NP -Completeness

For some of the hardness reductions presented in this chapter,NP-complete variants of the following SETCOVER(SC) problem are used.

SETCOVER(SC)

Input: A family of setsF={S₁, . . . ,S_`}over a universeU={u₁, . . . ,u_r} of elements and a positive integerh.

Question: Is there a size-h set cover, that is, a collection ofhsets inF whose union isU?

Note that even 3-SETCOVER(3-SC), where each set fromF is of size at most three, isNP-complete [Kar72]. In order to make the reduction work, we need to leth,|F|, and the number of occurrences of every element in the sets inF fulfill the following properties whose correctness is easy to see:

1. We add (multiple copies of) singletons to the familyF to ensure that

|F| −2h+1≥0 and to ensure that every element appears in at least hsets inF.

2. We also add a new elementu^∗to the universeU, addhcopies of the singleton {u^∗}to the familyF, and seth:=h+1 to ensure that each element appears in at most|F| −hsets inF.

A common starting point for each of our SC reductions is to transformF (andU) into a binary matrix in a similar way as Christian et al. [Chr+07]

transformed DOMINATINGSETinstances into binary matrices. The corre-sponding|F| × |U|-matrix is called SC-matrixand is defined as

M(F,U)=(x_i,j) withx_i,j:=

(1 ifu_j∈S_i, 0 otherwise.

6 Lobbying

U= {1, 2, 3, 4, 5}

F = {S1={2, 4}, S₂={3, 5}, S₃={1, 2, 5}, S4={1,2,3}, S₅={4,5} }

1 2 3 4 5 S1:

S2: S3: S4: S₅:

0 1 0 1 0 0 0 1 0 1 1 1 0 0 1 1 1 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0

6 7 0 0 0 0 0 0 0 0 0 0 1 0 0 1 k:= |F| −h=3 h=2

Figure 6.2: Illustration of the reduction from 3-SC to LOBBYING. Left: a 3-SC instance; Right: the constructed LOBBYINGinstance. The SC-matrix together with two dummy rows and two dummy columns (inside the dashed polygon) ensure that any set of rows which are not selected by some solution of size three corresponds to a set cover. The 3-SC solution (S₄andS₅) is highlighted in boldface on the left. The LOBBYINGsolution (modifying the first three rows) is highlighted by gray backgrounds.

Observe that each column j∈{1, . . . ,|U|}has 1s in at most (|F| −h) rows.

We will make use of the SC-matrix in Section6.2.3and in the following theorem.

An instance of LOBBYINGwithk≥ d(n+1)/2e(that is, more than half of the rows can be modified) is a yes-instance. Following general ideas of Mahajan and Raman [MR99] in the context of “above and below guarantee parameterization”, we investigate the complexity of LOBBYING in case the maximum number of 1s per row is bounded by a constant and kis slightly below the boundd(n+1)/2e. Such a parameterization is calledbelow guarantee.

Theorem 6.1. LOBBYINGremainsNP-complete for input matrices where the maximum number s of ones per row is three and for k⁰=1of the below guarantee parameter k⁰=(d(n+1)/2e −k).

6.2 Theoretical Results

Proof. Obviously, given a matrix Aand k= d(n+1)/2e −1 rows of A, we can check in polynomial time whether each row ofAhas at most three 1s and whether modifying thekgiven rows makes every column have more 1s than 0s. Hence, LOBBYINGwiths=3 andk⁰=1 remains inNP.

For theNP-hardness result, we describe a polynomial-time reduction from 3-SC to LOBBYINGwhere each row contains at most three 1s. The reduction is illustrated with an example inFigure 6.2. Let (F,U,h) denote a 3-SC instance. First, compute the SC-matrixM(F,U), which we refer to as original rows andoriginal columnsin the following. Second, add

|F| −2h+1 additionaldummy rowsand|F| −2h+1 additionaldummy columnscontaining only 0s. Recall that we can assume that|F| −2h+1≥0.

Finally, for 1≤i≤ |F| −2h+1, change the entry in theith dummy column of theith dummy row to 1. To complete the construction we setk:= |F| −h.

Each original row has at most three 1s since each set ofF has at most three elements. Each dummy row has exactly one 1. Further, the below guarantee parameterk⁰= d(n+1)/2e −k= d(2|F| −2h+1)/2e −(|F| −h)=1.

The quota for a column to have more 1s than 0s is|F| −h+1. Since every original column has 1s in at most|F| −horiginal rows and no 1s in the dummy rows, no original column has more 1s than 0s in the matrix. More precisely, the jth original column has gap value|F| −h+1− |{S_i|u_j∈S_i}|. Each dummy column has gap value |F| −h. Hence the maximum gap valuegis|F| −h.

Obviously, the reduction runs in polynomial time. Now, we show that (F,U,h) is a yes-instance of 3-SC if and only if the constructed matrix can be modified to have more 1s than 0s in every column by changing at mostk rows to all-1-rows.

For the “only if ” part, suppose that (F,U,h) is a yes-instance for 3-SC.

LetF⁰⊆F denote a set cover of sizeh. By modifying each original rowi that corresponds to a setS_i∉F⁰, every column has more 1s than 0s, that is, at least (|F| −h+1) 1s: Each jth dummy column gains (|F| −h) 1s from the modified original rows and another 1 from the jth dummy row.

Each original column gains (|F| −h) 1s from the modified original rows and another 1 from an unmodified original row, since the sets corresponding to the unmodified original rows form a set cover.

For the “if ” part, suppose that the constructed matrix has more 1s than 0s in every column by modifying at mostk= |F| −hrows. First, observe that the number of modified rows is exactly|F| −hsince the maximum gap value is|F| −h. Second, no dummy row is modified: Assume towards a

6 Lobbying

contradiction that thei⁰th dummy row is modified. Since thei⁰th dummy column has only one 1 in the matrix, namely, at thei⁰th dummy row, this column cannot get a majority of (|F| −h+1) 1s by the modification of any

|F|−hrows including thei⁰th dummy row—a contradiction. Third, we show that the sets corresponding to the unmodified original rows form a size-h set cover for (F,U,h). Each original columnj⁰gets exactly (|F| −h) 1s from the modified original rows and no 1 from any dummy row. To get a majority of (|F| −h+1) 1s, column j⁰must contain another 1 in some unmodified original row. Hence, the sets corresponding to the unmodified rows form a set cover of sizek. This shows the correctness of the construction.

Proposition 6.1. LOBBYINGremainsNP-complete for every constant inte-ger value k⁰>1of the below-guarantee parameter k⁰=(d(n+1)/2e −k).

Proof. TheNP-containment follows analogously toTheorem 6.1. To show theNP-hardness for any constantk⁰>1, we take the SC-matrixM(F,U) and addxadditional dummy columns andxk⁰additional dummy rows where x:=(|F| −2h−1)/k⁰+2. Note that we can add singletons to the familyF to make sure k⁰is a divisor of (|F| −2h−1). We fill the added entries as follows: For each j∈{1, . . . ,|U|}, set the entries ofk⁰−1 arbitrary dummy rows at the original column jto 1. For 1≤j≤x, set the entry in the jth dummy column of the [(j−1)k⁰+1]th, [(j−1)k⁰+2]th, . . ., [(j−1)k⁰+k⁰]th dummy rows to 1. Setk:= |F| −h.

In total, the constructed matrix has 2|F| −2h+2k⁰−1 rows and|U| +x columns. The quota for a column to have more 1s than 0s isq= |F| −h+k⁰. Each original column has gap valueq−(|{S_i|u_j∈S_i}| +k⁰−1)= |F| −h+ 1−(|{S_i|u_j∈S_i}|) and each dummy column has gap valueq−k⁰= |F| −h.

The reduction runs in polynomial time. Now, it remains to show the correctness.

Suppose that (F,U,h) has a set coverF⁰⊆F of sizeh. By modifying each original row i that corresponds to a setS_j∉F⁰, every column has more 1s than 0s.

Conversely, suppose that the constructed matrix has more 1s than 0s after modifying at mostkrows. Using the analogous reasoning as in the proof forTheorem 6.1, we can show that exactlykoriginal rows are modified and the original rows which are not modified correspond to the sets which can form a set cover for (F,U,h).

6.2 Theoretical Results