Graphs containing a cycle of length 6

We assume that all induced cycles inGare of length at most 6. SupposeCis an induced

cycle on 6 verticesv1, . . . , v6.

Ifxis a neighbour of the cycle, we know it cannot be of type 1 (by Claim C2). By

Claim C1, it cannot be of type≥4. If xis of type 3 and has two consecutive neighbours

in the cycle, then it's third neighbour must be one of the two vertices not adjacent to the rst two. We say that such type 3 vertices are of type 3a. If x does not have 2

consecutive neighbours in the cycle, then it must be adjacent to v1, v3, v5 or v2, v4, v6. We say that such vertices are of type 3b. Ifxis of type 2, and is adjacent to two vertices

of the cycle which are of distance 2 from each other in the cycle, sayx were adjacent to v1 and v3, thenG[v1, x, v2, v6, v5, v4], would be aS1,1,3, which would be a contradiction. Thus ifxis of type 2, it is adjacent to either two opposite or two consecutive vertices of

the cycle. We say such a vertex xis of type 2a and 2b respectively (see also Figure 6.6).

Suppose there is a vertex x of type 3a. Without loss of generality, x is adjacent

tov1, v2 andv4. If xis black then, becauseG[x, v1, v2, v4]is a paw,v4 must be white, so

v3 and v5 must be black, sov2 must be white, so v1 must be black, sov6 must be white. Ifx is white, then v1, v2 must be black, sov6 must be white. In both cases v6 is white, thus in any valid partition v6 must be white.

(a) Type 2a (b) Type 2b (c) Type 3a (d) Type 3b

Figure 6.6: The possible neighbours of a C6 in an S1,1,3-free graph

R18 If C=v1−v2−v3−v4−v5−v6−v1 is a cycle of length 6, and x is adjacent to 3 vertices on C, two of which are adjacent, sayv1, v2, v4, output(G, B, W ∪ {v6}). We may now assume there are no vertices of type 3a.

Suppose, for contradiction, that there is a vertexx of type 3b and a vertexy of

type 2b. Without loss of generality, we may assume x is adjacent to v1, v3, v5 and y is adjacent to v1 and v2. But then x must be adjacent to y otherwise G[v1, v6, y, x, v3, v4] would be an S1,1,3. But then G[v1, y, v2, x] would be a diamond. This contradiction implies that if there is a vertex of type 3b, the only neighbours to the cycle not of type 3b must be of type 2a.

Suppose there is a vertex x of type 3b, adjacent to v1, v3 and v5, say. Then

G[v1, v2, v3, x],G[v3, v4, v5, x]andG[v5, v6, v1, x]areC4's, so the only valid colourings are

v1, v3, v5 white,v2, v4, v6, xblack or vice-versa. This means that any type 2a neighbours must be black and there can be no two such neighbours with the same neighbourhood in the cycle. In fact, because all the black vertices in the cycle need to have a black neighbour, there must be exactly one of each of the three sorts of type 2a neighbours and they must be pairwise nonadjacent.

R19 Suppose C = v1−v2−v3−v4−v5−v6−v1 is a cycle in G of length 6, and x is adjacent to 3 vertices on C, none of which are adjacent, say v1, v3, v5. Check that there are exactly three vertices y1, y2, y3 such that yi's neighbours on the cycle are

xi and xi+3, otherwise output (G, V(G), V(G)).

Any type 2a vertices must be black and must be matched to a vertex on the cycle. Thus any other neighbour (i.e. a neighbour of type 3b or of type 0) of a type 2a vertex must be white.

R20 SupposeC is a cycle inGof length 6, if there is a type 3b or type 0 vertexxadjacent

Suppose there are type 0 verticesz, z0 such thatxand z0 are adjacent toz. Then x must be adjacent to z0 otherwise G[v1, v2, v6, x, z, z0] would be an S1,1,3. Thus every type 0 vertex must be adjacent to a vertex of type 3b. In other words, the graph is

dominated by the cycle, the type 3b vertices and all three type 2a vertices. However, there are at most 2 ways of colouring this set of vertices (depending on the colour of vertex v1). Thus in this case, the problem can be solved in polynomial time by the use of Lemma 69.

R21 SupposeCis a cycle inGof length 6, if there is a type 3b vertexx. Try both possible

colourings forC and apply Lemma 69. If a solution exists, output the optimal one,

otherwise output (G, V(G), V(G)).

So we have now reduced to the case where there are no vertices of type 3. Suppose there is a type 2a vertexy and a type 2b vertexxwith a common neighbour in the cycle.

Without loss of generality, we may assume x is adjacent to v1 and v2 and y is adjacent tov1 andv4. The verticesxand ycannot be adjacent, otherwiseG[v1, x, v2, y]would be a diamond. But then G[v1, v6, x, y, v4, v3], would be an S1,1,3. Thus no type 2a vertex can have a common neighbour in the cycle with a type 2b vertex.

No two type 2b vertices may have a common neighbour in the cycle (since G is

(diamond, butterf ly, K4)-free). If there are 3 type 2b vertices, without loss of generality

a1 adjacent to v1 and v2, a2 adjacent to v3 and v4 and a3 adjacent to v5 and v6, then there can be no type 2a vertices. Also, if a1 is white, then v1 and v2 must be black, sov3 and v6 must be white, so a2, a3, v4, v5 must all be black, which is a contradiction. Thus in this case all the ai must be black and the cycle must be coloured in alternating

colours.

R22 SupposeC is a cycle inGof length 6, if there are 3 verticesa1, a2, a3 of type 2b adja- cent tov1&v2,v3&v4andv5&v6respectively, check if(G, B∪{a1, a2, a3, v1, v3, v5}, W∪ {v2, v4, v6})or(G, B∪{a1, a2, a3, v2, v4, v6}, W∪{v1, v3, v5})violates Rule R2. If nei- ther of them does, pick the one that had the better weight. If one of them violates the rule, output the other. If both violate the rule, output (G, V(G), V(G)).

If there are 2 type 2b vertices which are not opposite, without loss of generality

a1adjacent tov1 andv2,a2 adjacent tov3 andv4, then becauseG[v2, a1, v1, v3]is a paw,

v2 and v3 must have dierent colours. Again, in this case we cannot have any type 2a vertices. Suppose without loss of generality, thatv2 is white. Thena1, v1 andv3 must be black, sov6is white, sov4andv5are black, which is a contradiction, sinceG[a2, v3, v4, v5] is a paw. Thus in this case there is no valid colouring.

R23 If C is a cycle in Gof length 6 and has two type 2b vertices that are not opposite,

output (G, V(G), V(G)).

Now suppose there are 2 type 2b vertices which are opposite, without loss of generalitya1 adjacent to v1 and v2,a2 adjacent tov4 and v5. Note that in this case we may (or may not) have some type 2a vertices adjacent to v3 and v6. Ifv3 is black then since G[v2, a1, v1, v3]is a paw, v2 must be white. Similarly ifv6 is black thenv1 must be white. v1 and v2 cannot both be white, so at least one ofv3 andv6 must be white.

If there are no vertices of type 2a, there is only one way to colour the cycle. R24 IfC is a cycle inGof length 6 and has two type 2b verticesa1, a2 that are opposite

and are adjacent tov1&v2 andv4&v5 respectively, butChas no type 2a neighbours, output (G, B∪ {v1, v2, v3, v4}, W∪ {v3, v6, a1, a2}).

If there is a vertex x of type 2a, it is adjacent to v3 and v6, at least one of which must be white, sox must be coloured black. If x is adjacent to a1, this uniquely determines the colouring ofC.

R25 IfC is a cycle inGof length 6 and has two type 2b verticesa1, a2 that are opposite and are adjacent tov1&v2 andv4&v5 respectively andxis a type 2a vertex adjacent to a1, then output(G, B∪ {v1, v2, v3, v4}, W ∪ {v3, v6, a1, a2}).

If there are two opposite type 2b vertices, there must be a type 2a vertex and no vertex of type 2a may be adjacent to to a vertex of type 2b. We now show that the cycle, together with the type 2 vertices dominates the graph. (Note that there are at most 3 ways to colour this set of vertices, depending on the colours of v3 and v6.) Let x be a type 2a vertex,a1 be a type 2b vertex adjacent to v1 and v2 and supposea1 has a type 0 neighbour y. Then y must be adjacent to x otherwise G[v6, x, v5, v1, a1, y] would be an S1,1,3. Now suppose thatz is a type 0 vertex adjacent tox and z0 is a type 0 vertex adjacent toz, but notx. ThenG[v6, v1, v5, x, z, z0]would be anS1,1,3. This contradiction shows that the cycle, together with the type 2 vertices dominates the graph.

R26 IfC is a cycle inGof length 6 and has two type 2b verticesa1, a2 that are opposite and are adjacent tov1&v2 andv4&v5, the cycle and the type 2 vertices dominate the graphG. Try all 3 of the possible colourings for the cycle and vertices of type 2 and

apply Lemma 69.

Now suppose there is exactly 1 type 2b vertex, without loss of generality a1 adjacent tov1 and v2. Note that in this case we can again have type 2a vertices, but as

before, only if they are adjacent tov3 andv6. If v1 is white, then a1, v2 and v6 must all be black, so v3 must be white, sov4 must be black, so v5 must be black (since v4 has no neighbours outside the cycle). This is a contradiction, sincev5 cannot be black if it has 2 black neighbours. Thus our original assumption must have been wrong andv1 must be coloured black. Similarly,v2 must be black. This yields a unique colouring of the cycle. R27 If C is a cycle in G of length 6 and has exactly one type 2b vertex a1, adjacent to

v1&v2, say, output (G, B∪ {v1, v2, v4, v5}, W∪ {a1, v3, v6}).

So we are now left with the case where there are only type 2a vertices. We claim that the cycle together with the type 2a vertices dominates the rest of the graph. Indeed, suppose not. Then there must be a type 2a vertexx and type 0 verticesz, z0 such that x, z, z0 form a path of length 3. Without loss of generality, assume that x is adjacent

to v1 and v4. Then G[v1, v2, v6, x, z, z0] is an S1,1,3, which is a contradiction. Thus the cycle, together with the vertices of type 2a does indeed dominate the whole graph. Now we need to look at how many ways we can colour this set. The cycle can be coloured in 5 ways: alternately black/white (2 ways) or black, black, white, black, black, white (3 ways). Each of these gives at most one valid way of colouring all the type 2a vertices. R28 If C is a cycle in Gof length 6 and all neighbours of the cycle are of type 2a, then

the cycle together with the type 2a vertices dominates the graph. Try all 5 of the possible colourings for the cycle and vertices of type 2a and apply Lemma 69.

After applying the above reduction rules, we end up with a graph which contains no induced cycles of length≥6.