FUNCTIONAL- COMPLETENESS CRITERIA FOR FINITE

FUNCTIONAL- COMPLETENESS CRITERIA

FOR FINITE DOMINS

by

PETER SCHOFIELD

A t h e s i s s u b m itte d to th e U n iv e r s i ty o f L e ic e s te r

f o r th e d e g re e o f D o c to r o f P h ilo so p h y

UMI Number: U296355

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SUMMARY

N ecessary and s u f f ic ie n t c o n d itio n s f o r the fu n c tio n a l

com pleteness o f a s e t F o f fu n c tio n s w ith v a r ia b le s and va lu es

ran g in g over N = [ 0 , 1 , . . . , n J , where n ^ 1, a re in v e s t ig a te d

and in p a r t i c u l a r , com pleteness c r i t e r i a f o r a s in g le fu n c tio n

a re determ ined#

Complete s o lu tio n s a re known in th e s p e c ia l c a se s n = 1 ,2 ,

and r e s u l t s about th e se s p e c ia l c ase s which a re o f use in

fo rm u la tin g g e n e ra l theorem s a re d iscussed#

Proceeding to the g en e ra l case some p re lim in a ry c r i t e r i a

(which presuppose t h a t c e r ta in 2 -p lace fu n c tio n s a re gen era ted

by F) fo r th e fu n c tio n a l com pleteness o f F a re d e r iv e d .

These r e s u l t s show th a t th e s e t c o n s is tin g of a l l 2 -p lace fu n c tio n s

i s complete# In th e s p e c ia l case n + 1 = p (a prim e number) th e

fu n c tio n s o f . F a re shown to have a s p e c ia l form , and t h i s i s

used in some i l l u s t r a t i o n s of complete subsets#

The v a lu e sequence o f a fu n c tio n s a t i s f y in g th e S-éupecki c o n d itio n s

( th a t i s , depending on a t l e a s t 2 o f i t s argument p la c e s , and ta k in g

a l l n + 1 v a lu es of N) i s now examined, and some p ro p e r t ie s

o f such a fu n c tio n a re found# These r e s u l t s a re th en used in

dem onstra ting th e com pleteness o f a s e t F which g e n e ra te s a l l

1-p la c e fu n c tio n s , to g e th e r w ith a fu n c tio n s a t i s fy in g th e

S-ôupecki cond itions*

Our main r e s u l t s g ive improved s u f f i c i e n t co n d itio n s f o r

th e com pleteness o f F . In p a r t i c u l a r a s e t F i s com plete

i f i t g e n e ra te s a t r i p l y t r a n s i t i v e group of perm u ta tio n s o f N

and co n ta in s e i t h e r ( i ) only a s in g le fu n c tio n o r ( i i ) a t l e a s t

one fu n c tio n s a t i s f y in g th e S-ftupecki c o n d itio n s , th e l a t t e r a p a r t

from c e r ta in e x c e p tio n a l c a s e s . A d e ta i le d in v e s t ig a t io n

shows t h a t th e se occur only when n = 2 o r when n + 1 i s

a power o f 2 and a l l fu n c tio n s o f F a re l in e a r in each

v a r ia b le , r e l a t i v e to some mapping of N a s a v e c to r space

over Zg.

F in a l ly a d i f f e r e n t mapping o f N in to i s co n sid e red ,

and i t i s shown th a t th e fu n c tio n s of . F can be g iven a unique

r e p re s e n ta t io n r e l a t i v e to t h i s mapping. This r e p re s e n ta t io n

i s th e n used to f in d some examples of complete subsets*

The work of th i s th e s i s i s claim ed a s o r ig in a l except

where re fe re n c e to a n o th e r 's work i s made.

I shou ld l ik e to thank P ro fe sso r R .L .G oodstein f o r h is

h e lp and encouragement and f o r su g g estin g th e problems

considered h e re . I am a ls o g r a t e f u l to Dr. Roy 0. D avies

f o r t r a n s la t in g v a rio u s R ussian papers on t h i s work, and f o r

h is h e lp in th e p re s e n ta tio n and c o r re c t io n o f th e ty p e s c r ip t .

CONTENTS

SUMMARY

PAGECHAPTER I . INTRODUCTION

1 .1 . Basic d e f in i t io n s 1

1.2* Composition and com pleteness 2.1.3* 1-p la c e fu n c tio n s A-

1 .4 , Ba ckground 8

CHAPTER I I . COMPLETENESS CRITERIA IN Ei AND Eg

2 .1 . In tro d u c tio n 11

2 .2 . Completeness in Ej 11

2.3« Pre-comple t e su b se ts 14

CHAPTER I I I . SIMPLE GENERAL THEOREMS

3 .1 , In tro d u c tio n 17

3 .2 . Simple Completeness Theorems 18

3.3» The s p e c ia l case n + 1 = p (a prim e number) 21

CHAPTER IV. SOME FUNDAMENTAL RESULTS

4 .1 , In tro d u c tio n 32-

4 .2 , Consequences o f th e S-êupecki c o n d itio n s 33

4*3. Fundamental com pleteness c r i t e r i a in En 39

CHAPTER V. MAIN RESULTS

5*1 • In tro d u c tio n

5*2, A u x ilia ry r e s u l t s

3 . 3, An ex ce p tio n a l case 5 6

5 «4* In tro d u c tio n o f v e c to r n o ta t io n 52

3. 3 , Main conclusions 73

82CHAPTER V I. A MAPPING CF En INTO E?

6.1» In tro d u c tio n

6 .2 , Permanent s e ts 8^

6 . 3 , Some com plete su b se ts o f En 9?

BIBLIOGRAPHY. 110

I - INTRODUCTION - I

1.1 B asic D e f in it io n s

L e t n be a n a tu ra l number n ^ 1 and l e t N = { 0 , 1 , . . . , n j .

Denote by Ep th e s e t o f a l l fu n c tio n s ^(xi ) , o f any number

o f v a r ia b le s , whose v a r ia b le s range over N and whose va lues a re

elem ents of N.

REMARK. Mary a l te r n a t iv e n o ta t io n s f o r th e se b a s ic s e ts e x i s t .

For example, Sâlomàa... [27] [18] [29] uses the s e ts N + 1 = [ l , 2 , . . .,n+1 j"A/ ^

and 1^4,1 and Y ab lonsk ii [^9] uses th e s e ts = |0 , 1 , . . . , n j and

Pn* i . We have chosen th e above f o r reaso n s o f s im p l ic i ty o f n o ta t io n

and the f a c t th a t th e zero elem ent p lays an im p o rtan t ro le in some of

ou r r e s u l t s . I t should be no ted th a t th e number o f elem ents in th e b a s ic

s e t N i s n + 1, whereas the s e t o f a l l fu n c tio n s over N i s denoted

by En.

We c a l l the sequence o f values tak en by a fu n c tio n f ( x i , . . . , 2 j < ) f 3m ;

and w r i t t e n in th e o rd e r f ( 0 , . . . , 0 , 0 ) , f ( 0 , . . . , 0 , l ) , . . . , f ( 0 , . . . , 0 , n ) ,

f ( 0 , . . . , 1 , 0 ) , , . . , f ( 0 , . . . , 1 , n ) , . . . , f ( n , . . . , n ) th e value sequence o f f .

We sometimes r e fe r to an ordered s e t of k values (% , . . . ) as a p o in t.

denoted by a s in g le l e t t e r /i, and we may then w rite f( /i) fo r f (di > • • •>/ k ) •

The t r a d i t i o n a l way o f w r it in g down th e va lu e sequence of f i s in

th e form o f a t r u t h ta b le , and i t i s w e ll known th a t ev ery k -p lace

fu n c tio n f (xi c % can be u n iq u e ly re p re se n te d by a t r u t h ta b le

o f le n g th (n + 1)^* Obviously th e re a re (n + l ) d i s t i n c t

k -p la c e fu n c tio n s f ( x i , . . . , x k ) f % . When k = 2 th e most convenient

way o f w r i t in g down th e va lue sequence o f a 2 -p lace fu n c tio n f (x ,y )e

i s in th e form o f an (n + 1) x (n + 1) m atrix*so :

i ‘( x , y ) 0 1 . . • n

0 Pel * • • Pen1 P i i • • • Pin

n Pni • • • Pnn

where f o r each p a i r o f in d ic é s i , j (O ^ i , j ^ n) = f ( i , j ) .

1 .2 . Com position and Completeness

We c a l l xj an e s s e n t ia l v a r ia b le of a k -p la c e fu n c tio nCLJ_1 ,

f (xi , . . . ) i f th e re e x is t numbers a i , • . . ,a j+ 1 , . . . ,ai< such th a tA

f ( a i , . . . , a j w i , % j , a j * i , . . . , % k ) i s n o t c o n s ta n t. Any v a r ia b le of f

which i s n o t an e s s e n t i a l v a r ia b le we c a l l an in e s s e n t ia l v a r ia b le o f

f . For example, th e 2 -p lace fu n c tio n f ( x ,y ) = g (x ) , where g(x) i s n o t a

c o n s ta n t 1-p la c e fu n c tio n , has one e s s e n t ia l v a r ia b le x , and one in e s s e n t ia l

v a r ia b le y .

L e t th e i n f i n i t e s e t o f v a r ia b le s a v a ila b le f o r the fu n c tio n s o f

En be enum erated as fo llo w s: % , % , . . .,Z|( ***» (we sometimes w r ite

x ,y f o r Xi,3fe r e s p e c t iv e ly ) . For each f i n i t e p o s i t iv e in te g e r k we s h a l l

agree to re p re s e n t every k -p lace f \in c tio n o f En by a fu n c tio n in th e

f i r s t k v a r ia b le s o f th e above enum eration; f o r example f ( x i , . . . , x k )

( th e f i r s t k v a r ia b le s can be re g a rd ed as the v a r ia b le p la ce s of f ) .

By a com position o f fu n c tio n s o f a su b se t F C i t i s p o ss ib le

to produce f u r th e r fu n c tio n s (n o t n e c e s s a r i ly d i s t i n c t from th e fu n c tio n s

o f F ) . I f we d e fin e a p ro je c t io n fu n c tio n g(%i , ) , fo r any

f ix e d in te g e r k ^ 1, to be any fu n c tio n o f the form

g(xi ) = XL, f o r some index i , 1 ^ i ^ k , then

we have the fo llow ing concise d e f in i t io n o f com position.

DEFINITION'. A su b se t o f fu n c tio n s F of ^ i s c lo sed under

com position i f f (gi f F whenever f e F and f o r each

i ( i = 1 , . . . , k ) gt f F o r gj, i s a p ro je c t io n fu n c t io n . The fu n c tio n s

g l a l l have the same number o f argument p la ce s and i f th i s number i s -6

^ (ê l f ~ > • • • , . . .,Q< (3Ql , . • • ,X^) ) .

By the closu re under com position o f a su b set F o f En we mean the

s m a lle s t s e t F say , such th a t F Ç F and F i s c lo se d under^ F g en era te s by com position a fu n c tio n ^

com position . We say a ls o t h a t a s u b s e t^ if and only i f ÿ belongs to

th e c lo su re o f F under com position .

The s u b se ts o f E whose c lo su res under com position are E i t s e l f

are of s p e c ia l i n t e r e s t , and we c a l l a su b se t F com plete i f and o n ly

i f the c lo su re o f F under com position is Ep . That i s , F i s complete

i f and on ly i f F g e n e ra te s every member o f E!n •

REMARK. In e f f e c t , the above d e f in i t io n o f com position re q u ire s F to

be c lo sed w ith re s p e c t to th e fo llo w in g types of o p e ra tio n s :

( i ) re -o rd e r in g and id e n t i fy in g th e v a r ia b le s o f a fu n c tio n f f F;

( i i ) in tro d u c in g (e lim in a tin g ) any number o f in e s s e n t ia l v a r ia b le s

in to (from) the argument p la ce s o f a fu n c tio n f € F;

( i i i ) s u b s t i tu t in g fu n c tio n s o f F in to any number o f argument p lace s

o f a fu n c tio n f f F .

1*3 1 - p lace fu n c tio n s

I f f ( x i , . . * ,x k ) € Efl has a t most one e s s e n t ia l v a r ia b le ,

x t say , th en th e re i s a 1 -p lace fu n c tio n g such th a t

f ( x i , . . . ,% k ) = g (x i)

f o r a l l v a lu e s o f . I t fo llow s t h e r e f o r e th a t th e c lo su re

under com position o f th e su b se t o f a l l 1-p la c e fu n c tio n s o f i s

th e su b se t o f Ep c o n s is tin g o f a l l fu n c tio n s w ith a t most one e s s e n t i a l

v a r ia b le . This su b se t i s n o t i t s e l f s in c e E co n ta in s fu n c tio n s

w ith a t l e a s t two e s s e n t i a l v a r ia b le s ; f o r example, f (xi ,3^ , . • .,:q< )= % +%

(modulo n + 1 ) . T herefo re the s u b se t o f a l l 1 -p lace fu n c tio n s o f Ep i s n o t

com plete.

REMARK. The q u estio n o f th e s m a lle s t s e t o f 1-p la c e fu n c tio n s vdiich

g en era tes every 1-p la c e fu n c tio n o f E has been in v e s t ig a te d by Salomaa

in [2 7 ], where he proves th a t i f n = 1 th e n a t l e a s t 2 1-p la c e-v-w

fu n c tio n s a re needed to g en e ra te a l l 1-p la c e fu n c tio n s , and i f n > 2

a t l e a s t 3#

The fo llow ing n o tio n s concerning 1-p la c e fu n c tio n s w i l l p la y an

e s s e n t i a l ro le in our r e s u l t s .

A 1 -p lace fu n c tio n o f i s s a id to be o f type

where th e a 's a re n a tu ra l numbers s a t i s fy in g a% + % + . . • + a^ = n+1,

i f i t takes ex a ctly t d is t in c t va lu es , v ^ , . . . , v ^ sa y , where v

i s tak en e x a c tly aj, tim e s . I t i s easy to see th a t a 1 -p la c e fu n c tio n

o f type [ l , 1 , . . . , l ] , w ith n + 1 u n i t s , i s a p e rm u ta tio n o f N, and

t h a t a 1 -p lace fu n c tio n o f type [n + l ] i s a c o n s ta n t fu n c tio n .

We r e c a l l th a t a s e t o f 1-p la c e fu n c tio n s G i s s a id to be i f

$-p lv t r a n s i t i v e on whenever n i , . . . , n ^ are d i s t i n c t elem ents o f

N, and ni*, ; a re d i s t i n c t elem ents o f N, th e re e x is t s a 1—p lace

fu n c tio n g(x) € G, tak in g a t l e a s t th e v a lu es ni', , . • .,n4,! , su ch th a ti s c lo sed and

g (n i) = ntV, fo r i = 1 , . . . , s . I f G^ c o n s i s t s e n t i r e ly of p e rm u ta tions

o f N, then G i s an s -p ly t r a n s i t i v e group of perm u ta tio n s o f N.

REMARK. We s h a l l make use o f v a rio u s w ell-known r e s u l t s on s -p ly

t r a n s i t i v e groups o f p e rm u ta tio n s .

For a f ix e d 1 -p la ce fu n ctio n s (x )a fu n ction f (x i , . . . , % ) €

i s sa id to be se lf-co n ju g a te under s i f and only i f

f(s(x i ) , . . . , s(xi<)) = s ( f ( x i , . . . ,%())

fo r a l l v a lu es of x i , . . . , x k *

REMARK. I f g ( x i , . . . , x ^ ) , f o r some in te g e r 6 !^ 1 , i s a p ro je c t io n

fu n c tio n , th en

g ( s ( x i ) , . . . , s ( x ^ ) ) = s (x t) f o r some in te g e r i , 1 ^ i ^ *6

= s (g(xi , . . . ,x^) ) .Hence, fo r any f ix e d 1-p la c e fu n c tio n s ev ery p r o je c t io n fu n c t io n o f

En i s s e lf -c o n ju g a te under s .

THEOREM 1 .1 , I f f (xi , . . . , 2}< ) c Ep i s se lf-c o n ju g a te under some fix ed

1-p la ce fu n ction s (x ) , then every fu n ction generated by f i s a lso

se lf-co n ju g a te under s ( x ) .

PROOF. I f 0 ( x i , . . . , x ^ ) i s g en e ra ted by f , then

= f (gl (3Q1. , . . .,X ^) , . • (xi , . . *,X^) ) ,

where each of th e -6-place fu n c tio n s gi ( i = 1 , . . . , k ) i s e i th e r

g en era ted by f o r i s a p ro je c t io n fu n c tio n . S ince <f> i s n e c e s s a r i ly

a com position in v o lv in g a f i n i t e number o f o c cu rren ces , m say , o f th e

fu n c tio n f , we s h a l l proceed by in d u c tio n on t h i s number. I f a l l th e

fu n c tio n s a re p ro je c t io n fu n c tio n s , th e n in view o f th e

above remark (f) i s s e lf -c o n ju g a te under s . This e s ta b l is h e s th e r e s u l t

f o r th e case m = 1. Suppose now th a t 1, and t h a t every fu n c tio n

g en era ted by f w ith a t most m occurrences o f f in i t s com position

sequence i s s e lf -c o n ju g a te under s . L e t (f> be any fu n c tio n generated

by f w ith m + 1 occurrences o f f in i t s com position sequence.

Then each o f th e fu n c tio n s i-s s e lf -c o n ju g a te under s , s in ce

g t,( i = 1, . . . , k ) i s e i th e r a fu n c tio n g e n e ra te d by f w ith a t most *

occurrences o f f in i t s com position sequence ( in view o f th e i n i t i a l

f in th e com position sequence o f o r a p r o je c t io n fu n c tio n .

T herefore <f> i s s e lf -c o n ju g a te under s and the r e s u l t fo llo w s .

For a f ix e d p erm u ta tion p(x) o f N a su b se t F^ o f i s s a id

to be con jugate to a su b se t F of Ep 3f and on ly i f to every fu n c tio n

f ( x i , . . . , 2 k ) € F th e re corresponds a fu n c tio n f ( ] % ,. . . , :% ) e F such th a tP P

(^1 , • • * ) = p ( f (p (^1 ) , • • *>P ) ) )>

and conversely*

I t fo llo w s from th e above d e f in i t io n t h a t F i s con jugate to

Fp under th e p e rm u ta tion p“ (x ) , and so we c a l l F , F^ a p a i r o f

con jugate su b se ts o f Ep .

THEOREM 1 .2 . I f . f o r some f ix e d perm uta tion p(x) o f N, F, F^

i s a p a i r o f con jugate su b se ts o f , th en F is com plete i f and only

i f Fp i s com plete .

PROOF. Suppose i f p o ss ib le t h a t F i s complete b u t F^ i s n o t .

Then th e re i s a fu n c tio n <^(xi , . . . ,x ^ ) e % n o t g en era ted by F^.

S ince F i s com plete, th e fu n c tio n

, . . . , x ^ ) = p“ (0 ( p ( 2i ) , . . . , p ( x ^ ) ) )

i s g en era ted by F , and so <p* a r is e s ffom a com position sequence

in v o lv in g a f i n i t e number o f fu n c tio n s , s a y , where each

g t ( i = 1 , . . . ,m ) i s e i th e r a fu n c tio n o f F o r a p r o je c t io n fu n c tio n .

I f , f o r i = 1 , . . . , m , we re p la c e each o f th e fu n c tio n s g i (xi , . • . ,x^)

by th e fu n c tio n gl * (xi , . . . ,x ^ ) = p(g(p“ (xi ) , . . . ,p “ (xj< ) )) in th e

com position sequence o f < f> ', th e n we o b ta in a com position sequence

in v o lv in g a f i n i t e number o f fu n c tio n s gi * , . . . , & *, where each o f th e

fu n c tio n s gi’( i = 1 , . . . ,m ) i s e i th e r a fu n c tio n o f F^ o r a p ro je c t io n

fu n c tio n (s in c e each p ro je c t io n fu n c tio n i s s e lf -c o n ju g a te under s ) .

The r e s u l t in g fu n c tio n i s th e re fo re gen era ted by F , b u t i t i s th eP

fu n c tio nP * (p ) , • • • > P (zk ))) = ÿ (xi , . . . , x k ) ,

8

which c o n tra d ic ts th e assum ption th a t (f> i s no t g en era ted by F^.

'T herefore F^ i s a ls o com plete. Since F i s con jugate to F^ under

(x) th e r e s u l t in the o th e r d i r e c t io n fo llow s by a s im ila r argum ent.

1 .4 - Background

In t h i s work we s h a l l be concerned m ainly w ith th e problem of f in d in g

co n v en ien t c r i t e r i a fo r a su b se t F o f Ep to be com plete, and i f

p o s s ib le deducing c r i t e r i a f o r a s in g le fu n c tio n to be com plete.

These problems a rose ou t o f th e w e ll known "S h e ffe r S troke F u n c tio n s”

o f th e P r e p o s i t io n a l C alculus ( th e se are th e 2—p lace com plete fu n c tio n s

o f E l ) . In [5 3 ], S h e ffe r dem onstrated th a t ev ery fu n c tio n of th e

P re p o s i t io n a l C alculus i s g en era ted by j u s t one o f th e se fu n c tio n s .

Follow ing th e work o f S h e ffe r , P o s t, in [2 4 ], in a complete survey o f a l l

th e d o s e d s e ts o f Ep fo r the s p e c ia l case n = 1, com ple te ly so lved the above

problem s in t h i s s p e c ia l c a se . However, such a survey has proved too la b o rio u s

to be c a r r ie d o u t in a l l b u t the s p e c ia l case n = 1.

Unaware o f P o s t 's r e s u l t s , Y ablonsk ii in [4 6 ] succeeded by a more d i r e c t

p ro o f in a t t a in in g an id e n t ic a l s e t of co n d itio n s to those o f P o s t fo r

a su b se t to be complete in Ei • He then went on to f in d in [47]

id e a l c o n d itio n s f o r a su b se t o f Ep to be com plete in th e s p e c ia l

case n = 2 . The corresponding problem fo r a s in g le fu n c tio n in th i s

s p e c ia l case was p a r t ly so lved ( f o r a 2 -p lace fu n c tio n only) by M artin in

[17] and r e c e n t ly has been com pletely so lved by W heeler in [4 3 ] .

The re le v a n t r e s u l t s o f th e se two s p e c ia l cases a re m entioned in

C hapter 2 .

P roceeding to th e g en era l case . P ost in [2 2 ] f i r s t dem onstrated th e

e x is te n c e o f complete su b se ts o f Ep fo r every f i n i t e in te g e r n ^ 1.

He proved th a t , i f n ^ 2, th e su b se t o f c o n s is t in g o f th e 1-p la c e

fu n c tio n x + 1 (modulo n + 1) to g e th e r w ith th e 2 -p lace fu n c tio n m ax(x,y),

i s com plete. In C hapter 6 we s h a l l p re se n t an unusual way o f g e n e ra lis in g

t h i s r e s u l t . Many o th e r examples o f complete su b se ts and fu n c tio n s have

s in c e been found; f o r example, by Evans and Hardy [ 5 ] , G o tlind [ 9 ] ,

Webb and Salomaa [ 2 7 ] #

However, in g en e ra l attem pts to f in d c r i t e r i a f o r a su b se t o f E

to be complete have met w ith on ly p a r t i a l su cc e ss . The f i r s t s ig n i f ic a n t

r e s u l t was due to S6upecki who proved in [3 5 ] "üie fo llo w in g :

i f n ^ 2, and a su b se t F of g en era tes th e s e t o f a l l 1-p la c e fu n c tio n s of

Ep, to g e th e r w ith a s in g le non-degenera te 2 -p lace fu n c tio n F (x ,y ) which

ta k es a l l n + 1 va lues o f N, th en F i s com plete.

He then used th i s r e s u l t in concluding th a t a s in g le 2 -p lace fu n c tio n

f ( x ,y ) which gen era tes the s e t o f a l l 1-p la c e fu n c tio n s o f Ep i s com plete.

In [4 9 ] Y ablonskii in tro d u ced th e fo llow ing g e n e r a l is a t io n of 86upecki*s

c o n d itio n s fo r th e fu n c tio n f ( x ,y ) : a fu n c tio n f ( x i , . . ) s a t i s f i e d

the Sûpecki co n d itio n s i f f has a t l e a s t two e s s e n t ia l v a r ia b le s and ta k es

a l l n + 1 v a lu e s . He th e reb y e s ta b lis h e d S-6upecki's r e s u l t s f o r th e more

g en era l case when f i s a k -p la c e (k ^ 2) fu n c tio n o f E . O ther

10

improvements on S-6upecki*s r e s u lt s have been concerned w ith reducing

the c la s s o f 1-p la ce fun ction s generated by F. Such improvements

have been obtained by Y ablonskii (Theorem of [ 4 9 ] ) and Salomaa

[ 2 7 ] [ 2 8 ] [ 2 9 ] * In p a r ticu la r Salomaa has proved (Theorem 1 o f [ 2 8 ] )A/ A/ 'V Ax XIXth at fo r n ^ 4 , a su b set F i s complete i f ; i t generates, both

a fu n ction s a t is fy in g th e Sôupecki conditions and a lso the a ltern a tin g

group Ap+i of permutations o f N. This r e s u lt was used in th e proof

o f the fo llow in g (Theorem 2 o f [ 2 8 ] ) • fo r n ^ 4 , a s in g le fu n ction

f ( x i , . . . , x k ) e Efl i s complete i f and only i f i t generates the a ltern a tin g

group Ap+i • In Chapter 5 we are concerned w ith improving these two

r e s u lt s o f Salomaa s t i l l fu rth er .

11

I I - COMPLETENESS CRITERIA IN Ei AND % - I I

2.1» In tro d u c tio n

In C hapter 1 we m entioned th a t th e problem o f f in d in g c r i t e r i a f o r

a su b se t (o r s in g le fu n c tio n ) to be com plete in Ep has been com pletely

so lved in th e s p e c ia l cases n = 1, 2 . However, s in ce our in te n t io n i s

to improve th e p a r t i a l r e s u l t s o f th e g en era l c a se , we s h a l l p re s e n t in

th i s c h ap te r only those r e s u l t s of th e se two s p e c ia l cases which a re

of use in fo rm u la tin g our G eneral Theorems.

In 0 2 .2 we s h a l l show t h a t every fu n c tio n f € Ei belongs to th e

r in g o f polynom ials over th e f i e ld o f re s id u e s modulo 2 , and we use th i s

r e p re s e n ta t io n in a s tra ig h tfo rw a rd account o f th e n ecessa ry and s u f f i c i e n t

co n d itio n s f o r a s in g le fu n c tio n to be complete in Ei . In § 2 .3 we

s h a l l in tro d u ce th e n o tio n o f a p re-com plete su b se t o f Ep and l i s t a l l the

p re-com plete su b se ts o f Ei and % in two theorems concerning th e

r e l a t i o n between complete and pre-com plete su b se ts in th e se s p e c ia l c a se s .

2 .2 , 'C om pleteness ' in Et

THEOREM 2 .1 , ^ E i s a su b se t o f E i , w hich g e n e ra te s th e two

2-p la c e fu n c tio n s x + y modulo 2, xy, to g e th e r w ith th e c o n s ta n ts 0, 1,

th en F i s com plete.

PRÛOF. I t i s s u f f i c i e n t to prove th a t f o r ev ery f i n i t e in te g e r k ^ 1

F g en era te s th e s e t o f a l l k -p la c e fu n c tio n s o f Ei .

n

We proceed by in d u c tio n on k s t a r t i n g w ith th e case k = 1,

which fo llo w s from th e f a c t th a t th e s e t o f a l l 1-p la c e fu n c tio n s of

El i s [ 0 ,1 ,x ,x + l] , and i t i s e a s i ly seen t h a t F g en e ra te s these

if fu n c t io n s . Assume t h a t k ^ 1, and t h a t f o r each f i n i t e in te g e r

K(1 ^ K ^ k) F g en era te s every \ p lace fu n c tio n o f Ei Then

fo r each f ix e d k + 1 - p lace fu n c tio n P ( x i , # « ,y ) % th e

fu n c tio n s f ( x i , . . . , x k , O), f ( x i , . . . , x k , l ) a re generated by F,

s in ce they have a t most k v a r ia b le s . But

f ( x i , . . . , x k ,y ) = f ( x i , . . . , x k ,0 ) ( y ) + f ( x i , • . .,xk , 1 ) (y + l) ,

vdiere a d d it io n i s c a r r ie d ou t modulo 2, and so F g en era tes every

k+1 - p la ce fu n c tio n of E i , which proves th e r e s u lt*

THEOREM 2 .2 . The neaessany and s u f f i c i e n t co n d itio n s f o r a s in g le

fu n c tio n f ( x i , . . .,%k) to be com plete in % a re th a t

^ i ) f ( x ^ . . . , x ) = X + 1.

( i i ) f (x i+ 1 , .. . ,x k -+ l) / f(x i , . . . ,X k ) + 1.

C ondition ( i i ) s t a t e s t h a t f i s n o t s e lf -c o n ju g a te under x + 1.

PROOF, (a) N ecess ity ; i f f o r some v a lu e i c iO ,lj f s a t i s f i e s

f ( i , . . . , i ) = i , th e n every 1 -p lace fu n c tio n 0 g en era ted by f must

have # ( i ) = i , and so f w i l l no t be complete fo r in p a r t i c u la r f

cannot g en era te x + 1. Now f ( x , . . . , x ) = 0 , 1 , x o r x + 1 , and the

on ly p o s s ib i l i t y i s x + 1 . T herefo re ( i ) i s n e ce ssa ry .

I f f i s s e lf -c o n ju g a te under x + 1 , th en by Theorem 1.1 eveiy

fu n c tio n gen era ted by f w i l l a lso be s e lf -c o n ju g a te under x + 1 .

15

C onsequently f i s n o t com plete, s in ce E% co n ta in s fu n c tio n s which

a re n o t s e lf -c o n ju g a te under x + 1 ; f o r example th e c o n sta n ts 0, 1•

T herefo re ( i i ) i s n e ce ssa ry .

(h) S u ff ic ie n c y ; we s h a l l show t h a t f g en era te s th e co nstan ts

0 ,1 , to g e th e r w ith th e two 2 -p lace fu n c tio n s x + y modulo 2, xy

and th e n th e r e s u l t w i l l fo llow by Theorem 2 .1 .

I f f i s n o t s e lf -c o n ju g a te under x + 1 , th en f o r some f ix e d

s e t o f v a lues « x , . . . , a |< , where fo r each index i ( l ^ i ^ k)

a i f ! 0 , l j ,

f (flti+1, . . . +1 ) ^ 1 *

C onsider th e 1-p la c e fu n c tio n

^ (x ) = f (x + K i , . . . , X + ttk ) .

By c o n d itio n ( i ) f g en era te s x + 1 , and so f g en era te s ^ , b u t

ijf s a t i s f i e s

^ (x + 1 ) / ijf ix) + 1.

Only (p = 0 o r 1 s a t i s f i e s t h i s c o n d itio n , and i t fo llow s th a t f

g en era te s th e co n stan ts 0,1 (th e y w i l l be ip, ÿ + l ) .

Suppose f i s l i n e a r , th enk

a d d itio n c a r r ie d ou t modulo 2 . I f a lso f ( x , . . . , x ) = x + 1 th e nk

n e c e s s a r i ly we have Z a& = b = 1, b u t i f t h i s was so f would beLsi

s e lf -c o n ju g a te under x + 1 c o n tra d ic t in g ( i i ) . T herefore f i s

n o n - lin e a r , and we can take th e p roduct term o f l e a s t degree to be

X i3 % ...x j. Hence f (x ,y , 1, . . . , 1 , 0 , . . .,0)j». w ith the z e r o 's a l lo c a te d

u

l a s tto th e ^ k - j p la c e s , i s a n o n - lin e a r polynom ial in x ,y , say

xy .+ A ix + Agy + Ag, and so f genera tes th e p ro d u c t xy by th e com position

xy = f (x + Ag , y + A i , 1 , , . . , 1 , 0 , . . . , 0 ) + (A@+ AiAg ) .

F in a l ly we have

X + y = ((x + l ) ( y + 1 )+ I)(x y + 1 ),

and) i t fo llow s th a t f generates x + y , which com pletes th e proof*

REMARK. Theorem 2 .2 was o r ig in a l ly proved by P o s t in [2^]* The

above p ro o f i s a s l ig h t s im p l i f ic a t io n o f th e one used by R.A. Cuninghamg -

'G reen in [3].2 -1 d is t in c t k -place

2^-'! ..-ICOROLLARY. There a re 2 ‘

complete fu n c tio n s in Ei •pk __2

IROOF. C ondition ( i ) of Theorem 2 .2 g iv e s 2 k -p la c e2^“ —1fu n c tio n s in E i , and o f th e se 2 a re s e lf -c o n ju g a te under x + di.

This g ives

k -p lace com plete fu n c tio n s in E i .

2 .3 . P re-com plete su b se ts

DEFINITION. A c lo sed su b se t F o f i s c a l le d p re-com plete

i f fo r every fu n c tio n h e E n o t con tained in F th e su b se t [h | U F

i s com plete.

This d e f in i t io n , due to K uznetsov, i s quoted by Y ablonsk ii in [49]*

Follow ing KuSnetsov we say th a t a fu n c tio n ^ (xj. , . . . , % )

conserves th e p re d ic a te P ( x i , . . . , x ^ ) i f th e fo rm ula

15

D P(0 (x^1**21*" '"**k1 ^ ^*12*^22* * *'**k2^* * *"*^^*1^*^26* * *

i s v a l id f o r a l l v a lu e s of the v a r ia b le s x^^ ( i = 1 , . . . , k , j = 1 , . . . , 6 ) .

The s e t o f a l l fu n c tio n s which conserve th e p re d ic a te P is c a l le d th e

co n se rv a tio n c-laas o f P . I t i s easy to see t h a t th e co n se rv a tio n

cLàss o f a f ix e d p re d ic a te P i s c lo se d under com position , and th is

n o ta t io n g iv es us a sim ple and compact form of re p re se n tin g pre-com plete

s u b se ts .

P o s t, in [2^, in a complete a n a ly s is o f th e c lo sed su b se ts of l a ,

subsequently , found a l l the p re-com plete su b se ts o f E%. There a re f iv e o f them,

and th ey a re l i s t e d in th e s ta tem e n t of Theorem 2.3* These f i v e pre-com plete

su b se ts were l a t e r found ind ep en d en tly by Y ablonsk ii in [4^, who then

went on to f in d in &l7] a l l th e p re-com plete su b se ts o f Eg , l i s t e d in

Theorem 2.4* These r e s u l t s le d to th e fo llo w in g theorem s on com pleteness

c r i t e r i a in Ei,Eg 1

THEOREM 2.3* The n ecessa ry and s u f f i c i e n t c o n d itio n f o r a su b se t

to be complete in Ei i s th a t i t s h a l l n o t be co n ta in ed in any one of

th e co n se rv a tio n c là s s e s o f the fo llo w in g 5 p re d ic a te s ;

X = 0 , X = 1, X / y , x + y = z + u , x ^ y .

PROOF. See e i th e r P o st [24] o r Y ab lonsk ii [46].

THEOREM 2. 4 * The n ecessa ry and s u f f i c i e n t c o n d itio n , f o r a su b se t

to be complete in Ea i s t h a t i t s h a l l n o t be co n ta in ed in any one of

th e c o n se rv a tio n c la s se s o f the fo llo w in g I 8 p re d ic a te s : '

16

x = i , x / i , x + 1 = y , x + y = z + u, x + i ^ y + i , (x = i ) f ^ ( y = i ) ,

x = y v x = i v y = i , x = y v x = z v y = Z, where i i s a parameter

taking the values 0 , 1, 2 , and add ition i s taken modulo 3»

PROOF. See Y ab lonsk ii [47] o r [49]./V'*'

REMARK. Theorems 2 .3 and 2 .4 a re th e s p e c ia l oases n = 1, 2

o f th e g e n e ra l r e s u l t :

a su b se t o f Ep i s com plete i f and only i f i t i s n o t e n t i r e ly con tained in

any one o f th e p re-com ple te su b se ts o f Ep .

A p ro o f o f t h i s r e s u l t (due to Kuznetsov) i s given by Y ablonskii in

■ [^ . However, in th e g en e ra l case , a ttem p ts to f in d a l l th e pre-com plete

su b se ts o f En have met w ith g re a t d i f f i c u l t i e s . We have e x p l ic i t e ly

s ta te d Theorems 2 .3 , 2 .4 above in view o f t h e i r a p p lic a t io n to

p rov ing "end c a se s” in l a t e r Theorems.

REMARK. In [44], W heeler d e riv e s n ecessa ry and s u f f i c i e n t co n d itio n s

f o r a s in g le fu n c tio n to be com plete in % .

17

I I I - SIMPLE GENERAL THEOREMS - I I I

3 .1 . In tro d u c tio n

Since th e s u b se t o f a l l fu n c tio n s o f En w ith a t most one e s s e n t ia l

v a r ia b le i s c lo sed under com position , a n ecessa ry c o n d itio n f o r a su b se t

o f En to be complete i s th a t i t s h a l l g en e ra te a fu n c tio n w ith a t

l e a s t two e s s e n t ia l v a r ia b le s .

In § 3 .2 we s h a l l p re se n t some p re lim in a ry c r i t e r i a f o r a su b se t F

o f En to be com plete. These c r i t e r i a presuppose th a t an e x ten s iv e

c la s s o f 2 -p lace fu n c tio n s g en era ted by F i s a v a i la b le , b u t th e

r e s u l t s proved are in te r e s t in g and v a lu a b le .

In § 3 .3 we s h a l l c o n s id e r th e s p e c ia l case when n + 1 = p

(a prime number). In th i s case we s h a l l show t h a t every fu n c tio n

f e E . belongs to th e ring : o f polynom ials over th e f i e l d of re s id u e sp-1modulo p , and we s h a l l m ention examples o f com plete and p re-com plete

fu n c tio n s and su b se ts in E . .p-1

18

3 .2 , Simple Completeness Theorems

DEFINITION# We d e fin e th e fo llo w in g p a i r o f 2 -p lace fu n c tio n s

each belonging to En :

X + y by

x + 0 = x = 0 + x ;

xy by

xO = 0 = Ox

x1 = X = 1x,

o r in term s o f th e m atrix n o ta tio n *

x + y 0 1 . . . n xy 0 1 2 . . . n

0 0 1 . . . n 0 0 0 0 . . . Ù

1 1 1^11 • • Min 1 0 1 2 . . . n

2 0 2 V2 2 • .• % , , ,

• • • • • • • • •n n Mn 1 • • Mnn n 0 n i nn

We may take as th e v a lu e o f P iJ (1 (2 ^ k , -6. ^ n)

any elem ent of N. This g iv e s us ( n + l) ^ ^ d i s t i n c t p a ir s of

2 -p lace fu n c tio n s x+y, xy.

EEMA.EK# x+y (modulo n + l) , xy (modulo n+1 ) i s a p a r t i c u la r

p a i r of the above ty p e , bu t u n le ss we s t a t e o th e rw ise , we s h a l l

take x+y, xy to mean th e fu n c tio n s d e fin e d above.

19

THEOREM 3*1* I f a su b se t F g e n e ra te s a f ix e d p a i r

o f 2-p la c e fu n c tio n s x+y, xy , t o g e th e r w ith th e s e t o f a l l 1-p la c e

fu n c tio n s of Ep, th e n F i s com ple te .

PROOF. I t i s s u f f i c i e n t to prove th a t f o r each f i n i t e

in te g e r k ^ 1 F g e n e ra te s ev ery k -p la c e fu n c t io n f (x^ , . . . ,X |< ) c Ep •

We s h a l l use in d u c tio n on k , s t a r t i n g w ith th e case k=1 w hich

fo llo w s by th e hypotheses of the Theorem. Assume th a t k ^ 1 and

th a t f o r each in te g e r K (1 ^ k) F g e n e ra te s every K -place

fu n c tio n of Ep. S ince F g e n e ra te s ev ery 1-p la c e fu n c tio n of Ep,

then in p a r t i c u la r F g en era te s th e n+1 1—p lace fu n c tio n s

Mo,Mi,*.*,Mn d e fin ed by

^ l(x ) = 0 , fo r X / i ;

M i(i) = 1,

f o r each index i (O ^ i ^ n ) .

Yve can e iç r e s s each f ix e d (k + l)-p la ce fu n c tio n f ( x ^ , . . . , X k , y ) e Ep

as fo llo w s;

f ( x i , . . . , X k , y ) = z_i f ( x i , . . . , X k , i ) ^ j^(y) (a )1=0

where a d d i t io n and m u l t ip l ic a t io n a re w ith r e s p e c t t o th e

o p e ra tio n s x+y, xy re s p e c t iv e ly ( in f a c t (A) i s ju s t a co n v en ien t

way o f w r it in g down the com position in v o lv in g x+y, xy , p ^ , . . . , ^ p ) .

By the induction hypothesis F generates the n+1 fu n ction s

f ( x i , . . . , X k , i ) , for i = 0 , 1 , . . . , n , s in ce each f ( x % , . . . , X k , i ) i s a t

most a k -p lace fu n ction o f Ep . Therefore F generates f ( x i , . . . , X k , y ) ,

and the r e s u lt fo llo w s .

20

THEOREM 3*2. Denote by the s e t o f a l l k-p la c e fu n c tio n s o f ;

th e n i s com plete i f and only i f k ^ 2 .

PROOF. I f k ^ 2 , th e n by id e n t i fy in g v a r ia b le s (o r e lim in a tin g

in e s s e n t ia l v a r ia b le s ) in s u i ta b le k -p lace fu n c tio n s o f E we can show

th a t g en era te s ^ and , and so th e r e s u l t fo llo w s by

Theorem 3*1* I f k = 1 , th e n the c lo su re o f under com position is the

s e t of a l l fu n c tio n s o f Ep w ith a t most one e s s e n t ia l v a r ia b le ,

which i s a p ro p e r su b se t o f % , and so ^ is not com plete.

REMARK. Theorem 3*2 i s due to P o s t , who p re se n te d i t in [22]in th e form; every fu n c tio n can be expressed aa a f i n i t e com position

o f th e two fu n c tio n s max(x,y) and x + 1 (modulo n + l ) . I t i s d f

fundam ental im portance in th e work and w i l l be r e f e r r e d to in l a t e r

c h a p te rs .

The fo llo w in g Theorems a re improvements on Theorem 3*1*

21

THEOREM 3»3. I f a su b se t F g e n e ra te s a f ix e d p a i r of

2-p la c e fu n c tio n s x + y , xy, to g e th e r w ith a l l c o n s ta n ts and

the n + 1 1-p la c e fu n c tio n s ^q, ju^, . . . ,jUp d efin ed by

m i x ) - O f fo r X / i j

M l(i) = 1 j

fo r each index i (O ^ i ^ n ) , th en F i s com plete.

PROOF. We s h a l l prove th a t - F generates the s e t o f a l l

1-p la ce fu n ction s of Ep , and t ie n th e r e s u lt fo lla v s by Theorem 3*1*

I f f ( x ) i s any f ix e d 1-p la ce fu n ctio n of Ep , then we can

express f in the form

f ( x ) = Z f ( i ) / u i ( x ) ,L= 0

where a d d itio n is w ith resp ec t to the operation x + y , and m u ltip lica tio n

i s w ith resp ect to xy . Therefore F generates f ( x ) and the

r e s u lt fo llo w s .

REI.ARK. I f F generates the p a r ticu la r pair o f 2-p lace

fu n c tio n s x + y (modulo n + 1 ) j xy (modulo n + 1 ) , th e n fo r

F to be complete i t i s s u f f ic ie n t only fo r F to generate a lso

the n + 1 1-p lace fu n ction s s in ce every con stan t,

c , say , can be expressed in the form

c = Z[/Jt(x) + . . . + jJi(x)] (modulo n + l ) .L-oi-------------------V ---------'

c tim es

Therefore F generates a l l constants, and i s complete by Theorem 3*3*

22

THEOREM 3*4. I f a su b se t F g e n e ra te s a f ix e d p a i r of

2-p la c e fu n c tio n s x+y, xy , to g e th e r w ith a doubly t r a n s i t i v e s e t of

1-p la c e fu n c tio n s o f N, then F i s com plete.

PROOF. We s h a l l prove t h a t F g en e ra te s th e n+1 1-p la c e

fu n c tio n s d e fin ed by

m i x ) = 0^ f o r X i j

M l(i) = 1,

f o r each index i (1 ^ i ^ n ) , to g e th e r w ith a l l c o n s ta n ts , and

th en th e r e s u l t w i l l fo llo w by Theorem 3*3*

G-iven any i (O ^ i ^ n ) , we choose n 1—p la ce fu n c tio n s

Skt where k = 0 , 1 , . . . , n (k / 1 ) , g en era ted by F such th a t

Ski (k) = 0 , S k i ( i ) = 1 .T h e n

jUj.(x) = n S k c ( x ) , Iq/i

where m u l t ip l ic a t io n i s in th e sense o f th e o p e ra tio n x ,y, and

so m i x ) i s gen era ted by F .

I t remains to show th a t F g en e ra te s a l l c o n s ta n ts .

Suppose u (x ) = u i s any c o n s ta n t fu n c tio n o f Ep, and l e t S be

a 1-p la c e fu n c tio n g e n e ra te d by F such th a t S(o) s u .

We have

Mo(m i(x )) = 0 , f o r a l l x ,u nd so

u (x ) = S im O tti(x ))) .

T herefo re F g e n e ra te s a l l c o n s ta n ts .

23

COROLLARY. I f a su b se t F g e n e ra te s a 2-p lao e fu n c tio n o f

type xy , to g e th e r w ith a doubly t r a n s i t i v e group of perm u ta tions

of N, then P i s com plete.

PROOF. The r e s u l t fo llo w s by Theorem 3*4- a f t e r we have

shown th a t F a ls o g e n e ra te s a fu n c tio n of type x+y.

Choose a perm uta tion r g en e ra ted by F such th a t

r ( 0 ) = 1 , r ( l ) = 0 ;

th en F g e n e ra te s th e 2—p lace fu n c tio n

h (x ,y ) = r - i ( r ( x ) , r ( y ) ) ,

where m u l t ip l ic a t io n i s w ith r e s p e c t to xy . We have

h (0 ,x ) = X a h ( x ,0 ) ,

and so h (x ,y ) i s o f type x+y and th e r e s u l t fo llo w s .

REMARK. In g en e ra l we canno t r ^ l a c e ’xy* by ’x+y’ in th e

C o ro lla ry , a s i s shown by th e example in w hich n = 3 , and F

c o n s is ts o f th e a l te r n a t in g group o f perm uta tions A* on | 0 , 1 , 2 , 3 j ,

a l l c o n s ta n ts , and th e p a r t i c u la r fu n c tio n o f type x+y d e fin ed

as fo llo w s ;x+y 0 1 2 3

0 0 1 2 31 1 0 3 22 2 3 0 13 3 2 1 0

L et T = [A4, a l l c o n s ta n ts ] . Then T co n ta in s a doubly t r a n s i t i v e

group o f perm utations of N, namely A 4 , and i t i s easy to v e r i f y

th a t i f r ( x ) , s ( x ) e T th e n a ls o r (x )+ s (x ) e T. This means t h a t

i f g i s any 1-p la c e fu n c t io n g e n e ra te d by com positions of x+y and

members cf T, th e n g (x ) € T. But T does n o t c o n ta in a l l 1-p la c e

fu n c tio n s of E 3, and so F i s n o t com plete.

24

3*3, The S p ec ia l Case n+1=p ( a pirime number)

THEOREM 3*6. I f n+1 = p ( a prim e number) , th en every

f unet io n f(x^ ,X|< ) e belongs to th e r in g of polynom ials

over the f i e l d o f re s id u e s modulo p#

PROOF. L et F be th e su b se t cf E^ c o n s is t in g o f the

p a r t i c u la r p a i r o f 2 -p lace fu n c tio n s x+y (modulo p ) , xy (modulo p ) ,

to g e th e r w ith a l l co n s tan ts and th e p 1-p la c e fu n c tio n s

d e fin ed by

m i x ) = 0 > f o r X / i )

M l(i) = 1,

f o r each i (O ^ i ^ p - l ) .

We can re p re s e n t Ml(x) as a polynom ial over th e f i e l d o f

re s id u e s modulo p a s fo llo w s;p—1

Ml(x) =(p - 1) n (x + a ) 5a=0

a /p -4

where a d d i t io n and m u l t ip l ic a t io n a re c a r r ie d ou t modulo p .

Hence ev ery fu n c tio n g en era ted by F belongs to the r in g of

polynom ials over th e f i e l d o f re s id u e s modulo p , b u t by Theorem 3*3

F i s com plete, and so th e r e s u l t fo llo w s .

25

REMARK. Theorem 3*6 g iv es us a conven ien t method of

e x p re ss in g fu n c tio n s belong ing to E^ We s h a l l m ention the

fo llow ing examples of com plete fu n c tio n s and su b se ts in E^ ^ .

EXAMPLE 3 .1 . ^ n+1 = p , and ÿ ( x i , . . . , x ) =P P

1 + XI , where a d d i t io n , s u b tra c t io n and

m u lt ip l ic a t io n a re c a r r ie d o u t modulo p , th e n <f) i s complete#

This simple r e s u l t i s proved by R.A.Cunningham Green in [ 3 ]•REIvlARK* Examples o f complete g e n e ra to rs over G alois

f i e l d s are to be found in [ 8 ] by R .L .G oodstein .

26

EXAMPLE 3*2. n+1 = p , and f o r f ix e d v a lu es

0? 0 (modulo p ) ) F c o n s is ts o f the p a i r of 2-p la c e

fu n c tio n s

# (x ,y ) = (x -a ) + (y -a ) + a ;

4 (x ,y ) = /9 (x -a )(y -a ) + y(^-cc) + ô (y -a ) + e ,

where a d d i t io n , s u b tra c t io n and m u lt ip l ic a t io n a re c a r r ie d o u t

modulo p , th en i f a ^ e (modulo p) F i s com plete, and i f

a s e (modulo p) F g en era te s th e p re-com plete su b se t o f Ep_^

which p re se rv e s th e p re d ic a te x = a .

We see t h a t ^ p re se rv e s th e p re d ic a te x =t a ( t h a t i s

<p{cL,a.) = a ) , and i f a s e th en ijf w i l l a ls o p re se rv e x = a .

We s h a l l make th e fo llo w in g change o f v a r ia b le ;

X = X - a .

We have

0 (x ,y ) = X + Y + a

^ (x ,y ) = /9XY + yX + SY + e ,

and i f ( x i , • • . ,Xk ) i s any fu n c tio n o f which p re se rv e s th e

p re d ic a te x = a , then

^ (x i , • • • ,Xk ) = ^ ’ (Xi , • • • , Xk )+ CL)

where Xj ‘ = x j -a f o r j = 1 , . • • ,k , and ^ ' (X , . , ]^ ) p re se rv e s

th e p re d ic a te X = 0 ( th a t i s ^ * ( 0 , . . . ,0 ) = O).

miAJîK ON SUBSTITUTION. I f ^ ( x i , . . . , X k ) , a re any

f ix e d k+1 fu n c tio n s of which p re se rv e the p re d ic a te x=a, th e n

* (^1 ) ■*" ^9

where f o r each index j ( l ^ j ^ k)

+ K «

We s h a l l need th e fo llo w in g two lemmas.

27

LEM'LA. 3*1. (j> g e n e ra te s ev ery l i n e a r fu n c tio n of E^_^

which -preserves th e p re d ic a te x=a«

PROOF. F o r each f ix e d k ^ 2 we have

» . . . 1 fXk ) #. ) ) = +Xg + . . . + Xk + oc,

where Xj = x j - a , f o r j = 1 , . . . , k , and i t i s easy to s e e , by

id e n t i fy in g s u i ta b le v a r ia b le s in th e se fu n c tio n s , t h a t 0 g en era te s

eveiy l in e a r fu n c tio n which p re se rv e s th e p re d ic a te x=a.

LEIvlIA. 3 .2 . F g en e ra te s th e 2 -p lace fu n c tio n XY + — [e -a ] + a

(Note t h a t s in c e n+1 i s prime 1_ € N.)

PROOF. Lemma 3.1 ^ g e n e ra te s th e fu n c tio n

T ( x ,y ,z ) = ^ [(p - y ) X + (p - j ) Y + Z] + a ,

and so F g e n e ra te s

T (x ,y ,^ ( x ,y ) )= X Y + ^ [e -a ] + oc .

28

PROCF OF EXAMPLE 3 .2 . P a r t ( i ) I f a ^ € (modulo p ) .

By Lemma 3.1 ^ g e n e ra te s th e c o n s ta n t oc, f o r

oc~ X + . . . + X + ct.I .............. I

p tim es

We have

f p( oL, a . ) = € cl) )

and so fo r j = 0 , 1 . , , , . p —1 F g e n e ra te s th e p c o n s ta n ts

j(€ -o :) + a . Since e-tt ^ 0 (modulo p ) , th e se a re the co n sta n ts

0 , 1 , . . . , p —1 in some o rd e r .

F a ls o g en era te s th e 1-p la c e fu n c tio n s

x+j = 0 (x ,j+ a ) f o r j= 0 , 1, . . . , p - 1 ,

and hence F g e n e ra te s the p a ir of 2—place fu n c tio n s

x + y (modulo p) = <p(x + a , y + a ) - a ;

xy (modulo p) = t ( x + a , y + a , ^ (x + a , y + a ) )

- - a ) - a ,

(see Lemma 3 .2 ) .

Therefore F g en e ra te s th e r in g o f polynom ials over tlê f i e ld

of re s id u e s modulo p , and by Theorem 3.6 t h i s i s a s u f f i c i e n t

c o n d itio n f o r F ô be com plete.

29

P a r t ( i i ) I f a s e (modulo p '), th en every fu n c tio n

g en era ted by F w i l l p re se rv e th e p re d ic a te x = a . C onversely

we s h a l l prove th e fo llo w in g .

LEMJ.1A 3*3* ^ ^ i s any f ix e d fu n c tio n of which

p re se rv e s the p re d ic a te x=a, th en F g e n e ra te s

PROOF. I f I i s l in e a r th e n the r e s u l t fo llo w s im m ediatly

by Lemma 3*1 • Suppose th a t ^ i s n o n - l in e a r ; th en f o r a f ix e d

index 6 (-6 depending on th e number of term s in ^ ) th e re e x i s t s

a l i n e a r fu n c tio n L(xJi , . . . ,x ^ ) , vh ich p re se rv e s th e p re d ic a te x=a,

and & n o n -lin e a r fu n c tio n s d e fin ed by

Mj ( x i , . . • ,Xk ) = n (]Qn ) + ot>m= 1

where f o r each p a i r of in te g e rs j,m ( 1^ j ^ 6 , 1 ^ m ^ k)

Xj = x j - a and € [ 0 , 1 . ,p—1 j , such t h a t

^ ( x i , . . . ,X k ) = L (M i,..« ,M ^)«

I t i s s u f f i c i e n t to prove th e re fo re th a t F g en era te s every

fu n c tio n of type Mj .

In Lemma 3*X l e t /c(x,y) = T ( x , y , ^ ( x , y ) ) , and s in ce

a s f we have

K (x,y) = XY + a .

T herefore f o r each f ix e d in te g e r k = 3 ^ 3 ,# . . , F g e n e ra te s

th e k -p la c e fu n c tio n

/c(xi ,/t(xq , » . . ,^(xk-. 1 ,Xk ) . . . ) ) = X Xg # . .X% + oc ,

I t is easy to s e e , by id e n t i fy in g s u i ta b le v a r ia b le s in th e se f u n c t io n s ,

t h a t F g en era te s ev ery fu n c tio n of ty p e Mj , and so the r e s u l t fo llow s*

30

We s h a l l now prove P a r t ( i i ) o f Example 3*2. Obviously

F is not com plete s in c e E __ C ontains fu n c tio n s w hich do not

p re se rv e th e p re d ic a te x=a. On the o th e r hand by Lemma 3*3

F g e n e ra te s every fu n c tio n o f E^ which p re se rv e s x=a.

That i s F* ( th e c lo su re o f F under com position ) i s th e

co n se rv a tio n c la s s of th e p re d ic a te x s a . I f h i s any f ix e d

fu n c tio n of Ep such t h a t h / , th en h does not p re se rv e

th e p re d ic a te x = a . T herefore h ( a , . . . , a ) ^ a and th e

su b se t F^ u fh ^ o f En w i l l be com plete by a s im ila r argum ent

to P a r t ( i ) . But th i s is th e co n d itio n f o r to be p re -co m p le te ,

and so th e r e s u l t fo llo w s .

31

REMARK. In connec tion w ith p re-com plete su b se ts , i t i s

of i n t e r e s t to determ ine fo r a g iven pre-com plete su b se t P o f Ep

w hether or no t th e re i s a s in g le f u n c t io n ‘d c P such t h a t d g en era te s

P . For th e p a r t i c u la r p re—com plete su b se t of E^ which p re se rv e s

th e p re d ic a te x = a , mentioned in Example 3*2, i f p = 2 th en

d ( x ,y ,z ) = XY + Z + a i s such a fu n c tio n because

d (x ,x ,y ) = X + Y + a ;

d ( x ,x ,d ( x ,y ,x ) ) = XY + a ,

end i f p ^ 3 th en d (^ i > • * * >^p^y) ® X%Xg + (p—l)XiXg + Y + ct

i s such a fu n c tio n because

d ( x , . . . , x , y ) = X + Y + a ;

(and so <p g e n e ra te s th e c o n s ta n t a)

d ( x ,y ,a , . . . ,a') = XY + a .

This i n tu rn le a d s to th e more g en e ra l q u e s tio n ; Given any

su b se t of fu n c tio n s $ = di****;#& o f Ep , what a re the n e ce ssa ry and

s u f f ic ie n t c o n d itio n s f o r $ to co n ta in a s in g le fu n c tio n , d l say ,

such th a t d l g e n e ra te s §? This q u estio n i s e a s i ly answered i f $

c o n s is ts of only 1-p la c e fu n c tio n s o f Ep, f o r th en d i ,* '* /& 6 must

a l l be powers of d l» but g e n e ra lly i t depends upon th e fundam ental

q u es tio n ; What a re th e necessary and s u f f ic ie n t c o n d itio n s f o r

a fu n c tio n f f En to gen era te a g iven fu n c tio n g e Ep? For n=1,2 ,

by Theorems 2 .} , 2 ,4 re s p e c tiv e ly , we may deduce t h a t a n ecessa ry

c o n d itio n i s th a t e i t h e r $ co n ta in s a com plete fu n c tio n , or # i s

co n ta in ed e n t i r e ly in one o f the p re-com plete s u b s e ts .

32

IV - SOME FUNDAMENTAL RESULTS - IV

4 .1 . In tro d u c tio n

In t h i s c h ap te r we s h a l l prove some fundam ental r e s u l t s

on com plete su b se ts and fu n c tio n s in En • In f a c t , the theorem s on

complete su b se ts (Theorems 4 .4 , 4 .6 ) a re e x p re s s ly d esig n ed to

y ie ld c r i t e r i a f o r a s in g le fu n c tio n to be com plete (Theorems 4 .5 ,

4 .7 ) .

We in tro d u ce th e fo llo w in g n ecessa ry p ro p e r t ie s o f a complete

fu n c tio n .

We c a l l x j an e s s e n t ia l v a r ia b le of a k—p la ce fu n c tio n

f ( x i , . . . , X k ) i f th e re e x i s t numbers a i , . . . , a j _ i , a j + i , . . . , a k

such t h a t the 1-p la c e fu n c tio n f ( a i , . . . , a j - 1 ,x j , a j + 1 , . . . ,a k ) i s

n o t c o n s ta n t .

A fu n c tio n f ( x i , . . . , X k ) e En i s s a id to s a t i s f y th e

Sûpecki c o n d itio n s i f i t ta k e s a l l n+1 va lu es of N, and

has a t l e a s t two e s s e n t i a l v a r ia b le s . I t i s easy to see th a t

a n ecessa ry co n d itio n f o r f to be com plete i s t h a t f s h a l l

s a t i s f y th e S-êupecki c o n d itio n s , fo r i f f does not take a l l

n+1 v a lu e s th en f cannot g en e ra te any fu n c tio n of En which

ta k es th e m issing v a lu e s , and i f f has a t most one e s s e n t ia l

v a r ia b le then f g e n e ra te s on ly fu n c tio n s o f Ep p o sse ss in g a t

most one e s s e n t ia l v a r ia b le .

33

4 .2 . Consequences of th e S-ê'Upecki co n d itio n s

THEOREM 4 .1 . I f n ^ 2 and f e En s a t i s f i e s th e Sûpecki

c o n d it io n s , w ith Xj. a s one e s s e n t ia l v a r ia b le , th e n th e re a re p o in ts

a = = Gî, • • • ,/?k) such t h a t th e th re e v a lu es

f(oc), fC 01 ,#3 ,.#* ,K k)*

a re d i s t i n c t .

We c a l l a,/9 S’êupecki p o in ts f o r f .

PROOF. S ince x^ i s e s s e n t ia l th e re e x i s t numbers

a,nd Ys J • • • >Yk such th a t

F(y i^jY 2 > • • • ^Yk ) u , ^*(Yi ^Ys > • • • *Yk ) — ^ >

viiere u^ / u® . We s h a l l now d is t in g u is h two c a s e s .

CASE 1. Vfe have f (y i * >Ys * • • • »Yk ) = € [u^,u®j fo r some

number Y i^ . Bince one of X2 , . . . ,X k i s e s s e n t i a l , th e re e x i s t

numbers Si and S g ^ , . . . ,# ^ ^ ; such th a t

f (S i ,Sg^, • . . , 5k^ ) ^ F (S i,S g ^ , • . • ,Sk^ ) .

One o f th e se two v a lu es must be d i f f e r e n t from f ( S i ,Y s> • • • >Yk ) i we

may suppose t h a t

f (Si ,5a^ , . . . ,Sk^ ) ^ i*(Si ,Ys > • • • >Yk ) •

One o f u^ ,u^ ,u® must be d i f f e r e n t from b o th of th e s e , u*’ say .

Then th e fo llo w in g th re e v a lu es a re d i s t i n c t ;

F ( y i ** »Y2 > • • • >Yk ) > F ( S i ,Y 2 J • • • >Yk ) > f ( S i , S g ,S k ^ ) •

CASE 2 . We have f (x i ,Y 2 , . . . ,Y k ) c [u^ ,u^ ] f o r a f ( X i., L SucH tia 't

Since f ta k es a t l e a s t 5 d i s t i n c t v a lu e s th e re e x i s t n u m b e r^ S i, . . . ,Sk )

u® / [u ^ ,u ^ j . L e t f ( 8i,Y 2 ,. .* ,Y k ) = u* '(i = 1 o r 2 ) . Then th e

fo llow ing th re e v a lu es a r e d i s é i n c t j

> Y2 ;*"*yYk); ^ (S i ,Y2 j • • • >Yk ) » ^*(§1 »S2 > • • • ^^k ) .

u

REMARK. Theorem 4.1 i s im p l ic i t in Y a b lo n sk ii 's p ro o f

o f h is ’fundam ental lemma’ ([4 9 ] p . 6 9 ). The above p ro o f of

Theorem 4.1 was shown to me by Dr, Roy 0. D avies; i t i s s im p le r

than my o r ig in a l proof based on Salomaa’s fo rm u la tio n of what is h e re

Theorem 4 .2 .

I t i s easy to see t h a t i f we weaken th e S-êupecki c o n d itio n s

by re p la c in g ’f assumes a l l n+1 v a lu e s ’ by *f assum es a t

l e a s t 3 d i s t i n c t v a lu e s ’ th e n Theorem 4.1 g iv es a co n d itio n

which i s s u f f ic ie n t a s w e ll as n e ce ssa ry .

DEFINIT ION. L e t G%, ...,G k be non-empty su b se ts ©f N.

Then we denote by f ( G i , . . . ,G k ) th e s e t of va lues assumed by

f ( x i , . . . , x k ) when fo r each index i , 1 ^ i ^ k , on ly values

belong ing to a re a ss ig n ed fo r x t .

35

THEOREM 4 .2 . ^ 2 and s a t i s f i e s th e

S-6upecki c o n d itio n s , w ith a s one e s s e n t ia l v a r ia b le , th en

fo r each in te g e r j 2 ^ j < n+1, th e re a r e s e t s , i= 1 , . . . , k ,

each co n sis tin g : of a t most j v a lu e s , such th a t th e s e t f(Gt, , • . . ,G|< )

c o n s is ts of a t l e a s t j+1 v a lu e s .

PiÔQP. By Theoreiî 4 .1 f w i l l have a p a i r of S-ûpecki p o in ts

a,/9; l e t th e co rrespond ing v a lu es be u^ ,u^ ,u® . We may suppose

th a t the n+1 d i s t i n c t v a lu es of N a re u^ ,u^ ,u^ , u f , . . . ,u"'’' , and

s in ce f assumes a l l n+1 v a lu e s , f o r each i ( l ^ i ^ n+ l) th e re

e x is t s numbers X i^ ,...,X k * ’ such th a t

f ( x i^ , . . . ,X k * ’) = u ^ .

L et j be a f ix e d in te g e r , 2 ^ j < n+1, and f o r each index

i (1 ^ i ^ k) l e t th e s e t Gl = [ a i , /9 i ,xl , . . . , xl' '’'^ Î .

The s e ts G i , . . . ,G k each c o n s is t of a t most j d i s t i n c t v a lu e s ,

and the s e t f (Gi , . . . ,Gj< ) c o n s is ts of a t l e a s t the v a lu es u^ ,u^ ,u® ,u^ , • . . ,

^ j+ i^ Tliat i s , f (G% , . . . ,Gk) c o n s is ts cf a t l e a s t j+1 e lem en ts .

REMARK*. Theorem 4 .2 i s a ls o proved by Y ab lonsk ii in [49]*

For o th e r r e s u l t s on th e value sequences o f fu n c tio n s w ith a t l e a s t

two e s s e n t i a l v a r ia b le s see papers by A.Salomaa [3 0 ] and

Roy 0 . Davies [ 4 ] .

36

IffiiVIARK. We a ls o in c lu d e in t h i s § th e fo llo w in g theorem

(Theorem 4«3)* Though no t a consequence of th e S6upecki

c o n d itio n s . Theorem 4 .3 g iv es c r i t e r i a (depending on th e 1-p la c e

fu n c tio n s g en era ted ) f o r a s in g le fu n c tio n f to s a t i s f y the

S6upecki c o n d itio n s . A r e s u l t of t h i s k in d i s o f te n used by

au th o rs in th i s work ; f o r example Salomaa [27] [28] [2.9]

and Y ablonskii [4 9 ] , b u t th e proof i s seldom g iv en .

37

THEOREM 4*5* I f a s in g le fu n c tio n f c Ep g e n e ra te s

1-p la c e fu n c tio n s <f>i >• • • (6 ^ 2) such Hi a t

( i ) each cf th e n+1 v a lues of N is tak en by a t l e a s t one of

th e fu n c tio n s d i , . . . and ( i i ) th e re i s no 1-p la c e

fu n c tio n d such th a t f o r each index i ( l ^ i ^ &)

d i ( x ) = d ^ ^ x ) f o r some power k i , where d ^ K ^ ) = d ( " " * (^ ( x lL . .)

(w ith d re p ea ted kj, t im e s ) , th e n f s a t i s f i e s th e Sûpecki

co n d itio n s .

PROOF. Suppose f does n o t s a t i s f y th e S-êupecki c o n d itio n s ;

i f f does no t tak e a l l n+1 v a lu e s , then f has a t l e a s t one

m issing v a lu e , u say . In t h i s case every fu n c tio n g en era ted by

f w i l l a ls o bave u a s a m issing v a lu e , in p a r t ic u la r

This c o n tra d ic ts a s s e r t io n ( i ) .

Suppose th e re fo re th a t f tak es a l l n+1 v a lu es and has a t

most one e s s e n t i a l v a r ia b le . L et f ( x , . . . , x ) = d(x)* ty a simple

in d u c tio n argument i t i s easy to show th a t every fu n c tio n g

g en era ted by f has a t most one e s s e n t i a l v a r ia b le ; and t h a t

th e 1-p la c e fu n c tio n g ( x , . . . , x ) i s a power o f d* I t fo llow s

th e re fo re t h a t every 1—p lace fu n c tio n g en era ted by f is a power

o f (p, c o n tra d ic t in g a s s e r t io n ( i i ) . The procf i s com plete.

3 8

COROLLARY 1 • I f f g e n e ra te s a l l c o n s ta n ts , th e n f

s a t i s f i e s th e S6upecki c o n d itio n s .

PROOF. To each value u e N th e re corresponds a c o n s tan t

u (x ) = u . I f d f En i s any s i n ^ e 1-p la c e fu n c tio n such

th a t d ^ (x ) = u , f o r some power k , th en n e c e s s a r i ly we must

have d (u ) = u ( i f n o t , th e n = 4>{u) / u , which i s

a c o n tra d ic t io n ) . T herefore no s in g le 1 -p lace fu n c tio n o f Ep

can g en e ra te more than one c o n s ta n t, and th e r e s u l t fo llo w s by

Theorem 4.3*

COROLLARY 2# ^ n ^ 2 and f g e n e ra te s a doubly

t r a n s i t i v e group P o f p e m u ta t io n s cf N, th en f s a t i s f i e s

th e S6upecki c o n d itio n s .

PROOF. Each member of P , being a p e im u ta tio n , ta k es

a l l n+1 v a lu es . I f d i s a s in g le p e rm u ta tion which g e n e ra te s

P th e n , s in ce P i s t r a n s i t i v e , d must be a c i r c u la r perm uta tion

of N. But ±f <p ±3 c i r c u l a r , i t canno t g e n e ra te any non—i

• id e n t ic a l perm uta tion f ix in g an elem ent o f N, and s in c e n ^ 2

P w i l l co n ta in perm uta tions o f t h i s ty p e . T herefore no s in g le

p e rm u ta tion d g a ie r a te s P , and so th e r e s u l t fo llo w s by

Theorem 4.3*

39

4*3* Fundamental oom pleteness c r i t e r i a in En

THEOREM 4 .4 . I f n ^ 2 %nd a su b se t F g e n e ra te s the s e t

o f a l l 1-p la c e fu n c tio n s of Ep , to g e th e r w ith a fu n c tio n s a t i s fy in g

th e S-êupecki c o n d itio n s , then F i s com plete#

PROOF. We may suppose t h a t i s e s s e n t ia l f o r f , and

th en by Theorem 4.1 th e re i s a p a i r o f Séupecki p o in ts l e t th e

correspond ing v a lu es be u^ ,u® ,u®. Since u^ ^ u® we must have

(%i / , and s in ce v? / u® we must have oli / f o r some index

i ^ 2 . We s h a l l need th e fo llo w in g two lemmas;

LEMMA 4 . 1. Under the hypotheses o f Theorem 4 .4 F g en e ra te s

a 2-p la c e fu n c t io n , which we s h a l l denote by x y , having

th e fo llow ing p r o p e r t ie s ;

X 0 0 = X = 0 ^ X ,

f o r a l l X e N.

PROOF. L et a (x ) be a p erm u ta tion such th a t a (u ^ ) = 2,

a(u^ ) = 0 , a(u®) = 1, and co n sid er th e k -p la c e fu n c tio n

gen era ted by F

h (x j , «. . ,Xk ) = a ( f (xji , . * * ) ) •

V/e have

h (a ) = 2 , h(/9i ,tt2 , ) = Of k(/9) = 1,

and s in ce f ta k e s a l l n+1 v a lu es o f N, f o r each in te g e r i

(3 ^ i ^ n ) , th e re e x i s t numbers Xi ,xk such th a t

h ( x i^ , .* . ,x k ^ ) = i .

^0

L et a i and a g , b e 1—p lace fu n c tio n s such th a t

Q>i ( 0 ) = ^ 1 f &i ) = (%i J &i ( 1 ) “ 3Ci f o r 1 = f 9 « • fiif

and f o r each index j (2 ^ j k)

a j (0) = ocj f a j ( 1 ) = /9j f a j ( i ) = xj f o r i =

C onsider th e 2—p lace fu n c tio n g en e ra ted by F

b (x ,y ) = h ( a i ( x ) , ag ( y ) , . . • ,a k ( y ) ) .

Y/e have

b (0 ,0 ) = 0 , b ( 0 , l ) = 1, b ( l ,0 ) = 2,

and fo r each in te g e r i ( 3 ^ i ^ n) b ( i , i ) = i .

On w ritin g b (x ,y ) in the m a trix n o ta t io n , and le a v in g any

undeterm ined v a lu e cf b b lan k , we have

We s h a l l proceed in d u c tiv e ly , denoting by (x ,y ) a fu n c tio n o f

En which has th e fo llo w in g p ro p e r t ie s :

Cfn(0 ,x ) = X = Cm(x,0 ) fo r each x (O ^ x ^ m).

Now f g e n e ra te s a fu n c tio n of ty p e C i ( x , y ) , namely d ( b ( x ,y ) ) ,

where d i s any l ^ l a c e fu n c tio n such t h a t d(o) = 0, d ( l ) = d(2)

and any fu n c tio n o f type Cn (x ,y ) can be taken a s x ^ y .

= 1 )

41

I t i s s u f f i c i e n t th e re fo re to prove th a t f o r each in te g e r m

( l i ^ m < n ) i f F g e n e ra te s a fu n c tio n o f type Cn,(x,y) th en

F g e n e ra te s a fu n c tio n o f type C m +i(x,y),

Suppose F g e n e ra te s a fu n c tio n of type Cm(x,y) (1 ^ m < n ) ,

and l e t @1,62 be the 1—p la ce fu n c tio n s d e fin e d by

e i ( o ) = 0 , 6i ( x ) = x - 1 , f o r x = 1 , . . . , n ;

63( 0 ) = 0 , 63( 1) = 1 , 63( 2) = 0 , 63 (x ) = x - 1 ,

f o r x=3 , . . . , n ;

63( 0 ) = 0 , 63( 1 ) = 1 , 63 (x ) = x+1, f o r x=2 , . . . , n ,

where a d d it io n and s u b tra c t io n a re c a r r ie d out modulo n+1 .

C onsider the two 2—p lace fu n c tio n s g en era ted by F

6 4 (x ,y ) = e s ( c m ( e i ( x ) ,e i ( y ) ) ) ;

@6 = eg(cm(e3 (x ) , e g ( y ) ) ) .

On w r it in g 64 , eg in the m a trix n o ta tio n and le av in g un im portan t

v a lu es b la n k , we have r e s p e c t iv e ly

64(x ,y )0123

m+1

n

0 1 2 5 *«ni+1 . . n0 0 1 3 . . m+1 013

m+1 m+1 m+1

n

I t i s noTf easy to v e r i fy t h a t the 2-p la c e fu n c tio n g en era ted by F

b (e 4 (x ,y ) , ©sCxjy))

is o f type C m + i(x ,y ), and so th e r e s u l t fo llo w s .

1 2

LEMMA. 4#2. Under the hypotheses o f Theorem 4 .3 , f o r

each t r i p l e o f f ix e d numbers i , j , k (O ^ i , j , k ^ n) Fk'Ugenerates th e 2-n la c e fu n c tio n rM (x ,y ) d e fin ed by

r*Lj(x,y) = 0 o th e rw ise .

PROOF. Suppose b ( l , l ) = v , say , and s e le c t a value

i € [0 ,1 ,2 j d i f f e r e n t from v . L et b(Yi ,Y2 ) = where

Yj = 0 o r 1 , f o r j = 1 ,2 . Choose 1-p la c e fu n c tio n s S i,S 2 ,S3

such th a t

5i ( l ) = Yi , 8 i(x ) = 1-^1 f o r X / 1 ;

83( 1) = Y 2 , 83 ( x ) = I-Y2 f o r X / 1 ;

s s ( i ) = 1 , 63 (x) = 0 f o r X / i .

V/e have

r i i ( x , y ) = 8 3 ( b ( s i ( x ) , S 2 ( y ) ) ) ,

and so F gen erate s r j i ( x , y ) .

For each t r i p l e of f ix e d numbers i , j , k (O ^ i , j , k ^ n)

choose 1-p la c e fu n c tio n s t i , t j , t k ' such th a t

t i ( i ) = 1, t i ( x ) / 1 fo r X / i ;

t j ( j ) = 1 , t j ( x ) / 1 fo r X / j ;

tk * (o )^ t% '( l ) = k .

Thenr i j ( x , y ) = tk’ ( r i \ ( t L ( x ) , t j ( y ) ) ) .

43

We s h a l l now r e tu r n to th e p roof of Theorem 4* 3*

I f <f>{x,y) i s any f ix e d 2—p lace fu n c t io n o f Ep then

n n0 ( x , y ) = Z Z r i j ( x , y )

1=0 J=o

where a d d i t io n i s w ith r e s p e c t to th e o p e ra tio n x y ,

and th e fu n c tio n s r y (x ,y ) a re a s d e fin ed in Lemma 4 .2 .

T herefore by Lemmas 4*1, 4 .2 F g e n e ra te s ev ery 2—p lace fu n c tio n

of En, and by Theorem 3 .2 of C hapter 3 th i s i s a s u f f i c i e n t

c o n d itio n f o r F to be com plete.

REIvIAEK. Theorem 4*3 was f i r s t proved f o r th e case k = 2

by S-êupecki in [55]» The c o n d itio n n ^ 2 in Theorem 4 .1 cannot

be re la x e d , as i s shown by th e example in which n = 1 and F

c o n s is ts o f th e s e t o f a l l l i n e a r polynom ials (see Theorem 1 .2 ) .

I t i s easy to v e r ify th a t F i s c losed under com position ,and

con tains a fu n c tio n s a t i s f y in g th e S lupeck i co n d itio n s and a lso th e

s e t of a l l 1 -p lace fu n c tio n s o f (namely 0 ,1 ,x ,x + l) , b u t

con ta in s no n o n - lin e a r polynom ial of Ei^. T herefore F i s no t

com plete.

44

THEO HEM 4 . 5 . I f a s in g le fu n c tio n ^ g en era te s the

se t o f a l l 1 -p la c e fu n c tio n s of Ep , th en f i s com plete,

PROOF, % Theorem 4 .3 C o ro lla ry 1, s in ce f g e n e ra te s

a l l c o n s ta n ts , f s a t i s f i e s th e Sûpecki c o n d itio n s . Hence

Theorem 4 .5 fo llo w s by Theorem 4 .4 i f n ^ 2 . I f n = 1,

th e n , by argum ents s im ila r to th o se in th e proof o f Theorem 2 .2 ,

f o r a s in g le fu n c tio n f to g en e ra te a l l 1—p lace fu n c tio n s of

Ej^ i t i s n ecessa ry th a t f ( x , . . . , x ) = x + 1 , and th a t

>i’ ) + 1 (otherwise f cannot generate the co n sta n ts).

But, by Theorem 2.2. these are s u f f ic ie n t con d ition s fo r f to

be complete.

REMARK. The proofs of Theorems 4 .3 , 4 .4 are based on

the proofs o f Theorems 5 .1 , 5*4 o f [ 2 7 ] by A.Salomaa, but

the reasoning i s more d ir e c t than Salom aa's.

CORÛLLAR.Y. I f f i s a s in g le fu n c tio n w hich g en era te s th e s e t o f

a l l .i-p la c e fu n c tio n s o f Ep ( j ^ I ) , th e n f i s com plete.

PROOF. For j = 1 th i s i s Theorem 4 .5 . I f j > 1, th en by

id e n t i fy in g v a r ia b le s in s u i ta b le j-p la c e fu n c tio n s we can show

t h a t f g en era tes the s e t o f a l l 1-p la c e fu n c tio n s o f Ep and th e

r e s u l t follow s by Theorem 4 .5 .

45

THE OEM 4 .6 . I f 2 and a su b se t P g e n e ra te s the

s e t o f a l l 1 —p lace fu n c tio n s which a re n o t perm uta tions o f N ,

to g e th e r w ith a fu n c tio n f s a t i s f y in g th e S-ûpecki c o n d it io n s ,

th e n F i s com plete .

PROOF. We s h a l l prove t h a t F a lso g e n e ra te s every

p erm u ta tion of N, and co nsequen tly F g e n e ra te s th e s e t of

a l l 1-p la c e fu n c tio n s of Ep . The r e s u l t th e n fo llo w s by

Theorem 4 .4 .

V/e m%r suppose t h a t i s e s s e n t i a l f o r f , and th en

b y Theorem 4.1 th e re is a p a i r o f S*êupeclci p o in ts a,/3 for f ;

l e t th e corresponding v a lu es be u^ ,u^ ,u® . We may suppose t h a t

the n 4- 1 d i s t i n c t e lem ents o f N a re u^ ,u^ ,u® ,u^ , . . . ,

and s in ce f assumes a l l n + 1 v a lu e s ,fo r each in te g e r i

( 4 i ^ n+1 ) th e re e x i s t numbers Xj_ ^, . . . ,xi< such th a t

f (X] . ,X|< ) = u .

L e t p(x) be any f ix e d perm uta tion o f N, and choose

k 1-p la c e fu n c tio n s q i , . . . , q k d e fin ed by

Ç h (p " i(u i) ) = %i, Ç b(p ri(u 2 )) = ^ 1 , qi(p"Îu® )) = /? i ,

fo r i = 4 , . . . , n+1;

and f o r each index 3 (2 ^ j ^ k)

qj(p"^(u^)) = a j , qj(p“ ^(u®)) = aj , qj (p“ (u® ) ) = /9j ,

qj = xj for i = 4 , . . . ,n + 1 .

4 &

Since q i , . . . , q k tak e a t most n values th e y a re n o t perm uta tions

o f N and so i h ^ a re g en era ted by P . T herefore P g en e ra te s

the 1-p la c e fu n c tio n

r ( x ) = f ( q i ( x ) ,q 2 ( x ) , . . . , q k ( x ) ) ,

and r ( x ) s a t i s f i e s

r(p ~ ^ (x ) ) = X

That i s

r ( x ) = p (x ) .

T herefore P g e n e ra te s p (x) and th e r e s u l t fo llo w s .

4 7

THEOREM 4*7* I f f is a s in g le fu n c tio n of Ep which

g e n e ra te s th e s e t of a l l 1-p la c e fu n c tio n s which a re n o t

perm utations o f N, th e n f i s com plete.

PROOF. ^ Theorem 4 .3 C o ro lla ry 1 , s in ce f g e n e ra te s

a l l c o n s ta n ts , f s a t i s f i e s th e S-ûpecki c o n d itio n s .

Hence Theorem 4 .7 fo llow s from Theorem 4 .6 i f n ^ 2 .

I f n = 1 , th e n th e two 1-p la c e fu n c tio n s of E ^ which a r e not

p erm uta tions of a re th e c o n s ta n ts 0 ,1 . N ecessary

co n d itio n s f o r f to g en era te th e c o n s ta n ts 0,1 a re th a t

f 1 , and Ih a t f ( x , . . . , x ) = x+1 ( f o r i f

f ( x , . , . x ) = i then f cannot g en era te 1 - i , and i f f ( x , . . . , x ) = x

then f g e n e ra te s no o th e r 1-p la c e f u n c t io n ) . Theorem 2 .2

th e se a re a ls o s u f f i c i e n t c o n d itio n s fo r f to be com plete.

REMARKS. Theorems 4 .6 , 4 .7 were o r ig in a l ly proved in [A-9 ]

by Y ablonslcii, However, the above p roo fs do not use h is re a so n in g .

The s e t ©f a l l 1-p la c e fu n c tio n s o f i s th e f u l l tra n s fo rm a tio nsemigroup of N in to i t s e l f , the semigroup o p e ra tio n being com position .The idem potents o f th i s semigroup a re th e 1-p la c e fu n c tio n s f € such

th a t f ( f ( x ) ) In [ l l j , Howie has proved th a t th e s e t of a l l 1-p la c e

fu n c tio n s o f Ep which a re n o t p e rm u ta tions o f N,. i s generated by th e

su b se t o f a l l idempo te n ts o f th e type [ 2 , 1 , . . . , 1 ] , w ith n u n i t s . It*

fo llow s th e re fo re by Theorem 4 .7 , t h a t i f a s in g le fu n c tio n f e Ep

g en e ra te s every idem potent o f "type [ 2 , 1 , . . . , l j , w ith n u n i t s , "then

f i s com plete.

48

V - MAIN RESULTS - V

5 . 1 • In tro d u c tio n

In Theorem 4 .6 we proved th a t a s in g le fu n c tio n o f Ep which

g en e ra te s the s e t o f a l l 1-p la c e fu n c tio n s o f Ep i s com plete. In th is

C hapter we s h a l l be concerned w ith com pleteness c r i t e r i a fo r a su b se t o f

Ep which g e n e ra te s perm u ta tio n s o f N. These c r i t e r i a a re p re sen te d in

Theorems 5 .1 , 5 .2 , 5 .3 o f § 5 .5 .

In § 5 .2 we c o l le c t some f a i r l y s tra ig h t- fo rw a rd a u x i l ia r y r e s u l t s ,

w h ile in § 5 .3 and § 5 .4 we in v e s t ig a te in d e t a i l an e x ce p tio n a l s i tu a t io n

which may a r is e when n+1 i s a power o f 2 . F in a l ly in ^ 5 .5 we s t a t e

and prove our main theorem s, and compare them w ith e a r l i e r r e s u l t s .

We s h a l l make use o f v a rio u s n o tio n s and d e f in i t io n s in tro d u ced in

p rev io u s c h a p te rs . F o r example; S -^peck i co n d itio n s (C hapter 4 ) ,

s -p ly t r a n s i t i v i t y (C hapter 1) and p ro p e r t ie s o f 1-p la c e fu n c tio n s

(C hapter I ) .

5 . 2 . A u x ilia ry r e s u l t s

LEMMA 5. 1. n ^ 3 , and a su b se t F g e n e ra te s a t r i p l y t r a n s i t i v e

group o f perm uta tions of- N, to g e th e r w ith a fu n c tio n f ( x i , . . . ,% k )

s a t i s f y in g th e S ^ p e c k i c o n d itio n s , th e n F g en era te s a 1-p la c e fu n c tio n

which tak es more th a n 1 and l e s s th a n n+1 d i s t i n c t v a lu e s .

PROOF, By Theorem 4 .2 th e re i s a p o in t y (= Yi>****Yk) su ch t h a t ,

f o r s e ts S i , . . . ,S k d e fin ed by Sj, = N / îy i l , Tor i = 1 , . . . , k ,

f ( ? i , . . . > ^ ) — N.

49

From th i s we deduce th a t th e re i s a p o in t y *(* Yi * , • • • ,Yk * ), where

YL* f St fo r i = such t h a t f(Y*) = T(y )* Choose k

perm uta tions g en era ted by F such t h a t (c ) =Y l>

r t ( l ) = Y t* , f o r i = 1, . . . , k . C onsider th e 1-p la c e fu n c tio n

g en era ted by F

f (x) = f (rj. ( x ) , • • . , 1 (x) ) •

I t i s easy to see t h a t f (0 ) = f ( l ) , and so f(x ) tak es l e s s th an n+1

v a lu e s .

To complete the p ro o f i t s u f f ic e s to c o n sid e r th e case when

f (x ) i s c o n s ta n t. We may suppose t h a t Xi i s e s s e n t ia l fo r f , and

th en by Theorem 4.1 th e re i s a p a i r o f Séupecki p o in ts a , l e t the

co rrespond ing values be u^ , i f , i f . Since f i s c o n s ta n t, i t

fo llo w s th a t F g en era te s a l l c o n s ta n ts , and so F g enera tes th e

1-p la c e fu n c tio n f (x,aa , . • .yxk ) • This fu n c tio n takes th e va lues

u , i f , and th e p roo f i s com pleted i f i t does n o t a ls o tak e th e value

if . Suppose then th a t

f(oC ,R2 ,.. * . ,(%k ) = U ,

where 4= cti , ^ i . Choose a f u r th e r p o in t f (=fi ) such t h a t

4= and f o r each index i ( 2 ^ i ^ k) th e 3 v a lu es i

a re c o in c id e n t i f a t = /3 l , o r d i s t i n c t i f a t 4 P i* C onsider th e

value o f f ( f ) . I f f ( c ) = if , th e n we choose k 1-p la c e fu n c tio n s

8 i , . . . , 8 k g en era ted by F such th a t s t ( 0 ) = a t , s t ( l ) = P t ,

S t (2) = f t . Tor i = 1 , . . . , k (accord ing as a t = P t o r a t 4 P i» 81

i s a c o n s ta n t o r a p e rm u ta tion r e s p e c t iv e ly ) . C onsider the 1-p la c e

fu n c tio n gen era ted by F

so

b(x) = f ( s i ( x ) , . . . , S k ( x ) ) .

We have b (0 ) = ^ b ( l ) = b (2 ) = i f , and so b ta k es more th a n 1

and l e s s th a n n+1 values#. I f f (f ) 4 > th e n we choose a f u r th e r

perm u ta tio n s* g en era ted by F such th a t S i '( o ) = , S i* ( l) = /3 i,

8i *(2) = f 1 # C onsider th e 1-p la c e fu n c tio n gen era ted by F

b*(x) = f ( s i * ( x ) ,% ( x ) , . . . ,S k ( x ) ) .

We have b '( 0 ) = b * (l) = u® 4 t ' ( 2 ) , and so b* tak es more th a n 1 and

le s s th an n+1 d i s t i n c t v a lu e s . The p ro o f i s com pleted,

REMARK, Using th e above p ro o f Salomaa ([20 ] Lemma 1,4) d e riv e s

th e same co n c lu sio n when F g en e ra te s the a l te r n a t in g group An-Ki on

N ( to g e th e r w ith a fu n c tio n s a t i s f y in g th e S6upecki c o n d it io n s ) ,

b u t as h is p ro o f uses only th e t r i p l e t r a n s i t i v i t y o f Aq^we have a p p lie d

i t to th i s more g e n e ra l case .

The d b n d itio n n ^ 3 in Lemma 5.1 cannot be re la x e d ,a s i s shown

by th e example in w hich n = 2, and F c o n s is ts o f a l l l i n e a r fu n c tio n s ,

th a t i s , a l l fu n c tio n s e x p re ss ib le in th e form

f ( x i , . . . , 2k) = lb + ihXi + . . . + (modulo 3 ) .

I t i s e asy to v e r i f y t h a t F i s c lo sed under com position and co n ta in s

th e sym m etric group o f perm uta tions % on N, to g e th e r w ith a fu n c tio n

s a t i s f y in g th e Sûpecki c o n d itio n s , fo r example x + y (modulo 3 ) , b u t

F co n ta in s no 1-p la c e fu n c tio n ta k in g e x a c tly 2 v a lu e s .

51

LEIÆMA. 5 .2 . I f n ^ 2 , and a su b se t F g e n e ra te s a t r i p l y

t r a n s i t i v e group o f p e rm u ta tio n s of N, to g e th e r w ith a 1—p la c e

fu n c tio n ta k in g e x a c tly t v a lu e s , where 1 < t < n + 1 , th e n

F g e n e ra te s a 1-p la c e fu n c tio n ta k in g e x a c tly 2 v a lu e s .

PROOF. Suppose F g e n e ra te s th e 1—p lace # (x ) ta k in g

t d i s t i n c t v a lu e s v ^ , . . . , v ^ , where 1 < t < n + 1 .

I f t = 2 th e re i s no th ing to p ro v e ; i t i s th e re fo re s u f f i c i e n t

to prove th a t i f t ^ 3 then th e hypotheses imply th a t F g e n e ra te s

a 1—p lace fu n c tio n ta k in g v a lu e s , •vdiere 1 < t* < t .

L et V*’ s [x : # (x ) = v* 'j. S ince t < n , a t l e a s t one o f

th e se s e t s , say , co n ta in s a t l e a s t two members, a,/9 say .

Let Y be any member of V®, l e t g (x ) be a p e rm u ta tion

g en era ted by F , such th a t

g ( v ^ ) = a , g(v®) = /9, g (v ® )= Y ,

and l e t 0 (x ) deno te the fu n c tio n ^ (g (< ^ (x ))). We have

e(vM = I v ^ j , e(v®) = JvM, e(v®) = [v®J,

w hile on each o f th e rem aining t —3 s e ts V * ,. . . ,V ^ th e fu n c tio n

6 ta k e s a c o n s ta n t v a lu e . Thus 6 ta k e s t* v a lu e s , where

1 < t* < t .

LEMMA 5*3. Under the c o n d itio n s o f Lemma 5 .2 , F g e n e ra te s

a l l c o n s ta n ts .

52

PROOF. Let T be any g iv en member of N* Lemma 5*2,

F g e n e ra te s a 1-p la c e fu n c tio n ^ (x ) ta k in g e x a c tly 2 v a lu e s ,

v^,v® say . At l e a s t one o f ih e se v a lu e s , v^ say , must be

assumed a t l e a s t tw ice, a t ,a® say . Let g ,h be

perm uta tions generated by F such t h a t g(v^ ) = v , and

h(v^ ) =s , h(v® ) = a® • Then th e 1 -p la c e fu n c tio n

g (^ ( h ( ^ ( x ) ) ) ) tak es th e c o n s ta n t v a lu e v ,

LSI&IA 5 .4 . ^ n ^ 1 , and a su b se t F g e n e ra te s

a doubly t r a n s i t i v e group o f perm uta tions o f N, to g e th e r

w ith a 1-p la c e fu n c tio n g ta k in g e x a c tly 2 v a lues and

a 2—p lace fu n c tio n h such th a t

h (0 ,0 ) = 0 ; h ( 0 , l ) = h ( l , l ) = h ( l ,0 ) = 1,

th en F g e n e ra te s every 1—p la c e fu n c tio n of type [ n . l l .

PROOF. Choose numbers a,/9 such th a t g (a ) ^ g (p ) .

Choose perm u ta tions r ( x ) and S i ( x ) , . . . ,Sp(x) g en era ted by F

such th a t r ( g ( a ) ) = 0 , r(g (/? )) = 1; s i (o ) = a , s i ( i ) =

f o r i = 1 , . . . , n . L et t i ( x ) = r ( g ( s ( , ( x ) ) . Then th e 1 -p la c e

fu n c tio n g en era ted by F

^ (x ) = h [ t i ( x ) , h [ t 2 ( x ) , . . • ,h [tn ^ 1 ( x ) , t p ( x ) ] • . , ] ]

assumes th e v a lu e 0 f o r x = 0 and th e v a lu e 1 o th e rw ise .

Now l e t ^ (x ) be any o th e r 1-p la c e fu n c tio n of ty p e [ n , l ] ,

assuming th e v a lu e v^ f o r x = y and th e value Vg / V;i o th e rw ise .

55

Choose p en in ita tio n s s , t gen era ted by F such th a t s ( l ) = v^ ,

s (2 ) = v^; t ( y ) “ 1 . Then *(x) = s( * ( t ( z ) ) ) , and th e re fo re

ip i s g en era ted by F*

LE&1MA 5 .5 . I T n ^ 2 , th en fo r any n a tu ra l numbers

t , b ^ , , . , , b ^ s a t i s f y in g 1 < t < n+ 1 , b^ + . . . + b^ * n ,

th e s e t o f a l l 1- p lace fu n c tio n s of ty p e , [ b ^ , . . , ,b ^ ] g e n e ra te s

every 1- p lace fu n c tio n of type [b^ + b ^ ,b ^ , , . , ,b^] ,

PROOF. L et 0 be an a r b i t r a r y f ix e d , fu n c tio n of ty p e

[b^ + b ^ ,b ^ , . . . , b ^ ] assuming th e value v^ e x a c tly b^ + b^ t im e s .g

Since t .< n + 1, 0 has a m issin g v a lu e , v say , and any fu n c tio n

(j)(x) d e fin ed by: 4>(x) = 0(x ) , excep t fo r b^ v a lu es o f x fo r

which 4»(x) = v^ ; 0(x ) = v^ , w i l l c le a r ly be o f type [b ^ ,b ^ , . , , ,b^] ,

Choose a fu n c tio n <t»* of type [ b ^ ,b ^ , . , . ,b^(| such th a t

4>*(0(x ) ) = 0(x ) ; * '(v * ) = v ^ .......................... ( i )

We can c o n s tru c t (j>' as fo llow s: s in ce t < n + 1 , fo r some

fix e d index i ( l ^ i ; ^ t ) b j ^ ^ 2 , Let th e t - 1 d i s t i n c t v a lu es

o f 0 be v ^ ,0* , . . . , 0^ ^ , and l e t (j)' ( s a t i s f y in g c o n d itio n ( i ) ) tak e

th e va lu e v^ e x a c tly b^ tim es and th e v a lu es v ^ ,0* , . . . , 0^ e x a c tly

b ^ , . . . ,b ^ _ ^ , t im e s ;re s p e c t iv e ly . S ince v i s a

m issing v a lu e o f 0, i s of re q u ire d ty p e .

F in a l ly we have

0(x ) = <j)»( * ( x ) ) ,

and th e r e s u l t fo llo w s ,

REMARK, This sim ple r e s u l t iff Lemma 1 .3 o f [gs] ,

54

LEMMA. 5*6* IT n ^ 2 , and a su b se t F g en era te s a t r i p l y

t r a n s i t i v e group o f p em iu ta tio n s of N, to g e th e r w ith a fu n c tio n

s a t i s f y in g th e S-êupecki co n d itio n s an d , f o r some

in te g e r m s a t i s f y in g 1 ^ m ^ n - 1, every 1 -p la c e fu n c tio n o f

type [n+1-m ,1, . • • ,1 ] (w ith m u n i t s ) , then F g e n e ra te s every

1-p la c e fu n c tio n o f ty p e [n -m ,1, . . . , l ] (w ith an a d d it io n a l u n i t ) .

PROOF* app ly ing Lemma 5*5 we can show th a t F g e n e ra te s

every 1-p la c e fu n c tio n which ta k e s some value a t l e a s t n+1-m tim es .

V / V V\For each v c N, th e re i s a p o in t x = (x^ , . . . ,Xk ) such

th a t f (x ^ ) = V. We may suppose w ith o u t lo s s o f g e n e r a l i ty th a t

Xi i s an e s s e n t i a l v a r ia b le f o r f , and th e n by Theorem 4*1 th e re

i s a p a ir o f S>ftupecki p o in ts a,/9; l e t th e co rrespond ing v a lu es

be u^ ,u® ,u® .

Let e (x ) be any f ix e d 1-p la c e fu n c tio n of ty p e [n -m ,1 , . . . , 1 ] ,

w ith

e ( a i ) =3 w° ( i= 1 , . . . , n - m ) ; e (b j ) = ( j = 1 , . . . ,m + l )

Choose a p e rm u ta tio n r g en era ted by F such th a t

r ( u ^ ) = w^, r(u® ) = w°, r(u® ) = w®.

C onsider th e 1-p la c e fu n c tio n s h^ and h^(& = 2 , . . . , k ) d e fin e d

a s fo llo w s;

55

h i(x ) = I

/?x fo r X rs

«1 f o r X = b i

r " i (wJ) f o r X = b a , • •

h^(x ) =

[

fo r X = 8-1 , • • • ,b%

f o r X = b2r - i (w J )

^6 f o r X = b3,*«»,bn»n*

We see t h a t each of th e se fu n c tio n s ta k e s one value a t l e a s t

n+1-m tim es, and i s th e re fo re gen era ted by F , b u t i t i s easy

to v e r i f y t h a t

e (x ) = r ( f ( h i ( x ) , h g ( x ) , . . . , h k ( x ) ) ) ,

and so e (x ) i s a lso g en e ra ted by F and th e lemma i s p roved .

COROLLARY. Under the c o n d itio n s o f Lemma 5*6 , F g e n e ra te s

every 1-p la c e fu n c t io n .

PROOF. re p e a te d a p p lic a tio n s o f Lemma 5*6, we can

deduce t h a t F g e n e ra te s every 1-p la c e fu n c tio n of type [1 ,1 ,

th a t i s , every pe rm u ta tio n , and ev e iy 1-p la c e fu n c tio n o f type

[ 2 , 1 , . , . , l ] . app ly ing Lemma 5*5, we can deduce th a t F a ls o

g e n e ra te s every 1—p lace fu n c tio n which i s n o t a p e rm u ta tio n .

LEMMA 5*7* I f n ^ 2 , th e s e t o f perm u ta tions o f N th a t

a re s e lf -c o n ju g a te under a g iven 1-p la c e fu n c tio n s (x ) ^ x i s not

doubly t r a n s i t i v e .

56

PROOF. I f Xq i s any elem ent of N fo r which s(xo ) / Xo,

and r i s any p erm u ta tio n s e lf -c o n ju g a te under s (x ) f o r which

r(xo ) = xb , th en a ls o r ( s ( x o ) ) = s ( r (x o ) ) = s(xq)*

5•3* An e x c e p tio n a l case

We s h a l l now show th a t i f n ^ 3 and F s a t i s f i e s th e

c o n d itio n s o f Lemmas 5.1 and 5*2, then i t w i l l a lso s a t i s f y

th e a d d it io n a l c o n d itio n of Lemma 5*4, u n le ss n+1 i s a power

of 2 and F has a very s p e c ia l form . We s h a l l f u r th e r

in v e s t ig a te t h i s s p e c ia l form in § 5 *4 , in p a r t i c u la r showing

in Lemma 5*10 C o ro lla ry 2 th a t i t can indeed occu r.

LEMMA. 5*8* Let n ^ 2 and suppose th a t F g e n e ra te s

a t r i p l y t r a n s i t i v e group o f perm u ta tions of N, to g e th e r w ith

a fu n c tio n f ( x ^ , . . . ,X k ) s a t i s f y in g th e S’êupecki co n d itio n s

and a 1-p la c e fu n c tio n g (x ) tak in g e x a c tly 2 v a lu e s , bu t F

does not g en era te any 2-p la c e fu n c tio n h (x ,y ) such th a t

h (0 , 0 ) = 0 ; h (0 , l ) = h ( l , l ) = h ( l , 0) = 1 . (a )

Then th e re e x i s t s a number v e N such t h a t ( i ) v / [0 ,1 ,2 }

and ( i i ) F does not g en e ra te any 1-p la c e fu n c tio n th a t on th e s e t

[ 0 , 1 , 2,v j ta k es one v a lu e e x a c tly 3 tiroes and an o th er va lue e x a c tly

once. Moreover F g e n e ra te s 1-p la c e fu n c tio n s g ^ ,* .* ,g ^ each

ta k in g th e v a lu es 0,1 and s a t i s f y in g g l(o ) = 0 ( i = 1 , . . . , j ) ,

such t h a t th e correspondence x (g^ ( x ) , . • . ,g^ (x) ) between elem ents

of N and sequences of j 0*s and 1*s i s o n e -to -o n e . In

p a r t i c u la r n+1 = 2'^.

57

*PROOF. We may suppose th a t i s an e s s e n t ia l v a r ia b le

f o r f , and th en by Theorem 4 .1 . th e re i s a p a i r of S-ôupecki

p o in ts a,/9; l e t th e co rrespond ing th re e v a lu es be u^ ,u® ,u® .

Since u^ / u® we must have a i ^ /9i and since u® / u® we

must have a t / ^ t Tor some index i ^ 2 .

Choose k fu n c tio n s r j ( x ) ( j = 1 , . . . , k ) gen era ted by F

such th a t r j ( l ) = a j ; r j (2 ) = /9j . We can always make t h i s

choice because i f a j then we may ta k e a s r j ( x ) a s u i ta b le

perm uta tion g en e ra ted by F, and i f aj = /9j th en s in c e , by

Lemma 5*3, F g e n e ra te s a l l c o n s ta n ts we may tak e r j (x) = a j .

Let s (x ) be a p erm uta tion g en era ted by F such th a t

s (u ^ ) = V, f o r V 3 0 , 1 , 2 .

C onsider the 2 -p lace fu n c tio n g e n e ra te d by F

T *(x,y) = s ( f ( r i ( x ) , r g ( y ) , . . . , r k ( y ) ) ) .

V/e have

f » ( 0 , 0 ) = 0 , f * ( l , 0 ) = 1 , = 2 , f » ( 0 , l ) = V ( s a y ) ,

and from th e re a so n in g so f a r on ly th e p resence o f such a fu n c tio n

need be r e ta in e d .

Since n ^ 2 th e 1-p la c e fu n c tio n g tak es one of i t s

2 v a lu es a t l e a s t tw ice , and so we may choose d i s t i n c t numbers

a,/5?Y N such th a t g (a ) / g(/9) = gCr)*

My o r ig in a l p ro o f was in ad eq u a te . I am g r a te f u l to Dr. Davies fo r p o in tin g t h i s o u t, and fo r h is h e lp in the p re se n t p to o f .

58

Suppose i f p o s s ib le t h a t v e [0 ,1 ,2 } , and l e t v = Kg,

where ,«2 a,re th e numbers 0 ,1 ,2 in some o rd e r. Choose

f [ 0,1 i so t h a t f*(/Ji,jUg) = Ko, and choose perm uta tions

b i , t g , t 3 , t 4 gen era ted by F such th a t

t i ( 0 ) = ^ 1 , t i ( l ) = 1 - tg (o ) = /ia , t g ( l ) = 1 - ^3 ;

taCao) = a, taCai) = /5, ta (ag ) = y; t4(g(a)) = 0, t 4 (g(^)) = 1.

Then i t i s easy to v e r i f y t h a t th e 2 -p lace fu n c tio n (g en e ra ted

by F)

h (x ,y ) = t 4 ( g ( t a ( f * ( t i ( x ) , t g ( y ) ) ) ) )

s a t i s f i e s ( a ) , c o n tra ry to h y p o th e s is . Thus a s s e r t io n ( i ) i s p roved .

Suppose i f p o s s ib le t h a t F g e n e ra te s a 1-p la c e fu n c tio n

d (x ) such th a t d(%o) / d ( % i ) = d ( a g ) = d ( % a ) , where Go,Gi,K2,%3

a re th e numbers 0 ,1 ,2 ,v in some o rd e r. Let S3 be a p erm u ta tion

g en era ted by F such th a t S a ( d ( a o ) ) = 0 and S a ( d ( a i ) ) = 1, choose

1 19^2 ^ [ 0 , 1 } so t h a t f *(jL/i ,/jg ) = Go, and choose p e rm u ta tions

S i ,Sg g en era ted by F such th a t

81( 0 ) = / i i , 81( 1 ) = 1 - ^ 1 ; Sg(0) = /Jg, S g (l) = 1 - /ig.

Then i t i s easy to v e r ify t h a t th e 2 -p lace fu n c tio n (g en e ra ted by F)

h (x ,y ) = 5 a (d ( f* (x ,y ) ) )

s a t i s f i e s (a ) c o n tra ry to h y p o th e s is . Thus a s s e r t io n ( i i ) i s proved.

For th e l a s t p a r t of th e p roo f i t i s convenient to fo rm u la te

an in te rm e d ia te r e s u l t .

59

ASSERTION ( i l l ) . For each in te g e r k such th a t 2 ^ ^ n+1,

F g e n e ra te s k 1-p la c e fu n c tio n s each ta k in g e x a c tly

th e v a lu es 0 ,1 and s a t i s f y in g g*’(0 ) = 0 ( i = 1 , . , , , k ) such

th a t (w ith i i , . . . , i k running over a l l sequences of k 0 ' s and 1 *s )^1 * * *

(a ) fo r each f ix e d K, 1 ^ K ^ k , the s e t s V

= [x ; g^(x) = i^ f o r 1 ^ 6 ^ K} a l l have th e same power,

(b) whenever r i s a perm u ta tio n g en era ted by F f o r which

r ( o ) , r ( l ) , r ( 2 ) e V we a ls o have r ( v ) e V ^

I t i s c l e a r t h a t th e rem ainder o f Lemma 5 w i l l fo llo w from

A sse rtio n ( i i i ) , s in c e i f j i s th e l a r g e s t in te g e r fo r which

2^ ^ n+1 th en we must have 2* = n+1.

PROOF OF ASSERTION ( i i i ) . We use in d u c tio n on k , s ta r t in g

w ith th e t r i v i a l case k = 0 , vhere i t i s understood th a t^1 • •

V = N f o r k = 0 . Suppose th e r e s u l t t r u e fo r k , and

l e t 2*^^ ^ n .

Since g^(o ) = 0 , f o r i = 1 , . . . , k , we have 0 e -(w ith k z e ro s ) . Denote by 1 ' any o th e r elem ent of , andl e t t 4 , t s be perm utations g en era ted by F such th a t

t4 (g (a ) ) = 0 , t4 (g (^ )) = 1; ts (o) = o t, tg ( 1 ' ) = ^ .

Let g' '*’^ (x ) = t 4 ( g ( t e ( x ) ) ; then g‘ '^^(x) i s g en e ra ted by F ,

tak es e x a c tly th e v a lu es 0 ,1 , and s a t i s f i e s gk+ i(o ) = 0 ,

gk+ 1, Let

y i i .* . i k ik + i _ ; g^(x ) = i^ fo r 1 < -6 ^ k+1 j .

6 0

I t w i l l be s u f f ic ie n t to prove th a t f o r a l l i i , . . . , i k th e

two s e t s jjave th e same power and th a t fo r

a l l i i , • , . ik , ik + 1 , whenever r i s a perm uta tion g en era ted by

P f o r which r ( l ) , r ( 2 ) , r ( 3 ) € we a ls o have

r ( v ) €

Let us f i r s t show th a t none o f th e s e ts

empty. Suppose i f p o s s ib le t h a t empty; then

g^* i s c o n s ta n t on Choose a perm u ta tion r i (x )

gen era ted by F such t h a t i*i(o) = 0 , r i ( l ) = 1* and i 'i ( 2 ) = u# * # ( # (

where u i s some elem ent of and. l e t r i ( v ) e v ^ ^

say . I f i ^ = i ^ fo r 1 ^ 6 ^ k then r i ( v ) c and

th e 1 -p lace fu n c tio n g* * (r^ (x ) ) ta k e s one va lue e x a c tly 3 tim es

on [ 0 ,1 ,2 ,v j , c o n tra d ic t in g ( i i ) . Hence i^ / i ^ f o r some C,

b u t th e n th e fu n c tio n g ^ ( r i ( x ) ) ta k es one value e x a c tly 3 tim es

on [ 0 ,1 ,2 ,v ] , ag a in c o n tra d ic t in g ( i i ) .

Next we s h a l l show th a t fo r a l l i i , . . . , i k th e two non-empty

s e ts ^ h ^ . . . i k i th e same power. Suppose i f

p o ss ib le th a t f o r example

= À > fi = > 0

(where c i s an a b b re v ia tio n f o r th e sequence i i . . . i k ) #

Let 0*', 1” be a r b i t r a r y elem ents of r e s p e c t iv e ly , and

denote th e elem ents of by Vq ,V i, . . .,v;\^^ where v° = 0 " .

For each i = 1 , . . . ,A ,—1 choose a p e m u ta t io n cj, g en era ted by F

such t h a t

61

c i ( 0 ) = 0", o i ( l ) = 1", c i ( 2 ) = VI.

In view o f th e in d u c tio n h y p o th esis ( b ) , we have

c i(v ) € f o r i = - 1 . Now f o r th e 1-p la c e

fu n c tio n ^(c j,(x ))w e have

g k + i ( o i ( 0 ) ) = g k + i ( c i ( 2 ) ) = 0; g k + i ( c i ( l ) ) = 1,

and we deduce from ( i i ) th at g * * ^ (c i(v )) = 1 , that i s , C[(v) €

for i = - 1 . On the other hand, c i ( l ) = 1” for a l l i ,

and s in ce X > ^ i t fo llow s th a t fo r some i ^ j we have

c t (v ) = c j (v ) = v say, iidiere v c

Choose a pe rm u ta tio n c , g en era ted by F , such t h a t

c (v ^ ) = 0” , o ( v j ) = V ', c(v®) = V®.

Consider th e two 1—p lace fu n c tio n s g e n e ra te d by F

a f * i ( o ( c i ( x ) ) ) ; g k + i ( c ( o j ( x ) ) ) .

They co incide in va lue a t th e p o in ts x = 0 ,1 ,v , b u t d i f f e r

f o r X = 2; co n seq u en tly , one of them ta k e s some value e x a c tly

3 tim es on [ 0 ,1 ,2 ,v j , co n trad ic ting ( i i ) .

F in a l ly we show th a t f o r a l l i i , . . . , i k , whenever r i s

a p erm u ta tion g en era ted by F such th a t r ( 0 ) , r ( l ) ,r-(2) e ' * *^k+1 ^

we a lso have r ( v ) € * * *^k+1 (This f i n a l c o n d itio n on ly a p p lie s

when I • • ‘^k+i I ^ ^ Suppose i f p o s s ib le th a t f o r example

r ( 0 ) , r ( l ) , r ( 2 ) e r ( v ) /

(where c" i s an a b b re v ia tio n f o r th e sequence i i # . . i k ) #

6 2

Since r ( v ) e by th e in d u c tio n hy p o th esis (b ) , we have

r ( v ) € ' ^ ^ 9 and th e 1-p la c e fu n c tio n g ^ + i( r (x ) ) ta k es one

va lue e x a c tly 3 tim es on } 0 ,1 ,2 ,v ] , con trad icting ( i i ) .

The proof i s com plete.

5 .4 . In tro d u c tio n of v e c to r n o ta tio n

In t h i s s e c tio n we assume th e conclusion of Lenma 5*8;

in p a r t i c u la r n+1 = 2^, and P g en e ra te s 1-p la c e fu n c tio n s

ta k in g the v a lu es Q,1 such th a t th e correspondence

X -* [ g ^ ( x ) , . • . ,gJ (x) j between elem ents of N and ro w -vec to rs

o f O 's and 1 's i s o n e -to -o n e . ( i t should always be c le a r

from th e co n tex t w hether [ a , b , c , . . . j deno tes a row v e c to r

o r th e s e t whose elem ents a re a , b , c , . . . . ) We s h a l l deno te by

X = [x^j =

th e row -vector corresponding to x , so t h a t x^ = g ^ (x ) .

We a ls o in tro d u ce a correspond ing n o ta t io n f o r fu n c tio n s

h ( x i , . . , ,Xk) € En ; thus h ( x i , . . . ,x k ) i s th e row -vecto r

[ h ( x i , . «. jXk) I 3 [ h ( x i , . . . ,Xk] , . . . ,h ^ (x ^ , • . • , x ^ ) ] , where

h ^ (x i , . . .,X k) = g” ( h ( x i , . . . ,X k ) ) . V/e can a ls o reg a rd h a s

a fu n c tio n h ( x i , . . . ,X k ) of th e corresponding k ro w -v ec to rs ;p

then each component h i s a fu n c tio n of th e k j v a r ia b le s

xj^(l ^ /J ^ k , 1 ^ A ^ j ) , where x^ = [3V , . T h e value0

o f each of th e se v a r ia b le s , and of h , i s 0 o r 1, and so h e Ei*

63

I t fo llow s by Theorem 2.1 o f C hapter 2 t h a t we may now w rite

h^ u n iq u e ly a s a polynom ial over th e f i e l d Zg s [ 0 , l j o f re s id u e s .

modulo 2 , th a t i s a s a sum (modulo 2 ) of d i s t i n c t term s each of

th e form

P i* '" jUr

(w ith d i s t i n c t f a c t o r s ) . Such a term i s n a tu r a l ly c a l le d l in e a r

i f r = 1 , a polynom ial i s l in e a r i f each of i t s te r n s i s l i n e a r ,

and h ( x i , . . . ,X k ) e Ep i s c a l le d l in e a r ^ i f each o f th e components

h ^ (x i , . . . ,Xk) i s a l in e a r po lynom ial. In th i s case we can w r ite

+ ho

In p a r t i c u la r i f h (x ) e En i s a 1 -p la c e l in e a r fu n c tio n then

h (x ) = ^ + ^ ,

where H 3 [h^] i s a j x j m a trix over Zg . Let us deno te

by 0 th e v e c to r [ 0 , . . . , 0 j , and b y e ^ ( l ^ ^ ^ j ) th e v e c to r

d i f f e r in g from 0 in having 1 in th e & -th p la c e ; by

0 , e i , . . . , e j we den o te th e co rrespond ing elem ents of N.

No co n fu sio n w i l l a r i s e from th i s n o ta t io n , s in ce in Lemma 5 .8

we have chosen our correspondence x -» [ g ^ (x ) , . . .,g^ ( x ) } such

th a t each g^(x) (6 s 1 , . . . , j ) s a t i s f i e s g^(o ) = 0 . Hence

th e zero elem ent of N s a t i s f i e s 0 [ 0 , . . . , 0 ] 3 0 .

t We s h a l l no t be u s in g th e word * lin e a r* in the sense o f a fu n c tio n o f th e form Zaj,xi + b modulo n , in Tdiich i t has been used f o r example in [Z9],[49].

64

I f th e 1 -p lace fu n c tio n h e En s a t i s f i e s h (o ) = 0 , then we have ^ = 0,

and in th i s case th e & -th row v ec to r of H i s sim ply th e v ec to r h (e ^ ) .

I f x ,y € N we denote by x 0 y th e elem ent z € N such t h a t z = x + jg,

th a t i s , g ^ (z ) =s g^(x ) + g^(y ) (modulo 2) f o r 6 = 1 , . . . , j . V/e s h a l l use

w ith o u t s p e c ia l m ention v a rio u s obvious p ro p e r t ie s o f t h i s a d d i t io n , f o r

example, th e com m utative, a s s o c ia t iv e and idempotency (x 0 x = O) law s.

F in a l ly we s h a l l deno te by Z th e s e t o f a l l quadruples [ft,/? y 8} of

d i s t i n c t elem ents of N f o r which a 0 /9 0 Y 0 S = 0 . Observe t h a t t h i s

i s e q u iv a le n t to 8 = a a.nd t o a + / 9 + ^ + 6 = 0 .

We now pass to th e proof o f v a rio u s r e s u l t s in whose fo rm u la tio n s

th e above n o tio n s p lay an e s s e n t i a l r o l e .

LEMT.IA. 5 •8 (a ) . Under th e hypotheses of Lemma 5*8 every 1 -p la c e

fu n c tio n s (x ) g en era ted by F s a t i s f i e s s(o) 0 s ( l ) 0 s (2 ) 0 s (v ) = 0 .

PROOF. In view of a s s e r t io n ( i i ) o f Lemma 5*8 each fu n c tio n

g ^ (s (x ) ) = 1 , . . . , j ) tak es each o f th e v a lu es 0,1 an even number

o f tim es on th e s e t [ 0 ,1 ,2 ,v ] , and so

g ^ (s (0 ) ) + g ^ ( s ( l ) ) + g^(s(2)) + g ^ (s (v ) ) = 0 (modulo 2) , 6 = 1 , . . . , j ,

which is e q u iv a le n t to th e c o n c lu s io n .

LEMIvIA 5 .8 (b ) . Under the hypo die ses o f Lan ma 5«8 every 1-p la c e

func t i on s (x ) g en era ted by F s a t i s f i e s s ( g ) f f l s(/9) s (y )(+ )8 (S ) = 0

f o r every quadruple [a ,/9 ,Y ,S l € Z .

PROOF. Choose a p erm u ta tion r , g e n e ra te d by F , such th a t r (o ) = a ,

r ( l ) = r ( 2 ) = Y* 7n view of Lemma 5 .8 (a ) we have r ( v ) = a© /90Y = 8 .

Applying Lemma 5 .8 (a ) to the fu n c tio n s ( r ( x ) ) we deduce th e re q u ire d

fo rm u la .

65

LEMMA 5*9• 1-p la c e fu n c tio n h e En i s l in e a r i f and only

i f h (g ) 0 h(j3) 0 h (y ) 0 h(S ) = 0 f o r every quadruple § j e Z .

PROOF. I f h i s l in e a r , say h(x) = xH + ho, then

h(%) + 6 % ) + & (l) + = (« + g + % + = 0 , a s re q u ire d .

•6I f h i s non—lin e a r , th e n some component h co n ta in s

a n o n - lin e a r te rm ; choose one co n ta in in g as few v a r ia b le s as

p o s s ib le . W ithout lo s s of g e n e r a l i ty we may suppose th a t i t

i s T(x ) = x^x®. . .x^ ^ 2 ) . D efine

a =

g =

Y = 01, B ■

Then [a.,/3,x,8] € Z , b u t we s h a l l show th a t

h'^(a) + h'^Cg) + h"^(^) + h'^Cg) s 1 (mod 2 ) .

In f a c t we v e r i f y a t once t h a t

T(a) + T(g) + T(y ) + T(8) 3 1 (mod 2 ) ;

b u t f o r every o th e r term S(x) o f h^ we have

S (a) + S(g) + S(;^) + S(S) = 0 (mod 2 ) .

Namely, i f S(x) i s l in e a r th i s fo llo w s from th e f i r s t p a r t of

the p ro o f , w h ile i f S(x) i s n o n - lin e a r then (by th e d e f in i t io n o f T)

66

i t co n ta in s a f a c to r x'" w ith m > 6 and th e re fo re S(x) s 0

(mod 2) f o r x = a,g,Y ,

LEMMA 5*10* The group G- of a l l 1-p la c e l in e a r fu n c tio n s

o f En which a r e a lso p e m u ta tio n s of N i s t r i p l y t r a n s i t i v e .

PROOF. A 1-p la c e l in e a r fu n c tio n h e , w ith

h (x ) = ^ , i s a perm uta tion i f and only i f th e mapping

X h (x ) i s o n e -to -o n e , and th e n ecessa ry and s u f f i c i e n t c o n d itio n

f o r t h i s i s t h a t th e m atrix H he n o n -s in g u la r . In o rd e r to

prove th e t r i p l e t r a n s i t i v i t y i t w i l l th e re fo re be s u f f ic ie n t to

show th a t g iv en any th re e d i s t i n c t numbers ao,cti, Gg € N we can

f in d ho € N and a n o n -s in g u la r m a trix H over Zg such t h a t

OH + ho = Go ; e^H + ho = gi ( i = 1 ,2 ) . fw# /V#These eq u a tio n s hold i f and on ly i f ho = Go and th e f i r s t two

rows o f H a re «i - Go and ag - Go # The l a t t e r two v e c to rs

a re l in e a r ly in d ependen t, because Go,Gi , Gg a re d i s t i n c t , and

th e re fo re we can s e le c t th e rem aining j - 2 rows so a s to c o n s tru c t

a n o n -s in g u la r m a trix H, a s re q u ire d .

COROLIARY 1 . The o rd e r o f G- i s 24^5^ (2-J - 2 I) . ---

PROOF. There a re 2^ p o ss ib le choices f o r ho, and

n ( 2J - 2 ^ ) cho ices fo r th e non ^sin g u la r m a trix H.1=0

REMARK. I have no t been ab le to dec id e f o r j > 2 w hether

G- can co n ta in any t r i p l y t r a n s i t i v e p ro p e r subgroups. I t fo llo w s

67

by a theorem of B urnside ( [ 1 ] , p . 177) t h a t th e o rd er of such

a p ro p e r subgroup would be d iv i s ib le by 2- (2^-1 )(2 ^ -2 ) • This

r e s u l t dem onstrates th e im p o s s ib i l i ty o f t r i p l y t r a n s i t i v e p ro p er

subgroups of G- f o r j = 2 , b u t g e n e ra l ly , s in c e th e o rd e r o f G-

i s g r e a te r th an 2- (2^-1 ) (2 ^ -2 ) , th e problem i s s t i l l open.

COROLIARY 2 ; The s e t L of a l l l i n e a r fu n c tio n s of Ep

s a t i s f i e s th e hypotheses of Lemma 5*8*

PROOF. L in c lu d e s a t r i p l y t r a n s i t i v e group o f

perm uta tions of N (by Lemma 5*10); a fu n c tio n s a t i s f y in g

th e S-êupecki c o n d itio n s , f o r example G-(x,y) = x 0 y ; and

a 1-p la c e fu n c tio n ta k in g e x a c tly 2 v a lu e s , f o r example th e

fu n c tio n s g ^ (x ) a re e a s i ly seen t o be l in e a r . Hov/ever, i t i s

easy to v e r i f y t h a t L i s c lo se d under com position , and th a t

no 1-p la c e l in e a r fu n c tio n i s o f ty p e [ n , l ] . I t th e re fo re

fo llo w s from Lemma 5*4 th a t L does n o t in c lu d e any fu n c t io n

h (x ,y ) s a t i s f y in g c o n d itio n s (A ), and t h i s com pletes th e

v e r i f i c a t io n o f th e hypotheses o f Lemma 5*8.

LEMMA 5*11" F C En generates a tr ip ly tr a n s it iv e

group of permutations of N, togeth er w ith a l l constants and

a non-linear fu nction h ( x i , . . . ,Xk) , then F generates

a 1-p la ce non -linear fu n c tio n .

PROOF. We may assume th a t k ^ 2 and t h a t a l l perm uta tions

gen era ted by F a re l i n e a r , because o therw ise th e re i s no th ing

68

to p rove . F or some C, h ( x i , . . . , x k ) must be a n o n - lin e a r

polynom ial, and so contain a non -lin ear term S = . . . .12

We co n sid e r two c a s e s .p

CASE 1. In h th e re i s a t l e a s t one term S in ^b ich

some two o f th e ju 's a re e q u a l. Let S re p re s e n t such a term

w ith th e l e a s t number o f f a c t o r s , and suppose f o r example th a t

=jU2 = = m = 1, w h ile jLis ^ 2 f o r s ^ i + 1. D efine

co n stan ts o (k = 2 , . . . , k ) a s fo llo w s : = 1 i f 8 co n ta in sKA A A * A *the f a c to r x ; o therw ise o = 0 . Let S* = x x . . .

be any term of h d i f f e r e n t from S b u t w ith p re c is e ly th e same

f a c to r s o f th e form x ^ : thus we may suppose t h a t = 1 and

A’s = As fo r s = 1 , . . . , i , w hile nl ^ 2 f o r s ^ i + 1 . Then

i t i s easy to see (by th e d e f in i t io n o f S) th a t i f th e co n stan ts

C3 , . . . ,C|< a re s u b s t i tu te d f o r X a , . . . , X k th e term S* v an ish es .

I t fo llo w s th e re fo re th a t th e r e s u l t in g fu n c tio n h ^ ( x ,^ , . . . ,Ck )

w i l l co n ta in th e n o n - lin e a r term x ^ i . . . x ^ ^ ( to g e th e r p o s s ib ly

with other d is t in c t non -lin ear term s), and hence h ( x , 0 2 , . . . , C k )

i s a 1-p la c e n o n - lin e a r fu n c tio n g en era ted by P.0

CASE 2 , For every term S in h , a l l the /Li’s a re

unequal. I f k ^ 3 th e n by s u b s t i tu t in g s u i ta b le co n stan ts

69

we can red u ce th i s case to one in which k=2. (This i s e a s i ly

proved by a method s im ila r to t h a t of Case 1 .) Thus we o b ta in

a n o n - lin e a r polynom ial o f degree 2 in of the form

h ^ ( x i , X g ) = XiAxs* + L ( x i , x s ) ,

■vhere L i s l i n e a r , 35® (denotes th e tra n sp o se (co lum n-vector)

o f , and A i s a non-zero m atrix over Zg.

Then = 1 fo r some Y ,8 ; and w ith o u t lo s s o f g e n e r a l i ty

we may suppose th a t th e lead in g elem ent a ^ = 1 , because i f

r , s a re ( l in e a r ) p e rm u ta tions g en e ra ted by P such th a t

r ( o ) = 0, r ( e ^ ) = e j s (o ) = 0, s(eg ) = e^

th en th e le ad in g elem ent in th e corresponding m a trix f o r th e fu n c tio n

h ( r ( x i ) , s (x g )) i s 1.

I f A is n o t symmetric then i t i s easy to v e r i f y t h a t the

1-p la c e fu n c tio n h (x ,x ) g en era ted by P i s n o n - lin e a r : i f

a . / a«,^ then h ^ (x ,x ) con ta in s the term x*^x^. Consequentlyyo o# /V isawe may suppose th a t A i s sym m etric, and in p a r t i c u la r a^g = &2 1 *

Choose a pe im u ta tio n t g en era ted by P such th a t

t(0) = 0, t(ei) = e% , t(j ) = i ^ 1 2 » j •

Then i t i s easy to v e r i f y t h a t f o r th e fu n c tio n h ( t ( x i ) ,x g ) th e

corresponding m a trix , say M, has m g = = a^g + 1*

C onsequently M is non-sym m etric, and thus h ( t ( x ) ,x ) i s n o n - l in e a r .

70

COROLLARY* Under th e hypotheses o f Lemma 3*8, F g en era tes only

l i n e a r fU n o tio n s*

PROOF. Immediate from Lemmaa 3 .8 (h ) , 5*9 3*11#

REMARK. I have n o t been ab le to decide w hether under th e hypotheses

o f Lemma 3*8 F must n e c e s s a r i ly g en e ra te a l l th e l i n e a r fu n c tio n s , t h a t

id , w hether any p roper su b se t o f L can a lso s a t i s f y th e hypotheses o f

Lemma 3*8.

LEMMA 3*12. n = 2^ and f i s a l i n e a r fu n c tio n o f % ,

th en f canno t g en e ra te a doubly t r a n s i t iv e group o f perm uta tions o f N.

PROOF. We s h a l l prove th a t f i s s e lf -c o n ju g a te under e itjier some n o n -id en tica l permutation S o f N o r some constant 1-p la ce fu n c tio n . I t i s easy to see t h a t ev ery fu n c tio n g en era ted by f th e n

has the same property, and the conclusion fo llow s by Lemma 3*7*

Since f i s l i n e a r , we can w r i te

J k= ! 2 2 X ^ i + b

A=1 ^=1

where each c o e f f ic ie n t a^ i s 0 o r 1 . We denote by B th e j x jK Im atrix w ith elem ents B .. = 2 a. , and we co n sid e r two c a se s .

^ /i=1CASE 1. The m a trix B -I i s s in g u la r . Then ih e re i s ^ ^ such

t h a t £B = o, and f i s s e lf -c o n ju g a te under th e p e rm u ta tio n de fin ed by

^ (x ) = X + jc because

s (x k )) = ! 2 2 ( / + o'^)j + b = + ^ BA fI ff=1 ^ ^

- £ (^1 ^ ~ ^ (£ (3 )* * * * ^ ))*

71

CASE 2 . The matrix B-I i s non -singu lar. Then "there i s d

such th a t dB - ,d =jb, and f i s se lf-co n ju g a te under th e constant

1-p la ce fu n ction eûal to d because

f ( d , . . . , d) = [ ^ 2 a% d^] + b = dB + b = d.~ ~ ^ A»i ju-i ^ ~ ^ ^ ~

LEMMA 5*13* I f r i s a n o n - lin e a r p erm u ta tion of N. th en

th e re e x is t s a l i n e a r p erm uta tioh h such t h a t th e fu n c tio n r h r “ i s

n o n - l in e a r .

PROOF. In view o f Lemma 5*9, fo r some quadruple ^

we have

r ‘ (a ) 0 r (ft) & r (y ) © r " " (S) / 0 . ( l )

Choose a lin e a r permutation s such "that

8 ( f ( a ) ) = 0 , s (r“ ^ ) ) = ei , s(r"^ (y ) ) = % .

Then

s(r"^ (5)) / e i g)eb (2)

because otherw ise, observing th a t 0 , ei , % , ei @ % are d is t in c t

and

0 0 ei 0 3 0 (ei % ) = 0 ,

we could deduce, s in ce s“ i s l in e a r , th a t ( l ) i s f a l s e .

Choose a lin e a r permutation h* such "that

h' (0) = 0 , h*(e i ) = e i , h ’ (% ) = %, h ' ( s ( r » i ( a ) ) / s ( r “ ( a ) ) .

We can alw ays make th i s choice because i f s ( r “ ^ ( s ) ) , say , th e n

i t w i l l be s u f f i c i e n t to a rrange t h a t h* (eg ) = s(r“^ ( s ) ) .

This can be done: we need h*(x) = xH + b , and th e requ irem en ts a re

72

met by ta k in g b = 0 , and th e f i r s t 3 rows o f H to be

ei , ^ , ^(r*“ (S )) r e s p e c t iv e ly . These a re independent by (2 ) ,

and we can choose j —3 f u r th e r independent rows to make H

n o n -s in g u la r . I f s (r* '^ (5 )) = % th en we can s im i la r ly arrange th a t

h * (ea ) = % 0 e t j , and so on.

D efine th e l i n e a r p e rm u ta tio n h as s“ ^h*s and c o n s id e r the

1-p la c e fu n c tio n k = r h r“ . We f in d th a t k (a) = a ,

k(j9 ) = )5 , kCy) = y , b u t k (ô) / 8 . Hence k (a ) k(/3 ) (^) kCy) ® k(8 ) / 0

and s in ce f Z i t fo llow s t h a t k i s nonr-1 in e a r .

REMARK. In t h i s s e c tio n our r e s u l t s have been concerned w ith

th e c la s s L o f fu n c tio n s o f En t h a t a re l i n e a r r e l a t iv e to a

p a r t i c u l a r mapping o f N onto , where nW = 2 (namely the mapping2

X -* { ^ ( x ) , . . . , ^ ( x ) i ) . There a re obv iously 2 ! such mappings, and

each determ ines a corresponding c la s s L o f l i n e a r fu n c tio n s . These

c la s s e s a re con jugate to one an o th er under th e perm uta tions of N, and

i t i s o f i n t e r e s t to determ ine how many o f them a re d i s t i n c t . By Lemma

5.13 a l l th e c la s s e s con jugate to a g iven c la s s L under a g iven p e rm u ta tion

which i s n o t l i n e a r in th e sense o f L, w i l l be d i s t i n c t from L, whereas

s in ce ' l i n e a r i t y ' i s conserved under com position , a c la s s con jugate to

L under a l i n e a r ( in th e sense o f L) perm u ta tio n w i l l be id e n t ic a lj - i

w ith L i t s e l f . By Lemma 5*10 C o ro lla ry 1 th e re a re 2' II (2^-2^)J - i

such c la s s e s , and th e re fo re th e re a re 2 1/2^ II (2^ - 2^) d i s t i n c t

c la s s e s L o f l i n e a r fu n c tio n s .

(T his e v a lu a tio n was made in c o lla b o ra t io n w ith Roy 0 . Davies,..)

73

5*5 Main conclusions

THEOREM 5*1. I f n ^ 5 and P generates a t r ip ly

tr a n s it iv e group o f permutations' on N, to g eth er w ith a fu nction

s a t is fy in g the S6upecki co n d itio n s , then P i s com plete, u n less

n+1 = 2 and F generates only fu n ction s th a t are lin e a r

r e la t iv e to some p a rticu la r mapping o f N onto

PROOF. Ey Lemmas 5*1 and 5*2, F generates a 1-p la ce

fu n ction tak ing e x a c t ly 2 v a lu e s . I f F a lso generates

a 2-p lace fu n ctio n h (x ,y ) such th a t

h(0 ,0) = 0; h ( 0 , l ) = h ( l , 0 ) = h ( l , l ) = 1,

then by Lemma 5*4, Lemma 5*5, Lemma 5*6, C orollary, and Theorem

4*4 (chapter 4 , p. 3 9 ) F i s complete. I f F does not

generate such a fu n ction then i t s a t i s f i e s th e hypotheses of

Lemma 5*8, and th erefore by Lemma 5*11 C o ro lia iy , F generates

only fu n ctio n s th a t are lin e a r r e la t iv e to a cer ta in mapping

o f N onto z j .

REMARK. The example g iven in the Remark fo llo w in g

Lemma 5*tshows th a t th e con d ition n ^ 5 in Theoron 5*1 cannot

be relaxed .

THEOREM 5 . 2 . I f n ^ 4 and F generates a quadruply

tr a n s it iv e group o f -permutations on N, together w ith a fu n ctio n

s a tis fy in g th e S6upecki con d itio n s, then F i s complete.

7 4

PROOF* The r e s u l t fo llo w s im m ediately from Theorem 5*1,

u n le ss n+1 = 2 and F g en era te s only l i n e a r fu n c tio n s

( r e la t iv e to some mapping o f N onto 2 ^ ) ; h u t f o r n+1 = 2J ,

j > 2 , th e group of l in e a r perm u ta tions of En i s no t quadruply

t r a n s i t i v e . This i s c le a r from Lemma 5$9, which shows th a t

i f a ,/? ,Y a re d i s t i n c t e lem ents of N th en th e v a lu e o f any

1 -p la c e l in e a r fu n c tio n h e a t th e p o in t a Q /9 0 y i s

b (a ) (5> h(/?) Q h ( y ) , and i s thus determ ined by th e v a lu es

h ( a ) , h(/9), h (y ) .

REMARK. The r e s u l t f a i l s f o r n=3, because in t h i s case

th e group of l in e a r p e rm u ta tions i s th e whole symmetric group

S4 and th u s quadruply t r a n s i t i v e , so t h a t the c la s s F of

l i n e a r fu n c tio n s s a t i s f i e s the hypotheses a lthough obviously

n o t complete*

THEOREM 5*3* I f n ^ 2 and f i s a s in g le fu n c tio n o f En

which g e n e ra te s a t r i p l y t r a n s i t i v e group o f perm uta tions of N,

then f i s com plete.

PROOF. I f n ^ 3 , th e n by Theorem 4*3 C o ro lla ry 2 (C hapter 4 ,

p* 3 8 ) f must s a t i s f y th e S'&upecki c o n d itio n s , and th e r e s u l t

fo llo w s from Theorem 5*1 u n le ss n+1 =s 2 J and f g e n e ra te s only

l in e a r fu n c tio n s ( r e l a t iv e to some mapping of N onto Zg);

b u t th e n f i t s e l f i s l i n e a r , and by Lemma I 5 cannot g e n e ra te

even a doubly t r a n s i t i v e group of perm utations*

75

I t rem ains to prove th e Theorem f o r n=2. In t h i s case

th e only t r i p l y t r a n s i t i v e group o f p e rm u ta tio n s on N i s

th e symmetric group Sg, and i t i s w e ll known th a t each member

r of 8 3 can be w r i t t e n in th e form

r ( x ) = ax + b modulo 3,

fo r some c o n s ta n ts a ,b .

Since a l l th e p re-com ple te su b se ts o f Eg a re known

(th e y a r e l i s t e d in Theorem 2 .4 o f C hapter 2 ) , i t i s easy to

v e r i fy t h a t Sg i s e n t i r e ly co n ta in ed in only two of them,

nam ely, th e c o n se rv a tio n c la s s e s of th e two p re d ic a te s

( i ) X + y = 2 + u , ( i i ) x « y v x = z v y = z . I f we denote

th e se by L* and L” r e s p e c t iv e ly , th en i t w i l l be s u f f ic ie n t to

prove t h a t no s in g le fu n c tio n f belonging to e i th e r L' o r L"

g e n e ra te s 83.

I t i s easy to v e r ify th a t L* i s the s e t of a l l lin ea r

(modulo 3 ) fu n ction s o f Eg, and so i f f e L* then

f ( x i , . . , , X k ) = ao + a^Xi + . . . + a^xk (modulo 3)

f o r some c o n s ta n ts a g , a ^ , . . . ,a p .

According a s + ag + . . . + a% ^ 1 or + ag + . . . + a% e 1

(modulo 3 ) f ( x i , . . . , X k ) i s s e l f con jugate under some c o n s ta n t

fu n c tio n a € [ 0 , 1 , 2 j o r th e c i r c u la r p e rm u ta tio n x+1,

r e s p e c t iv e ly , and so i t fo llo w s by Lemma 5*7 th a t f cannot

g en e ra te 83 in t h i s c a se .

76

I f f , . . . ,%k ) L", then f o r ev ery th r e e p o in ts

a = ( % i , . . . , a k ) , /3 = ( / ? i , • • • ,/?k) , Y = (yi , • • • ,Yk) , such t h a t

«1 = V a i = Y l V = Y l , f o r i = we have

f ( a ) = f (/?) V f(a ') = f ( y ) v f(/9) = f ( y ) . I t fo llo w s th e re fo re

t h a t th e re cannot e x i s t s e t s G% , . . . ,Gk each c o n ta in irg a t most

two elem ents of N such th a t th e s e t f (G% , # . . ,Gk ) = N, and so

by Theorem 4*2 (C hap ter 4 , p*35 ) f does n o t s a t i s f y th e

S-ftupecki c o n d itio n s . Hence f has a t most one e s s e n t i a l

v a r ia b le , or has a t l e a s t one m issing v a lu e , and in e ith e ÿ c a se

i t i s e a s i ly seen th a t f cannot g e n e ra te S 3.

This completes the proof cf Theorem 3*3*

77

REMARK. In [ “S, 8 ] Salomaa has proved the fo llow ing

two r e s u l t s ;

( i ) " I f n ^ 4 and F i s a su b se t of En which g e n e ra te s

th e a l t e r n a t in g group An of p e rm u ta tio n s of N, to g e th e r

w ith a fu n c tio n s a t i s fy in g th e S-èupecki c o n d itio n s , th en

F i s com plete".

( i i ) " I f n ^ 3 and f i s a s in g le fu n c tio n o f F which

g e n e ra te s th e a l te r n a t in g group An of perm uta tions of N,

to g e th e r w ith a fu n c tio n s a t i s f y in g th e S-ûpecki c o n d itio n s ,

then F i s com plete".

S ince the a l t e r n a t in g group An of p e rm u ta tio n s of N

i s (n-1 ) -p ly t r a n s i t i v e on N i t fo llo w s th a t f o r n ^ 5

Theorems 3*1, 3*3 a re improvements on th e se two r e s u l t s .

A ccording to M arshall H a ll ( [ 1 0 ] , p .6 8 ), a p a r t from th e

a l te r n a t in g and sym m etric groups o f p e rm u ta tions of N th e re

a re only fo u r groups known which a re quadruply t r a n s i t i v e .

These a r e th e M athieu groups on 11 ,12 ,23 and 24 l e t t e r s , and

i t i s no t known w hether t h ^ a re e x c e p tio n a l o r p a r t o f an

i n f i n i t e sequence o f groups w hich a re quadruply t r a n s i t i v e . ^

However, th e re e x i s t i n f i n i t e l y many t r i p l y t r a n s i t i v e groups

o f p e rm u ta tio n s ; Jo rdan [fj} showed th a t i f p i s prim e, th en

1* In [ 1 2 ] I to shows th a t i f p and q are both prime numbers,w ith p ^ 2q+1 and p > 11, then a n o n -so lv ab le t r a n s i t i v e p e rm u ta tion group on [ 0 , 1 , . . . , p - l j i s quadruply t r a n s i t i v e . However, in view o f Salom aa's improvements on Theorems 3*1,5*3 fo r the s p e c ia l case n+1=p (a prim e num ber), t h i s r e s u l t i s n o t re le v a n t t o th e p re se n t problbm."..

78

th e re e x is t s a t r i p l y t r a n s i t i v e group o f perm uta tions on

p^+1 l e t t e r s of o rd e r (p^+1 )p^(p^-1 ) , and th e groups G- of

Lemma 1 2 form a n o th e r i n f i n i t e sequence o f t r i p l y t r a n s i t iv e

g roups.

For th e s p e c ia l case n+1 = p (a prime number) i t i s

p o ss ib le to o b ta in b e t t e r com pleteness c r i t e r i a th a n Theorems

5 .1 , 3 . 3 . In [ 2. 9 ] Salomaa has proved th e fo llo w in g r e s u l t :

I f n+1 = p (a prime number ^ 3) and f is a s in g le

fu n c tio n which g en e ra te s (b u t i s no t s e lf -c o n ju g a te under)

a c i r c u la r p e rm u ta tion c (x ) on N, th e n f i s com plete.

THEOREM 3 . 4 . I f a c lo sed su b se t F ^ E r in c lu d e s

a t r i p l y t r a n s i t i v e group of perm uta tions of N and a ls o

a fu n c tio n s a t i s f y in g th e S6upecki c o n d it io n s , then F is

p re-com plete i f and only i f n+1 = 2* and F is th e s e t o f

a l l fu n c tio n s th a t a re l in e a r r e l a t i v e to some mapping N

onto Zg .

(For a d e f in i t io n of a p re-com plete su b se t o f Ep see

C hapter 2 g 3*-)

PROOF. I f F i s p re-com plete and s a t i s f i e s the o th er

co n d itio n s th e n , by Theorem 3 ,1 , n+1 = 2* and F in c lu d es

only fu n c tio n s l in e a r r e l a t iv e to some such mapping (o th erw ise

F would be com p le te ); and F must in c lu d e a l l th e l in e a r

79

fu n c tio n s because o therw ise a l in e a r fu n c tio n n o t in F

could be a d jo in e d to F w ith o u t making i t com plete (s in c e

s t i l l only l in e a r fu n c tio n s would be g e n e ra te d ) . The r e s u l t

in th e o th er d i r e c t io n fo llow s a t once from Theorem 3.1.

REMARK. F or each in te g e r n ^ 3) Y ab lonsk ii has

c o n s tru c te d in 1 ^ ,] an example o f a p re-com plete su b se t of

En which in c lu d es th e s e t of a l l 1-p la c e fu n c tio n s (and

consequen tly a t r i p l y t r a n s i t i v e group o f p e im u ta tio n s on n ) ,

but t h i s p a r t i c u la r su b se t c o n ta in s no fu n c tio n s a t i s f y in g th e

S-ûpecki c o n d itio n s . He a ls o c o n s tru c ts s e v e ra l c la s s e s o f

p re-com ple te s u b s e ts , b u t th e l in e a r s e t s o f th is paper a r e n o t

among them . T herefore by th e R.emark fo llo w in g Lemma 1 3 , f o rJ - 1

n = 2* we have an a d d i t io n a l 2'^l/2'^ II (2*^—2*’) examples o fl»opre-com plete s u b s e ts .

We should l ik e to m ention th a t many of our d e f in i t io n s

and r e s u l t s concern ing l i n e a r i i y can probably be g e n e ra l is e d

to th e case when n i s th e j^ ^ power o f any prime p , bu t

th e re would not be any d i r e c t a p p l ic a t io n to th e main s u b je c t

o f th is t h e s i s .

REIAARK. In C hapter 3 (p*31 ) m entioned t h a t each of

th e p re-com plete su b se ts g iv en in Example 3*2 was g en e ra ted by

a s i n ^ e in c luded fu n c tio n . The fo llo w in g r e s u l t shows th a t

th i s is not alw ays th e case :

80

THEOREM 5*5« I f n+1 = 2^, and P i s th e p re-com plete

su b se t o f En c o n s is tin g o f a l l fu n c tio n s th a t a re l in e a r

r e l a t i v e to some mapping o f N onto Zg, then P does n o t

co n ta in a s in g le fu n c tio n (f> such th a t <p g en e ra te s F •

PROOF. I f j ^ 2 then by Lemma 3 .1 0 , F c o n ta in s

a t r i p l y t r a n s i t i v e group of perm uta tions o f N and th e

r e s u l t fo llow s im m ediately from Lemma 3 .1 2 . I f j= 1 , th en

F i s the s e t o f a l l l i n e a r fu n c tio n s of E^ and consequently

in c lu d e s the s e t of a l l 1-p la c e fu n c tio n s of E]_, namely

[0 ,1 ,x ,x + lJ . ^ argum ents s im ila r to th o se cf Theoren 2 .2

(c h a p te r 2 ,p . 1 2 ) a n ecessa ry c o n d itio n f o r a s in g le fu n c tio n

4> to g en era te th e s e t of a l l 1-p la c e fu n c tio n s of E % i s

th a t # ( x , . . . , x ) = x+1. I f in a d d itio n (p i s l i n e a r , th en

(p w i l l be s e lf -c o n ju g a te under the c i r c u la r perm uta tion x+1,

and in p a r t i c u la r <p cannot g e n e ra te th e c o n s ta h t fu n c tio n s

0 ,1 . T herefore no s in g le l in e a r fu n c tio n g e n e ra te s th e s e t

of a l l 1-p la c e fu n c tio n s of E a n d th e r e s u l t fo llo w s .

REMARK. In [A 9 ] Y ab lonsk ii has advanced th e fo llo w in g

c o n je c tu re :

each c lo sed s e t F Ç Eq has a f i n i t e b a s is (vhere a b a s is o f F

i s any s e t F ' Ç F such th a t the c lo su re of F* i s F ) .

In P o s t proves t h i s co n jec tu re f o r th e ^ e c i a l

case n=1, b u t acco rd ing to Y ab lonsk ii th e re i s no known proof

f o r the c a se n 2 . We f u r th e r d e f in e a m inim al b a s is B

81

o f a c lo sed su b se t P to be a " sm a lle s t b a s is " ( th a t i s ,

th e re i s no o th e r b a s is B* o f F such t h a t | B '| < | b | ) and

say th a t a c lo se d su b se t F i s o f ty p e ( i ) i f i i s th e o rd er

of any minimal b a s is cf F . The p re-com plete su b se ts g iv en

in Example 3*2 a re th e re fo re o f ty p e (1 ) , whereas th o se of

Theorem 5 «5 a re of type g r e a te r th a n 1* The ta s k of

d ec id in g w hether o r not a g iv en p re-com plete su b se t cf En

i s of type (1 ) ought to be f a i r l y s t r a i^ t f o r w a r d * The

problem a lso a r i s e s o f d e te rm in in g f a r a g iven in te g e r n th e

number of p re-com plete su b se ts o f type (1 ) . For exam ple, i f

n=1, th en th e re a re e x a c tly 3 p re-com ple te su b se ts of ty p e (1)^

These no tions g iv e r i s e t o th e fo llo w in g r e s u l t .

THEOREM 3*6. I f a su b se t F (c En) g e n e ra te s a p re-com plete

su b se t of En o f type ( i ) , and i f |p | < i , then F i s com plete .

PR-GOFr, F g e n e ra te s th e p re-com plete su b se t b u t i s n o t

a b a s i s . Therefore F in c lu d es a fu n c tio n n o t in th e p re-com plete

su b se t, and so F i s com plete.

8 2

V I . - A MAPPING- OF En INTO E / - V I

6 .1 . I n t r o d u c t i o n

This f i n a l c h a p te r p r e s e n t s an u n u s u a l ap p ro ach to th e

problem of g e n e r a t i n g th e com ple te s e t E p . We b e g in i n

§ 6 . 2 by d e f i n i n g n - t u p l e s of f u n c t i o n s f ^ , f s , . . . , fn £ E *

w ith s p e c i a l p r o p e r t i e s . I n § 6 .3 t h e f u n c t i o n s t& be

of are^mapped onto the se s p e c i a l t y p e s of n - t u p l e s ,tbit-

and t h i s mapping i s u sed i n § 6 .4 t o p ro v e v a r io u s p a r t i c u l a r

s u b s e t s o f Ep a r e comp].ete.

We have found i t n e c e s s a r y to a l t e r t h e n o t a t i o n u sed in

p re v io u s c h a p te r s , d e n o t i r g by x ,y (w i th i n d i c e s ) th e v a r i a b l e s

ra n g in g o v e r [ 0,1 ] ;<jf),f , g , h ( w i th i n d i c e s ) f u n c t i o n s of E]_ ;X,Y

( w i th i n d i c e s ) v a r i a b l e s ra n g in g o v e r n ( = [ 0 , 1 , . . . , n | ) Jand

e t c . ( wi t h i n d i c e s ) f u n c t io n s of Ep . V7e s h a l l a l s o

den o te a n a r b i t r a r y s u b s e t of Ep by S .

8 3

6 .2 . Perm anent s e t ' s

DEFINITION. F o r e a c h i n t e g e r n ^ 1 an n - p la c e f u n c t i o n

f ( x i , . . . , X p ) e El i s c a l l e d ( l , n ) perm anen t i f i t s a t i s f i e s

th e c o n d i t i o n

I ( Xi , . . • , Xj , . . . , Xp ) = f ( Xi , . . , Xj , • . . , Xp ) , . . . (-!-)

where f o r e a c h in d e x j (1 ^ j ^ n) Xj d e n o te s t h e p ro d u c t

Xi Xg . . . Xj •

THEOREM 6 . 1 . f ( x i , . . . , Xp) i s ( l , n ) perm anent i f and o n ly

i f t h e r e e x i s t s a f u n c t i o n <p € such t h a t

l ( x i , , , . , Xj , • . . , Xp ) = < ( X i , . . • , Xj , • . • , Xp ) . . . . ( i i )

PROOF. I f f s a t i s f i e s ( i ) , th en f s a t i s f i e s ( i i ) w i th

( f ) = f , C o n v erse ly f o r any <p ± f f s a t i s f i e s ( i i ) th e n

f ( Xi , . . . , Xj , « . . , Xp J — (f)(^ Xi , . • . , Xi Xg . . . Xj , • . . , Xi . . . Xp )

— 0 ( ^ 1 f • • • ,Xij , . • . ,Xp )

t h e r e f o r e f s a t i s f i e s ( i ) .

DEFINITION. For each p a i r o f i n t e g e r s m,n 1 th e

m n-p lace f u n c t i o n f ( x n , . . . ,X[,j , . . . ) e Ei i s c a l l e d (m,n)

8 4

permanent i f f c o n ta in s m s e t s o f n v a r i a b l e s x^j

( l ^ i ï ^ m , 1 ^ j ^ n) and s a t i s f i e s t h e c o n d i t i o n

',Aere f o r each p a i r o f i n t e g e r s i , j ( l J ^ i ^ m , 1 ^ j ^ n)

x^ j d e n o te s th e p ro d u c t x^ i x^g • • • * LJ .

THEOREM 6 . 2 , f . ,X[,j , . . , ,x^n ) (m,n) perm anen t

i f an d o n ly i f t h e r e e x i s t s 0 c such t h a t

f (x i 1 , . . . ,xL j , . . , ,XjYin ) = (^(xi 1 , . . . , Xlj , . . . ,Xmn j .

PROOF. S im i la r to Theorem 6 . 1 .

DEFINITION. An (m,n) perm anent f u n c t i o n f £ Ei i s

c a l l e d n - r e m a n e n t ( f o r an y m).

REI'/IARK. We s h a l l u se w i th o u t ^ e c i a l m en tion the f a c t

t h a t th e c l a s s o f n -perm anen t f u n c t i o n s i s c lo s e d under

a d d i t i o n and m u l t i p l i c a t i o n (modulo 2 ) .

DEFINITION. An n - t u p l e o f f u n c t i o n s f ^ , . . . , f j , . . . , f p ,

where f o r each i n t e g e r j (1 ^ j ^ n) f j e E^ , i s c a l l e d

a perm anent n - t u p l e i f each f j s a t i s f i e s th e c o n d i t i o n

f j = f j , where f j = f^fg . . . f j .

THEOREM 6 .3 - ^ n - tu p le o f f u n c t io n s f , . . . , f j , . . . , fp

i s a perm anen t n- t u p l e i f and on ly i f t h e r e e x i s t , . . . ,0n ,

S5

where f o r e a c h inte^sier j (I ^ j ^ n) <j6j e such t h a t

f J = 0 j , where < j = 0 i < 2 • • *<Pj . ■ - .

PROOF, S im i la r t o Theorem 6 . 1 .

DEFINITION. An n - t u p l e of f u n c t i o n s f ^ , . . . , f j , . . . , fp

i s c a l l e d an (m, n) perm anent s e t i f f , . . . , f j , . . . ,fp

i s a perm anent n - t u p l e of (m, n) perm anent

f u n c t i o n s .

THEOREM 6 .4 . ^ n- t u p l e o f f u n c t io n s f , . . . , f j , . . . , fp

i s a n (m,n) p e rm a n m t s e t i f and o n ly i f t h e r e e x i s t

< i , . . . , ( p j , . . , ,0n , where f o r e ac h in d e x j ( l ^ j n )

(pj £ E l ) such t h a t

f J — (^11 , • • • j ^ l j , • • • ) •

PROOF-. B y Theorem 6 .2 a n d Theorem 6.3»

DEFINITION. An (m, n) pe rm anen t s e t i s c a l l e d

n -p e rm an en t ( f o r any m).

8 6

6 . 3 . A c o r re sp o n d e n c e between Ep and n - perm anent s e t s

C o n s id e r th e f o l lo w in g mapping

X -> Xi , . , , , x j , . . . , X p ( I )

where X e N (= [ 0 , 1 , . . , , n | ) , and f o r e ac h in d e x j

(1 ^ j ^ n) Xj = 1 i f j ^ X, Xj = 0 i f j > X.

NO'! ATI ON. Denote th e n - t u p l e o f v a lu e s x^ , . . . ,x j , . . . ,Xp

w i th t h e above p r o p e r t i e s by x . We s h a l l have n+1 such

n - t u p l e s x° , x^ , . . . ,x! .

DEFINITION. We c a l l an n - t u p l e o f v a lu e s x ^ , . . . , x j , . . .Xp,

\ w h e r e Xj ç [ 0 , l ] f o r j = 1 , . . . , n , an n -p e r n a n e a t p o i n t i f f o r

each in d e x j ( l ^ j ^ n) xj = xj .

XI t i s e a s y t o v e r i f y t h a t f o r e a c h X e N x i s an n -perm anent

p o i n t . C o n v e rse ly , i f x^ , . . . ,Xj , . . . ,Xp i s an F t-pem anen t p o in t

th e n t h e r e e x i s t s a u n iq u e X (O X ijC n) such t h a t Xj = 1

i f j ^ X, Xj = 0 i f j > X, t h a t i s x ^ , . . . , x j , . . . ,Xp i s

o f t h e f o r m x f o r some X e N. Hence we have p roved t h e

f o llovfing :

THEOiMvI 6 .5 . ( l ) i s a o n e - to -o n e mapping betw een e le m e n ts

o f N and n-perm anen t p o i n t s .

8 7

RElvIAxK. B y " th e v a lu e sequence cf a perm anent n - tu p le

^ 1 , # # #, j , m t h e v a n a h l e s i , * # * , x^j , # * #, x^ p ,

we s h a l l mean a sequence of n -perm anen t p o i n t s . F o r exam ple ,

th e v a lu e seq u en ce o f the p e r n a n e n t double d>i (x^ ,x^ ) ,^g ) IS

( 1 )

( 2 )

( 5 )

( 4 )

and i n v iew o f t h e f a c t t h a t 0g = each o f th e d o u b le s

( 1 ) , ( 2 ) , ( 3 ) , (4 ) w i l l be a 2 -perm anent p o i n t .

THEOREM 6 .6 . L e t <Pi, • • * , f j > • > • ,<Pn be any perm anent

n - t u p l e o f f u n c t i o n s i n m s e t s of n v a r i a b l e s x^j

( i = 1 , . . . , m , j = 1 , . . . , n ) and c o n s id e r th e v a lu e sequence

o f 0 1 , . . . , 0 j , . . . , 0 n . In p a r t i c u l a r c o n s id e r th e perm anent

•point (form ed by th e v a lu e s of 0 i , . . . ,0 j , . . . ,0p ) a t each o f

th e p o i n t sX, S i

(x i ...X]^ . . .Xgi } , .............. .................. \ A )

f o r a l l sequences X i , , . , ,X(,, . . . ,Xm w here e a c h X[ e N,Xr

and where X[ i s t h e n-p e rm an en t p o in t x ^ j , . . . , xlj , . . . ,X[,p

d e f in e d by x^ j = 1 i f j ^ X [, X ( , j = 0 i f j > Xq.

8 8

I f t h e pennanenb p o in t o f 0 i , . . . , 0j , . . . ,0p i s f i x e d a t ev e ry

p o i n t o f ty p e (A) , th e n th e r e e x i s t s an (m, n) perm anent s e t

f 1 , . . . , f j , . . . ,fp such t h a t f o r e a c h in d e x j = 1 , . . . , n

I j = 0 j a t e v e r y p o in t of ty p e ( a ) . M oreover, such an

(m, n) perm anent s e t i s u n iq u e .

PRO OB'. D efine th e n - tu p l e f , , f j , . . . , fp by

(^ 1 1 , • • • LJ , * • • ,^nin i — (^11 , • • • L j , • • • ) ,

f o r each in d e x j = 1 , . . . , n . S in ce 0 ^ , . . . , 0 j , . . . ,0p i s

a pe rm anen t n - t u p l e we have 0 j = 0 J f c r j = 1 , . . . , n ,

and so by Theorem 6 .4 f ^ . , f j , . . . , fp i s an (m,n) perm anent

s e t . A ls o , a t e a c h of t h e p o i n t s ( a ) , s in c e each x i ( f o r i = 1 , . . . , m )

i s an n -p e rm an en t p o in t , we have

( ^ 1 1 , • • *, J , • • • (^11 , • • • I j , • • • , i n ) ~

, / Xi Xl \— 0j (x i , . . . , x q J • • • / .

That i s

f j = 0 j , f o r j = 1 , . . . , n .

M oreover, f j i s u n iq u e , s i n c e t h e v a lu e of f j a t ev/€ry

p o in t Xi 1 . . . X^j . * .Xqip , w here each ^Lj ^ i ^ , 1 j ds g iv e n by

f \ / XCt X I Xrp \f \X ii , . . fXRj , • • • ,Xmn / — , • • • ,^L , • • • / ?

Xwhere f o r each in d e x i = 1 , . . . , m x [ i s the n -p e rm an en t p o in t

X

89

NOTATION* Denote by Ep th e s e t o f a l l n -perm anen t s e t s ,

THEOREM 6 . 7 . The mapping

J : F(Xi X l , . . . ,Xm ) ^*1, • • • 3 • • •

of Ep to Ep*, where f (Xi , . . . ,Xfj, ) e Ep and f . , f j , . . . ,fp

i s t h e (rn,n) perm anent s e t i n t h e v a r i a b l e s Xi 1 , • . . ,X(,j , . . . ,x^p

whose perm anent p o in t i s d e f in e d a t each o f th e p o i n t s

Xi^^ , . . . ,XL^*', . . . ,Xfn^ ( f o r a l l seq u en ces X , . . . ,X|_, . . . ,X^ where

each Xt e N) f j (x^^^ , . . . , x ^ ^ ^ , . . . , x ^ ^ ) = 1 i f j ^ P( X i , . . . ,Xm ) ;

f j (x i^ ^ , . . . ,X(,^'', . . . , Xm^ ) = 0 i f j > F(Xi , . . . ,X(D ) , i s o n e - to -o n e

and o n to .

PROOF* L e t F ( X i , . . , , X m ) be any f i x e d m -p lace f u n c t i o n

o f Ep and c o n s id e r th e c o rre sp o n d e n c e

J I F(Xi , * * . , X [ , . . . ,Xm ) -» 01, • * • ,0J , • * • >0n

where 0 i , . . . , 0j , . . , ,0p i s a perm anent n - t u p l e i n th e v a r i a b l e s

X i1 , . . . , X [ j , . . . ,Xmp whose perm anent p o i n t i s d e f in e d a t each of

th e p o i n t [ ^ 1^^ x^^^ ( f o r a l l Xj , . . , , X [ , . . . ,Xm e N* )

by 0j (x i . ,x^^ ^, • • • ,Xm ; = 1 i f j F(Xi , . . ,Xm J j

0 j (x i^^ , . . . , xl^ , X m ^ ) = 0 i f j > f (Xi , . . . ,Xm ) . Such a perm anent

n - t u p l e c e r t a i n l y e x i s t s , f o r i t can be fo n n e d s im p ly by f i l l i n g i n

a l l t h e u n d e f in e d v a lu e s i n th e v a lu e sequence o f 0 i , . . . , 0 j , . . . ,0p

by n-perinanen t p o in t s .

9 0

D efin e t h e (m,n) perm anent s e t f ^ . , f j , . . . , fp by

(^11 , * • • ,^LJ , • • * , ^ n ) “ (^11 , • • • j , • • • , ^ n )

f o r j = 1 , . . . , n .

Then, by Theorem 6 , 6 , f^ . , f j , . . . , fp i s u n iq u e and

J : F(Xi , . . . , X q , * , . ,Xm ) f ] _ , . » . , f j , . . , , f n »

C o n v e rse ly , th e n—perm anent p o i n t o f a g iv e n n—p e rn a n e n t s e t

f i , . . . , f j , . . , , fp a t e a c h o f th e p o i n t s o f i t s v a lu e sequence i s

u n iq u e ly d e f in e d . In p a r t i c u l a r , a t t h e p o i n t s . . . . . .

( f o r a l l s eq u en ces X , . . . ,X|_, . . . ,Xm e ) . Thus t h e r e e x i s t s

a u n ique f u n c t i o n F su ch t h a t

J I F(X], , . . . , X [ , . . . ,Xm ) f 1 , . , , f j , • • • , fn •

91

6 . 4 . Some co m p le te s u b s e t s o f En

We s h a l l now p ro ceed to a p p ly t h e o n e - to -o n e co rre sp o n d e n c e

J t o the p rob lem o f the g e n e r a t i o n of th e com ple te s e t E p .

F i r s t l y we have a re m ark on s u b s t i t u t i o n .

REMARK. Suppose t h a t f (Xi , . . . ) , G% , . . . a r e

f u n c t i o n s , each b e lo n g in g to Ep , a n d t h a t

J t E(Xi , . . . ,Xm ) f i , . . . , f j ( x i i , . . . , X L j , . . . ,Xmp ) , . . . , f p ,

and fo r each in d e x i = 1 , . . . , m ,

S l i > • «.j SLj > • • • ^SLn .

Then

J : F ( & i , . . . , G - j ) -> h i , . . . , h j , . . . , h p ,

where f o r j = 1 , . . . ,n

KSîvltlRK. I n f u t u r e , we s h a l l a lw ays assume t h a t th e symbol ->

d e n o te s t h e c o r re sp o n d e n c e J .

TERMINOLOGY. V/e say t h a t an a r b i t r a r y s u b s e t S of Ep

ge n e r a t e s a f i x e d s u b s e t T i f S g e n e r a t e s e v e r y s i n g l e f u n c t i o n

o f T.

DEFINITION. For e a c h f i x e d in d e x j (1 ^ j ^ n) we s h a l l

d e n o te th e s e t of a l l n -pe rm anen t s e t s f ^ , . . . , fp f o r w h ich

92

f j + i = . . . = f n = 0 by Cj*, and th e c o r re sp o n d in g s u b s e t

o f Sp by C j . That i s , Cj i s th e s e t of a l l f u n c t i o n s

F ( f Ep) such t h a t t h e n -perm anen t s e t f i , . . . , f p , where

F f 1 , . . . ,fp , b e lo n g s to C j* .

I t i s e a s y to v e r i f y t h a t £ j i s c lo s e d u r d e r c o m p o s i t io n

( f o r e a c h f i x e d j ) .

We s h a l l need t h e f o l lo w in g I n a u c t iv e C r i t e r i o n o f

C om ple teness .

INDUCTIVE CRITERION OF COMPLETENESS. I f a s u b s e t S

o f Ep g e n e r a te s C^, and f o r each in d e x j (1 ^ j < n) th e

s u b s e t S U Cj g e n e r a t e s th e s u b s e t Cj +i , th e n S i s co m p le te .

REMARK. B efore we p r e s e n t a p ra c td .c a l f or m o f t h i s C r i t e r i o n

(Lemma 6 . 1 ) v/e s h a l l -make some rem arks a b o u t tJrie ' g e n e r a t in g

p o w er’ cf t h e c i r c u l a r p e r m u ta t io n X + 1 modulo n + 1 .

I t i s e a s y t o v e r i f y t h a t

X + 1 Xp + 1 , Xp + Xi , . « . , Xp + Xp _

and t h a t , f o r j = 1 , . . . , n - 1 , t h e j + 1^^ power o f X + 1 i s

th e f u n c t i o n

( 9 ' ' ; f j y > f ’j + l 7 ^ j + 2 ) • • • 3 f*n )

X + j + 1 -> X + 1 - j +Xp_ j+1 , . . . +Xp_ j +1 ,X^_j+1 ,Xp-j +Xi , . . . ,Xp_ j +Xp_ (j + i)

93

I f F i s any f u n c t i o n o f Ep such t h a t

F-> f j , . « , fk f O j . . . j O j

v-i.th j -1 u n i t s , and n -k z e r o s , where 1 ^ j k ^ n^ then

f o r e a c h p a i r o f f i x e d i n t e g e r s i , -6. (O i j - 1 , 0 ^ ^ n -k )

we can move th e sequence f j , . . . , fk i p l a c e s t o t h e l e f t , o r

& pl& ces to the r i g h t , b y t h e r e s p e c t i v e s u b s t i t u t i o n s

( i ) F ' = ( f ) + n + 1 - i

, f j , « . . , f | ^ , 0 , « . . , 0 ,

w i t h j - ( i + l ) u n i t s , and n + i -k z e r o s ;

( i i ) F” = (F) + &

wi t h j+6-1 u n i t s , and n—(k+&) z e r o s .

vVe s h a . l l c a l l t h i s p ro c e s s s h u n t i n g .

LEHvttLfV 6 . 1 . I f a s u b s e t S En g e n e r a t e s , t o g e t h e r

;v i th t h e c i r c u l a r p e rm u ta t io n X+1 modulo n+1 , and f o r e a ch in d e x j

(1 (6 j < n) th e s u b s e t 3 U Cj g e n e r a t e s th e .1+1- p la c e f u n c t i o n. . » * *

G-J+ 1 ( X i , . . . ,Xj+1 ) -> X i , . . . ,x j ,xj+ 1 , 0 , . . . , 0 , \\faere f o r e a c h in d e x

i ( l i ^ j+1 ) Xj, d e n o te s t h e p ro d u c t i % g . . .X[ i , t h e n S

i s co m p le te .

PROOF. I t s u f f i c e s to p ro v e t h a t f c r e ac h in d e x j

(1 ^ j < n ) , t h e s u b s e t S U ^ g e n e r a te s t h e s u b s e t £ j + i > &nd

94

th e n t h e r e s u l t f o l l o w s by th e I n d u c t iv e C r i t e r i o n o f C o m p le ten ess .

Suppose F -> f i , . . . , f j + i , 0 , . . . , 0 i s any f i x e d f u n c t i o n

o f £ j + i and c o n s id e r th e j+1 f u n c t io n s d e f in e d

f o r i = 1 , . . . , j+1 by

F^ -» f L , 0 , . . . , 0 ,

w i th n-1 z e r o s .

Each F**, f o r i = 1 , . . . , n , b e lo n g s to C and i s t h e r e f o r e

g e n e r a te d by S. Hence, S U Cj g e n e r a t e s F , s i n c e

F = +

and t h e r e s u l t f o l lo v / s .

The n e x t Lemma i s co n ce rn e d w i th c r i t e r i a s a t i s f i e d by a n

a r b i t r a r y s u b s e t cf En w hich g e n e r a t e s C%.

LEI<IMA 6 , 2 . I f a s u b s e t S g e n e r a te s th e 4- f u n c t i o ns

a ( x ) -> 0 , 0 , . . . , 0 ;

B(x ) 1 , 0 , . . . , 0 ;

C(X,Y) -» Xi+yx , 0 , . . . , 0 ;

D(X,y ) -> y%

t o g e t h e r w i th th e n 1- p l a c e f u n c t i o n s H ' , . . . , H T d e f i n e d ,

f o r i = 1 , . . . , n , ^

H g x )

th e n S g e n e r a t e s C^.

9 5

PROOF. By Theorem 2.1 ( s e e C h ap te r 2 ) from A,B,C,D,

we can show t h a t S g e n e r a t e s e v e ry f u n c t i o n F ' e C of t h e

form

F -» f ( x x i , . . . , X | < ; x } j O , . . . , 0 ,

f o r any f i x e d in d e x k > 0.

We can t h e r e f o r e shov/ f rom A,B,C,D,H*^, . . . t h a t S

g e n e r a t e s e v e r y f u n c t i o n o f t h e form

F f ( x % 0 , . . . , 0 ,

f o r any f i x e d in d e x m > 0 ; t h a t i s , ev e ry f u n c t i o n of C .

For exam ple: l e t n=m=2. From A, B,C,D, S g e n e r a t e s e v e r y

f u n c t i o n o f th e fo rm

F ' (Xj ,Xg ,Xs ,^4 ) , an d

from A, . . . S g e n e r a t e s e v e ry f u n c t i o n o f t h e form

F ( X i , X s ) = F ' (H (X, ) ,H= (Xi ) , t f ( X s ) ,H= (Xs ) )

-*■ f ( X i i ,Xl2,X2 1, %g2) | 0* . ' ' . •

TIiFOREM 6 .8 . I f S i s a s u b s e t of Ep which g e n e r a t e s th e

c i r c u l a r p e rm u ta t io n X+1 modulo n+1 on N, t o g e t h e r w i th a 2 - p la c e*

f unet i o n X H Y d e f in e d a s f o l l o w s ;

d e f in e X A Y f o r f i x e d i n d i c e s r , s (1 r ^ s ^ n ) by

X A Y - > X x A Yi y •• • )Xr a ÿ r fYr+ i • fYs f iŸs+ i f >

9 5

where

A + y ^ , f o r 1 < ^ ^ r ,

f o r a f i x e d in d e x t ( l ^ t ^ r , t < n ) d e f in e

X* -> X i Xi_, X Xi., . where a ................. a iS any t a, . a -------- t+1 ' n ^t+1 nf i x e d i n c r e a s i n g sequence o f i n t e g e r s t+1 ^ ^ n ,

*Y -> , j^t+l " • • • ÏÉierË ( 3 i y - , / 3 ^ ± s _any

i n c r e a s i n g sequence o f i n t e g e r s 1 ^ / ? i ^ . . . ^ ^ t , a,nd

f o r f i x e d r , s , t ( l ^ t ^ r ^ s ^ n , t < n) we d e f i n e X A* Y

X A Y = X* A* Y.

Then S i s c o m p le te .

PROOF. We s h a l l p ro v e Theorem 6 . 8 i n two s t a g e s . F i r s t

we s h a l l shov/ t h a t S g e n e r a t e s 0% . Then we s h a l l show t h a t

f o r e a c h in d e x j ( l ^ j ^ n ) , the s u b s e t S g e n e r a t e s t h e* * »x

j+ 1 - p la c e f u n c t i o n G-j+i -» x% , . . . ,x j ,xj+ , 0 , . . . , 0 . The r e s u l t

c l e a r l y f o l lo w s by Lemma 6.1 from t h e s e two a s s e r t i o n s , a n d t h e

h y p o th e se s o f th e Theorem.

We s h a l l u se w i th o u t me n t i o n t h e f o l l o w i n g p r o p e r t i e s o f

th e f u n c t i o n s x y , x^y : i f , . . . i s an y sequence o f

f t in c t io n s each b e lo n g in g to and s a t i s f y i n g (pj = pj f o r

j = 1 , 2 , , . , , k , th e n f o r any p a i r o f i n d i c e s -6,m ( l ^ 6 ^ m ^ k )

h = h •

= h -

97

I n f a c t , x y , XaJ a r e t h e w e l l known 'minimum* and * maximum '

f u n c t io n s of r e s p e c t i v e l y .

S tage ( i ) . We s h a l l show t h a t 8 g e n e r a t e s t h e 4 f u n c t io n s

A(x ) -> 0 , 0 , . . « , 0 ;

B(x ) -> 1, 0 , . . . , 0 ;

C(X,y ) -> Xj^yi, 0 , . . . , 0 ;

D(X,Y) Xi+yi, 0 , . . . , 0)t o g e t h e r w i th the n 1- p l a c e f u n c t i o n s d e f in e d f o r each

i m e x i (1 i ^ n) by

H^ X) -> X(, , 0 , . . . , 0 .

I t wi 11 th e n f o l l o w by Lemma 6 .2 t h a t S g e n e r a te s 0% . We s h a l l

l a y the p ro o f o u t a s a s e r i e s o f Lemmas.

LEAft'IA 6 . 3 . Under t h e h y p o th e se s o f Theorem 6 .8 , 3 g e n e r a te s

th e two 1- p la c e f u n c t i o n s A,B d e f in e d by

a ( x) -> 0 , 0 , • • • , 0 ;

b( x ) -> 1 , 0 , . . . , 0 •

PROOF. F o r each i n t e g e r j = 1 , 2 , . . . , n—1 8 g e n e r a te s th e

j+ 1 ^^ power o f X+1, a n d so 8 g e n e r a t e s t h e 1- p l a c e f u n c t io n

i ( x ) = x + n n x + n - i n . . . n X + i n x

1, # . . , 1, Xp+ 1 f • • • f ^ > 0, . , . , 0.

98

C o n sid e r t h e seq u en ce o f 1 - p l a c e f u n c t i o n s , 1 ^ , . . . g e n e r a t e d

by S and d e f in e d by

I ^( X) = I (X) + n+1 - r ,

( s h u n t in g X r + i , . . . , x ^ r p l a c e s to th e l e f t ) and f o r e a c h

i n t e g e r u = 1 , . . . , n-1

i ^ \ x ) = i X i ^ ( x ) ) .

We have

a ( x ) = I"(X )

b ( x ) = i " ( x ) + 1 .

T h e re fo re S g e n e r a t e s A,B*

REMARK* I t i s more c o n v e n ie n t to show t h a t S g e n e r a t e s

, . . . b e f o r e C and D.

LEMMA 6 .4 . Under t h e h y p o th e se s o f Theorem 6 . 8 , S

g e n e r a t e s th e n 1- p la c e f u n c t i o n s , . . . ,H" d e f in e d f o r e ac h

in d e x i ( l ^ i h ) Uy

H'-(X) -> X L , 0 , . . . , 0 .

PROOF. V/e s h a l l p ro v e f i r s t l y t h a t S g e n e r a t e s H" ( X ) .

Suppose t h a t f o r some f i x e d in d e x i ( l < i $ C n ) S g e n e r a t e s

th e 1 - p l a ce f une t i on

dL (x) -> i , . . . , 1 , X [ _ , . . . , X p ,

w i th i -1 u n i t s . Then S g e n e r a t e s YP i n th e f o l l o w i n g way;

9 9

T 2 n + 1— Lc o n s id e r th e seq u en ce of f u n c t i o n s g m e r a t e d

by S , an d d e f in e d by

J l ( x ) = J | . ( X ) ,

and f o r each i n t e g e r u = - iu+i . . , u . .

J l .(X) = J t ( J | , ( x ) + n ) .

U + 1 -QThat i s , we o b t a in J [ from J|, by f i r s t s h u n t in g t h e sequence

xi+urt^ • • • 3 one p la c e to t h e l e f t , c o n s e q u e n t ly in t r o d u c in g

an e x t r a z e r o , a n d t h e n r e p l a c i n g xi+u.-iby a u n i t . We haven+ 1— L ^

J l ( X ) - > ,Xp , 0 , . « . , 0 ,

w i th i - 1 u n i t s ,

and S g e n e r a t e s im m ed ia te ly by the s u b s t i t u t i o n

tf’ (x) = X x ) ) + n + 1 - ( i - 1 )

( s h u n t in g Xp i - 1 p la c e s to t h e l e f t ) .

C o n s id e r now th e two 1- p l a ce f u n c t i o n s g e n e r a te d by S.

( Ai (x ) + t ) D X-> 1 , . . . , 1 ,x^^^ J • • • jXp ,Xp+ I f . ' . jXg , 0 , . . . , 0 ;

(A(x) + n) A X 1 , . . . , 1 , 1 , . . . , 1 ,Xp+ . ,Xg ,Xg+ I f . . . fXp *

A ccording a s r = n , o r r < n we have

(a (x ) + t ) a X = ( x ) , o r (a ( x ) + n) a X = Jp+ % (x ) r e s p e c t i v e l y .

I n v iew of Lemma 6 .3 , in e i t h e r c a s e , S g e n e r a t e s a f u n c t i o n o f

type J l , f o r some f i x e d index i ( l < i ^ n ) , and so ^ g e n e r a t e s

(x) by t h e above a rgum ent.

100

We shfî.ll now prove t h a t f o r each u = 0 , ' l , , , . , n —1 S

g e n e r a t e s t h e u+1 1 -p la c e f u n c t i o n s i f . , î f “

I t i s e a sy to s e e t h a t t h e Lemma f o l l a v s on t a k in g u = n - 1 .

vVe p ro c e e d by i n d u c t io n on u , s t a r t i n g w i th th e case u = 0

w hich we have a l r e a d y shown to be t r u e . Assume t h a t f o r some

f i x e d i n t e g e r u (O (C u ^ n—2) S g e n e r a te s th e u+1 1- p l a ce

f u n c t i o n s i f , i f " . , l f “ th e n i t s u f f i c e s t o p ro v e t h a t

S g e n e r a t e s t h e 1—p la c e f u n c t i o n pn- ( u+ i y ^

C o n s id e r t h e 1- p l a ce f u n c t i o n g e n e r a te d by S

J ' ( X) = I f (X + n)

">■ Xj_ + 1 , 0 , * . * , 0 *

Hence S g e n e r a t e s t h e 1- p l a c e f u n c t i o n

t f ( x ) = J ' ( J ' ( X ) ) ,

and th e 2—p la c e f u n c t i o n

J(X ,Y) = H^(X n Y)

Xi A y , o , . . . , o ««1

B y th e i n d u c t io n h y p o th e s i s S g e n e r a te s t h e 1—p la c e f u n c t i o n

J"(x) = H "-" (XS1) + « 1-1

w ith (%i-1 u n i t s , t h a t is ^ + X p _ ( ^ i ) o cc u rrin g in th e p la c e .

1 0 1

Noting t h a t a( ^ + ^ - ( u + i ) ) = ^ - ( u + i ) see tha.t S

g e n e ra te s Hn-(u+i) a s f o l l o w s :

I f - 1 1)0= J ( l f “ ^(X) n J ” (X) )

LEMIÆA 6*5. Under th e h y p o th e se s o f Theorem 6 .8 , S

g e n e r a te s th e 2—p la c e f u n c t i o n s .

6(X,Y) -> ^2ÿ i , 0 , . . . , 0 ;

f O f . ' . f O .

PROCF. Ey Lam ma 6 . 4 S g e n e r a t e s th e 1—p la c e f u n c t i o n

i f (X) -> x i , 0 , . . . , 0 ,

ana so S g e n e r a t e s th e 2—p la c e f u n c t i o n

j ( x , y ) = H ^ X n Y)

Xi A y , o , « . . , o ,

ana th e 1—p la c e f u n c t i o n

J ’ ” (X) = H^(X) + n

-> Xi+1 , . . . ,Xi + 1 ,

w i th Xi + 1 r e p e a t e d n t im e s .

We have

C(X,Y) = J' " ( X ) , J " ' (T) ) ) ) ,

and D(X,Y) = C( J " ' ( c ( J " ' (X) , J ' " ( x ) ) )> J " ' ( C ( X , y ) ) ) .

Hence S g e n e ra te s C, D.

This completes the p roo f cf the f i r s t s t a g e o f Theor’em 6 . 8 .

102

S tage ( i i ) . We s h a l l r e q u i r e t h e fo l lo w in g Lemma,

LMh.A 6 , 6 . Under t h e h y p o th e se s o f Theorem 6 . 8 , S

'•enera tes t h e 1 - u la c e f u n c t i o n s

K ( X ) x% , . . . , , 0 , . . . , 0 ,

w i th Xj r e treated t+1 t i m e s ,

L (x ) 1 , . . . , 1 , Xp , . . . , ,

w ith t u n i t s #

PROOF. We s h a l l p rove f i r s t l y t h a t S g e n e r a t e s K.

There a r e two c a s e s .

Case ( i ) . I f t = r : by Lemmas 6 .3 and 6 .4 , S g e n e r a t e s

th e 1- p l a c e f u n c t i o n s A(x ) and H^(x), r e s p e c t i v e l y , and so 3

g e n e r a te s t h e 1- p l a c e f u n c t i o n s

H^(x) + n -> Xi + 1 , . . . ,Xi + 1 ,

and

K' (X) = (A (x)+n) n (H^(x)+n)

1 , . . . ,1 ,Xi + 1 , . . . ,Xi + 1 ,

w i th t u n i t s #

We have

k ( x) = k ' ( x ) + 1,

and i n t h i s c a se 3 g e n e r a t e s K.

103

Case ( i i ) . I f t < r : a s b e f o r e , S g e n e r a te s A and

, and so 3 g e n e r a te s t h e 1- p l a ce f u n c t i o n s

K"(X) = H ^(hH x ) + n) + n

w i th r e p e a t e d n t im e s ;

K' ” (X) = K’’(X) n a ( x )

w ith Xi r e p e a t e d r t i m e s .

A ccord ing a s 2 t ^ r#-1or 2 t < r —)we have

K(x ) = ( (A (x )+ r -^ 4 l |n K"(X)) + n + l + t - r ,

o r

K(x ) = (K "(x ) n (A(x)+r-f:tlJ)f n + l + t - r ,

r e s p e c t i v e l y ( r e p l a c i n g th e f i r s t x ^ ' s by u n i t s , an d

sh u n t in g the re m a in in g sequence of t+? x ^ ’ s p la c e s t o t h e l e f t )

Hence i n t h i s c a s e S g e n e r a t e s K.

F in a l . ly we have

L(x ) = K ( î f (x ) + n ) + n ,

where

H^(X) -> X p , 0 , . . . , 0 .

% Lemma 6 . 4 S g e n e r a t e s H" , and so S g e n e r a te s L .

The p ro o f of the Lemma i s com ple ted .

104

I t rem a in s to show t h a t , f o r each i n i ex j ( l ^ j < n ) , th e

s u b s e t S U Cj g e n e r a t e s t h e j+ 1 -p la c e f u n c t i o n>St *

G-J+ 1 -> Xj^, . . • , xj , x j + 2 , 0 , . . . , 0 »

T here a r e two c a s e s .

Case ( i ) . I f j ^ t : th e n the f u n c t io n s. * *

and. *

^ , . . . , X j + j _ ) -> X j + i , 0 , . . . , 0

each be long t o £ j , and a r e t h e r e f o r e g e n e r a te d by S U Cj .

By Lemma 6 . 6 S g e n e r a t e s K( x ) , and so S U £ j g e n e r a te s

P ( Xi , . . . ,XjV 1 ) = K(P' ' X i , . . . ,X ji 1 ) )jje O

-> X j + i , . . . , X j + ] _ , 0 , . . . , 0 5

w ith Xj+ 1 r e p e a t e d t +1 t im e s .

I f t < s , t h e n S U £ j g e n e r a t e s Gj+i by th e s u b s t i t u t i o n

Gj+ 1 ( X i , . . . ,Xj+ 1 ) = (Cj ( X i , . • • ) + t—j n P ( X i , . . . ,Xj+ I ) ) + n+1 + j —t ,*

That i s , we s h u n t X i , . . . ,x j t - j p l a c e s t o tine r i g h t , r e p l a c e t h e* * « *

f i r s t z e ro by x j + i , and sh u n t th e new sequence x^ , . . . , xj ,Xj+^ t - j

p la c e s t o th e l e f t .

I f t = s , th e n the 1—pla.ce f u n c t i o n

Q' ( x ) X j+1 , x j + X i , . . . ,Xj+Xj_ 1 , 0 , . . . , 0

b e lo n g s to £ j , T h e re fo re S U Cj g e n e r a t e s t h e 1- p la c e f u n c t i o n .

Q"(X) = Q' (X) + n

-» X i , . . . ,Xj ) . . ; , Xj ,

w i th Xj r e p e a t e d n—j t im e s .

105

C o n s id e r th e sequence o f 1 - p la c e f u n c t i o n s ”

g e n e r a te d by S U £ j and d e f in e d by

Q°(X) = Q"(X),

and f o r each u=0 , 1 , , . . , t - j -1

Q "*\x) =(«"(x) n A(x)) + 1 .

T hat i s , if' j < t we form by r e p l a c i n g t h e l a s t n - s

Xj ' 3 by z e ro s and s h u n t in g th e re m a in in g sequence one p l a c e t o

th e r i g h t , we th e n fo rm from Q^, f o r u=1 , , . . , t - ( j +1 ) , by

r e p l a c i n g th e l a s t Xj by a z e ro and s h u n t in g th e r e m a in in g

sequence one p la c e t o th e r i g h t .

We have

Q "(x) -*■ 1 , . . . , X]_, . . • , Xj , Xj , 0 , . . . , 0 ,

w i th e x a c t l y 2 x j ' s and n—( t + 1 ) z e r o s .

Hence

G j i i ( X i , . . . , X j i i ) = ( q* " J ( s (Xi , . . . , X j ) ) * P(Xi , . . . , X j> i ) ) +

+ n+1+ j - t ,

and so i n t h i s c a s e S U £ j g e n e r a t e s Gj+i .

Case 2. I f t < j < n : i n t h i s c a s e we s h a l l p rove

t h a t S U ^ g e n e r a te s th e two f u n c t i o n s

106

* * » >/'- * *T1 , • * • , Xp ) -> , • • • , x ^ , Xp , • • • , Xp , Xp , , . • , Xp ;

* îi« »U(Xl , • . . ,Xj+ 1 ) -> 1 ) » * * )%t+ 1 ) ^ t + l 3 * ' ' 3 ^ + l ) ^ 3 . . . , 0 .

From T and U i t f o l lo w s t h a t S U £ j g e n e r a te s Gj+ ^ , s i n c e

^j+ 1 (Xi , • • • jXj+ 1 ) = T(Xi , . . . ,Xp ) n TJ ( X i , » . . ,Xj+ 1 ) •

We s h a l l show f i r s t l y t h a t S U ^ g e n e r a t e s th e r - p l a ce

f u n c t i o n T(Xi , • . . ,Xp ) , S in ce t < j , th e f u n c t i o n* * »

T ( X , • « *, Xp ) -> Xp+1, Xp +x^ , • • • , Xp + x ^ , 0 , , « « j 0 ,

b e lo n g s t o £ j , and so S U £ j g e n e r a te s T by t h e s u b s t i t u t i o n

T ( X i , . . . , X p ) = T ' ( X i , . . . , X r . ) + n .

I t rem ains to show t h a t S U Cj g e n e r a te s t h e j+ 1 - p la c e

f u n c t i o n U. B y Lemma 6 . 6 S g e n e r a t e s t h e 1- p l a ce f u n c t i o n

L ( X ) 1 , . « « , 1 , Xp , • • • , Xp f

wi oh t u n i t s ,

ana so S g e n e r a te s t h e 1- p l a c e f u n c t i o n

v ( x ) = L(X) n X

A . ♦ •,X.j.^ « « ,Xp .

C o n sid e r t h e sequence o f 1—p la c e f u n c t i o n s

g e n e r a te d by S a n d d e f in e d by

Vi ( x ) = V(X) + 1

3 n+1 > - * « , ^ + 1 , ^ t +1 5 • • • Xn- 1 J

w i th âq,+1 r e p e a t e d t +1 t im e s ,

107

and f o r e a c h i n t e g e r u = 1 , 2 , . . . , n - ( t -*1 )

V^^(X) = V^(V^(X) )

( 3 • • • ) ^t+ 1 3 • • • 3 i 't+U >* » . . « . » « .

^ + i - u + ^ - u + ' J • • • 3 ^ + i-U + ^ - u + 1 f ' 3%n + x ^ -11+ I )

^ t+u+ i > ' ^ t+ u +2 ' ' ' )

Xn-u+ ' ^ -u + XLy+ 1 , * . . ,Xn-u+ (u+i ) •

We have

V ( i f (x) ) - > x.j- 3 , • • • ,x;-5+ X ,x-t+ 2 , • • • ,Xp ,

ana s i n c e t ^ 1 , t h e j+ 1- p l a c e f u n c t i o n

1 / X * *U \Xx J • • • ,Xj+X ) i -fc+x 3 • • • ^x!j+X J 0 , • . . , 0

b e lo n g s to C j . T h e re fo re S U Cj g e n e r a t e s U by th e

s u b s t i t u t i o n

U ( X i , . . . , X j * i ) = V ^ ( V " - R u ' ( X i , . . . , X j i i ) + t ) ) .

This com ple tes t h e p ro o f o f t h e seco n d s t a g e and Theorem 6 . 8

f o l l o w s .

1 0 8

D e fin e X n Y f o r f i x e d i n d i c e s r , s ( 0 r s < n) by

y « # • • • • • • » •X A Y -» Xi AYi , * ' * ,Xr A^r *yr+ ! > • • • > Ys jX^+iYs+i i»*» jX/iYn j

and f o r a f i x e d i n t e g e r t ( s ^ t < n , t > O) d e f in e

“X -» x^ J • • • X , • • • jX|i, where ax, *» * , ^ 4.ux Kt ■ ^

i s a n y i n c r e a s i n g sequence o f i n t e g e r s 1 ^ (%x ^ . . . ^ a ^ ^ t ;

y Y i s • • • > y J • • • ,Yg 9 where /St+ 1 ? • • • >fn

i s an y i n c r e a s i n g sequence o f i n t e g e r s t +1 ^/9.J-+x K . . . K ^ n ,

th e n f o r f i x e d r , s , t (0 ^ r ^ s ^ t < n , t > 0 )

X Â Y = “X A y".

'THEOREM 6 . 9 - ^ ^ i s a s u b s e t of En w hich g e n e r a te s

th e c i r c u l a r p e r m u ta t io n X+1 (modulo n+1) t o g e t h e r w i th

a 2- p l a c e X A Y, th e n Z i s c o m p le te #

PROOF. C o n s id e r t h e p e r m u ta t io n R(x ) of N d e f i n e d by

R( X) -> Xn+ 1 , . • • , Xn + x— 1 , . «. , Xx + 1 •

vVe s h a l l p ro v e t h a t th e f u n c t i o n X A Y i s c o n ju g a te u n d e r R(x )*

to a s u i t a b l e f u n c t i o n X A Y a s d e f in e d in Theorem 6 . 8 . The

r e s u l t w i l l th e n fo l lo w by Theorem 6 . 8 and Theorem 1 . 1 o f

C h ap te r 1 a f t e r we have shown uhat ^ g e n e r a t e s a 1—p la c e

f u n c t i o n c o n ju g a te u n d e r R (x) to th e c i r c u l a r p e rm u ta t io n X+1

(modulo n+1 )#

1 09

F o r f i x e d i n d i c e s r , s , t ( 0 ^ r ^ s ^ t < n , t > O)*

c o n s id e r t h e p a r t i c u l a r f u n c t i o n X O Y d e f in e d a s f o l l o w s : d e f i n e

(X n y ) ' by

U • ♦ • « • • ' *n 1 ) -> XxaYi , • • • ,Xp_ s AYn-s ^yn+1- s > • • • >Yn-r jXn+

and d e f in e

Yi*A • • • #X -*■ X x , . . . , x ^ _ t , Xn+X-a^3 • • • jXp+X-«1 >

y Yn+ 1—/?n 3 • • • > y n + , Yn+ i —t 3 • • • fYn 3o+ 1

th e n

X n Y = ( x * n # Y ) \

I t i s e a sy t o v e r i f y t h a t

r ( i t ^ ( x ) n IT X Y ) ) = X n Y,

and so X A Y i s c o n ju g a te u n d e r R(x ) t o a s u i t a b l e f u n c t i o n

o f ty p e X A Y .

F i n a l l y S g e n e r a t e s t h e 1 -p la c e f u n c t i o n X+n (modulo n+1 )

and we have

r (R"^(X) + 1) = X + n .

T h e re fo re X + 1 i s a l s o c o n ju g a te u n d e r R(x ) t o a f u n c t io n

g e n e r a te d by S. Hence the r e s u l t f o l l o v / s .

110

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