1 Showing Recognizability

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1 CSCC63 Worksheet Recognizability and Decidability 1 1 Showing Recognizability 1.1 An Example - take 1 Let Σ be an alphabet. L = { M M is a T M and L(M) }, i.e., that M accepts some string from Σ. Prove that L is recognizable. Q. How do we show that a language is recognizable? A. Show that a TM R recognizes it. For R to recognize L, it needs to take it s input M, and run M on every possible string w and check if M accepts w. We use the notation O to indicate a string encoding of an object O. How the object is encoded doesn t really matter as the TM can be programmed to recognize it accordingly. So M means a string representation of a TM M. Let w 1, w 2, w 3,... be an effective enumeration of Σ. Comprehension Check. Do you know how to enumerate Σ? We give a TM R that recognizes L... R = On input M : 1 for i = 1 to 2 run M on input w i 3 if M accepts w i then 4 accept or does it? What might go wrong? A. If M loops on w i then M and therefore R never halt. Q. How can we fix it? A. We limit the number of steps M performs on each iteration - essentially splitting line 2 into two smaller steps. R = On input M : 1 for s = 1 to 2 for i = 1 to s 3 run M on input w i for s steps 4 if M accepts w i within s steps then 5 accept 1 Parts of this worksheet are based upon the notes of Nick Cheng. 1

2 Correctness : If M L, then there are numbers k and t such that M accepts w k in (exactly) t steps. So when s = max(t, k) and i = k, the condition in line 4 will be satisfied. Hence R accepts M as wanted. : If M L then there is no number k such that M accepts w k. So the conditon in line 4 cannot be satisfied. Hence R loops on M as wanted. 1.2 Example - take 2 - for you to read on your own. Let s prove L recognizable again by finding an enumerator. Why can we do this? A. Theorem 3.21 states that a language is recognizable iff some enumerator enumerates it. For a TM M, we need to check if there are numbers k and t such that M accepts w k within t steps. So our enumerator must check these infinite number of cases for an infinite number of TMs. Let M 1, M 2, M 3,... be an effective enumeration of the set of all TM descriptions. Let w 1, w 2, w 3,... be an effective enumeration of Σ. We give an enumerator that enumerates L. E = Ignore input: 1 for s = 1 to 2 for i = 1 to s 3 for j = 1 to s 4 run M i on input w j for s steps 5 if M i accepts w j within s steps then 6 print M i Why does this work? Write an explanation similar to above justifying the above algorithm. 2

3 2 Decidable and Recognization Languages Q. What is the relationship between decidability and recognizability? Theorem 4.22 on pp : Language L is decidable both L and L c are recognizable (L c is complement of L). ( ) This is easy since if L is decidable, then certainly L is recognizable and L is also decidable so L c is recognizable. For ( ), we can construct a machine M 1 that recognizes L and M 2 that recognizes L c. Create TM, M as follows: M = On input w: 1. Run both M 1 and M 2 in parallel. 2. If M 1 accepts w, accept; if M 2 accepts w, reject. The Acceptance Problem for DFAs. Test whether a particular DF A accepts a given string. Define A DF A to be the language that contains the encodings of all DF As together with the strings that they accept. Formally. A DF A = { B, w B is a DF A that accepts input w} Theorem 4.1 A DF A is a decidable language. A TM can check B is valid encoding of DF A (using any reasonable convention), then use tape to keep track of current state of B and position on string w while simulating B s transitions one by one (going back and forth to check description of B). At the end, easy to check if last state is accepting or rejecting. Using conventions above: 3

4 On input B, w : 1. Simulate B on w (use appropriate tape alphabet symbols to keep track of position of B on input w; use separate portion of tape to keep track of B s current state). 2. Accept if B accepts; reject if B rejects. Since a DFA always halts, TM above is a decider for A DF A. Define the following languages: A NF A = { B, w B is an NFA that accepts w} A RE = { R, w R is a RE that can generate w} Q. Are A NF A and A RE recognizable, decidable or neither? A. Decidable. Recall constructions for transforming NFA into corresponding DFA, and RE into corresponding DFA. These constructions are algorithmic so can be carried out by TMs (by Church-Turing thesis). This immediately gives that A NF A and A RE are decidable. Q. Is E DF A = { B B is a DFA that accepts no string } decidable? A. yes. the proof idea is to start at the start state, mark it as visited. Visit all states reachable from the start state (BFS). Mark them. If you even mark an accepting state, reject. If you mark all states and never reach an accepting state, accept. READ: Theorem 4.4 on p.196 for the proof. Theorem 4.1 A T M = { M, w M is a TM that accepts input w} is recognizable. There is a TM U (the universal TM ) that takes a reasonable encoding of M and its input w, and that carries out M s computation on w. U accepts if M accepts w, U rejects if M rejects w, and U loops if M loops on w. U = On input M, w : 1. Simulate M on input w (use portion of tape to represent M s configuration state and content of M s tape, including head position and move back-and-forth between M s description and M s configuration for each simulation step). 4

5 2. Accept if M accepts; reject if M rejects. Q. What is the difference between A DF A and A T M? A. Difference from A DF A : U could get stuck in infinite loop in stage 1 (if M never halts on w, U will simply keep simulating), so U recognizes A T M but U does not decide A T M this does not show that A T M undecidable, just that U is not a decider for A T M. Note: U is general-purpose computer : like all TMs, hard-wired to carry out exactly one task, but that task depends on instructions provided in input. 3 Countable Sets and Diagonalization Definition. An infinite set is countable if there is a way to list every member of the set. Determine whether each of the following sets is countable: {0, 1} :is countable because we can list strings in lexicographic order, i.e., use correspondence e, 0, 1, 00, 01, 10, 11, 000, 001,... (where e represents the empty string) Z: is countable because use correspondence Q: is countable because we can construct a zig-zag argument. 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/ R: is not countable. Proof by contradiction through diagonalization. Idea, suppose that there is a correspondence with the natural numbers. Then we can list the numbers as follows: f(1) = a 0.a 1 a 2 a 3 a 4... f(2) = b 0.b 1 b 2 b 3 b 4... f(3) = c 0.c 1 c 2 c 3 c 4... f(4) = d 0.d 1 d 2 d 3 d define a number x = 0.x 1 x 2 x 3 x 4... such that x 1 a 1, x 2 b 2, x 3 c 3 etc. Now we are guaranteed that x f(i) i N. These types of proofs are referred to as diagonalizations. 5

6 4 Diagonalization and Languages Q. What does diagonalization have to do with languages? Consider: Q. Are the number of T Ms countable? Yes. Each Turing machine can be encoded into a string over an alphabet Σ. For any alphabet, Σ, the set of strings Σ is countable. (left to you) Simply list all strings of length 0, followed by all strings of length 1, followed by all strings of length 2, etc. To list all Turing machines, take the listing of Σ and remove those strings that do not represent Turing machines. Q. Is the number of languages countable? no Let B be the set of all infinite binary sequences. B is uncountable (why??). There is a correspondence between B and the set of all languages, so these two sets are the same size. Let L be the set of all languages over an alphabet Σ. Let Σ = {s 1, s 2, s 3,...}. For each language A L, there is a unique sequence in B. Given a language A, A may consist of {s 1, s 4, s 5,..., s i,..., s j,...}. Consider the binary string where the i th bit is a 1 if s i A and a 0 otherwise. Then this infinite binary string uniquely represents A. Hence there is a correspondence between L and B. Since, there are a countable number of recognizable languages, (how do we know this?) since there are a countable number of turing machines There are some languages that are not unrecognizable! Q. But are there any natural unrecognizable languages? Next week... 6