Combinatorics, Probability and Computing (1996) 5, 297-306
Copyright © 1996 Cambridge University Press
Finding a Longest Alternating Cycle in a
2-edge-coloured Complete Graph is in RP
RACHID SAAD*
Cite Ibn Khaldun, Bat 68 A2, Boumerdes, Algeria
Received
15 December
1992; revised 15 December 1994
Jackson [10] gave a polynomial sufficient condition for a bipartite tournament to contain
a cycle of a given length. The question arises as to whether deciding on the maximum
length of a cycle in a bipartite tournament is polynomial. The problem was considered by
Manoussakis [12] in the slightly more general setting of 2-edge coloured complete graphs:
is it polynomial to find a longest alternating cycle in such coloured graphs? In this paper,
strong evidence is given that such an algorithm exists. In fact, using a reduction to the well
known exact matching problem, we prove that the problem is random polynomial.
1. Introduction
Apart from their interest as a possible framework for many applications (alternating
cycles are considered as transversals in latin squares in Andersen [7]; in Chen and
Daykin [6] a kind of travelling salesman problem with many means of communication
was considered in terms of coloured graphs), edge-coloured complete graphs extend
multipartite tournaments in a natural way, as pointed out by Benkouar, Manoussakis,
Paschos and Saad [2]. Many results presented by Benkouar, Manoussakis and Saad [3]
and Benkouar et al. [2] extend well known results in bipartite tournaments to 2-edge
coloured complete graphs. A necessary and sufficient condition for a 2-edge coloured
complete graph to contain an alternating Hamiltonian cycle was given by Bankfalvi and
Bankfalvi [1]. Based on this result, an O(n25)-time algorithm for finding an alternating
Hamiltonian cycle was derived by Benkouar et al. [2], which, in the case of bipartite
tournaments, was established by Haggkvist and Manoussakis [9]. More precisely, such a
problem has been reduced by Benkouar et al. [2] to finding a perfect alternating factor in
a 2-edge-coloured complete graph. Poljak [13] has reduced the perfect alternating factor
problem in a &-edge-coloured complete graph to the perfect matching problem. On the
other hand, Gutin [8] has proved that, for every strongly connected bipartite tournament,
the maximum length of a factor equals the maximum length of a cycle, but no efficient
+
This work was done when the author was at the Laboratoire de Recherche en Informatique, Bat 490,
Universite Paris-Sud, 91405 Orsay Cedex, France.
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298
R. Saad
way of deciding on the length of such a cycle or factor was given. The aim of this paper, is
to show a polynomial time reduction of the longest alternating cycle problem in a 2-edge
coloured complete graph to the well known exact matching problem.
2. Notation
The notion of an alternating cycle was originally introduced by Bollobas and Erdos [5].
However, in what follows, we shall revert to the notation of Benkouar et al. [2].
2.1. Alternating paths and cycles
1 Let Kn be the complete graph on n vertices, and suppose its edges are coloured with
two colours 1 and 2. Let c and K%, respectively, denote the function which assigns a
colour from {1,2} to each edge of Kn and the graph Kn, endowed with the colouring
c. Assume throughout colour 1 (resp. 2) to be the red (resp. blue) colour. If A and B
are two subsets of vertices of K£, AB denotes the set of edges having one vertex in A
and another in B; if, in addition, these edges have the same colour, this colour will be
denoted by c(AB). If, in addition, A = {x}, then the colour c(AB) will be denoted by
c(xB).
A path P (resp. a cycle C) of Kn is said to be alternating if and only if any two
consecutive edges of P (resp. C) have different colours. K% is said to be alternating
bipancyclic if it contains an alternating Hamiltonian cycle and an alternating cycle of
length k or k + 2 for every even integer k < n.
As pointed out by Haggkvist and Manoussakis [9], every bipartite tournament of
vertex classes A and B can be viewed as a 2-edge coloured complete graph: it suffices
to complete the graph by adding red (resp. blue) edges in A (resp. B) and to colour
every (A, B) arc blue and every (B,A) arc red so that to every cycle in the original
bipartite tournament there corresponds an alternating cycle in the coloured complete
graph, and vice versa. Every K% that arises from a bipartite tournament in this way is
said to be bipartite.
Let x and y be two vertices of K% and P be an alternating x-y path of K%. We denote
by P(x) (resp. P{y)) the colour of the only edge of P incident to x (resp. y).
2 For a graph G, a factor is a 2-regular subgraph of G (i.e. a disjoint union of cycles
of G). A perfect factor of G is a spanning factor of G. For an edge-coloured graph,
alternating factors and perfect alternating factors are defined likewise. Let A be a
complete edge-coloured induced subgraph of K£. We denote by af(A) (resp. ac(A))
the length of a largest alternating factor (resp. cycle) of A.
2.2. Decision problems and the RP class
The following decision problem is denoted by AC:
Instance: an edge-coloured complete graph K% and an integer k.
Question: does K% contain an alternating cycle of length / > k? An instance of AC will
be denoted by (Kcn,k).
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Finding a Longest Alternating Cycle
299
The problem of deciding on the maximum length of a bipartite tournament is denoted
by BC.
The following decision problem is called the exact matching problem:
Instance: a graph G = (V, E), a subset of edges F c E and an integer k.
Question: does there exist a perfect matching of G having exactly k edges in common
with F?
An instance of exact matching will be denoted by (G,F,k).
Similarly, the following decision problem is called exact factor
Instance: a graph G = (V,E), a subset of edges F c E and an integer k.
Question: does there exist a perfect factor of G having exactly k edges in common with
F?
An instance of exact factor will be denoted by (G,F,k).
The usual polynomial time reduction between problems is denoted by <. A random
algorithm is a Turing machine which uses coin flips during the computation. At each
step of the computation, a coin is tossed and then, from the outcome of the coin, the
state of the machine and the character read from the input tape, a decision is taken
(deterministically) concerning the new state of the machine and the direction in which the
head of the machine is to move. RP is the class of problems L which admit polynomial
random algorithms A such that on every instance / of L, the output 'yes' of A is always
correct, whereas the probability of the output 'no' being false is less than or equal to
1/2. For more details on the random complexity classes, see Johnson's chapter in Van
Leeuwen [14]. Every member of RP is said to be random polynomial.
The aim of this paper is to prove that AC is random polynomial by demonstrating a
polynomial reduction to the exact matching problem.
3. Preliminary results
In this section, some basic results on alternating cycles and the RP class are presented.
The following lemma was proved by Beineke and Little [4],
Lemma 3.1. Let B be a Hamiltonian bipartite tournament of order n. Then B contains
cycles of length k or k + 2 for all even k <n.
The 'alternating' version of this result yields the following:
Lemma 3.2. If Kcn is alternating Hamiltonian, then it is alternating bipancyclic.
Proof. Let xiX2...x n xi be an alternating Hamiltonian cycle of K% and A = {x,-,i even}
and B = {x,, i odd}. Next, consider every edge of colour 1 (resp. 2) of AB as directed
from A to B (resp. from B to A). The bipartite tournament so obtained is Hamiltonian
and contains, by the preceding lemma, a cycle of length k or k + 2 for every even k < n.
The conclusion of Lemma 3.2 follows immediately.
D
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R. Saad
Now let F = C\ U C2 U. •. U Cu (k > 2) be an alternating factor of Kcn with the maximum
size and the minimum number of cycles (among those of maximum size). For every i,
let us label the vertices of C,: C, = x'ox\x^...x'o so that c(x'ox\) = 1 and let us put
Xi = {x) I j even}, Y-, = {x) | j odd}.
Algorithm (2) in Benkouar et al. [2] describes an efficient way of obtaining such an F
and the following lemma is derived from it:
Lemma 3.3. Up to a permutation of the cycles C,, F satisfies:
1. For every i < j , c(XtXi) = c(XiCj).
2. For every i < j , c(y,T;) = c^C/).
3. c(XtXt) + ciYiYt).
The following theorem is due to Karp, Upfal and Wigderson [11], who designed an
efficient random algorithm to solve the exact matching problem:
Theorem 3.1. Exact matching is in RP.
Eventually, we shall need the following theorem, which is established as Theorem 2.6
of Benkouar, Manoussakis and Saad [3]:
Theorem 3.2. There is a O(n2)-time algorithm which eitherfindsa shortest alternating cycle
through two given vertices x and y in a 2-edge-coloured complete graph K%, or shows that
such a cycle does not exist.
4. Main result
Let K£ be an edge coloured complete graph, and consider the relation R on its vertices
defined as follows: xRy if and only if x = y or there exist two alternating x-y paths P
and P' such that:
P(x) = 1, P[y) = i and P'(x) = 2, P'(y) = j , j ± '•
O
Let us first observe the following facts about the relation R:
Fact 1. If C\, C2 are two alternating cycles of K.% having at least one common vertex and
if x e V(Ci), y G V(C2), then xRy.
Fact 2. If xRy then either x and y belong to an alternating cycle or there exist two
alternating x-y paths P and P' satisfying (*) and such that:
(i) V(P') c V(P).
(ii) The coloured subgraphs induced by V(P) — {x} and V(P) — {y} are both alternating
Hamiltonian.
Proof of Fact 1. Assume without loss of generality that |K(Ci)| < |K(C2)|. The proof is
by induction on |K(Ci)|, the length of the shortest cycle. Let us first prove that Fact 1 is
true for |K(C,)| = 4.
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Finding a Longest Alternating Cycle
301
First case: \V(Ci)\ = 4
Let us put C\ = XX1X2X3X and C2 = yy\y2---y, with c(yyi) = 1. If x e V(C2), there is
nothing to prove. Suppose that x £ V(C2). The case \V{C\) n F(C 2 )| = 1 is trivial. So let
us suppose that 2 < |F(Ci) n K(C2)| < 3. Let us put xi = }>,,, X2 = y,-2 and x 3 = yi} (with
the possibility that yi2 = y,3). We can suppose that (3 > i'i, otherwise relabel C2 in the
decreasing order of the }>,s and interchange colours 1 and 2. We shall also assume that
c(xxi) = 1.
Subcase a: c{xyh) = c(yhyh+1) and c{xyh) = c(yhyh+[)
is alternating and contains x
Suppose that c(xy) = 1. Then the cycle C = yyiy2---y^xy
and y. Similarly, if c(xy) = 2, the cycle xy^y^-i ...yx is alternating and contains x and y.
Thus xRy.
Subcase b: c(xyu) ± c(yhyh+l) or c(xyh) + c(yhyh+l)
Suppose that c(xyix) ^ c(yilyil+\) (the same proof will hold for the other subcase). Then
the x-y path P, = xyhyil+lyh+2...y
is alternating.
Thus, we can also suppose that c(xyi}) ^ c(_y,3y,3+i), for otherwise we have the alternating
x-y path P2 = xy,3y,3_i>';3_2...}' so that Pi and P2 satisfy (*).
Now if 12 is larger than i"i, the path P 3 = xy-^y^y^y^-xy-n-i...y
is alternating, so that
P 3 and Pi satisfy (*). So suppose i2 < i\.
Now if i2 is odd we obtain the alternating path P 4 = xy^y^y^-iy^^...y,
so that P4
and P2 satisfy (*). Similarly if i2 is even we obtain the path P 5 = xyilyilyi1-\yi2-2 • • • y so
that P5 and P\ satisfy (*). Thus xRy and the base of the induction is proved.
Second case: |K(Ci)| > 6
First suppose that C\ contains two consecutive vertices x,,x, + i such that:
{x,,x,+i} n V(C2) = 0 and c(x,x, +1 ) = j , with j e {1,2}.
Then give the colour m =£ j , m G {1,2} to the edge x,_ix,+2, delete the vertices x, and
x;+i and consider the cycles C[ = xx\x2...x,_ix,+2X,+3...x
and C2.
By the induction hypothesis, xRy in this new coloured graph, which implies that xRy
in the original coloured graph, since each alternating path using the edge x,_ix,+2 in the
new coloured graph corresponds to an alternating path in the original coloured graph
using the sequence x,-_ix,-x,-+ix,-+2So we can suppose that C\ has no two vertices x,,x, + i such that {x,,x, + i} n V(C2) = 0.
On the other hand, as the complete graph is 2-edge coloured, one of the cycles
XX1X2X3X and XX3X4X5...X is alternating. Let us denote it by C[. We know from the
preceding assumption that F(CJ)n V(C2) =/= 0 and so by the induction hypothesis we
conclude that xRy.
D
Proof of Fact 2. Let P and P' satisfy (*), and suppose that each of P and P' is inclusion
minimal. Assume \V(P)\ > \V(P')\.
If P(x) = P(y) = 1, then c{xy) = 2, for otherwise the edge xy is a shorter alternating
path than P. Hence, x and y lie on an alternating cycle.
If P(x) = 1, P(y) = 2 and \V(P)\ > 4:
Let us put P = xx]X2...X2i-i3'- Then from the minimality of P, we have for every 1
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R. Saad
such that i < k, c(xx2,-i) = 2 and c{x2i-\y) = 1, and hence the alternating path P" =
yx2i-ix2iX2i+ix so that P and P" satisfy (*) and V(P") <=. V(P). Moreover, we have two
alternating cycles spanning V(P) — {x} and V(P) — {y}, respectively:
xxxx2...X2k-\x
and yx2*-iX2*-2 • • • My-
This leaves only the case: \V(P)\ = 2. But then |V(P')I = 2 and x and y belong to an
alternating cycle.
D
Lemma 4.1. R is an equivalence relation.
Proof. It suffices to prove the transitivity of R. From Facts 1 and 2, we can suppose that
the paths involved in the relations xRy and yRz are internally vertex-disjoint. But then,
by concatenating them appropriately, we find x-z paths defining xRz.
Q
Notice that the above two facts yield the following alternative definition of the relation
R: xRy if and only if either x = y or there exist some vertex z e V(K°) and two
alternating cycles Cx and Cy (Cx may be identical to Cy) containing z such that x G V(CX)
and y e f(C y ). Indeed, Fact 1 shows that the existence of such Cx and Cy is sufficient to
conclude that xRy and Fact 2 shows that if xRy, either x and y belong to some alternating
cycle, or there exists an alternating x-y path P such that V(P) — {x} and V(P) — {y}
induce two alternating cycles containing, respectively, y and x, and having a nonempty
intersection. This yields the following (informal) algorithm for deciding whether two given
vertices x and j; satisfy xRy:
1. For every vertex z e V(Kn) — {x}, find an alternating cycle Cxz in Kcn containing x and
z by applying the O(n2 )-time algorithm of Theorem B.
2. If Cxz does not exist, conclude that xRy is not true else, by again applying the
algorithm of Theorem B, we may find an alternating cycle Cyz containing y and z.
3. Again, if Cyz does not exist, conclude that xRy is not true or else conclude that xRy.
Clearly, the time complexity of this algorithm is O(n3). In the sequel, any equivalence
class of R will be called a component. Computing the component of an x e V(Kn) reduces
to deciding for every vertex y whether xRy. Thus, by using the preceding algorithm to
recognise xRy for every y, we obtain an 0(n4)-time algorithm for building the component
of a vertex x.
Now, returning to the problem of the longest alternating cycle in Kcn, it is clear that
every alternating cycle C of K^ is included in a unique component. As the determination
of all the components of R is polynomial, we shall restrict the study to these components.
Lemma 4.2 given here is rather a straightforward consequence of Lemma 3.3. It was also
proved independently by Gutin [10] for the case of bipartite tournaments.
Lemma 4.2. If A is a component, then af(A) = ac(A).
Proof. Let F be a maximum alternating factor with the least number of alternating cycles
F = Ci U C2 U ... Q . Suppose k > 2.
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Finding a Longest Alternating Cycle
303
Here we return to the notation and labelling of Lemma 3.3.
In the sequel, we shall denote by a and b the respective lengths of C\ and Ck- The
parity of an integer will be taken to be 1 or 2 (0 is excluded).
From Lemma 3.3 above, we can suppose that:
1. For every i < j , c(X,X,) = c(XiCj).
2. For every i < j , dYft) = c(YtCj).
3. c[X,X,) + ciYM).
Suppose now, without loss of generality, that for every j < k, c{X\Xi) = c{X\Cj) = 1.
As A is a component, for every x\, i= l,...a, there is an x\-x\ alternating path P, with
Pi(xi) 7^ Parity of i. Let us take a smallest such P,: P, = x\y\y2...ymx\.
Observe first
that for every m, the edge coloured subgraph induced by V(Cm) is 'bipartite' in the sense
that the parities of the subscripts of any two consecutive vertices in an alternating path
of the complete edge-coloured subgraph induced by V(Cm) are distinct. Hence, if P is an
alternating x/-xj path included in V(Ci), with P(x/) ^ parity of /, then P{xXj) = parity
of/
As, for every j , c(xlj(F — C\)) = parity of j , there exists no x\-x\ alternating path
P a V(F) with P(xj) j= parity of i. That is, starting from any vertex x of V{Ci) with an
edge of colour different from the parity of x, we cannot find our way out of V(C\) by an
alternating path included in F. Moreover, from the minimality of P, and the preceding
parity argument, we know that y\ ^ V(F).
Let j be the first integer such that yj e V(F) and y ; _i g V(F) and let Cp be the
alternating cycle such that yj € Cp. In the sequel, the subscript q of the vertex x'q is always
taken modulo the length of the cycle C,.
Case p = 1
Let us denote by q the subscript such that xlq = yj.
From the minimality of P;, we must have c(xlqyj+\) = parity of q. Consequently
c(yj-iXq) ^ parity of q. Let a and /? denote the respective parities of i and q — i and
consider the alternating cycle:
C[ = xjyiy2...yj-iXqX1^
...x/+1x£+1x£+2...xka+bxq+^xlq+l3+1
...x\.
As C[ misses at most one vertex of C\ U Q (the vertex xq+1) and it contains at least a
vertex of Kn — F, we must have:
Now C{ U Ci U... Ck-i is a maximum alternating factor with fewer cycles than F, which
contradicts our assumptions on F.
Case p > 2
Let us again denote by q the subscript such that xpq = yj and by a the parity of i. Let cp
be the length of Cp and let r denote the integer such that r = cp — 1 if q and i do not
have the same parity and r — cp if q and i have the same parity.
If c{yj-\xpq) = parity of q, consider the alternating cycle:
C[ = x{yiy2...
yMxpqxpq+l...
xpq+rxi+lxi+2...
x,-.
Then C[ is again an alternating cycle of length at least |K(Ci| + | V(CP)|.
Now CJ U C2 U ... U Cu-i is a maximum alternating factor with fewer cycles than F,
which contradicts our assumptions on F.
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R. Saad
Similarly, if c{yj-\xpq) ^ parity of q, we have the following alternating cycle:
C\ = Xiy\yi...yj_\xpqxpq_x...xp_rxi+ixi+2...x,
and the same conclusion follows.
•
The following lemma is an easy generalisation of a known reduction from the perfect
factor problem to the perfect matching problem in graphs. Its proof will therefore be
omitted.
Lemma 4.3. Exact factor < exact matching.
Now, we can state our theorem.
Theorem 4.1. AC e RP.
Proof. From Theorem 3.1 and Lemma 3.3 we know that the exact factor problem is in
RP. We are going to reduce AC to the exact factor problem. Let (K^,k) be an instance
of AC. As every alternating cycle belongs to an equivalence class of R and finding such
classes is polynomial, we can suppose that K% consists of only one equivalence class.
Next, from Lemma 2.4 and Lemma 3.2, we know that K% contains an alternating cycle
of length / > k if and only if it contains an alternating factor of length k or k + 2.
Now construct a graph G = (V,E) as follows: First split each vertex x of K% into three
vertices labelled xi,X2,X3. Next do the following:
1. For every x € V(K^), let xi,X2,X3 induce K3.
2. Put Vi = {xi I x e V(Kcn)}, V2 = {x2 I x e V(Kl)}, K3 = { x 3 | x e V(Kcn)} and
V = Vi U V2 U V}.
3. Let V\ induce the graph G\ obtained from Kcn by removing the edges of colour 2,
V2 induce the graph G2 obtained from Kcn by removing the edges of colour 1 and F3
induce the independent graph on n vertices.
This describes the graph G (see Figure 1).
Next consider the subset of edges of G defined by F = [x\X2 e E(G) | x e V(K^)}.
We now claim that K% contains an alternating factor of length k or k + 2 if and only if
G contains a perfect factor with n — /corn — k — 2 edges in F.
Proof of our claim. If Kfj contains an alternating factor / of length /, then put all the red
edges o f / in Gi and the blue edges of/ in G2. Then add all the X1X3 and X3X2 edges, for
every x. Moreover, if a vertex x does not belong to / , add the edge xjX2- The subgraph
of G so obtained is a perfect factor of G with n — / edges in F.
Conversely, let g be a perfect factor of G having /' edges in F.
Notice that all the X1X3 and X2X3 edges belong to g, so that gfiGi and g n Gi contain
no path of length 2.
Now by removing the vertices xi, X2 such that x\x2 £ F, we obtain a matching in the
'rest' of G\ and another one in the the rest of G2, so that by 'gluing' them we obtain
an alternating factor of length n — V in K%- This proves the claim. Now, let A be a
random polynomial algorithm for the exact matching problem. Clearly, applying A on
the instances (G,F,n — k) and (G,F,n—k — 2) respectively and taking the decision of the
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Finding a Longest Alternating Cycle
305
Boolean sum of the decisions given by A, we obtain a random polynomial algorithm for
AC.
D
As a consequence of the theorem and the remark in the introduction that every bipartite
tournament can be viewed as a particular 2-edge-coloured complete graph, we obtain the
following corollary:
Corollary 4.1. BC is in RP.
/
//
w
\
\
Figure 1
5. Conclusion
Surprisingly enough, the problem of deciding on the maximum length of an alternating
cycle in a 2-edge-coloured complete graph is not known to be polynomial. We prove here
that it is random polynomial. Our proof is in fact a reduction to the exact matching
problem via a computation of the 'components' of an equivalence relation, the search for
a maximum alternating factor inside each component and finally a reduction to exact
factor and exact matching. Thus, our reduction is (roughly) in O(n4). As the search version
of the exact matching problem is in RP, the search version of our problem is also in
RP. Moreover, as exact matching is known to be in random NC, the question could be
raised as to whether our reduction could be efficiently parallelised. Indeed, the reduction
of Lemma 4.3 is in NC as is the reduction described in our main theorem. However,
we have not yet been able to efficiently parallelise the search of a maximum alternating
factor.
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R. Saad
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