The Church-Turing Thesis

Giorgi Japaridze Theory of Computability The Church-Turing Thesis Chapter 3

What is this course about? 3.0.a Giorgi JaparidzeTheory of Computability • Main questions that will be asked and answered: • What does computation mean? • What can be computed and what can not? • How quickly? • With how much memory?

Preliminaries 3.0.b Giorgi JaparidzeTheory of Computability Alphabet --- a finite set of objects called thesymbolsof the alphabet. E.g.:  = {a,b,…,z}  = {0,1}  = {0,1,$} Stringover--- a finite sequence of symbols from. E.g.:x = 01110 is a string over. |x|=5 ---“thelengthof x is 5”. Theempty stringis denoted. ||=0. Concatenationxyof the stringsxandy--- the result of appendingy at the end ofx. k xk --- xx…x Language --- a set of strings

3.1.a Giorgi JaparidzeTheory of Computability Components of a Turing machine (TM) a a b a b b - - - - - - (Q,,,,start,accept,reject) • Q is the set of states •  is the input alphabet not containing the blank symbol - •  is the tape alphabet, where - and  •  is the transition function of the type Q Q{L,R} • start,accept,rejectQ, where rejectaccept; the states • accept and reject are called halting states. • There are no transitions from the halting states, and they immediately take effect! xy,R q1 q2 If the current tape symbol is x, replace it with y, move the head right and go to q2. The label xx,R is simply written as xR. If we here have L instead of R, then the head is moved left, unless it was on the first cell, in which case it remains where it was.

3.1.b Giorgi JaparidzeTheory of Computability 0  L x  L How a Turing machine works q5 xR -  R -  L xR start q2 q3 0 -,R 0x,R •  R • x  R -  R 0 x,R 0R xR q4 reject accept • Configuration: • Current state; • Tape contents; • Head position -  R -xq50x - x 0 x - - - - - - -

3.1.c1 Giorgi JaparidzeTheory of Computability Definitions A TM accepts an input string iff, on this input, sooner or later it enters the accept state. Otherwise the string is considered rejected. • Thus, the input is rejected in two cases: • The machine enters the reject state at some point, or • The machine never halts (never enters a halting state). A Turing machine is said to be a decider iff it halts on every input. The language recognized by a TM --- the set the strings that TM accepts; If this machine is a decider, then we say that it not only recognizes, but also decides that language. A language is said to be Turing-recognizable iff some TM recognizes it. A language is said to be Turing-decidable iff some TM decides it.

3.1.c2 Giorgi JaparidzeTheory of Computability Recognizing vs. deciding accept 0  R -  R -  R •  R • 0 R 0  R reject What language does the above machine recognize? Does it decide that language? accept 0  R -  R 0  R -  R reject What language does the above machine recognize? Does it decide that language?

3.1.d Giorgi JaparidzeTheory of Computability Diagrams without the reject state 0  L x  L xR q5 -  R -  L start xR q2 q3 0 -,R 0x,R -  R 0R 0 x,R accept xR q4 The reject state can be safely removed. It will be understood that all the missing transitions lead to the reject state.

3.1.e1 Giorgi JaparidzeTheory of Computability Designing TM: Example 1 Design a TM that recognizes (better – decides) {#n =m | n=m} 1. Sweep left to right across the tape, testing if the input has the form #*=*; if not, reject; if yes, go back to the beginning of the tape And go to step 2 (state q3). L = L x L - L R R = R - L # R start q1 q2 q9 #  L q3

3.1.e2 Giorgi JaparidzeTheory of Computability Designing TM: Example 1 2. Keep going to the right, replace the first  you see before = with x andgo to step 3 (state q5); if you reach = without seeing a , go to step 4 (state q6). xR x,R q4 q3 q5 # R = R q6

3.1.e3 Giorgi JaparidzeTheory of Computability Designing TM: Example 1 3. Keep going to the right, pass = and then replace the first  you see with x andgo to the beginning of the tape, step 2 (this can be done by going to state q9); if you reach a blank without seeing a , reject. q9 x,R R = R xR q7 q5

3.1.e4 Giorgi JaparidzeTheory of Computability Designing TM: Example 1 4. Keep going to the right as long as you see x. If you see a  before reaching a blank, reject; otherwise accept. xR - R accept q6

3.1.f Giorgi JaparidzeTheory of Computability Designing TM: Example 2 Design a TM that decides {#n <m | n<m} L < L x L L < L x L - L R R < R - L # R start q1 q2 q9 #  L x,R xR R x,R < R xR q4 q7 q3 q5 # R xR < R R accept q6

3.1.g Giorgi JaparidzeTheory of Computability Designing TM: Example 3 Design a TM that decides {#n +m = k | n+m=k} + L L = L x L - L R R R = R - L + R # R start q1 q2 q9 q0 #  L x,R xR +R R +R x,R = R xR q4 q7 q3 q5 # R xR = R - R accept q6

3.1.h Giorgi JaparidzeTheory of Computability Designing TM: Example 4 Design a TM that decides {#n m = k | nm=k} Step 1: Check if the string has the form #* * = * . If not, reject; If yes, go back to the beginning of the tape, step 2. Step 2: Find the first  between # and, delete it and go to step 3; If no suchwas found, go to step 5. Step 3: Find the first  between and =,delete it and go to step 4; If no suchwas found, go to the beginning of the tape, restoring on the way back all the deletedbetween and=,and go to step 2. Step 4: Find the first  after=, delete it, go back to ,and go to step 3; If no suchwas found before seeing a blank, reject. Step 5: Go right past = ; if no  is found there before reaching a blank, Accept; otherwise reject.       =       - #

3.1.i Giorgi JaparidzeTheory of Computability Design a fragment of a TM that, from a state “Beg?”, goes to a state “Yes” or “No”, depending on whether you are at the beginning of the tape or not, without corrupting the contents of the tape. Testing whether the head is at the beginning of the tape Tape alphabet: {x1,…,xn} Add a new tape symbol: $ • Read the current symbol, remember it, type $, and move left; • Read the current symbol. • If it is $, restore the remembered symbol and go to “Yes”. • If it is not $, move right, restore the remembered symbol and go to “No”. x1,…,xnR a1 b1 $ x1,L x1,R x1$,L … L Beg? Yes Temp No xn$,L $ xn,L xn,R x3 x7 an bn x1,…,xnR Beg?

3.1.j Giorgi JaparidzeTheory of Computability Implementing the “go to the beginning of the tape” operation Design a fragment of a TM that, from a state “Go to the beginning”, goes to the beginning of the tape and state “Done”. Move left and test if you are at the beginning of the tape. If yes, go to “Done”. If not, repeat the step. R No L L Done Go to the beginning Beg? L Yes R

3.1.k Giorgi JaparidzeTheory of Computability Design a fragment of a TM that, beginning from the current cell, shifts the contents of the tape right, typing in the current cell a 0. Assume the tape alphabet is {0,1,-}. Shifting tape contents • If the current symbol is blank, type 0 and you are done. • Read the current symbol, remember it, type 0 and move right. • While the current symbol is not -, remember it, type • the previously remembered symbol and move right. • 4. Once you see a blank, type the remembered symbol • and you are done. 00,R Remembered 0 00,R - 0,R - 0,R Shift Done Done 01,R 10,R - 1,R 10,R Remembered 1 11,R

3.1.l Giorgi JaparidzeTheory of Computability A TM for the element distinctness problem Design a TM that decides the language E={#x1#x2#…#xn | each xi{0,1}* and xixj for each ij} See page 147 for a description of such a TM # x1 # x2 # x3 # x4 #x5

3.2.a Giorgi JaparidzeTheory of Computability Turing machines with the “stay put” option A transition of this type of machine may have either L (move left), Or R (move right), or S (stay put). ab,S q1 q2 “Replace a with b and go to state q2 without moving the head” This does not increase the power of the machine, because the above transition can be simulated with the ordinary TM as follows: ab,R L q1 q2

3.2.b Giorgi JaparidzeTheory of Computability Multitape Turing machines A multitape TM has k tapes, each with its own read/write head. Initially, the input is written on the first tape, and all the other tapes are blank, with each head at the beginning of the corresponding tape. For a 3-tape TM, a transition will look like 0,1,11,0,1,R,R,L q1 q2 • “If you are in state q1 and see 0 on Tape1, 1 on Tape2 and 1 on Tape3, • type 1 on Tape1, 0 on Tape2 and 1 on Tape3; • move Head1 right, Head2 right and Head3 left; • go to state q2.”

3.2.c Giorgi JaparidzeTheory of Computability Multitape Turing machines: Example Design a fragment of a 2-tape TM that swaps the contents of the tapes, from the position where the heads are at the beginning and there are no blanks followed by non-blank symbols. Tape alphabet: {0,1,-} 0,00,0,R,R 0,11,0,R,R 0,-  -,0,R,R 1,00,1,R,R 1,11,1,R,R 1,-  -,1,R,R -,00,-,R,R -,11,-,R,R -,- - ,-,L,L Done Swap Tape1 Tape2 1 1 0 - 0 0 - -

3.2.d Giorgi JaparidzeTheory of Computability Simulating multitape TM with ordinary TM Theorem 3.13:Every multitape TM M has an equivalent single-tape TM S. Proof idea: The tape contents and the head positions of M can be represented on the single tape of S and correspondingly updated as shown on the following example for a 3-tape M; S follows the steps of M and accepts iff M accepts. Tape1 M: Tape2 Tape3 S: 0 0 1 - q0 - - - - - - - - 0 0 1 - - - - - - - - - - - -

3.2.e Giorgi JaparidzeTheory of Computability Nondeterministic Turing machines A nondeterministic TM is allowed to have more than one transition for a given tape symbol: A string is accepted iff one of the possible branches of computation takes us to the accept state. q2 ab,R q1 ac,L q3 Theorem 3.16:Every nondeterministic TM has an equivalent deterministic TM. Proof omitted. Idea: Simulate every possible branch of computation in a breadth-first manner.

3.2.f Giorgi JaparidzeTheory of Computability Turing machines with an output(not in the textbook!) The only difference with ordinary TM is that a TM with an output (TMO) has a state halt instead of accept and reject; if and when such a machine reaches the halt state, the contents of the tape (up to the first blank cell) will be considered its output. Example: Design a machine that, for every input w{0,1}*, returns w0.

3.2.g Giorgi JaparidzeTheory of Computability Computable functions A function g: * * is said to be computable, iff there is a TMO M such that for every input w* M returns the output u with u=g(w). In this case we say that M computes g. Example: Let f: {0,1}* {0,1}* be the function defined by f(w)=w0, so that f()=0, f(0)=00, f(1)=10, f(00)=000, f(01)=010, etc. Then obviously f is computable. The graph of a function g: * * is the language {(w,u) | w*, u=g(w)} E.g., the graph of the above function f is {(,0), (0,00), (1,10), (00,000), (01,010), …}, i.e. {(w,u) | w{0,1}*, u=w0}

3.2.h Giorgi JaparidzeTheory of Computability Computability vs. decidability Theorem:A function is computable iff its graph is decidable. Proof sketch. Let g: * *be a function. (): Suppose C is a TMO that computes g. Construct a TM D that works as follows: D = “On input t: 1. If t does not have the form (w,u), where w,u*,thenreject. Otherwise, 2. Simulate C for input w. If it returns u, accept; otherwise reject.” (): Suppose a TM D decides the graph of g. Let s1,s2,s3,… be the lexicographic list of all strings over the alphabet . Construct a TMO C that works as follows: C =“On input t: Simulate D for each of the inputs (t,s1), (t,s2),(t,s3),… until you find si suct that (t,si) is accepted, and return this si”

3.3.a Giorgi JaparidzeTheory of Computability The intuitive notion of algorithms: An algorithm (procedure, recipe) is a collection of simple instructions for carrying out some task. Algorithms • Examples: • The elementary school algorithms for adding, subtracting, multiplying, • dividing, etc.; • Calculating students’ average grades; • Sorting algorithms; • Any other procedures that Professor Zhu DaMing has taught you. To carry out an algorithm, you only need to mechanically (blindly) follow its instructions. No understanding of the objects you are manipulating, no understanding of what you are doing is needed! We need a strict mathematical definition of an algorithm. The above “definition” obviously does not qualify.

3.3.b Giorgi JaparidzeTheory of Computability The structure of an algorithm: Input(s) Output or Input(s) Yes (accept) or No (reject) Algorithms should deal with strings The attempt to formalize the notion of algorithm should start from strictly specifying what kind of objects can be inputs/outputs. These are going to be only. strings No loss of generality: non-string objects can be easily represented with strings. A tuple of such objects can also be represented with a (one single) string. After all, computers exclusively deal with strings --- strings over the alphabet {0,1} !

3.3.c Giorgi JaparidzeTheory of Computability Encoding Encoding--- representing objects as strings in some standard way. There are usually many ways to encode; as one encoding can be easily (algorithmically) converted into another, it does not matter which encoding we choose. Encoding numbers:0,1,2,3,…,34,…, or 0,1,10,11,…,100010,… Encoding pairs of numbers:(4,2), (1,3), …, or 4#2, 1#3,…, or aaaabb, abbb, … Encoding people:“David Stewart”, “Dongfei Wei”, …, or “Stewart, David”, “Wei, Dungfei”, or 315-21-5541, 326-11-1521… Encoding assembly language instructions: Their machine language equivalents. If O is an object, <O> will stand for a (the) encoding of O.

3.3.d Giorgi JaparidzeTheory of Computability Encoding graphs 1 2 3 G = <G> = (1,2,3,4,5)((1,4),(1,5),(2,3),(4,5)) 4 5 Algorithm M for checking if a graph is connected: M = “On input <G>, the encoding of a graph G: 1. Select the first node of G and mark it. 2. Repeat the following stage until no new nodes are marked: 3. For each node in G, mark it if it is attached by an edge to a node that is already marked. 4. Scan all the nodes of G to determine whether they all are marked. If they are, accept; otherwise reject.”

3.3.e Giorgi JaparidzeTheory of Computability The Church-Turing thesis The Church-Turing thesis: (the intuitive notion of) Algorithm Turing machine = It can never be mathematically proven! Importance: Without the Church-Turing thesis, you can never prove that a certain problem has no algorithmic solution. Illustrative example Hilbert’s 10th problem (see page 154 of textbook).

The Church-Turing Thesis