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
Download Presentation The PPT/PDF document "Background, Reserve & Gandy Machines" is the property of its rightful owner. Permission is granted to download and print the materials on this web site for personal, non-commercial use only, and to display it on your personal computer provided you do not modify the materials and that you retain all copyright notices contained in the materials. By downloading content from our website, you accept the terms of this agreement.
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?