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Branch and cut algorithms for detecting critical nodes in undirected graphs

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Abstract

In this paper we deal with the critical node problem, where a given number of nodes has to be removed from an undirected graph in order to maximize the disconnections between the node pairs of the graph. We propose an integer linear programming model with a non-polynomial number of constraints but whose linear relaxation can be solved in polynomial time. We derive different valid inequalities and some theoretical results about them. We also propose an alternative model based on a quadratic reformulation of the problem. Finally, we perform many computational experiments and analyze the corresponding results.

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Acknowledgements

The authors would like to thank two anonymous referees, whose constructive comments helped to improve the initial version of the paper.

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Correspondence to Andrea Grosso.

Appendix: Convex hull description

Appendix: Convex hull description

Ignoring the cardinality constraint ∑ iV x i K, formulation (8)–(11) reads as follows:

(54)
(55)

When the underlying graph G is a path, the constraint matrix of system (54) is totally unimodular, as (ignoring unit columns) the 1s appear in consecutive positions on each row. Therefore in this case the linear relaxation of the above set provides a description of the convex hull of integer solutions. However, the linear relaxation of (54)–(55) has fractional extreme points already for G=K 3 or when G is a tree. In the following subsections we give a linear-inequality description of the convex hull of (54)–(55) for some special cases. We first discuss here the proof technique and give some results that will be used later.

First of all we observe that the convex hull of (54)–(55) is a full-dimensional polyhedron, as the following N+N(N−1)/2+1 feasible points are affinely independent (recall that N=|V| and the number of variables is N+N(N−1)/2):

  • for iV, the point defined by setting x i =1 and all other variables to 0;

  • for i,jV,i<j, the point defined by setting x i =y ij =1 and all other variables to 0;

  • the origin.

The convex hull of (54)–(55) will be described by some candidate linear system Ax+Byd. In order to prove that this system actually describes the convex hull of (54)–(55), we will use a well-known technique due to Lovász [10]: first we will observe that Ax+Byd is a valid relaxation for our set; then we will show that for every optimization problem of the form

(56)

with (p,q)≠0, there is an inequality in the system Ax+Byd that is satisfied at equality by all the optimal solutions of problem (56). Since our set is full-dimensional, this proves the result, as when the objective function is parallel to a facet of the convex hull of (54)–(55), an inequality that is tight for all optimal solutions must induce that facet; thus all the facets of the convex hull of (54)–(55) appear in Ax+Byd.

We now discuss some preliminary results. In the following we assume that an objective function (p,q)≠0 is fixed. Also, expressions such as “optimal solution” or “optimal value” will always implicitly refer to problem (56).

Lemma 12

If p i >0 for some iV, then all optimal solutions satisfy x i =1. If q ij <0 for some i,jV,i<j, then all optimal solutions satisfy y ij =0.

Proof

If a feasible solution satisfies x i =0, we can increase x i to 1: this gives a feasible solution with better objective value. Similarly, if a feasible solution satisfies y ij =1, we can decrease y ij to 0. □

Therefore, from now on we assume that p0 and q0.

Note that if (x,y) is an optimal solution, then

$$ y_{ij}=\min \biggl\{1,\min\biggl\{\sum_{r\in V(P)}x_r:P\in \mathcal{P}(i,j)\biggr\}\biggr\}$$
(57)

for every ij such that q ij >0. A feasible solution (x,y) with y ij satisfying (57) for all ij (including those for which q ij =0), will be called a standard solution. Since a standard solution is uniquely determined by its x-components, when constructing a standard solution we will only specify x.

Lemma 13

If p=0 or q=0, then one of the bounds 0≤x i ≤1 is tight for all optimal solutions.

Proof

If p=0, then, since (p,q)≠0, there exist i,j such that q ij >0. We claim that all optimal solutions satisfy y ij =1. Assume this is not true, i.e., there is an optimal solution with y ij =0. Then the standard solution x=1 has a larger objective value, a contradiction.

If q=0, then, since (p,q)≠0, there exists i such that p i <0. We claim that all optimal solutions satisfy x i =0. Assume this is not true, i.e., there is an optimal solution with x i =1. Then the standard solution x=0 has a larger objective value, a contradiction. □

Lemma 14

If iV is the only node such that p i <0, then x i =0 for all optimal solutions.

Proof

If a feasible solution satisfies x i =1, we can decrease x i to 0 and set all other components of x to 1 (without changing y): this gives a better feasible solution. □

Therefore, from now on we assume that q0 and p i <0 for at least two nodes.

For a given objective function as in (56), we denote by Q the set of pairs ij such that q ij >0. We also define graph G Q =(V,Q). The following result will be used several times.

Lemma 15

Let iV be a node of degree 1 in G Q and let j be its unique neighbor in G Q . If p i <0 and ijE, then all optimal solutions satisfy x i +x j =y ij .

Proof

Assume that there is an optimal solution (x,y) such that x i +x j >y ij . Since q ij >0, it follows that x i =x j =y ij =1. Then we find a better solution by decreasing x i to 0 and taking the corresponding standard solution. □

We now consider two configurations in detail.

1.1 A.1 Clique with three nodes

Here we consider the case G=K 3 and prove that the convex hull of (54)–(55) is obtained by adding the following inequality to the linear relaxation of (54)–(55):

$$ y_{12}+y_{23}+y_{13} \le x_1+x_2+x_3+1.$$
(58)

Note that inequality (58) is valid, as it is a clique inequality.

Let px+qy be an objective function satisfying the above assumptions. Recall that |Q|≥1. If 1≤|Q|≤2, we can use Lemma 15 (the existence of an index i as required in Lemma 15 is guaranteed by the fact that p i <0 for at least two nodes). So we only have to consider the case |Q|=3, i.e., q ij >0 for all ij.

If q ij >0 for all ij, every optimal solution is a standard solution. Thus we have eight candidates, one for each possible choice of x∈{0,1}3. It can be checked that the only candidates for which (58) is not tight are x=0 and x=1. However, x=1 cannot be optimal, as by choosing an index i such that p i <0 and decreasing x i to 0, a feasible solution with larger objective value is obtained. It follows that if x=0 is not optimal, then all optimal solutions satisfy (58) at equality.

It only remains to consider the case in which x=0 is an optimal solution. Clearly in this case the optimal value is 0. If all optimal solutions satisfy x 1+x 2=y 12, the proof is complete. So we assume that there is an optimal solution such that x 1+x 2>y 12. The only candidate with this property is the standard solution x=(1,1,0). The optimality of this solution implies

$$ p_1+p_2+q_{12}+q_{23}+q_{13}=0.$$
(59)

Now, if we consider the two standard solutions x=(1,0,0) and x=(0,1,0), since their objective values cannot exceed 0, we have:

$$p_1+q_{12}+q_{13}\le0,\qquad p_2+q_{12}+q_{23}\le0.$$

If we take the sum of these two inequalities and subtract equation (59), we obtain q 12≤0, a contradiction.

1.2 A.2 Cycle of length 4

Let V={1,2,3,4} and E={12,23,34,14}. We prove that in this case the convex hull of (54)–(55) is obtained by adding the following five inequalities to the linear relaxation of (54)–(55):

(60)
(61)

The validity of (60) and (61) has been already proven in Sect. 3.2.

Lemma 16

Suppose that p i <0 for exactly two nodes (i and j, say). (a) If ijE, then all optimal solutions satisfy x i +x j =y ij . (b) If ijE, then all optimal solutions satisfy x i =0.

Proof

(a) Without loss of generality, i=1 and j=2. If there is an optimal solution (x,y) such that x 1+x 2>y 12, then at least one of x 1,x 2 is equal to 1, say x 1=1. Note that y 12=1 only if x 2=1. Then we obtain a better solution by decreasing x 1 to 0 and setting (or leaving) x 3 and x 4 to 1 (other components unchanged).

(b) Without loss of generality, i=1 and j=3. If there is an optimal solution (x,y) with x 1=1, then the standard solution x=(0,1,0,1) would have a better objective value. □

Therefore, from now on we assume that p i <0 for at least three nodes.

Lemma 17

If there is an optimal solution with more than two x-components equal to 1, then there exists ijE such that all optimal solutions satisfy x i +x j =y ij .

Proof

First of all note that no optimal solution satisfies x=1 (otherwise one can obtain a better solution by choosing an index i such that p i <0 and decreasing x i to 0).

It is easy to see that there cannot be an optimal solution with three x-components equal to 1 if p i <0 for all i. Thus we assume that p i =0 for some i, say p 1=0 without loss of generality (thus p 2,p 3,p 4<0).

Now we claim that if q 23>0 then all optimal solutions satisfy x 2+x 3=y 23. If this is not true, there is an optimal solution with x 2=x 3=y 23=1. But then the standard solution x=(1,0,1,0) would have a better objective value.

It remains to consider the case in which p 1=q 23=0 and there is an optimal solution with three x-components equal to 1. Such a solution must satisfy x 1=x 2=x 4=1. However, we obtain a better feasible solution by decreasing x 2 to 0 and setting (or leaving) y 23 to 0. □

Therefore, from now on we assume that all optimal solutions have at most two x-components equal to 1.

Lemma 18

If q 13>0 and q 24>0, then all optimal solutions satisfy (61) at equality.

Proof

Since q 13>0 and q 24>0, for every optimal solution the components y 13,y 24 satisfy (57), i.e., their value is maximal. Now it can be checked that every feasible solution with at most two x-components equal to 1 and with y 13,y 24 maximal satisfies (61) at equality. □

Therefore, from now on we assume that q 13 q 24=0. We now analyze several cases, depending on the cardinality of Q. Note that 1≤|Q|≤5.

Case 1: |Q|=1

Let us assume that Q={ij}. If ijE, we can apply Lemma 15 (as at least one of p i ,p j is negative). So we assume that ijE, say ij=13 without loss of generality. Since, by our assumptions, at least three components of p are negative, we can assume without loss of generality that p 1,p 2<0. We claim that then x 1+x 2+x 3=y 13 for all optimal solutions. Assuming by contradiction that this is not true, consider an optimal solution such that x 1+x 2+x 3>y 13. Note that since q 13>0, variable y 13 satisfies condition (57). We now distinguish two possibilities.

If y 13=0, then x 1=x 3=0 (by (57)) and x 2=1. Then we can obtain a better feasible solution by decreasing x 2 to 0 and taking the corresponding standard solution.

If y 13=1, at least two of x 1,x 2,x 3 are equal to 1. If x 2=1, we can decrease x 2 and take the corresponding standard solution; otherwise we have x 1=x 3=1 and we can decrease x 1 and take the corresponding standard solution. In both cases we obtain a better feasible solution.

Case 2: |Q|=2

If Q is either a maximum matching with both edges in E or a path of length 2 with both edges in E, we can apply Lemma 15. Since Q≠{13,24} (as q 13 q 24=0), the only remaining case is when Q is a path with one edge in E and the other edge not in E, say Q={12,13} without loss of generality. If p 1<0, we can apply Lemma 15. So we assume p 1=0, which implies p 2,p 3,p 4<0. Now, proceeding exactly as in Case 1, one can show that x 1+x 2+x 3=y 13 for all optimal solutions.

Case 3: |Q|=3

The only case that cannot be treated with the above lemmas and assumptions is when Q is a 3-clique, say Q={12,23,13} without loss of generality.

If p 4=0, we claim that x 1+x 2=y 12 for every optimal solution. If this is not true, there is an optimal solution with x 1=x 2=y 12=1. We obtain a better solution by decreasing x 1 to 0 (note that p 1<0) and setting (or leaving) x 4 to 1. So we now assume p 4<0.

Assume that the origin is an optimal solution. If all optimal solutions satisfy x 1+x 2=y 12, the analysis of this case is complete. So we assume that there is an optimal solution such that x 1+x 2>y 12. Since, by our assumption, there is no optimal solution with more than two component of x equal to 1, all optimal solutions such that x 1+x 2>y 12 satisfy x=(1,1,0,0), with objective value

$$ p_1+p_2+q_{12}+q_{23}+q_{13}=0.$$
(62)

Now, if we consider the two standard solutions x=(1,0,0,0) and x=(0,1,0,0), since their objective values cannot exceed 0, we have:

$$p_1+q_{12}+q_{13}\le0,\qquad p_2+q_{12}+q_{23}\le0.$$

If we take the sum of these two inequalities and subtract equation (62), we obtain q 12≤0, a contradiction.

Thus we now assume that the origin is not optimal. We claim that then all optimal solutions satisfy (60) at equality (with i=1,j=2,k=3).

We first consider optimal solutions with exactly two x-components equal to 1. Up to symmetries, constraint (60) is not satisfied at equality only if x 4=1 and x 2=0. However, since one of x 1 and x 3 is equal to 1, we find a better feasible solution by decreasing x 4 to 0 and taking the corresponding standard solution.

We now consider optimal solutions with exactly one x-component equal to 1. Constraint (60) is not satisfied at equality only if x 4=1. However, in this case the origin would be a better solution.

Case 4: 4≤|Q|≤5

The only cases that cannot be treated with the above lemmas and assumptions are (up to symmetries) the following: (a) Q={12,23,34,14} and (b) Q={12,23,34,14,13}. For the most part, these two cases can be analyzed together.

First of all we show that if p i =0 for some i, then x i =1 for all optimal solutions. Assume that p i =0 and there is an optimal solution with x i =0. Note that x can have at most one x-component equal to 1: otherwise, by increasing x i we would obtain an optimal solution with more than two x-components equal to 1, a contradiction to our assumptions. So there is at most one x-component equal to 1. Now, if we increase x i to 1 and construct the corresponding standard solution, we obtain a better solution (as at least one y ij with q ij >0 can be increased from 0 to 1), a contradiction. Therefore, from now on we assume that p i <0 for all iV.

We will show that there is ijE such that all optimal solutions satisfy x i +x j =y ij . The proof is by contradiction, so from now on we assume that for every ijE there is an optimal solution such that x i +x j >y ij .

Note that there is always a standard solution that is optimal. In the following we consider all possible standard solutions. Recall that we can assume that the number of x-components equal to 1 is at most two in any optimal solution.

Assume that the standard solution x=0 is optimal (thus the optimal value is 0). Take an optimal solution with x 1+x 2>y 12. Such a solution satisfies x 1=x 2=1, thus

$$ p_1+p_2+q_{12}+q_{13}+q_{14}+q_{23}=0.$$
(63)

If we consider the two standard solutions x=(1,0,0,0) and x=(0,1,0,0), we have p 1+q 12+q 13+q 14≤0 and p 2+q 12+q 23≤0. If we sum these two inequalities and subtract (63), we find q 12≤0, a contradiction.

Assume that the standard solution x=(1,0,0,0) is optimal. Then the optimal value is α=p 1+q 12+q 13+q 14. Take an optimal solution such that x 1+x 2>y 12: its objective value is p 1+p 2+q 12+q 13+q 14+q 23=α, hence p 2+q 23=0. Similarly, considering an optimal solution such that x 1+x 4>y 14, we find p 4+q 34=0. Now, if we consider the standard solution x=(0,1,0,1), we have p 2+p 4+q 12+q 13+q 14+q 23+q 34α, which, together with the conditions obtained above, gives p 1≥0, a contradiction.

Note that the above case also covers the situation in which the standard solution x=(0,0,1,0) is optimal (by symmetry arguments); moreover, if 13∉Q, the above analysis also covers the cases in which x=(0,1,0,0) and x=(0,0,0,1) are optimal. However, if 13∈Q these two cases need a different argument. So let us assume that 13∈Q and the standard solution x=(0,1,0,0) is optimal (the other case is analogous). Then the optimal value is α=p 2+q 12+q 23. Take an optimal solution such that x 1+x 4>y 14: its objective value is p 1+p 4+q 12+q 13+q 14+q 34=α, hence p 1+p 4+q 13+q 14+q 34=p 2+q 23. Now, if we consider the standard solution x=(0,1,0,1), we find p 2+p 4+q 12+q 13+q 14+q 23+q 34α, thus p 2+q 23p 1 also holds. If we consider the standard solution x=(1,0,0,0), we find p 1+q 12+q 13+q 14α, thus p 1+q 13+q 14p 2+q 23p 1. We then obtain q 13+q 14≤0, a contradiction.

Assume that the standard solution x=(1,1,0,0) is optimal. Then the optimal value is α=p 1+p 2+q 12+q 13+q 14+q 23. Take an optimal solution such that x 1+x 4>y 14: its objective value is p 1+p 4+q 12+q 13+q 14+q 34=α, hence p 4+q 34=p 2+q 23. Now, if we consider the standard solution x=(0,1,0,1), we find p 4+q 34p 1. Finally, with x=(1,0,0,0) we find p 2+q 23≥0. The last three conditions together imply p 1≥0, a contradiction. Note that this case also covers the situation in which one of the following standard solutions is optimal: (0,1,1,0), (0,0,1,1), (1,0,0,1).

Assume that the standard solution x=(1,0,1,0) is optimal. Then the optimal value is α=p 1+p 3+q 12+q 13+q 14+q 23+q 34. Take an optimal solution such that x 1+x 2>y 12: its objective value is p 1+p 2+q 12+q 13+q 14+q 23=α, hence p 2=p 3+q 34. Similarly, if we consider an optimal solution such that x 3+x 4>y 34, we find p 4=p 1+q 12. Now, if we take the standard solution x=(0,1,0,1), we find p 2+p 4p 1+p 3. The last three conditions together imply q 12+q 34≤0, a contradiction.

If 13∉Q, the above case also covers the situation in which x=(0,1,0,1) is optimal. It only remains to consider the case when 13∈Q and the standard solution x=(0,1,0,1) is optimal. In this case the optimal value is α=p 2+p 4+q 12+q 13+q 14+q 23+q 34. Take an optimal solution such that x 1+x 2>y 12: its objective value is p 1+p 2+q 12+q 13+q 14+q 23=α, hence p 1=p 4+q 34. Similarly, if we consider an optimal solution such that x 1+x 4>y 14, we find p 1=p 2+q 23. Now, with the standard solution x=(1,0,0,0), we find p 1p 2+p 4+q 23+q 34. The last three conditions together imply p 1≥0, a contradiction.

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Di Summa, M., Grosso, A. & Locatelli, M. Branch and cut algorithms for detecting critical nodes in undirected graphs. Comput Optim Appl 53, 649–680 (2012). https://doi.org/10.1007/s10589-012-9458-y

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