Parallel Prefix addition

Transcription

1 Marcelo Kryger Sudent ID Parallel Prefx addton The parallel prefx adder presented next, performs the addton of two bnary numbers n tme of complexty O(log n) and lnear cost O(n). Lets notce the followng fact: If the carry bts c n,..., c 0 are known, performng the addton s smplfed; to get the fnal result <s n > would suffce to perform s = XOR( a, b, c ) for [ 0,, n 1] 1. And vce-versa, f the result s known, the carry bts can be easly obtaned by applyng the XOR functon: c = XOR( a, b, s ). Concluson: calculatng the result bts s almost as complex as calculatng the carry bts. We consder now the problem of computng the carry bts. The nput conssts of ab, { 01, } n, c IN { 01, }. The output conssts of the carry bts c n,, c 0 { 01, } Carry calculaton Defne σ 1, σ, n { 012,, } by: a + b n 0 σ = 2 = 1, c 0 = 1 0 = 1, c 0 = 0 Sometmes the values 0, 1 and 2 are named kll, propagate and generate respectvely. We ll use regular expresson notaton. In regular expressons the symbol * ndcates that the symbol precedng t can appear 0 or more tmes n the expresson. For example: 1 * 2 { 2, 12, 112, 1112,, , } Clam: For every [ 1,, n 1], c + 1 = 1 f and only f there exsts j such that ( σ, σ 1. Frst drecton s smple, for every, gven j such that ( σ, σ 1, t can be seen that the carry generated at σ j s propagated through, up to σ. The second drecton wll be proved by nducton, gven that = we ll fnd j such that ( σ, σ 1, σ, j Proof 1 (by nducton on ): Base: =-1. By defnton c 0 = 1 σ 1 = 2. c For =n t s just s n =c n Page 1

2 Marcelo Kryger Sudent ID Step: c 1 + = 1 a + b + c 2 We consder 3 cases: (1) σ = 2 : In ths case both nput bts are 1, and carry s generated. We choose j = (2) σ = 1 :In ths case one of the nput bts s 1. If = the nducton assumpton for leads the proof for +1: c If only one of the nput bts s one, and carry s generated, at bt, then the carry at bt -1, has to be 1. Therefore exst j< that comples wth the clam, accordng to the nducton assumpton + = 1 and σ = 1 c = 1 j< 1 ( σ 1 ( σ, σ 1 c 1 (3) σ = 0 c + 1 = 0 Parallel Prefx Calculaton Now we want to fnd out f there exsts j such that ( σ. We ll buld an automata (aka state machne - SM or fnte state machne - FSM) consstng of only two states (whch means only one bt s needed to mplement t). The nput of the FSM, at step, s. The ntal state s S 0. State S 1 ndcates, at step, that there exsts j such that ( σ ; state S 0 ndcates there s not such j. σ σ =0,1 σ =0 S 0 S 1 σ =1,2 σ =2 Fgure 1. Carry Calculaton State Machne Lets prove, by nducton on, that the FSM works correctly,.e, that after readng a gven sequence of nputs σ 1, σ, the FSM s n state S k f and only f c + 1 = k. It s smple to show the frst drecton, after readng the sequence σ 1, σ,, f c + 1 = 1 then the FSM s n state S () = S 1 : Lets consder the two cases: c 1 σ j (1) Assume + = 1. Accordng to proof 1, exsts j such that ( σ, σ, j. From Fgure 1 can be seen that when (=2) s read the FSM goes to state S 1 (regardless of ts prevous state), any further σ (=1) that s read, does not change the state, Page 2

3 Marcelo Kryger Sudent ID therefore S () = S 1 (2) Assume =. Usng the opposte argument than before, there exsts no j such c that ( σ (otherwse, accordng to proof 1, c + 1 = 1 ). The frst element of every sequence s σ 1, whch, by defnton can be ether 0 or 2. Then for every j: ( σ ) ( 1 * 2) ( σ ) ( 1 * 0) From Fgure 1 can be seen that when σ j (=0) s read the FSM goes to state S 0 (regardless of ts prevous state), any further σ (=1) that s read does not change the state, therefore S () = S 0. Gven the FSM n Fgure 1 we wll buld a network that receves, as nputs, for all, and calculates, n parallel, the sequence of states S n 1,, S 0 that the FSM wll go through for those nputs. Parallel Prefx Functon Lets defne the functon, and ts truth table: Table 1: truth table σ a a b b The operator s an assocatve operator,.e. a b c = ( a b) c = a ( b c), but t s not commutatve,.e. a b b a. Non commutatvty can be mmedately seen from the fact that the table above s not a symmetrc matrx. Table 2 proves assocatvty The symbol φ n the table represents don t care Page 3

4 Marcelo Kryger Sudent ID Table 2: Assocatvty of functon a b c ( a b) ( a b) c ( b c) a ( b c) 0 φ φ 0 0 φ 0 2 φ φ 2 2 φ φ φ Lets defne the followng notaton: σ σ 1 σ 0 σ 1 Clam: c 1 + = 1 = 2 We already proved that j ( σ, σ, j c + 1 = 1 Drecton 1: Gven that c + 1 = 1 lets prove that for k from -1 to, the multplcaton of all, has to have result 2. c 1 Applyng the truth table of : If = then exst j such that ( σ and by openng the regular expresson and multplyng ts elements, usng 1 2 = 2, 1 1 = 1 and assocatvty, we see that = 2, therefore can be dvded n two: up to element j and from element j down to -1: k = j + = 1 = 2 j 1 c = = ( 1 2) (?) = 2? = 2 Page 4

5 Marcelo Kryger Sudent ID Drecton 2: Gven that the multplcaton of all, for k from -1 to, has the result 2, lets prove that c + 1 = 1 Lets look at the gven prefx: σ j j = 1 σ j j = 1 We ll show that the above equaton s true f ( σ, σ, 1 ) ( 1 * 2. * ). Afterwards, by proof 1: j ( σ c + 1 = 1, we can conclude that c + 1 = 1 = 2 c + 1 = 1 = 2 σ σ 1 = 2 [ EQ - 1 ] Proof 2 (by nducton on ) of: ( σ, σ, 1 ) ( 1 * 2. * ) Base, k=-1: Accordng to EQ-1: = 2 ( 1 * 2. * ) σ 1 Step, lets assume t s true for k 1 and prove t for k = for all three values of : σ * 2. * ( σ, σ, 1 ) = ( σ, σ 1, σ, 1 ) = ( σ, 1 ) ( 21 * 2. * ) = 2 ( 1 * 2. * ) σ = 2 = ( 11 * 2. * ) ( 1 * 2. * ) σ = 1 contradcton wth EQ-1 σ = 0 The Problem of Parallel Prefx Computaton a n 1 Inputs: a 0 { a } Σ : Σ Σ Σ Page 5

6 Marcelo Kryger Sudent ID Outputs: The set of all prefxes of a 1 a 0 a n 1 a n 2, namely: For every : 0 n a j We transformed ths way, the problem of calculatng the carry bts nto the problem of calculatng all bts prefxes. j = 0 (Fast) Algorthm for Parallel Prefx Calculaton (PPC) Stage 1: Reducton. Transform a problem of sze n nto smaller problems of sze n/2 Stage 2. Combne the outputs of the smaller problems wth the orgnal nputs and get the fnal output a 7 a 6 a 5 a 4 a 3 a 2 a 1 a 0... a 7 a 6 a 5 a 4 a 3 a 2 a 1 a 0 PPC(n/2)... a 7... a 0 a 5... a 0 a 3 a 2 a 1 a 0 a 1 a 0... a 6... a 0 a 4... a 0 a 2 a 1 a 0 a 0 Fgure 2. PPC network From Fgure 2 t can be seen that the outputs of the PPC(n/2) network are the odd prefxes, and that, by combnng them wth the even nputs, we can obtan the even prefxes. Page 6

7 Marcelo Kryger Sudent ID Tme and Cost complexty Assumng that T ( ) = O( 1) and C ( ) = O( 1) T( n) = T n T ( ) = 2 log( n) T ( ) = O( logn) Cn ( ) = C n ( n 1) C ( ) n ( n 1) -- 2 n = C ( ) 2 n C ( ) = On ( ) Page 7