The interval constrained 3-coloring problem
TThe interval constrained 3-coloring problem
Jaroslaw Byrka Andreas Karrenbauer ∗ Laura Sanit`a † Institute of MathematicsEPFL, Lausanne, Switzerland { Jaroslaw.Byrka,Andreas.Karrenbauer,Laura.Sanita } @epfl.ch Abstract
In this paper, we settle the open complexity status of interval constrained coloring with a fixednumber of colors. We prove that the problem is already NP-complete if the number of differentcolors is 3. Previously, it has only been known that it is NP-complete, if the number of colors ispart of the input and that the problem is solvable in polynomial time, if the number of colors is atmost 2. We also show that it is hard to satisfy almost all of the constraints for a feasible instance.This implies APX-hardness of maximizing the number of simultaneously satisfiable intervals.
In the interval constrained 3-coloring problem, we are given a set I of intervals defined on [ n ] := { , . . . , n } and a requirement function r : I → Z ≥ , which maps each interval to a triple of non-negative integers. The objective is to determine a coloring χ : [ n ] → { , , } such that each intervalgets the proper colors as specified by the requirements, i.e. (cid:80) i ∈ I e χ ( i ) = r ( I ) where e , e , e are thethree unit vectors of Z .This problem is motivated by an application in biochemistry to investigate the tertiary structureof proteins as shown in the following illustration. More precisely, in Hydrogen-Deuterium-ExchangeFigure 1: Coloring of the residues of a protein chain according to their exchange rates.(HDX) experiments proteins are put into a solvent of heavy water ( D O ) for a certain time afterwhich the amount of residual hydrogen atoms, that have exchanged with deuterium atoms, is mea-sured [1]. Doing this experiment for several timesteps, one can determine the exchange rate of the ∗ supported by the Deutsche Forschungsgemeinschaft (DFG) within Priority Programme 1307 “Algorithm Engineering”. † supported by Swiss National Science Foundation within the project “Robust Network Design”. a r X i v : . [ c s . D M ] D ec esidues. These exchange rates indicate the solvent accessibility of the residues and hence they pro-vide information about the spatial structure of the protein. Mass spectroscopy is one of the methodsfor measuring these exchange rates. To this end, the proteins are digested, i.e. cut into parts whichcan be considered as intervals of the protein chain, and the mass uptake of each interval is measured.But thereby only bulk information about each interval can be obtained. Since there is not only oneprotein in the solvent but millions and they are not always cut in the same manner, we have this bulkinformation on overlapping fragments. That is, we are given the number of slow, medium, and fast ex-changing residues for each of these intervals and our goal is to find a feasible assignment of these threeexchange rates to residues such that for each interval the numbers match with the bulk information.Though the interval constrained 3-coloring problem is motivated by a particular application, itsmathematical abstraction appears quite simple and ostensibly more general. In terms of integer linearprogramming, the problem can be equivalently formulated as follows. Given a matrix A ∈ { , } m × n with the row-wise consecutive-ones property and three vectors b , , ∈ Z m ≥ , the constraints A A
00 0
AI I I · x x x = b b b (1)have a binary solution, i.e. x , , ∈ { , } n , if and only if the corresponding interval constrained 3-coloring problem has a feasible solution. We may assume w.l.o.g. that the requirements are consistentwith the interval lengths, i.e. A · b + b + b , since otherwise we can easily reject the instance asinfeasible. Hence, we could treat x as slack variables and reformulate the constraints as Ax = b , Ax = b , x + x ≤ . (2)It is known that if the matrix A has the column-wise consecutive ones property (instead of row-wise ),then there is a reduction from the two-commodity integral flow problem, which has been proven tobe NP-complete in [2]. However, the NP-completeness w.r.t. row-wise consecutive ones matrices hasbeen an open problem in a series of papers as outlined in the following subsection. The problem of assigning exchange rates to single residues has first been considered in [3]. In thatpaper, the authors presented a branch-and-bound framework for solving the corresponding coloringproblem with k color classes. They showed that there is a combinatorial polynomial time algorithmfor the case of k = 2 . Moreover, they asked the question about the complexity for k > . In [4],the problem has been called interval constrained coloring . It has been shown that the problem is NP-hard if the parameter k is part of the input. Moreover, approximation algorithms have been presentedthat allow violations of the requirements. That is, a quasi-polynomial time algorithm that computes asolution in which all constraints are (1 + ε ) -satisfied and a polynomial time rounding scheme, whichsatisfies every requirement within ± , based on a technique introduced in [5]. The latter implies thatif the LP relaxation of (1) is feasible, then there is a coloring satisfying at least of the requirements.APX-hardness of finding the maximum number of simultaneously satisfiable intervals has been shownin [6] for k ≥ provided that intervals may be counted with multiplicities. But still, the questionabout the complexity of the decision problem for fixed k ≥ has been left open. In [7], severalfixed parameter tractability results have been given. However, the authors state that they do not knowwhether the problem is tractable for fixed k . 2 .2 Our contribution In this paper, we prove the hardness of the interval constrained k -coloring problem for fixed parameter k . In fact, we completely settle the complexity status of the problem, since we show that alreadythe interval constrained 3-coloring problem is NP-hard by a reduction from 3-SAT. This hardnessresult holds more generally for any problem that can be formulated like (1). Moreover, we evenshow the stronger result, that it is still difficult to satisfy almost all of the constraints for a feasibleinstance. More precisely, we prove that there is a constant (cid:15) > such that it is NP-hard to distinguishbetween instances where all constraints can be satisfied and those where only a (1 − (cid:15) ) fraction ofconstraints can be simultaneously satisfied. To this end, we extend our reduction using expandergraphs. This gap hardness result implies APX-hardness of the problem of maximizing the number ofsatisfied constraints. It is important to note that our construction does neither rely on multiple copiesof intervals nor on inconsistent requirements for an interval, i.e. in our construction for every interval ( i, j ) we have unique requirements that sum up to the length of the interval. Theorem 1.
It is NP-hard to decide whether there exists a feasible coloring χ for an instance ( I , r ) ofthe interval constrained 3-coloring problem. Proof.
The proof is by reduction from the 3-SAT problem.Suppose to be given an instance of the 3-SAT problem, defined by q clauses C , . . . , C q and p variables x , . . . , x p . Each clause C i ( i = 1 , . . . , q ) contains 3 literals, namely y ( i ) , y ( i ) , y ( i ) . Each literal y h ( i ) ( i = 1 , . . . , q and h = 1 , , refers to a variable x j , that means, it is equal to either x j or ¯ x j for some j in , . . . , p . A truth assignment for the variables x , . . . , x p satisfies the 3-SAT instance ifand only if, for each clause, at least one literal takes the value true . . . . . . . . . . . . . . . . C ! C ! C ! C q ! n ! . . . . . . . . . y (i) ! y (i) ! y (i) ! C i ! Figure 2: The sequence of nodes in an instance of the interval constrained 3-coloring problem.We now construct an instance of the interval constrained 3-coloring problem. For each clause C i we introduce a sequence of consecutive nodes. This sequence is, in its turn, the union of threesubsequences, one for each of the three literals (see Fig. 2).3n the following, for the clarity of presentation, we drop the index i , if it is clear from the context.We denote color 1 by RED, color 2 by BLACK and color 3 by WHITE. Literal y ( i ) . The subsequence representing literal y is composed of 8 nodes. Among them, thereare three special nodes, namely t , f and a , that play a key role since they encode the informationabout the truth value of the literal and of the variable x j it refers to. The basic idea is to achieve thefollowing two goals: 1) given a feasible coloring, if χ ( t ) is BLACK, we want to be able to constructa truth assignment setting x j to true , while if χ ( f ) is BLACK, we want to be able to construct a truthassignment setting the variable x j to f alse ; 2) given a feasible coloring, if χ ( a ) is RED, we want tobe able to construct a truth assignment where y is true .To achieve the first goal, we will impose the following property: Property 2.
In any feasible coloring, exactly one among t and f will be BLACK. To achieve the second goal, and being consistent with the first one, we must have the property that:
Property 3.
In any feasible coloring, if χ ( a ) = RED , then χ ( t ) = BLACK if y = x j , while χ ( f ) = BLACK if y = ¯ x j . To guarantee properties (2) and (3), we introduce a suitable set I ( y ) of six intervals , shown inFig. 3a. ! " ! ! ! ! $ ! ! % ! ! & ! ! ’ ! ( " ! ) " ! * " ! + " ! (a) !" ! !$ ! !% ! ! " ! " ! $ " ! ! " ! " ! $ " ! ! " ! " ! $ " ! (b) Figure 3: Literal y . The picture on the right shows the three feasible colorings. On a black and whiteprintout red color appears as grey.The requirement function for such intervals changes whether y = x j or y = ¯ x j . If y = x j ,we let r ( I ) = (1 , , ; r ( I ) = (1 , , ; r ( I ) = (1 , , ; r ( I ) = (1 , , ; r ( I ) = (0 , , ; r ( I ) = (2 , , . For any feasible coloring there are only three possible outcomes for such sequence,reported in Fig. 3b. Observe that the properties (2) and (3) are enforced.Now suppose that y = ¯ x j : then we switch the requirement function with respect to WHITE andBLACK, i.e. define it as follows: r ( I ) = (1 , , ; r ( I ) = (1 , , ; r ( I ) = (1 , , ; r ( I ) = In principle, interval I and the node it contains are not needed. However, this allows to have the same number ofWHITE and BLACK colored nodes for the sake of exposition. , , ; r ( I ) = (0 , , ; r ( I ) = (2 , , . Trivially, the possible outcomes for such sequence areexactly the ones in Fig. 3b but exchanging the BLACK and WHITE colors. Literal y ( i ) . The sequence of nodes representing literal y is similar to the one representing y .We still have a sequence of 8 nodes, and three special nodes t , f and a . As before, we let t and f encode the truth value of the variable x j that is referred to by y , while a encodes the truth value ofthe literal y itself. Therefore, we introduce a set I ( y ) of intervals in order to enforce the followingproperties: Property 4.
In any feasible coloring, exactly one among t and f will receive color BLACK. Property 5.
In any feasible coloring, if χ ( a ) = RED , then χ ( t ) = BLACK if y = x j , while χ ( f ) = BLACK if y = ¯ x j . Fig. 4a shows the nodes and the six intervals that belong to I ( y ) : observe that the sequence is similarto the one representing y , but the position of node a and the intervals are now “mirrored”. If y = ¯ x j ,we let r ( I ) = (1 , , ; r ( I ) = (1 , , ; r ( I ) = (1 , , ; r ( I ) = (1 , , ; r ( I ) = (0 , , ; r ( I ) = (2 , , . Fig. 4b reports the three possible outcomes for such sequence in a feasible coloring.Note that properties (4) and (5) hold.Now suppose that y = x j : once again, we switch the requirement function with respect toWHITE and BLACK. ! " ! " ! $ " ! % & ! % ’ ! % " ! % ( ! % ) ! % * ! + " ! (a) !" ! !$ ! !% ! ! " ! " ! $ " ! ! " ! " ! $ " ! ! " ! " ! $ " ! (b) Figure 4: Literal y Literal y ( i ) . The sequence of nodes representing literal y is slightly more complicated. It iscomposed of 36 nodes, and among them there are 4 special nodes, namely t , f , a (cid:96) and a r (seeFig. 5). Still, we let t and f encode the truth value of the variable x j that is referred to by y , while a (cid:96) and a r encode the truth value of the literal.Similarly to the previous cases, we want to achieve the following goals: 1) given a feasible color-ing, if χ ( t ) is BLACK, we want to be able to construct a truth assignment setting the variable x j to5 . . . . . t ! f ! a l a !
13 nodes !
15 nodes ! I ! I ! I ! I ! I ! I ! I ! I ! I ! I ! I ! I ! I ! I !
18 nodes ! Figure 5: Literal y true , while if χ ( f ) is BLACK, we want to be able to construct a truth assignment setting the variable x j to f alse ; 2) given a feasible coloring, if χ ( a (cid:96) ) = χ ( a r ) = RED, we want to be able to construct atruth assignment where the literal y is true . We are therefore interested in the following properties: Property 6.
In any feasible coloring, exactly one among t and f will receive color BLACK. Property 7.
In any feasible coloring, if χ ( a (cid:96) ) = RED and χ ( a r ) = RED , then χ ( t ) = BLACK if y = x j , and χ ( f ) = BLACK if y = ¯ x j . In this case, we introduce a set I ( y ) of 14 suitable intervals, shown in Fig. 5. The requirements forthe case y = ¯ x j are given in the following table. I I I I I I I I I I I I I I RED
BLACK
W HIT E
Observe that the set of intervals { I , . . . , I } is defined exactly as the set I ( y ) , therefore the possibleoutcomes for the sequence of 8 nodes covered by such intervals are as in Fig. 4b. Similarly, the setof intervals { I , . . . , I } is defined exactly as the set I ( y ) , therefore the possible outcomes for thesequence of 8 nodes covered by such intervals are as in Fig. 3b. Combining r ( I ) and r ( I ) with r ( I ) , it follows that in any feasible coloring χ , exactly one node among t and f has WHITE (resp.BLACK) color, enforcing Property (6). Still, note that if χ ( a (cid:96) ) = RED and χ ( a r ) = RED , thenboth the leftmost node and the rightmost node covered by interval I have color BLACK, therefore t must have color WHITE otherwise r ( I ) is violated. Together with Property (6), this enforcesProperty (7).In case y = x j , once again we switch the requirement function with respect to WHITE andBLACK.It remains to describe the role played by the first 13 nodes and the last 15 nodes of the sequence, thatso far we did not consider. We are going to do it in the next paragraph. Intervals encoding truth values of literals.
For each clause C i , we add another set I ( C i ) of inter-vals, in order to link the nodes encoding the truth values of its three literals. The main goal we pursue6s the following: given a feasible coloring, we want to be able to construct a truth assignment suchthat at least one of the three literals is true . To achieve this, already having properties (3), (5) and (7),we only need the following property: Property 8.
For any feasible coloring, if χ ( a ) (cid:54) = RED and χ ( a ) (cid:54) = RED , then χ ( a (cid:96) ) = χ ( a r ) = RED . Fig. 6 shows the six intervals that belong to I ( C i ) . The requirement function is: r ( I ) = (1 , , r ( I ) = (1 , , r ( I ) = (1 , , r ( I ) = (1 , , r ( I ) = (1 , , r ( I ) = (1 , , . We nowshow that Property (8) holds. Suppose χ is a feasible coloring, and let v , . . . , v be the first 13 nodesof the sequence introduced for literal y . By construction, if χ ( a ) (cid:54) = RED , then there is a node v j : χ ( v j ) = RED and j ∈ { , , } , otherwise r ( I ) is violated. Similarly, if χ ( a (cid:96) ) (cid:54) = RED ,then there is a node v j : χ ( v j ) = RED and j ∈ { , , } , otherwise r ( I ) is violated. On theother hand, this subsequence contains exactly one node with RED color, otherwise r ( I ) is violated.It follows that at least one among a and a (cid:96) has RED color. The same conclusions can be stated fornodes a r and a . Putting all together, it follows that the Property (8) holds. y (i) ! C i ! . . .
13 nodes ! a t ! I ! . . . . . . a l ! I ! a f ! . . . I !
15 nodes ! I ! I ! . . . I ! y (i) ! y (i) ! Figure 6: Set of intervals I ( C i ) . Intervals encoding truth value of variables (later also called: variable intervals).
Our last set ofintervals will force different nodes to take the same color, if they encode the truth value of the samevariable. In particular, we aim at having the following property:
Property 9.
In any feasible coloring, χ ( t h ( i )) = χ ( t k ( i (cid:48) )) if both literals y h ( i ) and y k ( i (cid:48) ) refer to thesame variable x j . To achieve this, for each pair of such literals we add a big interval I ( y h ( i ) , y k ( i (cid:48) )) from f k ( i (cid:48) ) to t h ( i ) (assuming i (cid:48) < i without loss of generality). Note that, by construction, there is a subset of intervalsthat partitions all the internal nodes covered by the interval. That means, we know exactly the numberof such nodes that must be colored with color RED, BLACK and WHITE (say z , z , z respectively).Then, we let the requirement function be r ( I ( y h ( i ) , y k ( i (cid:48) ))) = ( z , z + 1 , z + 1) . Under theseassumptions, if χ is a feasible coloring then χ ( t h ( i )) (cid:54) = χ ( f k ( i (cid:48) )) , and in particular one node willhave WHITE color and the other one BLACK color. Combining this with properties (2),(4) and (6),the result follows.Notice that such an interval constrained 3-coloring instance can clearly be constructed in polyno-mial time. Now we discuss the following claim in more details.7 laim 10. There exists a truth assignment satisfying the 3-SAT instance if and only if there exists afeasible coloring χ for the interval constrained 3-coloring instance. First, suppose there exists a feasible coloring. We construct a truth assignment as follows. Weset a variable x j to true if χ ( t h ( i )) = BLACK , and to f alse otherwise, where y h ( i ) is any literalreferring to x j . Note that, by Property (9), the resulting truth value does not depend on the literal wetake. Still, combining Property (8) with properties (3),(5) and (7), we conclude that, for each clause, atleast one literal will be true . By construction, we therefore end up with a truth assignment satisfyingthe 3-SAT instance. The result follows.Now suppose that there is a truth assignment satisfying the 3-SAT instance. The basic idea, is toconstruct a coloring χ such that the following property holds for all literals: Property 11. χ ( t h ( i )) = BLACK (resp. WHITE) if and only if y h ( i ) refers to a true -variable (resp. f alse -variable). Consider the sequence of nodes representing literal y ( i ) , and suppose y ( i ) = x j for some j . Wecolor such nodes as in Fig. 3b- (1) if the literal is true in the truth assignment, and as in Fig. 3b -(3) otherwise. If y ( i ) = ¯ x j , switch BLACK and WHITE colors, in both previous cases. Now focus onthe sequence of nodes representing literal y ( i ) . If y ( i ) = ¯ x j for some j , we color such nodes as inFig. 4b- (1) if the literal is true , and as in Fig. 4b -(3) otherwise. If y ( i ) = x j , switch BLACK andWHITE colors, in both previous cases. Finally, consider the sequence of nodes representing literal y ( i ) . Suppose y ( i ) = ¯ x j . We color the 18 nodes in the middle of the sequence as in Fig. 7 -(1) if y ( i ) is true , as in Fig. 7 -(2) if both y ( i ) and y ( i ) are f alse , and as in Fig. 7 -(3) otherwise.Once again, if y ( i ) = x j , we switch BLACK and WHITE colors, in all the previous three cases.Notice that, by construction, Property (11) holds, and all requirements for the intervals in I ( y h ( i ))( i = 1 , . . . , q and h = 1 , , ) are not violated. . . . . . . t ! f ! a l a ! . . . . . . t ! f ! a l a ! . . . . . . t ! f ! a l a !
13 nodes !
15 nodes !
18 nodes ! (1) (2) (3) Figure 7: Coloring of nodes representing literal y Now we show how to color the first 13 nodes ( v , . . . , v ) and the last 15 nodes ( w , . . . , w ) ofthe sequence representing literal y ( i ) , in such a way that the requirements of the intervals I , . . . , I in I ( C i ) are not violated ( i = 1 , . . . , q ) . Note that, by construction, at least one node among a and a (cid:96) is colored with RED. In fact, if y ( i ) is true then χ ( a ) = RED, while if y ( i ) = f alse then a (cid:96) is colored with RED. Similarly, at least one node among a and a r is colored with RED, since χ ( a r ) (cid:54) = RED only if both literals y ( i ) and y ( i ) are f alse : then, necessarily y ( i ) is true , andtherefore χ ( a ) = RED . Let us focus on the nodes v , . . . , v , and let u be the node in between v a (cid:96) . In the following, we refer to WHITE as the opposite color of BLACK and vice versa. As wealready discuss, we can have only two cases:Case 1: χ ( a ) = χ ( a (cid:96) ) = RED . We color v with the opposite color of f , and the nodes v and v with BLACK and WHITE. Note that r ( I ) is not violated. We then color v , v , v with theopposite color of v , v , v respectively. Similarly, we color v with the opposite color of u . Then,we color v and v with BLACK and WHITE, so that r ( I ) is not violated. Once again, we assignto v , v , v the opposite color of v , v , v respectively. Finally, we let χ ( v ) = RED . Note that r ( I ) is not violated.Case 2: χ ( a ) (cid:54) = RED and χ ( a (cid:96) ) = RED , or vice versa. Suppose χ ( a ) (cid:54) = RED (the othercase is similar). Both nodes a and f can have only BLACK or WHITE colors. Then, we can color v and v with the opposite color of a and f respectively, and v with color RED, so that r ( I ) isnot violated. Still, we color v and v with the opposite color of v and v . Finally, we color v and v with BLACK and WHITE. To the remaining nodes v , . . . , v we assign the same colors as in Case1. One checks that requirements of intervals I and I are not violated.One can prove in a similar manner that nodes ( w , . . . , w ) can be properly colored, withoutviolating the requirements of intervals I , I , I .Finally, since Property (11) holds, it is easy to see that, for each couple of literals y h ( i ) , y k ( i (cid:48) ) , therequirement r ( I ( y h ( i ) , y k ( i (cid:48) ))) is also not violated. The result then follows. We will now argue that not only the interval constrained 3-coloring problem but also its gap versionis NP-hard, i.e., it is hard to distinguish between satisfiable instances and those where only up to a (1 − (cid:15) ) fraction of constraints may be simultaneously satisfied.For the purpose of our argument we will use the following, rather restricted, definition of gaphardness. We will only talk about maximization versions of constraint satisfaction problems. Thinkof an instance of the problem as being equipped with an additional parameter t called threshold. Weask for a polynomial time algorithm which given the instance answers: • “YES” if all the constraints can be satisfied, • “NO” if there is no solution satisfying more than t constraints.Note that for instances, where more than t but not all constraints can be simultaneously satisfied, anyanswer is acceptable. We will now restrict our attention to the case where the threshold is a fixedfraction of the total amount of constraints in the instance. We call problem A to be gap NP-hard ifthere exists a positive (cid:15) such that there is no polynomial time algorithm to separate feasible instancesfrom those where only at most a (1 − (cid:15) ) fraction of the constraint can be simultaneously satisfiedunless P = N P .Observe that gap NP-hardness implies APX-hardness, but not vice versa. For example the linearordering problem (also known as max-subdag) is APX-hard [8], but is not gap NP-hard, since feasibleinstances may be found by topological sorting.Let us first note that the 3-SAT problem, which we used in the reduction from the previous section,has the gap hardness property. It is the essence of the famous PCP theorems that problems with suchgap hardness exist. For a proof of the gap hardness of 3-SAT see [9].9efore we show how to modify our reduction to prove gap hardness of the interval constraintcoloring problem, we need to introduce the notion of expander graphs . For brevity we will only givethe following extract from [9].
Definition 12.
Let G = ( V, E ) be a d -regular graph. Let E ( S, S ) = | ( S × S ) ∩ E | equal the numberof edges from a subset S ⊆ V to its complement. The edge expansion of G is defined as h ( G ) = min S : | S |≤| V | / E ( S, S ) | S | . Lemma 13.
There exists d ∈ Z and h > , such that there is a polynomial-time constructiblefamily { X n } n ∈ Z of d -regular graphs X n on n vertices with h ( X n ) ≥ h . (Such graphs are calledexpanders). Let us now give a “gap preserving” reduction from gap 3-SAT to gap interval constrained 3-coloring. Consider the reduction from the previous section. Observe that the amount of intervals ineach literal gadget, and therefore also in each clause gadget, is constant. The remaining intervals arethe variable intervals. While it is sufficient for the NP-hardness proof to connect occurrences of thesame variable in a “clique” fashion with variable intervals, it produces a potentially quadratic numberof intervals. Alternatively, one could connect these occurrences in a “path” fashion, but it would givetoo little connectivity for the gap reduction. The path-like connection has the desired property of usingonly linear amount of intervals, since each occurrence of a variable is linked with at most two otherones. We aim at providing more connectivity while not increasing the amount of intervals too much.A perfect tool to achieve this goal is a family of expander graphs.Consider the instance of the interval coloring problem obtained by the reduction from the previoussection, but without any variable intervals yet. Consider literal gadgets corresponding to occurrencesof a particular variable x . Think of these occurrences as of vertices of a graph G . Take an expandergraph X | V ( G ) | and connect two occurrences of x if the corresponding vertices in the expander areconnected. For each such connection use a pair of intervals. These intervals should be the originalvariable interval and an interval that is one element shorter on each of the sides. We will call this pairof intervals a variable link. Repeat this procedure for each of the variables.Observe that the number of variable links that we added is linear since all the used expandergraphs are d -regular. By contrast to the simple path-like connection, we now have the property, thatdifferent occurrences of the same variable have high edge connectivity. This can be turned into highpenalty for inconsistent valuations of literals in an imperfect solution. Theorem 14.
Constrained interval 3-coloring is gap NP-hard.
Proof.
We will argue that the above described reduction is a gap-preserving reduction from the gap3-SAT problem to the gap interval 3-coloring problem. We need to prove that there exists a positive (cid:15) such that feasible instances are hard to separate from those less than (1 − (cid:15) ) satisfiable.Let (cid:15) be the constant in the gap hardness of gap 3-SAT. We need to show two properties: that the“yes” instances of the gap 3-SAT problem are mapped to “YES” instances of our problem, and alsothat the “NO” instances are mapped to “NO” instances.The first property is simple, already in the NP-hardness proof in the previous section it was shownthat feasible instances are mapped by our reduction into feasible ones. To show the second property,we will take the reverse direction and argue that an almost feasible solution to the coloring instancecan be transformed into an almost feasible solution to the SAT instance.10uppose we are given a coloring χ that violates at most (cid:15) fraction of the constraints. Suppose theoriginal 3-SAT instance has q clauses, then our interval coloring instance has at most c · q intervals forsome constant c . The number of unsatisfied intervals in the coloring χ is then at most (cid:15)qc .We will say that a clause is broken if at least one of the intervals encoding it is not satisfied by χ .We will say that a variable link is broken if one of its intervals is not satisfied or one of the clauses itconnects is broken. An unsatisfied variable link interval contributes a single broken link; an unsatisfiedinterval within a clause breaks at most d intervals connected to the clause. Therefore, there is atmost d (cid:15)qc broken variable links in total.Recall that each variable link that is not broken connects occurrences of the same variable in twodifferent not broken clauses. Moreover, by the construction of the variable link, these two occurrencesdisplay the same logical value of the variable.Consider the truth assignment φ obtained as follows. For each variable consider its occurrencesin the not broken clauses. Each occurrence associates a logical value to the variable. Take for thisvariable the value that is displayed in the bigger set of not broken clauses, break ties arbitrarily.We will now argue, that φ satisfies a big fraction of clauses. Call a clause bad if it is not broken,but it contains a literal such that in the coloring χ this literal was active, but φ evaluates this literal tofalse. Observe that if a clause is neither broken nor bad, then it is satisfied by φ . It remains to boundthe amount of bad clauses.Consider the clauses that become bad from the choice of a value that φ assigns to a particularvariable x . Let b x be the number of such clauses. By the connectivity property of expanders, theamount of variable links connecting these occurrences of x with other occurrences is at least h b x . Aswe observed above, all these variable links are broken. Since there are in total at most d (cid:15)qc brokenlinks, we obtain that there is at most h d (cid:15)qc bad clauses. Hence, there are at most ( h d + 1) (cid:15)qc clauses that are either bad or broken and they cover all the clauses not satisfied by φ .It remains to fix (cid:15) = h (3 d + h ) c (cid:15) to obtain the property, that more than (cid:15) unsatisfiable instancesof 3-SAT are mapped to more than (cid:15) unsatisfiable instances of the constrained interval 3-coloringproblem. Acknowledgment
We thank Steven Kelk for valuable discussion.
References [1] Lam, T., Lanman, J., Emmett, M., Hendrickson, C., A.G., M., Prevelige, P.: Mapping ofprotein:protein contact surfaces by hydrogen/deuterium exchange, followed by on-line high-performance liquid chromatography-electrospray ionization fourier-transform ion-cyclotron-resonance mass analysis. Journal of Chromatography A (2002) 85–95[2] Even, S., Itai, A., Shamir, A.: On the complexity of timetable and multicommodity flow problems.SIAM Journal on Computing (1976) 691–703[3] Althaus, E., Canzar, S., Emmett, M.R., Karrenbauer, A., Marshall, A.G., Meyer-Baese, A., Zhang,H.: Computing of H/D-Exchange Speeds of Single Residues form Data of Peptic Fragments.Proceedings of the 23rd Annual ACM Symposium on Applied Computing, 2008 (2008) 1273–1277 114] Althaus, E., Canzar, S., Elbassioni, K., Karrenbauer, A., Mestre, J.: Approximating the intervalconstrained coloring problem. Proceedings of the 11th Scandinavian Workshop on AlgorithmTheory (SWAT) (2008) 210–221[5] Gandhi, R., Khuller, S., Parthasarathy, S., Srinivasan, A.: Dependent rounding and its applicationsto approximation algorithms. J. ACM (2006) 324–360[6] Canzar, S.: Lagrangian Relaxation - Solving NP-hard problems in Computational Biology viaCombinatorial Optimization. PhD thesis, Universit¨at des Saarlandes (2008)[7] Komusiewicz, C., Niedermeier, R., Uhlmann, J.: Deconstructing Intractability - A Case Study forInterval Constrained Coloring. In: Proceedings of the 20th Annual Symposium on CombinatorialPattern Matching (CPM’09). Volume 5577., Springer (2009) 207–220[8] Papadimitriou, C.H., Yannakakis, M.: Optimization, approximation, and complexity classes. J.Comput. Syst. Sci. (1991) 425–440[9] Dinur, I.: The pcp theorem by gap amplification. J. ACM54