4 vs 7 sparse undirected unweighted Diameter is SETH-hard at time n^{4/3}
44 vs 7 sparse undirected unweighted Diameter isSETH-hard at time n / Univ Lyon, CNRS, ENS de Lyon, Université Claude Bernard Lyon 1, LIP UMR5668, [email protected]
Abstract
We show, assuming the Strong Exponential Time Hypothesis, that for every ε >
0, approximatingundirected unweighted
Diameter on n -vertex m -edge graphs within ratio 7 / − ε requires m / − o (1) time, even when m = ˜ O ( n ). This is the first result that conditionally rules out a near-linear time5 / Diameter . Theory of computation → Graph algorithms analysis
Keywords and phrases
Diameter, inapproximability, SETH lower bounds, k-Orthogonal Vectors
The diameter of a graph is the length of a longest shortest path between two of its vertices.We write
Diameter for the algorithmic task of computing the diameter of an input graph.Throughout the paper, n implicitly denotes the number of vertices of a graph, and m ,its number of edges. We will often prefix Diameter with undirected/directed to indicatewhether or not edges may be oriented , and unweighted/weighted to indicate whether or notnon-negative edge weights are allowed.A fairly recent and active line of work aims to determine the best runtime for an algorithmapproximating Diameter within a given ratio. First, there is an exact algorithm running intime ˜ O ( mn ), which computes n shortest-path trees from every vertex of the graph. Secondly,there is a 2-approximation running in time ˜ O ( m ), which computes a shortest-path tree froman arbitrary vertex and outputs the largest distance found. There are an ˜ O ( m / ) time3 / Diameter [1, 15, 6], and for every non-negativeinteger k , an ˜ O ( mn k +1 ) time (2 − − k )-approximation for undirected weighted Diameter [4].We refer the interested reader to the survey of Rubinstein and Vassilevska Williams [16].We will now focus on sparse graphs, for which m = ˜ O ( n ). This is because the current paperdeals with conditional lower bounds on approximating Diameter , and all such results evenwork with that restriction. Observe that, on sparse graphs, the first result of the previousparagraph is a near-quadratic 1-approximation, while the second result is a near-linear2-approximation. One can represent these ratio-runtime trade-offs in the two-dimensionalplane. The ultimate goal of fine-grained complexity, in that particular context, is to obtain acomplete curve of algorithms linking these two extreme points, matched by tight conditionallower bounds. We now present one way of deriving conditional lower bounds for polytimeproblems. In directed
Diameter , we are to compute the length of a longest shortest path taken from any vertexto any vertex. where ˜ O ( · ) suppresses the polylogarithmic factors with an extra additive factor depending on the weights a r X i v : . [ c s . D S ] J a n n / Lower bounds based on the Strong Exponential Time Hypothesis
The Strong Exponential Time Hypothesis (SETH, for short) asserts that for every ε > k such that k -SAT cannot be solved in time (2 − ε ) n on n -variableinstances [11]. At first glance, this assumption should only be useful to rule out somespecific running time for NP-hard problems which, like the satisfiability problem, seems torequire superpolynomial time. Such conditional lower bounds to classical [7] or parameterizedalgorithms [8] are overviewed in a survey [14] on the consequences of the SETH (as well asthe weaker assumption ETH) on solving computationally hard problems.Interestingly, using the SETH to rule out a given running time for a polynomial-timesolvable problem took more time. In a survey of fine-grained complexity [18], VassilevskaWilliams dates the first reduction (albeit used positively) from SAT to a problem in Pback to 2005 [17]. We will see that this reduction to , whereone wants to find two orthogonal 0 , Diameter . As it turns out, the first SETH-based lower bound fora polytime graph problem occurred almost a decade later, on the very unweighted undirected
Diameter [15].There might have been a psychological barrier in reducing a “hard” problem to an “easy”one, in order to derive a conditional lower bound. However this makes perfect sense. Let usgive an apropos example. Suppose (as it is actually the case) that one can create in time O ( n ) a list of n n = O (2 N/ ), from an N -variable SAT formula, such thatthere is pair of orthogonal vectors in the list if and only if the formula is satisfiable. Nowa truly subquadratic algorithm, that is in time n − ε for some ε >
0, for would enable to solve
SAT in time O (2 (1 − ε/ N ) = O ((2 − δ ) N ) for some δ > is SETH-hard attime n , and more generally a problem Π is SETH-hard at time T if it requires time T − o (1) under the SETH. SETH lower bounds for Diameter
There is a handful of SETH-hardness results on approximating
Diameter [15, 2, 3, 9, 12, 13].Unless the SETH fails, any 3 / − ε -approximation for sparse undirected unweighted Diameter ,with ε >
0, requires time n − o (1) [15] (this is the above-mentioned seminal result to the fine-grained complexity within P), whereas any 5 / − ε -approximation requires time n / − o (1) [12](an early version of [13]). Since a 5 / Diameter running in near-linear timewas consistent with the then knowledge (up until mid-August 2020, even in weighted directedgraphs) Rubinstein and Vassilevska Williams [16] and Li [12] ask for such an algorithm orsome lower bounds with a ratio closer to 2.In the last few months, there were several developments on directed graphs. The authorshowed that, under the SETH, 7 / − ε -approximating sparse directed weighted Diameter requires time n / − o (1) [3]. Then Wein and Dalirrooyfard [9], and independently, Li [13] (anupdated version of [12]) both show that not only this result holds on directed unweighted graphs but they generalize it in the following way: Unless the SETH fails, for every ε > k (cid:62) k − k − ε -approximating directed unweighted Diameter requirestime n kk − − o (1) .Despite these advances, a near-linear time 5 / undirected Dia-meter may still have existed. In this paper, we rule out this possibility by showing thefollowing (see Figure 1 for a visual summary of what is now known on approximatingundirected
Diameter ). . Bonnet 3 (cid:73) Theorem 1.
Unless the SETH fails, for any ε > , / − ε -approximating Diameter onundirected unweighted n -vertex ˜ O ( n ) -edge graphs requires n / − o (1) time. In particular we resolve [16, Open Question 2.2.], on the existence of a near-linear time5 / undirected Diameter , by the negative.1 k +1 k
32 53 74 − k − k − k approximation factor r un t i m ee x p o n e n t [15, RV13] [13, Li20] T h e o r e m [15, RV13][2, CLRSTV18][4, CGR16][13] Figure 1
Approximability of sparse undirected unweighted
Diameter . Blue areas are feasible,as witnessed by algorithms at bottom-left corners (blue dots). The red regions are SETH-hard, aswitnessed by reductions at top-right corners (red dots). Dotted cyan areas are not SETH-hard,unless the NSETH fails. The current landscape for the sparse undirected weighted
Diameter isthe same, except the middle red region is due to Backurs et al. [2] instead of [13]. The axis-parallelblack curve represents the tractability frontier as foreseen by Conjecture 2.
In light of the recent results (see in particular the paragraph on barriers to SETH-hardness),it is reasonable to conjecture that the four variants of sparse
Diameter (undirected/directedunweighted/weighted) are equally approximable. More precisely, we venture the followingoptimistic prediction. (cid:73)
Conjecture 2.
Sparse (un)directed (un)weighted
Diameter is − k -approximable in time ˜ O ( n k +1 k ) for every k ∈ N + ∪ {∞} . Unless the SETH fails, approximating sparse (un)directed(un)weighted Diameter within ratio better than − k +1 requires time n k +1 k − o (1) for every k ∈ N + . Conjecture 2 is naturally equivalent to obtaining the algorithms for the directed weighted
Diameter and the SETH-hardness for the undirected unweighted
Diameter . Settling theconjecture would give a complete landscape of the approximability of
Diameter , where ifone represents the results in the two-dimensional space of approximation factor vs runtimeexponent, the feasible and infeasible regions are separated by a rectilinear curve with infinitely n / many corners (the black curve drawn in Figure 1). In that respect, our contribution is to givethe third lower bound on the curve (i.e., North-East corner) after Roditty and VassilevskaWilliams gave the first [15], and Li, the second [12]. Hopefully our new ideas (together withthe recent constructions in the directed case of Wein and Dalirrooyfard [9], and Li [13])will also help in generalizing the lower bound predicted by Conjecture 2 to every positiveinteger k . Barriers to SETH lower bounds
Conjecture 2 is partly prompted by intriguing results due to Li [13]. To state them, we needto recall the definition of a strengthening of SETH introduced by Carmosino et al. [5]. Itis called NSETH for Nondeterministic SETH. NSETH asserts that for every ε >
0, thereis an integer k such that the k -Taut problem cannot be solved in non-deterministic time(2 − ε ) n , where k -Taut asks, given a k -DNF formula whether every truth assignment satisfiesit (in other words, if it is a tautology). Li shows, for all four variants of Diameter but thedirected weighted one, that no point positioned strictly above the rectilinear black curveof Figure 1 can be shown SETH-hard, under the NSETH (and, if randomized reductions arepermitted, under a stronger assumption, called NUNSETH for Non-Uniform NSETH).Conjecture 2 is very optimistic since it predicts that every such point will be explainedby an algorithm. There are many alternatives to that event. For instance NSETH couldbe false , or the intractability region could extend further North via a non SETH-basedreduction, or via a deterministic SETH-based reduction in the directed weighted case. Besidesit would require significant progress in approximating the sparse directed Diameter , whencurrently no algorithm running in time n − ε achieves approximation factor better than 2.The second half of Conjecture 2 shown for every k (cid:62) k = 1 , , Techniques
Like every mentioned
Diameter lower bound (for more details, see the paragraphs on k -OV and Diameter in the surveys [18, 16]), we reduce from k -Orthogonal Vectors , whereone seeks, in a given set of N , ‘ , k vectors such that at every index,at least one of these k vectors has a 0 entry. Under the SETH, k -Orthogonal Vectors requires time N k − o (1) [17], even when ‘ is polylogarithmic in N .Here we will reduce from . We thus wish to build a graph on˜ O ( N ) vertices and edges with diameter 7 if there is an orthogonal quadruple (i.e., a solutionto the the instance), and diameter 4 otherwise. Following areduction to ST - Diameter by Backurs et al. [2] (arguably also following [15]) most of thereductions (as in [3, 9, 13]) feature layers L , L , . . . , L k − , L k , with only (forward) edgesbetween two consecutive L i . The vertices within the same layer share the same number of“vector attributes” and “index attributes”. The interplay between vector and index attributesin defining the vertices and edges is made so that if there are no k orthogonal vectors, thenthere are paths of “optimal” length k between every pair in L × L k , whereas if there isset X of k orthogonal vectors, a pair in L × L k jointly encoding X is suddenly very farapart (usually and ideally at distance 2 k − If we are totally honest, even the weaker SETH does not gather such a wide consensus, and is false ifquantum computation is allowed. where one seeks the length of a longest shortest path from a vertex of S to a vertex of T . Bonnet 5 Then the challenge is to make sure that, on NO-instances, the other pairs (not in L × L k )are at distance at most k , without destroying the previous property. The core of our reductionis similar to our previous construction for directed weighted Diameter [3]. However wesimplify and streamline it in the following way. As in the first construction of Li [12], wecollapse some layers into one. We will have L = L (= T ) and L = L (= C ), while L iscalled P . This makes the case analyses simpler (fewer kinds of pairs to consider).At this point, we face the same issue as in [3]: There are pairs in T × P that are toofar apart. On directed graphs, this can be fixed by adding parallel layers and appropriate“back” edges [3, 9] or simply “back” edges [13]. This is no longer an option. Instead we adda set I of vertices with only index attributes. These vertices link the right pairs of T × P with path of length 4 (we are back to the first variation on the theme [15]). To emphasizethat the situation is somewhat delicate, we observe that not all the pairs of T × P can be atdistance 4, since otherwise every pair in T × T is at distance at most 6. We set I at distance3 of T (by initially putting edges of weight 3). This permits to cliquify I without creating T T -paths of length at most 6. In turn, this puts every pair involving I at distance at most 4,as well as pairs of ( C ∪ P ) × P . Note that as long as d ( T, X ) + d ( T, Y ) (cid:62) k − all the pairs of X × Y at distance 4 (or k ), without creating undesired T T -paths oflength at most 6 (or 2 k − T and I . This involves some vertex splitstransforming T into T, T , T , and a simpler echo of the idea of having the clique I , with aclique I connecting appropriately the pairs in T × T . Organization
In Section 2, we recall graph-theoretic notations, and give the relevant background on the
Or-thogonal Vectors problem. In Section 3, we present a simpler reduction with edge weights.It thus achieves the statement of Theorem 1 for sparse undirected weighted
Diameter .In Section 4, we tune this reduction to get rid of the edge weights, and establish Theorem 1.
We use standard graph-theoretic notations. If G is a graph, V ( G ) denotes its vertex set, and E ( G ), its edge set. We denote the edge set between X ⊆ V ( G ) and Y ⊆ V ( G ) by E ( X, Y ).If S ⊆ V ( G ), G [ S ] denotes the subgraph of G induced by S . Weighted graphs have positiveedge weights. (Throughout the paper, we will only need edges of weight 1 and 3.) Weexclusively deal with undirected graphs (for which the distance function is symmetric). For u, v ∈ V ( G ), d G ( u, v ) denotes the distance between u and v in G , that is, the number of edgesin a shortest path between u and v . For every positive integer r and every vertex u ∈ V ( G ), N rG [ u ] denotes the set of vertices v such that d G ( u, v ) (cid:54) r . In unweighted graphs, the closedneighborhood of u , denoted N G [ u ], coincides with N G [ u ]. However in a weighted graph N G [ u ] would for instance not contain the neighbors of u via an edge of weight greater than 1.This subtlety will arise only once, and we will remind the reader in due time. For everypositive integer r and every vertex S ⊆ V ( G ), N rG [ S ] denotes the set of vertices v ∈ V ( G )such that d G ( u, v ) (cid:54) r for some u ∈ S . We observe that, in unweighted graphs, N rG [ S ]coincides with N G [ N G [ · · · N G [ N G [ S ]] · · · ]] where N G [ · ] is applied r times and N G [ S ] is theclosed neighborhood of S . We drop the subscript in the above notations, if the graph G isclear from the context.We denote by diam( G ) the diameter of G , that is, max u,v ∈ V ( G ) d ( u, v ). The Diameter problem asks, given a graph G , for the value of diam( G ). We call uv -path, a path going n / from vertex u to vertex v , and ST -path (with possibly S = T ), any path going from somevertex u ∈ S to some vertex v ∈ T .If ‘ is a positive integer, [ ‘ ] denotes the set { , , . . . , ‘ } . If v is a vector and i is a positiveinteger, then v [ i ] denotes the i -th coordinate of v . We use maj( a , . . . , a h ) to denote thevalue with the largest number of occurrences in the tuple ( a , . . . , a h ).For every fixed positive integer k , the k -Orthogonal Vectors ( k -OV for short) problemis as follows. It asks, given a set S of 0,1-vectors in { , } ‘ , if there are k vectors v , . . . , v k ∈ S such that for every i ∈ [ ‘ ], Π h ∈ [ k ] v h [ i ] = 0, or equivalently, v [ i ] = v [ i ] = · · · = v k [ i ] = 1does not hold. Williams [17] showed that, assuming the SETH, k -OV requires N k − o (1) timewith N := | S | . Furthermore, using the Sparsification Lemma [10], this lower bound holdseven when, say, ‘ = d log N e . Here we will leverage this lower bound for k = 4. This is, inthe context of the SETH-hardness of approximating Diameter , a usual opening step: Forexample, Roditty and Vassilevska Williams [15] uses this lower bound for k = 2, Li [12], for k = 3 and general k (cid:62)
3, the author [3], for k = 4, Wein and Dalirrooyfard [9], for general k (cid:62) k = 4. From any set S of N vectors in { , } ‘ , we build an undirected weighted graph G = ρ ( S )(with edge weights 1 and 3, only) with O ( N + N ‘ + ‘ ) vertices and O ( N ‘ + N ‘ + ‘ )edges such that if S admits an orthogonal quadruple then the diameter of G is (at least) 7,whereas if S has no orthogonal quadruple then the diameter of G is (at most) 4. We recallthat requires N − o (1) time, unless the SETH fails, even when ‘ = d log N e [17]. In thatcase, the graph G has O ( N ) vertices and ˜ O ( N ) edges. Hence any algorithm approximatingsparse undirected weighted Diameter within ratio better than 7 / n / − δ , with δ >
0, would refute the SETH.
We first describe the vertex set of G , then its edge set, and finally check that the number ofvertices and edges are as announced. Vertex set
Every vertex of G is the concatenation of a possibly empty tuple of vectors of S , called vector tuple , followed by a possibly empty tuple of possibly equal indices of [ ‘ ], called indextuple . Each coordinate of the vector tuple is called a vector field , while each coordinate ofthe index tuple is called an index field . The set V ( G ) is partitioned into four sets: T (for t riples), C (for c ouples), P (for p airs), and I (for i ndices). The names behind T, C, P reflectthe number and the nature (ordered or unordered) of the vector fields. Each of these setscomprise vertices with up to three vector fields and five index fields. They are defined in thefollowing way. T : for every { a, b, c } ∈ (cid:0) S (cid:1) , we add vertex ( a, b, c ) to T . Thus vertices of T have threevector fields and no index field. C : for every { a, b } ∈ (cid:0) S (cid:1) and i, j, k ∈ [ ‘ ] such that a [ i ] = a [ j ] = a [ k ] = 1 andmaj( b [ i ] , b [ j ] , b [ k ]) = 1, we add vertex ( a, b, i, j, k ) to C . Thus vertices of C have twovector fields and three index fields. P : for every a, b ∈ S and i, j, k ∈ [ ‘ ] such that a [ i ] = a [ j ] = a [ k ] = 1 and b [ i ] = b [ j ] = b [ k ] = 1, we add vertex ( { a, b } , i, j, k ) to P . We will still see a and b as filling the two . Bonnet 7 vector fields of the vertex, without a first vector field and a second vector field. Contraryto vertices of C , ( { a, b } , i, j, k ) and ( { b, a } , i, j, k ) are two names for the same vertex(whereas ( a, b, i, j, k ) and ( b, a, i, j, k ) are two distinct vertices, whose existence impliesslightly different properties). Thus vertices of P also have two vector fields and threeindex fields. Note also that { a, b } is a multiset, since a may be equal to b . I : for every p , p , i, j, k ∈ [ ‘ ], we add vertex ( p , p , i, j, k ) to I . The chosen labels forthe five index fields anticipate that, to build the edge set, it is convenient to imaginea separation after the first two index fields of the tuple. The vertices of I have no vectorfield and five index fields. Edge set
We will put some edges between T and C , C and P , P and I , and I and T . In addition,we put index-switching edges within I and within C . An index-switching edge is betweentwo vertices of the same set ( I or C ) with the same vector tuple (which is always the casein I ) and distinct index tuples. The only edges with a weight different than 1 are the edgesbetween I and T , which all have weight 3. Thus, unless specified otherwise, an edge hasweight 1.The total list of edges is as follows.We add all the index-switching edges within I and C . Thus G [ I ] is a clique and G [ C ] isa disjoint union of at most (cid:0) | S | (cid:1) cliques (while G [ T ] and G [ P ] remain independent sets).More explicitly, we have an edge between every pair of distinct vertices ( p , p , i, j, k ) ∈ I and ( p , p , i , j , k ) ∈ I , and for every a = b ∈ S between every pair of distinct vertices( a, b, i, j, k ) ∈ C and ( a, b, i , j , k ) ∈ C . E ( T, C ): We add an edge between every ( a, b, c ) ∈ T and ( a, b, i, j, k ) ∈ C provided thatthere is an h ∈ { i, j, k } such that b [ h ] = c [ h ] = 1. E ( C, P ): We add an edge between every ( a, b, i, j, k ) ∈ C and ( { c, d } , i, j, k ) ∈ P whenever a ∈ { c, d } . E ( T, I ): We add an edge of weight 3 between every ( a, b, c ) ∈ T and ( p , p , i, j, k ) ∈ I whenever a [ p ] = b [ p ] = c [ p ] = a [ p ] = b [ p ] = c [ p ] = 1. E ( I, P ): We add an edge between every ( p , p , i, j, k ) ∈ I and ( { a, b } , i, j, k ) ∈ P whenever a [ p ] = b [ p ] = 1 or a [ p ] = b [ p ] = 1.This ends the construction. See Figure 2 for an illustration. Vertex and edge count
There are O ( N ) vertices in T , O ( N ‘ ), in C ∪ P , and ‘ , in I , hence O ( N + N ‘ + ‘ ) = O ( N ) in total. There are O ( N ‘ ) edges in E ( T, C ) ∪ E ( C, P ), O ( N ‘ ), in E ( C ), O ( N ‘ ),in E ( T, I ), O ( N ‘ ), in E ( I, P ), and O ( ‘ ) in E ( I ), hence O ( N ‘ + N ‘ + ‘ ) = ˜ O ( N )edges in total. Furthermore G can be built in time ˜ O ( N ). Assuming that there is no orthogonal quadruple, we show that every pair of vertices of G is at distance at most 4. For that we repeatedly use that, for every a, b, c, d ∈ S ,ind( a, b, c, d ) := min { i ∈ [ ‘ ] | a [ i ] = b [ i ] = c [ i ] = d [ i ] = 1 } is a well-defined index in [ ‘ ]. Weonly take the minimum index to have a deterministic notation, but there is nothing particularwith it, and any index of the non-empty { i ∈ [ ‘ ] | a [ i ] = b [ i ] = c [ i ] = d [ i ] = 1 } would work allthe same. n / ( a, b, c )( a, b, i, j, k ) ( a, b, i , j , k )( { d, e } , i, j, k )( p , p , i, j, k )( p , p , i , j , k ) a [ i ] = a [ j ] = a [ k ] = ( b [ i ] , b [ j ] , b [ k ]) = [ i ] = d [ j ] = d [ k ] = e [ i ] = e [ j ] = e [ k ] = ∃ h ∈ { i , j , k } , c [ h ] = b [ h ] = ∈ { d , e } a [ p ] = b [ p ] = c [ p ] = a [ p ] = b [ p ] = c [ p ] = d [ p ] = e [ p ] = [ p ] = e [ p ] = TCPI (3 , , , , Figure 2
The weighted construction G . In bold, the conditions for the existence of a vertex or ofan edge. The edge in blue, and more generally every edge of E ( T, I ), has weight 3, while all otheredges have weight 1. The pairs in red recall, for vertices of the corresponding set, the length of theirvector tuple followed by the length of their index tuple.
We first observe that every vertex is at distance at most 3 from I . (cid:73) Lemma 3. N [ I ] ⊇ I ∪ P , N [ I ] ⊇ I ∪ P ∪ C , and N [ I ] = V ( G ) . Proof.
The first and second inclusions are actually equalities but we will not need thosefacts. N [ I ] ⊇ I ∪ P since every ( { a, b } , i, j, k ) ∈ P is adjacent (with an edge of weight 1) to( i, i, i, j, k ) ∈ I . Then, N [ I ] ⊇ N [ I ∪ P ] ⊇ I ∪ P ∪ C since every ( a, b, i, j, k ) ∈ C is adjacentto ( { a, a } , i, j, k ) ∈ P . Finally, N [ I ] ⊇ N [ I ∪ P ∪ C ] = V ( G ) since every ( a, b, c ) ∈ T isadjacent to ( a, b, i, i, i ) ∈ C for some i ∈ [ ‘ ], for otherwise a, b, c is an orthogonal triple. (cid:74) We now exhibit paths of length at most 4 between every pair of vertices of G . For the casedisjunction, initially imagine the K with loops on vertices T, C, P, I , where edges correspondto kinds of pairs that are left to check. The following paragraphs remove all its edges in theorder: all edges incident to I , all remaining edges incident to P but T P , all remaining edgesincident to C , the loop on T , and finally the edge T P . Between u ∈ I and v ∈ V ( G ) As G [ I ] is a clique and, by Lemma 3, N [ I ] = V ( G ), every vertex u ∈ I is at distance atmost 4 from every vertex v ∈ V ( G ). Between u ∈ P and v ∈ P ∪ C For every u ∈ P , N [ u ] ⊃ I and so N [ u ] ⊃ P ∪ C , by Lemma 3. In particular there is apath of length at most 4 between u and any vertex v ∈ P ∪ C . . Bonnet 9Between u ∈ C and v ∈ T ∪ C Let ( a, b ) be the two vector fields of u ∈ C , ( c, d ) be the first two vector fields of v ∈ T ∪ C , and e be the third vector field of v if v ∈ T . Let i = ind( a, b, c, d ), j = ind( a, c, d, e ) if v ∈ T , and j = i if v ∈ C . We observe that ( a, b, i, i, j ) , ( { a, c } , i, i, j ) , ( c, d, i, i, j ) are (existing) verticesof C , P , and C , respectively, and that u − ( a, b, i, i, j ) − ( { a, c } , i, i, j ) − ( c, d, i, i, j ) is a pathof length 3 in G . The existence of these vertices is implied by a [ i ] = b [ i ] = c [ i ] = d [ i ] = 1, a [ j ] = c [ j ] = d [ j ] = 1. The first edge of the path is an index-switching edge within C . Theexistence of the other edges is implied by a ∈ { a, c } , c ∈ { a, c } , and the fact that the indextuple ( i, i, j ) does not change.Finally if v ∈ C , then the index-switching edge ( c, d, i, i, j ) − v completes the uv -path oflength 4. If instead v ∈ T , then the edge ( c, d, i, i, j ) − ( c, d, e ) = v completes the uv -path oflength 4. This edge exists since d [ j ] = e [ j ] = 1. Between u ∈ T and v ∈ T Let u = ( a, b, c ) , v = ( d, e, f ) ∈ T , i = ind( a, b, c, d ), j = ind( a, b, d, e ) and k = ind( a, d, e, f ).Then u = ( a, b, c ) − ( a, b, i, j, k ) − ( { a, d } , i, j, k ) − ( d, e, i, j, k ) − ( d, e, f ) = v is a path oflength 4 in G . These vertices exist since a and d have value 1 on indices i, j, k , b , on indices i, j , and e , on indices j, k . The first edge exists since b [ i ] = c [ i ] = 1, the next two edges existfor similar reasons as invoked in the previous paragraph, and the fourth edge exists since e [ k ] = f [ k ] = 1. Between u ∈ T and v ∈ P Let u = ( a, b, c ) ∈ T and v = ( { d, e } , i, j, k ) ∈ P . We set p = ind( a, b, c, d ), p =ind( a, b, c, e ), and exhibit a uv -path of length 4 via I . Indeed u = ( a, b, c ) − ( p , p , i, j, k ) − ( { d, e } , i, j, k ) = v is a path of length 4 in G (recall that the first edge of the path hasweight 3). Edge ( a, b, c ) − ( p , p , i, j, k ) ∈ E ( T, I ) exists since a [ p ] = b [ p ] = c [ p ] = a [ p ] = b [ p ] = c [ p ] = 1. Edge ( p , p , i, j, k ) − ( { d, e } , i, j, k ) ∈ E ( I, P ) exists since d [ p ] = e [ p ] = 1and the three last indices ( i, j, k ) remain unchanged. Let a, b, c, d ∈ S be an orthogonal quadruple, that is, such that there is no index i ∈ [ ‘ ]satisfying a [ i ] = b [ i ] = c [ i ] = d [ i ] = 1. We may further assume that a, b, c, d are all distinctsince checking for an orthogonal triple can be done in time ˜ O ( N ). We will now show thatthere is no path P of length at most 6 between u = ( a, b, c ) ∈ T and v = ( d, c, b ) ∈ T .Since the distance between every pair of vertices in T × I is at least 3, a T T -path oflength at most 6 cannot contain an edge of the clique G [ I ], nor more generally intersects I at least twice. We thus distinguish two cases: (case A) P visits I exactly once, and (case B) P remains within T ∪ C ∪ P . Before proving that no uv -path P of length at most 6 visits I ,thereby ruling out case A, we state a couple of useful observations. (cid:73) Observation 4.
There is at most one path of length 2 between ( { d, e } , i, j, k ) ∈ P and ( a, b, c ) ∈ T , namely ( { d, e } , i, j, k ) − ( a, b, i, j, k ) − ( a, b, c ) , which in particular implies that a ∈ { d, e } . More basically, the only neighbors of ( a, b, c ) ∈ T (at distance 1, so not in I ) are of theform ( a, b, i, j, k ) ∈ C . We can generalize this observation to paths contained in T ∪ C . n / (cid:73) Observation 5.
For every path within G [ T ∪ C ] , all the vertices of the path have the samefirst two vector fields. Case A: P visiting I As P cannot visit I twice, if it visits I then it has length exactly 6 and is one of the followingkinds: (case 1) T − I − T , (case 2) T − C − P − I − P − C − T , or (case 3) T − I − P − C − T (recall that the edges in E ( I, T ) have weight 3). An important feature of such paths is thatno index-switching edge can be used, thus the three last index fields (when they exist) haveto remain the same.
Case 1.
A path ( a, b, c ) − ( p , p , i, j, k ) − ( d, c, b ) would in particular imply that a [ p ] = b [ p ] = c [ p ] = d [ p ] = 1, contradicting the orthogonality of a, b, c, d . Case 2.
By Observation 4 applied to both ends of the path, P is of the form ( a, b, c ) − ( a, b, i, j, k ) − ( { a, e } , i, j, k ) − ( p , p , i, j, k ) − ( { d, f } , i, j, k ) − ( d, c, i, j, k ) − ( d, c, b ) withsome e, f ∈ S . The existence of the vertices ( a, b, i, j, k ) , ( d, c, i, j, k ) ∈ C implies that a [ i ] = a [ j ] = a [ k ] = d [ i ] = d [ j ] = d [ k ] = 1, and that b and c have value 1 on at least twoindices (with multiplicity) of multiset { i, j, k } . In particular, there is an h ∈ { i, j, k } suchthat a [ h ] = b [ h ] = c [ h ] = d [ h ] = 1, a contradiction to a, b, c, d being orthogonal. Case 3.
By Observation 4 applied to the second half of the path, P has then the form( a, b, c ) − ( p , p , i, j, k ) − ( { d, e } , i, j, k ) − ( d, c, i, j, k ) − ( d, c, b ). The first three vertices yielda contradiction. Indeed, the existence of edge ( p , p , i, j, k ) − ( { d, e } , i, j, k ) implies that d [ p z ] = 1 for some z ∈ { , } , while the existence of ( a, b, c ) − ( p , p , i, j, k ) implies that a [ p z ] = b [ p z ] = c [ p z ] = 1. Case B: paths P within T ∪ C ∪ P We now consider paths P in G [ T ∪ C ∪ P ]. Since a = d , P has to visit P , since otherwisethe first vector field cannot change, by Observation 5. We then observe that no shortest uv -path visits T a third time (one more time than the two endpoints u and v ). A T T -pathvisiting T a third time would contain a segment C − T − C that can be shortcut into C − C .Indeed, ( a, b, i, j, k ) − ( a, b, c ) − ( a, b, i , j , k ) has a chord ( a, b, i, j, k ) − ( a, b, i , j , k ) whichis an index-switching edge of C .We further distinguish two cases: (case 1) P does not contain any index-switching edge,or (case 2) P contains at least one index-switching edge. Case 1.
In that case, P is of the form T − C − P − C − T or T − C − P − C − P − C − T .Either way, we consider the unique neighbors of u and v in P . These neighbors have to be( a, b, i, j, k ) ∈ C and ( d, c, i, j, k ) ∈ C for some i, j, k ∈ [ ‘ ]. Indeed no index-switching edgenor return to T is allowed here. Thus we conclude as in case A.2. Case 2.
We now assume that P contains at least one index-switching edge (of C ). Inthat case, as P has length at most 6, it can visit P only once. Hence P is of the kind T − C − C − P − C − C − T , where one of the two edges C − C is optional. We consider thelast vertex u ∈ C before visiting P , and the first vertex v ∈ C after visiting P . There is,by design, no index-switching edge between u and v on path P . Thus by Observation 5,there are i, j, k ∈ [ ‘ ] such that u = ( a, b, i, j, k ) and v = ( d, c, i, j, k ). We then conclude asin case A.2. . Bonnet 11 So far we showed the announced lower bound for sparse undirected weighted
Diameter .We show how to tune the previous construction to get the same lower bound for sparseundirected unweighted
Diameter . The weighted graph G had only non-trivial edge weightsin E ( T, I ). We now describe how to replace these weighted edges, to get an unweightedgraph G = ρ ( S ). We start with a short summary of the changes. We will replace T by three copies T, T , T with an induced perfect matching between T and T , and between T and T . We link T to I as we linked T to I , and T and C , and T and C , as we linked T and C . We finally adda set I of vertices with empty vector tuple (like I ) that we link to T and I only. Addition to the vertex set
We add three sets to V ( G ) to get V ( G ): two identical copies of T , denoted by T and T ,and a set I isomorphic to [ ‘ ]. More precisely, for every i ∈ [ ‘ ], we add vertex ( i ) to I .Thus I has no vector field and a unique index field. We use a subscript to distinguish the homologous vertices in T, T , T . Vertices ( a, b, c ) T ∈ T, ( a, b, c ) T ∈ T , ( a, b, c ) T ∈ T arethe three vertices of G corresponding to the same vertex ( a, b, c ) of G . Edition of the edge set
We first remove the edges of G with weight 3 (between T and I ). For every { a, b, c } ∈ (cid:0) S (cid:1) , weadd the edges ( a, b, c ) T − ( a, b, c ) T and ( a, b, c ) T − ( a, b, c ) T . We also add edges between T and C , the same way we have defined edges between T and C . That is, ( a, b, c ) T − ( a, b, i, j, k )is an edge if and only if ( a, b, c ) T − ( a, b, i, j, k ) is an edge. Let us recall that the existence ofthis edge (and of its endpoint in C ) implies that a, b, c have value 1 on indices { i, j, k } , threetimes, at least twice, and at least once, respectively, and that there is an h ∈ { i, j, k } suchthat a [ h ] = b [ h ] = c [ h ].We add edges (of weight 1) between T and I , the same way we defined the weight-3edges of G between T and I . Thus there is an edge ( a, b, c ) T − ( p , p , i, j, k ) whenever a [ p ] = b [ p ] = c [ p ] = a [ p ] = b [ p ] = c [ p ] = 1. We further add an edge between ( i ) ∈ I and ( a, b, c ) T whenever a [ i ] = b [ i ] = 1. Finally we add all the index-switching edges in I ,and we make I and I fully adjacent, that is, we turn G [ I ∪ I ] into a clique.This finishes the edition to the unweighted construction. See Figure 3 for an illustration. New vertex and edge count
We added to V ( G ) O ( N ) vertices in T ∪ T , and ‘ , in I . Thus G has also O ( | V ( G ) | ) = O ( N + N ‘ + ‘ ) = O ( N ) vertices. We added to E ( G ) O ( N + N ‘ ) edges incidentto T , and O ( N ‘ + ‘ + ‘ ) edges incident to I . (The edges between T and I were alreadycounted in G between T and I .) Thus G has O ( | E ( G ) | ) = O ( N ‘ + N ‘ + ‘ ) = ˜ O ( N )edges. Again G can be computed in time ˜ O ( N ). In case S has no orthogonal quadruple, we use similar arguments as in G , to find paths oflength at most 4 between every pair of vertices in G . We first show that I is at distance at n / ( a, b, c ) T ( a, b, c ) T ( a, b, c ) T ( i )( i ) ( a, b, i, j, k ) ( a, b, i , j , k )( { d, e } , i, j, k )( p , p , i, j, k )( p , p , i , j , k ) a [ i ] = a [ j ] = a [ k ] = ( b [ i ] , b [ j ] , b [ k ]) = [ i ] = d [ j ] = d [ k ] = e [ i ] = e [ j ] = e [ k ] = ∃ h ∈ { i , j , k } , c [ h ] = b [ h ] = ∈ { d , e } a [ p ] = b [ p ] = c [ p ] = a [ p ] = b [ p ] = c [ p ] = [ i ] = b [ i ] = [ p ] = e [ p ] = [ p ] = e [ p ] = TT T I CPI (3 , , , ,
1) (2 , , , Figure 3
The unweighted construction G . In bold, the conditions for the existence of a vertexor of an edge. The pairs in red recall, for vertices of the corresponding set, the length of their vectortuple followed by the length of their index tuple. most 3 of every vertex of G . (cid:73) Lemma 6. N [ I ] ⊇ I ∪ I ∪ T , N [ I ] ⊇ I ∪ I ∪ T ∪ P ∪ T , and N [ I ] = V ( G ) . Proof.
The inclusions are actually equalities. N [ I ] ⊇ I ∪ I ∪ T since I is fully adjacent to I and every ( a, b, c ) T ∈ T is adjacent to some ( i ) ∈ I , for otherwise a, b is an orthogonalpair. N [ I ] ⊇ N [ I ∪ I ∪ T ] ⊇ I ∪ I ∪ T ∪ P ∪ T since every ( { a, b } , i, j, k ) ∈ P isadjacent to ( i, i, i, j, k ) ∈ I and every ( a, b, c ) T ∈ T is adjacent to ( a, b, c ) T ∈ T . Finally, N [ I ] ⊇ N [ I ∪ I ∪ T ∪ P ∪ T ] = V ( G ) since every ( a, b, i, j, k ) ∈ C is adjacent to( { a, a } , i, j, k ) ∈ P and every ( a, b, c ) T ∈ T is adjacent to ( a, b, c ) T ∈ T . (cid:74) We also show the following inclusions. (cid:73)
Lemma 7. N [ I ] ⊃ P ∪ T , and N [ I ] ⊃ P ∪ C ∪ T ∪ T . Proof. N [ I ] ⊃ P , N [ I ] ⊃ C , and N [ T ] ⊃ T have all been shown in Lemma 6. Thereforewe shall just prove that N [ I ] ⊃ T . Indeed every vertex ( a, b, c ) T ∈ T is adjacent to some( i, i, i, i, i ) ∈ I , since otherwise a, b, c is an orthogonal triple. (cid:74) For the case disjunction, initially imagine the K with loops on vertices T, T , T , C, P, I, I ,where edges correspond to the kinds of pairs that are left to check. The following paragraphsremove all its edges in the order: all edges incident to I and to I , all remaining edges incidentto P and to T but T P and
T T , all remaining edges incident to C , all remaining edgesincident to T as well the loop on T , the edge T P , and finally the edge
T T . Between u ∈ I ∪ I and v ∈ V ( G ) As G [ I ∪ I ] is a clique and, by Lemma 6, N [ I ] = V ( G ), then N [ u ] = V ( G ) holds forevery vertex u ∈ I ∪ I . . Bonnet 13Between u ∈ P ∪ T and v ∈ P ∪ C ∪ T ∪ T For every u ∈ P ∪ T , by Lemma 7 and the fact that G [ I ] is a clique, N [ u ] ⊃ I and, againby Lemma 7, N [ u ] ⊃ P ∪ C ∪ T ∪ T . In particular there is a path of length at most 4from u to any vertex v ∈ P ∪ C ∪ T ∪ T .The following two cases work as in G , since ( a, b, c ) T and ( a, b, c ) T are twins in G [ T ∪ T ∪ C ∪ P ]. Between u ∈ C and v ∈ T ∪ T ∪ C This holds by replacing the occurrence of ( c, d, e ) by ( c, d, e ) T or ( c, d, e ) T , and everyoccurrence of T by T ∪ T , in the paragraph Between u ∈ C and v ∈ T ∪ C of the weightedconstruction. Between u ∈ T ∪ T and v ∈ T ∪ T Again this holds by replacing occurrences of ( a, b, c ) (resp. ( c, d, e )) by ( a, b, c ) T or ( a, b, c ) T (resp. ( c, d, e ) T or ( c, d, e ) T ). Between u ∈ T and v ∈ P This works as in G by following three edges of weight 1 from T to I , instead of a singleedge of weight 3. For every u = ( a, b, c ) T ∈ T and v = ( { d, e } , i, j, k ) ∈ P , there is apath u = ( a, b, c ) T − ( a, b, c ) T − ( a, b, c ) T − ( p , p , i, j, k ) − ( { d, e } , i, j, k ) = v in G , with p = ind( a, b, c, d ) and p = ind( a, b, c, e ). Between u ∈ T and v ∈ T This case is the real novelty compared to G , and the reason for introducing I . For every u = ( a, b, c ) T ∈ T and v = ( d, e, f ) T ∈ T , there is a path u = ( a, b, c ) T − ( a, b, c ) T − ( a, b, c ) T − ( i ) − ( d, e, f ) T = v in G , with i = ind( a, b, d, e ). The last two edges exist since a [ i ] = b [ i ] = d [ i ] = e [ i ] = 1. Again we assume that there is an orthogonal quadruple a, b, c, d ∈ S such that a, b, c, d are pairwise distinct. We claim that there is no path of length at most 6 in G between u = ( a, b, c ) T and v = ( d, c, b ) T . Since the distance between T and I ∪ I is at least 3, any T T -path of length at most 6 visits I ∪ I at most once. For the sake of contradiction, let P be such a path that we further assume shortest (hence in particular chordless) and, amongshortest uv -paths, having the fewest edges in E ( T , C ). We will show that P cannot visit I ,nor use any edge of E ( T , C ). Finally we observe that T T -paths of length at most 6 in G respecting these two interdictions are in length-preserving one-to-one correspondence with T T -paths in G . P cannot visit I The only possible kind of a
T T -path of length at most 6 visiting I is T − T − T − I − T − T − T . This forces P to be of the form ( a, b, c ) T − ( a, b, c ) T − ( a, b, c ) T − ( i ) − ( d, c, b ) T − ( d, c, b ) T − ( d, c, b ) T for some i ∈ [ ‘ ]. However the third and fourth edges imply that thereis an i ∈ [ ‘ ] such that a [ i ] = b [ i ] = c [ i ] = d [ i ] = 1, a contradiction to the orthogonality of a, b, c, d . n / P cannot use any edge of E ( T , C ) Assuming that P contains at least one edge in E ( T , C ), we first show that it has to containa subpath C − T − T − I ∪ I or I ∪ I − T − T − C . Let w = ( a , b , c ) T ∈ T ∩ V ( P )be a vertex of P with one neighbor x ∈ C ∩ V ( P ) on P . The other neighbor y of w on P is necessarily in T . Indeed if y ∈ T , then y = ( a , b , c ) T , and xy ∈ E ( G ) is a chord. Ifinstead y ∈ C , then one can replace the subpath x − ( a , b , c ) T − y by x − ( a , b , c ) T − y ,contradicting the minimality of the number of used edges in E ( T , C ) (since this numberdecreases by 2).Thus the only possibility is that y ∈ T . Then the other neighbor of y on P (otherthan w ) has to be in I ∪ I , since otherwise P is not a simple path. Hence P contains asubpath of the kind C − T − T − I ∪ I (or the reverse, I ∪ I − T − T − C ). Now weobserve that C is at distance at least 1 from T , while I ∪ I is at distance at least 3 from T .Therefore such a path P would have length at least 7. Such a path P would also exist in G We can now assume that P does not use any vertex of I nor any edge of E ( T , C ). Every such simple T T -path (visiting I at most once) also exists in the weighted graph G , with the samelength. To see it, we notice that if P contains an edge ( a , b , c ) T − ( a , b , c ) T , then it hasto contain a subpath of the form ( a , b , c ) T − ( a , b , c ) T − ( a , b , c ) T − ( p , p , i, j, k ) ∈ I ,and is emulated in G by taking the weight-3 edge ( a , b , c ) − ( p , p , i, j, k ). However weshowed in the previous section that no uv -path of length at most 6 exists in G . References Donald Aingworth, Chandra Chekuri, Piotr Indyk, and Rajeev Motwani. Fast Estima-tion of Diameter and Shortest Paths (Without Matrix Multiplication).
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