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Chudnovsky, M, Lo, I, Maffray, F et al. (2 more authors) (2019) Coloring square-free Berge
graphs. Journal of Combinatorial Theory, Series B, 135. pp. 96-128. ISSN 0095-8956
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Coloring square-free Berge graphs
Maria Chudnovsky
∗Irene Lo
†Fr´
ed´
eric Maffray
‡Nicolas Trotignon
§Kristina Vuˇskovi´
c
¶August 8, 2018
Abstract
We consider the class of Berge graphs that do not contain an induced cycle of length four. We present a purely graph-theoretical algorithm that produces an optimal coloring in polynomial time for every graph in that class.
Keywords: Berge graph, square-free, coloring, algorithm
1
Introduction
A graphGisperfect if every induced subgraphH ofGsatisfiesχ(H) =ω(H), where χ(H) is the chromatic number of H and ω(H) is the maximum clique size inH. In a graph G, a hole is an induced cycle with at least four vertices and anantihole is the complement of a hole. We say that graphGcontains a graphF, ifF is isomorphic to an induced subgraph ofG. A graphGisF-free
if it does not containF, and for a family of graphsF,GisF-freeifGisF-free for every F ∈ F. Berge [2, 3, 4] introduced perfect graphs and conjectured that a graph is perfect if and only if it does not contain an odd hole or an odd antihole. ABerge graph is any graph that contains no odd hole and no
∗Mathematics Department, Princeton University, NJ, USA. Partially supported by NSF grants DMS-1265803 and DMS-1550991. This material is based upon work supported in part by the U. S. Army Research Laboratory and the U. S. Army Research Office under grant number W911NF-16-1-0404.
†Department of Industrial Engineering and Operations Research, Columbia University, New York, NY, USA.
‡CNRS, Laboratoire G-SCOP, University of Grenoble-Alpes, France. Partially supported by ANR project STINT under reference ANR-13-BS02-0007
§CNRS, LIP, ENS, Lyon, France. Partially supported by ANR project STINT under refer-ence ANR-13-BS02-0007, and by the LABEX MILYON (ANR-10-LABX-0070) of Universit´e de Lyon, within the program Investissements d’Avenir (ANR-11-IDEX-0007) operated by the French National Research Agency (ANR)
odd antihole. This famous question (the Strong Perfect Graph Conjecture) was solved by Chudnovsky, Robertson, Seymour and Thomas [7]: Every Berge graph is perfect. Moreover, Chudnovsky, Cornu´ejols, Liu, Seymour and Vuˇskovi´c [6] devised a polynomial-time algorithm that determines if a graph is Berge.
It is known that one can obtain an optimal coloring of a perfect graph in polynomial time due to the algorithm of Gr¨otschel, Lov´asz and Schrijver [12]. This algorithm however is not purely combinatorial and is usually considered impractical. No purely combinatorial algorithm exists for coloring all Berge graphs optimally and in polynomial time.
The length of a path or cycle is the number of its edges. In what follows we will use the term “path” to mean “induced (or chordless) path”. For a path
P with ends a, b, the interior of P is the setV(P)\ {a, b}; the interior of P
is denoted byP∗. We letC
k denote the hole of length k (k≥4). The graph
C4 is also referred to as a square. A graph is chordal if it is hole-free. It is
well-known that chordal graphs are perfect and that their chromatic number can be computed in linear time (see [11]).
Farber [10], and later Alekseev [1], proved that the number of maximal cliques in a square-free graph onnvertices isO(n2). Moreover it is known that
one can list all the maximal cliques in a graph G in time O(n3K), where K
is the number of maximal cliques; see [20, 18] among others. It follows that findingω(G) (the size of a maximum clique) can be done in polynomial time for any square-free graph, and in particular findingχ(G) can be done in polynomial time for a square-free Berge graph. Moreover, Parfenoff, Roussel and Rusu [19] proved that every square-free Berge graph has a vertex whose neigbhorhood is chordal, which yields another way to find all maximal cliques in polynomial time. However getting an exact coloring of a square-free Berge graph is still hard, and this is what we do. The main result of this paper is a purely graph-theoretical algorithm that produces an optimal coloring for every square-free Berge graph in polynomial time.
Theorem 1.1 There exists an algorithm which, given any square-free Berge graphGonnvertices, returns a coloring of Gwithω(G)colors in timeO(n9).
A prism is a graph that consists of two vertex-disjoint triangles (cliques of size 3) with three vertex-disjoint paths P1, P2, P3 between them, and with no
other edge than those in the two triangles and in the three paths. Note that if two ofP1, P2, P3 have lengths of different parities, then their union induces an
odd hole. So in a Berge graph, the three paths of a prism have the same parity. A prism iseven (resp.odd) if these three paths all have even length (resp. all have odd length).
Theorem 1.2 ([17]) There exists an algorithm which, given any graph G in classAon nvertices, returns a coloring of Gwith ω(G)colors and a clique of sizeω(G), in timeO(n6).
Note that every antihole of length at least 6 contains a square; so a square-free graph contains no such antihole.
Since Theorem 1.2 settles the case of graphs that have no prism, we may assume for our proof of Theorem 1.1 that we are dealing with a graph that contains a prism. The next sections focus on the study of such graphs. We will prove that whenever a square-free Berge graphGcontains a prism, it contains a cutset of a special type, and, consequently, thatG can be decomposed into two induced subgraphs G1 andG2 such that an optimal coloring ofG can be
obtained from optimal colorings ofG1 andG2.
Note that results from [16] show that finding an induced prism in a Berge graph can be done in polynomial time but that finding an induced prism in general is NP-complete.
In [15], it was proved that when a square-free Berge graph contains no odd prism, then either it is a clique or it has an “even pair”, as suggested by a conjecture of Everett and Reed (see [9]). However, this property does not carry over to all square-free Berge graphs; indeed it follows from [13] that the line-graph of any 3-connected square-free bipartite line-graph (for example the “Heawood graph”) is a square-free Berge graph with no even pair.
We finish this section with some notation and terminology. In a graph G, given a setT ⊂V(G), a vertex ofV(G)\T iscomplete toT if it is adjacent to all vertices ofT. A vertex ofV(G)\T isanticomplete toT if it is non-adjacent to every vertex ofT. Given two disjoint setsS, T⊂V(G),S iscompletetoT if every vertex ofSis complete toT, andS isanticompletetoT if every vertex of
S is anticomplete to T. Given a cycle, any edge between two vertices that are not consecutive along it is achord. A cycle that has no chord ischordless.
Theline-graph of a graphH is the graphL(H) with vertex-setE(H) where
e, f∈E(H) are adjacent inL(H) if they share an end inH.
In a graph J, subdividing an edge uv ∈ E(J) means removing the edge
uv and adding a new vertex w and two new edges uw, vw. Starting with a graphJ, the effect of repeatedly subdividing edges produces a graph H called asubdivision ofJ. Note thatV(J)⊆V(H). Abipartite subdivisionof a graph
J is any subdivision of J that is bipartite.
Lemma 1.3 Let Gbe square-free. LetK be a clique inG, possibly empty. Let
X1, X2, . . . , Xk be pairwise disjoint subsets ofV(G), also disjoint fromK, such that Xi is complete to Xj for all i=6 j, and let X =SiXi. Suppose that for everyv in K, there is an integeriso that v is complete to X\Xi. Then there is an integerisuch that (K∪X)\Xi is a clique in G.
Proof. First observe that there exists an integerjsuch thatX\Xjis a clique, for
otherwise two ofX1, . . . , Xk are not cliques and their union contains a square.
Now we claim thatK is complete to at leastk−1 of theXi’s. For suppose
on the contrary that K is not complete to any ofX1 andX2. Then there are
verticesv1, v2∈K,x1∈X1,x2∈X2such that fori∈ {1,2}andj∈ {1,2}\{i}, vi adjacent toxi and non-adjacent to xj. By the assumption, v1 6=v2. Then
{v1, x1, x2, v2} induces a square, contradiction. Hence there exists an index h
such thatKis complete to X\Xh.
Suppose that the lemma does not hold. Then j 6=hand there are vertices
x, x′ ∈X
j,v∈K,w∈Xhsuch thatxandx′are non-adjacent andvandware
non-adjacent. Then {x, v, x′, w} induces a square, contradiction. This proves
the lemma.
In a graph G, we say (as in [7]) that a vertexv can belinked to a triangle
{a1, a2, a3}(via pathsP1, P2, P3) when: the three pathsP1, P2, P3are mutually
vertex-disjoint; for eachi ∈ {1,2,3}, ai is an end of Pi; for all i, j ∈ {1,2,3}
withi 6=j, aiaj is the only edge between Pi and Pj; and v has a neighbor in
each ofP1, P2, P3.
Lemma 1.4 ((2.4) in [7]) In a Berge graph, if a vertex v can be linked to a triangle{a1, a2, a3}, then v is adjacent to at least two ofa1, a2, a3.
2
Good partitions
In a graphG, atriad is a set of three pairwise non-adjacent vertices.
Agood partition of a graphGis a partition (K1, K2, K3, L, R) ofV(G) such
that:
(i) LandR are not empty, andLis anticomplete toR;
(ii) K1∪K2andK2∪K3 are cliques;
(iii) IfP =p1− · · · −pk is a path withp1∈K1,pk∈K3,k≥3 andP∗⊆L,
thenp2is complete to K1.
(iv) EitherK1 is anticomplete toK3, or for every v∈Lthe set N(v)∩K1 is
complete toN(v)∩K3;
(v) For somex∈Landy∈R, there is a triad ofGthat contains{x, y}.
Theorem 2.1 LetGbe a square-free Berge graph. IfGcontains a prism, then
Ghas a good partition.
The proof of this theorem will be given in the following sections, depending on the presence inGof an even prism (Theorem 4.2), an odd prism (Theorem 5.2), or the line-graph of a bipartite subdivision ofK4 (Theorem 6.1).
Lemma 2.2 LetGbe a square-free Berge graph. Suppose thatV(G)has a good partition(K1, K2, K3, L, R). LetG1=G\R andG2=G\L, and for i= 1,2
letci be anω(Gi)-coloring of Gi. Then an ω(G)-coloring of Gcan be obtained in polynomial time.
Proof. We may assume (by makingK2 maximal) that no vertex ofK3is
com-plete toK1. SinceK1∪K2is a clique, by permuting colors we may assume that c1(x) =c2(x) holds for every vertexx∈K1∪K2.
Say that a vertex u in K3 is bad if c1(u) 6= c2(u), and let B be the set
of bad vertices. If B = ∅, we can merge c1 and c2 into a coloring of G and
the lemma holds. Therefore let us assume that B 6=∅. We will show that we can produce in polynomial time a pair (c′
1, c ′
2) ofω(G)-colorings ofG1 andG2,
respectively, that agree onK1∪K2 and have strictly fewer bad vertices than
(c1, c2). Repeating this argument at most|B| times will prove the lemma.
For each h∈ {1,2} and for any two distinct colorsi andj, let Gi,jh be the bipartite subgraph ofGhinduced by{v∈V(Gh)|ch(v)∈ {i, j}}; and for any
vertexu∈K3, letChi,j(u) be the component ofGi,jh that containsu.
Letu∈B, with i=c1(u) and j =c2(u). Then Chi,j(u)∩K2 =∅ for each h∈ {1,2} because uis complete toK2. Say thatuis free ifChi,j(u)∩K1=∅
holds for someh∈ {1,2}. In particularuis free whenever colorsiandjdo not appear inK1.
We may assume that there is no free vertex. (1)
Suppose thatuis a free vertex, withC1i,j(u)∩K1=∅say. Then we swap colors iandjonC1i,j(u). We obtain a coloringc
′
1ofG1where the color of every vertex
inK1∪K2 is unchanged, by the definition of a free vertex; soc′1 and c2 agree
on K1∪K2. For all v ∈ K3\B we have c1(v) 6= i, because c1(u) = i, and c1(v) 6=j, because c1(v) =c2(v)6=c2(u) =j; so the color of v is unchanged.
Moreover we havec′
1(u) = j =c2(u), soc′1 and c2 agree on u. Hence the pair
(c′
1, c2) has strictly fewer bad vertices than (c1, c2). Thus (1) holds.
Choosewin B with the largest number of neighbors inK1. Then:
Every vertexu∈B satisfiesN(u)∩K1⊆N(w)∩K1. (2)
For suppose that some vertex x∈ K1 is adjacent to uand not to w. By the
choice ofwthere is a vertex y∈K1 that is adjacent tow and not tou. Then
{x, y, u, w} induces a square, a contradiction. Thus (2) holds.
Up to relabelling, letc1(w) = 1 andc2(w) = 2. By (1)wis not a free vertex,
soC11,2(w)∩K16=∅ andC21,2(w)∩K1 6=∅. Hence for some i∈ {1,2} there is
a pathP =w-p1-· · ·-pk-ain C11,2(w), with k≥1,p1∈K3∪L,p2, . . . , pk ∈L
anda∈K1 with c1(a) =i; and for somei′ ∈ {1,2} there is a pathQ=w-q1
-· -· -·-qℓ-a′ in C21,2(w), with ℓ≥1,q1 ∈K3∪R, q2, . . . , qℓ∈R anda′ ∈K1 with c2(a′) =i′. It follows that at least one of the colors 1 and 2 appears inK1. We
claim that:
For suppose that there are verticesa1, a2∈K1 withc1(a1) = 1 andc1(a2) = 2.
We know thatwis anticomplete to{a1, a2}. SincePis bicolored byc1, it cannot
contain a vertex complete to{a1, a2}; so, by assumption (iii),P does not meet L. This implies that P = w-p1-a1 and p1 ∈ K3. Then c2(p1) 6= 2, because c2(w) = 2, and sop1 ∈B; but then (2) is contradicted sincewis non-adjacent
toa1. Thus (3) holds.
By (3) we havei=i′ anda=a′. Let j= 3−i. Note that ifi= 1 then P
has even length andQhas odd length, and ifi= 2 then P has odd length and
Qhas even length. So P andQ have different parities. Ifp1∈L andq1 ∈R,
thenV(P)∪V(Q) induces an odd hole, a contradiction. Hence,
At least one ofp1 andq1is in K3. (4)
We claim that:
There is no vertexy inK3 such thatc1(y) = 2 andc2(y) = 1. (5)
Suppose that there is such a vertexy. Ifp1∈K3andq1∈K3, thenp1=y=q1
and (V(P)∪V(Q))\ {w} induces an odd hole, a contradiction. So, by (4), exactly one of p1 and q1 is in K3. Suppose that p1 ∈ K3 and q1 ∈ R. So p1 =y, and in particularp1ais not an edge. If p1 has no neighbor onQ\w,
thenV(P)∪V(Q) induces an odd hole. So suppose thatp1 has a neighbor on Q\w. Then there is a path Q′ from p
1 to a′ with interior inQ\w, and since
it is bicolored by c2 the parity of Q′ is different from the parity of Q. Then
(V(P)\ {w})∪V(Q′
) induces an odd hole, a contradiction. Whenp1∈Land q1∈K3 the proof is similar. Thus (5) holds.
p1∈/ K3. (6)
For suppose thatp1 ∈K3. We havec2(p1)6= 1 by (5) and c2(p1)6= 2 because c2(w) = 2. Hence letc2(p1) = 3. So color 3 does not appear inK2. Note that
every vertex inK3\B has its color different from 1,2,3 because ofwandp1.
Suppose that color 3 does not appear inK1. Then, by (3),C2j,3(p1)∩K1=
∅. We swap colors j and 3 on C2j,3(p1). We obtain a coloring c′2 of G2 such
that the color of all vertices inK1∪K2 is unchanged, soc′2 agrees withc1 on K1∪K2. For every vertexv in K3\B we havec2(v)6= 3, because c2(p1) = 3,
and c2(v) ∈ {/ 1,2}, because c2(v) = c1(v) and {1,2} = {c1(w), c1(p1)}; so c′
2(v) =c2(v). Moreover,c′2(p1) =j. Ifj= 1, thenc′2(w) =c2(w) and the pair
(c1, c′2), contradicts (5) (withy=p1). Ifj= 2, thenc′2(p1) =c1(p1), so the pair
(c1, c′2) has strictly fewer bad vertices than (c1, c2). Therefore we may assume
that there is a vertexa3 inK1withc1(a3) = 3.
Vertexp1is not adjacent toa3becausec2(p1) =c2(a3), andp1is not adjacent
toaby (2) and becausewis not adjacent toa. This impliesk≥2, so the path
P\w meets L. Assumption (iii) implies that P \w contains a vertex that is complete to{a, a3}, and sinceP is a path, that vertex ispk.
Suppose that a3 has a neighborpg onP\ {w, pk}, and choose the smallest
contains no vertex that is complete to {a, a3} because a has no neighbor on P \pk, so assumption (iii) is contradicted. Therefore a3 has no neighbor on P\ {w, pk}.
Suppose that i= 1. ThenP has even length, and by (3) color 2 does not appear in K1. If w is adjacent to a3, then since k ≥2, we see that (V(P)\
{a})∪ {a3}induces an odd hole. So wis non-adjacent toa3. Hence {a, a3} is
anticomplete to{w, p1}. Since, by (1), p1 is not a free vertex, and color 2 does
not appear inK1, there is a pathSbetweenp1anda3inC22,3(p1), andShas even
length becausec2(p1) =c2(a3). If w∈V(S), thenV(S)∪ {p1, . . . , pk} induces
an odd hole. Ifw /∈V(S), then the interior ofSis inR, andV(S)∪ {p2, . . . , pk}
induces an odd hole.
Now suppose thati= 2. By (1), p1 is not a free vertex, so there is a path T from p1 to{a, a3}inC12,3(p1). SinceT is bicolored byc1it cannot contain a
vertex that is complete to {a, a3}, so assumption (iii) implies thatT does not
meetL. So we haveT =p1-x-a for some vertexx in K3 with c1(x) = 3. We
havec2(x)6= 3 because c2(p1) = 3; sox∈B. But the fact thatais adjacent to xand not towcontradicts (2). Thus (6) holds.
By (4) and (6) we havep16∈K3 andq1∈K3. In particular,P meetsL. We
havec1(q1)6= 1 becausec1(w) = 1, andc1(q1)6= 2 by (5). Hence letc1(q1) = 3.
Note that every vertex inK3\B has its color different from 1,2,3 because of wandq1.
Color 3 appears inK1. (7)
Assume the contrary. Then, by (3),C1j,3(q1)∩K1=∅. We swap colorsj and 3
onC1j,3(q1). We obtain a coloring c′1 of G1 such that the color of every vertex
in K1∪K2 is unchanged, so c′1 agrees withc2 onK1∪K2. Also every vertex v in K3\B satisfies c′1(v) = c1(v). Moreover, c′1(q1) = j. If j = 1, then c′
1(q1) =c2(q1), so the pair (c′1, c2) has strictly fewer bad vertices than (c1, c2).
If j = 2, then c′
1(w) = c1(w) = 1, so the pair (c′1, c2) contradicts (5) (with y=q1). Thus we may assume that (7) holds.
By (7) there is a vertexa3inK1 withc1(a3) = 3. By (2),q1is anticomplete
to{a, a3}. Vertexq1has a neighbor inP\w, for otherwiseV(P)∪V(Q) induces
an odd hole. So there is a pathP′ from q
1 toawith interior inP\w, andP′
meetsL because it containspk. By assumption (iii),P′ contains a vertex that
is complete to{a, a3}, and sinceP is a path that vertex ispk.
Suppose that C1i,3(q1)∩K1 =∅. Then we swap colors iand 3 on C1i,3(q1).
We obtain a coloringc′
1ofG1 such that the color of every vertex inK1∪K2is
unchanged, soc′
1agrees withc2onK1∪K2. Also every vertexvinK3\Bsatisfies c′
1(v) = c1(v). Moreover, c′1(q1) = i. If i = 1, the pair (c′1, c2) has strictly
fewer bad vertices than (c1, c2). If i= 2, then c′1(w) = c1(w) = 1, so (c′1, c2)
contradicts (5) (withy=q1). Therefore we may assume thatC1i,3(q1)∩K16=∅.
LetZ be a path fromq1 to {a, a3} in C1i,3(q1). Since Z is bicolored byc1,
no vertex ofZ can be complete to{a, a3}, and so assumption (iii) implies that Z does not meet L. This means that either i = 1 and Z =q1-w-a3, or i = 2
anticomplete toK3. Sincepkis complete toK1, and no vertex ofK3is complete
toK1, assumption (iv) implies thatpk is anticomplete toK3. In particular,pk
is non-adjacent toq1 andk≥2.
If a3 has a neighbor in P \ {w, pk}, then (sinceq1 also has a neighbor in P\ {w, pk}) there is a path from q1 to a3 with interior in V(P)\ {w, pk}, so,
by (iii), that path must contain a vertex that is complete to{a, a3}; but this
is impossible because a has no neighbor in P \pk. So a3 has no neighbor in P\ {w, pk}.
Now if i = 1, then w is adjacent to a3, and P has even length, hence
(V(P)\ {a})∪ {a3}induces an odd hole. Soi= 2, andZ=q1-z-a3withz∈K3
andc1(z) = 2. Recall thatpk is non-adjacent toz. The pathz-w-P-pk has odd
length, and it is bicolored byc1, so it contains an odd path P′′ from z to pk.
But thenV(P′′)∪ {a
3}induces an odd hole. This completes the proof.
3
Prisms and hyperprisms
In a graphGletR1, R2, R3 be three paths that form a prismK with triangles
{a1, a2, a3} and {b1, b2, b3}, where each Ri has ends ai and bi. A vertex of
V(G)\Kis amajor neighbor ofKif it has at least two neighbors in{a1, a2, a3}
and at least two neighbors in{b1, b2, b3}. A subsetX ofV(K) islocal if either X⊆ {a1, a2, a3}orX ⊆ {b1, b2, b3}or X ⊆V(Ri) for some i∈ {1,2,3}.
IfF, K are induced subgraphs ofGwithV(F)∩V(K) =∅, any vertex inK
that has a neighbor inF is called anattachment ofF inK, and whenever any such vertex exists we say thatF attaches toK.
Here are several theorems from the Strong Perfect Graph Theorem [7] that we will use.
Theorem 3.1 ((7.4) in [7]) In a Berge graphG, letR1, R2, R3 be three paths,
of even lengths, that form a prismK with triangles{a1, a2, a3}and{b1, b2, b3},
where eachRihas endsaiandbi. Assume thatR1, R2, R3all have length at least
2. Let R′
1 be a path from a′1 tob1, such thatR′1, R2, R3 also form a prism. Let y be a major neighbor ofK. Theny has at least two neighbors in{a′
1, a2, a3}. Theorem 3.2 ((10.1) in [7]) In a Berge graphG, letR1, R2, R3be three paths
that form a prismK with triangles {a1, a2, a3} and{b1, b2, b3}, where each Ri has ends ai and bi. Let F ⊆V(G)\V(K) be connected, such that its set of attachments inK is not local. Assume no vertex in F is major with respect to
K. Then there is a pathf1-. . .-fn inF withn≥1, such that (up to symmetry) either:
2. n ≥ 2, f1 is adjacent to a1, a2, a3, and fn is adjacent to b1, b2, b3, and
there are no other edges between{f1, . . . , fn}andV(K), or
3. n≥2,f1 is adjacent to a1, a2, and fn is adjacent to b1, b2, and there are
no other edges between {f1, . . . , fn} andV(K), or
4. f1 is adjacent to a1, a2, and there is at least one edge between fn and
V(R3)\{a3}, and there are no other edges between{f1, . . . , fn}andV(K)\
{a3}.
Hyperprisms
Ahyperprism is a graphH whose vertex-set can be partitioned into nine sets:
A1 C1 B1 A2 C2 B2 A3 C3 B3
with the following properties:
• Each ofA1, A2, A3, B1, B2, B3is non-empty.
• For distinct i, j ∈ {1,2,3}, Ai is complete to Aj, and Bi is complete to
Bj, and there are no other edges betweenAi∪Bi∪Ci andAj∪Bj∪Cj.
• For each i ∈ {1,2,3}, every vertex of Ai∪Bi ∪Ci belongs to a path
betweenAi andBi with interior in Ci.
For each i∈ {1,2,3}, any path from Ai to Bi with interior in Ci is called an
i-rung. The triple (Ai, Ci, Bi) is called a strip of the hyperprism. If we pick
any i-rung Ri for each i∈ {1,2,3}, we see that R1, R2, R3 form a prism; any
such prism is called aninstance of the hyperprism. IfH contains no odd hole, it is easy to see that all rungs have the same parity; then the hyperprism is called even or odd accordingly. Note that ifH is an even hyperprism, thenAi
is anticomplete toBi for eachi. On the other hand, ifH is an odd hyperprism,
there may be edges between Ai and Bi for any i; however, if H is square-free
there is at most one integerisuch that there is an edge betweenAi andBi.
Let G be a graph that contains a prism. Then G contains a hyperprism
H. Let (A1, . . . , B3) be a partition ofV(H) as in the definition of a hyperprism
above. A subsetX ⊆V(H) islocalif eitherX⊆A1∪A2∪A3orX ⊆B1∪B2∪B3
orX ⊆Ai∪Bi∪Cifor somei∈ {1,2,3}. A vertexxinV(G)\V(H) is amajor neighbor ofH ifxis a major neighbor of some instance ofH. The hyperprism
H is maximal if there is no hyperprismH′ such that V(H) is strictly included
inV(H′).
Lemma 3.3 LetGbe a Berge graph, letH be a hyperprism inG, and letM be the set of major neighbors ofH inG. LetF be a component ofG\(V(H)∪M)
• A path P, with ∅ 6= V(P) ⊆ V(F), such that V(H)∪V(P) induces a hyperprism (of the same parity asH).
• A path P, with∅ 6=V(P)⊆V(F), and for eachi∈ {1,2,3}an i-rungRi ofH, such thatV(P)∪V(R1)∪V(R2)∪V(R3)induces the line-graph of
a bipartite subdivision of K4.
Proof. WhenH is an even hyperprism, the proof of the lemma is identical to the proof of Claim (2) in the proof of Theorem 10.6 in [7], and we omit it. WhenH is an odd hyperprism, the proof of the lemma is similar to the proof of Claim (2), with the following adjustments: the case when the integer n in that proof is even and the case whennis odd are swapped, and the argument on page 126 of [7], lines 16–18, is replaced with the following argument:
Suppose thatfn is not adjacent tob1; sof1 is adjacent tob1,n≥2,
andfn is adjacent toa2. Let R3 be any 3-rung, with endsa3∈A3
and b3 ∈B3. Thena1b1 is an edge, for otherwisef1-a1-R1-b1-f1 is
an odd hole; and f1 has no neighbor in {a3, b3}, for otherwise we
would haven= 1. Likewise,a2b2is an edge, andfn has no neighbor
in{a3, b3}. But thenV(R1)∪V(R2)∪V(R3)∪ {f1, . . . , fn}induces
the line-graph of a bipartite subdivision ofK4, a contradiction.
This completes the proof of the lemma.
4
Even prisms
We need to analyze the behavior of major neighbors of an even hyperprism. The reader may note that in the following theorem we are not assuming that the graph is square-free.
Theorem 4.1 Let G be a Berge graph that contains an even prism and does not contain the line-graph of a bipartite subdivision of K4. Let H be an even
hyperprism inG, with partition(A1, . . . , B3)as in the definition of a hyperprism,
and letxbe a major neighbor ofH. Then either:
• xis complete to at least two ofA1, A2, A3 and at least two ofB1, B2, B3,
or
• V(H)∪ {x} induces a hyperprism.
Proof. Since xis a major neighbor of H, there exists for eachi ∈ {1,2,3} an
i-rung Wi of H such that xis a major neighbor of the prism KW formed by
W1, W2, W3. Suppose that the first item does not hold; so, up to symmetry, x
has a non-neighboru1∈A1and a non-neighboru2∈A2. For eachi∈ {1,2}let Ui be an i-rung with endui, and let U3 be any 3-rung. Thenxis not a major
neighbor of the prismKU formed byU1, U2, U3. We can turnKW intoKU by
two consecutive prismsKandK′ such thatxis a major neighbor ofKand not
a major neighbor ofK′. SinceKandK′ are consecutive they differ by exactly
one rung. LetKbe formed by rungsR1, R2, R3, where eachRihas endsai∈Ai
andbi ∈Bi (i= 1,2,3), and let A={a1, a2, a3} and B ={b1, b2, b3}; and let K′
be formed byP1, R2, R3 for some i-rungP1. LetP1 have endsa′1∈A1 and b′
1∈B1, and letA′={a′1, a2, a3}andB′ ={b′1, b2, b3}.
Let α = |N(x)∩A|, β = |N(x)∩B|, α′
= |N(x)∩A′|
, β′
= |N(x)∩
B′|. We know that α ≥2 and β ≥2 since xis a major neighbor of K, and
min{α′, β′} ≤ 1 sincex is not a major neighbor ofK′. Moreover, α′ ≥α−1
andβ′ ≥β−1 sinceK and K′ differ by only one rung. Up to the symmetry
onA, B, these conditions imply that the vector (α, β, α′, β′) is equal to either
(3,2,3,1), (3,2,2,1), (2,2,2,1) or (2,2,1,1). In either case we have β= 2 and
β′ = 1, soxis adjacent to b
1, non-adjacent to b′1, and adjacent to exactly one
ofb2, b3, say tob3.
Suppose that (α′, β′) is equal to (3,1) or (2,1). We can apply Theorem 3.2
to K′ and F ={x}, and it follows that xsatisfies item 4 of that theorem, so xis adjacent to a′
1, a2, b3 and has no neighbor in V(K′)\({a′1, a2} ∪V(R3)).
But then V(R2)∪ {x, b3} induces an odd hole, a contradiction. So we may
assume that (α, β, α′ , β′
) = (2,2,1,1), which restores the symmetry betweenA
andB. Sinceα= 2 and α′
= 1,x is adjacent toa1, non-adjacent toa′1, and
adjacent to exactly one ofa2, a3. Ifxis adjacent toa2, thenK′and{x}violate
Theorem 3.2. Sox is adjacent toa3 and not to a2, and Theorem 3.2 implies
thatxis a local neighbor ofK′ withN(x)∩K′ ⊆V(R
3), soxhas no neighbor
inP1or R2. Then we claim that:
For every 1-rung Q1, the ends of Q1 are either both adjacent tox
or both non-adjacent tox. (1)
For suppose the contrary. Thenxis not a major neighbor of the prism formed byQ1, R2, R3, and consequently that prism and the set F ={x}violate
Theo-rem 3.2. So (1) holds.
Let A′
1 = A1\ N(x) and A′′1 = A1 ∩N(x), and B1′ = B1\ N(x) and B′′
1 =B1∩N(x). By (1), every 1-rung is either betweenA′1 andB1′ or between A′′
1 andB ′′ 1. LetC
′
1be the set of vertices ofC1that lie on a 1-rung whose ends
are in A′ 1∪B
′
1, and let C ′′
1 be the set of vertices of C1 that lie on a 1-rung
whose ends are inA′′ 1∪B
′′
1. By (1),C ′ 1andC
′′
1 are disjoint and there is no edge
betweenA′ 1∪C
′ 1∪B
′ 1 andC
′′
1 or betweenA ′′ 1∪C
′′ 1 ∪B
′′ 1 and C
′ 1.
Pick any 1-rungP′
1 with ends in A ′ 1∪B
′
1. Then Theorem 3.2 implies (just
like forP1) thatxis a local neighbor of the prism formed by P1′, R2, R3, so x
has no neighbor inP′
1. It follows that: xhas no neighbor inA′
1∪C1′ ∪B1′. (2)
Moreover, we claim that:
A′
For suppose on the contrary, up to relabelling vertices and rungs, thata′ 1 and a1are non-adjacent. Then, by (2), V(P1)∪ {x, a1, a2, b3}induces an odd hole.
Thus (3) holds. LetA′
2=A2\N(x),A′′2 =A2∩N(x),B′2=B2\N(x) andB2′′=B2∩N(x).
LetC′
2be the set of vertices ofC2that lie on a 2-rung whose ends are inA′2∪B2′,
and let C′′
2 be the set of vertices ofC2 that lie on a 1-rung whose ends are in A′′
2∪B2′′. By the same arguments as for the 1-rungs, we see that every 2-rung is
either betweenA′
2 andB2′ or betweenA′′2 andB2′′, thatC2′ andC2′′ are disjoint
and that there is no edge betweenA′
2∪C2′∪B2′ andC2′′or betweenA′′2∪C2′′∪B2′′
andC′
2. Alsoxhas no neighbor inA′2∪C2′∪B2′, andA′2is complete toA′′2, and B′
2 is complete toB2′′. Note that, sincexa1′ andxa2are not edges, the setsA′1, B′
1,C1′,A′2,B′2, C2′ are all non-empty. It follows that the nine sets A′
1 C1′ B1′
A′
2 C2′ B2′
A′′
1∪A′′2∪A3 C1′′∪C2′′∪C3∪ {x} B1′′∪B2′′∪B3
form a hyperprism. So the second item of the theorem holds.
Theorem 4.2 Let Gbe a square-free Berge graph that contains an even prism and does not contain the line-graph of a bipartite subdivision of K4. Then G
has a good partition.
Proof. LetH be a maximal even hyperprism inG, with partition (A1, . . . , B3)
as in the definition of a hyperprism. Recall that, sinceH is an even hyperprism,
Ai is anticomplete toBifor each i. LetM be the set of major neighbors ofH.
LetZ be the set of vertices of the components ofV(G)\(V(H)∪M) that have no attachment inH. By Lemma 3.3 every component of G\(V(H)∪M ∪Z) attaches locally toH. For eachi= 1,2,3, letFi be the union of the vertex-sets
of the components ofG\(V(H)∪M∪Z) that attach toAi∪Bi∪Ci. LetFAbe
the union of the vertex-sets of the components ofG\(V(H)∪M∪Z∪F1∪F2∪F3)
that attach toA1∪A2∪A3, and defineFB similarly. By Lemma 3.3 the setsF1, F2, F3, FA, FB are well-defined and are pairwise anticomplete to each other,
andV(G) =V(H)∪M∪Z∪F1∪F2∪F3∪FA∪FB.
By Theorem 4.1, every vertex inM is complete to at least two ofA1, A2, A3
and at least two ofB1, B2, B3.
Suppose that M contains non-adjacent vertices x, y. By Theorem 4.1, x
andy have a common neighbora in Aand a common neighbor b in B. Then
{x, y, a, b} induces a square, a contradiction. Therefore M is a clique. By Lemma 1.3,M∪Ai is a clique for at least two values ofi, and similarlyM∪Bj
is a clique for at least two values ofj. Hence we may assume that bothM∪A1
andM∪B1are cliques.
Define setsK1=A1,K2=M,K3=B1,L=A2∪B2∪C2∪F2∪A3∪B3∪ C3∪F3∪FA∪FB andR=V(G)\(K1∪K2∪K3∪L). (SoR=C1∪F1∪Z.)
Every path fromK3toK1that meetsLcontains a vertex fromA2∪A3, which
toK3, and the setsC1, C2, C3are non-empty, so, picking any vertexxi∈Cifor
eachi ∈ {1,2,3}, we see that {x1, x2, x3} is a triad with a vertex in L and a
vertex inR. So (K1, K2, K3, L, R) is a good partition ofV(G).
5
Odd prisms
Now we analyze the behavior of major neighbors of an odd hyperprism. The fol-lowing theorem is the analogue of Theorem 4.1, but here we need the assumption that the graph is square-free, and there is an additional outcome.
Let H be an odd hyperprism in G, with partition (A1, . . . , B3) as in the
definition of a hyperprism. Fori= 1,2,3, letA∗
i ={x∈Ai |xhas a neighbor
in Bi} andBi∗={x∈Bi |xhas a neighbor in Ai}. So A∗i and B
∗
i are either
both empty or both non-empty. Moreover, sinceG is square-free, A∗
i ∪B
∗
i is
non-empty for at most one value ofi.
Theorem 5.1 Let G be a square-free Berge graph. Let H be a maximal odd hyperprism inG, with partition(A1, . . . , B3)as in the definition of a hyperprism,
and letm∈V(G)\V(H)be a major neighbor of H. Then either:
• mis complete to at least two ofA1, A2, A3 and at least two ofB1, B2, B3,
and for everyi∈ {1,2,3} mis complete to A∗
i ∪B
∗
i, or
• A∗ 1 and B
∗
1 are non-empty, m is complete to A ∗ 1∪B
∗
1, to at least one of A2, A3 and to at least one ofB2, B3. Further, suppose thatm has a
non-neighbor in at least two ofB1, B2, B3. Fori∈ {1,2,3}, letBi1be the set of non-neighbors ofm inBi, andCi1 be the set of all the vertices ofCi that belong to rungs between B1
i andAi, and letA1i be the set of all vertices of
Ai that are in rungs whose other end is inBi1. Then:
– A1
1⊆A∗1, and
– for every pathP fromB1
1∪C11to(C1\C11)∪(A1\A11), some vertex
ofV(H)\(A1∪B1∪C1)has a neighbor in P∗.
– for every pathP fromm toB1
1∪C11 some vertex ofV(H)\(A1∪ B1∪C1)has a neighbor in P∗.
– Let i∈ {2,3}. –A1
i is complete to Ai\A1i, andBi1 is complete toBi\Bi1, – for every pathP fromC1
i to(Ai∪Bi∪Ci)\(A1i∪Bi1∪Ci1), some vertex of V(H)\(Ai∪Bi∪Ci)has a neighbor in P∗;
– for every path P from Ci \Ci1 to A1i ∪Bi1∪Ci1, some vertex of
V(H)\(Ai∪Bi∪Ci) has a neighbor inP∗; –mis complete toA1
i and for every pathP frommtoBi1∪Ci1some vertex of V(H)\(Ai∪Bi∪Ci)has a neighbor in P∗.
Proof. We first observe that:
For suppose on the contrary, up to symmetry, that there is a 1-rungP1 that
contains no neighbor ofm. LetP1 have endsa1∈A1 andb1∈B1. Supposem
has neighborspandqsuch thatp∈A2∪A3,q∈B2∪B3, andpis non-adjacent
to q. Thenp-a1-P1-b1-q-m-p is an odd hole, contradiction. Hence, sincem is
major, we may assume thatm is anticomplete to A3∪B3 and has neighbors a′
1, b ′
1,a2, b2 such that a1′ ∈A1, b′1 ∈B1, a2∈A2,b2∈B2, anda2 and b2 are
adjacent. Note thatA1∪A3is anticomplete toB1∪B3sinceA∗2∪B ∗
2 6=∅. Pick
anya3∈A3andb3∈B3. Suppose that botha′1, b ′
1have a neighbor inP ∗ 1. Then
there is a 1-rungP′
1 with endsa′1, b′1 and interior inP1∗, and thenV(P1′)∪ {m}
induces an odd hole. Hence we may assume thatb′
1has no neighbor inP1∗. Then b1b′1 is not an edge, for otherwise m-a2-a1-P1-b1-b′1-m is an odd hole. Suppose
thata′
1 has a neighbor c1 in P1∗, and choose c1 as close to b1 as possible along P1. Thena′1-c1-P1-b1 is a 1-rung, so it is odd; but thenm-a′1-c1-P1-b1-b3-b′1-m
is an odd hole. Soa′
1 is also anticomplete toP1∗, and by symmetrya1a′1is not
an edge. But thenm-a′
1-a3-a1-P1-b1-b3-b′1-m is an odd hole. This proves (1).
For eachi,mis complete toA∗
i ∪Bi∗. (2)
For suppose the contrary. Then there are vertices ui ∈ A∗i and vi ∈ Bi∗ such
thatuivi is an edge andmhas a non-neighbor in {ui, vi}. Since mis a major
neighbor ofH, it has a neighbor ain (A1∪A2∪A3)\Ai and a neighborb in
(B1∪B2∪B3)\Bi. Then the subgraph induced by{m, a, b, ui, vi} contains a
square or aC5, a contradiction. Thus (2) holds.
In view of (2) we may assume thatmdoes not satisfy the property of being complete to at least two of B1, B2, B3 (for otherwise the first outcome of the
theorem holds). So we may assume thatmis not complete toB1, not complete
toB2, and (possibly exchanging the roles ofB2andB3), thatmhas a neighbor
inB3. Letb2∈B2 be a non-neighbor ofm andb3∈B3 be a neighbor ofm.
Let b1 be a non-neighbor of m in B1, and let a1-P1-b1 be a rung
throughb1. Thenm is adjacent toa1 and anticomplete toV(P1)\
{a1}. Moreover, let P be a path from m to V(P1)\ {a1}. Then
some vertex ofV(H)\(A1∪B1∪C1) has a neighbor inP∗.
(3)
Leta2-P2-b2 be a rung through b2. If there is a pathP violating (3), then by
(1)m has a neighbor in P2, and therefore we can link mto {b1, b2, b3} viaP, P2 and m-b3, a contradiction to Lemma 1.4. Thus no such path exists, and in
particularm is anticomplete to V(P1)\ {a1}. Now it follows from (1) thatm
For every symbolX in{A, B, C}there is a partition ofX1into two
sets X′ 1 andX
′′
1 such that:
– A′
1 is complete toA ′′ 1, andB
′
1 is complete toB ′′ 1;
– A′
1 is anticomplete toB ′ 1;
– for every path P from C′ 1 to A
′′ 1 ∪B
′′ 1 ∪ C
′′
1, some vertex of V(H)\ (A1∪B1 ∪C1) has a neighbor in P∗, and in particular C′
1 is anticomplete toA′′1∪B1′′∪C1′′;
– for every path P from C′′
1 to A′1 ∪B′1 ∪C1′, some vertex of V(H)\ (A1∪B1 ∪C1) has a neighbor in P∗, and in particular C′′
1 is anticomplete toA′1∪B1′ ∪C1′;
– m is complete to A′
1 and for every path P from m to B1′ ∪C1′
some vertex of V(H)\(A1∪B1∪C1) has a neighbor inP∗, and in
particularmis anticomplete toB′ 1∪C1′.
(4)
Pick rungsa2-P2-b2 anda3-P3-b3 containingb2 andb3 respectively. By (3), mis adjacent toa2.
LetB′
1={y∈B1\B1∗|y is non-adjacent tomand there exists a rung from A1\A∗1 to y}, and let A′1={x∈A1\A∗1|there is a rung fromxtoB1′}. Let C′
1 ={z ∈C1|z lies on a rung betweenB1′ andA′1}. So mis anticomplete to B′
1and, by (3),mis complete toA′1and anticomplete toC1′. LetB1′′=B1\B1′, A′′
1 =A1\A′1, and C1′′ =C1\C1′. Let Qbe any rung with ends x∈A′1 and y∈B′
1. We prove five claims (a)–(e) as follows.
(a)For every pathP fromB′′
1 toA′1∪C1′ some vertex ofV(H)\(A1∪B1∪C1)
has a neighbor inP∗.
We know thatB′′
1 is anticomplete to A′1 sinceA1′ ⊆A1\A∗1. Suppose up, to
relabelling, that there is a path P from some vertex b1 in B1′′ to a vertex of Q\ {y}, contradicting (a). Then there is a pathQ′fromxtob
1with interior in P∗∪Q∗. By the maximality ofH,Q′ ∗⊆C
1, andQ′ has odd length at least 3.
By the definition ofB′
1 and the existence ofQ ′
imply that b1 is adjacent tom.
Supposemhas a neighbor inP∗
. Since no vertex ofV(H)\(A1∪B1∪C1) has
a neighbor inP∗
, it follows from the second statement of (3) thatP is a path fromb1 to a1, andP∗∩V(Q) = ∅, and that P∗ is anticomplete toV(Q). Ify
is adjacent tob1, thenx−P−b1−y−Q−xis an odd hole, a contradiction.
Thusb1 is non-adjacent toy, and so b1−m−x−Q−y−b2−b1, is an odd
hole, again a contradiction. This proves thatm is anticomplete toP∗. Recall
that m is adjacent toxand anticomplete to V(Q)\ {x}. It follows that m is anticomplete toQ′ ∗. consequentlyV(Q′)∪ {m}induces an odd hole. Since this
holds for allQ, claim (a) is established. (b)B′′
1 is complete to B1′.
Suppose, up to relabelling, that someb1 inB1′′is not adjacent toy. Recall that mis anticomplete toV(Q)\ {x}. By (a) b1 has no neighbor inQ. Then b1 is
non-adjacent tom, for otherwise x-Q-y-b2-b1-m-xinduces an odd hole. Pick a
runga1-P1-b1. By (2),b1∈/ B1∗, hence, by the definition ofB1′, we havea1∈A∗1,
argument as forb1it follows thatb∗1is not adjacent tom, which contradicts (2).
Thereforeb∗
1 is adjacent to y and, by (2), to m. By (a)b∗1 has no neighbor in Q\y. We know thata1 is not adjacent toy sincey /∈B∗1. Moreovera1 has no
neighbor inQ\x, for otherwise we can linka1to {b3, y, b∗1}viaa3-P3-b3,Q\x
and a1-b∗1, a contradiction to Lemma 1.4. Then xa1 is an edge, for otherwise x-Q-y-b∗
1-a1-a3-x is an odd hole. There is no edge between Q and P1 except a1x, for otherwise there would be a rung fromxto b1, implyingb1 ∈B1′. But
thenb1-P1-a1-x-Q-y-b3-b1 is an odd hole. ThusB1′′ is complete toy, and since
this holds for allQ, the claim is established.
(c)For every path P fromA′′
1 toB1′, some vertex of V(H)\(A1∪B1∪C1)
has a neighbor inP∗.
For suppose up to relabelling that a1-P1-y is a path contradicting (c). By the
maximality ofH,P1is a rung ofH. By the definition ofB1′ we havea1∈A∗1, so a1has a neighborb∗1∈B1∗. By (a) and (b),b∗1 is adjacent toyand anticomplete
to Q\y. Since y 6∈ B∗
1, it follows that a1 is not adjacent to y. Then a1 has
no neighbor inQ\x, for otherwise we can linka1to {y, b∗1, b2}viaQ\x,a1-b∗1
anda2-P2-b2. Moreovera1 is adjacent tox, for otherwise a1-b∗1-y-Q-x-a2-a1 is
an odd hole. SinceV(Q)∪V(P1) does not induce an odd hole, there is an edge
betweenQ\y andP1\ {a1, y}. Since b1∗-y-P1-a1-b∗1 cannot be an odd hole, b ∗ 1
has a neighbor inP1\ {a1, y}. But this implies the existence of a rung between xandb∗
1, which contradicts (a).
(d)A′′
1 is complete to A ′ 1.
Suppose on the contrary that somea1 inA′′1 is non-adjacent tox. Leta1-P1-b1
be a rung. By (c),b1∈B′′1, and by (a)–(c), the only edge betweenP1andQis b1y. Thena1-P1-b1-y-Q-x-a3-a1 is an odd hole, a contradiction.
(e)For every pathP fromC′
1 toA′′1∪C1′′∪B1′′, some vertex ofV(H)\(A1∪ B1∪C1)has a neighbor in P∗, and for every path P fromC1′′ toA′1∪C1′ ∪B′1
some vertex ofV(H)\(A1∪B1∪C1) has a neighbor inP∗.
Indeed, in the opposite case there is a path that violates (a) or (c).
It follows from (3) and claims (a)–(e) that all the properties described in (4) are satisfied. So (4) holds.
By (4), ifB′
16=∅, then (A1, C1, B1) is a strip such thatmis complete toA′1
and anticomplete toB′ 1∪C
′
1. Likewise, ifB ′
26=∅, then (A ′ 2, C
′ 2, B
′
2) is a strip
such thatmis complete toA′
2and anticomplete toB ′ 2∪C
′
2. So if bothB ′ 16=∅
andB′
26=∅, using the properties described in (4), we obtain a hyperprism: A′
1 C1′ B1′ A′
2 C2′ B2′ A3∪A′′1∪A′′2∪ {m} C3∪C1′′∪C2′′ B3∪B1′′∪B′′2.
contrary to the maximality ofH. Thus we may assume that B′
1 =∅, and consequentlyA∗1, B1∗ are both
Let B11 be the set of non-neighbors of m in B1. Then every rung
with an end inB1
1 has its other end in A ∗
1. Moreover, letC11 be the
set of all the vertices of C1 that belong to rungs between B11 and A∗
1. Then for every pathP from B11∪C11 to (C1\C11)∪(A1\A∗1)
some vertex ofV(H)\(A1∪B1∪C1) has a neighbor inP∗.
(5)
Since B′
1 = ∅, it follows that every rung of H with an end in B11 has its
other end in A∗
1. Next suppose that there is a path P from c ∈ B11∪C11 to d∈(C1\C11)∪(A1\A∗1) violating (5). By the definition of a hyperprism, there
is a path Pc from c to b ∈ B11 with interior in C1, and a path Pd from d to
a∈A1\A∗1. Since P∗ violates (5), no vertex ofV(H)\(A1∪B1∪C1) has a
neighbor in P∗, and therefore P∗∩(A
1∪B1) = ∅. Since B11 is anticomplete
to A1, there is a path Q from a to b with non-empty interior contained in V(P)∪V(Pa)∪V(Pb)\ {a, b}. By the maximality of H, Q is a rung of H,
contrary to the fact that every rung with an end inB1
1 has its other end inA∗1.
This proves (5).
Since A∗
1, B1∗ are non-empty, it follows that B′2 6=∅, and so (A′2, C2′, B2′) is
a strip. By (2) m has a neighbor in B1. Suppose that m is not complete to B3. Then there is symmetry between B1 and B3, and thereforeB3′ 6=∅ and
(A′ 3, C
′ 3, B
′
3) is a strip. Applying (4) toA2, B2, C2 and toA3, B3, C3, we obtain
a hyperprism:
A′
2 C
′
2 B
′ 2 A′
3 C
′
3 B
′ 3 A1∪A′′2∪A′′3∪ {m} C1∪C2′′∪C3′′ B1∪B2′′∪B′′3.
contrary to the maximality ofH.
Thus we may assume thatm is complete toB3. Sincem has a neighbor in A∗
1 ⊆A1, reversing the roles of A and B we may assume that m is complete
to at least one ofA2, A3. Now by (3), (4) and (5), the second outcome of the
theorem holds.
Theorem 5.2 Let G be a square-free Berge graph that contains an odd prism and does not contain the line-graph of a bipartite subdivision of K4. Then G
admits a good partition.
Proof. SinceG contains an odd prism, it contains a maximal odd hyperprism (A1, C1, B1, A2, C2, B2, A3, C3, B3) which we callH. LetM be the set of major
neighbors ofH. LetMgoodbe the set of all the vertices ofM that are complete
to at least two of A1, A2, A3 and to at least two ofB1, B2, B3, and let Mbad
be the remaining major neighbors of H. Let Z be the set of vertices of the components ofV(G)\(V(H)∪M) that have no attachment inH. SinceH is maximal, by Lemma 3.3 there is a partition ofV(G)\(V(H)∪M∪Z) into sets
• Fori= 1,2,3,N(Fi)⊆Ai∪Ci∪Bi∪M;
• N(FA)⊆A1∪A2∪A3∪M andN(FB)⊆B1∪B2∪B3∪M;
• The setsZ, F1, F2, F3, FA, FB are pairwise anticomplete to each other.
We observe that:
At least two ofA1, A2, A3are cliques, and at least two ofB1, B2, B3
are cliques. (1)
This follows directly from Lemma 1.3 (withK=∅). SinceH is maximal, 5.1 implies that:
Let m ∈ M. For every i ∈ {1,2,3}, m is complete to A∗
i ∪B
∗
i.
Moreover, if i, j∈ {1,2,3}are distinct andA∗
i =A
∗
j =∅, thenmis
complete to at least one ofAi, Aj and to at least one ofBi, Bj.
(2)
We claim that:
M is a clique. (3)
Suppose that there are non-adjacent verticesm1, m2 inM. By (2),m1 andm2
have a common neighbor inA1∪A2∪A3. Therefore leta1be a common neighbor
ofm1andm2 inA1. Ifm1 andm2have a common neighborb∈B2∪B3, then
{m1, m2, a1, b}induces a square, a contradiction. In view of (2),A∗2=A∗3 =∅,
and we may assume up to symmetry that m1 is not complete to B2, so it is
complete toB3, and consequently that m2 is not complete to B3, and so it is
complete toB2. Pick a non-neighborb2ofm1inB2and a non-neighborb3ofm2
inB3. Then{m1, m2, a1, b2, b3}induces aC5, a contradiction. This proves (3). Mgood∪Ai is a clique for at least two values of i, and similarly
Mgood∪Bj is a clique for at least two values ofj. (4)
This follows directly from (2), (3) and Lemma 1.3. Thus (4) holds.
Let j, k ∈ {1,2,3} be distinct and such that A∗
j =A
∗
k = ∅. Then
either M ∪Aj or M ∪Ak is a clique, and similarly eitherM ∪Bj
or M∪Bk is a clique.
(5)
This follows directly from (2), (3) and Lemma 1.3 applied toM, Aj, Ak and to
M, Bj, Bk. Thus (5) holds.
SinceGis square-free, we may assume, up to symmetry, thatA1is
anticom-plete toB1, and thatA2is anticomplete toB2, and soC16=∅andC26=∅. Pick
anyx1∈C1,x2∈C2 anda3∈A3. So{x1, x2, a3}is a triad τ.
Suppose that bothM∪A1 andM∪B1are cliques. SetK1=A1,K2=M, K3 = B1, L = A2∪B2∪C2∪F2∪A3∪B3∪C3∪F3∪FA∪FB and R =
V(G)\(K1∪K2∪K3∪L). (So R = C1∪F1∪Z.) We observe that K1 is
vertex fromA2∪A3, which is complete toK1, andτ is a triad with a vertex in Land a vertex inR. Thus (K1, K2, K3, L, R) is a good partition ofV(G). The
same holds if bothM ∪A2 andM∪B2 are cliques.
By (5), we may assume thatA∗
36=∅,M ∪A1 andM ∪B2 are cliques, and
neither ofM∪A2andM∪B1is a clique. Suppose thatMbad=∅. Then by (4)
M∪A3 andM ∪B3 are both cliques. SetK1=A1, K2 =M, K3 =B2∪B3, L=A2∪C2∪F2∪A3∪C3∪F3∪FA and R=V(G)\(K1∪K2∪K3∪L).
(SoR=C1∪B1∪F1∪Z.) We observe thatK1 is anticomplete toK3, every
path fromK3 to K1 with interior in L contains a vertex from A2∪A3, which
is complete to K1, and τ is a triad with a vertex in L and a vertex in R.
Thus (K1, K2, K3, L, R) is a good partition ofV(G). Thus we may assume that Mbad6=∅. LetMA be the set of vertices ofM with a non-neighbor in bothA2
and A3, and letMB be the set of vertices of M with a non-neighbor in both
B1 andB3. SoMbad=MA∪MB. By Lemma 1.3 applied with K=M\MB,
X1 = B1 and X2 =B3, we deduce that either (M \MB)∪B3 is a clique, or
(M \MB)∪B1 is a clique.
Suppose first that (M \MB)∪B1 is a clique. Then, since M ∪B1 is not
a clique, it follows that MB 6= ∅. Let m ∈ MB be chosen with N(m)∩B1
maximal; letB′
1=B1∩N(m) andB1′′=B1\B′1. SinceB1, M are both cliques
andG has no C4, it follows from the choice of m that MB is anticomplete to
B′′ 1. LetC
′′
1 be the set of vertices ofC1that are in rungs with vertices ofB′′1, let A′′
1 be the set of vertices inA1that are in rungs with vertices ofB1′′, and letF ′′ 1
be the set of vertices ofF1such that there is a path from them toB′′1∪C ′′ 1 with
interior inF′′
1. Recall that every vertex ofMB has a non-neighbor inB1 and a
non-neighbor inB3, and so the second outcome of Theorem 5.1 holds. It follows
that the setC′′
1 ∪F1′′is anticomplete toMB∪(C1\C1′′)∪(F1\F1′′), andMB is
complete toA′′
1. Now setK1=A′′1,K2=M\MB,K3=B′′1,R=C1′′∪F1′′and L=V(G)\(K1∪K2∪K3∪R). By Theorem 5.1 the setLis anticomplete to R. We know thatK1∪K2 is a clique (becauseM∪A1is a clique) andK2∪K3
is a clique (because of the current assumption that (M \MB)∪B1is a clique).
Moreover, K1 is anticomplete to K3, and, again by Theorem 5.1, every path
fromK3toK1with interior inLcontains a vertex ofA2∪A3∪(A1\A′′1)∪MB,
which is complete toA′′
1. We may assume thatτ∩C1 ⊆C1′′, and thus τ is a
triad with a vertex inLand a vertex inR. Consequently, (K1, K2, K3, L, R) is
a good partition.
Hence we may assume that (M\MB)∪B1is not a clique, and so (M\MB)∪B3
is a clique. By symmetry we may assume that (M \MA)∪A3 is a clique.
Switching the roles ofA andB if necessary, we may assume thatMB 6=∅.
Let B′′
1 be the set of vertices in B1 that are not complete to MB, and let
B′
1 = B1\B1′′. So B1′′ 6= ∅. Let C1′′ be the set of vertices of C1 that are in
rungs with vertices ofB′′
1, letA′′1 be the set of vertices inA1 that are in rungs
with vertices ofB′′
1, and letF1′′ be the set of vertices of F1 such that there is
a path from them toB′′
1 ∪C1′′ with interior inF1′′. By 5.1 the setC1′′∪F1′′ is
anticomplete to (C1\C1′′)∪(F1\F1′′); recall thatM ∪A1 is a clique. LetA′′3
LetC′′
3 be the set of vertices ofC3 in rungs withA′′3, andF3′′ the set of vertices
ofF3 such that there is a path to them fromA′′3∪C3′′ with interior inF3′′. Let C′
3=C3\C3′′ andF3′=F3\F3′′.
We claim that M \MB is complete to B1′. Suppose m ′ ∈
M \MB has a
non-neighbor inb′ 1 ∈B
′
1. Choose m∈ MB. Since m∈ MB, someb3 ∈B3 is
non-adjacent tom. Now{m, m′ , b′
1, b3}induces aC4, a contradiction.
LetK1=A′′1∪A ′
3,K2=M,K3=B1′∪B2∪B3∗,R=B ′′ 1∪C
′′ 1∪F
′′ 1 ∪FB∪
(B3\B3∗)∪C ′ 3∪F
′
3 andL=V(G)\(K1∪K2∪K3∪R). By Theorem 5.1 the
setLis anticomplete toR, and every rung ofH with an end inA′′
3 has its other
end inB∗
3. Since M ∪A1 is a clique, A3∪(M\MA) is a clique, and since by
the definition ofA′
3,MAis complete to A′3, it follows thatK1∪K2 is a clique.
SinceM ∪B2 is a clique,M \MB is complete to B1′, by the definition ofB1′, MB is complete toB1′,B3∪(M \MB) is a clique, and by Theorem 5.1MB is
complete toB∗
3, it follows thatK2∪K3 is a clique.
We now check that condition (iii) in the definition of a good partition holds. Suppose thatP is a path from K3 to K1 with P∗ 6=∅ andP∗ ⊆L. We may
assume thatV(P) is disjoint from A2∪A3′′∪(A1\A′′1), which is complete to A′′
1∪A′3 (recall thatA1andA3are both cliques). It follows from Theorem 5.1
that the ends of P are in A′
3 and in B ∗
3, and P ∗ ⊆
C′′ 3 ∪F
′′
3. In particular, A′′
3 6=∅. We claim that some m∈MA is anticomplete toA′′3. Choosem∈MA
with N(m)∩A′′
3 minimal. We may assume thatA ′′
3∩N(m)6= ∅; let a ∈A ′′ 3
be a neighbor ofm. It follows from the definition of A′′
3 that somem ′ ∈
MA is
non-adjacent toa. By the choice ofm, there isa′ ∈A′′
3 adjacent tom
′ and not
tom. But now sinceA3 and MA are both cliques,{m, m′, a, a′}induces a C4,
a contradiction. This proves the claim; letm∈MAbe anticomplete toA′′3. By
Theorem 5.1 m is anticomplete to C′′
3 ∪F3′′, and so m is anticomplete to P∗.
But nowV(P)∪ {m} induces an odd hole, a contradiction. This proves that condition (iii) holds.
Next we check that condition (iv) in the definition of a good partition is satisfied. Suppose that some l ∈ L has a neighbor k1 ∈ K1 and a neighbor k3 ∈K3. Thenl ∈C3′′∪F3′′. SinceH is an odd prism,k1-l-k3 is not a rung of
length two, and thereforek1 is adjacent tok3. This proves that condition (iv)
holds. Finally, we may assume that τ∩C1 ⊆C1′′, and thus τ is a triad with
a vertex inLand a vertex inR. This proves that (K1, K2, K3, L, R) is a good
partition, and completes the proof of the theorem.
6
Line-graphs
The goal of this section will be to prove the following decomposition theorem.
Theorem 6.1 LetGbe a square-free Berge graph, and assume thatGcontains the line-graph of a bipartite subdivision ofK4. ThenGadmits a good partition.
In a graphH, abranch is a path whose interior vertices have degree 2 and whose ends have degree at least 3. Abranch-vertex is any vertex of degree at least 3.
In a graph G, an appearance of a graph J is any induced subgraph of G
that is isomorphic to the line-graph L(H) of a bipartite subdivision H of J. An appearance of J is degenerate if either (a) J = H = K3,3 (the complete
bipartite graph with three vertices on each side) or (b) J = K4 and the four
vertices ofJ form a square inH. Note that a degenerate appearance of a graph contains a square since in either case (a) or (b) the graphH contains a square. An appearanceL(H) ofJ in Gisovershadowed if there is a branchB ofH, of length at least 3, with endsb1, b2, such that some vertex ofG is non-adjacent
in Gto at most one vertex inδH(b1) and at most one in δH(b2), whereδH(b)
denotes the set of edges ofH (vertices ofL(H)) of whichb is an end.
Anenlargement of a 3-connected graph J (also called a J-enlargement) is any 3-connected graphJ′ such that there is a proper induced subgraph of J′
that is isomorphic to a subdivision ofJ.
To obtain a decomposition theorem for graphs containing line graphs of bipartite graphs, we first thicken the line graph into an object called a strip system, and then study how the components of the rest of the graph attach to the strip system.
LetJ be a 3-connected graph, and letGbe a Berge graph. AJ-strip system
(S, N)inGmeans
• for each edgeuvofJ, a subsetSuv =Svu ofV(G),
• for each vertexv ofJ, a subset Nv ofV(G),
• Nuv=Suv∩Nu,
satisfying the following conditions (where for uv ∈ E(J), a uv-rung means a pathRof Gwith endss, t, say, whereV(R)⊆Suv, and sis the unique vertex
ofRin Nu, and tis the unique vertex ofRinNv):
• The setsSuv (uv∈E(J)) are pairwise disjoint;
• For eachu∈V(J),Nu⊆
S
uv∈E(J)Suv;
• For eachuv∈E(J), every vertex ofSuv is in auv-rung;
• For any two edges uv, wx of J with u, v, w, x all distinct, there are no edges betweenSuv andSwx;
• Ifuv,uwinE(J) withv6=w, thenNuv is complete toNuwand there are
no other edges betweenSuv andSuw;
• For eachuv∈E(J) there is a specialuv-rung such that for every cycleC
The vertex set of (S, N), denoted byV(S, N), is the setSuv∈E(J)Suv.
Note thatNuvis in general different fromNvu. On the other hand,Suv and
Svu mean the same thing.
The following two properties follow from the definition of a strip system:
• For distinct u, v ∈ V(J), we have Nu∩Nv ⊆ Suv if uv ∈ E(J), and
Nu∩Nv =∅ ifuv6∈E(J).
• Foruv∈E(J) andw∈V(J), ifw6=u, v thenSuv∩Nw=∅.
In 8.1 from [7] it is shown that for everyuv∈E(J), alluv-rungs have lengths of the same parity. It follows that the final axiom is equivalent to:
• For every cycleC ofJ and every choice ofuv-rung for everyuv∈E(C), the sums of the lengths of theuv-rungs has the same parity as|V(C)|. In particular, for each edgeuv∈E(J), alluv-rungs have the same parity. By achoice of rungswe mean the choice of oneuv-rung for each edgeuvofJ. It follows from the preceding points that for every choice of rungs the subgraph ofGinduced by their union is the line-graph of a bipartite subdivision of J.
We say that a subset X of V(G) saturates the strip system if for every
u∈V(J) there is at most one neighbor v of usuch that Nuv 6⊆X. A vertex
v in V(G)\V(S, N) is major with respect to the strip system if the set of its neighbors saturates the strip system. A vertex v ∈ V(G)\V(S, N) is major with respect to some choice of rungsif theJ-strip system defined by this choice of rungs is saturated by the set of neighbors ofv.
A subset X of V(S, N) is local with respect to the strip system if either
X⊆Nv for some v∈V(J) orX ⊆Suv for some edgeuv∈E(J).
Lemma 6.2 Let G be a Berge graph, let J be a 3-connected graph, and let
(S, N)be a J-strip system inG. Assume moreover that if J =K4 then(S, N)
is non-degenerate and that no vertex is major for some choice of rungs and non-major for another choice of rungs. Let F ⊆V(G)\V(S, N)be connected and such that no member of F is major with respect to (S, N). If the set of attachments ofF inV(S, N)is not local, then one can find in polynomial time one of the following:
• A path P, with ∅ 6=V(P)⊆V(F), such that V(S, N)∪V(P) induces a
J-strip system.
• A path P, with ∅ 6=V(P) ⊆V(F), and for each edge uv ∈ E(J) a uv -rungRuv, such thatV(P)∪Suv∈E(J)Ruv is the line-graph of a bipartite subdivision of a J-enlargement.
We say that a K4-strip system (S, N) in a graph Gis special if it satisfies
the following properties, where for alli, j∈[4], Oij denotes the set of vertices
in V(G)\V(S, N) that are complete to (Ni∪Nj)\Sij and anticomplete to
V(S, N)\(Ni∪Nj∪Sij):
(a) N13=N31=S13 andN24=N42=S24.
(b) Every rung inS12 and S34 has even length at least 2, and every rung in S14 andS23 has odd length at least 3.
(c) O12 andO34are both non-empty and complete to each other.
(d) If some vertex ofV(G)\(V(S, N)∪O12∪O34) is major with respect to
some choice of rungs in (S, N), then it is major with respect to (S, N). In particular, there is no overshadowed appearance of (S, N) inG\(M∪
O12∪O34), whereM is the set of vertices that are major with respect to
(S, N).
(e) There is no enlargement of (S, N) inG\(O12∪O34), and (S, N) is maximal
inG\(O12∪O34).
Lemma 6.3 Let Gbe Berge and square-free, and let J be a 3-connected graph. Let (S, N) be a J-strip system in G. Let m∈ V(G)\V(S, N). If m is major for some choice of rungs in(S, N), then one of the following holds:
1. There is a J-enlargement with a non-degenerate appearance in G (and such an appearance can be found in polynomial time).
2. There is a J-strip system (S′, N′) such that V(S, N) ⊂ V(S′, N′) with
strict inclusion (and(S′, N′)can be found in polynomial time).
3. mis major with respect to(S, N).
4. Ghas a specialK4-strip system.
Proof. Letmbe major for some choice of rungs in (S, N). Suppose that there is noJ-enlargement with a non-degenerate appearance inG, and (S, N) is maximal in G, and that m is not major with respect to (S, N). Let X be the set of neighbors ofm. LetM be the set of vertices ofV(G)\V(S, N) that are major with respect to (S, N). LetM∗
be the set of vertices of V(G)\V(S, N) that are major with respect to some choice of rungs. Som∈M∗\M.
As noted earlier, every degenerate appearance of any 3-connected graph contains a square, soGcontains no degenerate appearances of any 3-connected graph. Hence, by 8.4 in [7], we must haveJ =K4. LetV(J) ={1,2,3,4}. Since mis major with respect to some choice of rungs and not major with respect to the strip system, we may choose rungsRij, R′ij(i6=j∈ {1,2,3,4}) forming line
graphsL(H) andL(H′) respectively, so thatX saturatesL(H) but notL(H′).
Let the ends of eachRij be rij and rji, where{rij |j ∈ {1,2,3,4} \ {i}}is
a triangleTi for eachi. Similarly, let the ends of eachR′ij ber
′
ij andr
′
ji, where
{r′
ij |j∈ {1,2,3,4} \ {i}}is a triangleTi′ for eachi.
SinceX saturates L(H), it has at least two members in each of T1, . . . , T4,
and sinceX does not saturate L(H′
), there is some T′
i that contains at most
one member ofX. SinceT3=T3′ andT4=T4′ we may assume that|X∩T1|= 2
and|X∩T′
1|= 1, sor12∈X, r12′ 6∈X, and exactly one ofr13, r14 is inX, say r13∈X andr146∈X. Also, at least two vertices ofT3are inX, and similarly for T4, so there are at least two branch-vertices ofH′incident inH′with more than
one edge in X. By 5.7 in [7] applied to H′, we deduce that 5.7.5 in [7] holds,
so (since odd branches ofH′ correspond to even rungs inL(H′) and vice-versa)
there is an edgeij ofJ such that
R′
ij is even and [X∩V(L(H
′))]\V(R′
ij) = (T
′
i ∪T
′
j)\V(R
′
ij). (1)
In particular,T′
i andT
′
jboth contain at least two vertices inX, soi, j≥2. Since
r13∈X, it follows that one ofi, j is equal to 3, sayj = 3, and sor13=r31, in
other wordsR13 has length 0. Hencei∈ {2,4}. We claim that:
i= 4. (2)
For suppose that i = 2. By (1) R23 is even and [X ∩V(L(H′))]\V(R23) =
{r′
21, r24, r31, r34}. Since at least two vertices of T4 are in X it follows that r42=r24andr43=r34 (andr41∈/ X). HenceR24andR34 both have length 0,
and sinceR23 is even this is a contradiction to the last axiom in the definition
of a strip system. Thus (2) holds.
Therefore we havei= 4 andj= 3. So (1) translates to:
R34 is even and [X∩V(L(H′))]\V(R34) ={r31, r32, r41, r42}. (3)
This implies that V(R′
12)∩X = ∅; moreover, if r23 ∈ X then r23 =r32, and
similarly ifr24∈X thenr24=r42.
One ofR23, R24has length 0, the other has odd length,R14has odd
length, andr21∈X. (4)
Since the pathr32-R23-r23-r24-R24-r42 can be completed to a hole viar42-r43 -R34-r34-r32, the first path is even, and so exactly one of R23, R24is odd. Since
the same path can be completed to a hole viar42-r41-R14-r14-r13-r32, it follows
thatR14is odd. Since one of R23, R24 is odd, they do not both have length 0,
and hence at most one ofr23, r24is inX. On the other hand, sinceX saturates L(H), the triangleT2 has at least two vertices fromX; it follows thatr21∈X
and that exactly one ofr23, r24 is inX (in other words exactly one of R23, R24
has length 0). Thus (4) holds.
R12 has length 0. (5)
For suppose thatr216=r12. Ifr21 has a neighbor inR′12, thenmcan be linked
onto the triangleT′
