CS F-15 Undecidiability 1

Transcription

1 CS F-15 Undecidiability : Univeral TM Turing Machine are Hard Wired Addition machine only add 0 n 1 n 2 n machine only determine if a tring i in the language0 n 1 n 2 n Have een one Programmable TM Random Acce Computer TM 15-1: Univeral TM We can create a Univeral Turing Machine Take a input a decription of a Turing Machine, and the input tring for the Turing Machine Simulate running the machine on the input tring Turing Machine Interpreter Writing a Java Interpreter in Java, for intance, i not all that trange eentially what we are doing with Turing Machine 15-2: Encoding a Turing Machine Our Univeral Turing Machine need to have a pecific, finite alphabet of tape ymbol We need to be able to imulate any Turing Machine with any tape alphabet Ue an encoding cheme 15-3: Encoding a Turing Machine Tape alphabet Σ for Univeral Turing Machine: q a 01,() Encoded tate: q001,q010,q011,q100, Encoded Tape ymbol a001,a010,a011,a100, 15-4: Encoding a Turing Machine Turing Machine that change all a tob, and allb toa 15-5: Encoding a Turing Machine b a R b ye a

2 CS F-15 Undecidiability 2 a b q 0 (q 2, ) (q 1,b) (q 1,a) q 1 (q 0, ) (q 0, ) (q 0, ) q : 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 ( alway ymbol 1, alway ymbol 10 a010 2) 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) 15-7: Encoding a Turing Machine Halting tate can be coded implicitly No outgoing edge = halting tate If we want a ye and no tate Firt halting tate i ye Second halting tate i no 15-8: Encoding a Turing Machine Given any Turing MachineM, we can create an encoding of the machine,e(m) Some machine will require more digit to repreent tate & ymbol Why uedq andaeparator We can actually encode any Turing Machine (and any tape) uing jut 0 and 1 (more on thi in a minute) 15-9: Univeral Turing Machine Take a input an encoding of a Turing Machinee(M), and an encoding of the input tapee(w) Simulate running M on w 3-Tape Machine: Simulated Tape Current State Tape Tranition Function Tape 15-10: Univeral Turing Machine Input Tape a a a a a a a State Tape q 0 0 Tranition 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 1 0 0

3 CS F-15 Undecidiability : Encoding a Turing Machine Encoding a Turing Machine uing jut 0 and 1 : If we knew how many tate there were, and how many ymbol in the input alphabet, we wouldn t need the eparator aq(), Each encoding tart with the # of digit ued for tate, and the # of digit ued for alphabet ymbol, 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) : Halting Problem Halting Machine take a input an encoding of a Turing Machinee(M) and an encoding of an input tringe(w), and return ye ifm halt onw, and no if M doe not halt onw. Like writing a Java program that pare a Java function, and determine if that function halt on a pecific input e(m) e(w) Halting Machine ye no 15-13: Language v. Problem Brief Interlude We will ue Language and Problem interchangeably Any Problem can be converted to a Language, and vice-vera Problem: Multiply two numberx andy Language: L = {x;y;z : z = x y} 15-14: Language v. Problem Brief Interlude We will ue Language and Problem interchangeably Any Problem can be converted to a Language, and vice-vera Problem: Determine if a number i prime Language: L = {p : p i prime} 15-15: Language v. Problem Brief Interlude We will ue Language and Problem interchangeably Any Problem can be converted to a Language, and vice-vera Problem: Determine if a Turing MachineM halt on an input tringw Language: L = {e(m),e(w) : M halt onw}

4 CS F-15 Undecidiability : Halting Problem Halting Machine take a input an encoding of a Turing Machinee(M) and an encoding of an input tringe(w), and return ye ifm halt onw, and no if M doe not halt onw. Like writing a Java program that pare a Java function, and determine if that function halt on a pecific input How might the Java verion work? 15-17: Halting Problem Halting Machine take a input an encoding of a Turing Machinee(M) and an encoding of an input tringe(w), and return ye ifm halt onw, and no if M doe not halt onw. Like writing a Java program that pare a Java function, and determine if that function halt on a pecific input How might the Java verion work? Check for loop while (<tet) <body Ue program verification technique to ee if tet can ever be fale, etc : Halting Problem The Halting Problem i Undecidable There exit no Turing Machine that decide it There i no Turing Machine that halt on all input, and alway ay ye if M halt on w, and alway ay no if M doe not halt onw Prove Halting Problem i Undecidable by Contradiction: 15-19: Halting Problem Prove Halting Problem i Undecidable by Contradiction: e(m) e(w) e(m) Aume that there i ome Turing Machine that olve the halting problem. Halting Machine ye no We can ue thi machine to create a new machineq: Q e(m) e(m) Halting Machine ye no ye run forever 15-20: Halting Problem

5 CS F-15 Undecidiability 5 e(m) Q e(m) e(m) Halting Machine ye no ye run forever ye R M DUPLICATE M HALT 15-21: Halting Problem no ye MachineQtake a input a Turing MachineM, and either halt, or run forever. What happen if we runqone(q)? IfM HALT ayqhould run forever one(q),qhalt IfM HALT ayqhould halt one(q), Q run forever Q mut not exit butqi eay to build ifm HALT exit, om HALT mut not exit 15-22: Halting Problem (Java) Quick ideline: Prove that there can be no Java program that take a input two tring, one containig ource code for a Java program, and one containing an input, and determine if that program will halt when run on the given input. boolean Halt(String SourceCode, String Input); 15-23: Halting Problem (Java) boolean Halt(String SourceCode, String Input); void Contrarian(String SourceCode) { if (Halt(SourceCode, SourceCode)) while (true); ele return; } 15-24: Halting Problem (Java) boolean Halt(String SourceCode, String Input); void Contrarian(String SourceCode) {

6 CS F-15 Undecidiability 6 if (Halt(SourceCode, SourceCode)) while (true); ele return; } Contrarian("void Contrarian(String SourceCode { \ if (Halt(SourceCode, SourceCode)) \ } "); What happen? 15-25: Halting Problem II What if we retrict the input language, to prohibit running machine on it own encoding? Blank-Tape Halting Problem Given a Turing MachineM, doem halt when run on the empty tape? 15-26: Halting Problem II What if we retrict the input language, to prohibit running machine on it own encoding? Blank-Tape Halting Problem Given a Turing MachineM, doem halt when run on the empty tape? Thi problem i alo undecidable Prove uing a reduction 15-27: Reduction Reduce Problem A to Problem B 15-28: Reduction Convert intance of Problem A to an intance of Problem B Problem A: Power x y Problem B: Multiplication x y If we can olve Problem B, we can olve Problem A If we can multiply two number, we can calculate the powerx y If we can reduce Problem A to Problem B, and Problem A i undecidable, then: Problem B mut alo be undecidable 15-29: Reduction Becaue, if we could olve B, we could olve A To prove a problem B i undecidable: Start with a an intance of a known undecidable problem (like the Halting Problem)

7 CS F-15 Undecidiability 7 Create an intance of Problem B, uch that the anwer to the intance of Problem B give the anwer to the undecidable problem If we could olve Problem B, we could olve the halting problem thu Problem B mut be undecidable 15-30: Halting Problem II Show that the Blank-Tape Halting Problem i undecidable, by reducing the Halting Problem to the Blank-Tape Halting Problem Given any machine/input pairm,w, create a machinem uch thatm halt on the empty tape if and only if M halt onw 15-31: Halting Problem II Given any machine/input pair M,w, create a machine M uch that M halt on the empty tape if and only if M halt onw MachineM : Erae input tape (ignore input) Write M,w on input tape Run Univeral Turing Machine No matter what the input to M i, the input i ignored, andm imulate runningm onw M halt on the empty tape iffm halt onw If we could olve the Empty-Tape Halting Problem, we could olve the tandard Halting Problem 15-32: Halting Problem II input M (ignored) e(m) e(w) Univeral Turing Machine 15-33: Halting Problem II (Java) boolean MPrime(String Input) { String code = " (any java ource)" String input = " (any tring)" jvm(javac(code), input); } 15-34: More Reduction

8 CS F-15 Undecidiability 8 What do we know about Problem A if we reduce it to the Halting Problem? That i, given an intance of Problem A, create an intance of the halting problem, uch that olution to the intance of the halting problem i the ame a the olution to the intance of Problem A 15-35: More Reduction What do we know about Problem A if we reduce it to the Halting Problem? That i, given an intance of Problem A, create an intance of the halting problem, uch that olution to the intance of the halting problem i the ame a the olution to the intance of Problem A We know that the Halting Problem i at leat a hard a Problem A If we could decide the Halting Problem, we could decide Problem A 15-36: More Reduction What do we know about Problem A if we reduce it to the Halting Problem? That i, given an intance of Problem A, create an intance of the halting problem, uch that olution to the intance of the halting problem i the ame a the olution to the intance of Problem A We know that the Halting Problem i at leat a hard a Problem A If we could decide the Halting Problem, we could decide Problem A Which tell u nothing about Problem A! 15-37: More Reduction Given two Turing MachineM 1,M 2, i L[M 1 ] = L[M 2 ]? 15-38: More Reduction Given two Turing MachineM 1,M 2, i L[M 1 ] = L[M 2 ]? Start with an intancem,w of the halting problem Create M 1, which accept everything Create M 2, which ignore it input, and run M,w through the Univeral Turing Machine. Accept if M halt onw. IfM halt onw, thenl[m 2 ] = Σ, andl[m 1 ] = L[M 2 ] IfM doe not halt onw, thenl[m 2 ] = {}, andl[m 1 ] L[M 2 ] 15-39: More Reduction Given two Turing MachineM 1,M 2, i L[M 1 ] = L[M 2 ]? input M 2 M 1 input (ignored) e(m) e(w) Univeral Turing Machine (ignored) ye 15-40: More Reduction

9 CS F-15 Undecidiability 9 If we had a machine M ame that took a input the encoding of two machine M 1 and M 2, and determined if L[M 1 ] = L[M 2 ], we could olve the halting problem for any pairm,w: Create a Machine that accept everything (eay!). Encode thi machine. Create a Machine that firt erae it input, then write e(m),e(w) on input, then run Univeral TM. Encode thi machine Feed encoded machine intom ame. IfM ame ay ye, thenm halt onw, otherwiem doe not halt onw 15-41: More Reduction I the language decribed by a TMM regular? Thi problem i alo undecidable Ue a reduction 15-42: More Reduction Recall: to how a problemp i undecidable: Pick a known undecidable problemp UND Create an intance ofp, uch that if we could olvep, we could olvep UND SinceP UND i known to be undecidable,p mut be undecidable, too : More Reduction I the language decribed by a TMM regular? Let M,w be an intance of the halting problem Create a new machine M, that firt run M on w. If that proce halt, the input tring i run though a machine that accept the languagea n b n 15-44: More Reduction input M e(m) e(w) Univeral Turing Machine an bn After M halt on w What il[m ]? 15-45: More Reduction

10 CS F-15 Undecidiability 10 input M e(m) e(w) Univeral Turing Machine an bn After M halt on w What il[m ]? IfM halt onw, thenl[m ] = a n b n, which i not regular IfM doe not halt onw, thenl[m ] = {}, which i regular 15-46: More Reduction So, if we have a machinem REG, that took a input a Turing machinem 1, and decided ifl[m 1 ] i regular, then: For any Turing machinem and tringw, we can decide ifm halt onw Create M fromm andw FeedM throughm REG IfM REG ay ye, thenm doe not halt onw. IfM REG ay no, thenm doe halt onw 15-47: More Reduction Given a Turing Machine M, i L[M] 0? That i, are there any tring accepted by M? 15-48: More Reduction Given a Turing Machine M, i L[M] 0? That i, are there any tring accepted by M? Undecidable, by reduction from the halting problem. Given any TMM and tringw, we create a TM M uch that: L[M ] = Σ ifm halt onw L[M ] = {} otherwie 15-49: More Reduction Given a Turing Machine M, i L[M] 0? That i, are there any tring accepted by M? ConiderM : Erae input Simulate running M on w Accept 15-50: Quetion about Grammar The following quetion about unretricted grammar are all undecidable:

11 CS F-15 Undecidiability 11 Given a GrammarGand tringw, iw L[G]? Given a GrammarG, iǫ L[G]? Given GrammarG 1 andg 2, il[g 1 ] = L[G 2 ]? Given a GrammarG, il[g] = {} 15-51: Quetion about Grammar The following quetion about unretricted grammar are all undecidable: Given a GrammarGand tringw, iw L[G]? By reduction from the Halting Problem: Given any MachineM, we can contruct an unretricted GrammarG, uch thatl[g] = L[M] w L[M] iffw L[G] 15-52: Quetion about Grammar The following quetion about Context-Free Grammar are decidable: Given a GrammarGand tringw, iw L[G]? Compiler would be hard to write, otherwie Given a GrammarG, iǫ L[G]? Thi i a pecial cae of determining if w L[G] 15-53: Quetion about Grammar However, there are ome problem about CFG that are not decidable: Given any CFG G, il[g] = Σ Given any two CFG G 1 andg 2, il[g 1 ] = L[G 2 ] Given two PDAM 1 andm 2, il[m 1 ] = L[M 2 ] Given a PDA M, find an equivalent PDA with the mallet poible number of tate 15-54: Quetion about Grammar Given any CFG G, il[g] = Σ? Prove thi problem i undecidable by reduction from the problem Given an unretricted grammar G, i L[G] = {} That i, given any unretricted grammarg, we will create a CFG G : L[G ] = Σ iffl[g] = {} 15-55: Quetion about Grammar Given any unretricted grammarg, we will create a CFG G, uch thatl[g ] = Σ iffl[g] = {} Firt, we will modifyg, to create an equivalent grammar S absc S X Ba ab BX Xb ax a 15-56: Quetion about Grammar

12 CS F-15 Undecidiability 12 Given any unretricted grammarg, we will create a CFG G, uch thatl[g ] = Σ iffl[g] = {} Firt, we will modifyg, 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 15-57: Quetion about Grammar A tandard derivation in thi new grammar i one in which each odd tep applie of rule of the formu i A i, and every even tep applie a rule of the forma i v i S A 1 absc aba 2 c abxc aa 4 c axbc A 5 bc abc S A 1 A 1 absc S A 2 A 2 X Ba A 3 A 3 ab BX A 4 A 4 Xb ax A 5 A 5 a 15-58: Quetion about Grammar Each tandard derivation of a tring generated from G can be conidered a tring over the alphabet V { } (recallv containσ a well a non-terminal) We can define a new langauged G, the et of all valid tandard derivation of tring generated byg. 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 15-59: Quetion about Grammar A boutrophedon verion of a derivation i one in which the odd numbered element of the derivation are revered: S x R 1 x 2 x R 3 xr n 1 x n Derivation S A 1 absc aba 2 c abxc aa 4 c axbc A 5 bc abc Boutrophedon verion of the derivation S A 1 absc ca 2 Ba abxc ca 4 a axbc cba 5 abc 15-60: Quetion about Grammar D G i the language of all tandard derivation of tring inl[g] BD G i the language of all boutrophedon verion of tandard derivation of tring inl[g]

13 CS F-15 Undecidiability 13 S A 1 absc ca 2 Ba abxc ca 4 a axbc cba 5 abc BD G 15-61: Quetion about Grammar Given any Unretricted Grammar G: L[G] i the et of all tring generated byg D G i the et of all tring that repreent tandard derivation of tring generated byg BD G i the et of all tring that repreent boutrophedon verion of tandard derivation of tring in L[G] BD G i the et of all tring that do not repreent boutrophedon verion of tandard derivation of tring in L[G]. w BD G if w repreent only a partial derivation, or an incorrect derivation, or a mal-formed derivation 15-62: Quetion about Grammar What doe it mean ifbd G = Σ? 15-63: Quetion about Grammar What doe it mean ifbd G = Σ? BD G = {} D G = {} L[G] = {} So, if we could build a CFG that generate BD G, for any unretricted grammar G, and we could determine if L[G ] = Σ for any CFG G 15-64: Quetion about Grammar What doe it mean ifbd G = Σ? BD G = {} D G = {} L[G] = {} So, if we could build a CFG that generate BD G, for any unretricted grammar G, and we could determine if L[G ] = Σ for any CFG G We could determine if L[G] = {} for any unretricted grammar G which mean we could olve the halting problem (why?) 15-65: Building a CFG forbd G When iw BD G? w doe not tart with S w doe not end with v,v Σ w contain an odd # of w i of the formu y v, oru y

14 CS F-15 Undecidiability 14 u contain an even number of y contain exactly one y i not of the formy = y 1 A i y 2 y R 2 β iy R 1 for omei R,y 1,y 2 V, whereβ i i the right-hand ide of theith rule ing (there more) 15-66: Building a CFG forbd G When iw BD G? w i of the formu y v u contain an odd number of y contain exactly one y i not of the formy = y 1 α i y 2 y R 2 A iy R 1 for omei R,y 1,y 2 V, whereα i i the right-hand ide of theith rule ing 15-67: Building a CFG forbd G w doe not tart with S We can create a CFG for all tringw (V ) that do not tart with S Thi language i regular, we could even creatre a DFA or regular expreion for it : Building a CFG forbd G w doe not end with v, v Σ We can create a CFG for all tringw (V ) that do not end with v, v Σ Thi language i regular, we could even create a DFA or regular expreion for it : Building a CFG forbd G w doe contain an odd # of We can create a CFG for all tringw (V ) that contain an odd # of Thi language i regular, we could even create a DFA or regular expreion for it : Building a CFG forbd G w i of the formu y v, oru y u contain an even number of y contain exactly one y i not of the form y = y 1 A i y 2 y R 2 β i y R 1 for ome i R, y 1,y 2 V, where β i i the right-hand ide of theith rule ing We can create a PDA which accept tringw of thi form 15-71: Building a CFG forbd G We can create a PDA which can accept all tringw of thi form

15 CS F-15 Undecidiability 15 Firt, check that the firt part of the tring i of the form: V ( V V ) Non-determinitically decide when to top Puh Symbol on tack until a Check the input againt the tack, making ure there i at leat one mimatch Jut like the PDA for non-palindrome 15-72: Building a CFG forbd G A 3 ABc (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 (B,ε,ε) (c,ε,ε) (V,ε,ε) (ε,γ,ε) 15-73: Building a CFG forbd G w i of the formu y v u contain an odd number of y contain exactly one y i not of the form y = y 1 α i y 2 y R 2 A iy R 1 for ome i R, y 1,y 2 V, where α i i the right-hand ide of theith rule ing We can create a PDA which accept all tringwof thi form : Quetion about Grammar If we could determine, for any CFG G, ifl[g] = Σ, we could olve the halting problem Given any TM M and tring w: Create a new TM M, that acceptσ if M halt onw, and{} otherwie Create an Unretricted Grammar, uch thatl[g] = L[M ] Create a CFG G forbd G L[G ] = Σ if and only if L[G] = {},L[M ] = {}, andm doe not halt onw 15-75: Quetion about Grammar Undecidable problem about CFG Given any CFG G, il[g] = Σ

16 CS F-15 Undecidiability 16 Jut proved Given any two CFG G 1 andg 2, il[g 1 ] = L[G 2 ] Given two PDAM 1 andm 2, il[m 1 ] = L[M 2 ] Given a PDA M, find an equivalent PDA with the mallet poible number of tate 15-76: Quetion about Grammar Given any two CFGG 1 andg 2, il[g 1 ] = L[G 2 ] We can eaily create a CFG G 2 which generateσ How? Note that the preceding proof did not ay we cannot decide if L[G] = Σ for any grammar G, but intead, we cannot decide if L[G] = Σ for every grammarg. For any CFG G 1, L[G 1 ] = L[G 2 ] if and only ifl[g 1 ] = Σ 15-77: Quetion about Grammar Given two PDA M 1 andm 2, il[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 CFG are equivalent We could determine if a CFG acceptedσ We could determine if an Unretricted Grammar generated any tring We could determine if a Turing Machine accepted any tring We could olve the halting problem 15-78: Quetion about PDA Given a PDAM, find an equivalent PDA with the mallet poible number of tate Firt, prove that it i decidable whether a PDA with one tate acceptσ 15-79: Quetion about PDA It i decidable whether a PDA with a ingle tate acceptσ A PDA that acceptσ mut acceptσ. We can tet whether a PDA acceptσ Iw L[M] for a PDA M i decidable Tet each of the Σ tring equentially We can decide if a PDA M acceptσ 15-80: Quetion about PDA If a Single State PDA M acceptσ, thenm acceptσ. Why? 15-81: Quetion about PDA

17 CS F-15 Undecidiability 17 If a Single State PDA M acceptσ, thenm acceptσ. Start in the initial tate (which i alo a final tate) with an empty tack After reading a ingle character, tack i empty, and we re (till!) in the initial (and final!) tate accept the tring We are in exactly the ame poition after reading one ymbol a we were before reading in anything After reading the next ymbol, we will be in a final tate with an empty tack accept the tring 15-82: Quetion about PDA Given a PDAM, find an equivalent PDA with the mallet poible number of tate It i decidable whether a PDA with one tate acceptσ So 15-83: Quetion about PDA Given a PDAM, find an equivalent PDA with the mallet poible number of tate It i decidable whether a PDA with one tate acceptσ If we could minimize the number of tate in a PDA, we could decide if a PDA acceptedσ Minimize the number of tate. PDA acceptσ iff minimized PDA ha a ingle tate, and minimized PDA acceptσ 15-84: Tiling Quetion There are ome problem which eem to have no relation to Turing Machine at all, which turn out to be undecidable Tiling Problem: Set of tile (infinite # of copie of each tile) Rule for which tile can be placed next to which other tile (think puzzle piece) Can we tile the entire plane? 15-85: Tiling Quetion origin tile 15-86: Tiling Quetion

18 CS F-15 Undecidiability 18 Tiling Sytem D = (D,d 0,H,V) D i a et of tile d D i the origin tile H D D lit of pair of which tile can be next to each other V D D lit of pair of which tile can be on top of each other 15-87: Tiling Quetion Tiling functionf : N N D Specifie which tile goe where f(0,0) = d 0 (f(m,n),f(m+1,n)) H (f(m,n),f(m,n+1)) V 15-88: Tiling Quetion Problem: Given a tiling ytem D = (D,d 0,H,V), doe a tiling exit? That i, i there a completely defined tiling functionf that atifie the requirement: f(0,0) = d 0 (f(m,n),f(m+1,n)) H (f(m,n),f(m,n+1)) V Thi problem i undecidable! 15-89: Tiling Quetion Tiling Problem i undecidable Proof by reduction from the empty tape halting problem Given any Turing Machine M Create a tiling ytem D A tiling will exit fordif and only if M doe not halt when run on the empty tape 15-90: Tiling Quetion Given any Turing MachineM, create a tiling ytem D Baic Idea: Each row of the tiling repreent a TM configuration Infinite row, infinite tape Create rule o that ucceive row i and j are only legal if TM can tranition form configuration i to configuration j in one tep Can tile the entire plane if and only if TM doe not halt 15-91: Tiling Quetion

19 CS F-15 Undecidiability 19 Label the edge of each tile Two tile can only be adjacent if the edge match Jut like a tandard jigaw puzzle 15-92: Tiling Quetion We will modify the Turing MachineM lightly to get M (make creating the tiling ytem a little eaier) Add a new ymbolto the tape ymbol ofm Add a new tart tate Add a tranition((,),(, )) M halt on empty tape if and only if M halt on the tape containing 15-93: Tiling Quetion Tile: For each ymbola that can appear on the tape of Turing MachineM, add the tile: a a Top and bottom edge labeled witha, left and right edge labeled withǫ 15-94: Tiling Quetion Tile: For each tranition((q,a),(p,b)) inδ M add the tile: (p,b) (q,a) 15-95: Tiling Quetion Tile: For each tranition((q,a),(p, )) in δ M add the tile: a (p,b) p p (q,a) b Add a copy of right-hand tile for every ymbolb

20 CS F-15 Undecidiability : Tiling Quetion Tile: For each tranition((q,a),(p, )) in δ M add the tile: (p,b) a p p b (q,a) Add a copy of left-hand tile for every ymbolb 15-97: Tiling Quetion Initial Tile: (,) 15-98: Tiling Quetion One final tile: 15-99: Tiling Quetion Thi tiling ytem ha a tiling if and only if M doe not halt on the empty tape. Example: (, ) (, ) : Tiling Quetion (, ) (, )

21 CS F-15 Undecidiability 21 (, ) (, ) (,) (, ) (,) (,) : Tiling Quetion (, ) (, ) (, ) (, ) (, ) (,) (,) : Tiling Quetion Example II: (, ) (p, ) p (, ) : Tiling Quetion

22 CS F-15 Undecidiability 22 (,) (, ) (,) (,) (p, ) (p,) p p p (, ) (, ) (,) (p, ) : Tiling Quetion (p, ) p p (, ) (, ) (p, ) (p, ) p p (, ) (, ) (,) (,) : Tiling Quetion Example III: (, ) (, ) : Tiling Quetion

23 CS F-15 Undecidiability 23 (,) (,) (, ) (,) (, ) (,) (, ) : Tiling Quetion (, ) (, ) (, ) (, ) (, ) (,) (,) (, ) (, ) : Tiling Quetion Example IV: a (, ) (p, ) p (, ) : Tiling Quetion

24 CS F-15 Undecidiability 24 a (,) a (,) (, ) (,) (,a) a (p, ) (p,) p p p (, ) a (, ) (,) (p,a) : Tiling Quetion (p,a) p a (,a) a Can t Place Any tile here (p, ) p p (, ) (, ) (,) (,)