Simple Reduction Rules

Given an instance(G, B, W) of the Minimum Efficient Edge Domination problem,

we dene the following potential function: ϕ(G, B, W) = 2|V(G)| − |B| − |W|. Note that ϕ(G, B, W) ≥ 0, with equality if and only if |V(G)| = |B|= |W|. The proofs in this chapter rely on a series of reduction rules. Applying these rules allows us to assume certain useful properties about the input instance. In this section we introduce some simple reduction rules that can be applied to general instances of the problem.

We apply the following reduction rules in the order given, i.e. before the applica- tion of any rule, we assume that all previous rules have been applied exhaustively. We will show that after any of these reduction rules is applied, the reduced problem will

have a valid solution if and only if the original one did. Moreover, the value of optimal solution of the reduced problem will be the same as that for the original problem, un- less otherwise stated (see comment after Rule R4). Each of these reduction rules runs in polynomial time. To see that this yields an algorithm that runs in polynomial time overall, note that each rule either solves the problem optimally or decreases the value of

ϕ(except Rule R1, which recurses on disjoint sets of vertices).

Note that if the problem instance (G, B, W) has no valid solution for some com-

ponent ofG, then the problem has no valid solution for G. If every component ofGhas

a valid solution, an optimal solution for G can be obtained by taking the union of the

optimum solutions for the components ofG.

Given an instance(G, B, W) of the problem withϕ(G, B, W)>0:

R1 IfGis disconnected, solve the problem componentwise. If the problem has no solution

on any of the components ofG, it has no solution onGitself. Note that ifD1, . . . , Dk

are the components ofG thenϕ(G, B, W) =ϕ(D1, V(D1)∩B, V(D1)∩W) +· · ·+

ϕ(Dk, V(Dk)∩B, V(Dk)∩W) and the weight of the optimal solution is the sum of

the weights of the optimal solutions on each of its components.

R2 IfB∩W 6=∅ then output(G, V(G), V(G)). Indeed, ifB∩W 6=∅then the problem has no valid solution, since no vertex can be both black and white. In this case we can output the instance (G, V(G), V(G)), which clearly has no valid solution and

which ϕevaluates to 0.

R3 Ifx∈W, output(G−x, B∪N(x), W\ {x}). Indeed, all neighbours of a white vertex

must be black. Once we force all the neighbours of a white vertex to be coloured black, we can remove the white vertex without loss of generality.

Note that Rule R3 allows us to assume that from now on,W =∅ in the input instance. R4 Ifx, y∈B, xy∈E(G), output (G− {x, y}, B\ {x, y}, W∪N({x, y})). Indeed, ifxy

is an edge where both x and y are black, then in any valid solution, xy must form

part of the matching, and so any other neighbours ofx and y must be white.

Rule R4 allows us to assume that from now on, B is an independent set in the

input instance. Note that this rule reduces the value of the optimal solution in G by w(xy).

R5 If N(x) = 0, output (G, B, W ∪ {x}). In this case, x cannot be matched with any

R6 IfV|(G)|= 2, output(G, V(G), W). In this case,Gmust consist of a single edge, so

both its vertices must be black.

This allows us to assume that Gis connected and contains at least 3 vertices.

R7 Ifx∈B, N(x) = 1, output (G, B∪N(x), W). In this case,x must be matched with

its unique uncoloured neighbour.

R8 If x, y ∈B, z∈ V(G), xz, yz ∈ E(G), Output (G, B, W ∪ {z}). If z has two black

neighbours, it cannot be coloured black.

R9 If G contains a diamond, colour it in the only possible way. If w, x, y, z ∈ V(G),

with wx, wy, wz, xz, yz∈E(G) andxy 6∈E(G), then in any valid solution w and z

must be black andxand ymust be white (see also Figure 6.3). We therefore output

(G, B∪ {w, z}, W ∪ {x, y}).

R10 If x1x2, x2x3, x3x4, x4x1 ∈ E(G), x1x3, x2x4 6∈ G, and x1 ∈ B output (G, B ∪ {x3}, W ∪ {x2, x4}). The vertices of an induced C4 must alternate in colour (see also Figure 6.2).

Figure 6.2: A C4 subgraph has two possible colourings

R11 IfGcontains aK4output(G, V(G), V(G))In this case, no valid colouring is possible (see also Figure 6.3).

R12 IfGcontains a buttery, colour it in the only way possible. Ifx, y1, y2, y3, y4 ∈V(G) with xy1, xy2, xy3, xy4, y1y2, y3y4 ∈ E(G) and y1y3, y1y4, y2y3, y2y4 6∈ E(G) then in any solution, x must be white and y1, y2, y3, y4 must be black (see also 6.3). We therefore output (G, B∪ {y1, y2, y3, y4}, W ∪ {x}).

R13 IfGcontains a paw, the leaf vertex and the central vertex must have dierent colours.

Ifw, x, y, z∈V(G)withwx, xy, xz, yz ∈E(G),wy, wz 6∈E(G), then in any solution,

w and x must have dierent colours (see also 6.4). Therefore ifx ∈B then output

(G, B, W ∪ {w})and if w∈B then output(G, B, W ∪ {x}).

We may now assume thatG is(butterf ly, diamond, K4)-free. Armed with these reduction rules, we are now ready to study the parameterized complexity of the problem.

(a) diamond (b) buttery (c)K4

Figure 6.3: A diamond or buttery subgraph has only one possible valid colouring. A

K4 subgraph has no valid colourings

x w

y

z

Figure 6.4: In a paw subgraph, wand x must have dierent colours.