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Научни трудове на Съюза на учените в България - Пловдив. Серия В. Техника и технологии. Том XVII, ISSN 1311 -9419 (Print); ISSN 2534-9384 (Online), 2019. Scientific Works of the Union of Scientists in Bulgaria - Plovdiv. Series C. Technics and Technologies. Vol. XVII., ISSN 1311 -9419 (Print); ISSN 2534-9384 (Online), 2019
A PROPERTY OF INTERVAL GRAPHS
Vasil Angelov Petrov Technical univers ity Sofia, Plovdia branch
Abstract
Sufficient conditions for the existence of an interval in a finite family of intervals on the real line intersecting all the intervals in the family are found.
Key words: family of intervals, real line
The starting point for this paper is the following simple assertion (which is more or less known)
Proposition 1. Let G be an undirected graph with 2n+ 1 vertices. Let for each n vertices there exists a vertex (different from them) adjacent to all the n vertices. Then there exists a vertex adjacent to all the vertices of the graph.
For the sake of completeness we shall give a proof.
Proof.Let if be a maximum clique. If |ff | > n + 1 then for the remaining vertices (which will be less or equal to n ) there exists a vertex v adjacent to all of them. But clearly v (E ff . So v is adjacent to all vertices of G and we are done. Now if |iT| <n there exists a vertex u £ K adjacent to all vertices in ff. But then ff U u form a clique which contradicts the maximallity of ff and the proof is finished.
Despite its simplicity Proposition 1 is in general the best we can assert. Let's consider some examples.
Example 1. We shall construct a 2n -regular graph on 2n + 2 vertices (compared with Pr. 1 we have increased the number of vertices by one) in the following way: let the vertices be v1,v2,v2n+2 and let vt is adjacent to all vertices except vi+n+1 (the indeces are mod 2n + 2 ). We shall show that for any n vertices there is a vertex (different from them) adjacent to them. Let the vertices be v^.v^.v^, if we consider also vii+n+1, vh+n+1,, viji+n+1 there still be two other vertices , say u and v and of course any of them will be adjacent to all vt ,j = 1,2, ,ti. On the other hand there is no vertex adjacent to all the vertices
of the graph and we are done.
Example 2. Now we shall go the other way, we shall consider a graph with 2n + 1 vertices (as in Pr. 1) but shall reduce n to n — 1. Let us consider the 2n — 2 -regular graph with vertices viri — 1,2,,2n constructed in exactly the same way as in Ex.1. We shall add one more vertex v2n+1, adjacent to all vertices
except v2n. Let vi1>vi2,,vin_1 be any n — 1 vertices of the graph. If v2n+1 is not among them, then like in Ex. 1 there is a vertex v in the 2n — 2 -regular graph adjacent to them. So let v2n+ ± be one of the chosen n — 1 vertices. For the n — 2 vertices (v2n+1 is excluded) there are in fact at least two vertices, adjacent to them (just like in Ex. 1). Thus one of them will be different from v2n and so it will be adjacent to v2n+1 and we are done. Therefore for any n — 1 vertices there is a vertex, adjacent to them, but obviously there is no vertex, adjacent to all the vertices in the graph.
Definition 1. A graph has the property Pk if for any k vertices of the graph there is a vertex (different from them) adjacent to all these k vertices.
In view of this definition Pr. 1 and the examples tell us that the minimum k such that { Pk is valid for a graph G with 2n + 1 vertices } implies the existence of a vertex, adjacent to all the 2n + 1 vertices of G is n. However if we consider not an arbitrary graph, but a graph belonging to a particular class, then k could be smaller than n. In what follows we are interested in finding the minimum k for the class of the interval graphs - a subclass of the so called intersection graphs([l]). This means that we are considering an arbitrary finite set of closed intervals on the real line. The vertices of our graph will be the intervals and two vertices will be adjacent if the corresponding intervals have a nonempty intersection.
Theorem 1. Let G be a family of n finite intervals on the real line. Let for any two of them there exists an interval (different from them) which intersects the two intervals. Then there exists an interval intersecting all the intervals in G.
Proof. The proof in fact is extremely easy. Let the intervals be [iiij&i] i = 1,2, ...,n. Let A = maxai and B = mini^. Let }p = \aprbp\ be any interval such that ap = A and let }q = \aq, bq\ be any interval such that bq = B. Suppose that B < A. There is an interval }s = [as, £>s] intersecting both Jv and Jq. Now let , ] be any interval of the family, different from Jp and Jq. Suppose that Js does not intersect it and [at, bj is situated to the left of Js (the other case is absolutely the same). Then bt < as < bq and we get a contradiction with the choise of Jp . Thus }s intersects all the intervals in the family.
Next suppose that B > A. Then ap < bq and obviously bq] [=]l for any i = 1,2, So in this case any interval intersects all other and we are done.
Remark 1. The fact that the intervals are closed is not at all essential. They could be open, halfopen, whatever.
Remark 2. It is interesting that Th. 1 could not be extended in general for infinite family of intervals. Even if we impose rather strong restrictions on the intervals (for example the family to be countable, the lengths of the intervals to be bounded away from 0 and go -this means that there exist positive constants p and q such that all lengths are in [p, <?]) the result does not follows as the next example shows.
Example 3. Let us consider the union of the following 3 sequences of 1111 11
intervals: {[^ — 1, -]}, {[—. 1 — —]} and {[1 — 2 — -]}. It is easy to check that
for any 2 intervals there is an interval intersecting them, but obviously there is no interval intersecting all of them, what's more - any interval is missing in fact infinitely many intervals.
Proposition 1, with which we started can be considered from a somewhat different angle. Since for any n vertices there exists a vertex, adjacent to them, then there should be in general a lot of vertices of degree at least n. So it is natural to ask- could we invert the statement of Pr.l, more precisely if all vertices are of degree at least n does this implies the existence of a vertex, adjacent to all other? Example 1 (and 2) tells us that this is not the case. Although all vertices are of degree 2n — 2 (this in fact only 1 less than the highest possible degree in a graph with 2n vertices) there isn't a vertex adjacent to all other. However if consider the class of interval graphs, we have such a result!
Theorem 2. Let G be a family of 2n + 1 intervals on the real line. If each of them intersects at least n other intervals of G, then there is an interval intersecting all intervals of G.
Proof. The proof is based on the following lemma.
Lemma 1. Under the conditions of Th.2 there exists a n + 1-clique or there is an interval intersecting all intervals of G.
Proof. We shall use induction on n. The assertion is valid for ?j = 1 because obviously there exists a 2-clique. Assuming the validity of the lemma for ?i — 1 we shall prove it for n. If for any two intervals there's an interval (different from them) intersecting them then Th. 1 implies that there's an interval intersecting all intervals of G and we are done. So let us assume that there is a couple of intervals such that no interval from the rest 2n — 1 intervals intersects both of them. We shall denote these intervals by }p — [ap,bp] and Jq — \_aq, bq] . Then each of the remaining 2ti - 1 intervals intersects at least n — 1 other intervals (Jp = [ap, bp] and Jq = [aq, bq\ are excuded). The induction hypothesis implies then that there is an interval intersecting the other 2n — 2 intervals or that there exists a n-clique. We shall start with the first possibility. Let }s = [a5 , &s] intersects the rest 2n — 2 intervals. At least one of the intervals Jp and Jq does not intersect /s. Let this be }p and let for the sake of definiteness Jp is situated to the right of Js, i.e. ap > bs. We shall note that Jp f\Jq + <p. Indeed if we assume the opposite, it will follow that each of them intersects n intervals from the rest, so we'll need 2n intervals but we have only 2n— 1. Let us denote by {/i = [a,,bl ]} the intervals which Jp intersects (we do not include ]q). There are at least n — 1 such intervals and each of them intersects also }s. So the inequalities at < bs < ap are valid. Let a{ — max^} Then [aJ p bs] a ]l for each i and [a,, &J l , so these intervals are part of an at least «-clique. Let us consider an arbitrary interval Jk, different from Jp,JqJs and not belonging to the group {Jt}. If
Ik ^ 4> we get a 7i + 1-clique and the assertion is proved. So let
Ik f l Jjs] = 4> Since Jk n Js <p, then bk < at. On the other hand Jq intersects at least n — 1 intervals different from Jp. These intervals could not be from the group {/¿j, hence }k intersects at least one of the Jks (as well as Jp), which implies [a;, &s] c Jq and we again have obtained a n + 1-clique. Thus the first possibility is fully investigated.
Now let us suppose that among the chosen 2n — 1 intervals there isn't an interval intersecting the other 2n — 2. Then the induction hypothesis implies that there is at least a Ti-clique among these 2?i — 1 intervals. By Helly theorem the intervals in this n-clique have a nonempty intersection and let us denote this interval (possibly degenerate) by K. If any of the other n + 1 intervals intersects K, we will have a ti + 1-clique. So we shall assume that neither of them intersects K. Since Jp n Jq ^ <p , they should be on one and the same side of K. Let for the sake of definiteness they are situated to the right of K. We shall also note that neither Jp nor jq c Jp. Otherwise if Jp c Jq for example and / n ]p ^ <p , then / n Jq ^ <p , which is impossible. Let for the sake of definiteness
Then for any interval /, different from Jp and Jq, we have / H \aq, bp\ = <p. Since Jq intersects at least ti intervals then there exist at least Ti—l intervals the left ends of which belong to (bp,bq]. Obviously neither of them is a part of the 7i-clique. Then the intervals with left end in (bp, bq) should be exactly n - 1. Any / from the clique does not intersect these intervals, Jp also does not intersect them. Then each of these intervals intersects no more than Ti — 2 intervals from this group plus Jq, i.e. n — 1 intervals, which contradicts the fact that each interval in G intersects at least n intervals. Having considered the second possibility, the proof of the lemma is finished.
Let us continue with the proof of the theorem. In view of Lemma 1 we may assume that there exists a n + 1-clique (the other option is trivial). So let Cn+S be a clique of maximum cardinality (s E [1, n + 1]). If s — n + 1 (or even s = n) there's nothing to prove. So let us assume that s E [1, n — 1]. By Helly theorem an interval K (possibly degenerate) exists, which is a subset of each interval in the clique. We shall split the intervals in G in three groups: group A consists of all intervals that do not intersect K and are situated to the left of it; the intervals of Cn+S form group B; group C consists of all intervals that do not intersect K and are situated to the right of it. In view of Th. 1 we shall also assume that there is at least one couple of intervals such that no interval in G intersects both of them. Due to symmetry we can reduce the possible locations of these two intervals (with respect to the 3 groups) to three cases:
case 1. One of the intervals is in A and the other is in C.
case 2. Both are in C.
case 3. We have one in B and one in C.
Case 1. Let Jp E A, Jq EC and no interval in G intersects both of them. Let \A | = I. Then |C| = n + 1 — s — I. Jp could intersect at most I — 1 intervals of A and jq could intersect at most n — s — I intervals of C . Then ]p intersects at least n — 1 + 1 intervals from Cn+S and Jq intersects at least s 1 intervals from Cn+S . Thus Jp and Jq intersect at least n + s + 1 intervals of Cn+S . Since |Cn+J| = + s there's an interval in Cn+S which intersects both Jp and Jq and we obtain a contradiction.
Case 2. Let Jp = \ap,bp\ and Jq = \aq, bq] are in C and no interval in G intersects both of them. As we noticed in the proof of Lemma 1 Jp n Jq ^ <p . On the other hand neither Jp l-Jq, nor Jq a ]p. Otherwise (suppose Jp ~]q) any interval intersecting Jp will also intersect Jq which is impossible. Let for the sake of definiteness the inequalities in (1) are valid. If / E B and / intersects Jq, then / will also intersect Jp. So there are no intervals in B, intersecting Jq and therefore only intervals from C could intersect Jq. But |C| < n and hence Jq could intersect only n - 1 intervals, which is a contradiction.
Case 3. The unique interval in B that Jq intersects is Jp (otherwise there will be an interval in B intersecting both Jp and Jq ) Thus Jq intersects at least 7i — 1 intervals in C and since \C\ < n then \C\ = n and any interval / in C intersects Jq. But then / could not intersect intervals in B. Since / could intersect at most n—1 intervals in C we get a contradiction.
Finally, having considered the three cases, the proof of Th. 2 is finished.
Remark 3. The result of Th. 2 is in general the best possible. If for example we increase the number of the intervals from 2n + 1 to 2n + 2 we cannot assert that there's an interval, intersecting all the others.
Example 4. Let us consider two families A ={[1-^l + k*= L2,,71 + 1} and B ={[4-i 4 + hk = l,2,,n+ 1}
rt rC iC iC
Obviously each interval in A and B intersects exactly n other intervals but still there isn't an interval intersecting all.
It is likely that the rusults in this paper can be extended to circular arc graphs. On the other hand it is very unlikely that we have something similar for intersection graphs generated by two or higher dimensional figures (disc graphs for example). There are many and many classes of intersection graphs and may be the most interesting question here is if these results are something rare, specific only for interval graphs or not. This is completely open problem.
REFERENCES
1. McKee, Terry A.; McMorris, F. R. (1999), Topics in Intersection Graph Theory, SIAM Monographs on Discrete Mathematics and Applications, 2, Philadelphia: Society for Industrial and Applied Mathematics
