TRANSPOSE ON VERTEX SYMMETRIC DIGRAPHS

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1 TRANSPOSE ON VERTEX SYMMETRIC DIGRAPHS Vance Faber Center for Computng Scences, Insttute for Defense Analyses Abstract. In [] (and earler n [3]), we defned several global communcaton tasks (unversal exchange, unversal broadcast, unversal summaton) on drected or vertex symmetrc networks. In that paper, we manly focused on unversal broadcast. In [4], we dscussed unversal sum. Here we contnue ths work and focus on unversal exchange (transpose) n drected vertex symmetrc graphs. Introducton. In unversal exchange, processor has a vector of data a j, where the element a j s a packet needed by processor j. Thus we start wth a matrx A where each processor has a separate row and we end the task wth each processor havng a separate column, n other words, we are computng the transpose. Ths gves rse to two fundamental questons: Gven a drected vertex symmetrc graph G (or any drected graph for that matter), how many tme steps τ (G) does t take to perform a transpose? Gven two drected vertex symmetrc graphs (or general drected graphs for that matter), how do we compare them wth respect to ther ablty to perform a transpose? Graph symmetres. One mportant property of vertex symmetrc graphs s that the number of edges drected nto and out of a gven vertex s a constant d. In our model of communcaton, we assume that on every tme step all the edges n the graph can be used smultaneously, that s, each processor can smultaneously exchange a sngle packet of data wth all of ts neghbors. From here on out, we consder only drected Cayley coset graphs (equvalently, drected vertex symmetrc graphs) G = ( Γ,, H ). Brefly, a Cayley coset graph G has P = [ Γ : H ] cosets of the subgroup H n the group Γ as vertces and degree d = where s a collecton of elements n Γ and H generates Γ. (Note that we allow mult-edges here but we nsst that the graph be connected.) The edges are gven by ( gh, gδ H ) for δ. A necessary and 1

2 suffcent condton for the edges to be well-defned between cosets n ths way s that the unon of the left and rght cosets H = H. In the case where H s the dentty subgroup, G s a drected Cayley graph. Let the number of vertces of dstance k from a gven vertex be n k. A smple fundamental lower bound on τ (G) s gven by the followng lemma. Lemma 1. The tme for transpose s at least as great as the average dameter of the graph: θ (G) Proof. Each element ( ) θ ( G) = τ G knk / d. (Equaton 1) a j travels on a path (task graph) j T from to j. Snce the graph s vertex symmetrc, the total length of these paths from a sngle vertex s at least k n k and that means that all these transfers occupy at least P knk edges. If we could use all Pd edges at once on each tme step, we would acheve a tme equal to θ (G). Problems achevng the optmum value. In order to gve an actual value for the transpose, we have to produce a communcaton graph, a schedule that lsts for each tme whch edges transfer whch data. Ths was dscussed n [, Defntons. and.13]. In the case of transpose, ths s a collecton of drected paths T j n G for each ordered par of vertces wth the propertes that the tmes ncrease on each T j and no edge s labeled twce wth the same tme. (Note that ths can also be thought of as embeddng the complete drected graph nto G.) If we want to acheve θ (G), the paths T j must be nearly the shortest paths from to j. (There s a bt of leeway n the case where θ (G) has been rounded up.) One problem s that any collecton of shortest paths mght have some edges represented more often than others. Another problem s that shortest paths from one vertex mght conflct wth shortest paths from another vertex. We dscuss these problems one at a tme. Paths wthout conflcts. Suppose that G ( Γ,, H ) = s a Cayley graph ( H s the dentty subgroup.) Suppose we express each element g n G by a word w(g) n (n other words, a product of) the members of. Then w dentfes a path from the dentty e to the element g n G. Each entry n w can be thought of as a sngle drected edge n G. Let δ ( j) be the generator n entry j of ths path.

3 Theorem. Suppose we label the edges of the paths w (g) wth ncreasng tmes so that no generator appears more than once wth a gven tme. Then the collecton of paths hw (g) for all h n G s a communcaton graph for transpose on G where the labels on the edges of the hw (g) are dentcal to the labels on w. (g) Proof. We have to show that no edge has the same tme twce. Suppose the same edge appears n h 1 w( g 1 ) and h w( g ) wth the same tme. Thus ths edge can be represented n two ways: ( 1 1 j h g, h gδ ( )) and h h, h hδ ( )). So we must have h1 g = hh and h1 gδ ( j) = hhδ ( k) ( k for some group elements g and h. Ths mples δ ( j ) = δ ( k) = δ. Now the tmes on the edge h g, h g ) n the path h w ) and the edge h h, h h ) n ( 1 1 δ 1 ( g 1 ( δ the path h w( g ) are the same as the tmes on the edge ( g, gδ ) n the word w ( g 1 ) and the edge ( h, hδ ) n the word w ( g ), respectvely, by constructon. Unless g = h and w ( g1) = w( g ), ths volates our assumpton that no two edges wth the same generator n the set of paths can have the same tme. Ths theorem allows us to schedule a transpose on G by creatng a consstent set of tmes on the paths from the dentty to the other elements. Once we have chosen a set of such paths (words) W, we can calculate a lower bound on τ (G) wth respect to these paths. Defnton 1. We let the regular bound on transpose be { δ : δ ( j) W, δ ( = δ } ψ ( G) = mn max j) W δ Remark. Snce we have forced each occurrence of the generator δ to have a dfferent tme, the tme that we obtan for transpose n ths way s at least as great as ψ (G). Also. 3

4 so ( G) ψ ( G) kn k { δ : δ ( j) W, δ ( j) = δ } / d ψ ( ) / d = mn G W δ θ. What s not clear s whether there s any provable relatonshp between the regular bound and the optmal transpose. The deal stuaton would be f each generator appears roughly the same number of tmes n the optmal W but often that cannot be arranged wthout ncreasng the average sze of the words n W beyond the average dameter. There s no obvous reason why the regular bound s a lower bound on the tme for a transpose on G because there s no necessty for tmes to be consstently assocated wth generators throughout the graph so the same tme could appear twce on the same generator n the paths from a fxed vertex to the other vertces. In addton, for ψ (G) to be the maxmum tme n the labelng, every tme must be assgned to one of the occurrences of the generator that appears most often. Ths mght not be possble. Queston 1. Is ( G) ψ ( G) τ =? Spannng factorzatons. The dffculty n routng on vertex symmetrc graphs that are not Cayley graphs s the falure to be able to label edges wth specfc generators. As an example, the problem s seen n the Petersen graph [7] that s vertex symmetrc but cannot be represented as a Cayley graph. The Petersen graph s the complement of the lne graph of K 5 ; the vertces can be represented as -element subsets and { u, v} and { x, y} are connected f they are dsjont. The automorphsm group s somorphc to S 5 ; each automorphsm α s a derved permutaton of vertces of K 5. (That s, gven an automorphsm α t s a map α { u, v} = { ˆ( α u), ˆ( α v)} for a permutaton αˆ.) The stablzer subgroup B of the vertex { 1,} s thus the set of twelve derved permutatons that fx the set { 1,}. One set of generators conssts of the permutatons δ 1, δ, δ derved from ( 13)(4), ( 13)(5), ( 14)(5). Let x = (34)(15). Then 3 xb B are two dstnct vertces of G. Also ( δ 1) xδ1 = (1)(345) B so xδ 1B = δ1b. Ths means that there are two dstnct edges gong out from the xδ 1B = δ1b, one to xb and the other to B, that can be thought to have δ δ. vertex the label δ 1 dependng on whch coset representatve s used, x 1 or 1 We now dscuss a possble soluton to ths problem. Ths secton s taken from [1]. A 1-factor of a drected graph G s a subgraph wth both n-degree and out-degree equal to 1. (Some authors have called ths a -factor. Our defnton seems more consstent wth the notaton n undrected graphs. For example, f the edges are all b-drectonal and the factor s a unon of -cycles, then ths 1 4

5 would be an ordnary 1-factor n an undrected graph.) It s known (Petersen 1891, see [7]) that every regular drected graph wth n-degree and out-degree d has a 1-factorng wth d 1-factors. For completeness, we gve the proof here. Fact 1. Every drected graph G where the n-degree and out-degree of every vertex s d has a edge dsjont decomposton nto d 1-factors. Proof. Form an auxlary graph B wth two new vertces u and u for each vertex u. The edges of B are the pars ( u, v ) where ( u, v) s a drected edge n G. The undrected graph B s bpartte and regular wth degree d so t can be decomposed nto d 1-factors. Each of these 1-factors corresponds to a drected 1-factor n G. In order to create global routng schemes on G, we consder regular graphs wth factorzatons wth addtonal propertes. Defnton. Let F 1, F, K, F d be the factors n a 1-factorng of G. We call a fnte strng of symbols from the set { F } a word. If v s a vertex and ω s a word, then v ω denotes the drected path (and ts endpont) n G startng at v and proceedng along the unque edge correspondng to each consecutve factor represented n the word ω. If G s a graph wth n vertces, we say that a 1- factorng and a set of n words W ω, ω, ω, K, ω }, ω =, s a = { 0 1 n 1 spannng factorzaton of G (wth word lst W ) f for every vertex v, the vertces vω are dstnct. Schedules. A schedule assocated wth a factorzaton s an assgnment of a tme (a label) to each occurrence of each factor n the words of W such that no tme s assgned more than once to a partcular factor and tmes assgned to the factors n a sngle word are ncreasng. The tme of a schedule s the largest tme assgned to any of the factors. If T s the total tme, the schedule can be thought of as a d T array where each row corresponds to a factor and an entry n that row ndcates whch occurrence of that factor has been assgned the correspondng tme. An entry n a row n the array can be empty ndcatng no occurrence of that factor has been assgned the gven tme. The power of a spannng factorzaton les n the fact that a schedule can be used to descrbe an algorthm for conflct free global exchange of nformaton between the vertces of the graph. Theorem 3. Suppose we have a schedule for a factorzaton of the graph G. Then the collecton of drected label-ncreasng paths vω for all v and nonempty ω have the property that no edge n the graph s assgned the same tme 0 5

6 twce. A schedule for a spannng factorzaton yelds a tme labeled drected path between every two vertces so that no edge s labeled wth the same tme twce. Proof. Each edge n the graph s assgned to a sngle one factor. Assume there s an edge n the one factor F that has been assgned the same tme twce. Snce every occurrence of F n the words n W has been assgned a unque tme, ths can only mean that there are two dfferent vertces u and v and an ntal subword ω of a word n W such that the edges ( uω, uω F) and ( vω, vω F) are the same edge. Then u ω and v ω must be the same vertex. Let us assume that ths s the shortest ω for whch ths happens. The word ω cannot be empty snce u and v are dfferent. But then the last factor n ω must also be the same edge, a contradcton. If we start wth a spannng factorzaton, then all the non-empty paths from v are unque, there are n 1 of them and none of them can return to v so they must reach to every other vertex n the graph. There are some addtonal propertes that a spannng factorzaton mght have. Defnton. We say a spannng factorzaton s balanced f each factor appears nearly as often n the schedule as any other. We say the factorzaton s short f the average number of tmes a factor appears s the same as the theoretcal lower bound θ based on the average dstance between any two vertces and the number of edges. We say the factorzaton s optmal f t s short and balanced. A schedule Σ s mnmum for a spannng factorzaton, f t has tme τ (Σ) equal to the theoretcal mnmum tme for the factorzaton based on maxmum number of tmes a factor appears. In mathematcal terms, we can wrte () balanced f max{ F : F W} = d = 1 { F : F W} / d ; () short f here s the dameter; d = 1 W} / d = kn D k { F : F, k= 1 nd N k s the number of tmes the dstance between two vertces s k and D () optmal f 6

7 D knk max { F : F W} = ; k= 1 nd (v) mnmum f Note that these parameters are ordered D k= 1 kn nd k d = 1 ( Σ) = max{ F : F W} τ. { F : F W} / d max{ F : F W} τ ( Σ) Creaton of schedules for spannng factorzatons are dscussed n [6], where the followng s proven. Fact. Every dameter two spannng factorzaton has a mnmum schedule unless the max belongs to a factor F whch s not n a word of length one and s entrely absent n words of length two n one poston, ether frst or second. In that case, the shortest tme for any schedule s one more than the theoretcal mnmum. Theorem 4. Every Cayley graph has a short factorzaton. Proof. Ths s a sketch. Take a tree T 1 of shortest paths from the dentty of the group. The factors consst of all the edges labeled wth a specfc generator. The words are just the paths n T 1, so the factorzaton s short. Queston. Does every vertex symmetrc graph have a factorzaton or even a short factorzaton? Graphs of dameter wth a spannng factorzaton. There s one type of graph for whch we can always construct an optmal transpose. Defnton. If G s a drected graph and v s a vertex n G, then for any postve nteger r, S r (v) s the set of all vertces u n G for whch there s a drected path from v to u of length less than or equal to r. Defnton 3. We call L S ( e) S 1( e) r = the r layer. r r. 7

8 Theorem 5. Let be a graph of dameter wth a spannng factorzaton and the elements of the layer can be represented by words x y n the generators. Let M (δ ) be the number of tmes that the generator δ appears as an x and N(δ ) be the number of tmes that the generator δ appears as a y. If M ( ) + N( ) ( G) 1 ( G) = ( G) ( G. δ δ θ for all δ, then τ θ = ψ ) Proof. To prove ths theorem, we need to show that we can fnd a schedule wth θ (G) columns. Frst note that f n s the number of vertces n the layer, then δ ( M ( ) + N( δ )) = n δ. Snce θ G) = ( d + n ) / d = n / d 1 ( + δ, then ( M ( δ ) + N( δ ))/ d = θ ( G) 1 and the hypothess says that the total number of tmes any generator appears s as close to the average as possble. In order to construct the schedule, we appeal to Corollary 4 n [6]. Here a generator s called a machne and a word s called a job and each generator appearng as an x s called a type 1 step and each generator appearng as a y s called a type step. Corollary 4 (from [6]). Gven s 1 jobs wth one step and s jobs wth two steps, T = s s )/ d ) f and only f there exsts an optmal schedule (n tme ( 1 + each machne s assgned at most T steps and no machne s assgned exclusvely steps of one type, ether type 1 or type from jobs wth steps. In our case, we have s = d 1 and the machne (generator) δ s assgned M ( δ ) + N( δ ) + 1 θ ( G ) = T steps whch by hypothess s at most. Snce every machne s assgned one job wth a sngle step (the members of the 1 layer), the exclusvty condton s satsfed so the hypotheses hold and the theorem guarantees the exstence of the requred schedule. In addton, a ccorollary of Theorem 1 n [6] yelds a more general statement. 8

9 Corollary 6. Let be a graph of dameter wth a spannng factorzaton and the elements of the layer can be represented by words x y n the generators. Let M (δ ) be the number of tmes that the generator δ appears as an x and y. Then N(δ ) be the number of tmes that the generator δ appears as a τ ( G ) 1 + max( M ( δ ) + N( δ )). δ Comparng transpose on two networks wth dfferent parameters. For ths comparson, we need to make some model assumptons. We nvestgated ths prevously n [5] wth dfferent model assumptons. Here we create a smple model problem utlzng a transpose. The dea s that we want to compare dfferent network archtectures wth the same overall cost and use the runnng tme of a model algorthm as a way to evaluate the archtecture. Frst, we examne the costs. We make the smplfyng assumpton that memory costs are the same between archtectures and only depend on the problem sze. We consder only two costs, one for the processors and one for the network. We let the cost of a processor be C. For the network, we assume that cost does p not depend upon how many wres are connected to a processng node but only to the overall number of wres n the network. We let the cost of one wre be C w. Then n ths model, the cost s C = PC + PdC = P( C + dcw ). C C p We can wrte ths as γ = = P + d. Cw Cw p w To descrbe the model algorthm, we start wth an N N matrx A wth column vectors v 1, v, L, vn. Our algorthm also uses a gven complcated functon f N N mappng R to R. Let F be the functon that maps the matrx A to the matrx F (A) wth columns f v ), f ( v ), L, f ( v ). Our algorthm s then the ( 1 N teratve algorthm wth M teratons gven by T j+ 1 A j ). A = F( We assume that the CPU tme for ths algorthm has the form NMα (N) for some functon α that represents the runnng tme of f. We make a smple p 9

10 assumpton that the CPU tme scales wth the number of processors so that the CPU tme wth P processors s then NMα( N) T p P =. Gven the P processors and a network G, we dstrbute the columns of A wth N / P columns n each processor. To form the transpose, each processor must send N / P rows from these columns to every other processor. Ths means the network must send ( / P) N peces of nformaton to every other processor. The total communcaton tme for the algorthm s then gven by T c N = M τ ( G). P We can optmstcally estmate τ by usng deal numbers. In the dscusson, we can assume that D s the average dameter of the network, that s, the sum. If the network has as many processors as possble for a certan kn k / P degree and dameter, t wll have about the transpose s able to use all the avalable band wdth on every transpose step, then each processor has to use D wres for each of the P 1 messages t has to transmt so the total band wdth requrement s approxmately N D M Pd =. T c If we assume that algorthm s T D P = d processors. If we assume that DP. Ths yelds aτ that s DP / d and a total communcaton tme of D = D+ d >> C p / C, then d D = Pd ~ γ T T NMα ( N) λ 1 P + c = + DN where λ = 1 / γ < 1. w λ + 1 so the total tme for the The term α (N) s proportonal to the number of operatons n the functon f. If f were a matrx multply, ths could be anywhere from N to N. Let us examne a regme where the two terms are of comparable order n N. We let ( N) N. Then we have to mnmze α = β 10

11 1 +1 β γ D + D. Clearly then, n ths regme we want D to be as large as possble because that allows more processors. However, f D gets too large, then d wll become small and the assumpton that d >> C p / C wll become nvald. In case the growth rate of α (N) s larger than N, then the P strongly and agan, we wll want D as large as possble. w T term domnates even more We can use these results to compare two dfferent networks wth a constraned cost. Frst, we see that constranng cost means constranng the number of wres, that s, C Pd < γ =. Assumng both networks satsfy ths constrant, the C w one wth the largest number of processors s the better one. Summary. We have dscussed transpose on vertex symmetrc networks. We have provded a method to compare the effcency of transpose schemes on two dfferent networks wth a cost functon based on the number processors and wres needed to complete a gven algorthm n a gven tme. REFERENCES 1. Randall Dougherty and Vance Faber, Network routng on regular drected graphs from spannng factorzatons, preprnt.. Vance Faber, "Global communcaton algorthms for Cayley graphs", Arxv.org: Vance Faber, "Global communcaton algorthms for hypercubes and other Cayley coset graphs," Techncal Report, LA-UR , Los Alamos Natonal Laboratory (1987). 4. Vance Faber, "Global sum on symmetrc networks," Arxv.org: Vance Faber, Latency and dameter n sparsely populated processor nterconnecton networks: a tme and space analyss, Techncal Report, LA-UR , Los Alamos Natonal Laboratory (1987). 6. Vance Faber, Amanda Streb and Noah Streb, Ideal unt-tme job shop schedulng, preprnt. 11

12 7. Wkpeda, en.wkpeda.org/wk/petersen_graph. 1