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CS 3240 – Chapter 11 They may not halt on every possible input! And not just because the creator of a specific TM was a doofus This is related to the major mathematical/computational discovery of the 20th century! There are propositions that cannot be decided (“proven”) A question is decidable if there is a TM that always halts and answers “yes” or “no” for each possible input The TM therefore constitutes an algorithm A language is decidable if there is a TM that always halts and answers “accept” or “reject” whenever an input string is in the language or not aka “Turing acceptable” A function is computable if there is a TM that always halts with the appropriate output for each possible input in the function domain aka “Total function” Let g(x,y) be some computable function Let f(x) = the smallest p where g(x, p) = 1, or 0, if such a p does not exist “Pseudo-algorithm” for f(x): m = 0; while (g(x, m) != 1) ++m; cout << m; If there is no m for a given g and x, then we hang! 3 possibilities in general when a TM processes an input string: Accepts (goes to an accepting halt state) Rejects (e.g., crashes, or gives a “no” answer) Hangs (infinite loop) A language for which there is a TM that always halts for and accepts strings in the language is recognizable “It knows one when it sees one” :-) But it may hang on strings not in the language A function that is defined for only some of its domain elements is a partial function It may hang (or “blow up”) on some inputs e.g., divide by zero This is the computational analogue to a recognizable language you’ll always get an answer with a valid input it may hang on invalid input Language Machine Grammar Regular Finite Automaton Regular Expression, Regular Grammar Context-Free Pushdown Automaton Context-Free Grammar Recursively Turing Machine Unrestricted Phrase- Enumerable Structure Grammar CS 3240 - Introduction 9 A language L over the alphabet is called recursively enumerable (aka “recognizable”) if there is a TM T that accepts every word in L and either rejects (crashes) or loops forever for every word in the language L', the complement of L: accept(T) = L reject(T) + loop(T) = L' recursively enumerable (r.e.) = recognizable We just saw one (slide 5) Let g(x,y) be some computable function Let f(x) = the smallest p where g(x, p) = 1, or 0, if such a p does not exist Let Lg be the set of functions, f, corresponding to all computable functions g, as explained above Such functions can be encoded as strings, and are therefore countable Lg can then be seen as a language (a set of strings) A language L over the alphabet is called recursive (aka “decidable”) if there is a TM T that accepts every word in L and rejects every word in L'; that is, accept(T) = L reject(T) = L' A TM that accepts a recursive set is a model of an algorithm. It always halts. recursive = decidable There exist languages that are r.e. but not recursive We just saw one There exist languages that aren't even r.e.! (You'll see one soon) All are “contrived” Languages generated by grammars are r.e. or “better” All r.e. languages are closed under union, intersection, concatenation, and Kleene* Everything but complement! Recursive languages are also closed under complement Also: If L and L' are r.e., then L is recursive Here come the Proofs… Let M be a machine that decides a recursive language, L Form the machine M' by inverting the acceptability output of M Goes to a reject state instead Then M' decides L' So L' is recursive Suppose L and L' are both r.e. Let M recognize L, and M' recognize L' M may hang on elements of L', but M' doesn't Form a new machine, M* that calls M and M' in parallel (non-deterministically) If M accepts w, so does M* If M' accepts w, reject w There are no other possibilities! (No hanging) Therefore, L is decidable/recursive, by definition TMs can recognize/accept strings from certain languages and/or compute functions If there is a TM, M, that accepts a language, L, and M always halts, then L is recursive If there is no such M for L, but there is instead a machine M that accepts every string in L, but M may hang on strings not in L, then L is recursively enumerable The complement of a recursive language is recursive In fact, recursive languages are closed under all operations, like regular languages are r.e. languages are closed under intersection The complement of a r.e. language may not be r.e. But if it is, then both languages are actually recursive! A Timely Interlude The real numbers in (0,1) are uncountable They cannot be mapped in a 1-to-1 fashion to the counting numbers Proof: Assume they can be: r0, r1, r2, … Arrange their digits in a table a[ ][ ] ▪ each row, a[i], contains the digits or ri The diagonal sequence (a[n][n]+1) mod 10, representing a valid real number, is not in the table! Contradiction! The power set of the natural numbers, N the set of all subsets (2N) Suppose it is countable The we can enumerate the sets: p0, p1, p2, … Now consider the set T = {i | i ∉ f(i) = pi } Certainly T is a set of integers, so T ∈ P(N) ▪ Call it pk Question: Is k in T = pk? The power set of any countably infinite set is uncountable Much bigger than a countable set! The number of countable sets is negligible compared to the number of uncountable sets Well? A language over an alphabet Σ is a subset of Σ* the latter being an infinite set The set of all languages over Σ is therefore the power set of Σ* (2∑*) which we just showed is uncountable So… the number of languages over any finite alphabet is uncountable The # of TMs are countable, but the # of languages is not Therefore, some languages cannot be recognized by a TM There aren't enough TMs to go around! Just like there are more reals than integers So, non-r.e. languages must exist! Just take the complement of any r.e. language that is not recursive Example (page 279): Consider all TMs, Mi with alphabet Σ={a}. Let X = {ai: ai ∈ L(Mi)} ▪ This is r.e., but not recursive ▪ Because we can construct a TM that carries out the computation, but it may not halt when ai ∉ L(Mi) Then X’ must be non-r.e.! Suppose X’ is r.e. Then there is a TM, Mk, that recognizes it Now ask the question, is ak ∈ X’ ? If it is, this means ak ∉ L(Mk) = X’, by definition Contradiction! If it isn’t, then this means that ak ∈ L(Mk) = X’ Again, by definition Again, a contradiction! ⇒ X’ is not r.e. What we have just described is the membership problem: “Given a r.e. language, L, and a string, w, is w in L?” We have just shown that the membership problem is, in general, undecidable Language Machine Grammar Regular Finite Automaton Regular Expression, Regular Grammar Context-Free Pushdown Automaton Context-Free Grammar Recursively Enumerable Turing Machine Unrestricted Grammar CS 3240 - Introduction 29 Left-hand side of the rule is a concatenation of one or more symbols There must be at least one variable on the left ▪ λ is not allowed on the left Any string is allowed on the right, including λ aka “Type 0” Grammar S → aAbc | abc | λ S ⇒aAbc Ab → bA ⇒abAc Ac → Bbcc ⇒abBbcc bB → Bb ⇒aBbbcc aB → aa | aaA ⇒aabbcc An A is created as the left-most variable. It travels to the right until it finds a 'c', then creates a new b and c, and becomes a B. The B moves back to create the extra needed a. The last rule allows the option to do it all over again (by introducing another A). This is similar to what the Turing machine for the same language does, except the TM marks instead of generates the letters. We have already seen S → XYS | λ CFGs for this XY → YX This unrestricted YX → XY grammar: X→a Introduces X’s and Y’s Y→b together Allows moving X’s and Y’s arbitrarily by swapping them Cannot use the decidability algorithms for CFGs e.g., CYK algorithm does not apply No “Normal Form” Non-null productions may create shorter strings Terminals can disappear! For every r.e. language, there is an unrestricted grammar that generates it For every unrestricted grammar, there is a TM that recognizes its language It may or may not decide it We will not prove this but the book does Like Unrestricted Grammars except: Right-hand side must be no shorter than left-hand side ▪ So strings never shrink Hence the name “non-contracting” or “monotonic” Cannot contain λ anywhere Context-sensitive languages don't need unlimited memory Since intermediate sentential forms never shrink, the largest memory requirement is proportional to |w| Accepting machine: Linear Bounded Automaton S → aAbc | abc Ab → bA Ac → Bbcc bB → Bb aB → aa | aaA Every context-sensitive language is recursive there is a TM that accepts it (i.e., always halts) But not all recursive languages are context- sensitive So Context Sensitive languages form a proper subset of Recursive Languages Just like Recursive languages are a proper subset of Recursively enumerable languages Type 0 grammar Type 1 grammar Type 2 grammar Type 3 grammar Language Grammar Machine Recursively Enumerable Unrestricted Phrase TM (that may not halt) (“Turing Recognizable”) Structure (Type 0) Recursive " TM (that always halts) (“Turing Decidable”) Context-Sensitive Context Sensitive (Type 1) Linear-bounded automata (monotonic/non- (bounded TM) contracting) Context Free Context Free (Type 2) PDA Deterministic Context Free " DPDA Regular Regular (Type 3) FA (Right-linear or Left-linear)