AAn Investigation into Graph Curvature’sAbility to Measure Congestion in NetworkFlow
Matthew P. Yancey ∗ December 7, 2015
Abstract
A recent trend in network research involves finding the appropriategeometric model for a given network, and to use features of the modelto infer information about the network. One piece of information thatthis paper will focus on is the severity and location of congestion inthe network’s traffic flow. To this end, many versions of “curved sur-faces” have been proposed, each with some set of parameters to allowthe model to fit the network, and then those parameters are linkedto congestion in some manner. These proposed spaces include Gro-mov’s hyperbolic spaces, a scaled variation of Gromov’s hyperbolicspace, curved 2-manifolds, additive metrics, and fixed points in theautomorphism group of the graph. However, in each case above, thelink between the parameters of the geometric space and the conges-tion of the network has been intuitive and non-rigorous. This paperis a thorough and rigorous treatment of each space’s ability to de-scribe congestion. Our conclusion is that Gromov’s hyperbolicity isthe unique space from which information can be extracted. ∗ Institute for Defense Analyses / Center for Computing Sciences (IDA / CCS),[email protected] a r X i v : . [ m a t h . C O ] D ec ur investigation’s wide scope led to the resolution of conjecturesand open problems from a wide variety of topics. Specifically, thispaper resolves • a conjecture of Dourisboure and Gavoille on a 2-approximationmethod for calculating tree-length, • an open problem from Narayan and Saniee on the amount ofcongestion in the Euclidean grid, • all of the conjectures from Jonckheere, Lou, Bonahon, and Barysh-nikov relating congestion to rotational symmetry, and • a rejection of the implication by Jonckheere, Lohsoonthorn, andBonahon that scaled hyperbolicity implies properties character-ized by hyperbolic spaces such as the confinement of quasi-geodesicsto a neighborhood of the geodesic.We also show that Buneman’s distance approximating tree is the bestpossible up to a constant both additively and multiplicatively. There has been an attempt recently to describe routing in computer networksby modeling the traffic as a set of shortest paths in non-Euclidean space.The comparison has deep qualitative significance. One of the non-Euclideanspaces considered is a hyperbolic plane. In a hyperbolic plane, paths betweentwo points are cheaper the closer they are to the center, and so all shortestpaths arc towards the center as they travel. In computer networks, bits ofinformation that need to travel far will be passed from regional hubs up tobackbone routers, a subnetwork designed to route a large volume of trafficcheaply across long distances. Each successive stage up is then consideredcloser to the “logical center” of the network. But in what tangible ways isthis metaphor useful?Shavitt and Tankel [61], and later Bogu˜n´a, Papadopoulos, and Krioukov[12], described a heuristic method to map vertices of a computer network into2 2-dimensional hyperbolic space. They started with placing vertices withhighest degree (a parameter known to correlate with being in the backbone)at the center of the hyperbolic space, and then moving on to lower degreevertices placed at larger radii from the center and at an angle that minimizesthe distance to adjacent vertices. The goal of the project was to see if such amapping of the graph interactions between telecommunication companies toa hyperbolic space could be used to efficiently route Internet traffic via thenatural shortest path derived from the geometric space. The experiment wasa success. Another example of using a curved space to solve a problem is thescheme by Gavoille and Ly [40] for distance-labeling networks.After the appearance of these applications various types of curved spaceshave been presented as alternative methods to model big data or big net-works. Most of these spaces have a parameter, usually called the curvatureof the space. Proposed applications of this line of thinking include commu-nity detection [37], network security [44] [42], quantum information [43, 48],influence in regulatory networks [3, 14], graph mining [1], and connectivity ofsensor networks [5]. We will focus on the application of describing the conges-tion in a network. Suggested models of networks that incorporate congestionhave involved hyperbolic spaces [30], curved 2-manifolds [55], or fixed pointsin the automorphism group of the graph [47].Traffic on a computer network is modeled using uniformly distributeddemand . In this paper we assume all of our graphs are connected. Let P ( u, v )denote the number of shortest paths from u to v in G , and P ( u, v ; w ) denotethe number of shortest paths from u to v that use w . The demand on avertex w is defined to be D ( w ) = (cid:80) u,v P ( u, v ; w ) /P ( u, v ). The betweenness of w is D ( w ) / (cid:0) n (cid:1) ; we choose the less-common notation of demand becauseit better illustrates asymptotic behavior. In most real world networks theaverage distance between two vertices is small [51], and therefore the averageamount of demand on a vertex is approximately cn for some constant c . A congested network is then a network with a skewed distribution of demand;in the extremal case we should see a non-empty set of vertices with demandat least c (cid:48) n for some constant c (cid:48) . We will investigate several of the spacesproposed to model congested graphs, and whether or not there is any relationbetween the curvature parameter and the level of congestion.Our results will show that the only parameter of interest is Gromov’s δ -3yperbolic curvature. Some work investigating the implication of Gromov’swork on graphs has already been done: the introduction to [63] contains anexcellent survey on results characterizing graphs with small hyperbolicity,and inequalities have been constructed relating hyperbolicity to the inde-pendence and dominating numbers [59]. Hyperbolicity has also been usedto construct random networks that feature scale free degree distributions[50] and tunable clustering coefficients [23]. The first half of this paper is athorough analysis of how Gromov’s work applies to congestion in networks.The most common application of Gromov’s work to network analysis isthat for any δ -hyperbolic graph G there is a map φ from the vertex set to atree T such that for all x, y ∈ V ( G ) d T ( φ ( x ) , φ ( y )) ≤ d G ( x, y ) ≤ d T ( φ ( x ) , φ ( y )) + 2 δ log ( | V ( G ) | ) . This result, and that trees are 0-hyperbolic, inspires the intuition “ δ -hyperbolicgraphs (for small δ ) are tree-like.” This intuition can be misleading, as ourtree-like graphs do not always behave like trees. For example, every tree with n + 1 vertices will have a vertex with demand at least n . On the other hand, K n +1 is also 0-hyperbolic, and has the least amount of congestion among allgraphs with n + 1 vertices. A second example is a cycle C n , which is n -hyperbolic, and therefore not hyperbolic at all as n → ∞ . However, everyvertex in C n has demand (1 + o (1)) n .In the first example above, the distance approximating tree T for K n +1 is a star, where each edge has cost 0 . G maps to a leaf.Therefore the only vertex with non-trivial demand in T is not associated witha vertex of G at all. In the second example, the distance approximating treeof C n is a path with vertex set in order t , t , . . . t n . For any vertex u ∈ C n ,Gromov’s theorem constructs a map φ u such that φ u ( v ) = t d ( u,v ) . We see thatthe link between δ -hyperbolic curvature and congestion has less to do with δ and more to do with the distance approximating tree. It is then a coy turnof events that we will demonstrate a rigorous connection between congestionand hyperbolic curvature, and we will do it without reference to distanceapproximating trees.We will also investigate other parameters related to distance approxi-mating trees, such as tree-length [26] and the length of the largest inducedcycle [25, 32], even though there has been no previously reported relation to4ongestion in a network. Distance approximating trees are a subject of theirown interest, with applications to routing schemes [28, 32], evolutionary trees[2, 36], graph mining [1], computational geometry [6], and graph visualiza-tion [33]. We will show that while the methods change drastically, Gromov’smethod to generate a distance approximating tree has re-appeared multipletimes. For a survey of other related parameters, see [35].In Section 2 we will translate a portion of Gromov’s seminal work [41] intothe language of graph theory. We will focus on the distance approximatingtree, and using additional work show that the tree produced by Gromov’swork is in fact a layering tree. In Section 3 we will investigate other parame-ters that imply the existence of a distance approximating tree with boundederror, and show relations between those parameters and δ -hyperbolicity. InSection 4 we will consider the algorithmic problem of determining the valueof δ for a given graph. In Section 5 we will prove a relationship betweenhyperbolicity and congestion.In the second half of this paper we will briefly discuss the other pro-posed geometric models for inferring information about congestion. Some ofthese models are built on an intuitive understanding that falls apart as werigorously analyze them. Others are models designed to overcome the limita-tions of Gromov’s hyperbolicity - but are done in such a way that all of theadvantages are lost.Sections 6 and 7 will investigate measures loosely based on the concept ofthe center of rotation as a nexus for where the highest levels of congestion in anetwork would be located. In Section 8 we will analyze the level of congestionin the Euclidean grid, which has been held up as the standard non-hyperbolicgraph. In Section 9 we will consider scaled versions of Gromov’s hyperbolicity. Our first set of results concern the relationship between Gromov hyperbol-icity, thin triangles, and slim triangles. Gromov’s [41] seminal work gave theoriginal connection between hyperbolicity and thin triangles - but the coeffi-cient involved had an arithmetic error. As far as we know, Gromov’s typo hasbeen used ubiquitously. Alternative proofs given in various surveys provide5eaker results. The bounds we present are the strongest known at this time.It is folklore that a graph is 0-hyperbolic if and only if each 2-connectedsubgraph is complete (see [63]). Blocks (maximally 2-connected subgraphs)of a graph form a tree with the separating vertices. The following result isnot repeated in the main text of this manuscript, as it clearly follows.
Theorem 1.1. -hyperbolic graphs with bounded clique size have congestion. Baryshnikov and Tucci [9, 10] approached the problem of congestion usingthe boundary of a hyperbolic space, which is defined on the behavior ofinfinite geodesics. Their approaches involved several unstated assumptions,such as the graph being a group action on the Poincar´e disk [9] or every vertexbeing a part of an infinite ray [10] (one implication of either assumptionis that the graph has no leaves). These assumptions are necessary to theirtechniques. We call a graph a locally finite if each ball of finite radius containsa finite number of vertices. Li and Tucci [52] give a beautiful proof of largedemand for locally finite graphs using randomized methods: if each ball ofradius 2 δ has at most M points and D is the diameter of the graph, thenthere exists a ball B of radius 2 δ such that (cid:80) w ∈ B D ( w ) ≥ ( nD − M − ) .Our results start with two statements about how traffic of certain typeswill cluster around a central point. Corollary 5.2 If B , B are disjoint balls in a graph with ˆ δ -thin triangles,then there exists a ball B with radius 2ˆ δ such that for every pair of vertices x ∈ B and y ∈ B we have that every shortest path from x to y crosses B . Theorem 5.5
Let G be δ -hyperbolic. For any t , there exists a ball B of radius t + 18 δ such that every geodesic between x and y intersects B if d ( x, y ) ≥ diam( G ) − t .If the graph is a geometric network from a bounded dimensional hy-perbolic space, then Corollary 5.2 will imply congestion. If the graph is ageometric network from an unbounded dimensional hyperbolic space, thenTheorem 5.5 can be used to imply congestion. We wish to strengthen Corol-lary 5.2 by altering our focus from balls to halfspaces. To do so, we define ananalogue for halfspaces in discrete geometries.6et α be a large geodesic from u to v , r be the midpoint of α , and 2 f α ( z ) = d ( z, v ) − d ( z, u ). In this language, we think of α as a straight line through ourgraph, r is our origin ( f α ( r ) = 0), and f α ( z ) is the (index of the) projectionof the vertex z onto α . Bowditch [16] gave three definitions of a projectiononto a geodesic. This projection is similar to definition (P2) in description.The standard definition, the vertex in α that minimizes the distance to z , isdefinition (P3). Fortunately, we do not need to be confused: Bowditch wenton to prove that all three definitions agree up to a small multiple of δ indistance for δ -hyperbolic graphs. We continue this terminology to say that H k = { z : 2 f α ( z ) = k } is the analogue of a hyperplane for each fixed k and α . We define a half-space as the set of points z such that f α ( z ) >
0. Ahyperbolic network G with n points and origin r is ( a, b, c )-balanced if1. for every maximal geodesic α of length at least diam( G ) − a , bothhalf-spaces defined by α contain at least cn points each, and2. if n k = |{ a : d ( a, r ) = k }| , then cb k ≤ n k ≤ c − b k for some b > k ≤ diam( G ). Theorem 5.9 If G is a ( a, b, c )-balanced network with a ≥ Ω (log( c − ) + δ ),then there exists a ball of radius O (log( c − ) + δ ) whose demand sums to atleast Ω ( n ).Many proposed examples of hyperbolic networks embed in two-dimensionalspace. To contrast hyperbolic networks against a “flat” network, the Eu-clidean grid is held up as the example non-congested, non-hyperbolic net-work [47, 45, 55, 30]. However, we have not seen a rigorous calculation of thedemand in an Euclidean grid. Theorem 8.1 If w is the center of a two-dimensional Euclidean grid with n vertices, then (1 + o (1)) n . < D ( w ) < (1 + o (1)) n . .What has not been mentioned (rigorously or not) is that the O ( n . )demand is best possible. Theorem 8.2 If G is a planar graph, then there exists a vertex w ∈ V ( G )such that D ( w ) ≥ √ (1 − o (1)) n . . 7ur study of hyperbolicity includes an investigation into Gromov’s claimthat hyperbolic spaces are similar to trees. The proof to his claim is con-structive; it gives a fast algorithm to create a distance approximating tree.We present a series of claims that leads to the following result. Theorem 2.16
The distance approximating tree constructed by Gro-mov’s algorithm is a layering tree (see Definition 2.17).Layering trees have had an incredible history - from historical literature togeometric algebra to biology to computer vision. Layering trees were first in-vented by Buneman [20, 21] as he studied the phylogeny of Canterbury Talesmanuscripts in the 1970’s. Next came Gromov’s work in the 1980’s. This wasfollowed by Agarwala, Bafna, Farach, Paterson, and Thorup’s work [2] on thephylogeny of biological specimens in the 1990’s. They appeared most recently(and were finally named by) Chepoi and Dragan [25] in the 2000’s to assist incomputer vision. These four projects were all totally independent - and theyused wildly different methods to create their distance approximating trees.Proving that all four algorithms produce the same tree is non-trivial, hencethe following theorem is half original research and half a survey of literature.
Theorem 3.4
For a given graph G , let T be a distance approximatingtree for G . Let u, v ∈ V ( G ), let (cid:15) u,v = (cid:107) d G ( u, v ) − d T ( T ( u ) , T ( v )) (cid:107) . We havethat • If for all u, v we have that 1 ≤ d T ( T ( u ) ,T ( v )) d G ( u,v ) ≤ γ , then finding theminimum possible value for γ , denoted γ G , is NP-complete. • Let γ (cid:48) G be the minimum of possible values for max u,v (cid:107) d G ( u, v ) − d T ( T ( u ) , T ( v )) (cid:107) .Finding a distance approximating tree T where max u,v (cid:107) d G ( u, v ) − d T ( T ( u ) , T ( v )) (cid:107) is less than 9 / γ (cid:48) G is NP-hard.On the other hand, there exists a single distance approximating tree T forwhich all of the following is true1. if G is δ -hyperbolic, then max u,v (cid:107) d G ( u, v ) − d T ( T ( u ) , T ( v )) (cid:107) ≤ δ log ( | V ( G ) | ),2. 1 ≤ d T ( T ( u ) ,T ( v )) d G ( u,v ) ≤ γ G , 8. max u,v (cid:107) d G ( u, v ) − d T ( T ( u ) , T ( v )) (cid:107) ≤ γ (cid:48) G , and4. T can be computed in O ( | E ( G ) | ) time.Layering trees are tightly connected with a graph parameter called tree-length. Chepoi, Dragan, Estellon, Habib, and Vax`es [26] proved that graphswith tree-length (cid:96) are (cid:96) -hyperbolic and graphs that are δ -hyperbolic havetree-length at most 2 δ log ( | V ( G ) | ). Their proof heavily relied on the layeringtree of a hyperbolic graph, which was later formally stated (see [35]). Thefollowing result is not repeated in the main text of this manuscript, as itfollows immediately from the definition of tree-length. However, we includeit here as motivation for our extended investigation into tree-length. Theorem 1.2.
Locally finite graphs with bounded tree-length will experiencecongestion.
The result of Chepoi, Dragan, Estellon, Habib, and Vax`es is tight: tree-length is a strictly stronger condition than hyperbolicity.
Claim 3.8
There exists an infinite family of graphs { G i } with uniformlybounded hyperbolicity and there exists a constant c > G i is at least c log( | V ( G i ) | ).Lokshtanov [54] showed that finding the optimal tree-length is NP-complete,which places an emphasis on constructing approximation algorithms for tree-length. Given the connection to layering trees, the work in Theorem 3.4 im-mediately implies a strong algorithm. Dourisboure and Gavoille [33] providea second algorithm called k -disk-tree. If k -disk-tree terminates, then it hasconstructed a tree decomposition of the graph with tree-width at most 2 k .Dourisboure and Gavoille [33] prove that k -disk-tree terminates when k is atleast three times the tree-length of the graph, giving a bound that equals theresult from layering trees. Dourisboure and Gavoille [33] conjectured that the k -disk-tree algorithm will terminate when k is at least the tree-length of thegraph, which is false. Theorem 3.10
Let G r,(cid:96) be the Cartesian product of a cycle C (cid:96) and apath P r(cid:96) for r >
3. The tree length of G r,(cid:96) is (cid:98) (cid:96) (cid:99) + 1, but the k -disk-treealgorithm fails to terminate when k < (cid:96) − δ such that a given graph is δ -hyperbolic. Cohen, Coudert, and Lancin [30] gavea heuristic algorithm that involves trimming simplicial vertices, which hasbeen used several times since [31, 1]. Their argument for removing simplicialvertices is flawed; a simple counterexample is a clique on four vertices withone edge removed. We provide a weaker version of the flawed statementthat is sharp. Moreover, it implies (and generalizes) the often used step ofreducing to 2-connected subgraphs, which is included in other algorithms tocalculate hyperbolicity [15] that have been independently verified throughgenuine application [14] as quick and efficient. Theorem 4.1
Let B be a separator for a a graph G , where the diameterof G [ B ] is k . Let G i be component i of G − B with B added back in. If G i is δ i -hyperbolic for each i , then G is (2 k + max i ˆ δ i )-hyperbolic.Sections 6, 7, and 9 largely contain counterexamples to various conjecturesand do not contain constructive results. This section is dedicated to investigating Gromov’s [41] distance approxima-tion tree for graphs with small δ -hyperbolicity. In Sections 2.1 and 2.2 wereview selected statements from Gromov’s work as a short proof of his resulton distance approximating trees (which do not appear until 80 pages intothe manuscript). While we do not attribute it, all results in those sectionsare from [41]. In Section 2.3 we investigate further how Gromov’s methodbehaves on graphs. Definition 2.1.
Let r be some fixed vertex and define ( x.y ) r = ( d ( x, r ) + d ( y, r ) − d ( x, y )) / . If X is δ -hyperbolic then for every three points x, y, z ∈ X , e have that ( x.z ) r ≥ min (( x.y ) r , ( y.z ) r ) − δ. The smallest such δ among all roots r is called the hyperbolic curvature ofthe space. There are several definitions of δ -hyperbolicity; we will primarily use theone above. The term ( x.y ) r is called the Gromov product , and in a hyperbolicspace it suffices as an estimate to the distance from r to the geodesic from x to y . It may also be helpful to think of ( x.y ) r as an inverse angle. In thiscase the center of the angle, r , is called the root of the graph. There is anequivalent definition of hyperbolicity that will be useful in several places. Definition 2.2 (Four Points condition) . A graph is
Four Points ˆ δ -hyperbolic if for every four vertices a, b, c, d ∈ V ( G ) , we have that d ( a, b ) + d ( c, d ) ≤ max (( d ( a, c ) + d ( b, d )) , ( d ( a, d ) + d ( b, c ))) + 2ˆ δ. Note that not all roots are created equal, but Gromov proved that noneare too bad. Our theoretical arguments in the rest of the paper assume thatwe have an optimal root, and the next result suggests that we do not needto concern ourselves too much about this requirement.
Theorem 2.3.
If a graph is δ -hyperbolic at a root r , then it is δ -hyperbolicat any root.Proof. Let b, c, d be arbitrary points and a be our desired new root. Because G is δ -hyperbolic, we have − δ ≤ ( a.b ) r − min (( a.c ) r , ( b.c ) r )and − δ ≤ ( c.d ) r − min (( b.d ) r , ( b.c ) r ) , so it follows that − δ ≤ ( a.b ) r + ( c.d ) r − min (( a.c ) r + ( b.d ) r , b.c ) r ) . By a symmetrical argument, we see that − δ ≤ ( a.b ) r + ( c.d ) r − min (( a.c ) r + ( b.d ) r , a.d ) r ) . − δ ≤ ( a.b ) r + ( c.d ) r − min (( a.c ) r + ( b.d ) r , ( a.d ) r + ( b.c ) r ) . If we expand by the definition of ( ◦ . ◦ ) r and cancel, the above becomes − δ ≤ − d ( a, b ) − d ( c, d ) − min ( − d ( a, c ) − d ( b, d ) , − d ( a, d ) − d ( b, c )) . Adding d ( a, b ) + d ( a, c ) + d ( a, d ) inside and outside the minimum operationon the right hand side above gives us − δ ≤ d ( a, c ) + d ( a, d ) − d ( c, d ) − min ( d ( a, b ) + d ( a, d ) − d ( b, d ) , d ( a, b ) + d ( a, c ) − d ( b, c ))= 2 (( c.d ) a − min (( b.d ) a , ( b.c ) a )) . There is another parameter related to hyperbolic curvature, called
ThinTriangles whose value is bounded above and below by some multiples of δ . Definition 2.4. [Thin Triangles Condition] A geodesic space G has ˆ δ -thintriangles if for every triple x , x , x ∈ V ( G ) with shortest paths p i,j from x i to x j , then the vertices y k , y (cid:48) k that are on p i,j and p i,(cid:96) and are distance k from x i , for all ≤ k ≤ ( | p i,j | + | p i,(cid:96) | − | p j,(cid:96) | ) / x j .x (cid:96) ) x i satisfy d ( y k , y (cid:48) k ) ≤ ˆ δ . The relationship between thin triangles and hyperbolicity is subtle. Ourapproach to this relationship may appear to be pedantic to those more famil-iar with the area. The fact that any relationship exists is sufficient to thosewho study infinite spaces. However, the coefficients matter when an idea isput into practice and so we take time to carefully outline just how closelyrelated these distinct features are.In Lemma 6.3.B, Gromov proved that a space with ˆ δ -thin triangles is also2ˆ δ -hyperbolic. For completeness, we recreate a proof of this statement below.The complex part comes during Lemma 6.3.A, which claims that δ -hyperbolicgeodesic spaces have 2 δ -thin triangles. The issue is that the proof relies onLemma 6.1.B in such a way that the bound is actually ˆ δ -thin triangles forˆ δ = 2 (cid:100) log (5 − (cid:101) δ = 4 δ . The distinction is largely ignored. Texts covering12yperbolicity usually avoid Gromov’s proof for a weaker bound proven usingelementary methods, such as the proof by Batty and Papasoglu [11] andAlonso et. al. [4] of 18 δ or Albert, DasGupta, and Mobasheri’s proof [3] of6 δ + 2. We will prove the 4 δ bound using elementary methods below.Some texts, such as Bowditch’s [16], skip thin triangles altogether for aweaker condition called slim triangles. A triangle x, y, z is called ˜ δ -slim if forevery point u on the geodesic from x to y , there exists a point w on eitherthe geodesic from x to z or the geodesic from y to z such that d ( u, w ) ≤ k .In fact, the proof of 18 δ above is actually a proof of two statements: that δ -hyperbolic geodesic spaces have 3 δ -slim triangles and that geodesic spaceswith ˜ δ -slim triangles have 6˜ δ -thin triangles. Bridson and H¨afliger [18] have aproof that geodesic spaces with ˜ δ -slim triangles have 4˜ δ -thin triangles; we willgive a cleaner proof of this result. Burago, Burago, and Ivanov [22] state thatgeodesic spaces with ˜ δ -slim triangles have 3˜ δ -thin triangles, but the proof isleft as an exercise. By considering S , the best possible bound is no betterthan 2˜ δ , but this remains an open problem in extremal combinatorics.In some applications we may not wish to think of our graphs as geodesicspaces. For example, in networks our space is a discrete set of points and theconstruction of edges is an artificial method to represent distances. In suchcases we can define an analogue of a geodesic as a sequence of points suchthat the distance between any two vertices equals 1 (assuming the graph isunweighted). It is not immediately clear that a connection between hyperbol-icity and thin triangles exists in such a setting. The very thing that allowedus to define an analogue of a geodesic will also save us here: the spaces de-fined with and without the edges are quasi-isometric with no multiplicativeerror and an additive error of 1 /
2. We will prove the 4 δ bound using elemen-tary methods below assuming that our space is an unweighted graph. Thisis sharp: C is 1 / Claim 2.5.
A graph with ˆ δ -thin triangles is δ -hyperbolic.Proof. We will use the four point condition of hyperbolicity, so let a, b, c, d be four arbitrary points. Let P a,b = u , u , . . . , u (cid:96) be a shortest path from a to b . Consider the triangle created by the points a, b, c using path P a,b .Let i ∗ = (cid:100) ( d ( a, c ) + d ( a, b ) − d ( b, c )) / (cid:101) . By the conditions of having ˆ δ -thintriangles, we know that u i ∗ is within ˆ δ of some vertex on the shortest path13rom b to c and within ˆ δ of some vertex on the shortest path from a to c .We do the same thing for the triangle involving vertices a, b, d , and find a j ∗ such that u j ∗ is close to a vertex on each of the shortest paths from a to d and b to d . By symmetry, assume that i ∗ ≤ j ∗ .Let x be the vertex on the shortest path from a to d such that d ( a, x ) = j ∗ ,and so d ( x, u j ∗ ) ≤ ˆ δ . Let y be the vertex on the shortest path from c to b suchthat d ( b, y ) = (cid:96) − i ∗ , and so d ( y, u i ∗ ) ≤ ˆ δ . Therefore d ( a, d ) = j ∗ + d ( x, d ) and d ( c, b ) = (cid:96) − i ∗ + d ( c, y ). Note that d ( a, b ) = (cid:96) , and by the triangle inequality, d ( c, d ) ≤ d ( c, y ) + d ( y, u i ∗ ) + d ( u i ∗ , u j ∗ ) + d ( u j ∗ , x ) + d ( x, d ) ≤ ≤ δ + j ∗ − i ∗ + d ( c, y ) + d ( x, d ) . We conclude that d ( a, d ) + d ( b, c ) − d ( a, d ) − d ( b, c ) ≤ δ . Claim 2.6.
Let G be a geodesic space or the vertex set of an unweightedgraph. If G is δ -hyperbolic, then G has δ -thin triangles.Proof. First, let us assume that G is a geodesic space. Let x, y, z be the threecorners of a triangle. Without loss of generality, assume that w is on thegeodesic from x to y . By symmetry we also assume that d ( x, z ) − d ( x, w ) ≥ d ( y, z ) − d ( y, w ), or equivalently that d ( x, w ) ≤ ( d ( x, z ) + d ( x, y ) − d ( y, z )) / x.z ) w ≤ ( y.z ) w and so0 = ( x.y ) w ≥ min { ( x.z ) w , ( y.z ) w } − δ = ( x.z ) w − δ. Let w (cid:48) and w (cid:48)(cid:48) be the points on the geodesic from x to z such that d ( x, w (cid:48) ) =( z.w ) x and d ( x, w ) = d ( x, w (cid:48)(cid:48) ). We will show that d ( w, w (cid:48)(cid:48) ) ≤ δ .Note that d ( x, w ) = ( z.w ) x + ( x.z ) w , d ( w, z ) = ( x.z ) w + ( x.w ) z , d ( x, z ) =( w.z ) x + ( w.x ) z , and d ( z, w (cid:48) ) = ( x.w ) z . If we let P be the perimeter of thetriangle xwz , then d ( x, w (cid:48) ) + d ( w, z ) = d ( x, w ) + d ( w (cid:48) , z ) = P/
2. By the fourpoint condition, d ( w, w (cid:48) )+ d ( x, z ) ≤ δ + P/
2, and so d ( w, w (cid:48) ) ≤ δ +( x.z ) w ≤ δ . Because w (cid:48) and w (cid:48)(cid:48) are on the same geodesic that starts at x , we havethat d ( w (cid:48) , w (cid:48)(cid:48) ) = d ( x, w (cid:48)(cid:48) ) − d ( x, w (cid:48) ) = ( x.z ) w . By the triangle inequality, d ( w, w (cid:48)(cid:48) ) ≤ d ( w, w (cid:48) ) + d ( w (cid:48) , w (cid:48)(cid:48) ) ≤ δ .Now suppose that G is a graph. If P is even, then the above proof is validwithout adaptation. Suppose P is odd, so that ( z.w ) x is not an integer. In this14ase, define w (cid:48) to be as above, but with d ( x, w (cid:48) ) = ( z.w ) x + 0 .
5. Now we havethat ( P + 1) / d ( x, w (cid:48) ) + d ( w, z ) = d ( x, w ) + d ( w (cid:48) , z ) + 1 and so d ( w, w (cid:48) ) ≤ δ + 0 .
5. Furthermore, d ( w (cid:48) , w (cid:48)(cid:48) ) = d ( x, w (cid:48)(cid:48) ) − d ( x, w (cid:48) ) = ( x.z ) w − .
5. Aftercancellation we arrive at the same conclusion.
Claim 2.7. If G is a ˜ δ -slim geodesic space, then it is also a δ -thin geodesicspace.Proof. Let xyz be a geodesic triangle in our space and w a point on the xy geodesic such that d ( x, w ) ≤ ( y.z ) x . Let w (cid:48)(cid:48) be the point on the xz geodesicsuch that d ( x, w (cid:48)(cid:48) ) = d ( x, w ). We will show that d ( w, w (cid:48)(cid:48) ) ≤ δ .By the assumption of slim triangles, there exists a point w (cid:48) on the geodesic xz or the geodesic yz such that d ( w, w (cid:48) ) ≤ ˜ δ . We say that when w (cid:48) is on thegeodesic xz we have the good case . If we have the good case, then by thetriangle inequality | d ( x, w (cid:48) ) − d ( x, w ) | ≤ d ( w, w (cid:48) ) ≤ ˜ δ . Because w (cid:48)(cid:48) and w (cid:48) are both on the geodesic xz , it follows that d ( w (cid:48) , w (cid:48)(cid:48) ) = | d ( x, w (cid:48) ) − d ( x, w (cid:48)(cid:48) ) | .Therefore the good case implies that d ( w, w (cid:48)(cid:48) ) ≤ d ( w, w (cid:48) ) + d ( w (cid:48) , w (cid:48)(cid:48) ) ≤ δ .We will show that we have the good case if d ( x, w ) < ( y.z ) x − ˜ δ . Because w is on the xy geodesic, this assumption implies that d ( w, y ) > ( x.z ) y + δ . By way of contradiction, suppose that w (cid:48) is on the geodesic yz . By thetriangle inequality d ( w (cid:48) , y ) ≥ d ( y, w ) − d ( w, w (cid:48) ) > ( x.z ) y . Because w (cid:48) is onthe yz geodesic, this means that d ( w (cid:48) , z ) < ( x.y ) z . By the triangle inequality d ( x, z ) ≤ d ( x, w ) + d ( w, w (cid:48) ) + d ( w (cid:48) , z ) < d ( x, z ), which is a contradiction.If ( y.z ) x ≤ ˜ δ , then d ( w, w (cid:48)(cid:48) ) ≤ d ( w, x ) + d ( x, w (cid:48)(cid:48) ) ≤ δ . The claim thenfollows from the fact that w is within ˜ δ + (cid:15) of a good case for all (cid:15) > .2 Constructing a Distance Approximating Tree Theorem 2.8. If G is δ -hyperbolic, then there exists a tree T with weightededges such that every vertex of G is a vertex in T (this mapping is possiblynot injective or surjective), and for every pair of vertices u, v ∈ V ( G ) , wehave that d G ( u, v ) ≥ d T ( u, v ) ≥ d G ( u, v ) − (cid:15) G , where (cid:15) G ≤ δlog ( n − . The rest of Section 2.2 is dedicated to proving Theorem 2.8. The proof isbroken up into smaller claims. We do this because it makes the proof easierto read and it highlights smaller true statements that interested readers maywish to consider. This paper is testament to the idea that several of theclaims can be strengthened when additional assumptions are considered.The first statement proves the theorem under the special case when δ = 0.This was actually proven by Buneman [20] over a decade earlier. Bunemanalso gave a generalized method for constructing a distance approximatingtree for any δ [21] that includes Gromov’s method as a special case, howeverthere have been no calculations of quality for this method nor has a reasonableimplementation ever been outlined except under very restrictive assumptions. Claim 2.9.
Theorem 2.8 is true when δ = 0 . This means there exists a tree T with weighted edges such that d T ( u, v ) = d G ( u, v ) for every pair of vertices.Proof. Let our root, r , be an arbitrary vertex of G .Let P be a partition of V ( G ). At first, P will have one part: all of V ( G ).At the end, P will have | V ( G ) | parts: each vertex in G will be its own partin P . Let P i be the partition at stage i . We intend to grow a partial tree T i at each stage, with T i − ≤ T i .Note that ( u.v ) r ≥ a for any constant a is an equivalence relation betweenthe vertices u and v , because G is 0-hyperbolic. Thus, we will slowly grow aparameter (cid:96) i , and at each stage the parts of the partitions are the equivalenceclasses where ( u.v ) r > (cid:96) i . If k < k (cid:48) , then any part in P k (cid:48) will be a subset of16 a b gf i hc de a,b,c,d,e,f,g,h,i, ja,b c,d,e,f,g,h,i,jL0 = 0 a b c,i d e,g,f,h,jL1 = 0.5 j L2 = 1 a b d g,f,h,jc i eL3 = 2 a b d g,jc i e hL4 = 2.5 a b d g jc i e hffL5 = infinity
Figure 1: Step 1 in constructing a distance approximating tree fora 0-hyperbolic graph. a,b,c,d,e,f,g,h,i, ja,b c,d,e,f,g,h,i,jL0 = 0 a b c,i d e,g,f,h,jL1 = 0.5L2 = 1 a b d g,f,h,jc i eL3 = 2 a b d g,jc i e hL4 = 2.5 a b d g jc i e hffL5 = infinity ra b c d e fg hi j a b c i d e h
Figure 2: Step 2 in constructing a distance approximating tree fora 0-hyperbolic graph. 17ome part in P k . We grow the (cid:96) i to be the first point after (cid:96) i − such that P i (cid:54) = P i − . Step 1:
We set T − to be a single vertex, which we will call r (cid:48) . Because d is a metric distance function, ( u.v ) r = d ( u, r ) + d ( v, r ) − d ( u, v ) ≥ (cid:96) = 0 and P to be theappropriate partition. We set T to be T − plus one vertex for each part in P , and each of these vertices is adjacent to r (cid:48) . Those edges have weight (cid:96) .At stage i , we will construct T i from T i − by adding one vertex for eachpart in P i . For each of these vertices in T i − T i − , we will add one edge fromthat new vertex to the unique part in P i − that contained the given part in P i . Each of the new edges will be weighted (cid:96) i +1 − (cid:96) i .Continue this process until every part is one vertex. See Figure ?? for anillustrated example of this step. Step 2
Suppose this process ends at stage k . We have built a tree, and wemust now determine how to map the vertices of G to the vertices of T := T k .First, we map the root r of G to the root r (cid:48) of T . Every other vertex u of G ismapped to the vertex that is distance d G ( r, u ) from r (cid:48) that is on the shortestpath from r (cid:48) to the unique part of P k that equals { u } . See Figure ?? for anillustrated example of this step.We claim that for all pairs of vertices u, v , we have that d T ( u, v ) = d G ( u, v ). Clearly this is true if u = r or v = r , so assume otherwise. Thereis a unique subtree of T that connects r (cid:48) , u , and v . It consists of three pathsthat meet at one vertex, and that vertex has distance ( u.v ) r from r (cid:48) . Thedistance between u and v in T is then the sum of the other two paths. So, d T ( u, v ) = d T ( u, r (cid:48) ) + d T ( v, r (cid:48) ) − u.v ) r = d G ( u, r ) + d G ( v, r ) −
22 ( d G ( u, r ) + d G ( v, r ) − d G ( u, v ))= d G ( u, v )So now we know that any 0-hyperbolic space can be represented as a treewith weighted edges. The rest of this section is devoted to proving that a18 -hyperbolic space can be approximated by a 0-hyperbolic space, where thedistance between any two points is shrunk by at most (cid:15) G . Definition 2.10.
Let f ( x, y ) = max x = w ,w ,...,w k = y min ≤ i ≤ k − ( w i .w i +1 ) r , for arbitrary k . Claim 2.11.
For all pairs of vertices x, y , f ( x, y ) − ( x.y ) r ≤ δlog ( n −
1) = (cid:15) G / . Proof.
Let x = w , w , . . . , w k = y be a shortest sequence of vertices thatmaximizes f ( x, y ). We will prove that f ( x, y ) − ( x.y ) r ≤ δlog ( k ) by inductionon k . We will also show that that w i (cid:54) = r for all i , and no w i is repeated,which implies that k ≤ n −
1. Once we have proven both of these statements,the claim follows.If w (cid:48) , w (cid:48) , . . . , w (cid:48) j is a sequence such that for all a there exists a b such that w (cid:48) a w (cid:48) a +1 = w b w b +1 , thenmin ≤ i ≤ k − ( w (cid:48) i .w (cid:48) i +1 ) r ≥ min ≤ i ≤ k − ( w i .w i +1 ) r . Because f ( x, y ) wants to maximize over such sequences, we may assume thatno w i is repeated, or else a shorter sequence could be taken without cost.Suppose there exists an i such that w i = r . Because d is a metric space,we have that( w i − .r ) r = 12 ( d ( w i − , r ) + d ( r, r ) − d ( w i − , r )) = 0 = ( r, w i +1 ) r . Also because d is a metric space, ( a.b ) r ≥ a, b . So the sequence w , w , . . . , w i − , w i +1 , . . . w k has to produce at least a large a value for f ( x, y )while being one vertex shorter. Therefore we may assume w i (cid:54) = r for all i ,and so k ≤ n − f ( x, y ) − ( x.y ) r ≤ δlog ( k ). We claimthat for every sequence w , . . . , w k , we have that min ≤ i ≤ k − ( w i .w i +1 ) r ≤ w .w k ) r + log ( k ) δ , not just the sequence that maximizes f ( x, y ). The basecase is k = 2 and follows trivially as we are only trying to minimize on oneelement. So suppose k >
2, and by induction the claim follows for all se-quences of vertices where k (cid:48) < k . Let i (cid:48) = (cid:98) k/ (cid:99) . By Definition 2.1, we knowthat ( x.y ) r ≥ min { ( x.w i (cid:48) ) r , ( y.w i (cid:48) ) r } − δ . By symmetry, assume without lossof generality that ( x.w i (cid:48) ) r ≤ ( x.y ) r + δ . By induction, min ≤ i ≤ i (cid:48) ( w i .w i +1 ) r ≤ ( x.w i (cid:48) ) r + log ( k/ δ , and so there exists an i between 1 and i (cid:48) where ( w i .w i +1 ) r ≤ ( x.y ) r + δ (log ( k ) −
1) + δ .As a corollary to the above proof, we also know that f ( x, x ) = ( x.x ) r = d ( x, r ). Also, f ( x, r ) = 0 for all x . We will use these facts in the proofto the next claim. We will also use the fact: If x = w , . . . w k = y and y = w (cid:48) , . . . w (cid:48) k (cid:48) = z are sequences that maximize f ( x, y ) and f ( y, z ), re-spectively, then w , . . . , w k , w (cid:48) , . . . w (cid:48) k (cid:48) is a sequence that may or may notmaximize f ( x, z ) and so f ( x, z ) ≥ min { f ( x, y ) , f ( y, z ) } . We also see that f ( x, y ) ≤ min { d ( x, r ) , d ( y, r ) } , because w = x , w k = y , and for all v wehave by the triangle inequality that ( x.v ) r ≤ d ( x, r ). Claim 2.12.
We define a distance function d (cid:48) ( x, y ) = d ( x, r ) + d ( y, r ) − f ( x, y ) . The following are true:1. d (cid:48) is a metric distance function, after equivalence classes defined by d ( x, x (cid:48) ) = 0 have been identified to one point.2. d (cid:48) ( x, y ) ≤ d ( x, y ) .3. d (cid:48) ( x, y ) ≥ d ( x, y ) − log ( n − δ .4. d (cid:48) is -hyperbolic.Proof. Part One: d (cid:48) is a metric distance function, after equivalence classesdefined by d ( x, x (cid:48) ) = 0 have been identified to one point.Clearly d (cid:48) is symmetric. We also have that d (cid:48) ( x, x ) = d ( x, r ) + d ( x, r ) − f ( x, x ) = 0. For the triangle inequality, we have that d (cid:48) ( x, z ) − ( d (cid:48) ( x, y )+ d (cid:48) ( y, z )) ≤ ( d ( x, r ) − d ( x, r ))+( d ( z, r ) − d ( z, r )) − d ( y, r )+2( f ( x, z ) − min { f ( x, y ) , f ( y, z ) } ) − { f ( x, y ) , f ( y, z ) } ≤ . d (cid:48) ( x, y ) ≥ x and y . By symmetry, let d ( x, r ) ≤ d ( y, r ). By the above discussion, f ( x, y ) ≤ d ( x, y ) ≤ d ( y, r ), so d ( x, r ) + d ( y, r ) − f ( x, y ) ≥ Part Two: d (cid:48) ( x, y ) ≤ d ( x, y ).Consider the sequence x = w , w = y . Because f maximizes over se-quences that start at x and end at y , we see that f ( x, y ) ≥ ( x.y ) r , so d (cid:48) ( x, y ) ≤ d ( x, r ) + d ( y, r ) − x.y ) r = d ( x, y ) . Part Three: d (cid:48) ( x, y ) ≥ d ( x, y ) − log ( n − δ This follows directly from Claim 2.11.
Part Four: d (cid:48) is 0-hyperbolic.What we want to show is that every triple of vertices x, y, z satisfies d (cid:48) ( x, r )+ d (cid:48) ( z, r ) − d (cid:48) ( x, z ) ≤ min { d (cid:48) ( x, r )+ d (cid:48) ( y, r ) − d (cid:48) ( x, y ) , d (cid:48) ( y, r )+ d (cid:48) ( z, r ) − d (cid:48) ( y, z ) } . Because f ( x, r ) = 0, we note that d (cid:48) ( x, r ) = d ( x, r ) for all x . So what weneed to show reduces to f ( x, z ) ≤ min { f ( x, y ) , f ( y, z ) } . This is exactly whatwe showed above to be true.The proof of Theorem 2.8 follows quickly now. Given vertex set V ( G )and distance metric d (cid:48) , build a distance approximating tree T by the methodin Claim 2.9. This tree gives the exact distances for d (cid:48) , and by Claim 2.12those distances are within 2 log ( n − δ the distances for d . Now, let’s assume G is an unweighted graph. Claim 2.13.
For every pair of vertices x, y , there exists a sequence of ver-tices x = w , . . . w k = y that maximizes f ( x, y ) that is a path. Let α =21in i d ( w i , r ) . Furthermore, f ( x, y ) = α − . if there exists an i such that d ( w i , r ) = d ( w i +1 , r ) = α , and f ( x, y ) = α otherwise.Proof. By way of contradiction, assume that w i w i +1 / ∈ E ( G ) for some given i . Let w i = s , s , . . . , s (cid:96) = w i +1 be a shortest path from w i to w i +1 . By ourassumption, (cid:96) ≥ d ( w i , r ) ≥ d ( w i +1 , r ). We claim that the sequence w , w , . . . , w i , s , w i +1 , . . . , w k produces a “score” that is at least as good as the sequence w , w , . . . , w k .The “score” we are trying to improve is to maximize the product ( ◦ . ◦ ) r be-tween successive members of the sequence. Thus, we only need to prove thatmin { ( w i .s ) r , ( s .w i +1 ) r } ≥ ( w i .w i +1 ) r , because all other successive membersare the same. Because s is on the shortest path from w i and w i +1 , we knowthat d ( w i , s ) , d ( s , w i +1 ) ≤ d ( w i , w i +1 ) −
1. Because s and w i are adjacent,we know that d ( s , r ) ≥ d ( w i , r ) − ≥ d ( w i +1 , r ) −
1. Therefore2( w i .s ) r = d ( w i , r ) + d ( s , r ) − d ( w i , s ) ≥ d ( w i , r ) + (( d ( w i +1 , r ) − − ( d ( w i , w i +1 ) − w i .w i +1 ) r and 2( w i +1 .s ) r = d ( w i +1 , r ) + d ( s , r ) − d ( w i +1 , s ) ≥ d ( w i +1 , r ) + (( d ( w i , r ) − − ( d ( w i , w i +1 ) − w i .w i +1 ) r . This concludes the case when d ( w i , r ) ≥ d ( w i +1 , r ). If d ( w i , r ) ≤ d ( w i +1 , r ),then insert s (cid:96) − instead of s , and the proof follows symmetrically. Hence ourproof that w , . . . , w k can be assumed to be a path is complete.Let w , . . . , w k be a path that maximizesmax x = w ,w ,...,w k = y min ≤ i ≤ k − ( w i .w i +1 ) r . Because w , . . . , w k is a path, we know that d ( w i , w i +1 ) = 1 and | d ( r, w i ) − d ( r, w i +1 ) | ≤ i . It follows that ( w i .w i +1 ) r = ( d ( r, w i ) + d ( r, w i +1 ) − / d ( r, w i ) + g and 22 g = 0 if d ( r, w i +1 ) = d ( r, w i ) + 1, • g = − / d ( r, w i +1 ) = d ( r, w i ), and • g = − d ( r, w i +1 ) = d ( r, w i ) −
1. Note that in this case i does notminimize d ( w i , r ).Let ( x.y ) (cid:48) r = ( d (cid:48) ( x, r ) + d (cid:48) ( y, r ) − d (cid:48) ( x, y )) /
2. That is, when we createGromov’s distance approximating tree for a graph G with non-zero hyperboliccurvature, ( x.y ) (cid:48) r is the adjusted angle using the adjusted distance metric d (cid:48) that has the same domain and zero hyperbolic curvature. Claim 2.14.
Let T be the tree generated by Gromov’s distance approximatingmethod.1. ( x.y ) (cid:48) r = f ( x, y ) .2. x and y are the same vertex in T if and only if d ( x, r ) = d ( y, r ) = k and there exists a path x = w , . . . w k = y such that d ( w i , r ) ≥ k and max { d ( w i , r ) , d ( w i +1 , r ) } > k for all i .Proof. The first statement is a direct calculation, and so we omit it.Suppose x and y are the same vertex in T . Equivalently, d (cid:48) ( x, y ) = 0. Let x = w , w , . . . , w k = y be a path that maximizes f ( x, y ).By way of contradiction, assume d ( x, r ) < d ( y, r ). By Claim 2.13, we knowthat f ( x, y ) ≤ d ( w , r ) = d ( x, r ) < d ( y, r ). Therefore d (cid:48) ( x, y ) = d ( x, r ) + d ( y, r ) − f ( x, y ) >
0, a contradiction. By symmetry, we conclude d ( x, r ) = d ( y, r ).Similarly, if there exists an i such that d ( w i , r ) < d ( x, r ) or d ( w i , r ) = d ( w i +1 , r ) = d ( x, r ), then f ( x, y ) < d ( x, r ) , d ( y, r ) and so d (cid:48) ( x, y ) >
0, acontradiction again.The other direction is clear, as we have already shown that d (cid:48) ( x, y ) ≥ S i ( r ) = { u : d ( r, u ) = i } . We call S i ( r ) the i -shell. Corollary 2.15.
Vertices x and y of G are in the same vertex of T if and onlyif d ( x, r ) = d ( y, r ) and x and y are in the same component of the subgraphof G induced on the vertex set V ( G ) − (cid:0) ∪ i ≤ d ( r,x ) S i ( r ) − { x, y } (cid:1) . Theorem 2.16.
There exists a fast method to construct T .For every i , we have that (cid:96) i ∈ Z . So for now, just let (cid:96) i = 0 . i .First, we perform all of the steps when i is even. The vertices at step i willbe the vertices in S i ( r ) , with identification based on necessary and sufficientconditions in Claim 2.14. Wait to add edges until we perform the steps when i is odd.If i is odd then create a graph H whose vertices are those vertices of T created in step i + 1 . Two vertices u T , v T of H form an edge if and only ifthere are vertices u G , v G of G such that u G ∈ u T , v G ∈ v T , and u G v G ∈ E ( G ) .Create a vertex in T for each connected component of H that contains at leastone edge of H . Add an edge from each new vertex to each vertex in T fromstep i + 1 that is in the corresponding component of H and to the appropriatevertex from step i − . Weight all of these edges / . For each componentof H that is an isolated vertex u T , add an edge in T from u T to the uniquevertex v T in layer i − such that there exists vertices u G , v G in G where u G ∈ u T , v G ∈ v T , and u G v G ∈ E ( G ) , and weight these edges . The vertices generated during step i for odd i are called Steiner points.Proof.
Each vertex in T is to have distance d (cid:48) ( x, r ) = d ( x, r ) from r in T . Itfollows that the steps when i is even are correct.Recall that the construction in Claim 2.9 only has a level at (cid:96) i = i/ V ( G ) has ( x.y ) (cid:48) r = i/
2. Also recall that ( x.y ) (cid:48) r = f ( x, y ). If i is odd, then this is only possible when the path that maximizes f ( x, y ) hastwo consecutive vertices w p and w p +1 each with distance ( i + 1) / r . Let p , p , . . . , p (cid:96) − be the indices where this happens and p = x , p (cid:96) = y . Becausethis is a path, the pairs of vertices w p j w p j +1 for an edge for 1 ≤ j ≤ (cid:96) − a b cd e fg h i j k l m no p rabcd e fgh i j k l mo np
11 1 11 1 Figure 3: An example distance approximating tree for a graphwith non-zero hyperbolic curvature. This example is very easy tocompute using Theorem 2.16.Those edges then form edges in H . Note that by Claim 2.9 w p j and w p j +1 will be in the same vertex of T during stage i + 1 for 0 ≤ j ≤ (cid:96) . Therefore x and y will be in the same component of H .Beyond this technical analysis of how the distance approximating tree isconstructed, we have learned certain features about the tree. For example,this tree is essentially the layering tree from other works. Definition 2.17.
Let G be a graph with vertex u . A layering tree T rootedat u is the graph created from G by identifying vertices x and y if and onlyif x and y are in the same component of the subgraph of G induced on thevertex set V ( G ) − (cid:0) ∪ i ≤ d ( r,x ) S i ( r ) − { x, y } (cid:1) . Layering trees are the distance approximating trees constructed indepen-dently by Chepoi and Dragan [25], which is also used in [28, 17, 32, 33].Gromov’s distance approximating tree is a layering tree with Steiner pointsadded. A Steiner point will only affect the distance between two vertices inthe tree by at most 1 additavely, so these trees are essentially the same. Wewill see in the next Section that layering trees are quite prevalent.25 .4 Short-cuts, Quasi-cycles, and Embeddings
Based on the computational complexity of determining the hyperbolicityof a graph, Rodr´ıguez began characterizing classes of graphs with boundedhyperbolicity. An R -shortcut in a graph G is a pair of vertices u, v in acycle C such that the distance between U and v in G is R less than thedistance between u and v in C . A shortcut is an R -shortcut for R > C is a cycle with noshortcuts, then 4 δ ≥ | C | . This is a weaker statement than the four pointscondition by choosing 4 equi-distance points along the cycle. An r -shortgraph has no shortcuts for cycle C that intersects C only at vertices x and y when d ( x, y ) > r . Rodr´ıguez [58] characterizes the set of r -short graphs withbounded hyperbolicity as those where a cycle has finite length if the cyclecontains a R -shortcut for R > r . If there are no long shortcuts and no cyclehas a path significantly longer than a shortcut, then this result is impliedby Wu and Zhang’s [63] result that δ ≤ λ ( G ) /
4. Rodr´ıguez’s work can bethought of as an additive variation of a multiplicative bound on shortcutsdescribed by Gromov.
Proposition 2.18 ([41], Proposition 7.2.E) . If X is a δ -hyperbolic space,then there exist numbers L ≥ L > L > (cid:15) > and λ > depend-ing only on δ such that for every cycle C = x , x , . . . , x L then for everypath x i , x i +1 , . . . , x j ⊆ C such that (cid:15) ≤ ( j − i ) M ODL ≤ L we have that λd ( x i , x j ) ≤ ( j − i ) M ODL . Verbeek and Suri [62] give a converse of Gromov’s result. An ( α, β )-quasi-cycle is a cycle C = x , x , . . . , x L such that αd C ( x i , x j ) ≤ d ( x i , x j ) for all1 ≤ i, j ≤ L when d C ( x i , x j ) ≥ βL . They show that there exists α (cid:48) , c > G is not δ -hyperbolic then there exists a ( α (cid:48) , / cδ . On the other hand, Verbeek and Suri [62] show that if G has an ( α, / L , then any embedding into hyperbolic space of G will have distortion at least Ω ( L/ log( L )). Moreover, locally finite graphs withbounded-sized quasi-cycles will embed into hyperbolic space with boundedadditive error. However, this last proof is based on the work of Bonk andSchramm [13], which is only informative for spaces with infinite diameter.26 Distance Approximating Trees
In this section we take particular care to distinguish when we talk about thedistance between two vertices in T versus in G . To do this, let T ( G ) be sometree, and then a distance approximating tree is a map T : V ( G ) → T ( G ) suchthat | d G ( u, v ) − d T ( G ) ( T ( u ) , T ( v )) | is small for all pairs of vertices u, v ∈ V ( G ).We will never use T again, its existence only implicitly assumed when wediscuss the structure of T ( V ( G )). We believe that this abuse of notation willhelp clarify our notation as we progress. We begin with a simple analysis of the construction of Gromov’s distanceapproximating tree T . The following claim almost immediately follows fromthe fact that T is a layering tree but would not be clear otherwise. Thisclaim has already been proven several times, and Chepoi, Dragan, Newman,Rabinovich, and Vax`es [29] provide a short proof. Claim 3.1.
Let T be Gromov’s distance approximating tree for a graph G .Let D = max T ( u )= T ( v ) d ( u, v ) . For all vertices u, v , d ( T ( x ) , T ( y )) ≤ d ( x, y ) ≤ d ( T ( x ) , T ( y )) + D. The point of Claim 2.11 is to prove that in Gromov’s distance approxi-mating tree, D ≤ δ log ( n − D has been an area of research on its own.The L ∞ -distance between a graph G and distance approximating tree T is (cid:107) T − G (cid:107) ∞ = max u,v ∈ V ( G ) | d G ( u, v ) − d ( T ( u ) , T ( v )) | . Let γ (cid:48) G = inf T (cid:107) T − G (cid:107) ∞ .Agarwala, Bafna, Farach, Paterson, and Thorup [2] show that finding a dis-tance approximating tree T such that (cid:107) T − G (cid:107) ∞ ≤ γ (cid:48) G / L ∞ to be a measure of additive error, and we will show that Gromov’sdistance approximating tree will have additive error at most 3 γ (cid:48) G . Agarwala,Bafna, Farach, Paterson, and Thorup [2] gave a polynomial-time construc-tion of a distance approximating tree T ∗ for general metric spaces such that (cid:107) T ∗ − G (cid:107) ∞ ≤ γ (cid:48) G . While they do briefly mention the Four Points Condition,27o reference to Gromov’s work is given. That their tree is equivalent to Gro-mov’s has already been noted [29], but we know of no instance where it hasbeen noted that both are layering trees. Due to length and technical details,we omit the proof of the equivalence between their distance approximatingtree and Gromov’s, except to provide the following comparisons: • the distance function ( D + C a ) on vertices u and v equals 2( m a − ( u.v ) a ), • a layering tree is a type of a -restricted additive metric, • finding the minimum spanning tree in the proof of Theorem 3.3 asapplied by Lemma 3.5 in [2] is equivalent to finding the path thatdefines the function f ( x, y ) in Definition 2.10, • – the process of removing the maximum weighted edge of the min-imum spanning tree and recursing on each subtree in the proofof Theorem 3.3 as applied by Lemma 3.5 in [2] is the same thingas finding a partition S ∪ S of the vertex set that maximizes α ,where α ≤ d ∗ ( u, v ) for all u ∈ S and v ∈ S and d ∗ is the distancefunction defined by ( D + C a ) – while Step 1 of Claim 2.9 of this paper using distance function d (cid:48) defined in Claim 2.12 in such a way that ( x, y ) (cid:48) r = f ( x, y ) findsa non-trivial partition P β of the vertex set with minimal β suchthat ( x, y ) (cid:48) r > β if and only if x and y are in the same part (so α = 2( m a − β )), as illustrated in Figure ?? , • and Step 2 of Claim 2.9 of this paper, as illustrated in Figure ?? , isequivalent to the process of transitioning from ultrametric U a ( D + C a )to a -restricted additive metric U a ( D + C a ) − C a .From this analysis, we actually get stronger claims. The work in [2] tell usthat Gromov’s tree is the best possible layering tree for the L ∞ metric, andthat there exists a layering tree T such that (cid:107) T − G (cid:107) ∞ ≤ γ (cid:48) G .We say that T approximates G with distortion γ if for all vertices u, v ∈ V ( G ) we have that γd ( T ( u ) , T ( v )) ≤ d ( u, v ) ≤ d ( T ( u ) , T ( v )). In this case,we call T non-contracting . Note that Gromov’s tree is non-expanding; theseare equivalent when d ( T ( u ) , T ( v )) (cid:54) = 0 for all u and v by uniformly scaling28he edges of T . For a fixed graph G , let γ G be the infimum of distortionsacross all approximations of G . Dragan and Yan [34] showed that determiningif γ G ≤ k is NP-complete when k ≥
25. B˘adoiu, Indyk, and Sidiropoulos[6] provided an algorithm that accepts an unweighted graph G as input,and produces a distance approximating tree with distortion at most 100 γ G .This was improved to 27 γ G by B˘adoiu, Demaine, Hajiaghayi, Sidiropoulos,Zadimoghaddam [7], and Chepoi, Dragan, Newman, Rabinovich, and Vax`es[29] further improved the bound to 6 γ G . Gromov’s tree almost works, exceptfor the issue when T ( u ) = T ( v ) for u (cid:54) = v . B˘adoiu, Indyk, and Sidiropoulosresolve this issue by creating a distance approximating tree T (cid:48) , Gromov’stree T is a sub-tree of T (cid:48) , and each u ∈ V ( G ) is mapped such that T (cid:48) ( u )is a leaf that is adjacent to T ( u ). This is another instance where a distanceapproximating tree in the literature is essentially the same as Gromov’s tree.The improvement by Chepoi, Dragan, Newman, Rabinovich, and Vax`es usesthe same tree, but with a tighter argument.To prove the O (1) bound on γ G , B˘adoiu, Indyk, and Sidiropoulos provedthat a good approximation tree can not exist if G contains a specific sub-structure, and that substructure closely resembles a “thin rectangles” varia-tion of Definition 2.4. A graph has k -thin rectangles if for any four vertices u , u , u , u and paths P u i ,u i +1 , let ˜ δ i = min { d ( v, w ) : v ∈ P u i ,u i +1 , w ∈ P u i +2 ,u i +3 } , then we have that k ≥ min ˜ δ i . Lemma 2.1 of [6] shows that everygraph has γ G -thin rectangles. The same proof implies the following strongerstatement. Claim 3.2.
A graph has k -edge thin rectangles if for any four vertices u , u , u , u and paths P u i ,u i +1 , let ˜ δ i = min { d ( v, w (cid:48) ) + d ( w, v (cid:48) ) : vw (cid:48) ∈ E ( P u i ,u i +1 ) , ww (cid:48) ∈ E ( P u i +2 ,u i +3 ) } , then we have that k ≥ min ˜ δ i , where arithmetic on vertexindices is performed modulo . If G does not have k -edge thin rectangles for k > and T is a non-contracting distance approximating tree for G , thenthere exists an edge uu (cid:48) ∈ ∪ i P u i ,u i +1 such that d ( T ( u ) , T ( u (cid:48) )) ≥ k . Therefore γ G ≥ k . The heart of the improvement by Chepoi, Dragan, Newman, Rabinovich,and Vax`es is the following claim. They prove it in one page without men-tioning thin rectangles; we will prove it in a few sentences. Recall that D = max T ( u )= T ( v ) d ( u, v ) for Gromov distance approximating tree T . Claim 3.3.
For all graphs G , we have that D ≤ γ G . roof. Let T be Gromov’s distance approximating tree with root r and let T ( x ) = T ( y ). If d ( x, r ) ≤ γ G , then d ( x, y ) ≤ d ( x, r ) + d ( r, y ), and we aredone. Let r = y , y , . . . , y k = y and r = x , x , . . . , x k = x be short-est paths, and let x = w , w , . . . , w (cid:96) = y be a path such that d ( w i , r ) ≥ d ( x, r ) + 1 for all 1 < i < (cid:96) . Let u = x , u = y , u = y k − γ G , u = x k − γ G such that P , = w , w , . . . , w (cid:96) , P , = y k , y k − , . . . , y k − γ , P , = y k − γ , y k − γ − , . . . , y , r, x , . . . , x k − γ , and P , = x k − γ , x k − γ − , . . . , x k . By thetriangle inequality, for all u ∈ P , and v ∈ P , we have that k = d ( x, r ) ≤ d ( x, u ) ≤ d ( r, v ) + d ( v, u ) ≤ k − γ G + d ( v, u ), so d ( v, u ) ≥ γ G . Let u (cid:48) ∈ N ( u ) ∩ P , and v (cid:48) ∈ N ( v ) ∩ P , . If y k − γ G x k − γ G ∈ E ( G ) then d ( x, y ) ≤ d ( x, x k − γ G ) + 1 + d ( y k − γ G , y ) ≤ γ G + 1 and we are done. We can also assumethat (cid:96) >
2, so it follows that d ( v, u ) + d ( u (cid:48) , v (cid:48) ) ≥ γ G + 2. By γ G -edge thinrectangles we see that there exists an x i and a y j such that k − γ ≤ i, j ≤ k and d ( x i , y j ) ≤ γ G . So d ( x, y ) ≤ d ( x, x i ) + d ( x i , y j ) + d ( y j , y ) = 3 γ G .We take care to distinguish δ , γ G , and γ (cid:48) G . The arithmetic on vertex in-dices in the following statements will be done modulo 4. Using Definition 2.2of hyperbolicity and letting the paths be the two shorter pairs of diagonals,we see that G having k -thin rectangles implies that G is δ -hyperbolic. As aconsequence of Lemma 2.1 in [6], we see that δ < γ G . Fournier, Ismail, andVigneron showed that violating the Four Points condition would lead to alarge γ (cid:48) G ([39], Section 2.3), and therefore γ (cid:48) G ≥ δ . The reverse of these state-ments are not true; we will see examples in Section 3.2 of a family of graphs, G ( i ) , with bounded hyperbolicity and γ G ( i ) , γ (cid:48) G ( i ) = Θ(log( n )). This is sharp,as by Theorem 2.8 we have that γ G ( i ) , γ (cid:48) G ( i ) ≤ δ log ( n − T be non-contracting in the definition of γ (cid:48) G and did in the definition of γ G . We have to use the fact that Gromov’stree T satisfies (cid:107) T − G (cid:107) ∞ ≤ γ (cid:48) G to get the bound γ G ≤ γ (cid:48) G .We summarize some of these results with the following theorem. Notethat a layering tree can be computed in O ( | E ( G ) | ) time (see Remark 1 of[25]). Theorem 3.4.
For a given graph G , let T be a distance approximating treefor G . Let u, v ∈ V ( G ) , let (cid:15) u,v = (cid:107) d G ( u, v ) − d T ( T ( u ) , T ( v )) (cid:107) . We have that • If for all u, v we have that ≤ d T ( T ( u ) ,T ( v )) d G ( u,v ) ≤ γ , then finding the mini- um possible value for γ , denoted γ G , is NP-complete. • Let γ (cid:48) G be the infemum of possible values for max u,v (cid:107) d G ( u, v ) − d T ( T ( u ) , T ( v )) (cid:107) .Finding a distance approximating tree T where max u,v (cid:107) d G ( u, v ) − d T ( T ( u ) , T ( v )) (cid:107) is less than / γ (cid:48) G is NP-hard.On the other hand, there exists a single distance approximating tree T forwhich all of the following is true1. if G is δ -hyperbolic, then max u,v (cid:107) d G ( u, v ) − d T ( T ( u ) , T ( v )) (cid:107) ≤ δ log ( | V ( G ) | ) ,2. ≤ d T ( T ( u ) ,T ( v )) d G ( u,v ) ≤ γ G ,3. max u,v (cid:107) d G ( u, v ) − d T ( T ( u ) , T ( v )) (cid:107) ≤ γ (cid:48) G , and4. T can be computed in O ( | E ( G ) | ) time. For a tree T , let S ( T ) be the set of all sub-trees of T . A tree-decomposition ofa graph G is a tree T G and a map φ : V ( G ) → S ( T G ), such that if xy ∈ E ( G )then φ ( x ) ∩ φ ( y ) (cid:54) = ∅ . Let φ − : V ( T ) → V ( G ) be a map that when given avertex t of T returns all vertices x of G such that t ∈ φ ( x ). This set-up is mostpopularly known for the graph parameter tree-width , which is defined to be tw ( G ) + 1 = min T,φ max t ∈ V ( T ) | φ − ( t ) | . We are more interested in tree-length ,which is defined to be tl ( G ) = min T,φ max φ ( x ) ∩ φ ( y ) (cid:54) = ∅ d G ( x, y ).Tree-length and tree-width are unrelated: cliques have unbounded tree-width and tree-length 1 versus cycles have unbounded tree-length and tree-width 1. These two examples prove that for any k , the family of graphswith tree-length at most k is not minor-closed. Therefore many of the niceproperties about graphs with bounded tree-width do not apply to graphswith bounded tree-length. We are interested in tree-length because Chepoi,Dragan, Estellon, Habib, and Vax`es found a link between δ -hyperbolicity andtree-length. 31igure 4: An example of a graph with bounded hyperbolic curva-ture and unbounded tree-length as the graph grows in the obviousway. It also has positive curvature by a 2-manifold measure andunbalanced values of inertia and demand. Theorem 3.5 ([26]) . A δ -hyperbolic graph G has tree-length at most (cid:15) G + 1 .Furthermore, a graph with tree-length at most D is D -hyperbolic. Their methods involve the layering tree, which was formally stated inlater works.
Corollary 3.6 ([35]) . If G has a layering tree T , where D T = max T ( x )= T ( y ) d ( x, y ) ,then the tree-length of G is at most D T + 1 . Because (cid:15) G depends on both δ and n , this result would imply that tree-length is stronger than hyperbolicity. Indeed, we present a construction ofa family of graphs in Figure 4 where the hyperbolicity is bounded but thetree-length is unbounded. The family of graphs demonstrated by Figure 4are called ringed trees and are known to have constant hyperbolicity [24, 38].We will prove in the following claim that the tree-length grows at a linearrate as the number of rings, or levels, in the graph grows. Because each ringgrows the number of vertices exponentially, we see that the tree-length growslike Θ(log( n )). Therefore the results of Chepoi et. al. comparing tree-lengthto hyperbolicity are tight. First, we need a lemma. Lemma 3.7 ([33], Lemma 5) . Let G and G be connected subgraphs of , and let ( T, φ ) be a tree decomposition of G . Let A = ∪ u ∈ V ( G ) φ ( u ) and B = ∪ v ∈ V ( G ) φ ( v ) . Then there exists a vertex x ∈ V ( T ) such that • x ∈ A ∩ B , or • every path in G from a vertex in G to a vertex in G crosses a vertexin φ − ( x ) . Using this lemma, the result is now easy.
Claim 3.8.
Let S be the set of vertices in the (cid:96) level of a ringed tree. Orderthe vertices along the cycle as u , u , . . . , u (cid:96) . Then in any tree-decomposition ( T, φ ) of the graph, there exists a w ∈ V ( T ) such that there exist vertices u i , u j ∈ φ − ( w ) with i < j where min { j − i, (2 (cid:96) + i − j ) } ≥ (cid:96) − .Proof. Let G be the path defined by u , u , . . . , u (cid:96) − and G be the pathdefined by u (cid:96) − , u (cid:96) − +1 , . . . , u (cid:96) − ) . Let our vertex w ∈ V ( T ) be the bag x guaranteed by Lemma 3.7. If w ∈ A ∩ B , then we are done. Otherwise, φ − ( w ) contains a vertex in the set u (cid:96) − +1 , . . . , u (cid:96) − − and a vertex in theset u (cid:96) − )+1 , . . . , u (cid:96) .While we know that the tree-length of a graph is then bounded aboveby the layering tree and bounded from below by the hyperbolicity, the issueof determining its exact value remains open. Dourisboure and Gavoille [33]proved that tl ( G ) ≥ ( D − /
3, where D is the term defined in Claim 3.1. Bythe results in the previous section, it follows that tl ( G ) is a 3-approximationof γ G . Lokshtanov [54] showed that finding the optimal tree-length is NP-complete, which places an emphasis on constructing approximation algo-rithms for tree-length. Dourisboure and Gavoille attempted to craft an algo-rithm that would be a 2-approximation.First, we will describe the algorithm, which we will call ( k, (cid:96) ) -Disk Tree .The algorithm progresses iteratively, let ( T i , φ i ) denote the tree constructedat the end of stage i (we begin with T = ∅ ). During stage i , we grow T i − into T i by adding one vertex and staying constant on the existing vertices (inother words, φ − i | V ( T i − ) = φ − i − ). Let H i be the graph induced on the vertexset φ − ( T i ), and let S i be the unique vertex defined by T i − T i − . At eachstage i the algorithm maintains the property P i , which is that33 if x and y are in H i and there exists a path from x to y in G − H i , then d ( x, y ) ≤ (cid:96) , and • for each connected component C of the graph G − H i there exists avertex w ∈ V ( T ) such that ∪ u ∈ C ( N ( u ) ∩ H i ) ⊆ φ − ( w ).To do so, we choose some connected component C of the graph G − H i − ,and let x i ∈ N ( C ) ∩ H i − (if i = 0, then let x i be any vertex of G chosenrandomly). We originally set S i = ( N ( C ) ∩ H i − ) ∪ ( B k ( x i ) − H i − ), andthen iteratively remove vertices from S i ∩ B k ( x i ) until property P i is true. If S i ⊆ H i − , then repeat stage i with a different pair ( C, x i ).Dourisboure and Gavoille [33] prove that when k = (cid:96) , each ( T i , φ i ) is adistance approximating tree of H i such that max φ i ( x ) ∩ φ i ( y ) (cid:54) = ∅ d G ( x, y ) ≤ k .Furthermore, they showed that if k = (cid:96) ≥ tl ( G ) −
2, then there existsa r such that H r = G , and so ( k, (cid:96) )-Disk Tree is a 6-approximation algo-rithm for constructing a minimum-length tree decomposition. Dourisboureand Gavoille [33] conjectured that with a refinement, the ( k, (cid:96) )-Disk Treealgorithm can be augmented into a 2-approximation algorithm. Conjecture 3.9 ([33], conjecture 1) . Suppose that k = (cid:96) = tl ( G ) , and when-ever there is an option between removing z or y while iteratively removingvertices from S i ∩ B k ( x ) until T i satisfies P i , we always choose to remove z if d ( z, x i ) > d ( y, x i ) . Under these conditions, there exits an r where H r = G . Unfortunately, the conjecture is not true. In a tree decomposition eachset φ − ( w ) is a separating set when w is not a leaf in T . Sometimes theseparating sets of a graph form long and thin subgraphs of G , instead offorming successively smaller balls as is suggested in the conjecture. This isexactly the situation in the following counterexample. Theorem 3.10.
Let G r,(cid:96) be the Cartesian product of a cycle C (cid:96) and a path P r(cid:96) for r > . The tree length of G r,(cid:96) is (cid:98) (cid:96) (cid:99) + 1 , but the k -disk-tree algorithmfails to terminate when k < (cid:96) − .Proof. Consider the Cartesian product of a cycle and a path - that is V ( G ) = { u i,j : 1 ≤ i ≤ (cid:96), ≤ j ≤ r(cid:96) } and E ( G ) = { x a,b x a,c : | b − c | = 1 } ∪ { x a,c x b,c : | b − a | ≡ mod (cid:96) ) } . We see that tl ( G ) ≤ (cid:98) (cid:96)/ (cid:99) + 1 by constructing a path34ecomposition P = w , . . . , w (cid:96) − where φ − ( w i ) = { x a,b : i ≤ b ≤ i + 1 } . Onthe other hand, this graph contains a subgraph H that is isometric to an (cid:98) (cid:96)/ (cid:99) by (cid:98) (cid:96)/ (cid:99) + 1 Euclidean grid, whose tree-length is known to be (cid:98) (cid:96)/ (cid:99) ([33], Theorem 3). Any tree-decomposition of G induces a tree-decompositionof H , and since the subgraph is isometric we see that the tree length of G isexactly (cid:98) (cid:96)/ (cid:99) .Now suppose we try to run ( h, h )-Disk Tree on G , where h = (cid:96) − (cid:96) . This radius is large enough to separate the graph, there exists a“left” and “right” H i (draw the copies of C (cid:96) vertically and the copies of P r(cid:96) horizontally). However, the diamond shape is a problem: the diameter of theboundary of the ball is greater than (cid:96) . The algorithm iteratively removespoints from the boundary of the ball until the diameter of the boundary isat most h . But it chooses the vertex to remove based on distance from thecenter: so the boundary never flattens into a column, but always maintainstwo points that are (cid:96)/ P r(cid:96) , andthe two components H i merge into one. With the two components merged,the ball continues to shrink until it has radius ( (cid:96) − /
2. The algorithm nowstalls: the next ball falls will shrink using the same logic until is is a subsetof the original ball.There is a parameter that is even stronger than tree-length. Let λ ( G )denote the length of the longest induced cycle in G (in the literature G iscalled k -chordal, where k ≥ λ ( G )). Chepoi and Dragan [25] prove that thereexists a distance approximating tree T of G such that for all x, y ∈ G , wehave that | d ( x, y ) − d ( T ( x ) , T ( y )) | ≤ λ ( G ) /
2. Furthermore, their methodproduces a layering tree as their distance approximating tree. By Corollary3.6, this implies tl ( G ) ≤ λ ( G ) /
2, and it follows from Theorem 3.5 that δ ≤ λ ( G ) /
2. Wu and Zhang [63] gave a direct proof that δ ≤ λ ( G ) / δ = (cid:100) λ ( G ) − (cid:101) . The converse is not true: thereexist graphs with large induced cycles and small tree-length. For example,consider a cycle C n = { u , . . . , u n } plus vertices { w , . . . , w n/ } such that N ( w i ) = { u i , u n − i } . This graph has a path decomposition with boundedlength while the length of the cycle tends to infinity. Wu and Zhang givea different example: for any graph G , create a graph G (cid:48) = G + v , where35 ( v ) = V ( G ). If G is not a tree, then tl ( G (cid:48) ) ≤ λ ( G (cid:48) ) = λ ( G ).Dourisboure and Gavoille [33] proved that outerplanar graphs have tree-length at most (cid:100) λ ( G ) / (cid:101) . They conjectured that this stronger bound extendsto all graphs. This would be sharp, as Dourisboure and Gavoille show that C n has tree-length (cid:100) n/ (cid:101) . Any (partial) results towards this conjecture wouldbe very interesting. Based on Theorem 2.2, calculating the value of δ in δ -hyperbolicity can bedone in O ( n ) time, and a 2-approximation can be made in O ( n ) time.Fournier, Ismail, and Vigneron [39] use matrix multiplication methods tocalculate the exact answer in O ( n . ) time, and a 2-approximation can bemade in O ( n . ) time. They also argue that ˆ δ ≤ γ G (see section 2.3 oftheir work). Using this argument, they provide an algorithm to approximate δ : calculate Gromov’s distance approximating tree T , and estimate ˆ δ to bemax a,b d ( a, b ) − d ( T ( a ) , T ( b )). Their analysis is that their algorithm runs in O ( n ) time and has a multiplicative error of at most 2 log ( n ).The layering tree constructed in [25] can be computed in O ( | E | ) time(see Remark 1 of [25]). To extend their algorithm to construct Gromov’stree, we only need to add a step to include the Steiner points. Their algo-rithm already constructs the vertex set of H in Theorem 2.16; to enhancetheir algorithm into finding Gromov’s tree only requires adding edges to H and finding the components. Let E (cid:96) be the set of edges to be addedto H in the (cid:96) step, so adding edges and finding components in step (cid:96) willtake O ( | E (cid:96) | log( E (cid:96) )) time. Because (cid:80) (cid:96) | E (cid:96) | < | E | , Gromov’s tree can becomputed in O ( | E | log( | E | )) time. This may be significantly faster than the O ( n ) time reported in [28] (see Section 1.2) and [2]. Claim 3.1 implies thatmax a,b d ( a, b ) − d ( T ( a ) , T ( b )) = max T ( a )= T ( b ) d ( a, b ) = D , reducing the num-ber of pairs that need to be examined. Note that for fixed a , calculating d a = max T ( a )= T ( b ) d ( a, b ) can be done by a breadth-first search tree. A sim-ple argument shows that max a ∈ A d a is a 2-approximation factor of D when T ( A ) = T ( V ( G )): if T ( a ) = y ∈ V ( T ), then by the triangle inequality2 d a ≥ max T ( b )= T ( c )= y d ( b, c ). Therefore this algorithm can be modified to an36 BCZ
Figure 5: The vertex list (
B, A, C, Z ) witness that this graph hasˆ δ = 8 /
2. When Z is removed, this graph satisfies ˆ δ = 7 / O ( | E | log( | E | )) time and has a multiplicative error ofat most 4 log ( n ).Cohen, Coudert, and Lancin [30] give an exact algorithm with heuristicimprovements by using a series of theoretical results to streamline the searchfor good quadruples of vertices. Note that their definition of hyperbolicitycomes from the Four Points condition (Definition 2.2). For a quadruple of ver-tices { u , u , u , u } , let S = d ( u , u ) + d ( u , u ), S = d ( u , u ) + d ( u , u ),and S = d ( u , u ) + d ( u , u ). If L is the largest value of the S i and L isthe second largest, then 2 δ = max u ,u ,u ,u L − L .One of the improvements involves clique separators. A separator of agraph is a vertex set B such that the graph G − B is disconnected. Cohen,Coudert, and Lancin [30] claim that if G [ B ] is a complete graph, δ >
1, andthe vertex list { u , u , u , u } witness the maximum of δ , then { u , u , u , u } can not be incident with two separate components of G − B . This is false.A counterexample is given if Figure 5. They use this result to pre-processthe graph by removing simplicial vertices; the reader should note that ourcounterexample involves a simplicial vertex. This claim is repeated by thesame group in 2015 [31] and is used by Abu-Ata and Dragan [1].A weaker condition is true, that would allow one to use clique separatorsto gain an approximate solution. We present this below. Borassi, Coudert,Crencenzi, and Marino [15] build off of the work of Cohen, Coudert, andLancin, but fortunately we did not find the claim in question utilized by theircode. Borassi, Coudert, Crencenzi, and Marino do restrict their attention to37locks (maximal 2-connected subgraphs). Because blocks are separated bya subgraph with diameter zero (a single vertex), our theorem justifies thisaction. Theorem 4.1.
Let B be a separator for a a graph G , where the diameter of G [ B ] is k . Let G i be component i of G − B with B added back in. Let ˆ δ i bethe four-points hyperbolic curvature of G i ( δ = δ ). Under these conditions,the four-points hyperbolic curvature of G is at most k + max ˆ δ i .Proof. The proof of this Theorem will use the version of hyperbolicity in Def-inition 2.2. We will show that for every quadruple of vertices { u , u , u , u } in G , there is some quadruple of vertices in G i for some i with a hyperbolicvalue at most k worse. In cases 1 and 2 below, we will actually show that L − L ≤ k for the given set of points. Because ˆ δ i ≥ i , this will besufficient.Let t i denote the distance between vertex u i and the closest vertex of B (if u i is in B , then t i = 0). By the triangle inequality, we see that d ( u i , u j ) ≤ t i + t j + k for all i and j . Let T = t + t + t + t , so that the aboveimplies S (cid:96) ≤ T + 2 k for all (cid:96) . We proceed with case analysis, based on how { u , u , u , u } is distributed among the components of G − B . Note that if u j and u i are in different components of G − B then d ( u i , u j ) ≥ t i + t j . Weassume that no component contains all four points, or else the bound max δ i clearly applies. CASE ONE:
No component of G − B contains more than one vertex from { u , u , u , u } . Under these conditions, it is clear that T ≤ S (cid:96) ≤ T + 2 k forall (cid:96) . CASE TWO:
Some component G i contains exactly two vertices of the list { u , u , u , u } . Without loss of generality, assume that { u , u } are the samecomponent (we do not distinguish when { u , u } are in the same component).Following the logic from the previous case, we know that T ≤ S , S ≤ T +2 k .Note that we do not have a lower bound on S anymore, but it is still truethat S ≤ T + 2 k . If S < T , then { L , L } = { S , S } , and the value of S will be irrelevant. Therefore we are done with this case. CASE THREE:
Some component G i contains exactly three vertices of thelist { u , u , u , u } . (Note that this is where Cohen, Coudert, and Lancin38ade their mistake). Without loss of generality, assume that { u , u , u } arein G (cid:96) and u is in G (cid:96) (cid:48) .Consider the tuple { u , u , u , b } for some arbitrary vertex b chosen in B .Using the above logic, we know that t i + k + t ≥ d ( u i , u ) ≥ t i + t and t i + k ≥ d ( u i , b ) ≥ t i . This implies that t + k ≥ d ( u i , u ) − d ( u i , b ) ≥ t − k .Let S (cid:48) i be the hyperbolicity parameters for this new quadtuple. We claim that | ( S i − S (cid:48) i ) − ( S j − S (cid:48) j ) | ≤ k for all i and j , which would conclude the proofof the theorem. Without loss of generality, let us bound | S − S (cid:48) − S + S (cid:48) | ,which by symmetry will bound δ overall.We have that | S − S (cid:48) − S + S (cid:48) | = | d ( u , u ) + d ( u , u ) − ( d ( u , b ) + d ( u , u )) − ( d ( u , u ) + d ( u , u )) + ( d ( u , u ) + d ( u , b )) | = | d ( u , u ) − d ( u , b ) − d ( u , u ) + d ( u , b ) | = | d ( u , u ) − d ( u , b ) − ( d ( u , u ) − d ( u , b )) |≤ k. The other improvement that one can make is to choose points such that d ( u , u ) ≥ d ( u , u ) ≥ k for some large k . It was independently shownby Rodr´ıguez, Sigarreta, Vilare, and Villeta [60] that the hyperbolicity of agraph is at most half the diameter. Cohen, Coudert, and Lancin extend thisby proving that if L = S , then L − L ≤ min( d ( u , u ) , d ( u , u )). Theiralgorithm then orders pairs of vertices by the distance between them andsaves time by using the above criteria to terminate early. We can strengthenthis idea for unweighted graphs in two ways. First, instead of just claimingthat d ( u , u ) and d ( u , u ) are large, we can claim that the values are max-imal based on some local criteria. This criteria can be verified for all pairsinvolving one fixed vertex v by a breadth-first-search tree rooted at v .During the preparation of this manuscript, Borassi, Coudert, Crencenzi,and Marino [15] independently published the results of Theorem 4.2 and 4.3.Those results were used to construct an algorithm which has been indepen-dently verified [14] as being fast enough for most practical purposes, and39heir code is publicly available. Theorem 4.2.
In the following statement, the arithmetic on the indexes isdone modulo . Let G be δ -hyperbolic. There exists four points u , u , u , u such that d ( u , u )+ d ( u , u ) = max (( d ( u , u ) + d ( u , u )) , ( d ( u , u ) + d ( u , u )))+2 δ, and for all i and all w ∈ N ( u i ) , we have that d ( w, u i +2 ) ≤ d ( u i , u i +2 ) .Proof. Let u , u , u , u be chosen so that d ( u , u )+ d ( u , u ) − max (( d ( u , u ) + d ( u , u )) , ( d ( u , u ) + d ( u , u ))) = 2 δ, and among all such sets of points, d ( u , u ) + d ( u , u ) is maximum. We willprove the claim by contradiction. By symmetry, assume that there exists a w ∈ N ( u ) such that d ( w, u ) = d ( u , u ) + 1. Because G is unweighted, itmust be that d ( u , u ) ≤ d ( w, u ) + 1 and d ( u , u ) ≤ d ( w, u ) + 1. Because G is δ -hyperbolic,2 δ ≥ d ( b, u ) + d ( u , u ) − max (( d ( b, u ) + d ( u , u )) , ( d ( b, u ) + d ( u , u ))) ≥ d ( u , u )+ d ( u , u ) − max (( d ( u , u ) + d ( u , u )) , ( d ( u , u ) + d ( u , u ))) = 2 δ. This contradicts the maximality of d ( u , u ) + d ( u , u ) < d ( b, u ) + d ( u , u ).The second method to improve the selection criteria is that all four pointsshould be far apart from each other, not just from their partner along thelarge diagonals. Theorem 4.3. If G is δ -hyperbolic, then δ ≤ max u ,u ,u ,u ∈ V ( G ) min i (cid:54) = j d ( u i , u j ) . Proof.
By symmetry, we will show that L − L ≤ d ( u , u ). Case 1: L = S = d ( u , u ) + d ( u , u ). In this case, L ≥ S = d ( u , u )+ d ( u , u ). By the triangle inequality, d ( u , u ) ≤ d ( u , u )+ d ( u , u )+ d ( u , u ) = S + d ( u , u ). Therefore L − L ≤ d ( u , u ).40 ase 2: L (cid:54) = S . By symmetry, assume that L = S , and so L ≥ S = d ( u , u ) + d ( u , u ). By the triangle inequality, d ( u , u ) ≤ d ( u , u ) + d ( u , u ) and d ( u , u ) ≤ d ( u , u ) + d ( u , u ). Therefore L − L ≤ d ( u , u ). Because a clique is 0-hyperbolic, clearly any connection between hyperbol-icity and congestion will require additional assumptions. Li and Tucci [52]give a beautiful proof of large demand for locally finite graphs using ran-domized methods: if each ball of radius 2 δ has at most M points and D isthe diameter of the graph, then there exists a ball B of radius 2 δ such that (cid:80) w ∈ B D ( w ) ≥ ( nD − M − ) . For graphs with slowly growing diameter (i.e.diam( G ) ≈ log( n ) is common), this result is just short of proving congestion.Baryshnikov and Tucci [9, 10] approached the problem of congestion usingthe boundary of a hyperbolic space, which is defined on the behavior ofinfinite geodesics. Their approaches involved several unstated assumptions,such as the graph being a group action on the Poincar´e disk [9] or every vertexbeing a part of an infinite ray [10]. These assumptions are necessary to theirtechniques. For example, one could begin with an infinite group action of thehyperbolic surface G that satisfies their assumptions and then choose a basepoint x and a ray γ = x , x , . . . . Augment G into a graph G (cid:48) as follows: foreach i add leaves y i, , . . . , y i,f ( i ) whose only neighbor is x i , where f is somefunction that grows asymptotically faster than | B i ( x ) | . The conclusions ofBaryshnikov and Tucci fail on G (cid:48) . This is a serious flaw - most real worldnetworks have many, many leaves!Next we will consider congestion under the uniform traffic rate, and in-vestigate how the congestion grows relative to the rate at which the graphgrows. Having a set S of vertices with Ω( n ) congestion suggests that thereis a constant probability p u = O (1) for each vertex in u ∈ S that the shortestpath between two randomly chosen vertices x and y will cross u . We willaim for something easier: there exists a “small” set S such that there is aconstant probability p S = O (1) that the shortest path between two randomlychosen vertices x and y will cross at least one vertex in S . Jonckheere, Lou,41onahon, and Baryshnikov [47] show that in hyperbolic space, such an S canbe chosen as a ball around the origin with finite radius. This will be discussedmore in Section 8. We will show that this is also true of a hyperbolic graphwith large but finite radius and an assumption on balance. Note that in agraph, we have no bounds on how large a ball is, even if it has finite radius.Our work will focus on showing that there is a single vertex that is withina close distance of the shortest path between almost any pair of vertices.That hyperbolic spaces have congestion has been implied as “intuitively”true because of results relating to exponential divergence and the MorseLemma. Albert, DasGupta, and Mobasheri [3] stated two theorems directlyalong these lines. We will begin with a result in this flavor: we will show oureventual result is true for pairs of points selected one each from two fixeddisjoint balls, where the center of congestion is a carefully chosen point onthe geodesic between the centers of the balls. For the rest of this section, wewill be working with the thin triangle definition of hyperbolicity. Lemma 5.1.
Let G be a space with ˆ δ -thin triangles. Let u, x, v, y be verticessuch that d ( u, x ) , d ( v, y ) < d ( u,v )2 , and let r be the midpoint on the shortestpath from u to v . Under these conditions, there exists a vertex w on theshortest path from x to y such that d ( r, w ) ≤ δ .Proof. Let u, x, v, y be as above, and let P a,b denote a shortest path betweenvertices a and b . By the triangle inequality, d ( x, v ) , d ( y, u ) > d ( u,v )2 .Consider the triangle formed by shortest paths between vertices u, v, x ,where r is on P u,v . By the thin triangles condition, r is within distanceˆ δ to a vertex r (cid:48) on P u,x ∪ P v,x . Furthermore, if r (cid:48) ∈ P u,x then d ( u, r (cid:48) ) = d ( u, r ) = d ( u, v ) / r (cid:48) ∈ P v,x , then d ( v, r (cid:48) ) = d ( v, r ) = d ( u, v ) /
2. Be-cause d ( u, x ) < d ( u, v ) / < d ( v, x ), it must be that r (cid:48) ∈ P v,x and d ( v, r (cid:48) ) = d ( u, v ) / v, x, y , where r (cid:48) ∈ P v,x . By the thin triangle condition, r (cid:48) is within distanceˆ δ of a vertex r (cid:48)(cid:48) ∈ P x,y ∪ P v,y . Furthermore, if r (cid:48)(cid:48) ∈ P v,y , then d ( v, r (cid:48)(cid:48) ) = d ( v, r (cid:48) ) = d ( u, v ) /
2, which is a contradiction because d ( v, y ) < d ( u, v ) /
2. So r (cid:48)(cid:48) ∈ P x,y . 42e are now done: set w := r (cid:48)(cid:48) , and by the triangle inequality d ( r, w ) ≤ d ( r, r (cid:48) ) + d ( r (cid:48) , r (cid:48)(cid:48) ) ≤ δ . Corollary 5.2.
Let u and v be arbitrary vertices in a graph with ˆ δ -thintriangles. For any fixed λ , there exists a ball B of radius δ such that forevery pair of vertices x and y that satisfy d ( x, u ) < λd ( u, v ) and d ( y, v ) < (1 − λ ) d ( u, v ) , then any shortest path from x to y crosses B .Proof. By Theorem 4.1, we can add two paths, P x,x (cid:48) and P y,y (cid:48) , to G to create G (cid:48) , where G (cid:48) = G ∪ P x,x (cid:48) ∪ P y,y (cid:48) , P x,x (cid:48) ∩ G = { x } , P y,y (cid:48) ∩ G = { y } , and G (cid:48) hasthe same δ -hyperbolicity as G (it is clear that it has the same thin trianglesproperty as well). Let B x be the ball in G with radius λd ( u, v ) centered at x and let B y be the ball in G with radius (1 − λ ) d ( u, v ). Let a = | P x,x (cid:48) | , so thatthe ball of radius a + λd ( u, v ) in G (cid:48) centered at x (cid:48) is B x ∪ P x,x (cid:48) . Similarly,let b = | P y,y (cid:48) | , so that the ball of radius b + λd ( u, v ) in G (cid:48) centered at y (cid:48) is B y ∪ P y,y (cid:48) .Choose a and b so that a − b = d ( u, v )(1 − λ ). Then the midpoint r between x (cid:48) and y (cid:48) is λd ( u, v ) away from x and (1 − λ ) d ( u, v ) away from y ,and so we can apply Lemma 5.1.One might hope that if we choose u and v correctly, then each ball ofradius d ( u, v ) / u and v would contain some constant proportionof all vertices. This assumption is far from true in high-dimensional space,especially when the high-dimensional space is hyperbolic. This demonstratesthe issue with a quasi-geodesic approach to this problem: there is no clearchoice of a singular geodesic that should act as our “center” that capturesthe behavior of the whole space. We will show that the choice of the geodesicdoes not matter too much. However, to prove that we will need to follow adifferent line of thought. What follows more closely relates to the applicationof a visual metric to the boundary of an infinite hyperbolic space. We willgive an argument for finite discrete spaces that is an analogue of finding pairsof points whose distance in the visual metric is close to 1.In order to find such pairs of vertices, we will give a series of “half-space”results. Let α be a large geodesic from u to v , r be the midpoint of α , and2 f α ( z ) = d ( z, v ) − d ( z, u ). In this language, we think of α as a straight linethrough our graph, r is our origin ( f α ( r ) = 0), and f α ( z ) is the (index of43he) projection of the vertex z onto α . Bowditch [16] gave three definitionsof a projection onto a geodesic. This projection is similar to definition (P2)in description, and will be similar to definition (P1) in use. The standarddefinition, the vertex in α that minimizes the distance to z , is definition(P3). Fortunately, we do not need to be confused: Bowditch went on to provethat all three definitions agree up to a bounded distance. That bound is tooimprecise for our use, but hopefully knowledge of this result will settle yourintuition about how a projection should behave. We continue this terminologyto say that H k = { z : 2 f α ( z ) = k } is the analogue of a hyperplane for eachfixed k . It should be clear that if 2 | k | < d ( u, v ), then H k ∪ H k +1 is a separatingset, and we define a half-space as the set of points z such that f α ( z ) > Lemma 5.3.
Let u, x, v, y be vertices such that α is a geodesic from u to v , f α ( y ) > , r is the midpoint of α , and d ( u, x ) < d ( u, v ) . Under these condi-tions, the shortest path from x to y includes a vertex w such that d ( r, w ) ≤ δ .Proof. Let u, x, v, y be as above, and let P a,b denote a shortest path betweenvertices a and b . Specifically let α = P u,v = z , z , . . . , z d ( u,v ) , where z = u , z d ( u,v ) / = r , and z d ( u,v ) = v . Consider the triangle formed by the vertices u, v, y . By the thin triangles condition, r is within distance ˆ δ to a vertex r (cid:48) on P u,y ∪ P v,y . By assumption f α ( y ) >
0, so d ( u, y ) > d ( y, v ), so d ( u, r ) < ( v.y ) u , and therefore r (cid:48) ∈ P u,y . The thin triangles condition also implies that d ( u, r (cid:48) ) = d ( u, r ) = d ( u, v ) / u, x, y . By the thin triangle condition, r (cid:48) is within distance ˆ δ to a vertex r (cid:48)(cid:48) on P x,y ∪ P u,x . If r (cid:48)(cid:48) ∈ P x,y , then set w = r (cid:48)(cid:48) and we are done. This is true if ( x.y ) u < d ( u, r (cid:48) ) = d ( u, v ) /
2. By thetriangle inequality, d ( x, y ) ≥ d ( u, y ) − d ( u, x ), so ( x.y ) u ≤ d ( u, x ), which isless than d ( u, v ) / α , all origins from sufficiently long lines areroughly the same. Lemma 5.4.
Let G be δ -hyperbolic and have ˆ δ -thin triangles. Let u, v and x, y be four points such that d ( u, v ) = diam( G ) and d ( x, y ) ≥ diam( G ) − t .Let r be the midpoint between u and v and r (cid:48) the midpoint between x and y .Under these conditions, d ( r, r (cid:48) ) ≤ t + 4ˆ δ + 2 δ . roof. By symmetry on u and v and the thin triangle property, the shortestpath from x to u contains a vertex s such that d ( r, s ) ≤ δ . Because s is ona shortest path from x to u , it follows that d ( x, s ) = d ( x, u ) − d ( u, s ). By thetriangle inequality, d ( s, u ) ≥ d ( r, u ) − d ( s, r ). Putting all of this together, d ( x, r ) ≤ d ( x, s ) + d ( s, r ) ≤ d ( x, u ) − d ( u, s ) + d ( s, r ) ≤ d ( x, u ) − d ( r, u ) + 2 d ( s, r ) ≤
12 diam( G ) + 4ˆ δ. Using the same argument as above, we see that d ( y, r ) ≤ diam( G ) + 4ˆ δ also.We will now apply the four points condition to r, r (cid:48) , x, y . The two inequal-ities are all that we need to bound two of the three terms:max { d ( x, r ) + d ( y, r (cid:48) ) , d ( x, r (cid:48) ) + d ( y, r ) } ≤ diam( G ) + 4ˆ δ − t . By the four points condition, we see that d ( x, y ) + d ( r, r (cid:48) ) ≤ diam( G ) + 4ˆ δ +2 δ − t .This result alone creates an interesting implication in Theorem 5.5. Forthe rest of this section, r is a fixed midpoint on a geodesic of length diam( G ).Note that Theorem 5.5 is unique to hyperbolic spaces: it clearly fails in S k for any k . Theorem 5.5.
Let G be δ -hyperbolic and have ˆ δ -thin triangles.There exists a vertex r ∈ V ( G ) such that for any two vertices x and y that lie on a geodesic from u to w where d ( u, w ) > max { diam( G ) − t, d ( x, u ) , d ( y, w ) } , we have that the shortest path from x to y containsa vertex r (cid:48) with d ( r, r (cid:48) ) ≤ t + 4ˆ δ + 2 δ . Furthermore, our arguments above are constructive and can be used tofind r , the center of the congestion. The results of [27] combined with Lemma5.4 imply that a vertex near r can be found in linear time in the number ofedges. It should be noted that artificial networks can be designed such that45he center of congestion is skewed away from the vertex r that our methodswill identify as the center of congestion. For example, consider the tree T ( k, (cid:96) )with vertex set { r (cid:48) , u , . . . , u k , v , . . . , v (cid:96) } , where each u i is a leaf adjacent to r (cid:48) and r (cid:48) , v , . . . , v (cid:96) is a path. Our method reports that r = v ( (cid:96) − / is thecenter of congestion, but if k >> (cid:96) then r (cid:48) is the center of congestion. Thisexample also illustrates that our theorem is tight: T ( k, (cid:96) ) is 0-hyperbolic,the vast majority of vertices are in paths of length (diam( G ) + 1 − (cid:96) ) (so z = (cid:96) + 2), and d ( r, r (cid:48) ) = ( (cid:96) − / maximal if it is not a proper subset of another geodesic. Definition 5.6.
A hyperbolic network G with n points and center r is ( a, b, c ) -balanced if1. for every maximal geodesic α of length at least diam( G ) − a , both half-spaces defined by α contain at least cn points each, and2. if n k = |{ a : d ( a, r ) = k }| , then cb k ≤ n k ≤ c − b k for some b > and k ≤ diam( G ) . Both criteria are met by at least one probabilistic model for hyperbolicnetworks [50]. The first criteria is quite necessary, as we saw above. Thesecond criteria is slightly stronger than needed, but we have chosen to stateit in a way that is generally agreed to be true for a natural hyperbolic network.What we need is actually that some positive proportion of all points lie onthe boundary. Specifically, the criteria as listed implies that if k ≥ b (2 /c )46nd d := diam( G ) /
2, then |{ a : d ( a, r ) > d − k }| /n ≥ c ( b d +1 − − c − ( b d − k +1 − c − ( b d +1 − ≥ b d +1 ( c − ( b k c ) − ) c − ( b d +1 − ≥ cb d +1 c − ( b d +1 − ≥ c / Lemma 5.7. If x is a point such that d ( x, r ) ≥ diam( G ) / − t inside a space G with ˆ δ -thin triangles, then x is in a geodesic that is at least diam( G ) − t − ˆ δ long.Proof. By definition, r is the midpoint on a geodesic between vertices u and v such that d ( u, v ) = diam( G ). Consider the thin triangle condition for thetriangle xuv , and by symmetry assume that r is within distance ˆ δ of a vertex r (cid:48) on the geodesic from x to v . Because r (cid:48) is on the geodesic, we know that d ( x, v ) = d ( x, r (cid:48) ) + d ( r (cid:48) , v ). By the definition of the thin triangles condition,diam( G ) / d ( v, r ) = d ( v, r (cid:48) ). By the triangle inequality, d ( x, r (cid:48) ) ≥ d ( x, r ) − d ( r, r (cid:48) ) ≥ diam( G ) / − t − ˆ δ . Remark 5.8.
Lemma 5.7 is unique to hyperbolic spaces. Let X be the spacedefined by the upper half of a circle of radius R in -dimensional space.Suppose we put the standard Euclidean geometry on this space. The longestgeodesic in X is from the point ( − R, to the point ( R, with a center atthe origin. So every point p = ( x, y ) such that x + y = R and y ≥ is in X and satisfies d ( p, (0 , X ) / . However, the longest geodesic in X that contains the point (0 , R ) has length √ R = diam( G ) / √ . Theorem 5.9.
Suppose that G is an (ˆ δ + 2 log b (2 /c ) , b, c ) -balanced networkthat is δ -hyperbolic and has ˆ δ -thin triangles. Under these conditions if diam( G ) > δ + 6 log b (2 /c ) , then there exists a vertex r and a set of c n pairs of ver-tices such that any geodesic between any pair of vertices in the set includes avertex w where d ( r, w ) ≤ log b (2 /c ) + 6 . δ + 2 δ ≤ log b (2 /c ) + min { δ, . δ } . oreover, if G is a ( a, b, c ) -balanced network with a ≥ Ω (log( c − ) + δ ) ,then there exists a ball of radius O (log( c − ) + δ ) whose demand sums to atleast Ω ( n ) .Proof. By our discussion above, G being (ˆ δ − b ( c ) , b, c )-balanced im-plies that there exists 3 c n/ x such that d ( x, r ) > diam( G ) / − b (2 /c ). By Lemma 5.7, this implies that x is in a geodesic α of length atleast diam( G ) − ˆ δ − b (2 /c ). Let β be a maximal geodesic that contains α ,and let r (cid:48) be the midpoint of β . By Lemma 5.4, d ( r (cid:48) , r ) ≤ log b (2 /c )+4 . δ +2 δ .Let β have endpoints w and z , and without loss of generality, assume that d ( x, w ) ≤ ˆ δ + 2 log b (2 /c ). By assumption on balance, the half-space definedby β that contains z has at least cn points. Moreover, because diam( G ) islarge, we have that2 d ( x, w ) ≤ δ + 4 log b (2 /c ) ≤ diam( G ) − ˆ δ − b (2 /c ) ≤ | α | ≤ | β | = d ( w, z ) , and so we may apply Lemma 5.3 to say that any geodesic from x to a vertex y in the half space contains a vertex within distance 2ˆ δ of r (cid:48) . By the triangleinequality, the geodesic from x to y is then within log b (2 /c ) + 6 . δ + 2 δ of r .We may have double counted our pairs in this fashion, so we divide by 2.To prove the “moreover” part, notice that if our diameter is not at least3ˆ δ + 6 log b (2 /c ), then we can just define our ball to be all of G . -manifold Jonckheere, Lou, Bonahon, and Baryshnikov ([47], Section 3) state two con-jectures, each with two parts. The purpose of this section and the next sectionis to answer all of them. The conjectures deal with two different concepts ofgraph curvature, and the conjectures relate them towards the amount of con-gestion in a network and the location of the highest amounts of congestionin a network. Both concepts can be thought of as variations of rotation. Thefirst concept involves rotational inertia, or just inertia, and we will investi-gate that concept in this section. The second concept attempts to identifythe center of a graph (implicitly identified as the point of highest congestion)48s the unique fixed point under multiple angles of reflective symmetry. Wewill discuss the second concept in the next section.Most of the conjectures involves locating the “center” of a graph (notto be confused with the technical definition of the center of a graph). Thecenter of a hyperbolic space is clearly the area of highest congestion, andJonckheere, Lou, Bonahon, and Baryshnikov rigorously prove this. The con-jectures are then attempts at comparing the center of a hyperbolic space toother definitions of centers - either minimizing the second moment of inertiaor a fixed point in several symmetries. The second moment of inertia of avertex w is defined to be (cid:80) v d ( w, v ) . The first sign that all of these def-initions of center may be distinct is that there is no given reason for whywe should consider the second moment instead of the first moment or theseventh moment.In the end, we conclude that these definitions give no more informationabout the network tomography than the set of vertex degrees, and the set ofvertex degrees give no information about the network tomography. The first two definitions require that the graph plus a set of closed diskscalled faces form a CW-complex isomorphic to an orientable 2-manifold. Let | f | denote the number of edges incident on the boundary of a given face, f .Let d ( v ) denote the degree of a vertex v , which is the number of edges that v is contained in.Let d G : V ( G ) × V ( G ) → R + be the standard graph distance withweighted edges (with weight w ) - also known as the hop distance . The gen-eral idea is to embed the vertex set into some space, and we will requirea distance metric d : V ( G ) × V ( G ) → R + that represents the distance inthe target space between the images of two vertices. We will use a weightfunction w : E ( G ) → R + , where each edge is given a weight based on thedistance provided by the mapping: w ( uv ) = d ( u, v ) for all uv ∈ E ( G ). If weassume that d ( u, v ) = 1 whenever uv ∈ E ( G ), then it is easy to see that d ( u, v ) ≤ d G ( u, v ) by the triangle inequality.49he measure of curvature is frequently based on angles. The main defini-tion of curvature in this section comes from Alexandrov Angles. Definition 6.1 (Alexandrov Angles) . For a given graph G and vertex v ∈ V ( G ) , let { u , u , . . . , u k } be the neighbors of v in cyclic order. The angle u i vu i +1 (where i is taken modulo k ) is defined to be α i = cos − (cid:18) d ( v, u i ) + d ( v, u i +1 ) − d ( u i , u i +1 ) d ( v, u i ) d ( v, u i +1 ) (cid:19) . Under these conditions, the curvature at a vertex v is k G ( v ) = 2 π − (cid:80) α i (cid:80) area ( vu i u i +1 ) , where the area of a triangle is defined using Heron’s formula: area ( abc ) = 14 (cid:112) ( a + b + c )( a + b − c )( b + c − a )( a + c − b ) . Remark 6.2.
If every edge has weight equal to and d ( u i , u i +1 ) = 1 for all i , then k G ( v ) = 2 π . (cid:18) d ( v ) − (cid:19) . As a comparison, a second definition of curvature comes from the genusof the space a graph embeds into, and from Euler’s formula. This seconddefinition was proven to not measure congestion well in [55]. Still, we willmention it in the notation of angles to compare definitions.
Definition 6.3 (Gaussian Curvature) . For a given graph G , the curvatureof a vertex v is k G ( v ) = 1 − . d ( v ) + (cid:88) v ∈ f | f | − . The curvature of the whole graph G is the sum of the curvatures of theindividual vertices in G . This equals the Euler Characteristic of the graph(which is the sum of the number of vertices plus the number of faces minusthe number of faces), which equals minus twice the genus of the -manifoldthe graph is embedded in. d = 7.Our solutions to the conjectures will involve the construction of severalgraphs. The metrics that we use on these graphs will all be the same: d ( u, v ) = (cid:100) d G ( u, v ) / (cid:101) , and the weight on each edge will be 1. One of the nice thingsabout this metric is that we will always be able to apply the formula inRemark 6.2.A graph is regular if every vertex has the same degree. Jonckheere, Lou,Bonahon, and Baryshnikov [47] generate a finite approximation of an infiniteregular graph with degree d using the following method.Start with a graph G such that d ( v ) ≤ d for all vertices v . The vertices u such that d ( u ) < d are called “boundary” vertices. Pick an arbitrary vertex w in the boundary and two neighbors u , u of w such that u wu is a pathon a face f and u , u are also on the boundary. Add k = d − d ( w ) vertices w , . . . w k , where the neighbors of w i are w i − , w i +1 , and w (let u = w and u = w k +1 ). We then have k + 1 new triangular faces w i w i +1 w , andinstead of u wu , the face f now has the sequence u w w . . . w k u . By carefulselection of the v , each vertex on the boundary should always have at least twoneighbors on the boundary. By maintaining this property, we can grow theinterior of the graph to unbounded size. We will also use this constructiontechnique. Using Remark 6.2, this construction will give us a graph withnegative curvature if d >
6, positive curvature if d <
6, and zero curvature if d = 6.In the next subsection, we will discuss some issues related to this construc-tion. Each subsection after that will state and answer one of the questionsposed in the conjectures. We will prove the first statement to be true; after51hat we will give counterexamples to all other statements. We assume thatthe graph curvature is based on Definition 6.1, unless stated otherwise. It is not immediately clear what the curvature of the whole graph is in Defi-nition 6.1. Context from Jonckheere, Lou, Bonahon, and Baryshnikov ([47],Section 4) implies that the curvature for the graph is a fixed constant, andthe curvature for each vertex should equal that constant value. This creates abit of a conflict with their desire to examine graphs with negative curvature.Jonckheere, Lou, Bonahon, and Baryshnikov ([47], Section 3) state con-jectures relating only to finite graphs and consider the simplified case wherethe 2-manifold is the plane. But if we assume that every pair of vertices in-duces at most one edge, then it is well known that the average degree of thevertices is strictly less than 6. If every face is a triangle and every edge hasweight equal to 1, then we may apply remark 6.2 to see that the curvatureof the graph must be positive!Clearly, one too many assumptions and simplifications have been madehere. Perhaps we should simply reduce the assumption that G is planar: afterall, other authors have not made this assumption previously. In this case, if g is the genus of the surface that the graph embeds into, we have that theaverage degree is 6 + g − | V ( G ) | .A second assumption we may want to discard is that the graph is finite.Jonckheere, Lou, Bonahon, and Baryshnikov ([47], Section 4) construct aplanar graph with negative curvature by making every face a triangle andevery vertex have degree d ≥
7. The problem is that they apply their iterativeconstruction finitely many times - which leaves them with a great plenty ofvertices with degree 3 or 4. Furthermore, certain parameters in their conjec-tures - such as the center of inertia - are not well-defined on infinite graphs.A third assumption we may wish to discard is that the curvature of thevertices is constant. This assumption is never explicitly mentioned. But theredoes not seem to be a natural method to determine what the curvature ofthe graph as a whole would be in this context. Part of the problem is that52f we continue to only consider finite planar graphs, then there will alwaysbe vertices with positive curvature. A stronger statement is also true: if thecurvature of the graph is the sum of the curvature on the vertices (as inDefinition 6.3), then every planar graph where every face is a triangle andevery edge has weight one will have positive curvature.Therefore we will always consider below that our graph is infinite or non-planar or both. We will attempt to cover the conjectures under all of theseconditions, sometimes also considering Definition 6.3 for curvature.
Conjecture 3.3.1 has two parts. We further separate them to simplify thearguments.
Conjecture 6.4 ([47], 3.3.1(a)) . Let G be a large but finite graph with neg-ative curvature. There are very few vertices with the highest demand. If we drop the assumption that our graph is finite, then our counterex-ample is an infinite hyperbolic tiling of the two-dimensional disk. Note thatthe demand is infinite in this situation. However, we argue that because thegraph is the same after an appropriate shift, then the demand across allvertices must be uniform.The second option is that our graph is not planar, and our second coun-terexample is a lexicographic product of a cycle and a clique. This graphhas vertex set { u i,j : 1 ≤ i ≤ k, ≤ j ≤ k (cid:48) } , with edge set { u i,j u i (cid:48) ,j (cid:48) : ( i = i (cid:48) ) or ( | i − i (cid:48) | = 1 and j = j (cid:48) ) } . This graph is vertex-transitive and finite, butnot planar. Furthermore, it satisfies both Definition 6.3 and 6.1 of negativecurvature.A third counterexample involves constructing a graph using the methodin Section 6.1. Begin by expanding the graph along one side so that it growslong and narrow. Then curl the graph around and connect the two ends. Thegraph now resembles a thick cycle, and the demand will be spread across aloop along that cycle. This process can be repeated as the graph grows sothat the demand will always be spread across a large loop.53 onjecture 6.5 ([47], 3.3.1(b)) . Let G be a large but finite graph with neg-ative curvature. The vertices with the highest demand are near the verticeswith the smallest inertia. We use the construction mentioned in Section 6.1 to generate the coun-terexample, but we choose the border vertices to expand in a very asym-metrical way. Our graph should be approximately ‘Y’ shaped. The boundaryvertices should be h apart (in other words, the letter is h vertices thick), thetwo short legs of the ‘Y’ are k vertices long (1 << h << k ), and the long legof the ‘Y’ is 3 k vertices long (all of this approximate because the graph isnot actually a grid). We will call the area around where the three legs meetthe base of the graph.We will first find the vertices with the lowest moment of inertia. Becauseour distance metric d is approximately proportional to the graph distance d G , without loss of generality we will interchangeably swap the two in thisdiscussion. Consider a vertex that is in the long leg of the graph and dis-tance i from the base of the graph. The moment of inertia for that vertex isapproximately 2 k + i (cid:88) j =1 ( hj ) − i (cid:88) j =1 ( hj ) + k − i (cid:88) j =1 ( hj ) . Recall that (cid:80) (cid:96)j =1 j = (cid:96) ( (cid:96) + 1)(2 (cid:96) + 1) /
6, so the above formula simplifies to h (cid:18) i (1 + 5 k ) − i ( k + 7 k ) + 5 k + 33 k + 58 k (cid:19) , which is minimized when i = ( k + 7 k ) / (2 + 10 k ) ≈ . k .We claim that the vertices with the highest demand will be right where allthree legs meet. Suppose v is distance i away from the base of the graph and inthe long leg, and i >
0. Consider the line of h vertices that separates the graphat that distance. Every shortest path that uses one of those separating vertexmust begin and end on opposite sides of that divide. That line separates h (3 k − i ) vertices from h (2 k + i ) vertices, meaning that the average vertex onthat line has demand roughly h (6 k + ik − i ). This maximizes when i = k/ . hk . Now consider what happens when i = 0. Thevertices in the base of the graph separate the long leg from the short legs,54ut they also separate the short legs from each other. Therefore these verticeshave demand 7 hk .Both of the neighborhoods with local maxima on demand are at least 0 . k away from where the vertices with lowest inertia are. Because k is unbound-edly large, we see that vertices with extreme values of inertia are infinitelyfar from vertices with the highest demand. Conjecture 6.6 ([47], 3.4.1) . Let G be a large but finite graph with positivecurvature. G has more balanced values for demand and inertia than a graphwith negative curvature. We have already discussed a graph with negative curvature with uniformdemand and inertia. Hence, this vaguely-defined inequality has already beencontradicted. We add emphasis to this statement by constructing a graphwith positive curvature whose values of demand and inertia are highly un-balanced.The standard example of a graph with positive curvature is a Euclideangrid. Our counterexample is a polar-coordinates analogue to the Euclideangrid, where the size of each ring grows exponentially. Specifically, our coun-terexample is an infinite ringed tree (see Figure 4 for an illustration). Notethat this example is degree 5 regular, so it has positive curvature accordingto Definition 6.1. Furthermore, it is planar, so it also has positive curvatureaccording to Definition 6.3. It is easy to see that with few exceptions, shortestpaths from the top hemisphere to the bottom hemisphere travels through thecenter of the graph, so the center experiences Θ( | V ( G ) | ) demand. However,ringed trees have curvature according to Definition 2.1.For an example that has positive curvature by all definitions in this paperwith unbalanced demand and inertia, consider a non-convex finite subset of aEuclidean grid. See Figure 7 for an illustration of this counterexample. Eachlarge convex subset of the graph violates any possible negative curvatureproperty, while the small degrees force positive curvature by definitions 6.1and 6.3. Furthermore, demand and inertia have extreme values at the smallbottle necks. 55igure 7: An example of a graph with positive curvature andunbalanced values of inertia and demand. There are two conjectures in this section,
Definition 7.1.
A graph isomorphism is a permutation f : V ( G ) → V ( G ) such that uv ∈ E ( G ) if and only if f ( u ) f ( v ) ∈ E ( G ) .The symmetric group of a graph G is the set of graph isomorphisms on G . The symmetry group is vertex-transitive if for every pair of vertices u, v ,there exists a graph isomorphism f such that f ( u ) = v . Conjecture 7.2 ([47], 3.4.2) . Let G be a large but finite graph with positivecurvature. If the graph has a vertex-transitive symmetry group, then both thedemand and the inertia are constant for all vertices. The truth of this statement has nothing to do with the curvature of thegraph. We can prove a much stronger statement below - both by avoiding theuse of curvature and generalizing the assumption about vertex-transitivity.
Theorem 7.3.
If there exists a graph isomorphism f such that u = f ( w ) ,then u and w have the same demand and inertia. Before we prove this statement, we will clarify one issue about the as-sumptions. The standard definition of a graph isomorphism implies certain56elationships about the graph topology, and only about the graph topology.However, in this case we are dealing with a second metric, d , which is onlyslightly restrained by the topology of the graph. If we do not enforce that d ( u, v ) = d ( f ( u ) , f ( v )) for f to be a graph isomorphism, then the conjectureis clearly false (consider any highly symmetric graph, set d = d G in all cases,except one pair of vertices where (cid:15) is subtracted). Therefore we will assumethat this additional property holds. Proof. If w, v , v , . . . , v k is a path in G , then f ( w ) , f ( v ) , f ( v ) , . . . , f ( v k ) isa path in f ( G ). Furthermore, if T is a shortest-paths minimum spanning treerooted at w , then f ( T ) is a tree in f ( G ) whose total weight is equal to T .Because G is finite, we also know that f ( T ) spans f ( G ). Therefore the inertiafor vertex u = f ( w ) is no greater than the inertia on w .On the other hand, if v , . . . v k , w, v (cid:48) , . . . , v (cid:48) k (cid:48) is a shortest path that crosses w , then f ( v ) , . . . f ( v k ) , f ( w ) , f ( v (cid:48) ) , . . . , f ( v (cid:48) k (cid:48) ) is a shortest path that crosses u = f ( w ). Therefore the demand at vertex w is no more than the demand at u = f ( w ).If f is a graph isomorphism, then so is f − . Hence both of the above twoinequalities are equalities. Conjecture 7.4 ([47], 3.3.2) . Let G be a large but finite graph with negativecurvature. If the graph has a symmetric group that fixes a unique point v ,then v is the unique point of minimum inertia and maximum demand. Consider a path of n vertices. It has one non-trivial isomorphism: a mapthat reverses the ordering of the vertices. Informally, one can consider thisisomorphism as a “flip” of the graph. Note that this flip fixes exactly onevertex if n is odd and zero vertices if n is even. The center vertex of apath - the one fixed by the flip - does have the lowest inertia and maximumdemand, as stated in the conjecture. A path is a one-dimensional structure;our next graph will be a 2-dimensional structure. Reconsider the ‘Y’ graphfrom the previous section, but with symmetry so that it has a non-trivial flip-isomorphism that maps one short leg to the other (and the long leg twistsaround with it). Moreover, this flip should fix exactly one vertex, v , that isin the large leg. Finally, v should have a large (but constant) distance from57he base of the graph. By the discussion in Section 6.3, we know that v willnot have the highest demand or lowest inertia. The Euclidean grid (also known as the square lattice or the graph Z ), is theinfinite graph with vertex set { ( x, y ) : x, y ∈ Z } and edge set { ( x, y )( x (cid:48) , y (cid:48) ) : | x − x (cid:48) | + | y − y (cid:48) | = 1 } . We will be considering the finite subgraph induced onvertex set V ( G ∗ ) = { ( x, y ) : −√ n < x, y ≤ √ n } . Multiple research groups[47] [45] [55] have championed the Euclidean grid as being characteristic of“flat,” or non-hyperbolic graphs. The term flat does not seem to have atechnical definition, but the intuition behind it is “approximately planar.” Itis easy to see that the Euclidean grid is not hyperbolic, as an m × n subgrid,with m ≤ n , has hyperbolicity δ = m [30].Jonckheere, Lou, Bonahon, and Baryshnikov [47] show that if B is a ball ofradius 1 in d -dimensional space centered at the origin, then the probabilitythat the shortest path between x and y intersects B , where x and y arerandomly chosen in a ball with radius R , is • Θ(1) if the space is hyperbolic, and • Θ( R − d ) if the space is Euclidean.There is a known connection [13] between graphs that are δ -hyperbolic andhyperbolic space, but as was discussed in Section 1, sometimes this connec-tion is degenerate or trivial. Furthermore, the attempt by Jonckheere, Lou,Bonahon, and Baryshnikov to associate Euclidean space with the Euclideangrid is misleading. There is little congestion in Euclidean space because thestraight line in Euclidean space does not arc inward like it does in hyperbolicspace. However, given two randomly chosen vertices in the Euclidean grid, itis probability 1 / ,
0) and moving right with prob-ability 1 / /
2. Among all points ( i, j ) suchthat i + j = 2 k , the point that the walk is most likely cross is ( k, k ). UsingSterling’s Formula, we see that the probability of the walk hitting the point( k, k ) is Θ( k − / ). By this alternative definition of demand, each vertex givesvery little demand to any vertex that is not close.Narayan and Saniee [55] give experimental evidence that the maximumdemand seen in a δ -hyperbolic network is θ ( n ), while the demand at anyvertex in a flat network is O ( n . ). They claim that the largest demand ata vertex in the Euclidean grid is θ ( n . ), and sketched the proof using thefollowing logic: traffic that flows from the left n/ n/ √ n nodes, and so each has demand θ ( n . ).The missing piece of this argument is proving that the traffic from the leftto the right is spread uniformly across the center. If we attempted to usethe same argument on the ringed tree in Figure 4, where the center linenow consists of log( n ) vertices, then the analogous conclusion would be thateach vertex along the prime meridian has demand Θ( n / log( n )). However,inspection clearly indicates that the vertex at the origin will have Θ( n )demand. We will give a rigorous proof that the demand in the Euclidean gridhas a Θ( n . ) maximum by carefully analyzing how pairs of points distributedemand among the vertices on a shortest path between them. Theorem 8.1.
The amount of demand that crosses the point (0 , in G ∗ ismore than (1 + o (1)) n . and less than (1 + o (1)) n . .Proof. We will only consider demand generated by shortest paths from points( x, y ) to ( x (cid:48) , y (cid:48) ), where x ≤ ≤ x (cid:48) and y ≤ ≤ y (cid:48) . By symmetry, thisaccounts for one half of all shortest paths that cross (0 , a ∈ ( b, c ) d asshorthand to denote that bd ≤ a ≤ cd . Using this shorthand, we will showthat the set of paths we are considering generate demand at (0 ,
0) inside therange of (1 / , / o (1)) n . . Let P k ,k denote the set of shortest pathsbetween vertices ( x, y ) and ( x (cid:48) , y (cid:48) ) such that x (cid:48) − x = k and y (cid:48) − y = k . Case 1: k ≤ √ n/ k steps to the right and k steps upward.59ccordingly, we may describe each path beginning at a fixed vertex ( x, y )as an ordered partition of determined size as follows: path p is a word s , s , . . . , s k ∈ { , , . . . , k } k +1 such that (cid:80) k i =0 s i = k . The bijection be-tween paths and words of this form is given by p stepping vertically s i timesbetween horizontal steps i and i + 1. If we place a uniform distribution on allpaths in P k ,k for some given fixed starting point ( x, y ), the expected valueof size of the intersection between the vertices of the path and the vertex set { (0 , k ) : k ∈ Z } is the expected value of s − x + 1. In the set of words of thistype, it becomes clear that there exists perfect symmetry between the valuesplaced on s i and s j for any 0 ≤ i, j ≤ k . By the symmetry, we see that thisexpected value is k k +1 + 1.Let D z,k ,k be the demand at vertex at vertex (0 ,
0) generated by theshortest paths from ( z, y ) to ( x (cid:48) , y (cid:48) ), where the pair of vertices are in P k ,k .Given any fixed walk from ( z, y ) to ( x (cid:48) , y (cid:48) ), the number of times that it crosses(0 ,
0) over all possible y values is s − z + 1. Therefore D z,k ,k = k k +1 + 1.Let D k ,k be the demand at vertex (0 ,
0) generated by the shortest pathsfrom ( x, y ) to ( x (cid:48) , y (cid:48) ), where the pair of vertices are in P k ,k . If k ≤ √ n/ D k ,k = (cid:80) z = − k D z,k ,k = k + k + 1. If k > √ n/
2, then D k ,k = (cid:80) − k z = −√ n/ D z,k ,k = (cid:16) k k +1 + 1 (cid:17) ( √ n − k ) ≤ k + √ n − k . Because (cid:80) (cid:96)i = (cid:96)/ ( (cid:96) − i ) = (cid:80) (cid:96)/ i =0 i , we have that √ n/ (cid:88) k =1 √ n (cid:88) k =1 D k ,k = √ n/ (cid:88) k =1 √ n/ (cid:88) k =1 D k ,k + √ n (cid:88) k = √ n/ D k ,k = √ n/ (cid:88) k =1 √ n/ (cid:88) k =1 ( k + k + 1) + √ n (cid:88) k = √ n/ (cid:18) k k + 1 + 1 (cid:19) ( √ n − k ) ∈ √ n/ (cid:88) k =1 (1 , √ n/ (cid:88) k =1 ( k + k + 1)= (1 / , / o (1)) n . . Case 2: k ≥ √ n/
2. This case follows almost identically. The only dis-tinction is that we can no longer sum across all possible values for y , andtherefore we get the weaker statement that D z,k ,k ≤ k k +1 + 1 instead of the60quality we had before. The same proof then yields (cid:80) √ nk = √ n/ (cid:80) √ nk =1 D k ,k ∈ (0 , / o (1)) n . .In the sense of congestion, we can think of the Euclidean grid as anextremal example of a flat network. Theorem 8.2. If G is a planar graph with n vertices, then there exists a w ∈ V ( G ) such that D ( w ) ≥ √ (1 − o (1)) n . .Proof. The Lipton-Tarjan Separator Theorem [53] states that for any planargraph with n vertices, there exists a partition into the vertex set into threesets A ∪ B ∪ C , such that no edge has one endpoint in A and the other in C , | B | ≤ √ n , and | A | , | B | ≤ n/
3. From this, we see that the total demandsummed across the vertices of B will be (1 − o (1)) n , and therefore somevertex in the network will have demand at least √ (1 − o (1)) n . .The Euclidean grid is considered the canonical example for the sharpnessof the Lipton-Tarjan Separator Theorem (see “Planar separator theorem”on Wikipedia). Recall that the maximum amount of demand will be be-tween Θ( n ) and Θ( n ). We conclude that the congestion of network that isa flat graph is not insignificant, and that the Euclidean grid has the small-est amount of congestion among all flat graphs because (a) it has no smallseparators and (b) it distributes traffic evenly across the separators.To emphasize the importance of (b) one last time, let us consider onemore example. Let G n be the graph with vertex set { , , . . . , n − } andedge set { ij : i < j, j − i = 2 k , k ∈ N } ∪ { ij : i < j, i − j + n = 2 k , k ∈ N } .This graph has diameter Θ(log( n )) and so (*) δ = O (log( n )) and (**) thetotal demand for the graph sums to O ( n log( n )). Because the graph hasa relatively small value for δ -hyperbolicity, we might expect to see a largeamount of congestion. However, because the graph is vertex transitive, byTheorem 7.3 we know that the demand is spread perfectly even, and so themaximum demand is O ( n log( n )), which is less than the Euclidean grid.61 Scaled Hyperbolicity
Pestana, Rodr´ıgues, Sigarreta, and Villeta [57] have proven that a finite graphis δ -hyperbolic for δ ≤ n/
4. An infinite graph is hyperbolic if it is δ -hyperbolicfor some finite δ , so how do we characterize when a finite graph deserves thelabel “hyperbolic?” Answering this question is the motivation behind thedefinitions of scaled hyperbolicity. As we investigate scaled hyperbolicity andpossible values for various parameters, we will show that scaled hyperbolicityis inherently different than δ -hyperbolicity.For three vertices a, b, c , let vdiam( a, b, c ) = max { d ( a, b ) , d ( b, c ) , d ( a, c ) } .Let P x,y be a shortest path from x to y , and let I ( a, b, c ) = sup P a,b ,P b,c ,P a,c inf { d ( u, v )+ d ( v, w ) + d ( u, w ) : u ∈ P a,b , v ∈ P b,c , w ∈ P a,c } . We then define R -scaled hy-perbolicity as H R ( G ) = sup vdiam( a,b,c ) >R I ( a,b,c )vdiam( a,b,c ) . If P a,b is the shortest pathof the three, then by choosing u, v = b and w = a , we see that H R ≤ H R above without thecondition on taking the supremum over all shortest paths. They did not takeinto account that there may be many shortest paths between two vertices ina graph, for example, between any three points a, b, c in the Euclidean gridthere exists shortest paths P a,b , P b,c , P a,c such that P a,c ∪ P a,b ∪ P b,c forms atree. Therefore if we do not include the condition that we want the supre-mum over all shortest paths between a given triple of vertices, we may callthe canonical non-curved graph hyperbolic! With this modified definition, agraph is then considered scaled hyperbolic if H R ( G ) < / R . The constant 3 / H R = 3 /
2, while negatively curved Riemannian manifoldssatisfy H R < / R >
1, because any three vertices a, b, c thatinduce three edges satisfy I ( a,b,c )vdiam( a,b,c ) = 2. Even if the graph has no triangles,a similar approach can be used with equal-distant points along the shortestcycle. By the thin triangles condition, if we choose R > δ , then any δ -hyperbolic graph G with satisfy H R ( G ) < /
2. Therefore if we choose R toolarge, then we are reduced to the same state that every finite graph will behyperbolic. 62e claim that for any fixed R there exist infinitely many graphs G thatare (2 R/ H R ( G ) = 2. A k -subdivision of a graph G is agraph G /k such that each edge uv of G is replaced by a disjoint path P u,v that starts at u , ends at v , and has length k . By the four points condition,it is easy to see that if G is δ -hyperbolic, then G /k is ( δ + 2 k )-hyperbolic. If k > R/ G is not a tree, then every cycle in G /k has length at least R ,and so by our argument above H R ( G /R ) = 2. See Figure ?? for an example ofa 0-hyperbolic graph; they have been characterized [8] as the graphs G suchthat if H is a subgraph of G that is maximal among 2-connected graphs,then H is a complete graph. By k -subdividing such a 0-hyperbolic graphwith k = (cid:100) R/ (cid:101) , we see that the claim is true. Because we have infinitelymany, this class contains graphs that are unbounded in size.So it must be that R grows as a function of G . Jonckheere, Lohsoonthorn,and Bonahon [46] suggest that R grows proportional to the diameter of thegraph. Consider the Euclidean grid on a vertex set { ( i, j ) : 0 ≤ i ≤ m R , ≤ j ≤ R/ } , where m R is any sequence that satisfies the desired growth rateof R . Suppose a = ( a , a ), b = ( b , b ), and c = ( c , c ) are three pointsin such a grid, and by symmetry assume that a ≤ b ≤ c . There existsa point ( x, y ) ∈ P a,c such that x = b , and clearly b ∈ P a,b ∩ P b,c . Because0 ≤ b , y ≤ R/
2, we see that I ( a, b, c ) ≤ R by choosing u = b , v = b , and w = ( x, y ). Therefore H R ( G ) ≤ RR < /
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