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1 NOTICE WARNING CONCERNING COPYRIGHT RESTRICTIONS: The copyight law of the United States (title 17, U.S. Code) govens the making of photocopies o othe epoductions of copyighted mateial. Any copying of this document without pemission of its autho may be pohibited by law.

2 PARALLEL NEIGHBOR-SORT (OR THE GLORY OF THE INDUCTION PRINCIPLE) Nico Habemann Canegie-Mellon Univesity Pittsbugh, Pa. August 1972 This wok was suppoted by the Advanced Reseach Pojects Agency of the Office of the Secetay of Defense (F C-0107) and is monitoed by the Ai Foce Office of Scientific Reseach. This document has been appoved fo public elease and sale; its distibution is unlimited.

3 ABSTRACT Aay A[1:N] (N ^ 2) is to be soted in ascending ode using pocedue NS(i) which aanges the values of an adjacent pai A[i], A[i+1] in the ight ode. It is shown that a paallel pocess P which can pefom at least N T 2 executions of pocedue NS in paallel will sot aay A in p steps whee p ^ N. The poof is based on the obsevation that the distance of the position of a numbe in aay A afte p steps and its final position is bounded by N - p.

4 1. INTRODUCTION Elements of a given aay A[1:N] (whee N ^ 2) ae to be soted in ascending ode assuming that nothing is known about the initial ode. Suppose aay A can only be eaanged by soting adjacent pais (A[i], A[i+1]) with pocedue NS declaed as: pocedue NS(i); intege i; if i < N do if A[i] > A[i+1] do begin intege w; w «- A[i]; A[i] «- A[i+1]; A[i+1] «- w end; This equiement elements of aay A that only allows opeations on neighboing pais of esticts of couse the ange of soting algoithms consideably. But this estiction is plausible if paallel pocesses ae consideed which ae capable of caying out many sot opeations on elements of A simultaneously. The paallel pocesses consideed hee ae supposed to be able to execute in paallel at least N - 2 instances of pocedue NS. Such a pocess could fo instance ode in one step all elements with index with egad to thei neighbo element with index. This pape addesses itself to the question of how many paallel steps would be needed to sot aay A[1:N] with pocedue NS. The answe tuns out to be that the numbe of paallel steps equied is less o equal to N. This seems plausible when thinking of the smallest o lagest numbe, but not so obvious when obseving how an "in between" numbe may wande about befoe aiving at its destination.

5 -2- Example paallel step p The numbes 2 and 7 ae stating off in the wong diection, the numbes 4 and 8 begin heading in the ight diection but pass thei destination. But it seems as if a numbe v cannot wande too fa fom its destination. The example shows that any numbe v is at most two positions off fom its destination fo p = N-2 and only one fo p = N-l. The genealization of this obsevation is: a numbe v is afte p steps at most N-p positions away fom its destination. The lage pat of the sequel deals with the coectness poof of this obsevation. The existence of the uppebound N fo the numbe of steps p needed to sot aay A can easily be deived fom it. 2. REPHRASING THE PROBLEM The soting of aay A[1:N] (N ^ 2) with abitay values can easily be tanslated into a mapping of I -* I whee I = U,2,...,Nj. The idea is to eplace each value v by the index of its destination which will be eached afte soting. Thus, to sot aay A means to exchange elements of A until fo all j 6 il,2,...,nj A[j] = j. A sequential pocess that sots aay A using pocedue NS is descibed by

6 -3- intege k,z; fo z «- N-l step -1 until 1 do fo k 1 step 1 until z do NS(i) It is well known that this pocess would cetainly not be the most efficient one if the estiction of "eaanging neighbos only 11 was emoved. It equies l/2 N(N-l) sequential executions of pocedue NS, wheeas Quick Sot pocedues equie in the ode of l/2 N log2 N compaisons. On the othe hand, paiwise compaison is attactive if many of those can be pefomed in paallel at no additional cost. But the sequential pocess descibed above cannot be tansfomed easily into a paallel pocess because of its iteative natue. Instead, paallel pocesses that will be consideed use two paallel statements, named S,, and S : s o d d : pabegin NS(1); NS(3); NS(1 + (N-l) -f 2*2) paend s e v e n. Pabegin NS(2); NS(4); NS(N - 2*2) paend The esult of executing a paallel statement should be unambiguously intepetable without equiing knowledge of the ode in which vaiables ae accessed. It would not make sense fo instance to use two consecutive indices as paametes to calls of pocedue NS in a paallel statement, e.g., the esult of pabegin NS(i); NS(i+l);... paend cannot be intepeted uniquely. Fo instance, (A[i], A[i+1], A[i+2]) =* (11, 7, 3) could be mapped by this paallel statement into

7 -4- (7, 3, 11) o (3, 7, 11) o (7, 11, 7)11 Paallel statements S., and S. do not have such ambiguities, because each 9 element A[i] is involved in not moe than one instance of pocedue NS. Let pocess P^^ be defined by intege p; fo p = 1 step 1 until N do if p - p 4-2*2 then S else S.. and pocess P by J intege p; fo p = 1 step 1 until N do if p = p T 2*2 then S,. else S Pocess P,, stats with S,,, pocess P stats with S The main ' theoem to pove is Theoem 1, Pocess P o pocess P sot aay A in ascending ode (o: aay A is soted by p altenating execution of S ^ and S e v e n no matte which is fist, and p ^ N). The poof is based on the following esult: Theoem 2. ABS(i^(p) - j) ^ N-p (1) fo all j g N, whee i_.(p) is the index of j in aay A afte p steps (i.e., p altenating execution of S and S, no matte which one is taken fist), The poof of theoem 1 follows immediately fom (1):

8 ABS(ij(n) - j) = 0 fo p = N and fo all j lf..»nj, i.e., the index of j afte N steps is j, o A[j] = j. 3. PROOF OF THEOREM 2 Theoem 2 puts an uppe limit on how fa a numbe j is out of balance so to speak. This depends on (*j(p) ::= how many numbes to the left of j ae geate than j afte p steps Pj(p) how many numbes to the ight of j ae less than j afte p steps. (The subscipt j will be omitted fom now on whee confusion is unlikely.) The numbes j ae togethe [i(p) - a(p)3 + P(p)> an< * so i(p) - j - *(p) - P(p) (2) and initially 1(6) - j - cto) -'0 (3) The goal is to pove that 0 ckp) N-p (4) and 0 P(p) N-p (5) The poof of theoem 2 follows then fom (2), (4) and (5). - (N-p) * -p(p) *c(p)-p(p) - i(p)-j * cf(p) * N-p Let B[1:N] (N 2> 2) be an aay of which initially B[k] = k fo k»l,...,m and 1 m N-l B[k] = 0 fo k=m+l,...,n

9 -6- Aay B consists of the sequence l,...,m followed by at least one 0. S S a PP l i e d t o a a Y B tansfom no othe pais than (k,0) into (0,k) fo k 6 >!&} Thus, the non-zeo numbes will not ovetake each othe and the zeos peseve also thei elative ode. The case [ P JJ) is ] o [P, m is ] is denoted by C ; the othe case J b y C - The pai (B[m], B[m+1]) is compaed in the fist step iff ^ e v e n applies. Lemma 1. The numbe m, initially in B[m], is in B[m+p] afte p steps of C o p+1 steps of C JJf fo 0 p N-m. Poof. The fist step of C ^ has no effect on aay B since no pai (1,0) is consideed. Afte the fist step C ^ has eached the same state as the initial state of C. Thus, whateve is achieved in p steps of C, is ' obtained in p+1 steps of C Conside C. The statement is tue fo p=0. Suppose it is tue fo p-pg and p^ is a numbe such that 0 p Q N-m-1. So, (B[m+pg], B[m+Pg+1]) - (m,0), since m is the ightmost non-zeo numbe. The pais consideed in the p*"* 1 step of C ae (B[m-*p-l+2*v], B[m+p+2*v]) fo all v such that 2 m+p+2*v N. Hence, one of the pais consideed in step Pg+1 i s (B[m + (p Q +l) - 1], B[m + (p Q +l)]) fo v - 0 [fo which value 2 m+p Q +l+2*v ^ N is tue]. But this pai equals (m,0) accoding to the supposition and it is subsequently tansfomed into (0,m). Hence, the assumption that B[m+p Q ] - m fo p = p Q implies that

10 -7- B[ni-pg+l] = m fo p = PQ+1» Thus, the statement is tue fo all p such that 0 ^ p ^ N-m accoding to the induction pinciple. Thee is obviously a duality in aay B of zeos and non-zeo numbes and so, lemma 1 about the ightmost non-zeo numbe caies ove into a statement about the leftmost 0: Lemma 2. of C The leftmost 0, initially in B[mfl], is in B[m-p+l] afte p steps o p+1 steps of C,, fo 0 p m Lemma 3. Numbe k 6 \1 9...,mj, initially in B[k], is in B[2*k+p-m] afte p steps of C o p+1 steps of C fo m k + v p N. Poof. As obseved in the poof of lemma 1, C ^ equied one step moe to achieve the same as C. Conside C. The statement is tue fo k^m (lemma 1). Suppose it is tue fo k = kg {l,...,ml, i.e., B[2*k Q + p-m] = k Q fo all p such that m kg-l+p * N. It is tue fo p - m+l-k 0, because B[m - (m+l-k Q ) + 1] = B[k Q ] contains the leftmost zeo accoding to lemma 2 and so B[k Q -l] = k^-l. S u PP o s e i t : is tue fo p - p Q and m k Q + p Q N, i.e., B[2*(k 0-1) + p Q -m] = k Q -l. But it is known that fo this value of p B[2*k Q + Pg-m] - k 0 a n d s o B[2*k Q - p Q -m] = 0. In step p Q B[2*k Q + P 0 " M ] Sot the value k Q, hence in step p Q +l the pai (B[2*k Q + p Q -2-m], B[2*k Q + P()-l-m]) is consideed. It equals (k Q -l,0) and is subsequently tansfomed into (0,k n -l). Thus,

11 -8- B[2*(k Q -l) + (p 0 +l) - m] = k Q -l fo p = p Q +l Applying the induction pinciple twice shows that the statement is tue. Let H (p) denote the numbe of non-zeos left of any zeo in aay B afte p steps of P^dd o P Lemma 4. I (p) N-P Poof. Numbe k Q {l,...,m} is moved fom B[k Q ] to B[N-m+k Q ] in p = N-k Q steps of C o p = N-k n +l steps of C 0 accoding to lemma 3. This numbe k Q has passed all zeos of aay B. But so have all numbes k fo which kg k m o N-p+1 k m. Thus, fo given p the numbes that possibly did not pass all zeos of aay B is less than N-p+1 o A (p) N-p. Let C[1:N] be an aay with the same numbe of non-zeo and zeo elements as aay B[1:N]. The non-zeo elements ae the numbes l,...,m (1 m N-l) fom left to ight. These numbes ae not necessaily stoed in consecutive elements of aay C. Thus, if ij(p) an( * *J(P) denote the indices of numbe j [l,...,mj afte p steps of P^dd o? e v e n in aays B and C espectively, i*(0) ^ ijco) fo all j e {1,...,m}. Lemma 5. ij(p) ij (p) fo all j {l,...,m} and p 6 io,...,n} Poof. Let j = m. Accoding to lemma 1 C (p) = i m ( 0 ) + P * i m ( 0 ) + P = i m ( p ) f o P * N " i m ( 0 ) (o v p N-i (0) + 1 fo C..) m and i (p) ^ i (p) - N fo N-i (0) + 1 p N. m m m

12 -9- Suppose the elation is tue fo j k+1. The objective is to pove that this implies that the elation is also tue fo j=k. Fo p=0 i k ( ) ^ is tue. Suppose it is tue fo p 83 p Q {0,...N-1}. It could not happen that, if i*(p Q ) = i k (p), B[i*(p Q ) + 1] - 0' and C[i k (p) + 1] + 0 because it would mean that WPO* > W + 1 = W + 1 = WV> and this is in contadiction with the supposition that the elation is tue fo all p {0,...,N} fo j = k+1. It can easily be deived that the elation is also tue fo p Pg + * i n a ^ othe cases (viz. eithe B[i k (p Q ) + 1] ^ 0 o both B[i k (p Q ) + 1] - C[i fc (p) + 1] = 0 o i k(p 0 ) < APP^ 1 1 * t h e i n " duction pinciple twice shows that i*(p) ij(p) fo all j 11,...,m} and p 6 [0,...,N] Let A(p) denote the numbe of non-zeos in aay C to the left of at least one zeo afte p steps Lemma 6. A(p) < & (p) fo all p 10,...,N] Poof. If thee ae not any non-zeos to the ight of all zeos then A(p) = A (p) = m. Let the numbes k+l,...,m be to the ight of all zeos in aay B and the numbes j+l,...,m to the ight of all zeos in aay C. Then B[n-m-+k] = 0, which is the ightmost zeo in aay B, C[N-m+j] = 0, which is the ightmost zeo in aay C and A(p) = j and I (p)» k. Suppose k < j.

13 -10- This implies that j has all zeos to its left in aay B and so i.(p) - N-m+j. But ij(p) < N-m+j in aay C and so ij(p) > ij(p). But this elation is in contadiction with lemma 5. Let us now etun to aay A[1:N]. The objective was to show that tfj(p) ^ N-p fo all j g llf.fnj. Thee ae no numbes in aay A geate than N and so the elation is tivial fo j * N: c^(p) - 0 N-p is tue. Conside a numbe j [l,...,n-lj and the mapping T (A) -» C defined by T j : intege i,m; m «- 0; fo i «- 1 step 1 until N do if A[i] > j then begin m «- m+1; C[i] *- m end else C[i] 0 The esult of the mapping is that aay C contains at least one numbe ^ 0 and at least one 0, The non-zeo numbes ae odeed in ascending ode fom left to ight, but not necessaily in contiguous locations. Lemma 7. Tj.S(A) - S.T^(A) fo all j 6 [l,...,n-ll, i.e., the index elation between numbes ^ j in aay A and zeos in aay C is invaiant fo paallel steps S.. and S Poof. Take a numbe j {1,...,N-H and let (k^k^ be a numbe pai in aay J A that is consideed in S o S... If k x > j and k 2 > j then T..S(k lf k 9 ) = T (k.,k ) J J 1 2 I (i,i+l) 1 2 o and S.TjC^,^) = S(i,i+1) = (i,i+l). If k x ^ j and k 2^ j, the esult is both ways (0,0).

14 -11- If > j and k 2 ^ j then Tj-SCk^) = TjC^,^) = (0,i) and S.TjCk^k^ = S(i,0) = (0,i) If k j and k 2 > j, the esult is both ways (0,i). The statement can be genealized to include an abitay sequence of steps S J J and S Lemma 8. c^(p) ^ JL(p) fo all j 6 {l,...,n-lj and p {0 f...,n} Poof. Let Vj(p) - lk k > j and i^p) < ij(p)}. a.(p) - numbe of elements of V.(p). J 3 T j ( k V j ( p ) ) = k ' > a n d i k, ( p ) = i k ( p ) < i j ( p ) a n d Tj(j) - 0 in C[i^.(p)] accoding to lemma 7. Hence, ckp) = numbe of non-zeos left of C[i^.(p)] - 0, wheeas (p) = numbe of non-zeos left of the ightmost 0. Thus, <y(p) A(p). The cucial elation a.(p) * N-p (4) fo all j U,...,N} and p {0,...,N'j can now be deived fom a N (p) = 0 fo j - N and all p 6 1.0,...,N} and fo all j 6 [1,...,N-1] and all p g 10,...,N] ij(p) * N-p (lemma 4) V P) * V P) (lemma 6) ffj(p) ^ JJj(p) (lemma 8)

15 -12- Fo the poof of the main theoem, that aay A[1:N] can be soted in N paallel steps, we also needed the elation Pj (P) * N-p It is obvious that the poof that this elation is tue can be constucted by systematic changes of left, ight and less than, geate than in the lemmas. The check is left to the eade. It is supising that a theoem which seemed quite obvious equied such an elaboate poof. ACKNOWLEDGEMENTS I am gateful to Anita Jones, Lay Snyde, and Tim Teitelbaum fo discussing the solution with me. It was a pleasue to obseve how E. W. Dijksta constucted vaiations and altenatives of the algoithms P Q( J < J and p e v e n- I a** 1 also gateful to R. Floyd who pointed out to me that an algoithm to sot the evese sequence, N, N-l,..., 1, also sots any othe pemutation of these numbes. This theoem is essentially used whee aays B and C ae discussed (p. 8) and poved fo algoithms P ^ a n d p e v e n *- n the fom of lemmas 5 and 6.

16 Secuity Classification DOCUMENT CONTROL DATA R&D (Secuity classificmtion of title, body of abstact and indexing annotation must be enteed when the oveall epot is classified) 1. ORIGINATING ACTIVITY (Copoate autho) Compute Science Depatment Canegie-Mellon Univesity Pittsbugh, Pa REPORT TITLE 2a. REPORT SECURITY CLASSIFICATION UNCLASSIFIED 2b. GROUP PARALLEL NEIGHBOR-SORT (OR THE GLORY OF THE INDUCTION PRINCIPLE) 4. DESCRIPTIVE NOTES (Type of epot and Inclusive dates) Scientific Final 3- AUTHOR(S) (Fist name, middle initial, last name) Nico Habemann 6. REPORT DATE August «. CONTRACT OR GRANT NO. F C-0107 b. PROJECT NO «61102F * DISTRIBUTION STATEMENT Appoved fo public elease; distibution unlimited. 7a. TOTAL NO. OF PAGES 7b. NO. OF REFS 15 9a. ORIGINATOR'S REPORT NUMBER(S) 9b. OTHER REPORT NO(S) (Any othe numbes that may be assigned this epot) 11. SUPPLEMENTARY NOTES TECH OTHER 12. SPONSORING MILITARY ACTIVITY Ai Foce Office of Scientific Rsch (NM) 1400 Wilson Blvd. Alington, Va ABSTRACT Aay A[1:N] (N ^ 2) is to be soted in ascending ode using pocedue NS(i) which aanges the values of an adjacent pai A[i], A[i+1] in the ight ode. It is shown that a paallel pocess P which can pefom at least N «f 2 executions of pocedue NS in paallel will sot aay A in p steps whee p ^ N. The poof is based on the obsevation that the distance of the position of a numbe in aay A afte p steps and its final position is bounded by N - p. DD, F N o R vm473 Secuity Classification