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Continual pattern replication Munro, James Ian 1969

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CONTINUAL PATTERN REPLICATION by JAMES IAN MUNRO B. A . , U n i v e r s i t y of New Brunswick, 1968 A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF SCIENCE i n the Department of Computer Science We accept t h i s thes i s as conforming to the required standard. The U n i v e r s i t y of B r i t i s h Columbia August, 1969 In p r e s e n t i n g t h i s t h e s i s i n p a r t i a l f u l f i l m e n t o f the r e q u i r e m e n t s f o r an advanced degree at the U n i v e r s i t y o f B r i t i s h C o lumbia, I agre e t h a t t h e L i b r a r y s h a l l make i t f r e e l y a v a i l a b l e f o r r e f e r e n c e and s t u d y . I f u r t h e r agree t h a p e r m i s s i o n f o r e x t e n s i v e c o p y i n g o f t h i s t h e s i s f o r s c h o l a r l y p u r p o s e s may be g r a n t e d by the Head o f my Department o r by h i s r e p r e s e n t a t i v e s . I t i s u n d e r s t o o d t h a t c o p y i n g o r p u b l i c a t i o n o f t h i s t h e s i s f o r f i n a n c i a l g a i n s h a l l not be a l l o w e d w i t h o u t my w r i t t e n p e r m i s s i o n . Department o f Computer S c i e n c e The U n i v e r s i t y o f B r i t i s h Columbia Vancouver 8, Canada Date September, 1969 ABSTRACT This thes i s continues the studies of A . Waksman (1969) i n the repeated generation of f i n i t e s t r i n g s i n a one dimensional array of f i -n i t e automata. Waksman handles thispproblem by the use of a "modulo ar i thmet ic" a lgor i thm. This i s shown to be very r e s t r i c t i v e with regard to the number of characters permitted i n the output s t r i n g . In f a c t , i t i s shown that unless the length of the s t r i n g which i s to be repeated i s oc of the form p , where p i s prime, only one output character i s permit ted . This of course makes the process qu i t e meaningless. For t h i s reason, a new algor i thm i s developed. This i s r e -f e r r e d to as the wheel a lgor i thm, s ince there i s an obvious analpgy between i t and a wheel, with the output s t r i n g on i t s circumference r o l l i n g alonggthe array and l eav ing the imprint of the characters i n the s t r i n g behind i t i n the same way that a wheel leaves t i r e t r a c k s . The number of s tates required for such an a lgor i thm i s large and so the b i -nary wheel a lgor i thm i s in troduced. By us ing t h i s a lgor i thm, i n which an output s ta te i s represented as a s t r i n g of b i t s i n s evera l c e l l s , the number of s tates r e q u i r e d , i n a d d i t i o n to the n output s ta tes , can be reduced to about 4 n . Both the wheel and b inary wheel algorithms are then extended to the two dimensional and f i n a l l y the d-dimensional cases. i i TABLE OF CONTENTS Page I In troduct ion 1 II A More D i r e c t Method of Determining {bO 7 I I I The Value of ^ . . . 12 IV An A l t e r n a t i v e Method of Cont inual R e p l i c a t i o n of L i n e a r Patterns — The Wheel Algor i thm . . . . 22 V Extension of the Wheel Algor i thm to the 2-dimensional Case . 31 VI Extension to the d-dimensional Space . . . . . . . 44 VII Conclus ion 46 i i i ACKNOWLEDGEMENTS I wish to thank D r . R. S. Rosenberg for the guidance and ass i s tance which he extended to me i n the preparat ion of t h i s thes i s and Dr. A. Mowshowitz for h i s h e l p f u l remarks i n prepar ing the f i n a l manuscript . The f i n a n c i a l ass i s tance of the Nat iona l Research Counc i l of Canada i s a lso g r a t e f u l l y acknowledged. 1. I INTRODUCTION In a recent paper, Waksman (1969) uses a one-dimensional array of f i n i t e s tate machines to model the cont inua l r e p l i c a t i o n of a sequence of k symbols represented by the states of the c e l l s of the a r r a y . By cont inua l r e p l i c a t i o n of a sequence of length k we mean that a f ter an appropriate length of t ime, dependent on m, the subsets of c e l l s {ik+1, i k + 2 , . . . ( i + l ) k } for i = l , 2 , . . . , m w i l l each be r e p l i c a s of the de-s i r e d sequence of k symbols. Thus i f our alphabet were {0,1} and we wanted to reproduce the s t r i n g of length 4 (0111) c o n t i n u a l l y , a f t e r an appropriate length of time we would have {0111011101110111...} The p r i n c i p a l r e s t r i c t i o n on the transformation to be performed i s that the s tate of any c e l l at time (t+1) i s a funct ion of the states of that c e l l and i t s two immediate neighbours at time t . Waksman shows a method of generating- c o n t i n u a l l y s t r i n g s of length k i n the prev ious ly described manner provided the characters from which the s t r i n g i s con-s truc ted are chosen from the r i n g of integers mod g^, ^§k> where g k = gcd{(^)} i - l , 2 k - 1 . (1.1) That i s , g^ i s the greatest common d i v i s i o n of the set of b inomial co-k e f f i c i e n t s (^) other than those which are 1. The problem i s formulated by Waksman as fo l lows: Suppose we want to generate c o n t i n u a l l y the s t r i n g (a^} i = l , 2 , . . . , k ; then, we s t a r t with the i n i t i a l conf igurat ion c o n s i s t i n g of c e l l 0 i n a t r a n s i t i o n s t a t e , P, the next k c e l l s i n s tates b^ i = l , 2 , . . . , k and a l l other c e l l s i n the quiescent s ta t e , Q. I t can be seen that the r e l a t i o n 2. between the s t r i n g s and b^ i s given by a. = I (T"h b . (1.2) The a c t u a l method of c a l c u l a t i n g {b^}given {a^} w i l l be ex-p la ined l a t e r . I t i s to be understood that a l l ar i thmet ic w i l l be modulo At each time step one and only one c e l l i s i n the t r a n s i t i o n s t a t e , P . At time t , c e l l t only w i l l be i n s tate P. The process for any c e l l i s b a s i c a l l y broken i n t o two segments, the' time before i t enters s ta te P and the time a f t e r i t enters statePP. At each time step the c e l l i n s tate P and i t s r i g h t hand neighbour exchange s ta te s . A f t e r t h i s has happened to a c e l l , i t r e ta ins th i s s t a t e . The i n t e r e s t i n g part of the process i s , then, what happens before a c e l l enters the t r a n s i t i o n s t a t e . I f c e l l i i s i n s ta te a . at time t , with the i n i t i a l cond i t ion a„ .=b. t , i o , i 1 for i=l,2,...^kY we def ine a , . by -' C T J , J X a t + l , i = a t , i + a t , i - l , (1.3) where a c e l l i n the quiescent s tate i s considered to be i n s tate 0. To redef ine the funct ion i n a more formal manner we may fol low Table 1 which i n d i c a t e s the transformation up to the time of entry in to the terminal s ta te . To r i g o r o u s l y define the funct ion we may th ink of the c e l l s as having an a d d i t i o n a l f l i p f lop which i s o f f u n t i l the r i g h t neighbour of a c e l l enters s tate P. aAt that time the f l i p f lop i s turned on and no fur ther change can occur i n the s tate of the c e l l . 3. •TABLE 1* F l i p f lop i n p o s i t i o n 0 Q = quiescent s ta te P = t r a n s i t i o n s tate $ = no c e l l present ( l e f t neighbour of c e l l 0) a. = an element of Ze, * The output given i n t h i s and a l l s i m i l a r tables i s at time t+1. c e l l s ta te at time t a. l Right neighbour L e f t a k a i + a k a i + \ neighbour P P P L e f t neighbour Q a. Right neighbour Q Q L e f t neighbour Right neighbour a. No other conf igurat ions can occur . Thus i f we i n i t i a l i z e the array as prev ious ly mentioned and fol low the t r a n s i t i o n of Table 1, c e l l 0 w i l l enter the terminal s tate a^, c e l l 1, h^+b2 = a^ and i n general c e l l i-1 enters terminal s tate i E (. ^) b . = a_^  (by equation 1.2) j = l 2 ~ 3 1 at time t = i . Out problem i s now to c a l c u l a t e { b } for a given set { a} . Waksman found that { b .} could be generated from { a.} i n the fo l lowing manner. Let a, .=a. i=l.,2 , . . . ,k and wr i t e a l T . a . „ . . . a 1 7 . Then l i I ' ' 11 12 l k 4. complete a k x k matrix by l e t t i n g a . = a + a m = 2 , 3 , . . . , k mj m-l,jv- m - l , j + l . , * ' * 1 and a , = a, , = a, (1.4) mk l k k The ar i thmet ic i s of course mod g^. Waksman shows that the process i s continued for m = k+1, k + 2 , . . . , and that the matrix i s r e -peated, or that a m ^ o c ] c j = a m j a = a n y tue p o s i t i v e i n t e g e r . Thi$ property i s due to the fact that the ar i thmet ic i s done modulo g^. This process i s e s s e n t i a l l y what happens i n the ac tua l generating machine before the P s tate i s entered. The only d i f f erence i s that the k - tup le remains s t a t i o n a r y instead of moving to the r i g h t . A f t e r the k x k matrix has been formed the f i r s t column w i l l be c a l l e d the f i r s t t r a n s p o s i t i o n column. This column i s taken as an i n i -t i a l i z i n g row for a second matrix formed i n the same manner, and hence the f i r s t column of t h i s matrix i s c a l l e d the second t r a n s p o s i t i o n column. Waksman shows that i f g^ such matrices are formed the g^th t r a n s p o s i t i o n column w i l l be the set {a.}. Therefore the f i r s t row of the g, th matr ix , X K. or the ( g , - l ) s t t r a n s p o s i t i o n column generates {a.} i n a manner c o r r e s -ponding to that of the generating f u n c t i o n . Hence the (g^- l )s t t r a n s -p o s i t i o n column may be taken as {b_^ }. A numerical example may make t h i s process somewhat c l e a r e r * Suppose we are fo generate {021} c o n t i n u a l l y , so k=3 and Then wr i t e 0 2 1 and fol low formula (1.4) 3-1=2 times 5. (1) 0 2 1 2 0 1 0 2 2 i s the f i r s t t ransposition column 2 1 1 0 2 0 i s the second transposition column and so 0,2,0 = b. ' ' l To check that {0,2,0} w i l l indeed produce {0,2,1} we s h a l l form the or t h i r d matrix. (3) 0 2 0 2 2 0 1 2 0 We see then, that the desired sequence {0,2,1} i s produced. To i l l u s t r a t e the previously mentioned fac t that i f t h i s process i s continued the e n t i r e matrix w i l l be repeated, and so the g^th transposition column w i l l be re-peated we s h a l l continue a few more time steps. • 1 2 0 0 2 0 2 2 0 1 2 0 0 2 0 2 2 0 1 2 0 02.2 0 2 2 0 (2) 0 2 2 2 1 2 0 0 2 6. We s h a l l now s t a r t the generating funct ion with {b^}={0,2,0} TABLE 2 0 1 2 3 4 5 6 7 8 9 10 0 P 0 2 0 Q Q 1 0 P 2 2 0 Q 2 0 2 P 1 2 0 Q - - - -3 0 2 1 P 0 2 0 Q - - -4 0 2 1 0 P 2 2 0 Q - -5 0 2 1 0 2 P 1 2 0 Q -6 0 2 1 0 2 1 P 0 2 0 Q 7 0 2 1 0 2 1 0 P 2 2 0 A f t e r time step 3 we note that the des ired sequence has been generated once and that the "generating bud" i s the same as i t was when i n i t i a l i z e d . Thus we see that the process w i l l work. An important cons iderat ion i n judging the merit of such a scheme i s the number of states r e q u i r e d . Waksman requires the fo l lowing s ta tes : 1 Q the quiescent s tate 1 P the t r a n s i t i o n s tate 2.g^ the integers from 0 to (g^-1) and a lso a f l i p f lop to i n d i c a t e whether a c e l l has entered i t s terminal s tate or not . This gives a t o t a l of 2(g^+l) s ta te s , that i s about twice as many states as output charac ters . Another method of cont inua l generation of se-quences w i l l be introduced i n Chapter 4 and at that time i t w i l l be useful to compare the number of states required for the present method and the one introduced at that time. II A More Direct Method of Determining {b_^ } Once {b_^ } has been determined and the generating function has been i n i t i a l i z e d the generation of a^'s is straight forward. The process occurs as fast as can be expected for such a structure; that i s , one new output c e l l per time step. The generation of {b^} by Waksman's method i s , however, very tedious, expecially i f g^ is f a i r l y large. Fortunately i t turns out that the method is somewhat inefficient and that {b_^ } may be determined directly, rather than by iterative procedures, from {a_^ }. Consider the following discussion: Recall equation 1.2 i . , a. = S (T~:) b . . 1 • 1 J-1 3 J=l J J Rewriting this in vector-matrix notation, we have: a l i o 1 (2.1) or a = A b We want to express b in terms of a. 2.1 to yield This may be done by rewriting equation 8. Our problem i s now simply to i n v e r t A . We s h a l l show that: A " 1 = A 1 = 1 0 ... -1 1 1 1 u j : i ) ( - D 1 + j } ( - l ) k + 1 k ( - l ) k + 2 -1 0 k ( - l ) 2 k _ 1 1 To express t h i s simply i n words, A has the same elements as A , however the ( k , j ) t h element has the s ign of (-1)"''+ .^ A more use fu l way of w r i t i n g t h i s r e l a t i o n would be as Theorem 2.1. Theorem 2.1 b . = E d " h ( - l ) i + J a , . 1 j-1 j To prove t h i s theorem we need the fo l lowing lemma: Lemma 2.11 m=j l E ( i - l ) ! (m-1) ! m=j (m- l ) ! ( i -m) ! ( j - l ) ! ( m - j ) ! ( - D m 9. ( i - l ) - ( j - l ) = i - j (i-j)-(m-j)=i-m l e t k = m-j h = i - j = <-l)j (J" 1) S ( h ( - D k  J " i k=0 K By expanding (1-1) we note that h . 2 C ) ( - l ) = 0 k=0 K and hence QED We are now ready to prove the theorem. Proof of Theorem 2.1 f 1-1 (. n) for i=l,2,...,k j=l,2,...,l Let a. . ="S 0 for 1=1,2,...,k j=l+l,...,k This defines the matrix A = {a..}. S i m i l a r l y we define f / i - l w _ i N i + : J <;_?<-i>-a'.. H l — l y 2 j • i«jk 2™* 1*^*1 > • • • >k and so define A v = {a1;.} 13 10. Then from the d e f i n i t i o n of {b }. i n equation 1.2, we have shown that a = A b . What we are to prove i s that b . = Z d " h ( - l ) i + j a . 1 j-1 2 2 or that b = A ' a . That means A - 1 = A ' or that AA'=I. Let AA' = { « . . } k then « . . = Z a. a ' . i i i im mi J m=l J We s h a l l show that <*.. =1 and that « . . = 0 i f i ; ^ j , by i i i j cons ider ing the three cases ( i ) i < j ( i i ) 1 = j ( i i i ) i > j ( i ) i < j A i s t r i a n g u l a r , therefore a. =0 for m = i + l , . . . , k • im A 1 i s t r i a n g u l a r , therefore a ^ = 0 for m=l,2 , . . . , j-1 Therefore a. a ' . = 0 for m=l,2 (j-1) im mj and for m=(i+1) , . . . ,k and as i < j a. a . =-0 for m=l,2, . . . ,k J im mj ' ' ' Hence « . . = 0 for i < j . ( i i ) i = j Again we have a. =0 for m = i + l , . . . , k im a ' . = 0 for m=l,2,...(j-1) 11, but as i = j we have a. a ' . = 0 for m ^ i im mj therefore . = a . , a ! . 1 1 i i i i but a . . - ( ? - b - l i i l - l a i i = ( i : i ) ( - D 2 i = i therefore <=. . = 1 f o r i = j ( i i i ) i > j again a. =0 for m=(i+1) , . . . ,k im a ' . = 0 for m = l , . . . , ( j - l ) therefore = 0 for m = l , . . . , j - l ' m m - ' and m= (i+1) , . . . ,k Hence « . . = E a. a . = E a. a . i i , ". im mi . im mi-J m=l J m=j • • m=j m _ 1 2 ~ 1 ( - D d E c b e ' i ^ - 1 ) 1 1 1 m—1 i-1 m=j 0 by lemma 2.11 Therefore . = 0 f or i > j . u j-l i f i = j Hence . . = { n i j 0 otherwise So AA' = I or A' = A - 1 thus b . = E ( T " b ( - l ) i + j a . 1 J - l J " 1 J QED 12. R e c a l l i n g the example at the end of chapter I we s h a l l now r e c a l c u l a t e {b.} given {a.} = {0,2,1} k=3 Su=^ By theorem 2.1 h - l ( J i i x - i ) 1 * . , 3 —1 so b l * a l - ° b 2 = a 2 - & 1 = 2 b 3 = a 3 - 2 a 2 + a 1 = 1 - 2(2) + 0 = 0 (mod 3) Hence we have {b^} = {0,2,0} which agrees with the value c a l c u l a t e d by Waksman's method. I t i s easy to see that when k i s f a i r l y large Waksman's method involves a great deal of c a l c u l a t i o n — (g, - l ) k ( k - l ) a d d i t i o n s . The method which we have j u s t developed i s k k much more d i r e c t , r e q u i r i n g l ess than E 2 ( i - l ) ar i thmet ic operat ions . i = l That i s less than 2* 1 * k = k ( k - l ) - ar i thmet ic operat ions . The method i s b e t t e r by a fac tor g^. For the e n t i r e process to have any meaning g^ must be at l eas t 2. I l l The Value of ^ g.^ i s e s s e n t i a l l y the number of characters permitted i n the alphabet over which the s t r i n g i s generated by the Waksman technique, s ince there are g^ elements i n ^g^* Waksman says nothing more about the s i ze of the alphabet , from which the characters to be generated may be chosen, other than to define g, as g .c .d{ ( k ) } i = l , 2 , . . . , k - l . As i t turns out , upon c l o s e r rC X 13. i n v e s t i g a t i o n , g^ i s 1 unless k i s a prime or a power of a prime. To put i t even more s imply , g^ i s i n general 1 and so the Waksman technique of generating sequences c o n t i n u a l l y i s meaningless except i n s p e c i a l cases of k. For to generate a s t r i n g of any length with only one character c o n t i n u a l l y i s merely to generate t h i s one character c o n t i n u a l l y . To prove the r e s u l t we have j u s t s tated we must f i r s t prove severa l lemmas. F i r s t , define Np'(x) as the number of times the prime p i s a f a c t o r of x. Hence N2(12)=2, N,-(17)=0 et c e t era , I t i s qui te c l e a r N p (xq)=N p (x)+N p (y) when x#), y#). Lemma 3.11 N (i+kp e)=N ( i ) for 1=1,2 ( p 3 - l ) P P or i f i = p 3 and p ^(k+l) . Proof Let i = mp p^m then 0<_a<3 so N (i+kp 3 ) = N ( p a ( m + k p e " ° C ) ) = <*+N (m+kp3"") P P P But as pVi» P^On+kp^ ) then N p ( i+kp 3 ) = cc = N p ( i ) for 1=1,2 (pS-l) I f on the other hand, i - p 3 and p. ^(k+1) we have N (i+kp B ) = N ( (k+l )p e ) = 3 + N (k+1) P P P But , as J?Vk+l) N ( i+kp 3 ) = 6 = N ( i ) P P QED 14. Lemma 3.12 cc N (ps!) = Z p j _ 1  P J-1 Proof: We s h a l l prove t h i s lemma by induc t ion on F i r s t , i t i s obvious ly true for «=0 and f o r r : a = l . We s h a l l assume the lemma i s true for " ^ B and prove i t for <*=8+l. Hence by induct ion the lemma w i l l be t r u e . 8 3 i — 1 8+1 Assuming N (p !) = E (p- ) wr i te out (p 1) i n f u l l and P j - 1 d i v i d e i t in to p sect ions as shown. p ^ + D , . 1 f 2 . m m t P ^ I (p.B+l) . . . 2 p 6 | . . . | ( ( p ' - l ) p e + l ) ' . . . p -Now by lemma 3.11 we know that N p ( i+kp B ) = N p ( i ) for i = l , 2 , . . . , p e when k = 0 , 1 , . . . <p -2) and a lso for i = l , 2 , . . . (p --1) when k=p- l . - 8 Furthermore when i=p and k=p-l N (i+kp B ) = N ( ( p - l + l ) p B ) = N ( p B + 1 ) = 8+1 = N(p P ) +1 P P P Thus p i s a fac tor of each of the sect ions shown, except the l a s t , the same number of times; and i s a f a c t o r of the l a s t one more time than of g the o thers . However, the f i r s t s ec t ion i s p ! , and so p i s a fac tor 3 1-1 E (p J ) t imes. j = l Thus N ( p ( 3 + 1 ) ! ) = P E ( p J _ 1 ) + 1 P j - 1 15. This lemma may be extended to give the value of N p ( r ! ) where r i s not n e c e s s a r i l y a power of p. Lemma 3.13 N ( (p 'x) ! ) = xN (p"!) + N (x!) P P P Proof : C l e a r l y the lemma i s true for x=l . We s h a l l therefore assume the lemma i s true for x=y and prove i t for x=y+l. Hence by induct ion the lemma w i l l be t r u e . cAy+D)! = (P°V)! • (i+P°V)- ... -(y+D" = (P°V)I * n (i+P°V) 1=1 By lemma 3.11 N p (i+p°V) = N p ( i ) for i = l , 2 ( p ' - l ) and N p (p^+p^y) = <* + N p(y+1) = N p ( p ° ) + N p(y+1) then N ( ( p ° ( y + l ) ! ) = N ((p°V)!) + N (p*!) + N (y+1) P P P P (using the i n d u c t i o n assumption) - y N p ( p a ! ) + N p ( y ! ) + N (p*!) + N (y+1) = (y+1) N p ( p a ! ) + -N ((y+1)!) [noting that N ((y+1)!) = N (y!) + N (y+1)] P P P QED Lemma 3.14 I f k = p x, where p 1 x, x > 1 and p i s prime, then P ^ g, . \ 16. Proof ; g k = g.c.d{(J)} 1=1,2 (k-1) . To show p ^ g^ we need only f i n d one i such that p \ (^). • cc Consider the case i n which i = p . P a x Let us evaluate N (( )) P P Since ( ? c c x ) - (p g x! ) V P « ! (( x - l ) p « ) ! ' . C we have N (<f- X )) - N (( p ' x ) ! ) - N (p*!) - N ( ( ( x - l ) p ! ) tr ^ r Jr tr but then " 1-1 N ((p x)! ) = x E p + N (x!) P J - l ? OC N (p^!) = E p i"" 1  P J - l C and N ( ( (x - l )p") ! ) = (x-1) E p j - 1 + N ((x-1)!) P J - l P from lemmas 3.12 and 3,13. Then N ( ( ? « X ) ) = x E p J - 1 + N (x!) P V j = 1 P - E p j - 1 - (x-1) E p ^ " 1 - N ((x-1)!) J - l j - l P N (x!) - N ((x-1)!) + (x-1 - (x -1) ) E p i " 1  P P J - l = N (x!) - N ((x-1)!) P P N p (x ) = 0 as p | x Hence p ^ g k QED 17. With the proof of t h i s lemma the des ired r e s u l t fol lows e a s i l y . Theorem 3.1 k I f k ^ p , g, = 1, or the elements of {(.)} i = l , 2 , . . . ( k - 1 ) are r e l a t i v e l y prime. Proof : k k k = C { ( i ) } » hence g f c | k. Then for every prime p which d iv ides k, apply lemma 3.14. Hence, no f a c t o r of k i s a fac tor of g^.. Then, s ince g^ | k , g^ = 1. QED cc Let us now consider the case i n which k = p and determine the s i z e of the alphabet permitted i n generating c o n t i n u a l l y a s t r i n g of length k by Waksman's method. We can see by the fo l lowing Theorem that i n t h i s case g^ = p. Theorem 3.2 I f k = p , where p i s prime and « >_ 1, then g^ = p. Proof : By the d e f i n i t i o n of g^, i t must be a fac tor of k. Hence, i n cc t h i s case, g^ = p , where 0 <_ 3 <_ We s h a l l show that g^ = p by showing f i r s t that 3 >_ 1 and secondly that 3 <_ 1. 1) We are to show 3 >_ 1, that i s , ,p | g^. Consider f i r s t the term ( P a - i ) * ( p a - i + l ) . . . (P^-l) which i s of the form of the product of i consecutive integers 18. d iv ided by i ! . By Theorem 74 of Hardy and Wright (1945), i ! d iv ides the product of any i consecutive p o s i t i v e i n t e g e r s . Hence cc cc N ( (p - i ) ( p -i+1) . . . (p -1) ) P - i ! i s defined and 0. cc cc I f we replace (p - i ) , i n the term, by p , c l e a r l y the cc < cc value of N w i l l be increased as N (p - i ) < N (p ) for P P P 1 < i < p". Then N ((?")) - N ( - ( p " - i + l ) - : ; - - ( p ^ l ) ' p " ) p i p l ! > N ( (p'-D (p'-l+D ' ( p ' - l ) \ p i ! > 0. Thus N ( ( ? ) ) > 1. P 1 QED We s h a l l now show that 6 <_ 1 and so that g^ = p. Consider now the case i n which i = p , 1 and so the member p (P ) = p !  (P ) ! (p - P ) ! Therefore , N [ ( P , ) ] - N (p " l ) - N (P"" 1 !) - N ((p"- P"" 1 ) ! ) P p ^ - l P P P 19. j - 1 where N (P !) = Z P P J - 1 cc_ l N ( p " _ 1 ! ) = Z P j - 1  P J - 1 N ((P."- p" X ) ! ) - N ( ( P ™ - 1 (P - l ) ) !) P P cc—2. cc = (P- l ) Z P j _ 1 + N (P- l ) = (P- l ) Z p j - 1 j - 1 P j - 1 as P ^ (P- l ) Therefore . cc a cc— 2_ cc—J_ N [ ( P )] = Z P j _ 1 - Z P j _ 1 - (P- l ) Z P j _ 1  P P " 1 j - 1 j = l j = l « - l . , = l + ( P - l -(P-D) s p J _ 1 j - 1 = 1 cc Hence = P where k = P QED We can now see the f u l l impl i ca t ions of Waksman's method of c o n t i n u a l r e p l i c a t i o n of s t r i n g s . I f the length of the des ired s t r i n g , k, i s a prime, P , or a power of P; then the alphabet from which the elements of k may be chosen i s b i j e c t i v e to Z^. Otherwise the alphabet cons i s t s of a s i n g l e character and hence no meaningful s t r i n g can be generated. As an example, l e t us generate the s t r i n g (120222101). We note that t h i s can be done using Waksman's technique as k=9 hence 8^-3. The c a l c u l a t i o n of {b.} i s c a r r i e d out i n Table 4. TABLE 3 Pasca l ' s T r i a n g l e The b inomial c o e f f i c i e n t s ("!") j -*• 0 1 2 3 k 5 6 7 0 1 1 1 1 2 1 2 1 3 1 3 3 1 1 4 6 4 1 5 1 5 10 10 5 1 6 1 6 15 20 15 6 1 7 1 7 21 35 35 21 7 1 8 1 8 28 56 70 56 28 8 21. TABLE 4 C a l c u l a t i o n of {b.} 1 Refer to TABLE 3 for C^""1) {a±} = {1 ,2 ,0 ,2 ,2 ,2 ,1 ,0 ,1} b-i = ~E ) ( - 1 ) 1 + ? a. ar i thmet ic i s mod;g, 1 j = i J x J ° k b l = 1 b 2 = 2 - 1 = 1 b 3 = 0 -2(2) + 1 = 0 b,4 l = 2 -3(0) + 3(2) -1 = 1 b 5 = 2 -4(2) + 6(0) - 4(2) +1 = 2 b,, = 2 -5(2) + 10(2) - 10(0) + 5(2) - 1 = 0 b b ? = 1 -6(2) + 15(2) - 20(2) + 15(0) - 6(2) + 1 = 1 b g ; = 0 -7(1) + 21(2) - 35(2) + 35(2) - 21(0) + 7(2) - 1 = 0 = 1 -8(0) + 28(1) - 56(2) + 70(2) - 56(2) + 28(0) - 8(2) + 1 = 2 With (b^} determined we may proceed with the cont inua l generation of the sequence under the ru le s of TABLE 1. 22. TABLE 5 Generation of the sequence (120222101) c e l l i -»• 0 1 2 3 4 5 6 7 -8 9 <!° 11 12 13 14 15 16 17 18 19 20 0 P 1 1 0 1 2 0 1 0 2 Q Q Q 1 1 P 2 1 1 0 2 1 1 2 2 Q Q 2 1 2 P 0 2 1 2 0 2 0 1 2 Q Q 3 1 2 0 P 2 0 0 2 2 2 1 0 2 Q Q 1 2 0 2 P 2 0 2 1 1 0 1 2 2 Q Q 5 1 2 0 2 2 P 2 2 0 2 1 1 0 1 2 Q Q - - — ' -6 1 2 0 2 2 2 P 1 2 2 0 2 1 1 0 2 Q Q - - -7 1 2 0 2 2 2 1 P 0 1 2 2 0 2 1 2 2 Q Q - -8 1. 2 0 2 2 2 1 0 P 1 0 1 2 2 0 0 1 2 Q Q -9 1 2 0 2 2 2 1 0 1 P 1 1 0 1 2 0 1 0 2 Q Q 10 1 2 0 2 2 2 1 0 1 1 P 2 1 1 0 2 1 1 2 2 Q A f t e r the 9th time step we not ice that the pa t t ern a^ has been generated once and that the i n i t i a l conf igura t ion b_^  i s now i n c e l l s 10-18. At t h i s point we see that the process w i l l c l e a r l y generate the des i red sequence c o n t i n u a l l y . IV An A l t e r n a t i v e Method of Cont inual R e p l i c a t i o n of L i n e a r Patterns - The Wheel Algor i thm I t i s c l e a r that Waksman's method i s very r e s t r i c t i v e with regards to the alphabet permitted i n cont inua l generation of a sequence 23. of a r b i t r a r y l ength . It i s a lso qui te c l ear that no simple modi f i ca t ion of h i s "modulo ar i thmet ic scheme" can be general ized to any s i g n i f i c a n t degree. For t h i s reason we now turn to a more general method of r e p l i -2 c a t i o n . This method, i n i t s elementary form, requires 0(n ) s ta te s , however, ra ther than 0(n) as d i d Waksman's method, where n i s the s i ze of the alphabet from which the characters are chosen. This method may be modified somewhat so that the number of states required i s about n + 4 log2 n and so l e ss than the 2 n+2 required i n Waksman's method for reasonably large n . The idea behind t h i s general scheme i s qui te s imple . Write the s ta te of each c e l l as a p a i r of elements of the des ired output 3. C e alphabet — say [^ ] [^ ] [^]. Then th ink of t h i s s i x - t u p l e of states as p o s i t i o n s on a wheel r o l l i n g to the r i g h t , and so c lockwise . Hence the a lgor i thm may be r e f e r r e d to as the wheel a lgor i thm. The next ID EL C p o s i t i o n of the wheel w i l l be [^ ] [^ ] [ ]• Now suppose b i s l e f t i n the p o s i t i o n ( i n i t i a l l y occupied by [^], that i s , the wheel leaves a track behind i t as i t r o l l s , the image of the bottom part which was l a s t i n that p o s i t i o n . a c e Using [^ ] [^ ] [^ ] as an i n i t i a l conf igura t ion we have TABLE 6. TABLE 6 Q # [ f ] * Q b [*] [ f l [°J Q b d [df] [\] [ a] Q b d f [ f ] [ d] [ b] Q e c a 24. From t h i s i t can be seen the sequence b , d , f , e , c , a w i l l be generated c o n t i n u a l l y . Thus to generate a , b , c , d , e , f the appropriate i n i t i a l conf igurat ion would be obtained by w r i t i n g t h i s sequence i n a f e d counter-c lockwise manner around the wheel — [ ] [ , ] [ ] • I t may be a b c noted that t h i s method i s d i r e c t l y a p p l i c a b l e to sequences of even l ength . I f a sequence of odd length i s to be r e p l i c a t e d i t may be w r i t t e n twice and considered a sequence of even l ength . Thus to generate c o n t i n u a l l y c b a [abcj the appropriate i n i t i a l conf igura t ion would be [ ] [ , ] [ ] . a. D c The general r u l e for determining the s tate of a c e l l at time t+1 i s that the upper h a l f of the c e l l moves two pos i t i ons to the r i g h t and the lower h a l f stays i n p o s i t i o n , except at the ends. The lower h a l f of the le f t -most p a i r stays i n p o s i t i o n , but a l so moves to occupy the top h a l f of the c e l l to i t s r i g h t , which then becomes the leftmost c e l l . The top h a l f of the r ight-most occupied c e l l does not move two pos i t ions to the r i g h t , but moves to occupy the bottom h a l f of the c e l l immediately to i t s r i g h t which was prev ious ly i n the quiescent s t a t e , but now becomes the right-most occupied c e l l . This process v i o l a t e s one of the ru les which Waksman had s ta ted . That i s wi th the wheel a lgor i thm the s tate of c e l l i at time t+1 w i l l depend, i n genera l , on the s tate of c e l l i - 2 (the top h a l f of the c e l l moves two places to the r i g h t ) at time t . To modify the scheme and avoid t h i s problem requires the i n t r o d u c t i o n of a time f l i p - f l o p and so e s s e n t i a l l y doubl ing the number of s ta t e s . This f l i p - f l o p w i l l a l t e r n a t e , being 0 at even time steps and 1 at odd t imes. The upper h a l f of the p a i r representat ion of a c e l l w i l l move 1 p o s i t i o n to the r i g h t at each 25. time s tep . The ends w i l l be handled by appropriate means depending on t ime. TABLE 7 formal ly describes the func t ion . TABLE 7 a , b , c 0 -Q -[ a] -L b J Present State L e f t Neighbour output characters an a r b i t r a r y character quiescent s tate generating p a i r s $ - — - - no c e l l present X - _ any character Right Neighbour X(t=l) X(t=0) [ 6] L b J 'Si g b $ b -a X $ a b a Q Q a a t=l ; .Q t=G 26. we have: V c t=0 a b Q 0 0 t= l a b c Q w \\ t=2 a V a b c Q loj loj t=3 a fo] a b c b W w t=4 b f -\ a a b c b 0 To i l l u s t r a t e t h i s , consider the fo l lowing example: It i s des ired to generate c o n t i n u a l l y the pat tern febcbj. Then I n i t i a l conf igurat ion (the bottom symbol i s a time f l i p - f l o p ) t=5 t-6 t=7 w 0" a b c b a w 1 f ' c b a l'/b c b a Loj loj t i c f \ 0 a b c b a b w w Note that any s ta te could be subst i tu ted for the dummy state 0 One s l i g h t drawback to t h i s method i s that 2 time steps are required to generate each new output charac ter . The number of states 2 requ ired can be seen to be 0(n ) as: states are required for a l l poss ib le p a i r s of outputs and time f l i p - f l o p . n output s tates 1 Q, the quiescent s tate 2 2n +n+l s tates i s the t o t a l number r e q u i r e d . I t should be noted that the number of states required i s t o t a l l y independent of the length of the s t r i n g to be r e p l i c a t e d . The next quest ion to be asked i s can the number of states r e -2 quired be reduced from 0(n ) to 0(n) , perhaps at the expense of the time required for generation of each output character? The b a s i c schemexof t h i s a lgor i thm i s to represent each output s tate as a b inary number and so only p a i r s of b inary d i g i t s are manipu-l a t e d . This mapping i s an a r b i t r a r y b i j e c t i o n from the output alphabet to ]L^. As i t takes l og n* b i t s to represent uniquely the integers of a time counter running from 0 to l og n w i l l be required as w e l l . Hence 4 ( log n) + 1 s tates are required to represent t h i s . The n output states and the quiescent s ta te are a l so r e q u i r e d . This gives a t o t a l of n + 4 ( log n) + 5 s tates which are needed for th i s method. Let F denote the mapping from the output alphabet to Z^. For example, i f the output alphabet i s ( a , b , c ) , then n=3 and we could define F(a)=0, F(b)=l ,F(c)=2 . TABLE 8 ind ica te s formal ly the workings of the b inary wheel a lgor i thm. * log n w i l l be used to denote the l eas t integer >_log- n . 28. TABLE 8 a , b , c b inary d i g i t s i f i n p a i r s , output s tates i f alone Q quiescent s tate 0 0 s tate X any a r b i t r a r y c e l l s tate At each step the time counter i s incremented by 1 modulo ( ( log N) +1). Output s tates are reached at time (t+1). Right Neighbour L e f t Neighbour 1P Q I f l ft ft $,g b b Q Q a X a Q [ a] L b J c Q ft c 0,b a t=l ,2 ( ( log n ) - l ) ft ft Q ft ft [ ° i g 2 t" 2 d — . intege . < : F(g) 4. O I f r i g h t neighbour i s Q then a=0 i s the only pos s ib l e value during these i n t e r -mediate time s teps . t=log n X X $,b 29. ft [ F _ 1 ( F ( a ) + 2 t " 1 . d ) ] ft ft Q ft ft ft g ft -ft As an example, consider the generation of the sequenceC02l|> where n=3 and so l og n=2. Then 0 = 00 2 1 = o i 2 2 = 10 nI n i t i a l i z e at time 0 w r i t i n g 021 i n b inary around the wheel i n a counter clockwise manner but with the order of b i t s of each characte i n v e r t e d . Since the product of l og n and the length of the sequence i s even only one copy of the b inary representat ion of the sequence i s r e -q u i r e d . Thus: t=0 0 0 0 0 This represents 0 2 30. t=l f *s 1 f°l 0 0 0 1 llj w llj t=2 fo] fo" fo' 0 0 1 1 UJ t=3 fo' fo] fo] 0 0 1 1 loj 0 V J l ° J t=4 fo] fo] fo] 0 0 1 1 0 w 1 V J t=5 fol 0 fo' 0 2 1 0 0 Uj w UJ t=6 fl] fo] 0 2 1 0 0 10J loj loj t=7 f f f f r \ 0 0 2 1 0 0 0 w llj l l j t=8 1 1 f \ 0 0 2 1 0 0 0 UJ 12J 12J t=9 fo] fl] f l] 0 2 1 0 0 0 loj loj l°J At t h i s point the s t r i n g has been completely generated once and the generating bud sec t ion ( c e l l s 4 ,5,6) i s the same as the i n i t i a l con-f i g u r a t i o n of the o r i g i n a l generating bud sec t ion ( c e l l s 1,2,3 at t=0). 31. A reminder on time considerat ions would appear to be i n order . Waksman's method i s undoubtedly the fas tes t of the three methods discussed as one c e l l i s generated at each time s tep . The simple wheel a lgor i thm requires two time steps for each symbol and the b inary wheel i s the s lowest , -requiring ( ( log n) +1) time steps for each character to be generated. The advantage of the wheel and b inary wheel algorithms i s that the length of the des ired sequence has no bear ing on the s i ze of alphabet which may be used. In fact the s i z e of alphabet and length of s t r i n g are both completely a r b i t r a r y . For a given s t r i n g length the s i z e of alphabet i s determined automat ica l ly i n Waksman's method. Furthermore, i t i s only i n s p e c i a l cases that the s i z e of t h i s a l p h a b e t , i s not 1. The advantage of the b inary wheel a lgor i thm over the simple wheel a lgor i thm i s the d r a s t i c reduct ion i n the number of s tates r e q u i r e d . V Extension of the Wheel Algor i thm to the 2-Dimensional Case. The next step i n the preceding l i n e of thought i s c l e a r l y to generate rec tangu lar , rather than j u s t l i n e a r pat terns . This can be approached i n much the same manner as the l i n e a r case. The problem can now be formulated i n the fo l lowing manner; Define a funct ion on a 2-dimensional array of i d e n t i c a l f i n i t e s tate automata, such that the s tate of c e l l ( i , j ) at time (t+1) i s a func t ion of the states of that c e l l and i t s four immediate neighbours t ( i , j ) , ( i - l , j ) , ( i + l , j ) , ( i , j - l ) , ( i , j + l ) ] and so that an a r b i t r a r y predetermined rectangular pat tern of c e l l s tates w i l l be generated 32. c o n t i n u a l l y throughout the space. This problem may be solved by apply ing the wheel a lgor i thm twice . F i r s t the a lgor i thm i s used, moving i n the h o r i z o n t a l d i r e c t i o n , to produce generators s i m i l a r to those used i n the one dimensional case. These then generate the pat tern v e r t i c a l l y . The generators i n the one dimensional case are p a i r s of output characters together with a time f l i p - f l o p . In the two dimensional case we s h a l l s t a r t with p a i r s of p a i r s and a time f l i p - f l o p . The f i r s t a p p l i c a t i o n of the wheel a lgori thm generates p a i r s of charac ters , and a l so sets a time f l i p - f l o p to zero . Consider the general case of rectangular pat tern r e p l i c a t i o n , that i s , generate a i a . ~ . . . . . . a m, l m, z m,n a 2 , l a, .. . . -a- „ • a, c o n t i n u a l l y . 1,1 1,2 l , n I t should be noted that c e l l s are numbered as points i n the f i r s t quadrant of the Cartes ian plane and not as' matrix elements. Hence a.. 1 i s i n the lower l e f t hand corner , i » i The f i r s t problem i s to decide what the i n i t i a l conf igura t ion should be. To generate the columns i n the upward d i r e c t i o n , we require that at some time column j + kn k = 0 , l , 2 , . . . , j = l , 2 , . . . , n -be of the 33. form 1.3 a t m » 3 j 2,1 a -t • m-i+1, j a. . I 1»J a„ . I 2>3 J m>3 a. . I x>3j with a l l c e l l s above c e l l (m,j) i n the quiescent s t a t e . Once t h i s conf igura t ion i s reached a s tra ight forward a p p l i c a t i o n of the wheel a lgor i thm w i l l generate the (j + kn)th column i n the proper manner. Therefore the generation process on the i t h row i=l,2,..., should y i e l d as output m-i+1,1 a. m-;+i,2 i a i,2 m-i+1, j a. . I i»3 m-f+l,n a. repeated ly . Thus the i n i t i a l conf igurat ion becomes ev ident . C e l l ( i 34. i = l , 2 , . . . , m , j = l , 2 , . , w i l l be i n s tate m-i+1, n-j+1 [ a i , n - j + l -m-i+1,j a. . I 1»3 0 (5.1) i n i t i a l l y . I t may be noted that i n c e r t a i n cases only part of t h i s i n i t i a l c o n f i g u r a t i o n must be present , although the i n c l u s i o n of the e n t i r e con-f i g u r a t i o n as s ta ted above w i l l c e r t a i n l y produce the correc t r e s u l t . I t may be noted that i f m i s even the f i r s t — rows are i d e n t i c a l to rows y + 1 to m; and hence, only the f i r s t — rows need be i n i t i a l i z e d . S i m i -l a r l y i f n i s even only the f i r s t — columns need be i n i t i a l i z e d . To i l l u s t r a t e t h i s process more f u l l y l e t us consider the c o n t i n u a l generation of the pat tern a b c d e f g h i j k m The time f l i p - f l o p w i l l be seen to be 0 at even times and 1 at odd t i m e s . f o r h o r i z o n t a l generation and reversed for v e r t i c a l generat ion. 35. The i n i t i a l conf igura t ion i s given by (5.1)aas t=0 Q Q ft ' f t ' ft ft Q 0 * ft' ' f t ' ft ft Q 0 . o . ft' ft ft Q 0 0 The development of the pa t t ern may be c a l c u l a t e d by fo l lowing a tab le s i m i l a r to TABLE 7, but i n which we consider p a i r s of the form [^ ] as one symbol and produce output symbols of the form ft 0 Af ter these symbols are produced we continue with them as i n TABLE 7, reading lower neighbour for l e f t neighbour, and upper neighbour for r i g h t ne igh-bour. Thus the des ired output symbols are propagated i n an upward d i r e c t i o n . 36. t=l t=2 ft ft ft ft ' ft ft [ [°] 1 ft c Q r-H 1 l h J \ ft} ft Q m [ c ] Q 1 Q Q Q ft ft ft ft ft 0 ft ft ft 0 ft ft ft 0 ft m [°] 0 « 37. The generation of the pat tern ean be seen c l e a r l y from t h i s p o i n t . The general form of the pat tern whi le being generated can be seen at time t=12. t=12 column 1 2 row 10 tt tt Q Q tt 1-tt l tt 1 f Q Q t°] tt tt 1 k Q Q tt tt tt m 1 d m 5 Q Q 6 Q Q tt Q 1 [*] [tt 1 tt [tt 1 l tt .7 8 9 Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q tt tt Q tt tt 0 ' t t 1 ' t t ' Q tt I ° J ' t t ' [ i ] Q L k J 10, { o J 38. This a lgor i thm works as qu ick ly as can be expected. The length of each column i s increased by one every second time step and the number of columns conta in ing f i n a l output symbols i s increased every second time s tep . However the number of states required for th i s opera-t i o n i s f a i r l y large s ince we are forced to deal with quadruples of out-put s tates i n the i n i t i a l generation process . The a c t u a l number of s tates may be c a l c u l a t e d as fo l lows: 1 quiescent s tate 4 2n a l l p o s s i b l e quadruples of output s tates i n each of the two poss ib le time pos i t ions 2 2n a l l pos s ib l e p a i r s i n each time p o s i t i o n n output s tates ° n 4 2 2n +2n +n+l C l e a r l y th i s i s qu i te an undes irable number of s tates to r e -quire e s p e c i a l l y when n i s qu i te l a r g e . A t • t h i s point we may look back to the one dimensional case and r e c a l l that the corresponding problem was solved by mapping b i j e c t i v e l y the n output charac ters , i n an a r b i t r a r y manner, onto the r i n g of integers modulo n , that i s Z^. Then only b i t s need be manipulated u n t i l the ac tua l character generating time s tep . There must however, be a time counter running from 0 to l og n assoc iated with the p a i r s of b i t s and a time f l i p - f l o p , associated with the quadruples. Therefore the number of states for t h i s method may be determined as fol lows 1 quiescent s ta te 4 2:2 poss ib le quadruples of b i t s and a time f l i p - f l o p 2 2 . ( ( l o g n)+l pos s ib l e p a i r s with time counter n output states n+4 log n+36 states are required 39. As i n the one dimensional case, the b inary representat ion has two drawbacks. F i r s t , i t i s more d i f f i c u l t to wr i t e down the required i n i t i a l c o n f i g u r a t i o n , which occupies log n times as many c e l l s i n each d i r e c t i o n as does the non-binary form, and secondly, the output characters are produced only once i n every ( log n) + 1 time steps i n each column. In genera l , however, the column generation processes w i l l be out of phase with each other due to the fact that column generation w i l l begin on a new column at every other time s tep . Symbols are produced i n a " t r i a n g u l a r " form as i n the non-binary case. Hence the rate of product ion of a new output ce l l s i s p r o p o r t i o n a l to the time t . A simple example of the use of the b inary wheel a lgor i thm i n two dimensions would probably make the workings of the general case much c l e a r e r . fa b Suppose we are to r e p l i c a t e c o n t i n u a l l y the square c ^ Then n=4, so log n=2. We can define a b i s e c t i o n F from the alphabet to by a «-»- 0 = 00 2 b -w 1 = 0 1 2 c 2 = 10 d -M- 3 = 1 1 2 Then using the non-binary method we could i n i t i a l i z e the Si process with c e l l (1,1) i n s tate [ ] 0 However, us ing the b inary technique the process i s not qui te as easy to i n i t i a l i z e . Let us f i r s t look at the sets of p a i r s which 40. must be generated by the quadruples i n order to generate the des ired p a t t e r n . For the sake of c l a r i t y , l e t us temporari ly abandon the funct ion F as defined and l e t F(a) = 2-a± + &2 F(b) = 2'b± + b 2 F(c) = 2'c± + c 2 F(d) = 2 - d 1 + d 2 where a , , a>2 9 • • • 3. ire e i ther 0 or 1. Then to generate the odd numbered columns, which have a {caca ...} form we must produce (row 2) (row 1) r \ a i C 2 0 as the elements i n the f i r s t two rows of these columns. S i m i l a r l y fb. (row 2) (row 1) ,0 , 0 must be generated i n the even numbered columns. Thus the output for the h o r i z o n t a l generator i n row 1 i s 41. a l C 2 10 J f \ a 2 C l and i n row 2 0 ,0 , fb. 0 must be generated c o n t i n u a l l y . Therefore the i n i t i a l conf igurat ion of row. 1 i s f and for row 2 a 2 , c l , 0 TABLE 10 traces the development of the pat tern through severa l time steps a f t er r e p l a c i n g a^, a ^ , . . . with t h e i r defined va lues . H o r i -zonta l propagation i n s t r a i g h t forward, v e r t i c a l propagation follows TABLE 8 i n Chapter IV . t=0 row (2) row(l) ft ft ' f t ' ft 0 t= l ft 0 ft-[°] 0 ft ft [°] ft ft 0 ' f t ' ft 0 42. TABLE 10 43. [°] [°] tt 0 tt 0 f0 1 [ ] L0 J [°] [°] f t 0 ] 1 tt' 0 1 V J ' t t ' t t ' 0 V. J c b tt 0 tt tt 0 [°] ft [°] tt ' t t ' f 0 •[°] 0 tt I : t t tt 44. From t h i s the general r e p l i c a t i o n pat tern may be seen. VI Extension to d-dimensional space The next step i n the development of t h i s theory i s the ex-tens ion to a r b i t r a r y (d) dimensional space. That i s , to def ine a funct ion on a d-dimensional array of i d e n t i c a l f i n i t e automata so tha t , given the appropriate f i n i t e i n i t i a l c o n f i g u r a t i o n , the e n t i r e p o s i t i v e region of d-space w i l l be f i l l e d with repeated images of an a r b i t r a r y predetermined d-dimensional hypercuboid . Again we have the condi t ion that the s tate of any c e l l at time (t+1) be a funct ion of the states of that c e l l and i t s 2^  immediate neighbours at time t . The technique used i s the obvious extension of that used i n the 2-dimensional case. The wheel a lgor i thm i s used on the dth l e v e l generators to form ( d - l ) s t l e v e l generators . These i n turn generate (d-2)nd l e v e l generators i n the same way that the v e r t i c a l or f i r s t l e v e l generators were produced by the h o r i z o n t a l , or second l e v e l generators i n the 2-dimensional case. In any case, a f t er d such t r a n s -formations the terminal or output s tates emerge. As i n the 2-dimensional case, when n i s large the number of s tates required becomes much l a r g e r . A c t u a l l y , as may be expected, when d i s large the number of states r e -quired becomes as tronomica l . Therefore , to keep the number of states w i t h i n reason s igna l s may be sent i n b i n a r y , as i n the 2-dimensional b inary wheel method. Only iiti the ac tua l generation of output symbols do non-binary forms have to be deal t w i th . Hence, i t should not be sur -p r i s i n g that the e f f ec t of dimension s i z e (d) and alphabet s i z e (n) upon 45. number of s tates required are t o t a l l y independent. That i s number states = F(d) + G(n) . Let us now determine the number of states required to generate c o n t i n u a l l y an a r b i t r a r y d-dimensional hypercuboid of elements of an alphabet of c a r d i n a l i t y n . F i r s t consider the non-binary representat ion . To apply the wheel a lgor i thm and move from 1 dimensional generators to the output s ta te we require the quiescent s ta t e , the n output states and 2 a l l pos s ib l e p a i r s of outputs together with a time f l i p - f l o p , or 2n s ta te s . To generate the p a i r s , quadruples are needed, and so on up to 2^ t u p l e s . Thus the number of s tates S(d,n) required w i l l be given by The quiescent s tate and n output s tates are required as are the 4 ( ( l og n)+l) states which are needed i n moving from binary p a i r s to output s t a t e s . At higher l e v e l s , however, the process i s a simple t r a n s f e r of 22 tup le s . In general the j t h l e v e l requires a l l poss ib le i 2 j 2 J tuples at both time p o s i t i o n s , or 2«,2 s ta t e s . Therefore the number of s tates required i n the d-dimensional a p p l i c a t i o n of the b inary wheel a lgor i thm (S, (d,n)) i s given by d S(d,n) = 1 + n + E n j - 1 (6.1) a large number for s u r p r i s i n g l y small d . In the b inary wheel representa t ion , the number i s much lower. S, (d,n) = 1 + n + 4 ( ( l og n)+l) + 2 E 2 b . „ J=2 d = n + 41og n + E 2 j = l - 3. (6.3) 46. As was mentioned be fore , t h i s can be viewed as the sum of a funct ion of n and a funct ion of d . VII Conclus ion The r e s t r i c t i o n s inherent i n Waksman's method of c o n t i n u a l r e p l i c a t i o n of a l i n e a r s t r i n g have been shown. I f the length of the C  des ired s t r i n g i s not of the form P where P i s a prime, only one character may be produced. This means that no meaningful s t r i n g may be generated. Furthermore i f the s tr ing length i s of the form P , only P output characters are permit ted . These r e s t r i c t i o n s cannot be overcome us ing a modulo a r i t h m e t i c a lgor i thm. For t h i s reason the wheel a lgor i thm was developed. Using t h i s a lgor i thm the number of characters i n the output alphabet i s completely independent of the length of the s t r i n g , and i n f a c t , both are a r b i t r a r y . The b inary wheel a lgor i thm was developed to reduce the number of s tates requ ired to produce a s t r i n g conta in ing n d i f f e r e n t characters to 0 (n) . I t was shown a l so that both the wheel a lgor i thm and the b inary wheel a lgor i thm can be general ized to produce c o n t i n u a l l y a d-dimensional hypercuboid . F i n a l l y , i t should again be noted that the number of s tates requ ired to generate patterns i n d-space us ing an alphabet of c a r d i n a l i t y n i s of the form F(d) + G(S) where F(d) = 0(2 Z ) and G(n) = n + 41og. n . BIBLIOGRAPHY Hardy, G. H . and Wright , E . M. An Introduct ion to the Theory of  Numbers. Oxford U n i v e r s i t y Press , London, 1945. Waksman, A . A Model of R e p l i c a t i o n . Journal of the A s s o c i a t i o n f o r Computing Machinery 16,1 (January 1969), pp. 178-188. 

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