Chapter 3 Church-Turing Thesis. CS 341: Foundations of CS II

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1 CS 341: Foundations of CS II Marvin K. Nakayama Computer Siene Department New Jersey Institute of Tehnology Newark, NJ CS 341: Chapter Contents Turing Mahines Turing-reognizable Turing-deidable Variants of Turing Mahines Algorithms Enoding input for TM Chapter 3 Churh-Turing Thesis CS 341: Chapter DFA PDA Previous Mahines Reads input from left to right Finite ontrol (i.e., transition funtion) based on urrent state, urrent input symbol read. Has stak for extra memory Reads input from left to right Can read/write to memory (stak) by popping/pushing Finite ontrol based on urrent state, what s read from input, what s popped from stak. CS 341: Chapter Turing mahine (TM) Infinitely long tape, divided into ells, for memory Tape initially ontains input string followed by all blanks Tape head ( ) an move both right and left Can read from and write to tape Finite ontrol based on urrent state, urrent symbol that head reads from tape. Mahine has one aept state and one rejet state. Mahine an run forever: infinite loop.

2 CS 341: Chapter Key Differene between TMs and Previous Mahines Turing mahine an both read from tape and write on it. CS 341: Chapter Example: Mahine for reognizing language A = { s#s s {0, 1} } Tape head an move both right and left. Idea: Zig-zag aross tape, rossing off mathing symbols. Tape is infinite and an be used for storage. Aept and rejet states take immediate effet. Consider string 01101#01101 A. Tape head starts over leftmost symbol # Reord symbol in ontrol and overwrite it with X X # San right: rejet if blank enountered before # CS 341: Chapter When # enountered, move right one ell. X # If urrent symbol doesn t math previously reorded symbol, rejet. Overwrite urrent symbol with X X # X San left, past # to X Move one ell right Reord symbol and overwrite it with X X X # X CS 341: Chapter After several more iterations of zigzagging, we have X X X X X # X X X X X... After all symbols left of # have been mathed to symbols right of #, hek for any remaining symbols to the right of #. If blank enountered, aept. If 0 or 1 enountered, rejet. X X X X X # X X X X X... The string that is aepted or not by our mahine is the original input string 01101# San right past # to (last) X and move one ell to right...

3 CS 341: Chapter Desription of TM M 1 for { s#s s {0, 1} } M 1 = On input string w: 1. San input to be sure that it ontains a single #. If not, rejet. 2. Zig-zag aross tape to orresponding positions on either side of the # to hek whether these positions ontain the same symbol. If they do not, rejet. Cross off symbols as they are heked off to keep trak of whih symbols orrespond. 3. When all symbols to the left of # have been rossed off along with the orresponding symbols to the right of #, hek for any remaining symbols to the right of the #. If any symbols remain, rejet; otherwise,aept. CS 341: Chapter Formal Definition of Turing Mahine Definition: A Turing mahine (TM) is a 7-tuple M =(Q, Σ, Γ, δ,q 0,qaept, q rejet ),where Q is a finite set of states Σ is the input alphabet not ontaining blank symbol Γ is tape alphabet with blank Γ and Σ Γ δ : Q Γ Q Γ {L, R} is the transition funtion, where L means move tape head one ell to left R means move tape head one ell to right q 0 Q is the start state qaept Q is the aept state q rejet Q is the rejet state, withq rejet qaept. CS 341: Chapter Transtion Funtion of TM Transition funtion δ : Q Γ Q Γ {L, R} δ(q, a) =(s, b, L) means if TM in state q Q, and tape head reads tape symbol a Γ, then TM moves to state s Q overwrites a with b Γ moves head left (i.e., L {L, R}) q a b, L read write, move s Before After a b a a a b b a CS 341: Chapter Start of TM Computation M =(Q, Σ, Γ,δ,q 0,qaept,q rejet ) begins omputation as follows: Given input string w = w 1 w 2 w n Σ with eah w i Σ, i.e., w is a string of length n for some n 0. TM begins in start state q 0 Input string is on n leftmost tape ells Rest of tape ontains blanks w 1 w 2 w 3 w n... Head starts on leftmost ell of tape Beause Σ, first blank denotes end of input string.

4 CS 341: Chapter TM Computation When omputation on TM M =(Q, Σ, Γ,δ,q 0,qaept,q rejet ) starts, TM M proeeds aording to transition funtion δ : Q Γ Q Γ {L, R} a b, L q s read write, move If M tries to move head off left end of tape, then head remains on first ell. Computation ontinues until qaept or q rejet is entered. Otherwise, M runs forever: infinite loop. In this ase, input string is neither aepted nor rejeted. CS 341: Chapter Example: Turing mahine M 2 reognizing language A = { 0 2n n 0 }, whih onsists of strings of 0s whose length is a power of 2. Idea: The number k of zeros is a power of 2 iff suessively halving k always results in a power of 2 (i.e., eah result > 1 is never odd). M 2 = On input string w: 1. Sweep left to right aross the tape, rossing off every other If in stage 1 the tape ontained a single 0, aept. 3. If in stage 1 the tape ontained more than a single 0 and the number of 0s was odd, rejet. 4. Return the head to the left end of the tape. 5. Go to stage 1. CS 341: Chapter Run TM M 2 with Input 0000 Tape initially ontains input Run stage 1: Sweep left to right aross tape, rossing off every other 0. x 0 x... (Put in first ell to mark beginning of tape.) CS 341: Chapter Diagram of TM for { 0 2n n 0 } 0 0,L x x, L q 5,L Run stage 4: Return head to left end of tape (marked by ). x 0 x... Run stage 1: Sweep left to right aross tape, rossing off every other 0. x x x... Run stages 4 and 1: Return head to left end and san tape. Run stage 2: If in stage 1 the tape ontained a single 0, aept. q 1 q 2 q 3 0,R 0 x, R 0 x, R q rejet q aept q 4 0 0,R

5 CS 341: Chapter 3 0 0,L 3-17 x x, L Run TM on input w = 0000 Step State Tape q 1 q 2 q 3 0,R 0 x, R 0 0,R 0 x, R q rejet 0 q q q 3 x q 4 x q aept q 5,L Step State Tape q 4 4 q 3 x 0 x... 5 q 5 x 0 x... 6 q 5 x 0 x CS 341: Chapter TM for { 0 2n n 0 } Turing mahine M 2 =(Q, Σ, Γ,δ,q 1,qaept,q rejet ),where Q = {q 1,q 2,q 3,q 4,q 5,qaept,q rejet } Σ={0} Γ={0,x, } q 1 is start state qaept is aept state q rejet is rejet state Transition funtion δ : Q Γ Q Γ {L, R} is speified in previous diagram. For example, δ(q 4, 0) = (q 3,x,R) δ(q 3, ) =(q 5,,L) CS 341: Chapter Computation hanges urrent state urrent head position tape ontents TM Configurations q 2 State Tape Configuration provides snapshot of TM at any point during omputation: urrent state q Q urrent tape ontents Γ urrent head loation CS 341: Chapter TM Configurations Configuration 1011q 2 01 means urrent state is q 2 LHS of tape is 1011 RHS of tape is 01 head is on RHS 0 q 2 State Tape Definition: a onfiguration of a TM M =(Q, Σ, Γ,δ,q 0,qaept,q rejet ) is a string uqv with u, v Γ and q Q, and speifies that urrently M is in state q tape ontains uv tape head is pointing to the ell ontaining the first symbol in v.

6 CS 341: Chapter TM Transitions Definition: Configuration C 1 yields onfiguration C 2 if the Turing mahine an legally go from C 1 to C 2 in a single step. Speifially, for TM M =(Σ, Γ,δ,q 0,qaept,q rejet ), suppose u, v Γ a, b, Γ q i,q j Q transition funtion δ : Q Γ Q Γ {L, R}. Then onfiguration uaq i bv yields onfiguration uaq j v if CS 341: Chapter TM Transitions Similarly, onfiguration uaq i bv yields onfiguration uq j av if δ(q i,b)=(q j,,l). q i b, L q j Before u a b v After u a v δ(q i,b)=(q j,,r). q i b, R q j Before u a b v After u a v CS 341: Chapter Speial ase: q i bv yields q j v if q i TM Transitions δ(q i,b)=(q j,,l) If head is on leftmost ell of tape and tries to move left, then it stays in same plae. b, L q j Before After b v v CS 341: Chapter Remarks on TM Configurations Consider TM M =(Q, Σ, Γ, δ,q 0,qaept, q rejet ). Starting onfiguration on input w Σ is q 0 w An aepting onfiguration is uqaeptv for some u, v Γ A rejeting onfiguration is uq rejet v for some u, v Γ Aepting and rejeting onfigurations are halting onfigurations. Configuration wq i is the same as wq i

7 CS 341: Chapter 3 0 0,L 3-25 x x, L q 1 q 2 q 3 0,R 0 x, R 0 0,R 0 x, R q rejet q aept q 5,L On input 0000, get following sequene of onfigurations: q 4 q , q 2 000, xq 3 00, x0q 4 0, x0xq 3, x0q 5 x, xq 5 0x, q 5 x0x, q 5 x0x, q 2 x0x, xq 2 0x, xxq 3 x, xxxq 3, xxq 5 x, xq 5 xx, q 5 xxx, q 5 xxx, q 2 xxx, xq 2 xx, xxq 2 x, xxxq 2, xxx qaept. CS 341: Chapter Formal Definition of TM Computation Turing mahine M =(Q, Σ, Γ,δ,q 0,qaept,q rejet ). Input string w Σ. Definition: M aepts input w if there is a finite sequene of onfigurations C 1,C 2,...,C k for some k 1 with C 1 is the starting onfiguration q 0 w C i yields C i+1 for all i =1,...,k 1 sequene of onfigurations obeys transition funtion δ C k is an aepting onfiguration uqaeptv for some u, v Γ. Definition: The set of all input strings aepted by TM M is the language reognized by M and is denoted by L(M). Note that L(M) Σ. CS 341: Chapter Turing-reognizable Definition: Language A is Turing-reognizable if there is a TM M suh that A = L(M). Remarks: Also alled a reursively enumerable or enumerable language. On an input w L(M), themahinem an either halt in a rejeting state, or it an loop indefinitely How do you distinguish between a very long omputation and one that will never halt? Turing-reognizable not pratial beause never know if TM will halt. CS 341: Chapter Turing-deidable Definition: A deider is TM that halts on all inputs, i.e., never loops. Definition: Language A = L(M) is deided by TM M if on eah possible input w Σ, the TM finishes in a halting onfiguration, i.e., M ends in qaept for eah w A M ends in q rejet for eah w A. Definition: Lang A is Turing-deidable if TM M that deides A. Remarks: Also alled a reursive or deidable language. Differenes between Turing-deidable language A and Turing-reognizable language B A has TM that halts on every string w Σ. TM for B maylooponstringsw B.

8 CS 341: Chapter Desribing TMs It is assumed that you are familiar with TMs and with programming omputers. Clarity above all: high-level desription of TMs is allowed M = On input string w: 1. San input... but it should not be used as a trik to hide the important details of the program. Standard tools: Expanding tape alphabet Γ with separator # dotted symbols 0, a, to indiate ativity, as we ll see later. Typial example: Γ={0, 1, #,, 0, 1} CS 341: Chapter Example: Turing mahine M 3 to deide language C = { a i b j k i j = k and i, j, k 1 }. Idea: If i olletions of j things eah, then i j things total. TM: for eah a, rossoffj s by mathing eah b with a. M 3 = On input string w: 1. San the input from left to right to make sure that it is amemberofl(a b ),andrejet if it isn t. 2. Return the head to the left-hand end of the tape 3. Cross off an a and san to the right until a b ours. Shuttle between the b s and the s, rossing off eah until all b s are gone. If all s have been rossed off and some b s remain, rejet. 4. Restore the rossed off b s and repeat stage 3 if there is another a to ross off. If all a s are rossed off, hek whether all s also are rossed off. If yes, aept; otherwise, rejet. CS 341: Chapter Running TM M 3 on Input a 3 b 2 6 C Tape head starts over leftmost symbol a a a b b... Stage 1: Mark leftmost symbol and san to see if input L(a b ) A a a b b... Stage 3: Cross off one a and ross off mathing b s and s A a a b b... Stage 4: Restore b sandreturnheadtofirsta not rossed off CS 341: Chapter Stage 3: Cross off one a and ross off mathing b s and s A a a b b... Stage 4: Restore b sandreturnheadtofirsta not rossed off A a a b b... Stage 3: Cross off one a and ross off mathing b s and s A a a b b... Stage 4: If all a s rossed off, hek if all s rossed off. aept A a a b b...

9 CS 341: Chapter TM Triks CS 341: Chapter Variant of TM: k-tape Question: How to tell when a TM is at the left end of the tape? One Approah: Mark it with a speial symbol. Alternative method: remember urrent symbol overwrite it with speial symbol move left if speial symbol still there, head is at start of tape otherwise, restore previous symbol and move left. 3-tape TM Eah tape has its own head. Transitions determined by urrent state, and what all the heads read. Tape Tape Tape Eah head writes and moves independently of other heads. CS 341: Chapter k-tape Turing Mahine Definition: A k-tape Turing mahine M =(Q, Σ, Γ, δ,q 0,qaept, q rejet ) has k different tapes and k different read/write heads: Q is finite set of states Σ is input alphabet (where Σ) Γ is tape alphabet with ({ } Σ) Γ q 0 is start state Q qaept is aept state Q q rejet is rejet state Q δ is transition funtion δ : Q Γ k Q Γ k {L, R} k where Γ k =Γ } Γ Γ {{}. k times CS 341: Chapter Transition funtion Suppose Multi-Tape TM δ : Q Γ k Q Γ k {L, R} k δ(q i,a 1,a 2,...,a k )=(q j,b 1,b 2,...,b k,l,r,...,l) Interpretation: If mahine is in state q i,and heads 1 through k read a 1,...a k, then mahine moves to state q j heads 1 through k write b 1,...,b k eah head moves left (L) orright(r) as speified.

10 CS 341: Chapter Multi-Tape TM Equivalent to 1-Tape TM Theorem 3.13 For every multi-tape TM M, there is a single-tape TM M suh that L(M) =L(M ). CS 341: Chapter Basi Idea of Proof of Theorem 3.13 Simulate k-tape TM using 1-tape TM Tape Remarks: In other words, for every multi-tape TM M, thereisanequivalent single-tape TM M. Proving and understanding this kind of robustness result is essential for appreiating the power of the TM model. 3-tape TM Tape Tape We will onsider different variants of TMs, and show eah has equivalent basi TM. Equivalent 1-tape TM Tape # # 0 0 # # CS 341: Chapter Proof of Theorem 3.13 Let M k =(Q, Σ, Γ,δ,q 0,qaept,q rejet ) be a k-tape TM. Initially, M k has input w = w 1 w n on tape 1 other tapes ontain only blanks eah head points to first ell. Tape 1 w 1 w 2 w n Tape 2 Tape 3 Construt 1-tape TM M 1 with expanded tape alphabet Γ = Γ Γ {#} Head positions are marked by dotted symbols. CS 341: Chapter Proof of Theorem 3.13 On input w = w 1 w n,the1-tape TM M 1 does the following: First M 1 prepares initial string on single tape: # w 1 w2 w n # # # For eah step of M k,tmm 1 sans tape twie 1. Sans its tape from first # (whih marks left end of tape) to (k +1)st # (whih marks right end of used part of tape) to read symbols under virtual heads 2. Resans to write new symbols and move heads If M 1 tries to move virtual head to the right onto #, then M k is trying to move head onto unused blank ell. So M 1 has to write blank on tape and shift rest of tape right one ell.

11 CS 341: Chapter Turing-reognizable k-tape TM From Theorem 3.13, we get the following: Corollary 3.15 Language L is TM-reognizable if and only if some multi-tape TM reognizes L. CS 341: Chapter Nondeterministi TM Definition: A nondeterministi Turing mahine (NTM) M an have several options at every step. It is defined by the 7-tuple where Q is finite set of states M =(Q, Σ, Γ, δ,q 0,qaept, q rejet ), Σ is input alphabet (without blank ) Γ is tape alphabet with { } Σ Γ q 0 is start state Q qaept is aept state Q q rejet is rejet state Q δ is transition funtion δ : Q Γ P(Q Γ {L, R}) CS 341: Chapter Transition Funtion δ of NTM δ : Q Γ P(Q Γ {L, R}) CS 341: Chapter Computing With NTMs On any input w, evolution of NTM represented by a tree of onfigurations (rather than a single path). If (at least) one aepting leaf, then NTM aepts. a, L q j C 1 t =1 q i, R q k C 2 C 3 C 4 t =2 a, L d, R q l C 5 C 6 C 7 C 8 C 9 t =3 Multiple hoies when in state q i and reading from tape: rejet C 10 aept t =4 δ(q i,)={ (q j,a,l), (q k,,r), (q l,a,l), (q l,d,r) }

12 CS 341: Chapter NTM Equivalent to TM Theorem 3.16 Every nondeterministi TM N has an equivalent deterministi TM D. Proof Idea: Build TM D to simulate NTM N on eah input w. D tries all possible branhes of N s tree of onfigurations. If D finds any aepting onfiguration, then it aepts input w. If all branhes rejet, then D rejets input w. If no branh aepts and at least one loops, then D loops on w. CS 341: Chapter Proof of Equivalene of NTM and TM On eah input w, NTMN s omputation is a tree Eah branh is branh of nondeterminism. Eah node is a onfiguration arising from running N on w. Root is starting onfiguration. TM D searhes through tree to see if it has an aepting onfiguration. Depth-first searh (DFS) doesn t work. Why? Breadth-first searh (BFS) works. Tree doesn t atually exist. So TM D needs to build tree as it searhes through it. CS 341: Chapter Proof of Equivalene of NTM and TM Simulating TM D has 3 tapes 1. Input tape ontains input string w never altered 2. Simulation tape used as N s tape when simulating N s exeution on some path in N s omputation tree. 3. Address tape keeps trak of urrent loation of BFS of N s omputation tree. CS 341: Chapter Address Tape Works as Follows Every node in the tree has at most b hildren. b is size of largest set of possible hoies for N s transition funtion. Everynodeintreehasanaddress that is a string over the alphabet To get to node with address 231 start at root take seond branh then take third branh then take first branh Ignore meaningless addresses. Γ b = {1, 2,...,b} Visit nodes in BFS order by listing addresses in Γ b in string order: ε, 1, 2,..., b, 11, 12,..., 1b, 21, 22,...

13 CS 341: Chapter Proof of Equivalene of NTM and TM aept onfiguration has address 231. Configuration C 6 has address 12. Configuration C 1 has address ε. Address 132 is meaningless. C 1 t =1 C 2 C 3 C 4 t =2 C 5 C 6 C 7 C 8 C 9 t =3 rejet C 10 aept t =4 CS 341: Chapter Simulating TM D Works as Follows 1. Initially, input tape ontains input string w. Simulation and address tapes are initially empty. 2. Copy input tape to simulation tape. 3. Use simulation tape to simulate NTM N on input w on path in tree from root to the address on address tape. At eah node, onsult next symbol on address tape to determine whih branh to take. Aept if aepting onfiguration reahed. Skip to next step if symbolsonaddresstapeexhausted nondeterministi hoie invalid rejeting onfiguration reahed 4. Replae string on address tape with next string in Γb in string order, andgotostage2. CS 341: Chapter Remarks on TM Variants Corollary 3.18 Language L is Turing-reognizable iff a nondeterministi TM reognizes it. Proof. Every nondeterministi TM has an equivalent 3-tape TM 1. input tape 2. simulation tape 3. address tape 3-tape TM, in turn, has an equivalent 1-tape TM by Theorem CS 341: Chapter TM Deidable NTM Deidable Definition: A nondeterministi TM is a deider if all branhes halt on all inputs. Remark: Can modify proof of previous theorem (3.16) so that if NTM N always halts on all branhes, then TM D will always halt. Corollary 3.19 A language is deidable iff some nondeterministi TM deides it. Remarks: k-tape TMs and NTMs are not more powerful than standard TMs: The Turing mahine model is extremely robust.

14 CS 341: Chapter Enumerators Remarks: Reall: a language is enumerable if some TM reognizes it. But why enumerable? Definition: An enumerator is a TM with a printer TM takes no input TM simply sends strings to printer may reate infinite list of strings dupliates may appear in list enumerates a language CS 341: Chapter Enumerators Theorem 3.21 Language A is Turing-reognizable iff some enumerator enumerates it. Proof. Must show 1. If E enumerates language A, thensometmm reognizes A. 2. If TM M reognizes A, then some enumerator E enumerates A. To show 1, given enumerator E, build TM M for A using E as blak box: M = On input string w, 1. Run E. 2. Every time E outputs a string, ompare it to w. 3. If w is output, aept. CS 341: Chapter Seond Half of Proof of Theorem 3.21 We now show 2: If TM M reognizes A, then some enumerator E enumerates A. Let s 1,s 2,s 3,... be a list of all strings in Σ Given TM M, definee using M as blak box as follows: Repeat the following for i =1, 2, 3,... Run M for i steps on eah input s 1,s 2,...,s i. If any omputation aepts, print out orresponding string s Note that dupliates may appear. CS 341: Chapter Algorithm is Independent of Computation Model All reasonable variants of TM models are equivalent to TM: k-tape TM nondeterministi TM enumerator random-aess TM: head an jump to any ell in one step. Similarly, all reasonable programming languages are equivalent. Can take program in LISP and onvert it into C, and vie versa. The notion of an algorithm is independent of the omputation model.

15 CS 341: Chapter Algorithms CS 341: Chapter Hilbert s 10th Problem What is an algorithm? Informally a reipe a proedure a omputer program Historially, algorithms have long history in mathematis but not preisely defined until 20th entury informal notions rarely questioned, but insuffiient to show a problem has no algorithm. Muḥammad ibn Mūsā al-khwārizmī ( ) soure: wikipedia David Hilbert ( ) soure: wikipedia In 1900, David Hilbert delivered a now-famous address Presented 23 open mathematial problems Challenge for the next entury 10th problem onerned algorithms and polynomials CS 341: Chapter Polynomials A term is a produt of variables and a onstant oeffiient: A polynomial is a sum of terms: 6x 3 yz 2 6x 3 yz 2 +3xy 2 x 3 10 A root of a polynomial is an assignment of values to variables so that the value of the polynomial is zero. The above polynomial has a root at (x, y, z) =(5, 3, 0). We are interested in integral roots. Some polynomials have integral roots; some don t. Neither 21x 2 81xy +1nor x 2 2 has an integral root. CS 341: Chapter Hilbert s 10th Problem Problem: Devise an algorithm that tests whether a polynomial has an integral root. In Hilbert s words: to devise a proess aording to whih it an be determined by a finite number of operations... Hilbert seemed to assume that suh an algorithm exists. However, Matijasevi proved in 1970 that no suh algorithm exists. Mathematiians in 1900 ouldn t have proved this. No formal notion of an algorithm existed. Informal notions work fine for onstruting algorithms. Formal notion needed to show no algorithm exists for a problem.

16 CS 341: Chapter Churh-Turing Thesis CS 341: Chapter Hilbert s 10th Problem Alonzo Churh ( ) soure: wikipedia Alan Turing ( ) soure: wikipedia Consider language D = { p p is a polynomial with an integral root }. Sine 6x 3 yz 2 +3xy 2 x 3 10 has an integral root at (x, y, z) =(5, 3, 0), Formal notion of algorithm developed in 1936 λ-alulus of Alonzo Churh Turing mahines of Alan Turing Definitions appear very different, but are equivalent. Churh-Turing Thesis The informal notion of an algorithm orresponds exatly to a Turing mahine that halts on all inputs. 6x 3 yz 2 +3xy 2 x 3 10 D. Sine 21x 2 81xy +1has no integral root, 21x 2 81xy +1 D. Hilbert s 10th problem asks whether this language is deidable. i.e., Is there a TM that deides D? D is not deidable, but it is Turing-reognizable. CS 341: Chapter Hilbert s 10th Problem Consider simpler language of polynomials over single variable: D 1 = { p p is a polynomial over x with an integral root } D 1 is reognized by following TM M 1 : On input p, whih is a polynomial over variable x 1. Evaluate p with x set suessively to values 0, 1, 1, 2, 2, 3, 3, If at any point the polynomial evaluates to 0, aept. CS 341: Chapter Hilbert s 10th Problem It turns out, though, that D 1 is deidable. Can show that the roots of p (over single variable x) lie between where ±k max 1 k is number of terms in polynomial max is maximum oeffiient 1 is oeffiient of highest-order term Note that Thus, only have to hek integers between k max 1 and k max 1. If p has an integral root, the mahine eventually aepts. If not, mahine loops. M 1 reognizes D 1, but does not deide D 1. Matijasevi proved suh bounds don t exist for multivariate polynomials.

17 CS 341: Chapter Enoding Input to a Turing mahine is a string of symbols over an alphabet. ButwewantTMs(algorithms)thatworkon graphs languages Turing mahines et. Need to enode an objet as a string of symbols over an alphabet. Can often do this in many reasonable ways. We sometimes distinguish between an objet X its enoding X. CS 341: Chapter Undireted graph G Enoding an Undireted Graph 1 One possible enoding 2 3 G =(1, 2, 3, 4) ((1, 2), (1, 3), (2, 3), (3, 4) ) }{{}}{{} nodes edges In this enoding sheme, G of graph G is string of symbols over some alphabet Σ, where the string starts with list of nodes followed by list of edges 4 CS 341: Chapter Conneted Graphs Definition: An undireted graph is onneted if every node an be reahed from any other node by travelling along edges CS 341: Chapter Deision Problems Deision problem: (omputational) question with YES/NO answer. Answer depends on partiular value of input to question. Example: Graph onnetedness problem: Is an undireted graph onneted? 3 3 Graph G 1 2 Graph G 2 2 Conneted graph G 1 Unonneted graph G Example: Let A be the language onsisting of strings representing onneted undireted graphs: A = { G G is a onneted undireted graph }. So G 1 A and G 2 A. 3 Input to question is undireted graph. For input G 1,answerisYES. For input G 2,answerisNO. 3

18 CS 341: Chapter Instane and Language of Deision Problem Instane of deision problem is speifi input value to question. Instane is enoded as string over some alphabet Σ. YES instane has answer YES. NO instane has answer NO. Universe Ω of a deision problem omprises all instanes. Language of a deision problem omprises all its YES instanes. Example: For graph onnetedness problem, Universe onsists of (enodings of) every undireted graph G: Ω={ G G is an undireted graph } Language A of deision problem CS 341: Chapter Proving a Language is Deidable Reall for graph onnetedness problem, Ω={ G G is an undireted graph }, A = { G G is a onneted undireted graph }. To prove A is deidable language, need to show TM that deides A. For a TM M to deide A, thetmmust take any instane G Ω as input halt and aept if G A halt and rejet if G A (i.e., never loops indefinitely) A = { G G is a onneted undireted graph } is subset of universe; i.e., A Ω CS 341: Chapter Graph G 1 1 TM to Deide if Graph is Conneted A = { G G is a onneted undireted graph } Graph G 2 1 M = On input G Ω, whereg is an undireted graph: 0. Chek if G is a valid graph enoding. If not, rejet. 1. Selet first node of G and mark it. 2. Repeat until no new nodes marked: 3. For eah node in G, mark it if it s attahed by an edge to a node already marked. 4. San all nodes of G to see whether they all are marked. If they are, aept; otherwise, rejet CS 341: Chapter TM M for Deiding Language A For TM M that deides A = { G G is a onneted undireted graph } Stage 0 heks that input G Ω is valid graph enoding, e.g., two lists first is a list of numbers seond is a list of pairs of numbers first list ontains no dupliates every node in seond list appears in first list Stages 1 4 then hek if G is onneted. When defining a TM, we often do not expliitly inlude stage 0 to hek if the input is a valid enoding. Instead, the hek is often only impliitly inluded.

19 CS 341: Chapter Hierarhy of Languages (so far) All languages??? Turing-reognizable TM, k-tape TM, NTM, enumerator,... Deidable Deider (deterministi, nondet, k-tape,...) Context-free CFG, PDA Regular DFA, NFA, Reg Exp Examples??? { 0 n 1 n 2 n n 0 } { 0 n 1 n n 0 } (0 1) Finite { 110, 01 } CS 341: Chapter Turing mahines Summary of Chapter 3 tape head an move right and left tape head an read and write TM omputation an be expressed as sequene of onfigurations Language is Turing-reognizable if some TM reognizes it But TM may loop forever on input string not in language Language is Turing-deidable if a TM deides it (must always halt) Variants of TM (k-tape, nondeterministi, et.) have equivalent TM Churh-Turing Thesis Informal notion of algorithm is same as deiding by TM. Hilbert s 10th problem undeidable. Enoding TM input and deision problems.