Two Arc Disjoint Paths in Eulerian Digraphs - CiteSeerX

Two Arc Disjoint Paths in Eulerian Digraphs Andr as FRANK, Toshihide IBARAKI and Hiroshi NAGAMOCHI

Andr as FRANK Department of Computer Science, Mathematical Institute, Eotvos University, M useum kor ut 6-8, Budapest VIII, Hungary 1088, email: [email protected] Toshihide IBARAKI Department of Applied Mathematics and Physics, Graduate School of Engineering, Kyoto University, Kyoto 606-01 Japan, fax 81-75-761-4866, email: [email protected] Hiroshi NAGAMOCHI Department of Applied Mathematics and Physics, Graduate School of Engineering, Kyoto University, Kyoto 606-01 Japan, fax 81-75-761-4866, email: [email protected]

Abstract:

Let G be an Eulerian digraph, and fx1 ; x2 g; fy1 ; y2 g be two pairs of vertices in G. A directed path from a vertex s to a vertex t is called a st-path. An instance (G; fx1 ; x2 g; fy1 ; y2 g) is called feasible if there is a choice of h; i; j; k with fh; ig = fj; k g = f1; 2g such that G has two arc-disjoint xh xi - and yj yk -paths. In this paper, we characterize the structure of minimal infeasible instances, based on which an O (m + n log n) time algorithm is presented to decide whether a given instance is feasible, where n and m are the number of vertices and arcs in the instance, respectively. If the instance is feasible, the corresponding two arc-disjoint paths can be computed in O (m(m + n log n)) time.

Keywords:

algorithm

Eulerian digraph, disjoint paths, minimum cut, planar graph, polynomial time

Abbreviated title: Two Paths in Eulerian Digraphs AMS subject classi cations: 05C35, 05C40

1

1 Introduction Finding a set of edge-disjoint paths connecting pairs of speci ed vertices (called terminals) in a graph or a digraph is one of the classical and fundamental problems in graph theory, which has a wide variety of applications. A path between terminals s and t (or a directed path from s to t) is called an st-path. If the graph is undirected, an important result by Robertson and Seymour [10] says that edge-disjoint paths for k pairs fsi ; ti g of terminals, i = 1; 2; . . . ; k can be obtained in polynomial time for a xed k. In case of k = 2, a complete characterization of undirected graphs G that do not have edge-disjoint s1t1- and s2t2-paths is available (Dinic and Karzanov [2, 3], Seymour [11] and Thomassen [12]). Such G can be reduced to a graph G that has a planar representation with following properties (see Fig. 1): (i) the four terminals have degree 2, and all other vertices are of degree 3, and (ii) the terminals are located on the outer face in the order of s1 ; s2; t1; t2. Contrary to this, the characterization of arc-disjoint path problems in digraphs seem much more dicult. For example, the weak 2-linking problem (i.e., to decide whether there are arcdisjoint s1t1- and s2t2 -paths) in a general digraph is shown by Fortune, Hopcroft, and Wyllie [4] to be NP-complete. However, if the digraph under consideration is Eulerian, the situation becomes slightly easier. For a given digraph G = (V ; E ) with ordered terminal pairs (si ; ti ), i = 1; 2; . . . ; k , call H = (V ; f(ti ; si ) j i = 1; 2; . . . ; kg) its demand digraph. The weak 2-linking problem in an Eulerian digraph G + H is known by Frank [5] to be polynomially solvable. Furthermore, Ibaraki and Poljak [9] showed that the weak 3-linking problem for an Eulerian digraph G + H can also be solved in polynomial time. It is based on the observations that weak 3-linking problem is equivalent to nding arc-disjoint x1x2-, x2x3- and x3 x1-paths in an Eulerian digraph with terminals x1; x2; x3, and that the resulting problem is infeasible if and only if it is reducible to a 2-connected Eulerian digraph G , which has a planar representation (see Fig. 2) such that (i) all terminals have degree 2, and all other vertices have degree 4, and (ii) every face is a directed cycle, and all the terminals are located on the outer face (which is also a directed cycle) in the order of x3 ; x2; x1. In this paper, we generalize the above result to the two arc-disjoint path problem in an Eulerian digraph G (but G + H is not Eulerian), which decides whether there are arc-disjoint paths connecting two unordered terminal pairs fx1; x2g and fy1; y2g (i.e., x x - and y y - paths, where either x x = x1x2 or x2x1, and either y y = y1y2 or y2y1). This problem includes the above weak 3-linking problem as a special case: for a given instance (G; (s1 ; t1 ); (s2; t2 ); (s3; t3)) of the 3-linking problem (where G + H is Eulerian), add four new vertices x1; x2 ; y1 and y2 together with seven new arcs (t1; y1); (y1; s2); (t2; y2); (y2; x1); (x1; s3 ); (t3; x2); (x2; s1 ) to obtain an instance (G ; fx1; x2g; fy1; y2g) of the two arc-disjoint path problem (where G is Eulerian), which is clearly feasible if and only if the instance (G; (s1 ; t1); (s2 ; t2 ); (s3; t3 )) is feasible. We show that the problem can be solved in O(m + n log n) time, where m and n are respectively the numbers of arcs and vertices in G, by deriving an analogue of the above structural characterization of infeasible instances: An Eulerian digraph G with four terminals x1; x2; y1; y2 is infeasible if and only if it is reducible to an Eulerian digraph G that has a planar representation (see Fig. 3(a),(b)) such that (i) all terminals have degree 2, and all other vertices have degree 4, (ii) there is at most one cut vertex, and (iii) every face is a directed cycle, and all terminals are located on the outer face in the order of x1; y ; x2 ; y , where fy ; y g = fy1; y2g. The proof for this is, however, substantially dierent from that of [9]. For a feasible instance, we also show that the corresponding two arc-disjoint paths can be computed in O(m(m + n log n)) 0

0

0

0

00

0

00

0

00

0

0

0

0

00

0

00

2

00

time.

2 Preliminaries Let

G = (V; E ) be a digraph which may have multiple arcs. Denote by deg (v ), indeg (v ) and outdeg (v ) the degree, indegree and outdegree of a vertex v in G, respectively, where the degree

of a vertex is the sum of its out- and indegrees. We call a digraph Eulerian if the outdegree and indegree of each vertex are equal. Under a path or a cycle, we always understand a directed path or cycle. Repetition of arcs is not allowed, but repetition of vertices is allowed. A cycle that visits every arc exactly once is called Eulerian. A path from s to t is called an st-path. If fP1; P2 ; . . . ; Pk g is a collection of arc-disjoint paths such that the last vertex of Pi coincides with the initial vertex of Pi+1 for each i = 1; 2; . . . ; k 0 1, we denote by P = hP1; P2; 1 1 1 ; Pk i the concatenation of the paths. In the following discussion, digraph G, path P , or cycle C may sometimes be treated either as a vertex set or an arc set, as far as its meaning is unambiguous from the context. If it is necessary to specify, we use E (G) and V (G) to mean the arc set and the vertex set of a digraph G, respectively. For a digraph G = (V; E ) and an arc set E 0  E , we denote the digraph (V; E 0 E 0 ) by G 0 E 0. For a vertex set Z  V , the subdigraph induced by Z is denoted by G[Z ] = (Z; EZ ) where EZ = f(u; v ) 2 E j u; v 2 Z g, and G[V 0 Z ] may be denoted by G 0 Z . For a subset Z of vertices, +(Z ) denotes the set of arcs from Z to V 0 Z , 0 (Z ) the set of arcs from V 0 Z to Z , and  (Z ) =  +(Z ) [ 0(Z ). If G is Eulerian, then j+ (Z )j = j 0(Z )j holds for every Z , where jAj denotes the cardinality of a set A, and therefore j(Z )j is always even. A set Z  V is called a k-cut if j (Z )j is k. For two disjoint S; T  V , we say that a cut Z separates S and T if S  Z and T  V 0 Z , and de ne  (S; T ) to be the set of arcs from S to T and arcs from T to S . Throughout this paper, a singleton set fv g may also be denoted as v . Two cuts Z1 and Z2 intersect each other if Z1 \ Z2 6= ;, Z1 0 Z2 6= ; and Z2 0 Z1 6= ;, and they cross each other if, in addition, V 0 (Z1 [ Z2 ) 6= ; holds. For two crossing cuts Z1 and Z2 , we easily see that

j (Z1)j + j(Z2 )j  j(Z1 \ Z2)j + j (Z1 [ Z2)j;

(2.1)

j (Z1)j + j(Z2 )j = j(Z1 0 Z2)j + j(Z2 0 Z1)j + 2j(Z1 \ Z2; V 0 (Z1 [ Z2))j

(2.2)

and hold. Some further notions, such as planarity, edge connectivity, and vertex connectivity, we refer to the unoriented graph G obtained from G by ignoring arc orientation. A digraph G is called connected if G is connected. For a connected digraph G = (V; E ), a vertex z is called a cut vertex if G 0 fz g has more than one connected component. We call an undirected path in G a chain. A chain with end vertices s and t is called an st-chain, and these s and t are said to be connected (by the chain). A concatenation of a collection of chains is also de ned analogously to paths. Consider an instance (G = (V; E ); X; Y ) with X = fx1; x2g  V and Y = fy1; y2g  V . Throughout this paper, when we refer to an instance (G; X; Y ), we assume that G is Eulerian and is connected (hence strongly connected since G is Eulerian). Each t 2 X [ Y is called a 0 00 0 00 terminal. We say that an instance (G; X; Y ) is feasible if it has two arc-disjoint x x - and y y paths such that fx0 ; x00 g = X and fy0 ; y00 g = Y ; otherwise it is infeasible. Lemma 2.1 Let in

G

PX

0 E (PX ), then

x0 x00 -path with x0 ; x00 (G; X; Y ) is feasible. be an

f

3

g = X in (G; X; Y ). If y1 and y2 are connected

Proof:

Since

G

components in

G

is Eulerian,

0

X 0

E (P

)

G

0

E (P

X)

E (P

X 0

has an

00

contained in the same connected component in

G

0

y1

and

H2 ,

y2

are connected in

0

X

Since

X

y1

and

-path

P

X, 0

and each of the connected

) is Eulerian (possibly, a single vertex). If

feasible. Assume therefore that respectively.

0

x x

y2

X ) 0 E (PX ), 0

E (P

y1

and

y2

then the instance is

are contained in two distinct components

0

G

0

X

E (P

are

H1 and ) but not connected in G

X

0

0

) E (P ), H1 and H2 must contain vertices v1 and v2 in V (P ), respectively. Without loss generality, assume that P 0 visits v1 before v2 . Then, H1 has a y1 v1 -path, P 0 contains a v1 v2 -path, and H2 has a v2 y2 -path. This implies that the instance is feasible. E (P

X

X

2

Lemma 2.2 If an instance (G; X; Y ) satis es X \ Y 6= ;, then it is feasible. Proof: Assume without loss of generality that x1 = y1 for X = fx1 ; x2g and Y = fy1 ; y2g. Since G is connected, there are arc-disjoint x1 x2 -path PX and x2 x1 -path PX . Consider the connected component containing y2 in G 0 E (PX ) 0 E (PX ). It contains a vertex in V (PX ) or V (PX ) (say, V (PX )), since G is connected. Then, y1 and y2 are connected in G 0 E (PX ). 0

0

0

0

2

Lemma 2.1 then implies that (G; X; Y ) is feasible. In the following, therefore, we assume

X

\

Y

=

; for an instance (

G; X; Y

).

with X = fx1; x2 g and Y = fy1; y2g has an infeasible planar representation (IPR, for short) if the following conditions hold (see Fig. 3(a); (b)).

De nition 2.1 We say that an instance

(i) (ii) (iii)

G

(G; X; Y )

is planar, and has at most one cut vertex.

All the terminals have degree 2, and all other vertices have degree 4. has a planar representation in which every face is a directed cycle (or equivalently, the arcs incident to a vertex are alternately oriented out and in), and all the terminals lie on the boundary of the outer face (which is also a directed cycle) in the order of x1 ; y ; x2; y , where Y = fy ; y g. 2

G

0

0

00

00

We then have the next lemma.

Lemma 2.3 Any (G; X; Y ) which has an IPR is infeasible. Proof: If an IPR has arc-disjoint x x - and y y -paths, where fx ; x 0

00

0

00

0

00

g=

X

and

f

0

y ;y

00

g=

Y

,

then these two paths must cross at some nonterminal vertex in the planar representation (since every terminal has degree 2, and is located on the boundary of the outer face). However, the two paths cannot cross at a nonterminal vertex, because the arcs incident to a vertex are alternately

2

oriented out and in.

It is easy to see that a feasible instance (G; X; Y ) never becomes infeasible by contracting any arc. We say that an instance (G; X; Y ) is (G0 ; X; Y ) obtained by contracting

any

minimal infeasible, if it is infeasible but the instance

arc becomes feasible. The main contribution of this

paper is to show that the converse of Lemma 2.3 holds for such minimal infeasible instances.

j j = 6, however, there is a minimal infeasible instance with = f 1 2 1 2 g f g (see Fig. 3(c)), which is

In case of and

E

V

V

x ; x ; y ; y ; v; w

= (v; x1 ); (x1 ; w ); (v; x2 ); (x2 ; w ); (w; y1 ); (y1 ; v ); (w; y2 ); (y2 ; v )

clearly infeasible but has no IPR. We shall see that this is the only exception.

4

3 Irreducible Instances Let us consider the following three types of reductions: (1) Let Z be a 2-cut, and Z \ (X [ Y ) = ;. Let u be the tail of the arc from V 0 Z to Z , and v the head of the arc from Z to V 0 Z . Delete Z , and if u 6= v then add the arc (u; v ) to G[V 0 Z ]. See Fig. 4(1). (2) Let Z be a 2-cut, jZ j  2, and jZ \ (X [ Y )j = 1. Then contract Z to the terminal t 2 Z , deleting any resulting loops. (The resulting terminal t has degree 2.) See Fig. 4(2). (3) Let Z be a 4-cut such that G[Z ] is connected, contract Z into a single vertex. See Fig. 4(3).

jZ j  2, and Z \ (X [ Y ) = ;.

Then

The next is immediate from the de nition of reductions. (G; X; Y ) (1),(2) and (3).

Lemma 3.1 An instance of the reductions

is feasible if and only if it is feasible after performing any

2

We say that an instance (G; X; Y ) is reducible if one of the above reductions (1)(3) can be applied (in this case, we also call such 2-cut or 4-cut reducible); otherwise irreducible. An irreducible instance cannot have a 4-cut W with W \ (X [ Y ) = ; even if W does not induce a connected subdigraph, because in such a case there is a 2-cut Z  W with Z \ (X [ Y ) = ; (i.e., it is reducible). As will be shown in Section 9, an irreducible instance of a given instance (G; X; Y ) can be obtained in polynomial time. In this section, we present some properties of infeasible irreducible instances. Lemma 3.2 Any minimal infeasible instance is irreducible.

Obvious because any of the reductions (1),(2) and (3) can be performed by an adequate sequence of arc contractions, during which any infeasible instance never becomes feasible by Lemma 3.1. 2 Proof:

Lemma 3.3 Let

(i) (ii)

(G = (V; E ); X; Y )

There is no 2-cut

Z

be an infeasible irreducible instance. Then

that separates

X

and

Y.

Each terminal has degree 2.

(iii)

Each nonterminal vertex has degree 4.

(iv)

For any pair of vertices one of

(u; v)

and

(v; u)

u; v

2 V , there is at most one arc connecting them (i.e., at most )

exists in E .

(i) Assume that a 2-cut Z separates X and Y , where  (Z ) = fe1 ; e2g. Since G is connected and Eulerian, there are arc-disjoint x1x2-path PX and x2x1-path PX . Clearly, one of them (say, PX ) contains no arcs from fe1; e2 g. Similarly G has a y y -path PY , where fy ; y g = Y , such that E (PY ) \fe1; e2g = ;. These PX and PY are arc-disjoint in G, and hence (G; X; Y ) is feasible. (ii) Assume deg(x1)  4 for a terminal x1 2 X , without loss of generality. If G has arcdisjoint x1y1-path P1 and x1y2-path P2, then G 0 E (P1 ) 0 E (P2) has arc-disjoint y1 x1 -path P3 and y2 x1 -path P4, since G is Eulerian. Let H be the connected component in G 0 E (P1) 0 E (P2 ) 0 E (P3 ) 0 E (P4 ) that contains x2 . Since G is connected, H must contain a vertex z in V (Pi ) for some i. Assume z 2 V (P1 ) [ V (P2 ) (the case of z 2 V (P3 ) [ V (P4 ) can be treated Proof:

0

0

0

00

5

00

similarly). Then E (P ) [ E (P ) [ E (H ) contains a path P from x to x via z . However, y and y are connected in G 0 E (P ) 0 E (P ) 0 E (H ), since Q = hP ; P i is a y y -chain, and the instance would be feasible by Lemma 2.1, contradicting the assumption. Therefore, at least one of the above P and P does not exist; i.e., by Menger's theorem, there must be a 2-cut W such that x 2 W and Y  V 0 W . Since deg(x )  4, we have jW j  2. From (i) of this lemma, x 2 V 0 W holds. This, however, implies that there is a reducible 2-cut W , a contradiction. (iii) Assume deg(u)  6 for a nonterminal vertex u. Let W be a cut that minimizes j(W )j among cuts W such that u 2 W and fx ; y ; y g  V 0 W . By the minimality of j(W )j, G[W ] is connected. By deg (u)  6, j (W )j = 2 would imply jW j  2, and W is reducible. Hence j(W )j  4. From this, either (a) j(W )j  6, or (b) j(W )j = 4 and x 2 W must hold (otherwise, W would be reducible). In case of (a), by Menger's theorem, G has three arc-disjoint paths P ; P and P from u to some vertices w ; w ; w 2 fx ; y ; y g. By (ii) of this lemma, every terminal has degree 2, and then we can assume that these paths P ; P and P are ux -, uy - and uy -paths, respectively. Since G is Eulerian, G 0 E (P ) 0 E (P ) 0 E (P ) has three arc-disjoint x u-path P , y u-path P and y u-path P . Let H be the connected S E (P ) that contains x . Since G is connected, H must contain component in G 0 a vertex z in V (P ) for some i. Assume z 2 V (P ) [ V (P ) [ V (P ) [ V (P ) (the case of z 2 V (P ) [ V (P ) [ V (P ) [ V (P ) can be treated analogously). Then E (P ) [ E (P ) [ E (P ) [ E (P ) [ E (H ) contains a path P from x to x via z . However, y and y is connected in G 0 E (P ) 0 E (P ) 0 E (P ) 0 E (P ) 0 E (H ), since Q = hP ; P i is a y y -chain, and the instance would be feasible by Lemma 2.1, contradicting the assumption. Therefore, we assume (b), i.e., there is a 4-cut W separating fu; x g and fx ; y ; y g. In this case, applying the above argument to u and fx ; y ; y g, we can conclude that there is also a 4-cut W such that fu; x g  W and fx ; y ; y g  V 0 W . These two cuts W and W cross each other, and from (2.1) we have 1

2

2

1

1

1

X

2

3

Y

2

4

1

1

2

2

1

1

2

1

1

2

2

1

2

3

1

2

3

1

1

2

1

1

1

2

1

1

4

1

5

2

i

i=1;...;6

1

5

4

4

2

6

3

1

X

1

6

1

4

1

2

2

4

1

3

5

Y

2

0

1

2

1

1

2

2

6

1

2

2

0

3

3

2

i

3

2

2

1

1

2

2

0

2

0

j (W )j + j(W )j  j(W \ W )j + j (W [ W )j: Here j(W )j = j(W )j = 4 and j(W [ W )j  4 by (i). This implies j(W \ W )j  4, i.e., W \ W with (W \ W ) \ (X [ Y ) = ; is a reducible 4-cut (or contains a reducible 2-cut W  W \ W ), a contradiction. 0

0

0

0

0

0

0

0

00

0

(iv) If there are multiple arcs (u; v ) and (u; v ) in E , then u and v are nonterminal vertices by (iii), and deg(u) = deg(v) = 4 holds. This means that Z = fu; v g is a reducible 2- or 4-cut, a contradiction. Similarly if there are two arcs (u; v ) and (v; u), it is also easy to show that there is a reducible 2- or 4-cut Z = fu; v g, a contradiction. 2 (G; X; Y ) be an irreducible infeasible instance. Then G has at most two 2-cuts = fy ; y g. If there are two such 2-cuts Z and Z , cut vertex z such that Z = Z [ fz g or Z = (V 0 Z ) [ fz g. Conversely, any cut

Lemma 3.4 Let that separate then

G

has a

fx ; y g and fx ; y g, where Y 0

1

00

2

0

00

0

0

0

vertex is obtained in this manner.

\

Let

[

Z

and

f

g

f

g

Z 0 be the two 2-cuts that separate x1 ; y 0 and x2 ; y 00 , where we assume Z (X Y ) = Z (X Y ) = x1 ; y0 without loss of generality. Choose Z (resp., Z 0 ) as the cut minimizing Z (resp., maximizing Z 0 ) among such 2-cuts. We rst show that any other 2-cut Z 00 that separates x1; y0 and x2; y00 satis es Proof:

\ j j 0

f

f

g

f

g j j

g

 Z ): (3.1) If Z crosses Z , we have j (Z \ Z )j  4 from the choice of Z , and j(Z [ Z )j  2 as Z [ Z separates fx ; y g and fx ; y g. This means j (Z )j + j(Z )j = 4 < 6  j(Z \ Z )j + j (Z [ Z )j; Z

Z Z

[

00

0

(and hence Z

00

0

00

1

0

2

00

00

00

00

00

6

00

a contradiction to (2.1). Similarly, we see that Z 00 also cannot cross Z 0, and hence (3.1) holds. Since j(Z 0 0 Z )j  j (Z )j + j(Z 0 )j (= 4), Z 0 0 Z is a 2- or 4-cut satisfying (Z 0 0 Z ) \ (X [ Y ) = ;. Then, by irreducibility, Z 0 0 Z must be a 4-cut consisting of a single vertex (say, z ). This implies that z is a cut vertex. From jZ 0 0 Z j = 1 and (3.1), (G; X; Y ) has at most two 2-cuts that separate fx1; y0g and fx2 ; y00g. Conversely, let z be a cut vertex in G. Since G is Eulerian, z cannot have degree 2. By Lemma 3.3(ii)(iii) and (iv), z is a nonterminal vertex, and there are four distinct vertices, say u1 ; u2 ; v1 ; v2 , adjacent to z . Again since G is Eulerian, G 0fz g has exactly two components. Let W1 ; W2  V be the vertex sets of these components, and assume without loss of generality that (u1 ; z ); (u2; z ); (z; v1 ); (z; v2) 2 E , u1 ; v1 2 W1 and u2; v2 2 W2 since G is Eulerian and at least one arc going out (resp., going in) each of W1 and W2. Clearly W1 and W2 = V 0 (W1 [ fz g) are both 2-cuts. By irreducibility and Lemma 3.3(i), these satisfy jW1 \ X j = jW1 \ Y j = j(W1 [ fzg) \ X j = j(W1 [ fzg) \ Y j = 1. This implies that a cut vertex z is obtained in the manner of the lemma statement. 2 (G = (V; E ); X; Y ) be an irreducible infeasible instance which has an IPR, B be the cycle of the outer face in the IPR. If V 5 and there is no 2-cut Z such that Z X = Z Y = 1 and min Z ; V Z 3, then V (X Y ) induces a connected digraph in G E (B ). Lemma 3.5 Let

j j

and let

j \ j j \ j fj j j 0 jg  0 [ 0 Proof: By jV j  5, V 0 (X [ Y ) 6= ;. By Lemma 3.3(ii) and the de nition of IPR, each terminal in X [ Y is an isolated vertex in G 0 E (B ). Assume that V 0 (X [ Y ) induces two connected components H1 and H2 in G 0 E (B ). By the planarity of the IPR, any vertices u1 ; v1 2 V (H1 ) \ V (B ) and any vertices u2 ; v2 2 V (H2 ) \ V (B ) cannot appear alternately (in the order of u1 ; u2 ; v1; v2 ) along cycle B . This means that there are two arcs (u; v); (u0 ; v0 ) 2 E (B )

such that V (H1) and V (H2 ) are contained in two distinct connected components H10 and H20 in G 0 f(u; v ); (u0; v 0)g, respectively. Hence Z = V (H10 ) is a 2-cut, and by the irreducibility, both V (H10 ) and V (H20 ) must contain two terminals, one from X and the other from Y by Lemma 3.3(i). Clearly, each of V (H10 ) and V (H20 ) contains a nonterminal, and has at least three vertices; i.e., minfjZ j; jV 0 Z jg  3. 2 The next can be shown by inspecting all possible irreducible and infeasible instances with jV j  7, based on Lemma 3.3. (G; X; Y ) be an irreducible infeasible instance with jV j  7. If jV j 2 f4; 5; 7g, (G; X; Y ) has an IPR. If jV j = 6, (G; X; Y ) is the instance shown in Fig. 3(c) (in this case there is no irreducible infeasible instance with jV j = 6 in which some two terminals are adjacent each other). 2

Lemma 3.6 Let then

In this paper, we prove the next result. Theorem 3.1 Let

(G; X; Y )

(G; X; Y )

be a minimal infeasible instance, and satisfy

has an IPR.

jV j 6=

6.

Then

2

We shall need Sections 4-8 to prove Theorem 3.1 for general jV j  8.

4 Outline of the proof This section describes an outline of how to prove Theorem 3.1 in Sections 5-8. We rst assume that there is a smallest counterexample (G3; X; Y ) to Theorem 3.1, i.e., 7

(G3; X; Y ) is a minimal infeasible instance with n 6= 6 vertices, but has no IPR,

(4.1)

where G3 minimizes the number n3 of vertices among such instances. By Lemma 3.6, n3  8 is assumed. In Sections 5 and 6, we characterize cut vertices, 2-cuts and 6-cuts in G3 . Then in Sections 7 and 8, as outlined below, we derive a contradiction from the existence of such G3 , which proves Theorem 3.1. For the subsequent discussion, we introduce two operations. Let w be a nonterminal vertex with four incident arcs (s0 ; w); (s1; w); (w; s2); (w; s3 ), where s0; s1; s2 and s3 are all distinct. We say that arcs (s0; w) and (w; s2) are split o at w when four arcs (s0; w); (s1 ; w); (w; s2); (w; s3) are replaced with new two arcs (s0 ; s2) and (s1; s3 ), after eliminating w. Conversely, we say that two arcs e = (u; v ) and e0 = (u0; v 0) are hooked up (with a new vertex w) when we replace these two arcs with the new arcs (u; w); (w; v); (u0 ; w) and (w; v0 ) after introducing a new vertex w. Now we choose a nonterminal vertex w adjacent to a terminal (say x2) by arc (x2; w) in G3 and split o two arcs at w (recall that deg(w) = 4). If the resulting instance (G3 ; X; Y ) remains connected and irreducible, w

then we call such splitting (or two arcs) admissible. Based on the properties obtained in Sections 5 and 6, we show in Section 7 that G3 always has an admissible splitting. Clearly (G3 ; X; Y ) is infeasible w

since (G3; X; Y ) is infeasible. Also if (G3 ; X; Y ) is irreducible, then w

(G3 ; X; Y ) has an IPR, w

by the assumption on G3. However, we shall show in Section 8 that, for the arc e = (x2 ; v) and any other arc e0 in an irreducible infeasible instance (G; X; Y ) that has an IPR, the instance (G ; X; Y ) obtained by hooking up e and e0 satis es one of the following properties: 0

e;e

(i) (G

; X; Y ) is reducible,

(ii) (G

; X; Y ) has an IPR,

e;e0

e;e0

(iii) (G

0

e;e

; X; Y ) is feasible.

Notice that G3 is obtained from G3 by hooking up two arcs in an IPR of G3 . However, this leads to a contradiction, because G3 = (G3 ) satis es none of (i)-(iii). Hence no such counterexample (G3; X; Y ) exists. w

w

w

5

Cut vertex and 2-cuts in

Lemma 5.1

e;e0

G3

The minimum counterexample (G3; X; Y ) in (4.1) has the following properties:

(i) There is no cut vertex. (ii) There is no 2-cut Z such that

j \ j = j \ j = 1 and minfj j j 0 jg  3. Z

X

Z

Y

Z ; V

Z

(i) Assume that G3 = (V ; E ) has a cut vertex z . Since no vertex with degree 2 is a cut vertex in a connected Eulerian digraph, z is nonterminal and has degree 4 by Lemma 3.3(ii),(iii). Let Z 0 and Z 00 be the vertex sets of the two components in G3 0 fz g. By Lemma 3.4, jZ 0 \ X j = jZ 0 \ Y j = 1 holds, and each of Z 0 [ fz g and Z 00 [ fz g is a 2-cut in G3. Let (u0 ; z ); (u00 ; z); (z; v 0); (z; v 00) 2 E be the four arcs incident to z . Without loss of generality Proof:

8

we can assume that u0; v 0 2 Z 0 and u00 ; v00 2 Z 00, Z 0 \ (X [ Y ) = fx1; y2g, and G3 has an Eulerian cycle which visits all the terminals in the order of y2 ; x1 ; y1 ; x2 (if there is an Eulerian cycle in the order of y2 ; x1 ; x2; y1, then the instance is feasible). We decompose (G3; X; Y ) into (G0 ; X 0; Y 0) and (G00 ; X 00 ; Y 00 ) as follows. Let G3[Z 0] (resp., G3[Z 00]) denote the subdigraph of G3 induced by Z 0 (resp., Z 00 ), and G0 (resp., G00 ) be the Eulerian digraph obtained by adding new vertices y10 ; x02 and new arcs (u0 ; y10 ); (y10 ; x02 ); (x02 ; v0 ) (resp., new vertices y200; x001 and new arcs (u00; y200); (y200 ; x001 ); (x001 ; v 00)) to G3 [Z 0 ] (resp., G3[Z 00]). Regard X 0 = fx1; x02 g, Y 0 = fy10 ; y2g, X 00 = fx001 ; x2 g and Y 00 = fy1 ; y200 g as the sets of new terminals. We show that (G0 ; X 0 ; Y 0 ) is irreducible. If (G0 ; X 0; Y 0) has a reducible cut W , then W must separate y10 and x02 (otherwise W would be reducible in (G3; X; Y )). Then W is a 2-cut such that jW j  2 and W \(X 0 [Y 0 ) = fy10 g or fx02 g. Since deg(y10 ) = deg(x02) = 2, W 0 fy10 g (or W 0 fx02g) is a 2-cut, which is reducible in (G3; X; Y ), a contradiction. Note that (G3 ; X; Y ) is infeasible only when both new instances (G0 ; X 0; Y 0) and (G00; X 00; Y 00) are infeasible. Therefore (G0 ; X 0; Y 0) must be irreducible and infeasible. Clearly instance (G0; X 0 ; Y 0 ) is smaller than G3 (since Z 00 [ fz g is replaced with two vertices in the new instance), and hence has an IPR by de nition of G3 (note that jV (G0)j 6= 6 by Lemma 3.6 since (G0; X 0 ; Y 0 ) has two adjacent terminals). Analogously, we can show that (G00 ; X 00 ; Y 00 ) also has an IPR. However, it is easy to see that G3 has an IPR if both instances (G0 ; X 0; Y 0) and (G00; X 00 ; Y 00) have IPRs, a contradiction. (ii) Let Z be such a 2-cut in (G3 ; X; Y ), where (Z ) = f(u0 ; v00); (u00; v 0 )g and u0 ; v0 2 Z and 00 u ; v 00 2 V 0 Z . Clearly, Z = V 0 Z is also such a 2-cut. Note that u0 and v 0 (resp., u00 and v 00 ) are distinct (otherwise it would be a cut vertex, contradicting the above (i)). Without loss of generality, assume that Z \ (X [ Y ) = fx1 ; y2 g, and G3 has an Eulerian cycle that visits terminals in the order of y2; x1; y1; x2. We decompose instance (G3; X; Y ) into two instances (G0 ; X 0; Y 0) and (G00; X 00; Y 00) as follows. Let G0 (resp., G00 ) be the digraph obtained by adding new vertices y10 ; x02 and new arcs (u0 ; y10 ); (y10 ; x02 ); (x02 ; v0 ) (resp., new vertices y200; x001 and new arcs (u00 ; y200); (y200; x001 ); (x001 ; v00 )) to G3[Z ] (resp., G3[Z ]). Regard X 0 = fx1; x02g, Y 0 = fy10 ; y2g, X 00 = fx001 ; x2 g and Y 00 = fy1 ; y200 g as the sets of new terminals. Analogously to (i), we see that each of the new instances is an irreducible infeasible instance. From the assumption of minfjZ j; jV 0 Z jg  3, each of the new instances is smaller than G3, and has an IPR by de nition of G3 or by Lemma 3.6. However, it is again clear that G3 has an IPR if these new instances have IPRs, a contradiction. 2 6

6-cuts

We rst observe a property of a 6-cut Z . Lemma 6.1 Let

(G = (V; E ); X; Y )

be an infeasible instance.

If there exists a 6-cut

\ (X [ Y ) = ; satisfying the following (i)-(iv), then (G; X; Y ) is irreducible. (i) jZ j  3. (ii) Any cut W with W  Z is irreducible. (iii) Any cut W with W  Z or W \ Z = ; is irreducible.

Z

(iv) (Z )

Z

with

contains no multiple arcs.

Proof: Let Z be such a 6-cut. From (ii) and (iii), it suces to show that any cut W which intersects Z is irreducible. Assume that a cut W intersecting Z is reducible (this implies j(W )j = 2 or 4). Since a reducible cut W contains at most one terminal, V 0 (W [ Z ) contains a terminal, and hence cuts W and Z cross each other. By (2.1),

9

j (W )j + j(Z )j  j(W \ Z )j + j (W [ Z )j;

(6.1)

and by (2.2),

j (W )j + j(Z )j = j(W 0 Z )j + j(Z 0 W )j + 2j(W \ Z; V 0 (W [ Z ))j: (6.2) Since Z contains no reducible cut by (ii), we have j (W \ Z )j  4. Also by (iii), we see that j (W [ Z )j  j (W )j + 2 holds (otherwise if j (W [ Z )j  j (W )j then W [ Z would be reducible by (W [ Z ) \ (X [ Y ) = W \ (X [ Y )). Therefore, j (Z )j = 6, j (W \ Z )j  4 and j(W [ Z )j  j(W )j + 2 imply that (6.1) holds by equality. Hence j (W \ Z )j = 4 holds, and this means jW \ Z j = 1 since Z contains no reducible cut. Then jZ 0 W j  2 by (i), and again by (ii), j(Z 0 W )j  6. By (iii), we have j(W 0 Z )j  j(W )j (otherwise, if j(W 0 Z )j  j(W )j 0 2 then W 0 Z would be reducible). By j (Z 0 W )j  6, j (Z )j = 6 and j (W 0 Z )j  j (W )j, (6.2) implies that j (Z 0 W )j = j (Z )j = 6, j (W 0 Z )j = j (W )j and j (W \ Z; V 0 (W [ Z ))j = 0. By (iii), jW 0 Z j = 1 must hold, since jW 0 Z j  2 and j (W 0 Z )j = j (W )j mean that W 0 Z is reducible. Now the vertex v 2 W \ Z has degree 4, and has no adjacent vertex in V 0 (W [ Z ) by j (W \ Z; V 0 (W [ Z ))j = 0. From j (Z 0 W )j = j (Z )j and j (W 0 Z )j = j (W )j, we then have j (fv g; W 0 Z )j = j (fv g; Z 0 W )j = 2. However, by jW 0 Z j = 1, the two arcs in  (fv g; Z 0 W ) are multiple, contradicting (iv). 2 6.1

Interchangeability

Let Z be a 6-cut in (G; X; Y ) such that

\ (X [ Y ) = ;; 0 0  0 (Z ) = fe0 1 ; e2 ; e3 g; Z

0

where zi

+ +  + (Z ) = e+ 1 ; e2 ; e3 ;

f

+

g

(6.3)

0

(resp., zi ) denote the head vertices of arcs ei

+

(resp., the tail vertices of arcs ei )

0 and z + may not be distinct. We say 0 0 0 + + + that Z is (e1 ; e2 ; e3 ; e 1 ; e 2 ; e 3 )-interchangeable, where fj1 ; j2 ; j3 g = f1; 2; 3g, if the subdigraph G[Z ] of G induced by Z has three arc-disjoint z 0 z +i -paths P i (i = 1; 2; 3). Z is called fully 0 0 0 + + + interchangeable, if it is (e1 ; e2 ; e3 ; e ; e ; e )-interchangeable for any choice of j1 ; j2 ; j3 with 1 2 3 fj1 ; j2 ; j3g = f1; 2; 3g. for i = 1; 2; 3 (see Fig. 5). Note that these vertices zi j

j

i

j

j

Z

\ (X [ Y ) = ;.

j

i;j

j

Lemma 6.2 An irreducible infeasible instance with

i

j

(G; X; Y )

has no fully interchangeable 6-cut

Z

Assume that there is such a 6-cut Z , and let GZ be the digraph obtained by contracting Z into a nonterminal vertex z . It is easy to see that

Proof:

(i) (G; X; Y ) is feasible if and only if (GZ ; X; Y ) is feasible, and (ii) (GZ ; X; Y ) is irreducible. Therefore, (GZ ; X; Y ) is also an irreducible infeasible instance, but deg (z ) = 6 contradicts

2

Lemma 3.3(iii). A directed cycle of length 3 is called a Lemma 6.3 Let

(i)

If

Z

(G; X; Y )

be an irreducible infeasible instance. Then:

is a 6-cut such that

connected, then

G[Z ]

triangle.

Z

\ (X [ Y ) = ;, jZ j = 3 and the induced subdigraph G[Z ] is

is a triangle.

10

(ii) If jZ j = 3 and the three vertices in Z are mutually adjacent, then the induced subdigraph G[Z ] is a triangle. (i) From Lemma 3.3(iv) and j(Z )j = 6, it is easy to see that the connected subdigraph G[Z ] contains exactly three arcs and such three arcs form an undirected cycle C of length 3 if the orientation is neglected. This C must be a directed cycle in G, because otherwise it is not dicult to see, by checking all possibilities, that Z is fully interchangeable, contradicting Lemma 6.2. (ii) If all vertices in Z are nonterminal, (ii) follows from (i). Therefore assume that Z contains a terminal. Z = fv1; v2; v3g can contain at most two terminals since any terminal has degree 2 from Lemma 3.3(ii). If Z contains two terminals, then clearly G[Z ] forms a triangle. Then assume that Z contains exactly one terminal, say Z \ (X [ Y ) = fv2g. If G[Z ] is not a triangle, we can assume without loss of generality that G[Z ] has arcs (v1; v2); (v2; v3); (v1; v3). Let G be the digraph obtained from G by contracting Z into terminal v2. It is easy to see that (G ; X; Y ) is also an irreducible infeasible instance. But v2 has degree 4, and contradicts Lemma 3.3(ii). Proof:

v2

v2

2

Let (Z ; G) denote (Z ) in a digraph G. Let (G; X; Y ) be an irreducible infeasible instance, and let Z be a 6-cut in (G; X; Y ) 0 0 + + + as de ned in (6.3). If Z is not (e0 1 ; e2 ; e3 ; e1 ; e2 ; e3 )-interchangeable, then properties (i)-(iv)

Lemma 6.4

hold.

(i) The induced subdigraph G[Z ] is connected. (ii) If G[Z ] has no z0z+-path for some i 2 f1; 2; 3g, then Z = fz0; z+g. (iii) z0 6= z+ for all i 2 f1; 2; 3g. (iv) If jZ j  3, then z0 6= z0 and z+ 6= z+ for 1  i < i0  3. Note that j0(fug; G[Z ])j = j+ (fug; G[Z ])j for all u 2 Z 0 fz 0; z + j i = 1; 2; 3g. Hence 0 + if G[Z ] has a z z -path P , then G[Z ] 0 E(P ) has arc-disjoint z0z+- and z0 z+ -paths for some i0 ; i00 ; j 0 ; j 00 with fi0 ; i00 g = f1; 2; 3g 0 fig and fj 0 ; j 00 g = f1; 2; 3g 0 fj g. (i) If the subdigraph G[Z ] of G consists of more than one connected component, then there would be a reducible 2-cut Z 0 with Z 0  Z . (ii) Assume without loss of generality that0G[Z ] has+no z10z1+-path. Then, by Menger's theorem, G[Z ] has a cut W  Z such that z1 2 W , z1 2 Z 0 W and j+(W ; G[Z ])j = 0. Here j0(W ; G[Z ])j  1 since G[Z ] is connected by (i). Let H denote the Eulerian digraph + 0 3 obtained by adding three new arcs e = (z ; z ) (i = 1; 2; 3) to G[Z ]. Now e31 2 0(W ; H ) and j0(W ; H )j  2 hold, and hence j+(W ; H )j  2 since H is Eulerian. Since j+(W ; G[Z ])j = 0, we see that e32; e33 2 +(W ; H ). Therefore, by j+(W ; H )j+= 02 =0 j0(W ; H )j, we have 0 + + 0 j (W ; G[Z ])j = 1. This implies that z1 ; z2 ; z3 2 W and z1 ; z2 ; z3 2 Z 0 W , and that j(W ; G)j = j(Z 0 W ; G)j = 4 holds. Hence the 4-cut W (resp., Z 0 W ) in G consists of a single vertex z10 (resp., z1+ ), respectively, from the irreducibility of G. 0 (iii) If z1 = z1+0, then G[Z ] has a z10 z1+ -path of null length. Since G is Eulerian, G[Z ] has + two arc-disjoint z2 z2 - and z30z3+-paths, because even if G[Z ] has arc-disjoint z20z3+- and z30z2+paths, the connectivity0of+G[Z ] (which follows from (i)) implies that these paths have a common vertex v , from which z2 z2 - and z30 z3+ -paths can be constructed. This contradicts that Z is not (e01 ; e02 ; e03 ; e+1; e+2; e+3)-interchangeable. (iv) From (ii), jZ j  3 means that0 G[Z0] has paths from z0 to z+ for all i = 1; 2; 3 (but 0they may not be arc-disjoint). Assume z1 = z2 since other cases are analogous, and choose a z3 z3+ path P3 in G[Z ]. Note that G[Z ] 0 E(P3) together with additional arcs (z1+; z10) and (z2+; z20) i

i

i

i

i

i

i

i0

i

i0

Proof:

i

i

i0

j

i

i

j0

i

i

11

i

i

i00

j 00

becomes Eulerian. This means that G[Z ] 0 E (P3 ) has arc-disjoint z10z1+- and z20z2+ -paths, where 0 0 + + + z10 = z20 . This contradicts that Z is not (e0 2 1 ; e2 ; e3 ; e1 ; e2 ; e3 )-interchangeable. 6.2

G3

Proper 6-cuts in

We call a 6-cut Z proper if (b)

jZ j  3, Z \ (X [ Y ) = ;,

(c)

Z

(a)

contains a vertex z such that (u; z) 2 0 (Z ) and (z; v) 2 + (Z ).

In this subsection, we prove that any proper 6-cut Z induces a triangle in the minimum counterexample (G3; X; Y ). 0 + 0 Let Z be a proper 6-cut in the minimum counterexample (G3; X; Y ), for which e+ i ; ei ; zi ; zi (i = 1; 2; 3) are de ned by (6.3). As Z is not fully interchangeable by Lemma 6.2, assume that 0 0 + + + Z is not (e0 1 ; e2 ; e3 ; e10; e2 ; +e3 )-interchangeable without loss of generality. From condition (c) and Lemma 6.4(iii), zi = zj holds for some i 6= j . Here we assume without loss of generality that z10 = z3+ (if necessary, exchange the indices i = 2; 3). By Lemma 6.4(iii) and (iv), we have the following four possible cases. Case-1. Case-2. Case-3. Case-4.

z20 ; z30 ; z2+ ; z3+ are all distinct (see Fig. 6(1)). z2+ = z30 and z20 = z1+ (or symmetrically, z20 = z1+ z1+ = z30 and z20 = z2+ (see Fig. 6(3)). z2+ = z30 and z20 = z1+ (see Fig. 6(4)).

6

6

and z2+ 6= z30) (see Fig. 6(2)).

Now let HZ3 be the Eulerian digraph obtained from G3[Z ] as follows (see Fig. 7):

1. For i = 1; 2, if zi+ and zi0+1 are distinct, add a new vertex ti+1 together with new arcs (zi+; ti+1) and (ti+1; zi0+1); if zi+ = zi0+1, then let ti+1 = zi0+1(= zi+).

2. Let t1 = z10(= z3+).

3. Replace the arc (t1 ; v) with two arcs (t1 ; t01 ) and (t01 ; v), inserting a new vertex t01 between t1 and v .

~ Y~ ) is feasible De ne sets of terminals X~ = ft1; t3 g and Y~ = ft01; t2g. Obviously, instance (HZ3 ; X; 0 0 0 + + + 3 ~ Y~ ) must be infeasible. if and only if Z is (e1 ; e2 ; e3 ; e1 ; e2 ; e3 )-interchangeable. Thus, (HZ ; X; 3 3 Note that HZ contains at most jZ j + 3 (< jZ j + jX [ Y j  n ) vertices, where n3 is the number of vertices in G3. The next lemma summarizes the properties of HZ3 . Let (G3 ; X; Y ) be the minimum counterexample, and let Z be a proper 6-cut in 0 0 + + + 3 ~ Y~ ) be the instance (G3 ; X; Y ), which is not (e0 1 ; e2 ; e3 ; e1 ; e2 ; e3 )-interchangeable, and (HZ ; X; de ned in the above. Then the following properties (i)-(iv) hold in all the above Cases-1,2,3,4. Lemma 6.5

~ Y~ ) is infeasible. (i) (HZ3 ; X;

~ Y~ ) is connected and irreducible, and has no 2-cut W such that jW \X~ j = jW \Y~ j = (ii) (HZ3 ; X; 1 and minfjW j; jV (HZ3 ) 0 W jg  3.

~ Y~ ) has an IPR. (iii) (HZ3 ; X; (iv)

0 0 + + + Z is (e0 1 ; e2 ; e3 ; ej1 ; ej2 ; ej3 )-interchangeable for any (j1; j2 ; j3 ) = (1; 2; 3). 12

choice of j1; j2; j3 from f1; 2; 3g except

In what follows, we consider all Cases-1,2,3 and 4 simultaneously. (i) Already proved. ~ Y~ ) has (ii) Since G3 [Z ] is connected by Lemma 6.4(i), HZ3 is connected. Assume that (HZ3 ; X; 3 ~ ~ a reducible cut W . If W \ (X [ Y ) = ;, then W would be a reducible cut also in (G ; X; Y ), a ~ Y~ ) such that jW j  2 and jW \ (X~ [ contradiction. Therefore, W must be a 2-cut in (HZ3 ; X; 3 )j = 2 as easily checked, we further ~ Y )j = 1. Since no such W with jW j = 2 attains j (W ; HZ 3 ~ ~ assume jW j  3. Let ft g = W \ (X [ Y ). Clearly, j(W ; HZ3 )j = 2 implies that j(W 0; G3)j  4 holds for W 0 = W 0 ft3g  V . This and jW 0j = jW j 0 1  2 mean that W 0 is a reducible ~ Y~ ) is irreducible. Assume that 2- or 4-cut in (G3; X; Y ), a contradiction. Therefore, (HZ3 ; X; 3 ~ Y~ ) has a 2-cut W with jW \ X~ j = jW \ Y~ j = 1 and minfjW j; jV (HZ3 ) 0 W jg  3, and (HZ ; X; let t01 2 W without loss of generality. Then jW j  3 implies that (W 0 (X~ [ Y~ )) [ fz3+ g is a reducible 4-cut in G3, contradicting irreducibility of (G3; X; Y ). ~ Y~ ) is infeasible by (i), and is connected and irreducible by (ii). (iii) The instance (HZ3 ; X; 3 ~ Y~ ) has an IPR by the Since HZ contains at most jZ j + 3 < n3 vertices, the instance (HZ3 ; X; 3 minimality assumption on G and by Lemma 3.6 (note that terminals t1 2 X~ and t01 2 Y~ are adjacent). ~ Y~ ), where B visits (iv) Let B be the directed cycle of the outer face in an IPR of (HZ3 ; X; 0 t1 ; t1 ; t3 ; t2 in this order, and let B (u; v ) denote the uv -path on B , where B (u; u) means a path of null length. Note that (j1 ; j2; j3) 6= (1; 2; 3) implies (a) j1 = 3, (b) j1 = 2, or (c) j1 = 1 and j2 = 3. If jV (HZ3 )j = 4, then only Case-4 can occur and the IPR is a cycle of length 4 visiting t1 ; t01 ; t3 ; t2 in this order. In this case, we can easily check by inspection that (iv) holds. We then ~ Y~ ) has no 2-cut W stated in the above (ii) and jV (HZ3 )j  5 assume jV (HZ3 )j  5. Since (HZ3 ; X; 3 holds, V (HZ ) 0 (X~ [ Y~ ) induces a connected component in HZ3 0 E (B ) by Lemma 3.5. (a) j1 = 3. We rst take a z10 z3+-path PA of null length in HZ3 . We then consider path B (z2+ ; z20 ) which contains a z30 z1+ -path PB = B (z30 z1+ ), and remove the arcs of E (B (z2+ ; z20 )) from HZ3 . Now indeg(u) = outdeg(u) holds for all u 2 V (HZ3 ) 0 fz2+ ; z20 g. Then, the entire set 3 ) 0 E (B (z +; z 0 )) of remaining arcs can be regarded as a z 0z +-path PC . Therefore, Z is E (HZ 2 2 + + 0 0 + +2 +2 (e1 ; e0 To show the (e0 ; e0 ; e0 ; e+ 2 ; e3 ; e3 ; e2 ; e1 )-interchangeable. 1 2 3 3 ; +e1 ; e2 )-interchangeability, 0 + 0 + 0 it suces to prove that the above z3 z1 -path PB = B (z3 z1 ) and z2 z2 -path PC have a common vertex, by which we can reconstruct arc-disjoint z30 z2+-path and z20z1+ -path. Now since V (HZ3 ) 0 (X~ [ Y~ ) induces a connected component in HZ3 0 E (B ), we obtain V (PB ) \ V (PC ) 6= ;. (b) j1 = 2. It is easy to see that z10z2+-path PA = B (z10 ; z2+ ) and path B (z2+ ; z20 ) (which contains a z30z1+ -path PB = B (z30z1+)) and z20z3+-path PC = E (HZ3 ) 0 E (B (z10; z20)) are arc-disjoint. Therefore, Z is (e0 ; e0 ; e0 ; e+ ; e+ ; e+ 1 2 3 2 3 1 )-interchangeable. By the connectedness of ~ [ Y~ ) in HZ3 0 E (B ), V (PB ) \ V (PC ) 6= ; holds and arc-disjoint z30z3+ -path and z20z1+V (HZ3 ) 0 (X path can be reconstructed from PB and PC , implying (e0 ; e0 ; e0 ; e+ ; e+ ; e+ 1 2 3 2 1 3 )-interchangeability. 0 + 0 + (c) j1 = 1 and j2 = 3. Clearly, a z2 z3 -path PA = B (z2 ; z3 ) and path B (z2+; z20) (which contains a z30 z1+ -path PB = B (z30 z1+ )), and z10 z2+-path PC = E (HZ3 ) 0 E (B (z2+ ; z10 )) are arcdisjoint. By the connectedness of V (HZ3 ) 0 (X~ [ Y~ ) in HZ3 0 E (B ), V (PB ) \ V (PC ) 6= ; holds and arc-disjoint z30z2+-path and z10z1+-path can be reconstructed from PB and PC , implying the 0 0 + + + (e0 2 1 ; e2 ; e3 ; e1 ; e3 ; e2 )-interchangeability. Proof:

Fig. 8 shows the smallest IPR of HZ3 in Cases-1,2,3,4. Let HZ3 [Z [ ft0 g] be the subdigraph induced by Z [ ft0 g from the smallest IPR of HZ3 in Fig. 8, and let HZ# be the digraph obtained from HZ3 [Z [ ft0g] by deleting the vertex t01 (merging the two arcs (z3+; t01); (t01 ; v ) into an arc 3 (z3+ ; v)). Let us consider the digraph G# HZ obtained from the minimum counterexample G by replacing G3[Z ] by HZ# , as shown Fig. 9. Let Z0 = V (HZ# ). Let Z00 be the set of vertices u 2 Z0 with j(fug; HZ# )j = 2, and z0 2 Z0 be the vertex with j (fug; HZ# )j = 4 in Cases-1 and 2. Note that each vertex in Z00 is either zi+ or zi0 for some i. 13

For a proper 6-cut Z in the minimum counterexample (G3; X; Y ), which is not # 0 0 0 + + + (e1 ; e2 ; e3 ; e1 ; e2 ; e3 )-interchangeable, let GZ be de ned as above. Then the following properties (i)-(iv) hold in all Cases-1,2,3,4 (de ned in the beginning of this subsection). Lemma 6.6

#

(i) (GZ ; X; Y )

#

(ii) (GZ ; X; Y )

(iii) (G3 ; X; Y )

is infeasible. is connected and irreducible.

has an IPR if (G# Z ; X; Y ) has an IPR.

# holds.

(iv) G3 [Z ] = HZ Proof:

0 0 0 + + +

(i) By Lemma 6.5(iv), the 6-cut Z in G3 is (e ; e ; e ; ej1 ; ej2 ; ej3 )-interchangeable for

1 2 3

any choice of j ; j ; j

from

1 2 3

f1; 2; 3g except (j1 ; j2; j3) = (1; 2; 3). # V (H )

Then it is easy to see that

# 0= Z also has the same interchangeability in GZ , implying # that (GZ ; X; Y ) is feasible if and only if (G3 ; X; Y ) is feasible. Since (G3 ; X; Y ) is infeasible, # (GZ ; X; Y ) is also infeasible. # # (ii) Clearly, (GZ ; X; Y ) is connected, since so is (G3 ; X; Y ). We apply Lemma 6.1 to (GZ ; X; Y ) # and 6-cut Z0 = V (H ). Clearly, Z0 \ (X [ Y ) = ; and jZ0 j  3, and  (Z0 ) contains no multiple the corresponding 6-cut Z

Z

arcs, satisfying conditions (i) and (iv) of Lemma 6.1. We see by inspection, that there is no

 Z0, and from the irreducibility of (G3; X; Y ), that there is no reducible cut  Z0 or W \ Z0 = ;, satisfying conditions (ii) and (iii) of Lemma 6.1. Therefore,

reducible cut W

W with W

#

(GZ ; X; Y ) is irreducible by Lemma 6.1.

# + 0 + 0 + 0 an IPR, in which we assume without loss of generality that arcs e3 ; e1 ; e2 ; e3 ; e1 ; e2 appear + 0 + 0 + 0 in this order along the cycle fz3 = z1 ; z2 ; z3 ; z1 ; z2 g (recall that, in an IPR, the arcs incident 3

(iii) Consider Case-1 (other cases can be treated analogously). Assume that (GZ ; X; Y ) has

~ Y ~ ) has an IPR, to each vertex are alternately oriented out and in). By Lemma 6.5(iii), (HZ ; X; in which we can assume without loss of generality that all terminals t ; t0 ; t ; t

1 1 3 2 appear in this + 0 + 0 + 0 order along the cycle of the outer face, and hence the arcs e3 ; e1 ; e2 ; e3 ; e1 ; e2 appear in the # # # same way as in the IPR of (G ; X; Y ). This implies that H in G can be replaced with H 3 Z

so that the resulting digraph G3 also has an IPR.

Z

Z

Z

#

(iv) From (i)-(iii) and the assumption on G3 , (GZ ; X; Y ) is an irreducible infeasible instance,

j  jZ0 j + jX [ Y j  7, (G#Z ; X; Y ) is also a counterexample to Theorem 3.1. Then by the minimality of G3 , jZ0 j = jZ j. By inspection, we see that Z with jZ j = jZ0 j can induce no other subdigraph than G3[Z ] = HZ# of Fig. 9 in all Cases-1,2,3,4. 2 j

#

but has no IPR. Clearly, by V (G )

In what follows, we strengthen Lemma 6.6(iv) and show that any proper 6-cut Z in G3 induces

a triangle, i.e., none of Cases-1,2 and 3 occurs. A proper 6-cut Z in (G3 ; X; Y ) is called if there is no proper 6-cut Z 0 with Z

 Z 0.

maximal

Let Z be a maximal proper 6-cut in the minimum counterexample (G3; X; Y ), de ned by (6.3). If Z satis es one of Case-1, Case-2 and Case-3 (i.e., G3[Z ] = HZ3 of Fig 9(1), (2) and (3), respectively), then the following properties (i)-(v) hold. 0 + 0 (i) In Case-1, there is no pair of terminals t; t0 2 X [ Y such that (t; z2 ); (z1 ; t); (t0 ; z3 ); + 0 + (z2 ; t0 ) 2  (Z ). In Case-2, there is no pair of terminals t; t0 2 X [Y such that (t; z2 ); (z3 ; t), 0 + 0 0 0 (t ; z3 ); (z1 ; t ) 2  (Z ). In Case-3, there is no pair of terminals t; t 2 X [ Y such that 0 + 0 + (t; z2 ); (z3 ; t), (t0 ; z3 ); (z2 ; t0 ) 2  (Z ). 0 + (ii) In Case-1, assume that there is no terminal t with (t; z2 ); (z1 ; t) 2  (Z ) (without loss of 0 generality by (i)). Then the instance (G ; X; Y ) obtained from (G3; X; Y ) by splitting o arcs e02 and (z20; z1+) at z20 is infeasible and irreducible. Lemma 6.7

14

(iii) In Case-2, assume that there is no terminal t with (t; z20 ); (z3+; t) 2 (Z ) (without loss of generality by (i)). Then the instance (G0; X; Y ) obtained from (G3; X; Y ) by splitting o arcs (z20; z3+) and e+3 at z3+ (= z10) is infeasible and irreducible.

(iv) In Case-3, assume that there is no terminal t with (t; z20 ); (z3+; t) 2  (Z ) or (t; z10 ); (z2+; t) 2  (Z ) (without loss of generality by (i)). Then the instance (G0 ; X; Y ) obtained from + 0 (G3; X; Y ) by splitting o arcs (z20 ; z3+) and e+ 3 at z3 (= z1 ) is infeasible and irreducible. (v) Let (G0; X; Y ) be the instance of (ii) of Case-1 (resp., (iii) of Case-2, and (iv) of Case-3). Then (G3; X; Y ) has an IPR if (G0 ; X; Y ) has an IPR.

(i) Assume that there are such terminals t and t0 in Cases-1,2,3. By Lemma 3.3(ii), terminals t and t0 have degree 2, and Z [ ft; t0 g is a 2-cut. In Case-1, G3 would have a cut vertex z10 = z3+ (see Fig. 9(1)), which contradicts Lemma 5.1(i). Then consider Cases-2 and 3. By Lemma 3.3(i), jft; t0 g \ X j = jft; t0 g \ Y j = 1 holds, and assume t = x1 and t0 = y1 without loss of generality. Furthermore, by Lemma 5.1(ii), we obtain V 0 (Z [ ft; t0 g) = fx2 ; y2 g. Now G3 has jZ j + 4 = 9 vertices in Case-2 and jZ j + 4 = 8 vertices in Case-3. By inspection, we see that (G3; X; Y ) in Case-2 is feasible or has an IPR and (G3; X; Y ) in Case-3 is feasible, a contradiction. (ii) Obviously (G0 ; X; Y ) is infeasible. We show that (G0; X; Y ) has no multiple arcs. If there are such multiple arcs, then they must be (u; z1+); (z1+ ; u) for some vertex u 2 V 0 Z , since G3 has no multiple arcs by Lemma 3.3(iv). This means that u is adjacent to both z20 and z1+ in G3 . By the assumption that there is no terminal t with (t; z20 ); (z1+; t) 2  (Z ), u is not a terminal. Then Z [ fug is a proper 6-cut, contradicting the maximality of jZ j. Now we apply Lemma 6.1 to (G0 ; X; Y ) and Z 0 = Z 0 fz20 g. Clearly, 6-cut Z 0 satis es Z 0 \ (X [ Y ) = ; and conditions (i) and (iv) of Lemma 6.1. From the irreducibility of G3 , condition (iii) of Lemma 6.1 holds for Z 0 . By inspection, we see that Z 0 satis es condition of (ii) of Lemma 6.1. Therefore, (G0 ; X; Y ) irreducible by Lemma 6.1. (iii) Analogous to (ii). (iv) The assumption that there is no terminal t with (t; z20); (z3+ ; t) 2 (Z ) or (t; z10 ); (z2+; t) 2  (Z ) in Case-3 does not lose generality, because if a terminal t is adjacent to both z20 and z3+ then no terminal is adjacent to both z30 and z2+ by (i) and two vertices z20 ; z30 cannot be adjacent to another terminal. This assumption ensures that (G0; X; Y ) has no multiple arcs. The rest of the proof is analogous to (ii). (v) It should be noted that, if an instance has an IPR, then any triangle (if any) in the instance gives rise to a face in its IPR. Let Z 0 = Z 0 fz20g in Case-1, and Z 0 = Z 0 fz10g in Cases-2 and 3. It is easy to see that (G0; X; Y ) still contains a triangle in G0[Z 0] in each of Cases-1, 2 and 3, and the vertices on these triangles are uniquely embedded in an IPR. Based on this, we can observe that if (G0; X; Y ) has an IPR then (G3; X; Y ) has an IPR. 2 Proof:

From this lemma, we can conclude that none of Cases-1, 2 and 3 can happen in (G3; X; Y ) as follows. If situations (ii), (iii) or (iv) occurs, then the instance (G0; X; Y ) is irreducible and infeasible, as shown in the lemma. Since the instance (G0; X; Y ) is smaller than (G3; X; Y ) and jV (G0)j = jV (G3)j 0 1  7, it has an IPR by the assumption on G3 and Lemma 3.6. Then, (G3 ; X; Y ) also has an IPR by Lemma 6.7(v). This is a contradiction. Therefore, only Case-4 is possible for a maximal proper 6-cut Z (note that Lemma 6.7(iv) no longer holds for Case-4, + since G3[Z ] has no triangle after splitting o arcs, say (z20; z3+ ) and e+ 3 at z3 ). This implies that any maximal proper 6-cut (and hence any proper 6-cut, which is not necessarily maximal) always induces a triangle. Lemma 6.8

Any proper 6-cut in (G3; X; Y ) induces a triangle. 15

2

7 Admissible Splitting In this section, we derive a condition for splitting two arcs at a nonterminal vertex to be admissible (de ned in Section 4), and then show that (G3; X; Y ) always has an admissible splitting.

Let (G3; X; Y ) be the minimum counterexample and let w be a nonterminal vertex in where (s0; w); (s1; w); (w; s2 ); (w; s3) are the four arcs incident with w. If s0 and s2 are not adjacent, and s1 and s3 are not adjacent, then the instance (G3w ; X; Y ) obtained by splitting o (s0; w) and (w; s2) at w is connected and irreducible (i.e., this splitting is admissible). Lemma 7.1

G3 ,

By Lemma 5.1(i), w is not a cut vertex in G3 , and hence (G3w ; X; Y ) is connected. Assume that (G3w ; X; Y ) has a reducible cut W  V 0 fwg. Since s0 ; s1; s2 ; s3 are distinct by Lemma 3.3(iv), let S = fs0; s1; s2 ; s3g. We see that W \ S = fs0 ; s2g or W \ S = fs1 ; s3g holds (otherwise W would be reducible in (G3; X; Y )). Without loss of generality, assume W \ S = fs0 ; s2 g (see Fig. 10). We consider three cases (a) j (W ; G3w )j = 2 and W \ (X [ Y ) = ;, (b) j(W ; G3w )j = 4, W \ (X [ Y ) = ; and jW j  2, and (c) j (W ; G3w )j = 2, jW \ (X [ Y )j = 1 and jW j  2. (a) In this case, W 0 = W [ fwg satis es j(W 0 ; G3)j = j(W ; G3w )j + 2 = 4 and jW 0 j  2, implying that W 0 was reducible in (G3 ; X; Y ), a contradiction. (b) W 0 = W [ fwg satis es j(W 0 ; G3 )j = j(W ; G3w )j + 2 = 6, and jW 0j  3. Since arcs (s1; w); (w; s3) are adjacent to w, W 0 is a proper 6-cut in G3, and by Lemma 6.8, it induces a triangle. However, this contradicts that s0 and s2 are not adjacent. (c) Let ftg = W \ (X [ Y ). We see that t 6= s0; s2 holds, because if t = s0 or t = s2 then W 0 ftg is a 4-cut in G3 and jW 0 ftgj = 1 must hold by irreducibility of G3 , contradicting that s0 and s2 are not adjacent. Therefore jW j  3. This means that W 0 = (W 0 ftg) [ fwg is a proper 6-cut in G3, which then induces a triangle by Lemma 6.8. However, this again contradicts that s0 and s2 are not adjacent. 2 Proof:

Lemma 7.2 Let w be a nonterminal vertex adjacent to a terminal t by arc (t; w ) in the minimum counterexample (G3 ; X; Y ), and (s1; w); (w; s2 ); (w; s3) be other three arcs incident with w. Then the following property (i) or (ii) holds.

and s3 are not adjacent, and s1 and s2 are not adjacent (i.e., splitting (t; w ) and (w; s3) at w is admissible by Lemma 7.1).

(i)

t

(ii)

t

and s2 are not adjacent, and s1 and s3 are not adjacent (i.e., splitting (t; w ) and (w; s2) at w is admissible by Lemma 7.1).

(a) Consider the case in which t is adjacent to s2 (i.e., G3 has arc (s2 ; t)). Clearly, t cannot be adjacent to s3 . Assume that s1 and s2 are adjacent (i.e., G3 has arc (s2 ; s1 ) by Lemma 6.3(ii)). Let u; v be other two vertices adjacent to s1 , where (u; s1); (s1; v ) 2 E (see Fig. 11). We will show that w (resp., s2) is not adjacent to u (resp., v ). Assume rst that w and u are adjacent (i.e., (w; u) 2 E by Lemma 6.3(ii), and hence u = s3 ). If u is a terminal, W = ft; u = s3; w; s1; s2g is a 2-cut with jW \ (X [ Y )j = 2. By Lemma 3.3(i), jW \ X j = jW \ Y j = 1, and by Lemma 5.1(ii), jV 0 W j = 2, implying jV 0 W j + jW j = 7 < n3 , a contradiction. (Recall that n3  8 holds by Lemma 3.6, as noted after (4.1).) Then u must be nonterminal. However, in this case Z = fu = s3; w; s1 ; s2g would be a proper 6-cut with four vertices, contradicting Lemma 6.8. Therefore, u is not adjacent to w. Similarly, we see that v is not adjacent to s2. In other words, splitting o (u; s1 ); (s1; w) at s1 is admissible by Lemma 7.1 (i.e.,, the resulting instance (G3s1 ; X; Y ) is irreducible by Lemma 7.1). (G3s1 ; X; Y ) Proof:

16

(containing n3 0 1  7) has an IPR by the assumption on G3 and Lemma 3.6. Since the arcs incident to any terminal lies on the outer face in such IPR, arcs (s2; t); (t; w); (w; s3) form part of the boundary of the outer face. Hence, arcs (u; w); (w; s2); (s2; v ) belong to the boundary of a face in the IPR. This implies that (G3; X; Y ) has also an IPR, which can be obtained by hooking up (u; w) and (s2 ; v ). This is a contradiction, implying that s1 and s2 are not adjacent. Therefore, in this case, we have (i). (b) If t is adjacent to s3, we can show that (ii) holds, by an analogous argument. (c) Finally, consider the case in which t is adjacent to none of s2 and s3. Assume that s1 and s3 are adjacent. We only have to show that s1 and s2 are not adjacent. However, if these are adjacent, Z = fw; s1; s2 ; s3g would be a proper 6-cut with four vertices, contradicting Lemma 6.8. 2 This lemma says that (G3; X; Y ) always has an admissible splitting at vertex w, which is adjacent to a terminal. We further characterize the digraph obtained by such splitting. Lemma 7.3 Let

t

be a terminal which is not adjacent to any other terminal,

terminal vertex adjacent to

(s1; w); (w; s2); (w; s3 ) stance obtained from

t

by arc

(t; w)

in the minimum counterexample

be other three arcs incident with

G3

by splitting arcs

w.

w

be a non-

(G3; X; Y ),

and

G3s2 (resp., G3s3 ) denote the in(t; w); (w; s3)) at w. Then one of

Let

(t; w); (w; s2) (resp.,

these instances is connected and irreducible, and has no cut vertex.

By Lemma 7.2, one of the instances G3s2 and G3s3 is connected and irreducible. Assume without loss of generality that G3s2 is connected and irreducible, i.e., Lemma 7.2(ii) holds. Then s1 and s3 are not adjacent, and t and s2 are not adjacent in G3 . If G3s2 does not have a cut vertex, then the lemma is shown. Therefore, assume that G3s2 has a cut vertex z (see Fig. 12). We rst show that w and z are not adjacent in G3 by contradiction. By Lemma 5.1(i), z is not a cut vertex in G3. Let Z and Z 0 = V 0 fw; z g 0 Z be the vertex sets of the two connected components in G3s2 0 fz g, where t 2 Z is assumed. Consider the three possible cases, in which z = s1 , z = s2 and z = s3 . First consider the case of z = s1 . Then Z 0 is a 2-cut in G3 , and jZ 0 j  2 holds since s1 and s3 are not adjacent. Then jZ \ (X [ Y )j = jZ 0 \ (X [ Y )j = 2 (otherwise Z 0 or V 0 Z 0 would be a reducible 2-cut in G3s2 ). Then, by Lemma 3.3(i) jZ 0 \ X j = jZ 0 \ Y j = 1. If Z = ft; s2 g  X [ Y then j (t; G3 )j = j (s2 ; G3 )j = 2 hold by Lemma 3.3(ii) and in this case we see that (G3; X; Y ) is feasible, a contradiction. Therefore Z 0 (X [ Y ) 6= ;. By this and Lemma 5.1(ii), we have Z 0 0 (X [ Y ) = ; and jZ 0 j = 2. Let t0 be the terminal in Z 0ftg. We see that W = Z [ fw; z g 0 ft; t0 g is a 6-cut in G3 (if W is a 2- or 4-cut, then it would be reducible). By n3  8, we have jW j = n3 0 4  4, and hence W is a proper 6-cut. However, jW j = 3 must hold by Lemma 6.8, a contradiction. Next consider the case of z = s3. In this case, we can observe jZ 0 j = 2 and jZ 0 \ X j = jZ 0 \ Y j = 1 in the similar manner as in the case of z = s1 . Let t0 be the terminal in Z 0 ftg, and Z 0 = ft00; t000g. We see that W = Z [fz g0ft; t0 g is a 6-cut in G3 (if W is a 2- or 4-cut, then it would be reducible). By n3  8, jW j = n3 0 5  3, and W is a proper 6-cut. By Lemma 6.8, W induces a triangle. By considering that t and s2 are not adjacent and G3 has no multiple arc, G3 is given as the instance shown in Fig. 13, where the triangle W = fz; s2 ; v g has two possible ways of orientations. For any choice of terminals ft; t0 ; t00; t000g from X [ Y and orientation of the triangle, we can check that the instance is always feasible, a contradiction. Finally, consider the case of z = s2. In this case, we can obtain j(Z )j = 2, jZ j = 2 and jZ \ X j = jZ \ Y j = 1 in the similar manner as in the above cases. This implies that t is adjacent to a terminal ft0g = Z 0 ftg, contradicting the assumption on t of this lemma. Therefore, w and z are not adjacent in G3. Proof:

17

Then, fs1; s3 g and ft; s2g are contained in distinct components in G32 0 fz g since z is a cut vertex in G32 , but not in G3 . That is, s1 and s2 (resp., t and s3) are not adjacent, and hence 3 is connected and irreducible by Lemma 7.2. G 3 We show that G33 has no cut vertex. Let G33 [Z ] (resp., G33 [Z 0 ]) denote the subdigraph of 3 induced by Z (resp., Z 0 ). Note that G3 [Z ] = G3 [Z ] and G3 [Z 0 ] = G3 [Z 0 ]. Clearly, all G 3 3 3 vertices in Z (resp., all vertices in Z 0) are connected in G33 [Z ] (resp., G33 [Z 0 ]), since otherwise 3 3 3 G would be reducible. This implies that z is no longer a cut vertex in G . Assume that G 3 3 has another cut vertex z 0 (6= z ). By a similar argument as above, z 0 is not equal to any of 3 0 fz 0 g. However, s1 ; s2 ; s3 , and fs1 ; s2 g and ft; s3 g are contained in distinct components in G 3 0 0 3 this is impossible, because if z 2 Z , then t and s2 are connected in G [Z ](= G33 [Z ]) (without using z 0 ), and otherwise if z 0 2 Z then s1 and s3 are connected in G3 [Z 0 ](= G33 [Z 0]) (without using z0 ). 2 s

s

s

s

s

s

s

s

s

s

s

s

s

s

s

s

8 Hooking up Arcs in IPR To complete the proof of Theorem 3.1, this section shows that, given an irreducible infeasible instance G that has an IPR, hooking up any two arcs in G cannot yield G3. More precisely, hooking up two arcs in G makes G satisfy at least one of the conditions in the following lemma, none of which G3 can satisfy. (G = (V ; E ); X; Y ) (where jV j  7) be cut vertex. For arc e = (x2 ; u) 2 E with

Lemma 8.1 Let

an irreducible instance which has an

IPR and has no

x2

let

(G

e;e

0

; X; Y )

0

X and any arc e

be the resulting instance obtained by hooking up e and

Then Ge;e0 is connected, and one of the following properties

(i)

2

0 e

(i)-(v) hold:

= (v; v0 ) 2 E ,

with a new vertex w .

Ge;e0 has a cut vertex.

(ii) (G

0

e;e

; X; Y )

has a 2-cut Z such that Z

j [ f g 0 j  3. V

w

(iii) (G

; X; Y )

is reducible.

(iv) (G

; X; Y )

has an IPR.

(v) (G

; X; Y )

is feasible.

e;e0

e;e0

e;e0



j \ j

V ,

Z

X

=

j \ j Z

Y

= 1,

j j Z

3

and

Z

Assuming that (G ; X; Y ) satisfy neither (i) nor (ii), we show that (G ; X; Y ) satis es one of (iii)-(v). Since G has no cut vertex, we only have to consider IPRs as illustrated in Figs. 14, 15 and 16, which respectively correspond the following three cases: Proof:

e;e0

e;e0

Case-1: (G; X; Y ) has no 2-cut W such that jW \ X j = jW \ Y j = 1. Case-2: (G; X; Y ) has a 2-cut Z such that jZ \ X j = jZ \ Y j = 1, Z 0 (X [ Y ) 6= ; and (V 0 Z ) 0 (X [ Y ) 6= ;, where x2 2 Z . Case-3: (G; X; Y ) has a 2-cut Z such that jZ \ X j = jZ \ Y j = 1, Z  X [ Y or V 0 Z  X [ Y , where x2 2 Z .

Let R be the IPR of (G; X; Y ), and let B denote the cycle of the outer face of R, which is a simple cycle since G has no cut vertex. Let v1; v2; . . . ; v (p = jV (B )j) be the vertices that appear along B clockwise, where v1 = x1 ; v = y0 ; v = x2 and v = y00 (1 < a < b < c), and fy0 ; y00g = Y are assumed without loss of generality. Let B (u; v) denote the subpath of B from u to v , where 0 B (u; u) means a path of null length. Let R denote the planar representation obtained from p

a

c

b

18

the IPR of G by eliminating the arcs in E (B ) together with X [ Y . We denote components of 0 00 0 00 G ; G ; 1 1 1, and the directed cycles representing their outer faces by B ; B ; 1 1 1. For these cycles, say B 0, we denote by B 0(u; v) the subpath of B 0 from u to v. The proof will be given separately for the above three cases.

R0 by

(Fig. 14): In this case, R0 consists of a single component G0 by Lemma 3.5. Also, no two terminals in R are adjacent on B . A more detailed illustration of such R is given in Fig. 17. Note that the directed cycle B 0 (which may not be simple) visits all vertices in V (B ) 0 X 0 Y in the order reverse to B (i.e., anticlockwise). Choose e = (x2 ; v +1), and partition the arc set E 0 e into three subsets: Case-1

b

1 = E2 = E3 =

E

f 00j 00 is adjacent to g ( ( +1 00)) [ ( 0( 0 0 10 2 e

e

e ;

E B vb E

e

;y

E B

E

+1; v 01)) 0 E1 ;

vb

b

E :

It is easy to see that (G ; X; Y ) satis es (iii) (resp., (iv)) for any e0 2 E1 (resp., e0 2 E2). We then show that (v) holds for all e0 2 E3. Since no two terminals in R are adjacent on B and R has no cut vertex, G has at least four nonterminal vertices in V (B 0 ). Case-1a: e0 = (v 01 ; y0 ) 2 E3 . Then G has a y0y00-path, where e and e0 are hooked up with vertex w. e;e

0

e;e0

a

PY

= hB (y0; v +1); B 0 (v +1; v 01 ); (v 01; w); (w; v +1); B (v +1; y00)i: a

a

a

a

b

b

Clearly G 0 E (P ) has an x1 x2-path P = h(x1; v2); B 0(v2; v 01 ); (v 01; x2)i, which implies that (G ; X; Y ) is feasible. Case-1b: e0 = (v ; v +1) 2 E (B (y0; v 01 ))  E3 . Then G has a y0 y00-path e;e0

Y

X

b

b

e;e0

k

k

e;e0

b

PY

= hB (y0 ; v ); (v

k;

k

) (w; v +1); B (v +1; y00)i:

w ;

b

b

It is also easy to see that x1 and x2 are still connected in G 0 E (P ), implying that (G ; X; Y ) is feasible by Lemma 2.1. Case-1c: e0 2 E3 0f(v 01 ; y0 )g0 E (B (y0 ; v 01)). In this case, consider the following y0y00 -chain in G: Q = hB (y0 ; v 01); B 0 (v +1; v 01); B (v +1; y00 )i e;e0

e;e

Y

0

a

b

Y

b

b

b

b

Clearly, e0 62 E (Q ) holds, and G still has y0y00 -chain Q . Let H be the IPR resulting from R by removing the arcs in E (B (v 01; v +1))[E (B 0 (v +1; v 01)) (see Fig. 18). We now claim that x1 is reachable in H from any vertex which is located on the boundary B 0 or in the area surrounded by B 0. By E (H) \ E (Q ) = ;, the claim will mean that, for any e0 = (u0; v 0) 2 E3 0 f(v 01; y0)g 0 E (B (y0 ; v 01)), G 0 E (Q ) has a v 0 x1 -path (hence, it has an x2x1-path). Then, by Lemma 2.1, this will complete the proof that (G ; X; Y ) is feasible. To prove the claim, it is sucient to show that x1 is reachable from any vertex on B 0 , since any vertex inside B 0 is clearly reachable to a vertex on B 0 . Partition set V (B 0) into two subsets V1 = V (B 0 (v 01 ; v +1)) and V2 = V (B 0 ) 0 V1. Since two paths B 0 (v 01 ; v +1) and B (v +1; v 01) remain in H, x1 is clearly reachable in H from any vertex v 2 V1 . We then show that x1 is reachable from any vertex v 2 V2 in H. Let us denote the vertex set V (B 0 (v +1; v 01)) (= V2 [ fv +1; v 01g) by fu0; u1; u2; . . . ; u ; u +1g, where B 0 visits these vertices u0 ; . . . ; u +1 in this order. Assume that there is a vertex u 2 V2 which cannot reach any vertex in V1, and that u has the smallest index among such vertices in V2. We follow the leftmost path P 3 from u in H until the path returns to u (note that P 3 must come back to u since indeg(v) = outdeg(v ) for all v 2 V 0 V1 in H). Clearly, jV (P 3)j  2. Let u be the rst vertex in V2 that P 3 visits. Note that P 3 visits no vertex u 2 V2 with i > h. Consider e;e0

Y

Y

a

c

c

b

Y

a

e;e0

b

Y

e;e0

b

b

c

c

c

c

a

c

b

q

b

q

q

k

k

k

k

k

h

i

19

the set V 3 of vertices in V 0 V (P 3 ) that are adjacent to a vertex of V (P 3 ) in H. Since the arcs incident to a nonterminal vertex are alternately oriented in and out from the de nition of IPR, all vertices in V 3 are located in the inside area surrounded by P 3. In other words, V1 and V (P 3 ) [ V 3 are disconnected in H. Thus, removal of the four arcs f(u 01 ; u ); e ; (u ; u +1 ); e g from G disconnects V1 and V (P 3) [ V 3, where e (resp., e ) is the arc in B (x2 ; y00) such that e and (u 01; u ) (resp., e and (u ; u +1)) belong to the same face in R. This contradicts that (G; X; Y ) has no reducible 4-cut. Therefore, x1 is reachable from any vertex in V1 [ V2 . k

k

k

k

h

h

k

k

h

h

h

h

k

h

(Fig. 15): In this case, R0 consists of two components G0 and G00, where G0 (resp., G00) is induced by V 0 = Z 0 (X [ Y ) (resp., V 00 = (V 0 Z ) 0 (X [ Y )). There are two cases; Case-2a: y 0 ; x2 2 Z and y 00 ; x1 2 V 0 Z (see Fig. 15(2a)), and Case-2b: x2 ; y 00 2 Z and x1 ; y 0 2 V 0 Z (see Fig. 15(2b)). In both Cases-2a and 2b, e0 must be chosen from E (G[V 0 Z ]) to avoid the case (ii) in the lemma statement. Case-2a(i): e0 2 E (B (x2; y00)) \ E (G[V 0 Z ]). It is easy to see that (G ; X; Y ) has an IPR. Case-2a(ii): e0 2 E (G[V 0 Z ]) 0 E (B (x2; y00)). Let v be the vertex in B (y0; y00) \ Z with the largest index, where b < h < c must hold (otherwise V 0 would be a reducible cut or a cut vertex). Then G has a y0y00-path Case-2

e;e0

h

e;e0

PY

= h(y0; v +1); B 0 (v +1; v ); B (v a

a

h

h

; y 00 ) :

i

Furthermore, it is also easy to see that x1 and x2 are connected in G 0 E (P ). Therefore, (G ; X; Y ) is feasible by Lemma 2.1. Case-2b(i): e0 = (v ; v +1) 2 E (B (y0; x2)) \ E (G[V 0 Z ]). Then G has a y0 y00-path P = hB (y0 ; v ); (v ; w); (w; v +1); B (v +1; y00)i. Obviously, x1 and x2 remain connected in G 0 E (P ), implying by Lemma 2.1 that (G ; X; Y ) is feasible. Case-2b(ii): e0 2 E (G[V 0 Z ]) 0 E (B (y0 ; x2 )). Let v be the vertex in B (y0 ; y00 ) \ Z with the smallest index, where a < h < b must hold. Then G has a y0 y00 -path P = hB (y0; v ); B 0 (v ; v 01); (v 01; y00)i. Since it is obvious that x1 and x2 are connected in G 0 E (P ), (G ; X; Y ) is feasible by Lemma 2.1. 0

Y

e;e

e;e0

k

k

k

e;e0

k

b

Y

e;e0

b

e;e0

Y

h

e;e0

Y

0

c

h

Y

e;e

h

c

0

e;e

(Fig. 16): In this case, there are two terminals x3 2 X and y3 2 Y which are adjacent on B , and R0 has a single component G0. Case-3a: Z = fy0 ; x2 g (i.e., (y0 ; x2 ) 2 E (B )) (see Fig. 16(3a)). This case can be treated in the same manner as in Case-1; the corresponding partition of E 0 e is de ned by E1 = fe00je00 is adjacent to eg, E2 = f(v 01; y0)g [ E (B 0 (v +1; v 01)) [ E (B (v +1; y00)) 0 E1 and E3 = E 0e0E1 0E2 and y0 y 00 -chain Q in Case-1c is chosen as Q = h(v 01 ; y 0 ); B 0 (v +1 ; v 01 ); B (v +1 ; y 00 )i. Case-3b: V 0 Z = fy00 ; x1g (i.e., (y00; x1) 2 E (B )) (see Fig. 16(3b)). This case can be treated in the same manner as in Case-2a. Case-3c: V 0 Z = fx1 ; y0 g (i.e., (x1; y0) 2 E (B )) (see Fig. 16(3c)). This case can be treated in the same manner as in Case-2b. Case-3d: Z = fx2 ; y00 g (i.e., (x2 ; y00) 2 E (B )) (see Fig. 16(3d)). This case can be treated in the same manner as in Case-1; the corresponding partition of E 0 e is de ned by E1 = fe00je00 is adjacent to eg, E2 = E (B 0 (v +1; v 01)) and E3 = E 0 e 0 E1 0 E2, and y0 y00-chain Q in Case-1c is chosen as Q = hB (y0; v 01 ); B 0 (v +1; v 01); B (y00; v +1)i. 2 Case-3

a

b

a

Y

Y

c

Y

b

b

a

b

a

b

Y

b

c

b

c

Now, we are ready to prove Theorem 3.1 by deriving a contradiction from the assumption that a minimum counterexample (G3; X; Y ) exists. G3 must have a nonterminal vertex adjacent to a terminal (otherwise, G3 consists of only terminals, contradicting n3  8). We can assume without loss of generality that (G3 ; X; Y ) has a terminal x2 which is not adjacent to any other terminal (because, if any terminal is adjacent to some other terminal, then j(V 0 (X [ Y ))j  4 holds, and hence jV 0 (X [ Y )j  1 is implied by irreducibility of G3 ). Then by Lemma 7.3, 20

two arcs (x2; w) and (w; s2) in (G3; X; Y ) can be split o at w so that the resulting instance (G3 ; X; Y ) is still connected and irreducible, and has no cut vertex, where (s1; w) and (w; s3) are the other arcs incident to w. In other words, G3 can be obtained from G3 by hooking up two arcs e = (x2; s2) and e0 = (s1; s3) after introducing w. We then apply Lemma 8.1 to G = G3 and G = G3. By Lemma 5.1, neither (i) nor (ii) of Lemma 8.1 holds for G3 . Furthermore, none of the remaining (iii){(v) of Lemma 8.1 is possible by the de nition of G3. This is a contradiction, and proves the next lemma. w

w

w

e;e0

Lemma 8.2 Let (G; X; Y ) (G; X; Y ) has an IPR.

be an infeasible irreducible instance, and satisfy

jV j 6= 6.

Then

2

Finally, this proves Theorem 3.1, since, by Lemma 3.2, this is a stronger statement than Theorem 3.1.

9 Complexity Results Based on Theorem 3.1 (or Lemma 8.2 to be more precise), we can test if a given instance (G; X; Y ) is feasible or not, in polynomial time. Lemma 9.1 Given an instance found in

O (m + n log n)

(G; X; Y ),

time, where

n

and

one of its irreducible instances

m

(G0 ; X; Y )

can be

denote the numbers of vertices and arcs in

respectively.

G,

2

Before describing the algorithm for computing an irreducible instance, let us review a cactus representation [1], a compact representation of all minimum cuts in an undirected graph. A connected undirected graph is called a cactus if, for each edge, there is exactly one simple cycle that contains it, where the cycle may be of length 2. Then, in a cactus, two cycles (if any) have at most one common vertex, which is a cut vertex. A vertex with degree 2 in a cactus is called a leaf vertex. Given an undirected graph G = (V; E ), we map it to a cactus 0 = (W; F ) by a mapping ' : V ! W , where ' may not be onto. The size of a minimum cut in G (resp., in 0) is de ned by (G) = minfj(Z ; G)j j ; 6= Z  V g (resp., (0) = minfj(S ; 0)j j ; 6= S  W g), where  (Z ; G) denotes the set of edges between Z and V 0 Z in G (similarly for (S ; 0)). Clearly, in a cactus 0 = (W; F ) with jW j  2, (0) = 2 holds. Let C (G) = fZ j ; 6= Z  V; j(Z ; G)j = (G)g and C (0) = fS j ; 6= S  W; j(S ; 0)j = (0)g denote the sets of all minimum cuts of G and 0, respectively. Note that S belongs to C (0) if and only if two arcs in (S ; 0) belongs to the same cycle. In the following description, we use the term \vertex" to denote an element in V , and the term \node" to denote an element in W . There may be a node x 2 W with '01(x) = ;, which is called an empty node. De ne '(Z ) 0 ' 1 (S )

 f'(v ) 2 W j v 2 Z g for Z  V; and  fv 2 V j '(v ) 2 S g for S  W:

A pair (0; ') of a cactus and a mapping ' is called a cactus representation for C (G) if it satis es (i) and (ii) below. (i) For any cut Z 2 C (G), there exists a cut S 2 C (0) such that Z = '01(S ) and V 0 Z = '01 (W 0 S ). (ii) Conversely, for any 2-cut S 2 C (0), Z = '01 (S ) satis es Z 2 C (G). 21

It is known [1] that G = (V; E ) always has such a cactus representation (0 = (W; F ); ') with jW j = O(jF j) = O(jV j), which can be constructed in O(jE j + (G)2jV j log jV j) time [7]. We say that a cut Z 2 C (G) and a cut S 2 C (0) correspond each other if Z = '01(S ) and V 0 Z = '01(W 0 S ). Note that if 0 has an empty node, a minimum cut in C (G) may correspond to more than one minimum cut in C (0), while any minimum cut in C (0) always corresponds to exactly one minimum cut in C (G). Obviously, any leaf node w 2 W corresponds to a minimum cut in C (G), and there are at least two leaf nodes in 0. Lemma 9.2 For an undirected graph

G = (V; E )

and a designated vertex

t3

2

V,

let

Z=

fZ1; Z2 ; . . . ; Z g be the set of all cuts Z such that (i) Z 2 C (G) and Z  V 0 ft3 g. (ii) jZ j is maximal subject to (i) (i.e., no cut Z 0 2 C (G) with Z  Z 0  V 0 ft3g). Then any two cuts Z ; Z 2 Z are mutually disjoint, and the set Z can be computed in O (jE j + (G)2jV j log jV j) time. 3 3 Proof: Consider a cactus representation (0 = (W; F ); ') for C (G). Let w = '(t ) 2 W , and let q

i

i

i

i

i

i

j

0 have p cycles passing through w3. In other words, removal of w3 from 0 creates p connected components with node sets Wi , i = 1; 2; . . . ; p. Let Zi = '01(Wi ), i = 1; . . . ; p. Since each Wi is a 2-cut in 0, we have Wi 2 C (0) and hence Zi 2 C (G) by the de nition of a cactus representation. Hence, each Zi satis es condition (i). If there is a cut Z 0 2 C (G) such that Zi  Z 0  V 0 ft3g, then there is a cut W 0 2 C (0) such that Z 0 = '01(W 0 ), Wi  W 0 and w3 2 W 0 W 0 . However, 0 cannot have any such 2-cut W 0 separating w3 and Wi by the de nition of W 0 . Therefore, each Zi satis es condition (ii). Obviously, Zi; . . . ; Zp are mutually disjoint by disjointness of W1 ; . . . ; Wp . The stated time complexity follows from the fact that a cactus representation (0 = (W; F ); ') with jW j + jF j = O(jV j) can be obtained in O(jE j + (G)2jV j log jV j) time [7], and computing connected components in 0 0 w3 can be done in O(jW j + jF j) = O(jV j) time.

2

Proof of Lemma 9.1: Given an instance (G; X; Y ), where n = jV j and m = jE j, the following algorithm applies all reductions of type (1), (2) and (3), de ned in the beginning of Section 3.

1. Type (1) reductions (i.e., 2-cuts Z such that jZ j  1 and Z \ (X [ Y ) = ;): We contract four terminals x1; x2; y1; y2 into a single vertex t3, and ignore arc orientation in G. Let G denote the resulting undirected graph. Clearly, (G)  2, since G is connected and Eulerian. It is easy to see that a cut Z  V 0 fx1; x2; y1; y2g is 2-cut in G if and only if (G) = 2, Z 2 C (G) and Z  V (G) 0 ft3 g. We can check if (G)  2 in O(m + n log n) time [7]. If (G) > 2 then there is no cut of type (1), and we go to 2. If (G) = 2, then by Lemma 9.2, the set fZ1; . . . ; Zpg of these cuts Z with maximal jZ j is uniquely determined, and obtained in O(m + n log n) time. Apply reduction (1) to all cuts Zi in (G; X; Y ). This can be done in O(m + n) time (since Zi \ Zj = ; for 1  i < j  p if p  2). Go to 2 after letting (G; X; Y ) be the resulting instance. 2. Type (2) reductions (i.e., 2-cuts Z such that jZ j  2 and jZ \ (X [ Y )j = 1): For each terminal t 2 X [ Y , let Gt denote the undirected graph obtained from G by contracting the other three terminals X [ Y 0ftg into a single vertex t, and ignoring arc orientations. We easily see that if G has a 2-cut Z with Z \ (X [ Y ) = ftg and jZ j  2, then (Gt ) = 2, Z 2 C (Gt ) and Z  V (G) 0 ftg hold. Then such Z contains t (otherwise, Z would be a cut of type (1), which must have been eliminated in the above 1) and is unique if it is 22

maximal (since at most one cut can contain t). Furthermore, such Z can be obtained in O(m + n log n) time, Lemma 9.2. We apply reduction (2) to the cut Z in (G; X; Y ) in O(m + n) time. This procedure for all four terminals can be done in O (m + n log n) time. Go to 3 after letting (G; X; Y ) be the resulting instance. 3. Type (3) reductions (i.e., 4-cuts Z such that G[Z ] is connected, jZ j  2 and Z \ (X [ Y ) = ;): Since the current (G; X; Y ) has no cut of type (1), any 4-cut Z  V 0 (X [ Y ) induces a connected subdigraph G[Z ]. Let G be the undirected graph obtained from (G; X; Y ) by contracting four terminals x1; x2; y1 ; y2 into a single vertex t3, and ignoring arc orientations. Clearly, (G)  4 (otherwise, (G; X; Y ) would have a reducible 2-cut). We easily see that a cut Z  V 0 (X [ Y ) is 4-cut in G if and only if (G) = 4, Z 2 C (G) and Z  V (G) 0 ft3g. We can check if (G)  4 in O(m + n log n) time. If (G) > 4 then there is no cut of type (3), and the current instance (G; X; Y ) is irreducible. If (G) = 4, then by Lemma 9.2, the set fZ1; . . . ; Zp g of these cuts Z with maximal jZ j is uniquely determined, and is obtained in O(m + n log n) time. Apply reduction (3) to all these cuts Zi in (G; X; Y ) to obtain an irreducible instance. This can be done in O (m + n) time. 2 Given an irreducible instance (G0 ; X; Y ), we can check if it is feasible or not in linear time as follows. If G0 has less than 7 vertices, its feasibility can be easily checked in O(1) time (since any irreducible infeasible digraph G0 with jV j < 7 has O(1) arcs). Otherwise test if the resulting irreducible instance (G0; X; Y ) has an IPR, which can be done in O(m + n) time by using a fast planar drawing algorithm [8]. If it has an IPR, then it is infeasible; otherwise it is feasible. Therefore, we have established the next theorem. Theorem 9.1 Given an instance

(G; X; Y ),

where

n

and

m

arcs, respectively, testing if it is feasible or not can be done in

are the numbers of vertices and

O (m + n log n)

time.

2

We now show that, if a given instance (G; X; Y ) is feasible, a solution (i.e., a pair of arcdisjoint x0 x00- and y0 y00-paths in G, where fx0; x00g = X and fy0; y00g = Y ) can be found in O (m(m + n log n)) time. Let (G = (V; E ); X; Y ) be an irreducible feasible instance. If V consists only of four terminals, then a solution is easily found in O(1) time. Otherwise, one of the following four cases A-D occurs, and we can nd a pair of arcs such that the instance remains feasible after splitting o them. A. There is a nonterminal vertex v with deg(v)  6 or a terminal v with deg(v ) = 4 in instance (G; X; Y ): Choose such a vertex v, and nd two arcs (v0 ; v) 2 0 (v ) and (v; v00 ) 2  +(v ) such that the instance obtained by splitting (v 0; v ) and (v; v00 ) at v remains feasible. Note that such a pair of arcs exists (since the instance is feasible), and by Theorem 9.1, it is found in O(m + n log n) time by checking feasibility among all (at most 9) possibilities. Split o such (v 0 ; v ) and (v; v00 ) at v and recompute an irreducible instance (G0 ; X; Y ) from the resulting instance in O(m + n log n) time (Lemma 9.1). B.

 8 for all v 2 V 0 (X [ Y ), deg(t) 6= 4 for all t 2 X [ Y , but there is a 6-cut  0 (X [ Y ): Then choose a minimal Z among such 6-cuts, and let v be a nonterminal

deg(v ) Z V

vertex in Z . Note that any nonempty cut W

 Z satis es j (W ; G)j  8 from the assumption on Z . 23

Since j0 (v )j = j+(v )j  4 by deg(v)  8 and j0 (Z )j = j+(Z )j = 3, there are arcs (v 0; v ) 2 0(v) 0 0(Z ) and (v; v00 ) 2 +(v) 0 +(Z ), where v 0; v 00 2 Z (possibly v 0 = v 00). Let (G0 ; X; Y ) be the instance obtained from (G; X; Y ) by splitting o these arcs (v 0 ; v ) and (v; v00) at v (in the case of v 0 = v 00, splitting simply means removal of those two arcs). We show that (G0 ; X; Y ) remains irreducible (hence feasible, because any irreducible instance having a nonterminal vertex v with deg(v )  6 is feasible by Lemma 3.3(iii)). Assume that (G0; X; Y ) has a reducible cut Z 0 . Since Z 0 was not reducible in (G; X; Y ), Z 0 must separate fvg and fv0 ; v00 g. Since Z 0  Z would imply j(Z 0 ; G)j  8 from the above assumption (i.e., j(Z 0 ; G0 )j  6) and hence Z 0 is not reducible in G0 , such Z 0 must intersect Z (and hence Z 0 and Z cross each other because (V 0 (Z 0 [ Z )) contains a terminal). In this case, we have j (Z \ Z 0; G)j  8, and j(Z \ Z 0 ; G0 )j  6 (= j(Z ; G0 )j): Also, we obtain

j(Z [ Z 0 ; G0)j  j(Z 0; G0)j + 2 (since otherwise j(Z [ Z 0 ; G0 )j  j(Z 0; G0)j implies that Z [ Z 0 is a reducible cut in G). However, these two inequalities contradicts (2.1) (i.e., j(Z ; G0)j + j(Z 0 ; G0)j  j(Z \ Z 0 ; G0 )j + j (Z [ Z 0 ; G0 )j). This shows that (G0 ; X; Y ) is irreducible, and hence feasible.

A minimal 6-cut Z in the above can be found in O(m + n log n) time as follows. Since such Z never intersects X [ Y , we contract the four terminals into a single vertex t3 , and ignore the arc orientation. Let Gt3 be the resulting undirected graph. Clearly, (Gt3 ) = 6 by the irreducibility of G and the assumption of case B. Find a cactus representation (0; ') for C (Gt3 ) in O(m + n log n) time [7]. Recall that 0 has at least two leaf nodes, and one of them, say z satis es t3 62 '01(z). By de nition of a cactus representation, Z = '01 (z ) is a minimal 6-cut in G. C.

 8 for all v 2 V 0 (X [ Y ), deg(t) 6= 4 for all t 2 X [ Y , and j(W )j  8 for all  0 (X [ Y ), but there is a 4-cut Z with Z \ (X [ Y ) = ftg for some terminal t: Then take a minimal Z among them. Since deg(t) = 6 4, we see that Z 0ftg =6 ; and deg(t)  6 (if deg(t) = 2, then Z 0ftg is a 6-cut with Z 0ftg  V 0 (X [ Y ), contradicting the assumption of case C). Since j 0(t)j = j+(t)j  3 by deg(t)  6 and j0(Z )j = j +(Z )j = 2, there are arcs (v 0; t) 2 0(t) 0  0(Z ) and (t; v00) 2 + (t) 0 +(Z ), where v0 ; v00 2 Z (possibly v 0 = v 00).

deg(v ) W V

Let (G0 ; X; Y ) be the instance obtained from (G; X; Y ) by splitting o these arcs (v 0 ; t) and (t; v00 ) at t (in the case of v0 = v00, splitting means removal of those two arcs). We show that (G0; X; Y ) remains irreducible (hence feasible, because any irreducible instance having a terminal vertex t with deg(t)  4 is feasible by Lemma 3.3(ii)). Assume that (G0 ; X; Y ) has a reducible cut Z 0. Since Z 0 was not reducible in (G; X; Y ), Z 0 must separate ftg and fv0 ; v00 g. Since Z 0  Z implies that j(Z 0)j  6 (if t 2 Z 0) by the minimality of Z and j (Z 0 )j  8 (if t 62 Z 0 ) by the assumption of case C, and hence Z 0 is not reducible in G0 , such Z 0 intersects Z (and hence Z 0 and Z cross each other since (V 0 (Z 0 [ Z )) contains a terminal). We see that Z 0 is not a reducible cut of type (1) in (G0; X; Y ), because otherwise Z 0 would be a reducible 4-cut in (G; X; Y ). If Z 0 is a reducible cut of type (3) in (G0 ; X; Y ), then Z 0 is a 6-cut Z 0  V 0 (X [ Y ) in (G; X; Y ), contradicting the assumption of case C. Therefore, Z 0 must be a reducible cut of type (2) in (G0; X; Y ). We rst consider the case of t 2 Z \ Z 0. We have j(Z \ Z 0 ; G)j  6 (by the minimality of Z ) and j(Z \ Z 0 ; G0 )j  4(= j(Z ; G0)j). Since Z 0 Z 0 contains no terminal, we obtain j (Z [ Z 0; G0 )j  j(Z 0; G0)j + 2 (otherwise j(Z [ Z 0; G)j = j (Z [ Z 0; G0)j  j(Z 0; G0)j implies that Z [ Z 0 is a reducible cut in G). However, these two inequalities contradicts 24

(2.1), as in case B. Then assume t 2 Z 0 Z 0 , implying that there is another terminal t0 in Z 0 0 Z . Clearly, j(Z 0 Z 0 ; G)j  6 (= j (Z ; G)j + 2) (from minimality of Z ). From j (Z 0; G0 )j = 2, we have j (Z 0 ; G)j = 4 and deg(t0 )  6 (if deg(t0 ) = 2, then Z 0 0 ft0g would be a 6-cut, contradicting the assumption of case C). From this and irreducibility of (G; X; Y ), j(Z 0 0 Z ; G)j  4 (= j (Z 0; G)j) holds. These inequalities contradicts (2.2). Consequently (G0 ; X; Y ) is irreducible, and hence is feasible. The above minimal 4-cut Z can be found in O(m + n log n) time as follows. Since such Z always separates ftg and X [ Y 0 ftg, contract the three other terminals in X [ Y 0 ftg into a single vertex t, and ignore the arc orientation. Let Gt be the resulting undirected graph. Clearly, (Gt ) = 4 (since case C does not occur for this t if (Gt )  6). Find a cactus representation (0; ') for C (Gt) in O(m + n log n) time [7]. By de nition of a cactus representation, 0 has a leaf node z with t 2 '01(z ), and Z = '01 (z ) is a desired minimal 4-cut in G. D.

 8 for all v 2 V 0 (X [ Y ), deg(t) 6= 4 for all t 2 X [ Y , j(W )j  8 for all  0 (X [ Y ) and j(Z )j  6 for all Z with jZ \ (X [ Y )j = 1: Then choose an

deg(v ) W V

arbitrary nonterminal vertex v and two arcs (v 0; v ) and (v; v 00). It is easy to see that the instance (G0 ; X; Y ) obtained from (G; X; Y ) by splitting o these arcs (v0 ; v) and (v; v00) at v remains irreducible (hence feasible, because any irreducible instance having a nonterminal vertex v with deg(v )  6 is feasible by Lemma 3.3(iii)).

Now repeat applying the above cases A-D as long as there is an applicable case. We note that exactly one of the cases A-D is always applicable unless we reach the instance having four terminals of degree 2 and no nonterminal vertex. In the latter case, we can easily nd a solution. The entire running time of this procedure is O(m(m + n log n)), since the number of arcs decreases at least by one after splitting o a pair of arcs. It is easy to see that a solution of the original instance (G; X; Y ) can be recovered in the same time complexity from the sequence such splittings. This establishes the next theorem. (G; X; Y ), where n and m are the numbers of vertices and edges, respectively, a solution of (G; X; Y ) can be computed in O (m(m + n log n)) time. 2

Theorem 9.2 Given a feasible instance

10 Discussion For the arc-disjoint path problems (G; Xi = fsi; ti g; i = 1; 2; . . . ; k) associated with Eulerian digraphs, dierent problem settings are conceivable depending upon the restrictions on G and the directions of the required paths: (i) either G + H is Eulerian, where H is the demand digraph, or G itself is Eulerian, and (ii) either si ti -paths are required for all i, or one of the si tiand ti si -paths is required for each i. The result in [9] shows that (G + H Eulerian, si ti-path, k = 3) can be solved in polynomial time, while our result here shows that (G Eulerian, one of the si ti - and ti si -paths, k = 2) can also be solved in polynomial time. By generalizing the proof in [4, 9], it is possible to prove that all types become NP-hard if k is considered as a part of input. Therefore, an interesting theoretical challenge, for each problem type, will be to nd out the maximum constant k that permits a polynomial time algorithm, or to show that any constant k permits a polynomial time algorithm.

Acknowledgements

This research was partially supported by the Scienti c Grant-in-Aid by the Ministry of Education, Science, Sports and Culture of Japan, and by the Inamori Foundation. 25

References [1] E. A. Dinic, A. V. Karzanov and M. V. Lomonosov, On the structure of a family of minimal weighted cuts in a graph, Studies in Discrete Optimization (in Russian), A.A. Fridman (Ed.), Nauka, Moscow, 1976, pp. 290{306.

On existence of two edge-disjoint chains in multi-graph connecting given pairs of its vertices, Graph Theory Newsletters, 8, 1979, pp. 2{3.

[2] E.A. Dinits and A.V. Karzanov,

[3] E.A. Dinits and A.V. Karzanov, On two integer ows of value 1 in a network, in: Combinatorial Methods for Network Flow Problems, A.V. Karzanov ed., Institute for System Studies, Moscow, 1979, pp. 127{137 (in Russian). [4] S. Fortune, J. E. Hopcroft, and J. Wyllie, The Theoret. Comput. Sci., 10, 1980, pp. 111{121.

directed subgraph homeomorphism problem,

[5] A. Frank, On connectivity properties of Eulerian digraphs, in Graph Theory in Memory of G.A.Dirac, Annals of Discrete Mathematics 41, Noth-Holland, Amsterdam, 1988, pp. 179{194. [6] A. Frank, Graph connectivity and network ows, Handbook of Combinatorics I, R. L. Graham, M. Gr otschel, and L. Lovasz, eds., North-Holland, 1995. [7] H. N. Gabow, Applications of a poset representation to edge connectivity and graph rigidity, Proc. 32nd IEEE Symp. Found. Comp. Sci., San Juan, Puerto Rico, 1991, pp. 812{821. [8] J. E. Hopcroft and R. E. Tarjan, 1974, pp. 549{568. [9] T. Ibaraki and S. Poljak, 4, 1991, pp. 84{98.

Ecient planarity testing, J. Assoc. Comput. Math., 21,

Weak three-linking in Eulerian digraphs, SIAM J. Disc. Math.,

[10] N. Robertson and P. D. Seymour, Graph binatorial Theory B, 63, pp. 65{110.

minors XIII: The disjoint path problem, J. Com-

[11] P. D. Seymour,

Disjoint paths in graphs, Disc. Math., 29, 1980, pp. 239{309.

[12] C. Thomassen,

2-linked graphs, Europ. J. Combinatorics, 1, 1980, pp. 371{378.

26

s1

s2

t2

t1

Figure 1: An infeasible instance for two edge-disjoint path problem in an undirected graph G.

x1

x

x3

2

Figure 2: An infeasible instance for weak 3-linking problem in an Eulerian digraph G + H .

27

x1

y″

y′

x1

x2

y″

y′

x2 (b) An example of IPR

(a) An example of IPR

v

x1

y2

x2

y1

w (c) The minimal infeasible instance which has no IPR Figure 3: Examples of infeasible instances for two arc-disjoint path problem in an Eulerian digraph

28

G.

Z

v

v

u

u

V-Z

V-Z (1)

t

t Z

V-Z

V-Z (2)

Z

z V-Z

V-Z (3)

Figure 4: Three types of reductions of irreducible cuts

29

Z.

e1-

e+ 3

z1-

z+ 3

e2-

z2-

Z z+ 1

z2+ z3e3-

e+ 1

V-Z Figure 5: A 6-cut

30

Z.

e+ 2

e1-

e+ 3

e1-

e+ 3

z+= z1-

z+= z1-

3

3

e2-

z2-

Z

z2+

e2-

e+ 2

z2-

Z

e+

1

e3-

e+

e3-

1

1

(2) Case-2

(1) Case-1

e1-

e+ 3

e1-

e+ 3

z+= z1-

z+= z1-

3

3

e2-

z2-

z2+

Z z+= z 1

e+ 1

e+ 2

z+

z3-

z+ 1

z2+= z3-

e+ 2

e2-

3

Z

z2-=z1+

z2+=z3-

2

e+

e3-

1

e3-

(4) Case-4

(3) Case-3 Figure 6: Four possible cases for a proper 6-cut

31

e+

Z

in (G3 ; X; Y ).

t1 (= z -, z+) 1 3

t1 (= z -, z+) 1 3 t′1 z2-

z2+ z+ 1

t2

t′1

z2-

t2

t3

z3-

z+

t3 (= z +, z - )

1

2 3

Z ∪ {t 1′ }

Z ∪ {t 1′ }

(2) Case-2

(1) Case-1

t1 (= z -, z+) 1 3

t1 (= z -, z+) 1 3

t′1

t′1

z2-

t2

z2+ t3 (= z +, z - )

t2 (= z2- ,z+)

t3

z+= z 1 3

2 3

1

Z ∪ {t 1′ }

Z ∪ {t 1′ }

(4) Case-4

(3) Case-3

3;X ~;Y ~ ) transformed from Figure 7: The instance (HZ

32

3 [Z ].

G

z +3 =z1- =t1 Z ∪ {t1′}

t1′ z0

z 2-

t2

z +1

z +3 =z1- =t1 t1′ z 2-

z 2+

t3

z 3-

Z ∪ {t1′}

z0

t2

z +1

(1) Case-1

(2) Case-2

z +3 =z1- =t1 t1′ z 2-

t2

z +2 =z3- =t3

z +3 =z1- =t1

Z ∪ {t1′}

t1′ Z ∪ {t ′} 1

z 2+

z +1 =z2- =t2

t3

z +1 =z 3(3) Case-3

z 2+=z3- =t3

(4) Case-4

Figure 8: The smallest IPRs of

33

HZ3

.

3

3

t

z1-= z+ 3

z1-= z+ 3 e2-

z2-

z2+

z0

z1+ t

e+ 1

z2-

z0

e2-

e+ 2

3

t

z3-

z2-

e+ 2

e+ 1

t′

e3-

e3-

t′

(2) Case-2

e1-

z2+

e1-

e+ 3

z1-= z+ 3 e2-

z2+= z3-

z1+

(1) Case-1 e+

e1-

e+

e1-

e+

z1-= z+ 3 e+ 2

e+

e2-

2

z2-= z1+

z1+= z3-

z2+= z3-

e+

e3-

1

e+ 1

t′

e3-

(4) Case-4

(3) Case-3 Figure 9: The instance

G#Z

obtained from

34

G3

by replacing

G3 Z [

] with

HZ#

.

s0

s

3

w

s2

s1

W Figure 10: Illustration for Lemma 7.1.

t

s

3

w

s2 s1 u

v Figure 11: Illustration for Lemma 7.2.

35

t

s

3

w

s1

s2

Z’

Z z

Figure 12: Illustration for Lemma 7.3.

t t″

w v

t′ t′′′ z=s3

s2

W Figure 13: The proof of Lemma 7.3.

36

x 1=v 1

y′=va

x 2=v b

y″=v c

(1) Figure 14: Case-1 in the proof of Lemma 8.1 (the shaded area indicates

R ; i.e., B 0

0

and its interior).

va-1 x 1=v 1

y′=va

x 1=v 1

y′=va B′

G″

Z G″

G′

B′

y″=v c

Z

G’

x 2=v b e

y″=v c

vh (2a)

e vb+1

vc-1 (2b)

Figure 15: Case-2 in the proof of Lemma 8.1.

37

vh x 2=v b

y′

x1

y′

x1 B′

B′

G′

G′

x

y″

e

x

y″

2

2

e

(a)

(b)

x1

y′

x1

y′

G’ B′ B′ G′ x

y″

e

2

x

y″ e (d)

(c)

Figure 16: Case-3 in the proof of Lemma 8.1.

38

2

v1 =x 1

v2

v3 =va-1

va =y′

vp

B va+1

B′

vc+1 v b-1 vc =y″ vc-1

e

vb =x 2

v b+1 Figure 17: Illustration of graph

G

39

used in the proof of Lemma 8.1.

E1 E2 E3

v2

v 3=va-1

v1 =x

1

va =y′

vp

va+1 u 0 = vc+1 u1

P*

V*

vc =y″ vc-1

ek

vb-1=uq+1

uk uh eh

e vb+1

vb =x 2 : vertices in V1 : vertices in V2

Figure 18: Illustration for the proof of Case-1c in Lemma 8.1.

40