Yi-Wu Chang
National Politics University, Taipei, Taiwan
Michael S. Jacobson and Jen}o Lehel
University of Louisville, Louisville, KY
msjaco01@ulkyvx.louisville.edu and j0lehe01@ulkyvx.louisville.edu
Douglas B. West
yUniversity of Illinois, Urbana, IL 61801-2975, west@math.uiuc.edu
Abstract.
We introduce the visibility number b(G) of a graph G, which is the minimum t such that G can be represented by assigning each vertex a union of at most t horizontal segments in the plane so that vertices u;v are adjacent if and only if some point assigned tou sees some point assigned tov via a vertical segment unobstructed by other assigned points. We prove the following:1) every planar graph has visibility number at most 2, which is sharp. 2) rb(Km;n)r+ 1, where r =d(mn+ 4)=(2m+ 2n)e.
3) dn=6eb(Kn)dn=6e+ 1.
4) When G has n vertices, b(G)dn=6e+ 2.
1. INTRODUCTION
Researchers in computational geometry have studied the use of graphs to model visi-bility relations in the plane. For example, in a polygon in the plane we say that two vertices \see" each other if the segment joining them lies inside the polygon. Letting vertices that see each other be adjacent denes the visibility graph of the polygon. Similarly, we can dene a visibility graph on a set of line segments in the plane, where two segments see each other if some segment joining them crosses no other segment. The literature on these models has dozens of papers, mostly concerning the computation and the recognition of visibility graphs. Also there are applications to search problems and motion planning.
We consider a simpler model in which visibility is vertical only. Let S be a family of horizontal bars in the plane. Tamassia and Tollis [5] dened the bar visibility graph of S
to be the graph with vertex setS in which two vertices are adjacent if and only if there is
y
Research supported in part by NSA/MSP Grant MDA904-93-H-3040. Running head: VISIBILITY NUMBER
AMS codes: 05C35, 05C10
Keywords: visibility representation, planar graph, interval number Written July 1994 and June 1998.
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some unobstructed vertical segment joining them. They characterized bar visibility graphs as the planar graphs having a planar embedding in which all cut-vertices lie on a common face. (Graphs generated by horizontal and vertical visibility of rectangles in the plane are studied in [2].)
Realistically, one would like visibility to occur along a channel of positive width. This enables two bars [(a;y);(b;y)] and [(b;z);(c;z)] to block visibility at bwithout seeing each other. We obtain this eect by letting bars be half-open segments of the form ((a;y);(b;y)]. We study problems for visibility graphs analogous to those studied for intersection graphs. The interval graph of a family S of intervals on the real line is the graph with vertex setS in which two vertices are adjacent if and only if as intervals they intersect. Bar visibility graphs provide a geometric analogue of interval graphs; visibility replaces inter-section as the adjacency relation, and we place the intervals at various heights. The models yield dierent families of graphs because intervening bars can block visibility, whereas in-tervals having a common point on the horizontal line are pairwise intersecting.
The interval graph model has been generalized to permit multi-interval representa-tions of all graphs. A t-interval is a union of (at most) t intervals on the real line. A
t-interval representation of G is an assignment of t-intervals to vertices of G so that ver-tices are adjacent if and only if their t-intervals intersect. The interval number i(G) of a graph G is the minimumt such that G has a t-interval representation.
Here we similarly generalize the bar visibility model. A t-bar is a union of (at most)
t horizontal bars in the plane. A t-bar representation of G is an assignment of t-bars to vertices of G so that vertices are adjacent if and only if some vertical segment links their
t-bars without intersecting any other t-bar in the representation. The visibility number
b(G) of a graph G is the minimum t such that G has a t-bar representation. When t is unspecied, we use the termmultibar.
For graphs without large cliques, visibility number tends to be smaller than interval number, because the upper and lower \sides" of a bar can be used independently to estab-lish edges. Using the result of [5], we show that every planar graph has visibility number at most two. (This compares with interval number at most three.)
For other families, our lower bounds arise from an easy lemma involving the maximum number of edges in N-vertex planar graphs. Combining this with constructions tells us (within 1) the visibility number for complete bipartite graphs (bicliques) and for cliques. The visibility number of a biclique Km;n is roughly half its interval number, but the clique
Kn has interval number 1 and visibility number roughly n=6.
We conjecture that, over graphs with n vertices, visibility number is maximized by
Kn. We provide a construction for arbitrary n-vertex graphs that always uses at most
dn=6e+ 2 bars for each vertex. This solves the extremal problem for n-vertex graphs with
an error of at most two. The construction uses the result of Lovasz [4] that everym-vertex graph can be decomposed into at most bm=2c paths and cycles.
2. PLANAR GRAPHS
We solve the extremal problem for planar graphs by expressing an arbitrary planar graph as the union of two bar visibility graphs.
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REMARK 1.
b(G[H)b(G) +b(H).Proof:
Bar visibility representations of G and H can be placed in disjoint vertical strips to represent G[H.THEOREM 2.
Every planar graph has a 2-bar representation in which all vertices other than cut-vertices are assigned 1-bars.Proof:
If H is a disjoint union of planar graphs having at most one cut-vertex in each component, then the result of Tamassia and Tollis [5] yields b(H) = 1. We express an arbitrary planar graph G as the union of two such graphs, which we call G0 and G1.Begin with G0 andG1 empty. Choose an arbitrary vertexv
2V(G) as a root. Place
the union of all blocks containing v into G0, and mark v nished. Proceed iteratively as
follows. For each unnished vertex added to Gi on the previous step, add to G1?i the
union of all blocks of G that have not yet been placed, and mark the vertex nished. Continuing in this breadth-rst manner through the blocks of G decomposes G into two subgraphs.
At each phase when a new subgraph consisting of pairwise disjoint \stars of blocks" is added to Gj, the new subgraph is disjoint from the earlier subgraphs added to Gj. Thus
each component of Gj has at most one cut-vertex, and the two graphs G0;G1 are bar
visibility graphs.
The minimal planar graphs that are not embeddable with every vertex on a single face are K4 and K2;3. Adding a pendant edge at each vertex of such a graph produces a
planar graph that is not a bar visibility graph. Thus Theorem 2 is sharp.
3. BICLIQUES (COMPLETE BIPARTITE GRAPHS)
Our subsequent lower bounds use an easy counting argument.
LEMMA 3.
The visibility number of a graph G with n vertices and e edges is at least (e+ 6)=(3n). If the graph is triangle-free, then b(G)(e+ 4)=(2n).Proof:
Consider a t-bar representation of G. The total number of bars used is N nt.In the plane, add one vertical segment joining each pair of bars that see each other. Now shrink each bar so that it becomes a single point. The added segments remain, covering the edges ofG. The result is a planar graphG0 with N vertices and at leasteedges. Since
it also has at most 3N ?6 edges, we have the desired bound.
If G is triangle-free, then the graph G0 will also be simple and triangle-free after we
contract all edges joining bars for the same vertex ofG. NowG0 has at most 2N
?4 edges,
and again these cover all edges of G. Lemma 3 yields b(Km;n) dmn
+4 2m+2n
e. Trotter and Harary [6] proved that i(Km;n) = dmn
+1
m+n
e. Our lower bound for b(Km;n) always equalsdi(Km;n)=2e ordi(Km;n)=2e+ 1. By
using the tops and bottoms of bars separately, we prove constructively that the visibility number of Km;n is within one of our lower bound. Our construction is motivated by the
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THEOREM 4.
If r=d mn +4 2m+2ne, then r b(Km;n)r+ 1.
Proof:
We construct an r+ 1-bar representation of Km;n. We may assume that m n,with partite sets X =fx
1;:::;xm
g and Y =fy
1;:::;yn
g. Let s =b(n?1)=2c?r.
Asmgrows, r increases to dn=2e. We construct a representation usingdn=2ebars for
each xi and one bar for each yj by arranging the bars for Y as a horizontal sequence of
vertical pairs and separating each pair vertically by a set of bars for X.
Therefore, we may assume thatrdn=2e?1 =b(n?1)=2c. Since (mn+4)=(2m+2n)
increases strictly with m (for n > 2) and equals (n?1)=2 when m=n 2
?n?4, we may
assume that m < n2
?n?3.
We construct a multibar representation using (up to) r+ 1 pairs of rows of bars for vertices ofY. In each pair of rows, the top row has bars for y1;:::;y
dn=2e, and the bottom
row has bars for the remainder of Y. In each row, the ith bar extends horizontally from
i?1 to i, except that when n is odd the bar foryn extends from (n?3)=2 to (n+ 1)=2.
The bars forX also form rows, with a row for some of X between successive rows for
Y. In Fig. 1, the bars have been shrunk for clarity within rows; the overlap of half-open bars is such that each bar sees only bars for vertices of the opposite partite set in the two nearest rows. Each bar for xi sees two bars above it and two below it, except that when
n is odd the rightmost bar in each row for X sees bars for only three vertices of Y.
x13 x11 x12 x11 x12 y11 y1 y11 y1 x8 x9 x10 x9 x10 x5 x6 x7 x8 x7 x4 x3 x4 x5 x6 x1 x2 x1 x2 x3 y11 y1 x13
Fig. 1. Part of a 4-bar representation of K13;11; s = 2 andr = 3.
Reading from left to right within successive rows forX from top to bottom, we alter-nate x1;x2 until we have s bars for each, then we alternate x3;x4, etc. These bars need
2sbm=2cpositions. In each of up to 2(r+1)?1 rows, there aredn=2e?1 positions available.
We thus require that 2sbm=2cb(n?1)=2c(2r+ 1). Proving that ms (s+r)(2r+ 1)
will show that there are enough locations for these bars.
Because mn + 4 2r(m+n), we have m(n=2?r) rn?2. When n is odd, this
inequality becomes m(s+1=2)r(2s+2r+1)?2. Sincer(2s+2r+1)<(s+r)(2r+1),
the desired inequality holds. When n is even, the known inequality becomes m(s+ 1)
2r(s+r+ 1)?2. Since 2r(s+r+ 1)?2 = (s+r)(2r+ 1) +r?s?2 and m > r, again
the desired inequality holds.
If the available postions are not all needed, we discard the later bars for Y to avoid visibilities between vertices of Y. If m is odd, we add one long bar for xm at the top and
another at the bottom; together these see all of Y. If m is odd and the bottom row of intervals for Y was deleted, then we can put it above the top bar for xm instead.
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For x1;:::;x2bm=2c, we have established visibility to 4s or to at least 4s
?1 (if n is
odd) vertices of Y. These visibilities involve distinct vertices of Y, because mnimplies
that r dn=4e, and hence s b(n?3)=4c. With r+ 1 bars allowed per vertex, we have
r+ 1?s = dn=2e?2s bars remaining to be assigned to xi. There remain n?4s (or at
most n+ 1?4s if nis odd) vertices in Y that xi must see. Picking up two of these with
each remaining bar completes the representation.
The visibilities established so far for xi consist of two strings A;B of consecutive
en-tries in the list y1;:::;yn, viewed cyclically. The lists may have length 2s each, or one
may have length 2s?1 if n is odd and it includes yn. The ending positions of A and B
are separated by dn=2e or bn=2c. If the last bar placed for xi sees yj and y 0
j at its right
end, then we add a small bar for xi by shrinking the next two bars for X between yj+1
andyj0
+1. Continuing in this fashion extendsAand B to create the remaining visibilities.
When n is odd, we need two visibilities for each new interval assigned to xi if and only if
xi already sees yn in the rst phase; otherwise xi starts with 4s visibilities and one of the
small intervals seeing yn picks up only one.
4. CLIQUES
A surprisingly simple construction forb(Kn) produces a representation using at most
one more bar per vertex than the counting bound from Lemma 3.
THEOREM 5.
dn=6eb(Kn)dn=6e+ 1.Proof:
Lemma 3 yieldsb(Kn)n?1 6 + 2 n
=dn=6e. For the upper bound, we rst reduce
to the case wheren is divisible by 6. In a visibility representation of a clique, deleting the bars for a vertex cannot introduce unwanted visibilities, since all visibilities are wanted. Deletion also cannot destroy visibilities. Hence b(Kn?1)
b(Kn).
Now let n= 6m. We partition the vertex set into three sets A1;A2;A3, each of size
2m. A clique with 2m vertices has a decomposition into m spanning paths, consisting of them rotations of a zig-zag path when the vertices are placed around a circle (see Fig. 2).
Fig. 2. Path decomposition ofKn.
Our visibility representation ofKnconsists of 3misomorphic modules. Each module is
a bar visibility representation of the joinPm_2K
1, using one bar per vertex. We represent
the path by a staircase of bars, each of which sees the bar before and after it. We add one long bar above and one long bar below; each sees the entire path (see Fig. 3).
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Fig. 3. Bar visibility representation of Pm_2K 1.
To each of the m paths in the decomposition of Ai, we assign two vertices in Ai+1
(indices modulo 3). This produces m pairwise edge-disjoint copies of Pm _2K
1. Their
union covers Ai and all edges fromAi to Ai+1. Doing this for each Ai yields 3m modules
whose union represents Kn.
A vertex of Ai appears in each path drawn fromAi, and it appears once as a top or
bottom bar in a module for a path in Ai?1. Thus each vertex is assigned m+ 1 bars.
Example 6.
b(K9) = 2. A graph represented with one bar per vertex must be planar,so the upper bound of Theorem 5 cannot be improved when 5 n 6. However, it can
be improved when 7n9.
When we allow only two bars per vertex, it is possible that we use the rst bar for each vertex in one vertical strip and the second bar in another. This would express the graph as the union of two planar subgraphs: thickness 2. The graph K9 has thickness 3.
However, the exibility of putting the representation for one subgraph above the other and thereby obtaining an extra edge yields b(K9) = 2 (see Fig. 4).
pq r s t w v u x r s q x t u p s v p q r s x w v u t s r x q t u p s v
Fig. 4. 2-bar representation of K9. 5. n-VERTEX GRAPHS
When bounding the visibility number of an n-vertex graph G, it is tempting to use Remark 1 and express Gas a union of planar graphs, since planar graphs have visibility at
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most 2. When n is not 9 or 10, every n-vertex graph is the union of at most b(n+ 7)=6c
planar graphs, with equality for cliques (see [1] for n6 4mod6, with the remaining case
settled through the work of many authors). This yields about n=3 as an upper bound on visibility number of n-vertex graphs. Using planar graphs whose cut-vertices lie on one face, which seems feasible, could eliminate the factor of 2 between this and the lower bound in Theorem 5.
Instead, we generalize the construction in Theorem 5 to prove directly that b(G) dn=6e+ 2. The construction in Theorem 5 establishes each edge once, but it is dicult to
modify it to delete an arbitrary set of edges. For example, letu;v;w appear consecutively on some path in the decomposition of A1, and let y;z be the vertices of A2 whose bars
surround this path. By extending u or w, it is possible to block v from seeing y or z. By deleting the bar for v and extending those for u and w to the same vertical line, we can delete all these edges. However, how can we delete vy;vz;uv and keep vw?
If all edges of the path in A were present, then we could delete arbitrary edges to y
andzby extending the bars for vertices on the path. Iftkis the maximum number of paths
needed to partition the edges of a k-vertex graph, we could thus obtain b(G) tn= 3 + 1.
Gallai [3] conjectured that tk =dk=2e, which would yield b(G)dn=6e+ 1.
We do almost as well by using the result of Lovasz [4] that everyk-vertex graph can be decomposed into bk=2c paths and cycles. Each vertex of odd degree must be an endpoint
of some path in such a decomposition. Thus the decomposition must consist entirely of paths when G has at most one vertex of even degree.
THEOREM 7.
If G has n vertices, thenb(G)dn=6e+ 2.Proof:
By adding isolated vertices, we may assume thatn is divisible by 6. Let n= 6m, and again partition V(G) into sets A1;A2;A3 of size 2m. To G[Ai], add one vertex wadjacent to all vertices with even degree in G[Ai]; call this graphG0
i. Since G0
i has at most
one vertex of even degree, G0
i has a decomposition into b(2m+ 1)=2c=m paths.
To each such path P, we assign two vertices of Ai+1. We design a module for our
representation that establishes all the edges of P contained in G and all the edges of G
joining Ai with these two vertices of Ai+1. In each such module, we use one bar for each
vertex of Ai and two bars for each of the two special vertices of Ai+1. Doing this for each
i and each P in the decomposition of G0
i produces a visibility representation with m+ 2
bars per vertex (see Fig. 5).
Let Iv be the bar (or two) to be assigned to v in such a module. Let y;z be the two
special vertices ofAi+1. We begin by representing P as a staircase of bars, with a bar for
y underneath and a bar for z above. The edges on P that don't involve the dummy vertex
w belong to G, so we never need to block these visibilities. Erasing Iw will produce a gap
that may cause us to break Iy and Iz.
Begining at the upper right end of P, which is also initially the right end of Iy, we
block visibilities between y and P as needed. When the current vertex v of P is not adjacent to y, we extend the bar for the next lower vertex of P to the right end of the current Iv. If the last vertex v before w is not adjacent to y, then we cannot extend a
lower bar to block Iv from Iy; instead, we break Iy and shorten the left end of the right
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Having arrived atIw, we delete it; the staircase was built so that the bar for the vertex
u after w on P does not see the bar for the vertex v before w on P. We now continue blocking vertices in the rest of P from seeing Iy as needed, in the same manner as before.
Visibilities up to Iz are corrected in the symmetric manner, working from the bottom left
end of P.
We still must consider the vertices of Athat do not belong to P. Each such vertex is adjacent to neither, one, or both of fy;zg. Vertices adjacent to neither can be ignored; we
need add no bar. For each vertex adjacent only to y, we add a bar at the right of P; none of these see each other, and Iu extends to the right to see them all. Similarly, the bars for
vertices of A?V(P) adjacent only to z can be added at the left of the bars for P.
For vertices of A?V(P) adjacent to both y and z, we add a bar in the gap between
the left and right portions ofP that was left by deleting w. Together they ll this gap so that Iy does not see Iz. The left portion of Iy and the right portion of Iz see these bars.
If there are no such vertices adjacent to both y and z, then we shorten the left portion of
Iy and the right portion ofz so that they won't see each other.
We have established and/or deleted all the desired adjacencies, using the desired num-ber of bars for each vertex.
y y
z z
Fig. 5. Module for multibar representation of general graph.
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[3] T. Gallai, Uber extreme Punkt- und Kantenmengen,Ann. Univ. Sci. Budapest, E}otv}os Sect. Math. 2 (1959), 133{138.
[4] L. Lovasz, On covering of graph, inTheory of Graphs(P. Erd}os and G. Katona, eds.). (Academic Press, 1968), 231{236.
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