Notes for Comp 497 (Comp 454) Week 12 4/19/05. Today we look at some variations on machines we have already seen. Chapter 21

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1 Notes for Comp 497 (Comp 454) Week 12 4/19/05 Today we look at some variations on machines we have already seen. Errata (Chapter 21): None known Chapter 21 So far we have seen the equivalence of Post Machines and Turing Machines. Consider the PDA. Do we gain anything if we add a second stack? What if we make it deterministic? Let us define a 2PDA as a deterministic PDA with 2 stacks (POP and PUSH now specify which stack.) It turns out that 2PDAs are more powerful than single-stack PDAs. They can accept any CFL and some non-cfl. An example of a recognizable language is a n b n a n. The 2PDA on page 481 recognizes a n b n a n Minsky showed that a 2PDA has the same power as a TM so now we have three equally powerful computing models: PM, TM, 2PDA. Formally, THEOREM 50 Any language accepted by some 2PDA can be accepted by some TM and vice versa. Proof by construction in two parts (1) Simulate a 2PDA using a TM. Seems clear that this should be possible given the more general nature of the TM. We just need to figure out how to map the input tape and the two stacks of the 2PDA onto the tape of the TM and then how to implement the READ, POP, and PUSH actions. Suppose the stacks (Y and Z) and tape (X) are as follows: Y1 Y2 X1 X2 X3 X4 Y3 Y4 Z1 Z2 We can map them onto different sections of the TM tape with separators, e.g.

2 X1... X4 # Y1... Y4 $ Z1 Z2 (We can use different markers if # and $ are in use) Now each of the 2PDA actions is a corresponding TM action and we can make use of the TM subroutines insert and delete. For example to simulate the READ operation, the TM performs the steps listed at the top of page 484. Similarly the POP operations on each stack will use the TM delete subroutines to remove a character from the tape though first branching on the character popped see page 485 and page 486 Similarly the PUSH operations on the stacks will use the TM insert subroutine. (2) Simulate a TM using a 2PDA This would seem to be more difficult because we only have two stacks to work with. We can simplify our task by showing that any language accepted by a Post Machine can be accepted by a 2PDA (and we already know the equivalence of PM and TM). We do this because the STORE of the Post Machine is similar to the stacks of the 2PDA Proof by construction. Given a PM with input in STORE, create using Stack2, a stack (Stack1) containing that input as in x y z z y x x y z STORE STACK2 STACK1 Now a READ operation in the PM is equivalent to a POP operation from STACK1. An ADD x operation in the PM is equivalent to (a) pushing x onto STACK2 then (b) creating STACK1 from STACK2 as we did with the initial input see pp. 487, 488.

3 An example 2PDA is shown on page 488 and on p. 489 is a TM that accepts the same language. At the foot of page 489 is a PM and on p. 490, a corresponding 2PDA. Thus we have the following set of equivalent machines (TM, PM, 2PDA). What if we add more stacks to a PDA do we get an even more powerful machine? No. THEOREM 51 Any language that can be accepted by a PDA with n stacks (n 2) can also be accepted by a TM. This is really an extension of our earlier mapping of a 2PDA onto a TM. We just need regions on the TM tape to represent the INPUT and each stack of the npda. In terms of power, here are the rankings to this point using the results of Kleene and the equivalencies we have seen recently. FA = TG = NFA < DPDA < PDA < 2PDA = npda = PM = TM Is there anything more powerful than our (deterministic) TM? Stay tuned.. Errata (Chapter 22): None known Chapter 22 In the previous chapter we looked at variations on the PDA. Next look at tweaking the TM. The variations we consider are: (1) Move-in-state TM (2) Stay-option TM (3) K-track tape TM (4) Two-way infinite tape TM (5) Non-deterministic TM (6) Read-only TM In the case of (2), (3), (4) and (5), we are interested in whether any of these will give us a more powerful machine than the TM we already have. In the case of (1) we want to know if the change of notation changes the power of the TM and in the case of (6) we want to identify the class of languages that such a restricted TM can recognize. (1) Move-in-state TM. This just means that the direction of movement along the tape is a property of the new state rather than the transition. Thus instead of

4 2 3 x,y,r we would have the movement direction be a property of state 3 (i.e. move-in-state) 2 3/R x,y Our original notation for TMs is not the only possible one. Authors other than Cohen sometimes use different ones including move-in-state. We can change our TM to a movein-state one. If transitions to a state have different directions as in a,b,l 2 x,y,r 3 then we can simply add new states a,b 2 3a/ L x,y 3b/ R So making the machine move-in-state does not make it more powerful it is just a notation convention. THEOREM 52 For every move-in-state TM M, there is an equivalent TM T that accepts the same language. Proof by construction, we just have to add the appropriate direction to all edges coming in to a particular state. THEOREM 53 For every TM T, there is an equivalent move-in-state machine M (same action on inputs, same final tape contents)

5 Proof by construction using the idea above which may, as we see, require additional states. So making the direction a property of a state rather than a transition does not add any power to the machine. (2) Stay-option TM This means giving the TM the option of not moving left or moving right, i.e. staying on the current square. This is not an increase in power either because a stay-option TM can be converted to a forced-movement TM as in the following example. 2 3 x,y,- converts to 2 3a 3 x,y,r *,*,L THEOREM 54 Stay-option machine = TM Proof: A normal TM is just a stay-option machine where we have chosen not to use the stay option. Any stay option machine can be converted to a normal TM using the technique above. (3) k-track TM (ktm) Would our TM be more powerful if the tape consisted of multiple parallel tracks? For example if k=3

6 # X a b b Y b a a a a A read operation reads from all tracks simultaneously and a write operation writes to all tracks simultaneously. THEOREM 55 ktm = TM We need to prove (1) Given a TM and positive integer k then is an equivalent ktm. (2) Given a ktm there is an equivalent TM (1) This is not just a case of adding k-1 tapes that are always blank but making an appropriate correspondence is not too difficult see pages 506 and 507. (2) See pages We might think immediately of encoding k symbols into a single symbol. If the ktm uses alphabet Σ then our new alphabet might be denoted Σ k At HALT we may unpack the composite characters in to single characters along the tape. (4) Two-way infinite tape Like the move-in-state idea this may just be a case of where Cohen s convention is different from that of other authors. Some texts assume that the TM tape is finite but unbounded in both directions. THEOREM 56 TMs with two-way infinite tapes are exactly as powerful as TMs with one-way infinite tapes. Proof is by construction (surprise!) Certainly anything a one-way infinite machine can do a two-way infinite machine can do also; just fix the problem of crashing at the left end. We can insert a marker symbol at

7 that point in the tape. Our machine will have no defined transitions for the symbol so will crash when it reads it. Can a one-way machine do everything that a two-way machine can? We can show the answer is yes by showing that anything a two-way machine can do can also be done by a 3TM and then use the result of Theorem 55. So how does the two-way infinite tape map onto three tracks in a one-way infinite tape? Two of the tracks hold the right-infinite and left-infinite parts of the tape and the third track is used to hold marker symbols so we know which track contains the current reading position. See diagram at the end of page 512. Relatively simple if movement is from one cell on the positive side of the tape to another cell on the positive side or from a cell on the negative side to another cell on the negative side. Trick part is where movement goes round the corner. But every movement has a mapping see pages 513 to 516 (5) Non-deterministic TM Would we gain any power by making the TM non-deterministic? A non-deterministic TM (NTM) would allow more than one transition from a state. For example, a,a,l 3 2 a,b,r 4 and a string on the tape is accepted by NTM if there is any path from START to HALT even if other paths crash. THEOREM 57 NTM = TM Contention is that making the TM non-deterministic does not add any power. Clearly anything recognized by TM can be recognized by NTM (that doesn t use its nondeterminism) so interesting question is how to show that any language accepted by an NTM is also accepted by some TM. Constructive proof shows that for a given NTM we can construct an equivalent 3TM and we already know that anything a 3TM can do, a TM can do.

8 Track 3 : Copy of input Track 2 : Which route through the TM we are trying (Cohen terms this mother s advice ) Track 1 : Working copy of input Systematically try all paths of length 1, length 2, length 3, copying track 3 to track 1 at the start of each new try. If input is accepted by the NTM then there will be a path and the TM will find it. Cohen shows on pages the algorithm for doing this. Following Theorem formalizes a useful result. THEOREM 58 Every CFL can be recognized by a TM True because CFL = PDA = NPDA = NPM = NTM = TM We needed Theorem 57 for this last step. (6) The read-only TM We can define a read-only TM as one that must write the same character that it reads, that is, transitions will be of the form (X,X,direction). It cannot write so it cannot compute a function but must merely act as a recognizer. What class of languages can it recognize? A read-only TM is sometimes known as a two-way FA because it is possible to go back and re-read inputs. THEOREM 59 A read-only TM accepts exclusively regular languages. Proof is by showing (as in the proof of Kleene s Theorem) that we can reduce the transition diagram labeled with (x,x,r) and (x,x,l) to a single transition with START Regular Expression HALT Consider (x,x,dir1) (y,y,dir2) A B C

9 Relatively straightforward to deal with if DIR1 is the same as DIR2 either rightwards (top of page 527) or leftwards (end of page 527). More of a problem if the directions are different.. We can devise a complex, but regular, expression to cover the various cases see discussion pp Net result is that we can reduce transition sequences to regular expressions. Read Chapters 21 and 22. Reading Assignment Homework #6 Here is Homework #6 due 4/26. Each question is worth 20 points. Covers Chapters 19, 20, 21 and Build a TM that accepts the language ODDPALINDROME. 2. Build a PM that takes in any string of a s and b s and leaves in its STORE the complement string that has the a s and b s switched. For example, if input is abaaabbaa then final STORE is babbbaabb. 3. Draw a 2PDA that accepts the language {a n b n a n b n }. 4. Outline the design of a 5TM that does decimal addition for three integers simultaneously. The numbers are on tracks 2, 3 and 4. The sum should be left on track 5. Track 1 is reserved for carries. 5. Outline the design of a pattern-matching 2TM. On track 1 is a long string (the text). On track 2 is a shorter string (the pattern). The 2TM halts only if the string on track 2 is a substring of the string on track 1.