Directory UMM :Data Elmu:jurnal:O:Operations Research Letters:Vol26.Issue3.Apr2000:
Operations Research Letters 26 (2000) 111–116
www.elsevier.com/locate/orms
On the structure and complexity of the 2-connected Steiner network
problem in the plane
Emily Larson Luebke, J. Scott Provan ∗
Department of Operations Research, University of North Carolina, Chapel Hill, NC 27599-3180, USA
Received 1 August 1999; received in revised form 1 September 1999
Abstract
We consider the problem of nding a minimum Euclidean length graph 2-connecting a set of points in the plane. We
show that the solution to this problem is an edge-disjoint union of full Steiner trees. This has three important corollaries.
The rst is a proof that the problem is NP-hard, even in the sense of nding a fully polynomial approximation scheme. The
second is a complete description of the solutions for 2SNPP for rectangular arrays of lattice points. The third is a linear-time
c 2000 Published by Elsevier
algorithm for constructing an optimal solution to 2SNPP given its topological description.
Science B.V. All rights reserved.
Keywords: Network graphs; Design; Survivability; Steiner tree; Polynomial approximation
1. Introduction
An important problem in the construction of survivable networks (see [3]) is the 2-Connected Steiner
Network Problem in the Plane (2SNPP). 2SNPP has
an input a set Z of k terminal points in the plane, and
as output the minimum Euclidean length 2-connected
network of straight line segments which contains all
of the points of Z. (The distinction between edge- and
vertex-connectivity is not important in this context,
since the optimal solution will be the same for both
versions.) Previous work on the 2-connected problem has concentrated primarily on the graphical version, where the 2-connected network is chosen from
∗ Corresponding author.
E-mail address: provan@imap.unc.edu (J.S. Provan)
a weighted graph (see [3]). This problem has been
shown to be NP-hard, the reduction coming directly
from the Hamiltonian Circuit problem. Some structural results are available for the Euclidean problem
[5,16], but the complexity of 2SNPP is still open.
The 1-connected analogue to this problem is the
well-known Steiner Tree Problem, which involves
nding the minimum length network (which is in fact
a tree) connecting all pairs of terminals. This problem
has enjoyed considerable attention (see [8]) and is
known to be NP-hard [2]. A complete set of solutions
to the problem has been found by Brazil et al. [1] when
the points of Z form a rectangular lattice of integer
points. Construction algorithms — most recently the
algorithm of Warme et al. [17] — solve the Steiner
tree problem by putting together sets of full Steiner
trees (FSTs). These are minimum length trees on a
c 2000 Published by Elsevier Science B.V. All rights reserved.
0167-6377/00/$ - see front matter
PII: S 0 1 6 7 - 6 3 7 7 ( 9 9 ) 0 0 0 6 8 - 1
112
E.L. Luebke, J.S. Provan / Operations Research Letters 26 (2000) 111–116
Fig. 1. A partition into full Steiner subnetworks.
subset of the terminals such that the terminals are all
end (degree 1) vertices of the tree. FSTs are easy
to construct geometrically and typically span a small
number of the terminals, and so are excellent building
blocks for Steiner tree solutions.
In this paper we give an interesting result concerning the structure of an optimal solution to 2SNPP,
namely, that the solution is made up of an edge-disjoint
union of FSTs. We then give three important consequences of this result. The rst is a proof that 2SNPP
is NP-hard, even in the sense of nding a “fully
polynomial approximation scheme”. The second is
a complete characterization of the 2SNPP solutions
for rectangular lattices, thus extending the results
of Brazil et al. The third shows that there exists a
linear-time algorithm for constructing the geometric
solution to the 2SNPP given the topological solution,
and we outline a rudimentary enumerative solution
algorithm for solving 2SNPP along the same lines as
those suggested for the Steiner tree problem.
2. The structural result
Consider optimal solution F associated with a
2SNPP instance Z. It follows immediately from the
results in [12] that every vertex of F has degree 2 or
3, and that the nonterminal vertices of F — which
we will call Steiner vertices — are of degree exactly
3. Now the edges F can be partitioned uniquely into
nonempty sets T1 ; : : : ; Tr with the property that each
Ti is maximal with respect to the fact that every pair
of edges in Ti can be connected by a path in F —
with possibly coincident endpoints — each of whose
interior vertices are Steiner vertices (see Fig. 1).
We call Ti a full Steiner subnetwork (FSS) of F.
It follows that two distinct FSSs intersect only at a
Fig. 2. Cycles intersecting in a simple path composed of Steiner
points.
set of terminal vertices. We can now state the main
structural result.
Proposition 1. For any optimal solution F to
2SNPP; each FSS of F is in fact a FST.
To prove this result, we rst give a lemma that is
interesting in its own right.
Lemma 1. If two cycles of F intersect in a simple
path; then this path must have an interior vertex that
is a terminal.
Proof. Let C1 and C2 be two cycles of F that intersect
in a simple path P all of whose interior vertices are
Steiner points, and choose these cycles so that the
number of edges in P is as small as possible (see
Fig. 2). Suppose there is at least one interior vertex s
on P. Since s is a Steiner point, we know that s must
be of degree exactly 3, and so there must be a unique
edge e adjacent to v that is not on P, and hence not
on either cycle.
Since F is a solution to the 2SNPP, then e must be
part of a cycle D, which must then contain an edge f
of P. Follow D starting with e away from f until the
rst point v at which D again intersects either C1 or
C2 , say C1 . Then by continuing on C1 in the appropriate direction back through f to s we obtain a cycle
C3 whose interaction with C2 is a simple path with
fewer edges than P, a contradiction to the choice of C1
and C2 .
E.L. Luebke, J.S. Provan / Operations Research Letters 26 (2000) 111–116
113
interior vertices Steiner points. It follows that the cycle
formed by paths P1 and P2 intersects C in P2 , violating
the conditions of Lemma 1. Thus such a cycle cannot
exist, and so all the FSSs are trees. By the de nition
of FSS these trees can have terminals only at the end
vertices, and the result follows.
3. NP-hardness of 2SNPP
Fig. 3. Cycle in full Steiner component.
The path P must therefore contain exactly one edge
(v1 ; v2 ). Remove (v1 ; v2 ) and consider all cuts (X; V \
X ) in F\{(v1 ; v2 )}. The only cuts that have changed
are those with v1 ∈ X and v2 ∈ V \ X , so examine
those cuts. Clearly v1 and v2 both lie in the cycle
C1 ∪C2 \{(v1 ; v2 )}, and so (X; V \X ) must have at least
two of the edges from this cycle in it. We conclude
that all cuts in F\{(v1 ; v2 )} have at least two edges in
them, and so F\{(v1 ; v2 )} is a solution to the 2SNPP.
This contradicts the fact that F is an optimal solution
to the 2SNPP, and the lemma follows.
Proof of Proposition 1. Let S be a FSS for F, and
suppose that S contains a cycle C. We can choose C
to contain at most one terminal, for if it contains at
least one terminal t, pick the two edges e1 and e2 on C
adjacent to t. By the de nition of FSS there must be
a cycle in S containing e1 and e2 and containing no
terminal vertices other than t. This will be our choice
for C.
Choose one of the Steiner points s on C, and consider the unique edge e adjacent to s that is not part of
C (see Fig. 3). Since F is a solution to the 2SNPP,
then e must be part of a cycle D, which must therefore also contain one of the edges of C adjacent to
e. Form the path P1 by following D starting at s and
going through e, to the rst point v that is again on C.
Now s and v partition the edges of C into paths P2
and P3 . Further, since C contains at most one terminal,
then at least one of P2 or P3 , say P2 , has all of its
In this section we prove that 2SNPP is NP-hard. In
particular, we reduce a special form of the Euclidean
Traveling Salesman Problem to it. The ETSP involves
nding the minimum Euclidean-length circuit through
a given set of points on the Euclidean plane. The Traveling Salesman Problem has always been closely associated with 2-connected network problems (see, for
example [4]), as a TSP circuit through a set of points
is also a feasible — though not necessarily optimal
— solution to 2SNPP for that set of points. We rst
show that under special conditions the TSP solution
is in fact the optimal solution to 2SNPP.
Lemma 2. Let Z be a set of k integer lattice points.
Then
• Every solution to 2SNPP is of length at least k.
• The only solutions to 2SNPP of length exactly k
are TSP circuits such that every two successive
points are adjacent in the lattice.
√
• All other solutions are of length at least k + 2−1.
Proof. Let F be an optimal solution to 2SNPP on Z.
From Proposition 1 we can partition F into a set of
edges e1 ; : : : ; en connecting vertices of Z, together with
a set T1 ; T2 ; : : : ; Tt of FSTs each of which spans at
least 3 terminals. For i=1; : : : ; t, let ki be the number of
terminals on Steiner tree Ti . Now the minimum length
spanning tree on this set of terminals is of length at
least ki − 1. The result of Du and Hwang [6] on the
Steiner ratio shows that the length Ti must be at least
√
3
(ki − 1) ¿ ki =2:
(∗)
2
This means that the total length of F is
l(F) =
n
X
i=1
l(ei ) +
t
X
j=1
l(Ti )
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E.L. Luebke, J.S. Provan / Operations Research Letters 26 (2000) 111–116
¿n+
√
t
X
3
(ki − 1)
2
j=1
1X
ki
2
t
¿n+
j=1
with the rst inequality being strict unless l(ei ) = 1
for all i; and the last inequality being strict unless
t = 0. Now each terminal must be contained in at least
two distinct
sets of e1 ; : : : ; en ; T1 ; T2 ; : : : ; Tt , so that
Pt
2n + i=1 ki = 2k. Thus
l(F)¿k
with equality holding only when t =0 and each l(ei )=
1, that is, when the solution is a Hamiltonian circuit
through the vertices of Z with each successive pair of
points being distance 1 apart.
Finally, if F is not of the above sort then one of
the following must hold:
Case 1: F contains an edge between
√ two
non-adjacent lattice points. Then l(F)¿k + 2 − 1.
Case 2: F contains a FST on four or more
√ terminals. Using (∗) we
√ have that l(F)¿k + ( 3=2)(4 −
1) − 4=2 ¿ k + 2 − 1.
Case 3: F contains two FST on at least 3 terminals√each. Again using (∗), we have
√ that l(F)¿k +
2(( 3=2)(3 − 1) − 3=2) ¿ k + 2 − 1.
Case 4: F contains exactly one FST on 3 terminals.
Since each terminal in a 2SNPP must be at least degree
2, then the number of edges in F, excluding those in
the FST, must be at least ⌈(2k − 3)=2⌉ = k − 1, with
each edge of length at least
√ ratio
√ 1. Using the Steiner
we have l(F)¿k − 1 + ( 3=2)(3 − 1) ¿ k + 2 − 1.
The result follows.
Corollary 1. 2SNPP is NP-hard.
Proof. Itai et al. [9] (see also [10]) show that the
following problem is NP-complete:
Input: Set Z integer lattice points in the plane.
Question: Does there exist a Hamiltonian circuit
through the points of Z such that the distance between each successive pair of
points is 1?
It follows immediately from Lemma 2 that for any
instance of HCGG the 2SNPP will be length k if and
only if HCGG has the required Hamiltonian circuit.
We note that the decision version of 2SNPP —
Does there exist a solution to 2SNPP for given set Z
that is of length at most a given rational number N ?
— is not known to be in NP, due to the problem of
comparing sums of irrational distances. An approach
for solving this class of problems that has been successful in several related papers [13–16] is to nd a
fully polynomial approximation scheme (FPAS) for
2SNPP. Here we wish to have an algorithm which,
given instance Z for 2SNPP and accuracy bound ,
will nd an -approximation to SNPP, that is, a feasible solution F for this instance whose length l̂ satis es (l̂ − l∗ )=l∗ ¡ , where l∗ is the length of the
optimal solution for this instance. Further, this algorithm must have a running time that is polynomial
in 1= and the length of the input describing Z. Hsu
and
√ Hu [5] give a polynomial algorithm to nd a
( 3 − 1)-approximation for 2SNPP. There is a limit
to how far these types of results can go, however, as
the next results shows.
Corollary 2. A FPAS for 2SNPP will not exist unless P = NP.
Proof. For any instance of HCGG, we have from
Lemma 2 that a solution to 2SNPP that is
√ not a HCGG
solution must have length at least k + 2 − 1 ¿ k +
0:414. Now for = 0:414=k, an -approximation to
2SNPP will have length less than k +0:414 if and only
if HCGG has a feasible solution. It follows that there
can be no algorithm that nds an -approximation in
time polynomial in 1= = k=0:414 unless P = NP.
4. The solution to 2SNPP for rectangular arrays of
lattice points
In this section we consider solutions of 2SNPP in
the case where Z forms an m × n rectangular array of integer lattice points on the Euclidean plane.
An elegant solution to the Steiner tree problem for
this class of instances was found by Brazil et al. [1].
Lemma 2 can be used to nd a simple set of corresponding solutions for the 2-connected case.
Corollary 3. Let Z be m × n rectangular array of
integer lattice points on the Euclidean plane. Then
a 2SNPP solution for this instance will always be a
E.L. Luebke, J.S. Provan / Operations Research Letters 26 (2000) 111–116
115
that the resulting plane network corresponds to the
optimal solution for 2SNPP. Proposition 1 shows that
by rst partitioning G into its FSSs we obtain a set
of edge-disjoint FSTs that will individually have the
same properties as they do for the Melzak construction. By using the linear-time Hwang implementation
of the Melzak construction to place each of the trees,
we obtain the following obvious corollary:
Fig. 4. Solutions to 2SNPP on a rectangular lattice.
single circuit through the points of Z of length
• k if either
m or n is even;
√
• k + 2 − 1 if both n and m are odd.
Proof. If either n or m is even, then it is easy to
construct a circuit consisting entirely of adjacent lattice points (see Fig. 4 (a), for example) which by
Lemma 2 is optimal. If both n and m are odd, then
by a simple parity argument there can be no circuit
consisting entirely of adjacent lattice points through
the odd number of vertices of Z. By Lemma 2 the optimal 2SNPP
√ solution must therefore have length at
least k + 2 − 1. It is also easy to construct a circuit
consisting entirely of adjacent lattice points except for
one diagonal lattice edge (see Fig.
√ 4(b), for example),
and this has length exactly k + 2 − 1. The corollary
follows.
Corollary 4. The optimal solution to 2SNPP for
given instance Z can be constructed from its topological description in linear time.
The real importance of Corollary 4 is that it gives
a basis for constructing a solution to 2SNPP by ecient enumeration of the appropriate graph topologies.
Proposition 1 shows that there can be at most 3(k − 2)
Steiner points in any optimal solution, since each FST
that spans a subset of ki of the terminals, can obtain at
most ki − 2 Steiner points, and each terminal is contained in at most 3 FSTs. Thus one method of solving
2SNPP could be to enumerate all possible 2-connected
graphs on the k terminals and at most 3(k − 2) proposed Steiner points and then to construct the associated geometric solutions, choosing the best as the optimal 2SNPP solution. This enumeration, however, is
not practical, and so it is necessary to extend ecient
FST enumeration and pruning techniques, like those
used in [17] for STP, to the 2SNPP problem. We leave
this problem for future research.
References
5. Finding a 2SNPP from topological structure
Results by Melzak [11] and Hwang [7] establish
a linear-time algorithm to construct a geometric realization for a solution to the Steiner tree problem
based on the topological description of the solution.
Proposition 1 allows these results to apply immediately to 2SNPP. In particular, suppose we are given
a topological description of a solution to 2SNPP, in
the form of a 2-connected graph G whose vertices are
designated as either labeled terminal vertices or
Steiner vertices. We wish to construct a geometric
solution corresponding to G, by placing the Steiner
points correctly in the plane and drawing straight
lines between Steiner points and terminal points so
[1] M. Brazil, J.H. Rubinstein, D.A. Thomas, J.F. Weng,
N.C. Wormald, Minimal Steiner trees for rectangular arrays
of lattice points, J. Combin. Theory Ser. A 79 (1997)
181–208.
[2] M.R. Garey, R.L. Graham, D.S. Johnson, The complexity of
computing Steiner minimal trees, SIAM J. Appl. Math. 32
(1977) 835–859.
[3] M. Grotschel, C.L. Monma, M. Stoer, Design of survivable
networks, in: M.O. Ball, T.L. Magnanti, C.L. Monma,
G.L. Nemhauser (Eds.), Network Models, Handbooks in
Operations Research and Managements Science Series,
Vol. 7, North-Holland, New York, 1995, pp. 617–672.
[4] G.N. Frederickson, J. Ja’Ja’, On the relationship between the
biconnectivity augmentation and traveling salesman problems,
Theoret. Comput. Sci. 9 (1982) 189–201.
[5] D.F. Hsu, X.-D. Hu, On shortest two-edge connected Steiner
networks with Euclidean distance, Networks 32 (1998)
133–140.
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[6] D.-Z. Hu, F.K. Hwang, A proof of Gilbert and Pollack’s
conjecture on the Steiner ratio, Algorithmica 7 (1992)
121–135.
[7] F.K. Hwang, A linear time algorithm for full Steiner trees,
Oper. Res. Lett. 4 (1986) 235–237.
[8] F.K. Hwang, D.S. Richards, P. Winter, The Steiner Tree
Problem, North-Holland, New York, 1992.
[9] A. Itai, C.H. Papadimitriou, J. Szwarc ter, Hamiltonian paths
in grid graphs, SIAM J. Comput. 11 (1982) 676–686.
[10] D. Johnson, C.H. Papadimitriou, Computational complexity,
in: E.L. Lawler, J.K. Lenstra, A.H.G. Rinnooy Kan,
D.B. Shmoys (Eds.), The Traveling Salesman Problem,
Wiley, New York, 1986, pp. 37–85 (Chapter 2).
[11] Z.A. Melzak, On the problem of Steiner, Canad. Math. Bull.
4 (1961) 143–148.
[12] C.L. Monma, B.S. Munson, W.R. Pulleyblank, Minimum
weight
two-connected
spanning
networks,
Math.
Programming 46 (1990) 153–171.
[13] J.S. Provan, Convexity and the Steiner tree problem,
Networks 18 (1988) 55–72.
[14] J.S. Provan, An approximation scheme for nding Steiner
trees with obstacles, SIAM J. Comput. 17 (1989)
920–934.
[15] J.S. Provan, A fully polynomial approximation scheme for
the Euclidean Steiner augmentation problem, in: D.-Z. Du,
J.M. Smith, J.H. Rubinstein (Eds.), Advances in Steiner
Trees, Kluwer Academic Publishers, Dordrecht, 2000, pp.
235–253.
[16] K.T. Talluri, Issues in the design and analysis of
survivable networks, Ph.D. Thesis, Massachusetts Institute of
Technology, Cambridge, MA, 1991.
[17] D.M. Warme, P. Winter, M. Zachariasen, Exact algorithms
for plane Steiner tree problems: a computational study, in:
D.-Z. Du, J.M. Smith, J.H. Rubinstein (Eds.), Advances in
Steiner trees, Kluwer Academic Publishers, Dordrecht, 2000,
pp. 81–116.
www.elsevier.com/locate/orms
On the structure and complexity of the 2-connected Steiner network
problem in the plane
Emily Larson Luebke, J. Scott Provan ∗
Department of Operations Research, University of North Carolina, Chapel Hill, NC 27599-3180, USA
Received 1 August 1999; received in revised form 1 September 1999
Abstract
We consider the problem of nding a minimum Euclidean length graph 2-connecting a set of points in the plane. We
show that the solution to this problem is an edge-disjoint union of full Steiner trees. This has three important corollaries.
The rst is a proof that the problem is NP-hard, even in the sense of nding a fully polynomial approximation scheme. The
second is a complete description of the solutions for 2SNPP for rectangular arrays of lattice points. The third is a linear-time
c 2000 Published by Elsevier
algorithm for constructing an optimal solution to 2SNPP given its topological description.
Science B.V. All rights reserved.
Keywords: Network graphs; Design; Survivability; Steiner tree; Polynomial approximation
1. Introduction
An important problem in the construction of survivable networks (see [3]) is the 2-Connected Steiner
Network Problem in the Plane (2SNPP). 2SNPP has
an input a set Z of k terminal points in the plane, and
as output the minimum Euclidean length 2-connected
network of straight line segments which contains all
of the points of Z. (The distinction between edge- and
vertex-connectivity is not important in this context,
since the optimal solution will be the same for both
versions.) Previous work on the 2-connected problem has concentrated primarily on the graphical version, where the 2-connected network is chosen from
∗ Corresponding author.
E-mail address: provan@imap.unc.edu (J.S. Provan)
a weighted graph (see [3]). This problem has been
shown to be NP-hard, the reduction coming directly
from the Hamiltonian Circuit problem. Some structural results are available for the Euclidean problem
[5,16], but the complexity of 2SNPP is still open.
The 1-connected analogue to this problem is the
well-known Steiner Tree Problem, which involves
nding the minimum length network (which is in fact
a tree) connecting all pairs of terminals. This problem
has enjoyed considerable attention (see [8]) and is
known to be NP-hard [2]. A complete set of solutions
to the problem has been found by Brazil et al. [1] when
the points of Z form a rectangular lattice of integer
points. Construction algorithms — most recently the
algorithm of Warme et al. [17] — solve the Steiner
tree problem by putting together sets of full Steiner
trees (FSTs). These are minimum length trees on a
c 2000 Published by Elsevier Science B.V. All rights reserved.
0167-6377/00/$ - see front matter
PII: S 0 1 6 7 - 6 3 7 7 ( 9 9 ) 0 0 0 6 8 - 1
112
E.L. Luebke, J.S. Provan / Operations Research Letters 26 (2000) 111–116
Fig. 1. A partition into full Steiner subnetworks.
subset of the terminals such that the terminals are all
end (degree 1) vertices of the tree. FSTs are easy
to construct geometrically and typically span a small
number of the terminals, and so are excellent building
blocks for Steiner tree solutions.
In this paper we give an interesting result concerning the structure of an optimal solution to 2SNPP,
namely, that the solution is made up of an edge-disjoint
union of FSTs. We then give three important consequences of this result. The rst is a proof that 2SNPP
is NP-hard, even in the sense of nding a “fully
polynomial approximation scheme”. The second is
a complete characterization of the 2SNPP solutions
for rectangular lattices, thus extending the results
of Brazil et al. The third shows that there exists a
linear-time algorithm for constructing the geometric
solution to the 2SNPP given the topological solution,
and we outline a rudimentary enumerative solution
algorithm for solving 2SNPP along the same lines as
those suggested for the Steiner tree problem.
2. The structural result
Consider optimal solution F associated with a
2SNPP instance Z. It follows immediately from the
results in [12] that every vertex of F has degree 2 or
3, and that the nonterminal vertices of F — which
we will call Steiner vertices — are of degree exactly
3. Now the edges F can be partitioned uniquely into
nonempty sets T1 ; : : : ; Tr with the property that each
Ti is maximal with respect to the fact that every pair
of edges in Ti can be connected by a path in F —
with possibly coincident endpoints — each of whose
interior vertices are Steiner vertices (see Fig. 1).
We call Ti a full Steiner subnetwork (FSS) of F.
It follows that two distinct FSSs intersect only at a
Fig. 2. Cycles intersecting in a simple path composed of Steiner
points.
set of terminal vertices. We can now state the main
structural result.
Proposition 1. For any optimal solution F to
2SNPP; each FSS of F is in fact a FST.
To prove this result, we rst give a lemma that is
interesting in its own right.
Lemma 1. If two cycles of F intersect in a simple
path; then this path must have an interior vertex that
is a terminal.
Proof. Let C1 and C2 be two cycles of F that intersect
in a simple path P all of whose interior vertices are
Steiner points, and choose these cycles so that the
number of edges in P is as small as possible (see
Fig. 2). Suppose there is at least one interior vertex s
on P. Since s is a Steiner point, we know that s must
be of degree exactly 3, and so there must be a unique
edge e adjacent to v that is not on P, and hence not
on either cycle.
Since F is a solution to the 2SNPP, then e must be
part of a cycle D, which must then contain an edge f
of P. Follow D starting with e away from f until the
rst point v at which D again intersects either C1 or
C2 , say C1 . Then by continuing on C1 in the appropriate direction back through f to s we obtain a cycle
C3 whose interaction with C2 is a simple path with
fewer edges than P, a contradiction to the choice of C1
and C2 .
E.L. Luebke, J.S. Provan / Operations Research Letters 26 (2000) 111–116
113
interior vertices Steiner points. It follows that the cycle
formed by paths P1 and P2 intersects C in P2 , violating
the conditions of Lemma 1. Thus such a cycle cannot
exist, and so all the FSSs are trees. By the de nition
of FSS these trees can have terminals only at the end
vertices, and the result follows.
3. NP-hardness of 2SNPP
Fig. 3. Cycle in full Steiner component.
The path P must therefore contain exactly one edge
(v1 ; v2 ). Remove (v1 ; v2 ) and consider all cuts (X; V \
X ) in F\{(v1 ; v2 )}. The only cuts that have changed
are those with v1 ∈ X and v2 ∈ V \ X , so examine
those cuts. Clearly v1 and v2 both lie in the cycle
C1 ∪C2 \{(v1 ; v2 )}, and so (X; V \X ) must have at least
two of the edges from this cycle in it. We conclude
that all cuts in F\{(v1 ; v2 )} have at least two edges in
them, and so F\{(v1 ; v2 )} is a solution to the 2SNPP.
This contradicts the fact that F is an optimal solution
to the 2SNPP, and the lemma follows.
Proof of Proposition 1. Let S be a FSS for F, and
suppose that S contains a cycle C. We can choose C
to contain at most one terminal, for if it contains at
least one terminal t, pick the two edges e1 and e2 on C
adjacent to t. By the de nition of FSS there must be
a cycle in S containing e1 and e2 and containing no
terminal vertices other than t. This will be our choice
for C.
Choose one of the Steiner points s on C, and consider the unique edge e adjacent to s that is not part of
C (see Fig. 3). Since F is a solution to the 2SNPP,
then e must be part of a cycle D, which must therefore also contain one of the edges of C adjacent to
e. Form the path P1 by following D starting at s and
going through e, to the rst point v that is again on C.
Now s and v partition the edges of C into paths P2
and P3 . Further, since C contains at most one terminal,
then at least one of P2 or P3 , say P2 , has all of its
In this section we prove that 2SNPP is NP-hard. In
particular, we reduce a special form of the Euclidean
Traveling Salesman Problem to it. The ETSP involves
nding the minimum Euclidean-length circuit through
a given set of points on the Euclidean plane. The Traveling Salesman Problem has always been closely associated with 2-connected network problems (see, for
example [4]), as a TSP circuit through a set of points
is also a feasible — though not necessarily optimal
— solution to 2SNPP for that set of points. We rst
show that under special conditions the TSP solution
is in fact the optimal solution to 2SNPP.
Lemma 2. Let Z be a set of k integer lattice points.
Then
• Every solution to 2SNPP is of length at least k.
• The only solutions to 2SNPP of length exactly k
are TSP circuits such that every two successive
points are adjacent in the lattice.
√
• All other solutions are of length at least k + 2−1.
Proof. Let F be an optimal solution to 2SNPP on Z.
From Proposition 1 we can partition F into a set of
edges e1 ; : : : ; en connecting vertices of Z, together with
a set T1 ; T2 ; : : : ; Tt of FSTs each of which spans at
least 3 terminals. For i=1; : : : ; t, let ki be the number of
terminals on Steiner tree Ti . Now the minimum length
spanning tree on this set of terminals is of length at
least ki − 1. The result of Du and Hwang [6] on the
Steiner ratio shows that the length Ti must be at least
√
3
(ki − 1) ¿ ki =2:
(∗)
2
This means that the total length of F is
l(F) =
n
X
i=1
l(ei ) +
t
X
j=1
l(Ti )
114
E.L. Luebke, J.S. Provan / Operations Research Letters 26 (2000) 111–116
¿n+
√
t
X
3
(ki − 1)
2
j=1
1X
ki
2
t
¿n+
j=1
with the rst inequality being strict unless l(ei ) = 1
for all i; and the last inequality being strict unless
t = 0. Now each terminal must be contained in at least
two distinct
sets of e1 ; : : : ; en ; T1 ; T2 ; : : : ; Tt , so that
Pt
2n + i=1 ki = 2k. Thus
l(F)¿k
with equality holding only when t =0 and each l(ei )=
1, that is, when the solution is a Hamiltonian circuit
through the vertices of Z with each successive pair of
points being distance 1 apart.
Finally, if F is not of the above sort then one of
the following must hold:
Case 1: F contains an edge between
√ two
non-adjacent lattice points. Then l(F)¿k + 2 − 1.
Case 2: F contains a FST on four or more
√ terminals. Using (∗) we
√ have that l(F)¿k + ( 3=2)(4 −
1) − 4=2 ¿ k + 2 − 1.
Case 3: F contains two FST on at least 3 terminals√each. Again using (∗), we have
√ that l(F)¿k +
2(( 3=2)(3 − 1) − 3=2) ¿ k + 2 − 1.
Case 4: F contains exactly one FST on 3 terminals.
Since each terminal in a 2SNPP must be at least degree
2, then the number of edges in F, excluding those in
the FST, must be at least ⌈(2k − 3)=2⌉ = k − 1, with
each edge of length at least
√ ratio
√ 1. Using the Steiner
we have l(F)¿k − 1 + ( 3=2)(3 − 1) ¿ k + 2 − 1.
The result follows.
Corollary 1. 2SNPP is NP-hard.
Proof. Itai et al. [9] (see also [10]) show that the
following problem is NP-complete:
Input: Set Z integer lattice points in the plane.
Question: Does there exist a Hamiltonian circuit
through the points of Z such that the distance between each successive pair of
points is 1?
It follows immediately from Lemma 2 that for any
instance of HCGG the 2SNPP will be length k if and
only if HCGG has the required Hamiltonian circuit.
We note that the decision version of 2SNPP —
Does there exist a solution to 2SNPP for given set Z
that is of length at most a given rational number N ?
— is not known to be in NP, due to the problem of
comparing sums of irrational distances. An approach
for solving this class of problems that has been successful in several related papers [13–16] is to nd a
fully polynomial approximation scheme (FPAS) for
2SNPP. Here we wish to have an algorithm which,
given instance Z for 2SNPP and accuracy bound ,
will nd an -approximation to SNPP, that is, a feasible solution F for this instance whose length l̂ satis es (l̂ − l∗ )=l∗ ¡ , where l∗ is the length of the
optimal solution for this instance. Further, this algorithm must have a running time that is polynomial
in 1= and the length of the input describing Z. Hsu
and
√ Hu [5] give a polynomial algorithm to nd a
( 3 − 1)-approximation for 2SNPP. There is a limit
to how far these types of results can go, however, as
the next results shows.
Corollary 2. A FPAS for 2SNPP will not exist unless P = NP.
Proof. For any instance of HCGG, we have from
Lemma 2 that a solution to 2SNPP that is
√ not a HCGG
solution must have length at least k + 2 − 1 ¿ k +
0:414. Now for = 0:414=k, an -approximation to
2SNPP will have length less than k +0:414 if and only
if HCGG has a feasible solution. It follows that there
can be no algorithm that nds an -approximation in
time polynomial in 1= = k=0:414 unless P = NP.
4. The solution to 2SNPP for rectangular arrays of
lattice points
In this section we consider solutions of 2SNPP in
the case where Z forms an m × n rectangular array of integer lattice points on the Euclidean plane.
An elegant solution to the Steiner tree problem for
this class of instances was found by Brazil et al. [1].
Lemma 2 can be used to nd a simple set of corresponding solutions for the 2-connected case.
Corollary 3. Let Z be m × n rectangular array of
integer lattice points on the Euclidean plane. Then
a 2SNPP solution for this instance will always be a
E.L. Luebke, J.S. Provan / Operations Research Letters 26 (2000) 111–116
115
that the resulting plane network corresponds to the
optimal solution for 2SNPP. Proposition 1 shows that
by rst partitioning G into its FSSs we obtain a set
of edge-disjoint FSTs that will individually have the
same properties as they do for the Melzak construction. By using the linear-time Hwang implementation
of the Melzak construction to place each of the trees,
we obtain the following obvious corollary:
Fig. 4. Solutions to 2SNPP on a rectangular lattice.
single circuit through the points of Z of length
• k if either
m or n is even;
√
• k + 2 − 1 if both n and m are odd.
Proof. If either n or m is even, then it is easy to
construct a circuit consisting entirely of adjacent lattice points (see Fig. 4 (a), for example) which by
Lemma 2 is optimal. If both n and m are odd, then
by a simple parity argument there can be no circuit
consisting entirely of adjacent lattice points through
the odd number of vertices of Z. By Lemma 2 the optimal 2SNPP
√ solution must therefore have length at
least k + 2 − 1. It is also easy to construct a circuit
consisting entirely of adjacent lattice points except for
one diagonal lattice edge (see Fig.
√ 4(b), for example),
and this has length exactly k + 2 − 1. The corollary
follows.
Corollary 4. The optimal solution to 2SNPP for
given instance Z can be constructed from its topological description in linear time.
The real importance of Corollary 4 is that it gives
a basis for constructing a solution to 2SNPP by ecient enumeration of the appropriate graph topologies.
Proposition 1 shows that there can be at most 3(k − 2)
Steiner points in any optimal solution, since each FST
that spans a subset of ki of the terminals, can obtain at
most ki − 2 Steiner points, and each terminal is contained in at most 3 FSTs. Thus one method of solving
2SNPP could be to enumerate all possible 2-connected
graphs on the k terminals and at most 3(k − 2) proposed Steiner points and then to construct the associated geometric solutions, choosing the best as the optimal 2SNPP solution. This enumeration, however, is
not practical, and so it is necessary to extend ecient
FST enumeration and pruning techniques, like those
used in [17] for STP, to the 2SNPP problem. We leave
this problem for future research.
References
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points correctly in the plane and drawing straight
lines between Steiner points and terminal points so
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