Mapping Reductions. Mapping Reductions. Introduction to Computability. Theory. Mapping Reductions. Lecture13: Mapping. Reductions

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1 Introduction to Computability Theory Lecture13: apping Reductions Prof. Amos Israeli 1 apping Reductions apping reductionsconstitute the generalization we are looking for. They are always in a specified form which: 1. Preserves the intrinsic difficulty of finding a reduction. 2. Ease the red tape requirements. 3 apping Reductions So far, we presented several reductions: From to, from to, A A from to, from to EQ, A REGULAR E EQ and several other. Each of these reductions had some special form not shared with the others. Can we generalize? 2 apping Reductions In other words: Assume we want to prove that B is undecidable by reduction from Athat is known to be undecidable. If our reduction has the form of a apping reductionthen the proof is complete. Nothing else is required. 4

2 apping Reductions The idea of a mapping reduction is very simple: If the instances (candidate elements) of one language, say A, are mapped to the instances of another language, say B, by a computable mapping in a way that I A iff ( I ) B, then a decider for Bcan be used to devise a decider for A. 5 apping Reductions This scheme works but there is one delicate point here. Can you see where it is? ( ) The mapping I must be computable. Given the input I, your ordinary Turing machine must be able to compute. For this we need the notion of Computable Functions. ( I ) 7 apping Reductions The sought for decider for Aworks as follows: Upon input Ido: ( ) 1. Compute I. 2. Use the decider for Bto determine whether ( I ) B. ( I ) B 3. If accept-otherwise - reject. 6 Computable Functions * * Let Σ be an alphabet: A function is a : Σ Σ computable function if there exists a, such that for every w Σ *, if computes ( w ) with input w, it halts with f on its tape. f 8

3 Examples of Computable Functions 1. Let mand nbe natural numbers, let m,n be a string encoding of mand n, and let m+n be the string representing their sum. The function ( m, n ) = m n f +, is computable. r g ( w ) = w 2. The function is a computable function. Can you devise -s to compute fand g? 9 Use of Comp. Func. for Reductions In the reduction from to - A from the input,w to was modified by S, the decider to s.t. 1 all endingcomputations on would become accepting computations on. f ( ) = 1 1 Since was carried out by S, it is obviously a computable function. 11 Examples of Computable Functions 3. Let be an encoding of and let be an encoding of another satisfying: ( ) L( ') 1.. L = ' 2. never makes four consecutive steps in the same head direction. The function t defined below is computable: t ( ) = ε ' if if is an ecodingof is not ecodingof any 10 Use of Comp. Func. for Reductions Another example is the reduction from to REGULAR A L ( ) 2 A - from the input to,w, was modified by S, to s.t. 2 is regular iff. Once again,, w A this is a computable function. Now we turn to define mapping reductions: 12

4 apping Reductions * Let Aand Bbe two languages over. A * * computable function f : Σ Σ is a mapping reductionfrom Ato Bif the following holds: * For each, I A iff f I. The I Σ ( ) B function f is called reductionof Ato B. The arguments of the reduction are often called instances. Σ 13 A Pictorial Illustration of B A f. A f ( A ) f A f ( A ) 15 apping Reductions If there exists a mapping reduction from Ato B, We say that Ais mapping reducible to Band denote it by B. A pictorial illustration of a mapping reduction appears in the next slide. 14 An Observation about and A B Assume that B. A f ( A ) f ( A ) A f Can we say something about the relationship between A and B? Answer: Yes B 16

5 Using apping Reductions If language Ais mapping reducible to language B, then a solution for B, can be used to derive a solution for A. This fact is made formal in the following theorem: 17 Let be a decider for Band let f be the reduction from Ato B. Consider N: N= On input w: ( w ) 1. Compute f. ( w ) 2. Run on f and output whatever outputs. Clearly Naccepts iff w B. QED 19 Theorem If B and Bis decidable, then Ais decidable. 18 Corollary If B and Ais undecidable, then Bis undecidable. 20

6 Discussion The previous corollary is our main tool for proving undecidability. Notice that in order to prove Bundecidable we reduce from Awhich is known to be undecidable to B. The reduction direction is often a source of errors. A similar tool is used in Complexity theory. 21 Revisited Theorem The language = Is undecidable. {, w is a that halts on w } 23 Revisited On the previous lecture we proved that = {, w is a that halts on w } is undecidable. Now, we prove this theorem once more by demonstrating a mapping reductionfrom A to. 22 Revisited The proof is by reduction from. In the next A slide we show that A. By the m HALT previous corollary it follows that is undecidable. 24

7 apping Reduction from A to The mapping reduction is presented by F : F= On input,w : 1. Construct. = On input x: 1. Run on x. 2. if accepts accept. 3. If rejects, enter a loop. 2. Output ',w. 25 The Halting Problem Revisited What happens if the input,w does not contain a valid description of a? By the specification of we know that in this A case. Therefore in this case, w F should output any string s satisfying s. A 27 (cont.) It is not hard to see that, w iff ', w A. In other words: A m. Since A is undecidable, the previous corollary implies that is undecidable too. 26 EQ Revisited Theorem The language {, N and N are - s and L ( ) L ( N ) } EQ = = Is undecidable. 28

8 Revisited The proof is by reduction from which is E known to be undecidable. In the next slide we show that. By the previous corollary it follows that is undecidable. E m EQ EQ 29 Theorem If B and Bis Turing recognizable, then Ais Turing recognizable. 31 apping Reduction from E to EQ Let be an encoding of a that rejects all r its inputs. The mapping f is defined as follows: For each instance of,,. It is not hard to verify that E, E f ( ) =, r r EQ ( ) ( ) ( ) eaning: L =φ L = L. QED r 30 Turing Recognizer Let be a decider for Band let f be the reduction from Ato B. Consider N: N= On input w: 1. Compute f. -if halts, 2. Run on f ( w ) and output whatever outputs. Clearly N accepts iff w B. QED N recognizes B. 32 ( w )

9 Corollary If B and Ais not Turing recognizable, then B is not Turing recognizable. 33 If B by a mapping f, then by the same mapping it holds that B. Since we know that A is not Turing recognizable, we can prove that language Ais not Turing recognizable by reducing to. A A 35 The Status of EQ Theorem EQ is neither Turing recognizable nor co- Turing recognizable. Reminder A language Lis co-turing recognizable if its complement, L, is Turing recognizable. 34 Now we use the previous remark and prove that EQ is not Turing recognizable, by showing that. A EQ Since, we conclude that A EQ and since we know that A is not Turing recognizable we conclude that so is. A EQ EQ 36

10 F= on input,w, where is a : 1. Construct that accepts on any input and 2 1 that accepts if accepts w. 2. Output. 1, 2 It is not hard to see that Fcomputes a reduction from to. A EQ 37 Since we conclude that is A EQ not Turing recognizable. EQ A EQ EQ Since we conclude that is not co-turing recognizable, hence EQ EQ = is not Turing recognizable. QED 39 G= on input,w, where is a : 1. Construct that rejects on any input and 2 1 that accepts if accepts w. 2. Output. 1, 2 It is not hard to see that Gcomputes a reduction from to. A EQ 38