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On the sum number and integral sum number of hypertrees and complete hypergraphs
Discrete Mathematics 236 (2001) 339–349
www.elsevier.com/locate/disc
On the sum number and integral sum number of hypertrees and complete hypergraphs Martin Sonntaga; ∗ , Hanns-Martin Teichertb a Faculty
of Mathematics and Computer Science, TU Bergakademie Freiberg, Bernhard-von-Cotta-Str. 2, D-09596 Freiberg, Germany b Institute of Mathematics, Medical University of L) ubeck, Wallstra+e 40, D-23560 L)ubeck, Germany Received 17 October 1997; revised 29 September 1998; accepted 5 March 2000
Abstract A hypergraph H is a sum hypergraph i0 there are a 1nite S ⊂ N+ and d1 ; d2 ∈ N+ with 1¡d1 6d2 ¡2d1 such that H is isomorphic to the hypergraph Hd+1 ;d2 (S) = (V; E) where V := S and
E :=
{x1 ; x2 ; : : : ; xk }: k ∈ {d1 ; d1 + 1; : : : ; d2 } ∧ (i=j ⇒ xi =xj ) ∧
k
xi ∈ S
:
i=1
For an arbitrary hypergraph H, the sum number = (H) is de1ned to be the minimum number of isolated vertices w1 ; : : : ; w ∈V such that H ∪ {w1 ; : : : ; w } is a sum hypergraph. For S ⊂ Z we obtain the concept of integral sum hypergraphs and the integral sum number
(H): In this paper, we determine (Knd ) and give bounds for (Knd ), where Knd denotes the d-uniform complete hypergraph on n vertices. Moreover, for d¿3 we prove that every d-uniform c 2001 Elsevier Science B.V. All hypertree H is an integral sum hypergraph, i.e. (H) = 0. rights reserved.
1. Introduction and denitions The concept of sum graphs and integral sum graphs was introduced by Harary [5,6]. Many authors investigated these kinds of graphs in the last years, for a brief summary, see for instance [8]. In the following, we generalize the concepts mentioned above to hypergraphs. All hypergraphs considered here are supposed to be nonempty and 1nite without loops and multiple edges. In standard terminology we follow [1]. A hypergraph H = (V; E) with vertex set V and edge set E ⊂ P(V )−{∅} is d-uniform i0 26d ∈ N and |e| = d (∀e ∈ E). For d6n we denote the d-uniform complete hypergraph on n vertices by Knd . ∗
Corresponding author. E-mail address: [email protected] (M. Sonntag).
c 2001 Elsevier Science B.V. All rights reserved. 0012-365X/01/$ - see front matter PII: S 0 0 1 2 - 3 6 5 X ( 0 0 ) 0 0 4 5 2 - 0
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Let S ⊂ N+ be 1nite and d1 ; d2 ∈ N such that 1¡d1 6d2 ¡2d1 . Hd+1 ; d2 (S) = (V; E) is called the (d1 ; d2 )-sum hypergraph of S i0 V = S and k xi ∈ S : E = {x1 ; x2 ; : : : ; xk }: k ∈ {d1 ; d1 + 1; : : : ; d2 } ∧ (i =j ⇒ xi =xj ) ∧ i=1
Furthermore, a hypergraph H = (V; E) is a sum hypergraph i0 there exist S ⊂ N+ and d1 ; d2 ∈ N such that H is isomorphic to Hd+1 ; d2 (S). In the following, in e0ect we will deal with d-uniform hypergraphs. For this end we de1ne Hd+ (S) := Hd;+d (S). Then, for d = 2 we obtain the known concept of sum graphs. For a hypergraph H, the sum number = (H) is de1ned to be the minimum number of isolated vertices w1 ; : : : ; w ∈V such that H ∪ {w1 ; : : : ; w } is a sum hypergraph. If also nonpositive integers are allowed as elements of S; i.e. S ⊂ Z, we obtain the de1nition of integral sum hypergraphs and the integral sum number = (H) in the same manner. One 1eld of investigation of sum graphs and integral sum graphs is the determination of (G) and (G), for several classes of graphs G. For the complete graphs Kn Bergstrand et al. [2] proved Theorem 1.1. For n¿4 the sum number of complete graphs is given by (Kn ) = 2n − 3: For (Kn ), it was shown by Chen [3] as well as by Sharary [7] that (Kn ) = (Kn ) is ful1lled for n¿4. In the 1rst part of this paper, we generalize Theorem 1.1 and determine (Knd ): Further we show (Knn ) = (Knn−1 ) = 0 and give bounds for (Knd ) if d6n − 2: In the second part of our article, we investigate connected hypergraphs with minimum number of edges, i.e. hypertrees. An edge sequence w = (v0 ; e1 ; v1 ; e2 ; : : : ; et ; vt = v0 ) with t¿2 pairwise distinct edges e1 ; e2 ; : : : ; et ∈ E and pairwise distinct vertices v0 ; v1 ; : : : ; vt−1 ∈ V is called a cycle of the hypergraph H = (V; E) i0 {vk−1 ; vk } ⊆ ek (∀k ∈ {1; 2; : : : ; t}). A hypergraph H = (V; E) is a hypertree i0 H is connected and cycle-free. We call the hypergraph H = (V; E) non-trivial i0 |V | ¿ 1. For graphs Ellingham [4] veri1ed Theorem 1.2. Every non-trivial tree G = (V; E) has sum number (G) = 1. In [8] this result was generalized to Theorem 1.3. If H = (V; E) is a non-trivial hypertree and max{|e|: e ∈ E}¡2 min{|e|: e ∈ E} − 1; then (H) = 1:
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In Section 3 we prove that d-uniform hypertrees are integral sum hypergraphs, if d¿3. 2. Complete hypergraphs In the following, let v1 ¡v2 ¡ · · · ¡vn be a labeling of the vertices of Knd and w1 ; : : : ; w a labeling of = (Knd ) isolated vertices such that H = Knd ∪ {w1 ; : : : ; w } can be recognized as a sum hypergraph. Further we choose the notation Vn ={v1 ; : : : ; vn } and W = {w1 ; : : : ; w }: Lemma 2.1. There are at least d(n − d) + 1 pairwise di7erent sums of vertices vˆ1 + · · · + vˆd ;
where vˆ1 ; : : : ; vˆd ∈ Vn and vˆ1 ¡ · · · ¡vˆd :
Proof. We consider the following sets of sums, each containing d summands: Sd = {v1 + · · · + vd−1 + vˆd : vˆd ∈ {vd ; : : : ; vn }}; Sk = {v1 + · · · + vk−1 + vˆk + vn+(k+1)−d + · · · + vn : vˆk ∈ {vk+1 ; : : : ; vn+k−d }}; k = d − 1; : : : ; 1: Clearly, each sum of Si is strictly larger than each sum of Si+1 for i = 1; : : : ; d − 1; thus Si ∩ Sj = ∅ for i =j. Further |Sd | = n − d + 1; |Sk | = n − d for k = d − 1; : : : ; 1: Hence, there are (n − d + 1) + (d − 1)(n − d) = d(n − d) + 1 pairwise di0erent sums vˆ1 + · · · + vˆd . Obviously, in case of d = n we have (Knn ) = 1 for n¿2: For d = n − 1 we know for graphs that (K3 ) = (K32 ) = 2 (see [2]). This result is generalized by Lemma 2.2. (Knn−1 ) = n − 1 for n¿3: Proof. There are exactly n edges in Knn−1 and only the maximum label vn ∈ Vn is chooseable as sum of the other labels v1 ; : : : ; vn−1 ; hence (Knn−1 )¿n − 1: Using a 1xed N ¿n−2 we label the vertices of Knn−1 in the following way: vi =N +i for i = 1; : : : ; n − 1; vn = N (n − 1) + (n(n − 1))=2 and choose for (n − 1) other vertices the labels w1 ¡ · · · ¡wn−1 as the remaining sums of (n − 1) labels vi (each containing the label vn ). This yields W = {N (2n − 3) + (n − 1)2 ; : : : ; N (2n − 3) + n(n − 1) − 1}. It remains to show that these vertices are isolates. For this end, we consider a sum s of d labels and suppose that at least one of these labels belongs to W . Using N ¿n − 2 it follows s ¿ (N (2n − 3) + (n − 1)2 ) + ((n − 2)N + ((n − 1)(n − 2))=2) ¿ (N (2n − 3) + (n − 1)2 ) + (N + 1)¿(N (2n − 3) + (n − 1)2 ) + (n − 1) = N (2n − 3) + n(n − 1) ¿ max{v: v ∈ Vn ∪ W };
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i.e. s ∈Vn ∪ W and the sum hypergraph property implies that all vertices of W are isolates. Now, we consider the number of possibilities to choose pairwise distinct vˆ1 ; vˆ2 ; : : : ; vˆd ; vj ∈ Vn such that vˆ1 + · · · + vˆd = vj . As mentioned in the proof above, for d = n − 1 only the sum v1 + · · · + vd = vn can occur. The question is more complicated for d6n − 2. We need the following interesting result for proving the main theorem. Lemma 2.3. For d6n − 2 there exist no labels {vˆ1 ; vˆ2 ; : : : ; vˆd ; vj } ⊂ Vn such that vˆ1 + vˆ2 + · · · + vˆd = vj : Proof. Assume, there are elements vˆ1 ¡vˆ2 ¡ · · · ¡vˆd ¡vj of Vn such that vˆ1 + vˆ2 + · · · + vˆd = vj is ful1lled. We consider the cases vj ¡vn ;
vj = vn ; vˆ1 ¿ v1 ;
vj = vn ; vˆ1 = v1
and derive a contradiction for each possibility. Let E = E(Knd ) denote the set of edges of Knd : Case 1: vj ¡vn : Then the vertices wJ 1 = vj + vˆ1 + · · · + vˆd−2 + vn and wJ 2 = vˆd + vˆ1 + · · · + vˆd−2 + vn are isolated vertices. We obtain wJ 1 = vj + vˆ1 + · · · + vˆd−2 + vn = (vˆ1 + · · · + vˆd ) + vˆ1 + · · · + vˆd−2 + vn = vˆ1 + · · · + vˆd−1 + (vˆd + vˆ1 + · · · + vˆd−2 + vn ) = vˆ1 + · · · + vˆd−1 + wJ 2 : Hence {vˆ1 ; : : : ; vˆd−1 ; wJ 2 } ∈ E, which is a contradiction because wJ 2 is an isolated vertex. Case 2: vj = vn ; vˆ1 ¿ v1 : We consider wJ 3 = vˆd + v1 + vˆ1 + · · · + vˆd−2 and distinguish two subcases. (a) wJ 3 is an isolated vertex. Clearly, wJ 4 = vn + v1 + vˆ1 + · · · + vˆd−2 is also an isolated vertex. We obtain wJ 4 = vn + v1 + vˆ1 + · · · + vˆd−2 = (vˆ1 + · · · + vˆd ) + v1 + vˆ1 + · · · + vˆd−2 = vˆ1 + · · · + vˆd−1 + (vˆd + v1 + vˆ1 + · · · + vˆd−2 ) = vˆ1 + · · · + vˆd−1 + wJ 3 : Hence {vˆ1 ; : : : ; vˆd−1 ; wJ 3 } ∈ E, which is a contradiction because wJ 3 is an isolated vertex. (b) wJ 3 = vr ∈ Vn − {vˆ1 ; : : : ; vˆd−2 ; vn }. As in subcase (a) wJ 4 and furthermore wJ 5 = vn + vr + vˆ1 + · · · + vˆd−2 are isolated vertices. We obtain wJ 5 = vn +vr +vˆ1 + · · · +vˆd−2 =vn +(vˆd +v1 +vˆ1 + · · · +vˆd−2 )+vˆ1 + · · · +vˆd−2 = (vn +v1 +vˆ1 + · · · +vˆd−2 )+vˆd +vˆ1 + · · · +vˆd−2 =wJ 4 +vˆd +vˆ1 + · · · +vˆd−2 : Hence {wJ 4 ; vˆd ; vˆ1 ; : : : ; vˆd−2 } ∈ E, which is a contradiction because wJ 4 is an isolated vertex. Case 3: vj = vn ; vˆ1 = v1 :
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Because of d6n−2 there exists a vt ∈ Vn −{vˆ1 ; : : : ; vˆd ; vj }: Further vˆ1 =v1 ¡vt implies that wJ 6 = vt + vˆ2 + · · · + vˆd is an isolated vertex and — because of vˆd ¡vn — it follows that wJ 7 = vt + vˆ2 + · · · + vˆd−1 + vn is another isolated vertex. We obtain wJ 7 = vt + vˆ2 + · · · + vˆd−1 + vn = vt + vˆ2 + · · · + vˆd−1 + (vˆ1 + · · · + vˆd ) = (vt + vˆ2 + · · · + vˆd−1 + vˆd ) + vˆ1 + · · · + vˆd−1 = wJ 6 + vˆ1 + · · · + vˆd−1 : Hence {wJ 6 ; vˆ1 ; : : : ; vˆd−1 } ∈ E, which is a contradiction because wJ 6 is an isolated vertex. Thus for each of the possible cases the assumption vˆ1 + · · · + vˆd = vj implies a contradiction and the proof is completed. As the main result, we determine the sum number of d-uniform complete hypergraphs for n¿d + 2; as a generalization of Theorem 1.1 we obtain: Theorem 2.4. For n¿d + 2 the sum number of complete d-uniform hypergraphs is given by (Knd ) = d(n − d) + 1: Proof. We use again the notations Vn = {v1 ; : : : ; vn } and W = {w1 ; : : : ; w } for the labelings of the vertices of Knd and the (Knd ) isolated vertices, resp., such that H = Knd ∪ {w1 ; : : : ; w } is a sum hypergraph. From Lemma 2.1, it follows that there are d(n − d) + 1 pairwise di0erent sums vˆ1 + · · ·+vˆd ∈ Vn ∪W ; {vˆ1 ; : : : ; vˆd } ⊂ Vn and by Lemma 2.3 even vˆ1 +· · ·+vˆd ∈ W is ful1lled. This consideration yields (Knd )¿d(n−d)+1: It remains to show (Knd )6d(n−d)+1; for this purpose we choose the following labeling: Let N ¿dn − d2 ¿n be a 1xed value and de1ne Vn = {i + N : i ∈ {1; : : : ; n}}; d(d + 1) d(d − 1) : W = dN + ; : : : ; dN + dn − 2 2 Obviously, Vn ∩ W = ∅ and each sum vˆ1 + · · · + vˆd ∈ W . Further a simple calculation yields |W | = d(n − d) + 1: The proof is completed, if we can show that all the labels wj ∈ W belong to isolated vertices. For this end we consider a sum s of d labels and suppose that at least one of these labels belongs to W . Using N ¿dn − d2 we obtain d(d − 1) d(d + 1) + (d − 1)N + s ¿ dN + 2 2 d(d + 1) ¿ dN + + (N + 1) 2 d(d + 1) d(d − 1) ¿ dN + + (dn − d2 + 1) = dN + dn − +1 2 2 ¿ max{v: v ∈ Vn ∪ W };
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i.e. s ∈Vn ∪ W and the sum hypergraph property implies that no vertex of W belongs to an edge. Remark 2.5. We can show that also the following two labelings, each containing the number 1 as minimum label, are sum hypergraph labelings of Knd ∪ {w1 ; : : : ; w }: 1. n¿d + 2 vi = 1 + (2d − b)(i − 1); wj = d + (2d − b)j; with b =
j=
i = 1; : : : ; n; d−1 k=0
n−1
k; : : : ;
k
k=n−d
0 if d even; 1 if d odd:
2. n = d + 1 vi = 10i−1 ; vn =
n−1
i = 1; : : : ; n − 1;
10i−1 ;
i=1
wi ;
i = 1; : : : ; n − 1
are the remaining sums of (n − 1) labels vi (each containing vn ). Chen [3] as well as Sharary [7] showed for complete graphs that (Kn ) = 0 for n63 and (Kn ) = (Kn ) for n¿4: We did not 1nd the exact integral sum number for d-uniform complete hypergraphs, but give a weaker result. Theorem 2.6. For the integral sum number of d-uniform complete hypergraphs holds
(Knd ) = 0
for d = n; n − 1;
(d − 1)(n − d − 1)6 (Knd )6d(n − d) + 1
for d6n − 2:
Proof. By Vn = {v1 ; : : : ; vn } we denote a labeling of the vertices of Knd . For 36d = n, the desired result follows from v1 = 0; v2 = −(v3 + · · · + vn ). For d = n − 1 we choose Vn = {1; : : : ; n=2; −1; : : : ; −n=2} if n is even and Vn = {0; 1; : : : ; (n − 1)=2; −1; : : : ; −(n − 1)=2} if n is odd. For the general case, we use the trivium (Knd )6 (Knd ) and the upper bound results from Theorem 2.4. Further, Lemma 2.1 is also valid for Vn ⊂ Z. Hence there are at least d(n − d) + 1 − n = (d − 1)(n − d − 1) labels necessary for isolated vertices. Conjecture 2.7. (Knd ) = (Knd ) for d6n − 2:
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3. Hypertrees In the case of graphs, i.e. 2-uniform hypergraphs, it is known (cf. [6,3]) that there exist trees with integral sum number 0 (e.g. the path Pn ) as well as trees with integral sum number 1. Because of Ellingham’s Theorem 1.2 we obtain (T ) = (T ) for in1nitely many trees T . For d¿3 we are able to show that all d-uniform hypertrees have integral sum number 0. To demonstrate this result we make use of a special labeling algorithm. Theorem 3.1. If H = (V; E) is a d-uniform hypertree and d¿3; then (H) = 0. Proof. It is suMcient to consider non-trivial hypertrees H = (V; E), i.e. |V | ¿ 1. We construct a labeling r: V → Z of the vertices of H such that the set S := {r(v): v ∈ V } de1nes an integral sum hypergraph H+ d (S) which is isomorphic to H. The labeling r ∗ induces a labeling r (e) := v∈e r(v) of the edges e ∈ E of H. At 1rst, we apply the following algorithm to the given hypertree H: Algorithm. 1. Let e ∈ E, let v ∈ e such that in case |E| ¿ 1 the degree of v is greater than 1, put r(v) := 2 and S := {v}. 2. If |S| ¿ 1, let v ∈ S; e ∈ E such that • e ∩ S = {v} and • r(v) = min{r(v ): v ∈ S ∧ ∃e ∈ E: e ∩ S = {v }}. Put {v1 = v; v2 ; : : : ; vd } := e. If |S| = 1 then put r(v2 ) := 3 and S := S ∪ {v2 }. (a) If there exists a completely labeled edge e , i.e. e ⊆ S, then put r(v2 ) := r ∗ (e) and S := S ∪ {v2 }, where e is the just completely labeled edge. (b) For j = 3; 4; : : : ; d label vj by the sum of the labels of all vertices of S plus |S|, i.e.: (i) j := 3; (ii) r(vj ) := v∈S (r(v) + 1); (iii) S := S ∪ {vj }; (iv) j := j + 1; (v) if j6d then go to (ii) 3. If V − S =∅ then go to 2. 4. Let {v1 ; v2 ; : : : ; vn } := V such that r(v1 )¡r(v2 )¡ · · · ¡r(vn ): Let e = {vi1 ; vi2 ; : : : ; vid } be the last completely labeled edge. If vi1 = v1 or vi1 = v3 ∧ d = 3, then r(vn ) := 3 −
v∈e v=vn
r(v);
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else r(vn ) := 2 −
r(v):
v∈e v=vn
5. Stop. We identify the vertices v1 ; v2 ; : : : ; vn−1 ; vn with their labels r(v1 ); r(v2 ); : : : ; r(vn−1 ); r(vn ): Obviously, the indices of v1 ; v2 ; : : : ; vn correspond to the order of the labeling of the vertices in the algorithm and vn ¡0¡v1 = 2¡v2 = 3¡v3 ¡ · · · ¡vn−1 : The feasibility of the algorithm follows from the absence of cycles in hypertrees. Moreover, it is easy to see that ∀e ∈ E ∃j ∈ {1; 2; : : : ; n − 1}: r ∗ (e) = vj :
(1)
It remains to show that ∀k ∈ {1; 2; : : : ; n} ∀J ⊆{1; 2; : : : ; n}: |J | = d ∧ vk =
vj ⇒ {vj : j ∈ J } ∈ E:
(2)
j∈J
Because (2) is trivial for |E| = 1, we suppose |E| ¿ 1. At 1rst, we consider the case n ∈J . Of course, k¡n, and vk has been labeled in step 2(a). Let J = {j1 ; j2 ; : : : ; jd } ⊆ {1; 2; : : : ; n − 1} with vk = j∈J vj and vj1 ¡vj2 ¡ · · · ¡vjd (¡vk ). Moreover, let e = {vi1 ; vi2 ; : : : ; vid } ∈ E be the edge used to label vk in step 2(a), where vi1 ¡vi2 ¡ · · · ¡vid (¡vk ). Because vi3 ; vi4 ; : : : ; vid have been labeled in step 2(b) and k = id + 1 = id−1 + 2 = · · · = i3 + d − 2 holds, we have vid = vjd ; vid−1 = vjd−1 ; : : : ; vi3 = vj3
and
vi1 + vi2 = vj1 + vj2 :
Owing to i2 = i3 − 1, vi2 ¿vj2 ¿ vj1 and — if i2 ¿ 2 — because of vi2 ¿ vi2 −1 + vi2 −2 we obtain vi2 = vj2 as well as vi1 = vj1 , i.e. {vj : j ∈ J } = e ∈ E. Consequently, it remains to verify (2) for the case n ∈ J . Let e# = {vi1 ; vi2 = vn−d+2 ; vi3 = vn−d+3 ; : : : ; vid = vn } be the last completely labeled edge in our algorithm. (Then, in (2) {vj : j ∈ J } ∈ E is equivalent to {vj : j ∈ J } = e# :) Now, assume J ⊆{1; 2; : : : ; n}, V := {vj : j ∈ J } ⊆ V and k6n − 1 with v ∈ V: (3) |V | = d ∧ vn ∈ V ∧ V ∈E ∧ vk = v∈V
Above all, we demonstrate vn−d+2 ; vn−d+3 ; : : : ; vn ∈ V :
(4)
Clearly, vn−d+3 ; vn−d+4 ; : : : ; vn−1 have been labeled in step 2(b) of the algorithm. Hence, every vertex vj (j ∈ {n − d + 3; n − d + 4; : : : ; n − 1}) has a label greater than the sum of all labels of the vertices v1 ; v2 ; : : : ; vj−1 . In connection with vn ∈ V and j∈J vj = vk ¿v1 = 2 = v∈e# v in case vi1 =v1 ∧ (vi1 =v3 ∨ d =3) we have vn−d+3 ; vn−d+4 ; : : : ; vn ∈ V :
(4 )
Since (4 ) is trivial for d = 3 and therefore also for vi1 = v3 ∧ d = 3, it remains to investigate the case vi1 = v1 ∧ d¿4.
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Then it holds 3 = v2 = v∈e# v, e# = {vi1 = 2; vn−d+2 ; vn−d+3 ; : : : ; vn } and vn−d+3 has been labeled in step 2(b). Assume, V = {vt1 ; vt2 ; vt3 ; vn−d+4 ; : : : ; vn } with vt1 ¡vt2 ¡vt3 ¡vn−d+3 : Because of vt1 + n−d+2 vt2 + vt3 ¡ i=1 (vi + 1) = vn−d+3 it follows that v∈V v¡(vi1 + vn−d+2 + vn−d+3 + vn−d+4 + · · · + vn ) − vi1 = v∈e# v − 2 = 3 − 2 = 1: Owing to v∈V v ¿ vn we obtain the contradiction v∈V v ∈V to (3). Analogously, the assumption that there is a j ∈ {n − d + 4; n − d + 5; : : : ; n − 1} with vj ∈V (instead of vn−d+3 ∈V ) provides the same contradiction. Hence, (4’) is valid in all possible cases. To 1nish the proof of (4) we assume vn−d+2 ∈V : In analogy with our preceding consideration it suMces to suppose V = {vt1 ; vt2 ; vn−d+3 ; : : : ; vn } with vt1 ¡vt2 ¡vn−d+2 : Obviously, vn−d+2 has got its label in step 2(a). Therefore, vn−d+2 ¿vn−d+1 + vn−d + vn−d−1 ¿ vt2 + vt1 holds and we get vn ¡ v∈V v = vt1 + vt2 + vn−d+3 + · · · + vn ¡(vi1 + vn−d+2 + vn−d+3 + · · · + vn ) − 2 = v∈e# v − 263 − 2 = 1. This contradicts again (3) and the proof of (4) is complete. Because of (3) and (4) there is a t ∈ {1; 2; : : : ; n − d + 1} − {i1 } such that
v=
v∈V
d−2
vn−l + vt :
(5)
l=0
Case 1: vi1 = v1 = 2: Step 2 of the algorithm provides ∀e; e ∈ E: e =e ⇒ e ∩ e = {v1 }:
d−2 Then, vi1 = 2 and vn = 3 − (vn−1 + vn−2 + · · · + vn−d+2 + vi1 ) imply l=0 vn−l = 1. Owing to (5) we obtain v = 1 + v ∈
V since v ¿ v + 1 holds for all t¿2. t t+1 t v∈V This contradicts (3) and Case 1 cannot occur. Case 2: vi1 ¿3: Because of (4), (5) and the de1nition of e# we see vt −vi1 = v∈V v− v∈e# v=vk −x, where 3 if i1 = 3 ∧ d = 3; x= 2 otherwise: Then
#
V =e implies vk =x and k =
2 if i1 = 3 ∧ d = 3; 1 otherwise:
Obviously, we have vk + vi1 = vt + x:
(6)
Assume t¡i1 . The case i1 = 3 ∧ d = 3 is impossible because t ∈ {1; 2} and vt + x = vt + 3 = vk + vi1 = vk + v3 = vk + 7 include vk + 7 ∈ {v1 + 3; v2 + 3} = {5; 6}, i.e. vk ∈ {−2; −1}. On the other hand, i1 =3∨d =3 provides vt +x =vt +2¡vi1 +26vi1 +vk incompatible with (6).
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Thus we have proved 26i1 ¡t:
(7)
Some simple considerations lead to 26k¡t:
(8)
Suppose k = 2. Then, because of 0 =vt − vi1 = vk − x, we have x = 2 and vt − vi1 = 1. But vt = vi1 + 1 contradicts t¿3 (cf. (7)). Hence we can sharpen (8) to 2¡k¡t:
(9)
Now, assume i1 = 2. d−2 d−2 On account of l=0 vn−l + vi1 = 2 and vi1 = v2 = 3 we get l=0 vn−l = −1. Then (3) in connection with (5) yields vt − 1 = vk and we obtain k = 1 contradictory to k¿3. So we have 2¡i1 ¡t:
(10)
Case 2.1: i1 =k: (9) and (10) imply t ¿ max{k; i1 }¿4 and from step 2 of the algorithm it follows vt ¿ vt−1 + vt−2 ¿vi1 + vk incompatible with (6). Case 2.2: i1 = k: Now we have t ¿ 3 and from (6) we obtain 2vk = vt + x:
(11)
The assumption k6t − 2 would result in the contradiction vt ¿ vt−1 + vt−2 ¿ vk + vk . Therefore, k = t − 1:
(12)
Case 2.2.1: vt was labeled in step 2(b). t−1 t−2 vt = i=1 (vi + 1) = (vt−1 + 1) + i=1 (vi + 1) ¿ vt−1 + vt−1 = 2vk contradictory to (11). Case 2.2.2: vt was labeled in step 2(a). t−1 There exists an l6t − d such that vt = i=t−d+1 vi + vl . Hence, 2vt−1 = vt + x = t−1 t−2 i=t−d+1 vi + vl + x and — using (11) — we see vt−1 = i=t−d+1 vi + vl + x. Because vt was labeled in step 2(a) the vertex vt−1 got its label in step 2(b) and it follows that {vl ; vt−d+1 ; vt−d+2 ; : : : ; vt−2 }={v1 ; v2 ; : : : ; vt−2 } as well as |{v1 ; v2 ; : : : ; vt−2 }|=x. The de1nition of x immediately provides 5 if i1 = 3 ∧ d = 3; t= 4 otherwise: Furthermore, in case i1 = 3 ∧ d = 3, i.e. t = 5, under consideration of (12) and Case 2.2 we obtain the contradiction 4 = t − 1 = k = i1 = 3. So i1 =3 ∨ d =3, i.e. t = 4, is valid. Since we are in Case 2.2.2 it follows from t = 4 that d = t − 1 = 3. Consequently, i1 ¿4 holds, contradictory to i1 = k = t − 1 = 3.
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This makes our proof complete. Having a look at the proof of Theorem 3.1 (especially at the labeling algorithm), we see that the d-uniformity as well as the absence of cycles in the hypergraphs under consideration are essential for the algorithm. It seems to be diMcult to use the given labeling algorithm (or a modi1cation of it) to determine the integral sum number of non-uniform hypertrees (cf. Theorem 1.3 for the sum number) or of hypergraphs which contain cycles. So the following problem arises: Problem 3.2. Is it possible to prove an analogue of Theorem 3.1 for non-uniform hypertrees? Acknowledgements The authors would like to thank the referees for their helpful suggestions. References [1] C. Berge, Hypergraphs, North-Holland, Amsterdam, New York, Oxford, Tokyo, 1989. [2] D. Bergstrand, F. Harary, K. Hodges, G. Jennings, L. Kuklinski, J. Wiener, The Sum Number of a Complete Graph, Bull. Malaysian Math. Soc. Ser. 2 12 (1989) 25–28. [3] Z. Chen, Harary’s conjectures on integral sum graphs, Discrete Math. 160 (1996) 241–244. [4] M.N. Ellingham, Sum graphs from trees, Ars Combin. 35 (1993) 335–349. [5] F. Harary, Sum graphs and di0erence graphs, Congr. Numer. 72 (1990) 101–108. [6] F. Harary, Sum graphs over all the integers, Discrete Math. 124 (1994) 99–105. [7] A. Sharary, Integral sum graphs from complete graphs, cycles and wheels, Arab Gulf J. Sci. Res. 14 (1) (1996) 1–14. [8] M. Sonntag, H.-M. Teichert, Sum numbers of hypertrees, Discrete Math. 214 (2000) 285–290.
www.elsevier.com/locate/disc
On the sum number and integral sum number of hypertrees and complete hypergraphs Martin Sonntaga; ∗ , Hanns-Martin Teichertb a Faculty
of Mathematics and Computer Science, TU Bergakademie Freiberg, Bernhard-von-Cotta-Str. 2, D-09596 Freiberg, Germany b Institute of Mathematics, Medical University of L) ubeck, Wallstra+e 40, D-23560 L)ubeck, Germany Received 17 October 1997; revised 29 September 1998; accepted 5 March 2000
Abstract A hypergraph H is a sum hypergraph i0 there are a 1nite S ⊂ N+ and d1 ; d2 ∈ N+ with 1¡d1 6d2 ¡2d1 such that H is isomorphic to the hypergraph Hd+1 ;d2 (S) = (V; E) where V := S and
E :=
{x1 ; x2 ; : : : ; xk }: k ∈ {d1 ; d1 + 1; : : : ; d2 } ∧ (i=j ⇒ xi =xj ) ∧
k
xi ∈ S
:
i=1
For an arbitrary hypergraph H, the sum number = (H) is de1ned to be the minimum number of isolated vertices w1 ; : : : ; w ∈V such that H ∪ {w1 ; : : : ; w } is a sum hypergraph. For S ⊂ Z we obtain the concept of integral sum hypergraphs and the integral sum number
(H): In this paper, we determine (Knd ) and give bounds for (Knd ), where Knd denotes the d-uniform complete hypergraph on n vertices. Moreover, for d¿3 we prove that every d-uniform c 2001 Elsevier Science B.V. All hypertree H is an integral sum hypergraph, i.e. (H) = 0. rights reserved.
1. Introduction and denitions The concept of sum graphs and integral sum graphs was introduced by Harary [5,6]. Many authors investigated these kinds of graphs in the last years, for a brief summary, see for instance [8]. In the following, we generalize the concepts mentioned above to hypergraphs. All hypergraphs considered here are supposed to be nonempty and 1nite without loops and multiple edges. In standard terminology we follow [1]. A hypergraph H = (V; E) with vertex set V and edge set E ⊂ P(V )−{∅} is d-uniform i0 26d ∈ N and |e| = d (∀e ∈ E). For d6n we denote the d-uniform complete hypergraph on n vertices by Knd . ∗
Corresponding author. E-mail address: [email protected] (M. Sonntag).
c 2001 Elsevier Science B.V. All rights reserved. 0012-365X/01/$ - see front matter PII: S 0 0 1 2 - 3 6 5 X ( 0 0 ) 0 0 4 5 2 - 0
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Let S ⊂ N+ be 1nite and d1 ; d2 ∈ N such that 1¡d1 6d2 ¡2d1 . Hd+1 ; d2 (S) = (V; E) is called the (d1 ; d2 )-sum hypergraph of S i0 V = S and k xi ∈ S : E = {x1 ; x2 ; : : : ; xk }: k ∈ {d1 ; d1 + 1; : : : ; d2 } ∧ (i =j ⇒ xi =xj ) ∧ i=1
Furthermore, a hypergraph H = (V; E) is a sum hypergraph i0 there exist S ⊂ N+ and d1 ; d2 ∈ N such that H is isomorphic to Hd+1 ; d2 (S). In the following, in e0ect we will deal with d-uniform hypergraphs. For this end we de1ne Hd+ (S) := Hd;+d (S). Then, for d = 2 we obtain the known concept of sum graphs. For a hypergraph H, the sum number = (H) is de1ned to be the minimum number of isolated vertices w1 ; : : : ; w ∈V such that H ∪ {w1 ; : : : ; w } is a sum hypergraph. If also nonpositive integers are allowed as elements of S; i.e. S ⊂ Z, we obtain the de1nition of integral sum hypergraphs and the integral sum number = (H) in the same manner. One 1eld of investigation of sum graphs and integral sum graphs is the determination of (G) and (G), for several classes of graphs G. For the complete graphs Kn Bergstrand et al. [2] proved Theorem 1.1. For n¿4 the sum number of complete graphs is given by (Kn ) = 2n − 3: For (Kn ), it was shown by Chen [3] as well as by Sharary [7] that (Kn ) = (Kn ) is ful1lled for n¿4. In the 1rst part of this paper, we generalize Theorem 1.1 and determine (Knd ): Further we show (Knn ) = (Knn−1 ) = 0 and give bounds for (Knd ) if d6n − 2: In the second part of our article, we investigate connected hypergraphs with minimum number of edges, i.e. hypertrees. An edge sequence w = (v0 ; e1 ; v1 ; e2 ; : : : ; et ; vt = v0 ) with t¿2 pairwise distinct edges e1 ; e2 ; : : : ; et ∈ E and pairwise distinct vertices v0 ; v1 ; : : : ; vt−1 ∈ V is called a cycle of the hypergraph H = (V; E) i0 {vk−1 ; vk } ⊆ ek (∀k ∈ {1; 2; : : : ; t}). A hypergraph H = (V; E) is a hypertree i0 H is connected and cycle-free. We call the hypergraph H = (V; E) non-trivial i0 |V | ¿ 1. For graphs Ellingham [4] veri1ed Theorem 1.2. Every non-trivial tree G = (V; E) has sum number (G) = 1. In [8] this result was generalized to Theorem 1.3. If H = (V; E) is a non-trivial hypertree and max{|e|: e ∈ E}¡2 min{|e|: e ∈ E} − 1; then (H) = 1:
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In Section 3 we prove that d-uniform hypertrees are integral sum hypergraphs, if d¿3. 2. Complete hypergraphs In the following, let v1 ¡v2 ¡ · · · ¡vn be a labeling of the vertices of Knd and w1 ; : : : ; w a labeling of = (Knd ) isolated vertices such that H = Knd ∪ {w1 ; : : : ; w } can be recognized as a sum hypergraph. Further we choose the notation Vn ={v1 ; : : : ; vn } and W = {w1 ; : : : ; w }: Lemma 2.1. There are at least d(n − d) + 1 pairwise di7erent sums of vertices vˆ1 + · · · + vˆd ;
where vˆ1 ; : : : ; vˆd ∈ Vn and vˆ1 ¡ · · · ¡vˆd :
Proof. We consider the following sets of sums, each containing d summands: Sd = {v1 + · · · + vd−1 + vˆd : vˆd ∈ {vd ; : : : ; vn }}; Sk = {v1 + · · · + vk−1 + vˆk + vn+(k+1)−d + · · · + vn : vˆk ∈ {vk+1 ; : : : ; vn+k−d }}; k = d − 1; : : : ; 1: Clearly, each sum of Si is strictly larger than each sum of Si+1 for i = 1; : : : ; d − 1; thus Si ∩ Sj = ∅ for i =j. Further |Sd | = n − d + 1; |Sk | = n − d for k = d − 1; : : : ; 1: Hence, there are (n − d + 1) + (d − 1)(n − d) = d(n − d) + 1 pairwise di0erent sums vˆ1 + · · · + vˆd . Obviously, in case of d = n we have (Knn ) = 1 for n¿2: For d = n − 1 we know for graphs that (K3 ) = (K32 ) = 2 (see [2]). This result is generalized by Lemma 2.2. (Knn−1 ) = n − 1 for n¿3: Proof. There are exactly n edges in Knn−1 and only the maximum label vn ∈ Vn is chooseable as sum of the other labels v1 ; : : : ; vn−1 ; hence (Knn−1 )¿n − 1: Using a 1xed N ¿n−2 we label the vertices of Knn−1 in the following way: vi =N +i for i = 1; : : : ; n − 1; vn = N (n − 1) + (n(n − 1))=2 and choose for (n − 1) other vertices the labels w1 ¡ · · · ¡wn−1 as the remaining sums of (n − 1) labels vi (each containing the label vn ). This yields W = {N (2n − 3) + (n − 1)2 ; : : : ; N (2n − 3) + n(n − 1) − 1}. It remains to show that these vertices are isolates. For this end, we consider a sum s of d labels and suppose that at least one of these labels belongs to W . Using N ¿n − 2 it follows s ¿ (N (2n − 3) + (n − 1)2 ) + ((n − 2)N + ((n − 1)(n − 2))=2) ¿ (N (2n − 3) + (n − 1)2 ) + (N + 1)¿(N (2n − 3) + (n − 1)2 ) + (n − 1) = N (2n − 3) + n(n − 1) ¿ max{v: v ∈ Vn ∪ W };
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i.e. s ∈Vn ∪ W and the sum hypergraph property implies that all vertices of W are isolates. Now, we consider the number of possibilities to choose pairwise distinct vˆ1 ; vˆ2 ; : : : ; vˆd ; vj ∈ Vn such that vˆ1 + · · · + vˆd = vj . As mentioned in the proof above, for d = n − 1 only the sum v1 + · · · + vd = vn can occur. The question is more complicated for d6n − 2. We need the following interesting result for proving the main theorem. Lemma 2.3. For d6n − 2 there exist no labels {vˆ1 ; vˆ2 ; : : : ; vˆd ; vj } ⊂ Vn such that vˆ1 + vˆ2 + · · · + vˆd = vj : Proof. Assume, there are elements vˆ1 ¡vˆ2 ¡ · · · ¡vˆd ¡vj of Vn such that vˆ1 + vˆ2 + · · · + vˆd = vj is ful1lled. We consider the cases vj ¡vn ;
vj = vn ; vˆ1 ¿ v1 ;
vj = vn ; vˆ1 = v1
and derive a contradiction for each possibility. Let E = E(Knd ) denote the set of edges of Knd : Case 1: vj ¡vn : Then the vertices wJ 1 = vj + vˆ1 + · · · + vˆd−2 + vn and wJ 2 = vˆd + vˆ1 + · · · + vˆd−2 + vn are isolated vertices. We obtain wJ 1 = vj + vˆ1 + · · · + vˆd−2 + vn = (vˆ1 + · · · + vˆd ) + vˆ1 + · · · + vˆd−2 + vn = vˆ1 + · · · + vˆd−1 + (vˆd + vˆ1 + · · · + vˆd−2 + vn ) = vˆ1 + · · · + vˆd−1 + wJ 2 : Hence {vˆ1 ; : : : ; vˆd−1 ; wJ 2 } ∈ E, which is a contradiction because wJ 2 is an isolated vertex. Case 2: vj = vn ; vˆ1 ¿ v1 : We consider wJ 3 = vˆd + v1 + vˆ1 + · · · + vˆd−2 and distinguish two subcases. (a) wJ 3 is an isolated vertex. Clearly, wJ 4 = vn + v1 + vˆ1 + · · · + vˆd−2 is also an isolated vertex. We obtain wJ 4 = vn + v1 + vˆ1 + · · · + vˆd−2 = (vˆ1 + · · · + vˆd ) + v1 + vˆ1 + · · · + vˆd−2 = vˆ1 + · · · + vˆd−1 + (vˆd + v1 + vˆ1 + · · · + vˆd−2 ) = vˆ1 + · · · + vˆd−1 + wJ 3 : Hence {vˆ1 ; : : : ; vˆd−1 ; wJ 3 } ∈ E, which is a contradiction because wJ 3 is an isolated vertex. (b) wJ 3 = vr ∈ Vn − {vˆ1 ; : : : ; vˆd−2 ; vn }. As in subcase (a) wJ 4 and furthermore wJ 5 = vn + vr + vˆ1 + · · · + vˆd−2 are isolated vertices. We obtain wJ 5 = vn +vr +vˆ1 + · · · +vˆd−2 =vn +(vˆd +v1 +vˆ1 + · · · +vˆd−2 )+vˆ1 + · · · +vˆd−2 = (vn +v1 +vˆ1 + · · · +vˆd−2 )+vˆd +vˆ1 + · · · +vˆd−2 =wJ 4 +vˆd +vˆ1 + · · · +vˆd−2 : Hence {wJ 4 ; vˆd ; vˆ1 ; : : : ; vˆd−2 } ∈ E, which is a contradiction because wJ 4 is an isolated vertex. Case 3: vj = vn ; vˆ1 = v1 :
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Because of d6n−2 there exists a vt ∈ Vn −{vˆ1 ; : : : ; vˆd ; vj }: Further vˆ1 =v1 ¡vt implies that wJ 6 = vt + vˆ2 + · · · + vˆd is an isolated vertex and — because of vˆd ¡vn — it follows that wJ 7 = vt + vˆ2 + · · · + vˆd−1 + vn is another isolated vertex. We obtain wJ 7 = vt + vˆ2 + · · · + vˆd−1 + vn = vt + vˆ2 + · · · + vˆd−1 + (vˆ1 + · · · + vˆd ) = (vt + vˆ2 + · · · + vˆd−1 + vˆd ) + vˆ1 + · · · + vˆd−1 = wJ 6 + vˆ1 + · · · + vˆd−1 : Hence {wJ 6 ; vˆ1 ; : : : ; vˆd−1 } ∈ E, which is a contradiction because wJ 6 is an isolated vertex. Thus for each of the possible cases the assumption vˆ1 + · · · + vˆd = vj implies a contradiction and the proof is completed. As the main result, we determine the sum number of d-uniform complete hypergraphs for n¿d + 2; as a generalization of Theorem 1.1 we obtain: Theorem 2.4. For n¿d + 2 the sum number of complete d-uniform hypergraphs is given by (Knd ) = d(n − d) + 1: Proof. We use again the notations Vn = {v1 ; : : : ; vn } and W = {w1 ; : : : ; w } for the labelings of the vertices of Knd and the (Knd ) isolated vertices, resp., such that H = Knd ∪ {w1 ; : : : ; w } is a sum hypergraph. From Lemma 2.1, it follows that there are d(n − d) + 1 pairwise di0erent sums vˆ1 + · · ·+vˆd ∈ Vn ∪W ; {vˆ1 ; : : : ; vˆd } ⊂ Vn and by Lemma 2.3 even vˆ1 +· · ·+vˆd ∈ W is ful1lled. This consideration yields (Knd )¿d(n−d)+1: It remains to show (Knd )6d(n−d)+1; for this purpose we choose the following labeling: Let N ¿dn − d2 ¿n be a 1xed value and de1ne Vn = {i + N : i ∈ {1; : : : ; n}}; d(d + 1) d(d − 1) : W = dN + ; : : : ; dN + dn − 2 2 Obviously, Vn ∩ W = ∅ and each sum vˆ1 + · · · + vˆd ∈ W . Further a simple calculation yields |W | = d(n − d) + 1: The proof is completed, if we can show that all the labels wj ∈ W belong to isolated vertices. For this end we consider a sum s of d labels and suppose that at least one of these labels belongs to W . Using N ¿dn − d2 we obtain d(d − 1) d(d + 1) + (d − 1)N + s ¿ dN + 2 2 d(d + 1) ¿ dN + + (N + 1) 2 d(d + 1) d(d − 1) ¿ dN + + (dn − d2 + 1) = dN + dn − +1 2 2 ¿ max{v: v ∈ Vn ∪ W };
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i.e. s ∈Vn ∪ W and the sum hypergraph property implies that no vertex of W belongs to an edge. Remark 2.5. We can show that also the following two labelings, each containing the number 1 as minimum label, are sum hypergraph labelings of Knd ∪ {w1 ; : : : ; w }: 1. n¿d + 2 vi = 1 + (2d − b)(i − 1); wj = d + (2d − b)j; with b =
j=
i = 1; : : : ; n; d−1 k=0
n−1
k; : : : ;
k
k=n−d
0 if d even; 1 if d odd:
2. n = d + 1 vi = 10i−1 ; vn =
n−1
i = 1; : : : ; n − 1;
10i−1 ;
i=1
wi ;
i = 1; : : : ; n − 1
are the remaining sums of (n − 1) labels vi (each containing vn ). Chen [3] as well as Sharary [7] showed for complete graphs that (Kn ) = 0 for n63 and (Kn ) = (Kn ) for n¿4: We did not 1nd the exact integral sum number for d-uniform complete hypergraphs, but give a weaker result. Theorem 2.6. For the integral sum number of d-uniform complete hypergraphs holds
(Knd ) = 0
for d = n; n − 1;
(d − 1)(n − d − 1)6 (Knd )6d(n − d) + 1
for d6n − 2:
Proof. By Vn = {v1 ; : : : ; vn } we denote a labeling of the vertices of Knd . For 36d = n, the desired result follows from v1 = 0; v2 = −(v3 + · · · + vn ). For d = n − 1 we choose Vn = {1; : : : ; n=2; −1; : : : ; −n=2} if n is even and Vn = {0; 1; : : : ; (n − 1)=2; −1; : : : ; −(n − 1)=2} if n is odd. For the general case, we use the trivium (Knd )6 (Knd ) and the upper bound results from Theorem 2.4. Further, Lemma 2.1 is also valid for Vn ⊂ Z. Hence there are at least d(n − d) + 1 − n = (d − 1)(n − d − 1) labels necessary for isolated vertices. Conjecture 2.7. (Knd ) = (Knd ) for d6n − 2:
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3. Hypertrees In the case of graphs, i.e. 2-uniform hypergraphs, it is known (cf. [6,3]) that there exist trees with integral sum number 0 (e.g. the path Pn ) as well as trees with integral sum number 1. Because of Ellingham’s Theorem 1.2 we obtain (T ) = (T ) for in1nitely many trees T . For d¿3 we are able to show that all d-uniform hypertrees have integral sum number 0. To demonstrate this result we make use of a special labeling algorithm. Theorem 3.1. If H = (V; E) is a d-uniform hypertree and d¿3; then (H) = 0. Proof. It is suMcient to consider non-trivial hypertrees H = (V; E), i.e. |V | ¿ 1. We construct a labeling r: V → Z of the vertices of H such that the set S := {r(v): v ∈ V } de1nes an integral sum hypergraph H+ d (S) which is isomorphic to H. The labeling r ∗ induces a labeling r (e) := v∈e r(v) of the edges e ∈ E of H. At 1rst, we apply the following algorithm to the given hypertree H: Algorithm. 1. Let e ∈ E, let v ∈ e such that in case |E| ¿ 1 the degree of v is greater than 1, put r(v) := 2 and S := {v}. 2. If |S| ¿ 1, let v ∈ S; e ∈ E such that • e ∩ S = {v} and • r(v) = min{r(v ): v ∈ S ∧ ∃e ∈ E: e ∩ S = {v }}. Put {v1 = v; v2 ; : : : ; vd } := e. If |S| = 1 then put r(v2 ) := 3 and S := S ∪ {v2 }. (a) If there exists a completely labeled edge e , i.e. e ⊆ S, then put r(v2 ) := r ∗ (e) and S := S ∪ {v2 }, where e is the just completely labeled edge. (b) For j = 3; 4; : : : ; d label vj by the sum of the labels of all vertices of S plus |S|, i.e.: (i) j := 3; (ii) r(vj ) := v∈S (r(v) + 1); (iii) S := S ∪ {vj }; (iv) j := j + 1; (v) if j6d then go to (ii) 3. If V − S =∅ then go to 2. 4. Let {v1 ; v2 ; : : : ; vn } := V such that r(v1 )¡r(v2 )¡ · · · ¡r(vn ): Let e = {vi1 ; vi2 ; : : : ; vid } be the last completely labeled edge. If vi1 = v1 or vi1 = v3 ∧ d = 3, then r(vn ) := 3 −
v∈e v=vn
r(v);
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else r(vn ) := 2 −
r(v):
v∈e v=vn
5. Stop. We identify the vertices v1 ; v2 ; : : : ; vn−1 ; vn with their labels r(v1 ); r(v2 ); : : : ; r(vn−1 ); r(vn ): Obviously, the indices of v1 ; v2 ; : : : ; vn correspond to the order of the labeling of the vertices in the algorithm and vn ¡0¡v1 = 2¡v2 = 3¡v3 ¡ · · · ¡vn−1 : The feasibility of the algorithm follows from the absence of cycles in hypertrees. Moreover, it is easy to see that ∀e ∈ E ∃j ∈ {1; 2; : : : ; n − 1}: r ∗ (e) = vj :
(1)
It remains to show that ∀k ∈ {1; 2; : : : ; n} ∀J ⊆{1; 2; : : : ; n}: |J | = d ∧ vk =
vj ⇒ {vj : j ∈ J } ∈ E:
(2)
j∈J
Because (2) is trivial for |E| = 1, we suppose |E| ¿ 1. At 1rst, we consider the case n ∈J . Of course, k¡n, and vk has been labeled in step 2(a). Let J = {j1 ; j2 ; : : : ; jd } ⊆ {1; 2; : : : ; n − 1} with vk = j∈J vj and vj1 ¡vj2 ¡ · · · ¡vjd (¡vk ). Moreover, let e = {vi1 ; vi2 ; : : : ; vid } ∈ E be the edge used to label vk in step 2(a), where vi1 ¡vi2 ¡ · · · ¡vid (¡vk ). Because vi3 ; vi4 ; : : : ; vid have been labeled in step 2(b) and k = id + 1 = id−1 + 2 = · · · = i3 + d − 2 holds, we have vid = vjd ; vid−1 = vjd−1 ; : : : ; vi3 = vj3
and
vi1 + vi2 = vj1 + vj2 :
Owing to i2 = i3 − 1, vi2 ¿vj2 ¿ vj1 and — if i2 ¿ 2 — because of vi2 ¿ vi2 −1 + vi2 −2 we obtain vi2 = vj2 as well as vi1 = vj1 , i.e. {vj : j ∈ J } = e ∈ E. Consequently, it remains to verify (2) for the case n ∈ J . Let e# = {vi1 ; vi2 = vn−d+2 ; vi3 = vn−d+3 ; : : : ; vid = vn } be the last completely labeled edge in our algorithm. (Then, in (2) {vj : j ∈ J } ∈ E is equivalent to {vj : j ∈ J } = e# :) Now, assume J ⊆{1; 2; : : : ; n}, V := {vj : j ∈ J } ⊆ V and k6n − 1 with v ∈ V: (3) |V | = d ∧ vn ∈ V ∧ V ∈E ∧ vk = v∈V
Above all, we demonstrate vn−d+2 ; vn−d+3 ; : : : ; vn ∈ V :
(4)
Clearly, vn−d+3 ; vn−d+4 ; : : : ; vn−1 have been labeled in step 2(b) of the algorithm. Hence, every vertex vj (j ∈ {n − d + 3; n − d + 4; : : : ; n − 1}) has a label greater than the sum of all labels of the vertices v1 ; v2 ; : : : ; vj−1 . In connection with vn ∈ V and j∈J vj = vk ¿v1 = 2 = v∈e# v in case vi1 =v1 ∧ (vi1 =v3 ∨ d =3) we have vn−d+3 ; vn−d+4 ; : : : ; vn ∈ V :
(4 )
Since (4 ) is trivial for d = 3 and therefore also for vi1 = v3 ∧ d = 3, it remains to investigate the case vi1 = v1 ∧ d¿4.
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Then it holds 3 = v2 = v∈e# v, e# = {vi1 = 2; vn−d+2 ; vn−d+3 ; : : : ; vn } and vn−d+3 has been labeled in step 2(b). Assume, V = {vt1 ; vt2 ; vt3 ; vn−d+4 ; : : : ; vn } with vt1 ¡vt2 ¡vt3 ¡vn−d+3 : Because of vt1 + n−d+2 vt2 + vt3 ¡ i=1 (vi + 1) = vn−d+3 it follows that v∈V v¡(vi1 + vn−d+2 + vn−d+3 + vn−d+4 + · · · + vn ) − vi1 = v∈e# v − 2 = 3 − 2 = 1: Owing to v∈V v ¿ vn we obtain the contradiction v∈V v ∈V to (3). Analogously, the assumption that there is a j ∈ {n − d + 4; n − d + 5; : : : ; n − 1} with vj ∈V (instead of vn−d+3 ∈V ) provides the same contradiction. Hence, (4’) is valid in all possible cases. To 1nish the proof of (4) we assume vn−d+2 ∈V : In analogy with our preceding consideration it suMces to suppose V = {vt1 ; vt2 ; vn−d+3 ; : : : ; vn } with vt1 ¡vt2 ¡vn−d+2 : Obviously, vn−d+2 has got its label in step 2(a). Therefore, vn−d+2 ¿vn−d+1 + vn−d + vn−d−1 ¿ vt2 + vt1 holds and we get vn ¡ v∈V v = vt1 + vt2 + vn−d+3 + · · · + vn ¡(vi1 + vn−d+2 + vn−d+3 + · · · + vn ) − 2 = v∈e# v − 263 − 2 = 1. This contradicts again (3) and the proof of (4) is complete. Because of (3) and (4) there is a t ∈ {1; 2; : : : ; n − d + 1} − {i1 } such that
v=
v∈V
d−2
vn−l + vt :
(5)
l=0
Case 1: vi1 = v1 = 2: Step 2 of the algorithm provides ∀e; e ∈ E: e =e ⇒ e ∩ e = {v1 }:
d−2 Then, vi1 = 2 and vn = 3 − (vn−1 + vn−2 + · · · + vn−d+2 + vi1 ) imply l=0 vn−l = 1. Owing to (5) we obtain v = 1 + v ∈
V since v ¿ v + 1 holds for all t¿2. t t+1 t v∈V This contradicts (3) and Case 1 cannot occur. Case 2: vi1 ¿3: Because of (4), (5) and the de1nition of e# we see vt −vi1 = v∈V v− v∈e# v=vk −x, where 3 if i1 = 3 ∧ d = 3; x= 2 otherwise: Then
#
V =e implies vk =x and k =
2 if i1 = 3 ∧ d = 3; 1 otherwise:
Obviously, we have vk + vi1 = vt + x:
(6)
Assume t¡i1 . The case i1 = 3 ∧ d = 3 is impossible because t ∈ {1; 2} and vt + x = vt + 3 = vk + vi1 = vk + v3 = vk + 7 include vk + 7 ∈ {v1 + 3; v2 + 3} = {5; 6}, i.e. vk ∈ {−2; −1}. On the other hand, i1 =3∨d =3 provides vt +x =vt +2¡vi1 +26vi1 +vk incompatible with (6).
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Thus we have proved 26i1 ¡t:
(7)
Some simple considerations lead to 26k¡t:
(8)
Suppose k = 2. Then, because of 0 =vt − vi1 = vk − x, we have x = 2 and vt − vi1 = 1. But vt = vi1 + 1 contradicts t¿3 (cf. (7)). Hence we can sharpen (8) to 2¡k¡t:
(9)
Now, assume i1 = 2. d−2 d−2 On account of l=0 vn−l + vi1 = 2 and vi1 = v2 = 3 we get l=0 vn−l = −1. Then (3) in connection with (5) yields vt − 1 = vk and we obtain k = 1 contradictory to k¿3. So we have 2¡i1 ¡t:
(10)
Case 2.1: i1 =k: (9) and (10) imply t ¿ max{k; i1 }¿4 and from step 2 of the algorithm it follows vt ¿ vt−1 + vt−2 ¿vi1 + vk incompatible with (6). Case 2.2: i1 = k: Now we have t ¿ 3 and from (6) we obtain 2vk = vt + x:
(11)
The assumption k6t − 2 would result in the contradiction vt ¿ vt−1 + vt−2 ¿ vk + vk . Therefore, k = t − 1:
(12)
Case 2.2.1: vt was labeled in step 2(b). t−1 t−2 vt = i=1 (vi + 1) = (vt−1 + 1) + i=1 (vi + 1) ¿ vt−1 + vt−1 = 2vk contradictory to (11). Case 2.2.2: vt was labeled in step 2(a). t−1 There exists an l6t − d such that vt = i=t−d+1 vi + vl . Hence, 2vt−1 = vt + x = t−1 t−2 i=t−d+1 vi + vl + x and — using (11) — we see vt−1 = i=t−d+1 vi + vl + x. Because vt was labeled in step 2(a) the vertex vt−1 got its label in step 2(b) and it follows that {vl ; vt−d+1 ; vt−d+2 ; : : : ; vt−2 }={v1 ; v2 ; : : : ; vt−2 } as well as |{v1 ; v2 ; : : : ; vt−2 }|=x. The de1nition of x immediately provides 5 if i1 = 3 ∧ d = 3; t= 4 otherwise: Furthermore, in case i1 = 3 ∧ d = 3, i.e. t = 5, under consideration of (12) and Case 2.2 we obtain the contradiction 4 = t − 1 = k = i1 = 3. So i1 =3 ∨ d =3, i.e. t = 4, is valid. Since we are in Case 2.2.2 it follows from t = 4 that d = t − 1 = 3. Consequently, i1 ¿4 holds, contradictory to i1 = k = t − 1 = 3.
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This makes our proof complete. Having a look at the proof of Theorem 3.1 (especially at the labeling algorithm), we see that the d-uniformity as well as the absence of cycles in the hypergraphs under consideration are essential for the algorithm. It seems to be diMcult to use the given labeling algorithm (or a modi1cation of it) to determine the integral sum number of non-uniform hypertrees (cf. Theorem 1.3 for the sum number) or of hypergraphs which contain cycles. So the following problem arises: Problem 3.2. Is it possible to prove an analogue of Theorem 3.1 for non-uniform hypertrees? Acknowledgements The authors would like to thank the referees for their helpful suggestions. References [1] C. Berge, Hypergraphs, North-Holland, Amsterdam, New York, Oxford, Tokyo, 1989. [2] D. Bergstrand, F. Harary, K. Hodges, G. Jennings, L. Kuklinski, J. Wiener, The Sum Number of a Complete Graph, Bull. Malaysian Math. Soc. Ser. 2 12 (1989) 25–28. [3] Z. Chen, Harary’s conjectures on integral sum graphs, Discrete Math. 160 (1996) 241–244. [4] M.N. Ellingham, Sum graphs from trees, Ars Combin. 35 (1993) 335–349. [5] F. Harary, Sum graphs and di0erence graphs, Congr. Numer. 72 (1990) 101–108. [6] F. Harary, Sum graphs over all the integers, Discrete Math. 124 (1994) 99–105. [7] A. Sharary, Integral sum graphs from complete graphs, cycles and wheels, Arab Gulf J. Sci. Res. 14 (1) (1996) 1–14. [8] M. Sonntag, H.-M. Teichert, Sum numbers of hypertrees, Discrete Math. 214 (2000) 285–290.