Connectedness of Unit Distance Subgraphs Induced by Closed Convex Sets
CConnectedness of Unit Distance SubgraphsInduced by Closed Convex Sets
Remie Janssen ∗ [email protected] Leonie van Steijn † [email protected] 26, 2021 Abstract
The unit distance graph G R d is the infinite graph whose nodes arepoints in R d , with an edge between two points if the Euclidean distancebetween these points is 1. The 2-dimensional version G R of this graphis typically studied for its chromatic number, as in the Hadwiger-Nelsonproblem. However, other properties of unit distance graphs are rarelystudied. Here, we consider the restriction of G R d to closed convex subsets X of R d . We show that the graph G R d [ X ] is connected precisely whenthe radius of r ( X ) of X is equal to 0, or when r ( X ) ≥ X is at least 2. For hyperrectangles, we give bounds for thegraph diameter in the critical case that the radius is exactly 1. In the Hadwiger-Nelson problem, the aim is to colour the plane with as fewcolours as possible, so that no pair of points at distance 1 from each other havethe same colour. Equivalently, it asks to find the chromatic number of G R , theunit distance graph of the plane which has node set R and an edge betweentwo points u, v ∈ R precisely when they are at distance 1 from each other.This problem is typically tackled by studying finite subgraphs of G R . However, G R has an interesting structure when restricted to (infinite) connected subsetsof the plane as well, such as a strip of the plane S r = R × [0 , r ]. It turnsout that, in the induced subgraph G R [ S r ] of G R , the existence of cycles ofa given length and the chromatic number both depend on the size r of thestrip [Bau98, Kru08, ACL + ∗ Delft Institute of Applied Mathematics, Delft University of Technology, Delft, The Nether-lands. Research funded by the Netherlands Organization for Scientific Research (NWO), Vidigrant 639.072.602 of dr. Leo van Iersel. † Mathematical Institute, Leiden University, Leiden, The Netherlands. a r X i v : . [ m a t h . C O ] F e b n this paper, we study subgraphs of G R d (the unit distance graph of R d )induced by closed and convex subsets of R d . However, instead of studying thechromatic number of these graphs, we investigate their connectedness and graphdiameters. In other words, we aim to characterize when there is a path betweenany two points in a closed convex set using unit distance steps. Additionally,we study bounds for the maximal number of unit steps needed to go from anypoint to any other point in a hyperrectangle. R d First, we define several basic subsets of R d that we will use throughout thispaper. We will use d ( x, y ) to denote the Euclidean distance, and x · y for thestandard inner product between x, y ∈ R d .A ball is defined by its center and its radius: Let d ∈ N , v ∈ R d , and r ∈ R ,then B d ( v, r ) = { x ∈ R d : d ( x, v ) ≤ r } is the d -dimensional ball of radius r around v . The sphere S d − ( v, r ) := { x ∈ R d : d ( x, v ) = r } is the boundary ∂ B d ( v, r ) of the ball B d ( v, r ).The d -dimensional hypercube with side lengths l is denoted C d ( l ). We al-ways use the embedding { x ∈ R d : x i ∈ [0 , l ] } of C d ( l ) in R d . Similarly, a d -dimensional hyperrectangle is defined by a d -tuple of non-negative real num-bers l = ( l , . . . , l d ) ∈ R d . We denote this hyperrectangle by R d ( l ) = { x ∈ R d : x i ∈ [0 , l i ] } .Let P, Q ∈ R d , then P Q denotes the line segment between P and Q . More-over, we write H ( P, Q ) = { x ∈ R d : ( Q − P ) · ( x − P ) ≤ } , which is theclosed half-space bounded by the hyperplane through P perpendicular to P Q that does not include Q .All of these subsets of R d , except for the sphere, are convex. In other words,for any pair of points in such a set (i.e., a ball, a hypercube, a hyperrectangle,a line segment, a hyperplane, or a halfspace) the line segment between them isalso contained in the set.To construct convex sets, we use the convex hull Conv( A ) of a set A ⊂ R d ,which is the smallest convex set containing A . The cone Cone( A ) of a set ofpoints A is the set { λx : x ∈ Conv( A ) , λ ≥ } . For a finite set of points A = ( a , . . . , a n ) in R d the convex hull is { λ a + · · · + λ n a n : (cid:80) ni =1 λ i = 1 } .If all a i are affinely independent, then S = Conv( A ) is an n -simplex. For eachsubset I ⊆ [ n ], the set Conv( { a i } i ∈ I ) is an | I | -dimensional face of Conv( A ).Finally, to construct additional subsets of R d , we use we use the Minkowskisum and scaling of a set. We denote the Minkowski sum by ⊕ , i.e., X ⊕ Y := { x + y : x ∈ X, y ∈ Y } . Scaling a set X ⊆ R d by a factor λ ∈ R ≥ is defined as λX := { λx ∈ R d : x ∈ X } . Note that the Minkowski sum of two convex sets isconvex, and λX is convex iff X is convex.2 .2 Enclosing balls Let X ⊆ R d , then any ball B d ( v, r ) ⊇ X containing X is an enclosing ball of X .For each bounded set X ⊂ R d , there is a unique minimal enclosing ball (m.e.b.),which is the enclosing ball of X with minimal radius. The radius r ( X ) of X isdefined as the radius of the m.e.b. of X .Note that the center of the m.e.b. of a convex set lies in the convex set. Arelated lemma is the following, which, according to [FGK03], is well-known andcan be traced back to Seidel (no reference given). Lemma 1 (Seidel) . Let X be a set of points on the boundary of some ball B with center C . Then B is the m.e.b. of X if and only if C ∈ Conv( X ) . For a simplex S ⊆ R d the center of the m.e.b of S is always contained in S . This center may lie on a face of S . Here, we are interested in so-calledwell-centered simplices, which are simplices where this is not the case. In otherwords, a simplex S is well-centered if the center of its m.e.b. is contained in itsrelative interior [VHGR10, VHG + S is the interiorof S when viewed as a subset of the affine span of S . We now define the main subject of this paper, the unit distance graph of asubset of R d . The unit distance graph in dimension d is denoted G R d ; it is theinfinite graph whose vertex set is R d , and two points x, y ∈ R d are connectedby an edge if d ( x, y ) = 1.We restrict this graph to subsets of R d by considering induced subgraphs.Let G = ( V, E ) be a graph, and V (cid:48) ⊆ V a subset of the vertices, then G [ V (cid:48) ]is the subgraph of G induced by V (cid:48) : the graph with vertices V (cid:48) and an edgebetween u, v ∈ V (cid:48) if { u, v } ∈ E .The graphs we are interested in are G R d [ X ], where X ⊂ R d is compact andconvex. In particular, we will study the connectedness and the diameter of thesegraphs, which are invariant under translation and rotation of X . The diameterdiam( G ) of a graph G is defined as the maximal graph distance between anypair of vertices of G . The graph distance between two nodes x and y of G is the minimum number of edges in a path between x and y through G . Toprevent confusion between the Euclidean distance and the graph distance, wewill henceforth reserve the term distance for the Euclidean distance, and theterm diameter for the graph diameter. The graph distance will always be givenin a number of steps. If there is a sequence of (unit distance) steps between twopoints u and v , then we say that v can be reached from u .3 P P P Q T (2) (3)
Figure 1: Wiggling through a rectangle from P = ( p,
0) to Q = ( q,
0) inLemma 1. The point Q must lie on the line segment P T (orange), where T = ( p + 2(1 − √ − h ) , P T can be reached from somepoint on the circle arc (3), which is simply the circle arc (2) translated to theright by distance 1. Moreover, (2) is the circle arc with points at distance onefrom the point P = ( p + 1 , Q from P via P , somepoint P on (2), and a point P on (3). All dotted lines have unit length. n -Simplices The following lemmas shows that each sufficiently large triangle has enough‘wiggle room’ to reach a significant part of one of the sides of the triangle usingsteps of length exactly one. This will be instrumental when proving that G R d [ S ]is connected for an n -simplex S . Lemma 2.
Let R = [0 , x ] × [0 , h ] ⊆ R for some h > and ≤ x ≤ . Then,for each pair of nodes u, v ∈ G R [ R ] that lie on the line segment [0 , x ] × { } ,there is a path between u and v in G R [ R ] .Moreover, this path consists of at most 2 steps if h ≥ , and at most (cid:108) x −√ − h ) (cid:109) steps if h < .Proof. To prove the lemma, we start in a point P ∈ [0 , x ] × { } and show thatwe can reach all points on a small line segment to the right of P in four steps.Obviously, if h ≥
1, each point on [0 , x + 1] × { } can be reached in at most twosteps from P , so we assume h < P = P = ( p,
0) with 0 ≤ p ≤ x , and let Q = ( p + q,
0) such that p + q/ ≤ x and 0 ≤ q ≤ − √ − h ) ≤
2. Then, by taking the following fourunit steps, we can reach the point Q from P (Figure 1). First, go a unit stepto the right to P = ( p + 1 , P = (cid:16) p + q/ , (cid:112) − (1 − q/ (cid:17) . Note that P ∈ S ( P , ∩ R . Indeed, P ∈ S ( P ,
1) because d ( P , P ) = 1,and P ∈ R because 0 ≤ p + q/ ≤ x and (cid:112) − (1 − q/ ≤ (cid:113) − (1 − (1 − (cid:112) − h )) = h. P = P + (1 , R because p + q/ ≤ x + 1. Lastly, take a unit step from P to Q , where the fact that d ( P , Q ) = 1 follows from a simple calculation.Now let u = ( u ,
0) and v = ( v ,
0) be two points on the line segment[0 , x ] × { } such that u ≤ v . By repeatedly using the sequence of four stepsgiven above, we can reach v from u in G R [ R ]. This takes at most4 (cid:38) v − u (cid:0) − √ − h (cid:1) (cid:39) steps if h ≤ u and v by 2 if h ≥ (cid:38) x (cid:0) − √ − h (cid:1) (cid:39) . Lemma 3.
Let T ∈ R d be the triangle Conv( { P , P , P } ) . Assume that d ( P , P ) = 1 + x > and that there is a point A in the relative interiorof T such that d ( P , A ) = d ( P , A ) = 1 . Then for each pair of points u, v ∈ P P ∩ B d ( P , x ) , there is a path from u to v in G R d [ T ] .Proof. Without loss of generality, we assume d = 2, P = (0 , P = (1 + x, P = ( P , P ) with P >
0. Finally, to simplify notation in the proof, weassume 0 ≤ P ≤ x . Note that this does not affect the conclusion: if P liesoutside this region, we can restrict to the triangle T (cid:48) = T ∩ [0 , x ] × R . Indeedthe result for T then follows because the assumptions of the lemma still hold for T (cid:48) , and G R d [ T (cid:48) ] is a subgraph of G R d [ T ] containing all u, v ∈ P P ∩ B d ( P , x ).We first prove that we can reach some small neighbourhoods of P and P in T using unit distance steps (Figure 2). Then, we extend the small regionaround P to the right by ‘wiggling’. Finally, translating the neighbourhood of P to the left by a distance of one, we reach the desired part of the line segment P P .Note that there is a small arc of the circle S ( P ,
1) containing A that lieswithin T , namely C = S ( P , ∩B ( A, (cid:15) ) for some (cid:15) > c of C to the right of A has distance d ( c, P ) = 1+ r > P ,so d ( Q, c ) ≥ Q ∈ B ( P , r ) ∩ T by the triangle inequality. Moreover,because P ≥
0, we have d ( Q, A ) ≤ Q ∈ B ( P , r ) ∩ T . These twodistance bounds, together with the continuity of the circle arc C from c to A and the intermediate value theorem imply that each Q ∈ B ( P , r ) ∩ T is atdistance 1 from some point on C . By symmetry in P and P , there is some0 < r (cid:48) ∈ R such that each Q (cid:48) ∈ B ( P , r (cid:48) ) ∩ T can be reached from A as well.Note that we now know we can reach [0 , r ] × { } and [1 + x − r (cid:48) , x ] × { } from A . To see that we can also reach [ x − r (cid:48) , x ] × { } , we note that this isjust the line segment [1 + x − r (cid:48) , x ] × { } translated to the left by distance1. The only remaining part of [0 , x ] × { } we need to reach is [ r, x − r (cid:48) ] × { } .5 C P = (0 , P = (1 + x, P ( r, c Figure 2: Reaching the corners of a triangle as in Lemma 3. To reach any point Q in the shaded are within T from A , take one step to P , then one to a pointon the arc C , and finally to Q . There exists a suitable point on C because theshaded area lies in B ( A, \ B ( c,
1) and C is a continuous path from c to A .If r + r (cid:48) ≥ x , then we are done, so suppose r + r (cid:48) ≤ x . The required resultfollows immediately from Lemma 2 when we observe that there is a rectangle R = [ r, x − r (cid:48) ] × [ h ] ⊆ T for some h >
0, whose base has width w = 1+ x − r (cid:48) − r .Indeed, the lemma then implies that we can reach [ r, r + ( w − × { } , with r + ( w −
1) = x − r (cid:48) .In this proof, we have used the existence of small balls, but these balls mayactually be reasonably large, so that the steps may make significant headwayalong P P . However, when the angle at P is very large, this wiggle space maybe limited. Lemma 4.
Let T ⊆ R be an obtuse triangle with radius r ( T ) = 1 , then G R [ T ] is connected.Proof. As T is obtuse, the radius of T is half of the longest side of T . Hence,without loss of generality, let the triangle be the convex hull of the points (0 , ,
0) and P = ( P , P ) with P , P >
0. We prove that there is a sequence of6teps between any two points in base B of the triangle, i.e., the line segmentbetween (0 ,
0) and (2 , B . To see this,let Q = ( Q , Q ) ∈ T be arbitrary, and assume without loss of generality that Q ≤
1, so that we have d ( Q, (2 , ≥
1. As (1 ,
0) is the center of the m.e.b. of T , we also have d ( Q, (1 , ≤
1. Applying the intermediate value theorem onthe function d ( Q, · ) along the line segment (0 , , b on B such that d ( b, Q ) = 1.To see that there is a sequence of steps between any two points in base B ofthe triangle we use a technique similar to the proof of Lemma 3. We first notethat we can reach some intervals [0 , r ] × { } and [2 − r (cid:48) , × { } from (1 ,
0) bygoing via the arcs above (1 ,
0) of the circles around (0 ,
0) and (2 ,
0) of radius one(similar to Figure 2, but with A on B instead of in the relative interior of T ).Then, by translating these line segments to the left and the right by a distanceof one, we can also reach the line segment [1 − r (cid:48) , r ] × { } . The remainingparts of B are [ r, − r (cid:48) ] × { } and [1 + r, − r (cid:48) ] × { } , which can both be reachedthrough a rectangle [ r, − r (cid:48) ] × [0 , h ] for some h > In this subsection, we will prove that G R d [ S ] is connected for any simplex S ofdimension at least 2 with radius r ( S ) = 1. To do this, we first show that eachpoint in S is at distance 1 from some point on a line segment CP ⊆ S where P is a vertex of S with d ( C, P ) = 1. Then, we show that we can reach eachpoint on such a line segment from C taking only unit steps. Together, thesefacts prove that G R d [ S ] is connected. Lemma 5.
Let S ⊆ R d be a simplex with m.e.b. B = B d ( C, and let P be avertex of S with d ( C, P ) = 1 . Then, each point in H ( C, P ) ∩ B is at distanceone from some point on CP .Proof. Let Q ∈ H ( C, P ) ∩ B be arbitrary. As Q ∈ H ( C, P ) and B d ( P, ∩ H ( C, P ) = { C } , we have d ( P, Q ) ≥
1. Furthermore, because Q ∈ B = B d ( C, d ( C, Q ) ≤
1. Combining this with continuity of the function d ( Q, · )along the line segment CP , the intermediate value theorem implies the existenceof a point U on CP such that d ( Q, U ) = 1.
Lemma 6.
Let S ⊆ R d be a simplex with m.e.b. B = B d ( C, r ) and let P , . . . , P n be the vertices of S where d ( C, P i ) = r , then (cid:83) i ∈ [ n ] H ( C, P i ) = R d .Proof. Without loss of generality, assume that C = (0 , . . . , Q ∈ R d such that Q (cid:54)∈ H ( C, P i ) for all i ∈ [ n ]. This implies that Q · P i > i ∈ [ n ], so P i ∈ R d \ H ( C, Q ) for all i ∈ [ n ]. In other words, all P i lie strictly on one side of the hyperplane perpendicular to CQ through C . Then,by Lemma 1, the center C must satisfy C ∈ Conv( { P . . . , P n } ) ⊆ R d \ H ( C, Q ).This contradicts the fact that C = (0 , . . . , (cid:54)∈ R d \ H ( C, Q ). We conclude thatthere is no Q ∈ R d such that Q (cid:54)∈ H ( C, P i ) for all i ∈ [ n ]; in other words, each Q ∈ R d lies in H ( C, P i ) for some i ∈ [ n ], so (cid:83) i ∈ [ n ] H ( C, P i ) = R d .7 emma 7. Let S ⊆ R d be a well-centered simplex of dimension at least withradius r ( S ) = 1 , then G R d [ S ] is connected.Proof. Let B = B d ( C,
1) be the m.e.b. of S = Conv( { V , . . . , V n } ), and let P be an arbitrary vertex of S . Because S is well-centered, we have d ( C, V i ) = 1for all i and therefore, by Lemma 6, H ( C, P ) ∩ { V , . . . , V n } (cid:54) = ∅ . Hence, S hasa vertex Q ∈ H ( C, P ) ∩ { V , . . . , V n } , and the 1-face P Q has length at least √
2. Because S is well-centered, C is in the relative interior of S , so there is apoint R ∈ S such that the triangle T = Conv( { P, Q, R } ) contains the center C of the m.e.b. of S .Let X be the point at distance | P Q | − Q on P Q . By Lemma 3, anypoint on
P Q can be reached from C through G R d [ T ]. Furthermore, there is acontinuous path from X to C through S d − ( P, ∩ T . All of the points on thispath are at distance 1 from P , so they can be reached from C via P in G R d [ T ].This path together with the line segment QX gives a continuous path P ( Q, C )from Q to C through T , and each of the points on this path can be reachedfrom C in G R d [ T ].Using continuity of d ( x, · ) along P ( Q, C ) and the fact that d ( Q, x ) ≥ d ( C, x ) ≤ x ∈ CP , the intermediate value theorem implies that eachpoint on CP can be reached from C in G R d [ T ] via some point on P ( Q, C ). Aswe chose an arbitrary vertex P of S , we conclude that, for each V i , all pointson CV i can be reached from C in G R d [ S ].Now note that each point in H ( C, V i ) ∩ B is at distance one from some pointon CV i , and (cid:83) i ∈ [ n ] H ( C, V i ) ∩ B = B (Lemma 6). Finally, as S ⊆ B , each pointin S can be reached from C in G R d [ S ]. In other words, G R d [ S ] is connected. Proposition 1.
Let S ⊆ R d be a simplex with radius r ( S ) = 1 of dimension atleast , then G R d [ S ] is connected.Proof. Let C be the center of the m.e.b. B of S , and note that C lies in therelative interior of a face F of S of dimension at least 1. Moreover, the m.e.b. of F is B . If F has dimension one, then there exists a point R ∈ S \ F because S hasdimension at least 2. Hence, S contains the obtuse triangle T = Conv( { R }∪ F ),and G R d [ T ] is connected by Lemma 4. In particular, F can be reached in itsentirety from C in G R d [ S ].If F has dimension at least 2, then G R d [ F ] is connected, because F is a well-centered simplex of radius 1 (Lemma 7). In both cases, G R d [ F ] is connected andeach point in B is at distance one from some point on F . Hence, we concludethat G R d [ S ] is connected.The following lemma can improve the diameter bound resulting from theproofs above. It states that in a well-centered simplex Conv( { V , . . . , V d +1 } ) ofdimension d , each point on a line segment CV i is at most d steps away from apoint on CV j for each j (cid:54) = i . Hence, its consequence is that, during a sequenceof unit steps between any two points in a well-centered simplex, we only have to‘wiggle’ once on one of the edges (i.e., apply the sequence of steps from Lemma 2at most once). As the wiggling part is the part that may take the most steps,this could make the diameter bound a lot smaller.8 emma 8. Let S ⊆ R d be a simplex with m.e.b. B d ( C, , and P , . . . , P n thevertices of S with d ( C, P i ) = 1 . Then, for each i ∈ [ n ] and u ∈ CP i , thereexists v ∈ CP such that there is a uv -path of at most n − steps in G R d [ S ] .Proof. By Lemma 5, we can reach each point on CP j from some point on CP i if P j ∈ H ( C, P i ). Hence, we define the graph G = ( V, E ), where V = { P , . . . , P n } and { P i , P j } ∈ E iff P j ∈ H ( C, P i ). To prove the lemma, we justhave to show that G is connected. For simplicity, we assume C = (0 , . . . , x ∈ H ( C, y ) precisely when x · y ≤ G is connected, we are done, so we assume that G is not connected,which implies it has a maximally connected subset A (cid:40) V . The proof that G is connected then uses the following strategy. We will first show that A mustbe contained in some half-space H ( C, Q ), then we will show that in fact allof V must lie in one such half-space, where V \ A cannot lie on the boundinghyperplane. This will then imply that C ∈ Conv( A ), which, by Lemma 1proves that the m.e.b. of A is B d ( C, (cid:83) a ∈ A H ( C, a ) = R d (Lemma 6), we conclude that each P i is in some H ( C, a ), so G is connected.Assume for the sake of contradiction that there is no x ∈ R d such that A ⊆ H ( C, x ). Then there is also no x such that A ∩ H ( C, x ) = ∅ , which impliesthat A ∩ H ( C, x ) (cid:54) = ∅ for all x ∈ R d . This holds in particular for all x ∈ V \ A ,which contradicts the maximal connectedness of A in G . Therefore, we mayconclude that A ⊆ H ( C, Q ) for some Q ∈ R d .Now let m be the center of the m.e.b. of A , and assume that m (cid:54) = C .This implies that the minimum min x ∈S d − ( C, max a ∈ A d ( x, a ) is attained atthe point m (cid:48) found by projecting m onto S d − ( C,
1) via the line from C to m .Moreover, the distance between two points on the sphere is monotone decreasingwith the inner product of the two points. Therefore, the fact that Q satisfiesmin a ∈ A − Q · a ≥
0, implies we also have min a ∈ A m (cid:48) · a ≥
0, and we can concludethat m, m (cid:48) ∈ Cone( A ) and A ⊆ H ( C, − m ).By maximal connectedness of A we know that A ⊆ R d \ (cid:83) b ∈ V \ A H ( C, b ),which implies m ∈ Cone( A ) \ { } ⊆ R d \ (cid:83) b ∈ V \ A H ( C, b ). In other words, V \ A ⊆ R d \ H ( C, m ) and A ⊆ H ( C, − m ). Now note that C must lie inConv( V ) by Lemma 1, and this convex combination cannot use any elementsfrom V \ A , as they all have a nonzero component in the Cm direction, whichcannot be compensated for by any other part of the convex combination. Hence, C ∈ Conv( A ), which implies that the B d ( C,
1) is the m.e.b. of A (Lemma 1).This leads us to the contradition that both m = C and m (cid:54) = C , so we concludethat the assumption m (cid:54) = C is false.Finally, we have that m = C , and we may use m = C ∈ Conv( A ). Hence,the m.e.b. of A is equal to the m.e.b. B d ( C,
1) of S (Lemma 1). Then, applyingLemma 6 to A and noting that A is connected in G proves the result.9 Closed Convex Sets
In this section, we use the connectedness results for simplices to characterizethe connectedness of closed convex sets. Clearly, the radius of such a connectedset must be at least one, because, otherwise, the center of the m.e.b. has noneighbours in the induced unit distance graph. We will show that this conditionis actually sufficient for a closed convex set to be connected, as long as the sethas affine dimension at least 2. We first focus on bounded sets, and considerunbounded sets afterwards.
Lemma 9.
Let X ⊆ R d be a bounded convex set such that G R d [ X ] is connected.Then for all λ ≥ , G R d [ λX ] is connected as well.Proof. Without loss of generality, suppose the m.e.b. of X is B = B d (0 , r ). Weprove that G R d [ λX ] is connected for all λ ≥ λr ≤ r + 1 (i.e., weincrease the radius by at most 1). The result then simply follows by repeatedapplication of this result.Let λ ≥ λr ≤ r + 1, and let x be any point in λX . Because X is convex and 0 ∈ X we have X ⊆ λS . Hence, we only have to prove that foreach x ∈ λX \ X , there is a point y ∈ X with d ( x, y ) = 1. To prove this, we findtwo points P, Q ∈ S at distance d ( P, x ) ≤ d ( Q, x ) ≥
0. The (continuous)line segment
P Q is contained in X , so, by the intermediate value theorem, thereis a point R ∈ X on P Q such that d ( R, x ) = 1.First we show that we can take P = x/λ . Because x ∈ λX , we have x/λ ∈ X ; and because | x | ≤ r ( λX ) ≤ r + 1 and 1 − /λ ≤ / ( r + 1), we have d ( x, x/λ ) = | x | (1 − /λ ) ≤
1. Now we show that there exists a Q ∈ X with d ( x, Q ) ≥
1. Suppose there weren’t such a Q , then X ⊆ B d ( x, r ( X ) < x is not the center of the m.e.b. of X . However, if r ( X ), then G R d [ X ] could not have been connected, which contradicts the conditions of thelemma. Hence, there exists a Q ∈ X with d ( x, Q ) ≥ Lemma 10.
Let X ⊆ R d be a compact and convex set and let B be its m.e.b. withcenter C , then C ∈ Conv( ∂B ∩ X ) .Proof. Suppose, to the contrary, that C (cid:54)∈ Conv( ∂B ∩ X ) and consider the sets X ∂ ( (cid:15) ) = Conv( ∂B ∩ X ) ⊕ B d (0 , (cid:15) ) parameterized by (cid:15) >
0. Then there mustexist such an (cid:15) > C (cid:54)∈ Conv( X ∂ ( (cid:15) )). Because X ∂ ( (cid:15) ) is convex, thehyperplane separation theorem implies the existence of a hyperplane H sepa-rating C from X ∂ ( (cid:15) ). Let L be the line segment between H to C perpendicularto H . From C to H along L , the distance to each point in X ∂ ( (cid:15) ) decreases.Furthermore, for some (cid:15) (cid:48) >
0, we have the distance bound d ( C, P ) < r ( X ) − (cid:15) (cid:48) for all P ∈ X \ X ∂ ( (cid:15) ). This follows from the fact that the closure of X \ X ∂ ( (cid:15) ) iscompact and does not contain any points at distance r ( B ) from C by definitionof X ∂ ( (cid:15) ). Hence, the triangle inequality implies that for any point Q ∈ B d ( C, (cid:15) (cid:48) )and any point P ∈ X \ X ∂ ( (cid:15) ), we have d ( Q, P ) ≤ d ( Q, C )+ d ( C, P ) < (cid:15) (cid:48) +( r ( X ) − (cid:15) (cid:48) ) = r ( X ). 10inally, combining these results, there is a point C (cid:48) (cid:54) = C on L ∩ B d ( C, (cid:15) (cid:48) )such that d ( C (cid:48) , P ) < r ( X ) for all P ∈ X . This contradicts the fact that B isthe m.e.b. of X , so we conclude that C ∈ Conv( ∂B ∩ X ). Theorem 1.
Let X ⊆ R d be closed and convex, then G R d [ X ] is connected iff r ( X ) = 0 , or r ( X ) ≥ and X contains at least three affinely independent points.Moreover, if X is bounded, then the diameter of G R d [ X ] is finite.Proof. The result is trivial when X contains at most two affinely independentpoints, so we assume X contains at least three affinely independent points.Without loss of generality, we also assume r ( X ) = 1; the full result then followsby applying Lemma 9 and, if r ( X ) is unbounded, by noticing that X can becovered by a set of compact convex sets which can be reached by taking stepsalong a ray of X .Let B be the m.e.b. of X with center C , then C ∈ Conv( ∂B ∩ X ) because X is closed (Lemma 10). In particular, there exist at most d +1 points S ⊆ ∂B ∩ X such that C ∈ Conv( S ) (Caratheodory’s theorem). By Lemma 1, the m.e.b. of S is B as well. As X contains at least three points, S contains at least twopoints. If | S | = 2, then Conv( S ) is a line segment of length two. Furthermore, X contains a third point P not in Conv( S ). Hence, X contains the triangle T formed by S and P , which is obtuse of radius one. Hence, G R d [ T ] is connected(Lemma 4), and each pair of points in Conv( S ) is connected by a path in G R d [ X ]. If | S | >
2, then Conv( S ) is a simplex of dimension at least two. Hence,by Proposition 1, G R d [Conv( S )] is connected. As each point on the m.e.b. of S can be reached from some point in Conv( S ) (Lemma 6), G R d [ X ] is connectedas well. The finiteness of the diameter follows directly from the constructionsof the paths in the lemmas.By a simple scaling argument, we obtain the following corollary for convexsets. Note that it is incomplete in the sense that connectedness of G R d [ X ] forconvex sets X is not completely characterized: the case where r ( X ) = 1 and X is not closed is still an open problem. We conjecture that the only extracondition for connectedness in that case is that the intersection of X and itsm.e.b. is non-empty. Corollary 1.
Let X ⊆ R d be convex, then G R d [ X ] is connected if r ( X ) = 0 , or r ( X ) > and X contains at least three affinely independent points. Although the connectedness of hyperrectangle-induced unit distance graphs ischaracterized completely by Theorem 1, the bound on the graph diameter basedon those proofs is quite large. In this section, we will show that the diameter ofthe unit distance graphs induced by hypercubes is constant in the dimension ofthe hypercube. For hyperrectangles, we show an upper bound of the diameterthat is a linear combination of the dimension and the graph diameter of the‘best’ two-dimensional rectangle contained in the hyperrectangle.11 roposition 2.
Let d > , then G R d [ C d ( l )] is connected iff l ≥ / √ d . More-over, for l = 2 / √ d , the diameter of this graph is at most .Proof. The connectedness follows directly from Theorem 1 and the fact that r ( C d ( l )) ≥ l ≥ / √ d . Hence, we focus on the diameter ofhypercubes C d (2 / √ d ). First we show that all points on 1-dimensional faces canbe reached from the center point M = (1 / √ d, . . . , / √ d ) in at most three steps,then we complete the proof by showing that each other point of the hypercubeis at distance one from a point in such a 1-dimensional face. This proves that G R d [ C d (2 / √ d )] is connected for all d > P be a point on a 1-dimensional face F . Without loss of general-ity, assume that P = ( P , , . . . , L P ( h ) = ( P / , h, . . . , h )lies within the hypercube and d (0 , L P ( h )) = d ( P, L P ( h )) for all h ∈ [0 , / √ d ].Moreover, we have the extremal distance bounds d (0 , L P (0)) ≤ / √ d < d (0 , L P (2 / √ d )) > (cid:113) ( d − / √ d ) >
1. By the intermediate value theorem,there is an h ∈ [0 , / √ d ] such that d (0 , L P ( h )) = d ( L P ( h ) , P ) = 1. Therefore,we can reach P from M in at most three steps in G R d [ C d (2 / √ d )] via the paththe points 0 and L P ( h ).Now, we show that each point in the hypercube is at distance one from apoint in a 1-dimensional face. Note that for each point P (cid:48) in the cube, thereis at least one vertex V at distance d ( P (cid:48) , V ) ≤ V (cid:48) such that d ( P (cid:48) , V (cid:48) ) ≥
1. Consider a path along the 1-dimensional faces of C d (2 / √ d )between V and V (cid:48) . By the intermediate value theorem, there is a point P (cid:48)(cid:48) onthis path such that d ( P, P (cid:48)(cid:48) ) = 1. As P (cid:48)(cid:48) lies on a path through 1-dimensionalfaces, it lies on a 1-dimensional face, and can thus be reached from M in atmost three steps. We conclude that the arbitrary point P (cid:48) ∈ C d (2 / √ d ) can bereached from M in at most four steps through G R d [ C d (2 / √ d )]. Lemma 11.
Let R ( l , l ) be a -dimensional rectangle with side lengths l ≥ l ,then G R d [ R ( l , l )] is connected iff (cid:112) l + l ≥ .Moreover, when (cid:112) l + l = 2 , the diameter of G R d [ R ( l , l )] is bounded by if l ≥ , and by (cid:24) l − − √ − l ) (cid:25) if l < .Proof. The connectedness result follows directly from Theorem 1, so we focus onthe diameter bound. If l ≥
1, then we can apply exactly the same argumentsas in Proposition 2, so we continue with the case l <
1. To prove the bound,we show that for any point P ∈ R ( l , l ), there is a path from M = ( l / , l / P in G R [ R ( l , l )] of at most 2 + 4 (cid:24) l − − √ − l ) (cid:25) steps.To see this, assume without loss of generality that P ≥ l / P ≥ l / P must then be at distance 1 from some point on any continuous pathfrom 0 to M by the intermediate value theorem. In particular, it is at distance1 from some point on the path consisting of the line segment L = [0 , l − × { } and the arc R ( l ) ∩ S (( l , , L can be reached in at most12 + 4 (cid:24) l − − √ − l ) (cid:25) steps from C (Lemma 2), and each point on the arc in 2steps. Hence, P can be reached from C in at most 2 + 4 (cid:24) l − − √ − l ) (cid:25) steps. Proposition 3.
Let R d ( l ) be a d -dimensional rectangle with side lengths l =( l , . . . , l d ) , then G R d [ R d ( l )] is connected iff | l | ≥ .Moreover, if | l | = 2 and l (cid:48) = (cid:113)(cid:80) i ∈ I l i and l (cid:48) = (cid:113)(cid:80) i (cid:54)∈ I l i for some ∅ (cid:54) = I (cid:40) [ d ] then the diameter is bounded by diam( G R d [ R d ( l )]) ≤ diam( G R [ R ( l (cid:48) , l (cid:48) )]) +2 .Proof. The connectedness result follows directly from Theorem 1, so we againfocus on the diameter bound.Note that for each ∅ (cid:54) = I (cid:40) [ d ], R d ( l ) contains a rectangle with side lengths l (cid:48) = (cid:113)(cid:80) i ∈ I l i and l (cid:48) = (cid:113)(cid:80) i (cid:54)∈ I l i . Such a rectangle necessarily contains a diag-onal D of R d ( l ). Hence, G R d [ R ( l (cid:48) , l (cid:48) )] is connected, which implies all points on D are connected in G R d [ R d ( l )] by a path of length at most diam( G R d [ R ( l (cid:48) , l (cid:48) )]).Finally, Lemmas 6 and 5 imply that each point in G R d [ R d ( l )] is at distance 1 fromsome point on D . Therefore, G R d [ R d ( l )] has diameter at most diam( G R [ R ( l (cid:48) , l (cid:48) )])+2. For hypercubes, this proposition together with Lemma 11 gives a proof fora diameter bound of diam( C d (2 / √ d )) ≤
10 independent of Proposition 2. Thisshows that the bound for hyperrectangles in Proposition 3 can possibly still beimproved.
In this paper, we have characterized the connectedness of unit distance graphsinduced by closed and convex subsets of R d . For compact convex sets, we haveshown that the diameter of this graph is bounded. Moreover, for the smallestconnected hypercubes in each dimension, this diameter stays constant with anincreasing dimension. Connected components of such graphs (i.e., for strips ofthe plane) have been considered superficially previously [ACL +
14, Bau98], butnever as the main subject of study, as we have done.There are some obvious problems that we have left open, such as the con-nectedness of the unit distance graphs induced by non-closed convex sets, oreven non-convex sets. The former is partially answered by Corollary 1, but theedge case of radius one is still open. This question can possibly be answeredusing techniques similar to the ones we used, but we made ample use of pathsthrough faces of simplices, which may not be present in the non-closed case. Thelatter question (finding a characterization of the connectedness for non-convexsets) is probably quite hard to answer in general, so it merits restriction to moretractable sets, such as polytopes. 13t would be interesting to see whether this translates to computational hard-ness of such questions as well. For example, we could consider the complexityof the following problem.Input: A (finite) polytope P ⊆ R d .Question: Is G R d [ P ] connected?Interestingly, it is not even immediately clear whether this is in NP; we prob-ably need some kind of characterization of connectedness to verify a certificate.Another interesting (computational) problem concerns the diameter: Givena set X ⊂ R d of radius 1, find a scaling factor λ > G R d [ λX ])is minimal. If λ is small, the diameter can be very large because of little wiggleroom; if λ is large, the diameter can be large because of the distance that needsto be bridged using steps of distance one.Another variation for all these problems is to consider the unit distance graphof Q d , which is common when studying the chromatic number as well (e.g.,[ACL + w n of a strip of the plane S w n so that the graph G R [ S w n ]contains a cycle of length n (where n is odd). R d We have focused on induced unit distance graphs that are connected, i.e., thathave one connected component. Obviously, we can relax this notion, and in-stead consider the number and ‘size’ of connected components when G R d [ X ] isdisconnected.For example, if the radius of a hyperrectangle R is smaller than 2, then themidpoint of the hyperrectangle has no neighbours in G R d [ R ], but this graphis not necessarily totally disconnected. Of course, this does happen when theradius is smaller than than 1. It is not immediately clear what the componentslook like for radii between 1 and 2.A quick illustration of this concept for squares makes it clear that thesecomponents can be quite interesting. If the square C ( l ) has side lengths l > √ , then it seems that G R d [ C ( l )] has one non-trivial connected componentconsisting of the points (cid:91) v ∈ V C ( l ) \ ˚ B ( v, , where V is the set of vertices of C ( l ), and ˚ B ( v,
1) is the open ball of radius1 around v ; all points on the interior of the shape bounded by these circle arcshave no neighbours (Figure 3). If l = √ , then G R d [ C ( l )] appears to have 314 ≥ √ l = √ √ ≤ l < √ × Figure 3: The non-trivial connected components of G R [ C ( l )] for varying l . Ineach square, the component consists of the shaded area, the thick black lines,and the filled nodes. Each of the remaining points in the middle of the square,which are in none of the displayed components, is a connected component of itsown.non-trivial connected components: (cid:0) C ( l ) \ B ((0 , , ∪ C ( l ) \ B (( l, l ) , (cid:1) \ { (0 , , ( l, l ) } , (cid:0) C ( l ) \ B ((0 , l ) , ∪ C ( l ) \ B (( l, , (cid:1) \ { (0 , l ) , ( l, } , { (0 , , (0 , l ) , ( l, , ( l, l ) } ∪ (cid:91) v, ∈ V C ( l ) ∩ S ( v, , where S ( v,
1) is the circle of radius 1 around v (Figure 3). For 1 / √ ≤ l < √ ,the last of these components falls apart into 4 separate components consistingof a vertex v and the arc C ( l ) ∩ S ( v, l = 1 / √ l = 2 / √ l = √ G R [ C ( l )] changes, and the lengths at whichthe chromatic number changes [Kru08]—only the critical value l = 8 / √
65 for thechromatic number does not show up here. It would be interesting to study thecase l = 8 / √
65 in more detail, and to investigate whether such a correspondencealso holds for hypercubes. 15 .2 Random walks
Our problem is obviously also related to random walk problems. In most of theseproblems, the step length has some continuous distribution (often exponential).Here, the step length is 1 in all cases, and only the new direction is chosen (e.g.,uniformly) at random. This leads to the question of determining the distributionfor the location after a fixed number of steps. Probabilistic problems of thisnature have been investigated thoroughly ever since Pearson asked the followingquestion [Pea05].A man starts from a point O and walks l yards in a straight line; hethen turns through any angle whatever and walks another l yards ina second straight line. He repeats this process n times. I require theprobability that after these n stretches he is at a distance between r and r + δr from his starting point, O .However, in related research, the space of the so called short uniform randomwalk is most often unbounded, i.e., R d (e.g., [BNSW11, BS16, Zho19]). There isat least some interest in bounded cases—as we have studied in a non probabilisticsetting—as well, but then, too the step size may vary, and the boundary maybe assumed to be reflexive (e.g., [CR87]).Although we do not venture into the probabilistic realm in this paper, thereare some obvious questions that this paper brings up. For example, we couldask essentially the same question as Pearson for bounded regions, but for a manwho is aware of his surroundings and walks stretches of length 1 without crossingthe boundary of the region. Note that the starting point is of great influence inthis problem, and a uniform distribution on the neighbourhood may either be acontinuous one (if there is an infinite number of neighbours), or discrete (if theneighbourhood is finite, such as for the midpoint of C ( √ G R d [ X ] isdisconnected, or is only connected via a node of finite degree. References [ACL +
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