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Decidable and undecidable problems deciding regular languages and CFL’s Undecidable problems

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Deciding CFLs Useful to have an efficient algorithm to decide whether string x is in given CFL –e.g. programming language often described by CFG. Determine if string is valid program. Can simulate NPDA, but this takes exponential time in the worst case.

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Deciding CFLs Convert CFG into Chomsky Normal form. parse tree for string x generated by nonterminal A: A BC x If A k x (k > 1) then there must be a way to split x: x = yz A → BC is a production and B i y and C j z for i, j < k yz

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Deciding CFLs An algorithm: IsGenerated(x, A) if |x| = 1, then return YES if A → x is a production, else return NO for all n-1 ways of splitting x = yz for all ≤ m productions of form A → BC if IsGenerated(y, B) and IsGenerated(z, C) return YES return NO worst case running time?

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Deciding CFLs worst case running time exp(n) Idea: avoid recursive calls –build table of YES/NO answers to calls to IsGenerated, in order of length of substring –example of general algorithmic strategy called dynamic programming –notation: x[i,j] = substring of x from i to j –table: T(i, j) contains {A: A nonterminal such that A * x[i,j]}

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Deciding CFLs IsGenerated(x = x 1 x 2 x 3 …x n, G) for i = 1 to n T[i, i] = {A: “A → x i ” is a production in G} for k = 1 to n - 1 for i = 1 to n - k for k splittings x[i, i+k] = x[i,i+j]x[i+j+1, i+k] T[i, i+k] = {A: “A → BC” is a production in G and B T[i,i+j] and C T[i+j+1,i+k] } output “YES” if S T[1, n] else output “NO”

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Emptiness problem for CFL’s L = { | L(G) is empty } We can show that L is decidable using the following algorithm: Input: A CFG G. Output: ‘yes’ if L(G) is empty, ‘no’ else. N1 = { }; N2 = { A | A -> w is a rule in G for some w in * } While (N1 != N2) { N1 = N2; N2 = { A | A -> for some in ( U N1)* } } if (S is included in N1) output yes else output no;

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Problems related to CFG Equivalence problem for CFG Given two CFG’s G1 and G2, determine if L(G1) = L(G2). It turns out that this problem is undecidable. This means, it is impossible to design an algorithm to solve this problem.

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13 Countable and Uncountable Sets Nthe natural numbers N = {1,2,3,…} are countable Definition: a set S is countable if it is finite, or it is infinite and there is a bijection N f: N → S

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14 Countable and Uncountable Sets Theorem: the positive rational numbers N Q = {m/n : m, n N } are countable. Proof: 1/1 1/2 1/3 1/4 1/5 1/6 … 2/1 2/2 2/3 2/4 2/5 2/6 … 3/1 3/2 3/3 3/4 3/5 3/6 … 4/1 4/2 4/3 4/4 4/5 4/6 … 5/1 … …

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Countable and Uncountable Sets R Theorem: the real numbers R are NOT countable (they are “uncountable”). How do you prove such a statement? –assume countable (so there exists bijection f) –derive contradiction (some element not mapped to by f) –technique is called diagonalization (Cantor)

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16 Countable and Uncountable Sets Proof: R –suppose R is countable R –list R according to the bijection f: nf(n) _ 13.14159… 25.55555… 30.12345… 40.50000… …

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17 Countable and Uncountable Sets Proof: R –suppose R is countable R –list R according to the bijection f: nf(n) _ 13.14159… 25.55555… 30.12345… 40.50000… … set x = 0.a 1 a 2 a 3 a 4 … where digit a i ≠ i th digit after decimal point of f(i) (not 0, 9) e.g. x = 0.2312… x cannot be in the list!

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18 non-RE languages Theorem: there exist languages that are not Recursively Enumerable. Proof outline: –the set of all TMs is countable –the set of all languages is uncountable –the function L:{TMs} →{languages} cannot be onto

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19 non-RE languages Lemma: the set of all TMs is countable. Proof: –each TM M can be described by a finite-length string N –can enumerate these strings, and give the natural bijection with N

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20 non-RE languages Lemma: the set of all languages is uncountable Proof: –fix an enumeration of all strings s 1, s 2, s 3, … (for example, lexicographic order) –a language L is described by its characteristic vector L whose i th element is 0 if s i is not in L and 1 if s i is in L

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21 non-RE languages –suppose the set of all languages is countable –list characteristic vectors of all languages according to the bijection f: nf(n) _ 10101010… 21010011… 31110001… 40100011… …

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22 non-RE languages –suppose the set of all languages is countable –list characteristic vectors of all languages according to the bijection f: nf(n) _ 10101010… 21010011… 31110001… 40100011… … set x = 1101… where i th digit ≠ i th digit of f(i) x cannot be in the list! therefore, the language with characteristic vector x is not in the list

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23 The Halting Problem Definition of the “Halting Problem”: HALT = { : TM M halts on input x } HALT is recursively enumerable. – proof? Is HALT decidable?

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24 The Halting Problem Theorem: HALT is not decidable (undecidable). Proof: –Suppose TM H decides HALT –Define new TM H’: on input if H accepts > then loop if H rejects > then halt

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25 The Halting Problem Proof: –define new TM H’: on input if H accepts > then loop if H rejects > then halt –consider H’ on input : if it halts, then H rejects >, which implies it cannot halt if it loops, then H accepts > which implies it must halt –contradiction.

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26 The Halting Problem Turing Machines inputs Y n Y n n Y n YnYYnnY H’ : box (M, x): does M halt on x? The existence of H which tells us yes/no for each box allows us to construct a TM H’ that cannot be in the table.

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27 So far… Can we exhibit a natural language that is non-RE? regular languages context free languages all languages decidable RE {a n b n : n ≥ 0 } {a n b n c n : n ≥ 0 } some language HALT

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28 RE and co-RE The complement of a RE language is called a co-RE language regular languages context free languages all languages decidable RE {a n b n : n ≥ 0 } {a n b n c n : n ≥ 0 } some language HALT co-RE

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29 RE and co-RE Theorem: a language L is decidable if and only if L is RE and L is co-RE. Proof: () we already know decidable implies RE –if L is decidable, then complement of L is decidable by flipping accept/reject. –so L is in co-RE.

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30 RE and co-RE Theorem: a language L is decidable if and only if L is RE and L is co-RE. Proof: () we have TM M that recognizes L, and TM M’ recognizes complement of L. –on input x, simulate M, M’ in parallel –if M accepts, accept; if M’ accepts, reject.

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31 A natural non-RE language Theorem: The complement of HALT is not recursively enumerable. Proof: –we know that HALT is RE –suppose complement of HALT is RE –then HALT is co-RE –implies HALT is decidable. Contradiction.

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Summary Main point: some problems have no algorithms, HALT in particular. regular languages context free languages all languages decidable RE {a n b n : n ≥ 0 } {a n b n c n : n ≥ 0 } some language HALT co-RE co-HALT

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