Automata-Theoretic Models for Computational Complementarityecalude/elenathesis.pdf ·...

Automata-Theoretic Models forComputational Complementarity

A thesis presented toThe University of Auckland

in fulfillment of the thesis requirementfor the degree of

Doctor of Philosophy

by

Elena Calude

The University of AucklandOctober 1997

Abstract

The purpose of this Thesis is to study, from the mathematical point of view, anew type of questions about finite automata, questions motivated by looking atautomata as toy models of physical particles.

Working along the line of research initiated by Moore, two computational com-plementarity principles are studied, (for finite–deterministic, complete or incom-plete, nondeterministic–automata with outputs but no initial states) both theo-retically and experimentally; they mimic the physical complementarity and theEinstein–Podolsky–Rosen effect.

Automata are studied via simulations (informally, the automaton A is simulatedby the automaton B if B can perform all computations A can execute and producesthe same outputs; two automata are equivalent in case they simulate each other).A new type of minimization problem will be solved and the solution is proved tobe unique up to an isomorphism; the minimal automaton equivalent to a givenautomaton can be constructed only in terms of outputs for deterministic completeor incomplete automata, but one needs the whole internal machinery for nondeter-ministic automata. It happens that minimal automata are exactly the automatawhich may feature computational complementarity.

Even if the original motivation will remain only metaphorical, the physical mo-tivation was good to suggest new definitions and constructions (simulation, univer-sality, complementarity) leading to new mathematical results (existence of universalfinite automaton, solving in a new way the minimalization problem for nondeter-ministic automata).

III

IV Abstract

Acknowledgments

I wish to warmly thank Dr. Hans Guesgen, my supervisor, for his continuous en-couragement and advice, especially during the period when my research reached adead end.

My thanks goes to Professor Karl Svozil for introducing me to computationalcomplementarity, and to Dr. Bakh Khoussainov, Dr. Marjo Lipponen, Dr. RaduNicolescu and Professor Sheng Yu for their scientific cooperation over the last years;I have learnt a lot from each of them.

I would like to express my deep gratitude to Professors Arto Salomaa and GeorgePaun for their helpful and constructive comments.

A word of thanks is due to Professors Bob Doran and Peter Gibbons who showedfaith in my research abilities from the early stages of my PhD programme (muchmore than myself).

I would like to thank Penny Barry, Dorothy Brown, Anita Lai, Neena Ranigaand Peter Shields for their kind and efficient support.

The financial support offered by the University of Auckland Research Committeeis acknowledged with thanks.

Not least, a warm thank you to Andreea and Cris for their encouragement,patience and faith in me.

V

VI Acknowledgments

Contents

1 Introduction 1

2 Physical Complementarity 3

3 Mathematical Background 7

4 Simulations 15

5 Computational Complementarity 255.1 Moore’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255.2 Moore’s Theorem Revisited . . . . . . . . . . . . . . . . . . . . . . 285.3 Two Complementarity Principles . . . . . . . . . . . . . . . . . . . 285.4 More About Moore’s Automaton . . . . . . . . . . . . . . . . . . . 365.5 Decidability and Complexity . . . . . . . . . . . . . . . . . . . . . . 37

6 Minimality and Universality 436.1 Minimality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436.2 Other Models of Minimality . . . . . . . . . . . . . . . . . . . . . . 466.3 Universality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

7 Testing Computational Complementarity 537.1 Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 537.2 Complementarity: Reversible Instances for Complete Deterministic

Automata . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 557.3 Complementarity: Non-Reversible Instances for Complete Determin-

istic Automata . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 647.4 A Glimpse into Programs . . . . . . . . . . . . . . . . . . . . . . . 677.5 Computational Complementarity Statistics . . . . . . . . . . . . . . 74

8 Open Questions 77

Bibliography 80

VII

VIII Table of Contents

Chapter 1

Introduction

Electronic computers perform computations by moving electrons around, so thephysical restrictions of an electron moving through matter determine how fast suchcomputers can run. Computer speed is limited by a) how fast the computer canmove information from where it is now to where it has to go next and b) by howfast that information can be processed once it gets there.1

This limitation is rather fundamental, because the fastest possible speed forinformation transmission is the speed of light, and the speed of an electron is alreadya substantial fraction of this. So, instead of improving the speed of computer devicesone can concentrate in the speed-up of computation. At a first glance, these maysound like the same thing, until one realizes that the number of operations requiredto perform a computation is determined not only by hardware performance, but bythe algorithm used to perform the computation. An efficient algorithm can performa computation much more quickly than can an inefficient algorithm, even if there isno change in the computer hardware. Even more, there are computable problemsfor which any change in the hardware will essentially make no difference for theiralgorithmic solution. So further improvement in the design of algorithms offers apossible route to continuing to make computers faster and a better exploitationof parallel operations2 is one of the most promising ways of increasing computingefficiency.

This Thesis is devoted to a mathematical study of some limitations imposed toclassical computers as a consequence of Bohr’s physical complementarity, a phe-nomenon which includes as a particular case Heisenberg’s uncertainty principle.The difficulties in understanding and conceptualizing the physical complementar-ity phenomenon, rather inevitable consequences of the quantum formalism, havemotivated the study of extremely simple models featuring complementarity.

Working along the line of research initiated by Moore, two computational

1Here is a nice example due to S. Lloyd: “Consider a garden hose: When you turn on thefaucet, how long does it take for water to come out the other end? If the hose is empty, thenthe amount of time is equal to the length of the hose divided by the velocity at which waterflows down the hose. If the hose is full, then the amount of time it takes for water to emerge isthe length of the hose divided by the velocity at which an impulse propagates down the hose, avelocity approximately equal to the speed of sound in water.”

2As a result of, for example, quantum-mechanical effects.

1

2 1. Introduction

complementarity principles3 are studied, both theoretically (by means of finite—deterministic, complete or incomplete, nondeterministic—automata with outputsbut no initial states) and experimentally (for the case of four-states complete de-terministic automata). Automata are studied via simulations (informally, the au-tomaton A is simulated by the automaton B if B can perform all computations Acan execute and produces the same outputs; two automata are equivalent in casethey simulate each other).

Motivated by physical considerations we discuss a new type of minimizationproblem. We solve this problem and the solution is proved to be unique up toan isomorphism; the minimal automaton equivalent to a given automaton can beconstructed only in terms of outputs for deterministic complete or incomplete au-tomata, but one needs the whole internal machinery for nondeterministic automata.It happens that minimal automata are exactly the automata which may featurecomputational complementarity.

The material is organized as follows. Chapters 2 and 3 are mainly introduc-tory; they contain a brief overview of the current understanding of Bohr’s phys-ical complementarity and a few mathematical facts necessary for our theoreticalconsiderations. In Chapter 4 we study various types of automata simulations inan attempt to model elementary particle behaviour via automata. Chapter 5 de-scribes two types of computational complementarity mainly by critically examiningMoore’s Theorem [56] termed the “uncertainty principle” by Conway [22, p. 21]and “computational complementarity” by Finkelstein and Finkelstein [29]. Chapter6 states and solves the minimization problem and relates minimality to universality.Chapter 7 discusses some experiments with four-states deterministic automata; tomake the programs relatively short we have developed some rudiments of a theoryof operators acting on an automaton space of states. Very complex four–states au-tomata, i.e. automata having an operator monoid size greater or equal than 58, arereversible. This supports a conjecture relating complexity and reversibility: highlycomplex automata are reversible.4 Finally, the last chapter enumerates some openproblems.

The Thesis contains some of the author’s contributions to a few joint papers: [13,12, 14, 15, 16]; however, I have included some results which have not been published.Finally I would like to notice that the organization of the Thesis is completelydifferent from the way the results were originally obtained: it goes from the generalcase (nondeterministic) to the most particular case (complete deterministic).

3They are abstract toy models featuring physical complementarity.4This fact might be interesting in the context of quantum computation.

Chapter 2

Physical Complementarity

Today scientists describe the universe in terms of two fundamental theories: thegeneral theory of relativity and quantum mechanics. The theory of relativity de-scribes the force of gravity and the large–scale structure of the universe. Quantummechanics deals with phenomena on extremely small scales. Relativity altered theclassical concept of physical objectivity but left open the possibility of a suprememathematician who, in Einstein’s view, neither cheats nor plays dice. Quantummechanics went one step further: it not only did situate the experimenter in theuniverse, but it stated that the experimenter can be modeled as a “sturdy, classicalentity” composed of a macroscopic number of microscopic objects. Nevertheless, theexperimenter can neither predict nor control certain “spontaneous” microphysicalevents: the observer is bound by complementarity1—that is, informally speaking,either experiences one certain type of observation, (exclusive) or a different, com-plementary one.

The unity of classical physics is replaced by a duality in quantum mechanics: weare forced to simultaneously hold in our minds two mutually inconsistent picturesof the same objects: “particle” and “wave”.2

In spite of the fact that physical complementarity appears to be a rather straight-forward consequence of the quantum formalism, the conceptualization of comple-mentarity is highly controversial causing considerable attention, concern, thoughtand neglect (see Jammer [45, 46]). Indeed, physical complementarity is tied upwith measurement, a highly nontrivial matter, as contemplations by Wigner [77],Wheeler [75], and Bell [4], among many others, show.

How subtle the issue may get can be best demonstrated by the fact that in certaininstances it is possible to “reconstruct” the quantum wave function after its alleged“collapse” (see Greenberger and Yasin [39]). Thereby, not a single (quantum) bitof information should remain available from the previous “measurement”. In sucha scenario, it is possible to “measure” complementary observables: the price to be

1A term invented by Bohr.2There are at least two philosophical attitudes which can be adopted: a) admit that our

knowledge is partial, b) claim that these seemingly contradictory descriptions do not simultaneouslyexist. In either case we would say, following Bohr, that position is complementary to momentumand energy is complementary to time.

3

4 2. Physical Complementarity

paid amounts to the total ignorance of the first “measurement outcome”.Recently, schemes for an “interaction-free wave function collapse” associated

with “interaction-free” measurement schemes have got renewed attention (see, forexample, Dicke [23], Elitzur and Vaidman [26], Vaidman[72], Kwiat, Weinfurter,Herzog, Zeilinger and Kasevich [48]). Compare Bohr’s statement (cf. [8], reprintedin Wheeler and Zurek [76, p. 89 and 103])

. . . the quantum postulate implies that any observation of atomic phe-nomena will involve an interaction with the agency of observation notto be neglected. . . . the impossibility of neglecting the interaction withthe agency of measurement means that every observation introduces anew uncontrollable element. Indeed, it follows . . . that the measurementof the positional co-ordinates of a particle is accompanied not only by afinite change in the dynamical variables, but also the fixation of its posi-tion means a complete rupture in the causal description of its dynamicalbehaviour, while the determination of its momentum always implies agap in the knowledge of its spatial propagation.

with Gabor’s statement [32, p. 124],

No observation can be made with less than one quantum passing throughthe observed object.

The “folklore” understanding of complementarity, in general, and the Heisen-berg uncertainty relation, in particular, is the existence of certain (complementary)features of a quantum system which cannot be measured and predicted simultane-ously with arbitrary accuracy. The first (but not last) attempt to overcome a certainvagueness in its definition (cf. Ballentine [1, pp. 364-367]) has been undertaken byPauli [57], who called two classical concepts complementary

if the applicability [[operationalizability]] of the one (e.g., position coor-dinate) stands in the relation of exclusion to that [[operationalizability]]of the other (e.g., momentum)

in the sense that any experimental setup for measuring one object interferes destruc-tively with any experimental setup for measuring the other object (see Jammer [45,p. 355]).

The “canonical” understanding of complementarity is expressed in Messiah [53,p. 154]

The description of properties of microscopic objects in classical termsrequires pairs of complementary variables; the accuracy in one mem-ber of the pair cannot be improved without a corresponding loss in theaccuracy of the other member.

...

It is impossible to perform measurements of position x and momentum pwith uncertainties (defined by the root-mean square deviations) ∆x and

2. Physical Complementarity 5

∆p such that the product of ∆x∆p is smaller than a constant unit ofaction h

2.

Complementarity maintains that wave and particle are mutually exclusive con-cepts: when a particular experiment shows one, the other is absent and the two cannever be known simultaneously. As a consequence, it is impossible to know every-thing about the world; indeed, we can only know half of everything. In Prigogine’swords [62, p. 51],

the world is richer than it is possible to express in any single language.

Is it possible to get around this limit somehow? Einstein has constantly soughta means to do so, but the debate continues to the present day.

6 2. Physical Complementarity

Chapter 3

Mathematical Background

We begin by introducing some notations and basic definitions.If S is a finite set, then |S| denotes the cardinality of S. A partial function

f : A→ B is a function defined for some elements from A. By 2A we denote the

power set of A, i.e., the set of all subsets of A. In case f is not defined on a ∈ Awe write f(a) = ∞. Let D(f) = a ∈ A | f(a) 6= ∞ denote the domain of f . IfD(f) = A, we say that f is total. Two partial functions f and g are equal1, whenD(f) = D(g) and f(a) = g(a), for every a ∈ D(f).

If Σ is a finite set, called alphabet, then Σ∗ stands for the set of all finite wordsover Σ; the empty word is denoted by λ. Put Σ+ = Σ∗ \ λ. By w+ we mean theset of all powers wi, i > 0, of the word w ∈ Σ? whereas w∗ = w+ ∪λ. The lengthof a word w is denoted by |w|.

A finite automaton, in the classical sense, reads words w over a fixed alphabetΣ, letter by letter in discrete steps. It has a finite number of internal states, aspecific initial state and a specific set of final states. At each moment of time, it isin one of these states. The state at the next moment depends on the state and thesymbol read at the present moment. In this sense, the reading of a letter causes astate transition. The automaton begins by reading the first letter of w in the initialstate. The word is accepted by the automaton if it causes some sequence of statetransitions leading to some final state after the whole word w has been read. Allaccepted words constitute the language accepted by the automaton. Formally (seeKuick and Salomaa [47]) a finite automaton over Σ is a 4-tuple A = (Q,M, q0, F )where

• Q is a finite nonempty set of states,

• M ∈ 2ΣQ×Q is a square matrix, specifing the state transitions,

• q0 ∈ Q is the initial state,

• F ⊆ Q is the set of final states.

In 1956 Moore was the first to study some experiments on finite deterministicautomata in an attempt to understand what kind of conclusions about the internal

1We write simply f = g.

7

8 3. Mathematical Background

conditions of a finite machine it is possible to draw from input-output experiments.The main conclusion of these studies is that it is impossible to determine the initialstate of an automaton, and consequently a discrete model of Heisenberg uncertaintyhas been suggested. For this aim the notion of automaton with initial state is notadequate so in what follows we will use a different definition for a finite automatonwhich does not include the initial states.

We fix two finite alphabets Σ and O: Σ contains input symbols, and O containsoutput symbols.2 A nondeterministic finite automaton over the alphabets Σ and Ois a triple A = (SA,∇A, FA), where

• SA is a finite nonempty set of states,

• ∇A is a total function from SA × Σ to the set 2SA of all subsets of SA, calledthe transition table,

• FA is a mapping from the set of states SA into the output alphabet O, calledoutput function.

As we will study only finite automata (deterministic or nondeterministic) in thisThesis, we will omit the word finite.

In drawing graph representations of automata, we denote states by and labelthem with symbols from the output alphabet.3 In Figure 3.1 we have pictured ameans that there is a transition σ from q to p, that is p ∈ ∇A(q, σ), and FA(q) =ν, FA(p) = µ.

-q p

σν| |µ

Figure 3.1

The function ∇A can be extended to a function ∇A : SA×Σ∗ → 2SA , as follows:

∇A(q, λ) = q, for all q ∈ SA,∇A(q, σw) =

⋃s∈∇A(q,σ)

∇A(s, w), for all q ∈ SA, σ ∈ Σ, w ∈ Σ?.

Let A = (SA,∇A, FA) be a nondeterministic automaton. There are several waysto introduce the notion of “response” of A to an input sequence of signals. Takew = σ1 . . . σn ∈ Σ? and s0 ∈ SA. A trajectory of A on s0 and w is a sequence

s0, s1, . . . , sn

of states such that si+1 ∈ ∇A(si, σi+1) for all 0 ≤ i ≤ n−1. A trajectory s0, s1, . . . , snemits the output FA(s0)FA(s1) · · ·FA(sn) ∈ O∗.

2In almost all examples we will be using the binary alphabet O = 0, 1.3Sometimes, we omit the name of the state.

3. Mathematical Background 9

The total response, denoted by RA, is a function which to any (s, w) ∈ SA ×Σ?

assigns the set RA(s, w) of all outputs emitted by all trajectories of A on s and w.The final response of A is a function fA which to any pair (s, w) ∈ SA ×Σ? assignsthe subset of all last symbols occurring in words in RA(s, w).

Formally, the total response of a nondeterministic automaton A is the totalfunction RA : SA × Σ?→2O∗,

RA(s, λ) = FA(s),RA(s, σw) =

⋃q∈∇A(s,σ)

FA(s) ·RA(q, w),

where q, s ∈ SA, σ ∈ Σ, w ∈ Σ? and for x ∈ O and Y ⊆ O?, x · Y = xy | y ∈ Y .Define the function last : O?→O by

last(λ) = λ,

last(wσ) = σ,

for σ ∈ O, w ∈ O?. Then the final response is the total function fA : SA ×Σ?→2O,fA(s, w) = last(u)|u ∈ RA(s, w).

In case ∇A(q, σ) has at most one element for every q ∈ SA and σ ∈ Σ, thenthe automaton will be called deterministic incomplete4. In this case we write ∆A

instead of ∇A. Since ∆A is a partial function, D(∆A) denotes the domain of ∆A.Thus D(∆A) is a subset of SA×Σ and needs not include every state–input pair. Fora current state q ∈ SA and a current input σ ∈ Σ, ∆A(q, σ) is the next state of A if(q, σ) ∈ D(∆A); otherwise, the next state is undefined. Formally, ∆A : SA×Σ

→ SAand for any q ∈ SA and σ ∈ Σ,

∆A(q, σ) = p iff ∇(p, σ) = p

and

∆A(q, σ) =∞ iff ∇(p, σ) = ∅.The extension of the transition function ∆A to a partial function, denoted by

∆A, ∆A : SA × Σ? → SA, acts as follows: for every s ∈ SA, w ∈ Σ? and σ ∈ Σ,

∆A(s, λ) = s,

∆A(s, σw) =

∆A(∆A(s, σ), w), if ∆A(s, σ) 6=∞,∞, otherwise.

For all p ∈ SA, the set

WA(p) = w ∈ Σ? | ∆A(p, w) 6=∞4Ginsburg [44] has studied Mealy machines with incomplete transitions.


consists of all words leading to full trajectories on state p. Following Ginsburg [36],we say that a word u is applicable to the state p if u ∈ WA(p).

The following lemma is a direct consequence of definitions.

Lemma 3.1 Let A = (SA,∆A, FA) be an incomplete automaton. Then

1) For any word w = σ1σ2 . . . σn ∈ Σ?, σi ∈ Σ, 1 ≤ i ≤ n, and anystate s ∈ SA, if ∆A(s, w) 6= ∞, then ∆A(s, σ1) 6= ∞, ∆A(s, σ1σ2) 6=∞, . . . , ∆A(s, σ1σ2 . . . σn−1) 6=∞.

2) For all p ∈ SA and u, v ∈ Σ?,

vu ∈ WA(p) iff v ∈ WA(p) and u ∈ WA(∆A(p, v)).

3) For all p ∈ SA, λ ∈ WA(p).

For an incomplete automaton, the total response is a sequence of outputs emittedby all the states that are visited in the entire computation of the input, whereasthe final response is the output emitted only by the last state. Formally, the totalresponse of A is the partial function RA : SA × Σ? → O∗,

RA(s, λ) = FA(s),

RA(s, σ1 . . . σn) = FA(s)FA(∆A(s, σ1))FA(∆A(s, σ1σ2))

. . . FA(∆A(s, σ1 . . . σn)),

for s ∈ SA, σ1 . . . σn ∈ WA(s), σi ∈ Σ, n ≥ 1 and 1 ≤ i ≤ n. The final response of Ais the partial function fA : SA × Σ? → O, fA(s, w) = FA(∆A(s, w)), for all s ∈ SAand w ∈ WA(s). Note that D(RA) = D(fA) and always |RA(s, w)| = |w|+ 1.

Example 3.2 Let Σ = a, b, O = 0, 1 and consider the three-state incompleteautomaton A presented in Figure 3.2. The output function is defined by FA(p) = 0and FA(q) = FA(r) = 1. Clearly, RA(p, aba) = 0111, RA(q, aba) = 1101, andRA(r, aba) =∞. We also have fA(p, aba) = 1, fA(q, aba) = 1, and fA(r, aba) =∞.

JJJJJ

p q

r

M %$

| a

|

|0

a

b 1

1

b

Figure 3.2


Further on, if ∇A(q, σ) has exactly one element for every q ∈ SA and σ ∈ Σ, thenthe automaton will be called deterministic (complete), and in this case we write δAinstead of ∇A.5 A deterministic automaton which is not complete will be sometimecalled strictly incomplete. Note that an automaton A is complete iff for all s ∈ SA,WA(s) = Σ?. The extension of δA to SA × Σ? is defined by the equations:

δA(q, λ) = q, for all q ∈ SA,

δA(q, σw) = δA(δA(q, σ), w), for all q ∈ SA, σ ∈ Σ, w ∈ Σ∗.

In case of deterministic complete automata we have one more type of response,the initial response. For completeness we list all definitions. The total response ofthe automaton A is the function RA : SA × Σ? → O? defined as follows:

RA(s, λ) = FA(s),

RA(s, σ1 . . . σn) = FA(s)FA(δA(s, σ1))FA(δA(s, σ1σ2))

. . . FA(δA(s, σ1 . . . σn)),

where σi ∈ Σ, s ∈ SA, 1 ≤ i ≤ n. The final response of A is the function fA :SA×Σ? → O defined, for all s ∈ SA and w ∈ Σ?, by fA(s, w) = FA(δA(s, w)). Theinitial response of A is the function iA : SA × Σ? → O defined, for all s ∈ SA andw ∈ Σ?, by iA(s, w) = FA(s).

Consider, for example, Moore’s automaton (given in [56]; see Figure 3.3), inwhich SM = 1, 2, 3, 4, Σ = a, b and the transition is given by the followingtables

q σ δ(q, σ)1 a 41 b 32 a 12 b 3

q σ δ(q, σ)3 a 43 b 44 a 24 b 2

and the output function is defined by FM(1) = FM(2) = FM(3) = 0, FM(4) = 1. Wehave RM(3, baab) = 01000, RM(4, aaa) = 1001, fM(3, baab) = 0, fM(4, aaa) = 1,iM(3, baab) = 0, and iM(4, aaa) = 1. Its graphical representation is presented inFigure 3.3.

5Notice that a complete deterministic automaton is an incomplete automaton with a totaltransition function.


6 6

@@@@@@@@@R

| |1

|0 0

|

b

04 3

21

a,b

a b

a

a,b

Figure 3.3 (Moore’s automaton)

Let A = (SA, δA, fA) be a deterministic automaton. The language generated by Aon the state q is L(A, q) = w ∈ Σ? | fA(δA(q, w)) = 1. The emptiness, inclusionand equality properties for regular languages are algorithmically decidable and theclass of regular languages is closed under union, intersection and complement (cf.Salomaa [65], Hopcroft and Ullman [43]).

In what follows we will omit the overline in the extensions of ∇A, ∆A, and δA,i.e., we write ∇A = ∇A, ∆A = ∆A, δA = δA.

An automaton A = (SA, δA, FA) is reversible if for every states p, q ∈ SA andu ∈ Σ? with δA(p, u) = q, there exists a word w ∈ Σ? such that δA(q, w) = p.

An automaton A is connected if for every pair of states s and p from SA, thereexists a sequence of states s1, s2, . . . , sm such that

1. s1 = s,

2. sm = p,

3. for any i ∈ 1, 2, . . . ,m − 1, there is σ ∈ Σ such that δA(si, σ) = si+1 orδA(si+1, σ) = si.

An automaton A is strongly connected if for every pair of states s and p fromSA, there is a word w ∈ Σ? such that δA(p, w) = q.

Clearly, any strongly connected automaton is connected, but the converse im-plication is false. Moore’s automaton (see Figure 3.3) is strongly connected andreversible. The automaton in Figure 3.4 is connected as any two states are corre-lated by a transition but is neither strongly connected nor reversible as no transitionto state 1 is possible.


@@@@@@@@@R

6

> 6

-

4 3

21

|

|

|

a|

0

0

1

ba

0

b,a

,

a b

b

Figure 3.4

The automaton from Figure 3.5 is reversible as the only transition out fromstate 1 is to state 1 itself and from states 2,3 and 4 we can reach any of the states2,3 or 4 on various inputs. The same automaton is not connected as state 1 is notlinked with any other state.

@@@@@@@@@R

6

4 3

21

6 %

$

|

a

|

|

|

0

0

1

0

b

a

a

Figure 3.5

Chapter 4

Simulations

Informally, an automaton A is simulated by an automaton B if B can perform allcomputations of A and produces the same outputs. We start with various modelsof simulation for the case of incomplete deterministic automata.

Let A = (SA,∆A, FA) and B = (SB,∆B, FB) be incomplete automata, and fixa mapping h : SA → SB. Consider the following conditions:

(i) For all s ∈ SA, WA(s) ⊆ WB(h(s)).

(ii) For all s ∈ SA, WA(s) = WB(h(s)).

(iii) For all s ∈ SA and σ ∈ Σ ∩WA(s), h(∆A(s, σ)) = ∆B(h(s), σ).

(iv) For all s ∈ SA and w ∈ WA(s), RA(s, w) = RB(h(s), w).

(v) For all s ∈ SA and w ∈ WA(s), fA(s, w) = fB(h(s), w).

We define two types of simulations: weak and strong. In the weak case, forevery state s ∈ SA there is a state h(s) ∈ SB such that h(s) does everything that sdoes (and possibly more) whereas in the strong simulation h(s) does exactly whats does (and nothing more). Formally, we say that

• A is weakly simulated by B, A ≺ B, if the mapping h satisfies (i), (iii) and(iv);

• A is weakly f–simulated by B, A ≺f B, if the mapping h satisfies (i), (iii)and (v);

• A is strongly simulated by B, A¿ B, if the mapping h satisfies (ii), (iii) and(iv);

• A is strongly f–simulated by B, A¿f B, if the mapping h satisfies (ii), (iii)and (v).

Clearly, strong simulation implies weak simulation, but the converse implicationis false as the following example shows.

15

16 4. Simulations

Example 4.1 The strictly incomplete automaton A, given in Figure 4.1, is weaklysimulated by the complete automaton B, given in Figure 4.2, via the mapping h :SA → SB, h(p) = s, h(q) = r; however A is not strongly simulated by B as there isno mapping from SA to SB which preserves the computational power of the state p.

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We will next study the connection between strong/weak simulation and f–strong/f–weak simulation. It turns out that it makes no difference whether wedefine simulation by total or final response.

Theorem 4.2 For any incomplete automata A and B:

1) A is strongly simulated by B iff A is strongly f–simulated by B.

2) A is weakly simulated by B iff A is weakly f–simulated by B.

Proof. We will prove 1), as for showing 2) the argument is similar.If A is strongly simulated by B it follows from the definitions of simulations that

A is strongly f–simulated by B. Now we assume that A is strongly f–simulated byB and we prove that A is strongly simulated by B. As we need to show that twopartial function are equal we have to prove that:

i) WA(s) = WB(h(s)) for all s ∈ SA and

ii) RA(s, w) = RB(h(s), w) for all s ∈ SA, w ∈ WA(s).

4. Simulations 17

As i) follows from the definition of the strong f–simulation we will prove ii) byinduction on |w|. For w = λ the equation RA(s, w) = RB(h(s), w) follows from thedefinition of the strong f–simulation. Suppose that RA(s, w) = RB(h(s), w) holdsfor all s ∈ SA and w ∈ WA(s) with |w| ≤ n. Let wσ ∈ WA(s), |w| = n and σ ∈ Σ.Then

RA(s, wσ) = RA(s, w)FA(∆A(∆A(s, w), σ)).

By induction hypothesis RA(s, w) = RB(h(s), w). Since ∆A(∆A(s, w), σ) =∆A(s, wσ), we have h(∆A(s, wσ)) = ∆B(h(s), wσ). Therefore

FA(∆A(s, wσ)) = FB(h(∆A(s, wσ))) = FB(∆B(h(s), wσ)).

Finally we have

RB(h(s), wσ) = RB(h(s), w)FB(∆B(h(s), wσ)) = RA(s, wσ). 2

Lemma 4.3 If h : SA → SB and B strongly/weakly (or f-strongly/f-weakly) simu-lates A via h, then h(∆A(s, w)) = ∆B(h(s), w), for all s ∈ SA, w ∈ WA(s).

Proof. Let assume that B f–weakly simulates A via h, and fix s ∈ SA.Notice that for all w ∈ WA(s), ∆A(s, w) 6= ∞, and h(∆A(s, w)) 6= ∞. To

prove that h(∆A(s, w)) = ∆B(h(s), w), for w ∈ WA(s), we will use inductionon |w|. When |w| = 0 it follows that h(∆A(s, w)) = h(∆A(s, λ)) = h(s) and∆B(h(s), w) = ∆B(h(s), λ) = h(s). So the above equality holds for |w| = 0. Assumethat it holds for all |w| ≤ n. Let now w ∈ WA(s) be uσ with σ ∈ Σ and |u| = n.Then h(∆A(s, w)) = h(∆A(s, uσ)) = h(∆A(∆A(s, u), σ))) = ∆B(h(∆A(s, u)), σ),= ∆B(∆B(h(s), u), σ) = ∆B(h(s), uσ) = ∆B(h(s), w). As f–strongly simula-tion implies f–weakly simulation, it follows that the above result is valid for f–strongly simulation. By Theorem 4.2 it follows that the equality is also valid forstrongly/weakly simulation. 2

Note that for strong simulations the equality h(∆A(s, w)) = ∆B(h(s), w) holdsactually true for all s ∈ SA and w ∈ Σ?, since ∆A(s, w) = ∆B(h(s), w) = ∞, forw /∈ WA(s). For weak simulations this is not the case; for instance, in Example 4.1,∆A(p, b) =∞ but ∆B(h(p), b) = ∆B(s, b) 6=∞.

The following result shows that in some sense all strong simulations preservethe completeness of automata.

Proposition 4.4 Let A and B be incomplete automata such that A¿ B. If B iscomplete, then A is also complete.

Proof. Let h : SA → SB be the mapping verifying properties (ii), (iii) and(iv).

Assume that B is a complete automaton. By Lemma 3.1, this is equivalent tothe fact that WB(q) = Σ?, for any q ∈ SB. Hence also WB(h(p)) = Σ?, for any

18 4. Simulations

p ∈ SA, and since WA(p) = WB(h(p)) by property (ii), the automaton A is alsocomplete. 2

Note that the converse implications in Proposition 4.4 fail to be true. Considerfor example automata A, given in Figure 4.4 and B, from Figure 4.3. ClearlyA¿ B via h : SA → SB where h(r) = q, but A is a complete automaton and B isa strictly incomplete one.

Corollary 4.5 Let A and B be incomplete automata. If A¿ B and B ¿ A, thenA and B are both either complete or strictly incomplete automata.

Proposition 4.4 and Corollary 4.5 are equally valid for strongly f–simulationbecause of Theorem 4.2. For weak simulations there is no corresponding result.In particular, Corollary 4.5 does not hold for weak simulations. For instance, theincomplete automata A and B, given in Figures 4.3 and 4.4 respectively, are weaklysimulating each other, even though A is strictly incomplete and B is complete.

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The behavioral simulation, which is weaker than all previous simulations, makesuse only of the outputs produced by the automaton, but not of the transition∆A (which cannot be measured under the physical interpretation of automata 1).To motivate the following formalization, consider the incomplete automata A andB from Figures 4.5 and 4.6, respectively.

1If the automaton is conceived as a black box, then its internal structure remains “hidden”;the experimenter can only check the output produced by the automaton, which can be thoughtas a measurement.

4. Simulations 19

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Neither A nor B strongly (nor weakly) simulates the other one as any possiblesimulation mapping h from SA to SB must associate p1 to q1; so h(p1) = q1, then,on one side

h(p2) = h(∆A(p1, a)) = ∆B(h(p1), a)) = ∆B(q1, a) = q2

and, on the other side

h(p2) = h(∆A(p1, b)) = ∆B(h(p1), b)) = ∆B(q1, b) = q3,

which is impossible.Nevertheless, if we consider the mapping h1 : SA → SB,

h1(p1) = q1, h1(p2) = q2, h1(p3) = q4, h1(p4) = q4,

we notice that for any state x ∈ SA there is a state h1(x) ∈ SB such that startingfrom this state, B responds to any word w ∈ WA(h1(x)), in the same way as Astarting from the state x. There is also a mapping h2 : SB → SA,

h2(q1) = p1, h2(q2) = p2, h2(q3) = p2, and h2(q4) = p3,

20 4. Simulations

having a similar behavior.We are now ready to give formal definitions for behavioral simulations (called

in what follows β-simulations). Let A = (SA,∆A, FA) and B = (SB,∆B, FB) beincomplete automata. We say that

• A is strongly β-simulated by B, A¿β B, if there is a mapping h : SA → SBwhich satisfies conditions (ii) and (iv).

• A is weakly β-simulated by B, A ≺β B, if there is a mapping h : SA → SBwhich satisfies conditions (i) and (iv).

We are not going to define behavioral f -simulation since the result analogue ofTheorem 4.2 shows that this definition brings nothing new.

We also notice that Proposition 4.4 (and hence also Corollary 4.5) is valid forstrongly β-simulation.

Finally, notice that strong simulation implies both weak simulation and strong β-simulation and weak simulation implies weak β-simulation, but weak simulation andstrong β-simulation are incomparable; for instance, in Example 4.1 the automatonA is weakly simulated but not strongly β-simulated by B, whereas A, from Figure4.5 and B, from Figure 4.6 are strongly β-simulating but not weakly simulatingeach other.

For nondeterministic automata we study only the behavioral (β) simulation.2

To this aim we need some rudiments of game theory. Informally, the behaviour ofa nondeterministic automaton A receiving an input w can be thought as a gamewith two players: Player 0 and Player 1. A move of any player consists of pickingup a state of A. Player 0 picks a state p. Player 1 tries to pick up a state q suchthat the observer cannot distinguish p and q using responses coming from p and q;Player 0 tries to prove the opposite. For the sake of completeness we include somesimple facts about finite games.

Let T be a finite tree, and W be a set of some paths from T . Nodes on evenpositions are positions of Player 0; the remaining nodes are positions of Player 1.A play is a finite sequence of nodes

x0y0 . . . xkyk

such that x0 is the root of T and the sequence x0y0 . . . xkyk is a path in T . A gameis the pair (T,W ).

A strategy for Player 0 (Player 1) is a function which maps every position xof Player 0 (Player 1) to a child (i.e., an immediate successor) of x. For instance,Player 0 can follow a strategy g and an initial play according to this strategy canbe:

g(x0)y0g(y0)y1g(y1)y2g(y2),

where x0 is the root of T .We say that Player 1 wins the game (T,W ) if there is a strategy g for Player 1

such that every play generated by g belongs to W ; otherwise Player 1 looses.

2The only simulation leading to the construction of the minimal automaton.

4. Simulations 21

Fact 4.6 In the game (T,W ) one of the players wins. If Player 1 does not win thisgame, then there is a strategy g for Player 0 such that every play induced by g doesnot belong to W .

Proof. Let CW be the set of all nodes in T which are the last elements of thepaths in W . We mark elements of T as follows:

Stage 0. Every element in CW is marked.Stage i+ 1. Consider a node x. If x is a position of Player 0, then x is marked

at this stage if all children of x are marked. Otherwise we do not mark x at thisstage (x may be marked at later stages). If x is a position of Player 1, then x ismarked if some child of x is marked. Otherwise x is not marked at this stage.

Clearly there is a stage after which no node will be marked. Thus, there are twocases:

Case 1. If the root is marked, then Player 1 wins. The winning strategy forPlayer 1 is the following: if x is marked and is a position for Player 1, then take amarked child of x.

Case 2. If the root is not marked, then Player 0 wins. The winning strategy forPlayer 0 is the following: if x is an unmarked position for Player 0, then take anunmarked child of x. 2

From the proof of this fact we get the following:

Corollary 4.7 Consider the game (T,W ). A strategy g is a winning strategy forPlayer 1 iff every play according to g goes through marked nodes.

Let A and B be two, not necessarily distinct, nondeterministic automata. Takestates p ∈ SA and q ∈ SB, and fix a positive integer n ≥ 1. We define a gameΓ(p, q, n) between two players: Player 0 and Player 1. Again, Player 0 tries to provethat outputs emitted by trajectories which begin in p are different from outputsemitted by trajectories originated in q. Player 1 tries to show the opposite. Thedifference from the previous games is that Player 0 (Player 1) is not restricted toconsider computations which begin from p (q) only. Player 0 (Player 1) is allowedto pick up any instance of a computation which begins from q (p) as well.

Here is a description of a play. Every play has at most n stages. Each stagebegins with a move of Player 0 and ends with a response of Player 1.

Stage 0. Player 0 picks up either p or q. Player 1 responds by picking up theother state.

Stage k + 1 ≤ n. At the end of stage k we have two sequences

p0p1 . . . pk

and

q0q1 . . . qk

22 4. Simulations

where p0 = p and q0 = q. Now Player 0 chooses a state either from⋃σ∈Σ∇A(pk, σ) or

from⋃σ∈Σ∇B(qk, σ). If Player 0 chooses a pk+1 from

⋃σ∈Σ∇A(pk, σ), then Player

1 responds by choosing a state qk+1 from⋃σ∈Σ∇B(qk, σ). If Player 0 chooses a

qk+1 from⋃σ∈Σ∇A(qk, σ), then Player 1 responds by choosing a state pk+1 from⋃

σ∈Σ∇B(pk, σ). This ends a description of stage k + 1 of a play.Let

p0p1 . . . pt

andq0q1 . . . qt

be sequences produced during a play. We say that Player 1 wins the play if for all0 < i ≤ t, σ ∈ Σ, we have pi ∈ ∇A(pi−1, σ) if and only if qi ∈ ∇B(qi−1, σ) andFA(pi) = FB(qi).

From the definition of the game Γ(p, q, n) we have the following lemma.

Lemma 4.8 If a player wins the game Γ(p, q, n), then he wins the game Γ(q, p, n).

Let A an B be nondeterministic automata. We say that the state p is equivalentto the state q, and we write p ≡ q, if Player 1 wins the game Γ(p, q, n), for allpositive integers n.

Suppose now that in the game Γ(p, q, n) the automata A and B coincide.

Lemma 4.9 The relation ≡ is an equivalence relation on SA.

Proof. It is clear that ≡ is symmetric and reflexive. For transitivity we supposethat p ≡ q and q ≡ s and we need to show that p ≡ s, that is Player 1 wins thegame Γ(p, s, n) for every n. Let g1 and g2 be winning strategies for Player 1 ingames Γ(p, q, n) and Γ(q, s, n), respectively. Then a winning strategy g for Player 1in the game Γ(p, s, n) can be described as follows. Suppose that at the end of stagek (k < n) of a play the players have produced two sequences

p0p1 . . . pk

ands0s1 . . . sk

where p0 = p and s0 = s. If at stage k + 1 Player 0 chooses a state pk+1 from⋃σ∈Σ∇A(pk, σ), then Player 1 follows the instructions bellow:

First, think of this move of Player 0 as a move in the game Γ(p, q, n).Secondly, using the strategy g1, respond to the move as you were in thegame Γ(p, q, n). Thirdly, consider this response of Player 1 as a moveof Player 0 in the game Γ(q, s, n). Finally respond, using the strategyg2, to the move as you were in the game Γ(q, s, n).

On the other hand, if Player 0 chooses a state sk+1 from⋃σ∈Σ∇A(sk, σ), then Player

1 follows the instructions:

4. Simulations 23

First, think of this move of Player 0 as a move in the game Γ(q, s, n).Secondly, using the strategy g2, respond to the move as you were in thegame Γ(q, s, n). Thirdly, consider this response of Player 1 as a moveof Player 0 in the game Γ(p, q, n). Finally respond, using the strategyg1, to the move as you were in the game Γ(p, q, n).

In both cases the strategy is clearly a winning strategy for Player 1. 2

We say that A is β-simulated by B if there is a mapping h : SA → SB such thatfor all s ∈ SA, s ≡ h(s). We denote this fact by A ≤β B.3 Thus, the function h inthis definition means that Player 1 wins the game Γ(p, h(p), n), for every n.

In Figure 4.7 there is an example of an automaton in which the states p and qare not ≡, in spite of the fact that they generate the same responses. Here is theargument.

A) The states p and q generate exactly the same responses, namely:00001, 00000.

B) The following play shows that p and q are not equivalent:

• Player 0 starts with p/0 and Player 1 responds with q/0.

• Player 0 moves to p1/0 and Player 1 responds with q1/0.

• Player 0 goes to q2/0 and Player 1 responds with p2/0.

• Player 0 moves to q3/0.

At this point, Player 1 has two options. If he chooses p3/0, then Player 0 moves toq5/0 and Player 1 has to go to p4/1. We have got two different series of outputs:00000 for Player 0 and 00001 for Player 1. If Player 1 chooses instead p5/0, thenPlayer 0 moves to p4/0 and Player 1 has to respond with p6/0. Again we have gottwo different series of outputs: 00001 for Player 0 and 00000 for Player 1.

3Note that the simulation relation defined above coincides with β-simulation of deterministicautomata, in case A and B are deterministic.

24 4. Simulations

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Chapter 5

Computational Complementarity

The difficulties in understanding and conceptualizing the complementarity phe-nomenon served as a motivation for considering extremely simple models featuringcomplementarity. The fact that ultimately complementarity refers to what an ex-periment can or cannot reveal makes experiments with automata relevant.1 It is theaim of this chapter to introduce and study two new computational complementarityprinciples expressed in terms of finite automata.

5.1 Moore’s Theorem

Moore’s paper [56] is the first attempt to study experiments on finite automatawith the aim of understanding what kind of conclusions about the internal statesof a finite automaton can be possible derived from input-output experiments.2

A (simple) Moore experiment can be described as follows: a copy of the au-tomaton will be experimentally observed, i.e. the experimenter will input a finitesequence of input symbols to the automaton and will observe the sequence of out-put symbols. The correspondence between input and output symbols depends onthe particular chosen automaton and on its initial state. The experimenter willstudy the sequences of input and output symbols and will try to conclude that “themachine being experimented on was in state q at the beginning of the experiment”.3

Moore’s experiments have been studied from a mathematical point of view byvarious researchers, notably by Ginsburg [35], Gill [34], Chaitin [17], Conway [22],and Brauer [10].

Consider now a deterministic complete automaton A = (SA, δA, fA). FollowingMoore [56] we shall say that a state q is “indistinguishable” from a state q′ (withrespect to A) if every experiment performed on A starting in state q produces the

1They can be conceived as toy models featuring complementarity.2To emphasize the conceptual nature of his experiments, Moore has borrowed from physics the

word “Gedanken”.3This is often referred to as a state identification experiment.

25

26 5. Computational Complementarity

same outcome as it would starting in state q′. Formally,

RA(q, x) = RA(q′, x),

for all words x ∈ Σ+.An equivalent way to express the indistinguishability of the states q and q′ is to

require, following Conway [22, p. 3], that for all w ∈ Σ?,

fA(δA(q, w)) = fA(δA(q′, w)).

Indeed,

RA(q, x1x2 . . . xn) = fA(q)fA(δA(q, x1))fA(δA(q, x1x2))

· · · fA(δA(q, x1x2 . . . xn)),

for all q ∈ Q, x1x2 . . . xn ∈ Σ?.A pair of states will be said to be “distinguishable” if they are not “indistin-

guishable”, i.e. if there exists a string x ∈ Σ+, such that RA(q, x) 6= RA(q′, x).

Moore [56] has proven the following important theorem:

Theorem 5.1 (Moore) There exists an automaton M such that any pair of itsdistinct states are distinguishable, but there is no experiment which can determinewhat state the machine was in at the beginning of the experiment.

Proof. Consider the automaton M displayed in Figure 3.3 and note that eachpair of distinct states can be distinguished by an experiment: 1, 2 by x = a, 1, 3 byx = b, 1, 4 by x = a, 2, 3 by x = a, 2, 4 by x = a, and 3, 4 by x = a.

However, there is no (unique) experiment capable to distinguish between everypair of arbitrary distinct states. Two cases have to be examined:

A) The experiment starts with b, i.e. x = bu, u ∈ Σ?.In this case RM(1, x) = RM(2, x), that is x cannot distinguish between the states

1, 2 as

RM(1, x) = RM(1, bu)

= fM(1)fM(δM(1, b))RM(δM(1, b), u)

= fM(1)fM(3)RM(3, u)

= 00RM(3, u)

and

RM(2, x) = RM(2, bu)

= fM(2)fM(δM(2, b))RM(δM(2, b), u)

= fM(2)fM(3)RM(3, u)

= 00RM(3, u).

5.1 Moore’s Theorem 27

B) The experiment starts with a, i.e. x = av, v ∈ Σ?.

In this case RM(1, x) = RM(3, x), that is x cannot distinguish between the states1, 3 as

RM(1, x) = RM(1, av)

= fM(1)fM(δM(1, a))RM(δM(1, a), v)

= fM(1)fM(4)RM(4, v)

= 01RM(4, v)

and

RM(3, x) = RM(3, av)

= fM(3)fM(δM(3, a))RM(δM(3, a), v)

= fM(3)fM(4)RM(4, v)

= 01RM(4, v). 2

Moore’s Theorem can be thought of as being a discrete analogue of the Heisen-berg uncertainty principle. Here is the analogy. The state of an electron E isconsidered specified if both its velocity and its position are known. Experimentscan be performed with the aim of answering either of the following:4

1. What was the position of E at the beginning of the experiment?

2. What was the velocity of E at the beginning of the experiment?

For Moore’s automaton, experiments can be performed with the aim of answer-ing either of the following:

1. Was the automaton in state 1 at the beginning of the experiment?

2. Was the automaton in state 2 (or 3) at the beginning of the experiment?

4The propositional system obtained from the Moore automaton is the modular lattice L12

(see Svozil [69, pp. 141-147]). An exact quantum mechanical analogue has been given by Foulisand Randall [30, Example III]: Consider a device which, from time to time, emits a particle andprojects it along a linear scale. We perform two experiments. In experiment A, the observerdetermines if there is a particle present. If there is not, the observer records the outcome of A asthe outcome 4. If there is, the observer measures its position coordinate x. If x ≥ 1, the observerrecords the outcome 2, otherwise 3. A similar procedure applies for experiment B: If there isno particle, the observer records the outcome of B as 4. If there is, the observer measures thex-component px of the particle’s momentum. If px ≥ 1, the observer records the outcome 1, 2,otherwise the outcome 1, 3. Still another quantum mechanical analogue has been proposed byGiuntini [37, pp. 159-162]. A pseudo-classical analogue has been proposed by Cohen [21] and byWright [78].


In either case, performing the experiment to answer question 1 changes the stateof the system, so that the answer to question 2 cannot be obtained. This meansthat it is only possible to gain partial information about the previous history of thesystem, since performing experiments causes the system to “forget” about its past.

Moore’s automaton is a simple model featuring an “uncertainty principle” (cf.Conway [22, p. 21]), later termed “computational complementarity” by Finkelsteinand Finkelstein [29]. These type of models have been intensively studied from thepoint of view of their experimental logical structure by Grib and Zapatrin [41, 42], aswell as by Svozil [69], Schaller and Svozil [66, 67, 68] and Dvurecenskij, Pulmannovaand Svozil [24]. See Svozil and Zapatrin [71] for a comparison of models.

5.2 Moore’s Theorem Revisited

In this section we prove a slightly stronger version of Moore’s Theorem. Thisresult will motivate the introduction, in the next section, of two new computationalcomplementarity principles.

Theorem 5.2 Moore’s automaton satisfies the following property: There exists astate q ∈ SM such that for every w ∈ Σ+ there exists a state p 6= q (depending uponw) such that RM(q, w) = RM(p, w).

Proof. Take q = 1 and notice that in case w = au,we have δM(1, a) = δM(3, a) =4 and fM(1) = fM(3) = 0, so we can take p = 3, and in case w = bv, thenδM(1, b) = δM(2, b) = 3 and fM(1) = fM(2) = 0, so we can take p = 2. 2

5.3 Two Complementarity Principles

Motivated by Theorem 5.2 we will introduce two new non-equivalent principles ofcomputational complementarity. Here is the basic idea. Consider the class of allelements of reality5 and consider the following properties:

A Any two distinct elements of reality can be mutually distinguished by a suit-ably chosen measurement procedure.

B For any element of reality, there exists a measurement–depending on the cho-sen element–which distinguishes between this element and all the others. Thatis, a distinction between any one of them and all the others is operational.

C There exists a universal measurement which distinguishes between any twoelements of reality. That is, a single pre-defined experiment operationallyexists to distinguish between an arbitrary pair of elements of reality. (Classicalcase.)

5The terms “elements of reality”, “properties”, and “observables” will be used as synonyms.

5.3 Two Complementarity Principles 29

We define first the notion of “distinguishability”. Assume that we want to“distinguish” between two states p and q of the deterministic incomplete automatonA by means of a “measurable experiment”, i.e. by the responses of the automatonto an input w ∈ Σ?. There are two basic possibilities:

(i) The experiment is not relevant in case ∆A(p, w) = ∆A(q, w) = ∞ (soRA(p, w) = RA(q, w) =∞); hence, another experiment is required.

(ii) The experiment is relevant in case w is applicable to at least one state p or q;in this case we have two further possibilities:

(a) w distinguishes between p and q in case w is applicable to both of themand RA(p, w) 6= RA(q, w), or w is applicable to either p or q, but not toboth.

(b) w does not distinguish between p and q in case w is applicable to bothof them and RA(p, w) = RA(q, w).

To summarize, w distinguishes between p and q if RA(p, w) 6= RA(q, w) (meaningthat either w is applicable to both p and q and the responses are different or w isapplicable to only one of the states). In the remaining cases, w may not distinguishor may not be relevant for distinguishing between p and q.

The above facts motivate the introduction of the following notions. LetA = (SA,∆A, FA) be an incomplete automaton. The states p, q ∈ SA are indis-tinguishable if and only if RA(p, w) = RA(q, w), for all w ∈ WA(p)6.

If the states p and q are not indistinguishable, we say that they are distinguish-able, and every word from the set

w ∈ WA(p) ∪WA(q) | RA(p, w) 6= RA(q, w) (5.1)

is said to distinguish between p and q. In the same way, a word w cannot distinguishbetween p and q if RA(p, w) = RA(q, w) or w 6∈ WA(p) ∪WA(q).

A natural questions arises: Does there exist automata having property C? Theanswer is affirmative and the automaton given in Figure 5.1 is an example of suchan automaton:7

6a) Note that WA(p) = WA(q). b) In case A is complete, p and q are indistinguishable iffthey are RA-equivalent. Hence, because of Lemma 6.11, we can define indistinguishability by finalresponse fA instead of total response RA.

7Note that the complete automaton in Figure 5.1 is strongly connected, i.e. every two statesare linked by some computation.


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Figure 5.1

The experiment ba distinguishes between any pair of distinct states.8

Complementarity corresponds to the following cases:

CI Property A but not property B (and therefore not C): The elements of real-ity can be mutually distinguished by experiments, but one of these elementscannot be distinguished from all the other ones by any single experiment.

CII Property B but not property C: Any element of reality can be distinguishedfrom all the other ones by a single experiment, but there does not exist a uni-versal experiment which distinguishes between any pair of distinct elements.

A deterministic incomplete automaton A satisfies the first principle of comple-mentarity if

1) for every distinct states p and q there exists an experiment wp,q (dependingupon p and q) such that RA(p, wp,q) 6= RA(q, wp,q), and

2) there exists a state r such that for every experiment w we can find a state qw(depending upon w) such that RA(p, w) = RA(qw, w).

The principle CI mimics Heisenberg’s uncertainty: indeed, any experiment willfail to distinguish between the exceptional state r and some distinct state. Anysuccessful experiment in determining that the automaton A was initially in stater fails to show that some other state, distinct from r is not the initial state of A(even, this state is probable not to be the initial state).

Moore’s automaton has been shown to satisfy the first principle of complemen-tarity (see Theorem 5.2).

A deterministic incomplete automaton A satisfies the second principle of com-plementarity if

8An automaton having property B but not C requiring a longer experiment is presented inFigure 5.12.


1) for every state q there exists an experiment wq (depending upon q) such thatfor every state q′ different from q, we have RA(q, wq) 6= RA(q′, wq),

9 and

2) for every experiment w there exists at least two distinct states q, q′ (dependingupon w) such that RA(q, w) = RA(q′, w).

In view of 2), each experiment “generates” a pair of distinct states which exercisea mutual influence, namely they cannot be separated by the experiment w; thisinfluence mimics, in a sense, the state of quantum entanglement.10 In a sense, CIImay be conceived as a toy model for the EPR effect (see Einstein, Podolsky, Rosen[25], Penrose [60, 61]), as well as for the Zou-Wang-Mandel effect (see Greenberger,Horne, Zeilinger [38], Wang, Zou, Mandel [79, 74]. Einstein, Podolsky and Rosenshowed how quantum physics requires that a property, such the polarization of aphoton, could be measured at a distance by simply measuring the polarization ofa second photon that had interacted with the first some time previously. Now,if we reject the idea that this measurement could have interfered with the distantobject, then we have to admit that the first photon must have possessed the measuredproperty before the measurement was actually carried on. In view of the fact that theexperimenter can adjust the distant apparatus it follows that the property measuredcan be varied. This is the reason why EPR concluded that all physical propertiesmust be “real” before they are measured, in open contrast with the Copenhageninterpretation of quantum physics.11 There is a huge literature discussing thesecontroversial matters; Bell [3], Feynman [27], Greenstein and Zajonc [40], Rae [64]are some particularly good references.

The natural reaction towards such bizarre ideas as uncertainty and the role ofthe experimenter in influencing the outcome of an experiment is to dismiss them (atleast to some extent), to assume that they are just “illusions”. Common sense willtell us that behind all these sophisticated scenarios there must be a clockwork ofNewtonian determinism which sits in the world of certainty and objective reality andcreates the appearance of uncertainty and subjectivity to our imperfect senses andinadequate measuring apparatus. This possibility was ruled out by the experimentscarried out in 1980s by a group headed by Alain Aspect working along the linessuggested twenty years before by Bell. Their conclusion is strong: there is nounderlying clockwork, the spooky action at a distance, that Einstein hated so much,is correct. We claim that CII reflects to some extent this state of affairs. Indeedunder CII , for each experiment w we have at least two states q, q′ (as distant as welike in terms of the emitting outputs f(q), f(q′)) which interact via the experimentw: any measurement of q is affecting q′ and, conversely, any measurement of q′ isaffecting q. It is interesting to note that this explanation supports Penrose’s view([61, p. 237]) that EPR effects are puzzle mysteries, that is genuinely puzzling,but directly experimentally supported. Greenberger, Horne and Zeilinger call similar

9Note that RA(q, ·) is a partial function having WA(q) as domain.10In particular, this influence cannot be used to send an actual message from a state to the

other.11It is reported, that Bohr replied “Don’t tell God what to do!” to Einstein’s famous words

“God does not play dice”.


experiences simply mindboggling Giuntini [38]. In fact, a second “reading” of thesephenomena could prove that their puzzling nature might “not be so puzzling”, afterall. Actually, we believe that a bit of the mystery associated with EPR effects, ingeneral, and with CII , in particular, might not be so “puzzling” at all, and hereare two arguments.

The basic building block of a quantum computer is a gate that has two inter-acting “qubits” 12: a control bit and a target bit. The control remains unchangedand its current state determines the target: in case the control is 0, then nothingis changed in the target, but if the control is 1, then the target undergoes a certaintransformation. In case the control is in some superposition of 0 and 1, the gate is“entangled”, i.e., the two qubits are correlated in a nonseparable state, a situationsimilar with EPR effects modeled by CII. The superposition and entranglementmake a quantum computer so different from a classical one.

A random (binary) sequence in Chaitin-Martin-Lof sense (see Chaitin [19, 18],Calude [11]) is the prototype of the ideal chaotic sequence, in which no predictionis possible to come true and no computation can be successful in approximatingmore than a finite part of the sequence. However, such a random sequence satisfiessome very interesting, deterministic laws. Here are two examples:

• a constructive property: each random sequence is Borel normal, in the sensethat every word of length n over the binary alphabet occurs in the sequencewith the exact expected probability, i.e. 2−n (see Calude [11]);

• a non-constructive property: in each random sequence at least one of the twosymbols 0 and 1 must occur in arithmetical progressions of every length (seevan der Waerden [73]).

With reference to normality, in a random sequence each bit has, in the long run,a “mysterious”, purely deterministic, influence on all digits. This “effect” can betested (of course, only on finite initial segments of the sequence); it can be proven“from outside”, i.e. at the level of the meta-language, but it is “un-reachable” forany observer from “inside”.

A language-theoretic version of the EPR effect is related to the so-called depthhypothesis. Psychologists have measured the “span of immediate memory”, i.e.the ability to memorize at a glance and repeat correctly random digits, nonsensewords, various items. It seems that the average ability is about seven (cf. Miller[55]). The depth hypothesis suggests that much of the syntactic complexity of anatural language can be understood in terms of this memory restriction.13 Froma mathematical point of view this restriction can be modeled by the property ofprojectivity (cf. Marcus [52, chapter 6]) which, in a sense, measures the “long run”syntactic subordination of words in natural languages. Again, the depth hypothesis,and more generally, any syntactic subordination, is “visible” from “outside” and notfrom “inside”.

12A two-level system can be presented as a quantum bit, or, qubit.13The syntax of English, for instance, has many devices for keeping utterances within the bounds

of this restriction; it has also resources to circumvent it, as to regain the loss of expressive power.


Coming back to our formal model, let us notice that property C implies propertyB, which, in turn, implies propertyA; none of the converse implications is true. Forthe first case see Theorem 5.2; the automaton presented in Figure 5.2 satisfies thesecond principle of complementarity.

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Indeed, the following pairs of states are distinguishable by every experiment:(1, 2), (1, 4), (2, 3), (3, 4). Accordingly, 1 is distinguishable from the other statesby w = a, 2 is distinguishable by w = b, 3 is distinguishable by w = a and 4is distinguishable by w = b, so the automaton has property B. It does not haveproperty C because:

• any experiment w which starts with b, i.e. w = bx, x ∈ Σ?, does not distin-guish between 1 and 3;

• any experiment w which starts with a, i.e. w = ay, y ∈ Σ?, does not distin-guish between 2 and 4.

Here are some more examples of incomplete automata satisfying CI and CII .The automaton, given in Figure 5.3, has property A since any two states are dis-tinguishable. For instance, the word w = b distinguishes between q1 and q2 sincew ∈ WA(q1) \WA(q2). But it does not have property B. If we choose w = ax orw = cx, x ∈ Σ?, then the states q1 and q2 are indistinguishable and if we choosew = bx then q1 and q4 are indistinguishable. Hence there is no experiment whichdistinguishes the state q1 from the other states.

On the other hand, the automaton, presented in Figure 5.4, has property B sinceby choosing the words a, b, c, and a for the states q1, q2, q3, and q4, respectively,each of them can be distinguished from the other states. There is, however, nosingle experiment which can distinguish between any two states. For w = ax,where x ∈ Σ?, the states q2 and q4, for w = bx the states q1 and q4, and finally forw = cx, the states q1 and q2, respectively, are indistinguishable.


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Figure 5.3

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Proposition 5.3 Complementarity properties CI, CII cannot appear for completeautomata with less than four states.14

Proof. Indeed, for the complete automaton A = (SA, δA, fA) with |S| ≤ 2 theabove statement is clear. So, let SA = 1, 2, 3. We only need to consider twocases:15

1. if fA(1) = fA(2) = fA(3), then no pair of states is distinguishable;

14This result was noticed by Conway [22, pp. 20-23], for property A.15It is worth noticing that this is true regardless the size of the alphabet. Indeed, assuming that

Σ has more than two symbols, the following analysis should be completed with one more case: iffA(1) 6= fA(2), fA(2) 6= fA(3), and fA(1) 6= fA(3), then A has property C.


2. if fA(1) = fA(2) and fA(1) 6= fA(3), and the states 1, 2 are distinguishable,then there exists an experiment w ∈ Σ+ such that E(1, w) 6= E(2, w). SincefA(1) = fA(2) 6= fA(3), RA(1, w) 6= RA(3, w) and RA(2, w) 6= RA(3, w).

Accordingly, the automaton A has property A iff it has property C.

2

For incomplete automata, complementary properties CI and CII can be foundalready for three-state automata.

Proposition 5.4 Properties CI and CII hold true for three-state incomplete au-tomata.

Proof. Consider the automaton in Figure 5.5.

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Figure 5.5

It has property A since, for instance, w = b distinguishes between p and r(RA(p, w) = 00 6= ∞ = RA(r, w)). But it does not have property B since thestate p does not have any single experiment which distinguishes it from the otherstates. Indeed, for w = ax, the states p and q, and for w = bx, the states p and r,respectively, are indistinguishable. Thus it has CI.

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Figure 5.6


In the same way, the automaton given in Figure 5.6, has CII, since now thewords b, a, and c can distinguish the states p, q, and r, respectively, from the otherstates, but there is no single experiment which can distinguish between any twogiven states. For w = ax the states r and q, for w = bx the states p and q, andfinally for w = cx the states p and r, respectively, are indistinguishable. We alsonotice that these properties exist even though the output alphabet has only oneletter. 2

5.4 More About Moore’s Automaton

Moore’s automaton can be used with different output functions to produce both CIand CII . The complement of the original output function leads to CI . To get CIIwe can use each of the output functions in Figures 5.7 and 5.8. Let us prove CIIfor the first choice of fM . Moore’s automaton has property B as for 1 we can usew = ab, for 2 we can use w = aab, for 4 we can use w = b (any w is good for 3).

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Figure 5.8

5.5 Decidability and Complexity 37

The automaton has not property C as:

• (1, 2) cannot be distinguished by any experiment w = by, y ∈ Σ?;

• if w = an, for some n ≥ 1, then 1 and 2 cannot be distinguished: RM(1, an) =0n+1 = RM(2, an), δM(1, a) = 4, δM(1, aa) = 2, δM(1, aaa) = 1, δM(2, a) =1, δM(2, aa) = 4, δM(2, aaa) = 2;

• if w = a3nby, for some y ∈ Σ?, n ≥ 1, then 1 and 2 cannot be distin-guished: RM(1, a3nb) = 03n+2 = RM(2, a3nb) and δM(1, a3nby) = δM(3, by) =δM(3, y) = δM(2, by) = δM(2, a3nby);

• if w = a3naby, for some y ∈ Σ?, n ≥ 1, then 2 and 4 cannot be distinguished:RM(2, a3nab) = 03n+3 = RM(4, a3nab) and δM(2, a3naby) = δM(2, aby) =δM(3, y) = δM(4, aby) = δM(4, a3naby);

• if w = a3naaby, for some y ∈ Σ?, n ≥ 1, then 1 and 4 cannot be distinguished:RM(1, a3naab) = 03n+4 = RM(1, a3naab) and δM(1, a3naaby) = δM(1, aaby) =δM(3, y) = δM(4, aaby) = δM(4, a3naaby).

5.5 Decidability and Complexity

Some natural mathematical questions arise: a) Is each of the properties A, B, Calgorithmically decidable?, b) How difficult is to test these properties?

From the work of Conway [22, chapter 2], it follows directly that for deterministiccomplete automata the problem of testing whether two states are or not distinguish-able is algorithmically decidable. This means that property A is algorithmicallydecidable. We will place ourselves in the case of incomplete automata. An incom-plete automaton A = (SA,∆A, FA) has property A if for each pair of its distinctstates p and q, there is a word w ∈ WA(p)∪WA(q) for which RA(p, w) 6= RA(q, w).In other words the distinguishing sets

Rp,q = w ∈ WA(p) ∪WA(q) | RA(p, w) 6= RA(q, w)

are nonempty for all p, q ∈ SA, p 6= q.Consider, for each state q ∈ SA, the finite (complete) deterministic automaton

with initial state q

Mq = (S ′,Σ′ ×O, q, δ,Λ),

where S ′ = SA ∪ # ∪ Λ and Σ′ = Σ∪ λ, q is the initial and Λ the final state.The transition function δ is defined, for p ∈ SA, σ ∈ Σ, τ ∈ O, by the equations:

δ(p, (σ, τ)) =

∆A(p, σ) if FA(p) = τ , σ 6= λ, and ∆A(p, σ) 6=∞,Λ if FA(p) = τ , and σ = λ,# otherwise,


and, for p ∈ #,Λ, σ ∈ Σ′, τ ∈ O,

δ(p, (σ, τ)) = #.

It follows immediately that for p ∈ SA and (σ1, τ1) · · · (σn, τn) ∈ (Σ′ ×O)∗,δ(p, (σ1, τ1) · · · (σn, τn)) = Λ iff

σ1, . . . , σn−1 ∈ Σ, σn = λ, and RA(p, σ1 . . . σn−1) = τ1 . . . τn,

and, hence, the language accepted by Mq is

L(Mq) = (wλ,RA(q, w)) | w ∈ WA(q).

We have obtained:

Theorem 5.5 Properties A, B and C are algorithmically decidable.

Proof. Indeed, we have:

A has property A iff Rp,q 6= ∅, for all p, q ∈ SA, p 6= q,

A has property B iff for all q ∈ SA,⋂p6=q Rp,q 6= ∅,

A has property C iff⋂q∈SA

⋂p6=q Rp,q 6= ∅,

and, for any states p and q,

Rp,q 6= ∅ iff (L(Mp) ∪ L(Mq)) \ (L(Mp) ∩ L(Mq)) 6= ∅. 2

The automaton A in Figure 5.9 satisfies CI.

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0 0 0 0q a b a

c b

s r p

Figure 5.9

Indeed, since all states are distinguishable but for the states p and s there is nosingle experiment which distinguishes them from all the other states. The applicablewords for each state are the following:

W (q) = λ, ac∗b∗, W (r) = c∗b∗, W (s) = b∗, W (p) = λ, ab∗,

and the distinguishing sets are:


Rq,r = ac∗b∗, c+b∗, c∗b+, Rq,s = ac∗b∗, b+, Rq,p = ac+b∗,Rr,p = ab∗, c+b∗, c∗b+, Rr,s = c+b∗, Rs,p = b+, ab∗.

Hence A cannot have property B since

Rq,p ∩Rr,p ∩Rs,p = ∅ = Rq,s ∩Rr,s ∩Rs,p,

but it has property A since every distinguishing set is nonempty.

We continue by showing that the decidability of the property A in the incompletecase can be deduced also from the known result of the complete case (see Conway[22] and Calude, Calude and Khoussainov [12]). Let A = (SA,∆A, FA) be anincomplete automaton. We will turn it complete by adding an extra state, denotedby α, which takes care of all undefined transitions of A. Formally, define Aα =(SAα ,∆Aα , FAα) such that SAα = SA ∪ α. The transition function ∆Aα is thesame as ∆A, except that ∆Aα(p, σ) = α if ∆A(p, σ) = ∞, and ∆Aα(α, σ) = α, forall σ ∈ Σ. In order to distinguish α from the other states we have to make surethat it produces a different output, FAα(α) = x where x 6∈ O.

Lemma 5.6 An incomplete automaton A has property A iff its completion automa-ton Aα has property A.

Proof. Assume first that A has property A, i.e., every pair of its distinct statesare distinguishable. Since the new state α produces a different output, it can bedistinguished from the other states by the empty word. On the other hand, for anypair of distinct states of A there is a word which is either applicable to both of thesestates or to one of them. In both of these cases w can distinguish the states also inAα.

Assume now that A does not have property A and let p and q be indistinguish-able, i.e., RA(p, w) = RA(q, w) for all w ∈ WA(p) = WA(q). Take any word w ∈ Σ?.We can write it in such a way that w = uσv where u ∈ WA(p) (u may also beempty) and uσ 6∈ WA(p). But this means that ∆Aα(p, uσ) = α = ∆Aα(q, uσ) and,hence, RAα(p, w) = RAα(q, w). So p and q must be indistinguishable also in Aα. 2

The following example shows that the completion automaton Aα does not pre-serve properties B and C. This is due to the fact that the states p and q can bedistinguished in A by the word w which is not applicable to both of these states inA but will become applicable in Aα.

Example 5.7 The completion map A 7→ Aα does not preserve properties B andC.

The automaton A in Figure 5.10 has property CI since there is no word whichdistinguishes p or s from the other states. However, the new automaton Aα inFigure 5.11 has property B since the word w = ac distinguishes p and the wordw = c distinguishes s, from the other states, respectively.


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An interesting problem is to evaluate how difficult it is to test properties A, B,C. A way to look at this problem is to evaluate the shortest length of experiments


needed to decide properties A, B, C. This problem has been studied for testingproperty A for complete deterministic automata by several authors, for instanceChaitin [17], Conway [22]. The main result (for a binary alphabet) can be stated asfollows: To test property A it is sufficient to test the condition RA(q, w) 6= RA(q′, w)for all words of length less than |(SA)| − 2. In fact, it is trivial to notice that weneed only to test the above condition for words of length equal to |(SA)| − 2.

This result is no longer true for properties B and C. Indeed, babababab isthe shortest word that distinguishes every pair of states in the automaton A =(1, 2, 3, 4, 5, 6, 7, δ, f) displayed in Figure 5.12.

Let us notice that in general, the shortest word that can distinguish every pairof states in the automaton (1, 2, . . . , n, δ, f) where δA(i, a) = i + 1, δA(i, b) = i,for 1 ≤ i ≤ n − 2, δA(n − 1, a) = n − 1, δA(n − 1, b) = δA(n, a) = δA(n, b) = n,fA(i) = 0, 1 ≤ i ≤ n− 1, fA(n) = 1) has length 2n− 5.

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Figure 5.12

Here is the argument. Each w ∈ a+ab+ab+ab+ab(a + b)∗ can distinguish everypair of states in SA. The shortest word in that set is w = babababab. No shorterexperiment can replace w:16

• every word w ∈ a(a+ b)∗ cannot distinguish between states 5, 6,

16We next use the classical notation for regular expressions (cf. Hopcroft and Ullman [43, pp.28-29].)


• every word w ∈ a+aa(a+ b)∗ cannot distinguish between states 4, 5,

• every word w ∈ a+ab+aa(a+ b)∗ cannot distinguish between states 3, 4,

• every word w ∈ b+ab+ab+aa(a+ b)∗ cannot distinguish between states 2, 3,

• every word w ∈ b+ab+ab+ab+aa(a+b)∗ cannot distinguish between states 1, 2.

Using Proposition 5.6 and the next result due to Conway[22], we derive the bestbound for testing property A for an incomplete automaton.

Theorem 5.8 (Conway) Let A = (SA, δA, fA) be a complete automaton in a bi-nary alphabet, |Σ| = 2. To test property A for A it is sufficient to test the condition

RA(q, w) 6= RA(p, w) (5.2)

for all words of length |SA| − 2.

Theorem 5.9 The length of the shortest experiment needed to decide property Afor an incomplete automaton A = (SA, δA, fA) in a binary alphabet is |SA| − 1.

Proof. Consider the complete automaton Aα = (SAα , δAα , fAα). By Theorem5.8 it is sufficient to test condition (5.2) for all words of length |SAα| − 2 in order todecide whether Aα has property A. On the other hand, by Proposition 5.6, this isequivalent for testing whether A has property A. Since |SAα| = |SA|+ 1, this givesus the bound |SA| − 1 for an automaton A.

Notice that the bound |SA| − 1 cannot be improved. For the automaton A inFigure 5.5 the shortest experiment distinguishing between the states p and q inautomaton A is w = ab. 2

Chapter 6

Minimality and Universality

In this chapter we state and solve a problem of minimization for automata withoutputs but without initial states and we relate the minimal automaton the theuniversal automaton. In opposition with current solutions for the minimizationproblem, in our model the minimal automaton is unique up to an isomorphism.

6.1 Minimality

To construct the minimal automaton “equivalent” to a given automaton A we haveto give a criterion for identifying those states of A which give the same response tothe same inputs. It is not difficult to see that any acceptable equivalence relationshould be “compatible” with all computations performed by A, i.e., it should be“well–behaved” with respect to the transition table ∇A. Here is the appropriatedefinition. A relation ≡ on SA is well–behaved if for all p ≡ q (p, q ∈ SA) and forevery σ ∈ Σ the following properties hold:

1. For every p′ ∈ ∇A(p, σ) there is a state q′ ∈ ∇A(q, σ) such that p′ ≡ q′.

2. For every q′ ∈ ∇A(q, σ) there is a state p′ ∈ ∇A(p, σ) such that q′ ≡ p′.

A well-behaved equivalence relation ≡ should guarantee that any two ≡ equiva-lent states simulate each other. Having a well–behaved equivalence relation ≡, onecan consider the factor automaton A/≡ and prove that it is minimal.

The first “natural” equivalence relation that comes to mind is the following: twostates p, q ∈ SA are ≡1 equivalent if for all w ∈ Σ? we have RA(p, w) = RA(q, w).Unfortunately, the next example shows that ≡1 is not well-behaved.1

Example 6.1 The automaton A in Figure 6.1 has the following property: thereexist two states p, q ∈ SA such that p ≡1 q, but for all p′ ∈ ∇A(p, σ) and q′ ∈∇A(q, σ), we have p′ 6≡1 q

′.

1Some other unsuccessful attempts to define the right well-behaved equivalence relation werereported in [14].

43

44 6. Minimality and Universality

Indeed, it is not hard to see that for all w ∈ Σ?, RA(p, w) = RA(q, w). It followsthat p ≡1 q. However, no p′ ∈ ∇A(p, a) is ≡1 equivalent to any q′ ∈ ∇A(q, a).

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a

Figure 6.1.

Fortunately, the equivalence relation ≡ defined in Chapter 3 is well-behaved.Recall that p ≡ q if Player 1 wins every game Γ(p, q, n), for all n > 0.

Theorem 6.2 The equivalence relation ≡ is well-behaved.

Proof. In view of Lemma 4.9, ≡ is an equivalence relation. Suppose thatp ≡ q. We need to show that for every σ ∈ Σ and every p′ ∈ ∇A(p, σ) there is aq′ ∈ ∇A(q, σ) such that p′ ≡ q′. Let q1, . . . , qs be all states belonging to ∇A(q, σ).Suppose that none of qi is ≡ equivalent to p′. Then for every qi there is an nisuch that Player 0 wins the game Γ(p′, qi, ni). Let hi be a strategy for Player 0 towin the game Γ(p′, qi, ni). Then Player 0 wins also any continuation of the game,Γ(p′, q′, ni + t), for every natural number t. Let n be the maximal number amongall n1,. . ., ns and consider the game Γ(p, q, n). Suppose that in this game the firstmove of Player 0 is p′. If Player 1 responses by not taking a state from q1, . . . , qs,then clearly Player 1 looses the game. On the other hand, if Player 1 chooses astate qi, then Player 0 simply follows the strategy hi. It is clear that in this casePlayer 0 wins the game Γ(p, q, n) which contradicts the fact that p ≡ q. 2

Let A be a nondeterministic automaton. We define the automaton M(A) asfollows:

1. The set of states SM(A) ofM(A) is [s] | s ∈ SA, where [s] = q ∈ SA | s ≡ q.

2. For all [q], [s] ∈ SM(A) and σ ∈ Σ, [q] ∈ ∇M(A)([s], σ) if and only if q ∈∇A(s, σ).

6.1 Minimality 45

3. FM(A)([s]) = FA(s).

The next lemma, concerning the relationship between A and M(A), is an exact ana-logue of the case for deterministic automata (see Calude, Calude and Khoussainov[12]).

Lemma 6.3 The automata A and M(A) simulate each other.

Proof. We prove that automaton A is simulated by M(A) via the mappings 7→ [s], for all s ∈ SA. We need to show that Player 1 has a strategy to win thegame Γ(s, [s], n), for each n. Suppose that at the end of stage k (k < n) of a playthe players have produced two sequences

s0s1 . . . sk

and[p0][p1] . . . [pk]

where s0 = s and [p0] = [s]. By induction, we can assume that pk ≡ sk. Supposethat at stage k + 1 Player 0 chooses a sk+1 from

⋃σ∈Σ∇A(sk, σ). Since sk ≡ pk, by

Theorem 6.2, there exists a pk+1 ∈⋃σ∈Σ∇A(pk, σ) such that sk+1 ≡ pk+1. Hence

Player 1 picks up this pk+1. Suppose that at stage k + 1 Player 0 chooses a [pk+1]from

⋃σ∈Σ∇M(A)([pk], σ). Again by Theorem 6.2 Player 1 can choose a sk+1 such

that sk+1 ≡ [pk+1].Similarly, one can prove that the automaton M(A) is simulated by A via the

mapping [s] 7→ min[s], where min[s] is the minimal element in [s] under some fixedlinear ordering in SA. 2

Say that two automata A and B are equivalent (and denote this by A ∼ B)if A ≤ B and B ≤ A. Clearly, the relation ∼ is an equivalence relation. Anondeterministic automaton A is minimal if for every nondeterministic automatonB such that A ∼ B one has |SA| ≤ |SB|.

Our goal is to prove that each class [A] = B| A ∼ B contains a minimalautomaton which is unique up to an isomorphism. We recall that two automataA and B are isomorphic if there is a bijective mapping h : SA → SB such thatfor all s, p ∈ SA, σ ∈ Σ, p ∈ ∇A(s, σ) if and only if h(p) ∈ ∇B(h(s), σ) andFA(s) = FB(h(s)).

Lemma 6.4 The automaton M(A) is minimal.

Proof. Suppose that B is minimal. Let h : SM(A) → SB be a mapping suchthat M(A) is simulated by B via h. Then h is one–to–one. Otherwise, there existtwo states [p] 6= [q] in SM(A) such that h([p]) = h([q]). Hence p ≡ h(p), h(p) = h(q),and h(q) ≡ q. It follows that [p] ≡ [q], and consequently, p ≡ q, i.e., [p] = [q]. Thisis a contradiction. Thus, |SM(A)| ≤ |SB|. 2


In the last step we show the unicity (up to an isomorphism) of the minimalautomaton.

Lemma 6.5 If B is minimal and A ∼ B, then B is isomorphic to M(A).

Proof. Suppose that B is minimal. There exists a mapping h : SM(A) → SBsuch that M(A) is simulated by B via h. From the proof of Lemma 6.4 we see thath must be a one–to–one mapping. Since the automaton B is minimal, h must beonto. Indeed, assume by contradiction that there is a mapping g : SB → SM(A)

such that B is simulated by M(A) via g and g(p) = g(q), for some p, q ∈ SB. Hencep ≡ q. Since M(B) ∼ B, B cannot be minimal, a contradiction. Consequently, his a bijection from SM(A) to SB.

We need to prove that h is an isomorphism. It is clear that FM(A)([s]) =FB(h([s])), for all s ∈ SA. Suppose that [s] ∈ ∇M(A)([p], σ). We need to showthat h[s] ∈ ∇B(h([p]), σ). Since [p] ≡ h([p]), there exists a q ∈ ∇B(h([p]), σ) suchthat q ≡ [s]. Hence q ≡ h([s]) since h establishes a simulation. If q 6= h([s]), thensince q ≡ h([s]), we have |SM(B)| < |SB|. This is again a contradiction with theassumption that B is minimal. Hence q = h([s]) and h([s]) ∈ ∇B(h([p]), σ). 2

The above lemmas prove the main theorem of this chapter:

Theorem 6.6 For every nondeterministic automaton A, the automaton M(A) sat-isfies the following properties:

1) The automata A and M(A) simulate each other.

2) The automaton M(A) is minimal.

3) The automaton M(A) is unique up to isomorphism.

6.2 Other Models of Minimality

In what follows we will compare the classical way of defining and constructingthe minimal incomplete automaton and the method presented in Section 6.1. Theproblem of determining and deleting “superfluous” states of a automaton with-out changing it, has been unraveled by many authors such as Ginsburg [36] andMikolajczak [54]. Solutions to this problem depend on the precise definition for“superfluous” states and “without changing it”, in other words they depend on therequirements we put on minimal realization. So, when an incomplete automatonM can be considered to be a “minimal representation” for a given incomplete au-tomaton A? The minimality condition is straightforward: M must have a minimalnumber of states. The condition of “representing” the initial automaton can havevarious interpretations:

6.2 Other Models of Minimality 47

• One way is to consider that all outputs produced by A can be produced byM on identical input words; in this interpretation it does not matter whetherM can produce outputs which cannot be obtained from A.

• Another way is to consider that automata A and M must generate exactlythe same output sequences in response to identical input sequences; in thiscase whatever output M can produce A must be able to produce and viceversa.

The classical way to solve the minimization problem for incomplete automata followsthe first approach. The solution consists in constructing a family of automata withminimal number of states, which represent a given automaton. We will consider anexample presented by Mikolajczak ([54], p.11). A complete Mealy automaton is a6-tuple (Q,Σ,Ω, δ, µ, q0) where

• Q is a finite set of states,

• Σ is a finite input alphabet,

• δ : Q× Σ→Q is the transition mapping,

• q0 is an initial sate (from Q)

• Ω is the output alphabet and

• µ : Q×Σ→Ω is the output mapping, that is µ(q, a) gives the output associatedwith the transition from state q on input a.

For the Mealy automaton A presented in Figure 6.2 two minimal Mealy covers 2

Amin1 and Amin2 are given in Figures 6.3 and 6.4. We notice two facts:

• The automaton A is not reversible (hence not strongly connected), but bothits minimal covers are reversible (even strongly connected). This is not thecase for the minimal automaton obtained using the method given in Section6.1.3

• There exist input words for which the output given by any of the minimalcovers cannot be obtained from the initial automaton A. For example, forw0 = bab, Amin1 produces 000 on p1 and Amin2 produces 001 on q1. A doesnot produce 000 and 001 on w0 for any state: indeed on w0 the automaton Aemits 100 on s3 and is undefined for s1 and s2.

2A detailed presentation involving covers can be found in Mikolajczak [54], Ginsburg [44].3See Proposition 7.4.


%$

.

JJJJJJJJ]JJJJJJJJ

s1 s2

a |0

0

W'

&

s3

ba 0|

a|

|b|–

0

b|1|

Figure 6.2

-

a

p1

Amin1

b|

?'

& 0

,p2

b|1|a 0

0|

Figure 6.3

O %

$

-

a|0

q1

'

& bq2

|0

Amin2

b|1a |0

Figure 6.4

Using the method presented in Section 6.1, one can see that the incompleteMealy automaton presented in Figure 6.2 is minimal.

To conclude: the classical minimal model does not preserve the whole informa-tion encoded in the initial automaton though it usually has less states and is clearlyoptimal in an weaker sense.

6.3 Universality 49

6.3 Universality

Consider the following class of incomplete automata with initial states: C =(A1, p1), (A1, p2), (A2, p3), (A3, p4), where A1, A2 and A3 are given in Figure6.5.

M !

QQQQQs

3

-1|a, ba

p | |3 1 1 p

M !

4

0

|a, b

1

a

a, b

a, br s

2 A :3A :

A :1

1p |0

p2|

Figure 6.5

-

a, b

q |M !

2a

1q |01

Figure 6.6

The incomplete automaton U from Figure 6.6 has the following properties:

1. Each incomplete automaton Ai is strongly β–simulated by U with a suitablychosen initial state from SU :

• WA1(p1) = WU(q1) and RA1(p1, w) = RU(q1, w), for all w ∈ WA1(p1),



• WA3(p4) = WU(q2) and RA3(p4, w) = RU(q2, w), for all w ∈ WA3(p4).

2. The automaton U , starting from any of its states, is strongly β–simulated bysome automaton (Ai, pj) from C:

• RU(q1, w) = RA1(p1, w), for all w ∈ WU(q1),

• RU(q2, w) = RA3(p4, w), for all w ∈ WU(q2).

We say that U is a “universal incomplete automaton” for the class C. Our aimis to define the notion of universal incomplete automaton and to prove that everyfinite class C can be embedded into a class having a universal incomplete automatonwhich is unique up to an isomorphism; finally, we show that universality is relatedto minimality.


Suppose that we have a finite class C containing pairs (Ai, qi) of incompleteautomata Ai = (Si,∆i, Fi) and initial states qi ∈ Si, i = 1, . . . , n. An incompleteautomaton UC = (SUC ,∆UC , FUC ) is universal for the class C if the following twoconditions hold:

1. For any 1 ≤ i ≤ n, there is a state s ∈ SUC such that WUC (s) = WAi(qi) andRUC (s, w) = RAi(qi, w), for all w ∈ WUC (s).

2. For any s ∈ SUC , there is an i, 1 ≤ i ≤ n, such that WUC (s) = WAi(qi) andRUC (s, w) = RAi(qi, w), for all w ∈ WAi(qi).

It is not hard to see that every incomplete automaton V (with no initial state)naturally defines a class C(V ) for which V itself is universal. Indeed, let q1, . . . , qn ∈SV be all the states of V , and for each i, define Ai = V . Clearly V is universal forthe class C(V ) = (A1, q1), . . . , (An, qn).

Not every finite class of finite incomplete automata with initial states has auniversal incomplete automaton; nevertheless, we can always enlarge it to a classwhich has this property.

Proposition 6.7 Every finite class of pairs of incomplete automata and initialstates can be embedded into a finite class which has at least one universal incompleteautomaton.

Proof. Let Γ = (Ai, qi) | 1 ≤ i ≤ n, where Ai = (Si,∆i, Fi). Assume thatall the state sets of these incomplete automata are pairwise disjoint. Consider theincomplete automaton U ,

U = (∪ni Si, ∪ni ∆i, ∪ni Fi).

Let C(U) be the class as defined above and take B = Γ ∪ C(U). It is easy to seethat Γ is contained in B and U is universal for B . 2

Theorem 6.8 The incomplete automata A and B strongly β–simulate each otheriff A and B are universal for the same class.

Proof. Suppose that A and B strongly β–simulate each other via h1 : SA → SBand h2 : SB → SA. Consider the class C(A), for which the incomplete automaton Ais universal. We show that B is universal for C(A). Suppose that (A1, q1) belongsto C(A). Then for all w ∈ WA1(q1), we have RA1(q1, w) = RB(h1(q1), w). For everyq ∈ SB there exists a state q′ = h2(q) ∈ SA such that for the pair (A, q′) we haveRA(q′, w) = RB(q, w), for all w ∈ WA(q′). Hence B is also universal for C(A).

Now assume that A and B are both universal for the class C =(A1, q1), (A2, q2), . . . , (An, qn). For every q ∈ SA there exists i ∈ 1, 2, . . . , n,such that RA(q, w) = RAi(qi, w), for all w ∈ WA(q). Since (Ai, qi) ∈ C and B isuniversal for C there is a state pi ∈ SB – say, the minimal one according to a fixed

6.3 Universality 51

linear order defined on the set of all states – such that RAi(qi, w) = RB(pi, w), forall w ∈ WAi(qi). Hence A is strongly β–simulated by B via the mapping qi 7→ pi.Similarly, B can be strongly β–simulated by A. 2

An incomplete automaton A which is universal for the class C is said to beminimal if it has the least number of states compared with all other incompleteautomata universal for the same class.

From this definition and Theorem 6.8 above we obtain:

Corollary 6.9 The following statements are equivalent:

1) The incomplete automaton A is a minimal universal automaton for a class C.

2) For every incomplete automaton B universal for the class C, if A ¿β B andB ¿β A, then |SA| ≤ |SB|.

Corollary 6.10 Let C be a finite class of incomplete automata with initial statesand M the unique (up to an isomorphism) universal minimal incomplete automatonfor C. Then M is a complete automaton iff all automata in C are complete.

In case of deterministic incomplete automata the equivalence relation ≡ can beexpressed in one of the following equivalent forms:

1) RA(p, w) = RA(q, w), for all w ∈ WA(p).

2) fA(p, w) = fA(q, w), for all w ∈ WA(p).

Lemma 6.11 Let p and q be any states of an incomplete automaton A =(SA,∆A, FA). Then

1) p ≡RA q iff p ≡fA q.

2) p ≡fA q implies ∆A(p, w) ≡fA ∆A(q, w), for all w ∈ WA(p).

3) p ≡fA q implies FA(p) = FA(q).

Proof. 1) If p ≡R q then p ≡f q by definition. So assume that p ≡f q. First,RA(p, λ) = FA(p) = fA(p, λ) = fA(q, λ) = FA(q) = RA(q, λ). Assume now thatRA(p, w) = RA(q, w) for all words w whose length is at most n. Then for anyσ ∈ Σ,

RA(p, wσ) = RA(p, w)FA(∆A(p, wσ))

= RA(p, w)fA(p, wσ)

= RA(q, w)fA(q, wσ)

= RA(q, w)FA(∆A(q, wσ))

= RA(q, wσ),


if wσ ∈ WA(p); otherwise, RA(p, wσ) =∞ = RA(q, wσ).2) Let w ∈ WA(p) = WA(q) and denote r = ∆A(p, w) and s = ∆A(q, w). The

applicable words of r and s can be obtained from the applicable words of p and q,

WA(r) = u | wu ∈ WA(p) and WA(s) = u | wu ∈ WA(q).

Since WA(p) and WA(q) are equal, also the sets WA(r) and WA(s) are equal. Onthe other hand, for any w ∈ WA(r),

fA(r, u) = fA(p, wu) = fA(q, wu) = fA(s, u).

Thus r and s are indistinguishable, too.3) The statement follows from the fact that FA(p) = fA(p, λ) = fA(q, λ) =

FA(q). 2

For any state s ∈ SA, let [s] denote the equivalence class of s under ≡, thatis, [s] = p ∈ SA | s ≡ p. The automaton M(A) = (SM(A),∆M(A), FM(A)) isconstructed as follows: SM(A) = [s] | s ∈ SA, and for all [s] ∈ SM(A),

FM(A)([s]) = FA(s), and

∆M(A)([s], σ) =

[∆A(s, σ)] , if σ ∈ Σ ∩WA(s),∞, otherwise.

From Theorem 6.6 we deduce:

Corollary 6.12 For any two minimal incomplete automata A and B the followingare equivalent:

1) A and B strongly simulate each other.

2) A and B strongly f-simulate each other.

3) A and B strongly β-simulate each other.

4) A and B are isomorphic.

Chapter 7

Testing ComputationalComplementarity

In this chapter a package of programs for testing CI and CII for complete determin-istic automata is described and the results of using these programs for four-statesautomata are discussed. We will have a particular interest in distinguishing re-versible from non–reversible automata.1 Very complex four–states automata, i.e.automata having an operator monoid size greater or equal than 58, are reversible.This supports a conjecture relating complexity and reversibility: highly complexautomata are reversible. To simplify the computational process we need a few the-oretical results which will be presented in the next section.

7.1 Operators

A good physical way to look at an automaton, which is equivalent to the classicalmathematical one, is to think of A in terms of the transformations (operators)Tσ : SA → SA, σ ∈ Σ, Tσ(q) = ∆A(q, σ), as “push buttons” allowing the automatonto change its states. Mathematically, we shall associate to A a class of operators(Tw)w∈Σ? ,

Tw : SA → SA, Tw(q) = ∆(q, w).

Clearly, for all u,w ∈ Σ?, Tu Tv = Tuv, so (Tw)w∈Σ? is a monoid (Tλ is the neutralelement).2 This monoid is finite. To get a system of independent names we definethe equivalence relation u ≈ v if Tu = Tv, pick from each equivalence class thesmallest word (in lexicographical order) and collect all these words into the finiteset S. Then each operator Tu has a “name” Tv with v ∈ S .

Sometimes the monoid of operators is a group, and in this special situation bothtypes of complementarity disappear. Here is the mathematical justification.

1As reversibility is intimately linked to both quantum physics and quantum computation.2This monoid is sometimes called the transition monoid, (cf. Clifford and Preston [20]); for

more details on the algebraic theory see also Gecseg and Peak [33].

53

54 7. Testing Computational Complementarity

Theorem 7.1 The following three statements are equivalent:

G1) The operators (Tw)w∈S form a group.

G2) Each operator Tσ, σ ∈ Σ, is bijective.3

G3) Every operator Tσ, σ ∈ Σ, has a right inverse.

Proof. For the implications ”G1)⇒ G2)” note that every Tσ is total Tσv = Tλ,for some v ∈ S, i.e., Tσv(q) = q, for all q ∈ SA, so Tσ is bijective. As the implication”G2)⇒ G3)” is trivial, let us assume that that G3) holds true, i.e. for every σ ∈ Σthere exists a word xσ (depending upon σ) such that Tσxσ = Tλ. Clearly, Tσ is total.

As each operator has a right inverse, it follows that for each word u therecorresponds a word u such that Tuu = Tλ. Indeed, if u = σ1σ2 . . . σn, then u =xσn . . . xσ1 does the job. Now take p = Tuu(q). As

∆A(p, u) = ∆A(∆A(q, uu), u) = ∆A(∆A(q, u), uu) = ∆A(q, u)

it follows that

∆A(p, u) = ∆A(q, u).

Using again the hypothesis and the above equality we get

p = ∆A(∆A(p, u), u) = ∆A(∆A(q, u), u) = q,

which tells us that Tu is the inverse of Tu. 2

Corollary 7.2 Each of the above equivalent conditions G1) - G3) implies that theautomaton A is complete and the equivalence of properties A, B and C.

Proof. Indeed, every operator Tw has an inverse Tw. If A has property A, thenfor each pair of distinct states q, q′ there exists a word wq,q′ such that RA(q, wq,q′) 6=RA(q′, wq,q′). To get an experiment which globally distinguishes between any twodistinct states we proceed as follows: we concatenate all words wq,q′wq,q′ (when q, q′

range in SA and are distinct) and we get the word w such that for all distinct statesq, q′ satisfies RA(q, w) 6= RA(q′, w). 2

Intuitively, the statement “G1) - G3) implies the absence of complementarity”can be understood as follows. Suppose an automaton has one of the equivalentproperties G1) - G3). Suppose further that an observer obtains a single copy ofit and adopts the following strategy. The observer runs an arbitrary number ofindependent experiments. After each experiment, the observer records its outcomeand steers the automaton back to its original (unknown) state.4 In this way, the

3As each operator is a function from the finite set SA into itself, it follows that an operator Tσis bijective if and only if it is injective if and only if it is surjective.

4Bennett, [5], has used a similar strategy for avoiding a huge memory overhead in reversiblecomputations.

7.2 Reversible Instances 55

observer can make sure that the experiment does not irreversibly destroy potentialinformation about the initial state of the automaton. Indeed, the setup is similarto Moore’s multi-automaton configuration, with the only difference that not allexperiments are performed simultaneously, but only one at a time. In that way,total and thus classical knowledge of the initial state is obtained.

This strategy fails for quantum systems. There, it is only possible to “reverse thewave function collapse” (“reconstruct the state”) if no knowledge of the measure-ment outcome is left over. All obtained information is needed in the reconstructionprocess itself. And, since any copying of proper q(u)bits of quantum informationis not allowed, the strategy fails to produce the classical “elements of reality.” Be-cause, stated pointedly, the observer either can make sure that he recovers theoriginal system, (exclusive) or records a single measurement outcome associatedwith one particular experiment.

From an observer “from inside” the automaton is reversible, i.e. each compu-tation can be reversed. To be more precise, an “inside observer” can reverse anycomputation, but the proof that each computation can be reversed can be achievedonly at the meta-level, i.e. at the level of a language “speaking about the automa-ton”. An external observer is “losing” information in the process of monitoring onlythe outcomes: for such an observer some computations cannot be reversed.

The size of the monoid of operators is a measure of the complexity of the automa-ton. In what follows we shall refer to this measure as the “size of the automaton”.

Experimental tests for the case of four-states automata show that each automa-ton satisfying CI or CII has a non-commutative monoid of operators. In physicalterms, non-commutativity is a mathematical form of complementarity, meaningthat “there is no state in which both measurables have well-defined values simulta-neously”.

How far could one go from the monoid structure associated with automata to theunitary transformations encountered in the evolution of quantum mechanical statesbetween measurements? A few warnings should be issued at the very beginning.Automaton logic, like quantum logic, is static. It is not concerned with dynamicalprocesses but with the inference of operational statements and their interrelation.Also, quantum systems are defined in the entire richness of finite/infinite dimen-sional Hilbert spaces and it can be expected that no finite structure can faithfullyrepresent such wealth of mathematics (see Svozil and Tkadlec[70]). Therefore, thereis little hope for complete “isomorphism”.

7.2 Complementarity: Reversible Instances for

Complete Deterministic Automata

Everyone is familiar with the strange effects produced by projecting a film backward.This “strangeness” is considered to be normal in classical dynamics, as was explicitly


stated by its founders Galileo or Huygens.5 Quantum mechanics, in which statepreparations and measurements are irreversible6, raised serious doubts related toreversibility and initiated the abolition of it.7

A new perspective on this issue came recently from computation theory.8 To-day’s computers erase a bit of information every time they perform a computationcorresponding to a many-to-one operation, (cf. Landauer [49], Bennett [5, 6], Fred-kin and Toffoli [31], Feynman [28]). Therefore, computational operations, such asthe explicit deletion of information or clearing some memory and that like, are “ir-reversible.” In spite of the fact that in the last 50 years the dissipated energy perbit of computational operation9 has decreased by roughly tenfold each five years(cf. Landauer [50]) the erasure is done very inefficiently, and much more than kTenergy is dissipated for each bit erased.10

In order to improve the computer hardware performance we have to continue toreduce the energy dissipated by each computational operation.

There are two alternative ways to approach this problem: a) improving by con-ventional methods, i.e. improving the efficiency with which we erase information;b) ultimately use computational operations that do not erase information, i.e. theso-called “reversible” computational operations, which can, in principle, dissipatearbitrarily little heat. As the energy dissipated per irreversible computational op-eration approaches the limit of ln 2×kT , the use of reversible operations is likely tobecome more attractive. If we are to reduce energy dissipation per logic operationbelow ln 2× kT we will be forced to use reversible logic.

The above facts show clearly how important is to model the idea of reversiblecomputational operation (for an excellent discussion see the paper by Bennett andLandauer [7]). Recall that the automaton (SA, δA, FA) is reversible11 if for everystates p, q ∈ SA and u ∈ Σ? with δA(p, u) = q, there exists a word w ∈ Σ? such thatδA(q, w) = p. In other words, every input state of a computation can be “reachedback” from the final state of the computation, by means of a suitable computation.A stronger definition was studied in the literature (see Bavel and Muller [2]).

5For instance, when they described the implications of the equivalence between cause and effectas an “axiom” for their mathematical model of motion.

6Up to instances where the wave function is reconstructed.7“Active science is thus, by definition, extraneous to the idealized, reversible world it is de-

scribing”, (see Prigogine and Stengers, [63, p. 61]).8More than 40 years ago Einstein had a similar point: irreversibility is an illusion, a subjective

impression. There is no irreversibility in the basic laws of physics.9A good analogy is with friction. In practice there is always friction. The loss of heat energy is

normally large. Carnot heat engines–in which heat energy is converted into work and back again,operate over a reversible closed cycle: they start off in a particular state to which they returnafter completing a cycle of operation. This exercise cannot be done with zero energy cost (thesecond law of thermodynamics) but it is—at least theoretically—possible to make the losses dueto friction as small as possible in case you run the engine infinitesimally slow.

10Here k is Boltzmann’s constant and T is the absolute temperature in Kelvin degrees, sokT ≈ 3× 10−21 Joule at room temperature.

11Condition G3) was used as a definition for “reversibility” by Bavel and Muller [2]; it seems tous to be too strong.


Proposition 7.3 The following two conditions are each equivalent to reversibility:

1. for all q ∈ Q and w ∈ Σ? there exists a word u ∈ Σ? such that δ(q, wu) = q;

2. for every state q ∈ Q and σ ∈ Σ there exists a word vqσ ∈ Σ? (depending uponσ and q) such that δ(q, σvqσ) = q.

Proof. We have to prove only the equivalence between the last condition andreversibility. Indeed, if u = σ1 . . . σn, then

δA(q, uvδA(q,σ1...σn−1)n . . . v

δA(q,σ1)2 vq1) = q. 2

Each of the above conditions is decidable, as we need to examine only wordsof length less than the size of SA, due to the Pumping Lemma (see Hopcroft andUllman [43, pp. 55-56]): for all q ∈ SA, u ∈ Σ?, δA(q, u) = δA(q, w), for somew ∈ Σ? with length less than the size of SA.

Experimental computations show that out of 359,040 reversible four-states au-tomata:

• 26,688 satisfy CI; the minimal size of an automaton displaying CI is 8 (see anexample in Figure 7.1); the maximal size is 79 (see an example in Figure 7.3);

• 16,128 satisfy CII; the minimal size of an automaton displaying CI is 9 (seean example in Figure 7.2); the maximal size is 145 (see an example in Figure7.4).

We start with examples of automata having CI, CII and minimal size. Theautomaton in Figure 7.1 has CI. Indeed, the automaton has property A: the pairsof states (1, 2), (1, 3), (1, 4) are distinguishable by any experiment, (2, 3) and (2, 4)can be distinguished by w = b, and (3, 4) can be distinguished by w = ab.

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66

?

4 3

21

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|

b

|

|

|

0

1 0

ba

a

b

0

b a

a

Figure 7.1

The automaton has not property B as for 4:


• every experiment w of the form ay, y ∈ Σ? does not distinguish between 4and 2, and

• every experiment w of the form by, y ∈ Σ? does not distinguish between 4and 3.

The monoid of operators has 8 elements

State Tλ Ta Tb Taa Tab Tba Tbb Taab1 1 4 1 2 1 4 1 32 2 2 3 2 3 4 1 33 3 4 1 2 1 4 1 34 4 2 1 2 3 4 1 3

and the induced table is the following:

Tλ Ta Tb Taa Tab Tba Tbb TaabTλ Tλ Ta Tb Taa Tab Tba Tbb TaabTa Ta Taa Tab Taa Taab Tba Tbb TaabTb Tb Tba Tbb Taa Tbb Tba Tbb TaabTaa Taa Taa Taab Taa Taab Tba Tbb TaabTab Tab Tba Tbb Taa Tbb Tba Tbb TaabTba Tba Taa Tbb Taa Taab Tba Tbb TaabTbb Tbb Tba Tbb Taa Tbb Tba Tbb TaabTaab Taab Tba Tbb Taa Tbb Tba Tbb Taab

The automaton in Figure 7.2 satisfies CII:

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66

?-

?

4 3

21

|1

|

|

|

0

1 0

ba

bb

ab

a a

Figure 7.2


The automaton has property B: the states (1, 2), (1, 4), (2, 3), (3, 4) are distin-guishable by any experiment, so the state 1 can be distinguished from any otherstate by w = b, 2 can be distinguished by w = a, 3 can be distinguished by w = b,and 4 can be distinguished by w = a. It has not property C as:

• if w = ay, y ∈ Σ?, then b1 and 3 are not distinguishable,

• if w = by, y ∈ Σ?, then 2 and 4 are not distinguishable.

In this case the monoid of operators has 9 elements

State Tλ Ta Tb Taa Tab Tba Tbb Tabb Tbaa1 1 2 4 3 1 2 1 4 32 2 3 1 2 1 2 4 4 33 3 2 1 3 1 2 4 4 34 4 2 1 3 1 2 4 4 3

and the induced table is the following:

Tλ Ta Tb Taa Tab Tba Tbb Tabb TbaaTe Te Ta Tb Taa Tab Tba Tbb Tabb TbaaTa Ta Taa Tab Ta Tab Tba Tabb Tabb TbaaTb Tb Tba Tbb Tbaa Tab Tba Tb Tabb TbaaTaa Taa Ta Tab Taa Tab Tba Tabb Tabb TbaaTab Tab Tba Tabb Tbaa Tab Tba Tab Tabb TbaaTba Tba Tbaa Tab Tba Tab Tba Tabb Tabb TbaaTbb Tbb Tba Tb Tbaa Tab Tba Tbb Tabb TbaaTabb Tabb Tba Tab Tbaa Tab Tba Tabb Tabb TbaaTbaa Tbaa Tba Tab Tbaa Tab Tba Tabb Tabb Tbaa

We continue with examples of automata having CI, CII and maximal size. Theautomaton in Figure 7.3 has CI and maximum size (i.e 79). The pair of states(1, 2), (1, 3), (1, 4) can be distinguished by any experiment. For (2, 3) and (2, 4) wecan use w = b; for (3, 4) we can use w = a. So, it has property A.

The automaton has not property B as (2, 4) cannot be distinguished by anyexperiment w = ay, y ∈ Σ? and (3, 4) cannot be distinguished by any experimentw = by, y ∈ Σ?.


6

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?

4 3

21

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|

a, b

|

|

|

0

1 0

0

a a

a

bb

b

Figure 7.3

The automaton in Figure 7.4 has CII maximum size (i.e 145):

?

6

->

=

4 3

21

%$

|

a

|

|

|

0 0

0 1

a, b

a

ab b

b

Figure 7.4

The pair of states (1, 2), (2, 3), (2, 4) can be distinguished by any experiment.The state 1 can be distinguished by any other state by w = b, 2 can be distinguishedby any experiment, 3 can be distinguished by w = aab, and 4 can be distinguishedby w = ab. So, it has B.

The automaton has not property C as

• (3, 4) cannot be distinguished by any experiment w = by, y ∈ Σ?;

• if w = an, for some n ≥ 1, then 1 and 3 cannot be distinguished: R(1, an) =an+1 = R(3, an);

• if w = a3nby, for some y ∈ Σ?, n ≥ 1, then 3 and 4 cannot be distinguished:R(1, a3nb) = a3n+2 = R(4, a3nb) and δ(3, a3nby) = δ(3, by) = δ(1, y) =δ(4, by) = δ(4, a3nby);


• if w = a3naby, for some y ∈ Σ?, n ≥ 1, then 1 and 3 cannot be distinguished:R(1, a3nab) = a3n+3 = R(4, a3nab) and δ(1, a3naby) = δ(1, aby) = δ(1, y) =δ(3, aby) = δ(3, a3naby);

• if w = a3naaby, for some y ∈ Σ?, n ≥ 1, then 1 and 4 cannot be distinguished:R(1, a3naab) = a3n+4 = R(1, a3naab) and δ(1, a3naaby) = δ(1, aaby) =δ(1, y) = δ(4, aaby) = δ(4, a3naaby).

The following proposition follows from the definitions of reversibility and con-nectivity, and the fact that for all states p and q in SA, and every string w ∈ Σ?,p = δA(q, w) iff [p] = δM(A)([q], w).

Proposition 7.4 . Let A be an incomplete automaton. Then the following condi-tions hold true:

1. A is reversible iff M(A) is reversible.

2. A is strongly connected iff M(A) is strongly connected.

3. A is connected iff M(A) is connected.

There are 3888 classes of isomorphic four states reversible automata, 2725 haveproperty A, 2445 have property B and 2273 have property C. Automata withmonoid operator of size greater than or equal to 80 do not have CI and automatawith monoid operator of size greater or equal to 146 do not have CII. Examiningthe following tables we notice a few interesting facts.

• It seems that automata with a large size operator monoid tend to be reversible(for our case, all automata with the monoid size greater or equal than 58 arereversible).

• Automata with operator monoid size 57 have a critical behaviour: none ofthem are reversible and none of them have CI nor CII. This means thatevery minimal automaton of this type has property C. There is no other classof four–state automata displaying these properties.

• Complementarity principles CI and CII appear mainly on automata withsmall operator monoid size: the larger the size, the less likely is CI or CII.

Size 1 2 3 4 5 6 7 8 9 10All 3 62 277 599 649 812 780 740 518 476Rev 3 18 22 41 44 82 34 156 66 48

Prop A 0 2 8 13 14 34 15 77 25 23CI 0 0 0 0 0 0 0 6 5 5CII 0 0 0 0 0 0 0 0 2 2


Size 11 1 2 13 14 15 16 17 18 19 20All 450 540 496 396 148 224 144 276 128 128Rev 90 188 110 80 32 104 0 108 64 64

Prop A 44 117 53 34 27 63 0 58 32 38CI 1 0 11 0 6 12 0 1 9 11CII 6 3 2 2 2 5 0 2 1 3

Size 21 22 23 24 25 26 27 28 29 30All 80 64 16 510 280 72 32 48 24 16Rev 16 0 16 366 104 40 16 32 8 16

Prop A 8 0 8 310 85 29 11 24 4 8CI 0 0 0 54 15 6 1 0 0 0CII 0 0 0 13 7 0 0 0 0 0

Size 31 32 33 34 35 36 37 38 39 40All 176 120 32 64 0 0 0 24 16 368Rev 8 120 0 64 0 0 0 24 0 224

Prop A 7 66 0 44 0 0 0 21 0 196CI 0 0 0 4 0 0 0 3 0 34CII 1 14 0 5 0 0 0 0 0 26

Size 41 42 43 44 45 46 47 48 49 50All 160 64 64 16 40 16 0 32 0 0Rev 144 48 64 16 24 16 0 32 0 0

Prop A 84 36 50 8 21 8 0 28 0 0CI 4 6 4 0 6 0 0 4 0 0CII 6 6 2 0 0 0 0 0 0 0

Size 51 52 53 54 55 56 57 58 59 60All 0 48 0 0 0 64 48 32 16 0Rev 0 48 0 0 0 64 0 32 16 0

Prop A 0 42 0 0 0 56 0 28 14 0CI 0 4 0 0 0 12 0 4 2 0CII 0 4 0 0 0 4 0 0 2 0

Size 61 62 63 64 65 66 67 68 69 70All 40 48 32 32 0 0 168 64 0 0Rev 40 48 32 32 0 0 168 64 0 0

Prop A 34 42 28 28 0 0 146 56 0 0CI 3 4 2 4 0 0 15 4 0 0CII 11 4 2 4 0 0 7 4 0 0


Size 71 72 73 74 75 76 77 78 79 80All 0 0 0 0 0 0 0 0 144 0Rev 0 0 0 0 0 0 0 0 144 0

Prop A 0 0 0 0 0 0 0 0 126 0CI 0 0 0 0 0 0 0 0 18 0CII 0 0 0 0 0 0 0 0 0 0

Size 81 82 83 84 85 86 87 88 89 90All 0 0 0 0 0 0 0 0 0 0Rev 0 0 0 0 0 0 0 0 0 0

Prop A 0 0 0 0 0 0 0 0 0 0CI 0 0 0 0 0 0 0 0 0 0CII 0 0 0 0 0 0 0 0 0 0

Size 91 92 93 94 95 96 97 98 99 100All 0 0 0 0 0 0 48 0 0 0Rev 0 0 0 0 0 0 48 0 0 0

Prop A 0 0 0 0 0 0 42 0 0 0CI 0 0 0 0 0 0 0 0 0 0CII 0 0 0 0 0 0 0 0 0 0

Size 101 102 103 104 105 106 107 108 109 110All 0 0 0 0 0 48 0 0 0 0Rev 0 0 0 0 0 48 0 0 0 0

Prop A 0 0 0 0 0 42 0 0 0 0CI 0 0 0 0 0 0 0 0 0 0CII 0 0 0 0 0 0 0 0 0 0

Size 111 112 113 114 115 116 117 118 119 120All 0 0 0 0 0 64 0 0 0 0Rev 0 0 0 0 0 64 0 0 0 0

Prop A 0 0 0 0 0 56 0 0 0 0CI 0 0 0 0 0 0 0 0 0 0CII 0 0 0 0 0 8 0 0 0 0

Size 121 122 123 124 125 126 127 128 129 130All 0 0 0 0 0 0 0 96 0 0Rev 0 0 0 0 0 0 0 96 0 0

Prop A 0 0 0 0 0 0 0 84 0 0CI 0 0 0 0 0 0 0 0 0 0CII 0 0 0 0 0 0 0 0 0 0


Size 131 132 133 134 135 136 137 138 139 140All 0 0 0 0 0 0 0 0 0 0Rev 0 0 0 0 0 0 0 0 0 0

Prop A 0 0 0 0 0 0 0 0 0 0CI 0 0 0 0 0 0 0 0 0 0CII 0 0 0 0 0 0 0 0 0 0

Size 141 142 143 144 145 146 147 148 149 150All 0 0 0 0 96 0 0 0 0 0Rev 0 0 0 0 96 0 0 0 0 0

Prop A 0 0 0 0 84 0 0 0 0 0CI 0 0 0 0 0 0 0 0 0 0CII 0 0 0 0 12 0 0 0 0 0

Size 151 152 153 154 155 156 157 158 159 160All 0 0 0 0 0 0 0 0 0 0Rev 0 0 0 0 0 0 0 0 0 0

Prop A 0 0 0 0 0 0 0 0 0 0CI 0 0 0 0 0 0 0 0 0 0CII 0 0 0 0 0 0 0 0 0 0

Size 161 162 163 164 165 166 167 168 169 170All 0 0 0 0 0 0 0 0 0 0Rev 0 0 0 0 0 0 0 0 0 0

Prop A 0 0 0 0 0 0 0 0 0 0CI 0 0 0 0 0 0 0 0 0 0CII 0 0 0 0 0 0 0 0 0 0

Size 171 172 173 174 175 176 177 178 179 180All 0 0 0 0 0 96 0 0 0 0Rev 0 0 0 0 0 96 0 0 0 0

Prop A 0 0 0 0 0 84 0 0 0 0CI 0 0 0 0 0 0 0 0 0 0CII 0 0 0 0 0 0 0 0 0 0

7.3 Complementarity: Non-Reversible Instances

for Complete Deterministic Automata

In case of non-reversible automata the sizes of automata having CI or CII decrease,in the minimal cases, to 3. Here are two examples.

The automaton in Figure 7.5 has CI . Indeed, the automaton has property A:the pairs (1, 2), (1, 3), (1, 4) are distinguishable by any experiment, the word w = bdistinguishes between 2, 3, and 2, 4, while w = a distinguishes between 3 and 4.The automaton has not property B as the state 3 cannot be distinguished from 2

7.3 Non-Reversible Instances 65

by any experiment starting with a, and 3 cannot be distinguished from 4 by anyexperiment starting with b.

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21

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|

b

|

|

|

0

1 0

0

a, ba, b

aa, b

Figure 7.5

The operators induced by the automaton in Figure 7.5 and their compositiontable are presented below:

State Tλ Ta Tb1 1 1 12 2 2 23 3 2 14 4 1 1

Tλ Ta TbTλ Tλ Ta TbTa Ta Ta TaTb Tb Tb Tb

The automaton in Figure 5.2 has CII (see the proof in Section 5.3). Thisautomaton has minimum size, i.e. 3, which makes the proof for non-C quite easy. Inthis case we have three operators (Tλ, Ta, Tb)—which are different from the operatorsof the automaton from Figure 7.3,

State Tλ Ta Tb1 1 1 12 2 2 23 3 2 14 4 2 1

but their composition tables do coincide.There are 7476 classes of isomorphic non-reversible four states automata; 3491

have property A, 500 verify CI and 156 verify CII. There is no four-state non-reversible automaton with the operator monoid size greater or equal to 59. PrincipleCI disappears from the size 47 and principle CII disappears from the size 43.


Size 1 2 3 4 5 6 7 8 9 10All 3 62 277 599 649 812 780 740 518 476

NonRev 0 44 255 558 605 730 746 584 452 428Prop A 0 2 33 133 179 266 300 223 208 221CI 0 0 2 17 37 37 47 27 41 42CII 0 0 1 4 11 13 16 14 7 14

Size 11 1 2 13 14 15 16 17 18 19 20All 450 540 496 396 148 224 144 276 128 128


Size 21 22 23 24 25 26 27 28 29 30All 80 64 16 510 280 72 32 48 24 16


Size 31 32 33 34 35 36 37 38 39 40All 176 120 32 64 0 0 0 24 16 368


Size 41 42 43 44 45 46 47 48 49 50All 160 64 64 16 40 16 0 32 0 0


Size 51 52 53 54 55 56 57 58 59 60All 0 48 0 0 0 64 48 32 16 0


7.4 A Glimpse into Programs 67

7.4 A Glimpse into Programs

Programs generating all equivalence classes of isomorphic automata and testingproperties A, CI, CII have been written in Pascal. In this section we are going topresent some routines.12

We start with the list of the main types used in the programs:

constlX = 6; lenght of input stringcS = 4; number of statescA = 2;number of letters in the input alphabetnrPair = cS * (cS - 1) div 2; total number of pairs of

distinct statessizeD = cA * cS;used for representing the transition

functiontypecS_vector = array[1..cS] of integer;used for the output

functioncS_cA_table = array[1..cS, 1..cA] of integer;

used for the transition functionlX_vector = array[1..lX] of integer;for input stringPairType = array[1..nrPair, 1..2] of integer;

digit = 0..cS;four_vector = array[1..cS] of digit;for fvalidT = 0..255;TSet = set of validT;TFunc = record

nr: four_vector;label: string[40];end;

TArrayType = array[1..255] of TFunc;four_two_table = array[1..cS, 1..2] of integer;six_two_table = array[1..6, 1..2] of integer;six_vector = array[1..6] of integer;cardArrayType = array[1..145] of integer;

The function GenerateX returns the representation of number in base cA. Ifthe representation has less than lX digits, then it fills with 0 to the left, to get lXdigits. For example for number = 15, cA = 3 and lX = 6, GenerateX is the vector0 0 0 1 2 0. This subroutine is used to generate w ∈ Σ? and to testCI and CII.

function GenerateX (number: integer): lX_vector;constbase = cA;

12The routines heavily reflect, for efficiency reasons, the state size. However, with minor modi-fications we can use these programs for the other cases as well.


vark: integer;quotient: integer;tempVal: lX_vector;

beginfor K := 1 to lX dotempVal[k] := 0;

if (number <> 0) thenbeginquotient := number;k := lX;while quotient >= base dobegintempVal[k] := quotient mod base;quotient := quotient div base;k := k - 1;

end;whiletempVal[k] := quotient;end;if..then

GenerateX := tempVal;end;

The procedure GeneratePairs generates and returns in the vector pair all distinctunordered pairs of states.

procedure GeneratePairs (var pair: PairType);cQ must be at least 2

vark, t, s: integer;

begink := 0;for t := 1 to cS - 1 dofor s := t + 1 to cS dobegink := k + 1;pair[k, 1] := t;pair[k, 2] := s;

end; forend; procedure

The function test returns true if the input word xValue distinguishes betweenthe states firstQ, secondQ and false otherwise.

function test (deltaValue: cS_cA_table; fValue: cS_vector;firstQ, secondQ:integer; xValue: lX_vector): boolean;

var


tempFc: boolean;a, b: integer;counter: integer;

beginif (fValue[firstQ] <> fValue[secondQ]) thentempFc := true

elsebegintempFc := false;a := firstQ;b := secondQ;counter := 1;while counter <= lX dobegina := deltaValue[a, xValue[counter] + 1];b := deltaValue[b, xValue[counter] + 1];if (fValue[a] <> fValue[b]) thenbegintempFc := true;counter := maxInt;endelsecounter := counter + 1;

end; while counterend; if..then..else

test := tempFc;end; function test

The function ACond returns true if the automaton having the transition map-ping given by deltaValue and the output mapping by fValue has property A andfalse otherwise.

function ACond (deltaValue: cS_cA_table; fValue: cS_vector): boolean;vartempAnswer, nowTest: boolean;k, t, qFirst, qSecond: integer;maxT: longInt;allPairs: PairType;currentX: lX_vector;

beginGeneratePairs(allPairs);tempAnswer := false;k := 1;while k <= nrPair do for all pairsbeginqFirst := allPairs[k, 1];


qSecond := allPairs[k, 2];t := 0;maxT := Power(2, lX) - 1;nowTest := false;while not (nowTest) and (t <= maxT) dobegincurrentX := GenerateX(t);nowTest := test(deltaValue, fValue, qFirst, qSecond, currentX);if not (nowTest) thent := t + 1;

end; while tK := K + 1;if t = Power(2, lX) thenK := maxInt

else if K = nrPair + 1 thentempAnswer := true;

end; while kACond := tempAnswer;

end; function ACond

The function BCond returns true if the automaton having the transition map-ping given by deltaValue and the output mapping by fValue has property B andfalse otherwise.

function BCond (deltaValue: cS_cA_table; fValue: cS_vector): boolean;varq1, q2, t, maxT: integer;tempAnswer, nowTest: boolean;currentX: lX_vector;

begintempAnswer := false;q1 := 1;maxT := Power(2, lX) - 1;while q1 <= cS do for all qbegint := 0;while t <= maxT dobegincurrentX := GenerateX(t);q2 := 1;while (q2 <= cS) dobeginnowTest := test(deltaValue, fValue, q1, q2, currentX);if (nowTest or (q1 = q2)) thenq2 := q2 + 1elseq2 := MaxInt - 2;


if (q2 = cS + 1) thent := maxInt - 2; next q1end;while q2.t := t + 1;if (t = maxT + 1) thenq1 := maxInt - 1;

end;whileq1 := q1 + 1;if (q1 = cS + 1) thentempAnswer := true;

end; while for q1BCond := tempAnswer;

end;end BCond

The function CCond returns true if the automaton having the transition map-ping given by deltaValue and the output mapping by fValue has property C andfalse otherwise.

function CCond (deltaValue: cS_cA_table; fValue: cS_vector): boolean;vart, k, q1, q2, maxT: integer;tempAnswer, nowTest: boolean;allPairs: PairType;currentX: lX_vector;

begintempAnswer := false;GeneratePairs(allPairs);maxT := Power(2, lX) - 1;t := 0;while t <= maxT dobegincurrentX := GenerateX(t);k := 1;while k <= nrPair dobeginq1 := allPairs[k, 1];current first stateq2 := allPairs[k, 2];current second statenowTest := test(deltaValue, fValue, q1, q2, currentX);if nowTest thenk := 1 + kelsek := maxInt - 1;if k = nrPair + 1 thenbegintempAnswer := true;t := maxInt - 1;end;


end;while kt := t + 1;end; while t

CCond := tempAnswer;end;conditionC

The function IsRev returns true if the automaton with the transition mappinggiven by deltaValue is reversible and false otherwise.

function IsRev (currentDelta: cS_cA_table): boolean;vartempProduct, base: cS_cS_table;i, j, k: integer;answer: boolean;

beginbase := BasicMatrix(currentDelta);tempProduct := base;for i := 1 to cS dotempProduct := productMatrices(tempProduct, base);

answer := true;i := 1;while (answer) & (i <= cS) dobeginj := i + 1;while (answer) & (j <= cS) dobeginanswer := (tempProduct[i, j] = tempProduct[j, i]);j := j + 1

end;i := i + 1;end;

IsRev := answer;end;

The following three functions are used to find the cardinality of the operatorsmonoid defined by every transition mapping of a complete four–state automaton.Compose2Trans, a four-dimensional vector, returns the composition of the firstTrand secondTr. The function transIntChange converts a four-dimensional vectorto an integer number. The function CardTX returns the size of operator monoidassociated with the automaton whose transition mapping is given by firstTr andsecondTr.

function Compose2Trans (firstTr, secondTr: four_vector):


four_vector;var

index: integer;tempFc: four_vector;

beginfor index := 1 to cS dotempFc[index] := secondTr[firstTr[index]];

Compose2Trans := tempFc;end;

function transIntChange (transFc: four_vector): integer;varvec1, vec2, vec3, vec4: integer;

beginvec1 := cS * cS * cS * (transFc[1] - 1);vec2 := cS * cS * (transFc[2] - 1);vec3 := cS * (transFc[3] - 1);vec4 := transFc[4] - 1;transIntChange := vec1 + vec2 + vec3 + vec4;

end;

function cardTX (myT1, myT2: four_vector): integer;varlocalSet: TSet;local: TArrayType;finalL, initialL: integer;currentT: four_vector;i, j,id: integer;done: boolean;

beginid:=(cS-1)+(cS-2)*cS+(cS-3)*cS*cS;localSet := [id];initialL := 0;finalL := 0;done := false;if not (transIntChange(myT1) in localSet) thenbeginlocalSet := localSet + [transIntChange(myT1)];finalL := finalL + 1;local[finalL].nr := myT1;local[finalL].label := ’*’;end;

if not (transIntChange(myT2) in localSet) thenbeginlocalSet := localSet + [transIntChange(myT2)];


finalL := finalL + 1;local[finalL].nr := myT2;local[finalL].label := ’1’;end;

done := (initialL = finalL);while not done dobegininitialL := finalL;for i := 1 to initialL dofor j := 1 to initialL dobegincurrentT := Compose2Trans(local[i].nr, local[j].nr);if not (transIntChange(currentT) in localSet) thenbeginlocalSet := localSet + [transIntChange(currentT)];finalL := finalL + 1;local[finalL].nr := currentT;local[finalL].label := local[i].label + local[j].label;end;end;

done := (initialL = finalL);end;

cardTX := finalL;end;

7.5 Computational Complementarity Statistics

The experimental results displayed in Sections 7.2 and 7.3 can be visualized in thefollowing four graphics. Figures 7.6, 7.7 and 7.8 deal with the reversible case whileFigure 7.9 is devoted to non-reversible automata.

400

300

200

100

0

Nu

mber of

Auto

mata Monoid Size

1 1 0 2 0 3 0 4 0 5 0 5 9

Reversible Automata

Prop A CI CII

200

100

0

Nu

mber of

Auto

mata

Monoid Size6 0 7 0 8 0 9 0 100 110 119

Reversible Automata

Prop A CI CII

7.4 Computational Complementarity Statistics 75

Figure 7.1: Reversible Automata (size 1-59)


9 0

6 0

3 0

0

Nu

mber of

Auto

mata

Monoid Size

120 130 140 150 160 170 180

Reversible Automata

Prop A CI CII

300

250

200

150

100

5 0

0

Nu

mber of

Auto

mata

Monoid Size1 1 0 2 0 3 0 4 0 5 0 5 9

Non-Reversible Automata

Prop A CI CII



Figure 7.4: Non Reversible (size 1-59)

Chapter 8

Open Questions

Many problems remain for further work. We mention here only a few of them:

• Explore the relations between universality and minimality for nondeterminis-tic automata.

For deterministic (complete and incomplete) automata there exist close re-lations between universality (defined in this Thesis) and minimality: The-orem 6.8, Corollaries 6.9, 6.10 and 6.12. Are these relations true for non-deterministic automata?

• Find better upper bounds for testing properties B, C.

More precisely, are properties B and C essentially more difficult to test thanproperty A? Considering the automaton presented in Figure 5.12 it seemsthat the answer to this questions is affirmative.

• Describe CI, CII for Mealy automata and for nondeterministic finite au-tomata.

Classically, a Moore–type automaton1 is a particular Mealy–type automaton2.In the theory developed in this Thesis this relation is not true. Indeed, theMoore–type automaton in Figure 8.1 is minimal in the class of Moore-type

1A Moore–type automaton is a 6-tuple (Q,Σ,Ω, δ, µ, q0) where

– Q is a finite set of states,

– Σ is a finite input alphabet,

– Ω is the output alphabet,

– δ : Q× Σ→Q is the transition mapping,

– q0 is an initial sate (from Q) and

– µ : Q→Ω is the output mapping, that is µ(q) gives the output associated with the state q.

2A Mealy-type automaton is a 6-tuple (Q,Σ,Ω, δ, µ, q0) where Q,Σ,Ω, δ and q0 have the samemeaning as in the definition of Moore–type automaton, but µ(q, σ) gives the output associatedwith the state q and the input letter σ, where µ : Q× Σ→Ω.

77

78 8. Open Questions

automata. When considered in the class of Mealy-type automata (see Fig-ure 8.2) it is not a minimal Mealy-type automaton; the equivalent minimalMealy–type automaton, see Figure 8.3, in the larger class of Mealy automata,is not anymore a Moore automaton.

6& %6& %

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a

b

p|0 q|1

Figure 8.1

6& %6& %

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$'p

a|0

a|0 b|1

q

b|1

Figure 8.2

6& %

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b|1a|0

Figure 8.3.

Svozil (see [69]) has shown that Moore’s Theorem holds true for a three-stateMealy automaton. It would be interesting to formalize CI and CII in thiscontext.

• Investigate the influence of the size of the underlying alphabet.

In this Thesis we have explored mainly three and four states automata. Itwould be extremely interesting to study the case of six states automata (asan elementary physical particle is described by six parameters, three positionsand three momenta).

8. Open Questions 79

• Explore the relations between CI, CII and automata/quantum logics.

The automaton logic (see Svozil [69]), a logic very close to quantum logic, de-scribes the logical relations determined by running experiments on automata.Can CI and CII be expressed in terms of the automaton logic?

• Investigate the distribution of CI and CII in relation to the number of 1’s inthe range of the output functions.

The simplest automata have the same input and output alphabet: Σ = O =0, 1. Preliminary experiments indicate that the distribution of 1’s in therange of the output function plays a crucial role in determining CI, CII.

• Intuitively, “reversibility” should exclude both CI and CII. The definitionused in this Thesis is compatible with both complementarity principles, sothe definition used may not be adequate. An alternative definition is thefollowing. An automaton A is reversible if the function Γ : SA × Σ→SA × Σ,given by Γ(q, σ) = (δA(q, σ), FA(q)) is injective (equivalently: surjective). Isthis reversibility notion incompatible with CI and CII?

80 Bibliography

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