Metric Dimension Parameterized by Treewidth
MMetric Dimension Parameterized by Treewidth
Édouard Bonnet
Univ Lyon, CNRS, ENS de Lyon, Université Claude Bernard Lyon 1, LIP UMR5668, [email protected]
Nidhi Purohit
Univ Lyon, CNRS, ENS de Lyon, Université Claude Bernard Lyon 1, LIP UMR5668, [email protected]
Abstract
A resolving set S of a graph G is a subset of its vertices such that no two vertices of G have thesame distance vector to S . The Metric Dimension problem asks for a resolving set of minimumsize, and in its decision form, a resolving set of size at most some specified integer. This problem isNP-complete, and remains so in very restricted classes of graphs. It is also W[2]-complete with respectto the size of the solution.
Metric Dimension has proven elusive on graphs of bounded treewidth.On the algorithmic side, a polytime algorithm is known for trees, and even for outerplanar graphs,but the general case of treewidth at most two is open. On the complexity side, no parameterizedhardness is known. This has led several papers on the topic to ask for the parameterized complexityof
Metric Dimension with respect to treewidth.We provide a first answer to the question. We show that
Metric Dimension parameterized bythe treewidth of the input graph is W[1]-hard. More refinedly we prove that, unless the ExponentialTime Hypothesis fails, there is no algorithm solving
Metric Dimension in time f (pw) n o (pw) on n -vertex graphs of constant degree, with pw the pathwidth of the input graph, and f any computablefunction. This is in stark contrast with an FPT algorithm of Belmonte et al. [SIAM J. DiscreteMath. ’17] with respect to the combined parameter tl + ∆, where tl is the tree-length and ∆ themaximum-degree of the input graph. Theory of computation → Graph algorithms analysis; Theory ofcomputation → Fixed parameter tractability
Keywords and phrases
Metric Dimension, Treewidth, Parameterized Hardness
Digital Object Identifier
The
Metric Dimension problem has been introduced in the 1970s independently by Slater[22] and by Harary and Melter [13]. Given a graph G and an integer k , Metric Dimension asks for a subset S of vertices of G of size at most k such that every vertex of G is uniquelydetermined by its distances to the vertices of S . Such a set S is called a resolving set , anda resolving set of minimum-cardinality is called a metric basis . The metric dimension ofgraphs finds application in various areas including network verification [1], chemistry [3],robot navigation [18], and solving the Mastermind game [4]. Metric Dimension is an entry of the celebrated book on intractability by Garey andJohnson [12] where the authors show that it is NP-complete. In fact
Metric Dimension remains NP-complete in many restricted classes of graphs such as planar graphs [6], split,bipartite, co-bipartite graphs, and line graphs of bipartite graphs [9], graphs that are bothinterval graphs of diameter two and permutation graphs [11], and in a subclass of unit diskgraphs [16]. On the positive side, the problem is polynomial-time solvable on trees [22, 13, 18].Diaz et al. [6] generalize this result to outerplanar graphs. Fernau et al. [10] give a polynomial-time algorithm on chain graphs. Epstein et al. [9] show that
Metric Dimension (andeven its vertex-weighted variant) can be solved in polynomial time on co-graphs and forests © Édouard Bonnet, Nidhi Purohit;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. 23:1–23:23Leibniz International Proceedings in InformaticsSchloss Dagstuhl – Leibniz-Zentrum für Informatik, Dagstuhl Publishing, Germany a r X i v : . [ c s . CC ] J u l augmented by a constant number of edges. Hoffmann et al. [15] obtain a linear algorithm oncactus block graphs.Hartung and Nichterlein [14] prove that Metric Dimension is W[2]-complete (paramet-erized by the size of the solution k ) even on subcubic graphs. Therefore an FPT algorithmsolving the problem is unlikely. However Foucaud et al. [11] give an FPT algorithm withrespect to k on interval graphs. This result is later generalized by Belmonte et al. [2] whoobtain an FPT algorithm with respect to tl + ∆ (where tl is the tree-length and ∆ is themaximum-degree of the input graph), implying one for parameter tl + k . Indeed intervalgraphs, and even chordal graphs, have constant tree-length. Hartung and Nichterlein [14]presents an FPT algorithm parameterized by the vertex cover number, Eppstein [8], by themax leaf number, and Belmonte et al. [2], by the modular-width (a larger parameter thanclique-width).The complexity of Metric Dimension parameterized by treewidth is quite elusive. Itis discussed [8] or raised as an open problem in several papers [2, 6]. On the one hand,it was not known, prior to our paper, if this problem is W[1]-hard. On the other hand,the complexity of
Metric Dimension in graphs of treewidth at most two is still an openquestion.
We settle the parameterized complexity of
Metric Dimension with respect to treewidth.We show that this problem is W[1]-hard, and we rule out, under the Exponential TimeHypothesis (ETH), an algorithm running in f (tw) | V ( G ) | o (tw) , where G is the input graph, twits treewidth, and f any computable function. Our reduction even shows that an algorithmin time f (pw) | V ( G ) | o (pw) is unlikely on constant-degree graphs, for the larger parameterpathwidth pw. This is in stark contrast with the FPT algorithm of Belmonte et al. [2] forthe parameter tl + ∆ where tl is the tree-length and ∆ is the maximum-degree of the graph.We observe that this readily gives an FPT algorithm for ctw + ∆ where ctw is the connectedtreewidth, since ctw (cid:62) tl. This unravels an interesting behavior of Metric Dimension ,at least on bounded-degree graphs: usual tree-decompositions are not enough for efficientsolving. Instead one needs tree-decompositions with an additional guarantee that the verticesof a same bag are at a bounded distance from each other.As our construction is quite technical, we chose to introduce an intermediate problemdubbed k -Multicolored Resolving Set in the reduction from k -Multicolored Inde-pendent Set to Metric Dimension . The first half of the reduction, from k -MulticoloredIndependent Set to k -Multicolored Resolving Set , follows a generic and standardrecipe to design parameterized hardness with respect to treewidth. The main difficulty isto design an effective propagation gadget with a constant-size left-right cut. The secondhalf brings some new local attachments to the produced graph, to bridge the gap between k -Multicolored Resolving Set and Metric Dimension . Along the way, we introducea number of gadgets: edge, propagation, forced set, forced vertex. They are quite stream-lined and effective. Therefore, we believe these building blocks may help in designing newreductions for
Metric Dimension . In Section 2 we introduce the definitions, notations, and terminology used throughout thepaper. In Section 3 we present the high-level ideas to establish our result. We definethe k -Multicolored Resolving Set problem which serves as an intermediate step for . Bonnet, N. Purohit 23:3 our reduction. In Section 4 we design a parameterized reduction from the W[1]-complete k -Multicolored Independent Set to k -Multicolored Resolving Set parameter-ized by treewidth. In Section 5 we show how to transform the produced instances of k -Multicolored Resolving Set to Metric Dimension -instances (while maintainingbounded treewidth). In Section 6 we conclude with some open questions.
We denote by [ i, j ] the set of integers { i, i + 1 , . . . , j − , j } , and by [ i ] the set of integers [1 , i ].If X is a set of sets, we denote by ∪X the union of them. All our graphs are undirected and simple (no multiple edge nor self-loop). We denote by V ( G ), respectively E ( G ), the set of vertices, respectively of edges, of the graph G . For S ⊆ V ( G ), we denote the open neighborhood (or simply neighborhood ) of S by N G ( S ), i.e.,the set of neighbors of S deprived of S , and the closed neighborhood of S by N G [ S ], i.e., theset N G ( S ) ∪ S . For singletons, we simplify N G ( { v } ) into N G ( v ), and N G [ { v } ] into N G [ v ]. Wedenote by G [ S ] the subgraph of G induced by S , and G − S := G [ V ( G ) \ S ]. For S ⊆ V ( G )we denote by S the complement V ( G ) \ S . For A, B ⊆ V ( G ), E ( A, B ) denotes the set ofedges in E ( G ) with one endpoint in A and the other one in B .The length of a path in an unweighted graph is simply the number of edges of the path.For two vertices u, v ∈ V ( G ), we denote by dist G ( u, v ), the distance between u and v in G ,that is the length of the shortest path between u and v . The diameter of a graph is thelongest distance between a pair of its vertices. The diameter of a subset S ⊆ V ( G ), denotedby diam G ( S ), is the longest distance between a pair of vertices in S . Note that the distanceis taken in G , not in G [ S ]. In particular, when G is connected, diam G ( S ) is finite for every S . A pendant vertex is a vertex with degree one. A vertex u is pendant to v if v is the onlyneighbor of u . Two distinct vertices u, v such that N ( u ) = N ( v ) are called true twins , and false twins if N [ u ] = N [ v ]. In particular, false twins are adjacent. In all the above notationswith a subscript, we omit it whenever the graph is implicit from the context. A tree-decomposition of a graph G , is a tree T whose nodes are labeled by subsets of V ( G ),called bags , such that for each vertex v ∈ V ( G ), the bags containing v induce a non-emptysubtree of T , and for each edge e ∈ E ( G ), there is at least one bag containing both endpointsof e . A connected tree-decomposition further requires that each bag induces a connectedsubgraph in G . The width of a (connected) tree-decomposition is the size of its largestbag minus one. The treewidth (resp. connected treewidth) of a graph G is the minimumwidth of a tree-decomposition (resp. a connected tree-decomposition) of G . The length of atree-decomposition is the maximum diameter of its bags in G . The tree-length of a graph G is the minimum length of a tree-decomposition of G . We denote the treewidth, connectedtreewidth, and tree-length of a graph by tw, ctw, and tl respectively. Since a connectedgraph on n vertices has diameter at most n −
1, it holds that ctw (cid:62) tl.The pathwidth is the same as treewidth except the tree T is now required to be a path,and hence is called a path-decomposition. In particular pathwidth is always larger thantreewidth. Later we will need to upper bound the pathwidth of our constructed graph.Since writing down a path-decomposition is a bit cumbersome, we will rely on the following C V I T 2 0 1 6 characterization of pathwidth. Kirousis and Papadimitriou [19] show the equality betweenthe interval thickness number, which is known to be pathwidth plus one, and the nodesearching number . Thus we will only need to show that the number of searchers required towin the following one-player game is bounded by a suitable function. We imagine the edges ofa graph to be contaminated by a gas. The task is to move around a team of searchers, placedat the vertices, in order to clean all the edges. A move consists of removing a searcher fromthe graph, adding a searcher at an unoccupied vertex, or displacing a searcher from a vertexto any other vertex (not necessarily adjacent). An edge is cleaned when both its endpointsare occupied by a searcher. However after each move, all the cleaned edges admitting afree-of-searchers path from one of its endpoints to the endpoint of a contaminated edge arerecontaminated. The node searching number is the minimum number of searchers requiredto win the game.
Parameterized complexity aims to solve hard problems in time f ( k ) |I| O (1) , where k is aparameter of the instance I which is hopefully (much) smaller than the total size of I . Moreformally, a parameterized problem is a pair (Π , κ ) where Π ⊆ L for some language L ⊆ Σ ∗ over a finite alphabet Σ (e.g., the set of words, graphs, etc.), and κ is a mapping from L to N . An element I ∈ L is called an instance (or input ). The mapping κ associates eachinstance to an integer called parameter . An instance is said positive if I ∈
Π, and a negative otherwise. We denote by |I| the size of I , that can be thought of as the length of the word I .An FPT algorithm is an algorithm which solves a parameterized problem (Π , κ ), i.e., decideswhether or not an input
I ∈ L is positive, in time f ( κ ( I )) |I| O (1) for some computablefunction f . We refer the interested reader to recent textbooks in parameterized algorithmsand complexity [7, 5]. The
Exponential Time Hypothesis (ETH) is a conjecture by Impagliazzo et al. [17] assertingthat there is no 2 o ( n ) -time algorithm for on instances with n variables. Lokshtanovet al. [20] survey conditional lower bounds under the ETH.An FPT reduction from a parameterized problem (Π ⊆ L, κ ) to a parameterized problem(Π ⊆ L , κ ) is a mapping ρ : L L such that for every I ∈ L :(1) I ∈ Π ⇔ ρ ( I ) ∈ Π ,(2) | ρ ( I ) | (cid:54) f ( κ ( I )) |I| O (1) for some computable function f , and(3) κ ( ρ ( I )) (cid:54) g ( κ ( I )) for some computable function g .We further require that for every I , we can compute ρ ( I ) in FPT time h ( κ ( I )) |I| O (1) forsome computable function h . Condition (1) makes ρ a valid reduction, condition (2) togetherwith the further requirement on the time to compute ρ ( I ) make the mapping ρ FPT , andcondition (3) controls that the new parameter κ ( ρ ( I )) is bounded by a function of theoriginal parameter κ ( I ). One can therefore observe that using ρ in combination with anFPT algorithm solving (Π , κ ) yields an FPT procedure to solve the initial problem (Π , κ ).A standard use of an FPT reduction is to derive conditional lower bounds: if a problem(Π , κ ) is thought not to admit an FPT algorithm, then an FPT reduction from (Π , κ ) to(Π , κ ) indicates that (Π , κ ) is also unlikely to admit an FPT algorithm. We refer thereader to the textbooks [7, 5] for a formal definition of W[1]-hardness. For the purpose ofthis paper, we will just state that W[1]-hard are parameterized problems that are unlikely . Bonnet, N. Purohit 23:5 to be FPT, and that the following problem is W[1]-complete even when all the V i have thesame number of elements, say t (see for instance [21]). k -Multicolored Independent Set ( k -MIS ) Parameter: k Input:
An undirected graph G , an integer k , and ( V , . . . , V k ) a partition of V ( G ). Question:
Is there a set I ⊆ V ( G ) such that | I ∩ V i | = 1 for every i ∈ [ k ], and G [ I ] isedgeless?Every parameterized problem that k -Multicolored Independent Set FPT-reducesto is W[1]-hard. Our paper is thus devoted to designing an FPT reduction from k -Multicolored Independent Set to Metric Dimension parameterized by tw. Letus observe that the ETH implies that one (equivalently, every) W[1]-hard problem is not inthe class of problems solvable in FPT time (FPT =W[1]). Thus if we admit that there is nosubexponential algorithm solving , then k -Multicolored Independent Set is notsolvable in time f ( k ) | V ( G ) | O (1) . Actually under this stronger assumption, k -MulticoloredIndependent Set is not solvable in time f ( k ) | V ( G ) | o ( k ) . A concise proof of that fact canbe found in the survey on the consequences of ETH [20]. A pair of vertices { u, v } ⊆ V ( G ) is said to be resolved by a set S if there is a vertex w ∈ S such that dist( w, u ) = dist( w, v ). A vertex u is said to be distinguished by a set S if for any w ∈ V ( G ) \ { u } , there is a vertex v ∈ S such that dist( v, u ) = dist( v, w ). A resolving set ofa graph G is a set S ⊆ V ( G ) such that every two distinct vertices u, v ∈ V ( G ) are resolvedby S . Equivalently, a resolving set is a set S such that every vertex of G is distinguishedby S . Then Metric Dimension asks for a resolving set of size at most some threshold k .Note that a resolving set of minimum size is sometimes called a metric basis for G . Metric Dimension ( MD ) Parameter: tw( G ) Input:
An undirected graph G and an integer k . Question:
Does G admit a resolving set of size at most k ?Here we anticipate on the fact that we will mainly consider Metric Dimension paramet-erized by treewidth. Henceforth we sometimes use the notation Π / tw to emphasize that Π isnot parameterized by the natural parameter (size of the resolving set) but by the treewidthof the input graph. We will show the following. (cid:73)
Theorem 1.
Unless the ETH fails, there is no computable function f such that MetricDimension can be solved in time f ( pw ) n o ( pw ) on constant-degree n -vertex graphs. We first prove that the following variant of
Metric Dimension is W[1]-hard. k -Multicolored Resolving Set ( k -MRS ) Parameter: tw( G ) Input:
An undirected graph G , an integer k , a set X of q disjoint subsets of V ( G ): X , . . . , X q , and a set P of pairs of vertices of G : { x , y } , . . . , { x h , y h } . Question:
Is there a set S ⊆ V ( G ) of size q such that(i) for every i ∈ [ q ], | S ∩ X i | = 1, and(ii) for every p ∈ [ h ], there is an s ∈ S satisfying dist G ( s, x p ) = dist G ( s, y p )? C V I T 2 0 1 6
In words, in this variant the resolving set is made by picking exactly one vertex in eachset of X , and not all the pairs should be resolved but only the ones in a prescribed set P .We call critical pair a pair of P . In the context of k -Multicolored Resolving Set , wecall legal set a set which satisfies the former condition, and resolving set a set which satisfiesthe latter. Thus a solution for k -Multicolored Resolving Set is a legal resolving set.The reduction from k -Multicolored Independent Set starts with a well-establishedtrick to show parameterized hardness by treewidth. We create m “empty copies” of the k -MIS -instance ( G, k, ( V , . . . , V k )), where m := | E ( G ) | and t := | V i | . We force exactly onevertex in each color class of each copy to be in the resolving set, using the set X . In eachcopy, we introduce an edge gadget for a single (distinct) edge of G . Encoding an edge of k -MIS in the k -MRS -instance is fairly simple: we build a pair (of P ) which is resolved byevery choice but the one selecting both its endpoints in the resolving set. We now need toforce a consistent choice of the vertex chosen in V i over all the copies. We thus design apropagation gadget. A crucial property of the propagation gadget, for the pathwidth of theconstructed graph to be bounded, is that it admits a cut of size O ( k ) disconnecting one copyfrom the other. Encoding a choice in V i in the distances to four special vertices, called gates ,we manage to build such a gadget with constant-size “left-right” separator per color class.This works by introducing t pairs (of P ) which are resolved by the south-west and north-eastgates but not by the south-east and north-west ones. Then we link the vertices of a copyof V i in a way that the higher their index, the more pairs they resolve in the propagationgadget to their left, and the fewer pairs they resolve in the propagation gadget to their right.We then turn to the actual Metric Dimension problem. We design a gadget whichsimulates requirement (i) by forcing a vertex of a specific set X in the resolving set. Thisworks by introducing two pairs that are only resolved by vertices of X . We attach this newgadget, called forcing set gadget, to all the k color classes of the m copies. Finally we have tomake sure that a candidate solution resolves all the pairs, and not only the ones prescribedby P . For that we attach two adjacent “pendant” vertices to strategically chosen vertices.One of these two vertices have to be in the resolving set since they are false twins, hence notresolved by any other vertex. Then everything is as if the unique common neighbor v of thefalse twins was added to the resolving set. Therefore we can perform this operation as longas v does not resolve any of the pairs of P .To facilitate the task of the reader, henceforth we stick to the following conventions:Index i ∈ [ k ] ranges over the k rows of the k -MRS/MD -instance or color classes of k -MIS .Index j ∈ [ m ] ranges over the m columns of the k -MRS/MD -instance or edges of k -MIS .Index γ ∈ [ t ], ranges over the t vertices of a color class.We invite the reader to look up Table 1 when in doubt about a notation/symbol relative tothe construction. k -Multicolored Resolving Set/tw In this section, we give an FPT reduction from the W[1]-complete k -MulticoloredIndependent Set to k -Multicolored Resolving Set parameterized by treewidth.More precisely, given a k -Multicolored Independent Set -instance ( G, k, ( V , . . . , V k ))we produce in polynomial-time an equivalent k -Multicolored Resolving Set -instance( G , k , X , P ) where G has pathwidth (hence treewidth) O ( k ). . Bonnet, N. Purohit 23:7 Let (
G, k, ( V , . . . , V k )) be an instance of k -Multicolored Independent Set where( V , . . . , V k ) is a partition of V ( G ) and V i := { v i,γ | (cid:54) γ (cid:54) t } . We arbitrarily number e , . . . , e j , . . . , e m the m edges of G . We start with a high-level description of the k -MRS -instance ( G , k , X , P ). For each colorclass V i , we introduce m copies V i , . . . , V ji , . . . , V mi of a selector gadget to G . Each set V ji is added to X , so a solution has to pick exactly one vertex within each selector gadget. Onecan imagine the vertex-sets V i , . . . , V mi to be aligned on the i -th row , with V ji occupyingthe j -th column (see Figure 1). Each V ji has t vertices denoted by v ji, , v ji, , . . . , v ji,t , whereeach v ji,γ “corresponds” to v i,γ ∈ V i . We make v ji, v ji, . . . v ji,t a path with t − e j ∈ E ( G ), we insert an edge gadget G ( e j ) containing a pair of vertices { c j , c j } that we add to P . Gadget G ( e j ) is attached to V ji and V ji , where e j ∈ E ( V i , V i ).The edge gadget is designed in a way that the only legal sets that do not resolve { c j , c j } are the ones that precisely pick v ji,γ ∈ V ji and v ji ,γ ∈ V ji such that e j = v i,γ v i ,γ . We add a propagation gadget P j,j +1 i between two consecutive copies V ji and V j +1 i , where the indicesin the superscript are taken modulo m . The role of the propagation gadget is to ensure thatthe choices in each V ji ( j ∈ [ m ]) corresponds to the same vertex in V i . V V V V V V V V V V V V V V V V V V P , P , P , P , P , P , P , P , P , P , P , P , P , P , P , P , P , P , G ( e ) G ( e ) G ( e ) G ( e ) G ( e ) G ( e ) Figure 1
The overall picture with k = 3 color classes, t = 5 vertices per color class, m = 6 edges, e = v , v , , e = v , v , , e = v , v , , etc. The dashed lines on the left and right symbolize thatthe construction is cylindrical. The intuitive idea of the reduction is the following. We say that a vertex of G is selected if it is put in the resolving set of G , a tentative solution. The propagation gadget P j,j +1 i ensures a consistent choice among the m copies V i , . . . , V mi . The edge gadget ensures thatthe selected vertices of G correspond to an independent set in the original graph G . If both C V I T 2 0 1 6 the endpoints of an edge e j are selected, then the pair { c j , c j } is not resolved. We now detailthe construction. For each i ∈ [ k ] and j ∈ [ m ], we add to G a path on t − v ji, , v ji, , . . . , v ji,t , and denotethis set of vertices by V ji . Each v ji,γ corresponds to v i,γ ∈ V i . We call j -th column the set S i ∈ [ k ] V ji , and i -th row , the set S j ∈ [ m ] V ji . We set X := { V ji } i ∈ [ k ] ,j ∈ [ m ] . By definition of k -Multicolored Resolving Set , a solution S has to satisfy that for every i ∈ [ k ] , j ∈ [ m ], | S ∩ V ji | = 1. We call legal set a set S of size k = km that satisfies this property. We call consistent set a legal set S which takes the “same” vertex in each row, that is, for every i ∈ [ k ], for every pair ( v ji,γ , v j i,γ ) ∈ ( S ∩ V ji ) × ( S ∩ V j i ), then γ = γ . For each edge e j = v i,γ v i ,γ ∈ E ( G ), we add an edge gadget G ( e j ) in the j -th column of G . G ( e j ) consists of a path on three vertices: c j g j c j . The pair { c j , c j } is added to the list ofcritical pairs P . We link both v ji,γ and v ji ,γ to g j by a private path of length t + 2. Welink the at least two and at most four vertices v ji,γ − , v ji,γ +1 , v ji ,γ − , v ji ,γ +1 (whenever theyexist) to c j by a private path of length t + 2. This defines at most six paths from V ji ∪ V ji to G ( e j ). Let us denote by W j the at most six endpoints of these paths in V ji ∪ V ji . For each v ∈ W j , we denote by P ( v, j ) the path from v to G ( e j ). We set E ji := S v ∈ W j ∩ V ji P ( v, j ) and E ji := S v ∈ W j ∩ V ji P ( v, j ). We denote by X j the set of the at most six neighbors of W j onthe paths to G ( e j ). Henceforth we may refer to the vertices in some X j as the cyan vertices .Individually we denote by e ji,γ the cyan vertex neighbor of v ji,γ in P ( v ji,γ , j ). We observe thatfor fixed i and j , e ji,γ exists for at most three values of γ . We add an edge between two cyanvertices if their respective neighbors in V ji are also linked by an edge (or equivalently, if theyhave consecutive “indices γ ”). These extra edges are useless in the k -MRS -instance, but willturn out useful in the MD -instance. See Figure 2 for an illustration of the edge gadget.The rest of the construction will preserve that for every v ∈ ( V ji ∪ V ji ) \ { v ji,γ , v ji ,γ } ,dist( v, c j ) = dist( v, c j ) + 2, and for each v ∈ { v ji,γ , v ji ,γ } , dist( v, c j ) = dist( v, g j ) + 1 =dist( v, c j ). In other words, the only two vertices of V ji ∪ V ji not resolving the critical pair { c j , c j } are v ji,γ and v ji ,γ , corresponding to the endpoints of e j . Between each pair ( V ji , V j +1 i ), where j + 1 is taken modulo m , we insert an identical copy ofthe propagation gadget, and we denote it by P j,j +1 i . It ensures that if the vertex v ji,γ is ina legal resolving set S , then the vertex of S ∩ V j +1 i should be some v j +1 i,γ with γ (cid:54) γ . Thecylindricity of the construction and the fact that exactly one vertex of V ji is selected, willtherefore impose that the set S is consistent. P j,j +1 i, comprises four vertices sw ji , se ji , nw ji , ne ji , called gates , and a set A ji of 2 t vertices a ji, , . . . , a ji,t , α ji, , . . . , α ji,t . We make both a ji, a ji, . . . a ji,t and α ji, α ji, . . . α ji,t a path with t − We use the expression private path to emphasize that the different sources get a pairwise internallyvertex-disjoint path to the target. . Bonnet, N. Purohit 23:9 V V V v , v , v , v , v , e , e , g c c G ( e ) 66666 Figure 2
The edge gadget G ( e ) with e = v , v , . Weighted edges are short-hands for subdivi-sions of the corresponding length. The edges between the cyan vertices will not be useful for the k -MRS -instance, but will later simplify the construction of the MD -instance. edges. For each γ ∈ [ t ], we add the pair { a ji,γ , α ji,γ } to the set of critical pairs P . Removingthe gates disconnects A ji from the rest of the graph.We now describe how we link the gates to V ji , V j +1 i , and A ji . We link v ji, (the “top”vertex of V ji ) to sw ji and v ji,t (the “bottom” vertex of V ji ) to nw ji both by a path of length 2.We also link v j +1 i, to se ji by a path of length 3, and v j +1 i,t to ne ji by a path of length 2. Thenwe make nw ji adjacent to a ji, and α ji, , while we make ne ji adjacent to α ji, only. We makese ji adjacent to a ji,t and α ji,t , while we make sw ji adjacent to a ji,t only. Finally, we add anedge between ne ji and nw ji , and between sw ji and se ji . See Figure 3 for an illustration of thepropagation gadget P j,j +1 i with t = 5.Let us motivate the gadget P j,j +1 i . One can observe that the gates ne ji and sw ji resolvethe critical pairs of the propagation gadget, while the gates nw ji and se ji do not. Considerthat the vertex added to the resolving set in V ji is v ji,γ . Its shortest paths to critical pairs below it (that is, with index γ > γ ) go through the gate sw ji , whereas its shortest paths tocritical pairs at its level or above (that is, with index γ (cid:54) γ ) go through the gate nw ji . Thus v ji,γ only resolves the critical pairs { a ji,γ , α i,γ } with γ > γ . On the contrary, the vertex ofthe resolving set in V j +1 i only resolves the critical pairs { a ji,γ , α ji,γ } at its level or above.This will force that its level is γ or below. Hence the vertices of the resolving in V ji and V j +1 i should be such that γ (cid:62) γ . Since there is also a propagation gadget between V mi and V i , this circular chain of inequalities forces a global equality. We put the pieces together as described in the previous subsections. At this point, it isconvenient to give names to the neighbors of V ji in the propagation gadgets P j − ,ji and C V I T 2 0 1 6 v ji, v ji, v ji, v ji, v ji, v j +1 i, v j +1 i, v j +1 i, v j +1 i, v j +1 i, V ji V j +1 i sw ji se ji nw ji ne ji a ji, α ji, a ji, α ji, a ji, α ji, a ji, α ji, a ji, α ji, Figure 3
The propagation gadget P j,j +1 i . The critical pairs { a ji,γ , α ji,γ } are surrounded by thindashed lines. The blue (resp. red) integer on a vertex of A ji is its distance to the blue (resp. red)vertex in V ji (resp. V j +1 i ). Note that the blue vertex distinguishes the critical pairs below it, whilethe red vertex distinguishes critical pairs at its level or above. P j,j +1 i . We may refer to them as blue vertices (as they appear in Figure 4). We denote bytl ji the neighbor of v ji, in P j − ,ji , tr ji , the neighbor of v ji, in P j,j +1 i , bl ji , the neighbor of v ji,t in P j − ,ji , and br ji , the neighbor of v ji,t in P j,j +1 i . We add the following edges and paths.For any pair i, j such that e j has an endpoint in V i , the vertices tl ji , tr ji , bl ji , br ji are linkedto g j by a private path of length the distance of their unique neighbor in V ji to c j . We add anedge between se ji and se j +1 i , and between nw ji and nw j +1 i (where j + 1 is modulo m ). Finally,for every e j ∈ E ( V i , V i ), we add four paths between se ji , se ji , nw ji , nw ji and g j ∈ G ( e j ). Moreprecisely, for each i ∈ { i, i } , we add a path from g j to se ji of length dist( g j , sw ji ) −
4, anda path from g j to nw ji of length dist( g j , nw ji ) −
4. These distances are taken in the graphbefore we introduced the new paths, and one can observe that the length of these paths is atleast t . This finishes the construction.We recall that, by a slight abuse of language, a resolving set in the context of k -Multicolored Resolving Set is a set which resolves all the critical pairs of P . Inparticular, it is not necessarily a resolving set in the sense of Metric Dimension . Withthat terminology, a solution for k -Multicolored Resolving Set is a legal resolving set. We now check that the reduction is correct. We start with the following technical lemma. Ifa set X contains a pair that no vertex of N ( X ) (that is N [ X ] \ X ) resolves, then no vertexoutside X can distinguish the pair. (cid:73) Lemma 2.
Let X be a subset of vertices, and a, b ∈ X be two distinct vertices. If for everyvertex v ∈ N ( X ) , dist ( v, a ) = dist ( v, b ) , then for every vertex v / ∈ X , dist ( v, a ) = dist ( v, b ) . Proof.
Let v be a vertex outside of X . We further assume that v is not in N ( X ), otherwisewe can already conclude that it does not distinguish { a, b } . A shortest path from v to a , has to go through N ( X ). Let w a be the first vertex of N ( X ) met in this shortest . Bonnet, N. Purohit 23:11 path from v to a . Similarly, let w b be the first vertex of N ( X ) met in a shortest pathfrom v to b . Since w a , w b ∈ N ( X ), they satisfy dist( w a , a ) = dist( w a , b ) and dist( w b , a ) =dist( w b , b ). Then, dist( v, a ) (cid:54) dist( v, w b ) + dist( w b , a ) = dist( v, w b ) + dist( w b , b ) = dist( v, b ),and dist( v, b ) (cid:54) dist( v, w a ) + dist( w a , b ) = dist( v, w a ) + dist( w a , a ) = dist( v, a ). Thusdist( v, a ) = dist( v, b ). (cid:74) We use the previous lemma to show that every vertex of a V ji only resolves critical pairsin gadgets it is attached to. This will be useful in the two subsequent lemmas. (cid:73) Lemma 3.
For any i ∈ [ k ] , j ∈ [ m ] , and v ∈ V ji , v does not resolve any critical pairoutside of P j − ,ji , P j,j +1 i (where indices in the superscript are taken modulo m ), and { c j , c j } .Furthermore, if e j ∈ E ( G ) has no endpoint in V i ⊆ V ( G ) , then v does not resolve { c j , c j } . Proof.
We first show that v ∈ V ji does not resolve any critical pair in propagation gadgetsthat are not P j − ,ji and P j,j +1 i . Let { a j i ,γ , α j i ,γ } be a critical pair in a propagation gadgetdifferent from P j − ,ji and P j,j +1 i . Let X be the connected component containing P j ,j +1 i of G − ( { nw j − i , se j − i , nw j +1 i , se j +1 i } ∪ C e ), where C e comprises { c j , g j } if e j has an endpointin V i and { c j +1 , g j +1 } if e j +1 has an endpoint in V i . Thus C e has size 0, 2, or 4. Onecan observe that N ( X ) = { nw j − i , se j − i , nw j +1 i , se j +1 i } ∪ C e , that V j i ∪ V j +1 i ⊆ X , andthat no “other V ji ” intersects X . In particular V ji is fully contained in G − X . We nowcheck that no vertex of N ( X ) resolves the pair { a j i ,γ , α j i ,γ } (which is inside X ). For each u ∈ { nw j − i , nw j +1 i } , it holds that dist( u, a j i ,γ ) = γ + 1 = dist( u, a j i ,γ ) (the shortest pathsgo through nw j i ), while for each u ∈ { se j − i , se j +1 i , it holds that dist( u, a j i ,γ ) = t − γ + 2 =dist( u, a j i ,γ ) (the shortest paths go through se j i ). If they are part of C e , g j and c j alsodo not resolve { a j i ,γ , α j i ,γ } , the shortest paths going through the gates nw j i or se j i , andrespectively g j and then the gates nw j i or se j i . For the same reason, g j +1 and c j +1 do notresolve { a j i ,γ , α j i ,γ } . Then we conclude by Lemma 2 that no vertex of V ji (in particular v )resolves { a j i ,γ , α j i ,γ } , or any critical pair in P j i .Let us now show that the pair { c j , c j } is not resolved by any vertex of ∪X \ ( V ji ∪ V ji ) suchthat e j ∈ E ( V i , V i ). Let Y := { tl ji , tr ji , bl ji , br ji , tl ji , tr ji , bl ji , br ji , nw ji , se ji , nw ji , se ji } ,and X be the connected component containing g j in G − Y . Again one can observe that N ( X ) = Y , X contains V ji ∪ V ji but does not intersect any “other V ji ”. We therefore showthat no vertex of Y resolves { c j , c j } , and conclude with Lemma 2. All the vertices of Y have a private path to g j whose length is such that they have a shortest path to c j goingthrough g j . Therefore ∀ u ∈ Y , dist( u, c j ) = dist( u, g j ) + 1 = dist( u, c j ). (cid:74) The two following lemmas show the equivalences relative to the expected use of the edgeand propagation gadgets. They will be useful in Sections 4.2.1 and 4.2.2. (cid:73)
Lemma 4.
A legal set S resolves the critical pair { c j , c j } with e j = v i,γ v i ,γ if and only ifthe vertex v ji,γ i in V ji ∩ S and the vertex v ji ,γ i in V ji ∩ S satisfy ( γ, γ ) = ( γ i , γ i ) . Proof.
By Lemma 3, no vertex of S \ { v ji,γ i , v ji ,γ i } resolves { c j , c j } . By construction of G , v ji,γ (resp. v ji ,γ ) is the only vertex of V ji (resp. V ji ) that does not resolve { c j , c j } . Indeedthe shortest paths of v ji,γ , for γ (cid:62) γ + 1, to { c j , c j } go through v ji,γ +1 which resolves thepair. Note that a shortest path between V ji and V ji has length at least 2 t + 4, so a shortestpath from v ji,γ to { c j , c j } cannot go through V ji . Similarly the shortest paths of v ji,γ ,for γ (cid:54) γ −
1, to { c j , c j } go through v ji,γ − which also resolves the pair. Thus only v ji,γ C V I T 2 0 1 6 (resp. v ji ,γ ), whose shortest paths to { c j , c j } go via g j , does not resolve this pair among V ji (resp. V ji ). Hence, the critical pair { c j , c j } is not resolved by S if and only if v ji,γ i = v ji,γ and v ji ,γ i = v ji ,γ . (cid:74)(cid:73) Lemma 5.
A legal set S resolves all the critical pairs of P j,j +1 i if and only if the vertex v ji,γ in V ji ∩ S and the vertex v j +1 i,γ in V j +1 i ∩ S satisfy γ (cid:54) γ . Proof.
By Lemma 3, no vertex of S \{ v ji,γ , v j +1 i,γ } resolves a critical pair of P j,j +1 i . Let us showthat the critical pairs that v ji,γ resolves in A ji are exactly the pairs { a ji,z , α ji,z } with z > γ . Forany z ∈ [ t ], it holds that dist( v ji,γ , a ji,z ) = min( t +2+ z − γ, t +2+ γ − z ) = t +2+min( z − γ, γ − z ),and dist( v ji,γ , α ji,z ) = min( t + 2 + z − γ, t + 3 + γ − z ) = t + 2 + min( z − γ, γ − z + 1). So if z > γ , dist( v ji,γ , a ji,z ) = t + 2 + γ − z = t + 2 + γ − z + 1 = dist( v ji,γ , α ji,z ). Whereas if z (cid:54) γ ,dist( v ji,γ , a ji,z ) = t + 2 + z − γ = dist( v ji,γ , α ji,z ).Similarly, we show that the critical pairs that v j +1 i,γ resolves in A ji are exactly the pairs { a ji,z , α ji,z } with z (cid:54) γ . For every z ∈ [ t ], it holds that dist( v j +1 i,γ , a ji,z ) = min( t + 3 + z − γ , t +3+ γ − z ) = t +3+min( z − γ , γ − z ), and dist( v j +1 i,γ , α ji,z ) = min( t +2+ z − γ , t +3+ γ − z ) = t + 2 + min( z − γ , γ − z + 1). So if z (cid:54) γ , dist( v j +1 i,γ , a ji,z ) = t + 3 + z − γ = t + 2 + z − γ =dist( v j +1 i,γ , α ji,z ). Whereas if z > γ , dist( v j +1 i,γ , a ji,z ) = t + 3 + γ − z = dist( v j +1 i,γ , α ji,z ). Thisimplies that all the critical pairs of A ji are resolved by S if and only if γ (cid:54) γ . (cid:74) We can now prove the correctness of the reduction. The construction can be computedin polynomial time in | V ( G ) | , and G itself has size bounded by a polynomial in | V ( G ) | . Wepostpone checking that the pathwidth is bounded by O ( k ) to the end of the second step,where we produce an instance of MD whose graph G admits G as an induced subgraph. k -Multicolored Independent Set in G ⇒ legal resolving set in G . Let { v ,γ , . . . , v k,γ k } be a k -multicolored independent set in G . We claim that S := S j ∈ [ m ] { v j ,γ , . . . , v jk,γ k } is a legal resolving set in G (of size km ). The set S is legal byconstruction. Since for every i ∈ [ k ], and j ∈ [ m ], v ji,γ i and v j +1 i,γ i are in S ( j + 1 is modulo m ), all the critical pairs in the propagation gadgets are resolved by S , by Lemma 5. Since { v ,γ , . . . , v k,γ k } is an independent set in G , there is no e j = v i,γ v i ,γ ∈ E ( G ), such that( γ, γ ) = ( γ i , γ i ). Thus every critical pair { c j , c j } is resolved by S , by Lemma 4. G ⇒ k -Multicolored Independent Set in G . Assume that there is a legal resolving set S in G . For every i ∈ [ k ], for every j ∈ [ m ], thevertex v ji,γ ( i,j ) in V ji ∩ S and the vertex v j +1 i,γ ( i,j +1) in V j +1 i ∩ S ( j +1 is modulo m ) are such that γ ( i, j ) (cid:54) γ ( i, j +1), by Lemma 5. Thus γ ( i, (cid:54) γ ( i, (cid:54) . . . (cid:54) γ ( i, m − (cid:54) γ ( i, m ) (cid:54) γ ( i, γ i := γ ( i,
1) = γ ( i,
2) = . . . = γ ( i, m −
1) = γ ( i, m ). We claim that { v ,γ , . . . , v k,γ k } is a k -multicolored independent set in G . Indeed, there cannot be an edge e j = v i,γ i v i ,γ i ∈ E ( G ),since otherwise the critical pair { c j , c j } is not resolved, by Lemma 4. In this section, we produce in polynomial time an instance ( G , k ) of Metric Dimension equivalent to ( G , X , km, P ) of k -Multicolored Resolving Set . The graph G has alsopathwidth O ( k ). Now, an instance is just a graph and an integer. There is no longer X and . Bonnet, N. Purohit 23:13 P to constrain and respectively loosen the “resolving set” at our convenience. This createstwo issues: (1) the vertices outside the former set X can now be put in the resolving set,potentially yielding undesired solutions and (2) our candidate solution (when there is a k -multicolored independent set in G ) may not distinguish all the vertices. We settle both issues by attaching new gadgets to G . Eventually the new graph G willcontain G as an induced subgraph. To settle the issue (1), we design a forced set gadget. Aforced set gadget attached to V ji contains two pairs of vertices which are only resolved byvertices of V ji . Thus the gadget simulates the action of X .There are a few pairs which are not resolved by a solution of k -Multicolored ResolvingSet . To make sure that all pairs are resolved, we add vertices which need be selected in theresolving set. Technically we could use the previous gadget on a singleton set. But we canmake it simpler: we just attach two pendant neighbors, that we then make adjacent, to somechosen vertices. A pair of pendant neighbors are false twins in the whole graph. So we knowthat at least one of these two vertices have to be in the resolving set. Hence we call that the forced vertex gadget, and one of the false twins, a forced vertex . It is important that theseforced vertices do not resolve any pair of P . So we can only add pendant twins to verticesthemselves not resolving any pair of P . To deal with the issue (1), we introduce two new pairs of vertices for each V ji . The intentionis that the only vertices resolving both these pairs simultaneously are precisely the verticesof V ji . For any i ∈ [ k ] and j ∈ [ m ], we add to G two pairs of vertices { p ji , q ji } and { r ji , s ji } ,and two gates π ji and ρ ji . Vertex π ji is adjacent to p ji and q ji , and vertex ρ ji is adjacent to r ji and s ji .We link v ji, to p ji , and v ji,t to r ji , each by a path of length t . It introduces two newneighbors of v ji, and v ji,t (the brown vertices in Figure 4). We denote them by tb ji and bb ji ,respectively. The blue and brown vertices are linked to π ji and ρ ji in the following way. Welink tl ji and tr ji to π ji by a private path of length t , and to ρ ji by a private path of length 2 t − ji and br ji to π ji by a private path of length 2 t −
1, and to ρ ji by a private path oflength t . (Let us clarify that the names of the blue vertices bl ji and br ji are for “bottom-left”and “bottom-right”, and not for “blue” and “brown”.) We link tb ji (neighbor of v ji, ) to ρ ji by a private path of length 2 t −
1. We link bb ji (neighbor of v ji,t ) to π ji by a private pathof length 2 t −
1. Note that the general rule to set the path length is to match the distancebetween the neighbor in V ji and p ji (resp. r ji ). With that in mind we link, if it exists, the topcyan vertex tc ji (the one with smallest index γ ) neighboring V ji to π ji with a path of lengthdist( v ji,γ , p ji ) = t + γ − v ji,γ is the unique vertex in N (tc ji ) ∩ V ji . Observe that withthe notations of the previous section tc ji = e ji,γ . We also link, if it exists, the bottom cyanvertex bc ji (the one with largest index γ ) to ρ ji with a path of length dist( v, r ji ) where v isagain the unique neighbor of bc ji in V ji .It can be observed that we only have two paths (and not all six) from the at most threecyan vertices to the gates π ji and ρ ji . This is where the edges between the cyan vertices will Also, it is now possible to put two or more vertices of the same V ji in the resolving set S C V I T 2 0 1 6 become relevant. See Figure 4 for an illustration of the forced vertex gadget, keeping in mindthat, for the sake of legibility, four paths to { π ji , ρ ji } are not represented. We now deal with the issue (2). By we add (or attach ) a forced vertex to an already presentvertex v , we mean that we add two adjacent neighbors to v , and that these two verticesremain of degree 2 in the whole graph G . Hence one of the two neighbors will have to beselected in the resolving set since they are false twins. We call forced vertex one of these twovertices (picking arbitrarily).For every i ∈ [ k ] and j ∈ [ m ], we add a forced vertex to the gates nw ji and se ji of P j,j +1 i .We also add a forced vertex to each vertex in N ( { π ji , ρ ji } ) \ { p ji , q ji , r ji , s ji } . This represents atotal of 12 vertices (6 neighbors of π ji and 6 neighbors of ρ ji ). For every j ∈ [ m ], we attach aforced vertex to each vertex in N ( g j ) \ { c j , c j } . This constitutes 14 neighbors (hence 14 newforced vertices). Therefore we set k := km + 12 km + 2 km + 14 m = 15 km + 14 m . V ji G ( e j ) g j c j c j ji se ji nw ji ne ji sw j − i se j − i nw j − i ne j − i p ji π ji q ji r ji ρ ji s ji Figure 4
Vertices tl ji , tr ji , bl ji , br ji (blue vertices) are linked to π ji , ρ ji by paths of appropriatelengths (see Section 5.1.1). Vertex tb ji is linked by a path to ρ ji , while bb ji is linked by a path to π ji .To avoid cluttering the figure, we did not represent four paths: from tl ji and bc ji to ρ ji , and frombl ji and tc ji to π ji . We also did not represent the paths already in the k -MRS -instance from theblue vertices to g j . Black vertices are forced vertices. Gray edges are the edges in the propagationgadgets already depicted in Figure 3. Not represented on the figure, we add a forced vertex toeach neighbor of the red vertices, except p ji , q ji , r ji , s ji , c j , c j . Finally we add four more paths andpotentially two edges (see Section 5.1.3). . Bonnet, N. Purohit 23:15 We use the convention that P ( u, v ) denotes the path from u to v which was specificallybuilt from u to v . In other words, for P ( u, v ) to make sense, there should be a point in theconstruction where we say that we add a (private) path between u and v . For the sake oflegibility, P ( u, v ) may denote either the set of vertices or the induced subgraph. We alsodenote by ν ( u, v ) the neighbor of u in the path P ( u, v ). Observe that P ( u, v ) is a symmetricnotation but not ν ( u, v ).We add a path of length dist( ν ( π ji , tr ji ) , sw ji ) = t between ν ( π ji , tr ji ) and se ji , and a pathof length dist( ν ( π ji , bl ji ) , ne j − i ) = 2 t − ν ( π ji , bl ji ) and nw j − i . Similarly, we add apath of length dist( ν ( ρ ji , tr ji ) , sw ji ) = 2 t − ν ( ρ ji , tr ji ) and se ji , and a path of lengthdist( ν ( ρ ji , bl ji ) , ne j − i ) = t between ν ( ρ ji , bl ji ) and nw j − i . We added these four paths so thatno forced vertex resolves any critical pair in the propagation gadgets P j − ,ji and P j,j +1 i .Finally we add an edge between ν ( g j , nw ji ) and ν ( c j , bc ji ) whenever V ji have exactly threecyan vertices. We do that to resolve the pair { ν ( c j , tc ji ) , ν ( c j , bc ji ) } , and more generallyevery pair { x, y } ∈ P ( c j , tc ji ) × P ( c j , bc ji ) such that dist( c j , x ) = dist( c j , y ). This finishesthe construction of the instance ( G , k := 15 km + 14 m ) of Metric Dimension . The two next lemmas will be crucial in Section 5.2.1. The first lemma shows how the forcingset gadget simulates the action of former set X . (cid:73) Lemma 6.
For every i ∈ [ k ] and j ∈ [ m ] , ∀ v ∈ V ji , v resolves both pairs { p ji , q ji } and { r ji , s ji } , ∀ v / ∈ V ji , v resolves at most one pair of { p ji , q ji } and { r ji , s ji } , ∀ v / ∈ V ji ∪ P ( v ji, , p ji ) ∪ P ( v ji,t , r ji ) ∪ { q ji , s ji } , v does not resolve { p ji , q ji } nor { r ji , s ji } . Proof.
Let Y := { tl ji , tr ji , bl ji , br ji } ∪ ( X j ∩ N ( V ji )) ∪ ( N ( { π ji , ρ ji } ) \ { p ji , q ji , r ji , s ji } ), and recallthat X j ∩ N ( V ji ) is the set of cyan vertices neighbors of V ji (if they exist). Let us assumethat these cyan vertices exist (otherwise the proof is just simpler). In particular, there areat least two cyan neighbors tc ji , bc ji ∈ X j ∩ N ( V ji ). Let X be the connected componentof G − Y containing { π ji , ρ ji } . For every vertex u ∈ { tl ji , tr ji , bl ji , br ji , tc ji , bc ji } , by the waywe chose the length of P ( u, π ji ) (resp. P ( u, ρ ji )), there is a shortest path from u to p ji (resp. r ji ) that goes through π ji (resp. ρ ji ). Thus dist( u, p ji ) = dist( u, π ji ) + 1 = dist( u, q ji ) anddist( u, r ji ) = dist( u, ρ ji ) + 1 = dist( u, s ji ).Let mc ji be the middle cyan vertex if it exists (the one which is not the top nor the bottomone). There is shortest path from mc ji to p ji (resp. r ji ) going via tc ji (resp. bc ji ) and then π ji (resp. ρ ji ). This is where the edges mc ji tc ji and mc ji bc ji are useful. Hence mc ji does not resolve { p ji , q ji } nor { r ji , s ji } , either. It is direct that no vertex of N ( { π ji , ρ ji } ) \ { p ji , q ji , r ji , s ji } resolves { p ji , q ji } nor { r ji , s ji } . Thus no vertex of Y resolves any of { p ji , q ji } and { r ji , s ji } . Therefore byLemma 2, no vertex outside X resolves any of { p ji , q ji } and { r ji , s ji } .We observe that X = V ji ∪ P ( v ji, , p ji ) ∪ P ( v ji,t , r ji ) ∪ { π ji , q ji , ρ ji , s ji } . Because of the pathfrom the top brown vertex to ρ ji , vertices of P ( v ji, , p ji ) \{ v ji, }∪{ q ji } , which do resolve { p ji , q ji } ,do not resolve { r ji , s ji } . Similarly because of the path from the bottom brown vertex to π ji ,vertices of P ( v ji,t , r ji ) \ { v ji,t } ∪ { s ji } , which do resolve { r ji , s ji } , do not resolve { p ji , q ji } . Finallyfor every u ∈ V ji , dist( u, q ji ) = dist( u, p ji ) + 2 and dist( u, r ji ) = dist( u, s ji ) + 2. Thereforevertices of V ji are the only ones resolving both { p ji , q ji } and { r ji , s ji } , while no vertex of G − X resolves any of these pairs. (cid:74) C V I T 2 0 1 6
We denote by f ( v ) the forced vertex attached to a vertex v . For Section 5.2.1, we alsoneed the following lemma, which states that the forced vertices do not resolve critical pairs. (cid:73) Lemma 7.
No forced vertex resolves a pair of P . Proof.
We first show that no critical pair in some P j,j +1 i is resolved by a forced vertex. Weuse a similar plan as for the proof of Lemma 3. Let Y := { nw j − i , se j − i , nw j +1 i , se j +1 i } ∪ C e ,where C e comprises { c j , g j } if e j has an endpoint in V i and { c j +1 , g j +1 } if e j +1 has anendpoint in V i . Let X be the connected component of G − Y containing P j,j +1 i . Note thatthe distances between the vertices of Y and the critical pairs in P j,j +1 i are the same between G and G . Hence as we showed in Lemma 3, no vertex of Y resolves a critical pair in P j,j +1 i .Thus by Lemma 2 no vertex outside X resolves a critical pair in P j,j +1 i .We now check that no forced vertex in X resolves a critical pair in P j,j +1 i . We show thatevery forced vertex in X has a shortest path to { nw ji , ne ji } ending in nw ji , and a shortest pathto { sw ji , se ji } ending in se ji . It is clear for f (nw ji ) and for f (se ji ), as well as for all the forcedvertices attached to neighbors of g j (in case e j has an endpoint in V i ). Indeed recall thatthe length of P ( g j , nw ji ) (resp. P ( g j , se ji )) is four less than the distance to nw ji (resp. sw ji )ignoring the path P ( g j , nw ji ) (resp. P ( g j , se ji )). So the shortest paths from the latter forcedvertices go to g j and then to nw ji (resp. se ji ). Similarly in case e j +1 has an endpoint in V i ,the shortest paths from the forced vertices attached to the neighbors of c j +1 to { nw ji , ne ji } (resp. { sw ji , se ji } ) go to g j +1 , then to nw j +1 i and nw ji (resp. then to se j +1 i and se ji ).Note that all the forced vertices attached to neighbors of π ji and ρ ji (resp. π j +1 i and ρ j +1 i )have a shortest path to { nw ji , ne ji } ending in nw ji (resp. to { sw ji , se ji } ending in se ji ). Finallydue to the paths P ( ν ( π ji , tr ji ) , se ji ) and P ( ν ( ρ ji , tr ji ) , se ji ), all the forced vertices attached toneighbors of π ji and ρ ji have a shortest path to { sw ji , se ji } ending in se ji . And due to thepaths P ( ν ( π j +1 i , bl j +1 i ) , nw ji ) and P ( ν ( ρ j +1 i , bl j +1 i ) , nw ji ), all the forced vertices attached toneighbors of π j +1 i and ρ j +1 i have a shortest path to { nw ji , ne ji } ending in nw ji .We now show that no critical pair { c j , c j } is resolved by a forced vertex. We set Y := { tl ji , tr ji , bl ji , br ji , tl ji , tr ji , bl ji , br ji , nw ji , se ji , nw ji , se ji , π ji , ρ ji , π ji , ρ ji } , with e j ∈ E ( V i , V i ), and X be the connected component of G − Y containing g j . We showed in Lemma 3, andit remains true in G , that no vertex of Y \ { π ji , ρ ji , π ji , ρ ji } resolves { c j , c j } . We observethat π ji and ρ ji have shortest paths to c j going through g j (via a vertex of { tl ji , tr ji , bl ji , br ji } ).Similarly π ji and ρ ji have shortest paths to c j going through g j . Therefore no vertex of { π ji , ρ ji , π ji , ρ ji } resolves the pair { c j , c j } . Hence by Lemma 2, no vertex outside X resolves { c j , c j } . The only forced vertices in X are attached to neighbors of g j , thus they do notresolve { c j , c j } . (cid:74) ⇒ k -MRS-instance has a solution. Let S be a resolving set for the Metric Dimension -instance. We show that S := S ∩ S i ∈ [ k ] ,j ∈ [ m ] V ji is a solution for k -Multicolored Resolving Set . The set S \ S is made of14 km + 14 m forced vertices, none of which is in some V ji ∪ P ( v ji, , p ji ) ∪ { q ji } ∪ P ( v ji,t , r ji ) ∪ { s ji } .Thus by Lemma 6, S \ S does not resolve any pair { p ji , q ji } or { r ji , s ji } . Now S is a set of k − (14 km + 14 m ) = km vertices resolving all the 2 km pairs { p ji , q ji } and { r ji , s ji } . Againby Lemma 6, this is only possible if | S ∩ V ji | = 1. Thus S is a legal set of size k = km . Letus now check that S resolves every pair of P in the graph G .By Lemma 7, S \ S does not resolve any pair of P in the graph G . Thus S resolves allthe pairs of P in G . Since the distances between V ji and the critical pairs in the edge andpropagation gadgets V ji is attached to are the same in G and in G , S also resolves everypair of P in G . Thus S is a solution for the k -MRS -instance. . Bonnet, N. Purohit 23:17 k -MRS-instance has a solution ⇒ MD-instance has a solution.
For every i ∈ [ k ], j ∈ [ m ], let F ji := [ u ∈{ nw ji , se ji }∪ N ( { π ji ,ρ ji } ) \{ p ji ,q ji ,r ji ,s ji } { f ( u ) } , and F j := [ u ∈ N ( g j ) \{ c j ,c j } { f ( u ) } . Let S be a solution for k -Multicolored Resolving Set . Thus | S | = km . Let F := S i ∈ [ k ] ,j ∈ [ m ] F ji ∪ S j ∈ [ m ] F j . We show that S := S ∪ F is a solution of Metric Dimension .First we observe that | S | = km + 14 km + 14 m = k . Since the distances between the sets V ji and the critical pairs (of P ) are the same in G and in G , the pairs of P are resolvedby S . In what follows, we show that F resolves all the other pairs. For every i ∈ [ k ], j ∈ [ m ],we define the subset of vertices:Π ji := [ u ∈{ tr ji , tl ji , br ji , bl ji , bb ji , tc ji } P ( π ji , u ) ∪ P ( v ji, , p ji ) ∪ { q ji } ,R ji := [ u ∈{ tr ji , tl ji , br ji , bl ji , tb ji , bc ji } P ( ρ ji , u ) ∪ P ( v ji,t , r ji ) ∪ { s ji } , and G j := [ u ∈{ tr ji , tl ji , br ji , bl ji , tl ji , tr ji , bl ji , br ji , nw ji , se ji , nw ji , se ji } P ( g j , u ) ∪ E ji ∪ E ji ∪ { c j } . Informally Π ji ( R ji , G j , respectively) consists of the vertices on the paths incident to π ji ( ρ ji , g j , respectively). Our objective is the following result. (cid:73) Lemma 8.
Every vertex in G is distinguished by S . We start with the forced vertices and their false twin. We denote by f ( v ) the false twinof the forced vertex f ( v ). (cid:73) Lemma 9.
All the vertices f ( v ) and f ( v ) are distinguished by F . Proof.
Any vertex f ( v ) is distinguished by being the only vertex at distance 0 of itself f ( v ) ∈ F . Since f ( v ) has only two neighbors f ( v ) and v , it also resolves every pair { f ( v ) , w } where w is not v . The pair { f ( v ) , v } is resolved by any vertex f ∈ F \ { f ( v ) } .Indeed dist( f, f ( v )) = dist( f, v ) + 1. Thus f ( v ) is distinguished. (cid:74) In general, to show that all the vertices in a set X are distinguished, we proceed in twosteps. First we show that every internal pair of X is resolved. Then, we prove that everypair of X × X is also resolved. Let us recall that X is the complement of x , here V ( G ) \ X .For instance, the two following lemmas show that every vertex of Π ji is distinguished by S . (cid:73) Lemma 10.
Every pair of distinct vertices x, y ∈ Π ji is resolved by S . Proof.
Let U ji be the set { tl ji , tr ji , bl ji , br ji , tc ji , bb ji } . We first consider two vertices x = y ∈ P ( π ji , u ), for some u ∈ U ji . As dist G ( π ji , u ) is equal to the length of P ( π ji , u ), it holdsthat dist G ( π ji , x ) = dist P ( π ji ,u ) ( π ji , x ) = dist P ( π ji ,u ) ( π ji , y ) = dist G ( π ji , y ). Without loss ofgenerality, we assume that dist( π ji , x ) < dist( π ji , y ). If x = π ji , then x and y have distinctdistances to ν ( π ji , u ). Hence dist( f ( ν ( π ji , u )) , x ) = dist( f ( ν ( π ji , u )) , y ) and S resolves { x, y } .Now if x = π ji , then f ( ν ( π ji , u )) resolves { x, y } for any u ∈ U ji \ { u } . C V I T 2 0 1 6
Secondly we consider x ∈ P ( π ji , u ) and y ∈ P ( π ji , u ), for some u = u ∈ U ji . Ifdist( π ji , x ) = 2 + dist( π ji , y ), then f ( ν ( π ji , x )) resolves { x, y } . Indeed dist( f ( ν ( π ji , x )) , x ) =dist( π ji , x ) = 2 + dist( π ji , y ) = dist( f ( ν ( π ji , x )) , y ). Else if dist( π ji , x ) = 2 + dist( π ji , y ), then f ( ν ( π ji , y )) resolves { x, y } (since dist( π ji , y ) = 2 + dist( π ji , x )).Two distinct vertices on P ( v ji, , p ji ) are resolved by, say, f ( ν ( π ji , br ji )) ∈ F . A vertex of P ( v ji, , p ji ) and a vertex of P ( π ji , u ), for some u ∈ U ji , are resolved by either f ( ν ( π ji , u )) or f ( ν ( π ji , u )) for a u ∈ U ji \ { u } . Finally q ji and a vertex in P ( v ji, , p ji ) \ { p ji } are resolved by,say, f ( ν ( π ji , br ji )), whereas q ji and a vertex in P ( p ji , u ) is resolved by either f ( ν ( π ji , u )) or f ( ν ( π ji , u )) for a u ∈ U ji \ { u } . Therefore every pair of distinct vertices in Π ji is resolvedby F , except { p ji , q ji } which is resolved by S . (cid:74)(cid:73) Lemma 11.
Every pair { x, y } ∈ Π ji × Π ji is resolved by F . Proof.
Again let U ji be the set { tl ji , tr ji , bl ji , br ji , tc ji , bb ji } . We first assume x is in P ( π ji , u )for some u ∈ U ji \ { tr ji , bl ji } . Let y be a vertex of Π ji such that dist( f ( ν ( π ji , u )) , x ) =dist( f ( ν ( π ji , u )) , y ), otherwise f ( ν ( π ji , u )) already resolves { x, y } . Every shortest path from f ( ν ( π ji , u )) to y go through π ji . One can observe that there is a u ∈ U ji \ { u } such that f ( ν ( π ji , u )) has a shortest path also going through π ji . Hence f ( ν ( π ji , u )) has the samedistance to y (as f ( ν ( π ji , u ))) but a larger distance to x . Hence f ( ν ( π ji , u )) resolves { x, y } .We now consider an x ∈ P ( π ji , u ) for some u ∈ { tr ji , bl ji } . Again let y be a vertex of Π ji suchthat dist( f ( ν ( π ji , u )) , x ) = dist( f ( ν ( π ji , u )) , y ). If all the shortest paths of f ( ν ( π ji , u )) to y goes through π ji , we conclude as in the previous paragraph. So they go through P ( ν ( π ji , u ) , se ji )(if u = tr ji ) or P ( ν ( π ji , u ) , nw j − i ) (if u = bl ji ). Since dist( f ( ν ( π ji , u )) , x ) (cid:54) t −
1, it alsoholds that dist( f ( ν ( π ji , u )) , y ) (cid:54) t −
1. The path P ( ν ( π ji , tr ji ) , se ji ) has length t and the path P ( ν ( π ji , bl ji ) , nw j − i ) has length 2 t −
1. Therefore one of f (se ji ), f (se j − i ), f (nw ji ), f (nw j − i )resolves { x, y } .We now assume x is in P ( v ji, , p ji ) ∪ { q ji } and y ∈ Π ji . Then f ( ν ( π ji , br ji )) resolves { x, y } if y is not in the path P ( ν ( π ji , tr ji ) , se ji ) or P ( ν ( π ji , u ) , nw j − i ). Otherwise at least one of f ( ν ( π ji , br ji )), f ( ν ( π ji , tr ji )), f ( ν ( π ji , bl ji )) resolves { x, y } . In conclusion, every pair of vertices { x, y } ∈ Π ji × Π ji is resolved by F . (cid:74) Lemmas 10 and 11 prove that every vertex in Π ji is distinguished by S . Using the samearguments, we get symmetrically that every vertex of R ji is distinguished by S . (cid:73) Lemma 12.
All the vertices in the paths P ( ν ( π ji , tr ji ) , se ji ) , P ( ν ( ρ ji , tr ji ) , se ji ) , P ( ν ( π ji , bl ji ) , nw j − i ) , P ( ν ( ρ ji , bl ji ) , nw j − i ) are distinguished by F . Proof.
Any vertex x ∈ P ( ν ( π ji , tr ji ) , se ji ) is uniquely determined by its distances to f (se ji ), f (se j − i ), and ν ( π ji , tr ji ). Any vertex x ∈ P ( ν ( π ji , bl ji ) , nw j − i ) is uniquely determined by itsdistances to f (nw ji ), f (nw j − i ), and ν ( π ji , bl ji ). The two other cases are symmetric. (cid:74) So far we showed that the vertices added in the forced set and forced vertex gadgetsare all distinguished. We now focus on the vertices in propagation gadgets. Let ∆ i := A ji ∪ { nw ji , ne ji , sw ji , se ji } . (cid:73) Lemma 13.
Every pair of distinct vertices x, y ∈ ∆ ji is resolved by S . Proof.
Since the distances between vertices of V ji and vertices of ∆ ji are the same between G and G , S resolves all the critical pairs { a ji,γ , α ji,γ } . Thus we turn our attention to thepairs which are not critical pairs. Since dist(nw ji , a ji,γ ) = γ and dist(nw ji , α ji,γ ) = γ , everypair { a ji,γ , a ji,γ } , { a ji,γ , α ji,γ } , or { α ji,γ , α ji,γ } , with γ = γ is resolved by f (nw ji ). . Bonnet, N. Purohit 23:19 Gate nw ji (resp. se ji ) and any other vertex in ∆ ji is resolved by f (nw ji ) (resp. f (se ji )).Gate ne ji (resp. sw ji ) is resolved from any vertex of ∆ ji \ { a ji, , α ji, } (resp. ∆ ji \ { a ji,t , α ji,t } ) by f (nw ji ) (resp. f (se ji )). Finally, ne ji (resp. sw ji ) and a vertex of { a ji, , α ji, } (resp. { a ji,t , α ji,t } )is resolved by f (se ji ) (resp. f (nw ji )). (cid:74) Now when we check that a pair made of a vertex in ∆ ji and a vertex outside ∆ ji is resolved,we can further assume that the second vertex is not in some Π ji ∪ R ji since we already showedthat these vertices were distinguished. (cid:73) Lemma 14.
Every pair { x, y } ∈ ∆ ji × ∆ ji is resolved by S . Proof.
We may assume that y is not a vertex that was previously shown distinguished. Thus y is not in some Π ji ∪ R ji nor in a path of Lemma 12. Then we claim that the pair { x, y } is resolved by at least one of f (se ji ), f (se j − i ), f (se j +1 i ), f (nw ji ). Indeed assume that f (se ji )does not resolve { x, y } , and consider a shortest path from f (se ji ) to y . Either this shortestpath goes through se j − i (resp. se j +1 i ), and in that case f (se j − i ) (resp. f (se j +1 i )) resolves { x, y } . Either it takes the path to g j (if e j has an endpoint in V i ) or to tl j +1 i , and then f (nw ji ) resolves { x, y } . Or it takes a path to V ji , and then f (se j − i ) resolves { x, y } . (cid:74) Lemmas 13 and 14 show that that every vertex in ∆ ji is distinguished by S . The commonneighbor of se j − i and tl ji is distinguished by { f (se j − i ) , f ( ν ( π ji , tl ji )) } . We are now left withshowing that the vertices in the edge gadgets, in the sets V ji , and in the paths incident tothe edge gadgets, are distinguished. (cid:73) Lemma 15.
Every pair of distinct vertices x, y ∈ G j is resolved by S . Proof.
Let v i,γ and v i ,γ be the two endpoints of e j , and U ji := { tl ji , tr ji , bl ji , br ji , tl ji , tr ji , bl ji , bl ji , nw ji , se ji , nw ji , se ji , v ji,γ , v ji ,γ } . Every pair in S u ∈ U ji P ( g j , u ) is resolved. Indeed, similarlyto Lemma 10, two distinct vertices x, y on a path P ( g j , u ) ( u ∈ U ji ) are resolved by f ( ν ( g j , u )),while two vertices on distinct paths P ( g j , u ) and P ( g j , u ) ( u = u ∈ U ji ) are resolved by atleast one of f ( ν ( g j , u )) and f ( ν ( g j , u )).We now show that any pair in Γ ji := E ji ∪ E ji \ { P ( g j , v ji,γ ) , P ( g j , v ji,γ ) } is resolved. Twodistinct vertices x, y ∈ Γ ji are resolved by, say, f ( ν ( g j , se ji )) if they are on the same path, ormore generally if they have different distances to c j . Thus let us assume that x and y areat the same distance from c j . If x ∈ E ji and y ∈ E ji (or vice versa) then the pair { x, y } isresolved by the vertex in S ∩ V ji or the vertex in S ∩ V ji . If x = y ∈ E ji (resp. ∈ E ji ), then { x, y } is resolved by f ( ν ( g j , nw ji )) (resp. f ( ν ( g j , nw ji ))). This is the reason why we addedan edge between ν ( g j , nw ji ) and ν ( c j , bc ji ) (recall Section 5.1.3).We now consider pairs { x, y } of S u ∈ U ji P ( g j , u ) × Γ ji . Any of these pairs are resolved byat least one of f ( ν ( g j , u )), f ( ν ( g j , u )), f ( ν ( g j , nw ji )), f ( ν ( g j , nw ji )), where x is on the path P ( c j , u ) and u is any vertex in U ji \ { u, nw ji , nw ji } . Finally c j is distinguished from all theother vertices in G but c j by the forced vertices attached to the neighbors of g j .Thus every pair { x, y } in G j is resolved by F , except { c j , c j } which is resolved by S . (cid:74)(cid:73) Lemma 16.
Every pair { x, y } ∈ G j × G j is resolved by F . Proof.
Consider an arbitrary pair { x, y } ∈ G j × G j . We can assume that x is not c j ,and that y is in one different G j or in one V j i (since we already showed that the othervertices are distinguished). Again let v i,γ and v i ,γ be the two endpoints of e j , and U ji := { tl ji , tr ji , bl ji , br ji , tl ji , tr ji , bl ji , bl ji , nw ji , se ji , nw ji , se ji , v ji,γ , v ji ,γ } . If x is on a path P ( g j , u ),then at least one of f ( ν ( g j , u )) and f ( ν ( g j , u )), with u being any vertex in U ji \ { u } , resolves C V I T 2 0 1 6 { x, y } . If instead x is on a path P ( c j , u ) with u ∈ { v ji,γ − , v ji,γ +1 , v ji ,γ − , v ji ,γ +1 } , thenat least one of f ( ν ( g j , nw ji )), f ( ν ( g j , nw ji )), f ( ν ( g j , u )), with u being any vertex in U ji ,resolves { x, y } . (cid:74) Lemmas 15 and 16 show that every vertex in G j is distinguished by S . We finally showthat the vertices in V ji are distinguished. A pair of distinct vertices x, y ∈ V ji is resolvedby f (nw ji ). We thus consider a pair { x, y } ∈ V ji × V ji . We can further assume that y is insome V j i , since all the other vertices have already been shown distinguished. Then { x, y } isresolved by at least one of f (nw ji ), f (nw j i ), the vertex in S ∩ V ji , and the vertex in S ∩ V j i .This finishes the proof of Lemma 8. Thus S is a solution of the Metric Dimension -instance.The reduction is correct and it takes polynomial-time in | V ( G ) | to compute G . Themaximum degree of G is 16. It is the degree of the vertices g j (nw ji and se ji have degreeat most 11, π ji and ρ ji have degree 8, and the other vertices have degree at most 5). Thelast element to establish Theorem 1 is to show that pw( G ) is in O ( k ). Then solving Metric Dimension on constant-degree graphs in time f (pw) n o (pw) could be used to solve k -Multicolored Independent Set in time f ( k ) n o ( k ) , disproving the ETH. G has pathwidth O ( k ) We use the pathwidth characterization of Kirousis and Papadimitriou [19] mentioned in thepreliminaries, and give a strategy with O ( k ) searchers cleaning all the edges of G . A basicand useful fact is that the searching number of a path is two. (cid:73) Lemma 17.
Two searchers are enough to clean a path u u . . . u n . Proof.
We place two searchers at u and u . This cleans the edge u u . Then we movethe searcher in u to u . This cleans u u (while u u remains clean). Then we move thesearcher in u to u , and so on. (cid:74)(cid:73) Lemma 18. pw ( G ) (cid:54) k + 83 . Proof.
For every j ∈ [ m ], let S j := N [ g j ] ∪ X j ∪ S i ∈ [ k ] N [ { v ji, , v ji,t , π ji , ρ ji } ] ∪{ nw ji , ne ji , sw ji , se ji } . We notice that | S j | (cid:54)
17 + 6 + 30 k + 4 = 30 k + 27. Another important observation isthat S ∪ S j disconnects the first j columns of G from the rest of G . Finally the connectedcomponents G − ( S j ∪ S j +1 ) that are not the main component (i.e., containing more thanhalf of the graph if m (cid:62)
4) are all paths.We now suggest the following cleaning strategy with at most 90 k + 83 searchers. We placeone searcher at each vertex of S ∪ S ∪ S . This requires 90 k + 81 searchers. By Lemma 17,with two additional searchers we clean all the connected components of G − ( S ∪ S ∪ S )that are paths. We then move all the searchers from S to S , and clean all the connectedcomponents of G − ( S ∪ S ∪ S ) that are paths. Since S ∪ S is a separator, the edgesthat were cleaned during the first phase are not recontaminated when we move from S to S . We then move the searchers of S to S , and so on. Eventually the searchers reach S ∪ S m − ∪ S m , and the last contaminated edges are cleaned. (cid:74) The main remaining open question is whether or not
Metric Dimension is polytime solvableon graphs with constant treewidth. In the parameterized complexity language, now we knowthat MD /tw is W[1]-hard, is it in XP or paraNP-hard? We believe that the tools and ideas . Bonnet, N. Purohit 23:21 developed in this paper could help answering this question negatively. The FPT algorithm ofBelmonte et al. [2] also implies that Metric Dimension is FPT with respect to tl + k were k is the size of the resolving set, due to the bound ∆ (cid:54) k + k − Metric Dimension with respect to tw + k ? We conjecture thatthis problem is W[1]-hard as well, and once again, treewidth will contrast with tree-length.It appears that bounded connected treewidth or tree-length is significantly more helpfulthan the mere bounded treewidth when it comes to solving MD . We wish to ask for theparameterized complexity of Metric Dimension with respect to ctw only (on graphs witharbitrarily large degree). Finally, it would be interesting to determine if planarity cansometimes help to compute a metric basis. Therefore we also ask all the above questions inplanar graphs.
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Symbol/term Definition/action { a ji,γ , α ji,γ } critical pair of the propagation gadget P j,j +1 i A ji set of vertices S γ ∈ [ t ] { a ji,γ , α ji,γ } bb ji bottom brown vertex, ν ( v ji,t , r ji )bc ji bottom cyan vertex (smallest index γ )bl ji neighbor of v ji,t in P j − ,ji blue vertex one of the four neighbors of V ji in the propagation gadgetsbr ji neighbor of v ji,t in P j,j +1 i brown vertex vertices ν ( v ji, , p ji ) and ν ( v ji,t , r ji ) { c j , c j } critical pair of the edge gadget G ( e j )cyan vertex neighbor of V ji in the paths to G ( e j ) E ji vertices in the paths from V ji to G ( e j ) e ji,γ alternative labeling of the cyan vertices, neighbor of v ji,γ F set of all forced vertices, S i ∈ [ k ] ,j ∈ [ m ] F ji ∪ S j ∈ [ m ] F j F ji set of forced vertices attached to neighbors of { π ji , ρ ji , nw ji , se ji } F j set of forced vertices attached to neighbors of g j f ( v ) forced vertex attached to a vertex vf ( v ) false twin of f ( v ) G ( e j ) edge gadget on { g j , c j , c j } between V ji and V ji , where e j ∈ E ( V i , V i )mc ji middle cyan vertex (not top nor bottom)ne ji north-east gate of P j,j +1 i nw ji north-west gate of P j,j +1 i ne ji , sw ji resolve the critical pairs of P j,j +1 i nw ji , se ji do not resolve the critical pairs of P j,j +1 i ν ( u, v ) neighbor of u in the path P ( u, v ) P list of critical pairs { p ji , q ji } pair only resolved by vertices in V ji ∪ P ( v ji, , p ji ) ∪ { q ji } π ji gate of { p ji , q ji } , linked by paths to most neighbors of V ji P j,j +1 i propagation gadget between V ji and V j +1 i P ( u, v ) path added in the construction expressly between u and v { r ji , s ji } pair only resolved by vertices in V ji ∪ P ( v ji,t , r ji ) ∪ { s ji } ρ ji gate of { r ji , s ji } , linked by paths to most neighbors of V ji se ji south-east gate of P j,j +1 i sw ji south-west gate of P j,j +1 i t size of each V i tb ji top brown vertex, ν ( v ji, , p ji )tc ji top cyan vertex (largest index γ )tl ji neighbor of v ji, in P j − ,ji tr ji neighbor of v ji, in P j,j +1 i V i partite set of GV ji “copy of V i ”, stringed by a path, in G and G v ji,γ vertex of V ji representing v i,γ ∈ V ( G ) W j endpoints in V ji ∪ V ji of paths from V ji ∪ V ji to G ( e j ) X set containing all the sets V ji for i ∈ [ k ] and j ∈ [ m ] X j neighbors of W j on the paths to G ( e j ) (cyan vertices) Table 1