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9. Undecidability

9. Undecidability. CIS 5513 - Automata and Formal Languages – Pei Wang. What can a computer do?. What language can be defined by any computational device whatsoever?

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9. Undecidability

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  1. 9. Undecidability CIS 5513 - Automata and Formal Languages – Pei Wang

  2. What can a computer do? What language can be defined by any computational device whatsoever? “… recognizing the strings in a language is a formal way of expressing any problem, and solving a problem is a reasonable surrogate for what it is that computers do.”  textbook page 315

  3. Problems and languages (review) In mathematics and computer science, a “problem” is often defined as a function from input to output In principle, a “function” problem can be converted into a “decision” problem that only has “yes/no” answers So to solve a problem means to decide whether a given string belongs to a language

  4. Decidable problems A problem is decidable if and only if there is an algorithm that always tells correctly whether an instance of the problem has answer “yes” or “no”; If no such algorithm exists, the problem is undecidable Decidability corresponds to a universal proposition on problem instances, while undecidability to an existential proposition

  5. Decidability and computability A function is computable if and only if there is an algorithm (or TM) that computes the output of the function with any valid input; If no such algorithm exists, the function is uncomputable The computability of a functionproblem can be converted to the decidability of a corresponding decisionproblem

  6. Recursive enumerability A language is recursively enumerable (RE) if its sentences are accepted by a TM A language L is recursive if a TM not only accepts its sentences, but also rejects its non-sentences in finite steps, that is, both L and its complement are RE, and L is decidable If a language L is not recursive, then it is undecidable

  7. Reduction and decidability A problem P1 is reducible to a problem P2 if an algorithm for solving problem P2 could also be used to solve problem P1 “P1 is reducible to a problem P2” means P1 corresponds to a special type of P2. In this case, if P2 is decidable, so is P1; if P1 is undecidable, so is P2 A problem’s undecidability is usually proved by reducing a known undecidable problem to it

  8. Undecidable problems A problem is undecidable if there cannot be any algorithm (with guaranteed halting) for it That is different from a problem for which no algorithm has been found Intuitively, since the algorithms (TMs) are countable, but the problems (languages) are uncountable, there are undecidable problems Countability and Cantor's diagonal argument

  9. A binary word as a TM Every TM can be coded as a finite binary string Every finite binary string can be interpreted as a TM according to the above coding method, or as a default TM that accepts { } So all finite binary strings can be enumerated in a table, in which each entry corresponds to (1) a word, (2) a TM, and (3) a RE language All TMs are in this infinitely long table

  10. A language that is not RE In the table of {0, 1}*, for every index i, word wi can be seen as a TM Mi, which accepts L(Mi) Define Ld = {wi | wi is not in L(Mi)} If there is a j such that Ld = L(Mj) Then wj is in L(Mj)  wj is in Ld  wj is not in L(Mj) So Ld cannot be in the table, so cannot be RE

  11. A RE language that is not recursive The universal language Lu = {(M, w) | w is in L(M)} is RE, as it can be accepted by the universalTMU, which simulates M’s reaction to w Lu is not recursive, otherwise its complement Lucwould be recursive, too, but Ldcan be reduced to Luc by mapping its element w into (w, w) Membership of a RE language is undecidable

  12. Language hierarchy with examples Regular: {0*1*} Context-free: {0n1n} Recursive (decidable): {0n1n2n} Recursively Enumerable: Lu = {(M, w) | w is in L(M)} Non-RE: Luc = {(M, w) | w is not in L(M)}

  13. Undecidable problems of RE Every nontrivial (i.e., neither none nor all) property of RE languages is undecidable, e.g. • Whether a RE language is empty • Whether a RE language is finite • Whether a RE language is regular • Whether a RE language is context-free Proof: Lucan be reduced to each of them Whether a CFG is ambiguous is undecidable, too

  14. The halting problem If all problems are decidable, then there is a program P(x, y) that can judge whether any program x halts on any input y In that case another program Q(x) can be built to halt if and only if P(x, x) returns No However, Q(Q) halt  P(Q, Q) returns No  Q(Q) does not halt Therefore, no such a P(x, y) can exist

  15. Motivation and implication Hilbert’s program: to find axioms for all mathematics, and algorithms to decide the truth-value of any mathematical statement Gödel's incompleteness theorem: no consistent system of axioms can prove all truths about the relations of natural numbers Turing's undecidability theorem: no general algorithm can solve the halting problem for all possible program-input pairs

  16. Self-reference with negation • Liar paradox • Cantor's diagonal argument • Russell's paradox • Gödel's incompleteness theorem • Turing’s halting problem For a general discussion, see Gödel, Escher, Bach: An Eternal Golden Braid

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