Chapter 2 The Submodular Cost Allocation Problem
2.5 Hardness of MSCA
Ifk = 2, the MSCA problem can be reduced to submodular function minimization as follows. Let f : 2V → R+be the function such thatf (S) = f1(S)+f2(V−S). As shown in the following proposition,
f is submodular. A submodular function can be minimized in polynomial time [61,74,104,142,154], and therefore MSCA is in P whenk = 2.
Proposition 2.5.1. Letf1 : 2V → R and f2: 2V → R be two submodular functions. Let f : 2V → R
be the function such that f (S) = f1(S) + f2(V − S) for all S ⊆ V . The function f is submodular.
Proof. Let g2 be the function such that g2(S) = f2(V − S) for all S. We can verify that g2 is
submodular as follows. Consider any two setsA and B. Since f2 is submodular, we have
g2(A) + g2(B) = f2(V − A) + f2(V − B)
≥ f2((V − A) ∩ (V − B)) + f2((V − A) ∪ (V − B))
=f2(V − (A ∪ B)) + f2(V − (A ∩ B))
=g2(A∪ B) + g2(A∩ B)
Since f1 and g2 are submodular, their sum is also submodular.
In this section, we prove that, for any k ≥ 3, the MSCA problem does not admit any finite approximation factor. The reason is that for k≥ 3 it is NP-hard to decide whether the optimal
solution has value zero or nonzero.
Theorem 2.5.2. It is NP-hard to decide whether the optimal value for an instance of MSCA is zero, even for k = 3 and functions of the form fi(S) = ci(S) + δGi(S) where ci is a linear function
and δGi is the cut function of a directed graph Gi.
We start with the following known fact. For any non-negative submodular function, the collection of sets of zero value forms a lattice. (We recall that a boolean lattice is a family of sets L ⊆ 2V closed under unions and intersections, i.e. ∀A, B ∈ L; A ∩ B ∈ L, A ∪ B ∈ L.) This observation suggests that the question of checking for solutions of zero value is related to the following problem that we call Lattice Matching.
Lattice Matching. Given k boolean lattices L1,L2, . . . ,Lk ⊂ 2V (by a suitable compact repre-
sentation), findk disjoint sets S1∈ L1, . . . , Sk ∈ Lk such that Ski=1Si =V .
We show that this problem is NP-complete fork≥ 3, by a reduction from 1-in-3 SAT [153]. A technical issue is the question of representing a lattice compactly on the input. For this purpose we use a representation of lattices by directed graphs that goes back to Birkhoff [20]. In fact this construction also provides explicit submodular functions whose zeros are exactly the points of the respective lattice, and hence we prove Theorem 2.5.2.
A special case of a lattice is the collection of all unions of sets from some partition (collection of disjoint sets)Pi. Thus the Lattice Matching problem has the following natural special case.
Partition Matching. Given k collections of sets P1, . . . ,Pk⊂ 2V, where each Pi is a partition
of a subset of V (the sets in Pi are disjoint), find disjoint sets S1, . . . , S` ∈ Ski=1Pi such that
S`
i=1Si=V .
Just like the problems above, the Partition Matching problem is in P for k≤ 2. We prove that this problem is NP-complete fork = 5. We leave it as an open question whether it is NP-complete fork = 3 and k = 4.
2.5.1 Representation of lattices by directed graphs
Lemma 2.5.3. For every non-negative submodular function f : 2V → R
+, the set L =
{S ⊆ V : f(S) = 0} is a lattice, i.e., it is closed under unions and intersections.
In the following, we describe how to encode a boolean lattice using a directed graph, and how this connects MSCA and Lattice Matching. We follow the construction of [78].
Definition 2.5.4. Given a lattice L ⊆ 2V such that ∅ ∈ L, let V
L = S {S : S ∈ L}. For each
v ∈ VL, let D(v) = T {S : S ∈ L, v ∈ S}. We define a directed graph GL = (VL, A) where A =
{(v, w) : v ∈ V1, w∈ D(v), w 6= v}.
The following lemma is implicit in [20, 78]. We include a simple proof for completeness.
Lemma 2.5.5. For every lattice L ⊆ 2V such that ∅ ∈ L, the directed graph GL encodes the lattice
in the sense that S ∈ L if and only if S ⊆ VL and GL has no arcs from S to VL\ S.
Proof. If S ∈ L, then S ⊆ VL by the properties of the lattice. Also, for each each v ∈ S,
D(v) =T{S0 ∈ L : v ∈ S0} ⊆ S and hence all arcs originating at v stay within S.
Conversely, if S ⊆ VL and there are no arcs leavingS, we know that for each v∈ S, D(v) ⊆ S.
By the properties of lattices,D(v) =T{S0 ∈ L, v ∈ S0} ∈ L. If S 6= ∅, we get that S =S
v∈SD(v)∈
L. If S = ∅, then S ∈ L by assumption.
Thus the directed graphGLencodes the latticeL in a compact way: its description has size O(n2),
where n = |V |. Furthermore, we observe that this description provides a submodular function whose zeros are exactly the sets inL.
Lemma 2.5.6. For a lattice L ⊂ 2V defined by the directed graph GL, the following function is
submodular and its zeros are exactly the sets in L:
fL(S) =|S \ VL| + δGL(S)
where δGL(S) =| {(v, w) ∈ A(GL) :v∈ S, w /∈ S} | is the directed cut function of GL.
Proof. By Lemma 2.5.5,S ∈ L if and only if S ⊆ VL and there are no arcs fromS to outside of S
It follows that the Lattice Matching problem — where the lattices are given by the associated directed graphs — is equivalent to checking whether the MSCA instance in which the functions are {fLi | i ∈ [k]} has zero cost. To prove Theorem 2.5.2, it remains to prove the NP-completeness of
Lattice Matching under this encoding.
2.5.2 NP-completeness of Partition Matching and Lattice Matching
In this section, we prove that the Lattice Matching problem is NP-complete for k = 3. First, as a warm-up, let us prove that its special case, the Partition Matching problem, is NP-complete for k = 7. We reduce from the following NP-complete problem [81].
3-bounded 3-set Packing. Given a system of triples T ⊆ 2V such that each element of V is
contained in at most 3 triples, it is NP-complete to decide whether there exists a collection of disjoint triples covering V .
Theorem 2.5.7. Partition Matching is NP-complete for k = 7.
Proof. Given an instance of 3-bounded 3-set Packing, we observe that each triple intersects at most 6 other triples (2 for each of its elements). Thus we can greedily color the triples with 7 colors in such a way that intersecting triples get different colors. We define Pi to be the collection of
all triples of color i. We obtain an instance of Partition Matching with k = 7, for which it is NP-complete to decide whether there exists a collection of disjoint triples covering V .
For lower values ofk, we use more careful reductions from the Monotone 1-in-3 SAT problem [153].
Monotone 1-in-3 SAT. Given a formula Vm
i=1(xi1 ∨ xi2 ∨ xi3) (without negations), it is NP-
complete to find a boolean assignment such that in each clause (xi1∨ xi2∨ xi3), exactly one variable
is True and two variables are False.
Theorem 2.5.8. Partition Matching is NP-complete for k = 5.
Proof. Given an instance of Monotone 1-in-3 SAT, we produce an instance of Partition Matching as follows. We define a ground setV consisting of
• Two elements xi
j,¬xij for each occurrence of a variablexj in clausei.
On this ground set, we define the following 5 collections of sets:
• P1 contains for each variablexj a set{vj} ∪ {xij :∀clause i containing variable xj}.
• P2 contains for each variablexj a set{vj} ∪ {¬xij :∀clause i containing variable xj}.
• P3 contains for each clausexi1∨ xi2∨ xi3 a set{xii1,¬x
i i2,¬x
i i3}.
• P4 contains for each clausexi1∨ xi2∨ xi3 a set{¬x
i i1, x i i2,¬x i i3}.
• P5 contains for each clausexi1∨ xi2∨ xi3 a set{¬x
i i1,¬x i i2, x i i3}.
We call the sets inP1∪P2variable-assignment sets, and the sets inP3∪P4∪cP5 clause-assignment
sets. Observe that in each collectionPi, the sets are pairwise disjoint. I.e., we have an instance of
Partition Matching withk = 5.
Now, if there is a boolean assignment such that exactly one variable is each clause is satisfied, we produce a solution of Partition Matching as follows. For each variable xj = True, we choose
the variable-assignment set containing vj that is in P2 (i.e. containing all the elements ¬xij).
For each variable xj = False, we choose the variable-assignment set containing vj that is in P1
(i.e. containing all the elements xij). Finally, for each clause xi1 ∨ xi2 ∨ xi3, we choose the set
corresponding to its assignment, from either P3, P4 or P5. It is easy to verify that these sets are
disjoint and cover the entire ground set V .
Conversely, let us assume that there is a collection of disjoint sets F ⊂ P1∪ P2∪ P3∪ P4∪ P5
that covers the ground setV . SinceF must cover the element vj for each variable, it must contain
a variable-assignment set in either P1 or P2 (no other sets contain vj). This choice determines a
boolean assignment: we set xj = True if vj is covered by a set from P2, and xj = False if vj is
covered by a set from P1. Now, consider the 6 elements xii1,¬x
i i1, x i i2,¬x i i2, x i i3,¬x i i3 for clause i.
Exactly 3 of these elements are covered by sets fromP1andP2, hence the remaining 3 elements must
form a clause-assignment set inP3∪ P4∪ P5. These clause-assignment sets correspond to satisfying
assignments and hence the formula is satisfied by the boolean assignment that we defined. Theorem 2.5.9. Lattice Matching is NP-complete for k = 3.
Proof. We use the same reduction as in the proof of Theorem 2.5.8, but we combine the 5 partitions into 3 lattices. Specifically, using the notation from the proof above, we define
• L1 = cl(P1∪ P3)
• L2 = cl(P2)
• L3 = cl(P4∪ P5)
where cl(P) means all the sets that can be generated from P by taking unions and intersections. By construction,L1,L2,L3 are lattices that contain all the sets that were contained in P1, . . . ,P5
(as well as some additional sets). In the case of a satisfiable formula, we can still choose disjoint sets coveringV as above. Let Si be the union of those of these sets that are contained in Li (i.e.,
Si is also inLi), andS1∪ S2∪ S3 =V is a feasible solution of the Lattice Matching problem.
The potential issue with this construction is that we might have created a feasible solution of the Lattice Matching problem in case the formula is not satisfiable. Let us argue that this is not the case. InL1, we get new sets obtained by intersecting sets in P1 and P3: specifically, this is the
singleton{xi
i1} for each clause xi1∨ xi2∨ xi3 (and sets obtained by taking unions of these singletons
with other sets inP1∪ P3). InL2, we obtain only the unions of sets inP2 (which are disjoint). In
L3, we obtain by intersection the singletons {¬xii1} for each clause xi1 ∨ xi2 ∨ xi3, and again sets
obtained by unions with other sets inP4∪ P5.
However, observe that for each variable, the element vj is still contained only in the variable-
assignment sets arising fromP1 and P2 (and larger sets inL1,L2). As before, vj must be covered
by a variable-assignment set and this defines a boolean assignment. This also covers either all the elements xij for variable xj, or all the elements ¬xij. Moreover, for each clause, the elements
x2i2 and ¬x2i2 can be covered only by triples corresponding to satisfying assignments of the clause (singletons are present only for the first variable in each clause). Thus we still need to use a triple corresponding to a satisfying assignment for each clause, and this is possible only if a globally satisfying assignment exists.