On k-Ary Spanning Trees
of Tournaments
Xiaoyun Lu,1 Da–Wei Wang,2* Gerard J. Chang,3† In–Jen Lin,4 C. K. Wong5‡ 1SYNNEX INFORMATION TECHNOLOGY
3797 SPINNAKER COURT FREMONT, CA 94538 E-mail: [email protected]
2INSTITUTE OF INFORMATION SCIENCE
ACADEMIA SINICA NANKANG, TAIPEI 11529, TAIWAN E-mail: [email protected]
3DEPARTMENT OF APPLIED MATHEMATICS
NATIONAL CHIAO TUNG UNIVERSITY HSINCHU 30050, TAIWAN E-mail: [email protected]
4DEPARTMENT OF COMPUTER SCIENCE
NATIONAL TAIWAN OCEAN UNIVERSITY KEELONG, TAIWAN E-mail: [email protected]
5DEPARTMENT OF COMPUTER SCIENCE AND ENGINEERING
THE CHINESE UNIVERSITY OF HONG KONG SHATIN, HONG KONG E-mail: [email protected]
Received January 8, 1998; revised August 25, 1998
*Part of this work was done while visiting the Chinese University of Hong Kong †Research by Dr. Gerard J. Chang supported in part by the National Science Council under grant NSC87-2115-M009-007
‡On leave from IBM T. J. Watson Research Center, Yorktown Heights, NY 10598, U.S.A.
c
Abstract: It is well known that every tournament contains a Hamiltonian path, which can be restated as that every tournament contains a unary spanning tree. The purpose of this article is to study the general problem of whether a tournament contains ak-ary spanning tree. In particular, we prove that, for any fixed positive integerk, there exists a minimum number h(k) such that every tournament of order at leasth(k)contains ak-ary spanning tree. The existence of a Hamiltonian path for any tournament is the same ash(1) = 1. We then show thath(2) = 4 andh(3) = 8. The values ofh(k)remain unknown fork ≥ 4. c 1999 John Wiley & Sons, Inc. J Graph Theory 30: 167–176, 1999
Keywords: tournament; spanning tree; neighbor; Hamiltonian path; rooted tree; parent; child; depth;
height
1. INTRODUCTION
A tournament T = (V, E) of order n is an orientation of the complete graph of order n, where V is the vertex set and E the (directed) edge set of T . We sometimes use n-tournament to denote a tournament of order n. For an edge xy ∈ E, we also write x → y. If x → y, we say that x beats y. For A ⊆ V and B ⊆ V with
A ∩ B = ∅, we write [A, B]T = {xy ∈ E : x ∈ Aand y ∈ B}. We also use
A ⇒ Bto denote that every vertex in A beats every vertex in B; x ⇒ B stands for
{x} ⇒ B, and A ⇒ y for A ⇒ {y}.
Let G = (V, E) be a directed graph with vertex set V and edge set E. If x → y, we call y an out-neighbor of x, and x an in-neighbor of y. The out-neighborhood of a vertex x is the set N+
G(x) = {y : x → yin G} and the in-neighborhood of x is NG−(x) = {y : y → xin G}. The out-degree of x is d+G(x) = |NG+(x)|, and the
in-degree of x is d−
G(x) = |NG−(x)|.
A rooted tree is a directed tree with a special vertex, called the root, such that there exists a unique (directed) path from the root to any other vertex. A leaf is a vertex of out-degree zero. If u → v is an edge, then u is called the parent p(v) of v and v a child of u. Note that any vertex other than the root has exactly one parent, and a vertex may have many children. Descendants of a vertex are those vertices who can be reached from it by a path. The depth of a vertex is the length of the unique path from the root to it. The height of a rooted tree is the length of a longest path from the root.
A rooted tree is called a k-ary tree, if all nonleaf vertices have exactly k children, except possibly one nonleaf vertex has at most k−1 children. If all nonleaf vertices have exactly k children, then we call it a full k-ary tree. When k = 1, 2, 3, the corresponding k-ary trees are called unary, binary, and ternary trees, respectively. Note that a k-ary tree of order km + 1 is a full k-ary tree. In particular, a unary tree is always full. A star is a rooted tree of height 1. A k-star is a star with k vertices, and, thus, consists of k − 1 edges with a common tail.
It is a fundamental and important problem in the study of tournaments to deter-mine if a tournament contains a special structure. There are many results of this
kind in the literature. For example, it is well known that every tournament contains a Hamiltonian path ([11]), and every tournament contains a spanning tree of height at most 2 ([3]). Further results can be found in [1–2, 4–10, 13–14].
The fact that every tournament contains a Hamiltonian path can be restated as that every tournament contains a unary spanning tree. The purpose of this article is to study the general problem of whether a tournament contains a k-ary spanning tree. We first establish that, for any positive integer k, there exists a number
h0(k)such that every n-tournament contains a k-ary spanning tree if n ≥ h0(k). According to this result, we may define h(k) as the least number such that every
n-tournament contains a k-ary spanning tree if n ≥ h(k). The well-known result for the Hamiltonian path is then the same as that h(1) = 1. We then show that
h(2) = 4and h(3) = 8. The exact values of h(k) remain unknown for k ≥ 4. 2. k-ARY SPANNING TREES
The goal of this section is to establish a fundamental result that, for any positive integer k, there exists a number h0(k)such that every n-tournament contains a k-ary spanning tree if n ≥ h0(k). For this purpose, we need the following two theorems. A tournament is transitive if x → y and y → z imply that x → z, in other words, its vertices can be linearly ordered such that each vertex beats all later vertices.
Given a positive integer m and an n-tournament T , Erd˝os (see [12]) proved the following result concerning the number of vertex-disjoint transitive sub-tournaments of order mT can have.
Theorem 1. For any fixed positive integer m, there exists a number f(m) such that every n-tournament contains bn/mc vertex-disjoint transitive sub-tournaments of order m if n ≥ f(m).
It is not hard to see that, for n sufficiently large, every n-tournament contains all oriented trees of order m, where m is a fixed positive integer. Sumner (see [2]) conjectured that this is true for n ≥ 2m − 2. Although this conjecture is still open, H¨aggkvist and Thomason [2] proved the following result.
Theorem 2. Every oriented tree of order m is contained in every tournament of order 12m.
The following two lemmas are frequently used in this article.
Lemma 1 (Extension Lemma). For any fixed positive integer k, there exists a number g(k) ≤ 2(k + 1)k−2such that, for any full k-ary tree R of order at least
g(k)and any transitive sub-tournament S of order s ≤ k, where R and S are vertex disjoint, R can be extended to a k-ary tree R0with V (R0) = V (R) ∪ V (S).(Note
that, if s = k, then R0 is full.)
The proof of this lemma heavily depends on the idea used in the proof of Lemma 6 in Section 4 and so is given at the end of this article.
Lemma 2 (Induction Lemma). Let R be a k-ary tree of T and S a k-star of T with root u, where R and S are vertex disjoint. If |[u, V (R)]| ≥ 1, then T contains a k-ary tree R0with V (R0) = V (R) ∪ V (S).Moreover, if u is an out-neighbor of
the root of R, then R0can be chosen to have the same root as R.
Proof. Let v be the root of R. If u → v, then E(R) ∪ E(S) ∪ {uv} induces a
k-ary tree. Thus, we may assume that v → u. Since |[u, V (R)]| ≥ 1, there exists a vertex x such that u → x. Let P be the path of R from v to x, and y the first vertex in P such that u → y. Then p(y) → u, and, thus, (E(R) − {p(y)y}) ∪ E(S) ∪
{p(y)u, uy}induces a desired k-ary tree.
The ‘‘moreover’’ part is evident by the argument above.
Note that the Induction Lemma is also true, if we assume that R is a full k-ary tree and S an s-star of T with s ≤ k.
Now, we are ready to prove the following fundamental theorem for the existence of a k-ary spanning tree of a tournament, which has a natural similarity to Theorems 1 and 2.
Theorem 3. For any fixed positive integer k, there exists a number h0(k)such
that every n-tournament contains a k-ary spanning tree if n ≥ h0(k).
Proof. Let h0(k) = (g(k) + k) + max{f(k) + k, 12(g(k) + k)} + k, where
f(k)and g(k) are the numbers in Theorem 1 and the Extension Lemma, respec-tively. For n ≥ h0(k), let a = d(g(k) − 1)/ke, b = b(n − 2)/kc − a, and c = n − (ak + 1) − bk. Then ak + 1 ≥ g(k), bk ≥ f(k), 1 ≤ c ≤ k, n = (ak + 1) + bk + c, and n − c ≥ 12(ak + 1).
We first choose a transitive sub-tournament Q of order c of T . Note that the existence of such Q follows from Theorem 1. Since n − c ≥ 12(ak + 1), by Theorem 2, T − Q contains all oriented trees of order ak + 1, especially a full
k-ary tree R of order ak + 1. Since bk ≥ f(k), by Theorem 1, the sub-tournament
T − R − Qcontains b vertex-disjoint transitive sub-tournaments of order k, say,
S1, S2, . . . , Sb. Since R has at least g(k) vertices, by the Extension Lemma, R can be extended to a full k-ary tree R1 with V (R1) = V (R) ∪ V (S1). By the
same lemma, we can extend R1 to a full k-ary tree R2 with V (R2) = V (R1) ∪ V (S2). Continuing this process, we can finally obtain a full k-ary tree Rb with
V (Rb) = V (R) ∪ V (S1) ∪ · · · ∪ V (Sb). Using the Extension Lemma again for
Rband Q, we obtain a k-ary spanning tree of T .
According to this result, we may define h(k) as the least number such that every
n-tournament contains a k-ary spanning tree as long as n ≥ h(k). It is well known that h(1) = 1. We shall show that h(2) = 4 and h(3) = 8 in the next two sections, respectively. The values of h(k) remain unknown for k ≥ 4.
3. BINARY SPANNING TREES
The purpose of this section is to determine the value of h(2). The following theorem is the main result.
Theorem 4. Every tournament T of order n ≥ 4 contains a binary spanning tree.
Let T3 be the tournament with V (T3) = {0, 1, 2}and E(T ) = {01, 12, 20}.
Then T3 does not contain a binary spanning tree. Therefore, Theorem 4 implies
that h(2) = 4.
We first establish two lemmas before proving Theorem 4.
Lemma 3. If every (km + 1)-tournament has a k-ary spanning tree, then so does every km-tournament.
Proof. Suppose that T is a km-tournament. Construct a (km + 1)-tournament
T0from T by adding a new vertex w such that V (T ) ⇒ w. If R0is a k-ary spanning tree of T0, then R = R0− wis a k-ary spanning tree of T .
For convenience, we use (u, v, w) to denote a transitive 3-tournament with u beats v and w and v beats w. The following fact is evident.
Fact 1. Any tournament of order at least 4 contains a transitive 3-tournament.
Lemma 4. Every 5-tournament T contains a binary spanning tree.
Proof. Let x be a vertex of T with a maximum out-degree, and T0 the sub-tournament induced by NT+(x). We consider the following three cases.
Case 1. d+T(x) = 4. By Fact 1, T0contains a transitive 3-tournament (u, v, w). Let y be the remaining vertex of N+
T (x). Then {xy, xu, uv, uw} induces a binary spanning tree.
Case 2. d+T(x) = 3. Say, NT+(x) = {u, v, w} and NT−(x) = {t}. If
|[t, {u, v, w}]| ≥ 1, say, t → u, then {tu, tx, xv, xw} induces a binary span-ning tree. So we may assume that {u, v, w} ⇒ t. Without loss of generality, we may assume that u → v. Then {xw, xu, uv, ut} induces a binary spanning tree.
Case 3. d+
T(x) = 2. In this case, T is regular. Say, NT+(x) = {u, v} and
NT−(x) = {s, t}with s → t. Then {st, sx, xu, xv} induces a binary spanning tree.
Proof of Theorem 4. According to Lemma 3, we need to prove only the theorem
for n = 2m + 1 ≥ 5. We use an induction on m. The base case m = 2 is proved as Lemma 4. We now assume that m ≥ 3, and the theorem holds for all m0 < m.
Let x be a vertex with a maximum out-degree in T , and y one of its out-neighbors. Note that d+T(x) ≥ 2. Now the tournament T − {x, y} has order 2(m − 1) + 1. By the induction hypothesis, it contains a binary spanning tree R. By the choice of x, we must have |[x, V (R)]| ≥ 1. According to the Induction Lemma, the tournament
4. TERNARY SPANNING TREES
The purpose of this section is to determine the value of h(3). The following theorem is the main result.
Theorem 5. Every tournament T of order n ≥ 8 contains a ternary spanning tree.
Let T7be the 7-tournament with V (T7) = {0, 1, 2, 3, 4, 5, 6}and E(T7) = {ij : i − j ≡ 1, 2or 4 (mod 7)}. It is straightforward to check that T7does not contain
a ternary spanning tree. Therefore, h(3) ≥ 8, and so Theorem 5 implies that
h(3) = 8.
For proving Theorem 5, we first establish the following five lemmas.
Although there exists a 7-tournament containing no ternary spanning tree, most 7-tournaments do contain ternary spanning trees.
Lemma 5. Any nonregular 7-tournament T contains a ternary spanning tree.
Proof. Let u be a vertex of T with a maximum out-degree, and T0 the sub-tournament of T induced by NT+(u). Note that 3 ≤ d+T(u) ≤ 6. We consider the following four cases.
Case 1. d+T(u) = 6. Let v be a vertex of T0with a maximum out-degree. Then
d+
T0(v) ≥ 3. Say, v ⇒ {x, y, z}. Let the remaining vertices of T0 be s and t. Then {us, ut, uv, vx, vy, vz}induces a ternary spanning tree.
Case 2. d+
T(u) = 5. Say, NT+(u) = {s, t, x, y, z} and NT−(u) = {v}. If
|[v, NT+(u)]| ≥ 2, say, v ⇒ {s, t}, then {vs, vt, vu, ux, uy, uz} induces a ternary spanning tree. We may, therefore, assume that {s, t, x, y} ⇒ v. In the sub-tournament induced by {s, t, x, y}, there is a transitive 3-sub-tournament, say, (t, x, y). Then {uz, us, ut, tx, ty, tv} induces a ternary spanning tree.
Case 3. d+T(u) = 4. Say, NT+(u) = {v, x, y, z}and NT−(u) = {s, t} with
s → t. If |[s, {v, x, y, z}]| ≥ 1, say, s → v, then {st, sv, su, ux, uy, uz} in-duces a ternary spanning tree. So we may assume that {v, x, y, z} ⇒ s. If
|[{v, x, y, z}, t]| ≥ 2, say, {v, x} ⇒ t with v → x, then {uy, uz, uv, vx, vs, vt} induces a ternary spanning tree. We, thus, may assume that |[{v, x, y, z}, t]| ≤ 1. From d+T(t) ≤ d+T(u), we have |[{v, x, y, z}, t]| ≥ 1, and so |[{v, x, y, z}, t]| = 1. Say, v → t. If |[v, {x, y, z}]| ≥ 1, say, v → x, then {uy, uz, uv, vx, vs, vt} induces a ternary spanning tree. So we may assume that {x, y, z} ⇒ v. Without loss of gen-erality, we may assume that y → x. Then {tu, tz, ty, yx, yv, ys} induces a ternary spanning tree.
Case 4. d+
T(u) = 3. In this case, the tournament T is regular.
Let H be a subgraph and x a vertex not in H. The vertex x is said to be good
with respect to H, if x ⇒ V (H). Let R be a tree that is vertex-disjoint from H.
A vertex x of R is minimally good with respect to H, if x is good but none of its descendants in R is good with respect to H. The following lemma is useful in
proving that every tournament of order 10 contains a ternary spanning tree, and can also be used as an induction step in the ternary case. The proof technique is also useful in the proof of the Extension Lemma in the next section.
Lemma 6. Let (x, y, z) be a transitive 3-tournament in a tournament T. If R is a full ternary tree in T − {x, y, z} with root g such that g is minimally good with respect to {x, y, z}, then R can be extended to a ternary tree R0 with root g and
vertex set V (R) ∪ {x, y, z}.
Proof. Suppose that |[x, V (R)]| ≥ 1. Then, by the Induction Lemma, the
desired R0 exists. Therefore, we may assume |[x, V (R)]| = 0, i.e., V (R) ⇒ x. Suppose that |[y, V (R)]| ≥ 2. Choose a vertex u ∈ R with a smallest depth such that y → u, and another vertex v of R such that y → v. Since y → u and g → y, it must be the case that u /= g. Note that p(u) → y by the choice of u. Then, (E(R) − {p(u)u, p(v)v}) ∪ {yu, yv, yz, p(u)y, p(v)x}induces a desired ternary tree. Therefore, we may assume that |[y, V (R)]| ≤ 1.
Suppose that |[z, V (R)]| ≥ 5. We then choose 5 distinct vertices u, v1, v2, v3, v4
such that z ⇒ {u, v1, v2, v3, v4}. Note that u is not the root g of R, since z → u but g → z. We may assume that u is the vertex with a smallest depth such that z → u and so p(u) → z. Since a vertex of R has at most three children, {p(v1), p(v2), p(v3), p(v4)} contains at least two distinct vertices, one
of them must not be an out-neighbor of y, say, p(v2) → y. Then (E(R) − {p(u)u, p(v1)v1, p(v2)v2}) ∪ {zu, zv1, zv2, p(u)z, p(v1)x, p(v2)y}induces a
de-sired ternary tree. Therefore, we may assume that |[z, V (R)]| ≤ 4.
Note that, to be a full ternary tree, the order of R must be 1 (mod 3). If R has at least 7 vertices, there must be a vertex u /= gin R that beats all three vertices x, y, and z, a contradiction to the assumption that g is minimally good. In the case that
Rhas 4 vertices, let u, v, and w be the children of g. Then g ⇒ {u, v, w, x, y, z}. Then the sub-tournament induced by {u, v, w, x, y, z} has a vertex t of out-degree at least 3. One can find a desired ternary tree with t as the other nonleaf vertex. The case when R has just one vertex is trivial.
Therefore, the lemma holds.
Lemma 7. Every 10-tournament T contains a ternary spanning tree.
Proof. A transitive 3-tournament (u, v, w) is called a nice transitive 3-tournament
if u, v, and w have a common in-neighbor in T and T − {u, v, w} is not regular. Since T − {u, v, w} is not regular, it has a k-ary spanning tree R by Lemma 5. Since u, v, w has a common in-neighbor, we can choose a vertex x in R that is minimally good with respect to {u, v, w}. Let R1 be the subtree of R induced by x and its descendants. Then, by Lemma 6, R1 can be extended to a ternary tree R0
1with root x and vertex set V (R1) ∪ {u, v, w}. Let R0be the tree obtained from R by replacing R1 with R01. Then R0 is a desired ternary tree of T . Therefore,
the existence of a nice transitive 3-tournament implies the existence of a ternary spanning tree.
If T has a vertex x with an out-degree in the set {4, 5, 7, 8, 9}, then x has an out-neighbor u with two out-neighbors v and w, which are also in N+
loss of generality, we may assume that v → w. As x has an out-degree that differs from 3 in T − {u, v, w}, we have that (u, v, w) is a nice transitive 3-tournament.
So we may assume that for any u ∈ V (T ), d+
T(u) ∈ {0, 1, 2, 3, 6}. As the sum of the out-degrees of all vertices in T is 45, there exist two vertices u and v of out-degree 6 such that u → v. Let x and y be two out-neighbors of u that differ from v. Then T − {u, x, y} is a 7-tournament in which v is of out-degree at least 4. So, T − {u, x, y} is not regular. By Lemma 5, T − {u, x, y} has a ternary spanning tree. Then, by the Induction Lemma, T has a ternary spanning tree.
Lemma 8. Every 9-tournament contains a ternary spanning tree.
Proof. The lemma follows immediately from Lemmas 7 and 3.
Lemma 9. Every 8-tournament T contains a ternary spanning tree.
Proof. Choose a vertex x in the tournament T with an out-degree not equal to
4. If d+T(x) /= 0, then T − u is not regular for any u ∈ NT+(x). If d+T(x) = 0, then
T − uis not regular for any u /= x. In any case, we can find a vertex u such that
T − uis not regular. By Lemma 5, T − u has a ternary spanning tree. One can easily extend this tree to a ternary spanning tree of T .
Proof of Theorem 5. Lemmas 9, 8, and 7 prove the cases for n = 8, 9, 10.
Suppose that n ≥ 11 and the theorem is true for all n0 < n. By first choosing a vertex u of out-degree at least 3, it is not difficult to see that there is a transitive 3-tournament (u, v, w) such that u has an out-neighbor in T0 = T − {u, v, w}. By the induction hypothesis, T0has a ternary spanning tree. By the Induction Lemma,
T has a ternary spanning tree.
5. PROOF OF THE EXTENSION LEMMA
This section gives a proof of the Extension Lemma by using the idea in the proof of Lemma 6.
Let R be a full k-ary tree of T with root u, and S a transitive sub-tournament of T of order s ≤ k, where R and S are vertex disjoint. Write V (S) = {x1, x2, . . . , xs} such that xi beats xj whenever i < j. Set Ai = NT+(xi) ∩ V (R) and Bi =
A1∪ A2∪ · · · ∪ Aifor 1 ≤ i ≤ s. For convenience, let B0 = ∅.
Lemma 10. If |Bi− Bi−1| ≥ |Bi−1|k + ifor some i, then R can be extended to
cover S. In other words, there exists a k-ary tree with vertex set V (R) ∪ V (S).
Proof. Note that |Ai| ≥ |Bi − Bi−1| ≥ |Bi−1|k + i ≥ 1. If u → xi, let
yi be a vertex in R such that xi → yi and p(yi) → xi (such a vertex clearly exists); if xi → u, let yi = u. We next observe that it follows from the above inequality and the fact that the set Bi−1 has at most |Bi−1|k children, Ai has
i − 1vertices y1, y2, . . . , yi−1(distinct from yi) whose parents are not in Bi−1, i.e.,
xi ⇒ {y1, y2, . . . , yi−1}and {p(y1), p(y2), . . . , p(yi−1)} ⇒ {x1, x2, . . . , xi−1}. If u → xi, then
(E(R) − {p(yj)yj : 1 ≤ j ≤ i}) ∪ {p(yj)xj, xiyj : 1 ≤ j ≤ i}
∪ {xixi+1, xixi+2, . . . , xixs} determines a k-ary tree with vertex set V (R) ∪ V (S). And if xi → u, then
(E(R) − {p(yj)yj : 1 ≤ j ≤ i − 1}) ∪ {p(yj)xj, xiyj : 1 ≤ j ≤ i − 1}
∪ {xiu, xixi+1, xixi+2, . . . , xixs} determines a k-ary tree with vertex set V (R) ∪ V (S).
Lemma 11. If there is a leaf y ∈ V (R) − Bs, then R can be extended to
cover S.
Proof. In this case, y ⇒ {x1, x2, . . . , xs}. Then E(R) ∪ [y, V (S)] induces a desired spanning tree.
Proof of the Extension Lemma. Suppose that there exists no k-ary tree with
vertex set V (R) ∪ V (S). Then, by Lemma 10, |Bi− Bi−1| ≤ |Bi−1|k + i − 1and so |Bi| ≤ |Bi−1|(k + 1) + i − 1for 1 ≤ i ≤ s. Using |B0| = 0and an induction, we have |Bs| ≤Ps−1i=1i(k + 1)s−1−i < 2(k + 1)k−2. Certainly, if R has at least 2(k + 1)k−2 vertices, we can find a leaf y in V (R) − Bs and use Lemma 11 to complete the proof.
ACKNOWLEDGMENTS
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