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Background, Reserve & Gandy Machines Background, Reserve & Gandy Machines

Background, Reserve & Gandy Machines - PowerPoint Presentation

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Background, Reserve & Gandy Machines - PPT Presentation

Andres Blass amp Yuri Gurevich The Reserve An ASM requires a Reserve of elements that can be imported into the set of currently used elements In previous talks this was defined as a naked set ID: 316788

background set structure atoms set background atoms structure reserve function elements sets active lemma states functions atom structural finite

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Slide1

Background, Reserve & Gandy Machines

Andres Blass & Yuri

GurevichSlide2

The Reserve

An ASM requires a Reserve of elements that can be imported into the set of currently used elements

In previous talks this was defined as a “

naked set” – a set with no structure on itIn application it is convinient to define structure ahead of time and we will focus on this in this talkSlide3

Introduction

2 ways for an algorithm to increase it’s working space:

The space was there all along and just wasn’t used

A genuinely new space is createdExample:In a Turing machine: First view: it has infinite tape

Second view: tape is finite but it’s size can be increased at any timeSlide4

In ASMs

We adopt the

first view

for an infinite ReserveReserve elements as input to functions (except equality) result in default values. No reserve elements are outputedIt is often desirable to define a structure on the reserve- This way when a new element is imported, sets that involve it already exist

and there’s no need to define

є

on itSlide5

Introducing Backgrond class of structures

Exists

above the set of Atoms

without imposing structure on themSpecifies the constructions (like finite sets) available for the algoImporting

from the Reserve:

Non- deterministically using Inessential non- determinism

Using Gandy- style determinism

What is the Relationship between these two methods ?Slide6

Structures

Syntax

First order logic structure includes:

vocabulary: finite collection of function names, can be marked static=, TRUE, FALSE, undef, Boole(), operatorsTerms- defined by induction, choosing the Nullary functions for the base caseSlide7

Structures

Semantics (meaning)

A structure X for a vocabulary

γ:Non- empty set S: the base set of X

Interpretations of function names in S

Nullary functions identified with its value, true ≠ false

A j-ary function f:

Domain and range defined to be all elements different from undef

Val(t,X) = Val(t’,X) implies that the interpretations of t and t’ in X are equalSlide8

Sequential time & AS postulates

For an algorithm A we define:

A set

S(A)- states of AA set I(A)

of S(A)- initial states of A

For

deterministic

: a map called the one- step transformation

For

non- deterministic

: a map - all possible transforms

States of A are first order structures

All states share the same vocabulary

If then they share the

same base set

and the basic functions

The map works identically

under isom

.Conclusion: Symmetry preservation: Every automrphism of X is autom. of Slide9

Hereditarily Finite Sets

Defenitions:

a set X is

transitive if it contains all elements of it’s elementsTC(x) is the least transitive set containing x

A set x is

hereditarily finite

if TC(x) is finite

Elements of x which are not sets are called

atoms

The collection of atoms in TC(x) is called the

atomic support

of x or

Sup(x)

Let U be a set of atoms, then

HF(U)

or the

hereditarily finite sets over U

is the collection of all HF sets x such that Sup(x) is in U. Those are all the HF sets which contain atoms only from UCorollary: For a family of subsets of U we have:

Proof: A set x belongs to the intersection if Sup(x) is in for all i, which is the same as saying that

□Slide10

Background Classes

Definitions:

The predicate Atomic() and the logical symbols are called

obligatoryX is explictly atom- generated if the smallest sub- structure of X that includes all atoms is X itselfFor 2 structures X, Y of the same vocabulary

X

≤Y

means that X is a

sub- structure

of Y

If X also has property K we say that X is a

K- substructure

of YSlide11

Background Classes

K is a

background class

if (def. 4.1):BC0 K is closed under isomorphism

BC1

For every set U, there’s a structure X

є

K with Atoms(X)=U

BC2

For all X,Y

є

K and for every embedding of sets there’s a unique embedding of structures that extends

BC3

For all X

є

K and every x

є

Base(X) there is a smallest K-substructure of

X that contains xSlide12

The Envelope

K is a background class, X

є

K, S is in Base(X). Let F be the set of substructures Y≤X such that Y belongs to K and Includes S. The smallest member of F, if exists, is called the envelope of S in X and Atoms(Y) is the support

of S in X

i.e., the smallest background sub- structure of X containing x

є

X is

By

BC3

, every singleton subset has an envelope

Definition

: a background class K is

finitary

if in every background structure the support of every sigleton set is finiteSlide13

Analysis

Lemma 4.4

: In

BC2, if ζ is onto then so is η

Proof

:

Lemma 4.5

: Suppose Z is a background structure, X,Y are background substructures of Z, U=Atoms(X), V=Atoms(Y). Then:

If then the identity on X is the unique embedding of X into Y that is the identity on Y

If then X<Y (X sub-structure of Y)

Proof

: 1 follows from a similar uniqueness argument. Since the identity is an embedding, 2 follows.Slide14

Analysis

Lemma 4.6

: In a background structure X, every set U of atoms has an envelope

Proof: By BC1, exists Y, Atoms(Y)=UThe identity on U extends toBy BC0 Z is a background structureBy Lemma 4.5 it includes every other that includes U, thus it is the smallest-> envelope

Corollary

: has an envelope for every atom a. This is weaker than

BC3

, which requires that every singleton subset has an envelope

Does BC3 follow from BC0- BC2, and therefore can be dropped?Slide15

Counter example

Let K be the class of structures X satisfying the following:

TRUE, FALSE and UNDEF are distinct

If Atoms(X) is non- empty then X contains no non- logic elementsOtherwise there’s exactly one non- logic element and the atomsK satisfies

BC0- BC2

but if there’s more than 1 atom and x is the unique non- logic non- atomic element then {x} doesn’t have an envelopeSlide16

Alternative requirements

Alternative requirements for

BC3

: Lemma 4.8(BC3’): In a background structure X, every has

an envelope

Proof

: Define . By 4.6, U has an envelope. This is also S’s envelope.

Lemma 4.9(BC3’’)

: For all X

є

K, the intersection of any family of K-substructures of X is a K-substructure of X

Proof

: For a family F, we show that , U being the intersection of all atoms in F Slide17

Alternative requirements

Corollary

: For all i, let , then

Lemma 4.11: In defenition 4.1, BC3 can be replaced with BC3’’Proof

: Assume BC3’’ and let X

є

K, x

є

X.

Let

F be the collection of substructures Y of X that contain

x

By

BC3’’ , which is clearly the smallest K-substructure with xSlide18

And now for some examplesSlide19

Set background (SB)

The non-logic part: the hereditarily finite sets over the atoms U

Non- obligatory basic function:

єOther optional vocabulary elements: , Singleton(x)={x}, BinaryUnion(x,y)

, Pair(x,y), UnaryUnion(X)- union of elements in x, TheUnique(X)- if X is singelton return it’s value

Both are explictly atom- generated

Can you spot the error in this example??Slide20

Set background refined

In the previous basic definition, the uniqueness requirement fails:

Consider X and Y HF sets over {1,2} and {1,2,3}

repectively

Consider the identity

It can be extended naturally by sending each set to itself

But it can also be extended in many other ways, for example send to {3}

If we add the optional elements the problem doesn’t occur because of the additional constraintsSlide21

Set background refined

Solution?

One possibility is to equip our Set background with enough functions so that the structure is preserved

This may lead to low levels of abstractions, so instead…

We refine our model and specify a special embedding which will be called

standard

. The requirement now is that every embedding has a unique extension to a standard embeddingSlide22

String background

The set of non- logic elements is the set of strings of elements of the set of atoms U

Nil function (empty string)

Unary function to convert atoms to strings

Concatenation of 2 strings

Other optional vocabulary elements:

Head(x), Tail(x)

Also explictly atom- generatedSlide23

List background

Non- logic part: lists over U

Differs from strings in that nesting is allowed

Basic functions: Nil (empty list), Append(x,y), optional: Head, TailExplictly atom- generatedSlide24

Set/ List background

Non- logic part is the least set V s.t.

In this representation, lists (“<>”) and sets (“{}”) are regarded as independent basic constructionsExample of a

non- finitary

background class: take the string class allowing infinite strings this timeSlide25

Background structures & the Reserve

Fix a background class

BC

, vocabulary of BC is the background vocabulary, function names in -

background function names

, members of BC-

background structures

Def(6.1): For an algorithm A, BC is the

background

of A if:

The vocabulary of A, , includes . Every background function is static in

For every state X of A, the - reduct of X (forget functions from - ) is a background structureSlide26

Even more definitions

The basic functions of A with names in are the

background basic functions

Other functions will be called the foreground basic functionsDef(6.2): Let X be a state of A. We call an element x from Base(X) exposed if x belongs to the range of the foreground function or x occurs in a tuple that belongs to the domain of a foreground function

Def(6.3): The

Active

part of a state X is the envelope of the set of exposed elements- Active(X). The

Reserve

is the set of atoms which don’t belong to the active partSlide27

Analysis

Lemma 6.4

: Every permutation of the reserve of X gives rise to a unique automorphism of X that is the identity on the active part of X

Proof: Let π be a permutation of the Reserve.

Extend it to the Active part of X using

π

(x)=x.

Since Active(X) and Reserve(X) are disjoint, this is an

automorphism

of the entire structure

Remark

: Any isomorphism between states X and Y gives rise to an isomorphism between Active(X) and Active(Y)Slide28

Inessential non-determinism (IND)

Def7.1

: For a non- deterministic algorithm A and background BC, A is

inessentially non- deterministic if for all states X of A: If (X,X’) and (X,X’’) belong to , then there is an isomorphism from X’ onto X’’ that coincides with the identity on Active(X)

Corollary7.2

: if (X,X’) and (Y,Y’) belong to , isom. From X to Y, the restriction to Active(x). Then extends to an isom. From X’ to Y’

Proof

: From IND we have Slide29

Application:

Nondeterministic choice problem and Gandy machinesSlide30

Gandy machines

For a fixed infinitely countable set of atoms U, define

Every permutation

π of U naturally extends to an automorphism on G: if xєHF(U) then πx={πy:yє

x}

A subset S of HF(U) is

structural

if it is closed under automorphisms, which in this case means closed under

π as defined above

Def(8.1):

function

F:S->HF(U) is

structural

if for every x

є

S and for every perm.

π there’s a perm.

ρ

of U that pointwise fixes Sup({πx}) and ρ

πFx=FπxSlide31

Lemma and main definition

Lemma 8.2

:

A structural function (SF) F over a structural set S extends to a SF over HF(U)F SF Over HF(U), S structural subset of HF(U). Them the restriction of F|S is a SF over S

Def(8.3):

a

Gandy machine

M is a pair (S,F) s.t.:

S is a structural subset of HF(U) (intuitively: the set of states of M)

F is a SF from

S

into S (the one step transformation function)

Some additional constraints irrelevant to usSlide32

Example

M=(S,F),

π

a permutation of US is the collection of all finite subsets of U. Obviously it is structural , aєU-x (add new element)

So the required

premutation

ρ

will transpose

π

b to c and leave everything else intact

Thus, M satisfies both requirementsSlide33

The Nondeterministic choice problem

Think of an arbitrary Gandy machine M=(S,F), x

є

S being the “current state” and Fx the next state of MIt is possible that Sup({Fx}) has new atoms that are not in Sup({x})The choice of such new atoms shouldn’t matter, thus the structurality requirement

1 .Is the structurality requiremnt correctlly captures this irrelevance?

2. Is there a better solution to the Nondeterministi choice problem?Slide34

Some answers ahead...

We claim that the answer to the

second

question is positive by proposing a nondeterministic formalization to Gandy machinesIf we consider only deterministic machines, the answer to the first question is positive as well: we will see that the structurality requirement is equivalent to inessential non- determinism as defined earlierSlide35

Nondeterministic specification

S is structural subset of HF(U),

F:S->S a unary operation over S. Define a nondeterministic algorithm

All states of A have the same base set and also TRUE, FALSE and UNDEFNon- logic basic functions: Atomic, є and Core(X) that returns M’s current state

consists of pairs (X,Y) s.t. Core(Y)=F(Core(X))Slide36

In our example

Sup(X)=Core(X) for all states

For a particular state X, what are the states Y that A can continue to?

Those are the states Y s.t. Sup(Y)=Sup(X)+{a}, where a is a new additional atomAny atom from U-Sup(X) will do!The Algorithm is completely oblivious to which atom is chosenSlide37

Analysis

The algorithm is a nondeterministic algoritm with background SB (sets)

The only

exposed element of state X of A is Core(X)The active part of X is Sup(X)UHF(Sup(X)) + TRUE,FALSE, UNDEF

Hence,

Reserve

(X)= U- Sup(X)

A

permutation of Reserve

(X) fixes pointwise Sup(X) and agrees with π on Reserve(X)

Corollary 10.3: (X,Y)

є

iff there’s a premutation π of Reserve(X) s.t. Core(Y)=πF(Core(X))Slide38

Inessential nondeterminism and structurality

Let S and F be defined as in the previous section, A the nondeterministic specification. then:

Theorem 11.1

: The following are equivalent:A is inessentially nondeterministicF is structural over S

Proof

:

Hanc

marginis

exiguitas

non

caperet

..Slide39

Fin

Questions?