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

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 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 □

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: 1.If then the identity on X is the unique embedding of X into Y that is the identity on Y 2.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

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: 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

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 x єHF(U) then πx={πy:y єx} 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 єS 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: 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

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 premutation ρ 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

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: Hanc marginis exiguitas non caperet..

39. Fin Questions?