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Background, Reserve & Gandy Machines. Andres Blass & 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 ” – a set with no structure on it

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Background, Reserve & Gandy Machines

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    1. Background, Reserve & Gandy Machines Andres Blass & Yuri Gurevich

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

    3. 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 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

    4. In ASMs • We adopt the first view for an infinite Reserve • Reserve 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

    5. Introducing Backgrond class of structures • Exists 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 ?

    6. Structures Syntax • First 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

    7. 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 equal

    8. Sequential time & AS postulates For an algorithm A we define: • A set S(A)- states of A • A 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

    9. Hereditarily Finite Sets Defenitions: • a set X is transitiveif 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 Uis 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 □

    10. Background Classes Definitions: • The predicate Atomic() and the logical symbols are called obligatory • X 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

    11. 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 XєK with Atoms(X)=U • BC2 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

    12. The Envelope • K 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

    13. 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.

    14. Analysis • Lemma 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 Does BC3 follow from BC0- BC2, and therefore can be dropped?

    15. 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 elements • Otherwise there’s exactly one non- logic element and the atoms • K 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

    16. 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 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

    17. 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 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

    18. And now for some examples

    19. Set background (SB) • The non-logic part: the hereditarily finite sets over the atoms U • Non- obligatory basic function: є Other 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??

    20. 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 constraints

    21. 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 embedding

    22. 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- generated

    23. 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, Tail • Explictly atom- generated

    24. Set/ List background • Non- 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

    25. 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 structure

    26. Even more definitions • The basic functions of A with names in are the background basic functions • Other functions will be called the foreground basic functions • Def(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

    27. 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)

    28. 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

    29. Application: Nondeterministic choice problem and Gandy machines

    30. 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: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

    31. 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 Sinto S (the one step transformation function) • Some additional constraints irrelevant to us

    32. Example M=(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 premutationρ will transpose πb to c and leave everything else intact • Thus, M satisfies both requirements

    33. The Nondeterministic choice problem • Think 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?

    34. 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

    35. 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 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))

    36. 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 atom • Any atom from U-Sup(X) will do! • The Algorithm is completely oblivious to which atom is chosen

    37. 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))

    38. 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: Hancmarginisexiguitas non caperet..

    39. Fin Questions?