Complexity of Maximum Cut on Interval Graphs
Ranendu Adhikary, Kaustav Bose, Satwik Mukherjee, Bodhayan Roy
CComplexity of Maximum Cut on Interval Graphs
Ranendu Adhikary
Department of Mathematics, Jadavpur University, [email protected]
Kaustav Bose
Department of Mathematics, Jadavpur University, [email protected]
Satwik Mukherjee
Department of Mathematics, Jadavpur University, [email protected]
Bodhayan Roy
Department of Mathematics, Indian Institute of Technology Kharagpur, [email protected]
Abstract
We resolve the longstanding open problem concerning the computational complexity of
Max Cut oninterval graphs by showing that it is NP-complete.
For a graph G = ( V, E ) , a cut is a partition of V into two disjoint subsets. Any cut determinesa cut set which is the set of all edges that have one endpoint in one partition and the otherendpoint in the other partition. The size of a cut is the cardinality of its cut set. The maximumcut problem or Max Cut asks for a cut of maximum size.
Max Cut is a fundamental andwell-known NP-complete problem [17]. The weighted version of the problem is one of Karp’soriginal 21 NP-complete problems [23]. Besides its theoretical importance, it has applicationsin VLSI circuit design [11], statistical physics [3] etc.
Max Cut remains NP-hard evenfor cubic graphs [4], split graphs [7], co-bipartite graphs [7], unit disk graphs [15] and totalgraphs [19]. On the positive side, polynomial time algorithms are known for planar graphs [20],line graphs [19], graphs not contractible to K [2] and graphs with bounded treewidth [7].It is well known that many classical NP-complete problems like colourability [18], Hamil-tonian cycle [24], minimum dominating set [12], minimum feedback vertex set [26], minimumvertex cover [27] and maximum clique [21] are polynomial time solvable for interval graphs.This is because interval graphs are well structured graphs with many nice properties anddecomposition models that are often exploited to design efficient dynamic programming or1 a r X i v : . [ c s . CC ] J u l reedy algorithms. Few problems that are known to be NP-hard in interval graphs includeoptimal linear arrangement [14], achromatic number [5], harmonious colouring [1], geodeticset [10], minimum sum colouring [28], metric dimension [16], identifying code [16] and locating-dominating set [16]. The class of interval graphs is widely regarded as an important graphclass with many real-world applications. Interval graphs arise naturally in modelling problemsthat involve temporal reasoning, e.g scheduling problems. Interval graphs are also extensivelyused in bioinformatics (e.g. DNA mapping [29], protein sequencing [22]) and mathematicalbiology (e.g. food webs in population biology [13]).Surprisingly, the computational complexity of Max Cut for interval graphs is a long-standing open problem. No polynomial time algorithm is known even for the subclass of unitinterval graphs. There are two previous works [8, 9] reporting polynomial time algorithms solv-ing
Max Cut for unit interval graphs. However, both algorithms were later reported to beincorrect [6, 25]. In this paper, we show that
Max Cut is NP-complete for interval graphs.
For any simple undirected graph G = ( V, E ) , a cut is a partition of V into two disjointsubsets A and B , i.e., V = A ∪ B and A ∩ B = ∅ . The corresponding cut set is theset of all edges that have one endpoint in A and the other endpoint in B , i.e., the set { ( u, v ) ∈ E | ( u ∈ A, v ∈ B ) ∨ ( u ∈ B, v ∈ A ) } . The size of the cut is the cardinality of itscut set. A typical instance of the decision version of Max Cut consists of a simple undirectedgraph G = ( V, E ) and an integer k such that ≤ k ≤ | E | . ( G, k ) is an yes-instance of MaxCut if and only if G has a cut of size at least k .Interval graphs are the intersection graphs of intervals on the real line. Formally, G = ( V, E ) is said to be an interval graph if there is a set S of intervals on the real line and a bijection ϕ : V −→ S such that u, v ∈ V are adjacent if and only if ϕ ( u ) ∩ ϕ ( v ) = ∅ . In this section, we show that
Max Cut is NP-complete on interval graphs.
Max Cut isknown to be NP-complete on cubic graphs [4]. We reduce
Max Cut on cubic graphs to
Max Cut on interval graphs.
Let ( G, x ) be an instance of Max Cut where G = ( V, E ) is a cubic graph. Let | V | = n and hence | E | = n . We shall reduce it to an equivalent instance ( G , f ( x )) of Max Cut where G = ( V , E ) is an interval graph. The construction of G is outlined in the following. G = ( V , E ) is described as the intersection graph of a set of intervals on the real line andthe vertices of G are referred to as intervals.1. Fix an arbitrary ordering of the vertices and edges of G = ( V, E ) as v , v , . . . , v n , e , e ,. . . , e m . We shall write any edge e ∈ E as an ordered pair of vertices that respects thefollowing convention. If e is an edge between v i and v j , where i < j , then we shall write e = ( v i , v j ) (not e = ( v j , v i ) ). 2. For each vertex v ∈ V , we construct a V-gadget G ( v ) and for each edge e ∈ E , weconstruct an E-gadget G ( e ) . They are shown in Fig. 1. The structure of a V-gadgetis identical to that of an E-gadget, the only difference is their size. Each V-gadget(E-gadget) consists of q (resp. q ) left long intervals, p (resp. p ) left short intervals, q (resp. q ) right long intervals and p (resp. p ) right short intervals. The left longintervals and the right long intervals of a V-gadget (E-gadget) all intersect each otherto form a clique of size q (resp. q ). All left short intervals of a V-gadget (E-gadget)are mutually disjoint and each of them intersect only the q (resp. q ) left long intervals.Similarly all right short intervals of a V-gadget (E-gadget) are mutually disjoint and eachof them intersect only the q (resp. q ) right long intervals. Therefore the number ofedges in each V-gadget (E-gadget) is q (2 q −
1) + 2 pq (resp. q (2 q −
1) + 2 p q ). p left shortintervals q l e f t l o n g i n t e r v a l s q r i g h t l o n g i n t e r v a l s p right shortintervals (a) p left shortintervals q l e f t l o n g i n t e r v a l s q r i g h t l o n g i n t e r v a l s p right shortintervals (b) Figure 1: a) A V-gadget. b) An E-gadget.3. We set q = 200 n + 1 , p = 2 q + 7 n , q = 10 n + 1 , p = 2 q + 7 n , where n is thenumber of vertices in G . Note that, due to these values of p , q , p and q , the followinginequalities hold:(a) p > q > p > q > n (b) q > ( p − q )6 n (c) q > ( p − q )6 n (d) q > p + q ) n + 9 n
4. There are a total of n V-gadgets, and n/ E-gadgets. All n/ gadgets are arranged inthe following order as shown in Fig. 2 : G ( v ) , G ( v ) , . . . , G ( v n ) , G ( e ) , G ( e ) , . . . , G ( e n/ ) .No two intervals belonging to different gadgets intersect each other.5. To establish relationships between the V-gadgets and E-gadgets we introduce n linkintervals (See Fig. 2). Link intervals connect V-gadgets to E-gadgets. This will bedescribed in the next point. A link interval can intersect a gadget in four different waysas described in the following. • A link interval is said to cover a gadget if it intersects all intervals of the gadget.(See Fig. 3a) 3
A link interval is said to intersect a V-gadget in the first manner if it intersectsonly the q right long intervals of the V-gadget. (See Fig. 3b). • A link interval is said to intersect an E-gadget in the second manner if it intersectsonly the p left long intervals of the gadget. (See Fig. 3c). • A link interval is said to intersect an E-gadget in the third manner if it intersectsonly the q left long intervals and the p left short intervals of the gadget. (See Fig.3d).6. For each edge e = ( v i , v j ) ∈ E , we introduce four link intervals: 1) a pair intersecting G ( v i ) in the first manner and G ( e ) in the second manner, and 2) another pair intersecting G ( v j ) in the first manner and G ( e ) in the third manner (See Fig. 4). Note that since G is cubic, due to our construction the total number of link intervals covering a V-gadgetis k for some integer k , where k may vary from to n − . Similarly the total numberof link intervals covering a V-gadget is k for some integer k , where k may vary from to n/ − . Also, the total number of link intervals intersecting a V-gadget in the firstmanner is . L i n k i n t e r v a l s G ( v ) G ( v ) G ( v n ) G ( e ) G ( e ) G ( e ) G ( e m ) Figure 2: Arrangement of the gadgets and the link intervals.
In this section, we study some properties of the interval graph G constructed from G in theprevious section. We consider a partition P ( A, B ) of G that yields a maximum cut. To provethat P satisfies some properties, in general we show that if P does not satisfy those properties,then the size of the cut can be increased, contradicting the maximality of the cut induced by P . Then we have the following lemmas. Lemma 1.
If a partition of G yields a maximum cut, then for any V-gadget G ( v i ) , all of itsleft short intervals lie in the same subset. Similarly, all of its right short intervals lie in thesame subset.Proof. Let P ( A, B ) be a partition of G that divides its vertices into two disjoint subsets A and B , and gives a maximum cut. Let LL Ai and LL Bi denote the subset of left long intervals of V i in A and B respectively. Denote by OL Ai (resp. OL Bi ) the set of all link intervals that cover4 a) A gadget is covered by a link interval. (b) A link interval intersects a V-gadgetin the first manner.(c) A link interval intersects an E-gadgetin the second manner. (d) A link interval intersects an E-gadgetin the third manner. Figure 3: Illustrations showing the four different ways a link interval can intersect a gadget. G ( v i ) G ( v j ) G ( v i , v j ) Figure 4: link intervals connecting an E-gadget G (( v i , v j )) with V-gadgets G ( v i ) and G ( v j ) . G ( v i ) and lie in subset A (resp. B ). With out loss of generality, let the following direction ofinequality hold: | LL Ai | + | OL Ai | > | LL Bi | + | OL Bi | Note that the inequality must be strict since we have set q to be odd, and the total numberof link intervals covering a V-gadget is even. Suppose that a left short interval of V i is in A .Then due to the above inequality, moving it to B increases the number of cut edges. Thiscontradicts the fact that the partition yields a maximum cut. Hence, all left short intervalsof G ( v i ) must be in B . Using similar arguments we can show that all right short intervals of5 ( v i ) must be same subset. Lemma 2.
If a partition of G yields a maximum cut, then in any V-gadget, all its right longintervals belong to the same subset and all its left long intervals belong to the same subset.Proof. Consider a partition P ( A, B ) of G that yields a maximum cut. By Lemma 1, all of theshort intervals on the same side of G ( v i ) belong to the same subset. Without loss of generality,we consider two cases, where (a) all the left short intervals of G ( v i ) are in A , and all the rightshort intervals of G ( v i ) are in B , and (b) all the short intervals of G ( v i ) are in A .First consider Case (a), where all the left short intervals of G ( v i ) belong to A , and all theright short intervals of G ( v i ) belong to B . Suppose a left long interval of G ( v i ) is in A . Thenmoving it to B results in losing at most q − cut edges due to its intersections with otherlong intervals of G ( v i ) , and n cut edges due to its intersections with the link intervals of G ( v i ) . However, we gain p cut edges. Since p = 2 q + 7 n , the quantity p − (2 q − n ) ispositive, hence the size of the cut increases. This contradicts the fact that the partition yieldsa maximum cut. Hence, all left long intervals of G ( v i ) must be in B .Now consider Case (b), where all the short intervals of G ( v i ) belong to A . It can be seen thatthe above argument is also applicable in this case, and the claim holds. Lemma 3.
If a partition of G yields a maximum cut, then for any V-gadget G ( v i ) , all the leftlong and right short intervals are in one subset, while all the right long and left short intervalsare in another.Proof. Let P ( A, B ) be a partition of G that gives a maximum cut. Then without loss ofgenerality, by Lemma 1, either (a) all left short intervals of V i are in A and all right shortintervals of V i are in B , or (b) all the short intervals of V i are in A .First consider Case (a), i.e. V i has all its left short intervals in A and right short intervals in B . Then it follows from the proof of Lemma 2 that all left long intervals of G ( v i ) are be in B and all right long intervals of G ( v i ) must be in A , as claimed.Now consider Case (b), i.e. V i has all its short intervals in A . Since all the short intervals of V i are in A , it implies from the proof of 2 that all the long intervals of V i are in B . We moveall the right short intervals of G ( v i ) to B and all right long intervals of G ( v i ) to A . Due totheir intersections with link intervals, this removes at most ( p − q )6 n edges from the cut. Butdue to the intersections among the left and right long intervals, it also adds at least q edgesto the cut. Since by our choice of q and p , we have q − ( p − q )6 n > , the total number ofedges in the cut increases. This contradicts the fact that the partition yields a maximum cutand hence this case is impossible. Lemma 4.
Lemma 3 holds for E-gadgets of G as well.Proof. Let P ( A, B ) be a partition of G that divides its vertices into two disjoint subsets A and B , and gives a maximum cut. We modify the proof of Lemma 1 a little, so that Lemma 1holds for E-gadgets as well. Consider an E-gadget G ( e i ) of G . Observe that the proof holds forthe right short intervals of G ( e i ) , since any link interval that intersects the right short intervals6f an E-gadget, must also cover the E-gadget. But the left short intervals of each E-gadgetare intersected by two link intervals in the third manner. Then denote by OL Ai (resp. OL Bi )the set of all link intervals that cover V i or intersect V i in the third manner, and lie in subset A (resp. B ). Let LL Ai and LL Bi denote the subset of left long intervals of V i in A and B respectively, as before. Again, without loss of generality we have the following inequality. | LL Ai | + | OL Ai | > | LL Bi | + | OL Bi | The rest of the proof is similar to that of 1, and it can be seen that the claim holds.The proof of Lemma 2 for E-gadgets remains the same as for V-gadgets. Lemmas 1 and 2along with the choice of p and q , imply Lemma 3 for E-gadgets as well. Lemma 5. G has a cut of size at least x if and only if G has a cut of size at least (2 pq + q ) n + (2 p q + q ) n + 3 n ( n − p + q ) + 3 n ( n − p + q ) + 6 nq + 3 np + 2 xq .Proof. First suppose that G has a cut of size at least x . Denote the subsets in the partition ofthe vertices of G by C and D . We partition the vertices of G as follows. If a vertex v i of G is in C , then in the corresponding V-gadget G ( v i ) of G , all left short intervals and right longintervals are placed in A , all right short intervals and left long intervals are placed in B . Finally,all link intervals of V i are placed in B . If v i is in D instead, then all the above placements ofintervals are swapped. Recall that for each E-gadget exactly two link intervals intersect it inthe second manner and exactly two link intervals intersect it in the third manner. If the linkintervals that intersect an E-gadget in the third manner is in A , then we place the left shortintervals and right long intervals of the E-gadget in B , and the left long intervals and rightshort intervals in A . If the link intervals are in B , then the placements of the intervals areswapped.Due to the above placement of intervals in A and B , the number of cut edges obtainedinternally from all the V-gadgets and E-gadgets of G are (2 pq + q ) n and (2 p q + q ) n respectively. The number of cut edges formed by the V-gadgets and the link intervals thatcover them is n ( n − p + q ) . The number of cut edges formed by the E-gadgets and thelink intervals covering them is n ( n − p + q ) . For each V-gadget, the link intervalsintersecting it in the first manner give q cut edges, resulting in a total of nq cut edges. Eachlink interval that intersects an E-gadget in the third manner gives p cut edges, thus we have np in total. However, a link interval that intersects an E-gadget in the second manner canproduce cut edges from the E-gadget only when the other link interval mentioned above is ina different subset, i.e. the vertices of G corresponding to the V-gadgets of these link intervalsare in C and D , and produce a cut edge. This means that such link intervals produce xq cut edges in total, proving the forward direction of the claim.Now we prove the backward direction of the claim. Assume that G has a cut of size at least (2 pq + q ) n + (2 p q + q ) n + 3 n ( n − p + q ) + 3 n ( n − p + q ) + 6 nq + 3 np + 2 xq .So the size of a maximum cut of G is at least (2 pq + q ) n + (2 p q + q ) n + 3 n ( n − p + q ) + 3 n ( n − p + q ) + 6 nq + 3 np + 2 xq . Consider a maximum cut of G that divides itsintervals into two disjoint subsets A and B . Due to Lemmas 3 and 4, the placement of intervals7f V-gadgets and E-gadgets in A and B is the same as above. So, the internal cut edges ofV-gadgets and E-gadgets, and the cut edges formed between gadgets and the link intervalsthat cover them total to (2 pq + q ) n + (2 p q + q ) n + 3 n ( n − p + q ) + 3 n ( n − p + q ) .The maximum number of cut edges among the link intervals is (3 n ) = 9 n . But we have p > q > p > q > n . This means that the remaining nq + 3 np + 2 xq cut edges areobtained from the partial intersections of the link intervals with the V-gadgets and E-gadgets.The partial intersections between link intervals and V-gadgets can contribute at most nq cut edges. Note that the partial intersections between link intervals and E-gadgets, andintersects among the link intervals can not give more than p + q ) n + 9 n cut edges. Since q > p + q ) n + 9 n , it implies that exactly nq of the remaining cut edges are obtained fromlink intervals intersecting V-gadgets in the first manner. This happens when for each E-gadget,the link intervals intersecting it in the first manner are all in the subset which contains the leftlong and right short intervals of the gadget. Now for the remaining np + 2 xq cut edges,observe that each E-gadget gives only p cut edges from its partial intersections with linkintervals if they belong to the same subset, and p + q ) cut edges if they belong to differentsubsets. If x number of E-gadgets give p + q ) such cut edges, then their corresponding V-gadgets must have had opposite placements of intervals in A and B . Then the correspondingvertices of G can be placed in different subsets to give a cut edge. Hence the total number ofcut edges in G is at least x . Theorem 1.
Max Cut is NP-complete on interval graphs.Proof.
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