Throughput in Asynchronous Networks
aa r X i v : . [ c s . N I] O c t Throughput in Asyn hronous NetworksPaul Bunn Rafail OstrovskyNovember 20, 2018Abstra tWe introdu e a new, (cid:16)worst- ase(cid:17) model for an asyn hronous ommuni ation network andinvestigate the simplest (yet entral) task in this model, namely the feasibility of end-to-endrouting. Motivated by the question of how su essful a proto ol an hope to perform in anetwork whose reliability is guaranteed by as few assumptions as possible, we ombine the main(cid:16)unreliability(cid:17) features en ountered in network models in the literature, allowing our model toexhibit all of these hara teristi s simultaneously. In parti ular, our model aptures networksthat exhibit the following properties: • On-line • Dynami Topology • Distributed/Lo al Control • Asyn hronous Communi ation • (Polynomially) Bounded Memory • No Minimal Conne tivity AssumptionsIn the on(cid:28)nes of this network, we evaluate throughput performan e and prove mat hing upperand lower bounds. In parti ular, using ompetitive analysis (perhaps somewhat surprisingly) weprove that the optimal ompetitive ratio of any on-line proto ol is /n (where n is the number ofnodes in the network), and then we des ribe a spe i(cid:28) proto ol and prove that it is n - ompetitive.The model we des ribe in the paper and for whi h we a hieve the above mat hing upperand lower bounds for throughput represents the (cid:16)worst- ase(cid:17) network, in that it makes no reli-ability assumptions. In many pra ti al appli ations, the optimal ompetitive ratio of /n maybe una eptable, and onsequently stronger assumptions must be imposed on the network toimprove performan e. However, we believe that a fundamental starting point to understandingwhi h assumptions are ne essary to impose on a network model, given some desired throughputperforman e, is to understand what is a hievable in the worst ase for the simplest task (namelyend-to-end routing). Additionally, our work may also serve as a framework to study additionalquestions (beyond end-to-end ommuni ation) within this (cid:16)worst- ase(cid:17) ase network setting.Keywords. Network Routing, ompetitive analysis, throughput, asyn hronous proto ols.0 Introdu tionWith the immense range of appli ations and the multitude of networks en ountered in pra ti e,there has been an enormous e(cid:27)ort to study routing in various settings. In terms of understandingwhat routing standards are possible, resear hers atta k the problem from two ends: understandingwhat is optimal, and developing and evaluating proto ols. Not surprisingly, provable results at bothends rely heavily on the model hosen to apture the features of the network.Typi ally, networks are modelled as a graph with verti es representing nodes (pro essors, routers,et .) and edges representing the onne tions between them. Beyond this basi stru ture, additionalassumptions and restri tions are then made in attempt to apture various features that real-worldnetworks may display. In parti ular, a network model must spe ify a hoi e between ea h of thefollowing hara teristi s: Weak Assumption Strong AssumptionView of Network: On-line O(cid:27)-lineEdge-Reliability: Dynami Topology Fixed TopologyControl Me hanism: Distributed/Lo al Control Centralized/Global ControlTiming/Coordination: Asyn hronous Communi ation Syn hronous Communi ationResour es: Bounded Memory Unlimited Memory(cid:16)Liveness(cid:17): No Conne tivity Assumptions Minimal Conne tivity GuaranteesNoti e that in ea h option above there is an inherent trade-o(cid:27) between generality/appli ability of themodel verses optimal performan e within the model. For instan e, a proto ol that assumes a (cid:28)xednetwork topology will likely out-perform a proto ol designed for a dynami topology setting, butthe former proto ol may not work in networks subje t to edge-failures. From both a theoreti al anda pra ti al standpoint, it is important to understand how ea h ( ombination) of the above fa torsa(cid:27)e ts routing performan e.In this paper, we study the feasibility of routing in an (cid:16)unrestri ted(cid:17) network model: one thatsimultaneously onsiders all of the more general features from the above list. Admittedly, in this(cid:16)worst- ase(cid:17) model it is unlikely that any proto ol will perform well, and one (or more) strongerassumptions must be made to a hieve a reasonable level of performan e. However, understandingbehavior in the worst ase, even with respe t to the most basi task of end-to-end ommuni ation, isimportant to determine how mu h (if any) the addition of ea h strong assumption improves optimalproto ol performan e.1.1 Previous WorkTo date, all network models have made at least one of the (and more ommonly multiple) strongassumptions listed above. The amount of resear h regarding network routing and analysis of routingproto ols is extensive, and as su h we in lude only a sket h of the most related works, indi atinghow their models di(cid:27)er from ours and providing referen es that o(cid:27)er more detailed des riptions.Competitive Analysis Competitive Analysis was (cid:28)rst introdu ed by Sleator and Tarjan [34℄ as ame hanism to analyze a spe i(cid:28) proto ol in terms of how vulnerable the proto ol is to being out-performed by alternative proto ols. Noti e that with respe t to the four hoi es for modeling anetwork, three hoi es an be represented by imperfe t information available to any on-line proto olthat operates within the network. In parti ular, in a dynami network, a proto ol does not know thefuture topology of the network; in a distributed network, nodes do not know the state of the othernodes and/or the links of the network; and in an asyn hronous network, a proto ol does not knowhow long pa kets will take to traverse a link. Re all that a given proto ol has ompetitive ratio /λ (or is λ - ompetitive) if an ideal o(cid:27)-line proto ol has advantage over the given proto ol by at mosta fa tor of λ . Competitive analysis may be applied to any of the possible network models, but ourwork is the (cid:28)rst to utilize this analysis in the unrestri ted network model onsidered in this paper.For a thorough des ription of ompetitive analysis, see [16℄.1nd-to-End Communi ation In this paper, we fo us on the task of performing throughput-e(cid:30) ientEnd-to-End Communi ation. Although there is an enormous amount of work in developing andanalyzing proto ols for end-to-end- ommuni ation, analysis in terms of throughput performan e hasbeen restri ted to syn hronous networks.1Max-Flow and Multi-Commodity Flow The Max-(cid:29)ow problem assumes networks that are: (cid:28)xedtopology, global- ontrol, and syn hronous. This is far more restri tive a network model than the onewe onsider in this paper.The multi- ommodity (cid:29)ow model (see for example Leighton et al . [30℄), is a generalization of themax-(cid:29)ow model, in whi h the network being onsidered may be dynami and distributed. However,even if the network is dynami , at every round the topology of the network is assumed to be adequateto meet the demands of all ommodities.2In the work of Awerbu h and Leighton [13℄, ompetitive analysis is utilized to demonstrate aproto ol for multi- ommodity (cid:29)ow whi h is (1 + ǫ ) - ompetitive.3 In parti ular, they guarantee thatif the ideal o(cid:27)-line proto ol an meet z (1 + 3 ǫ ) per ent of the demand of every ommodity at alltimes, then their proto ol will meet z (1 + ǫ ) per ent of the demand of every ommodity. Sin e alinear ompetitive ratio is the best any proto ol an hope to a hieve, their result is very lose tooptimal (in this network model).Admission Control and Route Sele tion For an extensive dis ussion about resear h in the area ofAdmission Control and Route Sele tion, see [31℄ and referen es therein. The admission ontrol/routesele tion model di(cid:27)ers from the multi- ommodity (cid:29)ow model in that the goal of a proto ol is notto meet the demand of all ordered pairs of nodes ( s, t ) , but rather the proto ol must de ide whi hrequests it an/should honor, and then designate a path for honored requests.There are numerous variants to the basi admission ontrol/route sele tion problem, in ludingProbabilisti Analysis (requests are des ribed via distributions) [28℄ and [29℄, Routing with Edge-Congestion [21℄, Multi- ast Routing [9℄, [15℄, and [25℄, Unsplittable Flow [20℄, Load Control Problems[5℄, Non-Blo king Networks [7℄, and Ad Ho Networks [12℄ and [32℄. Competitive analysis was (cid:28)rstutilized in the admission ontrol/route sele tion model to analyze throughput by Garay and Gopal[23℄ and Garay et al . [24℄ for sparse networks, and later for more general networks by a series ofauthors, in luding [10℄, [8℄, and [26℄. Competitive analysis has also been used to measure theoreti albounds on throughput performan e for multi- asting in syn hronous networks in [11℄.In the Asyn hronous Transfer Model (ATM), the ( ommodity, demand) requests from pairs ofnodes ( s, t ) are not available at the outset of the proto ol, and furthermore these requests aretemporary, with a given duration of time for whi h s wishes to transfer the ommodity to t . Weemphasize that the de(cid:28)nition of asyn hroni ity in ATM is di(cid:27)erent than the one onsidered in thispaper. In parti ular, (cid:16)asyn hroni ity(cid:17) in ATM literature is meant to emphasize the fa t that therequests are not known ahead of time, and thus proto ols fa e the added hallenge of handling newrequests adaptively. As in the admission ontrol/route sele tion model, the goal of a routing proto olin the ATM is to de ide whi h requests to honor (and how to route the honored requests) based onthe urrent status of the network, su h that all previously honored requests are not deleted. In thework of Awerbu h, Azar, and Plotkin [10℄, ompetitive analysis is utilized to demonstrate a proto olthat is log( nT ) - ompetitive4 in terms of throughput, where n is the number of nodes in the networkand T is an upper-bound on the all duration of any request.1There are several message-driven proto ols designed to work in asyn hronous networks, for example the Slideproto ol introdu ed by Afek and Gafni [2℄ (and further developed by [14℄, [1℄, and [27℄). While these proto ols workwell in pra ti e and have proven results in syn hronous networks, to date there has been no rigorous throughput-analysis of these proto ols in asyn hronous networks.2Alternatively, the on urrent (cid:29)ow problem onsiders networks whose topology may not be adequate to simultane-ously meet the demands of all quantities. In this ase the problem seeks to maximize the per entage z su h that atleast z per ent of the demand of every ommodity is met.3Sin e the model assumes a dynami and distributed network, the perfe t information available to the o(cid:27)-lineproto ol in ludes knowledge of the future topology (whi h links will be available ea h round) as well as a global-viewof the network.4The perfe t information available to the o(cid:27)-line proto ol in ludes knowledge of all future requests.2he admission ontrol/route sele tion model assumes a (cid:28)xed topology, syn hronous network (andis also often onsidered to be global- ontrol). In addition, when a request ( s, t ) is honored, often itis demanded that a routing path must be spe i(cid:28)ed upon a eptan e of the request, and this path isnot permitted to hange in an adaptive manner (e.g . a di(cid:27)erent path may not be utilized in order tooptimize throughput as a result of a future requests).Queuing Theory A model related to the ATM is the Queuing Theory (QT) model (see e.g . [17℄ and[6℄). In this model (as in the ATM), the pattern of requests ( hosen from some known distribution)is not known ahead of time, but unlike ATM, all requests must be honored. Queuing Theory asksthe question: given a distribution from whi h the requests will be hosen, how mu h memory doesa given proto ol require of ea h pro essor in order to guarantee all requests an be honored?In alternate instantiations, the queuing theory model may either allow adaptive routing, or requirethat paths be (cid:28)xed upon re eipt of a request; also, there are works onsidering both global- ontroland distributed networks. Most analysis in the QT model has onsidered (cid:28)xed topology networks,and in all models, the network is assumed to be syn hronous.Adversarial Queuing Theory Adversarial Queuing Theory (AQT) is similar to QT, ex ept the pat-tern of requests is ontrolled by an adversary who may wish to disrupt a given proto ol as mu h aspossible, and then ompetitive analysis is used to analyze performan e. There has been a lot of workin the AQT model (e.g . [18℄ and [3℄) that onsiders networks allowing dynami topology (althoughthe network is always assumed to be syn hronous).Competitive Analysis of Distributed Algorithms So far, none of the above models have onsideredasyn hronous networks. However, there has been a tremendous amount of e(cid:27)ort to analyze dis-tributed algorithms in asyn hronous5 shared memory omputation, in luding the work of Ajtai etal . [4℄. This line of work has a di(cid:27)erent (cid:29)avor than the problem onsidered in the present paperdue to the nature of the algorithm being analyzed ( omputation algorithm verses network routingproto ol). In parti ular, network topology is not a onsideration in this line of work.1.2 MotivationOne of the primary goals of all routing proto ols is to a hieve high throughput between a Senderand a Re eiver in a network, subje t to resour e onstraints. As seen above, to date almost all ofthe resear h analyzing the throughput-performan e of routing proto ols has fo used on syn hronousnetworks, and the majority of works also assume (cid:28)xed topology. However, in many pra ti al settings,su h as the Internet, the networks en ountered are fully asyn hronous, and network topology istypi ally unstable, with edges that may go up an down in an unpredi table manner. In the next fewparagraphs, we dis uss some of the hallenges in designing and analyzing throughput performan eof proto ols in our (cid:16)unrestri ted(cid:17) network model.The metri in whi h we will wish to analyze and ompare proto ols is in the rate of pa ket delivery,or throughput. Intuitively, the throughput of a proto ol measures the amount of information deliveredas a fun tion of time. However, apturing this intuition formally in the ontext of our networkmodel en ounters two di(cid:30) ulties. First, the asyn hronous nature of the network makes it di(cid:30) ultto formalize the notion of (cid:16)time.(cid:17) In a syn hronous network, throughput is typi ally measured by(cid:28)rst de(cid:28)ning dis rete (cid:16)rounds,(cid:17) during whi h every onne tion an transfer one unit of information.6Throughput in the syn hronous setting is then de(cid:28)ned to measure the amount of information re eivedas a fun tion of the number of rounds that have passed. However, in an asyn hronous network, thereis no a priori notion of a round, as onne tions may be transferring information at di(cid:27)erent timesand rates.7 Therefore, apturing the intuition of what it means for a proto ol to be e(cid:30) ient requires5The terminology of (cid:16)asyn hroni ity(cid:17) here is onsistent with the one de(cid:28)ned in this paper, as opposed to theasyn hronous transfer model (ATM) des ribed above.6Depending on the model, this unit may be the same for all onne tions, or it may vary for the di(cid:27)erent onne tions(in whi h ase the terms (cid:16)bandwidth(cid:17) or (cid:16)edge- apa ity(cid:17) are used). Also, the bandwidth for a single onne tion maybe (cid:28)xed for all time or may hange ea h round.7Our model not only allows the transmission rate of a onne tion to di(cid:27)er a ross the various onne tions, but also3 areful approa h in formally de(cid:28)ning throughput (see Se tion 2 for our pre ise de(cid:28)nition).A se ond di(cid:30) ulty in evaluating a proto ol's throughput performan e in the network model onsidered in this paper arises from the dynami topology nature of the network. Indeed, for themodel we will be onsidering, not only is the topology of the network not (cid:28)xed, but we will notimpose minimal onne tivity requirements. In parti ular, sin e we are onsidering ommuni ationbetween two designated nodes through a network, there is no assumption made about how often (oreven if ) these two nodes are onne ted by a path. Indeed, we will model the dynami nature ofour network by introdu ing an edge-s heduling adversary who ontrols all the links in the network,and whose goal may be to disrupt ommuni ation as mu h as possible between the Sender and theRe eiver. Sin e we make no onne tivity assumptions, the adversary an simply hoose to leave allthe links ina tive, rendering all ommuni ation impossible. We will handle this se ond di(cid:30) ulty byemploying ompetitive analysis.1.3 Our ResultsIn this paper, we lay the foundation for evaluating routing proto ols in an (cid:16)unrestri ted(cid:17) networkmodel: on-line, distributed, asyn hronous, dynami , with bounded memory and no minimal on-ne tivity assumptions. As demonstrated by the dis ussion above, network models that demonstratesome (but not all) of the weaker assumption hoi es have been onsidered by multiple authors, butto date no e(cid:27)ort has been made to understand what (if any) performan e guarantees are possiblefor extremely unreliable networks that are subje t to all of the hallenges modelled by the weakassumptions.We fo us on the task of end-to-end ommuni ation, and using ompetitive analysis we analyze theoptimal throughput that an be a hieved in this network model. We (cid:28)rst demonstrate that the bestpossible ompetitive ratio that any proto ol an hope to a hieve is /n . We then des ribe an expli itproto ol that realizes this optimal ompetitive ratio. Informally, our results an be summarized:Theorem 1 (Informal) The best ompetitive-ratio that any proto ol an a hieve in a distributedasyn hronous network with bounded-memory and dynami topology (and no onne tivity assumptions)is /n . In parti ular, given any proto ol P , there exists an alternative proto ol P ′ , su h that P ′ willout-perform P by a fa tor of at least n .Theorem 2 (Informal) There exists a proto ol that an a hieve a ompetitive ratio of /n in adistributed asyn hronous network with bounded-memory and dynami topology (and no onne tivityassumptions).Theorem 1 states that given any proto ol P , there exists an o(cid:27)-line proto ol P ′ and a s hedulingadversary su h that P ′ will out-perform P by at least a fa tor of n . The proof of Theorem 1 (i.e.the lower bound) is highly nontrivial and relies on a deli ate ombinatorial argument together witha nonstandard potential fun tion analysis (see further intuition in se tion 3).Meanwhile, (the proof of) Theorem 2 exhibits a proto ol (cid:16)Slide + (cid:17) that guarantees that evenagainst an omnis ient o(cid:27)-line proto ol and against any adversary, Slide + will never be out-performedby more than a fa tor of n . In other words, Slide + is within a fa tor of n of the highest possiblethroughput a hievable by an omnis ient algorithm that makes optimal routing de isions based ona omplete view of the future onditions of the network and its topology. Moreover, by Theorem1, this is the optimal guarantee for throughput that a proto ol an hope to enjoy, i.e . no on-linestrategy an a hieve a better ompetitive ratio.2 The ModelIn this se tion, we des ribe formally the model in whi h we will be analyzing routing proto ols.We begin by modeling the network as a graph G with n verti es (or nodes). Two of these nodes areallows the same onne tion to have di(cid:27)erent transmission rates for ea h pa ket it transfers.4esignated as the Sender and Re eiver, and the Sender has a stream of pa kets { p , p , . . . } that itwishes to transmit through the network to the Re eiver.Asyn hronous ommuni ation networks vary from syn hronous networks in that the transmissiontime a ross an edge in the network is not (cid:28)xed (even along the same edge, from one message trans-mission to the next). Sin e there is no ommon global lo k or me hanism to syn hronize events, anasyn hronous network is often said to be (cid:16)message driven,(cid:17) in that the a tions of the nodes in thenetwork o urs exa tly (and only) when they have just sent/re eived a message.For this reason, asyn hronous networks are ommonly modelled by introdu ing an edge-s hedulingadversary that ontrols the edges of the network as follows. A round is de(cid:28)ned to be a single edge E ( u, v ) in the network hosen by the adversary in whi h two sequential events o ur: 1) Amongthe pa kets from u to v (and vi e-versa) that the adversary is storing, it will hoose one (in anymanner it likes) and deliver it to v (resp. to u );8 2) After seeing the delivered pa ket, u (resp. v )sends requests of the form ( u, w, p ) = (sending node, target node, pa ket) to the adversary, whi hwill be stored by the adversary and may be delivered the next time u (resp. v ) is part of a round.Modelling asyn hroni ity in this manner aptures both the intuition that a node has no idea howlong a message that it (cid:16)sends(cid:17) to adja ent node v will take to arrive, and it also onsiders (cid:16)worst- ase(cid:17)asyn hroni ity in that a (potentially deliberately adversarial) adversary ontrols the s heduling ofrounds/edges.Aside from obeying the above spe i(cid:28)ed rules, we pla e no restri tion on the s heduling adversary.In other words, it may honor whatever edges it likes (this models the fa t our network makes no onne tivity assumptions), wait inde(cid:28)nitely long between honoring the same edge twi e (modelingboth the dynami and asyn hronous features of our network), and do anything else it likes (whilerespe ting the guidelines) in attempt to hinder the performan e of a routing proto ol.Note that our network model is on-line and distributed, in that we do not assume that the nodeshave a ess to any information (in luding future knowledge of the adversary's s hedule) aside fromthe pa kets they re eive during a round they are a part of. Finally, we insist that nodes have boundedmemory9 C (at most polynomial in n ).The goal of this paper is to analyze the performan e of routing proto ols in a network modelthat is: on-line, distributed, asyn hronous, dynami , with bounded memory and no onne tivityassumptions. Our me hanism for evaluating proto ols will be to measure their throughput, a notionwe an now de(cid:28)ne formally in the ontext of rounds and the edge-s heduling adversary. In parti ular,let f AP : N → N be a fun tion that measures, for a given proto ol P and adversary A , the numberof pa kets that the Re eiver has re eived as a fun tion of the number of rounds that have passed.10The fun tion f AP formalizes our notion of throughput.As mentioned in the Introdu tion, we utilize ompetitive analysis to gauge the performan e (withrespe t to throughput) of a given proto ol against all possible ompeting proto ols. In parti ular,for any (cid:28)xed adversary A , we may onsider the ideal (cid:16)o(cid:27)-line(cid:17) proto ol P ′ whi h has perfe t infor-mation.11 That is, for any (cid:28)xed round x , there exists an ideal o(cid:27)-line proto ol P ′ ( A , x ) su h that f AP ′ ( x ) is maximal.De(cid:28)nition 2.1. We say that a proto ol P has ompetitive ratio /λ P (respe tively is λ P - ompetitive)if there exists a onstant k and fun tion g(n,C) su h that for all possible adversaries A and for all x ∈ N : f AP ′ ( x ) ≤ ( k λ P ) · f AP ( x ) + g ( n, C ) (1)Note that while g may depend on the size of the network n and the bounds pla ed on pro essormemory C , both g and k are independent of the round x and the hoi e of adversary A . We maynow restate our two main results formally:8For ease of dis ussion, we assume that all edges in the network have a (cid:28)xed bandwidth/ apa ity, and that thisquantity is the same for all edges in the network. We emphasize that this assumption does not restri t the validity ofour laims in a more general model allowing varying bandwidths, but is only made for ease of exposition.9For simpli ity, we assume that all nodes have the same memory bound, although our argument an be readilyextended to handle the more general ase.10In this paper, we will onsider only deterministi proto ols, so f AP is well-de(cid:28)ned.11Here, perfe t information means that the o(cid:27)-line proto ol has knowledge of all future de isions of the adversary.5heorem 2.2. For any proto ol P operating in a distributed, asyn hronous, bounded memory net-work with dynami topology (and no onne tivity assumptions), the ompetitive ratio of P is: λ P ≥ n Theorem 2.3. In a distributed, asyn hronous, bounded memory network with dynami topology (andno minimal onne tivity assumptions), there exists a proto ol P that is n - ompetitive ( λ P = n ).We prove Theorem 2.2 in the next se tion, and then go on to demonstrate a proto ol in Se tion 4that a hieves ompetitive ratio /n .3 Optimal Competitive Ratio in Unrestri ted NetworksDue to spa e onstraints and the omplexity of the argument, we will only be able to sket hthe proof of Theorem 2.2 in this se tion. At a high level, the idea is to des ribe an adversarythat s hedules edges based on the given proto ol's a tions su h that the pa kets of the proto olget (cid:16)spread out(cid:17) among the nodes of the network. Meanwhile, with knowledge of the adversary'ss hedule, an o(cid:31)ine proto ol an hoose to only move pa kets along edges leading to the re eiver. Ashort des ription is below; the full proof an be found in Appendix A.The network model assumes that nodes have bounded memory, so let C denote the maximalnumber of pa kets that any node an store at any time. We will show that for any deterministi proto ol P , there exists an adversary A , a proto ol P ′ , and a sequen e of stri tly positive integers { m , m , . . . } su h that for any α > , by round x = P αi =1 m i C : f AP ′ ( x ) = αC and f AP ( x ) ≤ αC ( n − ≈ αCn , (2)from whi h we on lude that the ompetitive ratio of P is at best /n .We begin by des ribing the adversary, i.e . a s hedule (or order) of edges that will be honored.The s hedule will pro eed in y les, with the i th y le lasting m i C rounds. For the (cid:28)rst C rounds,the adversary (cid:28)nds the internal node A that is urrently storing the most pa kets (ties are brokenarbitrarily), and honors edge E ( S, A ) for C rounds (here S denotes the Sender). The proto ol thenpro eeds indu tively, starting with j = 2 and b A = A :1. The adversary (cid:28)nds node A j , where A j is the node in the network losest in height (but smaller)to b A j − . If there is no su h node, set A j to the Re eiver R .2. The adversary honors edge E ( b A j − , A j ) for C rounds3. The adversary sets b A j to be whi hever node ( b A j − or A j ) has fewer pa kets after the C roundsof edge E ( b A j − , A j ) has just passed.The above three steps are ontinued until the end of the C rounds for whi h A j = R .Noti e a few features of the adversarial strategy: 1) The Sender's ability to insert pa kets ishindered by the fa t the adversary is hoosing to honor edge E ( S, N ) for the node N with thesmallest apa ity to store more pa kets; 2) By sele ting in Step 2 the node storing fewer pa kets, theadversary is attempting to minimize the number of pa kets that make progress towards the Re eiver;indeed 3) Among all nodes in the network, the node N that is urrently storing the fewest pa ketswill be the one onne ted to the Re eiver in the (cid:28)nal C rounds of the y le. Also, it is lear that ano(cid:27)-line proto ol P ′ with knowledge of all future rounds will be able to deliver C pa kets every y le.Sin e a y le onsists of C ∗ m rounds for some positive integer m , we an generate a sequen e ofpositive integers { m i } oming from the i th y le, yielding the (cid:28)rst equality of (2), so it remains toprove the se ond bound in (2).Fix any on-line proto ol P we wish to analyze. If we ould demonstrate that P delivers at most C/ ( n − pa kets per y le, then (2) would be immediate. Unfortunately, one an imagine e.g . the6tate of the network at the beginning of some y le being su h that all internal nodes are storing themaximum C allowed pa kets. In this ase, P will be able to deliver C pa kets this round. Therefore,we instead need to argue that if P ever rea hes a state where it is able to deliver more than C/ ( n − pa kets in some y le (e.g . all nodes are full), then it must be that P has delivered fewer than anaverage of C/ ( n − pa kets per y le in the past.With this ounter-example in mind, we de(cid:28)ne a potential fun tion Ψ α , whi h intuitively measuresthe ability of P to deliver pa kets in the α th y le. We will show that whenever P delivers more than C/ ( n − pa kets, the di(cid:27)eren e Ψ α − Ψ α +1 will be positive and (cid:16)su(cid:30) iently large.(cid:17) Conversely,any time Ψ α +1 > Ψ α , we will show that ne essarily P delivered (cid:16)signi(cid:28) antly fewer(cid:17) than C/ ( n − pa kets in the α th y le.Formally, for any α ∈ N , let H αi denote the number of pa kets that node N αi is storing at theoutset of α , and then de(cid:28)ne: Ψ α = n − X i =1 (cid:18) (cid:19) n − i − · max (cid:18) , H αi − ( n − i − Cn − (cid:19) (3)Let Z α denote the number of pa kets the Re eiver re eives in the α th y le. Our main te hni alresult for this se tion is then:Theorem 3.1. For all α ∈ N : Z α + (Ψ α +1 − Ψ α ) ≤ Cn − (4)Proof. See the proof of Theorem A.11 in the Appendix.With Theorem 3.1 in hand, we obtain the se ond inequality of (2) as an immediate orollary:Theorem 3.2. For any α ∈ N and x = ( n − αC : f AP ( x ) ≤ αCn − (5)Proof. Consider the string of inequalities: f AP ( x ) = X β ≤ α Z β ≤ X β ≤ α (cid:18) C ( n − − (Ψ β +1 − Ψ β ) (cid:19) = 7 αCn − − Ψ α +1 ≤ αCn − , (6)where the last inequality follows from the fa t that Ψ α +1 ≥ and Ψ = 0 (the latter is true sin e atthe outset of the proto ol, all nodes are not storing any pa kets).4 Optimal On-line Lo al Control Proto olIn this se tion we present an on-line proto ol that enjoys ompetitive ratio /n . The proto olis a basi implementation of the (cid:16)Slide(cid:17) proto ol (or gravitational-(cid:29)ow), whi h was (cid:28)rst introdu edby Afek and Gafni [2℄, and further developed in a series of work [14℄, [1℄, and [27℄. We hose toanalyze the performan e of this proto ol in our (cid:16)unrestri ted(cid:17) network model be ause its inherentmessage-driven proto ol is well-suited for the asyn hronous network, and it has also been shownto out-perform more naive andidates for asyn hronous routing proto ols (e.g. broad ast) whenstronger network assumptions are made [19℄.In the following sub-se tion, we outline the basi Slide proto ol, and then in Se tion 4.2 wesket h the proof that guarantees that the basi version of Slide is at least n - ompetitive in a (cid:16)semi-asyn hronous(cid:17) network (de(cid:28)ned below). In Appendix B, we present a modi(cid:28) ation of the Slideproto ol that a hieves the optimal n - ompetitive ratio in the fully asyn hronous model of Se tion 2.Re all that in the asyn hronous model de(cid:28)ned in Se tion 2, the adversary maintains a bu(cid:27)er ofrequests ( s, t, p ) that it has re eived so far, and then the proto ol pro eeds in rounds where an edge E ( u, v ) is honored, during whi h: 7. From its bu(cid:27)er, the adversary hooses one pa ket ( u, v, p i ) to deliver to v , and one pa ket ( v, u, p j ) to deliver to u
2. Based only on pa kets ea h node u and v has re eived thus far (and the instru tions from theproto ol), u and v send requests to the adversaryThe basi Slide proto ol di tates that a pa ket should be sent from u to v if u urrently has morepa kets stored than v . Therefore, a request ( u, v, p ) that a node u submits to the adversary will haveadditional information piggy-ba ked onto the pa ket p . Namely, u will append its height (a number H representing the number of pa kets u urrently holds) to the pa ket when submitting a requestto the adversary, thus requests will have form ( u, v, ( p, H )) .Unfortunately, in the model above, when u submits a request ( u, v, ( p, H )) to the adversary, thevalue for H may be ome out-dated by the time the adversary honors edge E ( u, v ) (for instan e if theadversary honors edges E ( u, v ′ ) for v ′ = v in between honors to E ( u, v ) ). In the Appendix, we handlethis problem by having nodes ommuni ate (cid:16)approximate(cid:17) heights, and show that this approximationto the Slide proto ol (whi h, together with several other te hni al modi(cid:28) ations, we all Slide + ) willnot perform mu h di(cid:27)erently than a proto ol in whi h nodes have perfe t information about theirneighbors' heights when they are making a de ision to send/re eive a pa ket.12 In the remainder ofthis se tion, we assume a semi-asyn hronous model, de(cid:28)ned as follows: ′ The adversary does not maintain a bu(cid:27)er of requests of pa kets from nodes and must insteadsatisfy them immediately as spe i(cid:28)ed in ′ below ′ The adversary pro eeds in the same manner as before, by sele ting an edge E ( u, v ) to honora ording to the same guidelines as in Se tion 2 ′ During a round E ( u, v ) , the adversary (cid:28)rst (cid:16)awakens(cid:17) u and v to alert them they are a part ofthe urrent round. Nodes u and v may now submit their request, onsisting only of a pa ketplus ontrol information, to the adversary who must dire tly deliver the pa ket p to v duringthis round (similarly the pa ket p ′ that v submitted is delivered to u ).We note that in terms of ′ , it will not be important for the Slide proto ol whether or not u isaware of whi h edge the adversary is honoring, i.e . whi h node v will be re eiving u 's pa ket. Notethat in the semi-asyn hronous model, the problem of out-dated height information being transferredbetween two nodes is avoided. This modi(cid:28) ation is ne essary to prove that the basi Slide proto ola hieves ompetitive ratio /n , but we emphasize that our result in Theorem 2.3 remains valid in thefull asyn hronous network model of Se tion 2. In the Appendix, we present the Slide + proto ol, anddemonstrate how to extend the proof of ompetitive ratio /n for basi Slide in the semi-asyn hronousmodel to an equivalent ratio for Slide + in the fully asyn hronous network model of Se tion 2.4.1 Des ription of the Proto olThere are numerous instantiations of the Slide proto ol that vary slightly between one another,but the basi prin iple is always the same. Due to spa e onstraints, we will not provide a detaileddes ription of the proto ol, but refer the reader to [2℄ for the original proto ol, and [14℄, [1℄, [27℄,and [19℄ for various modi(cid:28) ations. Below, we present a basi implementation of the Slide proto ol,and then go on to prove that the basi Slide proto ol a hieves ompetitive ratio /n in the restri tedsemi-asyn hronous model of ′ − ′ des ribed above.13The network model assumes that nodes have bounded memory, so let C denote the maximalnumber of pa kets that any node an store at any time.14 Also, we will assume C/n ∈ N and inparti ular that C/n ≥ (the former assumption is not ne essary but will make the exposition easier;12Although the approximated proto ol demands that nodes have bu(cid:27)er sizes at least O ( n ) , whereas normal Slideonly requires bu(cid:27)ers of size O ( n ) ′ − ′ above), we des ribe the request that a node u will make to the adversarywhen it is (cid:16)awakened,(cid:17) and also how this node u will respond to the pa ket it re eives from v :1. If u is the Sender, then u (cid:28)nds the next pa ket p i ∈ { p , p , . . . } that has not yet been deleted(see 1a below), and forms: p := ( p i , C + Cn − . Meanwhile, when u re eives (in the sameround) the pa ket ( p j , h ) :(a) If h < C , then u deletes pa ket p i from his input stream { p , p , . . . } (and ignores there eived pa ket p j )(b) If h ≥ C , then u keeps p i (and ignores the re eived pa ket p j )2. If u is the Re eiver, then u forms the pa ket to send p := ( ⊥ , − Cn ) . Meanwhile, when u re eivesa pa ket of form ( p j , h ) , if p j = ⊥ , u stores/outputs p j as a pa ket su essfully re eived.3. If u is an internal node (not Sender or Re eiver) and u urrently has height H , then u (cid:28)ndsthe last pa ket p i that it has re eived, and sets: p := ( p i , H ) (if H = 0 , then set p i = ⊥ ).Meanwhile, when u re eives (in the same round) a pa ket of form ( p j , h ) :(a) If H ≥ h + C/n , then u will delete p i (and ignore the pa ket p j )(b) If H ≤ h − C/n , then u will keep p i , and also store p j (as the most re ent pa ket re eived)( ) If | H − h | < C/n , then u will keep p i and ignore pa ket p j Noti e that rules 1-3 essentially state that internal nodes will always a ept pa kets from the Sender(if they have room), always send pa kets to the Re eiver (if they have any to send), and will transfera pa ket to a neighboring internal node if and only if they are urrently storing at least
C/n morepa kets than that neighbor.4.2 Competitive Analysis of Slide in the Semi-Asyn hronous ModelDue to spa e onstraints, we provide here only a sket h of the proof that the above des ribedSlide proto ol enjoys ompetitive ratio /n . The full proof an be found in Appendix B.Re all that we wish to show that there exists a onstant k and fun tion g ( n, C ) su h that for anyround x and against any adversary A (see (1)): f AP ′ ( x ) ≤ ( kn ) · f AP ( x ) + g ( n, C ) (7)Above (and through the remainder of this se tion), P will denote the Slide proto ol, and for (cid:28)xed hoi e of adversary A and round x , P ′ ( A , x ) will denote the ideal o(cid:27)-line proto ol (sin e we will be(cid:28)xing x and A , we will usually write simply P ′ ). We will show that (7) will be true for all rounds x and all adversaries A for k = 4 and g ( n, C ) = 4 n C . We pro eed by (cid:28)xing an arbitrary adversary A and round x ∈ N , and showing that for these (arbitrary) hoi es, (7) will be satis(cid:28)ed.We begin with some notation and terminology. Any time the Sender rea hes Step 1a, we willsay pa ket p i was inserted. Similarly, anytime the Re eiver stores/outputs pa ket p j , we will say thepa ket has been re eived. Noti e that anytime an internal node u rea hes Step 3a, the node v at theother end has ne essarily rea hed Step 3b. In this ase, we will say pa ket p i was transferred from u to v . Let Y P denote the number of pa kets inserted by P as of round x ( Y P ′ is de(cid:28)ned analogously).The starting point of the proof is the following trivial observation:Observation 1. As of round x , there have been at most nY P pa ket transfers in P .The observation is a produ t of the fa t that pa kets are transferred in a FILO manner, so rules3a-3 state that a pa ket must drop in height by at least C/n − when it is transferred. Sin e anode's memory is C ≥ n , a single pa ket an be transferred at most n times.Let Z P ′ denote the pa kets that have been re eived by the Re eiver for proto ol P ′ as of round x (de(cid:28)ne Z P analogously). Noti e that f AP ′ ( x ) , the left-hand-side of (7), is equal to | Z P ′ | (we will9 asionally write Z P ′ when we really mean | Z P ′ | ; the meaning will be lear from ontext). We split Z P ′ into two disjoint subsets Z P ′ = Z P ′ ∪ Z P ′ , whi h we now des ribe.We an view the adversary A as simply a s hedule (or order) of edges that the adversary willhonor. We will imagine a virtual world, in whi h the two proto ols (Slide and the ideal o(cid:27)-lineproto ol) are run simultaneously in the same network. De(cid:28)ne Z P ′ to be the subset of Z P ′ onsistingof pa kets p ′ for whi h there exists at least one round E ( u, v ) su h that both p ′ and some pa ket p ∈ Y P were both transferred this round.15 Set Z P ′ = Z P ′ \ Z P ′ .Lemma 4.1. | Z P ′ | ≤ n | Z P | + 2 n C Proof. This follows from Observation 1 together with the fa t | Y P − Z P | ≤ nC .It remains to bound | Z P ′ | ≤ n | Z P | + 2 n C . To this end, we (cid:28)rst observe that when any pa ket p ′ ∈ Z P ′ was (cid:28)rst inserted, it was ne essarily inserted into some node u su h that with respe t to P , u had height C (otherwise P would have also inserted a pa ket this round, and then p ′ ∈ Z P ′ ).Similarly, when p ′ ∈ Z P ′ is re eived by the Re eiver from some node v , then with respe t to P , v hadheight zero. The idea will then be to assign a potential fun tion ϕ p ′ to every pa ket p ′ ∈ Z P ′ thatrepresents the urrent height with respe t to P of the node in whi h p ′ is urrently stored. Thus,when a pa ket p ′ ∈ Z P ′ is (cid:28)rst inserted, ϕ p ′ = C , and when p ′ ∈ Z P ′ is re eived, ϕ p ′ = 0 .Next, we de(cid:28)ne a se ond potential fun tion Φ t , whi h will obey:1. Φ = 0 at the outset of the proto ol, and every time there is a pa ket transfer in P , ∆Φ = C
2. For any pa ket p ′ ∈ Z P ′ , anytime ϕ p ′ hanges (aside from when ϕ p ′ is initialized to C uponinsertion of p ′ ), Φ hanges by an equivalent amountWith these de(cid:28)nitions, we have:Lemma 4.2. For any round t ≤ x , let Z t ⊆ Z P ′ denote the set of pa kets in Z P ′ that have beeninserted by round t . Then: Φ t = C ∗ ( No. of transfers in P as of t ) − X p ′ ∈Z t ( C − ϕ p ′ ) ≤ C ∗ h ( No. of transfers in P as of t ) − | Z P ′ | i ≤ C ∗ (2 nY P − | Z P ′ | ) (8)Consequently, if we an show that at all times Φ ≥ , (8) implies: | Z P ′ | ≤ nY P ≤ n | Z P | + 2 n C (9)Thus, provided Φ ≥ , we an on lude | Z P ′ | = | Z P ′ | + | Z P ′ | ≤ n | Z P | + 4 n C , as required. Themain te hni al hallenge is to argue why Φ ≥ , whi h is the ontent of Lemma B.17, and will requirea bit of work, all of whi h an be found in Appendix B.5 Con lusionIn this paper, we investigated the feasibility of routing in a distributed, asyn hronous, boundedmemory, network with dynami topology and no minimal assumptions on onne tivity. In parti ular,we used ompetitive analysis to evaluate optimal throughput performan e of end-to-end ommuni- ation routing proto ols in this general network. Within this setting, our (cid:28)rst result was to prove abound of /n as the best-possible throughput of any deterministi proto ol. That is, for any proto ol,15Note that we make no ondition that the two pa kets traveled in the same dire tion.10here exists a ompeting proto ol together with a s hedule of a tive edges, su h that the ompetingproto ol will out-perform the given proto ol (in terms of throughput) by at least a fa tor of n .We then went on to demonstrate that in the semi-asyn hronous network model, the Slide proto ola hieves the optimal ompetitive ratio of /n . In Appendix C, we present the (cid:16)Slide + (cid:17) proto ol,whi h is a modi(cid:28) ation of the standard Slide proto ol that allows us to a hieve the same optimal ompetitive ratio of /n in the fully asyn hronous model. By Theorem 2.2, this is the optimalguarantee for throughput that a proto ol an hope to enjoy in the general network model presentedin Se tion 2.Referen es[1℄ Y. Afek, B. Awerbu h, E. Gafni, Y. Mansour, A. Rosen, N. Shavit. 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(cid:16)Log-Spa e Polynomial End-to-End Communi a-tion.(cid:17) SIAM Journal of Computing 27(6): 1531-1549. 1998.[28℄ F. Kelly. (cid:16)Blo king Probabilities in Large Cir uit-Swit hed Networks.(cid:17) Advan ed Appl. Probab.,Vol. 18, pp. 473-505. 1986.[29℄ A. Kamath, O. Palmon, and S. Plotkin. (cid:16)Routing and Admission Control in General TopologyNetworks with Poisson Arrivals.(cid:17) Pro . 7th ACM-SIAM Symp. on Dis rete Algorithms, pp. 269-278. 1996.[30℄ T. Leighton, F. Makedon, S. Plotkin, C. Stein, É. Tardos, and S. Tragoudas. (cid:16)Fast Approxima-tion Algorithms for Multi ommodity Flow Problem.(cid:17) Pro . 23rd ACM Symp. on the Theory ofComputing, pp. 101-111. 1991.[31℄ S. Plotkin. (cid:16)Competitive Routing of Virtual Cir uits in ATM Networks.(cid:17) IEEE J. on Sele tedAreas in Communi ations, Vol. 13, No. 6, pp. 1128-1136. 1995.1232℄ L. Jia, R. Rajaraman, and C. S heideler. (cid:16)On Lo al Algorithms for Topology Control andRouting in Ad Ho Networks.(cid:17) Pro . 15th ACM Symp. on Parallel Algorithms and Ar hite tures,pp. 220-229. 2003.[33℄ F. Shahrokhi and D. Matula. (cid:16)The Maximum Con urrent Flow Problem.(cid:17) J. of ACM, Vol. 37,pp. 318-334. 1990.[34℄ D. Sleator and R. Tarjan. (cid:16)Amortized E(cid:30) ien y of List Update and Paging Rules.(cid:17) Commun.ACM, Vol. 28, No. 2, pp. 202-208. 1985.[35℄ A. Yao. (cid:16)New Algorithms for Bin Pa king.(cid:17) J. of ACM, Vol. 27, pp. 207-227. 1980.AppendixA Formal Proof of Throughput BoundIn this se tion, we go through the rigorous details of the proof of Theorem 2.2, whi h was sket hedin Se tion 3. We will use the same notation introdu ed there for the remainder of this se tion. Inparti ular, re all that there is some (cid:28)xed proto ol P that we wish to analyze, and we are onsideringa s heduling adversary A that pro eeds in y les.We begin with a redu tion of the given proto ol P to a virtual proto ol P ′ , whi h will be operatingwith respe t to a di(cid:27)erent s heduling adversary A ′ than P . The s hedule of edges honored by A ′ will be (in general) di(cid:27)erent than those honored by A , but A ′ will also pro eed in y les. For any y le α in P ′ 's world, de(cid:28)ne Ψ ′ α and Z ′ α analogous to Ψ α and Z α that were de(cid:28)ned for P in Se tion3. We emphasize that the two worlds of P and P ′ are di(cid:27)erent, and we are not attempting to apply ompetitive analysis to these two proto ols. Rather, the property that P ′ will satisfy is: ∀ α ∈ N : Ψ α = Ψ ′ α and Z α = Z ′ α (10)Then given that (10) holds for all y les α , if we an show for all α (subje t to A ′ 's s hedule): Z ′ α + (Ψ ′ α +1 − Ψ ′ α ) ≤ Cn − , (11)then the equivalent statement will be true for P , whi h is Theorem 3.1 in Se tion 3, and thus theproof will be omplete.We now explain the alternate s heduling adversary A ′ , whi h will be de(cid:28)ned in terms of anyarbitrary proto ol attempting to route in a network ontrolled by A ′ . As mentioned above, thes hedule of A ′ will pro eed in y les, ea h of whi h will last ( n − C rounds. At the beginning ofany y le α , A ′ labels the internal nodes by { N α , N α , . . . , N αn − } , so that for all ≤ i ≤ n − ,node N αi is storing more pa kets than N αi +1 at the outset of y le α (note that the labels/indi es ofthe internal nodes will hange every y le). For the (cid:28)rst C rounds of the y le, the adversary willhonor edge E ( S, N ) (here S denotes the Sender). We des ribe the remaining rounds in this y leindu tively (starting below for i = 1 , and e N α = N α ):1. The adversary honors edge E ( e N αi , N αi +1 ) for C rounds2. After the (cid:28)rst ( i + 1) C rounds of y le α have passed (i.e . edge E ( e N αi , N αi +1 ) has just beenhonored C times), let e N αi +1 ∈ { e N αi , N αi +1 } denote the node storing fewer pa kets than the other.Steps 1-2 are repeated through i = n − , so that E ( e N αn − , N αn − ) has just ompleted, and e N αn − hasbeen de(cid:28)ned. Then for the last C rounds of y le α , the adversary honors edge E ( e N αn − , R ) .13emma A.1. Given proto ol P routing in a network ontrolled by A (whose s hedule was des ribedin Se tion 3), there exists a proto ol P ′ ompeting against A ′ , su h that with respe t to ea h proto ol'sown y le, (10) is valid.Proof. Sin e we are onsidering only deterministi proto ols, we an de(cid:28)ne what P ′ will do in anyround based on what P is doing. We will a tually demonstrate something slightly stronger than(10), that is:Indu tion Hypothesis. Up to permutation of the internal nodes, the heights of ea hof the internal nodes in both worlds is the same at the start/end of any y le, as is thenumber of pa kets delivered in any y le.We pro eed by indu tion on the y le. In parti ular, (cid:28)x some y le α , and assume that the indu tionhypothesis is true for all y les β < α . In the (cid:28)rst C rounds of α in P 's world, A opens edge E ( S, A ) ,where A is the internal node urrently storing the most pa kets. Similarly, in the (cid:28)rst C rounds, A ′ opens edge E ( S, A ′ ) , where A ′ is the internal node urrently storing the most pa kets in P ′ 's world.By the indu tion hypothesis, although the labels of node A verses A ′ may be di(cid:27)erent, the nodethat label represents will have the same height in the two worlds, and we de(cid:28)ne P ′ to do the samething that P does in these (cid:28)rst C rounds.Let A denote the node for whi h the adversary A will honor edge E ( A , A ) for the next C rounds, and similarly for A ′ with respe t to A ′ . Note that by the indu tion hypothesis together withthe de(cid:28)nition of P ′ (so far) for the (cid:28)rst C rounds of y le α , we have that the height of A equals theheight of A ′ , and similarly the heights of A and A ′ mat h. Now de(cid:28)ne P ′ to do in the C rounds E ( A ′ , A ′ ) whatever P does in the C rounds E ( A , A ) .16 Thus, after C rounds have passed, thetwo networks are still identi al (up to permutation of the nodes).Let e A denote the node among { A , A } that is storing fewer pa kets after the C rounds of E ( A , A ) . Now in P 's world, the adversary will sear h for the node A with height losest to (butsmaller than) e A , and the adversary A will next honor edge E ( e A , A ) for C rounds. Noti e that,if e.g . P had A transfer all its pa kets to A during the C rounds of E ( A , A ) , it is possible that A is not the node that had the third highest height at the start of y le α (indeed, its even possiblethat A = R ).By the indu tion hypothesis, there is some node A ′ i ( i ≥ ) in P ′ 's world su h that at the start of α , the height of A equals the height of A ′ i (if A = R , then i = n − , i.e . set A ′ i = R ). Noti e thatin ontrast to P 's world, the s hedule of A ′ will ne essarily go through every internal node at leaston e. Indeed, for any ≤ m ≤ n − , the node in P ′ 's world that started y le α as the m th fullestnode will ne essarily be a part of rounds mC through ( m + 1) C − . Therefore, for ea h ≤ m ≤ i ,di tate that during rounds mC through ( m + 1) C − , proto ol P ′ will have the two nodes swap(cid:28)nal states. In parti ular, for any ≤ m ≤ i , if H ′ m denotes the height of A ′ m at the start of y le α , then we di tate that P ′ has transfers enough pa kets from A ′ m to A ′ m − during the C rounds of E ( A ′ m − , A ′ m ) su h that the height of A ′ m − at the end of the C rounds is equal to H ′ m . In thismanner, it is lear that by the time the virtual world of P ′ rea hes the end of iC y les (re all that i is de(cid:28)ned so that the height of A equals the height of A ′ i ), the state of the networks in the twoworlds will be identi al (up to permutation of the nodes). Furthermore, during the next C roundsof ea h y le, the adversaries A and A ′ will honor an edge between two nodes ( E ( A , A ) verses E ( A ′ i − , A ′ i ) ) su h that at the moment the C rounds start, the height of A equals A ′ i − , and theheight of A equals A ′ i . Therefore, this pro ess may be repeated iteratively through the end of the y le in ea h respe tive world, and it is lear that the indu tion hypothesis will remain valid by theend of y le α . (cid:4) For the remainder of the se tion, we will seek to prove (11) for the proto ol P ′ . To simplifynotation, it will be onvenient to de(cid:28)ne m = n − . At the outset of every y le α , we label the16In order to preserve Fa t 1 below, we demand that after the C rounds of E ( A ′ , A ′ ) , A ′ is storing fewer pa ketsthan A ′ . Therefore, if this is not the ase for E ( A , A ) , then de(cid:28)ne P ′ to end in a symmetri state as P , i.e . so thatthe pair of nodes ( A , A ) have the same height as the pair of nodes ( A ′ , A ′ ) , but in the latter pair, ne essarily A ′ isstoring at least as many pa kets as A ′ after the C rounds of E ( A ′ , A ′ ) .14nternal (i.e . ex luding the Sender and Re eiver) nodes { N α , N α , . . . , N αm } , su h that if i < j , thennode N αi is storing more (or an equal number of) pa kets at the start of y le α than N αj . For all α , let N α = S and N αn − = R . For any ≤ i ≤ n − , let H αi denote the height the node had at theoutset of α . We emphasize that while the heights of nodes may hange through the ourse of y le α , the labeling { N αi } and the quantities { H αi } will remain (cid:28)xed throughout the y le. Indeed, thefollowing fa t implies that the labeling of nodes is independent of α (and in fa t is (cid:28)xed for all time):Fa t 1. For all α ∈ N and all ≤ i ≤ m : N αi = N α +1 i Fa t 2. For any y le α , node N i is a part of C rounds of the y le: (cid:28)rst for C roundswith E ( N i − , N i ) , and then for C rounds with E ( N i , N i +1 ) These fa ts, along with the following observations, all follow from the de(cid:28)nition/ onstru tion of P ′ in the proof of Lemma A.1 above. To (cid:28)x notation, for ea h ≤ i ≤ m let A αi denote the numberof pa kets sent from A i to A i +1 during the C rounds E ( N i , N i +1 ) of y le α . Note that A αi may benegative if the net pa ket (cid:29)ow during E ( N i , N i +1 ) was towards N i .Lemma A.2. For any y le α and for any ≤ i ≤ m : A αi ≤ A αi − + H αi − H αi +1 (12) A αi ≤ H α +1 i − H αi +1 (13)Proof. Statement 1 follows from the two fa ts above as follows. Note that after the C rounds E ( N i − , N i ) but before the next C rounds, node N i will have height A αi − + H αi . Now by de(cid:28)nitionof proto ol P ′ , at the end of the C rounds of E ( N i , N i +1 ) , N αi will have a greater (or equal) numberof pa kets than N αi +1 . In parti ular, sin e there are A αi − + H αi + H αi +1 total pa kets between the twonodes at the start of the C rounds E ( N αi , N αi +1 ) , it must be that at the end of these C rounds, N αi is storing at least half of these. Sin e the number of pa kets stored by N αi after the C rounds of E ( N αi , N αi +1 ) is given by A αi − + H αi − A αi , Statement 1 follows.Also, again sin e proto ol P ′ spe i(cid:28)es that N αi must have more (or an equal number of) pa ketsas N αi +1 immediately after the C rounds of E ( N αi , N αi +1 ) , and by Fa t 2 the height of N αi will not hange through the remainder of y le α , Statement 2 follows. (cid:4) We are interested in the potential fun tion: Ψ ′ α = m X i =1 (cid:18) (cid:19) m − i · max (cid:18) , H αi − ( m − i ) Cm (cid:19) (14)For ea h ≤ i ≤ m , de(cid:28)ne: δ αi = (cid:26) if the nd term of the max statement in (14) dominates otherwise (15)Also, for any pair of indi es ≤ j < k ≤ m , de(cid:28)ne: (Ψ ′ α +1 − Ψ ′ α ) i,j = j X k = i (cid:18) (cid:19) m − k · (cid:20) max (cid:18) , H α +1 k − ( m − k ) Cm (cid:19) − max (cid:18) , H αk − ( m − k ) Cm (cid:19)(cid:21) (16)Claim A.3. For any index ≤ i ≤ m and any y le α : H α +1 i = H αi + A αi − − A αi (17)15roof. Noti e N α +1 i = N α (Fa t 1) and N i is a part of exa tly C rounds for the α th y le (Fa t 2).In the (cid:28)rst C rounds, H i hanges by A αi − , and in the se ond C rounds it hanges by − A αi . Sin e N i began the y le with height H αi , we have that its height at the start of the ( α + 1) th y le will be H αi + A αi − − A αi . (cid:4) It will be onvenient to introdu e the following notation:De(cid:28)nition A.4. For any ≤ i ≤ m and any y le α , de(cid:28)ne: ω αi := min (cid:18) , H αi − ( m − i ) Cm (cid:19) (18)Claim A.5. For any index ≤ i ≤ m and any y le α : If δ α +1 = 1 , then: (Ψ ′ α +1 − Ψ ′ α ) i,i = 12 m − i ( A αi − − A αi + ω αi )2) If δ α +1 = 0 , then: (Ψ ′ α +1 − Ψ ′ α ) i,i = 12 m − i ω αi (19)Proof. If δ α +1 = 1 , then onsider the equalities: (Ψ ′ α +1 − Ψ ′ α ) i,i = 12 m − i (cid:20) max (cid:18) , H α +1 i − ( m − i ) Cm (cid:19) − max (cid:18) , H αi − ( m − i ) Cm (cid:19)(cid:21) = 12 m − i (cid:20) ( A αi − − A αi + H αi ) − ( m − i ) Cm − max (cid:18) , H αi − ( m − i ) Cm (cid:19)(cid:21) = 12 m − i ( A αi − − A αi ) + ( if H αi ≥ ( m − i ) Cm m − i (cid:16) H αi − ( m − i ) Cm (cid:17) if H αi < ( m − i ) Cm = 12 m − i ( A αi − − A αi + ω αi ) where the se ond equality is from Claim A.3 together with the assumption that δ α +1 = 1 . Otherwise,if δ α +1 = 0 , then Statement 2 is immediate. (cid:4) Lemma A.6. For any pair of indi es ≤ i < j < m for whi h δ α +1 k = 1 for every i ≤ k ≤ j :17 (Ψ ′ α +1 (cid:21) Ψ ′ α ) i,j − j X k = i ω k m − k + A j m − j − ≤ A i − m − i + ( j − i + 1)2 m − i +1 ( A i − + H i ) − H j +1 m − j +1 + j − X k = i +1 ( j − k )2 m − k +2 H k (20)Proof. This follows via an indu tive argument on j − i together with Lemma A.2 and Claim A.5:Base Case: j = i + 1 : First onsider the right-hand-side of (20) with j = i + 1 :RHS (20) = A i − m − i + 22 m − i +1 ( A i − + H i ) − H i +2 m − i = A i − m − i + 12 m − i ( A i − + H i ) − H i +2 m − i = A i − m − i − + 12 m − i H i − H i +2 m − i (21)17On the right-hand side of (20), all supers ripts are α , whi h we have suppressed for notational onvenien e.16eanwhile, for j = i + 1 , the left-hand-side of (20) is:LHS (20) = (Ψ ′ α +1 (cid:21) Ψ ′ α ) i,i +1 − i +1 X k = i ω i m − i + A i +1 m − i − = (Ψ ′ α +1 (cid:21) Ψ ′ α ) i,i + (Ψ ′ α +1 (cid:21) Ψ ′ α ) i +1 ,i +1 − i +1 X k = i ω i m − i + A i +1 m − i − = 12 m − i ( A i − − A i + ω i ) + 12 m − i − ( A i − A i +1 + ω i +1 ) − i +1 X k = i ω i m − i + A i +1 m − i − = 12 m − i − A i +1 + 12 m − i ( A i + A i − ) ≤ m − i − ( A i + H i +1 − H i +2 ) + 12 m − i ( A i + A i − ) ≤ m − i − ( H i + A i − − A i − H i +2 ) + 12 m − i ( A i + A i − )= A i − m − i − + 12 m − i H i − H i +2 m − i (22)where the third equality is due to Claim A.5, the (cid:28)rst inequality is Statement 1 of Lemma A.2, andthe se ond inequality is Claim A.3. Noti e (21) mat hes (22), as required.Indu tion Step: Consider the string of inequalities: (Ψ ′ α +1 (cid:21) Ψ ′ α ) i,j − j X k = i ω i m − i + A j m − j − = (Ψ ′ α +1 (cid:21) Ψ ′ α ) i,i + (Ψ ′ α +1 (cid:21) Ψ ′ α ) i +1 ,j − j X k = i ω i m − i + A j m − j − ≤ m − i ( A i − − A i ) + A i m − i − + ( j − i )2 m − i ( A i + H i +1 ) − H j +1 m − j +1 + j − X k = i +2 ( j − k )2 m − k +2 H k ≤ A i − m − i + ( j − i + 1)2 m − i +1 ( A i − + H i ) − H j +1 m − j +1 + j − X k = i +1 ( j − k )2 m − k +2 H k where the (cid:28)rst inequality is by the indu tion hypothesis together with Claim A.5 and the lastinequality is by Statement 2 of Lemma A.2. (cid:4) Lemma A.7. For any pair of indi es ≤ i < i + 1 < j ≤ m for whi h δ α +1 j = 1 but δ α +1 k = 0 forevery i < k < j :18 (Ψ ′ α +1 − Ψ ′ α ) i +1 ,j − − j − X k = i +1 ω k m − k + A j − m − j ≤ A i m − i − + H i +1 m − i − H j m − j +1 + j − X k = i +1 H k m − k +1 (23)Proof. This follows via an indu tive argument on j − i together with Lemma A.2:Base Case: j − i = 2 : Looking at the right-hand-side of (23) for j = i + 2 :RHS (23) = A i m − i − + H i +1 m − i − H i +2 m − i − + H i +1 m − i = A i + H i +1 − H i +2 m − i − (24)18On the right-hand side of (20), all supers ripts are α , whi h we have suppressed for notational onvenien e.17eanwhile, looking at the left-hand-side of (23) for j = i + 2 :LHS (23) = (Ψ ′ α +1 − Ψ ′ α ) i +1 ,i +1 − i +1 X k = i +1 ω i m − i + A i +1 m − i − = A i +1 m − i − ≤ A i + H i +1 − H i +2 m − i − , (25)where the se ond equality is from Claim A.5 (sin e δ α +1 i +1 = 0 ) and the inequality is Statement 1 ofLemma A.2.Indu tion Step: Consider the string of inequalities: (Ψ ′ α +1 (cid:21) Ψ ′ α ) i +1 ,j − − j X k = i ω i m − i + A j − m − j = (Ψ ′ α +1 (cid:21) Ψ ′ α ) i +1 ,i +1 + (Ψ ′ α +1 (cid:21) Ψ ′ α ) i +2 ,j − − j X k = i ω i m − i + A j − m − j ≤ A i +1 m − i − + H i +2 m − i − − H j m − j +1 + j − X k = i +2 H k m − k +1 ≤ A i m − i − + H i +1 m − i − H j m − j +1 + j − X k = i +1 H k m − k +1 (26)where the (cid:28)rst inequality is by the indu tion hypothesis together with Claim A.5 and the lastinequality is by Statement 1 of Lemma A.2. (cid:4) Lemma A.8. For any y le α and any index ≤ i < m − , if δ α +1 i = 1 , δ α +1 i +1 = 0 , and δ α +1 i +2 = 1 ,then: (Ψ ′ α +1 − Ψ ′ α ) i +1 ,i +1 − i +1 X k = i +1 ω k m − k + A i m − i − ≤ A i m − i − + 12 m − i − Cm (27)Proof. Consider: (Ψ ′ α +1 − Ψ ′ α ) i +1 ,i +1 − i +1 X k = i +1 ω k m − k + A i m − i − = A i +1 m − i − ≤ A i + H i − H i +1 m − i − ≤ A i m − i − + 12 m − i − Cm where the (cid:28)rst equality is Statement 2 of Lemma A.5, the (cid:28)rst inequality is Statement 1 of A.2,and the last inequality follows from the fa t that δ α +1 i = 1 , δ α +1 i +1 = 0 , and δ α +1 i +2 = 1 implies that H i − H i +1 m − i − ≤ m − i − Cm . (cid:4) Lemma A.9. For any y le α and any index ≤ i < m − , if δ α +1 i = 0 , δ α +1 i +1 = 1 , and δ α +1 i +2 = 0 ,then: (Ψ ′ α +1 − Ψ ′ α ) i +1 ,i +1 − i +1 X k = i +1 ω k m − k + A i m − i − ≤ A i m − i − + 12 m − i − Cm (28)18roof. Consider: (Ψ ′ α +1 − Ψ ′ α ) i +1 ,i +1 − i +1 X k = i +1 ω k m − k + A i m − i − = A i m − i − + A i +1 m − i − ≤ A i + H α +1 i +1 − H αi +2 m − i − ≤ A i m − i − + 12 m − i − Cm where the (cid:28)rst equality is Statement 1 of Lemma A.5, the (cid:28)rst inequality is Statement 2 of A.2,and the last inequality follows from the fa t that δ α +1 i = 0 , δ α +1 i +1 = 1 , and δ α +1 i +2 = 0 implies that H α +1 i +1 − H αi +2 m − i − ≤ m − i − Cm . (cid:4) Claim A.10. For any y le α , we have: Z α + (Ψ ′ α +1 − Ψ ′ α ) m,m ≤ A αm − (29)Proof. Sin e ( H α +1 m − ( m − m ) Cm ) = H α +1 m ≥ , we have that the se ond term of min(0 , H α +1 m − ( m − m ) Cm ) always dominates, and hen e for all y les, δ α +1 m = 1 . Therefore, applying Claim A.5(for i = m ): (Ψ ′ α +1 − Ψ ′ α ) m,m = A αm − − A αm + ω αm ≤ A αm − − A αm = A αm − − Z α (30)where the inequality follows sin e ω αi ≤ for all y les α and nodes i , and the last equality is be ause N m is the node that will be onne ted to the Re eiver in the last C rounds of α , so by de(cid:28)nition A αm = Z α . (cid:4) We are now ready to prove the main result of this se tion, namely that (11) is satis(cid:28)ed for all y les α :Theorem A.11. For all y les α , the following is always true: Z ′ α + (Ψ ′ α +1 − Ψ ′ α ) ≤ Cm ,
Proof. Fix y le α , and onsider the string of bits { δ α +1 i } mi =1 : ( δ α +11 , δ α +12 , . . . , δ α +1 m − , δ α +1 m ) (31)By Claim A.10, we have: Z α + Ψ ′ α +1 − Ψ ′ α = Z α + (Ψ ′ α +1 − Ψ ′ α ) ,m ≤ (Ψ ′ α +1 − Ψ ′ α ) ,m − + A αm − (32)We now use Lemmas A.6, A.7, A.8, and A.9 on the appropriate indi es (based on the form of { δ α +1 i } ),whi h yields:191. For the smallest index i su h that δ α +1 i = 1 , we have leading term: A i m − (33)19We ombine these lemmas by starting at the far right index i = m − , and working our way down through smallerindi es by using the appropriate lemma. Noti e that the (cid:28)rst term on the RHS of the inequality of ea h lemma isexa tly the term needed on the LHS of the next lemma. 19. For any indi es i < j falling under Lemma A.6, we have ontributions: j − i + 12 m − i +1 ( A i − + H i ) + j − X k = i +1 ( j − k + 1)( m − i )2 m − k +2 (34)3. For any indi es i < j falling under Lemma A.7, we have ontribution: j X k = i m − i m − k +1 (35)4. For any indi es i + 1 falling under Lemma A.8 or A.9, we have ontribution: m − i − Cm (36)Noti e that in terms of the ontributions from (34), ( A i − + H i ) ≤ ( m − i − Cm by Statement 2 ofLemma A.2 together with the fa t that δ α +1 i − = 0 implies H α +1 i − < ( m − i − Cm . The theorem nowfollows immediately from the fa ts:1. For any ≤ i < j < ∞ , P jk = i n ≤ P ∞ k =1 12 n = 1
2. For any ≤ i < j < ∞ , P jk = i n n ≤ P ∞ k =1 n n = 2
3. For any ≤ i < j < ∞ , P jk = i n ( n − n ≤ P ∞ k =1 n ( n − n = 4 (cid:4) The remainder of the proof that the optimal ompetitive ratio is /n was presented in Se tion 3.B Rigorous Proof of Competitive Ratio of SlideThe high-level ideas of the proof of Theorem 2.2 were sket hed in Se tion 4.2, and we en ouragethe reader to re-read that se tion before pro eeding here. In this Se tion, we begin by providing inSe tion B.1 a deeper explanation of the proof than was provided in Se tion 4.2, but still does notgo into the details of the proofs. Then in Se tions B.2-B.5 we rigorously prove all the lemmas andtheorems.B.1 Motivation and De(cid:28)nitionsIn what follows, unless stated otherwise, all notation is as de(cid:28)ned in Se tion 4.2. Re all fromSe tion 4.2 that we wish to onstru t two potential fun tions. The (cid:28)rst one, denoted by ϕ p ′ , will beasso iated to every pa ket p ′ ∈ Z P ′ . However, ϕ p ′ will not be exa tly as de(cid:28)ned in Se tion 4.2, sowe provide now the motivation to explain how ϕ p ′ is a tually de(cid:28)ned, and why we need to slightly hange what it represents.Our (cid:28)rst attempt employed in Se tion 4.2 was to de(cid:28)ne ϕ p ′ to be the height, with respe t to P , of the node in whi h p ′ was urrently being stored. We state on e-and-for-all that whenreferen ing the height of a node, we will mean its height with respe t to the Slide proto ol P . As noted in Se tion 4.2, if we de(cid:28)ne ϕ p ′ this way, then for every p ′ ∈ Z P ′ , ϕ p ′ will be initiallyset to C (when P ′ (cid:28)rst inserts p ′ ), and ϕ p ′ will be zero when p ′ is delivered to the Re eiver. Thus,there is a net hange of − C to ϕ p ′ from the time of insertion by the Sender to the time of re eptionby the Re eiver. The goal was then to de(cid:28)ne a se ond overall network potential fun tion Φ , whi h20n reases by C every time P transfers a pa ket, and su h that any time ϕ p ′ hanges for any p ′ ∈ Z P ′ ,the umulative hanges of P p ′ ∈ Z P′ ϕ p ′ will be mimi ked by Φ . Sin e Φ in reases by C when thereis a pa ket transfer in P , one (good) way to think of this approa h is that for ea h drop in ϕ p ′ , wewould like to (cid:28)nd a pa ket transfer in P that an be (cid:16) harged,(cid:17) i.e . this pa ket transfer (cid:16)allowed(cid:17) ϕ p ′ to de rease.Unfortunately, with the simplisti de(cid:28)nition of ϕ p ′ equal to the height of the node it is urrentlystored in, we en ounter a problem. To larify the problem, as well as to set notation, at the verybeginning of ea h round x , we will label the internal nodes (i.e . not the Sender or Re eiver) as: { N x , N x , . . . , N xn − } , where the labeling respe ts heights, so that at the start of the round x , N xi +1 is storing at least as many pa kets as N x (ties are broken arbitrarily). Letting H xi denote the heightof N xi at the start of x (i.e . the number of pa kets N xi is storing with respe t to P ), we may restatethe riterion for labeling nodes at the start of ea h round by writing: H x ≤ H x ≤ · · · ≤ H xn − . Notethat nodes may hange labels from one round to the next, i.e . we may have N xi = N x +1 i . When theround is unimportant, we will suppress the supers ript x . Let S denote the Sender and R denotethe Re eiver.We may now explain why the simplisti de(cid:28)nition of ϕ p ′ above will not be adequate. De(cid:28)ne Q := C − nn , and onsider the following two s enarios that may be present at the start of some round x : S enario 1: H n − = C H n − = C . . . H = C H = C H = ( n − Q S enario 2: H n − = ( n − Q H n − = ( n − Q . . . H = 2 Q H = Q H = 0 In S enario 1, onsider a pa ket p ′ ∈ Z P ′ that begins round x in node N , so that ϕ p ′ = ( n − Q .Noti e that if the adversary honors the edge E ( N , R ) , the Slide proto ol will transfer a pa ket tothe Re eiver (Rules 2 and 3a of Se tion 4.1). Now by de(cid:28)nition of being in the set Z P ′ , in order for p ′ to be delivered to the Re eiver via node20 N , node N must have height zero when the adversaryhonors edge E ( N , R ) . Therefore, there must be exa tly ( n − Q transfers in P (to drain N ) before p ′ an be delivered to R via N . Thus, loosely speaking, we an (cid:16) harge(cid:17) the resulting drop in ϕ p ′ from ( n − Q to to these ( n − Q transfers in P .Now instead imagine we are in S enario 2, and again (cid:28)x a pa ket p ′ ∈ Z P ′ su h that ϕ p ′ = ( n − Q at the start of round x , so p ′ ∈ N n − . In this ase, noti e that p ′ has a way to rea h R withoutany pa kets being transferred in P . In parti ular, the adversary ould honor edge E ( N n − , N n − ) in round x , and then E ( N n − , N n − ) in round x + 1 , and so forth. Sin e the di(cid:27)eren e in heightsbetween adja ent nodes is less than C/n , the Slide proto ol will not transfer any pa kets during theserounds. Meanwhile, proto ol P ′ may di tate that p ′ is transferred ea h of these rounds, all the wayto the Re eiver. Thus, in this s enario, ϕ p ′ was able to de rease from ( n − Q to zero without anypa kets being transferred in P . Be ause we are trying to asso iate drops in ϕ p ′ to pa ket transfers in P , this is problemati .Noti e that the problem in S enario 2 is that there exists a (cid:16)bridge(cid:17) between N n − and R . Thatis, even though N n − has a relatively large height, there is still a way for pa kets p ′ ∈ Z P ′ that arein N n − to rea h R without P being able to transfer any pa kets. In ontrast, in S enario 1, p ′ ∈ N will also have ϕ p ′ = ( n − Q , but now there must be ( n − Q transfers in P before p ′ an rea h R (again, sin e p ′ ∈ Z P ′ requires that p ′ is never transferred at the same time as a pa ket in P ). Insummary, one might say that even though node N in S enario 1 has the same height as node N n − from S enario 2, these two nodes have di(cid:27)erent (cid:16)e(cid:27)e tual(cid:17) heights.Considering the above two S enarios, we were en ouraged to modify our de(cid:28)nition of ϕ p ′ asfollows:- For node N i , de(cid:28)ne the node's e(cid:27)e tual height:21 e H i := max(0 , H i − ( i − Cn ) p ′ must be transferred to R via N , but for the sake of the example,we imagine this is the ase.21The (cid:16)maximum(cid:17) is added to prevent the e(cid:27)e tual height of a node from being negative.21 For any p ′ ∈ Z P ′ that is urrently in N i , de(cid:28)ne its potential: ϕ p ′ := e H i This is almost the a tual de(cid:28)nition we eventually make for ϕ , but we will need to (cid:28)rst (cid:16)smooth-out(cid:17)this de(cid:28)nition. To motivate the need to smooth the de(cid:28)nition, onsider the following events, whi hrepresent the only ways that ϕ p ′ an hange (based on the new de(cid:28)nition of ϕ p ′ ):Case 1. p ′ is transferred from N i to N j in some round E ( N i , N j ) Case 2. p ′ ∈ N i when N i hanges height due to a pa ket transfer in P , but this pa ket transferdoes not ause a re-indexing of nodesCase 3. p ′ is in some node N i when a pa ket transfer in P auses N i to hange index to N j (i.e . this node moves from the i th fullest node to the j th fullest node)Sin e we are only on erned with p ′ ∈ Z P ′ , we note that whenever ϕ p ′ hanges as by 1) above,ne essarily P did not transfer a pa ket this round. In parti ular, this means that | H i − H j | < C/n .In order to ontrol hanges to ϕ p ′ that are a result of Case 1, we would therefore like for e H i ≈ e H j whenever H i ≈ H j . Although the de(cid:28)nition of e(cid:27)e tual height e H i above almost aptures this, thereis ne essarily a (cid:16)jump(cid:17) of C/n between the values e H i and e H j . This is one of the reasons we will wantto (cid:16)smooth-out(cid:17) the de(cid:28)nition of ϕ p ′ .Changes to ϕ p ′ that ome from Case 2 above are okay, sin e in su h ases ϕ p ′ will hange by one,and this an be (cid:16) harged(cid:17) to the fa t that there has been a pa ket transfer in P . Lastly, noti e that ϕ p ′ an only hange as in Case 3 above if there are two nodes at the outset of some round x , N i and N i +1 , su h that a pa ket transfer during round x auses them to swit h pla es (e.g . before thetransfer, H i = H i +1 , and then N i re eives a pa ket in round x ). Be ause there has been a pa kettransfer in P , we an (cid:16) harge(cid:17) some of the hanges in ϕ p ′ to this pa ket transfer, but again the fa tthat there will be a (cid:16)jump(cid:17) of C/n to hanges in ϕ will en ourage a (cid:16)smoothing(cid:17) of the de(cid:28)nition of ϕ . This leads to the notion of a family of nodes. In parti ular, we will partition the internal nodesinto families. Intuitively, two nodes will be in the same family if they are relatively lose to ea h otherin height (or more generally, if there is a (cid:16)bridge(cid:17) onne ting them, as in S enario 2 above). Thenwithin ea h family, we will distribute the umulative e(cid:27)e tual height of the nodes in that familyevenly among all nodes in the family. Formally, for a family of nodes22 F = { N i , N i +1 , . . . , N j } ,de(cid:28)ne the umulative e(cid:27)e tual height H F of the family F by: e H F := j X k = i e H k = j X k = i max (cid:18) , H k − ( k − Cn (cid:19) For any p ′ ∈ Z P ′ su h that p ′ is urrently in some node of family F , we will de(cid:28)ne ϕ p ′ to be theaverage e(cid:27)e tual height of the family, i.e . : ϕ p ′ := e H F |F | Of ourse, e H F may not divide evenly among the nodes in the family F , and then to for e ϕ p ′ ∈ N ,we will distribute the ex ess weight (the remainder) to the nodes with higher indi es. Based on thisde(cid:28)nition of ϕ p ′ , note that if p ′ transfers between two nodes of the same family, ϕ p ′ an hange byat most one.We re-visit the three ways ϕ p ′ may hange, explaining in ea h ase how we an (cid:28)nd a pa kettransfer in P to (cid:16) harge(cid:17) for the hange in ϕ p ′ . In terms of hanges to ϕ p ′ resulting from Case 1 above,we re all that ne essarily | H i − H j | < C/n . We show in Lemma B.12 that anytime | H i − H j | < C/n ,22We will show in the next se tion that nodes within the same family will always have adja ent indi es.22 i and N j are ne essarily in the same family, in whi h ase our de(cid:28)nition of ϕ now guarantees that ϕ p ′ an hange by at most one when p ′ is transferred between nodes. Changes to ϕ p ′ due to Case 2will be at most one (sin e the umulative e(cid:27)e tual height of the family will hange by at most one,and this hange will be distributed among nodes in the family), and we an (cid:16) harge(cid:17) su h hanges tothe pa ket transfer in P that aused Case 2 to o ur. Finally, for Case 3, if p ′ ∈ N i when N i 's index hanges but N i remains in the same family, than sin e ϕ is distributed evenly among nodes in thefamily, the hange in index will be irrelevant (i.e . this will not ause ϕ p ′ to hange). On the otherhand, we will show that whenever a node N i swit hes families as a result of a pa ket transfer in P ,the average e(cid:27)e tual height of its new family will di(cid:27)er by at most one from the average e(cid:27)e tualheight of its old family. Thus, in this ase the hange in ϕ p ′ is also bounded by one, and we an(cid:16) harge(cid:17) this hange to the pa ket transfer that aused families to re-align.De(cid:28)ning how to partition nodes into families so that the families behave the way we want (e.g . so that: 1) nodes with height within C/n of ea h other are in the same family; 2) Families an onlyre-align during a round in whi h P transfers a pa ket; and 3) When families re-align, the averagee(cid:27)e tual height of any node before and after the re-alignment di(cid:27)ers by at most one) requires a littlethought, and it is done pre isely in the following se tion. On e we have the formal de(cid:28)nition of afamily, we would like to formalize the notion of (cid:16) harging a hange in ϕ p ′ to a pa ket transfer in P .(cid:17)Namely, as mentioned in Se tion 4.2, we de(cid:28)ne a se ond network potential Φ that will in rease by C every time there is a pa ket transfer in P , and that will also mirror the umulative hanges of ϕ p ′ forea h p ′ ∈ Z P ′ . In order to prove Φ is always positive, we will distribute the total network potentialbetween the families: Φ = Φ F + · · · + Φ F l (37)and then show in Lemma B.17 that within ea h family F : Φ F ≥ . (38)The areful de(cid:28)nition of families and the pre ise de(cid:28)nition of the potential ϕ and the networkpotential Φ is presented below in Se tion B.2. The main lemma and proof of the fa t that at alltimes Φ ≥ an be found in Se tion B.5.B.2 Formal De(cid:28)nition of (cid:16)Family(cid:17) and Potential of a Pa ket ( ϕ p ′ )We begin by de(cid:28)ning formally the notion of a family introdu ed in the previous se tion. Notethat families will in general re-align during a round when there is a pa ket transfer in P , so we usethe notation F x to denote some family F that was in existen e at the start of round x . Re all thatat the start of ea h round x , the internal nodes are indexed a ording to their heights with respe tto P : { N , N , . . . , N n − } , so that H i ≤ H j if i < j (ties are broken arbitrarily). Also re all fromthe previous se tion the de(cid:28)nition of the e(cid:27)e tual height e H i of node N i : e H i := max (cid:18) , H i − ( i − Cn (cid:19) (39)At the start of ea h round, we will partition the internal nodes into families indu tively (starting fromthe emptiest nodes), so that the average e(cid:27)e tual height of ea h family is minimized. In parti ular:De(cid:28)nition B.1. At the start of round x , internal nodes will be partitioned into families {F xi } asfollows. Starting at i = 1 and k = 0 :F1 Find index k i − < k i ≤ n − su h that the following quantity is minimal: k i X j = k ( i − +1 e H j ( k i − k i − ) (40)23n ase there are multiple values for k i that a hieve the same minimum, de(cid:28)ne k i to be thelargest of all possibilities. Then de(cid:28)ne23 family F xi := { N xk ( i − +1 , . . . , N xk i } .F2 Set i = i + 1 and repeat Step F1 until all internal nodes are in some family.F3 The Sender and Re eiver will form their own, separate, families. Denote the Sender's familyby F n and the Re eiver's family by F .24De(cid:28)nition B.2. The umulative e(cid:27)e tual height e H F of a family F is the sum of the e(cid:27)e tual heightsof ea h of the nodes in the family. The average e(cid:27)e tual height h e H F i of a family is the umulativee(cid:27)e tual height divided by the size of the family. Su in tly, if F := { N i , N i +1 , . . . , N j } : e H F := j X k = i e H k and h e H F i := e H F |F | = j X k = i e H k j − i + 1 (41)Noti e that by onstru tion (see Rules F1 and F2), families are reated so that the averagee(cid:27)e tual height of (the lowest indexed) families is minimized.With the formal de(cid:28)nition of families in hand, we are ready to formally de(cid:28)ne the (cid:28)rst kind ofpotential, ϕ . Re all that this potential will be asso iated to pa kets p ′ ∈ Z P ′ , and if p ′ ∈ N i ∈ F atthe start of some round, then ϕ p ′ will (roughly) represent the average e(cid:27)e tual height h e H F i . Morepre isely, we will as ribe to ea h node N i ∈ F a potential ϕ i equal to the average e(cid:27)e tual height,ex ept that the potential for some nodes in the family will be one bigger to a ount for the ase that e H F |F| / ∈ Z . Formally:De(cid:28)nition B.3. Let F = { N i , N i +1 , . . . , N j } . Then the potential ϕ k of a node N k ∈ F will be either h e H F i or h e H F i + 1 . More pre isely, writing: e H F = ⌊h e H F i⌋ ∗ |F | + r (42)Then de(cid:28)ne subsets of F : F − := { N i , N i +1 , . . . , N j − r } and F + := { N j − r +1 , . . . , N j } (43)Then for nodes N k ∈ F + , de(cid:28)ne ϕ k = ⌊h e H F i⌋ + 1 . For nodes N k ∈ F − , de(cid:28)ne ϕ k = ⌊h e H F i⌋ . Finally,if p ′ ∈ Z P ′ and p ′ is urrently being stored in N k , then de(cid:28)ne the potential ϕ p ′ to be the potential of N k , i.e . ϕ p ′ := ϕ k .One immediate onsequen e of the above de(cid:28)nition that we will need later is:Lemma B.4. At the beginning of any round x and for any family F x , the sum of the potentials forthe nodes in F equals the umulative e(cid:27)e tual height of the family: X N ∈F ϕ N = e H F (44)De(cid:28)nition B.5. The network potential Φ is an integer satisfying the following properties:1. Φ begins the proto ol equal to zero.2. Φ in reases by C every time a pa ket is transferred in proto ol P
3. For any pa ket p ′ ∈ Z P ′ , any time ϕ p ′ hanges, Φ hanges by the same amount.23When the round x is unimportant, we will suppress the supers ript in our notation.24The only reason we pla e the Sender and Re eiver in a family at all is to make the terminology easier in thelemmas that follow. In parti ular, the notation we use for the Sender's family ensures that it will have a higher indexthan all other nodes (there will be a gap between the index of the largest indexed family of internal nodes and theSender's family, whi h is unimportant), and onversely the Re eiver's family will have a smaller index than all othernodes. 24.3 Preliminary LemmasIn this se tion, we state and prove the basi properties that follow from the de(cid:28)nitions of theprevious se tion.Lemma B.6. At all times, all families onsist of nodes with adja ent indi es. In parti ular, if atthe start of any round x there are l families, then there exist indi es k < k < · · · < k l − su h that: F = { N , . . . , N k } , F = { N k +1 , . . . , N k } , . . . , F l = { N k l − +1 , . . . , N n − } (45)Proof. This follows immediately from the rules regarding the onstru tion of families (see F1 and F2in the previous se tion). (cid:4) Lemma B.7. Fix some round x and some pair of nodes N xi and N xj for i < j . Then:1. If H xi ≥ H xj − C/n , then e H xi ≥ e H xj .2. If H xi < H xj − ( j − i ) C/n and e H j > , then e H xi < e H xj .Proof. Consider the following string of inequalities: e H i − e H j = max(0 , H i − ( i − C/n ) − max(0 , H j − ( j − C/n ) ≥ max(0 , H i − ( i − C/n ) − max(0 , ( H i + C/n ) − ( j − C/n ) ≥ max(0 , H i − ( i − C/n ) − max(0 , ( H i + C/n ) − (( i + 1) − C/n )= max(0 , H i − ( i − C/n ) − max(0 , ( H i − ( i − C/n )= 0
This proves Statement 1. For Statement 2, if e H i = 0 , then it is immediate. Otherwise, onsider theinequalities: e H j − e H i = H j − ( j − C/n − ( H i − ( i − C/n )= H j − H i + (( i − − ( j − C/n> ( j − i ) C/n + ( i − j ) C/n = 0 (cid:4)
We state a trivial observation regarding fra tions of positive numbers that will be useful in provingthe lemmas below.Observation 2. For any positive numbers a, b, c, d ∈ N :1. ab < cd ⇒ ab < a + cb + d < cd ab = cd ⇒ ab = a + cb + d = cd Lemma B.8. Let x be any round, and suppose that at the outset of the round there is some family F xα = { N i , N i +1 , . . . , N j } . Then the following statements are all true at the outset of round x : For any i ≤ k < j : P km = i e H m k − i + 1 ≥ h e H F α i ≥ P jm = k +1 e H m j − k For any j < k ≤ n − h e H F α i < P km = j +1 e H m k − j h e H F α i < h e H F α +1 i P km = i e H m k − i +1 ≥ P jm = k +1 e H m j − k follows immediately from Observation 2 together withthe rules regarding the onstru tion of families (see Rule F1 from the previous se tion), and in par-ti ular the fa t that indi es are found by minimizing (40). Statement 1 now follows from Observation2. Statement 2 also follows immediately from Rule F1 and Observation 2, and Statement 3 followsimmediately from Statement 2. (cid:4) Statement 3 of Lemma B.8 an be immediately extended:Corollary B.9. Let x be any round, and suppose that at the outset of the round there are l families.Then: h e H F i < h e H F i < · · · < h e H F l i Lemma B.10. Let x be any round, and suppose that at the outset of the round there is some family F xα = { N i , N i +1 , . . . , N j } . Then:For any ≤ k < i : P i − m = k e H m i − k < h e H F α i (46)Proof. Sin e k < i , ne essarily N k is in some family F β with index β < α . Then: P i − m = k e H m i − k ≤ h e H F β i < h e H F β +1 i < . . . < h e H F α − i < h e H F α i , (47)where the (cid:28)rst inequality is from Statement 1 of Lemma B.8 and the other inequalities are fromCorollary B.9. (cid:4) Lemma B.11. If at the start of some round x we have that e H xj +1 ≤ e H xj , then N j and N j +1 are inthe same family at the start of round x .Proof. Suppose for the sake of ontradi tion that they are not in the same family at the start ofround x . Let F x denote N j 's family at the start of the round. By Lemma B.6 and the fa t that j and j + 1 are adja ent indi es, we must have that F x = { N i , N i +1 , . . . , N j } for some i ≤ j . The keyobservation is that: e H j +1 ≤ e H j ⇒ e H j +1 ≤ e H j +1 + e H j ≤ e H j (48)If i = j , then (48) ontradi ts Statement 2 of Lemma B.8 (set k = j + 1 ). If i < j , then de(cid:28)ne: A : = j − X l = i e H l and B := j − i (49)Then by Lemma B.8: e H j +1 ≤ e H j ≤ AB ⇒ e H j +1 + e H j + AB + 2 ≤ e H j + AB + 1 = h e H F i , (50)whi h ontradi ts Statement 1 of Lemma B.8. (cid:4) Lemma B.12. If at the outset of any round x , we have that | H xi − H xj | ≤ C/n for any pair of nodes N xi and N xj , then ne essarily the nodes are in the same family at the start of round x .Proof. Suppose for the sake of ontradi tion that there exists some round x and some pair of nodes N xi and N xj for whi h | H xi − H xj | ≤ C/n , but these nodes are in di(cid:27)erent families. Sin e families onsist of adja ent indi es (Lemma B.6) and nodes are indexed a ording to their heights at thestart of the round, we may assume without loss of generality that i and j are adja ent (i.e. that j = i + 1 ). By de(cid:28)nition of indexing, we must have H i ≤ H i +1 , whi h ombined with the hypothesisof the lemma implies that H i +1 − C/n ≤ H i . But then e H i ≥ e H i +1 by Lemma B.7, and then N xi and N xi +1 in di(cid:27)erent families ontradi ts Lemma B.11. (cid:4) E ( N a , N b ) during whi h there is a pa kettransfer in P from N a to N b .Proof. This is immediate from the rules regarding onstru ting families, sin e the values of { e H i } (39) an only hange if there is a pa ket transfer in P , and thus the analysis in Rule F1 (40) will not hange if there has been no pa ket transfer in P . (cid:4) Lemma B.14. Suppose that in some round x = E ( N a , N b ) , the Slide proto ol transfers a pa ket from N a to N b . Let F α := { N e , . . . , N a , . . . , N f } denote N a 's family at the start of round x ( e ≤ a ≤ f ),and F β := { N c , . . . , N b , . . . , N d } denote N b 's family25 at the start of x ( c ≤ b ≤ d ). The followingdes ribes all possible hanges to the way families are organized between the start of round x and thenext round:Case 1: e H a and e H b do not hange. Then the families at the start of round x + 1 are iden-ti al the arrangement of families at the start of x .Case 2: e H a does not hange, and e H b in reases by one. Then:(a) Families F δ to the left of F β (i.e. δ < β ) do not hange(b) For any node N m with b ≤ m ≤ d , N m will be in the same family as N b at the start ofround x + 1 ( ) For any node N m with d < m , letting F xµ denote N m 's family at the start of round x , oneof the following happens:i. F xµ does not hangeii. Every node in F xµ is in the same family as N b at the start of x + 1 Case 3: e H a de reases by one, and e H b does not hange. Then:(a) Families F δ to the right of F α (i.e. δ > α ) do not hange(b) For any node N m with e ≤ m ≤ a , N m will be in the same family as N a at the start ofround x + 1 ( ) For any node N m with m < e , letting F xµ denote N m 's family at the start of round x , oneof the following happens:i. F xµ does not hangeii. Every node in F xµ is in the same family as N a at the start of x + 1 Case 4: e H a de reases by one, and e H b in reases by one. Then:(a) Families F δ to the right of F α (i.e. δ > α ) and to the left of F β (i.e. δ < β ) do not hange(b) For any node N m with e ≤ m ≤ a , N m will be in the same family as N a at the start ofround x + 1 ( ) For any node N m with b ≤ m ≤ d , N m will be in the same family as N b at the start ofround x + 1 β ≤ α , as if both N a and N b are internal nodes, then Rule 3 of the Slide proto ol (togetherwith the de(cid:28)nition of how nodes are indexed) guarantees that b < a , and then β ≤ α by Lemma B.6. If N a is theSender and/or N b is the Re eiver, then β ≤ α omes from our hoi e to denote the Sender's family by F n and theRe eiver's family by F (see Rule F3 regarding the formation of families).27d) For any node N m with d < m < e , letting F xµ denote N m 's family at the start of round x ,one of the following happens:i. F xµ does not hangeii. Every node in F xµ is in the same family as N a at the start of x + 1 iii. Every node in F xµ is in the same family as N b at the start of x + 1 iv. Every node in F xµ is in the same family as N a AND N b at the start of x + 1 Proof. That the four ases stated in the lemma over all possibilities is immediate from the de(cid:28)nitionof e(cid:27)e tive height e H (see De(cid:28)nition (39)). Case 1 follows immediately from the rules F1-F2 forforming families (see De(cid:28)nition B.1) sin e the e(cid:27)e tive heights have not hanged. We go throughea h of the other ases, and prove ea h Statement.Suppose that we are in Case 2, so that e H a does not hange, and e H b in reases by one. For δ < β , onsider a family F δ := { N i , . . . , N j } , and for the sake of ontradi tion, suppose that F δ hanges insome way from the start of round x to the start of round x + 1 . Without loss of generality, we willsuppose that δ < β is the minimal index for whi h F δ hanges.Case A: F δ Splits. In other words, N i and N j are not in the same family at the start of round x + 1 . Let F x +1 ι := { N i , . . . , N k } denote N i 's new family at the start of x + 1 , where k < j byassumption.26 Noti e that for all i ≤ m ≤ j , the e(cid:27)e tive height e H m will not hange betweenthe start of x and x + 1 (sin e j < b < a ). Therefore: P jl = k +1 e H l j − k ≤ P kl = i e H l k − i + 1 = h e H F x +1 ι i < P jl = k +1 e H l j − k , (51)where the (cid:28)rst inequality is Statement 1 of Lemma B.8 and the last inequality is Statement 2of Lemma B.8. Clearly (51) is impossible, yielding the desired ontradi tion.Case B: F δ Grows. In other words, at the start of round x + 1 there is some family F x +1 ι := { N i , . . . , N k } for k > j . If k < b , then for all i ≤ m ≤ k , the e(cid:27)e tive height e H m will not hange between the start of x and x + 1 , so: P jl = i e H l j − i + 1 < P kl = j +1 e H l k − j ≤ P jl = i e H l j − i + 1 , (52)where the (cid:28)rst inequality is Statement 2 of Lemma B.8 and the se ond inequality is Statement1 of Lemma B.8. Clearly (52) is impossible, yielding the desired ontradi tion. On the otherhand, if k ≥ b , then for all i ≤ m ≤ k and m = b , the e(cid:27)e tive height e H m will not hangebetween the start of x and x + 1 , but the e(cid:27)e tive height e H b in reases by one from the start of x and x + 1 . Therefore (using supers ripts only when ne essary to spe ify the round): P jl = i e H l j − i + 1 < P kl = j +1 e H xl k − j < P kl = j +1 e H x +1 l k − j ≤ P jl = i e H l j − i + 1 , (53)where the (cid:28)rst inequality is Statement 2 of Lemma B.8 and the last inequality is Statement 1of Lemma B.8. Clearly (53) is impossible, yielding the desired ontradi tion.This proves Statement (a) of Case 2. For Statement (b), (cid:28)x index m ∈ [ b, d ] (Statement (b) istrivially true for m = b , so assume b < m ≤ d ). For the sake of ontradi tion, suppose that N m isnot in the same family as N b at the start of x + 1 . Let F x +1 β := { N i , . . . , N b , . . . , N j } denote N b 'snew family at the start of x + 1 , so by assumption j < m ≤ d , and also c ≤ i by Statement (a) of26Ne essarily N i is the smallest-indexed node in F ι by our hoi e of minimality for δ .28ase 2. Noti e that e H xb + 1 = e H x +1 b , but that for all other i ≤ l ≤ m , e H l does not hange from thestart of x and x + 1 . If i = c (using supers ripts only when ne essary to spe ify the round): P dl = j +1 e H l d − j ≤ P jl = c e H xl j − c + 1 < P jl = c e H x +1 l j − c + 1 < P dl = j +1 e H l d − j , (54)where the (cid:28)rst inequality is Statement 1 of Lemma B.8 and the last inequality is Statement 2 ofLemma B.8. Clearly (54) is impossible, yielding the desired ontradi tion. If on the other hand c < i , then (using supers ripts only when ne essary to spe ify the round): P dl = j +1 e H l d − j ≤ P jl = c e H xl j − c + 1 = h e H F xβ i≤ P i − l = c e H xl i − c< P i − l = c e H x +1 l i − c< h e H F x +1 ι i = P jl = i e H x +1 l j − i + 1 < P dl = j +1 e H l d − j , (55)where the (cid:28)rst and se ond inequalities are both Statement 1 of Lemma B.8, the fourth inequalityis Lemma B.10, and the last inequality is Statement 2 of Lemma B.8. Clearly (55) is impossible,yielding the desired ontradi tion.This proves Statement (b) of Case 2. It remains to prove Statement ( ). Fix some m > d , andlet F wµ = { N w , . . . , N m , . . . , N y } denote N m 's family at the start of x . We prove Statement ( ) viathe following two sub laims:Sub laim 1. F µ does not Split. In other words, N w and N y will be in the same family at thestart of round x + 1 .Proof. Suppose not. Let F x +1 ω = { N i , . . . , N w , . . . , N j } denote N w 's family at the start ofround x + 1 , so c ≤ i ≤ w ≤ j < y (where the (cid:28)rst inequality is due to Statement (a)). Noti ethat for every i ≤ l ≤ y , the only possible e(cid:27)e tive height e H l that an possibly hange in round x is for l = b , in whi h ase e H xb + 1 = e H x +1 b . If i = w , then (using supers ripts only whenne essary to spe ify the round): P jl = w e H l j − w + 1 < P yl = j +1 e H l y − j ≤ P jl = w e H l j − w + 1 , (56)where the (cid:28)rst inequality is Statement 2 of Lemma B.8 and the se ond is Statement 1 of LemmaB.8. Clearly, (56) is impossible, yielding the desired ontradi tion. If on the other hand i < w ,then (using supers ripts only when ne essary to spe ify the round): P jl = w e H l j − w + 1 ≤ P w − l = i e H x +1 l + P jl = w e H xl j − i + 1 ≤ P yl = j +1 e H l y − j ≤ P jl = w e H l j − w + 1 , (57)where the se ond inequality is Statement 2 of Lemma B.8, the third is Statement 1 of LemmaB.8, and the (cid:28)rst omes from: P jl = w e H l j − w + 1 ≤ P w − l = i e H x +1 l w − i ⇒ P jl = w e H l j − w + 1 ≤ P w − l = i e H x +1 l + P jl = w e H xl j − i + 1 , (58)29here the (cid:28)rst inequality is Statement 1 of Lemma B.8. Clearly, (57) is impossible, yieldingthe desired ontradi tion.Sub laim 2. If F µ gets larger, then ne essarily N b will be in the same family as N w and N y at the start of round x + 1 .Proof. Suppose not. Let F x +1 ω = { N i , . . . , N w , . . . , N j } denote N w 's family at the start ofround x + 1 , so b < i ≤ w ≤ y ≤ j . Noti e that for every i ≤ l ≤ y , sin e b < i , the e(cid:27)e tiveheight e H l does not hange. If i = w , then sin e we are assuming F µ grows, we have j > y ,and: P yl = w e H l y − w + 1 < P jl = y +1 e H l j − y ≤ P yl = w e H l y − w + 1 , (59)where the (cid:28)rst inequality is Statement 2 of Lemma B.8 and the se ond is Statement 1 of LemmaB.8. Clearly, (59) is impossible, yielding the desired ontradi tion. If on the other hand i < w and j > y , then: P w − l = i e H l w − i < P yl = w e H l y − w + 1 < P jl = y +1 e H l j − y , (60)where the (cid:28)rst inequality is from Lemma B.10, and the se ond is from Statement 1 of LemmaB.8. But then (60) implies: P w − l = i e H l + P yl = w e H xl y − i + 1 < P jl = y +1 e H l j − y , (61)whi h ontradi ts Statement 1 of Lemma B.8. Finally, if i < w and j = y , then: P w − l = i e H l w − i < P yl = w e H l y − w + 1 , (62)whi h ontradi ts Statement 1 of Lemma B.8.Cases 3 and 4 follow analogous arguments. (cid:4) B.5 Statement and Proof of Fa t that Slide has Competitive Ratio /n Lemma B.15. Suppose at the start of round x , there exists nodes { N xi , N xi +1 , . . . , N xj } su h that H xi = · · · = H xj . Then under any permutation of the indi es σ : { i, i + 1 , . . . , j } → { i, i + 1 , . . . , j } ,we have that: j X k = i e H xk = j X k = i max(0 , H xk − ( k − C/n ) = j X k = i max(0 , H xσ ( k ) − ( k − C/n ) (63)In parti ular, the value for P jk = i e H xk will not hange if we re-index the nodes { N i , . . . , N j } in anyarbitrary manner.Proof. This is immediate from the hypothesis that H xi = H xi +1 = · · · = H xj . (cid:4) Lemma B.16. Suppose that in some round x , N a transfers a pa ket to N b in the Slide proto ol. Let F β denote N b 's family and F α denote N a 's family. Then either there is exa tly one node N b ′ ∈ F β su h that ϕ b ′ in reases by one, or ϕ N does not hange for every N ∈ F β . Similarly, either there isexa tly one node N a ′ ∈ F α su h that ϕ a ′ de reases by one, or ϕ N does not hange for every N ∈ F α .No other node N ∈ G will have ϕ N hange as a result of this pa ket transfer.30roof. If N b 's e(cid:27)e tual height e H b does not in rease as a result of the pa ket transfer (e.g . the `0' inthe maximum statement of (39) dominates), then F β 's umulative e(cid:27)e tual height does not hange,and as a result, the potential ϕ of all nodes in F β remains un hanged. If on the other hand B 'se(cid:27)e tual height does in rease, then this will raise the umulative e(cid:27)e tual height e H F β by one, andthis will be absorbed by some node in F − . A similar argument works with respe t to N a in F α . Thelast statement of the lemma follows from Lemma B.4. (cid:4) We are now ready to prove the main lemma that will allow us to argue that the Slide proto olhas ompetitive ratio /n . To (cid:28)x notation, for any internal node N , let H P ′ N denote the number ofpa kets p ′ ∈ Z P ′ that N is urrently storing. Re all the de(cid:28)nition of Φ (see De(cid:28)nition B.5); we willdistribute the overall potential Φ between all the families, and show that with the rules regarding hanges in Φ , the potential of a family is always positive. Namely:Lemma B.17. For every round x and for all families F that are present at the start of x : Φ ≥ X F max X N ∈F − C − H P ′ N , X N ∈F + H P ′ N ≥ (64)Proof. We prove this based on indu tion on the round x . The lemma is learly true at the outset ofthe proto ol, when Φ = Φ F = 0 , and all nodes are in the same family, sin e all nodes have heightzero. Suppose that at the start of round x = E ( N a , N b ) , (64) is satis(cid:28)ed. We show that no matterwhat happens in round x , (64) will remain satis(cid:28)ed at the start of round x + 1 .Case 1: Neither P nor P ′ transfer a pa ket. In this ase, families will not hange (Lemma B.13), andno pa kets in Z P ′ move, so there will be no hanges to either side of (64).Case 2: P ′ transfers a pa ket during x , but P does not. If the pa ket p ′ transferred by P ′ is in Z P ′ ,then neither side of (64) will hange. So suppose p ′ ∈ Z P ′ . Note that in Case 1, N a and N b are inthe same family, all it F (Sin e Slide does not transfer a pa ket, we have | H a − H b | < C/n , and seeLemma B.12). • If N a and N b are in F + , then ϕ a = ϕ b , so ϕ p ′ does not hange. In parti ular, neither side of(64) hanges in this ase. The same is true if N a and N b are both in F − • If N a ∈ F + and N b ∈ F − , then the hange on the left-hand side of (64) is -1 (sin e ∆ ϕ p ′ = − ),whi h mat hes the hange on the right-hand side of (64) (sin e H P ′ b in reases by one, and H P ′ a de reases by one). If instead N a ∈ F − and N b ∈ F + , then similar reasoning shows that the hange of both sides of (64) is +1 .Case 3: P transfers a pa ket from N a to N b in round x . Noti e that this ase is not on erned withwhether or not P ′ also transfers a pa ket, as su h a pa ket would ne essarily be in Z P ′ (by de(cid:28)nition),and hen e this pa ket movement in P ′ will not a(cid:27)e t either side of (64). Also, without loss ofgenerality N a is the sending node and N b is the re eiving node. By Lemma B.14, there are 4 aseswe must onsider:Case 3A: e H b and e H a do not hange. Then by Lemma B.14, there will be no re-stru turing of familiesbetween rounds x and x + 1 . Consequently, if F β denotes N b 's family and F α denotes N a 's family(possible α = β ), then for all other families, (64) will remain valid. Also, ϕ N does not hange forany N ∈ F β (similarly for N ∈ F α ) sin e e H b and e H a do not hange. Therefore, the right-hand sideof (64) also will not hange for F β and F α , and the only hange in the left-hand side omes fromthe in rease of C to Φ (see Rule 2 of De(cid:28)nition B.5), whi h an be divided arbitrarily among thefamilies {F } , and this will only help (64).Case 3B: e H b in reases by one, but e H a does not hange. Let F β = { N c , . . . , N b , . . . , N d } for some c ≤ ≤ d . By Lemma B.14, there exist integers r, s ≥ and indi es { k , . . . , k r } and { l , . . . , l s } su hthat c ≤ k < · · · < k r ≤ b ≤ d < l < · · · < l s and:Families at the start of x Families at the start of x F β = { N c , . . . , N b , . . . , N d } b F β = { N c , . . . , N k − }F β +1 = { N d +1 , . . . , N l − } b F β +1 = { N k , . . . , N k − }F β +2 = { N l , . . . , N l − } b F β +2 = { N k , . . . , N k − } ... ... F β + s = { N l s − , . . . , N l s − } b F β + r − = { N k r − , . . . , N k r − } b F β + r = { N k r , . . . , N l s − } and no other families hange.By Lemma B.16, there is only one node N ∈ F − β for whi h ϕ N in reases by one as a resultof the pa ket transfer. Although F β will hange in the manner des ribed by the table above, byLemma B.4, the number of nodes N ∈ G with ϕ N = ⌊h e H F β i⌋ (respe tively ϕ N = ⌊h e H F β i⌋ ) will not hange (aside from the single node N ′ for whi h ϕ N ′ in reases by one, as guaranteed by LemmaB.16), although the spe i(cid:28) nodes in F + and F − may vary. A simple omputation ensures that theright-hand side of (64) hanges in the exa t same way as the left-hand side of (64) whenever any twonodes in F swap pla es (in F + and F − ). Therefore, we may assume without loss of generality thatthere is exa tly one node N ′ ∈ F − β for whi h ϕ N ′ in reases by one as a result of the pa ket transfer,and for all other nodes N ∈ G , ϕ N does not hange between the start of x and x + 1 .For ea h ≤ i ≤ r and ≤ j ≤ s , de(cid:28)ne the following quantities:Families at the start of x Families at the start of xX i = P N ∈ b F − β + i ( C − H P ′ N ) X j = P N ∈F − β + j ( C − H P ′ N ) Y i = P N ∈ b F + β + i H P ′ N Y j = P N ∈F + β + j H P ′ N A i = | b F + β + i | A i = |F + β + i | B i = | b F − β + i | B i = |F − β + i | (65)Also de(cid:28)ne F ∗ = b F β + r ∪ F β , and: µ = X N ∈ b F −∗ ( C − H P ′ N ) ν = X N ∈F + ∗ H P ′ N α = | b F + ∗ | and β = |F −∗ | (66)By the indu tion hypothesis, we have that at the start of round x : s X j =0 Φ F β + j ≥ s X j =0 (cid:18) A j X j + B j Y j A j + B j (cid:19) (67)In addition to the above potential, we also have that Φ in reases by C as a result of the pa kettransfer in Slide. Meanwhile, the goal is to show that at the start of round x + 1 : r X i =0 Φ b F β + i ≥ r X i =0 (cid:18) A i X i + B i Y i A i + B i (cid:19) (68)Putting all these fa ts together, we want to show that: C + s X j =0 (cid:18) A j X j + B j Y j A j + B j (cid:19) ≥ r X i =0 (cid:18) A i X i + B i Y i A i + B i (cid:19) (69)32e demonstrate in the remainder of the proof how to show (69) is satis(cid:28)ed.First look at the term i = r for the right-hand side of (69): A r X r + B r Y r A r + B r = ( α + 1 + P sj =1 A j )( µ + P sj =1 X j − ( C − H P ′ N ′ )) A r + B r + ( β − P sj =1 B j )( ν + H P ′ N ′ + P sj =1 Y j ) A r + B r = α + 1 α + β ( µ − ( C − H P ′ N ′ )) + s X j =1 X j A j A j + B j + β − α + β ( ν + H P ′ N ′ ) + s X j =1 Y j B j A j + B j + ( Y − X ) (cid:18) α P B j − β P A j ( α + β )( A r + B r ) (cid:19) + · · · + ( Y s − X s ) (cid:18) A s ( β + P B j ) − B s ( α + P A j )( A s + B s )( A r + B r ) (cid:19) < C + α + 1 α + β ( µ − ( C − H P ′ N ′ )) + s X j =1 X j A j A j + B j + β − α + β ( ν + H P ′ N ′ ) + s X j =1 Y j B j A j + B j We have used above that (by Lemmas B.8 and Corollary B.9): αα + β < A A + B < · · · < A s A s + B s < α + A + · · · + A s α + β + P sj =1 ( A j + B j ) (70)Meanwhile, we look at the left-hand side of (69) for the j = 0 term: A X + B Y A + B = ( α + P r − i =0 A i )( µ + P r − i =0 X i A + B + ( β + P r − i =0 B i )( ν + P r − i =0 Y i ) A + B ≥ µ (cid:18) αα + β (cid:19) + ν (cid:18) βα + β (cid:19) − µ + P r − i =0 X i A + B + sum r − i =0 A i X i + B i Y i A i + B i , (71)where we have used for the inequality above: A A + B < A A + B < A A + B < · · · < A r − A r − + B r − < α + P r − i =0 A i α + β + P r − i =0 ( A i + B i ) , (72)with the inequalities following from Lemma B.8 and Corollary B.9. Putting this all together, wehave that: C + s X j =0 (cid:18) A j X j + B j Y j A j + B j (cid:19) ≥ r X i =0 (cid:18) A i X i + B i Y i A i + B i (cid:19) whi h is (69).The other ases are proven similarly. (cid:4) We state as an immediate onsequen e the lemma we needed in the dis ussion of Se tion 4:Lemma B.18. At all times: | Z P ′ | ≤ nY P ≤ n | Z P | + 2 n C (73)33 Competitive Analysis of the Slide + Proto olC.1 Des ription of Slide + Re all that we model an asyn hronous network via a s heduling adversary that maintains a bu(cid:27)erof requests of the form ( u, v, p ) , whi h is a request from node u to send pa ket p to node v . Thes heduling adversary pro eeds in a sequen e of honored edges ( alled rounds), whereby we will meanthe following when we talk about an edge E ( u, v ) being honored by the adversary:Step 1. From its bu(cid:27)er of requests, the adversary sele ts one request of form E ( u, v, p ) anddelivers p to v , and also sele ts one request of form E ( v, u, p ′ ) and delivers p ′ to u . If there areno requests ( u, v, p ) (resp. ( v, u, p ′ ) ), then the adversary sets p (resp. p ′ ) to ⊥ .Step 2. Node u (resp. v ) sends new requests to the adversary of form ( u, v, p ) (resp. ( v, u, p ′ ) ).Note that the two above-mentioned a tions take pla e sequentially, so that the requests queued tothe adversary in Step 2 an depend on the pa kets re eived in Step 1, but requests formulated duringStep 2 of some round E ( u, v ) will not be delivered until edge E ( u, v ) is honored again (at the earliest).Sin e nodes in the network only send/re eive pa kets when they are at one end of an edge urrentlybeing honored, nodes will not do anything ex ept when they are a part of an honored edge. Thus,in des ribing Slide + , we need only des ribe what an internal node u will do when it is part of anhonored edge E ( u, v ) . Re all that C denotes the size of ea h node's memory27, and for simpli itywe will assume that C/n ∈ N , and also for Slide + , we will require C ≥ n .Slide + Proto ol Des ription.During honored edge E ( u, v ) , let ( v, u, ( p ′ , h )) denote the message that u re eives from v in Step1 of the round (via the s heduling adversary). Also, u has re orded the request ( u, v, ( p, H )) thatit made during Step 2 of the previous round in whi h E ( u, v ) was honored; note that v will bere eiving this message during Step 1 of the urrent round.1. If u is the Sender, then:(a) If h < C , then u deletes pa ket p from his input stream { p , p , . . . } (and ignores there eived pa ket p ′ ), and then pro eeds to Step ( ).(b) If h ≥ C , then u keeps p i (and ignores the re eived pa ket p j ), and pro eeds to Step ( ).( ) The Sender (cid:28)nds the next pa ket p i ∈ { p , p , . . . } that has not been deleted and is not urrently an outstanding request already sent to the adversary, and sends the request ( u, v, ( p i , C + Cn − to the adversary. Also, u will update the fa t that the urrentmessage request sent to v is ( u, v, ( p i , C + Cn − .2. If u is the Re eiver, then u sends the request ( u, v, ( ⊥ , − Cn )) to the adversary. Meanwhile, if p ′ = ⊥ , then u stores/outputs p ′ as a pa ket su essfully re eived.3. If u is any internal node, then:(a) If H ≥ h + ( C/n − n ) , then u will ignore p ′ , delete p and the (cid:16)ghost pa ket asso iated to p (cid:17) (see Step 3d below), and slide down any pa kets/ghost pa kets to (cid:28)ll any gaps reated.Also, u will update his height H = H − , and pro eed to Step 3d below.(b) If H ≤ h − ( C/n − n ) , then u will keep p , and also store p ′ in the sta k lo ation that u had been storing the (cid:16)ghost pa ket(cid:17) for p (see Step 3d below), deleting the ghost pa ketin the pro ess. Also, u will update his height H = H + 1 , and pro eed to Step 3d below.( ) If | H − h | < C/n − n , then u will ignore pa ket p ′ and keep p , but delete the (cid:16)ghostpa ket(cid:17) asso iated to p , and then pro eed to Step 3d.27For simpli ity, we assume that all nodes have the same memory bound, although our argument an be readilyextended to handle the more general ase. 34d) Node u will sear h its sta k for the highest pa ket p ′′ (not in luding ghost pa kets) thatit has not already ommitted in an outstanding request to the adversary. It then sendsthe request ( u, v, ( p ′′ , H )) to the adversary. Additionally, u will reate a (cid:16)ghost pa ketasso iated to the pa ket/request p ′′ (cid:17) that it has just sent the adversary. This (cid:16)ghostpa ket(cid:17) will assume the (cid:28)rst un-(cid:28)lled spot in u 's memory sta k. Finally, u will updatethe fa t that the urrent message request sent to v is ( u, v, ( p ′′ , H )) .In the following se tion, we will prove that the above routing rules are ompatible with memoryrequirements (e.g . that Steps 3b and 3d do not require a node to store more than C (ghost) pa kets),as well as prove that Slide + enjoys ompetitive ratio /n .C.2 Analysis of Slide + Before providing the full details of the proof that Slide + enjoys ompetitive ratio /n , we willprovide a brief high-level des ription of how the proof works. First, noti e that the main te hni al hallenge in moving from the semi-asyn hronous model of Se tion 4 to the fully asyn hronous modelis that nodes an no longer make routing de isions based on urrent information. Indeed, the urrentstate of a node may hange drasti ally from the time it makes a request in Step 2 of some round E ( u, v ) and the time the request is (cid:28)nally sent by the adversary in Step 1 of the next round inwhi h E ( u, v ) is honored. Sin e the Slide proto ol uses the urrent height of a node to make routingde isions, the fa t that the height of a node may hange substantially between the time a pa ketrequest is made and the time the re eiving node re eives the pa ket is an issue that must be resolved.The above des ribed proto ol handles this issue by allotting (cid:16)ghost pa kets(cid:17) in Step 3d (thiswill ensure there is always room to store a pa ket sent from an honest neighbor), as well as havingnodes make routing de isions based on old height onsiderations. In parti ular, Steps 1-3 abovedi tate what u should do based on the height that u and v had during the last time E ( u, v ) washonored. Therefore, although this information may have be ome outdated sin e the last time u and v ommuni ated with ea h other, at least the de isions will be made onsistently, both in the sensethat the heights being ompared are syn hronized (i.e . they are from the same time as ea h other,although possible now out-dated), and in the sense that the nodes will know what the other will do(assuming both are honest) in terms of whether or not it will keep the pa ket just sent/re eived.This last fa t is ru ial to prevent pa ket deletion and dupli ation from o urring (at least amongsthonest nodes).The proof will follow the main stru ture of the proof provided for the semi-asyn hronous Slideproto ol, with one additional ategory to a ount for pa ket transferring de isions that were basedon signi(cid:28) antly outdated height information.Theorem C.1. The Slide + proto ol a hieves ompetitive ratio /n in any distributed, asyn hronous,bounded memory network with dynami topology (and no minimal onne tivity assumptions). Morespe i(cid:28) ally, for any adversary/o(cid:27)-line proto ol pair ( A , P ′ ) , if P denotes the Slide + proto ol, C denotes the apa ity (memory bound) of ea h node, and Z P x (resp. Z P ′ x ) denotes the number ofpa kets re eived by proto ol P (resp. P ′ ) as of round x , then for all rounds x : Z P ′ x ≤ nZ P + 8 n C (74)Proof. Fix any adversary/o(cid:27)-line proto ol pair ( A , P ′ ) , and let P denote the Slide + proto ol and Z P x and Z P ′ x as in the statement of the theorem. Motivated by the proof in the semi-asyn hronous setting,we imagine a virtual world in whi h the two proto ols are run simultaneously in the same network.We split Z P ′ x into the following three subsets (we will hen eforth suppress the index referen ing theround x ):1. Z P ′ onsists of pa kets p ′ ∈ Z P ′ for whi h there exists at least one round E ( u, v ) su h thatboth p ′ was transferred by P ′ and some pa ket p was transferred by P .2828Note that we make no ondition that the two pa kets traveled in the same dire tion.35. Z P ′ onsists of pa kets p ′ ∈ Z P ′ that were never transferred alongside a pa ket in P as in 1above, and su h that every time p ′ was transferred between two nodes u and v during a round E ( u, v ) , the heights H and h that were used by u and v in determining whether to store/deletethe pa kets delivered by the adversary during Step 1 of E ( u, v ) (see proto ol des ription above)were ea h within n of the urrent heights of u and v .3. Z P ′ = Z P ′ \ ( Z P ′ ∪ Z P ′ ) .Clearly, | Z P ′ | = | Z P ′ | + | Z P ′ | + | Z P ′ | , and hen e the Theorem follows from Lemmas C.3, C.4, andC.5 below. (cid:4) We will need the following trivial observation, whi h follows immediately from the des ription ofthe Slide + proto ol in Se tion C.1.Observation 3. At all times, an internal node u has at most n ghost pa kets and at most n out-standing requests (one for ea h of its edges v ).Proof. Rules 1( ) and 3(d) only allow a node to submit a single request for ea h round the node ispart of an honored edge, and this request is then delivered by the adversary in Step 1 of the nextround in whi h the edge is honored. Also, Rules 3(a- ) guarantee that the ghost pa ket orrespondingto the urrent honored edge will be deleted before another one is reated in Rule 3(d). (cid:4) In order to bound | Z P ′ | , we will need to bound the number of times any pa ket p an be transferredby the Slide + proto ol. In the asyn hronous Slide proto ol of Se tion 4, we showed that any pa ket p ould be transferred at most n times, as during every pa ket transfer in Slide, the pa ket must dropin height by at least C/n − . At (cid:28)rst glan e, it might seem that we annot make the same argumentin the fully asyn hronous setting sin e the Slide + proto ol is making routing de isions based on(potentially) outdated height information. However, the introdu tion of (cid:16)ghost pa kets(cid:17) will allowus to retain this quality. Indeed, the purpose of utilizing ghost pa kets is to anti ipate future pa kettransfers and reserve spots in a node's memory sta k at the appropriate height, allowing us to arguethat even if nodes nodes are using out-dated height information, pa kets will still (cid:16)(cid:29)ow downhill(cid:17)from Sender to Re eiver. This is aptured in the following lemma.Lemma C.2. Let Y P x denote the the set of pa kets inserted by P as of round x . Also let T P x denotethe set of pa ket transfers that have o urred in P as of round x . Then any pa ket in the Slide + proto ol is transferred at most n times.29 In parti ular, | T P x | ≤ n | Y P x | ≤ n ( | Z P x | + nC ) .Proof. We show that anytime a pa ket is transferred in the Slide + proto ol, the pa ket's height inthe new bu(cid:27)er is ne essarily at least C/n − n lower than its height in the old bu(cid:27)er. Sin e pa ketsonly move within bu(cid:27)ers when they are re eived or sent (or when they slide down as in 3(a)), andsin e30 n ( C/n − n ) > C , the lemma will follow. Fix a pa ket p , and onsider a round x = E ( u, v ) in whi h p is transferred from u to v . In parti ular, it must have been that the previous round x ′ < x in whi h E ( u, v ) was honored, u sent some request of form ( u, v, ( p, H )) to the adversary inStep 2. Noti e that when u sele ted p to form a part of its request as in 3(d), sin e u had height H and u has at most n − pa kets already ommitted as an outstanding request (Observation 3), p must have height at least H − n in u 's bu(cid:27)er. Meanwhile, let ( v, u, ( p ′ , h )) denote the requestthat v sent to the adversary in Step 2 of round x ′ . Noti e that in 3(d), v reserved a position in itsbu(cid:27)er (the (cid:16)ghost pa ket(cid:17)), into whi h p will be inserted when it is re eived in round x . Sin e theghost pa ket is assigned the topmost uno upied (by pa ket or ghost pa ket) position in v 's bu(cid:27)er,we have that p will have height no bigger than h + n . Therefore, p will drop in height by at least ( H − n ) − ( h + n ) = H − h − n when it is transferred from u to v . Sin e the riterion for a eptinga new pa ket (see 3(d)) demands that H − h ≥ C/n − n , we have that p will ne essarily drop inheight by at least C/n − n when it is transferred. (cid:4) + , we have demanded that C > n . 36oti e that Lemma C.2 is valid regardless of how long a request ( u, v, ( p, H )) has been queuedin the adversary's bu(cid:27)er, and also of how u and v 's sta ks may have hanged in the meantime. Weare now ready to state and prove the (cid:28)rst requisite bound:Lemma C.3. | Z P ′ | ≤ n | Z P | + 2 n C Proof. By de(cid:28)nition, | Z P ′ | ≤ | T P | , and the latter is bounded by n | Z P | + 2 n C by Lemma C.2. (cid:4) Lemma C.4. | Z P ′ | ≤ n | Z P | + 2 n C Proof. This bound follows the same reasoning as the proof of Lemma B.18. Suppose that pa ket p ′ ∈ Z P ′ is transferred by P ′ from u to v in round x . By de(cid:28)nition of Z P ′ , Slide + did not transfer apa ket, and thus (with the notation as in Rule 3(d) for Slide + ) | H − h | < C/n − n . Also by de(cid:28)nitionof Z P ′ , we have that v 's height in round x is within n of h , and u 's height in round x is within n of H . Consequently, u 's height in round x must be within C/n of v 's height. Then if we de(cid:28)ne familiesthe same way as in the proof for the semi-syn hronous Slide proto ol (see Se tion B), by LemmaB.12, u and v must be in the same family at the start of x . Indeed, all the lemmas and proofs ofSe tion B will remain valid31, and hen e Lemma B.18, whi h states that | Z P ′ | ≤ n | Z P | + 2 n C ,remains valid. (cid:4) Lemma C.5. | Z P ′ | ≤ n | Z P | + 4 n C Proof. Fix a pa ket p ′ ∈ Z P ′ . By de(cid:28)nition of Z P ′ , there exists some round x p ′ = E ( u, v ) in whi h p ′ was transferred from u to v , where either u 's height or v 's height has hanged by at least n sin ethe previous round x ′ p ′ < x in whi h E ( u, v ) was honored. Let S p ′ ⊆ T P denote n of these pa kettransfers, where ea h pa ket transfer in S p ′ orresponds to a pa ket sent (or re eived) by u (or v ),and took pla e between x ′ p ′ and x p ′ .Observation. For any pa ket transfer in Slide + , there are at most n pa kets p ′ ∈ Z P ′ forwhi h the pa ket transfer appears in S p ′ .Proof. Consider any round x ′ = E ( u, v ) in whi h a pa ket is transferred from u to v by Slide + ,and refer to this spe i(cid:28) pa ket transfer as T x ′ . Then for ea h edge of u and ea h edge of v andfor any p ′ ∈ Z P ′ , there an be at most one round x p ′ > x ′ for whi h T x ′ ∈ S p ′ . After all, on ea given edge of u or v , say for example E ( u, w ) , transfers a pa ket p ′ ∈ Z P ′ in round x p ′ > x ′ ,the heights of both u and w are updated, and there an never be another p ′′ ∈ Z P ′ and laterround x p ′′ > x p ′ su h that x p ′′ = E ( u, w ) and T x ′ ∈ S p ′′ . Therefore, T x ′ an appear in at most n sets of form S p ′ .Sin e |S p ′ | = n for ea h p ′ ∈ Z P ′ , we have that: X p ′ ∈ Z P′ |S p ′ | = n | Z P ′ | (75)Now sin e for any given pa ket transfer T x ∈ T P there an be at most n di(cid:27)erent values of p ′ ∈ Z P ′ su h that T x ∈ S p ′ , we have that: [ p ′ ∈ Z P′ S p ′ ≥ n | Z P ′ | n (76)But ∪ p ′ ∈ Z P′ S p ′ ⊆ T P , so: | T P | ≥ | ∪ p ′ ∈ Z P′ S p ′ | ≥ | Z P ′ | (77)In parti ular, | Z P ′ | ≤ | T P | ≤ nZ P + 4 n C , where the se ond inequality is Lemma C.2. (cid:4) Z P ′2