Butterfly Network Analysis and The Beneˇ s Network


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1 6.895 Theoy of Paallel Systems Lectue 17 Buttefly Netwok Analysis and The Beneˇ s Netwok Lectue: Chales Leiseson Lectue Summay 1. Netwok with N Nodes This section poves pat of the lowe bound on expected outing time fo an abitay N node netwok. 2. The Beneš Netwok This section motivates, intoduces, and analyzes the Beneˇ s connection netwok. 3. Routing on Buttefly Netwoks This section establishes that most outing poblems take only O(lg N ) time on buttefly netwoks. 1 Netwok with N Nodes In this section, we pove a lemma that completes the poof of the Netwok with N Nodes theoem given in lectue 16. Theoem 1 Fo an abitay netwok with N nodes, the expected outing time to simultaneously send a single message fom each pocesso to a andom pocesso is whee BW is the minimum bisection width. Poof E[Routing Time] =Ω(N/BW + diamete) Lemma 2 The expected outing time fo the N andom messages poblem laid out in theoem 1 is Ω(N/BW ). Poof Refe to the notes fo lectue 16 fo poof. Lemma 3 Each node v in a N node netwok chooses a destination dest(v) unifomly at andom. Then, E[Routing Time] =Ω(d), whee d is the diamete. Poof Find node x such that P{δ(x; dest(x)) d/2} 1/2 whee δ(x, y) is the shotest distance fom node x to node y. Then, E[time to oute x to its destination dest(x)] = kp {δ(x; dest(x)) = k} k kp {δ(x; dest(x)) = k} k d/2 d/2 P{δ(x; dest(x)) = k} k d/2 = d/2p{δ(x, dest(x)) d/2} = d/2 =Ω(d) 171
2 Figue 1: Example of a Beneˇ s Netwok. Nodes aligned vetically ae in the same level and nodes aligned hoizontally ae in the same ow. To find such an x, let δ(x, x )= d, and define S = { v : δ(x, v) < d/2} and S = { v : δ(x,v) < d/2}. We know S S = since fo all z S S, δ(x, x ) δ(x, z)+ δ(z, x ) < d/2+ d/2 < d gives a contadiction, implying no such z can exist. Theefoe, at least one of S and S has N/2 nodes. Without loss of geneality, assume that S N/2. Thus, dest(x) S with pobability 1/2, i.e., P{ δ(x, dest(x)) d/2} 1/2. Fom lemma 2 and lemma 3, we know that E[Routing Time] = Ω(N/BW ) and E[Routing Time] = Ω(diamete). Theefoe, we can conclude that E[Routing Time] = Ω(N/BW + diamete). 2 The Beneš Netwok 2.1 Wost Case Poblems fo the Buttefly Netwok We saw last lectue that the wost case congestion fo the buttefly netwok is n fo an n input netwok. This was motivated by consideing the poblem whee inputs x 1 x 2 x 3 x outputs 0000x 1 x 2 x 3 x 4. Thee ae many such bad inputs that cause n congestion fo the buttefly netwok. Fo example, conside the poblem of computing the matix tanspose whee each node epesents an element of a matix and we wish to compute (i, j) (j, i). Matix tanspose and othe opeations that elicit the wost case congestion in a buttefly netwok ae common in pogams witten fo supecomputes. It is theefoe a eal poblem to build supecomputes with buttefly netwoks because typical applications suffe seious pefomance penalties. 3 The Beneš Netwok This section intoduces the Beneš netwok, a netwok that allows povably congestionfee communication fom inputs to a pemutation of the outputs. A Beneš netwok is constucted by ovelapping the lowode 172
3 Figue 2: An example Beneˇ s Netwok with the middle d 2 levels eplaced with two halfsized Beneˇ s netwoks. cycles of two buttefly netwoks. An example is shown in figue 1. Theoem 4 Any npemutation can be outed (offline) on an ninput Beneš netwok with nodedisjoint paths. Poof The poof uses the fact that Beneˇ s netwoks ae decomposable into smalle Beneˇ s netwoks. Let n = 2 d. The poof is based on induction on d. Fo d = 1, the netwok is tivial and the theoem is obvious on examination. As the inductive step, assume that the theoem holds fo a n = 2 d input Beneš netwok. Obseve that the middle d 2 levels of the Beneˇ s netwok fom two Beneˇ s netwoks, one on top and one on bottom, each with n/2 =2 d 1 inputs, as shown in figue 2. By assumption, any input to one of the subbeneš netwoks can be outed to any output of the same subnetwok without congestion. Theefoe, all we need to do is detemine whethe each input is to be outed though the top o the bottom subnetwok and show that the subnetwok outputs can be coectly outed to the final outputs without congestion. The only constaint we must obseve to avoid congestion is that connected pais at level 1 and at level d do not ovelap. That is, each buttefly at the fist and last level must eithe oute packets staight o cossed: butteflies cannot send both input packets to a single output node. Theefoe, by obsevation, the two inputs compising a buttefly at the fist level must be outed into diffeent subnetwoks and the two inputs compising a buttefly at the last level must come fom diffeent subnetwoks. It is this final constaint that is impotant. Define a buttefly pai as { i + n/2 (i n/2), bfly(i) = i n/2 (i>n/2). The buttefly pai is the othe node (eithe input o output) that defines a buttefly in the netwok, bfly (i) = j and bfly (j) = i. Define π(i) as the pemutation that maps inputs to outputs: π(i) = j denotes that input i should be outed to output j and π 1 (j) = i denotes that output j should be outed fom input i. We now show how to oute inputs into the uppe and lowe subnetwoks in a way that satisfies ou constaints. We stat by outing the fist input though the uppe subnetwok and connecting it to the coect output, π(1). Next, we fulfill the constaint at the last level by outing the fist path s output 173
4 buttefly pai, bfly (π(1)), though the lowe netwok and back to its coect input, π 1 (bfly(π(1))). We satisfy the new constaint at the input by outing the appopiate input, π 1 (bfly(π(1))), though the uppe subnetwok and to its coect output. We continue going back and foth though the uppe subnetwok when connecting an input to an output and though the lowe subnetwok when connecting an output to an input until we have defined a completed loop. Since the fist path went though the uppe subnetwok and the final etun path went though the lowe subnetwok, the input constaint on the oiginal path is satisfied. If the fist cycle of paths doesn t include all nodes, we pick an unouted node and epeat this pocess until all inputs ae outed. Following this algoithm, half of the inputs ae outed though the top subnetwok and half of the inputs ae outed though the bottom subnetwok without congestion. We have satisfied the inductive hypothesis by showing that given a n/2 =2 d 1 input congestion fee Beneˇ s netwok, we can constuct a n =2 d input congestion fee Beneˇ s netwok. By induction, we have poved the theoem. As a side note, since we always have a choice of how to oute the fist input, we can emove a switch fom each inductive level of the buttefly netwok without affecting its pefomance chaacteistics. Consequently, fo a n = 2 d input netwok, we can emove d switches fom the netwok without loss of geneality. Coollay 5 An ninput Beneš netwok can simulate any nnode, degeed netwok in O(d lg n) time. Poof Let G be an nnode netwok with maximum degee d. Let H be an ninput Beneš netwok. We pove the coollay by showing how to simulate G on H. Identify the ith node of G with the ith input of H to simulate node computation. The ticky pat is showing how to simulate communication on edges in G using H s netwok. We do this by identifying d + 1 subsets of the edges in G such that fo each subset, we can oute packets fom input i to output j fo each edge (i, j) in the subset of G with at most O(lg n) congestion. Each subset is outed in a sepaate ound. d + 1 subsets with O(lg n) congestion each gives us a total of O(d lg n) time to simulate G. We now constuct d + 1 disjoint and spanning subsets of edges in G. Constuct a bipatite gaph Γ G =(U, V, E) whee U = {u 1,u 2,...u n }, V = {v 1,v 2,...v n },and E = {(u i,v j ) (i, j) is an edge of G}. Communication fom i to j in G is epesented by edge (u i,v j )inγ G. Since the maximum degee in G is d, the maximum degee in Γ G is also d, making Γ G a degula bipatite gaph. Theefoe, we can constuct an edgecoloing of Γ G using d + 1 colos (see Leighton s Intoduction to Paallel Algoithms and Achitectues fo poof). Each set of edges with a given colo foms one of the d + 1 subsets. Let S be the kth of the d + 1 subsets. Fo each edge (i, j) in S, we simulate communication on (i, j) by sending a packet fom input i to output j on H duing ound k. Since the edges incident to i and the edges incident to j ae all coloed diffeently, we know that no two packets oiginate fom o ae deliveed to the same node duing the same ound. Theefoe, S can be outed with ou O(lg n) bound. One ound fo each of the d + 1 subsets with O(lg n) outing time each completes the poof. 4 Routing on Buttefly Netwoks Even though the buttefly netwok has a outing time of Ω( n) on cetain pemutations (including many inteesting pemutations), most outing poblems take only O(lg n) time. The following theoem coves abitay Npacket outing poblems on buttefly netwoks, not just those that oute inputs to outputs. Theoem 6 Conside the N N Npacket outing poblems on an Nnode (ninput, whee n = Θ(N/ lg N)) buttefly netwok. At least N N (1 1/N Ω(1) ) of these poblems can be outed in O(lg N) time. 174
5 Figue 3: Depiction of a buttefly netwok. hoizontally ae in the same ow. Nodes aligned vetically ae in the same level and nodes aligned Poof We will see the poof of a weake esult than is needed to establish the theoem. The poof establishes a congestion bound that leads to an O(lg 2 n) time esult. WLOG, we oute packets in thee phases (figue 3 shows the ows and levels in a buttefly netwok): 1. Route packets staight acoss ows in the netwok to the coesponding outputs (the ightmost node in each ow). 2. Use geedy input to output outing, to get each packet to the coect ow in the netwok. 3. Route packets staight acoss to the final destination. We analyze the congestion at the output nodes at each ow of the netwok in Phase 1. All packets fom a given ow end up at the output node of that ow at the end of the phase. Each ow contains lg n nodes, and lg n = O(lg N ), so the congestion in Phase 1 is O(lg n). Fo Phase 2, we conside a node x at the kth level of the netwok (whee the inputs ae the 0th level and the outputs ae the (lg n)th level). By symmety, the x is equivalent to any othe node at the kth level. The numbe of packets that can each x duing Phase 2 is 2 k lg n since thee is a path to each node at the kth level of a buttefly netwok fom 2 k input nodes and each input node has at most lg n packets at the beginning of Phase 2. Thus, wost case congestion at a node x at the kth level = 2 k lg n. The pobability that a given packet passes though node x is at most 2 k. This esult follows fom the binay tee popety of buttefly netwoks and the fact that the destination of each packet is chosen independently and unifomly at andom fom all the nodes in the netwok. (See figue 4 fo a depiction of the binaay tee popety). Thus, P [a given packet passes though node x] 2 k. Conside any set of specific packets. Since the destinations of the packets ae independent of one anothe, the pobability that all the packets pass though x is given by P [all packets pass though x] (2 k ) =2 k. 175
6 Figue 4: Binay tee popety of a buttefly netwok. Each choice on a path fom an input node to an output node is the head of a binay tee. The pobability that at least packets pass though x is bounded by the numbe of ways to choose packets fom the total numbe of packets multiplied by the pobability that all packets pass though x: 2 k lg n P [ packets pass though x] 2 k. + Note: This agument ovecounts. If + packets pass though x, this event is counted times within ) the ( 2 k lg n ways. a ea Using the fomula b b,wehave If we choose = 2e lg N, then we obtain b ( e2 k lg n ) P [ packets pass though x] 2 k ( e lg n ) =. P [ packets pass though x] ( 1 ) 2e lg N 2 N 2e 1. N 5.4 By Boole s inequality, the pobability that any node has moe than =2e lg N packets is N times the pobability that a given node has moe than packets: 1 P [ packets pass though some node] N N 5.4 = N
7 Theefoe, N N (1 1/N 4.4 ) outing poblems see 2e lg N congestion. Since thee ae O(lg N ) levels and each node at a level has at most O(lg N ) congestion, the total congestion fo Phase 2 is O(lg 2 N ). At the end of Phase 2, thee ae O(lg N ) packets at each output node with high pobability. Theefoe, at each node in Phase 3 thee is O(lg N ) congestion with high pobability. Thus, the oveall time bound is O(lg 2 N ) foatleast N N (1 1/N Ω(1) ) outing poblems. Coollay 7 E[outing time] = O(lg N )(1 1/N 4.4 )+ O(N )(1/N 4.4 ) = O(lg N ). 7
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