Background, Reserve & Gandy

MachinesAndres Blass & Yuri Gurevich

The ReserveAn 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 it

In application it is convinient to define structure ahead of time and we will focus on this in this talk

Introduction2 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 created

Example:In a Turing machine:

◦First view: it has infinite tape◦Second view: tape is finite but it’s size

can be increased at any time

In ASMsWe adopt the first view for an infinite

ReserveReserve elements as input to functions

(except equality) result in default values. No reserve elements are outputed

It 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 it

Introducing Backgrond class of structuresExists above the set of Atoms

without imposing structure on them

Specifies the constructions (like finite sets) available for the algo

Importing from the Reserve:◦Non- deterministically using

Inessential non- determinism◦Using Gandy- style determinism

What is the Relationship between

these two methods?

StructuresSyntaxFirst order logic structure

includes:◦vocabulary: finite collection of

function names, can be marked static

◦=, TRUE, FALSE, undef, Boole(), operators

◦Terms- defined by induction, choosing the Nullary functions for the base case

StructuresSemantics (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 equal

SS j

Sequential time & AS postulatesFor 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 structuresAll states share the same vocabularyIf then they share the same base set

and the basic functionsThe map works identically under isom.Conclusion: Symmetry preservation: Every

automrphism of X is autom. of

)()(:A ASAS

)()( ASASA

AXX )',(

A)(XA

Hereditarily Finite SetsDefenitions: a set X is transitive if it contains all elements of it’s elements TC(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 U

Corollary: 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 □

iUi

ii

i UHFUHF )()(

iUi

iUxSup )(

Background ClassesDefinitions: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 itself

For 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 Y

Background ClassesK is a background class if (def. 4.1):BC0 K is closed under isomorphismBC1 For every set U, there’s a

structure XєK with Atoms(X)=UBC2 For all X,Y єK and for every

embedding of sets there’s a unique embedding of structures that extends

BC3 For all XєK and every xєBase(X) there is a smallest K-substructure of X that contains x

)()(: YAtomsXAtoms

YX :

The EnvelopeK is a background class, XєK, 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 xєX 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 finite

)(SEX)(SSupX

)( xEX

AnalysisLemma 4.4: In BC2, if ζ is onto then so is ηProof:

Lemma 4.5: Suppose Z is a background structure, X,Y are background substructures of Z, U=Atoms(X), V=Atoms(Y). Then:1. If then the identity on X is the unique

embedding of X into Y that is the identity on Y2. 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.

VU

VU

onto identity is uniqunessBy . toextends (identity) X.Y: toextends 11

AnalysisLemma 4.6: In a background structure X,

every set U of atoms has an envelope◦ Proof: By BC1, exists Y, Atoms(Y)=U◦ The identity on U extends to◦ By BC0 Z is a background structure◦ By 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

a

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

)(,: YZXY

Counter exampleLet 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 elements◦Otherwise 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 envelope

Alternative requirementsAlternative requirements for BC3: Lemma 4.8(BC3’): In a background structure

X, every has an envelopeProof: Define . By 4.6, U has

an envelope. This is also S’s envelope.Lemma 4.9(BC3’’): For all XєK, 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

)(XBaseS

}:})({{ SxxSupU

)(UEF

Alternative requirementsCorollary: For all i, let ,

then Lemma 4.11: In defenition 4.1, BC3

can be replaced with BC3’’Proof: Assume BC3’’ and let XєK,

xєX.◦Let F be the collection of substructures

Y of X that contain x◦By BC3’’ , which is clearly the

smallest K-substructure with x

)(XAtomsU i

)()( i

ii

i UEUE

)(UEF

AND NOW FOR SOME EXAMPLES

Set background (SB)The non-logic part: the hereditarily

finite sets over the atoms UNon- obligatory basic function: єOther optional vocabulary elements:1. , Singleton(x)={x}, BinaryUnion(x,y)2. , 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??

Set background refinedIn 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 constraints

}3,2,1{}2,1{:

Set background refinedSolution?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 embedding

String backgroundThe 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- generated

List backgroundNon- logic part: lists over UDiffers from strings in that

nesting is allowedBasic functions: Nil (empty list),

Append(x,y), optional: Head, TailExplictly atom- generated

Set/ List backgroundNon- logic part is the least set V s.t.

◦ ◦

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

Example of a non- finitary background class: take the string class allowing infinite strings this time

VU VxxVxxVxxn nnn ,...,,},...,{,...,, 111

Background structures & the ReserveFix 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 structure

0

0

0

00

Even more definitionsThe basic functions of A with names in

are the background basic functionsOther 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 part

0

AnalysisLemma 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 structureRemark: Any isomorphism between states

X and Y gives rise to an isomorphism between Active(X) and Active(Y)

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

A

A 0

0

'''': YYX

Application:

Nondeterministic choice problem and Gandy machines

Gandy machinesFor 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:yєx}

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 xєS and for every perm. π there’s a perm. ρ of U that pointwise fixes Sup({πx}) and ρπFx=Fπx

),,)(( UUHFUG

Lemma and main definitionLemma 8.2:

1. A structural function (SF) F over a structural set S extends to a SF over HF(U)

2. 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.:

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

2. F is a SF from S into S (the one step transformation function)

3. Some additional constraints irrelevant to us

ExampleM=(S,F), π a permutation of U

◦S 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 πb to c and leave everything else intact

◦Thus, M satisfies both requirements

},,...,{},,...,{

},...,{

1

1

1

caaxFbaaFx

aax

n

n

n

}{)( axxF

The Nondeterministic choice problemThink of an arbitrary Gandy machine

M=(S,F), xєS being the “current state” and Fx the next state of M

It 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?

Some answers ahead...We claim that the answer to the

second question is positive by proposing a nondeterministic formalization to Gandy machines

If 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 earlier

Nondeterministic specificationS is structural subset of HF(U), F:S->S a unary operation over S.

Define a nondeterministic algorithmAll states of A have the same base

set and also TRUE, FALSE and UNDEF

Non- 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))

MA

)(UHFU

A

In our exampleSup(X)=Core(X) for all statesFor 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 atom

Any atom from U-Sup(X) will do!The Algorithm is completely

oblivious to which atom is chosen

AnalysisThe 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, UNDEFHence, 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))

A

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 nondeterministic◦F is structural over S

Proof: Hanc marginis exiguitas non caperet..

FinQuestions?