VIEWS: 5 PAGES: 50 POSTED ON: 12/7/2012 Public Domain
Introduction to Computability Theory Lecture11: The Halting Problem Prof. Amos Israeli 1 The Halting Problem In this lecture we present an undecidable language. The language that we prove to be undecidable is a very natural language namely the language consisting of pairs of the form M , w where M is a TM accepting string w: ATM M , w M is a TM accepting w 2 The Halting Problem Since this language requires to decide whether the computation of TM M halts on input w, it is often called The Halting Problem. Theorem The halting problem is Turing Recognizable. 3 Proof Consider a TM U that gets a pair M , w as input and simulates the run of M on input w. If M accepts or rejects so does U. Otherwise, U loops. Note: U recognizes ATM ,since it accepts any pair M , w L , that is: any pair in which M accepts input w. 4 Simulating an Input TM On the previous lecture, we detailed the simulation of a DFA by a TM. Simulating one TM by another, using the encoding of the first TM is a very similar process. In the next slide we review the main characteristics of TM N simulating TM M, using M’s encoding <M>. 5 Simulating an Input TM TM N works as follows: 1. Mark M’s initial state and w’s initial symbol as the “current state” and “current head location”. 2. Look for M’s next transition on the description of its transition function. 3. Execute M’s transition. 6 Simulating an Input TM 4. Move M’s “current state” and “current head location” to their new places. 5. If M’s new state is a deciding state decide according to the state, otherwise – repeat stages 2-5. 7 Diagonalization Our goal in this lecture is to prove that ATM is not decidable. The proof uses a common mathematical technique known as Diagonalization. Diagonalization was first used by Cantor when he found a way to distinguish between several types of infinite sets. 8 Cardinality Cantor dealt with questions like: How many natural numbers are there? Infinity! How many real numbers are there? Infinity! Does the amount of natural numbers equal to the amount of real numbers? How is the size of infinite sets measured? 9 Cardinality Cantor’s answer to these question was the notion of Cardinality. The cardinality of a set is a property marking its size. Two sets has the same cardinality if there is a correspondence between their elements 10 Intuitive Notion of Correspondence At this point of the lecture, think about a correspondence between sets A and B as 2 lists: A list of A’s elements and in parallel a list of B’s elements. These 2 lists are juxtaposed so that each element of A corresponds to a unique element of B. 11 Example A ,2,3,4 B 2,4,6,8 A 1 2 3 4 1 B 2 4 6 8 Clearly, the cardinality of A is equal to the cardinality of B. How about the cardinality of infinite sets? 12 Example How about the cardinality of infinite sets? Is the cardinality of natural numbers larger than the cardinality of even natural numbers? Intuitively, the cardinality of any set should be larger that the cardinality of any of its proper subsets. Alas, our intuition of sets is driven by our daily experience with finite sets. 13 Example So let us try to create a correspondence between the natural numbers the even natural numbers? N ,2,3,4,...,n,... 1 N 1 2 3 ... n ... EN 2,4,6,8,...,2n,... EN 2 4 6 ... 2n ... Indeed f n 2n defines the wanted correspondence between the 2 sets. 14 Example N ,2,3,4,...,n,... 1 N 1 2 3 ... n ... EN 2,4,6,8,...,2n,... EN 2 4 6 ... 2n ... So the cardinality of N is equal to the cardinality of EN. 15 Countable Sets This last example suggests the notion of Countable Sets: A set A is countable if it is either finite or its cardinality is equal to the cardinality of N. A cool way of looking at countable sets is: “A set is countable if a list of its elements can be created”. 16 Countable Sets “A set is countable if a list of its elements can be created”. Note: This list does not have to be finite, but for each natural number i, one should be able to specify the i-th element on the list. For example, for the set EN the i-th element on the list is 2i . 17 Countable Sets We just proved that EN, the set of even natural numbers is countable. What about the set of rational numbers? Is the set Q of rational numbers countable? Can its elements be listed? 18 The set of Rationals is Countable Theorem The set of rational numbers is countable. Proof In order to prove this theorem we have to show how a complete list of the rational numbers can be formed. 19 The set of Rationals is Countable Recall that each natural number is defined by a pair of natural numbers. One way to look 1 / 1 1 / 2 1 / 3 1 / 4 1 / 5 ……… at the Rationals 2 / 1 2 / 2 2 / 3 2 / 4 2 / 5 ……… is by listing them 3 / 1 3 / 2 3 / 3 3 / 4 3 / 5 ……… 4 / 1 4 / 2 4 / 3 4 / 4 4 / 5 ……… in an infinite 5 / 1 5 / 2 5 / 3 5 / 4 5 / 5 ……… Rectangle. ………………………………………………... 20 The set of Rationals is Countable How can we form a list including all these numbers? If we first list 1 / 1 1 / 2 1 / 3 1 / 4 1 / 5 ……… The first row – 2 / 1 2 / 2 2 / 3 2 / 4 2 / 5 ……… We will never 3 / 1 3 / 2 3 / 3 3 / 4 3 / 5 ……… 4 / 1 4 / 2 4 / 3 4 / 4 4 / 5 ……… reach the second. 5 / 1 5 / 2 5 / 3 5 / 4 5 / 5 ……… ………………………………………………... 21 The set of Rationals is Countable 1/1 1/ 2 1/ 3 1/ 4 1/ 5 2 /1 2 / 2 2 / 3 2 / 4 2 / 5 3 /1 3 / 2 3 / 3 3 / 4 3 / 5 4 /1 4 / 2 4 / 3 4 / 4 4 / 5 5 /1 5 / 2 5 / 3 5 / 4 5 / 5 One way to do it is to start from the upper left corner, and continue in this fashion 22 The set of Rationals is Countable Note that some rational numbers appear more than once. For example: all numbers on the main diagonal are equal to 1, so this list is not final. In order to compute the actual place of a given rational, we need to erase all duplicates, but this is a technicality… 23 So perhaps all sets are countable Can you think of any infinite set whose elements cannot be listed in one after the other? Well, there are many: Theorem The set of infinite binary sequences is not countable. 24 Uncountable Sets Assume that there exists a list of all binary sequences. Such a list may look like this: 1 0 1 1 0 ……… 1 1 0 0 1 ……… 0 0 0 0 1 ……… 1 1 1 0 1 ……… 1 0 0 0 1 ……… ……………………................. 25 Uncountable Sets But can you be sure that all sequences are in this list? In fact, There exist 1 0 1 1 0 ……… many sequences 1 1 0 0 1 ……… that are not on 0 0 0 0 1 ……… 1 1 1 0 1 ……… the list: ……… 1 0 0 0 1 ……………………................. 26 Uncountable Sets Consider for example S=0,0,1,1,0,… . The sequence S is formed so that S1 1st elt. Of 1st seq. 1 0 1 1 0 ……… S2 2nd elt. Of 2nd seq. 1 1 0 0 1 ……… S3 0 0 0 0 1 ……… 3rd elt. Of 3rd seq. 1 1 1 0 1 ……… And so on … 1 0 0 0 1 ……… ……………………........... 27 Uncountable Sets In general: The i-th element of S, S i differs from the i-th element of the i-th sequence in the list. Note: the i-th element of the i-th sequence in the list is always the element on the diagonal. Can the sequence S appear on the list? 28 Uncountable Sets Assume there exists an index j S1 1 0 1 1 0 ... ... such that S S j S2 1 1 0 0 1 ... ... In this case, S j j S j S3 0 0 0 0 1 ... ... S4 1 1 1 0 1 ... ... But by definition: S j S j j S5 1 0 0 0 1 ... ... ... ... ... ... ... ... ... ... Contradiction!! S 0 0 1 1 0 ... ? 29 Uncountable Sets For any sequence on the list, S j , the sequence S differs from S j by its element on the diagonal, that is: S j S j j , so S cannot be on the list. For obvious reasons, this technique is called Diagonalization. 30 Uncountable Sets We just used diagonalization to prove that the set of infinite binary sequences is uncountable. Can a a similar proof for the set of real numbers? 31 Turing Unrecognizable Languages Corollary Some Languages are not Turing-recognizable. Proof For any (finite) alphabet, , the set of (finite) strings * , is countable. A list of all elements in * is obtained by first listing strings of length 1, then 2, …, then n… 32 Proof (cont.) The set of all TM-s is also countable because every TM, M , can be described by its encoding M , which is a string over . So the set of TM-s corresponds to a subset of *. Note: Here we use the (unproven but correct) fact that the cardinality of a set is always not greater then the cardinality of any of its supersets. 33 Proof (cont.) Since each TM recognizes exactly a single language, a list of all TM-s can be used as a list of all recognizable languages. If we show that the set of languages over is uncountable, we can deduce that at least a single language is not on the list, that is: it is not recognized by any TM. 34 Proof (cont.) We have already seen that the set of infinite binary sequences is uncountable. If we form a correspondence between the set of languages and the set of infinite binary sequences we will show that the set of languages is uncountable. 35 Proof (cont.) Consider a fixed list l of all words * . The correspondence is formed as follows: For every infinite binary sequence S, corresponds the language: LS i | where si 1 l QED 36 The Language ___ Is Undecidable ATM So far we proved the existence of a language which is not Turing recognizable. Now we continue our quest to prove: Theorem The language ATM M , w M is a TM accepting w is undecidable. 37 The Language ___ Is Undecidable ATM Before we start the proof let us consider two ancient questions: Question1: Can god create a boulder so heavy that god cannot lift? 38 The Language ___ Is Undecidable ATM Question2: In the small town of L.J. there is a single barber: Over the barber’s chair there is a note saying: “I will shave you on one condition: Thaw shall never shave thyself!!!” Who Shaves the Barber? 39 Proof Assume, by way of contradiction, that ATM is decidable and let H be a TM deciding ATM . That is H M , w accept if M accepts w reject if M rejects w Define now another TM new D that uses H as a subroutine as follows: 40 Proof Define now another TM new D that uses H as a subroutine as follows: D=“On input M where N is a TM: 1. Run H on input M , M . 2. Output the opposite of H’s output namely: If H accepts reject, otherwise accept.“ 41 Proof Note: What we do here is taking advantage of the two facts: Fact1: TM M should be able to compute with any string as input. Fact2: The encoding of M, M , is a string. 42 Proof Running a machine on its encoding is analogous to using a compiler for the computer language Pascal to, that is written in Pascal, to compile itself. As we recall from the two questions self- reference is sometimes means trouble (god forbid…) 43 Proof What we got now is: accept if M rejects M D M reject if M accepts M Consider now the result of running D with input D . What we get is: accept if D rejects D D D reject if D accepts D 44 Proof accept if D rejects D D D reject if D accepts D So if D accepts, it rejects wand if it rejects it accepts. Double Trouble. And it all caused by our assumption that TM H exists!!! 45 Proof Review 1. Define ATM M , w M is a TM accepting w . 2. Assume that ATM id decidable and let H be a TM deciding it. 3. Use H to build TM D that gets a TM encoding M and behaves exactly opposite to H’s behavior, accept if M rejects M D M namely: reject if M accepts M 46 Proof Review 4. Run TM D on its encoding D and conclude: accept if D rejects D D D reject if D accepts D Contradiction. 47 So Where is the Diagonalization? The following table describes the behavior of each machine on its encoding: M1 M2 M3 M4 M1 accept accept M2 accept accept accept accept M3 M4 accept accept 48 So Where is the Diagonalization? This table describes the behavior of TM H. Note: TM H rejects where M i loops. M1 M2 M3 M4 M1 accept reject accept reject M2 accept accept accept accept M3 reject reject reject reject M4 accept accept reject reject 49 Proof Review Now TM D is added to the table… M1 M2 M3 M4 D M1 accept reject accept reject accept M2 accept accept accept accept accept M3 reject reject reject reject reject M4 accept accept reject reject accept D reject reject accept accept ??? 50