Spanning Tree Problem

Theorem 5.7. For the minimum spanning-tree problem on a graph G with at least k nodes and for a discrete uncertainty set U = {c1, . . . , cm}, we can

polynomially reduce Problem (M2) to Problem (M3) with at least m + k − 1 scenarios, for any fixed k < m.

Proof. Instead of (M2_{) we consider the equivalent Problem (M}2

0) (see The-

orem 3.1). The proof is similar to the proof of Theorem 5.4. Let a graph G = (V, E) with |V | ≥ k and an uncertainty set U = {c1, . . . , cm} be

given. We then define the graph Gst

k := (Vkst, Ekst) with Vkst := V ∪ {w}

and Est

k := E ∪ {fi = {vi, w} : i = 0 . . . , k − 1} where the vi are arbitrary

pairwise different nodes in G. In fact we add one node and connect it to k different nodes with exactly one edge. The latter construction is shown in Figure 5.2. Again the idea of the proof is to define scenarios on Gst

k that force v0 v2 v1 w G f0 f1 f2

Figure 5.2: The graph Gst

k for k = 3.

the min-max-min problem to choose solutions which use exactly one of the new edges each. Note that a feasible solution could use all of the new edges simultaneously. To this end, we define scenarios ¯c1, . . . , ¯cm on Gstk by extend-

ing every scenario ci ∈ U by zero on f0 and by 2M on the edges f1, . . . , fk−1,

where M can be chosen as

M := m X i=1 |E| X j=1 |(ci)j| + 1.

Then we add scenarios d1, . . . dk−1 such that in scenario di all edges in G have

cost M and all edges f0, . . . , fk−1 have cost (|V | + 1) M , except for edge fi

min-max-min spanning-tree problem on Gst

k, where x(i) uses edge fi and none

of the other new edges to connect node w, then for the objective value holds max c∈{¯c1,...,¯cm,d1,...dk−1} min i=0,...,k−1c > x(i) < M,

since on ¯ci the minimum is attained by x(0) and therefore, by the definition

of M , it is strictly lower than M . On the other hand on di the minimum is

attained by x(i) _{and is exactly −M since any spanning-tree in G uses exactly}

|V | − 1 edges. Therefore we know that every optimal solution of the min- max-min problem on the graph Gst_k must contain for each edge fi a solution

which only uses edge fi under the new edges. Otherwise, if for any fi0 this is not true, then, if i0 ≥ 1, in scenario di0 the minimum mini=1,...,kc

>_x(i) _is

greater or equal to M and therefore max c∈{¯c1,...,¯cm,d1,...dk−1} min i=0,...,k−1c > x(i) ≥ M

which cannot be optimal by the observation above. If i0 = 0 by the same

reasoning applied to the scenarios ¯ci every optimal solution must contain

a solution which only uses edge f0. So let x(0), . . . , x(k−1) be an optimal

solution like above. Then we have max

c∈{d1,...,dk−1}

min

i=0,...,k−1c

>_x(i) _{= −M}

by definition of the scenarios {d1, . . . , dk−1}. On the other hand

max c∈{¯c1,...,¯cm} min i=0,...,k−1c > x(i) = max c∈{c1,...,cm} c>x(0) > −M and therefore min

x(1)_,...,x(k)_{∈Xc∈{¯c1,...,¯}max_{cm,d1,...dk−1}}i=1,...,kmin c >

x(i) = min

x∈Xc∈{cmax1,...,cm} c>x.

and the min-max optimal solution must be contained in x(0)_{, . . . , x}(k−1) _.

Note that the spanning-tree problem on graphs with less than k nodes can be extended to graphs with at least k nodes by adding extra nodes and connecting them with exactly one edge to one of the nodes of the original graph. A solution for the original problem can be obtained by projecting the solution on the extended graph to the original graph. Therefore it is no restriction to consider only graphs with at least k nodes.

Corollary 5.8. For any fixed k ∈ N and fixed m > k, Problem (M3_{) is}

NP -hard for the minimum spanning-tree problem for uncertainty sets U with |U | = m.

Proof. Problem (M2_{) for the minimum spanning-tree problem is NP -hard}

for m ≥ 2 (see Table 3.1). From Theorem 5.7 we derive that the min-max- min variant of the same problem is NP -hard if the number of scenarios is at least k + 1.

Note that by Proposition 5.3 for k ≥ m Problem (M3_{) can be solved in}

polynomial time for the minimum spanning-tree problem. Therefore the re- striction to m > k is necessary.

Corollary 5.9. For any fixed k ∈ N, Problem (M3_{) with discrete U is}

strongly NP -hard for the minimum spanning-tree problem.

Proof. The min-max variant of the minimum spanning-tree problem is strong- ly NP -hard if the number of scenarios is not constant (see Table 3.1). The result follows since all numbers in the construction in the proof of Theo- rem 5.7 are polynomial in |U |.