Computer Language Theory Chapter 4: Decidability Last modified

Computer Language Theory Chapter 4: Decidability Last modified 4/7/21 1

Limitations of Algorithmic Solvability In this chapter we investigate the power of algorithms to solve problems Some can be solved algorithmically and some cannot Why we study unsolvability Useful because then can realize that searching for an algorithmic solution is a waste of time Perhaps the problem can be simplified Gain an perspective on computability and its limits In my view also related to complexity (Chapter 7) First we study whether there is an algorithmic solution and then we study whether there is an “efficient” (polynomialtime) one 2

Chapter 4.1 Decidable Languages 3

Decidable Languages We start with problems that are decidable We first look at problems concerning regular languages and then those for context-free languages 4

Decidable Problems for Regular Languages We give algorithms for testing whether a finite automaton accepts a string, whether the language of a finite automaton is empty, and whether two finite automata are equivalent We represent the problems by languages (not FAs) Let ADFA {(B, w) B is a DFA that accepts string w} The problem of testing whether a DFA B accepts a specific input w is the same as testing whether (B,w) is a member of the language ADFA. Showing that the language is decidable is the same thing as showing that the computational problem is decidable So do you understand what ADFA represents? If you had to list the elements of ADFA what would they be? 5

ADFA is a Decidable Language Theorem: ADFA is a decidable language Proof Idea: Present a TM M that decides ADFA M On input (B,w), where B is a DFA and w is a string: 1. 2. Simulate B on input w If the simulation ends in an accept state, then accept; else reject 6

Outline of Proof Must take B as input, described as a string, and then simulate it This means the algorithm for simulating any DFA must be embodied in the TM’s state transitions Think about this. Given a current state and input symbol, scan the tape for the encoded transition function and then use that info to determine new state The actual proof would describe how a TM simulates a DFA Can assume B is represented by its 5 components and then we have w Keep track of current state and position in w by writing on the tape Initially current state is q0 and current position is leftmost symbol of w The states and position are updated using the transition function δ Note that the TM must be able to handle any DFA, not just this one TM M’s δ not the same as DFA B’s δ When M finishes processing, accept if in an accept state; else reject. The implementation will make it clear that will complete in finite time. 7

ANFA is a Decidable Language Proof Idea: Because we have proven decidability for DFAs, all we need to do is convert the NFA to a DFA. N On input (B,w) where B is an NFA and w is a string 1. 2. 3. Convert NFA B to an equivalent DFA C, using the procedure for conversion given in Theorem 1.39 Run TM M on input (C,w) using the theorem we just proved If M accepts, then accept; else reject Running TM M in step 2 means incorporating M into the design of N as a subroutine Note that these proofs allow the TM to be described at the highest of the 3 levels we discussed in Chapter 3 (and even then, without most of the details!). 8

Computing whether a DFA accepts any String EDFA { A A is a DFA and L(A) ) is a decidable language Proof: A DFA accepts some string iff it is possible to reach the accept state from the start state. How can we check this? We can use a marking algorithm similar to the one used in Chapter 3. T On input (A) where A is a DFA: 1. 2. Mark the start state of A Repeat until no new states get marked: 3. 4. Mark any state that has a transition coming into it from any state already marked If no accept state is marked, accept; otherwise reject In my opinion this proof is clearer than most of the previous ones because the pseudo-code above specifies enough details to make it clear how to implement it 9

EQDFA is a Decidable Language EQDFA {(A,B) A and B are DFAs and L(A) L(B)} Proof idea Construct a DFA C from A and B, where C accepts only those strings accepted by either A or B but not both (symmetric difference) If A and B accept the same language, then C will accept nothing and we can use the previous proof (for EDFA) to check for this. So, the proof is: F On input (A,B) where A and B are DFAs: 1. Construct DFA C that is the symmetric difference of A and B (details on how to do this on next slide) 2. Run TM T from the proof from last slide on input (C) 3. If T accepts (sym. diff ) then accept. If T rejects then reject 10

How to Construct C L(A ) L(B) Complement symbol L(C) (L(A) L(B)’) (L(A)’ L(B)) We used proofs by construction that regular languages are closed under , , and complement We can use those constructions to construct a FA that accepts L(C) Wait a minute! The book is quite cavalier! We never proved regular languages are closed under 11

Regular Languages Closed under Intersection If L and M are regular languages, then so is L M Proof: Let A and B be DFAs whose regular languages are L and M, respectively Construct C, the “product automation” of A and B More on this in a minute, but essentially C tracks the states in A and B (just like when we did the proof of union without using NFAs) Make the final states of C be the pairs consisting of final states of both A and B In the union case we the final state any state with a final state in A or B 12

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ACFG is a Decidable Language Proof Idea: For CFG G and string w want to determine whether G generates w. One idea is to use G to go through all derivations. This will not work, why? Because this method a best will yield a TM that is a recognizer, not a decider. Can generate infinite strings and if not in the language, will never know it. But since we know the length of w, we can exploit this. How? A string w of length n will have a derivation that uses 2n-1 steps if the CFG is in Chomsky-Normal Form. So first convert to Chomsky-Normal Form Then list all derivations of length 2n-1 steps. If any generates w, then accept, else reject. This is a variant of breadth first search, but instead of extended the depth 1 at a time we allow it to go 2n-1 at a time. As long as finite depth extension, we are okay 14

ECFG is a Decidable Language How can you do this? What is the brute force approach? Try all possible strings w. Will this work? The number is not bounded, so this would not be decidable Instead, think of this as a graph problem where you want to know if you can reach a string of terminals from the start state Do you think it is easier to work forward or backwards? Answer: backwards 15

ECFG is a Decidable Language (cont) Proof Idea: Can the start variable generate a string of terminals? Determine for each variable if it can generate any string of terminals and if so, mark it Keep working backwards so that if the right-side of any rule has only marked items, then mark the LHS For example, if X YZ and Y and Z are marked, then mark X If you mark S, then done; if nothing else to mark and S not marked, then reject You start by marking all terminal symbols 16

EQCFG is not a Decidable Language We cannot reuse the reasoning to show that EQDFA is a decidable language since CFGs are not closed under complement and intersection As it turns out, EQCFG is not decidable! We will learn in Chapter 5 how to prove things undecidable 17

Every Context-Free Language is Decidable Note that a few slides back we showed ACFG is decidable. This is almost the same thing We want to know if A, which is a CFL, is decidable. A will have some CFG G that generates it When we proved that ACFG is decidable, we constructed a TM S that would tell us if any CFG accepts a particular input w. Now we use this TM and run it on input G,w and if it accepts, we accept, and if it rejects, we reject. This is so close to the prior proof it is confusing. It comes from the fact that a CFL is defined by a CFG. This leads us to the following picture of the hierarchy of languages 18

Hierarchy of Classes of Languages We proved Regular Contextfree since we can convert a FA into a CFG We just proved that every Context-free language is decidable From the definitions in Chapter 3 it is clear that every Decidable language is trivially Turing-recognizable. We hinted that not every Turingrecognizable language is Decidable. Next we prove that! Regular ContextFree Decidable Turingrecognizable 19

Chapter 4.2 The Halting Problem 20

The Halting Problem One of the most philosophically important theorems in the theory of computation There is a specific problem that is algorithmically unsolvable. In fact, ordinary/practical problems may be unsolvable Software verification Given a computer program and a precise specification of what the program is supposed to do (e.g., sort a list of numbers) Come up with an algorithm to prove the program works as required This cannot be done! But wait, can’t we prove a sorting algorithm works? Note: the input has two parts: specification and task. The proof is not only to prove it works for a specific task, like sorting numbers. Our first undecidable problem: Does a TM accept a given input string? Note: we have shown that a CFL is decidable and a CFG can be simulated by a TM. This does not yield a contradiction. TMs are more expressive than CFGs. 21

Halting Problem II ATM {(M,w) M is a TM and M accepts w} ATM is undecidable It can only be undecidable due to a loop of M on w. If we could determine if it will loop forever, then could reject. Hence ATM is often called the halting problem. Note that this is Turing recognizable: As we will show, it is impossible to determine if a TM will always halt (i.e., on every possible input). Simulate M on input w and if it accept, then accept; if it ever rejects, then reject We start with the diagonalization method 22

Diagonalization Method In 1873 mathematician Cantor was concerned with the problem of measuring the sizes of infinite sets. How can we tell if one infinite set is bigger than another or if they are the same size? We cannot use the counting method that we would use for finite sets. Example: how many even integers are there? What is larger: the set of even integers or the set of all strings over {0,1} (which is the set of all integers) Cantor observed that two finite sets have the same size if each element in one set can be paired with the element in the other This can work for infinite sets 23

Function Property Definitions From basic discrete math (e.g., CS 1100) Given a set A and B and a function f from A to B f is one-to-one if it never maps two elements in A to the same element in B f is onto if every item in B is reached from some value in a (i.e., f(a) b for every b B). The function add-two is one-to-one whereas absolutevalue is not For example, if A and B are the set of integers, then addtwo is onto but if A and B are the positive integers, then it is not onto since b 1 is never hit. A function that is one-to-one and onto has a (oneto-one) correspondence This allows all items in each set to be paired 24

An Example of Pairing Set Items Let N be the set of natural numbers {1, 2, 3, } and let E be the set of even natural numbers {2, 4, 6, }. Using Cantor’s definition of size we can see that N and E have the same size. The correspondence f from N to E is f(n) 2n. This may seem bizarre since E is a proper subset of N, but it is possible to pair all items, since f(n) is a 1:1 correspondence, so we say they are the same size. Definition: A set is countable if either it is finite or it has the same size as N, the set of natural numbers 25

Example: Rational Numbers Let Q {m/n: m,n N}, the set of positive Rational Numbers Q seems much larger than N, but according to our definition, they are the same size. Here is the 1:1 correspondence between Q and N We need to list all of the elements of Q and then label the first with 1, the second with 2, etc. We need to make sure each element in Q is listed only once 26

Correspondence between N and Q matrix To get our list, we make an infinite containing all the positive rational numbers. Bad way is to make the list by going row-to-row. Since 1st row is infinite, would never get to the second row Instead use the diagonals, not adding the values that are equivalent So the order is 1/1, 2/1, ½, 3/1, 1/3, This yields a correspondence between Q and N That is, N 1 corresponds to 1/1, N 2 corresponds to 2/1, N 3 corresponds to ½ etc. 1/1 2/1 3/1 4/1 5/1 1/2 2/2 3/2 4/2 5/2 1/3 2/3 3/3 4/3 5/3 1/4 2/4 3/4 4/4 5/4 1/5 2/5 3/5 4/5 5/5 27

Theorem: R is Uncountable A real number is one that has a decimal representation and R is set of Real Numbers Includes those that cannot be represented with a finite number of digits, like Pi and square root of 2 Will show that there can be no pairing of elements between R and N Will find some x that is always not in the pairings and thus a proof by contradiction 28

Finding a New Value x To the right is an example mapping I now describe a method that will be guaranteed to generate a value x not already in the infinite list Generate x to be a real number between 0 and 1 as follows Assume that it is complete To ensure that x f(1), pick a digit not equal to the first digit after the decimal point. Any value not equal to 1 will work. Pick 4 so we have .4 To x f(2), pick a digit not equal to the second digit. Any value not equal to 5 will work. Pick 6. We have .46 Continue, choosing values along the “diagonal” of digits (i.e., if we took the f(n) column and put one digit in each column of a new table). n f(n) 1 3.1415 9 2 55.555 5 3 0.1234 5 4 0.5000 00 . . When done, we are guaranteed to have a value x not already in the list since it differs in at least one position with every other number in the list. 29

Implications The theorem we just proved about R being uncountable has an important application in the theory of computation It shows that some languages are not decidable or even Turing-recognizable, because there are uncountably many languages yet only countably many Turing Machines. Because each Turing machine can recognize a single language and there are more languages than Turing machines, some languages are not recognized by any Turing machine. Corollary: some languages are not Turing-recognizable 30

Some Languages are Not Turingrecognizable I The set of all strings * is countable A finite number of strings of each length, so we can list them by increasing length and hence they are countable The set of all Turing Machines M is countable since each TM M has an encoding into a string M Order by length and omit strings that do not represent valid TM’s and we have a countable list of Turing Machines 31

Some Languages are Not Turingrecognizable II The set of all languages L over is uncountable Recall a language is made up of a set of strings, so different from what we just counted on last slide. Each language is represented by an infinite binary sequence B, where each position in the sequence corresponds to a string Assume * {s1, s2, s3 }. We can encode any language as a characteristic binary sequence, where the bit indicates whether the corresponding s i is a member of the language. Thus, there is a 1:1 mapping. The set of all infinite binary sequences B is uncountable Can prove uncountable using same proof used to prove real numbers not countable L is uncountable because it has a correspondence with B Since B is uncountable and L and B are of equal size, L is uncountable So set of TMs is countable and the set of languages is not Means we cannot put set of languages into a correspondence with set of TMs. Therefore some languages do not have a corresponding Turing machine Thus some languages not Turing-Recognizable 32

Common Sense Explanation Comparing languages, a potentially infinite set of strings, versus number of strings Each language is represented by a sequence of infinite length whereas each individual string is of finite (but unbounded) length String is to Language as Natural number is to Real Number 33

Halting Problem is Undecidable Prove that halting problem is undecidable We started this a while ago Let A TM { M,w M is a TM and accepts w} Proof Technique: Assume ATM is decidable and obtain a contradiction A diagonalization proof 34

Proof: Halting Problem is Undecidable Assume ATM is decidable Let H be a decider for ATM On input M,w , where M is a TM and w is a string, H halts and accepts if M accepts w; otherwise it rejects Construct a TM D using H as a subroutine D calls H to determine what M does when input string is its own description M . Like running a C program where input is the program represented as a string D then outputs the opposite of H’s answer D( M ) accepts if M does not accept M and rejects if M accepts M Now run D on its own description D( D ) accept if D does not accept D and reject if D accepts D No matter what D does it is forced to do the opposite, which is a contradiction. Thus, neither TM D or TM H can exist. See next slide. 35

The Diagonalization Proof M1 M2 M3 M4 D M1 M2 M3 M4 Acce pt Acce pt Reje ct Acce pt Reje ct Acce pt Reje ct Acce pt Acce pt Acce pt Reje ct Reje ct Reje ct Acce pt Reje ct Reje ct Acce pt Acce pt Reje ct Acce pt . The TM D must invert the value on the diagonal. It can do this for M1 , D etc, but Reje Reje was accept ? then it M2 , not for D . IfAcce the entryAcce for D( D ) needs to be reject, and reject then ct ctif it waspt pt it needs to be accept. Contradiction. Similar to proof that Real numbers not countable. 36

A More Satisfying Proof for CS Majors The last proof uses some mathematical tricks and is not very intuitive Computer programs appear more concrete to most of us The halting problem naturally is about programs and infinite loops After teaching this many times, I developed a proof that most of you will find less arbitrary since it focuses programs and infinite loops. But it is nonetheless follows the same steps as the prior proof 37

Slightly more Concrete Version You write a program, halts(P, X) in C that takes as input any C program, P, and the input to that program, X Your program halts(P, X) analyzes P and returns “yes” if P will halt on X and “no” if P will not halt on X You now write a short procedure foo(X): foo(X) {a: if halts(X,X) then goto a; else halt} This program does not halt if P halts on X (infinite loop via goto) and it does if P does not halt on X Does foo(foo) halt? It halts if and only if halts(foo,foo) returns no It halts if and only if it does not halt. Contradiction. Thus we have proven that you cannot write a program to determine if any arbitrary program will halt or loop 38

What does this mean? Recall what was said earlier The halting problem is not some contrived problem The halting problem asks whether we can tell if some TM M will accept an input string We are asking if the language below is decidable ATM {(M,w) M is a TM and M accepts w} It is not decidable But as I keep emphasizing, M is an input variable too! Halting problem is Turing-recognizable (we discussed this) Of course, some algorithms are decidable, like sorting algorithms Simulate the TM on w and if it accepts/rejects, then accept/reject. The halting problem is special because it gets at the heart of the matter (it is related to ATM in general) 39

Co-Turing Recognizable A language is co-Turing recognizable if it is the complement of a Turing-recognizable language Theorem: A language is decidable if and only if it is Turing-recognizable and co-Turingrecognizable Why? To be Turing-recognizable, we must accept in finite time. If we don’t accept, we may reject or loop (it which case it is not decidable). Since we can invert any “question” by taking the complement, taking the complement flips the accept and reject answers. Thus, if we invert the question and it is Turing-recognizable, then that means that we would get the answer to the original reject question in finite time. 40

More Formal Proof Theorem: A language is decidable iff it is Turingrecognizable and co-Turing-recognizable Proof (2 directions) Forward direction easy. If it is decidable, then both it and its complement are Turing-recognizable Other direction: Assume A and A’ are Turing-recognizable and let M1 recognize A and M2 recognize A’ The following TM will decide A M On input w 1. 2. Run both M1 and M2 on input w in parallel If M1 accepts, accept; if M2 accepts, then reject Every string is in either A or A’ so every string w must be accepted by either M1 or M2. Because M halts whenever M1 or M2 accepts, M always halts and so is a decider. Furthermore, it accepts all strings in A and rejects all not in A, so M is also a decider for A and thus A is decidable 41

Implication For any undecidable language, either the language or its complement is not Turingrecognizable 42

Complement of ATM is not Turingrecognizable ATM’ is not Turing-recognizable Proof: We know that ATM is Turing-recognizable but not decidable If A ’ were also Turing-recognizable, then TM ATM would be decidable, which it is not Thus A ’ is not Turing-recognizable TM This should not be too surprising. It is harder to determine that something is not in the language 43

Computer Language Theory Chapter 5: Reducibility Due to time constraints we are only going to cover the first 3 pages of this chapter. However, we cover the notion of reducibility in depth when we cover Chapter 7. 44

What is Reducibility? A reduction is a way of converting one problem to another such that the solution to the second can be used to solve the first We say that problem A is reducible to problem B Example: finding your way around NY City is reducible to the problem of finding and reading a map If A reduces to B, what can we say about the relative difficulty of problem A and B? A can be no harder than B since the solution to B solves A A could be easier (the reduction is “inefficient” in a sense) In example above, A is easier than B since B can solve any routing problem 45

Practice on Reducibility In our previous class work, did we reduce NFAs to DFAs or DFAs to NFAs? We reduced NFAs to DFAs We showed that an NFA can be reduced (i.e., converted) to a DFA via a set of simple steps NFA can not be any more powerful than a DFA Based only on the reduction, NFA could be less powerful But since we know this is not possible, since an DFA is a degenerate form of an NFA, we showed they have the same expressive power 46

How Reducibility is used to Prove Languages Undecidable If A is reducible to B and B is decidable, what can we say? If A is reducible to B and A is decidable, what can we say? A is decidable (since A can only be “easier”) Also, B, which is decidable, can be used to solve A Nothing– B may not be decidable (so this is not useful for us) If A is undecidable and reducible to B, then what can we say about B? B must be undecidable (B can only be harder than A) This is the most useful part for Chapter 5, since this is how we can prove a language undecidable We can leverage past proofs and not start from scratch To show something undecidable, show an undecidable problem can be reduced to it. 47

Example: Prove HALTTM is Undecidable I Need to reduce ATM to HALTTM, where ATM already proven to be undecidable Can use HALTTM to solve ATM Proof by contradiction Assume HALTTM is decidable and show this implies ATM is decidable Assume TM R that decides HALTTM Use R to construct S a TM that decides ATM Pretend you are S and need to decide ATM so if given input M, w must output accept if M accepts w and reject if M loops on w or rejects w. First try: simulate M on w and if it accepts then accept and if rejects then reject. But in trouble if it loops. This is bad because we need to be a decider 48

Example: Prove HALTTM is Undecidable II Instead, use assumption that have TM R that decides HALTTM Now can test if M halts on w If R indicates that M does halt on w, you can use the simulation and output the same answer If R indicates that M does not halt, then reject since infinite looping on w means it will never accept The formal solution on next slide We already discussed this case when we informally discussed how the halting problem is related to ATM 49

Solution: HALTTM is Undecidable Assume TM R decides HALTTM Construct TM S to decide ATM as follows S “On input M, w , an encoding of a TM M and a string w: 1. 2. 3. 4. Run TM R on input M, w If R rejects (doesn’t halt), reject If R accepts, simulate M on w until it halts If M has accepted, accept; If M has rejected, reject” 50