# Automata Theory CS F-15 Undecidiability

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1 Automata Theory CS F-15 Undecidiability David Galles Department of Computer Science University of San Francisco

2 15-0: Universal TM Turing Machines are Hard Wired Addition machine only adds 0 n 1 n 2 n machine only determines if a string is in the language 0 n 1 n 2 n Have seen one Programmable TM Random Access Computer TM

3 15-1: Universal TM We can create a Universal Turing Machine Takes as input a description of a Turing Machine, and the input string for the Turing Machine Simulates running the machine on the input string Turing Machine Interpreter Writing a Java Interpreter in Java, for instance, is not all that strange essentially what we are doing with Turing Machines

4 15-2: Encoding a Turing Machine Our Universal Turing Machine needs to have a specific, finite alphabet of tape symbols We need to be able to simulate any Turing Machine with any tape alphabet Use an encoding scheme

5 15-3: Encoding a Turing Machine Tape alphabet Σ for Universal Turing Machine: q a 0 1, ( ) Encoded states: q001, q010, q011, q100,... Encoded Tape symbols a001, a010, a011, a100,...

6 15-4: Encoding a Turing Machine Turing Machine that changes all a s to b s, and all b s to a s

7 15-5: Encoding a Turing Machine b a R b yes a q 0 q 1 a b (q 2, ) (q 1,b) (q 1,a) (q 0, ) (q 0, ) (q 0, ) q 3

8 15-6: Encoding a Turing Machine a b q 0 (q 2, ) (q 1,b) (q 1,a) q 1 (q 0, ) (q 0, ) (q 0, ) q 3 Symbol Encoding a000 a001 ( always symbol 1, always symbol 10 2 ) a010 a a011 b a100 (q00, a000, q10, a000)(q00, a011, q01, a100)(q00, a100, q01, a011) (q01, a000, q00, a010)(q01, a011, q00, a100)(q01, a100, q00, a100)

9 15-7: Encoding a Turing Machine Halting states can be coded implicitly No outgoing edges = halting state If we want a yes and no state First halting state is yes Second halting state is no

10 15-8: Encoding a Turing Machine Given any Turing Machine M, we can create an encoding of the machine, e(m) Some machines will require more digits to represent states & symbols Why used q and a separators We can actually encode any Turing Machine (and any tape) using just 0 s and 1 s (more on this in a minute)

11 15-9: Universal Turing Machine Takes as input an encoding of a Turing Machine e(m), and an encoding of the input tape e(w) Simulates running M on w 3-Tape Machine: Simulated Tape Current State Tape Transition Function Tape

12 15-10: Universal Turing Machine Input Tape a a a a a a a State Tape q Transition Function Tape ( q 0 0, a 0 0 0, q 1 0 a ) ( q 0 0, a 0 1 1, q 0 1, a ) ( q 0 0, a

13 15-11: Encoding a Turing Machine Encoding a Turing Machine using just 0 s and 1 s: If we knew how many states there were, and how many symbols in the input alphabet, we wouldn t need the separators a q ( ), Each encoding start with the # of digits used for states, and the # of digits used for alphabet symbols, in unary. (q00, a000, q10, a000) (q00, a011, q01, a100) (q00, a100, q01, a011), (q01, a000, q00, a010) (q01, a011, q00, a100) (q01, a100, q00, a100)

14 15-12: Halting Problem e(m) e(w) Halting Machine takes as input an encoding of a Turing Machine e(m) and an encoding of an input string e(w), and returns yes if M halts on w, and no if M does not halt on w. Like writing a Java program that parses a Java function, and determines if that function halts on a specific input Halting Machine yes no

15 15-13: Language vs. Problem Brief Interlude We will use Language and Problem interchangeably Any Problem can be converted to a Language, and vice-versa Problem: Multiply two numbers x and y Language: L = {x;y;z : z = x y}

16 15-14: Language vs. Problem Brief Interlude We will use Language and Problem interchangeably Any Problem can be converted to a Language, and vice-versa Problem: Determine if a number is prime Language: L = {p : p is prime }

17 15-15: Language vs. Problem Brief Interlude We will use Language and Problem interchangeably Any Problem can be converted to a Language, and vice-versa Problem: Determine if a Turing Machine M halts on an input string w Language: L = {e(m),e(w) : M halts on w}

18 15-16: Halting Problem Halting Machine takes as input an encoding of a Turing Machine e(m) and an encoding of an input string e(w), and returns yes if M halts on w, and no if M does not halt on w. Like writing a Java program that parses a Java function, and determines if that function halts on a specific input How might the Java version work?

19 15-17: Halting Problem Halting Machine takes as input an encoding of a Turing Machine e(m) and an encoding of an input string e(w), and returns yes if M halts on w, and no if M does not halt on w. Like writing a Java program that parses a Java function, and determines if that function halts on a specific input How might the Java version work? Check for loops while (<test>) <body> Use program verification techniques to see if test can ever be false, etc.

20 15-18: Halting Problem The Halting Problem is Undecidable There exists no Turing Machine that decides it There is no Turing Machine that halts on all inputs, and always says yes if M halts on w, and always says no if M does not halt on w Prove Halting Problem is Undecidable by Contradiction:

21 15-19: Halting Problem e(m) e(w) e(m) Prove Halting Problem is Undecidable by Contradiction: Assume that there is some Turing Machine that solves the halting problem. Halting Machine yes no We can use this machine to create a new machine Q: Q e(m) e(m) Halting Machine yes no yes runs forever

22 15-20: Halting Problem e(m) Q e(m) e(m) Halting Machine yes no yes runs forever yes R M DUPLICATE M HALT no yes

23 15-21: Halting Problem Machine Q takes as input a Turing Machine M, and either halts, or runs forever. What happens if we run Q on e(q)? If M HALT says Q should run forever on e(q), Q halts If M HALT says Q should halt on e(q), Q runs forever Q must not exist but Q is easy to build if M HALT exists, so M HALT must not exist

24 15-22: Halting Problem (Java) Quick sideline: Prove that there can be no Java program that takes as input two strings, one containig source code for a Java program, and one containing an input, and determines if that program will halt when run on the given input. boolean Halts(String SourceCode, String Input);

25 15-23: Halting Problem (Java) boolean Halts(String SourceCode, String Input); void Contrarian(String SourceCode) { if (Halts(SourceCode, SourceCode)) while (true); else return; }

26 15-24: Halting Problem (Java) boolean Halts(String SourceCode, String Input); void Contrarian(String SourceCode) { } if (Halts(SourceCode, SourceCode)) else while (true); return; Contrarian("void Contrarian(String SourceCode { \ } "); What happens? if (Halts(SourceCode, SourceCode)) \...

27 15-25: Halting Problem II What if we restrict the input language, to prohibit running machine on its own encoding? Blank-Tape Halting Problem Given a Turing Machine M, does M halt when run on the empty tape?

28 15-26: Halting Problem II What if we restrict the input language, to prohibit running machine on its own encoding? Blank-Tape Halting Problem Given a Turing Machine M, does M halt when run on the empty tape? This problem is also undecidable Prove using a reduction

29 15-27: Reduction Reduce Problem A to Problem B Convert instance of Problem A to an instance of Problem B Problem A: Power x y Problem B: Multiplication x y If we can solve Problem B, we can solve Problem A If we can multiply two numbers, we can calculate the power x y

30 15-28: Reduction If we can reduce Problem A to Problem B, and Problem A is undecidable, then: Problem B must also be undecidable Because, if we could solve B, we could solve A

31 15-29: Reduction To prove a problem B is undecidable: Start with a an instance of a known undecidable problem (like the Halting Problem) Create an instance of Problem B, such that the answer to the instance of Problem B gives the answer to the undecidable problem If we could solve Problem B, we could solve the halting problem thus Problem B must be undecidable

32 15-30: Halting Problem II Show that the Blank-Tape Halting Problem is undecidable, by reducing the Halting Problem to the Blank-Tape Halting Problem Given any machine/input pair M,w, create a machine M such that M halts on the empty tape if and only if M halts on w

33 15-31: Halting Problem II Given any machine/input pair M,w, create a machine M such that M halts on the empty tape if and only if M halts on w Machine M : Erase input tape (ignore input) Write M,w on input tape Run Universal Turing Machine No matter what the input to M is, the input is ignored, and M simulates running M on w M halts on the empty tape iff M halts on w If we could solve the Empty-Tape Halting Problem, we could solve the standard Halting Problem

34 15-32: Halting Problem II input M (ignored) e(m) e(w) Universal Turing Machine

35 15-33: Halting Problem II (Java) boolean MPrime(String Input) { String code = "... (any java source)" String input = "... (any string)" jvm(javac(code), input); }

36 15-34: More Reductions... What do we know about Problem A if we reduce it to the Halting Problem? That is, given an instance of Problem A, create an instance of the halting problem, such that solution to the instance of the halting problem is the same as the solution to the instance of Problem A

37 15-35: More Reductions... What do we know about Problem A if we reduce it to the Halting Problem? That is, given an instance of Problem A, create an instance of the halting problem, such that solution to the instance of the halting problem is the same as the solution to the instance of Problem A We know that the Halting Problem is at least as hard as Problem A If we could decide the Halting Problem, we could decide Problem A

38 15-36: More Reductions... What do we know about Problem A if we reduce it to the Halting Problem? That is, given an instance of Problem A, create an instance of the halting problem, such that solution to the instance of the halting problem is the same as the solution to the instance of Problem A We know that the Halting Problem is at least as hard as Problem A If we could decide the Halting Problem, we could decide Problem A... Which tells us nothing about Problem A!

39 15-37: More Reductions... Given two Turing Machines M 1, M 2, is L[M 1 ] = L[M 2 ]?

40 15-38: More Reductions... Given two Turing Machines M 1, M 2, is L[M 1 ] = L[M 2 ]? Start with an instance M,w of the halting problem Create M 1, which accepts everything Create M 2, which ignores its input, and runs M,w through the Universal Turing Machine. Accept if M halts on w. If M halts on w, then L[M 2 ] = Σ, and L[M 1 ] = L[M 2 ] If M does not halt on w, then L[M 2 ] = {}, and L[M 1 ] L[M 2 ]

41 15-39: More Reductions... Given two Turing Machines M 1, M 2, is L[M 1 ] = L[M 2 ]? input M 2 M 1 input (ignored) e(m) e(w) Universal Turing Machine (ignored) yes

42 15-40: More Reductions... If we had a machine M same that took as input the encoding of two machines M 1 and M 2, and determined if L[M 1 ] = L[M 2 ], we could solve the halting problem for any pair M,w: Create a Machine that accepts everything (easy!). Encode this machine. Create a Machine that first erases its input, then writes e(m),e(w) on input, then runs Universal TM. Encode this machine Feed encoded machines into M same. If M same says yes, then M halts on w, otherwise M does not halt on w

43 15-41: More Reductions... Is the language described by a TM M regular? This problem is also undecidable Use a reduction

44 15-42: More Reductions... Recall: to show a problem P is undecidable: Pick a known undecidable problem P UND Create an instance of P, such that if we could solve P, we could solve P UND Since P UND is known to be undecidable, P must be undecidable, too.

45 15-43: More Reductions... Is the language described by a TM M regular? Let M,w be an instance of the halting problem Create a new machine M, that first runs M on w. If that process halts, the input string is run though a machine that accepts the language a n b n

46 15-44: More Reductions... input M e(m) e(w) Universal Turing Machine an bn After M halts on w What is L[M ]?

47 15-45: More Reductions... input M e(m) e(w) Universal Turing Machine an bn After M halts on w What is L[M ]? If M halts on w, then L[M ] = a n b n, which is not regular If M does not halt on w, then L[M ] = {}, which is regular

48 15-46: More Reductions... So, if we have a machine M REG, that took as input a Turing machine M 1, and decided if L[M 1 ] is regular, then: For any Turing machine M and string w, we can decide if M halts on w Create M from M and w Feed M through M REG If M REG says yes, then M does not halt on w. If M REG says no, then M does halt on w

49 15-47: More Reductions... Given a Turing Machine M, is L[M] > 0? That is, are there any strings accepted by M?

50 15-48: More Reductions... Given a Turing Machine M, is L[M] > 0? That is, are there any strings accepted by M? Undecidable, by reduction from the halting problem. Given any TM M and string w, we create a TM M such that: L[M ] = Σ if M halts on w L[M ] = {} otherwise

51 15-49: More Reductions... Given a Turing Machine M, is L[M] > 0? That is, are there any strings accepted by M? Consider M : Erases input Simulates running M on w Accepts

52 15-50: Questions about Grammars The following questions about unrestricted grammars are all undecidable: Given a Grammar G and string w, is w L[G]? Given a Grammar G, is ǫ L[G]? Given Grammars G 1 and G 2, is L[G 1 ] = L[G 2 ]? Given a Grammar G, is L[G] = {}

53 15-51: Questions about Grammars The following questions about unrestricted grammars are all undecidable: Given a Grammar G and string w, is w L[G]? By reduction from the Halting Problem: Given any Machine M, we can construct an unrestricted Grammar G, such that L[G] = L[M] w L[M] iff w L[G]

54 15-52: Questions about Grammars The following questions about Context-Free Grammars are decidable: Given a Grammar G and string w, is w L[G]? Compilers would be hard to write, otherwise Given a Grammar G, is ǫ L[G]? This is a special case of determining if w L[G]

55 15-53: Questions about Grammars However, there are some problems about CFGs that are not decidable: Given any CFG G, is L[G] = Σ Given any two CFGs G 1 and G 2, is L[G 1 ] = L[G 2 ] Given two PDA M 1 and M 2, is L[M 1 ] = L[M 2 ] Given a PDA M, find an equivalent PDA with the smallest possible number of states

56 15-54: Questions about Grammars Given any CFG G, is L[G] = Σ? Prove this problem is undecidable by reduction from the problem Given an unrestricted grammar G, is L[G] = {} That is, given any unrestricted grammar G, we will create a CFG G : L[G ] = Σ iff L[G] = {}

57 15-55: Questions about Grammars Given any unrestricted grammar G, we will create a CFG G, such that L[G ] = Σ iff L[G] = {} First, we will modify G, to create an equivalent grammar S absc S X Ba ab BX Xb ax a

58 15-56: Questions about Grammars Given any unrestricted grammar G, we will create a CFG G, such that L[G ] = Σ iff L[G] = {} First, we will modify G, to create an equivalent grammar S A 1 S A 2 Ba A 3 BX A 4 ax A 5 A 1 absc A 2 X A 3 ab A 4 Xb A 5 a

59 15-57: Questions about Grammars A standard derivation in this new grammar is one in which each odd step applies of rule of the form u i A i, and every even step applies a rule of the form A i v i S A 1 absc aba 2 c abxc aa 4 c axbc A 5 bc abc S A 1 S A 2 Ba A 3 BX A 4 ax A 5 A 1 absc A 2 X A 3 ab A 4 Xb A 5 a

60 15-58: Questions about Grammars Each standard derivation of a string generated from G can be considered a string over the alphabet V { } (recall V contains Σ as well as non-terminals) We can define a new langauge D G, the set of all valid standard derivations of string generated by G. S A 1 absc aba 2 c abxc aa 4 c axbc A 5 bc abc D G S A 1 absc aba 1 c ababscc ababa 2 cc ababxcc aa 3 BXcc aabbxcc aaba 4 cc aabxbcc aaa 4 bcc aaxbbcc aa 5 bbcc aabbcc D G

61 15-59: Questions about Grammars A boustrophedon version of a derivation is one in which the odd numbered elements of the derivation are reversed: S x R 1 x 2 x R 3... x R n 1 x n Derivation S A 1 absc aba 2 c abxc aa 4 c axbc A 5 bc abc Boustrophedon version of the derivation S A 1 absc ca 2 Ba abxc ca 4 a axbc cba 5 abc

62 15-60: Questions about Grammars D G is the language of all standard derivations of strings in L[G] BD G is the language of all boustrophedon versions of standard derivations of strings in L[G] S A 1 absc ca 2 Ba abxc ca 4 a axbc cba 5 abc BD G

63 15-61: Questions about Grammars Given any Unrestricted Grammar G: L[G] is the set of all strings generated by G D G is the set of all strings that represent standard derivations of strings generated by G BD G is the set of all strings that represent boustrophedon versions of standard derivations of strings in L[G] BD G is the set of all strings that do not represent boustrophedon versions of standard derivations of strings in L[G]. w BD G if w represents only a partial derivation, or an incorrect derivation, or a mal-formed derivation

64 15-62: Questions about Grammars What does it mean if BD G = Σ?

65 15-63: Questions about Grammars What does it mean if BD G = Σ? BD G = {} D G = {} L[G] = {} So, if we could build a CFG that generates BD G, for any unrestricted grammar G, and we could determine if L[G ] = Σ for any CFG G...

66 15-64: Questions about Grammars What does it mean if BD G = Σ? BD G = {} D G = {} L[G] = {} So, if we could build a CFG that generates BD G, for any unrestricted grammar G, and we could determine if L[G ] = Σ for any CFG G... We could determine if L[G] = {} for any unrestricted grammar G which means we could solve the halting problem (why?)

67 15-65: Building a CFG for BD G When is w BD G? w does not start with S w does not end with v, v Σ w contains an odd # of s w is of the form u y v, or u y u contains an even number of s y contains exactly one y is not of the form y = y 1 A i y 2 y R 2β i y R 1 for some i R, y 1,y 2 V, where β i is the right-hand side of the ith rule in G... (there s more)

68 15-66: Building a CFG for BD G When is w BD G?... w is of the form u y v u contains an odd number of s y contains exactly one y is not of the form y = y 1 α i y 2 y R 2 A iy R 1 for some i R, y 1,y 2 V, where α i is the right-hand side of the ith rule in G

69 15-67: Building a CFG for BD G w does not start with S We can create a CFG for all strings w (V ) that do not start with S This language is regular, we could even creatre a DFA or regular expression for it.

70 15-68: Building a CFG for BD G w does not end with v, v Σ We can create a CFG for all strings w (V ) that do not end with v, v Σ This language is regular, we could even create a DFA or regular expression for it.

71 15-69: Building a CFG for BD G w does contains an odd # of s We can create a CFG for all strings w (V ) that contain an odd # of s This language is regular, we could even create a DFA or regular expression for it.

72 15-70: Building a CFG for BD G w is of the form u y v, or u y u contains an even number of s y contains exactly one y is not of the form y = y 1 A i y 2 y R 2β i y R 1 for some i R, y 1,y 2 V, where β i is the right-hand side of the ith rule in G We can create a PDA which accepts strings w of this form

73 15-71: Building a CFG for BD G We can create a PDA which can accept all strings w of this form First, check that the first part of the string is of the form: V ( V V ) Non-deterministically decide when to stop Push Symbols on stack until a Check the input against the stack, making sure there is at least one mismatch Just like the PDA for non-palindromes

74 15-72: Building a CFG for BD G A ABc 3 (V,ε,ε) (V,ε,ε) (V,ε,ε) (a,ε,a) (b,ε,b) (c,ε,c)... (A,ε,A)... (A,A ε) 3 (B,ε,ε) (A,A ε) 3 (V,ε,ε) (ε,γ,ε) (B,ε,ε) (c,ε,ε)

75 15-73: Building a CFG for BD G w is of the form u y v u contains an odd number of s y contains exactly one y is not of the form y = y 1 α i y 2 y R 2A i y R 1 for some i R, y 1,y 2 V, where α i is the right-hand side of the ith rule in G We can create a PDA which accepts all strings w of this form.

76 15-74: Questions about Grammars If we could determine, for any CFG G, if L[G] = Σ, we could solve the halting problem Given any TM M and string w: Create a new TM M, that accepts Σ if M halts on w, and {} otherwise Create an Unrestricted Grammar, such that L[G] = L[M ] Create a CFG G for BD G L[G ] = Σ if and only if L[G] = {}, L[M ] = {}, and M does not halt on w

77 15-75: Questions about Grammars Undecidable problems about CFGs Given any CFG G, is L[G] = Σ Just proved Given any two CFGs G 1 and G 2, is L[G 1 ] = L[G 2 ] Given two PDA M 1 and M 2, is L[M 1 ] = L[M 2 ] Given a PDA M, find an equivalent PDA with the smallest possible number of states

78 15-76: Questions about Grammars Given any two CFGs G 1 and G 2, is L[G 1 ] = L[G 2 ] We can easily create a CFG G 2 which generates Σ How? Note that the preceding proof did not say we cannot decide if L[G] = Σ for any grammar G, but instead, we cannot decide if L[G] = Σ for every grammar G. For any CFG G 1, L[G 1 ] = L[G 2 ] if and only if L[G 1 ] = Σ

79 15-77: Questions about Grammars Given two PDA M 1 and M 2, is L[M 1 ] = L[M 2 ] We can convert a CFG to an equivalent PDA If we could determine if two PDA are equivalent: We could determine if two CFGs are equivalent We could determine if a CFG accepted Σ We could determine if an Unrestricted Grammar generated any strings We could determine if a Turing Machine accepted any strings We could solve the halting problem

80 15-78: Questions about PDA Given a PDA M, find an equivalent PDA with the smallest possible number of states First, prove that it is decidable whether a PDA with one state accepts Σ

81 15-79: Questions about PDA It is decidable whether a PDA with a single state accepts Σ A PDA that accepts Σ must accept Σ. We can test whether a PDA accepts Σ Is w L[M] for a PDA M is decidable Test each of the Σ strings sequentially We can decide if a PDA M accepts Σ

82 15-80: Questions about PDA If a Single State PDA M accepts Σ, then M accepts Σ. Why?

83 15-81: Questions about PDA If a Single State PDA M accepts Σ, then M accepts Σ. Start in the initial state (which is also a final state) with an empty stack After reading a single character, stack is empty, and we re (still!) in the initial (and final!) state accept the string We are in exactly the same position after reading one symbol as we were before reading in anything After reading the next symbol, we will be in a final state with an empty stack accept the string

84 15-82: Questions about PDA Given a PDA M, find an equivalent PDA with the smallest possible number of states It is decidable whether a PDA with one state accepts Σ So...

85 15-83: Questions about PDA Given a PDA M, find an equivalent PDA with the smallest possible number of states It is decidable whether a PDA with one state accepts Σ If we could minimize the number of states in a PDA, we could decide if a PDA accepted Σ Minimize the number of states. PDA accepts Σ iff minimized PDA has a single state, and minimized PDA accepts Σ

86 15-84: Tiling Question There are some problems which seem to have no relation to Turing Machines at all, which turn out to be undecidable Tiling Problem: Set of tiles (infinite # of copies of each tile) Rules for which tiles can be placed next to which other tiles (think puzzle pieces) Can we tile the entire plane?

87 15-85: Tiling Question origin tile...

88 15-86: Tiling Question Tiling System D = (D,d 0,H,V) D is a set of tiles d D is the origin tile H D D list of pairs of which tiles can be next to each other V D D list of pairs of which tiles can be on top of each other

89 15-87: Tiling Question Tiling function f : N N D Specifies which tile goes where f(0,0) = d 0 (f(m,n),f(m+1,n)) H (f(m,n),f(m,n+1)) V

90 15-88: Tiling Question Problem: Given a tiling system D = (D,d 0,H,V), does a tiling exists? That is, is there a completely defined tiling function f that satisfies the requirements: f(0,0) = d 0 (f(m,n),f(m+1,n)) H (f(m,n),f(m,n+1)) V This problem is undecidable!

91 15-89: Tiling Question Tiling Problem is undecidable Proof by reduction from the empty tape halting problem Given any Turing Machine M Create a tiling system D A tiling will exist for D if and only if M does not halt when run on the empty tape

92 15-90: Tiling Question Given any Turing Machine M, create a tiling system D Basic Idea: Each row of the tiling represents a TM configuration Infinite row, infinite tape Create rules so that successive rows i and j are only legal if TM can transition form configuration i to configuration j in one step Can tile the entire plane if and only if TM does not halt

93 15-91: Tiling Question Label the edges of each tile Two tiles can only be adjacent if the edges match Just like a standard jigsaw puzzle

94 15-92: Tiling Question We will modify the Turing Machine M slightly to get M (makes creating the tiling system a little easier) Add a new symbol > to the tape symbols of M Add a new start state s Add a transition ((s,>),(s, )) M halts on empty tape if and only if M halts on the tape containing >

95 15-93: Tiling Question Tiles: For each symbol a that can appear on the tape of Turing Machine M, add the tile: a a Top and bottom edges labeled with a, left and right edges labeled with ǫ

96 15-94: Tiling Question Tiles: For each transition ((q,a),(p,b)) in δ M add the tile: (p,b) (q,a)

97 15-95: Tiling Question Tiles: For each transition((q,a),(p, )) inδ M add the tiles: a (p,b) p p (q,a) b Add a copy of right-hand tile for every symbol b

98 15-96: Tiling Question Tiles: For each transition((q,a),(p, )) inδ M add the tiles: (p,b) a p p b (q,a) Add a copy of left-hand tile for every symbol b

99 15-97: Tiling Question Initial Tile: (s,>)

100 15-98: Tiling Question One final tile:

101 15-99: Tiling Question This tiling system has a tiling if and only if M does not halt on the empty tape. Example: > s (s, ) s (s, )

102 15-100: Tiling Question > s (s, ) s (s, ) > (s, ) > (s, ) > s (s,>) (s,>) (s, ) s (s,>) s >

103 15-101: Tiling Question > (s, )... > (s, ) > (s, ) > (s, ) > (s, ) s s (s,>) (s,>)

104 15-102: Tiling Question Example II: > s (s, ) s (p, ) p (s, )

105 15-103: Tiling Question > > > (s, ) (s,>) s (s,>) s s > (s,>) (p, ) (p,>) p p p (s, ) > (s, ) (s,>) s s s (p, ) >

106 15-104: Tiling Question > (p, ) p p > (s, ) > > (s, ) s s (p, ) > (p, ) p p > (s, ) > (s, ) s s (s,>) (s,>)

107 15-105: Tiling Question Example III: > s (s, ) s (s, )

108 15-106: Tiling Question > (s,>) > > s (s,>) (s, ) s s (s,>) > (s, ) (s, ) s s s (s,>) >

109 15-107: Tiling Question > > > > (s, ) s s (s, ) > (s, ) s s > (s, ) > (s, ) s s (s,>) (s,>) (s, ) s s (s, )

110 15-108: Tiling Question Example IV: > a s (s, ) s (p, ) p (s, )

111 15-109: Tiling Question > a (s,>) > a > s (s, ) s s (s,>) s (s,a) (s,>) > a (p, ) (p,>) p p p (s, ) > p (p,a) a a (s, ) (s,>) s s s (p,a) > (s,a) a s

112 15-110: Tiling Question Can t Place Any tile here... > (p, ) p p > (s, ) > (s, ) s s (s,>) (s,>)

### The Halting Problem is Undecidable

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