Ecient Parallel Algorithms for Computing All Pair. University of Kentucky. Duke University

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1 Ecient Parallel Algorithm for Computing All Pair Shortet Path in Directed Graph 1 Yijie Han 2 * Victor Y. Pan 3 ** John H. Reif*** *Department of Computer Science Univerity of Kentucky Lexington, KY **Department of Mathematic and Computer Science Lehman College, CUNY Bronx, NY ***Department of Computer Science Duke Univerity Durham, NC Abtract We preent parallel algorithm for computing all pair hortet path in directed graph. Our algorithm ha time complexity O(f(n)=p + I(n) log n) on the PRAM uing p proceor, where I(n) i log n on the EREW PRAM, log log n on the CRCW PRAM, f(n) i o(n 3 ). On the randomized CRCW PRAM we are able to achieve time complexity O(n 3 =p + log n) uing p proceor. Keyword: Analyi of algorithm, deign of algorithm, parallel algorithm, graph algorithm, hortet path. 1 Introduction A number of known algorithm compute the all pair hortet path in graph and digraph with n vertice by uing O(n 3 ) operation [Di][Fl][J]. All thee algorithm, however, ue at leae n 1 recurive tep in the wort cae and thu require at leat the order of n time in their parallel implementation, even if the number of available proceor i not bounded. O(n) time and n 2 proceor bound can indeed be achieved, for intance, in the traightforward parallelization of the algorithm of [Fl]. (Here and hereafter we aume the cutomary PRAM model of parallel computing [KR].) 1 Preliminary verion of thi paper wa preented on the 4th Annual ACM Sympoium on Parallel Algorithm and Architecture, June, Current addre: Electronic Data Sytem, Inc., Ecore Rd., Romulu, MI Supported by NSF Grant CCR and PSC CUNY Award # and #

2 NC algorithm are alo available for thi problem. However, they either need (n 3 log n) operation or only work for the more narrow cla of the input graph and/or digraph (which have the edge weight bounded, ay, by a contant or have a family of mall eparator available) [AGM][DNS][DS][GM][L][PK][PR1][PR2][S]. The recent algorithm of [PP1][PP2] ue O(log 2:5 n) parallel time and O(n 3 ) operation in the cae of a general graph with n vertice. In thi paper we improve the latter time bound to the new record value of O(I(n) log n), till uing O(n 3 ) operation. Here I(n) i the time complexity of computing the minimum of n element uing n proceor. I(n) i O(log n) under the EREW (Excluive Read Excluive Write) PRAM model, O(log log n) under the CRCW (Concurrent Read Concurrent Write) PRAM model [V], and a contant under the randomized CRCW PRAM model [R]. Moreover, the total number of operation involved in our O(I(n) log n) time algorithm can be decreaed to o(n 3 ) on the EREW PRAM and CRCW PRAM, by applying the reult in [Fr][T]. To build our algorithm, we incorporated the well known technique of [Fl] and of the reduction of the hortet path computation to matrix multiplication over the emiring, but added ome new nontrivial technique of tudying path in graph and digraph. We preent our algorithm in three tage to illutrate our intuition behind the algorithm. We rt give a imple parallel algorithm with time complexity O(n 3 =p+i(n) log 1:5 n) uing p proceor. Thi algorithm take O(n 3 ) operation when we ue no more than O(n 3 =(I(n) log 1:5 n)) proceor. We then how how to peed up thi algorithm to achieve time O(I(n) log n) uing O(n 3 =(I(n) log 2=3 n)) proceor. By uing more ophiticated idea we how the time complexity O(n 3 =p + I(n) log n) for the all pair hortet path problem. Straightforward application of the reult in [Fr][T] yield time complexity O(f(n)=p + I(n) log n), where f(n) = o(n 3 ). 2 Computing All Pair Shortet Path We ue number 0; 1; ::: ; n 1 to repreent input vertice, and a matrix A uch that it entry a ij repreent the weight of the arc from i to j. We will ue the emiring (A; min; +), o that A n 1 repreent the hortet ditance between all pair of vertice of the input graph. An arc i an ordered pair of vertice. A path i a nite equence of vertice. We may aume that there i an arc between every pair of vertice, ome of them may have weight 1. Our algorithm ue matrix multiplication to compute hortet path. The computation can therefore be viewed a contracting each hortet path to a ingle arc. For example, for an input matrix A the operation A := AAA contract the length of every hortet path by at leat one half. Thu, if our algorithm contract every path of length l < n to a ingle arc, then it compute all pair hortet path correctly. For a path p, we ue (p) to denote the cot (the number of tep or the number of iteration of a loop) for contracting p to a ingle arc. We alo ue the following pecial denition. Denition 1: For a given integer k, 1. [i] denote a vertex u uch that ik u (i + 1)k 1. [i]; 0 i < n=k, form a partition of vertice. 2

3 2. [i; j] denote a vertex u uch that ik u (j + 1)k 1, [i; j] i empty if j < i. 3. [i; j][g; h] denote an arc (u; v) uch that ik u (j + 1)k 1 and gk v (h + 1)k [i; j] denote an empty path or a path [i; j][i; j] [i; j] of length < n. 2.1 A Simple Parallel Algorithm The input matrix A i divided into ubmatrice. Each ubmatrix i a k k matrix. For convenience, aume that k divide n, and imilarly we aume that the value of all logarithm, power and ratio below are integer where thi i needed. There are a total of n 2 =k 2 ubmatrice A ij ; 0 i; j < n=k. Submatrix A ij contain element in row ik to (i+1)k 1 and column jk to (j + 1)k 1 of A (ee Fig. 1). Our next algorithm combine the technique of Floyd' algorithm and of path computation by mean of matrix multiplication. Algorithm APSP1 for t := 0 to n=k 1 do begin /*Compute the tranitive cloure of A tt uing matrix multiplication. */ A tt := A tt; for all i; j; 0 i; j < n=k; do in parallel A ij := minfa ij ; A it A tt A tj g; end Let (p) denote the number of iteration of the loop indexed by t in algorithm APSP1 that i needed to contract p to a ingle arc, where p i a path of length le than n. Fixing k in the denition 1, we have: Lemma 1: ([0; n=k 1][0; t] [0; n=k 1]) t + 1, 0 t < n=k. Proof: By induction. Before the iteration t < 0, [0; n=k 1][0; t] [0; n=k 1] = [0; n=k 1][0; n=k 1], which i a ingle arc. Now auming that the lemma i true for t, we how that it i true for t + 1. We conider three cae for a path p = [0; n=k 1][0; t + 1] [0; n=k 1]. (1) The interior vertice of p do not contain a vertex in [t + 1]. That i, p i of the form [0; n=k 1][0; t] [0; n=k 1]. In thi cae (p) t + 1, by the induction hypothei. (2) The interior vertice of p contain one vertex in [t + 1]. That i, p i of the form [0; n=k 1][0; t] [t + 1][0; t] [0; n=k 1]. In thi cae p = p 1 p 2, where p 1 = [0; n=k 1][0; t] [t + 1], p 2 = [t + 1][0; t] [0; n=k 1]. By the induction hypothei, (p 1 ) t + 1, (p 2 ) t + 1. Therefore, immediately after the t-th iteration (note that t tart at 0), p 1 ha been contracted to [0; n=k 1][t + 1], and p 2 ha been contracted to [t + 1][0; n=k 1]. Thu p = p 1 p 2 ha been contracted to [0; n=k 1][t + 1][0; n=k 1]. In the (t + 1)-t iteration, intruction A ij := minfa ij ; A i;t+1 A t+1;t+1 A t+1;j g, 0 i; j < n=k, contract the path to a ingle arc [0; n=k 1][0; n=k 1]. Therefore, (p) t + 2. (3) The interior vertice of p contain more than one vertex in [t+1]. That i, p i of the form [0; n=k 1][0; t] [t + 1][0; t + 1] [t + 1][0; t] [0; n=k 1]. In thi cae p = p 1 p 2 p 3, where p 1 = [0; n=k 1][0; t] [t + 1], p 2 = [t + 1][0; t + 1] [t + 1], and p 3 = [t + 1][0; t] [0; n=k 1]. p 2 can 3

4 further be decompoed into block, where each block i of the form [t + 1][0; t] [t + 1]. That i, each block tart at a vertex in [t+1], end with a vertex in [t+1], but goe through only vertice in [0; t]. By the induction hypothei, (p 1 ) t + 1, (p 3 ) t + 1. Let p 0 be any block of p 2. We alo have that (p 0 ) t + 1, by the induction hypothei. Therefore, immediately after the t-th iteration, p 1 ha been contracted to [0; n=k 1][t+1], p 3 ha been contracted to [t+1][0; n=k 1], and each block of p 2 ha been contracted to [t+1][t+1]. Thu p 2 ha been contracted to [t+1]. In the (t+1)-t iteration, intruction A t+1;t+1 := A t+1;t+1 contract p 2 to [t+1][t+1]. Therefore, p ha been contracted to [0; n=k 1][t+1][t+1][0; n=k 1]. Finally, in the (t+1)-t iteration, intruction A ij := minfa ij ; A i;t+1 A t+1;t+1 A t+1;j g, 0 i; j < n=k, contracte p to a ingle arc [0; n=k 1][0; n=k 1]. Therefore, (p) t Theorem 2: Algorithm APSP1 correctly compute all pair hortet path. Proof: Setting t = n=k 1 in Lemma 1 prove thi theorem. 2 Theorem 3: The time complexity of algorithm APSP1 i O((n 3 +nk 2 log k)=p+(ni(n) log n)=k). Proof: Each iteration of the loop indexed by t execute a tranitive cloure operation for A tt, and n 2 =k 2 matrix multiplication and minimization A ij := minfa ij ; A it A tt A tj g. In each iteration the tranitive cloure require O(k 3 log k) operation and n 2 =k 2 matrix multiplication and minimization take O(n 2 =k 2 k 3 ) = O(n 2 k) operation. Therefore, the whole algorithm take O(n 3 + nk 2 log k) operation. Each iteration can be done in time O(I(n) log n) if enough proceor are available. Thu the theorem i proved. 2 Setting k to n= log 0:5 n, we obtain a parallel algorithm with time complexity O(n 3 =p + I(n) log 1:5 n). 2.2 Speeding Up the Algorithm So far we were unable to decreae the time complexity below O(I(n) log 1:5 n) with O(n 3 ) operation becaue we do the tranitive cloure of A tt equentially for t = 0; 1; ::: ; n=k 1. The loop indexed by t in algorithm APSP1 repreent the erialim of the algorithm. We now peed up our algorithm by adopting a new deign. In the following algorithm we let k = n= log 1=3 n and x = log 2=3 n. Algorithm APSP2 for t := 1 to 5 log n do begin for all, 0 < n=k; do in parallel A := A A A ; if t mod x = 0 then A := AAA; end The intuition behind the deign of APSP2 i a follow. We multiply all the diagonal matrice A, 0 < n=k, imultaneouly in each iteration. By doing o we hope to eliminate the erialim in APSP1. We alo replace the intruction for all i; j; 0 i; j < n=k; do in parallel A ij := minfa ij ; A it A tt A tj g; 4

5 in APSP1 with intruction if t mod x = 0 then A := AAA; APSP2 ha 5 log n iteration. Each iteration execute a contant number of matrix multiplication. Therefore, if a ucient number of proceor are available APSP2 can be executed in O(I(n) log n) time. Alo note that the number of operation executed by APSP2 cannot be bounded by O(n 3 ) becaue intruction A := AAA i executed more than a contant number of time. Let (p) be the number of iteration of the loop indexed by t in algorithm APSP2 that i needed to contract p to a ingle arc. Fixing k in denition 1, we have: Lemma 4: If p = [i] for any 0 i < n=k, then (p) log jpj, where jpj i the length of p. Proof: Each iteration reduce the length of uch a path by at leat one half until the path ha been contracted to a ingle arc. 2 In the following we will prove inequalitie of the form (p 1 ) (p 2 ) + n, where p 1 and p 2 are path and n i a number. In proving uch an inequality we alway aume that jp 2 j jp 1 j. In fact, in mot ituation p 2 i a ubpath of p 1. Lemma 5: ([0; n=k 1][0; i] [0; n=k 1]) ([0; i] ) + x, 1 i n=k. Proof: After [0; i] ha been contracted to a ingle arc, the intruction A := AAA will be executed within the next x iteration to contract the path to a ingle arc. 2 Lemma 6: If ([i][0; i 1] [i]) z(l) + log l, where l i the length of the path (i.e. there i a term log l in the expreion upbounding ([i][0; i 1] [i])) and z(l) i a nondecreaing function of l, then ([i][0; i] [i]) z(l) + log l + log log l + 3, 1 i < n=k. Proof: Let p = [i][0; i] [i]. Let jpj = L. We will partition p into block which tart from a vertex in [i], end with a vertex in [i], but only go through vertice in [0; i 1]. A block mut go through at leat one vertex in [0; i 1]. Thu a block can be denoted by [i][0; i 1][0; i 1] [i]. p ha no more than L=2 t block of length 2 t. We bound N, the number of arc of the form [i][i] in p. We note that when a block i contracted to a ingle arc [i][i] (we ay thi block i removed or eliminated), it contribute 1 to N. Example: Let p = [i][0][2][i][i][i 1][3][4][i][i][i][2][i][3][4][i]. p ha 4 block. N = 3. Aume that p i contracted to p 1 = [i][0][i][i][i][i][i][2][i][i]. Then p 1 ha 2 block and N = 5. Aume now that p 1 i contracted to p 2 = [i][0][i][i][i]. Then p 2 ha 1 block and N = 2. We note that a a path p i being contracted, the length of p i decreaed, the number of block i decreaed, N may increae. There are two factor aect N. When a block i contracted to a ingle arc, the block i removed, but it contribute 1 to N. On the other hand, the contraction of p alo contract ubpath of the form [i], thu decreaing N. Let P = [i][0; i 1] [i], jp j = l L. By the aumption of the lemma we have (P ) z(l) + log l z(l) + log l. All block of length 2 1 in path p are removed after z(l) +1 iteration. There are at mot L block removed. Therefore after z(l) +1 iteration, thee block contribute at mot L to N. All block of length 2 2 in path p are removed after z(l) + 2 iteration. There are at mot L=2 block removed. Therefore after z(l) + 2 5

6 iteration, thee block contribute at mot L=2 to N. In general, all block of length 2 t+1 in path p are removed after z(l) + t + 1 iteration. There are at mot L=2 t block removed. Therefore after z(l) + t + 1 iteration, thee block contribute at mot L=2 t to N. Now we count how many arc [i][i] are removed after each iteration. Becaue intruction for all, 0 < n=k; do in parallel A := A A A ; i executed in each iteration, the length of a ubpath [i] i cut by at leat half after each iteration. Thu it eem that we can reduce N by half in each iteration imply becaue of thi intruction. Such counting i not accurate. Conider a path p = [i][2][3][i][i][4][5][i][i][2][0][i][i][3][i 1][i]. N i 3 for p. After an iteration (auming that intruction A := AAA i not executed in thi iteration), N i not changed becaue a ingle arc [i][i] cannot be removed by the iteration. However, when an arc [i][i] doe not contribute to the decreaing of N, it i becaue the [i][i] i adjacent to a block. After z(l) + t iteration, all block of length maller than 2 t are removed. Therefore, after z(l) + t iteration there are at mot L=2 t block left, and there are at mot L=2 t arc [i][i] which do not contribute to the decreaing of N. After z(l) iteration, N L. After z(l) + 1 iteration, we do N N=2 = Becaue the length of each ubpath [i] i reduced by half = +L = Becaue there are at mot L arc [i][i] not contributing to the decreaing of N: = +L = Becaue there are at mot L block of length 2 1 removed: = Therefore N 5L=2. After z(l) + 2 iteration N N=2 + L=2 + L=2. Therefore N 9L=4. In general, after z(l) + t iteration N N=2 + L=2 t 1 + L=2 t 1, we have N (4t + 1)L=2 t. After z(l) + log L + 1 iteration, all block are removed, and N (4(log L + 1) + 1)L=2L < 2 log L + 3. At thi moment p i contracted to a path of the form [i] and of length N. It take at mot log(2 log L + 3) < log log L + 2 more iteration to contract the path to a ingle arc. Thu (p) z(l) + log L + log log L Lemma 7: ([0; i] ) log l + i(3x + log log l + 3), 0 i < n=k, where l i the length of the path. Proof: By induction. For i = 0 we have ([0; 0] ) log l by Lemma 4. Now, auming that the lemma hold for i 1, we how that it hold for i. A path p = [0; i] may take the following form: (1). p contain no vertex in [i]. p i of the form [0; i 1]. In thi cae the lemma i true by the induction hypothei. (2). p contain exactly one vertex in [i]. p i of the form [0; i 1] [i][0; i 1]. Aume that [i] i neither the tarting vertex nor the ending vertex of p (we leave to the reader the cae where [i] i the tarting or the ending vertex). p = p 1 p 2 p 3, where p 1 = [0; i 1], p 2 = [0; i 1][i][0; i 1], p 3 = [0; i 1]. We aume that both p 1 and p 2 are not empty 6

7 and leave to the reader the cae when they are empty. After ([0; i 1] ) iteration p 1 i contracted to [0; i 1][0; i 1], p 3 i contracted to [0; i 1][0; i 1]. Therefore p i contracted to [0; i 1][0; i 1][i][0; i 1][0; i 1]. It take at mot 2x more iteration for the intruction A := AAA to be executed twice to contract the whole path to a ingle arc. Therefore (p) ([0; i 1] ) + 2x. Therefore, the lemma i true. (3). p contain more than one vertex in [i]. p i of the form [0; i 1] [i][0; i] [i][0; i 1]. p = p 1 p 2 p 3 p 4 p 5, where p 1 = [0; i 1], p 2 = [0; i 1][i], p 3 = [i][0; i] [i], p 4 = [i][0; i 1], p 5 = [0; i 1]. We aume that p 1 ; p 2 ; p 3 ; p 4 ; p 5 are not empty and leave to the reader the cae where ome of them are empty. After ([0; i 1] ) iteration p 1 i contracted to [0; i 1][0; i 1], p 5 i contracted to [0; i 1][0; i 1]. Now ([i][0; i 1] [i]) ([0; i 1] )+x (by Lemma 5) log l + (i 1)(3x + loglog l + 3) + x (by induction hypothei) = z(l) + log l. Therefore by Lemma 6, (p 3 ) log l + (i 1)(3x + log log l + 3) + x + log log l + 3 = T. Therefore, after T iteration p i contracted to [0; i 1][0; i 1][i][i][0; i 1][0; i 1]. It take at mot 2x more iteration for the intruction A := AAA to be executed twice to contract the whole path to a ingle arc. Therefore, (p) T +2x = log l +i(3x+log log l +3). 2 Theorem 8: Algorithm APSP2 correctly compute all pair hortet path in time O(n 3 log 1=3 n=p + I(n) log n). Proof: Setting i = n=k 1 and l = n in Lemma 7 prove that algorithm APSP2 compute all pair hortet path in log n + (n=k 1)(3x + log log n + 3) 5 log n iteration. If enough proceor are available each iteration can be executed in O(I(n)) time. Intruction A := A A A i executed O(log n) time for all n=k = log 1=3 n ubmatrice, which take O(k 3 (n=k) log n) = O(n 3 log 1=3 n) operation. Intruction A := AAA i executed O(log n=x) time, which take O(n 3 log 1=3 n) operation. 2 If we ue APSP2 for the purpoe of computing A tt in APSP1, we get Corollary: All pair hortet path can be computed in time O(n 3 =p + I(n) log 7=6 n) uing p proceor. Remark: Algorithm APSP1 and APSP2 equentially evaluate the tranitive cloure of the matrice on the diagonal. Thi property i ueful for reducing the complexity of computing the all pair hortet path in graph with a family of precomputed eparator[pr1] [PR2]. 2.3 A Fater Algorithm We divide matrix A into level (ee Fig. 2). The 0-th level i A (0) 0 = A. For each matrix B = A (j) at level j we divide it into four ubmatrice of equal ize, B i 0;0 ; B 0;1 ; B 1;0 ; B 1;1, and dene A (j+1) = B 2i 0;0 and A (j+1) = 2i+1 B 1;1. Thu there i a total of 2 j matrice A (j) at i level j, each of ize n=2 j n=2 j. We involve the matrice up to level L = (log log n)=2, and there are 2 L = log 1=2 n matrice at that level, each of ize n=2 L n=2 L. Let K be the larget number, which i both a power of 3 and le than or equal to log n. Algorithm APSP3 for t := 1 to 8 log n do begin 7

8 for all, 0 < 2 L ; do in parallel /* Do matrix multiplication for all matrice at level L.*/ := A (L) A (L) A (L) ; end A (L) if 0 i L 1 and i i the mallet j uch that t mod (K=3 j ) = 0 then /* Do matrix multiplication for all matrice at level i. */ for all, 0 < 2 i ; do in parallel A (i) := A (i) A (i) A (i) ; APSP3 execute A (L) := A (L) A (L) A (L) in every iteration. For matrice at other level, they are being multiplied a follow. For the time interval from 1 to 8 log n, we let T 0 = ft j t mod K = 0g. Intruction A (0) := A (0) A (0) A (0) i executed for all t 2 T 0. Note that becaue of the way K i dened, T 0 contain only a contant number of element. Let T 1 = ft j t mod (K=3) = 0g T 0. Intruction A (1) := A (1) A (1) A (1) i executed for all t 2 T 1. The reaon T 0 i being ubtracted i that A (1) i a ubmatrix of A (0), thu operation b=2c A (0) := A(0) b=2c b=2c A(0) b=2c A(0) \contain" operation A(1) b=2c := A (1) A (1) A (1). Thu we may view that intruction A (1) := A (1) A (1) A (1) i executed every K=3 iteration. In general. let P T i = ft j t mod (K=3 i i 1 ) = 0g j=0 T j, i < L. Intruction A (i) for all t 2 T i. We may view that intruction A (i) iteration. := A (i) A (i) A (i) := A (i) A (i) A (i) i executed i executed every K=3 i Let k = n=2 L. By denition 1, [i] denote any vertex in fin=2 L ; in=2 L + 1; :::; (i + 1)n=2 L 1g. Denote K=3 L i 1 by t i, 0 i L 1. Then intruction A (L i 1) := A (L i 1) can be viewed a being executed every t i iteration, and intruc- can be viewed a being executed every t 0 iteration. tion A (L 1) A (L i 1) A (L i 1) := A (L 1) A (L 1) A (L 1) To analyze path being contracted by APSP3, we dene function cot(x; l) a follow. cot(0; l) = log l; cot(1; l) = log l + 2t 0 + log log l + 3; If x 6= 0 and x i even, cot(x; l) = cot(x 1; l) + t i, where i i the larget integer j uch that x=2 j i odd; P i 1 If x 6= 1 and x i odd, cot(x; l) = cot(x 1; l) + t 0 + 2(log log l + 3) + j=0 t j, where i i the larget integer j uch that (x + 1)=2 j i odd. Let p be a path, let (p) be the number of iteration needed in APSP3 to contract p to a ingle arc. Let l be the maximum length of the path under conideration. We dene function COST (x; l) a follow. COST (0; l) = ([0] ); COST (1; l) = ([0; 1] ); If x 6= 0 and x i even, COST (x; l) = ([x; x + 2 i 1][0; x 1] [x; x + 2 i 1]), where i i the larget integer j uch that x=2 j i odd; If x 6= 1 and x i odd, COST (x; l) = ([x 2 i + 1; x][0; x] [x 2 i + 1; x]), where i i the larget integer j uch that (x + 1)=2 j i odd. The denition of COST i for path of length l. For example, COST (0; l) i obtained 8

9 by taking the maximum of the number of iteration needed to contract a path over the path of the form [0] and of length l. We give a imple way to gure out the formula we are evaluating. For even x, x 6= 0, we dene COST (x; l) = ([a; b][0; x 1] [a; b]), where a and b can be found out in Fig. 2. We rt nd the larget quare in which x i at the top left corner. For x = 2 thi quare i the quare containing 2, 3. For x = 4 thi quare i the quare containing 4,5,6,7. For x = 6 thi quare i the quare containing 6,7. a i the mallet number in the quare, that i, the number at the top left of the quare. b i the larget number in the quare, that i, the number at the bottom right of the quare. For even x, x 6= 0, we alo dene cot(x; l) = cot(x 1; l) + t i, where t i can be found out in Fig. 2. We rt nd the larget quare in which x i at the top left corner. Then t i i the label in the immediate left neighbor quare. Note that for all even x, x 6= 0, the quare containing t i ' do not overlap. If we um all t i for all even x, x 6= 0, we are in fact umming all t i ' except t 0 ' in Fig. 2. For odd x, x 6= 1, we dene COST (x; l) = ([a; b][0; x] [a; b]), where a and b can be found out in Fig. 2. We rt nd the larget quare in which x i at the bottom right corner. For x = 3 thi quare i the quare containing 0,1,2,3. For x = 5 thi quare i the quare containing 4,5. For x = 7 thi quare i the quare containing 0 to 7. a i the mallet number in the quare. b i the larget number in the quare. For odd x, x 6= 1, we P P alo dene cot(x; l) = cot(x 1; l) + t 0 + 2(log log l + 3) + t j, where t j can be found out in Fig. P 2. We rt nd the larget quare in which x i at the bottom right corner. The t j ' in t j are the t j ' in P the bottom \row" of the quare. P Note that for all odd x, x 6= 0, the quare containing t j ' do not overlap. If we um all t j for all odd x, x 6= 0, we are in fact umming all t i ' in Fig. 2 except the t 0 in the top left quare containing 0 and 1. Now conider the extra t 0. Conidering the fact that x 0 ' are not preent in the formula for cot(x; l), x i even and x 6= 0, if we um all t i for all x except x = 0; 1, we are in fact umming all t i in Fig. 2 twice except the t 0 in the top left quare containing 0 and 1. However, t 0 i counted twice in the formula for cot(1; l). Therefore, if we um t i for all x, we are in fact umming all t i in Fig. 2 exactly twice. We ue everal propertie of APSP3 in the analyi of path. One property we ue i ymmetry. We note that vertice in [2i] and vertice in [2i + 1] are ymmetrical with repect to APSP3. For example, path p 1 = [1][0; 1] [1] i ymmetrical to path p 2 = [0][0; 1] [0]. That i to ay, if we replace [1] in p 1 with [0] and replace [0] with [1] we obtain p 2 a the ymmetrical path of p 1. Becaue a matrix multiplication at level L 1 in APSP3 either involve both [0] and [1] or involve neither [0] nor [1], and matrix multiplication at level L involve [0] and [1]. Therefore, the way that p 1 and p 2 are contracted i the ame a far a APSP3 i ued to contract p 1 and p 2. Therefore, (p 1 ) = (p 2 ). We alo have that p 1 = [2][1] [2] i ymmetrical to p 2 = [2][0] [2], and p 3 = [4; 5][0; 7] [4; 5] i ymmetrical to p 4 = [6; 7][0; 7] [6; 7] (refer to Fig. 2). We alo ue ymmetry in a omewhat generalized ene. For example, p 5 = [6][0; 5] [7] i ymmetrical to p 6 = [6][0; 5] [6] becaue matrix multiplication for contracting p 5 and p 6 proceed in the ame pattern. Alo, p 7 = [6; 7][0; 3] [4; 5] i ymmetrical to p 8 = [6; 7][0; 3] [6; 7]. We have (p 1 ) = (p 2 ) if p 1 and p 2 are ymmetrical. 9

10 The econd property we will ue i revere. The revered path of p 1 = [1][0][2][3] i p 2 = [3][2][0][1]. That i, the revered path i obtained by revering the vertice of the original path. It i not dicult to ee that if p 1 i the revered path of p 2 then (p 1 ) = (p 2 ). The third property we will ue i proper ubet. If the et of path repreented by p 1 i contained in the et of path repreented by path p 2 then we ay that p 1 i a ubet of p 2. If the containment i proper then we ay p 1 i a proper ubet of p 2. For example, [0; 1][5; 6] [0; 1] i a proper ubet of [0; 3][5; 6] [0; 3]. If p 1 i a proper ubet of p 2 then (p 1 ) (p 2 ). We ue four lemma below to prove COST (x; l) cot(x; l). We ue an implied induction. That i, when we are proving COST (x; l) cot(x; l) we aume that COST (y; l) cot(y; l) i true for all 0 y < x. To help undertand the following proof, we ugget that the reader try everal particular value for x and refer to Fig. 2 for each cae analyzed. Lemma 9: COST (0; l) cot(0; l). Proof: Each iteration contract uch a path by at leat half. Therefore, the Lemma i true. 2 Lemma 10: COST (1; l) cot(1; l). Proof: We have the following cae for a path p = [0; 1] : (a1). p i of the form [0] or [1], then (p) log l becaue intruction A (L) i executed in each iteration.. := A (L) A (L) A (L) (a2). p i of the form [1][0] [1]. After [0] i contracted to a ingle arc, it take at mot t 0 iteration for the intruction A (L 1) := A (L 1) A (L 1) A (L 1) to be executed in order to contract the path to a ingle arc. Therefore, (p) ([0] ) + t 0 log l + t 0 (by (a1)). (a3). p i of the form [1][0; 1] [1], then (p) log l + t 0 + log log l + 3 (by (a2) and Lemma 6). (a4). p i of the form [0][0; 1] [0]. Thi cae i a ymmetrical cae of (a3). By ymmetry we have (p) log l + t 0 + log log l + 3. (a5). p i of the form [0][0; 1] [1]. Aume that p ha at leat two vertice in [0] (we leave to the reader the cae that p ha only one vertex in [0]). p can be written a p 1 p 2 p 3, where p 1 = [0][0; 1] [0], p 2 = [0][1], p 3 = [1]. By (a4), (p 1 ) log l + t 0 + log log l + 3. By (a1), (p 3 ) log l. Thu after log l + t 0 + log log l + 3 iteration, p 1 i contracted to a ingle arc [0][0], p 3 i contracted to a ingle arc [1][1] (it i poible p 3 i empty). Therefore p i contracted to [0][0][1][1]. Within t 0 iteration intruction A (L 1) := A (L 1) A (L 1) A (L 1) will be executed, which contract p to a ingle arc. Therefore (p) log l + 2t 0 + log log l + 3. (a6). p i of the form [1][0; 1] [0]. Thi cae i ymmetrical to (a5). Now a path p = [0; 1] can be in only one of the following two ituation. (1). p only contain vertice in [0], or p only contain vertice in [1]. Thi ituation i taken care of in (a1). (2). p contain vertice both in [0] and in [1]. We look at the tarting and ending vertice of p. If the tarting ending vertex pair < a; b > i < [1]; [1] >, the ituation i calculated 10

11 in (a3). If the pair i < [0]; [0] >, it i calculated in (a4). If the pair i < [0]; [1] >, it i calculated in (a5). Finally, if the pair i < [1]; [0] >, it i calculated in (a6). 2 Lemma 11: For even x, x 6= 0, COST (x; l) = ([x; x + 2 i 1][0; x 1] [x; x + 2 i 1]) cot(x; l), where i i the larget integer j uch that x=2 j i odd. Proof: For x 6= 0 and x even, and for p = [x; x + 2 i 1][0; x 1] [x; x + 2 i 1] we have the following cae. (b1). x i a power of 2. By the denition of i, x 2 i = 0. When [0; x 1] i contracted to a ingle arc, p i contracted to [x; x + 2 i 1][0; x 1][0; x 1][x; x + 2 i 1]. Vertice in [0; x 1] and [x; x + 2 i 1] are in A (L i 1) 0, therefore it take at mot t i more iteration to multiply the matrice at level L i 1, in order to contract the path to a ingle arc. Therefore COST (x; l) = ([x; x + 2 i 1][0; x 1] [x; x + 2 i 1]) ([0; x 1] ) + t i = COST (x 1; l) + t i cot(x 1; l) + t i = cot(x; l). (b2). x i not a power of 2. There are three ubcae. (b2.1). p doe not have any vertex in [x 2 i ; x 1]. Thu p i of the form [x; x + 2 i 1][0; x 2 i 1] [x; x + 2 i 1]. We have (p) = ([x; x + 2 i 1][0; x 2 i 1] [x; x + 2 i 1]) ([x 2 i ; x + 2 i 1][0; x 2 i 1] [x 2 i ; x + 2 i 1]) (by proper ubet) = COST (x 2 i ; l) cot(x 2 i ; l) cot(x; l). (b2.2). p contain one vertex in [x 2 i ; x 1]. p i of the form [x; x + 2 i 1][0; x 2 i 1] [x 2 i ; x 1][0; x 2 i 1] [x; x + 2 i 1]. p = p 1 p 2, where p 1 = [x; x + 2 i 1][0; x 2 i 1] [x 2 i ; x 1], p 2 = [x 2 i ; x 1][0; x 2 i 1] [x; x + 2 i 1]. (p 1 ) = ([x; x+2 i 1][0; x 2 i 1] [x 2 i ; x 1]) ([x 2 i ; x+2 i 1][0; x 2 i 1] [x 2 i ; x+2 i 1]) (by proper ubet) = COST (x 2 i ; l) cot(x 2 i ; l) cot(x 1; l). After (p 1 ) iteration p 1 i contracted to a ingle arc. Becaue p 2 i the revere of p 1, p 2 i alo contracted to a ingle arc after (p 1 ) iteration. Therefore, after (p 1 ) iteration p i contracted to p 3 = [x; x+2 i 1][x 2 i ; x 1][x; x+2 i 1]. Vertice in p 3 are all in a ubmatrix A (L i 1). Therefore, it take at mot t i more iteration with an execution of matrix multiplication at level L i 1 to contract p to a ingle arc. Thu, (p) cot(x 1; l) + t i = cot(x; l). (b2.3). p contain more than one vertex in [x 2 i ; x 1]. p i of the form [x; x+2 i 1][0; x 2 i 1] [x 2 i ; x 1][0; x 1] [x 2 i ; x 1][0; x 2 i 1] [x; x + 2 i 1]. p = p 1 p 2 p 3, where p 1 = [x; x + 2 i 1][0; x 2 i 1] [x 2 i ; x 1], p 2 = [x 2 i ; x 1][0; x 1] [x 2 i ; x 1], p 3 = [x 2 i ; x 1][0; x 2 i 1] [x; x + 2 i 1]. We have (p 1 ) = ([x; x + 2 i 1][0; x 2 i 1] [x 2 i ; x 1]) ([x 2 i ; x + 2 i 1][0; x 2 i 1] [x 2 i ; x + 2 i 1]) (by proper ubet) = COST (x 2 i ; l) cot(x 2 i ; l) cot(x 1; l). We alo have (p 3 ) = (p 1 ) (by revere), and (p 2 ) = COST (x 1; l) cot(x 1; l). Therefore, after cot(x 1; l) iteration p i contracted to p 4 = [x; x + 2 i 1][x 2 i ; x 1][x 2 i ; x 1][x; x + 2 i 1]. Vertice in p 4 are all in a ubmatrix A (L i 1). Therefore, it take at mot t i more iteration with an execution of matrix multiplication at level L i 1 to contract p to a ingle arc. Thu, (p) cot(x 1; l) + t i = cot(x; l). 2 Lemma 12: For odd x, x 6= 1, COST (x; l) = ([x 2 i +1; x][0; x] [x 2 i +1; x]) cot(x; l), where i i the larget integer j uch that (x + 1)=2 j i odd. Proof: Let i be the larget integer j uch that (x + 1)=2 j i odd, Let y = x 1. Let i 0 be the larget integer j uch that y=2 j i odd. We have the following cae: 11

12 (c1). ([x][0; y 1] [x]) ([y; y + 2 i0 1][0; y 1] [y; y + 2 i0 1]) = COST (x 1; l) cot(x 1; l). ([x][0; y 1] [y]) ([y; y+2 i0 1][0; y 1] [y; y+2 i0 1]) = COST (x 1; l) cot(x 1; l). ([y][0; y 1] [x]) ([y; y+2 i0 1][0; y 1] [y; y+2 i0 1]) = COST (x 1; l) cot(x 1; l). ([y][0; y 1] [y]) ([y; y+2 i0 1][0; y 1] [y; y+2 i0 1]) = COST (x 1; l) cot(x 1; l). All by proper ubet. (c2). ([y][0; y] [y]) cot(x 1; l) + log log l + 3 (by Lemma 6 and (c1), becaue cot(x 1; l) = z(l) + log l). (c3). [x][0; y] [x] can have three ubcae. (c3.1). [x][0; y] [x] doe not contain any vertex in [y]. Thu it i of the form [x][0; y 1] [x]. Thi reduce to (c1). (c3.2). [x][0; y] [x] contain one vertex in [y]. It i of the form p = [x][0; y 1] [y][0; y 1] [x]. p = p 1 p 2, where p 1 = [x][0; y 1] [y] and p 2 = [y][0; y 1] [x]. We have (p 1 ) = (p 2 ) (by revere). After (p 1 ) iteration p i contracted to [x][y][x]. Becaue [x] and [y] are in a ubmatrix A (L 1), it take t 0 more iteration to contract p to a ingle arc. Therefore ([x][0; y] [x]) ([x][0; y 1] [y]) + t 0. Thi reduce to cae (c1). (c3.3). [x][0; y] [x] contain more than one vertex in [y]. It i of the form p = [x][0; y 1] [y][0; y] [y][0; y 1] [x]. p = p 1 p 2 p 3, where p 1 = [x][0; y 1] [y], p 2 = [y][0; y] [y], p 3 = [y][0; y 1] [x]. We have (p 1 ) = (p 3 ) (by revere). After maxf(p 1 ); (p 2 )g iteration p i contracted to [x][y][y][x]. Becaue [x] and [y] are in a ubmatrix A (L 1), it take at mot t 0 more iteration to contract p to a ingle arc. Thu (p) maxf(p 1 ); (p 2 )g + t 0. Now (p 1 ) i evaluated in cae (c1) and (p 2 ) i evaluated in cae (c2). Summarizing three ubcae for (c3) and uing (c1) and (c2), we obtain that ([x][0; y] [x]) cot(x 1; l) + t 0 + log log l + 3. (c4). ([x][0; x] [x]) cot(x 1; l) + t 0 + log log l log log l + 3 (by Lemma 6 and (c3)). (c5). We ue i tep to evaluate ([x 2 i +1; x][0; x] [x 2 i +1; x]). We dene cot(x; l; 0) = cot(x 1; l) + t 0 + 2(log log l + 3) and cot(x; l; j) = cot(x; l; j 1) + t j 1, 1 j i. In tep 1 j i we evaluate ([x 2 j + 1; x][0; x] [x 2 j + 1; x]). We how that ([x 2 j + 1; x][0; x] [x 2 j + 1; x]) cot(x; l; j). We denote ([x 2 j + 1; x][0; x] [x 2 j + 1; x]) by COST (x; l; j). Therefore we need etablih COST (x; l; j) cot(x; l; j), 1 j i. We have the following tep: (c5.1). ([x 2 j 1 + 1; x][0; x] [x 2 j 1 + 1; x]) = ([x 2 j + 1; x 2 j 1 ][0; x] [x 2 j + 1; x 2 j 1 ]), by ymmetry. (c5.2). Evaluate ([x 2 j + 1; x 2 j 1 ][0; x] [x 2 j 1 + 1; x]). For uch a path p we have the following cae: (c5.2.1). p contain only one vertex in [x 2 j +1; x 2 j 1 ] and one vertex in [x 2 j 1 +1; x]. Then p i of the form [x 2 j + 1; x 2 j 1 ][0; x 2 j ] [x 2 j 1 + 1; x]. We have (p) = ([x 2 j + 1; x 2 j 1 ][0; x 2 j ] [x 2 j 1 + 1; x]) ([x 2 j + 1; x][0; x 2 j ] [x 2 j + 1; x]) (by proper ubet) = COST (x 2 j + 1; l) cot(x 2 j + 1; l) cot(x; l; j). (c5.2.2). p contain more than one vertex in [x 2 j + 1; x 2 j 1 ] but only one vertex in [x 2 j 1 +1; x]. Let p = p 1 p 2, where p 1 contain all vertice [x 2 j +1; x 2 j 1 ] in p and end at the lat vertex [x 2 j +1; x 2 j 1 ] in p. p 1 and p 2 can be written a p 1 = [x 2 j +1; x 12

13 2 j 1 ][0; x 2 j 1 ] [x 2 j + 1; x 2 j 1 ], p 2 = [x 2 j + 1; x 2 j 1 ][0; x 2 j ] [x 2 j 1 + 1; x]. (p 1 ) = ([x 2 j + 1; x 2 j 1 ][0; x 2 j 1 ] [x 2 j + 1; x 2 j 1 ]) ([x 2 j + 1; x 2 j 1 ][0; x] [x 2 j +1; x 2 j 1 ]) (by proper ubet) = ([x 2 j 1 +1; x][0; x] [x 2 j 1 +1; x]) (by ymmetry, ee cae (c5.1)) = COST (x; l; j 1) cot(x; l; j 1). (p 2 ) = ([x 2 j + 1; x 2 j 1 ][0; x 2 j ] [x 2 j 1 + 1; x]) ([x 2 j + 1; x][0; x 2 j ] [x 2 j + 1; x]) (by proper ubet) = COST (x 2 j + 1; l) cot(x 2 j + 1; l) cot(x; l; j 1). Therefore, after cot(x; l; j 1) iteration p 1 i contracted to [x 2 j + 1; x 2 j 1 ][x 2 j + 1; x 2 j 1 ], p 2 i contracted to [x 2 j + 1; x 2 j 1 ][x 2 j 1 + 1; x]. Therefore p i contracted to [x 2 j + 1; x 2 j 1 ][x 2 j + 1; x 2 j 1 ][x 2 j 1 + 1; x]. It take at mot t j 1 more := A (L j) iteration for the intruction A (L j) contract p to a ingle arc. Therefore, (p) cot(x; l; j 1) + t j 1 = cot(x; l; j). A (L j) A (L j) to be executed in order to (c5.2.3). p contain more than one vertex in [x 2 j 1 + 1; x] but only one vertex in [x 2 j + 1; x 2 j 1 ]. Thi cae i ymmetrical to cae (c5.2.2). By ymmetry we have (p) cot(x; l; j). (c5.2.4). p contain more than one vertex in [x 2 j + 1; x 2 j 1 ] and more than one vertex in [x 2 j 1 + 1; x]. Let p = p 1 p 2 p 3, where p 1 contain all vertice [x 2 j + 1; x 2 j 1 ] in p and end at the lat vertex [x 2 j + 1; x 2 j 1 ] in p. p 1, p 2 and p 3 can be written a p 1 = [x 2 j +1; x 2 j 1 ][0; x] [x 2 j +1; x 2 j 1 ], p 2 = [x 2 j +1; x 2 j 1 ][0; x 2 j ] [x 2 j 1 +1; x], p 3 = [x 2 j 1 +1; x][0; x] [x 2 j 1 +1; x]. (p 1 ) = ([x 2 j +1; x 2 j 1 ][0; x] [x 2 j + 1; x 2 j 1 ]) = ([x 2 j 1 + 1; x][0; x] [x 2 j 1 + 1; x]) (by ymmetry, ee cae (c5.1)) = COST (x; l; j 1) cot(x; l; j 1). (p 2 ) = ([x 2 j + 1; x 2 j 1 ][0; x 2 j ] [x 2 j 1 + 1; x]) ([x 2 j +1; x][0; x 2 j ] [x 2 j +1; x]) (by proper ubet) = COST (x 2 j +1; l) cot(x 2 j + 1; l) cot(x; l; j 1). (p 3 ) = ([x 2 j 1 + 1; x][0; x] [x 2 j 1 + 1; x]) = COST (x; l; j 1) cot(x; l; j 1). Therefore, after cot(x; l; j 1) iteration p 1 i contracted to [x 2 j + 1; x 2 j 1 ][x 2 j + 1; x 2 j 1 ], p 2 i contracted to [x 2 j + 1; x 2 j 1 ][x 2 j 1 +1; x], p 3 i contracted to [x 2 j 1 +1; x][x 2 j 1 +1; x]. Thu p i contracted to [x 2 j + 1; x 2 j 1 ][x 2 j + 1; x 2 j 1 ][x 2 j 1 + 1; x][x 2 j 1 + 1; x]. It take at mot t j 1 more iteration for the intruction A (L j) := A (L j) A (L j) A (L j) to be executed in order to contract p to a ingle arc. Therefore, (p) cot(x; l; j 1)+t j 1 = cot(x; l; j). Summarizing (c5.2.1), (c5.2.2), (c5.2.3) and (c5.2.4) we have (p) = ([x 2 j + 1; x 2 j 1 ][0; x] [x 2 j 1 + 1; x]) cot(x; l; j). (c5.3). Evaluating ([x 2 j 1 + 1; x][0; x] [x 2 j + 1; x 2 j 1 ]). Thi i the revered path of the path analyzed in cae (c5.2). Therefore ([x 2 j 1 + 1; x][0; x] [x 2 j + 1; x 2 j 1 ]) cot(x; l; j). (c5.4). Now evaluating ([x 2 j + 1; x][0; x] [x 2 j + 1; x]) = COST (x; l; j). The tarting ending vertex pair < a; b > of path p = [x 2 j + 1; x][0; x] [x 2 j + 1; x] i either < [x 2 j +1; x 2 j 1 ]; [x 2 j +1; x 2 j 1 ] > or < [x 2 j 1 +1; x]; [x 2 j 1 +1; x] > or < [x 2 j +1; x 2 j 1 ]; [x 2 j 1 +1; x] > or < [x 2 j 1 +1; x]; [x 2 j +1; x 2 j 1 ] >. Therefore p i one of the four path p 1, p 2, p 3, p 4, where p 1 = [x 2 j +1; x 2 j 1 ][0; x] [x 2 j +1; x 2 j 1 ], p 2 = [x 2 j 1 +1; x][0; x] [x 2 j 1 +1; x], p 3 = [x 2 j +1; x 2 j 1 ][0; x] [x 2 j 1 +1; x], and p 4 = [x 2 j 1 + 1; x][0; x] [x 2 j + 1; x 2 j 1 ]. We have (p 2 ) = COST (x; l; j 1) cot(x; l; j 1) cot(x; l; j), (p 1 ) = (p 2 ) (by ymmetry), (p 3 ) cot(x; l; j) (by (c5.2)), (p 4 ) cot(x; l; j) (by (c5.3)). Therefore we have (p) cot(x; l; j). 13

14 Summarizing (c1) to (c5) we have COST (x; l) = ([x 2 i + 1; x][0; x] [x 2 i + 1; x]) cot(x; l; i) = cot(x; l; 0) + P i 1 j=0 t j = cot(x 1; l) + t 0 + 2(log log n + 3) + P i 1 j=0 t j = cot(x; l). 2 Theorem 13: COST (x; l) cot(x; l), 0 x < 2 L. Proof: By Lemma 9 to Theorem 14: The total number of iteration needed to compute all pair hortet path in APSP3 i no more than 8 log n. Proof: By Theorem 13, cot(2 L 1; n) upper bound the number of iteration needed to contract any path of length le than n to a ingle arc. We have cot(2 L 1; n) log n+2 L (log log n+3)+ P L j=1 t j 12 L j+1 = log n+2 L (log log n+3)+ P L j=1 K2L j+1 =3 L j 2 log n+6 log n = 8 log n. A we have explained before, when P we um all t i ' for all x we are in L fact umming all t i ' in Fig. 2 twice. We thu get the term j=1 t j 12 L j+1. 2(log log n+3) i included once for each odd x except x = 1 where the term i only log log n + 3. Thu we get the term 2 L (log log n + 3). The term log n come from cot(1; n). 2 Theorem 15: Algorithm APSP3 compute all pair hortet path in time O(n 3 =p + I(n) log n). Proof: Each iteration can be executed in I(n) time if enough proceor are available. Thu we only need to how that the total number of operation executed i O(n 3 ). Matrice at level i, 0 i < L, are multiplied with themelve 8 log n=t L i 1 8 log n3 i =K 24 3 i time. There are 2 i uch matrice and each matrix multiplication take O((n=2 i ) 3 ) operation. Thu the total number of operation for thee matrix multiplication i O( P L 1 i=0 3i 2 i (n=2 i ) 3 ) = O(n 3 ). There are 2 L ubmatrice at level L. Each of them i multiplied to itelf 8 log n time. The total number of operation for thee matrix multiplication i O(2 L (n=2 L ) 3 log n) = O(n 3 ). 2 If we multiply two matrice by uing the reult in [Fr][T], we can reduce the number of operation to o(n 3 ). Corollary: Algorithm APSP3 compute all pair hortet path in time O(f(n)=p+I(n) log n) on the EREW and CRCW PRAM, where f(n) = o(n 3 ). 2 Thi corollary doe not apply to the randomized CRCW PRAM. Becaue we multiply two matrice in contant time uing n 3 proceor on the randomized PRAM, Fredman and Takaoka' technique cannot be applied in contant time to reult in aving. Reference [AGM] N. Alon, Z. Galil, O. Margalit. On the exponent of all pair hortet path problem. Proc. 32nd Ann. IEEE Symp. FOCS, , [DNS] E. Dekel, D. Naimi and S. Sahni. Parallel matrix and graph algorithm. SIAM J. Comput. 10, , [DS] E. Dekel and S. Sahni. Binary tree and parallel cheduling algorithm. IEEE Tran. Comput. 32, ,

15 [Di] E. W. Dijktra. A note on two problem in connextion with graph. Numeriche Mathematik 1, , [Fl] R. N. Floyd. Algorithm 97, Shortet path. Comm. ACM 5, 345, [Fr] [GM] [J] [KR] [L] [PK] [PP1] [PP2] [PR1] [PR2] M. L. Fredman. New bound on the complexity of the hortet path problem. SIAM J. Comput., Vol. 5, No. 1, 83-89(March 1976). H. Gazit and G. L. Miller. An improved parallel algorithm that compute the BFS numbering of a directed graph. Information Proceing Letter 28:1, 61-65, D. B. Johnon. Ecient algorithm for hortet path in pare network. J. ACM, 24, 1, 1-13, M. Karp, V. Ramachandran. A urvey of parallel algorithm for hared memory machine. \Handbook of Theoretical Computer Science," , North-Holland, Amterdam, A. Linga. Ecient parallel algorithm for path problem in planar directed graph. Proc. SIGAL'90, Lecture Note in Computer Science 450, R. C. Paige and C. P. Krukal. Parallel algorithm for hortet path problem. Proc Int. Conf. on Parallel Proceing, St. Charle, Illinoi, V. Y. Pan and F. P. Preparata. Supereective low-down of parallel algorithm. Proc ACM Symp. on Parallel Algorithm and Architecture, San Diego, California, V. Y. Pan and F. P. Preparata. Work-preerving peed-up of parallel matrix computation. SIAM J. Comput., Vol. 24, No. 4, V. Y. Pan and J. H. Reif. Fat and ecient olution of path algebra problem. Journal of Computer and Sytem Science, 38(3): , June V. Y. Pan and J. H. Reif. The parallel computation of minimum cot path in graph by tream contraction. Information Proceing Letter 40, 79-83, [R] R. Reichuk. Probabilitic parallel algorithm for orting and election. SIAM J. Comput., Vol. 14, No. 2, (May 1985). [S] [T] R. Seidel. On the all-pair-hortet-path problem. Proc. 24th ACM STOC, ACM Pre, (1992). T. Takaoka. A new upper bound on the complexity of the all pair hortet path problem. Info. Proce. Lett. 43, (1992). [V] L. G. Valiant. Parallelim in comparion problem. SIAM J. Comp., Vol. 4, No. 3, (Sept. 1975). 15

16 A 00 A 01 A 02 A 03 A 10 A 11 A 12 A 13 A 22 A 20 A 21 A 23 A 30 A 31 A 32 A 33 Fig t 0 1 t 1 2 t 0 t t 0 t t 0 7 Fig