. In [l], they proved the following result. 
THEOREM 5. Let T be a nonstar tree of order n, with n > 3. Then there exists a two-placement o 
of T such that a(T) c T*. 
,4nd they posed the following conjecture. 
CONJECTURE 6. Let T be a nonstar tree of order n, with n > 3. Then there exists a twO- 
placement CT of T such that a(T) c T3. 
Note that the square of a path of order n contains only (n-2) more edges. So, it is not possible 
in general to pack two copies of a tree T into T2. 
For the placement in T3, the problem seems more difficult. It is due to some vertices which 
remain fixed by any possible permutation. Note that, for instance, the vertex x4 of P7 = 
(a,22,. . . , ~7) remains always fixed for all possible permutations in P;. It is clear that is 
not possible to make a placement of the tree T, gotten by two paths of order 7 joined by an edge 
between their fourth vertices, in T3 by composition of two path permutations. In any case, the 
image of such an edge (that joins the fourth two vertices) remains itself. 
In this note, we prove Conjecture 6 for some families of tree for which the placement is not very 
difficult to do. In particular, we give the placement of a path, a star-path-star and a tree without 
vertices of degree two. During this modest study we noticed that some vertices of degree 2 tend 
to remain fixed and block the placement in the third power! 
We shall need some additional definitions and notations in order to formulate our results. 
Let IC be a vertex of a tree T. The components of T-z are called neighbor trees of 2. If y is any 
neighbor of x in T, we denote by Tv the neighbor tree of x which contains y. Consequently, if we 
delete an edge e = xy of T, we obtain two components of T-e, respectively, the neighbor tree T, 
of y and the neighbor tree T3, of x. For each vertex x E V(T), if c&-(z) > 2 then x is an internal 
vertex. Consider two distinct rooted trees T’ and T”, together with the path (x1, x2,. . . , xp). 
We denote T = T’ . (x1, x2,. . . , xP) e T” the tree obtained by identifying x1 with the root of the 
first tree, and xp with the root of the second tree (see Figure 1). By doing this, the first tree 
becomes the neighbor tree T,, of x2, and the second tree becomes the neighbor tree T%, of ~~-1. 
In the particular case where the first tree of this construction is reduced to a single vertex (thus, 
identified with xl), we simplify the notation into (x1, x2,. . . , xP) . Tr. A special case of this 
construction is exemplified by the star-path-star S’ . (xl,. . . , xP) . S”, where S’ and S” are two 
stars with centers, respectively, x1 and xp. For p = 2 we call this tree a double-star. 
Packing of Two Copies of Some ?‘rees 1117 
(a) A tree T rooted on a. (b) A star S rooted on the (c) A path P with end-vertices 21 and q 
center b. 
ygg;;2 
(d) A tree T. P. S obtained by identifying a with ZEN and b with 27. 
Figure 1 
Let P be a path of end-vertices x and y. A permutation u on V(P) is said to be P-nice iff the 
following four conditions are satisfied: 
(1) c is a tweplacement bf P; 
(2) c7(P) c P3; 
(3) distp(x,a(x)) = 1; 
(4) distdy, a(y)) E (0, 1). 
The path P is said to be nice if there exists a P-nice permutation. 
Let T be a tree of order at least three. Let x E V(T) be an internal vertex. A permutation o 
on V(T) is said to be (T, x)-good iff the following four conditions are satisfied: 
(1) u is a two-placement of T; 
(2) a(T) c T3; 
(3) distT(x,o(x)) = 1; 
(4) for every end-vertex y of T, distT(y, a(y)) = 2. 
The tree itself is said to be x-good if there exists a (T, x)-good permutation. Finally, T is called 
good if it is x-good for every internal vertex x E V(T). 
l3y using this terminology, we shall prove the following propositions. 
PROPOSITION 7. AlI paths of order n, with n > 3, are nice. 
In the case of star-path-star we prove the following result. 
PROPOSITION 8. Let T = & . P. Sz be a star-path-star such that Si, with i = 1,2, has at least 
one edge and P is at least of length 1. Then there exists a two-placement IS of T such that 
a(T) c T3. 
The main result of this paper is the following. 
PROPOSITION 9. Let T be a nonstar tree of order n, with n > 5, having no vertex of degree 2. 
Then T is good. 
PROPOSITION 10. Let T be a nonstar tree of order n rooted on a vertex r or r’, with rr’ E E(T). 
If d(r), d(r’) 2 2 and for each x E V(T) \ {r,~‘}, d(x) # 2’, then T is r-good and r/-good. 
We call a k-ary tree (k 2 2) a rooted tree on a vertex of degree k and all other vertices are of 
degree 1 or k + 1. Proposition 10 implies the following corollary. 
1118 H. KHEDDOUCI 
COROLLARY 11. Let T be a nonstar k-ary tree, with k 1 2, rooted on r. Then T is r-good. 
In the following result, we accept more vertices of degree 2. 
PROPOSITION 12. For i = 1,2, let Ti be a rooted tree on an internal vertex such that the size 
of Ti is at least three and dTi (x) # 2 for each x E V(Ti). Let P be a path of order p, with p 2 2. 
Then there exists a two-placement c of T = Tl . P. T2, such that a(T) c T3. 
We start with a sequence of lemmas (Section 2) that we use in the main part of the proof given 
in Section 3. 
2. LEMMAS 
In this section, we show some basic lemmas needed to prove our propositions. 
LEMMA 13. The paths of order 4, 5, 6, and 7 are nice. 
PROOF. Let n E {4,5,6,7}. Let P = (x1,x2,. . . , x,) be a path of order n. A nice permutation cn 
associated to this path is given below (we use standard notation for its decomposition into cycles): 
u4=(21 22 x4 x3); 
05 = (Xl x2 x4 x3)(25); 
06 = (xl x2 25 x4)(x3)(x6); 
ff7 = (Xl 22 55)(x3 27 x6)(24); 
Note that dist(xl,a,(xl)) = 1 and dist(x,,a,,(x,)) E (0, l}, for each 4 5 n 5 7. I 
LEMMA 14. Let S be a star at least of size two centered on x1 and P = (x1, x2,. . . , xp), with 
3 < p 5 5. Let T = S. P. Then there exists a two-placement cs of T such that a(T) c T3 and 
dist(x,,a(x,)) = 1. 
PROOF. Let {z1,z2,. . . , zt} be the end-vertices of S. By hypothesis t 2 2, since p 2 3, then T is 
not a star. If p = 3, let P’ = (~1, x1, x2, x3). By Lemma 13, there exists a PI-nice permutation a’ 
such that dist(zl, o’(zl)) = 1, dist(x1, a/(x1)) = 2, and dist(xs, a/(x3)) = 1. We may extend this 
permutation to the required permutation u’ of V(T) by putting o(z~) = zi, for each 2 5 i 5 t. It 
is easy to see that dist(a(zi), ~(21)) 5 3, for each 1 < i 5 t. 
For p = 4, by Lemma 13, P is nice and there exists a P-nice permutation ~7’ such that 
dist(x4, a’(x4)) = 1. So the same corresponding permutation may be extended to T by the cycle 
(w2,. . G). 
Incasep=5, let P’= (Zl,xl,... ,x5). By Lemma 13, there exists a P/-nice permutation 0’ 
such that dist(zl,o’(z1)) = 0 and dist(xb,a’(xs)) = 1. Therefore, 1 < dist(xl,a’(xl)) 5 2. We 
extend this permutation to T by putting D(z~) = pi, for each 2 5 i 2 t. I 
LEMMA 15. Let T be a double-star. If the degree of each internal vertex is at least two, then T 
is good. 
PROOF. Let x and y be the two internal vertices T. Let N(y) = {x,yl,y2,. . . ,yp} and N(x) = 
{Y,51,52,. . . , xpr}, with p, p’ 2 2 (see Figure 2). We give u by the following cycle (x x1 x2 . . . X,,J 
Y Yl Y2... yp). Thus, dist(x, C(X)) = dist(y, o(y)) = 1, dist(xi, o(xi)) = 2 for each 1 5 i 5 p’ and 
dist(yi, g(yi)) = 2 f or each 1 5 i 5 p. Therefore, T is x-good and y-good, so T is good. I 
X Y 
Figure 2. A double-star with centers I and y. 
Packing of Two Copies of Some Trees 1119 
3. PROOFS OF PROPOSITIONS 
In the following proofs, the verification for any two-placement (T of a tree T that distT(g(x), 
u(y)) 2 3, for any xy E E(T), is easy and left to the reader to do. 
3.1. Proof of Proposition 7 
Let P= (x1,52,... , x,) be a path of order n. The proof is by induction on n. For n = 4,5,6, 
and 7, by Lemma 13, P is nice. 
Suppose that the proposition holds for all n’ < n and n > 8. Let xjxj+i be an edge of P 
such that 4 5 j 5 n - 4. The neighbor paths Pzj and P,,+, of P - XjXj+l are, respectively, of 
order j and n - j. By the induction hypothesis, Pzi and Pzj+l are nice. Let 17 be a Pzj-nice 
permutation and g’ be a Pzj+,-nice permutation. SO distpj(xir o(xi)) = 1, distpj(xj,g(xj)) E 
(0, l}, and distpj+l(xj+i,g(xj+i)) = 1, distpj+l(xn,o(x,)) E {O,l}. Then the composition of 
these two permutations (not commutative) gives a P-nice permutation C” = o o C’ such that 
distp(xi, a”(xi)) = 1 and distp(x,, a”(~,)) E (0, 1). I 
3.2. Proof of Proposition 8 
Let 5’1 = Kl,, be of center xi and end-vertices .q,zz, . . . , zq, with q 2 1 and 5’s = K~,,J of 
center xp and of end-vertices yr,y2,. . . yqjr with q’ 2 1. Let P = (xi, x2,. . . ,xp), with p > 2. 
According to Proposition 7, we may assume that Si has at least two end-vertices. 
CASE 1. q’ = 1. 
In case 2 5 p 5 4, by Lemma 14, T has a two-placement into T3. If p = 5, by Lemma 13, P. S 
is nice and xi is not a fixed vertex. To obtain a two-placement u of T into T3 in this case, we 
put g(zi) = zi, for 1 5 i 5 q. 
For p 2 6, from T - 23x4, the neighbor trees T,, and T,, are, respectively, a star-path of 
diameter three and a path of length at least three. By Lemma 14, there exists a permutation 
such that dist(xs, a~=~ (x3)) = 1, and by Proposition 7, T,, is nice. So we obtain a two-placement 
of T in T3. 
CASE 2. q’ > 2. 
If p = 2, T is a double-star and a two-placement is done by Lemma 15. In case p = 3, let 
P’ = (.zr,xi,x2,xa). By Lemma 13, P’ is nice, dist(xi,op(xi)) = 2 and dist(xa,upt(xs)) = 1. 
So by putting o(q) = zi, for 2 5 i 5 q and a(yi) = yi, for 1 5 i 5 q’, we obtain a two-placement 
of T in T3. 
In case p = 4, by Lemma 13, P is nice and distp(xi,op(xi)) = distp(xd,up(x4)) = 1. So 
put a(zi) = zi, for 1 < i 5 q and a(yi) = yi, for 1 5 i 5 q’. Thus, Si . P. S2 admits a two- 
placement o in its third power. If p = 5, by Lemma 13, the path P’ = (xl,. . . ,x5, ~1) is nice 
such that dist(xi,ap(xr)) = 1, dist(yi, apt(yi)) = 0, and dist(xs, opt) # 0. Then we put 
I = zi, for 1 5 i 5 q and a(yi) = yi, for 2 5 i 5 q’. 
If p = 6,7,8,9, we consider Tj and Tj+i the neighbor trees obtained by deletion of the edge 
xjxj+i, with j = p/2 if p is even and j = (p + 1)/2 otherwise. By Lemma 14, Tj and Tj+l admit 
two-placements into their third power such that dist(qj (xj), a~~+~ (xj+i)) < 3. 
For p 2 10, let Vi = &.(x1, x2, x3) and Us = (~~-2, xp-i, xp).&. Define (T on (x4,25,. . . , xp--3) 
according to Proposition 7 and on Vi and Us according to Lemma 14. So dist(a(xs), ~(23)) < 3 
and dist(a(x,-i), c(xp-s)) < 3. I 
3.3. Proof of Proposition 9 
We prove for any internal vertex x E V(T) that T is x-good. Let x be any internal vertex 
of V(T). Let i(T) be the number of internal vertices in T. The proof is by induction on i(T). 
Since T is not a star, then i(T) > 1. If i(T) = 2, then T is a double-star. Therefore, by Lemma 15, 
T is good. 
1120 H. KHEDDOUCI 
Assume for any nonstar T’ C T, with 2 as an internal vertex in T’ and i(T’) < i(T), that T’ 
is x-good. 
Let v E V(T) (# x) b e an internal vertex whose neighbors are all end-vertices except one. Call 
n>vz,..., ‘uk the end-vertices adjacent to e, with k 2 2. Let T’ = T - (~~1,772,. . . , vk}. Observe 
that T’ is not a star. Moreover, e becomes an end-vertex in T’. So i(T’) = i(T) - 1 and by 
hypothesis T’ is x-good. Let g’ be a (T/,x)-good permutation. We construct a permutation u 
on T by combining g’ and the cycle (VI, ~2,. . . ,vk). Observe that distT(a(vi),a(vj)) = 2, for 
each 1 5 i # j 5 k. distT(g(v),g(vi)) = distT(g’(v), g(wi)) = 3. Moreover, for each vertex y 
of T’, cr(y) = a’(y). Therefore, T is x-good. Finally, since it is true for each internal vertex, 
then T is good. II 
3.4. Proof of Proposition 10 
The proofs of T is r-good (and T is r/-good) are similar to that given in the proof of Proposi- 
tion 9. I 
3.5. Proof of Proposition 12 
Let P = (XI, 22,. . . , xp), with p 2 2. If both Tl and Tz are stars, Proposition 8 is applied. 
Assume that T2 is not a star. It is easy to see that if Ti has, at least, three edges, it is not 
a star and has no vertex of degree 2 then Ti is good, for i = 1,2. Let T[ = Tl . (x1, x2) and 
Ti = (~~-1, xP) . T2. Note that T,’ is x,-good. We have four cases. 
CASE 1. p = 2. 
The tree T = Tl . (21, x2) . Tz has no vertex of degree two. So by Proposition 9, T is good. 
CASE 2. p = 3. 
So x2 is the unique vertex of degree 2 in T. By Proposition 10, T is x2-good. 
CASE 3. p = 4. 
If Tl is a star, then by Lemma 14, Tl . ( xi, x2, xs) admits a permutation 0’ such that a’(Tl . 
(x1,x2,x3)) c @‘I.( xi, x2, ~3))~ and dist(xs, a’(xs)) = 1. Since T2 is not a star, then it is x4-good 
and any x4-good permutation of T2 provides an extension of cr’ to T. 
If Tl is not a star, then Ti satisfies Proposition 9. Let 0 be an xl-good permutation of this 
tree. We may extend it to T by letting ~(23) = 5s and taking for ~~~ any x4-good permutation. 
CASE 4. p = 5. 
If Tl is a star, a permutation is obtained by considering Tl. (xi, x2, x3, x4) as in Case 3 and T2. 
In case Tl is not a star, use an xi-good permutation for Ti, and xs-good permutation for Ti and 
put 4x3) = x3. 
CASE 5. p > 6. 
In case Tl is a star, let T,,-, and Tzp = T2 be the two components of T - xp-lxp. As 
for Case 1 of the proof of Proposition 8, Tzpml has a packing CT into its third power such that 
dist(x,-1, ~~~~~~ (~~-1)) = 0 or 1. S o any x,-good permutation of Tz gives an admissible exten- 
sion of u to T. 
If Tl is not a star, let P’ = P \ {x~,xp}. Proposition 7 provides a permutation of P’ such 
that dist(xg,op~(xs)) E {O,l} and dist(xp-r,apt(x,-1)) = 1. Extend it to T using any xi-good 
permutation of Tl, and any x,-good permutation of Tz. I 
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A note on the packing of two copies of some trees into their third power

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