The complexity of L(p,q)-Edge-Labelling
TThe complexity of L(p,q)-Edge-Labelling
Gaétan Berthe
ENS de Lyon, Lyon, [email protected]
Barnaby Martin
Department of Computer Science, Durham University, Durham, United [email protected]
Daniël Paulusma
Department of Computer Science, Durham University, Durham United [email protected]
Siani Smith
Department of Computer Science, Durham University, Durham, United [email protected]
Abstract
We classify the complexity of L ( p, q ) -Edge- k -Labelling in the sense that for all positive integers p and q we exhibit k so that we can show L ( p, q ) -Edge- k -Labelling is NP-complete. Mathematics of computing → Graph theory
Keywords and phrases
L(p,q)-labeling, dichotomy, NP-hard
Digital Object Identifier
Given any fixed nonnegative integer values p and q , the L ( p, q )- k -labelling problem consists inan assignment of labels from { , . . . , k − } to the nodes of a graph such that adjacent nodesreceive values which differ by at least p , and nodes connected by a path of length 2 receivevalues which differ by at least q [2]. Some authors instead define the latter condition as beingnodes at distance 2 receive values which differ by at least q (e.g. [3]). These definitions arethe same so long as p ≥ q and much of the literature considers only this case (e.g. [6]). If q > p , the definitions diverge. For example, the vertices of a triangle K need labels { , , } in the second definition but { , , } in the first. We use the first definition, in line with [2]. L ( p, q )-Labelling has been heavily studied, both from the combinatorial and computationalcomplexity perspectives. L (1 , L (1 , L (2 , L ( p, q ) -Edge- k -Labelling , considers an assignment ofthe labels to the edges instead of the vertices, and now the corresponding distance constraintsare placed also on the edges. In [7], the complexity of L ( p, q ) -Edge- k -Labelling is classified.It is in P for k < k ≥
6. In [9], the complexity of L (1 , -Edge- k -Labelling is classified. It is in P for k < k ≥
4. In this paper wecomplete the classification of the complexity of L ( p, q ) -Edge- k -Labelling in the sense thatfor all positive integers p and q we exhibit k so that we can show L ( p, q ) -Edge- k -Labelling is NP-complete. That is, we do not exhibit the border for k where the problem transitions We refer the reader to the excellent survey [2] which actually appears continually updated at . © Author: Please provide a copyright holder;licensed under Creative Commons License CC-BY42nd Conference on Very Important Topics (CVIT 2016).Editors: John Q. Open and Joan R. Access; Article No. 23; pp. 23:1–23:21Leibniz International Proceedings in InformaticsSchloss Dagstuhl – Leibniz-Zentrum für Informatik, Dagstuhl Publishing, Germany a r X i v : . [ c s . D M ] A ug Regime Reduction from Place in article2 ≤ q / p NAE-3-SAT Section 3.21 < q / p ≤ q / p = 1 5-COL Section 4 ([9]) / < q / p ≤ q / p = / / < q / p < / < q / p ≤ / NAE-3-SAT Appendix ([7])
Table 1
Table of results and where to find their proofs. from P to NP-complete (indeed, we do not even prove the existence of such a border). Theauthors of [7] were looking for a more general result, similar to ours, but found the case( p, q ) = (2 ,
1) laborious enough to fill one paper [10]. In fact, their proof settles for us allcases where p ≥ q . We now give our main result formally. (cid:73) Theorem 1.
For all p, q > , there exists k so that L ( p, q ) -Edge- k -Labelling is NP-complete. The proof follows by case analysis as per Table 1, where the corresponding section foreach the subresults is specified. Each section begins with a theorem detailing the relevantNP-completeness.
We use the terms colouring and labelling interchangeably. A special role will be played bywhat we term the extended n -star (especially for n = 4). This is a graph built from an n -star K ,n by subdividing each edge (so it becomes a path of length 2). It is common in theliterature to refer to the problem as L ( p, q )-labelling (or L ( h, k )-labelling) but henceforth wewill use L ( a, b )-labelling to free these other letters for alternative uses.The following lemma applies equally to the vertex- or edge-labelling problem. (cid:73) Lemma 2.
Let gcd( a, b ) = d > . Then the identity represents a polynomial time reductionfrom L ( a / d , b / d ) − k − Labelling to L ( a, b ) − k · d − Labelling . Proof.
Let G be a graph. We claim that G is L ( a / d , b / d ) − k − Labellable iff G is L ( a, b ) − k · d − Labellable. One direction is trivial, for if we have a L ( a / d , b / d ) − k − Labelling λ of G , then λ = d · λ is a L ( a, b ) − k · d − Labelling of G . Suppose now, we have a L ( a, b ) − k · d − Labelling λ of G .Define λ = d · b λ / d c , that is the rounding down of λ ( d ) to the nearest multiple of d . Weclaim that λ is also a L ( a, b ) − k · d − Labelling of G . Let us claim that | λ ( x ) − λ ( y ) | < c implies | λ ( x ) − λ ( y ) | < c , for both c ∈ { a, b } . Since c , λ ( x ) and λ ( y ) are multiples of d , wededuce that | λ ( x ) − λ ( y ) | < c − d + 1 which gives immediately | λ ( x ) − λ ( y ) | < c . Recalling c , λ ( x ) and λ ( y ) are multiples of d , we deduce that λ = λ / d is a L ( a, b ) − k − Labelling of G . (cid:74) ba > We will begin with the cases where the proofs are simplest. . Berthe, B. Martin, D. Paulusma and S. Smith 23:3 < ba ≤ (cid:73) Theorem 3. If < ba ≤ , the problem L ( a, b ) -Edge- (5 a + 1) -Labelling NP-complete.
This case is relatively simple as the variable gadget is built from a series of extended 4-starschained together, where each has a pendant 5-star to enforce some benign property. We willuse colours from the set { , . . . , a } . (cid:73) Lemma 4.
Let < ba ≤ . In any valid L ( a, b ) -edge- (5 a + 1) -labelling of the extended -star, if one leaf is coloured then all leaves are coloured in the interval { , . . . , a } ; and ifone leaf is coloured a then all leaves are coloured in the interval { a, . . . , a } . Proof.
Suppose some leaf is coloured by 0 and another leaf is coloured by l / ∈ { , . . . , a } .There are four inner edges of the star that are at distance 1 or 2 from these, and one another.If l < a , then at least 2 a labels are ruled out, which does not leave enough possibilitiesfor the inner edges to be labelled in (at best) { a + 1 , . . . , a } . If l ≥ a , then it is notpossible to use labels for the inner edges that are all strictly above l . It is also not possibleto use labels for the inner edges that are all strictly below l . In both cases, at least 2 a labelsare ruled out. Thus the labels, read in ascending order, must start no lower than a andhave a jump of 2 a at some point. It follows they are one of: a, a, a, a ; or a, a, a, a ; or a, a, a, a . This implies that l is itself a multiple of a (whichever one was omitted in thegiven sequence). But now, since b > a , there must be a violation of a distance 2 constraintfrom l . (cid:74) We would like to chain extended 4-stars together to build our variable gadgets, where theleaf edges represent variables (and enter into clause gadgets) and we interpret one of theregimes { , . . . , a } and { a, . . . , a } as true, and the other as false. However, the extended4-star can be validly L(a,b)-edge-(5 a + 1)-labelled in other ways that we did not yet consider.We can only use Lemma 4 if we can force one leaf in each extended 4-star to be either 0 or5 a . Fortunately, this is straightforward: take a 5-star and add a new edge to one of the edgesof the 5-star creating a path of length 2 from the centre of the star. This new edge can onlybe coloured 0 or 5 a . In Figure 1 we show how to chain together copies of the extended 4-star,together with pendant 5-star gadgets at the bottom, to produce many copies of exactly oneof the regimes { , . . . , a } and { a, . . . , a } . Note that the manner in which we attach thependant 5-star only produces a valid L(a,b)-edge-(5 a + 1)-labelling because 2 a ≥ b (otherwisesome distance 2 constraints would fail). So long as precisely one leaf per extended 4-staris used to encode a variable, then each encoding can realise all labels within each of theseregimes, and again this can be seen by considering the leaves drawn top-most in Figure 1,which can all be coloured anywhere in { a, . . . , a } . Let us recap, a variable gadget (to beused for a variable that appears in an instance of NAE-3-SAT m times) is built from chainingtogether m extended 4-stars, each with a pendant 5-star, exactly as is depicted in Figure 1for m = 3. The following is clear from our construction (the designation top is with referenceto the drawing in Figure 1). (cid:73) Lemma 5.
Any valid L(a,b)- (5 a + 1) -labelling of a variable gadget is such that the topleaves are all coloured from precisely one of the sets { , . . . , a } and { a, . . . , a } . Moreover,any colouring of the top leaves from one of these sets is valid. The clause gadget will be nothing more than a 3-star (a claw) which is formed from a newvertex uniting three (top) leaf edges from their respective variable gadgets. The following isclear.
C V I T 2 0 1 6 • { a,..., a } • { a,..., a } • { a,..., a } • a • a • a • a • a • • a • a • • a • a • • a •• a • a • a • a • a • a • • a • a • • a • a • • a • a • a • a • a • a • a • a Figure 1
Three extended 4-stars chained together, each with a pendant 5-star below, to form avariable gadget for Theorem 3. The leaf edges drawn on the top will be involved in clauses gadgetand each of these three edges can be coloured with anything from { a, . . . , a } . If the top leaf edgeis coloured 5 a it may be necessary that the inner star edge below it is coloured not 3 a but 2 a (cf.Figure 2). This is fine, the chaining construction works when swapping 2 a and 3 a . (cid:73) Lemma 6.
A clause gadget is in a valid L(a,b)- (5 a + 1) -labelling in the case where two ofits edges are coloured , a and the third a ; or two of its edges are coloured a, a and thethird . If all three edges come from only one of the regimes { , . . . , a } and { a, . . . , a } , itcan not be in a valid L(a,b)- (5 a + 1) -labelling. We are now ready to prove Theorem 3.1.
Proof of Theorem 3.1.
We reduce from NAE-3-SAT. Let Φ be an instance of NAE-3-SATinvolving n occurrences of (not necessarily distinct) variables and m clauses. Let us explainhow to build an instance G for L ( a, b ) -Edge- (5 a + 1) -Labelling . Each particular variablemay only appear at most n times, so for each variable we take a copy of the variable gadgetwhich is n extended 4-stars, each with a pendant 5-star, chained together. Each particularinstance of the variable belongs to one of the free (top) leaves of the variable gadget. Foreach clause of Φ we use a 3-star to unite an instance of these free (top) leaves from thecorresponding variable gadgets. Thus, we add a single vertex for each clause, but no newedges (they already existed in the variable gadgets). We claim that Φ is a yes-instance ofNAE-3-SAT iff G is a yes-instance of L ( a, b ) -Edge- (5 a + 1) -Labelling .(Forwards.) Take a satisfying assignment for Φ. Let the range { , . . . , a } represent trueand the range { a, . . . , a } represent false. This gives a valid labelling of the inner vertices inthe extended 4-stars, as exemplified in Figure 1. In each clause, either there are two instancesof true and one of false; or the converse. Let us explain the case where the first two variableinstances are true and the third is false (the general case can easily be garnered from this).Colour the (top) leaf associated with the first variable as 0, the second variable a and thethird variable 5 a . Plainly these can be consistently united in a claw by the new vertex that . Berthe, B. Martin, D. Paulusma and S. Smith 23:5 • a a • a • a • a • • • • a a • a • a • a • • • Figure 2
The clause gadget and its interface with the variable gadgets (where we must considerdistance 2 constraints). Both possible evaluations for not-all-equal are depicted. appeared in the clause gadget. We draw the situation in Figure 2 to demonstrate that this willnot introduce problems at distance 2. Thus, we can see this is a valid L(a,b)-(5 a + 1)-labellingof G .(Backwards.) For a valid L(a,b)-(5 a + 1)-labelling of G , we infer an assignment Φ byreading, in the variable gadget, the range { , . . . , a } as true and the range { a, . . . , a } asfalse. The consistent valuation of each variable follows from Lemma 5 and the fact that it isin fact not-all-equal follows from Lemma 6. (cid:74) ≤ ba In the case 2 ≤ ba , we can no longer get away with just an extended 4-star on which to baseour variable gadget. We need to move to higher degree. On the other hand, we will be ableto dispense with the pendant 5-stars. (cid:73) Theorem 7. If ≤ ba , let n ≥ be such that ( n − a ≥ b then problem L ( a, b ) -Edge- ( n − a + b + 1 -Labelling NP-complete.
We will need the following lemma. (cid:73)
Lemma 8.
Let ≤ ba and let n ≥ be such that ( n − a ≥ b . In any valid L ( a, b ) -edge- ( n − a + b + 1 -labelling of the extended n -star, either all leaves are coloured in the interval { ( n − a + b, . . . , ( n − a + b } or all leaves are coloured in the interval { , . . . , a } . Proof.
Suppose some leaf is coloured by l in { a + 1 , . . . , ( n − a + b − } . Consider the n − a + b + 1 at some point. But now we have run out of labels, because( n − a + ( a + b + 1) > ( n − a + b which is the last label.Suppose now that some leaf is coloured by l in { , . . . , a } and another leaf is colouredby l in { ( n − a + b, . . . , ( n − a + b } . It is now not possible to choose n − l and l together must remove more than b ≥ a possibilities for labels at both the top and the bottom of the order. Using 2 b > b + 2 a , thisleaves no more than ( n − a which is not enough space for n − a in the n − , . . . , ( n − a for the inner edges of theextended n -star, with { ( n − a + b, . . . , ( n − a + b } enforced on the leaves (and the wholerange from { ( n − a + b, . . . , ( n − a + b } possible adjacent to the label ( n − a ). Theother regime comes from order-inverting the colours. (cid:74) The stipulation ( n − a ≥ b plays no role in the previous lemma. It is needed in orderto chain together extended n -stars to form the variable gadget whose construction we now C V I T 2 0 1 6 L ( a, b ) − ( n − a + b + 1 ( n − a + b ( n − a ( n − a + b ( n − a ( n − a + b ( n − a λ ( n − a λ ( n − a a ≤ b ( n − a ≥ b λ λλλ λλ λ λλ n-2 repetitions n-3 repetitions n-2 repetitions Figure 3
The variable gadget for Theorem 7. The leaf edges drawn on the top will be involvedin clauses gadget and each of these three edges can be coloured with anything from { ( n − a + b, . . . , ( n − a + b } . explain. The variable gadget is made from a series of extended n -stars joined in a chain.They can join to one another in a path running from one’s inner star edge labelled 0 toanother’s inner star edge labelled ( n − a . In this fashion, the inner star edge labelled( n − a is free for the (top) leaf edge that acts as the point of contact for clauses. This innerstar edge may sometimes need to be labelled ( n − a (cf. Figure 4) in which case the otherinner star edge labelled ( n − a will be needed to perform the chaining. In the followinglemma, the designation top is with reference to the drawing in Figure 3. (cid:73) Lemma 9.
Any valid L(a,b)-edge- ( n − a + b + 1 -labelling of a variable gadget is suchthat the top leaves are all coloured from precisely one of the sets { , . . . , a } and { ( n − a + b, . . . , ( n − a + b } . Moreover, any colouring of the top leaves from one of these sets is valid. The clause gadget will be nothing more than a 3-star (a claw) which is formed from a newvertex uniting three (top) leaf edges from their respective variable gadgets. The following isclear. (cid:73)
Lemma 10.
A clause gadget is in a valid L(a,b)- ( n − a + b + 1 -labelling in the casewhere two of its edges are coloured , a and the third ( n − a + b ; or two of its edgesare coloured ( n − a + b, ( n − a + b and the third . If all three edges come from onlyone of the regimes { , . . . , a } and { ( n − a + b, . . . , ( n − a + b } , it can not be in a validL(a,b)- ( n − a + b + 1 -labelling. We are now ready to prove Theorem 7. . Berthe, B. Martin, D. Paulusma and S. Smith 23:7 • a ( n − a + b • b + a • b • ( n − a • • • • n − a + b ( n − a + b • a • ( n − a • ( n − a • • • Figure 4
The clause gadget and its interface with the variable gadgets (where we must considerdistance 2 constraints). Both possible evaluations for not-all-equal are depicted. Note the difference( n − a + b − ( n − a = b − a > a . Proof of Theorem 7.
We reduce from NAE-3-SAT. Choose n such that ( n − a ≥ b . Let Φbe an instance of NAE-3-SAT involving N occurrences of (not necessarily distinct) variablesand m clauses. Let us explain how to build an instance G for L ( a, b ) -Edge- ( n − a + b + 1 -Labelling . Each particular variable may only appear at most N times, so for each variablewe take a copy of the variable gadget which is N extended n -stars chained together. Eachparticular instance of the variable belongs to one of the free (top) leaves of the variablegadget. For each clause of Φ we use a 3-star to unite an instance of these free (top) leavesfrom the corresponding variable gadgets. Thus, we add a single vertex for each clause, but nonew edges (they already existed in the variable gadgets). We claim that Φ is a yes-instanceof NAE-3-SAT iff G is a yes-instance of L ( a, b ) -Edge- ( n − a + b + 1 -Labelling .(Forwards.) Take a satisfying assignment for Φ. Let the range { , . . . , a } represent trueand the range { ( n − a + b, . . . , ( n − a + b } represent false. This gives a valid labelling ofthe inner vertices in the extended n -stars, as exemplified in Figure 3. In each clause, eitherthere are two instances of true and one of false; or the converse. Let us explain the casewhere the first two variable instances are true and the third is false (the general case caneasily be garnered from this). Colour the (top) leaf associated with the first variable as 0, thesecond variable a and the third variable ( n − a + b . Plainly these can be consistently unitedin a claw by the new vertex that appeared in the clause gadget. We draw the situation inFigure 4 to demonstrate that this will not introduce problems at distance 2. Thus, we cansee this is a valid L(a,b)-( n − a + b + 1-labelling of G .(Backwards.) For a valid L(a,b)-( n − a + b + 1-labelling of G , we infer an assignment Φby reading, in the variable gadget, the range { , . . . , a } as true and the range { ( n − a + b, . . . , ( n − a + b } as false. The consistent valuation of each variable follows from Lemma 9and the fact that it is in fact not-all-equal follows from Lemma 10. (cid:74) ba = 1 This case has been settled in [9]. We give an independent proof, that is nonetheless close tothat in [9], for completeness. In light of Lemma 2, it suffices to find k so that L (1 , -Edge- k -Labelling is NP-hard. (cid:73) Theorem 11.
The problem L (1 , -Edge- -Labelling problem is NP-Complete. This case is relatively straightforward. Any valid L(1,1)-edge-5-labelling of the extended4-star has the property that the leaves have the same colour. Moreover, all 5 colours are
C V I T 2 0 1 6 • • • • • • • • • • • • • • • • •• • • • • • Figure 5
Three extended 4-stars chained together, to form a variable gadget for Theorem 11.The leaf edges drawn on the top will be involved in clauses gadget. Suppose the top leaves arecoloured 0 (as is drawn). In order to fulfill distance 2 constraints in the clause gadget, we may needthe inner star vertices adjacent to them to be coloured not always 1 (for example, if that edge leaf 0is adjacent in a clause gadget to another edge coloured 1). This is fine, the chaining constructionworks when swapping inner edges 1 and 2 wherever necessary. possible. The variable gadget may be taken as a series of extended 4-stars chained together.In the following, the “top” leaves refer to one of the two free leaves in each extended 4-star(not involved in the chaining together). (cid:73) Lemma 12.
Any valid L(1,1)-edge- -labelling of a variable gadget is such that the topleaves are all coloured from precisely one of the sets { } , { } , { } , { } and { } . Moreover,any colouring of the top leaves from one of these sets is valid. The clause gadget will be nothing more than a 2-star (a path) which is formed from a newvertex uniting two (top) leaf edges from their respective variable gadgets. The following isclear. (cid:73)
Lemma 13.
A clause gadget is in a valid L(1,1)-edge- -labelling in the case where itsedges are coloured distinctly. If they are coloured the same, then it can not be in a validL(1,1)-edge- -labelling. We are now ready to prove Theorem 11.
Proof of Theorem 11.
We reduce from 5-COL. Let G be an instance of 5-COL involving n vertices and m edges. Let us explain how to build an instance G for L (1 , -Edge- -Labelling . Each particular vertex may only appear in at most m edges (its degree), sofor each vertex we take a copy of the variable gadget which is m extended 4-stars chainedtogether. Each particular instance of the vertex belongs to one of the free (top) leaves of thevariable gadget. For each edge of G we use a 2-star to unite an instance of these free (top)leaves from the corresponding two variable gadgets. Thus, we add a single vertex for eachedge of G , but no new edges in G (they already existed in the variable gadgets). We claimthat G is a yes-instance of 5-COL iff G is a yes-instance of L (1 , -Edge- -Labelling . The final reduction will be from a colouring problem, not from a satisfiability problem. However, wepersist in reference to variable and clause gadgets, in the name of consistency. . Berthe, B. Martin, D. Paulusma and S. Smith 23:9 (Forwards.) Take a proper 5-colouring of G and induce these leaf labels on the corres-ponding variable gadgets. Plainly distinct leaf labels can be consistently united in a 2-clawby the new vertex that appeared in the clause gadget. Thus, we can see this is a validL(1,1)-5-labelling of G .(Backwards.) For a valid L(1,1)-5-labelling of G , we infer a 5-colouring of G by readingthe leaf edge labels from the variable gadget of the corresponding vertex. The consistentvaluation of each variable follows from Lemma 12 and the fact that it is proper (not-all-equal)follows from Lemma 13. (cid:74) < ba < (cid:73) Theorem 14. If < ba < , then the problem L ( a, b ) -Edge- (3 a + b + 1) -Labelling isNP-complete. The regimes of the following lemma are drawn in Figure 6. (cid:73)
Lemma 15.
Let < ab < . In a L ( a, b ) -edge- a + b + 1 -labelling c of the extended 4-cross,there are three regimes for the leaves. The first is { b, . . . , a } , the second is { a + b, . . . , a } ,and the third is { a + b, . . . , a } . Proof.
In a valid L ( a, b )-edge-3 a + b + 1-labelling, we note c < c < c < c the colors ofthe 4 edges in the middle of the cross, and l , l , l , l the colors of the leaves such that l i isthe color of the leaf connected to the edge of color c i . Claim 1 . ∀ i, c < l i < c . We only have to prove one inequality, as the other one is obtained by symmetry.If l i ≤ c (bearing in mind also b < a ), we have:3 a + b ≥ c − l i = ( c − l i ) + ( c − c ) + ( c − c ) + ( c − c ) ≥ a + b. So ( c , c , c , c ) = ( b, a + b, a + b, a + b ), but a > b so there is no possible value for l .Absurd!So c < l i , and by symmetry l i < c . Claim 2 . There exists i ∈ { , , } such that c i +1 − c i ≥ a + b .We suppose the contrary.We have proved c < l , l < c .If l < c , c − c = c − l + l − c ≥ a + b . Impossible.If c < l < c , c − c = c − l + l − c ≥ a + b . Impossible.So c < l < c .Symmetrically, we obtain c < l < c .So c < l < c < c < l < c , we get: c − c ≥ ( l − c ) + ( c − l ) + ( c − c ) + ( l − c ) + ( c − l ) ≥ b + a> a + b C V I T 2 0 1 6 L ( a, b ) − a + b + 1 coloring < ba < b ≤ x, y ≤ a a a + b a + b a + bx y b Case 3 a + b ≤ x, y ≤ a x a a + b a + by a + b Case 22 a + b ≤ x, y ≤ a x a a a + b a + b a y Figure 6
The regimes of Theorem 14. . Berthe, B. Martin, D. Paulusma and S. Smith 23:11 • b • b • b • a + b • a + b • a + b • b • • a + b • b • • a + b • b • • a + b • b •• a + b • a + b • a + b • b • b • b Figure 7
Three extended 4-stars chained together, to form a variable gadget for Theorem 14.The leaf edges drawn on the top will be involved in clauses gadget. Suppose the top leaves arecoloured b (as is drawn). In order to fulfill distance 2 constraints in the clause gadget, we mayneed the inner star vertices adjacent to them to be coloured not always a + b (for example, if thatedge leaf b is adjacent in a clause gadget to another edge coloured a + b ). This is fine, the chainingconstruction works when swapping inner edges a + b and 2 a + b wherever necessary. Impossible.Now we are in a position to derive the lemma, with the three regimes coming from the threepossibilities of Claim 2. If i = 1, then the inner edges of the star are 0 , a + b, a + b, a + b andthe leaves come from { b, . . . , a } . If i = 2, then the inner edges of the star are 0 , a, a + b, a + b and the leaves come from { a + b, . . . , a } . If i = 3, then the inner edges of the star are0 , a, a + b, a + b and the leaves come from { a + b, . . . , a } . (cid:74) The variable gadget may be taken as a series of extended 4-stars chained together. In thefollowing, the “top” leaves refer to one of the two free leaves in each extended 4-star (notinvolved in the chaining together). The following is a simple consequence of Lemma 15 andis depicted in Figure 7. (cid:73)
Lemma 16.
Any valid L(a,b)-edge- (3 a + b + 1) -labelling of a variable gadget is such thatthe top leaves are all coloured from precisely one of the sets { b, . . . , a } , { a + b, . . . , a } or { a + b, . . . , a } . Moreover, any colouring of the top leaves from one of these sets is valid. The clause gadget will be nothing more than a 2-star (a path) which is formed from a newvertex uniting two (top) leaf edges from their respective variable gadgets. The following isclear. (cid:73)
Lemma 17.
A clause gadget is in a valid L(a,b)-edge- (3 a + b + 1) -labelling in the casewhere its edges are coloured distinctly. If they are coloured the same, then it can not be in avalid L(a,b)-edge- (3 a + b + 1) -labelling. We are now ready to prove Theorem 11.
Proof of Theorem 14.
We reduce from 3-COL. Let G be an instance of 3-COL involving n vertices and m edges. Let us explain how to build an instance G for L ( a, b ) -Edge- (3 a + b +1) -Labelling . Each particular vertex may only appear in at most m edges (its degree), sofor each vertex we take a copy of the variable gadget which is m extended 4-stars chained C V I T 2 0 1 6 together. Each particular instance of the vertex belongs to one of the free (top) leaves of thevariable gadget. For each edge of G we use a 2-star to unite an instance of these free (top)leaves from the corresponding two variable gadgets. Thus, we add a single vertex for each edgeof G , but no new edges in G (they already existed in the variable gadgets). We claim that G is a yes-instance of 3-COL iff G is a yes-instance of L ( a, b ) -Edge- (3 a + b + 1) -Labelling .(Forwards.) Take a proper 3-colouring of G and induce these leaf labels on the corres-ponding variable gadgets according to the three regimes of Lemma 15. For example, mapcolours 1, 2, 3 to b, a + b, a + b . Plainly distinct leaf labels can be consistently united in a2-claw by the new vertex that appeared in the clause gadget. Thus, we can see this is a validL(a,b)-(3 a + b + 1)-labelling of G .(Backwards.) For a valid L(a,b)-(3 a + b + 1)-labelling of G , we infer a 3-colouring of G byreading the leaf edge labels from the variable gadget of the corresponding vertex and mappingthese to their corresponding regime. The consistent valuation of each variable follows fromLemma 16 and the fact that it is proper (not-all-equal) follows from Lemma 17. (cid:74) ba = In light of Lemma 2, it suffices to find k so that L (3 , -Edge- k -Labelling is NP-hard. (cid:73) Theorem 18.
The problem L (3 , -Edge- -Labelling problem is NP-Complete. We use the colours { , . . . , } . The following can be verified by hand (although we havedone so by computer). (cid:73) Lemma 19.
In a valid L(3,2)-edge-11-labelling of the extended 4-star, the possibles labelof the three other leaves after one label is fixed are given in the following dictionary: { , , } { , } { , } { , } { , } { , , } The variable gadget may be taken as a series of extended 4-stars chained together. In thefollowing, the “top” leaves refer to one of the two free leaves in each extended 4-star (notinvolved in the chaining together). (cid:73)
Lemma 20.
Any valid L(3,2)-edge- -labelling of a variable gadget is such that the topleaves are all coloured from precisely one of the sets { , } or { , , , } . Moreover, anycolouring of the top leaves from one of these sets is valid. The clause gadget will be nothing more than a 3-star (a path) which is formed from a newvertex uniting three (top) leaf edges from their respective variable gadgets. The following isclear. The Python program for checking this can be found at https://perso.ens-lyon.fr/gaetan.berthe/edge-labelling/labelling-code.py . Use, e.g.: extended_four_star=[(0,1),(0,2),(0,3),(0,4),(1,5),(2,6),(3,7),(4,8)] with plotPoss(extended_four_star, 3, 2, 12, {(1,5):[2]}) to find 2 : { , , } . . Berthe, B. Martin, D. Paulusma and S. Smith 23:13 • • • • • • • • • • • • • • • • •• • • • • • Figure 8
Three extended 4-stars chained together, to form a variable gadget for Theorem 18.The leaf edges drawn on the top will be involved in clauses gadget. We show in this case how bothsides of the regime representing false can be achieved (2 and 9). (cid:73)
Lemma 21.
A clause gadget is in a valid L(3,2)-edge- -labelling precisely in the casewhere its edges are coloured one from { , } and two from { , , , } . Proof of Theorem 18.
We reduce from 1-in-3-SAT. Let Φ be an instance of 1-in-3-SATinvolving n occurrences of (not necessarily distinct) variables and m clauses. Let us explainhow to build an instance G for L (3 , -Edge- -Labelling . Each particular variable mayonly appear at most n times, so for each variable we take a copy of the variable gadget whichis n extended 4-stars chained together. Each particular instance of the variable belongs toone of the free (top) leaves of the variable gadget. For each clause of Φ we use a 3-starto unite an instance of these free (top) leaves from the corresponding variable gadgets.Thus, we add a single vertex for each clause, but no new edges (they already existed in thevariable gadgets). We claim that Φ is a yes-instance of 1-in-3-SAT iff G is a yes-instance of L (3 , -Edge- -Labelling .(Forwards.) Take a satisfying assignment for Φ. Let the range { , } represent true andthe range { , , , } represent false. In particular, every clause has two false and one shouldbe chosen as (e.g.) 2 and the other 9. Thus, where a variable is false, some of top leaves arelabelled 2 and others 9 (and this is shown in Figure 8). In each clause, we will have (say)2 , ,
5. Plainly these can be consistently united in a claw by the new vertex that appearedin the clause gadget. We draw the situation in Figure 8 to demonstrate that this will notintroduce problems at distance 2. Thus, we can see this is a valid L(3,2)-12-labelling of G .(Backwards.) For a valid L(3,2)-12-labelling of G , we infer an assignment Φ by reading,in the variable gadget, range { , } as true and the range { , , , } as false. The consistentvaluation of each variable follows from Lemma 20 and the fact that it is in fact not-all-equalfollows from Lemma 21. (cid:74) < ba < (cid:73) Theorem 22. If < ba < , then the problem L ( a, b ) -Edge- (4 b + a + 1) -Labelling isNP-complete. C V I T 2 0 1 6 • • • • • • • Figure 9
The clause gadget and its interface with the variable gadgets (where we must considerdistance 2 constraints).
This is probably the most involved case in terms of the sophistication of the proof. We needsome lemmas before we can specify our gadgets. (cid:73)
Lemma 23. If < b < a and λ < a + b , with k = λ + 1 , any edge k -labelling of theextended -star must involve inner edge labels of (0 ≤ ) p < q < r < s ( < k ) so that both q − p ≥ b and s − r ≥ b . Proof.
The assumption λ < a + b forces: λ − s < b , p < b , q − p, r − q, s − r < a + b .Consider colouring the edge beside that edge which is coloured by r . This can’t be colouredby anything other than something between p and q , forcing q − p ≥ b . Similarly, considercolouring the edge beside that edge which is coloured by q . This can’t be coloured byanything other than something between r and s , forcing s − r ≥ b . (cid:74)(cid:73) Corollary 24.
Let a ≤ b . The minimal k so that the extended -star gadget can be edge k -labelled is b + a + 1 . Proof.
We know it is at least 4 b + a + 1 from the previous lemma. Further, the colouringalluded to in the previous proof extends to a valid colouring. Set labels ( p, q, r, s ) to(0 , b, b + a, b + a ). Then, the edges next to p and q can be coloured 3 b + a , and the edgesnext to r and s can be coloured b . (cid:74)(cid:73) Lemma 25.
Let < ba < and k = 4 b + a + 1 . The extended -star gadget can be labelledonly such that two leaves are b and the other two are b + a . Proof.
The inequality ≤ ba proves it is a correct labelling.We have λ = 4 b + a < a + b so from the previous lemma we deduce the inner edgelabels are 0 , b, b + a, b + a . Adjacent to 2 b + a must be b and the same is true for 4 b + a .Adjacent to 2 b must be 3 b + a and the same is true of 0. (cid:74) The gadget
Note that colours are in the set { , . . . , b + a } . Below, in Figure 10, we give two gadgetsfor the variables, the end gadget and the (basic part of the) variable gadget . The variablegadget admits a number of (4 b + a + 1)-edge-labellings, but we want the only possibilities tobe that drawn and one that swaps 3 b + a and b . This we enforce by attaching an end gadgetat the end (e.g. the left-hand end). For example, one may join it by adding new edges (inthe present colouring of the end gadget, that would force the other colouring of the variablegadget). That is, we join the end gadget using the two edges drawn at the bottom below tothe (basic part of the) variable gadget using the two edges drawn (say) to the left below.The join is accomplished by adding two new edges, one for each position (left/bottom andright/top). In the variable gadget, the variables will extend from the 10-cycles, but this is . Berthe, B. Martin, D. Paulusma and S. Smith 23:15 • b + a... • b + a ... • b • b • b • b + a • b + a • b • b + a • b + a • b •• • • b + a • b + a • b + a • • b + a • • b + a • b { b + a,..., b } • b + a • b { b + a,..., b } • b + a • b • b • b + a • b • b + a • b • b + a • • •• • b + a • • b + a • • b + a • b + a • b { b + a,..., b } • b + a • b { b + a,..., b } • b + a • b Figure 10
End gadget (above) and basic part of variable gadget (below). possible only on one side. We now meet, in Figure 11, a full variable gadget drawn with avariable protrusion, in this case a b variable (the symmetric form gives a 3 b + a variable).Summing up, we derive the following lemma. (cid:73) Lemma 26.
In a full variable gadget complete with an end gadget, any valid edge- (4 b + a +1) -labelling has the property that the leaves on the protrusions from the edge gadgets either allcontain b and do not contain b + a ; or all contain b + a and do not contain b . The clause gadget will be nothing other than an extended 4-star, whose properties we derivealready in Lemma 25.
Proof of Theorem 22.
We reduce from monotone 2-in-4-SAT. Let Φ be an instance of 2-in-4-SAT involving n occurrences of (not necessarily distinct) variables and m clauses. Letus explain how to build an instance G for L ( a, b ) -Edge- (4 b + a + 1) -Labelling . Eachparticular variable may only appear at most n times, so for each variable we take a a fullvariable gadget with n protrusions. Each particular instance of the variable belongs to one ofthe free (top) leaves of the variable gadget. For each clause of Φ we use an extended 4-starto unite an instance of these free (top) leaves from the corresponding variable gadgets. Thisinvolves adding new vertices and new edges making a 4-star (the other edges of the extended4-star already existed in the variable gadgets). We claim that Φ is a yes-instance of 2-4-SATiff G is a yes-instance of L ( a, b ) -Edge- (4 b + a + 1) -Labelling .(Forwards.) Take a satisfying assignment for Φ. Let b represent true and 3 b + a representfalse. In particular, every clause has two true and two false literals these can be consistentlyunited in an extended 4-star as in Figure 12. Thus, we can see this is a valid L(a,b)-(4 b + a +1)-labelling of G .(Backwards.) For a valid L(a,b)-(4 b + a + 1)-labelling of G , we infer an assignment Φ byreading, in the variable gadget, b as true and 3 b + a as false. The consistent valuation of eachvariable follows from Lemma 26 and the fact that it is 2-in-4 follows from Lemma 25. (cid:74) C V I T 2 0 1 6 • b not 3 b + a • b • b b + a • • b + a • • b + a • b b • b + a • b { b + a,..., b } • b + a • b • b • b + a • b • b + a • b • b + a • • •• • b + a • • b + a • • b + a • b + a • b { b + a,..., b } • b + a • b { b + a,..., b } • b + a • b Figure 11
A full variable gadget drawn with a variable protrusion. • b + a b + a b • b • b • b + a • b + a • b • b • a • a • • • • Figure 12
The clause gadget and its interface with the variable gadgets (where we must considerdistance 2 constraints). . Berthe, B. Martin, D. Paulusma and S. Smith 23:17
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A NP-hardness of the L(a,b)-edge- (3 a + 1) -labelling problem, for a ≥ b We follow the exposition of [7], which addresses the L(2,1)-edge-7-labelling problem. Withpermission we have used (an adaptation) of their diagrams. Note that in [7], they would callthe problem we address the L(a,b)-edge- -labelling problem as, in their exposition 3 a refersto the set { , . . . , a } . (cid:73) Theorem 27.
The L(a,b)-edge-(3a+1)-labelling problem, for a ≥ b , is NP-hard. Proof.
By reduction from NAE-3-SAT using the gadgets and properties detailed in Lemmas 28and 29, below. (cid:74)
For 1 ≤ k ≤
3, we define the sets k = (cid:74) ( k − a + b, ka − b (cid:75) . The edges of a 4-star have tobe coloured 0 , a, a, a , in any valid L(a,b)-edge-(3a+1)-labelling. Then any neighbouringedge of the star has to be in one these sets, of the form k . These properties we will nowuse without further comment.A variable is represented by the variable gadget of Figure 13. (cid:73) Lemma 28.
In any valid L(a,b)-edge-(3a+1)-labelling of the variable gadget, the threeedges free in the top of a -star at the top of a repeatable section must be coloured (in allrepeatable and initial parts) by either { a, a, a } or { , a, a } . C V I T 2 0 1 6 e a e e ′ a e ′ e ′ a e The unique labeling of the cycle a a a a a a { a, a, a } { a, a, a } { a, a, a } Repeatable part Initial partInitial part e v w Variable gadget Figure 13
Variable gadget (adapted from [7]). . Berthe, B. Martin, D. Paulusma and S. Smith 23:19
Proof.
Let us consider various possibilities for the colouring of { e , e } and { e , e } (up tothe order inverting map that takes (0 , . . . , a ) to (3 a, . . . , { e , e } e e e e e e e I. a a a in 1 a, a impossible in the cycleII. a a in 22 a a, a impossible in the cycleIII. 0 a impossible3 a a — —IV. 0 a impossible2 a a — —V. 0 3 a in 1 a a impossible —VI. 0 2 a in 1 a a in 2 0 Table 2
Variable gadget table (adapted from [7]). can not be { a, a } . In this case, the cycle must continue (bearing in mind that everyelement 0 and 1 mod 3 must be from { , a, a, a } ) in a certain way. To the right itmust continue: ( a, a, , a, a, , a, a, . . . ). However, to the left it must continue:(2 a, a, , , a, , , a, . . . ). These paths can now never join together in a cycle. This rulesout Cases I and II.The remaining labellings, Case VI and its various symmetries, are possible and result inthe claimed behaviour. (cid:74) The clause is represented by the clause gadget of Figure 14. (cid:73)
Lemma 29.
Consider any valid L(a,b)-edge-(3a+1)-labelling of the clause gadget, such thatthe input parts of the variable gadgets satisfy the previous lemma. Two of the input variablegadget parts must come from one of the regimes { a, a, a } or { , a, a } , and the other inputpart from the other regime. In particular, if all three input variable gadget parts come fromonly one of the regimes, then this can not be extended to a valid L(a,b)-edge-(3a+1)-labelling. Proof.
Let us consider various possibilities for the colouring of e and { e , e } (up to theorder inverting map that takes (0 , . . . , a ) to (3 a, . . . , { a, a, a } or { , a, a } (namely, { , a } and { a, a } , respectively). The claimed behaviour isclear. (cid:74) C V I T 2 0 1 6
First variable e a e a e ′ e
21 0 2 a e a a Second variableThird variable a e ′ a e ′′ aa aa a Clause gadget
Figure 14
Clause gadget (adapted from [7]). . Berthe, B. Martin, D. Paulusma and S. Smith 23:21 e e e e e e e I. 0 2 a or 3 a a or 2 aa Both in 3 . ImpossibleBoth in 3 . Impossible —II. 0 a or 3 a a or 2 a a Both in 1 . ImpossibleBoth in 1 . Impossible —III. 0 2 a or aa or 2 a a In 1In 2 0 or 3IV. 0 2 a or aa or 2 a a In 2In 1 2 a or 3 Table 3