W. S. C HOU
Y. M ANOUSSAKIS
O. M EGALAKAKI M. S PYRATOS Z S . T UZA
Paths through fixed vertices in edge-colored graphs
Mathématiques et sciences humaines, tome 127 (1994), p. 49-58
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PATHS THROUGH FIXED
VERTICES
IN EDGE-COLOREDGRAPHS
W.S.
CHOU1,2,
Y.MANOUSSAKIS3,
O.MEGALAKAKI4
M.
SPYRAT4S5,
and Zs.TUZA1,6
RÉSUMÉ 2014
Chaînes alternées passant par des sommets donnés dans desgraphes
arêtes-colorés.Nous étudions le
problème
de trouver dans ungraphe
arêtes-coloré une chaîne alternéejoignant
deuxsommets donnés et passant par des sommets donnés (une chaîne est alternée si deux arêtes
adjacentes
arbitraires ont des couleurs
différentes).
Plusprécisément
nous démontrons que ceproblème
est NP-complet
dans le cas degraphes
2-arêtes-colorés.Ensuite nous montrons que le
problème
de l’existence d’une telle chaîne estpolynomial
dans le casoù l’on se restreint aux
graphes complets
2-arêtes-colorés.Nous étudions
également
leproblème
de trouver une (s,t)-chaîne(c’est-à-dire
une chaîne delongueur
s+t
qui
se partage en deux sous-chaînesmonochromatiques
de couleursdifférentes) joignant
deuxsommets donnés et passant par des sommets donnés, dans un
graphe complet
arêtes-coloré.ABSTRACT2014 We
study
theproblem of finding
analternating path having given endpoints
and passingthrough
agiven
setof
vertices inedge-colored graphs
(apath
isalternating if
any two consecutiveedges
are indifferent
colors). Inparticular,
we show that thisproblem
inNP-complete for
2-edge-colored graphs.
Then we
give
apolynomial
characterization when we restrict ourselves to2-edge-colored complete graphs.
We also
investigate
on(s,t)-paths through fixed
vertices, i.e.paths of length
s+t such that sconsecutive
edges
are in one color and t consecutiveedges
are in another color.1 This
work was done while the authors werevisiting
theUniversity
ofOrsay.
2 Institute of Mathematics, Academia Sinica,
Nankang, Taipei
11529, Taiwan, R.O.C.3 Université Paris XI
(Orsay),
LRI, Bât.490, 91405Orsay
Cédex, France, e-mail :[email protected].
4
Department
of Social Sciences,University
of Crete, 74100Rethymnon,
Crete, Greece.5 Université Paris XII,
Département d’Informatique,
61, avenue du Général de Gaulle, 94100 Créteil, France, e-mail :spyratos @ univ-paris 12.fr.
6
Computer
and Automation Institute,Hungarian Academy
of Science, H-1111Budapest,
Kende u.13- 17,Hungary,
e-mail : [email protected].1. INTRODUCTION AND TERMINOLOGY
Research in social sciences often deals with relations of
opposite
content, e.g., "love" -"hatred",
"likes" -"dislikes",
"tells the truth to" - "lies to" etc. Agood
model for repre-senting
such relations is a so-calledsigned graph (a graph
in which we associate to eachedge
one of thesigns "+"
or"-"), i.e.,
a2-edge-colored graph.
In this work we deal withproblems directly
linked to the existence of two relationalpatterns
in a2-edge-colored graph:
thealternating paths (cycles)
and the(s, t)-paths (cycles). Pictorially
an alterna-ting path (cycle)
is apath (cycle)
any twoadjacent edges
of which are in distinct colors.An
(s, t)-path (cycle)
is apath (cycle) of length s
+ t such that s consecutiveedges
arein one color and t consecutive
edges
are in another color. The notion of an(s, t)-path (cycle)
isdirectly
related to the balance of agraph
introducedby Cartwright
andHarary
[7] (and originated
inpsychology [7, 9, 13, 17, 19]).
We recall that a2-edge-colored graph
is balanced
if,
andonly if,
in eachcycle
the number ofedges
with color "-" is even. Ina recent work
[16],
it has been shown that a2-edge-colored complete graph
contains an(s, t)-hamiltonian cycle (with s and t non-fixed) if,
andonly if,
thegraph
is unbalanced.In a further result the same authors showed
that,
for s and tfixed,
asufficiently large signed complete graph
contains an(s, t)-cycle if,
andonly if,
thegraph
is unbalanced.In this work we
study
theproblem of finding alternating
as well as(s, t)-paths having given endpoints
andpassing through
agiven
set of vertices in2-edge-colored graphs.
For fur-ther results on the
subject
the reader isencouraged
to consult[1-6, 8, 11,12, 14-16, 18, 20].
Formally,
in whatfollows,
unless otherwisespecified,
we denote the vertex-set, theedge-
set and the order of a
graph
Gby V(G), E(G)
andn(G), respectively.
Whenjust
onegraph
is underdiscussion,
weusually
writeV, E
and n instead ofV (G), E(G)
andn(G), respectively.
Let
A,
B denotenon-empty
subsets of V. Thegraph
induced in Gby
A is denotedby G[A].
The set of alledges
that have oneendpoint
in A and the other one in B is denotedby
AB. If A ={x},
then forsimplicity
we may write xB instead of{x}B.
A
k-edge-coloring (or,
forsimplicity,
ak-coloring)
of G is amapping
c from E onto the set of"colorsn
{l, 2, ..., k}.
If e EE(G),
thenc(e)
is the color of theedge
e. For any v E V andany color
i,
let the color-ineighborhood
of v be defined asNs(v) _ {a
EV1{v} ~ c(va)
=i}.
For any
non-empty
subset A ofV,
we defineNi(A)
=U N;(a).
We let Gc denote agraph
aEA
G colored
by
ak-edge-coloring
c. Acomplete graph .~1 n
coloredby
ak-edge-coloring
c isdenoted
by Kc
Let x and y be two distinct vertices of Gc and let s be a subset of
V (G~) 1 {x, y}.
Anarbitrary simple path
between xand y
in Gcpassing through
all vertices of S is denotedby
~r,
s, y. Whenever S containsonly
a fewvertices,
say ,~ ={zl,
z2,z3},
then forsimplicity,
we write
Px,
zl, z2,x3, ~ instead
ofPx, S,
y,replacing
S in the notationby
a sequence of its elements in any order.Let us note that all
paths
andcycles
considered in this paper aresupposed
to be elemen-tary.
Definition of an
(si,
s2, ~ ~ ~,8a+l )-path.
Let P = be a
path
in Gc.Suppose
that P is non-monochromatic and let{Xi- j 1 j = 1, 2, ..., al
denote the set of itsalternating vertices, where 1; ~+1) j = 1, 2, ..., a - 1.
The alternation sequence of P is defined to be the sequence of a + 1 terms, and P is called a If P is monochromatic its alterna- tion sequence is defined to be the one-term sequence 1 >.Observe that an
(sl,
s2, ~ ~ ~ , iSa+1)- path
P in Gc is:(1)
Monochromatic iff a = 0 and(2) alternating
iff a > 1and si = 1,
for aIl i= 1, 2,...,
a + 1.Here we
investigate
thefollowing problem:
PROBLEM 1.1. Let S =
f xi,
X2, ...,Xl} be a
setof 1 specified
vertices in ak-edge-colored graph
Gc. Let xand y
be two distinctfixed
vertices in S. Under which conditions does there exist a sequence(si,
s2, ,8a+1)
such that there exists an(si,
s2, ,8a+1)- path
between x and ycontaining
the verticesof
S ?In the sections that
follow,
westudy
thecomplexity
of the aboveproblem
for small values of~,
and for the twospecial
cases:2. NP-COMPLETENESS RESULTS
The
following
theorem of[15]
is used in this section.THEOREM 2.1. The
following problem
II isNP-complete.
Instance. A
complete graph K n ,
a set C= ~ 1, 2, ..., 1~} of k >
4colors,
ak-edge-coloring
c : C
four
distinct vertices a*i, X2, Yl, Y2 inIlg, a fixed permutation
q =
(ci,
c2, ~ ~ ~,ck ) of
the colorsof
C.Question.
Does contain twovertex-disjoint alternating paths f rom
xl to yiand from
X2to Y2
respectively,
such that the sequenceof
colorsof
each is a concatenationof
a numberof copies
We start with an
NP-completeness
result inedge-colored graphs.
THEOREM 2.2. Let G~ be a
2-edge-colored graph.
Let X, y and z be three distinct verticesDeciding
whether there exists analternating path from
xto y through z
inNP-complete.
PROOF. Our
problem obviously belongs
to NP. To prove that it isNP-complete,
wetransform the
following
so-called localpath problem (LPP) [10],
into an instance of ourproblem:
Given three distinct vertices x, y and z in a directedgraph D, deciding
if thereis a directed
path
from zto y through z
in D isNP-complete.
Consider now an
arbitrary
instance of LPP in a directedgraph
D. Let Gc denote the2-edge-colored graph
obtained from D as follows :Split
each arc of D into twoparts,
the firstpart (containing
the tail of thearc) being
colored 1 and the otherpart being
colored2;
furthermore add a new vertex x’ andjoin
this vertex with xby
anedge
in color 2.Clearly,
this construction can be done inpolynomial
time.Now,
if there is a directedpath Px, z,
y inD,
thenclearly
there is analternating path
y in Gc.
Conversely,
if there is analternating path Px’, z, y
inGc,
then we caneasily
deduce the existence of a directed
path Px, z,
y in D. Thiscompletes
theproof.
oTHEOREM 2.3. The
following problem
isNP-complete.
Instance. A
complete graph K,, ,
a set C =f 1, 2, ..., k} of
k > 4colors ,
ak-edge- coloring c :
Cof K,,,
three distinctvertices x, z and y
afixed permutation
q =
(cl,
c2, ~ ~ ~,ck) of
the colorsof
C.Question.
Does~in
contain analternating path Px,
z, y whose sequenceof
colors is theconcatenation
o f a
numberof copies of q ?
PROOF. The transformation is established from the
problem
fl of Theorem 2.1 above.Consider an
arbitrary
instanceof H, by fixing
four vertices xl, x2, yi , y2in Ilg
andby
as-suming,
without lossof generality,
that therequired
sequence of colors is q =(1, 2, ,
,k).
Let now denote a
k-edge-colored complete graph
obtained fromKc by adding
k -1new vertices wi, W2, ..., wk_i and the
corresponding edges
and thencoloring
theedges by
c* : C which is an extension of c defined as follows:(1) maz((1, j}) if 1 i - j 1=
1 andc*(wiwj) =
1if 1 i - j 1> 1, i, j =
(4)
Theedges
between Wl andY(n)-{yl},
and betweeneach wi
andV(K in), 2 i k-2,
are colored k.
(5)
Theedges
between and{~2}
are colored 1.Clearly,
the above transformations can be done inpolynomial
time.Fix now three vertices
x’, y, Z
inby xl, y’ - y2, z’ -
W2. Itis easy to see that contains an
alternating path Pxr,
zl, y, whose sequence of colors is the concatenation of a number ofcopies of q if,
andonly if,
contains two vertex-disjoint alternating paths
from x1 to yi and from X2 to Y2respectively,
such that the sequence of colors of each is a concatenation of a number ofcopies
of q. o PROBLEM 2.4. Is thefollowing problem NP-complete?
Instance. A
complete graph
a set C ={ l, 2, ..., k} of
k > 4colors,
ak-edge- coloring
c : Cof J(n’
three distinct vertices x, z and y in apositive integer
t.
Question.
Doesh’n
contain analternating path ~c,
z, y such that each color appears at least t times on3.
POLYNOMIAL CHARACTERIZATIONS
In the last section we have shown that the
alternating Pp
z,y-path problem
with a par-ticular sequence of colors is
NP-complete
fork-edge-colored complete graphs, k
> 4. In Theorem 3.2 andCorollary
3.3 of thissection,
we show that thealternating Px, z, y-path
problem
is nolonger NP-complete,
if we restrict ourselves to the case of2-edge-colored complete graphs. However,
we note that in the case of threecolors,
theproblem
is stillopen.
In view of the
proof
of Theorem 3.2below,
we prove thefollowing
lemma.LEMMA 3.1. Let x, z, y be three
specified
distinct vertices in ak-edge-colored
com-plete graph
> 2. There exists analternating path Px,
z, yif,
andonly if,
there existsan
alternating path containing
at least oneof
theedges
xz, zy in such thatz, y)’
PROOF. The existence of
Px,
z, yclearly implies
the existence of in Conver-sely,
suppose that contains analternating path .Px
z, y and suppose it to be of shortestlength.
SetP~,
z, y = z; ... xp y, where for the sake ofhomogeneity
weidentify z
with Xi on this
path.
Assumeby
contradiction that 1 i p. Let q, r denote the colors of theedges
Xi-l Xi andrespectively.
Since y isalternating, q =1=
r. If theedge
xxi is colored otherwise than r, then is shorter than
7~,
z, y, a contradiction to theminimality property
ofPx,
z, y.Similarly
we obtain a contradiction if we assumethat the color of the
edge
YXi is other than q. It follows that thepath
xxiy is alterna-ting
and shorter than y, a final contradiction. Thiscompletes
theproof
of the lemma.THEOREM 3.2. Let x, y, z be three distinct vertices in a
2-edge-colored complete graph
There is an
alternating path
ini f,
andonly i f,
oneo f
thefollowing
threeconditions holds:
(2) c(xz)
=c(yz)
=1,
say, and eitherN1(x)
fl orN1(y)
flN2(z) =1= 0.
(3) c(xz)
=c(yz)
=1,
say,N1(x)
nN2(z) _ ~
=N1(y)
nN2(z),
and in thefollowing
sequence
of
subsetsof
there is apositive integer t
= 2s -b, b
= 0 or1,
so thatCt =1= 0
and eitherCt
is not contained inN1+6(X)
n or thecomplete subgraph of
~in generated by
U ’ " UC2s-6
is not monochromatic:PROOF.
SuiHciency:
If(1) holds,
thenclearly
thepath
xzy isalternating.
If(2) holds,
thentake u E
Nl(x)
f1N2(z) =f 0 (respectively u
ENl(y)
f1N2(Z) 0 0)
and thepath
xuzy(respectively
thepath xzuy)
isalternating. Suppose
that(3)
holds. Let t = 28 - 6(with
b = 0 or
1)
be the smallestpositive integer satisfying (3).
ThenCi = 0
for all 1 i t.Since
Ci
CNl(x)
flNl(y)
if 1 i t is even andCi
CN2(x)
flN2(y)
if 1 i t isodd,
it follows that the setsCl, C2, ~ ~ ~ ,
?Ct
arepairwise disjoint.
IfCt
is not containedin
NI+6(X)
flNl+b(y),
take zt inCt 1 (NI+6(X)
flNl+a(y)) (we
take zt = x if x ECt,
or zt = y if y e
Ct).
Take E so thatc(zt_izt)
= 1 +b,
Zt-2 ECt-2
so that= 2 - à and so on, and zi E
Ci
so thate(ziz2)
= 1. If zt EN2-6(X),
thenzi zy is the desired
path. If zt
E then zt y is the desiredpath. Finally,
suppose thatCt
CNI+6(X)
nNl+b(y)
but thecomplete subgraph
ofKc generated by
U ~ ~ ~U C2s-b
is not monochromatic. Take zt, in U ~ ~ ~UC2s-6
so that = 2 - b. Since the
complete graph
U ~ ~ ~ U is coloredby
1
+ b
andCt
CNl+a(x)
nNl+b(y),
we have zt, eCt. Now,
take zt-1, zl asabove. Then is the desired
path.
Necessity:
LetPx, z, y
denote analternating path
of shortestlength
from x to ythrough
z in
Kc.
From Lemma3.1, Px, z,
y is either of the form XZZ1 ... ZkY or of the form xzk...zlzy.Without loss of
generality,
we may assume that = XZZ1 ... ZkY- IfPx,
z, y is of theform xzy, then
(1)
holds. IfPx,z,y
is of the form xzzly, then(2)
holds.Now,
sup-pose k
> 2 and let us assume, without loss ofgenerality,
thatc(xz)
= 1. In this case,Ni (y)
flN2 (Z) = 0
=Nl(x)
f1N2(z),
since otherwise there would be analternating path
of
length
3 from x to ythrough
z.So,
bothNl(x)
andNl(y)
are contained inNl(z).
Construct the finite sequence of subsets of
Kc
as in the statement of thistheorem. Set k - 1
= 2s + ~, ~ = 0 or 1.
From thesufficiency part
and the fact that k is the smallest among allpossible lengths
ofalternating paths
from x to ythrough
z,Cl+b
U ~ ~ ~ UC2i-I+6
is contained in and is monochromatic for all 1 i s, andC2-6
U ... U is contained inNl+b(x)
nNI+6(y)
and is monochro-matic for all 1 i s.
Moreover,
U ... U UC2s+a
is also contained inis not monochromatic we are done from the
sufficiency part. So,
we assume that U ~ ~ ~ UC2.,-b
UC2s+b
is monochromatic.Construct
Ck
as in the statement of the theorem.Then, zk E Ck.
Note that zi ECi
for all1 i k - 1. Also note that
c(Zk- 1 Zk)
= 2 - ~.If Ck
were contained inN2- 6(Y)
the
path
zzzi ... would not bealternating, contradicting
ourassumption. Therefore, Ck
is not contained in and theproof is complete.
ilCOROLLARY 3.3. There is an
algorithm of complexity 0(n3) for finding
an alterna-ting path
y(if any) in
a2-edge-colored complete graph.
PROOF.
Clearly,
Condition(2)
of Theorem 3.2requires 0(n’)
time. Condition(3)
needs0(n3) time,
since thelength
of the sequence ofCi’s
is0(n),
whilespecifying
eachCi
re-quires
at most0 (n 2) operations;
therefore the decisionalgorithm
costs0(n3) operations
in the worst case. The
algorithm
forfinding
the desiredpath (if any)
followsdirectly
fromthe
proof
of Theorem 3.2. ilWe shall finish this section with a result on
(s, t)-paths through
fixed vertices. In or-der to state the next theorem we define a
2-edge-colored complete graph Go
as follows(see Figure 1):
The
edges
xyof
both(?i
andG2,
omittedhere,
are coloredarbitrarily
either 1 or 2.Dashed lines
represent
color1,
while solid linesrepresent
color 2.Figure
1Partition the vertex set of a
complete (non-colored) graph
of order n > 5 into three subsetsVi, V2
andV3
suchthat 1 V3 1 == 3,
sayV3
={.r,
y,w}.
Then color theedges between Vi
and
V2 by 1,
theedges
betweenV3
andVi
UV2 by
2 and color one of theedges
wx, wyby
color 1. AU other
edges
are coloredarbitrarily. Clearly
the obtained2-edge-colored graph Go
has no(8, t)-path
between x and ycontaining
a vertex Zl ofVi
and a vertex Z2 ofV2.
THEOREM 3.4. Let be a
2-edge-colored complete graph, n >
4. Let x, y, zi , Z2 bedistinct fixed
vertices(i)For
some sand t,
there exists an(s, t)-path Px, zl,
y in~in if,
andonly if, Iin
containstwo distinct
edges
el and e2, other than x,y, such thatel is adjacent to
x,e2 is adjacent to
y
and c( el) =1 c(e2).
For some s
and t,
there exists an(s, t)-path ~.~
y in~~n if,
andonly if,
thereexist the
edges
el and e2of (i~
and in additionIin is
notisomorphic to
anyof Go, GI, G2 of Figure 3 (isomorphism
here is considered in the usual sence andby taking
into accountthe colors
of edges.
PROOF.
Necessity being obvious,
let us prove the "if’ case.Proof of
(i).
Assume without loss ofgenerality
that theedge
XZI is in color 1. If theedge
yzi is in color2,
we have finished. Otherwiseby
thehypothesis
there is a vertex w insuch that at least one of the
edges
wy, wx is in color 2.Now, independently
ofthe color of the
edge
wzi , ifc(wy)
= 2 thenPx, xl,
y = otherwiseP~,
Zl, y = xwzly.Proof of
(ii).
Let us assume without loss ofgenerality
thatC(ZIZ2)
= 1.c(yz2),
or
clearly
thepath
or XZ2ZlY is the desired one.Consequently,
in what follows assume
C(XZI)
=C(YZ2)
andc(xz2)
=c(yzl).
Assume first
c(yzi)
=C(XZ2)
=c(xzl)
=c(yz2)
= 1.By
thehypothesis,
there exists avertex w in
{.r,
y, zi ,z2}
such that either wy or wx is in color 2.Now, independently
of the color of the
edge
WZ2, either thepath
or thepath
XWZIZ2Y is the desiredone.
Assume next
C(XZ1)
= 1 = andC(XZ2)
= 2 =c( yzl ).
If n =4,
thenKc
isisomorphic
to
G1.
In thesequel suppose n >
5. If there is a vertex w in R =Ilg - {x,
y, Zl,Z21
such that
c(wy)
=2,
thenclearly
thepath
xzlz2wy is the desired one. Assume therefore that alledges between y
andV(R)
are in color 1.Similarly
we may also assume that alledges
between x andV(R)
are in color 1. Now if there is a color-1edge
between z2(or zl )
and a vertex say w ofR,
then thepath (or
thepath XZ2Z1WY)
is the desiredone.
Otherwise, i.e.,
if alledges
betweenz2}
andV(R)
are in color 2then,
in casen =
5, Kg
isisomorphic
toG2
and in case n > 6 thepath xz2wziw’y
is the desired one, where w and w’ are twoarbitrarily
chosen vertices of R.Assume
finally c(xzl)
=C(YZ2)
=C(XZ2)
=c(yzi)
= 2. Let S denote thesubgraph
inducedby
the color -1edges
ofKc
and letV1
denote the component of S which contains both zl and z2(z,
and Z2belong
to a samecomponent
ofS,
sinceC(Zl Z2)
=1, by assumption).
Set
V2 = V B Vi. Clearly,
ifV2
is not theempty
set, alledges
betweenV1
andV2
are incolor 2 in If at least one of x, y
belongs
toVl,
say x EVl,
then there exists a color-1path having
x as oneendpoint
and z, or Z2 as the otherendpoint
andcontaining
both z,and z2.
Indeed,
sinceVI
is a connectedcomponent
ofS,
there exists a color-1path,
sayP,
between x and z, inVI (and
therefore inK").
If Z2 is an internal vertex ofP,
then Pis the desired
path. Otherwise,
we extend P to a new color-1path P’
between x and z2containg
zl,by adding
theedge
ziz2. Thepath
P(or P’)
can beeasily
transformed toan
(s, t)-path
between x, ypassing through
both zl and z2by adding (the color-2) edge
zly
(or Z2Y). Suppose
therefore thaty} C V2. By
thehypothesis
of case(ii)
thereexists w E
V B {x, y~
such that one of theedges
wx, wy is in color 1. Now if w were a vertex inVI
then one of the vertices x, y wouldbelong
toVi,
a contradiction to ourassumption.
Thus w EV2
and thereforeV2
contains at least 3 vertices. Now ifV2
has atleast 4
vertices,
then thepath
YWZlZZ2X(or
thepath XWZlZZ2Y)
is the desired one, wherez denotes any vertex in
V2 B lx,
y,w}. Consequently,
in thesequel
assume thatV2
hasprecisely
3vertices, namely
x, y and w.Now,
if there is a monochromatic color-2path
between z, and z2, say ZI t1 ... in
Vl,
then satisfies the conclusion of the theorem.Otherwise,
the vertices ofV1
can bepartitioned
into 2 subsetsTl
andT2
such that zl E
Tl,
z2E T2
and alledges
betweenTl
andT2
are in color 1. In this finalcase
Kc
isisomorphic
toGo.
ciLet us notice that the
proof
of the above theorem can beeasily
transformed to analgo-
rithm for
finding,
the desiredpaths (if any).
Moreprecisely,
theproof
of(i)
hascomplexity 0(n),
while theproof of(ii)
hascomplexity 0(n2).
The factor0(n2)
in(ii)
comes from thefact that we have to check if the
graph
inducedby
the color-1(color-2) edges
is connected.It also comes from the determination of a color-2
path
between zi and Z2 inVi B
w,y}.
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