Distributed Maximum Matching Verification in CONGEST
DDistributed Maximum Matching Verification in CONGEST
Mohamad Ahmadi
University of [email protected]
Fabian Kuhn
University of [email protected]
Abstract
We study the maximum cardinality matching problem in a standard distributed setting,where the nodes V of a given n -node network graph G = ( V, E ) communicate over the edges E in synchronous rounds. More specifically, we consider the distributed CONGEST model, wherein each round, each node of G can send an O (log n )-bit message to each of its neighbors. We showthat for every graph G and a matching M of G , there is a randomized CONGEST algorithm to verify M being a maximum matching of G in time O ( | M | ) and disprove it in time O ( D + (cid:96) ),where D is the diameter of G and (cid:96) is the length of a shortest augmenting path. We hope that ouralgorithm constitutes a significant step towards developing a CONGEST algorithm to compute a maximum matching in time ˜ O ( s ∗ ), where s ∗ is the size of a maximum matching. For a graph G = ( V, E ), a matching M ⊆ E is a set of pairwise disjoint edges and the maximummatching problem asks for a matching M of maximum possible cardinality (or of maximum possibleweight if the edges are weighted). Matchings have been at the center of the attention in graph theoryfor more than a century (see, e.g., [LP86]). Algorithmic problems dealing with the computation ofmatchings are among the most extensively studied problems in algorithmic graph theory. Theproblem of finding a maximum matching is on the one hand simple enough so that it can be solvedefficiently [Edm65a, Edm65b], on the other hand the problem has a rich mathematical structureand led to many important insights in graph theory and theoretical computer science. Apart fromwork in the standard sequential setting, the problem has been studied in a variety of other settingsand computational models. Exact or approximate algorithms have been developed in areas suchas online algorithms (e.g., [KVV90, EKW16]), streaming algorithms (e.g., [McG05]), sublinear-timealgorithms (e.g., [YYI09, MV13]), classic parallel algorithms (e.g., [KUW85, FGHP93]), as well asalso the recently popular massively parallel computation model (e.g., [CLM + + Distributed maximum matching:
In the distributed context, the maximum matching problemis mostly studied for networks in the following synchronous message passing model. The networkis modeled as an undirected n -node graph G = ( V, E ), where each node hosts a distributed processand the processes communicate with each other over the edges of G . As it is common practice,we identify the nodes with their processes and think of the nodes themselves as the distributedagents. Time is divided into synchronous rounds and in each round, each node v ∈ V can performsome arbitrary internal computation, send a message to each of its neighbors in G , and receive themessages of the neighbors (round r is assumed to start at time r − r ). If themessages can be of arbitrary size, this model is known as the LOCAL model [Lin92, Pel00]. In themore realistic
CONGEST model [Pel00], in each round, each node can send an arbitrary O (log n )-bitmessage to each of its neighbors. a r X i v : . [ c s . D C ] F e b ur contribution: As the main result of our paper, we give a distributed maximum matchingverification algorithm.
Theorem 1.1.
Given an undirected graph G = ( V, E ) and a matching M of G , there is a randomizeddistributed CONGEST model algorithm to test whether M is a maximum matching. If M is amaximum matching, the algorithm verifies this in time O ( | M | ) , otherwise, the algorithm disprovesit in time O ( D + (cid:96) ) , where D is the diameter of G and (cid:96) is the length of a shortest augmenting path. Our main technical contribution is a distributed algorithm that, given a matching M and aparameter x , determines if there is an augmenting path of length at most x in O ( x ) rounds of the CONGEST model. If there is an augmenting path of length at most x , the algorithm identifiestwo free (i.e., unmatched) nodes u and v between which such a path exists. We note that if thealgorithm can be extended to also construct an augmenting path of length at most x between u and v in time ˜ O ( x ), it would directly lead to an ˜ O ( s ∗ )-round algorithm for computing a maximummatching, where s ∗ is the size of a maximum matching. The reason for this follows from the classicframework of Hopcroft and Karp [HK73]. It is well-known that if we are given a matching M of size s ∗ − k for some integer k ≥
1, there is an augmenting path of length less than 2 s ∗ /k [HK73]. Hence,if we can find such a path and augment along it in time linear in the length of the path, we get atotal time of O ( s ∗ log s ∗ ) by summing over all values of k from 1 to s ∗ . The same approach has beenused in [AKO18] to compute a maximum matching in time O ( s ∗ log s ∗ ) in bipartite graphs. Whilefinding a shortest augmenting path is quite straightforward in bipartite graphs, getting an efficient CONGEST model algorithm for general graphs turns out to be much more involved. We thereforehope that our algorithm for finding the length and the endpoints of some shortest augmenting pathprovides a significant step towards also efficiently constructing a shortest augmenting path in the
CONGEST model and therefore to obtaining an ˜ O ( s ∗ )-time CONGEST algorithm to find a maximummatching. Before we discuss our algorithm, the underlying techniques and the main challenges inmore detail in Section 2, we next give a brief summary of the history of the distributed maximummatching problem.
Distributed maximal matching algorithms:
While except for [AKO18], there is no previouswork on exact solutions for the distributed maximum matching problem, there is a very extensiveand rich literature on computing approximate solutions for the problem. The most basic wayto approximate maximum matching is by computing a maximal matching , which provides a 1 / O (log n )-round algorithms in the CONGEST model. It was later shown by Ha´n´ckowiak, Karo´nski, and Panconesi [HKP98, HKP99] that maximalmatching can also be solved deterministically in polylogarithmic time in the distributed setting.The current best deterministic algorithm in the
CONGEST model (and also in the
LOCAL model) isby Fischer [Fis17] and it computes a maximal matching in O (log ∆ log n ) rounds, where ∆ is themaximum degree of the network graph G . At the cost of a higher dependency on ∆, the dependencyon n can be reduced and it was shown by Panconesi and Rizzi [PR01] that a maximal matchingcan be computed in O (∆ + log ∗ n ) rounds. The best known randomized algorithm is by Barenboimet al. [BEPS12] and it shows that (by combining with the result of [Fis17]) a maximal matchingcan be computed in O (log ∆) + O (log log n ) rounds in the CONGEST model. The known boundsin the
CONGEST model are close to optimal even when using large messages. It is known thatthere is no randomized o (cid:0) log ∆log log ∆ + (cid:113) log n log log n (cid:1) -round maximal matching algorithm in the LOCAL model [KMW16]. A very recent result further shows that there are also no randomized o (cid:0) ∆ +1 og log n log log log n (cid:1) -round algorithm and no deterministic o (cid:0) ∆ + log n log log n (cid:1) -round algorithms to compute amaximal matching [BBH + Distributed matching approximation algorithms:
There is a series of papers that target thedistributed maximum matching problem directly and that provide results that go beyond the 1 / O (1 /ε ) iterations of augmenting along a (nearly) maximalset of vertex-disjoint short augmenting paths, one is guaranteed to have a (1 − ε )-approximate so-lution for the maximum matching problem. The first distributed algorithms to use this approachare an O (log O (1 /ε ) n )-time deterministic LOCAL algorithm for computing a (1 − ε )-approximation ingraphs of girth at least 2 /ε − O (log n )-time deterministic LOCAL algorithm for com-puting a 2 / CONGEST model are by Lotker et al. [LPP08], who give a randomized algorithm to compute a (1 − ε )-approximate maximum matching in time O (log n ) for every constant ε >
0. For bipartite graphs,the running time of the algorithm depends polynomially on 1 /ε , whereas for general graphs it de-pends exponentially on 1 /ε . The algorithm was recently improved by Bar Yehuda et al. [BCGS17],who give an algorithm with time complexity O (cid:0) log ∆log log ∆ (cid:1) for computing a (1 − ε )-approximation.As in [LPP08], the time depends polynomially on 1 /ε in bipartite graphs and exponentially on 1 /ε in general graphs. Note that the time dependency on ∆ in [BCGS17] matches the lower boundof [KMW16]. In [AKO18], Ahmadi et al. give a deterministic O (cid:0) log ∆ ε + log ∆ ε (cid:1) -round CONGEST maximum matching algorithm that has an approximation factor of (1 − ε ) in bipartite graphs andan approximation factor of (2 / − ε ) in general graphs. Unlike the previous algorithms, the algo-rithm of [AKO18] is not based on the framework of Hopcroft and Karp. Instead, the algorithm firstcomputes an almost optimal fractional matching and it then rounds the fractional solution to aninteger solution by adapting an algorithm of [Fis17]. There also exist deterministic distributed algo-rithms to (1 − ε )-approximate maximum matching in polylogairhtmic time [EMR15,FGK17,GHK18],these algorithm however require the LOCAL model. The algorithms of [AKO18, FGK17, GHK18] di-rectly also work for the maximum weighted matching problem. Other distributed algorithms thatcompute constant-factor approximations for the weighted maximum matching problem appearedin [WW04, HKL06, LPP08, LPR09, Fis17, BCGS17]. We note that none of the existing approxima-tion algorithms can be used to solve the exact maximum matching problem in time o ( | E | ) in the CONGEST model.
Additional related work:
Our result can also be seen in the context of some recent interest in thecomplexity of computing exact solutions to distributed optimization problems. In particular, it wasrecently shown that several problems that are closely related to the maximum matching problemhave near-quadratic lower bounds in the
CONGEST model. In [CHKP17], it is shown that computingan optimal solutions to the maximum independent set and the minimum vertex cover problem bothrequire time ˜Ω( n ) in the CONGEST model. In [BCD + n ) lower bounds are provenfor other problems, in particular for computing an optimal solution to the minimum dominating setproblem and for computing a (7 / ε )-approximation for maximum independent set. Consequently,for maximum independent set, minimum vertex cover and minimum dominating set, the trivial O ( | E | )-time CONGEST model algorithm is almost optimal. If our result can be extended to actuallyfind the maximum matching in almost linear time, it would show that this is not true for themaximum matching problem.
Mathematical notation:
Before giving an outline of our algorithm in Section 2, we introduce some In the
LOCAL model, the algorithm can be implemented in time O (log( n ) / poly( ε )) also for general graphs. Thiswas independently also shown in a concurrent paper by Nieberg [Nie08]. W fromnode u to a node v in a graph G = ( V, E ) is a sequence of nodes (cid:104) u = v , v , . . . , v k = v (cid:105) such thatfor all j < k , { v j , v j +1 } ∈ E . A path P is a walk that is cycle-free, i.e., a walk where the nodes arepairwise distinct. Let V ( W ) denote the multi-set of the nodes in a walk W and let | W | denote thelength of the walk W , i.e., | W | = | V ( W ) |−
1. For simplicity, we write v ∈ W if v ∈ V ( W ). Moreover,we say an edge e is on walk W and write e ∈ W if e is an edge between two consecutive nodes in W . For two walks W = (cid:104) u , . . . , u s (cid:105) and W = (cid:104) v , . . . , v t (cid:105) with u s = v , we use W ◦ W to denotethe concatenation of the walks W and W . Further, for a path P = (cid:104) u , u , . . . , u i , . . . , u j , . . . , u k (cid:105) ,we use P [ u i , u j ] to denote the consecutive subsequence of P starting at node u i and ending at node u j , i.e., the subpath of P from u i to u j . We use parentheses instead of square brackets to excludethe starting or ending node from the subpath, e.g., P ( u i , u j ], P [ u i , u j ) or P ( u i , u j ). For a graph G , it is well-known that a matching M is a maximum matching of G if and only if thereis no augmenting path in G w.r.t. M . By performing a broadcast/convergecast, the size of the givenmatching can be learnt by all nodes in the graph in time linear in D , the diameter of G . After allnodes learn the size of the given matching, the algorithm looks for an augmenting path of lengthat most r in phases, where r is initially set to D and it doubles from each phase to the next. Thealgorithm stops as soon as either r > | M | or it detects a shortest augmenting path of length atmost r . Note that the length of an augmenting path cannot be more than 2 | M | + 1. Therefore, ifthe algorithm does not find a shortest augmenting path, then there is no augmenting path in G withrespect to M . The efficiency of the algorithm depends on how fast one can detect the existence ofa shortest augmenting path of length at most (cid:96) in G for an integer (cid:96) . The following main technicalresult states that this central challenging task can be accomplish efficiently. Lemma 2.1.
Given an arbitrary graph G and a matching M of G , there is a randomized algorithmto detect whether there exists an augmenting path of length at most (cid:96) in O ( (cid:96) ) rounds of the CONGEST model, with high probability.
Considering the above lemma, let us now study the time complexity of the algorithm. First, letus assume that the given matching M is a maximum matching. Then, it takes O ( D ) rounds to learnthe size of the given matching and O ( | M | ) rounds to look for shortest augmenting paths in phasessince (cid:80) | M | i =log D O (2 i ) ≤ O ( | M | ). However, since Ω( D ) is a lower bound for the size of a maximummatching, the overall time complexity of the algorithm to verify that M is a maximum matching is O ( | M | ). Now let us assume that M is not a maximum matching and (cid:96) is the length of a shortestaugmenting path. Then, it takes O ( D ) rounds to learn the size of the given matching and O ( (cid:96) )rounds for the algorithm to look for and eventually detect a shortest augmenting path in phases.Hence, the overall time complexity of the algorithm to disprove M being a maximum matching is O ( (cid:96) + D ). This overall implies Theorem 1.1. Let G = ( V, E ) be a graph, let M ⊆ E be a matching of G , and let f ∈ V be a free node (i.e., anunmatched node). Assume that we want to find a shortest augmenting path P connecting f withanother free node f (cid:48) . If the graph G is bipartite, such a path can be found by doing a breadthfirst search (BFS) along alternating paths from f . This works because in bipartite graphs, for everynode v on a shortest augmenting path P connecting f with another free node f (cid:48) , the subpath P [ f, v ]connecting f and v is also a shortest alternating path between f and v . In [Vaz12], Vazirani calls thisproperty, which holds in bipartite graphs, the BFS-honesty property . If the BFS-honesty property3olds, to find a shortest alternating path from a free node f to a node v , it suffices to know shortestalternating paths from f to all the nodes along this path. The BFS-honesty property does not holdin general graphs. A simple example that shows this is given in Figure 1. The shortest alternatingpath connecting node f with u is of length 3. The shortest alternating path connecting f with v is of length 5 and it contains node u , however the subpath connecting f with u on the alternatingpath to v is of length 4. f u vw Figure 1: The BFS-honesty property does not hold in general graphs (solid lines depict edges in thematching, dotted lines depict edges not in the matching).To show the use of the DFS-honesty property in the distributed setting more clearly, we nextsketch the algorithm of [AKO18] for finding a shortest augmenting path in a bipartite graph. Thealgorithm essentially works as follows. Every free node f ∈ V in parallel starts its own BFSexploration of G along alternating paths. The exploration of a free node f is done by propagatingits ID (i.e., f ) along alternating paths from f , where the ID is propagated by one more hop in eachsynchronous round. Whenever a node u receives the IDs of two different free nodes f and f (cid:48) in thesame round, it only forwards the ID of one of them. Note that each node only forwards a single freenode ID and it only forwards this ID once (in the round after it first receives it). This is sufficient ifthe BFS-honesty property holds. Moreover, this guarantees that IDs only traverse alternating pathsand avoid traversing cycles. Assume that the shortest augmenting path in the graph is of length (cid:96) = 2 k + 1. let P be such a path and assume that { u, v } is the middle edge of P . Note that thisimplies that the shortest alternating paths of nodes u and v are both of length k . Hence, u and v receive the ID of a free node exactly in round k and they will both forward that ID along edge { u, v } in round k + 1. When this happens, u and v learn about the fact that they are in the middleof an augmenting path of length 2 k + 1 and that path can be constructed simply by following backthe edges on which the alternating BFS traversals reached nodes u and v .Let us now discuss some of the challenges when adapting this ID dissemination protocol togeneral graphs. For simplicity, assume that we are only doing the BFS exploration from a single freenode f . Consider again the example in Figure 1. We have seen that the shortest alternating pathfrom f to v passes through node u , however the subpath from f to u is not the shortest alternatingpath from f to u . In fact, while the shortest alternating path from f to u reaches u on an unmatchededge, in order to reach node v , we have to use the shortest one of the alternating paths from f to u that reach u on a matched edge. This suggests that each node v should keep track of both kindsof shortest paths from node f and that v should forward f twice. A natural generalization of theprotocol would thus be the following: After receiving f on a shortest alternating path ending inan unmatched edge of v , v forwards f on its matched edge and after receiving f on a shortestalternating path ending in the matched edge of v , v forwards f on its unmatched edges. One wouldhope that this lets each node detect both kinds of shortest alternating paths from node f . However,as Figure 2 shows, this is not necessarily true. While in the Figure 2a, when v receives f over itsmatched edge, the ID was indeed forwarded on a shortest alternating path from f to v . However, inFigure 2b, the exploration passes through an odd cycle and node v is only reached on an alternatingwalk instead of an alternating path. In the example of Figure 2b, node w should detect that the BFStraversal passed through the odd cycle and w should therefore not forward f over its matched edge.However, it is not clear how w should distinguish between the cases in Figure 2a and Figure 2b.Note that in the BFS traversal of Figure 2a, f is not only forwarded on the alternating path to w ,4 vwz (a) f vw (b) Figure 2: Main challenge: Nodes need to be able to distinguish whether the BFS exploration reachesthem on an alternating path or only on alternating walks.but it is also forwarded through the odd cycle as in Figure 2b. In fact, the example of Figure 2 isstill a relatively simple case as odd cycles can be nested, and closed odd walks can look much morecomplicated than just passing through a single odd cycle. Detecting whether and when to forwardthe ID of a free node is the main algorithmic challenge that we face.A second challenge come from the fact that we need to do alternating BFS explorations fromall free nodes and it is not obvious how to coordinate these parallel BFS explorations while keepingthe message size small. In the bipartite case, it was enough for each node v to only participate inthe BFS exploration of a single free node f and to discard all other BFS traversals that reach node v . It is not clear whether the same thing can also be done in general graph. Luckily it turns out tostill be sufficient if each node v only participates in the BFS exploration of the first free node thatreaches v . Proving that this is sufficient is however more involved than that in the bipartite case. We start the outline of our algorithm to detect a shortest augmenting path by describing the requiredoutcome of the alternating BFS exploration in general graphs in more detail. As mentioned above,we intend to in parallel perform BFS explorations starting from all the free nodes f , . . . , f ρ . Wewill show that it is sufficient for each node v ∈ V to participate in the BFS exploration for exactlyone free node f i . This implies that at each point in time, the BFS explorations of the different freenodes f , . . . , f ρ induce a clustering of the nodes in V . There is a cluster for each free node f i , andeach node v ∈ V is either contained in exactly one of the ρ clusters or it is not contained in anycluster (i.e., has not been reached by any of the explorations). We call this induced clustering the free node clustering . The clustering is computed in synchronous rounds and we will guarantee thatit satisfies the following properties.(C1) Consider some node v ∈ V and some round number r ≥
1. If v has not joined any cluster inthe first r − v has an alternating path P of length r to a cluster center f (i.e.,a free node f ), and if all nodes of P except node v are in the cluster of node f after r − v joins the cluster of f or some other cluster with the same property. That is, v joins a cluster in round r such that afterwards, it has an alternating path P (cid:48) of length r to itscluster center such that P (cid:48) is completely contained in the cluster that v joined. Let C be thecluster that v joins and let U ⊆ C be the set of neighbors u of v such that the cluster containsan alternating path of length r from v through u to the cluster center. The set U is calledthe predecessors of v . If v joins a cluster in round r , we say that v is r -reachable (i.e., v ’sshortest alternating paths to its cluster center that are completely contained in the cluster areof length r ). If v ’s adjacent edge on the shortest alternating path of v is an unmatched edge(i.e., if r is odd), we say that v is r - -reachable and otherwise, we say that it is r - -reachable .(C2) Assume that v is r v -reachable. Let r > r v be the first round after which the cluster of v P of length r connecting v with f such that if v is r v -0-reachable, P starts with a matched edge at node v and if v is r v -1-reachable, P starts with an unmatchededge at node v (if such a round r exists). Then, after r rounds of the construction, v is awareof the existence of such a path. If v is r v -0-reachable, we say it is also r - -reachable and if itis r v -1-reachable, we say that it is also r - -reachable .To put it differently, a node in a cluster is called r -0-reachable ( r -1-reachable) if there is a shortestodd-length (even-length) alternating path of length r from the cluster center to the node that iscompletely contained in the cluster. The clustering after r rounds of the construction will be calledthe r -radius free node clustering . For the precise definition of the clustering and of the relatedterminology, we refer to Section 3.1. We will see that the r -radius free node clustering can beconstructed in r rounds in the CONGEST model. We give an outline of the distributed constructionof the clustering in the following Section 2.3. The details of the distributed construction and itsanalysis appear in Sections 3 and 4. Before discussing the distributed construction, we next sketchhow the free node clustering can be used to detect an augmenting path and why it is sufficient fordetecting a shortest augmenting path.
Detecting augmenting paths:
After computing the r -radius free node clustering for a sufficientlylarge radius r , we can use it to find an augmenting path as follows. Let u and v be two neighborsin G such that u and v are in different clusters (say for free nodes f and f (cid:48) ). Assume that for twointegers (cid:96) u , (cid:96) v ≥ { u, v } is in the matching, u is (cid:96) u -0-reachable (in its cluster), and v is (cid:96) v -0-reachable(in its cluster).2. The edge { u, v } is not in the matching, u is (cid:96) u -1-reachable (in its cluster), and v is (cid:96) v -1-reachable (in its cluster).In both cases the matching directly implies that there exists an augmenting path of length (cid:96) u + (cid:96) v + 1between the free nodes f and f (cid:48) . Further, after max { (cid:96) u , (cid:96) v } + 1 rounds, u and v are aware of theexistence of this path. Detectability of a shortest augmenting path:
It remains to show that the free node clusteringallows to find some shortest augmenting path. Assume that the length of a shortest augmentingpath in G with respect to the given matching M is 2 k + 1 for some integer k ≥
0. For an augmentingpath P = (cid:104) f = v , . . . , v (cid:96) = f (cid:48) (cid:105) of length (cid:96) = 2 k + 1 between two free nodes f and f (cid:48) , we let i ≥ j ≥ i is the largest integer such that all nodes in P [ v , v i ] are inthe cluster of f and such that j is the largest integer such that all the nodes in P [ v (cid:96) − j , v (cid:96) ] are inthe cluster of f (cid:48) . We define the rank of the augmenting path P as i + j . Note that the path P is detectable if and only if it has rank 2 k . We therefore need to show that there exists a shortestaugmenting path of rank 2 k .To prove that there is a shortest augmenting path of rank 2 k , we assume that P is a shortestaugmenting path of maximal rank and that the rank of P is less than 2 k and we show that thisleads to a contradition: this either allows to construct an augmenting path of length less than 2 k + 1or it allows to construct an augmenting path of length 2 k + 1 of larger rank. The actual proofis somewhat technical. It consists of two steps. If we assume, w.l.o.g., that i ≤ j , we first showinductively that all the nodes v i +1 , . . . , v max { i +1 ,j } are in the cluster of f (cid:48) . If j ≥ (cid:96) − j −
1, we haveproven that v (cid:96) − j − is in the cluster of f (cid:48) , which is a contradiction to the choice of j . Otherwise,node v (cid:96) − j − is in a cluster f (cid:48)(cid:48) (cid:54) = f (cid:48) (it is however possible that f (cid:48)(cid:48) = f ). We can now derive thedesired contradiction by a careful concatenation of parts of the paths connecting f (cid:48)(cid:48) with v (cid:96) − j − ,parts of the augmenting path between f and f (cid:48) , and parts of a path between f (cid:48) and v i +1 that wasconstructed in the earlier inductive argument. The details of the arguments appear in Section 3.2.6 .3 Distributed Construction of the Free Node Clustering We focus on a single step (round) of the the distributed construction of the free node clustering. Tothat end, consider graph G and matching M , and assume that the first r − t < r and ϑ ∈ { , } , every t - ϑ -reachable node correctlydetects the fact that it is t - ϑ -reachable and knows its predecessors. Then, let us explain the outlineof the approach towards implementing the r th step of the distributed construction of the clustering.Let us first focus on maintaining property (C1). To satisfy (C1), every ( r − ϑ -reachable nodesends its cluster ID over its adjacent matched edge if ϑ = 0 and over its adjacent unmatched edgesif ϑ = 1. This way, for every node that receives a cluster ID over its adjacent edge, by joining thecorresponding cluster, there would be an alternating path of length r from the cluster center tothe node such that the path is completely contained in the cluster. This maintains property (C1)for r steps of the clustering and it can be achieved in a distributed setting as explained. However,maintaining property (C2) is the main challenge as we try to elaborate in the sequel.Let us consider a node v in the cluster centered at some free node f after r − r (cid:48) - ϑ -reachable for some integers r (cid:48) < r and ϑ ∈ { , } . Let usthen assume that after the nodes have joined their corresponding clusters in the r th step, there isan alternating path of length r from f to v that has completely fallen into the cluster centered at f such that the path contains an adjacent matched edge of v if ϑ = 0 and an adjacent unmatched edgeof v otherwise. Therefore, v should learn about the existence of such a path to maintain property(C2). Nodes like v can be reached within the cluster from their corresponding cluster center in twoways; first through an alternating path from the cluster center to v such that the path is completelycontained in the cluster and contains a matched edge of v , and second through a similar path butcontaining an unmatched edge of v . Let us call these nodes that can be reached via both kinds ofpaths bireachable nodes .Let us define an odd cycle to be an alternating walk of odd length that is completely containedin a cluster and starts and ends at the same node. Node v that is the first and last node of anodd cycle is called the stem of the odd cycle. An odd cycle is said to be minimal if it has noconsecutive subsequence that is an odd cycle. Note that a minimal odd cycle can still have aconsecutive subsequence that is an even-length cycle. An odd cycle is moreover said to be reachable if either the stem is the cluster center or there is an alternating path from the cluster center to thestem of the odd cycle such that (1) it is completely contained in the cluster, (2) it is edge-disjointfrom the odd cycle, and (3) it includes the matched edge of the stem of the odd cycle. You cansee examples of reachable minimal odd cycles in Figure 2a, one with stem w and another one withstem z . All the nodes of an odd cycle except the stem are said to be strictly inside the odd cycle.Then, one can show that a node is bireachable if and only if it is strictly inside a reachable minimalodd cycle. We only need this simple observation to explain the intuition behind our approach formaintaining property (C2) and in Section 4 we formally prove the correctness of the approach. Asan example, node w is strictly inside the reachable minimal odd cycle with stem z in Figure 2a andhence bireachable, but w is not bireachable in Figure 2b.To help the nodes to distinguish whether they are strictly inside a reachable minimal odd cycleor not, we define a flow circulation protocol throughout each cluster. Let us consider the very simpleexample of a reachable minimal odd cycle in Figure 3a. When the cluster ID is sent over the middleedge of this odd cycle (i.e., e ) in both directions in the same round, we consider a flow generationof unit size over the edge and we call it the flow of e . Then, half of the generated flow is sent backtowards the stem of the odd cycle on each of the two paths. When the stem receives the whole unitflow of edge e in a single round, it learns that it is the stem of an odd cycle for which the flow isgenerated and discards the flow (it avoids sending the flow further). Whereas all the other nodes7 a) (b) Figure 3: Flow circulation in reachable minimal odd cyclesinside the odd cycle receive a flow of value less than 1. They interpret this incomplete flow receiptas being strictly inside a reachable minimal odd cycle. Moreover, they interpret the round in whichthey receive an incomplete flow for the first time as the length of an existing alternating path fromthe cluster center. Therefore, to maintain property (C2), a node detects the length of its shortestalternating path through its matched (unmatched) edge by receiving an incomplete flow for the firsttime if its shortest alternating path contains its unmatched (matched) edge.Many such odd cycles might share a common middle edge as the cycles in Figure 3b that shareedge e as their middle edge. Then, it is enough that every node divides the value of the receivedflow of e and sends them backwards until the cycle’s stem, i.e., node s , receives the whole unit flowof e . However, in case of having many interconnected and nested reachable minimal odd cycles thatdo not share a single middle edge, an edge might carry the flows of many different edges in the sameround. This is a problem when implementing the idea in the CONGEST model as we cannot boundthe number of flows that have to be sent concurrently over an edge. Instead of separately sendingflows generated at different edges e , we therefore sum all flows that have to go over the same edgeand only send aggregate values. Ideally, we would like to have the following desired differentiation; anode that receives an aggregated flow whose size is not an integer, learns that it is strictly inside atleast one reachable minimal odd cycle, and a node that receives an aggregated flow whose size is aninteger learns that it is the stem of at least one reachable minimal odd cycle but not strictly insideany such cycle and it discards the flow. To avoid that the sum of a set of fractional flows for differentedges sums to an integer, we can use randomization. Instead of always equally splitting a flow thathas to be sent over several edges, we randomly split the flow. This guarantees that w.h.p., flows onlysum up to an integer if they consist of all parts of all involved separate flows. Unfortunately, this isstill not directly implementable in the CONGEST model because we might need to split a single flowa polynomial in n many times and O (log n ) bits then are not sufficient to forward the flow value withsufficient accuracy. In order to apply the idea in the CONGEST model, we instead use flow valuesfrom a sufficiently large (polynomial size) finite field. In Section 4.4, we show that this suffices tow.h.p. obtain the same behavior as if flows for each edge were sent separately. Aggregating flowsthus allows to satisfy the congestion requirement, it however causes a number of further challengingproblems, which we present and discuss next.Let us consider the rather basic example of having only two nested reachable minimal odd cyclesin Figure 4. Let C denote the odd cycle with stem s and C (cid:48) denote the odd cycle with stem s (cid:48) .Observe that e is the middle edge of C and e (cid:48) is the middle edge of C (cid:48) . The received flow of e by x must be sent to y whereas the received flow of e (cid:48) by x must be sent to y , which requires node x totreat the two flows differently. That is, node x must recognize that the received flow of e correspondsto the odd cycle containing the alternating path ending at edge { y , x } , and the received flow of e (cid:48) corresponds to the odd cycle containing the alternating path ending at edge { y , x } . This cannotbe achieved due to the flow aggregation enforced by the congestion restriction. Therefore, node x isnot capable of correctly directing the flows along the right paths so that the flows of e only traverse8igure 4: Nested odd cycles: flow simulation of edge e (cid:48) on edge e (cid:48)(cid:48) the paths of cycle C and the flows of e (cid:48) only traverse the paths of cycle C (cid:48) .To resolve this issue and be able to still aggregate the flows, nodes should be able to treat allflows in the same way. Therefore, since every node knows its predecessors, we would like to establishthe generic regulation of always sending flows only towards all predecessors no matter what the flowis. However, by letting node x send the received flow of e (cid:48) to its only predecessor y , the nodesin the alternating path between x and s (cid:48) through y do not anymore receive any flow of e (cid:48) . To fixthis and keep node x free of treating flows differently, we eliminate the flow generation over e (cid:48) andsimulate it by generating a flow over e (cid:48)(cid:48) . That is, we shift the flow generation of cycle C (cid:48) from e (cid:48) to e (cid:48)(cid:48) . Then, half of the flow of e (cid:48)(cid:48) is sent by y to its predecessor y , and half of it is sent by x to itspredecessor y .Previously, we let the flow generation only occur over the middle edge of an odd cycle, that canbe easily recognized when an edge carries the same cluster ID in opposite directions in the sameround. Now by having flow generation over both middle edges like e as well as non-middle edges like e (cid:48)(cid:48) , we need a more involved flow generation regulation. Let every node send its cluster ID in at mostone round over its matched edge and in at most one round over its unmatched edges. A node sendsits cluster ID over its matched edges in round r if it is ( r − r (cid:48) if it is ( r (cid:48) − e nor those of e (cid:48)(cid:48) are each other’s predecessors while the endpoints send f to each other over e and e (cid:48)(cid:48) .Therefore, flow generation occur over both e and e (cid:48)(cid:48) , where e is an example of a middle edge thatthe endpoints send cluster IDs in the same round, and e (cid:48)(cid:48) is an example of a non-middle edge for ashifted flow generation that the endpoints send cluster IDs in different rounds. Also note that since y is the predecessor of y , a flow is not anymore generated over e (cid:48) within this new regulation.Now to see that shifting flow generation maintains the desired effects and avoids any side effects,let us compare the two cases of flow generation over e (cid:48) and its simulation over e (cid:48)(cid:48) . Each half of theflow of e (cid:48) is sent towards s (cid:48) , one along the path between y and s (cid:48) through y and one along the pathbetween y and s (cid:48) through y as depicted by arrows in Figure 4. In the simulation, each half of theflow of e (cid:48)(cid:48) is also sent towards s (cid:48) , one along the path between y and s (cid:48) through y and one along thepath between x and s (cid:48) through y again as depicted by arrows in Figure 4. We need the simulationto serve the purposes of the flow generation of e (cid:48) . However, there are two crucial differences thatmight question the desired effects of flow e (cid:48) if we run the simulation instead. To explain the firstdifference, consider the nodes inside odd cycle C . Nodes like y do not receive flows of e (cid:48) but receiveflows of e (cid:48)(cid:48) in the simulation. Moreover, nodes like y receive flows of e (cid:48) but not flows of e (cid:48)(cid:48) in thesimulation. The second difference is that node s (cid:48) as the stem of C (cid:48) receives the whole unit flow of e (cid:48) in a single round as desired to perceive the fact that it is the corresponding stem and discards theflow. However, in the simulation since e (cid:48)(cid:48) is not the middle edge of C (cid:48) and the flows are sent alongpaths with different lengths, s (cid:48) does not receive the whole unit flow of e (cid:48)(cid:48) in a single round. Thisavoids node s (cid:48) to perceive the fact that it is the stem of an odd cycle and the flow is further sent by9 (cid:48) . Let us see how crucial these differences are and how we can resolve them.Regarding the first difference, the decisive observation is that whenever a node receives a properfraction of a flow for the first time in round r , it detects the existence of an alternating path oflength r . Therefore, only the first receipt of such flow is important and must be at the right time fora node. All those nodes in C that differ in receiving the corresponding flows in the flow circulationof e (cid:48) and e (cid:48)(cid:48) already have received a proper fraction of flow e , and hence the receipt and the time ofreceiving later flows are irrelevant to them.However, the second difference is crucial and needs to be resolved. Note that the half flow of e (cid:48)(cid:48) is sent along the path between y and s (cid:48) through y that is the same path traversed by the half flowof e (cid:48) . Therefore, if y sends the half flow of e (cid:48)(cid:48) to y immediately in the next round of receiving thecluster ID from x , it reaches s (cid:48) at exactly the same time as the half flow of e (cid:48) would have reached s (cid:48) .Now assume that x would also have sent the other half flow of e (cid:48)(cid:48) along the path between x and s (cid:48) through y . If x would have sent the flow in the very next round of receiving the cluster ID from y ,then the flow of e (cid:48)(cid:48) would also have reached s (cid:48) in exactly the same round as the flow of e (cid:48) would havereached s (cid:48) . However, since x actually sends this flow along the path between x and s (cid:48) through y ,it reaches s (cid:48) sooner. This time difference is the difference of the length of the shortest alternatingpath between x and f ending in the matched edge of x and such a path ending in the unmatchededge of x . This difference is known by x , and x can therefore delay sending the flow by this numberof rounds and repair the unwanted side effects of the simulation. Note that x cannot send the flowin the very next round of receiving the cluster ID from y since it has not yet decided at that timeto send its cluster ID to y and hence cannot yet recognize the flow generation over e (cid:48)(cid:48) . Therefore,it has to anyway send the flow along a shorter path, e.g., the path through y .This discussed simple example inevitably abstracts away some details. In the example, flows areonly sent over alternating paths. However, if nodes always send flows to their predecessors, flowsdo not necessarily traverse alternating paths and the paths along which flows are sent might haveconsecutive unmatched edges. Then, along a path that a flow is forwarded, every node that has twoadjacent unmatched edges on the path delays forwarding the flow. We postpone further details toSections 3 and 4. In this section, we present the algorithm to detect a shortest augmenting path in time linear inthe length of the path in the
CONGEST model. The organization of this section is as follows. InSection 3.1, we formally define the free node clustering that was described in Section 2.2. We definethe clustering by giving a deterministic sequential algorithm that constructs the clustering in astep-by-step manner. Note that this deterministic algorithm is only for the purpose of providinga precise definition of the clustering. Then, in Section 3.2, we show that given such a clustering,at least one shortest augmenting path can be detected in a single round of the
CONGEST model.In Section 3.3, we provide a distributed algorithm to construct the free node clustering. Due tolack of space, the analysis of the distributed free node clustering algorithm appears in in Section 4.For the sake of simplicity, we first consider no restriction on the size of sent messages when wedescribe the algorithm and present its analysis in Section 4. We postpone dealing with the messagesize restriction to Section 4.4, where we then show how one can employ randomness to adapt thealgorithm to the
CONGEST model (along the lines described in Section 2.2). r -Radius Free Node Clustering To define the r -radius free node clustering of a graph G w.r.t. a given matching M of G , weintroduce a deterministic r -step algorithm, which we henceforth call the FNC algorithm,. The free10odes f , . . . , f ρ are the cluster centers. For all i , let C i denote the cluster that is centered at freenode f i . Initially every cluster C i only contains f i and during the execution of the algorithm morenodes potentially join the cluster. For consistency, we assume that there exists a step 0 in whichevery free node joins the cluster centered at itself, i.e., initially ∀ i ∈ [1 , ρ ] : C i = { f i } . Then, in everystep t ≥
1, every node that has not yet joined any cluster, concurrently joins the cluster centered at f i if and only if f i is the minimum-ID free node from which v has an alternating path P of length t such that V ( P ) \ { v } ⊆ C i . Throughout, let C i ( t ) denote the set of nodes in cluster C i after t stepsof the FNC algorithm. We define C ( r ) := { C ( r ) , . . . , C ρ ( r ) } to be the r -radius free node clusteringof G . See Algorithm 1 for the details of the FNC algorithm. Algorithm 1: r -Radius Free Node Clustering (FNC) Algorithm Input :
Graph G = ( V, E ), matching M of G , and integer r ≥ Output: r -Radius free node clustering of G V (cid:48) ← V \ { f , . . . , f ρ } ; forall i ∈ { , . . . , ρ } do C i ← { f i } ; forall t ∈ { , , . . . , r } do forall i ∈ { , . . . , ρ } do B i := ∅ ; forall v ∈ V (cid:48) do X ← { j | v has an alternating path P of length t s.t. V ( P ) \ { v } ⊆ C j } ; if X (cid:54) = ∅ then j (cid:48) ← argmin f jj ∈ X ; B j (cid:48) ← B j (cid:48) ∪ { v } ; V (cid:48) ← V (cid:48) \ { v } ; forall i ∈ { , . . . , ρ } do C i := C i ∪ B i ; return C ( r ) ← { C , . . . , C ρ } ;To simplify the discussions, we introduce the following definitions and terminology. In the followingdefinitions, v is an arbitrary node in G and ϑ is an arbitrary integer in { , } . We say that P is a path of v or v has a path P if P is a path starting at a free node and ending at node v . Definition 3.1 (Uniform Paths) . We say that a path P is uniform at time t ≥ if V ( P ) ⊆ C i ( t ) for some i ∈ { , . . . , ρ } . When the time t is clear from the context, we just say that P is uniform. The following lemma is a simple observation about uniform alternating paths.
Lemma 3.1.
Let P be an alternating path of length r from any free node to any node. If there isany time (possibly larger than r ) at which P is uniform, then P is uniform at time r .Proof. For a free node f i and a node v , let (cid:104) v = v r , . . . , v = f i (cid:105) be the given path P . Let t be aninteger such that P is completely contained in cluster C i after t steps of the FNC algorithm. Thatis, all nodes of P eventually join the cluster centered at f i . Thus, since nodes commit to the clustersthey join, the nodes of P cannot join any other clusters. Let us now by induction show that forall r (cid:48) ≤ r , v r (cid:48) ∈ C i ( r (cid:48) ). Since v ∈ C i (0), v joins C i in the first step of the FNC algorithm, i.e., v ∈ C i (1). Now fix an arbitrary integer r (cid:48) ≤ r such that for all r (cid:48)(cid:48) < r (cid:48) , v r (cid:48)(cid:48) ∈ C i ( r (cid:48)(cid:48) ). Node v r (cid:48) can Note that there might be some nodes in G that never join any cluster in any step of the FNC algorithm. C i . If it already joined the cluster in the first r (cid:48) − v r (cid:48) ∈ C i ( r (cid:48) ) since for all t (cid:48) ≥ C i ( t (cid:48) ) ⊆ C i ( t (cid:48) + 1). Otherwise, node v r (cid:48) must join cluster C i in step r (cid:48) of the FNC algorithm, i.e., v r (cid:48) ∈ C i ( r (cid:48) ). Hence, we can conclude that for all r (cid:48) ≤ r , v r (cid:48) ∈ C i ( r (cid:48) ). That is, for all r (cid:48) ≤ r , v r (cid:48) ∈ C i ( r ), and consequently P is uniform at time r . Definition 3.2 (Almost Uniform Paths) . We say that a path P of v is almost uniform (at time t )if V ( P ) \ { v } is uniform (at time t ). Note that every uniform path of v is also almost uniform. Definition 3.3 ( ϑ -Edges) . A free (i.e., an unmatched) edge is called a -edge, and a matched edgeis called a -edge. Definition 3.4 ( ϑ -Paths) . An alternating path P of v is called a ϑ -path of v if P contains a ϑ -edgeadjacent to v . Definition 3.5 (Predecessors) . We say u is a predecessor of v if u is the neighbor of v on a shortestuniform alternating path of v . Definition 3.6 (Reachability) . For an integer r ≥ , we say that v is r - ϑ -reachable if v has ashortest uniform ϑ -path of length r . Moreover, we say that v is r -reachable if it has a shortestuniform alternating path of length r . For the sake of consistency, we assume that every free node is 0-0-reachable and 0-1-reachable.
In this section, we show how the existence of a shortest augmenting path of length at most (cid:96) can bedetected in a single round of the
CONGEST model if the nodes of G are provided with the (cid:96) -radiusfree node clustering with respect to a given matching M of G and if in addition the (cid:96) -radius freenode clustering is well-formed in the following sense. Definition 3.7 (Well-Formed Clustering) . The r -radius free node clustering C ( r ) is said to be well-formed in a distributed setting if for all r (cid:48) ≤ r , ϑ ∈ { , } and i , every r (cid:48) - ϑ -reachable node v ∈ C i ( r ) ,beyond knowing its cluster ID, knows its predecessors and the fact of being r (cid:48) - ϑ -reachable. Before proving the aforementioned claim that is stated in the next lemma, consider the followingdefinition.
Definition 3.8 (Rank of an Augmenting Path) . For an integer (cid:96) , consider an arbitrary augmentingpath P of length (cid:96) between any pair of free nodes f s and f t . Let us name the nodes of P as (cid:104) f s = u , . . . , u (cid:96) = f t (cid:105) and let i and j be the largest integers such that the subpaths P [ f s , u i ] and P [ u (cid:96) − j , f t ] are uniform. Then, we define the rank of P to be i + j . Lemma 3.2.
Let all nodes of a given graph G be provided with the well-formed r -radius free nodeclustering with respect to a given matching M of G . If there is a shortest augmenting path of length (cid:96) ≤ r , then a shortest augmenting path can be detected in a single round of the CONGEST model.Proof.
Let us assume that there is a shortest augmenting path that is partitioned into two consec-utive subpaths such that each subpath is completely contained in a different cluster. Then, sinceall the nodes are provided with the well-formed r -radius free node clustering, the two neighboringnodes at the end of the subpaths can detect the existence of the shortest augmenting path in justa single round of the CONGEST model. To do so, they only need to inform each other about theircluster IDs and their reachabilities in a single round. Hence, to prove the lemma it is sufficient to12rove that there is a shortest augmenting path that is partitioned into two consecutive subpathssuch that each subpath is completely contained in a different cluster.Let us assume that the shortest augmenting path is of length 2 k + 1. Throughout the proof, weconsider that the graph is provided with an r -radius free clustering for some r ≥ k + 1. Then, toprove the lemma, it is enough to show that there is a shortest augmenting path of rank 2 k . Let usfirst prove the following helping claim. Claim 3.3.
If an arbitrary node v has an almost uniform ϑ -path of length k that is not uniform,then v has a uniform ϑ -path of length at most k .Proof of Claim 3.3. Let P be a path from a free node f i to v that is almost uniform butnot uniform. At time r , P [ f i , w ) is uniform since P is almost uniform. Moreover, P [ f i , w )is of length k −
1. Therefore, based on Lemma 3.1, P [ f i , v ) is uniform at time k −
1, i.e., V ( P [ f, v )) ⊆ C i ( k − v has an almost uniform alternating path, it joins some clusterin step k of the FNC algorithm if it has not yet joined any cluster. Hence, we can say that v joins some cluster C j in the first k steps of the FNC algorithm, i.e., v ∈ C j ( k ) for some j .Therefore, v has a shortest uniform alternating path P of length at most k . Note that P [ f i , v )is completely contained in C i , and P in C j . Hence P [ f i , v ) and P are disjoint. Thus, if P is a (1 − ϑ )-path of v , then the concatenation of P and P would be an augmenting path oflength less than 2 k + 1, which contradicts the assumed length of the shortest augmenting path.Therefore, P is a uniform ϑ -path of v , whose length is at most k .For the sake of contradiction, let us assume that the highest rank of any shortest augmenting path in G is T < k . Let P = (cid:104) f s = v , v , . . . , v k +1 = f t (cid:105) be an arbitrary shortest augmenting path of rank T . Then, let i and j be the largest integers such that all the nodes of P [ f s , v i ] and P [ v k +1 − j , f t ] arerespectively in clusters C s and C t , which implies i + j = T < k . Without loss of generality, let usassume that i ≤ j . To show a contradiction to our assumption on the highest rank of any shortestaugmenting path, let us first prove the following claim. Claim 3.4. . For ψ := max { i + 1 , min { j, k − j }} , all the nodes of P [ v i +1 , v ψ ] are in C t .Proof of Claim 3.4. We prove the claim by induction on z ∈ [ i + 1 , ψ ]. Let us first showthat v i +1 ∈ C t as the induction base. Due to the choice of i , node v i +1 is not in cluster C s .Therefore, P [ f s , v i +1 ] is an almost uniform ( i mod 2)-path of v i +1 from f s while v i +1 is in acluster centered at a different free node than f s . Based on Claim 3.3, v i +1 thus has a shortestuniform ( i mod 2)-path P i +1 of length at most i + 1. f s f t v ψ v i v i +1 v k − j +1 ∈ C s ∈ C t . . . . . .. . . . . . For the sake of contradiction, let us assume that v i +1 is in cluster C m for m (cid:54) = t . Moreover,path P [ v i +1 , f t ] is an (( i + 1) mod 2)-path of length 2 k − i of v i +1 . Therefore, P (cid:48) , the con-catenation of P i +1 and P [ v i +1 , f t ], is an alternating walk of length at most 2 k + 1 between f m and f t . Note that P (cid:48) cannot be a path since otherwise it would be an augmenting path thatis of length less than 2 k + 1 or rank more than i + j . It then contradicts at least one of thetwo choices of k or T . Therefore, P i +1 and P [ v i +1 , f t ] must have a common node in additionto v i +1 . Any common node of P i +1 and P [ v i +1 , f t ] must be in P [ v i +1 , v k − j ]. That is because P [ v k − j +1 , f t ] ∈ C t and P i +1 ∈ C m , where m (cid:54) = t . Let v x ∈ P [ v i +2 , v k − j ] be the closest nodeto f m on P i +1 . Note that max {| P [ f s , v x ] | , | P [ v x , f t ] |} ≤ k − i since i + 1 < x ≤ k − j and13 ≤ j . Moreover, since | P i +1 | ≤ i + 1 and v x (cid:54) = v i +1 , path P i +1 [ v x , f m ] is of length at most i . Hence, since m (cid:54) = t and m (cid:54) = s , the concatenation of P i +1 [ v x , f m ] and either P [ f s , v x ] or P [ v x , f t ] is an augmenting path of length less than 2 k + 1, which contradicts the choice of k .Therefore, v i +1 ∈ C t , i.e., m = t . Note that if ψ = i + 1, the proof of the Claim 3.4 is alreadycompleted. Therefore, for the remainder of the claim’s proof, let us assume ψ (cid:54) = i + 1, i.e., ψ = min { j, k − j } , and conclude the proof by showing the induction step.Regarding the induction step, for an arbitrary integer z ∈ [ i + 1 , ψ − P [ v i +1 , v z ] are in cluster C t and prove that node v z +1 is in C t too. Let usfirst show that v z +1 has an almost uniform ( z mod 2)-path of length at most z + 1 from f t .Let v y ∈ P [ v i +1 , v z ] be the closest node to f t on P i +1 . P [ f s , v y ] is of length less than j since y ≤ z < ψ ≤ j . Moreover, since | P i +1 | ≤ i + 1, path P i +1 [ v y , f t ] is of length at most i + 1.Hence, if P i +1 [ v y , f t ] is a (( y + 1) mod 2)-path of v y , then the concatenation of P [ f s , v y ] and P i +1 [ v y , f t ] would be an augmenting path of length j + i + 1 < k + 1, contradicting the choice of k . Therefore, P i +1 [ v y , f t ] is a ( y mod 2)-path of v y . Hence, P (cid:48) , the concatenation of P i +1 [ v y , f t ]and P [ v y , v z +1 ] is an almost uniform ( z mod 2)-path of v z +1 from f t whose length is at most z + 1.For the sake of contradiction, let us assume that v z +1 is in cluster C q for q (cid:54) = t . Then, v z +1 is in a different cluster than C t and it has an almost uniform ( z mod 2)-path of length at most z + 1 ≤ k from f t . Hence, based on Claim 3.3, v z +1 has a uniform ( z mod 2)-path of lengthat most z + 1 from f q , denoted by P z +1 . Now let v y (cid:48) ∈ P [ v z +1 , v k − j ] be the closest node to f q on P z +1 . P z +1 [ v y (cid:48) , f q ] must be a ( y (cid:48) mod 2)-path of v y (cid:48) since otherwise the concatenation of P [ v y (cid:48) , f t ] and P z +1 [ v y (cid:48) , f q ] would be an augmenting path that is of length less than 2 k + 1 orrank more than T . Moreover, let v y (cid:48)(cid:48) ∈ P [ v i +1 , v y (cid:48) ] be the closest node to f t on P i +1 . Then, P i +1 [ v y (cid:48)(cid:48) , f t ] is a (( y (cid:48)(cid:48) + 1) mod 2)-path of v y (cid:48)(cid:48) since otherwise the concatenation of P i +1 [ v y (cid:48)(cid:48) , f t ]and P [ f s , v y (cid:48)(cid:48) ] would be an augmenting path of length less than 2 k + 1 or rank more than T .Note that | P i +1 [ v y (cid:48)(cid:48) , f t ] | ≤ i + 1 and | P z +1 [ v y (cid:48) , f q ] | ≤ z + 1 ≤ ψ ≤ j . Therefore, since P [ v y (cid:48)(cid:48) , v y (cid:48) ]is of length less than 2 k − i − j , the concatenation of P i +1 [ v y (cid:48)(cid:48) , f t ], P [ v y (cid:48)(cid:48) , v y (cid:48) ] and P z +1 [ v y (cid:48) , f q ]is an augmenting path of length less than 2 k + 1 or rank more than T , contradicting the choiceof k or T .Now let us first consider the case j ≥ k . This implies that 2 k − j = min { j, k − j } . Moreover, since i + j < k , it holds that i + 1 ≤ k − j . Therefore, ψ = 2 k − j . Then, based on Claim 2, v k − j ∈ C t ,which contradicts the choice of j and concludes the proof. Let us then assume that j < k . Due tothe choice of j , let us assume that v k − j is in cluster C w for w (cid:54) = t . Moreover, path P [ v k − j , f t ] isan almost uniform ((2 k − j ) mod 2)-path of v k − j from f t whose length is j + 1. Therefore, basedon Claim 1, v k − j has a uniform ((2 k − j ) mod 2)-path P k − j of length at most j + 1 from f w . Let v x ∈ P [ v ψ +1 , v k − j ] be the closest node to f w on P k − j . Then, P k − j [ v x , f w ] is a ( x mod 2)-pathof v x since otherwise the concatenation of P k − j [ v x , f w ] and P [ v x , f t ] would be an augmenting pathof length less than 2 k + 1. Based on Claim 2, node v i +1 is in cluster C t but not C s . Therefore, P [ f s , v i +1 ] is an almost uniform but not uniform ( i mod 2)-path of v i +1 from f s . Based on Claim 1, v i +1 thus has a shortest uniform ( i mod 2)-path P i +1 of length at most i + 1. Let v x (cid:48) ∈ P [ v i +1 , v x ]be the closest node to f t on P i +1 . Then, P i +1 [ v x (cid:48) , f t ] is a uniform (( x (cid:48) + 1) mod 2)-path of v x (cid:48) whose length is at most i + 1. Then, the concatenation of P i +1 [ v x (cid:48) , f t ], P [ v x (cid:48) , v x ] and P k − j [ v x , f w ]is an augmenting path that is of length less than 2 k + 1 or rank more than i + j between f t and f w ,contradicting the choice of k or T . 14 .3 Distributed Free Node Clustering In this section, we present a distributed deterministic algorithm whose r -round execution providesthe well-formed r -radius free node clustering. This algorithm uses large messages. However, inSection 4.4, we show how to use randomness to adapt this algorithm to the CONGEST model. Thealgorithm makes all the free nodes (cluster centers) propagate their own IDs along their shortestalternating paths. It can essentially be seen as a multi-source breadth first search graph exploration.To correctly develop the well-formed free node clustering, it is crucial that the free node IDs, that wecall tokens , only traverse paths but not walks with cycles. The algorithm succeeds in preventing thetokens to traverse odd-length cycles by a technique of generating and circulating flows throughoutthe network as we see in the sequel.
Distributed r -Radius Free Node Clustering: DFNC Algorithm The algorithm is run for r rounds. Let the following variables be maintained by the nodes duringthe execution; r (0) v and r (1) v keep track of the v ’s reachabilities, cid v holds the cluster ID of v , and pred v holds the set of v ’s predecessors. At the beginning of the execution, for every free node v ,variables r (0) v and r (1) v are set to 0, variable cid v is set to v , and pred v is set to ∅ . Moreover, for everymatched node, all these variables are initially undefined and set to ⊥ . Every node v participates inthe token dissemination based on the following simple rule. For an arbitrary integer t ≥
1, in round t : • If r (0) v = t −
1, then v sends cid v over its adjacent 1-edge (if any). Otherwise, if r (1) v = t − v sends cid v over all its adjacent 0-edges (if any).Then, based on the above simple rule, every node sends tokens to its neighbors in at most tworounds, at most once over its 1-edge and at most once over its 0-edges. Token forwarding for a nodedepends on its variables considering the above simple rule. We already explained how the variablesare set for a free node. Therefore, in the first round of the execution every free node sends its IDto all its neighbors. Now let us explain how a matched node sets its variables, i.e., which clusterit joins and how it detects its reachabilities and its predecessors. Let round t be the first round inwhich a node v receives tokens. Let τ , . . . , τ j be the tokens that v receives from its neighbors inround t . Then, v sets cid v to min i τ i , and subsequently sets pred v to the set of all its neighbors thatsent cid v to v in round t . Now let us explain how node v sets its variables r (1) v and r (0) v . There aretwo types of messages sent by the nodes throughout the execution; tokens (i.e., free node IDs) and flow messages . Node v sets r (1) v and r (0) v based on the received tokens and flow messages. Beforewe explain how these variables are set by v , let us first define the flow messages by explaining flowgeneration and circulation throughout the network. Flow Generation: A flow is a key-value pair, where the key is an edge and the value is a real numberin [0 , e is simply called a flow of edge e . A flow message is thendefined to be a set of flows (i.e., key-value pairs) that are sent by a node to its neighbor in a round.A flow generation is an event that can only happen over an edge for which both endpoints belong tothe same cluster. Let us fix an arbitrary edge e = { u, w } where both endpoints belong to the samecluster. Then, we say that a flow is generated over edge e if and only if (1) none of u or w is theother one’s predecessor, and (2) both u and w send tokens to each other. Note that we only considerat most one flow generation for every edge throughout the whole execution. Let us assume that u and w are not each other’s predecessors, u sends token to w in round r u , and w sends token to u in round r w . Then, the flow generation over e is defined as an event in which u receives a singletonflow message { ( e, / } over e in round r w and w receives a singleton flow message { ( e, / } over e in round r u . It is important to note that nodes u and w might send tokens to each other in differentrounds, i.e., r u (cid:54) = r w . However, they cannot perceive the flow generation over e before they make15 owdirection ... ∗∗∗ u (a) ... ∗ u (b) ... ... ∗∗∗ u (c) Figure 5: The 3 possibilities of flow forwarding. The u ’s predecessors are marked by asterisks.sure that e carries tokens in both directions. In particular, if r w < r u , node u cannot in round r w perceive the flow receipt over e since it does not yet know whether it will send a token to w . Hence, u will perceive this flow receipt of round r w later in round r u in which it decides to send token over e and then knows that the edge carries tokens in both directions. However, we will see that u doesnot need to know about the flow receipt of round r w before round r u . Flow Circulation:
No matter if it receives a flow over its adjacent 0-edges or its adjacent 1-edge,every node always forwards the received flow to its predecessors by equally splitting the flow valueamong them. When the edge over which v receives a flow and the edges connected to its predecessorsare not all 0-edges (see Figure 5a and 5b), v forwards the flow immediately in the next round afterreceiving the flow. Otherwise (see Figure 5c), it delays forwarding the flow for r (1) v − r (0) v rounds. Anode furthermore avoids forwarding the whole flow of a single edge e in a single round (i.e., a flowof value 1 of e ). Let us see the details of the flow circulation in the following.Let I v ( t ) denote the set of all the flows that a node v receives in a round t , i.e., the union ofall the received flow messages by v in round t . Moreover, let O v ( t, e ) denote the output buffer ofa node v for its adjacent edge e in a round t , which is initially an empty set and eventually sentas a flow message over e by v in round t . Now let us fix an arbitrary node v , where E v is the setof v ’s adjacent edges that are connected to its predecessors. Node v updates its output buffers intwo ways; (1) it updates them with respect to the received flows and (2) it updates them to avoidforwarding the whole unit flow of a specific edge in a single round. Regarding the former case, fixan arbitrary round t in which v receives flows. If the edges over which v receives flows and the edgesin E v are all 0-edges (Figure 5c), let t (cid:48) := t + r (1) v − r (0) v + 1. Otherwise (see Figure 5a and 5b), let t (cid:48) := t + 1. Then, for every ( e, f ) ∈ I v ( t ), v updates its output buffers as follows: ∀ e (cid:48) ∈ E v : O v ( t (cid:48) , e (cid:48) ) ← O v ( t (cid:48) , e (cid:48) ) ∪ ( e, f | E v | ) .Now regarding the latter way of output buffers update, fix an arbitrary round t (cid:48) . At the beginningof round t (cid:48) , let O v ( t (cid:48) ) be the set of all the flows in the output buffers of v for round t (cid:48) , i.e., O v ( t (cid:48) ) := (cid:83) e (cid:48) ∈ E v O v ( t (cid:48) , e (cid:48) ). Then, let S v ( e, t (cid:48) ) be the sum of flow values of a specific edge e that are sent by v in round t (cid:48) , i.e., S v ( e, t (cid:48) ) := (cid:80) ( e,f ) ∈ O v ( t (cid:48) ) f . For every edge e , if S v ( e, t (cid:48) ) = 1, node v removes allflows of edge e from all its output buffers of round t (cid:48) . That is, v removes all flows of e and we saythat v discards the flow of e . After discarding all such flows, for every e (cid:48) ∈ E v , v forwards O v ( t (cid:48) , e (cid:48) )over edge e (cid:48) in round t (cid:48) if O v ( t (cid:48) , e (cid:48) ) (cid:54) = ∅ . Setting Variables r (1) v and r (0) v (Reachability Detection): We say that round t is an incomplete round for v if node v sends flow in round t + 1. Let t be the first round in which v receives tokens or bethe first incomplete round for v . If t is an even integer, v assigns t to r (1) v , otherwise, v assigns t to r (0) v . Note that the first round that a node receives a token (if any) is before its first incompleteround (if any) since it receives flows from the nodes it has already sent tokens to.16 The Analysis of the DFNC Algorithm
In this section, we show that an r -round execution of the DFNC algorithm provides the well-formed r -radius free node clustering as stated in the following lemma. Lemma 4.1.
For any integer r , an r -round execution of the DFNC algorithm on a graph G and amatching M of G provides the well-formed r -radius free node clustering of G with respect to M . We show the correctness of this lemma by induction on r . For the entire Section 4, we fix an arbitrarygraph G = ( V G , E G ) and an arbitrary matching M of G , where { f , . . . , f ρ } ⊆ V G are the free nodes.Thus, throughout this section, when we refer to unmatched (free) nodes, matched nodes, augmentingpaths, etc., they are always considered in G with respect to M . Let E also be an arbitrary executionof the DFNC algorithm on G and M . For all i and t , let C i ( t ) denote the set of the nodes in thecluster centered at free node f i in the t -radius free node clustering of G with respect to M . For all i and t , let D i ( t ) also denote the set of the nodes that join the cluster centered at free node f i inthe first t rounds of E . Then, for all t , C ( t ) := { C ( t ) , . . . , C ρ ( t ) } and D ( t ) := { D ( t ) , . . . , D ρ ( t ) } arerespectively the t -radius free node clustering and the clustering provided by the first t rounds of E .We provide the necessary arguments for the induction proof in three sections. In Section I,we start with proving the induction base, i.e., D (0) = C (0). Thereafter, we consider the followingassumption as the induction hypothesis of the proof: I.H.
For every t < r , the first t rounds of E provides the well-formed t -radius free nodeclustering of G with respect to M .We will then show that the clustering provided by the first r rounds of E is the r -radius free nodeclustering, i.e., D ( r ) = C ( r ). Moreover, it will be shown that the variables holding the set ofpredecessors of the nodes that join clusters in round r of E are correctly set. We continue to provethat the provided clustering is also well-formed in the next two sections. In Section II, we show thatfor every node v , r ( ϑ ) v = r if v is r - ϑ -reachable. In Section III, we show that for every node v , v is r - ϑ -reachable if r ( ϑ ) v = r . Putting all these pieces together completes the induction argument andproves Lemma 4.1. Let us start with the proof of the induction base in following lemma.
Lemma 4.2. D (0) is the well-formed -radius free node clustering of G , i.e., C (0) = D (0) .Proof. Consider the state of the nodes right at the beginning of E , i.e., at time 0. Then, the providedclustering is the set of singleton sets containing the free nodes, which is the same as the 0-radius freenode clustering of G , i.e., C (0) = D (0). Every free node v is 0-0-reachable and 0-1-reachable while r (0) v = 0 and r (1) v = 0. Moreover, we have no other 0-0-reachable or 0-1-reachable nodes except thefree nodes in G . Furthermore, every free node v has no predecessors while pred v = ∅ . Therefore,the clustering provided by the DFNC algorithm at the beginning of the execution, i.e., at time 0, isthe well-formed 0-radius free node clustering of G .Now let us show that the provided clustering after r rounds of E is actually the r -radius freenode clustering. Lemma 4.3.
Assuming that I.H. holds, the clustering provided by the first r rounds of E is the r -radius free node clustering, i.e., D ( r ) = C ( r ) . roof. In addition to having for all t < r , D ( t ) = C ( t ), the provided clustering by the first r − E (i.e. D ( r − i , C i ( r ) \ C i ( r −
1) = D i ( r ) \ D i ( r − i ∈ [1 , ρ ], fix an arbitrary node c ∈ C i ( r ) \ C i ( r −
1) and an arbitrary node d ∈ D i ( r ) \ D i ( r − c ∈ D i ( r ) \ D i ( r −
1) and d ∈ C i ( r ) \ C i ( r − c ∈ D i ( r ) \ D i ( r − c joins the cluster centered at f i in the r th round of E . Considering the DFNC algorithm, it is then enough to show the followingthree points; (1) c does not receive any token before round r , (2) c receives token f i in round r , and(3) f i is the minimum one among all the tokens that c receives in round r . Regarding the first point,since c ∈ C i ( r ) \ C i ( r − c joins cluster C i in the r th step of the FNC algorithm. Therefore, c is not in any cluster of C ( r −
1) and hence not in any cluster of D ( r − c does not joinany cluster in the first r − E . Hence, node c does not receive any token before round r asevery node joins a cluster in the first round of receiving a token. Regarding the second point, notethat c joins cluster C i in the r th step of the FNC algorithm. Therefore, c has an almost uniformalternating path P c of length r from f i in clustering C ( r − D ( r −
1) of G .Let w be c ’s neighbor on P c and connected to c by a ϑ -edge for some ϑ ∈ { , } . Let us show that P c [ f i , w ] is a shortest uniform (1 − ϑ )-path of w in D ( r − w has a uniform (1 − ϑ )-path P (cid:48) c of length t < r −
1. Then, based on Lemma 3.1, P (cid:48) c isuniform at time t and hence V ( P (cid:48) c ) ⊆ C i ( t ). Node c is not in any cluster of C ( t ) and thus not in P (cid:48) c .Thus, the concatenation of P (cid:48) c and (cid:104) w, c (cid:105) is an almost uniform alternating path of length less than r of c at time t . Therefore, c would join a cluster at latest in round t of the FNC algorithm, whichcontradicts c not being in any cluster of C ( r − P c [ f i , w ] is a shortest uniform (1 − ϑ )-pathof w in D ( r − w is ( r − − ϑ )-reachable and hence r (1 − ϑ ) w = r −
1. Therefore, w sendstoken cid w = f i to c in round r . Now regarding the last point, for the sake of contradiction, let usassume that c receives a token f j < f i over a ϑ -edge, say { c, w (cid:48) } , in round r . Then, r (1 − ϑ ) w (cid:48) = r − E . Then, since D ( r −
1) is well-formed, w (cid:48) is ( r − − ϑ )-reachable in clustering D ( r −
1) andconsequently in C ( r −
1) of G . Let P (cid:48)(cid:48) c be a shortest uniform (1 − ϑ )-path of w (cid:48) in clustering C ( r − G . Based on Lemma 3.1, P (cid:48)(cid:48) c is uniform at time r −
1, i.e., V ( P (cid:48)(cid:48) c ) ⊆ C j ( r − c (cid:54)∈ P (cid:48)(cid:48) c because c ∈ C i ( r ) \ C i ( r −
1) and hence c (cid:54)∈ C j ( r − P (cid:48)(cid:48) c and (cid:104) w (cid:48) , c (cid:105) isan almost uniform alternating path of length r of c from f j in C ( r − f j < f i . Therefore, c must have not joined C i in the r th step of the FNC algorithm, which contradicts the choice of c .Now let us show that d ∈ C i ( r ) \ C i ( r − d joins the cluster centeredat f i in the r th step of the FNC algorithm execution on G . Therefore, it is enough to show thefollowing three points; (1) d does not join any cluster in the first r − d has an almost uniform alternating path of length r in clustering C ( r −
1) and (3) f i is theminimum-ID free node from which d has an almost uniform alternating path of length r in clustering C ( r − d is not in D ( r −
1) and consequently not in C ( r − d does not join any cluster in the first r − d ∈ D i ( r ) \ D i ( r − d joins the cluster centered at f i in the r th round of E . Therefore, for some node z and integer ϕ ∈ { , } , d receives token f i from z over its ϕ -edgein round r , and hence r (1 − ϕ ) z = r −
1. As a result, since the first r − E provides thewell-formed ( r − z must have a shortest uniform (1 − ϕ )-path P d oflength r − D ( r −
1) and consequently in C ( r − P d must beuniform at time r −
1, whereas d is not in any cluster of D ( r − d (cid:54)∈ P d and hence theconcatenation of P d and (cid:104) z, d (cid:105) is an almost uniform ϕ -path of length r in C ( r − d has an almost uniform alternatingpath P (cid:48) d of length r from a free node f j (cid:48) < f i in C ( r −
1) and consequently in D ( r − z (cid:48) be d ’sneighbor on P (cid:48) d . Let us show that P (cid:48) d [ f j (cid:48) , z (cid:48) ] is a shortest uniform (1 − ϕ )-path of z (cid:48) . Let z (cid:48) have a18horter uniform (1 − ϕ )-path P (cid:48)(cid:48) d of length t (cid:48) < r −
1. Then, based on Lemma 3.1, P (cid:48)(cid:48) d is uniform attime t (cid:48) in clustering D ( t (cid:48) ). Then, since d is not in any cluster of D ( t (cid:48) ), the concatenation of P (cid:48)(cid:48) d and (cid:104) z (cid:48) , d (cid:105) would be an almost uniform alternating path of length less than r , and hence d would joinsome cluster before round r of E , which is contradictory. Therefore, P (cid:48) d [ f j (cid:48) , z (cid:48) ] is a shortest uniform(1 − ϕ )-path of length r − z (cid:48) . Therefore, r (1 − ϕ ) z (cid:48) = r −
1, and hence z (cid:48) sends token f j (cid:48) to d inround r . This contradicts node d joining C i in round r since f j (cid:48) < f j .As a final step of this section, we state in the following lemma that predecessors are correctlyset in the execution of the DFNC algorithm. Lemma 4.4.
Assuming that I.H. holds, for all i and every node v ∈ D i ( r ) , pred v is properly set to v ’s predecessors in round r of E .Proof. In addition to having for all t < r , D ( t ) = C ( t ), the provided clustering by the first r − E (i.e. D ( r − i and every node v ∈ D i ( r ) \ D i ( r − pred v is properly set to the set of v ’s predecessors in round r of E .Fix an arbitrary node v ∈ D i ( r ) \ D i ( r −
1) for any i . We only need to show that v receives token f i from node w in round r if and only if w is a predecessor of v .First, let us fix an arbitrary predecessor w of v , and show that w sends token f i to v in round r .Let P be a shortest uniform alternating path of v on which w is v ’s neighbor. Note that | P | = r . Let P be a ϑ -path for some ϑ ∈ { , } . Now let us show that P [ f i , w ] is a shortest uniform (1 − ϑ )-pathof w . For the sake of contradiction, let us assume that w has uniform (1 − ϑ )-path P (cid:48) of length t < r −
1. Based on Lemma 3.1, path P (cid:48) is uniform at time t , i.e., V ( P (cid:48) ) ⊆ C i ( t ). However, v / ∈ C i ( t )since v ∈ D i ( r ) \ D i ( r − t < r −
1. Therefore, v (cid:54)∈ P (cid:48) , and hence the concatenation of P (cid:48) and (cid:104) w , v (cid:105) would be an almost uniform ϑ -path of length less than r of v . Hence, v must havejoined some cluster before round r , which is contradictory. Therefore, P [ f i , w ] is a shortest uniform(1 − ϑ )-path of w . Hence, w is ( r − − ϕ )-reachable. Since D ( r −
1) is well-formed, it thusholds that r (1 − ϕ ) w = r −
1. As a result, w sends f i to v in round r .Second, let us fix an arbitrary node w that sends token f i to v in round r over a ϑ -edge forsome ϑ ∈ { , } . Thus, cid w = f i and r (1 − ϑ ) w = r −
1. Therefore, since D ( r −
1) is the well-formed( r − G , w has a shortest uniform alternating path P (cid:48)(cid:48) of length r −
1, which is a (1 − ϑ )-path. Based on Lemma 3.1, P (cid:48)(cid:48) is uniform at time r − v has not yet joined any cluster. Hence, v (cid:54)∈ P (cid:48)(cid:48) and therefore the concatenation of P (cid:48)(cid:48) and (cid:104) w , v (cid:105) isanalternating path. It will become a uniform ϑ -path of length r of v after node v joins D i in round r of E . Therefore, w is the neighbor of v on a uniform ϑ -path of v , and hence a predecessor of v . In this section, we show that after r rounds of the DFNC algorithm execution, for all ϑ ∈ { , } ,every r - ϑ -reachable node properly detects its r - ϑ -reachability as stated in the following Lemma: Lemma 4.5.
Assuming that I.H. holds, for every node v and integer ϑ ∈ { , } , r ( ϑ ) v = r in execution E if v is r - ϑ -reachable. To prove this Lemma, we need to first point out a few observations about flow circulationthroughout the network while running the DFNC algorithm. To that end, we provide a series ofnecessary definitions and helper lemmas in the sequel. Since every node forwards the received flowsto their predecessors, the flows do not necessarily traverse alternating paths. We call the paths along19hich a node sends flows towards the cluster center the node’s shortcuts . Consider the followingdefinition for a more precise description of a shortcut.
Definition 4.1 (Shortcuts) . For any matched node v and free node f i , a uniform path P := (cid:104) v = v , v , . . . , v (cid:96) = f i (cid:105) is called a shortcut of v if for all j < (cid:96) , v j +1 is the predecessor of v j . Lemma 4.6.
Assuming that I.H. holds, let distinct nodes u and v respectively be r u -reachable and r v -reachable for r v ≤ r u ≤ r . Then, v has no shortcut containing u .Proof. For the sake of contradiction, let us assume that v has a shortcut S that contains u . Let usname the nodes in S [ v, u ] as (cid:104) v = w , . . . , w m = u (cid:105) for some integer m . To show a contradiction,we prove by induction that r v > r u . To that end, we prove that for all j < m , the reachability of w j is strictly greater than that of w j +1 . For the induction base, we prove that the reachability of v = w is strictly greater than that of w . Considering I.H. and Lemma 4.3, D ( r ) = C ( r ). Therefore,since v is r v -reachable, v must receive a token for the first time in round r v . Hence, since w is thepredecessor of v , w must send a token to v . Therefore, w must have a received a token beforeround r v . Let t < r v ≤ r be the first round that w receives a token. Due to I.H. , w is t -reachable.Therefore, the reachability of v = w is strictly greater than that of w . Now since the reachabilityof w is less than that of v and hence less than r , we can inductively employ I.H. and prove thatthe reachability of w m = u is less than that of w = v . That is, r v > r u , which is contradictory. Lemma 4.7.
Assuming that I.H. holds, let a t -reachable node v have a shortcut S that contains a t (cid:48) - ϑ -reachable node u for any integers t (cid:48) , t ≤ r and ϑ ∈ { , } . If u ’s adjacent edge on S [ v, u ] is a (1 − ϑ ) -edge, then t (cid:48) < t .Proof. Let v be in cluster C i , and hence S is a path between v and f i . Let w be u ’s neighbor in S [ v, u ], where consequently { w, u } is a (1 − ϑ )-edge. Node u is the predecessor of w . Therefore, u is w ’s neighbor in a shortest uniform alternating path P of w . Since { w, u } is a (1 − ϑ )-edge and w is connected to its predecessors, namely u , with (1 − ϑ )-edges, every shortest uniform alternatingpath of w must be a (1 − ϑ )-path. Thus, P is a (1 − ϑ )-path. Hence P [ u, f i ] is a uniform ϑ -path of u . Since u is t (cid:48) - ϑ -reachable, it holds that | P [ u, f i ] | ≥ t (cid:48) . Therefore, | P | > t (cid:48) , that is w is t (cid:48)(cid:48) -reachablefor some t (cid:48) < t (cid:48)(cid:48) . If w = v , the proof is concluded. Otherwise, based on Lemma 4.6, the reachabilityof w is strictly less than that of v , which concludes t (cid:48) < t .To prove Lemma 4.5, we benefit from some specific way of marking some of the shortcuts andaccordingly labeling some of the nodes. Shortcut marking.
Let P be an arbitrary uniform ϑ -path of any node v in the cluster centeredat some free node f i . Then, we mark a shortcut S of a node z with respect to P when for everyedge { u, w } ∈ P that u is the predecessor of w , if S contains w then w ’s neighbor on S [ w, f i ] is u . Node labeling.
Considering path P , we label a node with P + v when it has a marked shortcutwith respect to P that contains v . Further, we label a node with P − v when it has no marked shortcutwith respect to P that contains v . Note that each node can have either label P + v or P − v , but notboth. Lemma 4.8.
Assuming that I.H. holds, let an arbitrary node v in cluster C i be r - ϑ -reachable and r (cid:48) - (1 − ϑ ) -reachable, where r (cid:48) < r . There is an edge { w, u } on every shortest uniform ϑ -path P of v such that all the nodes in P [ u, v ] are labeled P + v , node w is labeled P − v and | P [ f i , w ] | ≥ r (cid:48) . roof. Throughout the proof consider all marking and labeling with respect to path P . Let us namethe nodes in P as (cid:104) v = v , . . . , v r = f i (cid:105) . Then, let t be the integer such that | P [ v t , f i ] | = r (cid:48) . Since r (cid:48) < r , P [ v, v t ] is of length at least 1. Since P [ v t , f i ] is of length r (cid:48) , node v t is r (cid:48)(cid:48) -reachable for some r (cid:48)(cid:48) ≤ r (cid:48) whereas v is r (cid:48) -reachable. Then, based on Lemma 4.6, v t has no shortcut containing v and isconsequently labeled P − v . Note that every node in P has a marked shortcut. Therefore, since everyshortcut of v obviously contains v , node v has a marked shortcut containing itself and thus labeled P + v . Path P [ v, v t ] has one endpoint labeled P + v and one endpoint labeled P − v . Therefore, since allthe nodes in P [ v, v t ] have either label P + v or P − v , there is an edge in P [ v, v t ] whose endpoints havedifferent labels. Therefore, edge { u, w } is the closest edge to v on P whose endpoints have differentlabels.Let us next study the time it takes for a flow to traverse shortcuts. To do so, we define the promoted length of a shortcut (or a consecutive subpath of a shortcut) as the time it takes for a flowto traverse the path. The traversal time of a path by a flow is actually the sum of the length ofthe path and all the delays caused by the inner nodes in flow forwarding along the path. However,we would like to have a generalized definition for any walks rather than having the definition onlyfor shortcuts. Let us first formally define the delay by a node along a walk as follows. For ease ofdiscussion, for any ϕ ∈ { , } , we say that a node is ∞ - ϕ -reachable when it has no uniform ϕ -pathin G . Definition 4.2 (Delay) . Consider any walk P and node v ∈ P , where v is r - -reachable and r - -reachable. The delay by node v along walk P , denoted by d ( P, v ) , is defined to be r − r if v hastwo adjacent -edges on P and otherwise. Definition 4.3 (Promoted Length) . The promoted length of walk P is denoted and defined by (cid:107) P (cid:107) := | P | + (cid:80) v ∈ P d ( P, v ) . Lemma 4.9.
Assuming that I.H. holds, for any (cid:96) ≤ r , the promoted length of a shortcut of an (cid:96) -reachable node equals (cid:96) .Proof. Fix an arbitrary (cid:96) -reachable node u . Without loss of generality, let u be (cid:96) -0-reachable. Let S be an arbitrary shortcut of u . Further, let (cid:96) (cid:48) be the promoted length of S , i.e., (cid:107) S (cid:107) = (cid:96) (cid:48) . Then,the goal is to show that (cid:96) = (cid:96) (cid:48) . Let us first assume that S is an alternating path, and hence, S is ashortest uniform 0-path of u . Therefore, since the promoted length of an alternating path is equalto the length of the path, it holds that (cid:96) (cid:48) = (cid:96) , concluding the proof. Now let us assume that S isnot an alternating path. Let u , . . . , u t be the nodes in S that have two adjacent 0-edges on S suchthat for all j , u j is closer to u on S than u j − . For every node u j that has two adjacent 0-edgesin S , there are integers o j and z j < o j such that u j is z j -0-reachable and o j -1-reachable. For everynode u j , if it has two adjacent 0-edges in S , let d j := o j − z j , and otherwise let d j := 0. To provethat (cid:96) = (cid:96) (cid:48) , it is enough to show that there is a shortest uniform 0-path of length (cid:96) (cid:48) of u . To do so,we construct one in t phases.Let P initially be S , which is of promoted length (cid:96) (cid:48) . Then, in t phases we gradually transform P to a shortest uniform 0-path of u while the promoted length of P remains the same. To do so,in each phase, we update P in a way that the number of nodes with two adjacent 0-edges on P reduces by 1. In every phase j , 1 ≤ j ≤ t , we change P by replacing P [ u j , f i ] with a shortest uniform1-path of length | P [ u j , f i ] | + d j of u j . Note that every node in P [ u, u j ) has a shortcut containing u j such that u j ’s adjacent edge in P [ u, u j ] is a 0-edge. Therefore, based on Lemma 4.7, for every nodein P [ u, u j ), there is an integer (cid:96) (cid:48)(cid:48) > o j such that the node is (cid:96) (cid:48)(cid:48) -reachable. Moreover, it is easy tosee that for every node in P [ u j , f i ], there is an integer (cid:96) (cid:48)(cid:48) ≤ o j such that the node is (cid:96) (cid:48)(cid:48) -reachable.Therefore, P [ u, u j ) and P [ u j , f i ] have no common node, and consequently, P is still a path after its21hange in phase j . Furthermore, since u j does not have two adjacent 0-edges in P anymore, thepromoted length of P remains the same.In the last phase, we replace P [ u t , f i ] with a shortest uniform 1-path of u t . Moreover, every nodeis the next node’s successor in path P [ u, u t ]. Therefore, P is a shortest uniform alternating pathof u whose promoted length remained (cid:96) (cid:48) throughout the t phases as we argued. Since P is now analternating path, its length is the same as its promoted length, i.e., (cid:96) (cid:48) . Therefore, since a shortestuniform alternating path of v must be of length (cid:96) , we can conclude that (cid:96) = (cid:96) (cid:48) . Lemma 4.10.
Assuming that I.H. holds, let any node v assigns a flow of an arbitrary edge e to besent in round t ≤ r + 1 . Then, v does not assign any flow of e to be sent in any round except round t .Proof. Let u and w be the two endpoints of e . Let r u be the round in which u sends token to w ,and r w be the round in which w sends token to u . We first show that both r u and r w are at most r .Since v receives a flow of e , the flow must have traversed a shortcut of u or w and reached v (notethat as a special case when v is one of e ’s endpoints, the shortcut has length 0). Without loss ofgenerality, let us assume that a flow of e has traversed a shortcut S of u and reached v . Hence, theflow is assigned to be sent by w after round r u . Therefore, node v also assigns the flow to be sentafter round r u . That is r u < t and hence r u ≤ r . Let us show that it also holds that r w ≤ r . Tostudy the non-trivial case, let us assume that r u < r w . (1)Moreover, let us assume that w is z (cid:48) - ϑ -reachable and z -(1 − ϑ )-reachable for some integers ϑ ∈{ , } , z and z (cid:48) < z . A flow is generated over edge e . Hence, node u is not a predecessor of w , andtherefore w can only receive a token from u after round z (cid:48) , i.e., z (cid:48) < r u . (2)Therefore, since r u ≤ r , it holds that z (cid:48) < r . Hence, considering I.H. , node w has set its variable r ( ϑ ) w to z (cid:48) and must send token over its (1 − ϑ )-edges in round z (cid:48) + 1. However, based on (1) and(2), z (cid:48) + 1 (cid:54) = r w . Node w does not therefore send token to u over e in round z (cid:48) + 1. This concludesthat e is a ϑ -edge (and hence ϑ = 0). Therefore, node w must send a token to u in round z + 1, i.e., r w = z + 1 . (3)Moreover, node w must delay flow of e by z − z (cid:48) rounds. This means that t > r u + z − z (cid:48) . Considering(2), it results in having t > z + 1. Together with (3), it concludes that r w < t and hence r w ≤ r .We just proved that both r u and r w are at most r , i.e., r u ≤ r and r w ≤ r . (4)Let S u be the set of all the shortcuts of u containing v , and S w be the set of all the shortcutsof w containing v . A flow of e that reaches v should either traverse a shortcut in S w or S u . Hence,in both cases we show that the flow is assigned by v to be sent in the same round, i.e., t . Withoutloss of generality, let the flow that is assigned to be sent by v in round t has traversed a shortcut S ∈ S w to reach v . Let P be the concatenation of (cid:104) w, u (cid:105) and S . Then, considering the time that ittakes for the flow to traverse S and be sent by v in round t , it holds that t = r u + d ( P, w ) + (cid:107) S [ w, v ] (cid:107) + d ( S, v ) + 1 . (5)22ote that as a special case when v = w , it holds that (cid:107) S [ w, v ] (cid:107) = 0 and d ( S, v ) = 0, and the equation(5) properly shows the correct calculation of the round in which the flow is assigned to be sent by v .First let us show that every flow that traverses a shortcut in S w and reaches v is assigned to besent by v in round t . To that end, let us fix an arbitrary shortcut S w ∈ S w and an arbitrary flowthat traverses S w . Let P w be the concatenation of (cid:104) u, w (cid:105) and S w . Let us assume that this flow isassigned to be sent by v in round t w := r u + d ( P w , w ) + (cid:107) S w (cid:107) + d ( S w , v ) + 1 and show that t = t w : t (5) = r u + d ( P, w ) + (cid:107) S [ w, v ] (cid:107) + d ( S, v ) + 1= r u + d ( P w , w ) + (cid:107) S [ w, v ] (cid:107) + d ( S, v ) + 1= r u + d ( P w , w ) + (cid:107) S (cid:107) − (cid:107) S [ v, f i ] (cid:107) + 1= r u + d ( P w , w ) + (cid:107) S w (cid:107) − (cid:107) S w [ v, f i ] (cid:107) + 1= r u + d ( P w , w ) + (cid:107) S w (cid:107) + d ( S w , v ) + 1= t w The second equality above comes from the fact that a node is either connected to all its predecessorsby 0-edges or by a 1-edge. The forth equality above is because the promoted length of all shortcutsof a node have the same length, due to Lemma 4.9.Second let us show that every flow that traverses a shortcut in S u and reaches v is assigned tobe sent by v in round t . To that end, let us fix an arbitrary shortcut S u ∈ S u and an arbitraryflow that traverses S u . Let P u be the concatenation of (cid:104) w, u (cid:105) and S u . Let us assume that this flowis assigned to be sent by v in round t u := r w + d ( P u , u ) + (cid:107) S u [ u, v ] (cid:107) + d ( S u , v ) + 1 and show that t = t u . Let e be a ϕ -edge for some ϕ ∈ { , } . Based on equation (5) and I.H. , node u must be( r u − − ϕ )-reachable. Moreover, the length of a shortest uniform (1 − ϕ )-path of u is of length (cid:107) S u (cid:107) + d ( P u , u ). Therefore, it holds that r u = (cid:107) S u (cid:107) + d ( P u , u ) + 1 . (6)Considering w ’s shortcut S and the path P := S ◦ (cid:104) u, w (cid:105) , it similarly holds that r w = (cid:107) S (cid:107) + d ( P, w ) + 1 . (7)Now let us show that t = t u as follows. t (5) = r u + d ( P, w ) + (cid:107) S [ w, v ] (cid:107) + d ( S, v ) + 1= r u + d ( P, w ) + (cid:107) S (cid:107) − (cid:107) S [ v, f i ] (cid:107) + 1 (7) = r u + r w − (cid:107) S [ v, f i ] (cid:107) (6) = (cid:107) S u (cid:107) + d ( P u , u ) + 1 + r w − (cid:107) S [ v, f i ] (cid:107) = (cid:107) S u (cid:107) + d ( P u , u ) + 1 + r w − (cid:107) S u [ v, f i ] (cid:107) = r w + d ( P u , u ) + (cid:107) S u [ u, v ] (cid:107) + d ( S u , v ) + 1= t u Now let us at the end of this section provide the proof of Lemma 4.5 below:
Proof of Lemma 4.5.
Considering an r - ϑ -reachable node v , we show that r ( ϑ ) v = r after r rounds of E . Since v is r - ϑ -reachable, it has a shortest uniform alternating path of length at most r . Let us23rst consider the case when v is r -reachable, i.e., the shortest uniform alternating path of v is oflength r . Since D ( r −
1) is the well-formed ( r − v does not join anycluster in the first r − E . Moreover, based on Lemma 4.3, D ( r ) is the r -radius free nodeclustering. Therefore, v must join its cluster in round r of E . Hence, r is the first round in which v receives a token. Note that the length of any 0-path is odd, and the length of any 1-path is even.Therefore, r is even if ϑ = 1, and it is odd otherwise. Thus, based on the DFNC algorithm, node v sets r ( ϑ ) v to r in round r of E .For the remainder of the proof, let us consider the case when v is r (cid:48) -reachable for r (cid:48) < r , andhence, v joins its cluster before round r . That is, v is r - ϑ -reachable and r (cid:48) -(1 − ϑ )-reachable. Then,to show that r ( ϑ ) v = r , we need to show that r is the first incomplete round for v . Since the first r − E provides the well-formed ( r − G and v is r - ϑ -reachable, v cannot have an incomplete round before round r . Therefore, it is enough to show that v has anincomplete round in the first r rounds of E .Let P be an arbitrary shortest uniform ϑ -path of v . Then, based on Lemma 4.8, let e := { u, w } bethe edge on P where all the nodes in P [ u, v ] are labeled P + v , node w is labeled P − v and | P [ w, f i ] | ≥ r (cid:48) .Let us first show that a flow is generated over edge e . To this end, we need to show that u and w send tokens to each other and none of them is the other one’s predecessor.Let us first show that u and w are not each other’s predecessor. For the sake of contradiction,let us assume otherwise. Let us first consider the case when u is a predecessor of w . Let S u be amarked shortcut of u that contains v . S u does not contain w , and hence the concatenation of S u and (cid:104) w, u (cid:105) is a marked shortcut of w that contains v . Therefore, w must have been labeled P + v , whichis contradictory. Now let us consider the case when w is a predecessor of u . Then, u ’s neighbor on S u must be w . Hence, S u [ w, f i ] is a marked shortcut of w that contains v . Therefore, w must havebeen labeled P + v , which is again contradictory.Next we show that u and w send tokens to each other. Let { u, w } be a ϕ -edge for an integer ϕ ∈ { , } . First, let us show that w sends token to u . Since P [ w, f i ] is a uniform (1 − ϕ )-path oflength less than r of w , node w is (cid:96) w -(1 − ϕ )-reachable for some (cid:96) w < r . Therefore, since D ( r − r -radius free node clustering, r (1 − ϕ ) w < r . Hence, w sends token to u at latestin round | P [ w, f i ] | + 1 ≤ r . Now, let us show that u sends token to w . Every node in P [ v, u ] islabeled P + v and hence has a shortcut containing v . Therefore, based on Lemma 4.6, for every nodein P ( v, u ], there is an integer (cid:96) > r (cid:48) such that the node is (cid:96) -reachable. Let P (cid:48) be a shortest uniform(1 − ϑ )-path of v , which is of length r (cid:48) . Then, every node in P (cid:48) ( v, f i ] (cid:96) (cid:48) -reachable for some (cid:96) (cid:48) < r (cid:48) .Therefore, P [ v, u ] and P (cid:48) have no common node except v . Then, the concatenation of P [ v, u ] and P (cid:48) is a uniform (1 − ϑ )-path of u . Note that this concatenated path is of length less than r sincebased on Lemma 4.8, | P [ w, f i ] | ≥ r (cid:48) . Therefore, since D ( r −
1) is the well-formed r -radius free nodeclustering, r (1 − ϕ ) u < r . Hence, u sends token to w in the first r rounds of E .We show that r is the first incomplete round for v in which v receives a flow of e but not thewhole flow. Let S + be a marked shortcut of u containing v . Since every node has a marked shortcutand w has no marked shortcut containing v , let S − be a marked shortcut of w that does not contain v . Since a flow of e is sent along S − and this shortcut does not contain v , node v does not receivethe whole flow of e . Hence, to show that v receives a proper fraction of e in round r , it is enoughto show the following two facts; (1) None of the nodes in S + [ u, v ] discards the whole flow of e , andhence, v receives a flow of e , (2) Node v receives a flow of e in the first r rounds of E .For the sake of contradiction, let z ∈ S + [ u, v ] be the node that discards the whole flow of e .Therefore, all the flows of e are received by z , and hence, S − must contain z . Then, the concatenationof S − [ w, z ] and S + [ z, f i ] is a marked shortcut of w that contains v , which contradicts w being labeled P − v . As a result, there is no node in S + [ u, v ] that discards the whole flow of e . Thus, since v receivesa flow of e over S + [ u, v ]. 24ow let u be (cid:96) u - ϕ -reachable and (cid:96) (cid:48) u -(1 − ϕ )-reachable (recall that e was considered to be a ϕ -edge for a ϕ ∈ { , } ). Every node in P [ v, u ] is labeled P + v and thus has a shortcut containing v .Therefore, based on Lemma 4.6, for every node in P ( v, u ], there is an integer (cid:96) > r (cid:48) such that thenode is (cid:96) -reachable. Moreover, it is easy to see that for every node in P (cid:48) , there is an integer (cid:96) ≤ r (cid:48) such that the node is (cid:96) -reachable. Therefore, the concatenation of P (cid:48) and P [ v, u ] is a path, and inparticular a uniform (1 − ϕ )-path of length r (cid:48) + | P [ v, u ] | of u . Hence, (cid:96) (cid:48) u ≤ r (cid:48) + | P [ v, u ] | . (8)Let P + v := (cid:104) w, u (cid:105) ◦ S + . If S + is a (1 − ϕ )-path of u , u is (cid:96) (cid:48) u -reachable, and therefore, (cid:107) S + (cid:107) = (cid:96) (cid:48) u dueto Lemma 4.9. Otherwise, u is (cid:96) u -reachable and (cid:107) S + (cid:107) = (cid:96) u = (cid:96) (cid:48) u − d ( P + v , u ). Since in case S + is a(1 − ϕ )-path, it holds that d ( P + v , u ) = 0, we can overall conclude that (cid:107) S + (cid:107) = (cid:96) (cid:48) u − d ( P + v , u ) (8) ≤ r (cid:48) + | P [ v, u ] | − d ( P + v , u ) . (9)Moreover, letting r w be the round in which w sends token to u , r w ≤ | P [ w, f i ] | + 1 . (10)Note that the flow that traverse S + reaches v in round t := r w + d ( P + v , u ) + (cid:107) S + [ u, v ] (cid:107) . Let us showthat t ≤ r in the following: t := r w + d ( P + v , u ) + (cid:107) S + [ u, v ] (cid:107) (10) ≤ | P [ w, f i ] | + 1 + d ( P + v , u ) + (cid:107) S + [ u, v ] (cid:107) = | P [ w, f i ] | + 1 + d ( P + v , u ) + (cid:107) S + (cid:107) − (cid:107) S + [ v, f i ] (cid:107) (9) ≤ | P [ w, f i ] | + 1 + d ( P + v , u ) + r (cid:48) + | P [ v, u ] | − d ( P + v , u ) − (cid:107) S + [ v, f i ] (cid:107) = | P [ w, f i ] | + 1 + r (cid:48) + | P [ v, u ] | − (cid:107) S + [ v, f i ] (cid:107) = | P [ w, f i ] | + 1 + r (cid:48) + | P [ v, u ] | − r (cid:48) = | P [ w, f i ] | + 1 + | P [ v, u ] | = | P | = r In this section, we show that after r rounds of the DFNC algorithm execution, every node thatdetects r - ϑ -reachability is actually an r - ϑ -reachable node as stated in the following Lemma: Lemma 4.11.
Assuming that I.H. holds, for arbitrary node v and integer ϑ ∈ { , } , node v is r - ϑ -reachable if r ( ϑ ) v = r in execution E . A node v sets its variable r ( ϑ ) v to an integer r when round r is either the first round in which v receives a token or the first incomplete round for v . In both cases we need to show that there actuallyexists a uniform alternating path of length r for v . We will see that the challenging case is when v sets its variable r ( ϑ ) v to r because of having round r as its first incomplete round. Let us first present25 = u w vf i x x x x x x x P u P w Figure 6: P u and P w are initially the concatenation of the shortcuts of u and w with (cid:104) u, w (cid:105) . Nodes x , . . . , x are the nodes with two adjacent 0-edges on P u and P w in ascending order by their 1-reachabilities.the outline of proving the claim in this challenging case. If node v sets its variable r ( ϑ ) v to r becauseof having round r as its first incomplete round, then there must clearly exist some edge { u, w } suchthat a proper fraction of the flow of { u, w } reaches v in round r . In this case, we show that there isactually a uniform alternating path of length r for v that contains { u, w } . Let us assume that v is r (cid:48) -reachable. We first show that there are two shortcuts of u and w such that they have no common (cid:96) -reachable node for (cid:96) ≥ r (cid:48) and only one of them contains v (see Figure 6). We show this claimby presenting Algorithm 2 that actually constructs the shortcuts (stated in Lemma 4.12). Then,we transform each of the shortcuts to an alternating path. To do so, let P u be the concatenationof the shortcut of u and (cid:104) u, w (cid:105) , and let P w be the concatenation of the shortcut of w and (cid:104) u, w (cid:105) .Then, let x , . . . , x be the nodes having two adjacent 0-edges on P u or P w in Figure 6. Note that u and w might also have two adjacent 0 edges on the paths, e.g., x . Let nodes x , . . . , x be inascending order by their 1-reachabilities. Then, we present Algorithm 3 to transform the shortcutsto alternating paths in phases. In phase i , if x i ∈ P w , the algorithm replace P w [ f, x i ] with a uniform1-path of x i that is disjoint from P u [ f, u ] and contains v . Otherwise, the algorithm replace P u [ f, x i ]with a uniform 1-path of x i that is disjoint from P w [ v, w ]. Doing that, x i does not anymore have twoadjacent 0-edges on the paths, and hence we reduce the number of nodes with two adjacent 0-edgeson the paths by 1 in each phase. Therefore, eventually P u and P w become alternating paths, andthe concatenation of P u [ f, u ], (cid:104) u, w (cid:105) and P w [ f, w ] becomes a uniform alternating path of length r of v . We present the described procedure to change the shortcuts to the corresponding alternatingpaths in Algorithm 3.Let us now start providing a few definitions and a few observations through a set of helperlemmas in the sequel, and finally at the end of this section we present the proof of Lemma 4.11. Forease of discussion throughout this section, for every node v in a cluster of the free node clustering,let R ( v ) denote the reachability of v , i.e., R ( v ) := l if v is l -reachable. Lemma 4.12.
Assuming that I.H. holds, let v be an arbitrary r (cid:48) -reachable node, and let { u, w } bean arbitrary edge for which there exists a shortcut of u and a shortcut of w such that only one ofthem contains v . Moreover, assume that there does not exist a node x such that x is l -reachable for l ≥ r (cid:48) and such that all shortcuts of u and w contain x . Then, there exists a shortcut of u and ashortcut of w with no common l -reachable node for any l ≥ r (cid:48) such that only one of them contains v .Proof. Let us prove this lemma by constructing the desired shortcuts of u and w . We present analgorithm to actually construct the shortcuts (the pseudocode is given in Algorithm 2). Before we26iscuss the algorithm in detail, let us present the outline of the algorithm in the following. Letus initially define paths P u := (cid:104) u (cid:105) and P w := (cid:104) w (cid:105) . We update P u and P w in phases such that P u always remains a consecutive subpath of a shortcut of u starting at u and P w always remains aconsecutive subpath of a shortcut of w starting at w . Let t u and t w be the variables that respectivelymaintain the last nodes of P u and P w . Variables t u and t w are initially set to u and w respectively,and they get updated after updating the paths in each phase. Initially P u and P w are clearlydisjoint. Moreover, since at least one of u or w has a shortcut containing v , it initially holds thatmax { R ( t u ) , R ( t w ) } > r (cid:48) due to Lemma 4.6. In every phase, P u and P w are updated such thatmax { R ( t u ) , R ( t w ) } decreases while P w and P u remain disjoint. We will show that the phases arerun in a way such that eventually max { R ( t u ) , R ( t w ) } = r (cid:48) and exactly one of paths P u or P w contains v . Then, we construct the desired shortcuts S u and S w as follows. S u is the concatenationof P u and a shortcut of t u , and S w is the concatenation of P w and a shortcut of t w . It is easy to seethat S u and S w are shortcuts of u and w respectively. Paths S u and S w have no common l -reachablenode for any l ≥ r (cid:48) since P u and P w are disjoint and also every node of the shortcuts of t u and t w ,except t u and t w , is l -reachable for some l < r (cid:48) . Hence, further considering that exactly one of P u or P w contains v , it holds that exactly one of S u or S w contains v . In the rest of the proof, we will showhow to update P u and P w in phases so that eventually after some phase max { R ( t u ) , R ( t w ) } = r (cid:48) and exactly one of paths P u or P w contains v .Let us label all nodes in V G as follows. Every node that has a shortcut containing v is labeled v (1) , and every node that has a shortcut not containing v is labeled v (0) . Note that a node can haveboth labels v (1) and v (0) . At the beginning of the execution, t u has label v ( σ ) and t w has label v (1 − σ ) for some integer σ ∈ { , } since there exist a shortcut of u and a shortcut of w such that only oneof them contains v . Therefore, to eventually have the paths P u and P w as desired, it is enough todecrease max { R ( t u ) , R ( t w ) } in every phase while maintaining the following invariant: Invariant A : There exists an integer σ ∈ { , } s.t. t u is labeled v ( σ ) , t w is labeled v (1 − σ ) ,and P u ∩ P w = ∅ . That is because at the beginning of the path construction execution, max { R ( t u ) , R ( t w ) } > r (cid:48) andinvariant A holds. Moreover, based on Lemma 4.6, the only node that is l -reachable for l ≤ r (cid:48) andhas a shortcut containing v is node v itself. Therefore, one can observe that if one can always updatethe paths while maintaining invariant A , eventually max { R ( t u ) , R ( t w ) } = r (cid:48) after some updatingphase.A single phase of updating P u and P w performs the following four steps. However, it is crucial tonote that a successful update in the second or third steps concludes the phase and avoids executingthe next steps. Without loss of generality, let us assume that at the beginning of the phase, it holdsthat R ( t u ) ≤ R ( t w ). By symmetry, we can similarly state everything in the sequel for the case ofhaving R ( t w ) ≤ R ( t u ). • Step 1: If R ( t u ) = R ( t w ), P w ← P w ◦ (cid:104) t w , z (cid:105) , where z is a predecessor of t w that has label v (1 − σ ) . • Step 2: If t u has a predecessor t (cid:54)∈ P w that is labeled v ( σ ) , then P u ← P u ◦ (cid:104) t u , t (cid:105) . • Step 3:
Else, let t ∈ P w be a predecessor of t u that is labeled v ( σ ) . If there is a shortcut S of u that is disjoint from P w [ w, t ], let t (cid:48) ∈ S be the closest node to u on S with R ( t (cid:48) ) < R ( t u ),and then P u ← S [ u, t (cid:48) ] and P w ← P w [ w, t ]. • Step 4:
Else, let S (cid:48) be any shortcut of u or w that does not contain t . Let t ∈ P w [ w, t ] ∩ S (cid:48) be the closets node to t on P w . Moreover, let t be the closest node to t in S (cid:48) [ t , f ] such that R ( t ) < R ( t u ). Then, P u ← P u ◦ (cid:104) t u , t (cid:105) and P w ← P w [ w, t ] ◦ S (cid:48) [ t , t ].Now let us show by induction that every phase successfully updates paths P u and P w , where27ax { R ( t u ) , R ( t w ) } decreases while invariant A is maintained. To do so, consider an arbitraryphase of the path construction such that at the beginning of the phase, invariant A holds andmax { R ( t u ) , R ( t w ) } > r (cid:48) . Let ˆ t be max { R ( t u ) , R ( t w ) } at the beginning of the phase. We will showthat after this updating phase, max { R ( t u ) , R ( t w ) } < ˆ t while invariant A is maintained.Consider an arbitrary node s , where R ( s ) > R ( v ). If s has label v (1) , then its neighbor onits shortcut that contains v is labeled v (1) (note that s (cid:54) = v since R ( s ) > R ( v )). Hence, s has apredecessor labeled v (1) . If s has label v (0) , then its neighbor on its shortcut that does not contain v is labeled v (0) , and hence, s has a predecessor labeled v (0) . Therefore, we can say that every node s that is labeled v ( x ) , for any x ∈ { , } , has a predecessor labeled v ( x ) if R ( s ) > R ( v ).Now let us consider the first step. In case R ( t w ) = R ( t u ), it updates P w such that R ( t w ) becomesless than R ( t u ). Let us assume that at the beginning of the phase, R ( t u ) = R ( t w ). Every node in P u is l -reachable for some l ≥ R ( t u ). However, every predecessor of t w is l (cid:48) -reachable for some l (cid:48) < R ( t w ). Therefore, no predecessor of t w is in P u . Moreover, since R ( t w ) > R ( v ) and t w is labeled v (1 − σ ) , t w has a predecessor labeled v (1 − σ ) . Hence, P w can be successfully updated such that R ( t w )decreases while invariant A remains true.After Step 1 , we are sure that R ( t w ) (cid:54) = R ( t u ) and hence R ( t w ) < R ( t u ). Now let us show that atleast one of the three steps, from Step 2 to Step 4 , is successful in updating the paths. If the secondstep is successful to update P u , after the update, R ( t u ) = R ( t ) < ˆ t while t w is still labeled v (1 − σ ) as P w is not changed. Therefore, since t (cid:54)∈ P w , invariant A remains true and max { R ( t u ) , R ( t w ) } decreases. Now let us assume that the second step is not successful to update P u . Since R ( t u ) > R ( v )and t u is labeled v ( σ ) , node t u must have a predecessor t that is labeled v ( σ ) . Since the second stepis not successful, t ∈ P w . Node t is the predecessor of t u , and hence, R ( t ) < R ( t u ) ≤ ˆ t . Moreover,due to the choice of t (cid:48) , R ( t (cid:48) ) < R ( t u ) ≤ ˆ t . Therefore, if the third step is successful, by updatingthe paths, t u and t w are updated by setting t u ← t (cid:48) and t w ← t . Hence, R ( t u ) = R ( t (cid:48) ) < ˆ t and R ( t w ) = R ( t ) < ˆ t while P u ∩ P w = ∅ .Now let us assume that the second and third steps are not successful in updating the paths.Let us show that the fourth step is then guaranteed to be successful in updating the paths. Since R ( t ) > R ( v ), based on the lemma’s assumption, t cannot be a common node of all the shortcuts of u and w . Therefore, there must be a shortcut S (cid:48) of u or w that does not contain t . If S (cid:48) is a shortcut of w , it is easy to see that it has a common node with P w [ w, t ], e.g. w . Otherwise, if S (cid:48) is a shortcut of u , then S (cid:48) must have a common node with P w [ w, t ] since the second step was not successful. Hence,there exists the closest node t ∈ P w [ w, t ] ∩ S (cid:48) to t on P w . Thus, P w [ t , t ] ∩ S (cid:48) [ t , t ] = { t } . Now letus show that S (cid:48) [ t , t ] ∩ P u = ∅ . For the sake of contradiction, let us assume that t (cid:48)(cid:48) ∈ S (cid:48) [ t , t ] ∩ P u is the closest node to t on S (cid:48) . Then, the concatenation P u [ u, t (cid:48)(cid:48) ] and S (cid:48) [ t (cid:48)(cid:48) , t ] is a shortcut of u that is disjoint from P w [ w, t ], which contradicts the fact that the third step was not successful.Therefore, P u and P w remain disjoint after the update while max { R ( t u ) , R ( t w ) } decreases. Notethat since in Step 3 and
Step 4 , we respectively choose t (cid:48) and t as the closest nodes to u and t with reachability smaller than R ( t u ), we definitely have a phase in which max { R ( t u ) , R ( t w ) } = r (cid:48) and v is the endpoint of one of the paths.In the following, we borrow a few notations and definitions from [Vaz12]. Let us henceforthconsider an implicit direction on shortest 0-paths and 1-paths of the nodes from the cluster centerstowards the nodes. Let us consider a shortest uniform ϑ -path P of any node u and a shortest uniform ϕ -path Q of any node w for ϑ, ϕ ∈ { , } to define the following notions. Definition 4.4 (Common Edge) . An edge e on both paths P and Q is called a common edge of P and Q . Definition 4.5 (Forward/Backward Edge) . Considering the implicit direction on P and Q , if both P and Q traverse e in the same direction, e is called a forward edge and otherwise a backward edge. lgorithm 2: The Shortcuts Construction Algorithm. Shortcut-Construction(
G, u, w, r (cid:48) ) t w ← w t u ← u while max { R ( t u ) , R ( t w ) } (cid:54) = r (cid:48) do Update ( P w , P u , t w , t u ) T u ← a shortcut of t u T w ← a shortcut of t w S u ← P u ◦ T u S w ← P w ◦ T w return S u , S w Update( P w , P u , t w , t u ) Step 1: if R ( t u ) = R ( t w ) then P w ← P w ◦ (cid:104) t w , z (cid:105) for a predecessor z of t w that is labeled v (1 − σ ) t w ← z Step 2: if t u has a predecessor t (cid:54)∈ P w labeled v ( σ ) then P u ← P u ◦ (cid:104) t u , t (cid:105) t u ← t return Step 3: let t ∈ P w be a predecessor of t u that is labeled v ( σ ) if ∃ shortcut S of u s.t. S ∩ P w [ w, t ] = ∅ then let t (cid:48) ∈ S be the closest node to u on S with R ( t (cid:48) ) < R ( t u ) P u ← S [ u, t (cid:48) ] t u ← t (cid:48) P w ← P w [ w, t ] t w ← t return Step 4: else let S (cid:48) be any shortcut of u or w that does not contain t let t ∈ P w [ w, t ] ∩ S (cid:48) be the closets node to t on P w let t be the closest node to t in S (cid:48) [ t , f ] such that R ( t ) < R ( t u ) P u ← P u ◦ (cid:104) t u , t (cid:105) t u ← t P w ← P w [ w, t ] ◦ S (cid:48) [ t , t ] t w ← t efinition 4.6 (Separator) . If P and Q have a common edge and the induced graph by paths P and Q gets disconnected by removing edge e , then e is called a separator. Note that a separator edge of P and Q must be a forward edge. Definition 4.7 (Tenacity) . For a r (cid:48) - ϑ -reachable and r - (1 − ϑ ) -reachable node v , the tenacity of v is defined to be r + r (cid:48) , denoted by tn ( v ) . For a ϕ -edge e = { u, w } that u is (cid:96) - (1 − ϕ ) -reachable and (cid:96) - (1 − ϕ ) -reachable, the tenacity of e is defined to be (cid:96) + (cid:96) + 1 , denoted by tn ( e ) . Lemma 4.13.
Assuming that I.H. holds, let P be a shortest uniform ψ -path of any node w , and Q be a shortest uniform ϕ -path of any node u for ψ, ϕ ∈ { , } . Let us assume that there exists a node v ∈ P of tenacity greater than | P | + | Q | such that v (cid:54)∈ Q . If u (cid:54) = w or ϕ (cid:54) = ψ , the closest commonedge of Q and P [ v, w ] to the cluster center on Q is a separator.Proof. Let f be the cluster center of u and w , and let P [ f, v ] be a (1 − ϑ )-path of v for any ϑ ∈ { , } .Let e := { x (cid:48) , x } be the closest common edge of Q and P [ v, w ] to f on Q such that x (cid:48) is closer to f than x on Q .Let us first show that e is a forward edge of P and Q . For the sake of contradiction, let us assumethat e is a backward edge of P and Q . Let L be the concatenation of Q [ f, x (cid:48) ] and P [ v, x (cid:48) ]. Since e is abackward edge, L is an alternating walk. Due to the choice of e , paths Q [ f, x (cid:48) ] and P [ v, x ] are disjointand hence L is a path. Moreover, since P is alternating and P [ f, v ] is a (1 − ϑ )-path, v ’s adjacentedge on L is a ϑ -edge. Hence, L is a uniform ϑ -path of v . Consequently, the shortest uniform ϑ -pathof v is of length at most | Q [ f, x (cid:48) ] | + | P [ v, x (cid:48) ] | while the shortest uniform (1 − ϑ )-path of v is of lengthat most | P [ f, v ] | . Hence, tenacity of v should be at most | P [ f, v ] | + | Q [ f, x (cid:48) ] | + | P [ v, x (cid:48) ] | . Since | P [ f, v ] | + | P [ v, x (cid:48) ] | ≤ | P | and | Q [ f, x (cid:48) ] | < | Q | , tenacity of v should therefore be at most | P | + | Q | ,which contradicts the assumed tenacity of v . It concludes that e is a forward edge of P and Q .To show that e is a separator of P and Q , it is enough to show that Q [ f, x (cid:48) ] ∩ P [ x, w ] = ∅ and P [ f, x (cid:48) ] ∩ Q [ x, u ] = ∅ . The former equality holds as a direct implication of the choice of e . Let usthen show that P [ f, x (cid:48) ] ∩ Q [ x, u ] = ∅ . Let us first show that P [ f, v ] ∩ Q [ x, u ] = ∅ . For the sake ofcontradiction, let us assume that P [ f, v ] and Q [ x, u ] have a common edge. Let e (cid:48) = { y (cid:48) , y } be theclosest such edge to f on P such that y (cid:48) is closer to f than y on P . Let also e (cid:48)(cid:48) = { z (cid:48) , z } be theclosest common edge of Q [ x (cid:48) , y ] and P [ v, x ] to v on P such that z (cid:48) is closer to v than z on P . Notethat if Q [ x, y ] and P [ v, x (cid:48) ] are disjoint, then e and e (cid:48)(cid:48) are the same, i.e., z (cid:48) = x (cid:48) and z = x . If e (cid:54) = e (cid:48)(cid:48) ,then by applying a similar argument to that in the second paragraph of this proof, we can showthat e (cid:48)(cid:48) is also a forward edge.Now let us show that e (cid:48) is also a forward edge. For the sake of contradiction let us assume that e (cid:48) is a backward edge. To show a contradiction, let us prove that tn ( v ) < | P | + | Q | . Since e (cid:48) isassumed to be a backward edge, path Y , the concatenation of P [ v, z (cid:48) ], Q [ z (cid:48) , y (cid:48) ] and P [ f, y (cid:48) ] is auniform ϑ -path of v . Then, to show that tn ( v ) < | P | + | Q | , it is enough to show that | Y | ≤ | Q | .Note that | P [ v, z (cid:48) ] | + | P [ f, y (cid:48) ] | < | P [ f, x (cid:48) ] | and | Q [ z (cid:48) , y (cid:48) ] | < | Q [ x (cid:48) , u ] | . Hence, to show that | Y | ≤ | Q | ,it suffices to prove that | P [ f, x (cid:48) ] | ≤ | Q [ f, x (cid:48) ] | . Since Q [ f, x (cid:48) ] and P [ x (cid:48) , w ] are disjoint and e = { x (cid:48) , x } is a forward edge of P and Q , the concatenation of Q [ f, x (cid:48) ] and P [ x (cid:48) , w ] is a uniform ψ -path of w . Moreover, since P is a shortest uniform ψ -path of w , | Q [ f, x (cid:48) ] | + | P [ x (cid:48) , w ] | ≥ | P | . Therefore, | P [ f, x (cid:48) ] | ≤ | Q [ f, x (cid:48) ] | . This concludes that e (cid:48) is a forward edge.Considering the choice of e (cid:48) and the fact that e (cid:48) is a forward edge, the concatenation of P [ f, y ]and Q [ y, u ], must be a uniform ϕ -path of u . We already proved that | P [ f, x (cid:48) ] | ≤ | Q [ f, x (cid:48) ] | . Therefore,since | P [ f, y ] | < | P [ f, x (cid:48) ] | , it holds that | P [ f, y ] | ≤ | Q [ f, x (cid:48) ] | . Moreover, | Q [ y, u ] | < | Q [ x (cid:48) , u ] | . Hence, | P [ f, y ] | + | Q [ y, u ] | < | Q [ f, x (cid:48) ] | + | Q [ x (cid:48) , u ] | = | Q | . This contradicts the assumption that Q is a shortestuniform ϕ -path of u . Hence, Q [ x, u ] and P [ f, v ] have no common edge.It is left to show that Q [ x, u ] and P [ v, x (cid:48) ] have no common edge. For the sake of contradiction,let us assume that edge ˆ e = { s, s (cid:48) } is the closest common edge of Q [ x, u ] and P [ v, x (cid:48) ] to v on P ,30here s (cid:48) is closer to v than s on P . With a similar argument to that in the second paragraph ofthis proof, one can show that ˆ e is a forward edge. Therefore, path Z , the concatenation of P [ f, s ]and Q [ s, u ] is uniform ϕ -path of u . We already proved that | P [ f, x (cid:48) ] | ≤ | Q [ f, x (cid:48) ] | . Therefore, since | P [ f, s ] | < | P [ f, x (cid:48) ] | , it holds that | P [ f, s ] | < | Q [ f, x (cid:48) ] | . Moreover, | Q [ s, u ] | < | Q [ x (cid:48) , u ] | . Therefore, | P [ f, s ] | + | Q [ s, u ] | < | Q [ f, x (cid:48) ] | + | Q [ x (cid:48) , u ] | = | Q | . Therefore, Z is of length less than | Q | and a uniform ϕ -path of u . This contradicts the assumption on the length of Q as the shortest ϕ -path of u . Thisleads to having Q [ x, u ] and P [ f, x (cid:48) ] with no common edge, which concludes the proof. Lemma 4.14.
Assuming that I.H. holds, let an arbitrary node w send a token to its neighbor u inround r w ≤ r of execution E . Letting S w be a shortcut of w , it holds that r w = (cid:107) S w ◦ (cid:104) w, u (cid:105)(cid:107) .Proof. Let w be (cid:96) (cid:48) - ϑ -reachable and (cid:96) -(1 − ϑ )-reachable for some integers ϑ ∈ { , } and (cid:96) (cid:48) < (cid:96) . Basedon Lemma 4.9, it holds that (cid:107) S w (cid:107) = (cid:96) (cid:48) . (11)Let us consider the following two possibilities separately:(a) { w, u } is a (1 − ϑ )-edge.Due to I.H. , w sends the token to u in round (cid:96) (cid:48) + 1. Moreover, since w ’s adjacent edge on S w is a ϑ -edge and { w, u } is a (1 − ϑ )-edge, it holds that d ( S w ◦ (cid:104) w, u (cid:105) , w ) = 0. Hence, r w = (cid:96) (cid:48) + 1 (11) = (cid:107) S w (cid:107) + 1= (cid:107) S w ◦ (cid:104) w, u (cid:105)(cid:107) − d ( S w ◦ (cid:104) w, u (cid:105) , w ) − (cid:107) S w ◦ (cid:104) w, u (cid:105)(cid:107) .(b) { w, u } is a ϑ -edge.Due to I.H. , w sends the token to u in round (cid:96) + 1. However, since both w ’s adjacent edge on S w and { w, u } are ϑ -edges (i.e., ϑ = 0), it holds that d ( S w ◦ (cid:104) w, u (cid:105) , w ) = (cid:96) − (cid:96) (cid:48) . Hence, r w = (cid:96) + 1= (cid:96) − (cid:96) (cid:48) + (cid:96) (cid:48) + 1= d ( S w ◦ (cid:104) w, u (cid:105) , w ) + (cid:96) (cid:48) + 1 (11) = d ( S w ◦ (cid:104) w, u (cid:105) , w ) + (cid:107) S w (cid:107) + 1= (cid:107) S w ◦ (cid:104) w, u (cid:105)(cid:107) . Lemma 4.15.
Assuming that I.H. holds, let v be an arbitrary r (cid:48) -reachable node for which round r is the first incomplete round. If v receives a proper fraction of the flow of an edge e in round r ofexecution E , then tn ( e ) ≤ tn ( v ) .Proof. To prove the lemma, we show that tenacity of v is at least r + r (cid:48) and tenacity of e is r + r (cid:48) .Let f be the cluster center of node v .Let us first show that tenacity of v is at least r + r (cid:48) . For the sake of contradiction, let us assumeotherwise. Node v is r (cid:48) -reachable. Let ϑ ∈ { , } be the integer such that v is r (cid:48) -(1 − ϑ )-reachable.Therefore, v is r (cid:48)(cid:48) - ϑ -reachable for some r (cid:48)(cid:48) < r . Hence, r (cid:48)(cid:48) must be an incomplete round for v , whichcontradicts the assumption on r as the first incomplete round v . Hence, tn ( v ) ≥ r + r (cid:48) . (12)31ext, we show that tenacity of e is r + r (cid:48) . Let u and w be the two endpoints of e such that w has a shortcut S along which a proper fraction of the flow of e is sent and received by v in round r . Let r u be the round in which u sends a token to w , and r w be the round in which w sends atoken to u . Let e be a ψ -edge for any ψ ∈ { , } . Therefore, since the clustering is the well-formed( r − G , u and w are respectively ( r u − − ψ )-reachable and( r w − − ψ )-reachable. Therefore, tn ( e ) = r u + r w − d ( S, v ) = 0. Let v and v be v ’s neighbors on S such that v is the predecessorof v . If ϑ = 0, then v is connected to v by a 1-edge. Then, v must be connected to v by a 0-edgeas it can only have one adjacent 1-edge. Therefore, it holds that d ( S, v ) = 0. Now let us considerthe possibility of ϑ = 1. Hence, v is connected to v by a 0-edge. Now we show that v is connectedto v by a 1-edge. For the sake of contradiction, let us assume otherwise. Then, since v sends aflow of e to v in round r , node v must have send a token to v (over a 0-edge) in some round r (cid:48)(cid:48) < r .Therefore, due to I.H. , v must be a ( r (cid:48)(cid:48) − r (cid:48)(cid:48) − v , which contradicts round r being the first incomplete round of v . Therefore, we canconclude that all cases, it holds that d ( S, v ) = 0 . (14)Now let us conclude the proof as follows: tn ( e ) (13) = r u + r w − (4 . = r u + (cid:107)(cid:104) u, w (cid:105) ◦ S w (cid:107) − r u + (cid:107)(cid:104) u, w (cid:105) ◦ S w [ w, v ] (cid:107) + d ( S, w ) + (cid:107) S [ v, f ] (cid:107) − (14) = r u + (cid:107)(cid:104) u, w (cid:105) ◦ S w [ w, v ] (cid:107) − (cid:107) S [ v, f ] (cid:107) = r u + (cid:107)(cid:104) u, w (cid:105) ◦ S w [ w, v ] (cid:107) − r (cid:48) = r + r (cid:48) (15)Equation (12) and Equation (15) conclude the proof. Lemma 4.16.
Assuming that I.H. holds, let e = { u, w } be an arbitrary edge over which a flowis generated, where u sends a token to w in round r u , and w sends a token to u in round r w inexecution E . If max { r u , r w } < r , tenacity of every node that has two adjacent free edges on theconcatenation of e and a shortcut of u or w is less than tn ( e ) .Proof. Let e be a ψ -edge for any ψ ∈ { , } . The nodes are provided with the well-formed ( r − G and max { r u , r w } < r . Moreover, u and w send tokens over theadjacent ψ -edge e in rounds r u and r w respectively. Therefore, u is ( r u − − ψ )-reachable, and w is ( r w − − ψ )-reachable. Hence, tn ( e ) = r u + r w −
1. Without loss of generality, consider path S , the concatenation of e and an arbitrary shortcut of u (the argument for w is symmetric). Let x be an arbitrary node with two adjacent 0-edges on S . Let x be (cid:96) -0-reachable and (cid:96) -1-reachable.For simplicity, let us separately study the cases where u = x and u (cid:54) = x . First consider the casewhere u = x . It is easy to see that (cid:96) + 1 = r u . Moreover, since w is not a predecessor of u , itmust hold that (cid:96) , that is the first round of receiving any token for w is less than r w . Therefore, (cid:96) + (cid:96) < r u + r w − tn ( e ).Now let us consider the case when u (cid:54) = x . Then, (cid:96) < R ( u ) < r u . Moreover, (cid:96) < R ( u ) < r w since w is not a predecessor u and hence u must not receive a token from w in the first round ofreceiving tokens. Therefore, (cid:96) + (cid:96) < r u + r w − tn ( e ).32ere we present the proof of Lemma 4.11 as the final step of this section: Proof of Lemma 4.11.
We show that an arbitrary node is r - ϑ -reachable if r ( ϑ ) v = r after r roundsof E . Let us assume that v is in the cluster centered at f . There are two ways that r ( ϑ ) v can be setto r ; either r is the first round in which v receives a token or it is the first incomplete round for v .Let us first consider the former case, and let v (cid:48) be a neighbor of v that sends token f to v in round r . Then, r (1 − ϑ ) v (cid:48) = r −
1, and since D ( r −
1) is the well-formed ( r − G , v (cid:48) is ( r − − ϑ )-reachable. Let P v (cid:48) be a shortest uniform (1 − ϑ )-path of length r − v (cid:48) .Then, v (cid:54)∈ P v (cid:48) since otherwise v should have received a token before round r , which is contradictory.Therefore, path P v , the concatenation of P v (cid:48) and (cid:104) v (cid:48) , v (cid:105) , is a uniform ϑ -path of length r of v . Path P v is a shortest uniform ϑ -path of v , since otherwise, r ( ϑ ) v would have been set to an integer smallerthan r . Hence, v is r - ϑ -reachable.For the rest of the proof, we consider the latter case where r is the first incomplete round for v .Let v be r (cid:48) -reachable for some integer r (cid:48) < r . Let e be an edge such that v receives an incomplete flowof edge e in round r . Since v receives a flow of edge e , all the shortcuts of the two endpoints of e haveno l -reachable common node for any l ≥ r (cid:48) . That is because otherwise based on Lemma 4.10, thecommon node of all the shortcuts of the two endpoints of e that has maximum reachability discardsthe whole flow of e , and v never receives a flow of e , which contradicts the fact that v receives a flowof e . Therefore, based on Lemma 4.12, there exist a shortcut T u of u and a shortcut T w of w withno common l -reachable node for any l ≥ r (cid:48) such that exactly one of them contains v . Without lossof generality, let us assume that T w contains v . Let e be a ϕ -edge for an integer ϕ ∈ { , } . Then,we run the path construction algorithm , whose pseudocode is given by Algorithm 3, to construct ashortest uniform ϑ -path of length r of v . Algorithm 3:
The Path Construction Algorithm. Path-Discovery(
G, T u , T w , v, ϑ ) P v ← a shortest (1 − ϑ )-path of v ; P u ← (cid:104) w, u (cid:105) ◦ T u ; P w ← (cid:104) u, w (cid:105) ◦ T w [ w, v ] ◦ P v ; Let U be the set of nodes with two adjacent 0-edges on P u Let W be the set of nodes with two adjacent 0-edges on P w Let T ← U ∪ W while T (cid:54) = ∅ do Let s ∈ T be the node with the smallest 1-reachability Resolve ( P w , P u , s ) T ← T \ { s } return P u [ u, f i ] ◦ (cid:104) u, w (cid:105) ◦ P w [ w, v ] Resolve( P u , P w , s ) if s ∈ P u then if s has a shortest uniform -path P (cid:48) disjoint from P w [ v, w ] then P u ← P (cid:48) ◦ P u [ s, u ] else if s ∈ P w then if s has a shortest uniform -path P (cid:48) containing v and disjoint from P u [ f, u ] then P w ← P (cid:48) ◦ P w [ s, w ] 33he algorithm gradually constructs a shortest uniform (1 − ϕ )-path of u , i.e., P u [ f, u ], and ashortest uniform (1 − ϕ )-path of w , i.e., P w [ f, w ], such that P w [ f, w ] contains v , P u [ f, u ] does notcontain v , P u [ f, u ] and P w [ v, w ] have no common node, and | P u [ f, u ] | + | P w [ f, w ] | = r + r (cid:48) − P u [ f, u ], (cid:104) u, w (cid:105) and P w [ v, w ] is a uniform ϑ -path of length r of v . Everyiteration of the while-loop in the algorithm is called successful if the procedure Resolve successfullyupdates one of the paths P w or P u . Then, to prove that the algorithm returns a shortest uniform ϑ -path of length r of v , we show that all iterations of the while-loop are successful, and hence theconcatenation of P u [ f, u ], (cid:104) u, w (cid:105) , and P w [ w, v ] is a uniform ϑ -path of length r of v . Initially, P u [ f, u ]and P w [ v, w ] are disjoint. If one iteration of the while is successful in updating one of the paths, P u [ f, u ] and P w [ v, w ] remain disjoint. Considering an arbitrary iteration j of the while loop, we showthat if all previous iterations were successful in updating P u [ f, u ] and P w [ v, w ], iteration j is alsosuccessful in updating P u [ f, u ] or P w [ v, w ]. Now let us consider the following two cases separatelywhen procedure Resolve ( T w , T u , s ) is called.(I) s ∈ P u Here we show that there exists some shortest uniform 1-path of s that is disjoint from P w [ v, w ].Let us first show that there is a shortest uniform 1-path of s that does not contain v . To do sowe search and find such a path in a procedure explained as follows. First observe that s musthave a shortest uniform 1-path. Let Z be an arbitrary shortest uniform 1-path of s . If Z doesnot contain v , we are done with the search and Z is one of such paths. Let us thus assumethat Z contains v . We will first show that the adjacent edge of v on Z [ v, s ] is a ϑ -edge. Wewill then show that the concatenation of Z [ v, s ] and P u [ f, s ], that is an alternating walk, is oflength less than r . Thus, Z [ v, s ] and P u [ f, s ] must have a common edge. Then, we will arguehow having such a common edge leads to the existence of a shortest uniform 1-path of s thatdoes not contain v .Let us show that the adjacent edge of v on Z [ v, s ] is a ϑ -edge. Node s sends a token to itsneighbor in P u [ s, w ] in round | Z | + 1. Moreover, considering Lemma 4.7, all the nodes in P u [ s, u ] excluding s have reachability greater than | Z | , and hence they can only send tokensin rounds greater than | Z | . Therefore, node u that is in P u [ s, u ] sends a token to w in a roundgreater than | Z | , i.e., r u > | Z | . We also know that r u ≤ r since T w contains v and v receives aflow of e along T w in round r . This overall concludes that | Z | < r . Then, since | Z [ f, v ] | < | Z | ,path Z [ f, v ] is an alternating path of length less than r of v . Hence, the adjacent edge of v on Z [ f, v ] is not a ϑ -edge, since otherwise v would have a uniform ϑ -path of length less than r and hence have received an incomplete flow in a round before round r .Now let us show that the concatenation of Z [ v, s ] and P u [ f, s ] is of length less than r . Path P u [ f, s ] is a shortest uniform 0-path of s . Thus, tn ( s ) = | P u [ f, s ] | + | Z | . Based on Lemma 4.16, tn ( s ) < tn ( e ), and based on Lemma 4.15, tn ( e ) < tn ( v ). Moreover, tn ( v ) = r + r (cid:48) . Therefore, | P u [ f, s ] | + | Z | < r + r (cid:48) . Note that since Z [ f, v ] is a uniform alternating path of v , | Z [ f, v ] | ≥ r (cid:48) .Hence, | Z [ v, s ] | + | P u [ f, s ] | < r . This concludes that Z [ v, s ] and P u [ f, s ] must have a commonedge as otherwise v would have a uniform ϑ -path of length less than r .The last step is to show that this common edge leads to the existence of a shortest uniform1-path of s that does not contain v . Due to Lemma 4.13, the closest common edge of P u [ f, s ]and Z [ v, s ] to f on P u is a separator of Z and P u [ f, s ]. Let this edge be { t , t } . Then, theconcatenation of P u [ f, t ] and Z [ t , s ] is a shortest uniform 1-path of s that does not contain v . Let S be the set of all shortest uniform 1-paths of s that do not contain v .Let us now consider the following two cases separately: (recall that T is the set of nodes thathave two adjacent 0-edges on current P w or P u .)34a) P w ∩ T = ∅ Here we show that there is some path in S that is disjoint from P w [ v, w ]. We first showthat there is a path in S that does not have a common edge with P w [ v, w ]. This impliesthat the path dose not contain any node in P w ( v, w ) since the path is an alternating path.Moreover, since the path is in S , it does not contain v . Then, at the end we show thatthe path does not also contain w .For the sake of contradiction, let us assume that there is no path in S that has nocommon edge with path P w [ v, w ]. Considering Q an alternating path starting at f , letthe closest common edge of Q and any path Q (cid:48) to f on Q be denoted by ∂ ( Q on Q (cid:48) ).Let e := { s , s (cid:48) } ∈ P w [ v, w ] be the closest edge to w such that e is ∂ ( S on P w [ v, w ])for some S ∈ S . Path P w [ f, w ] is a shortest uniform (1 − ϕ )-path of w . Moreover, S is a shortest uniform 1-path of s with length less than the length of a shortest (1 − ϕ )-path of u . Hence, | P w [ f, w ] | + | S | < tn ( e ) < tn ( v ). Then, based on Lemma 4.13, e isa separator. Therefore, L := P w [ f, s ] ◦ S [ s , s ] is a shortest uniform 1-path of s thatcontains v . Since | L | + | P u [ f, u ] | = tn ( s ) < tn ( e ) < tn ( v ) = r + r (cid:48) and | L [ f, v ] | ≥ r (cid:48) , theconcatenation of L [ v, s ] and P u [ f, s ] is an alternating walk of length less than r from f to v that contains a ϑ -edge of v . Therefore, this concatenation cannot be a path. Hence, P u [ f, s ] and L [ v, s ] must have a common edge. Then, based on Lemma 4.13, the closestcommon edge of P u [ f, s ] and L [ v, s ] to f on P u is a separator of L and P u [ f, s ]. Let e = { s , s (cid:48) } be the edge. Then, L := P u [ f, s ] ◦ L [ s , s ] is a shortest uniform 1-pathof s that does not contain v . Note that L [ s , s ] is the same as L [ s , s ] and the sameas S [ s , s ]. Therefore, since e is a separator of S and P w , L [ s , s ] has no common edgewith P w [ f, s ]. Moreover, L [ s , s ] has no common edge with P w [ s , w ] as otherwise itcontradicts the choice of e . Therefore, L is a shortest uniform 1-path of s that does notcontain v and has no common edge with P w [ v, w ].It is left to show that L does not also contain w . Since P w ∩ T = ∅ , it holds that P w [ f, u ]is alternating, and hence w has an adjacent 1-edge on P w . For the sake of contradiction,let us assume that L contains w , and consequently it must contain the adjacent 1-edgeof w . All the nodes in L [ f, s ) have reachability of less than R ( s ), and hence less than R ( u ). Therefore, L [ f, s ) cannot contain u , and consequently cannot also contain { u, w } .Therefore, the 1-edge of w is on P w [ f, w ], and hence L must contain an edge on P w [ v, w ],which contradicts L having no common edge with P w [ v, w ]. As a result, L does notcontain w .(b) P w ∩ T (cid:54) = ∅ Let node s (cid:48) ∈ P w ∩ T be the closest node to f on P w . Note that the reachability of everynode in P w ( s (cid:48) , w ] is greater than the 1-reachability of s (cid:48) . Moreover, every node in anyshortest uniform 1-path of s has reachability at most the 1-reachability of s . Therefore,since the 1-reachability of s is smaller than that of s (cid:48) , any shortest uniform 1-path of s has no node in P w ( s (cid:48) , w ]. We need to show that there exists a path in S that has no nodein P w [ v, s (cid:48) ].Let s (cid:48)(cid:48) be the matched neighbor of s (cid:48) . Then the concatenation of P w [ f, s (cid:48) ] and (cid:104) s (cid:48) , s (cid:48)(cid:48) (cid:105) is a shortest uniform 1-path of s (cid:48)(cid:48) . Observe that s (cid:48)(cid:48) cannot be the same node as s sinceotherwise P u [ f, s ] ◦ (cid:104) s, s (cid:48) (cid:105) ◦ P w [ v, s (cid:48) ] would be a uniform ϑ -path of length less than r of v . Then, we argue similarly to part (a) to show that there is a path in S that does nothave a common edge with P w [ f, s (cid:48) ] ◦ (cid:104) s (cid:48) , s (cid:48)(cid:48) (cid:105) . This concludes that that there is a path in S that is disjoint from P w [ v, s (cid:48) ] and hence P w [ v, w ].35II) s ∈ P w Here we show that there exists some shortest uniform 1-path of s that is disjoint from P u [ f, u ]and contains v . Let us first show that there is a shortest uniform 1-path of s that contains v . To do so we search and find such a path as follows. First observe that s must have ashortest uniform 1-path. Let Z be an arbitrary shortest uniform 1-path of s . If Z contains v ,we are done with the search and Z is one of such paths. Let us thus assume that Z does notcontain v . Since Z is a shortest uniform 1-path and P w [ f, s ] is a shortest uniform 0-path of s , tn ( s ) = | Z | + | P w [ f, s ] | . Based on Lemma 4.16, tn ( s ) < tn ( e ), and based on Lemma 4.15, tn ( e ) < tn ( v ). Moreover, tn ( v ) = r + r (cid:48) . Therefore, | Z | + | P w [ f, s ] | < r + r (cid:48) . Hence, since | P w [ w, v ] | = r (cid:48) , | Z | + | P w [ v, s ] | < r . This concludes that the concatenation of Z and P w [ v, s ] isan alternating walk of length less than r from f to v that contains a ϑ -edge of v . Therefore, Z and P w [ v, s ] must have a common edge since otherwise v would have a uniform ϑ -path oflength less than r . Let g = { h , h (cid:48) } be ∂ ( Z on P w [ v, s ]). Then, based on Lemma 4.13, g isa separator of Z and P w [ f, s ]. Hence, the concatenation of P w [ f, h ] and Z [ h , s ] is a shortestuniform 1-path of s that contains v . Let S (cid:48) be the set of all shortest uniform 1-paths of s thatcontain v .Let us now consider the following two cases separately: (recall that T is the set of nodes thathave two adjacent 0-edges on current P w or P u .)(a (cid:48) ) P u ∩ T = ∅ Here we show that there is some path in S (cid:48) that is disjoint from P u [ f, u ]. We first showthat there is a path in S (cid:48) that does not have a common edge with P u [ f, u ]. This impliesthat the path dose not contain any node in P u [ f, u ) since the path is an alternating path.Then, we will show that the path does not also contain u .For the sake of contradiction, let us assume that there is no path in S (cid:48) that has no commonedge with path P u [ f, u ]. Let g = { h , h (cid:48) } ∈ P u [ f, u ] be the closest edge to u such that g is ∂ ( S (cid:48) on P u [ f, u ]) for some S (cid:48) ∈ S (cid:48) . Path P u [ f, u ] is a shortest uniform (1 − ϕ )-path of u .Moreover, S (cid:48) is a shortest uniform 1-path of s with length less than the length of a shortest(1 − ϕ )-path of w . Hence, | P u [ f, u ] | + | S (cid:48) | < tn ( e ) < tn ( v ). Then, based on Lemma 4.13, g is a separator. Therefore, L (cid:48) := P u [ f, h ] ◦ S (cid:48) [ h , s ] is a shortest uniform 1-path of s that does not contain v . Since | L (cid:48) | + | P w [ f, s ] | = tn ( s ) < tn ( e ) < tn ( v ) = r + r (cid:48) and | P w [ f, v ] | = r (cid:48) , the concatenation of L (cid:48) and P w [ v, s ] is an alternating walk of length lessthan r from f to v that contains a ϑ -edge of v . Therefore, this concatenation cannot bea path. Hence, P w [ v, s ] and L (cid:48) must have a common edge. Then, based on Lemma 4.13,the closest common edge of P w [ v, s ] and L (cid:48) to f on L (cid:48) is a separator of L (cid:48) and P w [ f, s ].Let g = { h , h (cid:48) } be the edge. Then, L (cid:48) := P w [ f, h ] ◦ L (cid:48) [ h , s ] is a shortest uniform1-path of s that contains v . Note that L (cid:48) [ h , s ] is the same as L (cid:48) [ h , s ] and the same as S (cid:48) [ h , s ]. Therefore, since g is a separator of S (cid:48) and P u , L (cid:48) [ h , s ] has no common edgewith P u [ f, h ]. Moreover, L (cid:48) [ h , s ] has no common edge with P u [ h , u ] as otherwise itcontradicts the choice of g . Therefore, L (cid:48) is a shortest uniform 1-path of s that contains v and has no common edge with P u [ f, u ].It is left to show that L (cid:48) does not also contain u . Observe that since P u [ f, w ] is alternating, u has an adjacent 1-edge on P u . For the sake of contradiction, let us assume that L (cid:48) contains u , and consequently it must contain the adjacent 1-edge of u . All the nodes in L (cid:48) [ f, s ) have reachability of less than R ( s ), and hence less than R ( w ). Therefore, L (cid:48) [ f, s )cannot contain w , and consequently cannot also contain { u, w } . Therefore, the 1-edge of u is on P u [ f, u ], and hence L (cid:48) must contain an edge on P u [ f, u ], which is contradictory.As a result, L (cid:48) does not contain u . 36b (cid:48) ) P u ∩ T (cid:54) = ∅ Let node s (cid:48) ∈ P u ∩ T be the closest node to f on P u . Note that reachability of everynode in P u ( s (cid:48) , u ] is greater than 1-reachability of s (cid:48) . Moreover, every node in any shortestuniform 1-path of s has reachability at most the 1-reachability of s . Therefore, since the1-reachability of s is smaller than that of s (cid:48) , any shortest uniform 1-path of s has no nodein P u ( s (cid:48) , u ]. We need to show that there exists a path in S (cid:48) that has no node in P u [ f, s (cid:48) ].Let s (cid:48)(cid:48) be the matched neighbor of s (cid:48) . Then the concatenation of P u [ f, s (cid:48) ] and (cid:104) s (cid:48) , s (cid:48)(cid:48) (cid:105) is a shortest uniform 1-path of s (cid:48)(cid:48) . Observe that s (cid:48)(cid:48) cannot be the same node as s sinceotherwise P u [ f, s (cid:48) ] ◦ (cid:104) s, s (cid:48) (cid:105) ◦ P w [ v, s ] would be a uniform ϑ -path of length less than r of v . Then, we argue similarly to part (a (cid:48) ) to show that there is a path in S (cid:48) that does nothave a common edge with P u [ f, s (cid:48) ] ◦ (cid:104) s (cid:48) , s (cid:48)(cid:48) (cid:105) . This concludes that that there is a path in S (cid:48) that is disjoint from P u [ f, s (cid:48) ]. In this section, we employ randomness to adapt the DFNC algorithm to the
CONGEST model andprove Lemma 2.1.
Proof of Lemma 2.1.
Two types of messages are sent in the DFNC algorithm execution; the tokensand the flow messages. Over every edge, at most one token is sent in each direction, which is theID of some free node in the network. Therefore, the dissemination of the tokens does not violatethe congestion restriction of the
CONGEST model. However, the described DFNC algorithm utilizeslarge flow messages. First note that a flow message might contain a large number of flows (i.e., key-value pairs). Moreover, since every time a node receives a flow, it divides its value in partitioningand forwarding the flow to its predecessors, the value of a flow might become a very small realnumber that needs a large number of bits to be represented. In this section, we modify the DFNCalgorithm to resolve this problem and only use O (log n )-bit flow messages, and we show that thismodification does not influence the desired effects of the DFNC algorithm.We only modify the content of the flow messages, but all other rules and regulations regardingtoken dissemination and flow forwarding remain the same. In the modified DFNC algorithm, theflow message that is sent by a node to its neighbor is only an integer in the ring of integers modulo γ ,i.e., Z /γ Z , where γ := n c for some large enough constant c . That is, all flows of different edges thatare sent over an edge in a single round are together aggregated and replaced with a single integer.At the beginning of the execution, every node v sends a distinct uniformly at random chosen integerin Z /γ Z , denoted by τ v,e , over each of its adjacent edges e . Now let us consider a flow generationevent over an arbitrary edge e = { u, w } , where u and w send tokens to each other in rounds r u and r w respectively. Without loss of generality, let u < v . Then, we define the flow generation as node w receiving flow τ u,e in round r u and u receiving flow ( − τ u,e ) mod γ in round r w .Regarding flow forwarding, let us consider an arbitrary node v . We explain in the following how anode processes its incoming flows and how it sends the processed flows. Node v has an output buffer O v ( t ) for every round t which is initially set to 0. Regarding processing the incoming flows, let usassume that v receives flows in an arbitrary round r . Let z be the sum modulo γ of all the receivedflows by v in round r . If z (cid:54) = 0 mod γ , then it updates its output buffers as follows. If the edgesover which v receives flows in round r and the edges connecting v to its predecessors are all 0-edges,then v sets O v ( r + d + 1) to (cid:0) O v ( r + d + 1) + z (cid:1) mod γ , where d is the difference of v ’s shortestuniform 0-paths and its shortest uniform 1-paths. Otherwise, v sets O v ( r + 1) to (cid:0) O v ( r + 1) + z (cid:1) mod γ . Now regarding forwarding the received and processed flows, let us consider an arbitraryround r (cid:48) for v . If O v ( r (cid:48) ) (cid:54) = 0, then v forwards flows to its predecessors as follows. Let p be the37umber of v ’s predecessors. If p = 1, v just forwards O v ( r (cid:48) ) to its only predecessor. Otherwise if p > v selects one of its predecessors uniformly at random which we call the poor predecessor of v .It independently chooses p − α , . . . , α p − from Z /γ Z and considers α p to be ( O v ( r (cid:48) ) − (cid:80) p − j =1 α j ) mod γ . It forwards α p to its poor predecessor and forwards α , . . . , α p − to the rest of its predecessors, one to each.Recall that E is the DFNC algorithm execution on G and M . Let R be the execution of themodified DFNC algorithm on G and M . Considering this flow circulation, when the sum modulo γ of some flows is not 0 in R , it corresponds to a flow with positive value of less than 1 in E andotherwise to a flow with value 1 in E . Fix an arbitrary edge e = { u, w } over which a flow is generated.Let us define layers L ( e ) , L ( e ) , . . . on e as follows. L ( e ) := { u, w } , and for all t > L t is theset of predecessors of the nodes in L t − . Moreover, let s ( e ) denote the smallest integer such that | L s ( e ) | = 1. Note that for all edges e , there exists such integer s ( e ) since the cluster center of e ’sendpoints is the single element of the last layer of e . Note that the single node in layer L s ( e ) doesnot send any integer that is influenced by the flow of e since the sum modulo γ of the generatedintegers regarding the flow of e is 0 and it is all received and assigned to be sent by the node in asingle round due to Lemma 4.10. Therefore, it is “discarded” by the node (actually cancelled out)when the sum is 0 mod γ .Now let us define L ( e ) := (cid:83) j
1, then despiteother received flows, the modulo sum of the received flows by v (cid:48) is a uniformly random number in Z /γ Z . Now let us consider the case when all the incoming edges to v (cid:48) from layer t − L t − that has a dotted edge to v (cid:48) definitely has a solid edge to some othernode in L t . Therefore, the sum of the flows of e received by the nodes in L t \ v (cid:48) from the nodes in L t − is a uniformly random number in Z /γ Z . Moreover, it is easy to see that the sum of all thereceived flows of e by the nodes in L t − is 0 (mod γ ). Hence, the total sum of the received flows38f e by the nodes in L t is also 0 mod γ . Therefore, since the sum of the received flows of e by thenodes in L t \ { v (cid:48) } is a uniformly random number in Z /γ Z , the sum of the received flows of e by node v (cid:48) from the nodes in L t − must also be a uniformly random number in Z /γ Z .We showed that the modified DFNC algorithm that is represented in this proof maintains thedesired effects of the flow circulation in the DFNC algorithm by just changing the flows content.Considering Lemma 4.1, it thus concludes the proof of Lemma 2.1. References [ABB +
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