Minimum Cut Problem

Theorem 5.13. For the minimum s-t-cut problem on a connected graph G and for a discrete uncertainty set U = {c1, . . . , cm}, we can polynomially

reduce Problem (M2_{) to Problem (M}3_{) 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 Theo-

rem 3.2). The proof is similar to the proofs in the previous sections. Let a con- nected graph G = (V, E) and an uncertainty set U = {c1, . . . , cm} be given.

We then define the graph Gcut

k := (Vkcut, Ekcut) with Vkcut := V ∪ {v0, . . . , vk−1}

and

E_kcut := E ∪ {li = (s, vi) , ri = (vi, t) : i = 0, . . . , k − 1} .

Note that the graph is constructed analogously to the graph Gsp in the proof

of Theorem 5.4 but we add one more path. The latter construction is shown in Figure 5.4. Note that for any feasible solution δ(S) either the edge li or the

v0

v1

v2

s t

G

Figure 5.4: The graph Gcut

k for k = 3.

edge ri is contained in δ(S) for each i = 0, . . . , k −1, but not both. The idea of

the proof is to define scenarios on Gcut

k that force the min-max-min problem

to choose k solutions which use only the edge ri for each i = 0, . . . , k − 1.

Note that a feasible solution could use all of the edges ri simultaneously. To

this end, we define scenarios ¯c1, . . . , ¯cm on Gcutk by extending every scenario

ci ∈ U by zero on l1, . . . , lk−1 and r0 and by 2M on r1, . . . , rk−1 and l0. Here

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

and all edges l0, . . . , lk−1 have cost 0 except edge li which has cost 2M .

Furthermore all edges r0, . . . , rk−1 have cost 2M , except for edge ri which

has cost −M . Now if we choose a solution x(0)_{, . . . x}(k−1) _{of the min-max-}

min s-t-cut problem on Gcut_k , where x(i) uses edge ri and all edges lj with

j 6= i, 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 . Therefore we know that every optimal solution of the min-max-min problem on the graph Gcut

k must contain for each

edge ri a solution which only uses edge ri under the new edges r0, . . . , rk−1.

Otherwise, if for any ri0 this is not true, then edge li0 must be used. In the case if i0 ≥ 1, in scenario di0 the minimum mini=1,...,kc

>_x(i) _{is then 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. In the case if i0 = 0 by

the same reasoning applied to the scenarios ¯ci, in every optimal solution there

must be contained a solution which only uses edge r0. So letx(0), . . . , x(k−1)

be an optimal solution with the properties like above. Then we have max c∈{d1,...,dk−1} min i=0,...,k−1c > x(i) = −M by the reasoning above. 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,...d k−1} min i=1,...,kc > x(i) = min x∈Xc∈{cmax1,...,cm} c>x.

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

Corollary 5.14. For any fixed k ∈ N and fixed m > k, Problem (M3_{) is}

strongly NP -hard for the minimum s-t-cut problem for uncertainty sets U with |U | = m.

Proof. Problem (M2_{) for the minimum s-t-cut problem is strongly NP -hard}

for m ≥ 2 (see Table 3.1). From Theorem 5.13 and since all numbers in the proof remain of polynomial size, we derive that the min-max-min variant of the same problem is strongly 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 s-t-cut problem. Therefore the restriction to m > k is necessary.

As we have seen in Table 3.1 Problem (M2_{) for the normal minimum cut}

problem is strongly NP -hard if we assume the number of scenarios to be part of the input. In the following we show that this also holds for Problem (M3_).

Theorem 5.15. For the minimum cut problem on a connected graph G and for a discrete uncertainty set U = {c1, . . . , cm}, we can polynomially reduce

Problem (M2_{) to Problem (M}3_{) with at least m + k − 1 scenarios.}

Proof. The result can be proved analogously to the proof of Theorem 5.13 by using the graph ¯Gcut

k := ¯Vkcut, ¯Ekcut
with ¯Vkcut := V ∪ {v0, . . . , vk−1} and

¯

E_kcut := E ∪ {fi = (w, vi) : i = 0, . . . , k − 1}

for an arbitrary node w ∈ V . The latter construction is shown in Figure 5.5.

w

v1

v0 v2

G

f0 f1 f2

Figure 5.5: The graph ¯Gcut_k for k = 3.

Corollary 5.16. For any fixed k ∈ N, Problem (M3_{) is strongly NP -hard}

Proof. Problem (M2_{) for the minimum cut problem is strongly NP -hard if the}

number of scenarios is part of the input (see Table 3.1). From Theorem 5.13 and since all numbers in the proof remain of polynomial size, we derive that the min-max-min variant of the same problem is strongly NP -hard if the number of scenarios is at least k + 1.