Computing minimum geodetic sets
in proper interval graphs?
Tınaz Ekim1 , Aysel Erey1 , Pinar Heggernes2 , Pim van ’t Hof2 , and
Daniel Meister3
1
Boğaziçi University, Istanbul, Turkey, {tinaz.ekim,aysel.erey}@boun.edu.tr
University of Bergen, Norway, {pinar.heggernes,pim.vanthof}@ii.uib.no
3
University of Trier, Germany, [email protected]
2
Abstract. We show that the geodetic number of proper interval graphs
can be computed in polynomial time. This problem is NP-hard on chordal
graphs and on bipartite weakly chordal graphs. Only an upper bound on
the geodetic number of proper interval graphs has been known prior to
our result.
1
Introduction
The notion of geodetic sets was introduced by Harary et al. [11] and it has applications in game theory [3, 12, 17]. It is closely related to convexity and convex
hulls in graphs, which have applications in telephone switching centres, facility
location, distributed computing, information retrieval, and communication networks [9, 14, 16, 18, 19, 22]. Given a graph G and a set D of vertices of G, the
geodetic closure of D, denoted by IG [D], is the set containing every vertex of
G that lies on a shortest path between some pair of vertices of D. The set D
is a geodetic set of G if IG [D] contains all vertices of G. The geodetic number
of G, g(G), is the cardinality of a minimum geodetic set of G. Computing the
geodetic number is NP-hard on chordal graphs and on bipartite weakly chordal
graphs [7]. It can be done in polynomial time on cographs [7], split graphs [7],
and ptolemaic graphs [8].
The main result of this paper is a polynomial-time algorithm for computing
the geodetic number of proper interval graphs. Our algorithm can be implemented to also output a geodetic set of minimum size. The computational complexity of computing the geodetic number of proper interval graphs has been
open since Dourado et al. [7] gave a tight upper bound on the geodetic number
of proper interval graphs. Interestingly, the related notion of hull number has
been known to be computable in polynomial time on proper interval graphs [6].
The hull number and the geodetic number problems are both defined through
convexity, but they require quite different computation methods. The difference
between these two parameters can be arbitrarily large [13].
Proper interval graphs have been subject to extensive study (see, e.g., the
books [2, 10]) since their introduction [20], and they can be recognised in linear
?
This work is supported by the Research Council of Norway.
2
time. In this paper, in addition to our main result on proper interval graphs,
we also report on the following results: a polynomial-time algorithm to compute the geodetic number of block-cactus graphs, a polynomial-time algorithm
to approximate the geodetic number of bipartite permutation graphs with an
additive factor 1, and a proof of NP-hardness of computing the geodetic number of cobipartite graphs. Two variants of the geodetic number of block-cactus
graphs have been studied before [23], but we are not aware of an algorithm for
the exact computation of the geodetic number of such graphs.
2
Definitions and notation
We consider simple finite undirected graphs. For a graph G, its set of vertices is
denoted by V (G) and its set of edges is denoted by E(G). The neighbourhood of
a vertex v of G, denoted by NG (v), is the set of vertices of G that are adjacent
to v. For a set S of vertices of G, G[S] denotes the subgraph of G induced by S.
We write G−v to denote the graph G[V (G) \ {v}]. A vertex ordering for G is
an ordered tuple that corresponds to a permutation of V (G). For a given vertex
ordering σ, we write u ≺σ v if u appears before v in σ. The first position in σ
will be referred to as the left end of σ, and the last position as the right end.
We will use the expressions to the left of, to the right of, leftmost, and rightmost
accordingly.
A sequence (y0 , . . . , yr ) of distinct vertices of G is called a y0 , yr -path of
length r in G if yi yi+1 ∈ E(G) for every 0 ≤ i ≤ r − 1. For a vertex pair u, v
of G, the distance between u and v in G, denoted by dG (u, v), is the smallest
integer k such that G has a u, v-path of length k; if no such path exists then
dG (u, v) = ∞. G is connected if G has a u, v-path for every vertex pair u, v;
otherwise, G is disconnected. A connected component of G is a maximal connected
induced subgraph of G. A path is called a cycle if the first and the last vertex in
the sequence are also adjacent. A chordless cycle is a connected graph that has
a cycle containing all edges.
For a vertex triple u, v, x of G, x ∈ IG [{u, v}] if and only if dG (u, v) =
dG (u, x) + dG (x, v) [7]. It directly follows that a geodetic set of a disconnected
graph is the union of geodetic sets of its connected components. Hence, the
geodetic number of a disconnected graph is the sum of the geodetic numbers
of the connected components. It therefore suffices to study geodetic sets of connected graphs, and we will assume all input graphs to be connected in the paper.
A vertex v is called simplicial if NG (v) is a clique of G. Since a simplicial vertex
cannot lie on a shortest path between any two other vertices, it is easy to see
that every geodetic set of G contains all simplicial vertices of G.
3
Minimum geodetic sets for proper interval graphs
Proper interval graphs are equivalent to the intersection graphs of intervals of
unit length of the real line. A vertex ordering σ for a graph G is called a proper
interval ordering if the following is true for every vertex triple u, v, w of G:
3
u ≺σ v ≺σ w and uw ∈ E(G) implies uv ∈ E(G) and vw ∈ E(G). A graph is a
proper interval graph if and only if it has a proper interval ordering [15]. Note
that the first and the last vertex in a proper interval ordering are simplicial.
We fix some definitions, that will be valid throughout this section. We consider an arbitrary but fixed connected proper interval graph G with proper interval ordering σ. Let a be the first vertex in σ. For k ≥ 0, let Lk = {x ∈ V (G) :
dG (a, x) = k} be the vertices of G at distance k to a. The height of a is the
largest integer k with Lk 6= ∅; let h be the height of a. Let Λ = hL0 , . . . , Lh i.
We call Λ the BFS (breadth first search) partition of G with root vertex a. For
every vertex pair u, v of G and every integer i with 1 ≤ i ≤ h, if u, v ∈ Li and
u ≺σ v then NG (v) ∩ Li−1 ⊆ NG (u), and each of L1 , . . . , Lh is a clique of G [4].
These neighbourhood inclusion and clique properties will be central throughout
this section. Let ci be the rightmost vertex from Li with respect to σ for every
0 ≤ i ≤ h; clearly, c0 = a, and ch is the last vertex in σ. Note that Li ⊆ NG (ci−1 )
for every 1 ≤ i ≤ h.
3.1
Minimum geodetic sets with desirable properties
We show that G has a geodetic set of smallest size that satisfies very restrictive
properties. Observe for a vertex u ∈ Lp and a vertex v ∈ Lq with p < q:
q − p ≤ dG (u, v) ≤ q − p + 1, since a shortest path between u and v contains a
vertex from every level between Lp and Lq and may contain two vertices from
at most one level. This follows from the neighbourhood inclusion and clique
properties.
Let D ⊆ V (G). We define the range sets R0 (D), . . . , Rh (D) and R(D) of D
on Λ, where 0 ≤ i < h:
Rh (D) =def D ∩ Lh
and
Ri (D) =def (D ∩ Li ) ∪
[
(NG (v) ∩ Li )
v∈Ri+1 (D)
R(D) =def
[
Ri (D) .
1≤i≤h
For convenience, we write R(x) instead of R({x}) and, analogously, IG [u, v]
instead of IG [{u, v}]. As a simple consequence of the properties of Λ, we obtain
the following lemma.
Lemma 1. Let u, v be a vertex pair of G with u ≺σ v. If u ∈ R(v), then
R(u) ⊆ R(v). If u ∈
6 R(v) and u ∈ Lp with 0 ≤ p ≤ h, then Ri (v) ⊆ Ri (u) for
every 0 ≤ i < p.
It follows from Lemma 1 that, for every vertex pair u, v with u ∈ Lp and
v ∈ Lq and u ≺σ v, u ∈ R(v) if and only if dG (u, v) = q − p. We use this result
to characterise the sets IG [u, v] in the next lemma, whose proof follows from
Lemma 1.
Lemma 2. Let u, v be a vertex pair of G with a ≺σ u ≺σ v. If u ∈ R(v), then
IG [u, v] ⊆ R(v). If u 6∈ R(v), then for every vertex x of G, x ∈ IG [u, v] \ R(v) if
and only if u ∈ R(x) and x 6∈ R(v) and u 4σ x ≺σ v.
4
We say that a vertex x has a below-neighbour if there are an integer i and
a neighbour y of x such that 0 ≤ i < h and x ∈ Li and y ∈ Li+1 . For a
set D ⊆ V (G), we denote by Υ ∗ (D) the set of ordered vertex pairs (u, v) from
D that satisfy the following conditions:
v ∈ {c1 , . . . , ch }
there is 1 ≤ i ≤ h with a ≺σ u ≺σ ci ≺σ v
u has a below-neighbour and u 6∈ R(v).
P1)
P2)
P3)
Let D ⊆ V (G) and let D be a set of ordered vertex pairs from D. We call (D, D)
a geodetic pair for G if D ⊆ Υ ∗ (D).
Lemma 3. There is a geodetic pair (F,SF) for G such that F ∪{a} is a minimum
geodetic set of G and V (G) ⊆ R(F ) ∪ (u,v)∈F IG [u, v].
Proof. Let D ⊆ V (G) be a minimum geodetic set of G. Since a and ch are
simplicial vertices, D contains both a and ch . Let D be the set of all ordered
vertex pairs (u, v) from D with a ≺σ u ≺σ v and uv 6∈ E(G) and u 6∈ R(v)
and u has
S a below-neighbour. Using Lemma 2, it can be shown that V (G) ⊆
R(D) ∪ (u,v)∈D IG [u, v].
Let E be the set of indices i with 1 ≤ i ≤ h such that Ri (D) ⊂ Li . Let
i ∈ E , and let bi be the vertex from Li \ Ri (D) that is rightmost with respect to
σ. Since bi ∈ IG [D], there is a vertex pair (ui , vi ) in D with bi ∈ IG [ui , vi ]. Let
D0 =def {(ui , vi ) : i ∈ E } and J =def {j : vi ∈ Lj for some i ∈ E }. Let ψ be the
mapping: for every i ∈ E and 1 ≤ j ≤ h, if vi ∈ Lj then ψ(vi ) =def cj . Let
F =def
n
o
D \ {vi : i ∈ E } ∪ cj : j ∈ J
and
F =def
n
o
(u, ψ(v)) : (u, v) ∈ D0 .
Observe that |F | ≤ |D|. By carefully analysing F, we can prove that F ⊆ Υ ∗ (F ),
which
means that (F, F) is a geodetic pair for G, and that V (G) ⊆ R(F ) ∪
S
t
u
(u,v)∈F IG [u, v]. It follows that (F, F) satisfies the claim of the lemma.
Let (D, D) be a geodetic pair for G. A vertex x from D appears in D if there
is (u, v) ∈ D such that x ∈ {u, v}. We call (D, D) a normal geodetic pair if the
following conditions are satisfied:
for every vertex u from D that does not appear in D:
u has no below-neighbour
N2) for every u, u0 , v, v 0 ∈ D with (u, v) ∈ D and (u0 , v 0 ) ∈ D:
if u = u0 then v = v 0 , and
if u ≺σ u0 ≺σ v ≺σ v 0 then there is 1 ≤ i < h such that u0 , v ∈ Li .
N1)
Theorem 1. G has a normal geodetic pair S
(F, F) such that F ∪ {a} is a minimum geodetic set of G and V (G) ⊆ R(F ) ∪ (u,v)∈F IG [u, v].
Proof. Suppose for a contradiction that G does not have a normal geodetic pair
that satisfies the claim. Let (D, D) be a geodetic pair for G satisfying Lemma 3
such that the number of violations of conditions N1 and N2 is a small as possible.
We can assume that (D, D) satisfies condition N1 and the uniqueness part of
5
condition N2. Hence, there are pairs (d, c) and (d0 , c0 ) in D with d ≺σ d0 ≺σ c ≺σ
c0 and indices l0 and m with 1 ≤ l0 < m < h and d0 ∈ Ll0 and c ∈ Lm . Then one
of the three cases below must apply. Due to the space restrictions, we only give
the construction. The correctness of the arguments follows from a sequence of
results about properties of D.
Case 1: d ∈ Ll0
Let D0 =def (D \ {(d, c)}) ∪ {(d0 , c)}. There is F ⊆ D0 so that (D, F) is a geodetic
pair for G satisfying Lemma 3 that has a smaller number of conflicting pairs and
therefore contradicts the choice of (D, D).
Case 2: d ∈ L1 ∪ · · · ∪ Ll0 −1 and d 6∈ R(d0 )
Let F =def (D \ {(d, c)}) ∪ {(d, c0 )}. Then, (D, F) is a geodetic pair for G
satisfying Lemma 3 that has a smaller number of conflicting pairs and therefore
contradicts the choice of (D, D).
Case 3: d ∈ L1 ∪ · · · ∪ Ll0 −1 and d ∈ R(d0 )
Let w be the rightmost vertex of G with respect to σ satisfying: w ≺σ c and
d0 ∈ R(w). It can be shown that IG [d0 , c0 ] ⊆ R({w, c0 }) ∪ IG [d, c] ∪ IG [w, c0 ]. Let
F =def (D \ {d0 }) ∪ {w} and F =def (D \ {(d0 , c0 )}) ∪ {(w, c0 )}. It is important to
observe that w 6= d0 , particularly since d0 has a below-neighbour. It follows that
(F, F) is a geodetic pair for G satisfying Lemma 3 that has a smaller number of
conflicting pairs and therefore contradicts the choice of (D, D).
t
u
3.2
Computing the geodetic number in polynomial time
We are ready to derive a polynomial-time algorithm for computing a minimum
geodetic set for an input proper interval graph.
The size of a geodetic pair (D, D) is |D|. The normal geodetic number of
G is theSsmallest size of a normal geodetic pair (D, D) that satisfies V (G) ⊆
R(D) ∪ (u,v)∈D IG [u, v]. Theorem 1 shows that the normal geodetic number
of G plus 1 is equal to the geodetic number of G. We present an algorithm to
compute the normal geodetic number of G. The algorithm is mainly based on
a the condition N2 for normal geodetic pairs. This condition allows to split a
normal geodetic pair into two smaller parts or to reduce a normal geodetic pair.
These two reduction operations will be formalised later.
For 0 ≤ i ≤ h, let ai be the leftmost vertex from Li with respect to σ; clearly,
a0 = a. Let (D, D) be a geodetic pair for G. We call (D, D) a minimal geodetic
pair if the following conditions are satisfied:
M1) R(D) ∪ IG [C] ⊂ R(D) ∪ IG [D] for every C ⊂ D
M2) for every vertex u from D \ {a1 , . . . , ah } that does not appear in D:
R(C) ∪ IG [D] ⊂ R(D) ∪ IG [D] where C =def (D \ {u}) ∪ {ai } and
i is the integer with u ∈ Li .
Condition M1 means that every pair in D is necessary and condition M2 means
that every vertex from D that does not appear in D and that is not in {a1 , . . . , ah }
must cover itself.
6
Our algorithm to compute the normal geodetic number of G is based on the
idea of incrementally computing a normal geodetic set by extending an already
covered part of V (G). Such a covered part can be described by parameters. A
descript is an extended (9 + 2)-tuple [p, q; d, e, e0 ; b0 ; b, c, c0 ] + [s, t] where p, q are
integers with 0 ≤ p < q ≤ h and:
– either s ∈ V (G) and t ∈ {c1 , . . . , ch } or s = t = ×, and
d, e, e0 ∈ Lp ∪ {×} and b0 ∈ Lp+1 and c, c0 ∈ Lq ∪ {×} and b ∈ Lq+1 ∪ {×}
– if d 6= × then e, e0 6= × and d 4σ e 4σ e0 , and
if s 6= × then s ≺σ cp ≺σ cq 4σ t, and
if s 6= × and d 6= × then s ≺σ d, and
if c 6= × then c0 6= × and c 4σ c0 , and
if b 6= × then q ≤ h − 1.
We employ a special symbol ×, that will have the meaning of non-existing vertex:
{×} =def {(×, ×)} =def [× . . . ×] =def ∅. For a vertex pair u, v of G with u ≺σ v
and i the integer with u ∈ Li , let
[u . . . v] =def {x : u 4σ x 4σ v}
and
[[u . . . v] =def {x : ai+1 4σ x 4σ v}.
Let (D, D) be a normal geodetic pair for G, and let (u, v) ∈ D. A (u, v)-field of D
is a pair (x, y) from D with u ≺σ x ≺σ y 4σ v such that there is no pair (x0 , y 0 )
in D with x0 ≺σ x ≺σ y 4σ y 0 . We can say that a (u, v)-field is a maximal pair
from D inside of (u, v). Let A and B be the vertices from D that respectively do
and do not appear in D. We call (D, D) a realizer for [p, q; d, e, e0 ; b0 ; b, c, c0 ]+[s, t]
if (D, D) is a minimal normal geodetic pair for G with D ⊆ Lp ∪ · · · ∪ Lq and
B ∩ Lp = ∅ and D ∪ {b} 6= ∅ such that the following conditions are satisfied,
where C =def D ∪ {b} and C =def D ∪ {(s, t)}:
–
–
–
–
d is the leftmost vertex from A ∩ Lp with respect to σ and [e . . . e0 ] ⊆ A
b0 is the leftmost vertex from Rp+1 (C) with respect to σ
[c . . . c0 ] ∩ (R(C) ∪ IG [C]) = ∅ and [ap+1 . . . cq ] ⊆ R(C) ∪ IG [C] ∪ [c . . . c0 ]
if s 6= × then
for every (s, t)-field (u, v) of D: [[u . . . v] ⊆ R(C) ∪ IG [D] ∪ [c . . . c0 ].
For the condition on d, if A ∩ Lp is empty then d = ×. Note that there are
descripts that have no realizer, for instance, if b 6= × and c ∈ R(b).
The normal geodetic number of G is equal to the smallest size of a realizer for
[0, h; ×, ×, ×; a1 ; ×, ×, ×]+[×, ×]. We give an algorithm that computes the smallest size of a realizer for an arbitrary descript by using already computed values for
“smaller” descripts. We define a function Γ over the set of descripts that yields
two integer values as follows: for a descript A = [p, q; d, e, e0 ; b0 ; b, c, c0 ] + [s, t]
for G, Γ (A ) =def (g1 , g2 ), where g1 is the size of a smallest realizer for A , and
g2 is the size of a realizer for A containing cq that is of smallest possible size.
If A has a realizer, then g1 and g2 exist, and g1 ≤ g2 ≤ g1 + 1. The algorithm
for computing (g1 , g2 ) is presented in Figure 1. The algorithm itself is simple.
The difficulty lies in proving the correctness of the algorithm, that g1 and g2 are
indeed the optimal values for the sizes of the two desired realizers. It needs to be
7
Algorithm SizeOfRealizer
begin
let F1 ⊆ V (G) \ {a} and F2 ⊆ V (G) \ {a, cq } be of smallest size such that
(F1 , ∅) and (F2 ∪ {cq }, ∅) are realizers for A ;
for every r with p < r < q and d0 , e00 , e000 ∈ Lr ∪ {×} and b00 ∈ Lr+1 do
if B =def [p, r; d, e, e0 ; b0 ; b00 , e00 , e000 ] + [s, t] and
C =def [r, q; d0 , e00 , e000 ; b00 ; b, c, c0 ] + [s, t] are descripts then
let Γ (B) = (m, m0 ) and Γ (C ) = (n, n0 );
0
let kB,C =def m + n and kB,C
=def m + n0
end if end for;
if s 6= × then
let B =def [p, q; d, e, e0 ; b0 ; b, c, c0 ] + [×, ×];
0
let kB
=def m0 where Γ (B) = (m, m0 )
end if;
if s = × and d 6= × then
for every d0 , e00 , e000 ∈ ([d . . . cp ] \ {d}) ∪ {×} do
if B =def [p, q; d0 , e00 , e000 ; b0 ; b, c, c0 ] + [d, cq ] is a descript then
0
let kB
=def m0 + 1 where Γ (B) = (m, m0 )
end if end for end if;
0
let k1 and k10 be the smallest values of respectively kB,C and kB,C
;
0
0
let k2 be the smallest value of kB ;
let Γ (A ) =def (g1 , g2 )
where g1 =def min{|F1 |, k1 , k20 } and g2 =def min{|F2 | + 1, k10 , k20 })
end.
Fig. 1. The presented algorithm takes as input a descript A , and Γ (A ), that is computed, is the sizes of a smallest realizer for A and a smallest realizer for A that contains
vertex cq .
shown that every realizer admits a reduction that yields a realizer for “smaller”
descripts, and realizers for “smaller” descripts can be extended to realizers for
larger descripts, and the executability of both types of operations can be determined from only considering the descripts, in particular, without the knowledge
of the actual realizers. The following three lemmas show that the extension of a
realizer is indeed possible; the three lemmas implicitly also define what we want
to mean by “smaller” descript, namely B and C are smaller than A .
Lemma 4 (Realizer extension 1). Let A = [p, q; d, e, e0 ; b0 ; b, c, c0 ] + [s, t] and
B = [p, r; d, e, e0 ; b0 ; b00 , c00 , c000 ] + [s, t] and C = [r, q; d0 , c00 , c000 ; b00 ; b, c, c0 ] + [s, t] be
descripts. Let Γ (A ) = (k, k 0 ) and Γ (B) = (m, m0 ) and Γ (C ) = (n, n0 ). Then,
k ≤ m + n and k 0 ≤ m + n0 .
Proof. Let (D, D) and (E, E) be realizers for respectively B and C . We show
that (D ∪ E, D ∪ E) is a realizer for A and D ∩ E = ∅. The cardinality results
of the lemma then directly follow by choosing the two realizers for B and C
appropriately.
8
Let A1 and A2 be the vertices from respectively D and E that appear in
respectively D and E, and let B1 and B2 be the vertices from respectively D and
E that do not appear in respectively D and E. We first show that D ∩ E = ∅.
Due to the definition, D ⊆ Lp ∪ · · · ∪ Lr and E ⊆ Lr ∪ · · · ∪ Lq . Since p < r < q
due to the definition of descripts, it follows that D ∩ E ⊆ Lr . It is important to
note that A1 ∩ Lr ⊆ {cr }. Since cr 6∈ E, it holds that D ∩ E ⊆ B1 ; in particular,
no vertex in D ∩ E has a below-neighbour. Thus, (D ∩ E) ∩ A2 = ∅, and so,
D ∩ E = ∅.
Next, observe that A1 ∪ A2 is the set of vertices that appear in D ∪ E and
B1 ∪ B2 is the set of vertices from D ∪ E that do not appear in D ∪ E. Let
A =def A1 ∪ A2 and B =def B1 ∪ B2 . Note that A ⊆ Lp ∪ · · · ∪ Lq and B ∩ Lp = ∅.
We have to show that (D ∪E, D ∪E) is a minimal normal geodetic pair for G and
that the four conditions for realizers are satisfied. We begin with the conditions
for minimal normal geodetic pairs. It is easy to see that (D∪E, D∪E) is a geodetic
pair for G, and since no vertex from B has a below-neighbour, condition N1 is
satisfied. The uniqueness part of condition N2 is clearly satisfied, so that the
satisfaction of condition N2 follows. This shows that (D ∪ E, D ∪ E) is a normal
geodetic pair for G. For the minimality conditions, the definition of (D∪E, D∪E)
implies that the violation of any of the conditions M1 and M2 means the violation
of that condition by (D, D) or (E, E). We conclude that (D ∪ E, D ∪ E) is indeed
a minimal normal geodetic pair for G.
We verify the four conditions for realizers. We begin with the first three
conditions:
– since A ∩ Lp = A1 ∩ Lp , it follows from the assumptions about (D, D) that
d is the leftmost vertex from A ∩ Lp with respect to σ and [e . . . e0 ] ⊆ A
– b0 is the leftmost vertex from Rp+1 (D ∪ {b00 }) with respect to σ, and b00 is the
leftmost vertex from Rr+1 (E ∪{b}) with respect to σ, so that Rr (D∪{b00 }) =
Rr (D ∪ E ∪ {b}), and thus, b0 is the leftmost vertex from Rp+1 (D ∪ E ∪ {b})
with respect to σ
– [c . . . c0 ] ∩ (R(E ∪ {b}) ∪ IG [E ∪ {(s, t)}]) = ∅ due to assumptions, and
[c . . . c0 ] ∩ (R(D ∪ {b}) ∪ IG [D ∪ {(s, t)}]) = ∅, so that
[c . . . c0 ] ∩ (R(D ∪ E ∪ {b}) ∪ IG [D ∪ E ∪ {(s, t)}]) = ∅, and
[ap+1 . . . cr ] ⊆ R(D ∪ {b00 }) ∪ IG [D ∪ {(s, t)}] ∪ [c00 . . . c000 ] and
[ar+1 . . . cq ] ⊆ R(E ∪ {b}) ∪ IG [E ∪ {(s, t)}] ∪ [c . . . c0 ] and
[c00 . . . c000 ] ⊆ A2 ⊆ R(A) due to assumptions, so that
[ap+1 . . . cq ] =
[ap+1 . . . cr ] ∪ [ar+1 . . . cq ] ⊆ R(D ∪ E ∪ {b}) ∪ IG [D ∪ E ∪ {(s, t)}] ∪ [c . . . c0 ].
For the fourth condition for realizers, the field condition, we assume that s 6= ×.
Let (u, v) be an (s, t)-field in D ∪ E. Then, (u, v) is an (s, t)-field in D or (u, v)
is an (s, t)-field in E. If the latter then
[[u, v] ⊆ R(E ∪ {b}) ∪ IG [E] ∪ [c . . . c0 ]
⊆ R(D ∪ E ∪ {b}) ∪ IG [D ∪ E] ∪ [c . . . c0 ] ,
9
if the former then
[[u, v] ⊆ R(D ∪ {b00 }) ∪ IG [D] ∪ [c00 . . . c000 ]
⊆ R(D ∪ E ∪ {b00 }) ∪ IG [D] ⊆ R(D ∪ E ∪ {b}) ∪ IG [D ∪ E] .
We have seen that all conditions are indeed satisfied. Thus, we conclude that
(D ∪ E, D ∪ E) is a realizer for A .
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Lemma 5 (Realizer extension 2). Let A = [p, q; d, e, e0 ; b0 ; b, c, c0 ] + [s, t]
be a descript with s 6= × and d 6= × and [c . . . c0 ] ∩ IG [s, t] = ∅. Let B =
[p, q; d, e, e0 ; b0 ; b, c, c0 ] + [×, ×]. Let Γ (A ) = (k, k 0 ) and Γ (B) = (m, m0 ). Then,
k 0 ≤ m0 .
The proof of Lemma 5 is given in the appendix.
Lemma 6 (Realizer extension 3). Let A = [p, q; d, e, e0 ; b0 ; b, c, c0 ] + [×, ×]
be a descript with d 6= ×. Let B = [p, q; d0 , e00 , e000 ; b0 ; b, c, c0 ] + [d, cq ] where
d0 , e00 , e000 ∈ ([d . . . cp ] \ {d}) ∪ {×}. Let Γ (A ) = (k, k 0 ) and Γ (B) = (m, m0 ).
Then, k 0 ≤ m0 + 1.
Proof. Let (D, D) be a realizer for B that contains cq . Let C =def D ∪ {d} and
C =def D ∪ {(d, cq )}. It holds that |C| = |D| + 1. We show that (C, C) is a realizer
for A . Since d ∈ Lp , it holds that Ri (C ∪ {b}) = Ri (D ∪ {b}) for every p < i ≤ q.
Furthermore, IG [D ∪{(d, cq )}] = IG [C], and thus, the claim directly follows. Note
here that the condition about (s, t)-fields is trivially satisfied.
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The proof of each of the three lemmas above strongly relied on the properties of realizers and the definitions of descripts. Note that the assumptions of
Lemma 4 directly require p < r < q.
We show next that the converse of the Lemmas 4, 5 and 6 is also true. If
we say that the realizer extension provides an upper bound on the optimal sizes
of realizers, then we can say that the converse operation of reducing realizers
provides lower bounds on the sizes of realizers. The correctness of the following
results heavily relies on the properties of minimal normal geodetic pairs.
Lemma 7 (Realizer reduction 1). Let A = [p, q; d, e, e0 ; b0 ; b, c, c0 ] + [s, t] be
a descript with p ≤ q − 2, let Γ (A ) = (k, k 0 ), and assume that k < k 0 . Then,
there are descripts B = [p, r; d, e, e0 ; b0 ; b00 , c00 , c000 ] + [s, t] and
C = [r, q; d0 , c00 , c000 ; b00 ; b, c, c0 ]+[s, t] such that k ≥ m+n, where Γ (B) = (m, m0 )
and Γ (C ) = (n, n0 ).
The proof of Lemma 7 is given in the appendix.
Lemma 8 (Realizer reduction 2). Let A = [p, q; d, e, e0 ; b0 ; b, c, c0 ] + [s, t] be
a descript, and let Γ (A ) = (k, k 0 ). Then, one of the three cases applies:
1) It holds that p ≤ q−2, and there are descripts B = [p, r; d, e, e0 ; b0 ; b00 , c00 , c000 ]+
[s, t] and C = [r, q; d0 , c00 , c000 ; b00 ; b, c, c0 ] + [s, t] such that k 0 ≥ m + n0 , where
Γ (B) = (m, m0 ) and Γ (C ) = (n, n0 ).
10
2) It holds that s 6= ×, and for B = [p, q; d, e, e0 ; b0 ; b, c, c0 ] + [×, ×] and Γ (B) =
(m, m0 ), it holds that k 0 ≥ m0 .
3) It holds that s = ×, and there are d0 , e00 , e000 ∈ ([d . . . cp ] \ {d}) ∪ {×} such
that B = [p, q; d0 , e00 , e000 ; b0 ; b, c, c0 ] + [d, cq ] is a descript and k 0 ≥ m0 + 1,
where Γ (B) = (m, m0 ).
Proof. Let (D, D) be a realizer for A that contains cq . If q = p + 1 then the first
case trivially applies. So, we can assume that q ≥ p + 2. If cq does not appear
in D, which is only possible in case q = h, then the construction in the proof
of Lemma 7 is applicable, and the first case of the lemma applies. So, assume
now that cq appears in D. Let u be the leftmost vertex from G with respect to
σ such that (u, cq ) ∈ D. Due to the assumption, u exists. If u ∈ Lp+1 ∪ · · · ∪ Lq
then, analogous to the previous case, the construction in the proof of Lemma 7
is applicable, and thus, the first case of the lemma applies.
As the remaining case, assume that u ∈ Lp . Note that this particularly
implies d 6= ×, and (d, cq ) ∈ D due to the definition of vertex d in realizers, and
thus, u = d due to the choice of u. If s 6= × then (d, cq ) is an (s, t)-field, and
(D, D) is a realizer for B of the second case. Finally, assume that s = ×. It can
be shown that the third case is applicable and that (D \ {d}, D \ {(d, cq )}) or
(D \ {d, cq }, D \ {(d, cq )}) is a realizer for B. If the latter geodetic pair is the
realizer, we have to add a new pair to make it formally fit. This is always possible
as long as D contains at least two pairs. A special case would be D = {(d, cq )}.
Note also that care needs to be taken for the newly created (d, cq )-fields.
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Note that the third case of Lemma 8 implicitly assumes d 6= ×. We can
only remark here that d = × would directly imply that the first case must be
applicable.
The combination of all results established in this section immediately leads to
the main result of our paper, given as Theorem 2 below. The algorithm orders
the descripts and iteratively applies Algorithm SizeOfRealizer of Figure 1.
The running time of a single application of the algorithm and the number of
applications of the algorithm are polynomial in the number of descripts, which
is a polynomial in the number of vertices of G.
Theorem 2. There is a polynomial-time algorithm that, given a connected proper
interval graph G and a proper interval ordering σ for G, computes the normal
geodetic number for G with respect to σ.
Corollary 1. The geodetic number of proper interval graphs can be computed
in polynomial time.
The proofs of Lemmas 4, 5 and 6 are constructive and show how to obtain a
realizer for the “bigger” descript from the “smaller” descripts. These constructions can be used to extend the algorithm of Theorem 2, and also Corollary 1,
to compute a minimum geodetic set of G.
11
4
Concluding remarks and further results
The algorithms for computing the geodetic number and a minimum geodetic
set of proper interval graphs (Theorem 2 and Corollary 1) have been our major
challenge and main results. In this section, we report shortly on results for other
graph classes. Due to limited space, we give only the main ideas behind the
results.
A graph is a cobipartite graph if it is the complement of a bipartite graph,
i.e., the vertex set of the complement admits a partition into two independent
sets. The famous NP-complete Dominating Set problem is known to be NPcomplete on connected bipartite graphs [1, 5]. Using a reduction from Dominating Set on bipartite graphs to Geodetic Set on cobipartite graphs, we are
able to show the following result.
Theorem 3. Given a cobipartite graph G and an integer k, it is NP-complete
to decide whether the geodetic number of G is at most k.
A cut vertex in a graph G is a vertex whose removal disconnects G. A block of
G is a maximal connected induced subgraph of G that itself has no cut vertex.
The proof of the following theorem relies on the fact that no cut-vertex of a
graph belongs to a minimum geodetic set.
Theorem 4. Let G be a connected graph, let A be the set of cut-vertices of G,
and let B1 , . . . , Bt be the blocks of G. For 1 ≤ i ≤ t, let Di be a geodetic set for
Bi of smallest possible size satisfying A ∩ V (Bi ) ⊆ Di . Then, (D1 ∪ · · · ∪ Dt ) \ A
is a minimum geodetic set for G.
A block-cactus graph is a graph whose blocks are either complete graphs or
chordless cycles. For complete graphs and chordless cycles, the special geodetic
set problem of Theorem 4 is efficiently solvable, which proves the following result.
Theorem 5. A minimum geodetic set for a block-cactus graph can be computed
in polynomial time.
Finally, we come to bipartite permutation graphs. Let G be a bipartite graph
with bi-partition (A, B), and let σA and σB be orderings for respectively A and
B. We say that (σA , σB ) is a strong ordering for G if for every vertex quadruple u, v, x, y of G with u, v ∈ A and x, y ∈ B and u ≺σA v and x ≺σB y,
uy ∈ E(G) and vx ∈ E(G) implies ux ∈ E(G) and vy ∈ E(G). G is a bipartite
permutation graph if and only if it has a strong ordering for bi-partition (A, B)
[21]. Notice the resemblance of strong orderings and proper interval orderings.
Using this similarity, the results of Section 3 can be applied to bipartite permutation graphs with small modifications. The main difficulty for bipartite permutation graphs is the possible lack of simplicial vertices. In proper interval
graphs, the simplicial vertices are important anchor points. In bipartite permutation graphs, we lose this property. Nevertheless, the results of Section 3, when
adapted to bipartite permutation graphs, give the following result.
Theorem 6. Let G be a bipartite permutation graph. A geodetic set for G of
size at most g(G) + 1 can be computed in polynomial time.
12
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13
Appendix
Here we give the proofs Lemmas 5 and 7.
Proof of Lemma 5. Let (D, D) be a realizer for B. We show that (D, D) is
a realizer for A , from which the claimed bound follows. Since IG [D] ⊆ IG [D ∪
{(s, t)}], it remains to verify two conditions about realizers. Note that
[ap+1 . . . cq ] ⊆ R(D ∪ {b}) ∪ IG [D] ∪ [c . . . c0 ] ,
so that it directly follows for s 6= × and every (s, t)-field (u, v) in D that
[[u . . . v] ⊆ R(D ∪ {b}) ∪ IG [D] ∪ [c . . . c0 ].
We consider [c . . . c0 ]. It holds that [c . . . c0 ]∩R(D∪{b}) = ∅ due to the assumptions about (D, D). If [c . . . c0 ] ∩ IG [D ∪ {(s, t)}] 6= ∅ then [c . . . c0 ] ∩ IG [s, t] 6= ∅,
a contradiction to the assumptions of the lemma.
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Proof of Lemma 7. Let (D, D) be a realizer for A of size k. Since k < k 0
due to the assumption of the lemma, D does not contain cq . We constructively
determine B and C from A and (D, D). Let A and B be the sets of vertices from
D that respectively appear and do not appear in D. If A is empty then D is empty,
and d = × must hold. In this case, B and C can be defined straightforwardly.
Henceforth, we assume that A, and thus D, is non-empty. We distinguish between
two cases about d.
Case 1: d = ×
Let d0 be the leftmost vertex from A with respect to σ. Let r be the integer with
d0 ∈ Lr . Note here that p < r < q. Let D1 =def B1 =def {x ∈ D : x ≺σ d0 } and
let A1 =def ∅. Note that D1 ⊆ B. Let D2 =def D \ D1 , and let A2 =def A and
B2 =def B \ B1 , and let D1 =def ∅ and D2 =def D. Note that (A1 , A2 , B1 , B2 ) is
a partition of D and no vertex in B = B1 ∪ B2 has a below-neighbour.
Case 2: d 6= ×
Let v be the vertex from D with (d, v) ∈ D. Note that v indeed exists and is
unique due to the uniqueness part of condition N2 and v 6= cq . Let r be the integer
with v ∈ Lr . Due to condition P1, v = cr . Due to the assumptions of the lemma,
p < r < q. Let D1 be the set of pairs (u, z) from D with d 4σ u ≺σ z 4σ v,
and let D2 =def D \ D1 . Let d0 be the leftmost vertex from D with respect to
σ such that d0 appears in D2 . Let D1 be the union of {x ∈ D : x 4σ cr−1 }
and x ∈ D ∩ Lr that do not appear in D2 . Note that v ∈ D1 and all vertices in
(D1 ∩ Lr ) \ {v} have no below-neighbour. Let D2 =def D \ D1 . Note that every
vertex from D that appears in D1 is contained in D1 , and every vertex from D
that appears in D2 is contained in D2 .
Observe that D1 ⊆ Lp ∪ · · · ∪ Lr and D2 ⊆ Lr ∪ · · · ∪ Lq . We continue the
proof by defining the two descripts B and C . Let b00 be the leftmost vertex from
Rr+1 (D ∪ {b}) with respect to σ. It is important to note that Rr+1 (D ∪ {b}) =
14
Rr+1 (D2 ∪ {b}). We define c00 and c000 . If Lr ⊆ R(D1 ∪ {b00 }) ∪ IG [D1 ∪ {(s, t)}]
then let c00 =def c000 =def ×, and if Lr 6⊆ R(D1 ∪ {b00 }) ∪ IG [D1 ∪ {(s, t)}] then
let c00 and c000 be the respectively leftmost and rightmost vertex from Lr with
respect to σ that are not contained in Rr (D1 ∪ {b00 }) ∪ IG [D1 ∪ {(s, t)}]. We
show that [c00 . . . c000 ] ∩ (R(D1 ∪ {b00 }) ∪ IG [D1 ∪ {(s, t)}]) = ∅. Suppose that
there is a vertex y in D1 ∪ {b00 } with x ∈ R(y). Then, x0 ∈ R(y) for every
vertex x0 ∈ Lr with x 4σ x0 , in particular, c000 ∈ R(y), a contradiction to the
definition of c000 . Suppose that x ∈ IG [u, z] for some (u, z) ∈ D1 ∪ {(s, t)}. The
results of Lemma 2 and Lemma 1 show that c00 ∈ IG [u, z], a contradiction. Thus,
[c00 . . . c000 ] ∩ (R(D1 ∪ {b00 }) ∪ IG [D1 ∪ {(s, t)}]) is indeed empty. This completes
the definitions of B and C .
We show that (D1 , D1 ) and (D2 , D2 ) are realizers for respectively B and C .
Since D1 ∩ D2 = ∅ and D1 ∪ D2 , the claim of the lemma then directly follows.
We verify the conditions for realizers. From the assumptions about (D, D), it
directly follows that (D1 , D1 ) and (D2 , D2 ) are normal geodetic pairs for G. The
minimality conditions M1 and M2 are also satisfied, as a consequence of (D, D)
being a minimal normal geodetic pair for G. We verify the four conditions for
being realizers:
– since A ∩ Lp = A1 ∩ Lp , it holds that d is the leftmost vertex from A1 ∩ Lp
with respect to σ; d0 is the leftmost vertex from A2 ∩ Lr with respect to σ
by definition
– b0 is the leftmost vertex from Rp+1 (D1 ∪ {b00 }) with respect to σ, and
b00 is the leftmost vertex from Rr+1 (D2 ∪ {b}) with respect to σ;
the latter follows from the choice of b00 and the shown properties, the former
follows from the fact that b0 is the leftmost vertex from Rp+1 (D ∪ {b}) with
respect to σ due to the assumptions about (D, D) and that Rr (D ∪ {b}) =
Rr (D1 ∪ {b00 }), which yields Rp+1 (D ∪ {b}) = Rp+1 (D1 ∪ {b00 })
– [c00 . . . c000 ] ∩ (R(D1 ∪ {b00 }) ∪ IG [D1 ∪ {(s, t)}]) = ∅ as shown above;
[c . . . c0 ] ∩ (R(D2 ∪ {b}) ∪ IG [D2 ∪ {(s, t)}]) = ∅ follows from the assumptions
about (D, D);
[ap+1 . . . cr ] ∪ [ar+1 . . . cq ] ⊆ R(D ∪ {b}) ∪ IG [D ∪ {(s, t)}] ∪ [c . . . c0 ] due to
the assumptions, so that
[ap+1 . . . cr ] ⊆ R(D1 ∪ {b00 }) ∪ IG [D ∪ {(s, t)}] and
[ar+1 . . . cq ] ⊆ R(D2 ∪ {b}) ∪ IG [D ∪ {(s, t)}] ∪ [c . . . c0 ], and thus, due to the
choices of r, c00 , c000 ,
[ap+1 . . . cr ] ⊆ R(D1 ∪ {b00 }) ∪ IG [D1 ∪ {(s, t)}] ∪ [c00 . . . c000 ] and
[ar+1 . . . cq ] ⊆ R(D2 ∪ {b}) ∪ IG [D2 ∪ {(s, t)}] ∪ [c . . . c0 ]
– assume that s 6= ×
let (u, v) be an (s, t)-field in D;
(u, v) is an (s, t)-field either in D1 or in D2 ; it follows that either
[[u, v] ⊆ R(D1 ∪ {b00 }) ∪ IG [D1 ] ∪ [c00 . . . c000 ] or
[[u, v] ⊆ R(D2 ∪ {b}) ∪ IG [D2 ] ∪ [c . . . c0 ].
Thus, (D1 , D1 ) and (D2 , D2 ) are indeed realizers for respectively B and C , and
the claim of the lemma holds.
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