Warm-Up Problem. Please fill out your Teaching Evaluation Survey! Please comment on the warm-up problems if you haven t filled in your survey yet.

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1 Warm-Up Problem Please fill out your Teaching Evaluation Survey! Please comment on the warm-up problems if you haven t filled in your survey yet Warm up: Given a program that accepts input, is there an input where the program runs forever? Theorem: The Exists-Input-Runs-Forever Problem is undecidable 1/31

2 Decidability Introduction and Reductions to the Halting Problem Carmen Bruni Lecture 23 Based on slides by Jonathan Buss, Lila Kari, Anna Lubiw and Steve Wolfman with thanks to B Bonakdarpour, A Gao, D Maftuleac, C Roberts, R Trefler, and P Van Beek 2/31

3 Last Time Define a decidable problem Describe the Halting Problem Show that problems are decidable Give reductions to prove undecidability 3/31

4 Learning Goals Prove that the Halting Problem is Undecidable Prove other problems are undecidable based on reductions to the Halting Problem 4/31

5 The Halting problem The decision problem: Given a program and an input, will halt when run with input? Halts means terminates or does not get stuck One of the first known undecidable problems The Halting problem is undecidable There does not exist an algorithm, which gives the correct answer for the Halting problem for every program and every input Exercise: Translate the above statement into a Predicate formula 5/31

6 The Halting Problem is Undecidable A proof by video 6/31

7 Common questions about the video Why can we feed a program as an input to itself? That is, why does make sense? We can convert any program to a string, then we can feed the string of the program to itself as input What does the negator do? It negates the behaviour of the machine If H predicts that the program halts, then the negator goes into an infinite loop and does not halt If H predicts that the program does not halt, then the negator halts The negator is designed to make H fail at its prediction task Why do we need the photocopier? In the video, H takes two inputs We need to make two copies of the input In code, we do not need the photocopier We simply need to call 7/31

8 The Halting Problem is Undecidable Theorem: The Halting problem is undecidable Proof by contradiction Assume that there exists an algorithm, which decides the Halting problem for every program and every input We will construct an algorithm, which takes program as input We will show that gives the wrong answer when predicting whether the program halts when run with input This contradicts the fact that decides the Halting problem for every program and every input Therefore, does not exist 8/31

9 The Halting Problem is Undecidable The algorithm : Takes a program as input Makes two copies of Runs : If outputs yes, then goes into an infinite loop and runs forever (ie it doesn t halt) If outputs no, then halts and returns 0 Now, we ask What happens if we try? 9/31

10 Proof that Does Halt Is Impossible Suppose that halted Then by construction must have outputted no, that is, does not halt when given itself as input This contradicts our assumption that did halt when given itself as input 10/31

11 Proof that Does Not Halt Is Impossible Suppose that did not halt Then by construction must have outputted yes, that is, does halt when given itself as input This contradicts our assumption that did not halt when given itself as input The combination of this slide and the next shows that this machine cannot exist However, every part of the machine does exist except for possibly the machine Hence cannot exist and we have a contradiction 11/31

12 Diagonalization The above method is related to Cantor s diagonalization method which will discuss at the end of today s lecture and the start of tomorrow s lecture 12/31

13 Specific Input Run Forever Problem Problem: Given a program, does it run forever on input? Theorem: The Specific Input Run Forever Problem is undecidable 13/31

14 Specific Input Run Forever Problem Problem: Given a program, does it run forever on input? Theorem: The Specific Input Run Forever Problem is undecidable Proof: Assume towards a contradiction that the specific input problem is decidable That is, there is an algorithm such that when I give it a program, it returns yes if and only if it runs forever on the input Given a program and an input, construct a program that: Ignores the input and runs on If when run on halts, then halt and return 0 Note that this halts at 0 [in fact it halts on all inputs!] if and only if halts on We now solve the Halting Problem (Continued on next slide) 13/31

15 Specific Input Problem Problem: Given a program, does it run forever on input? Theorem: The Specific Input Problem is undecidable Proof: (Continued) We now solve the Halting Problem Consider an algorithm : It consumes and It creates It runs algorithm on and then negates the output Now, algorithm on returns yes if and only if runs forever on input But, this can happen by construction if and only if does not halt on So negating the final answer means that if we get no from the above algorithm, then we have that does not halt on input, a contradiction Hence algorithm cannot exist, that is, the Specific Input Run Forever Problem is undecidable 14/31

16 Partial Correctness Problem Problem: Given a Hoare triple, does satisfy the triple under partial correctness? Theorem: The Partial Correctness Problem is undecidable 15/31

17 Partial Correctness Problem Problem: Given a Hoare triple, does satisfy the triple under partial correctness? Theorem: The Partial Correctness Problem is undecidable Proof: Assume towards a contradiction that there is an algorithm such that when, is satisfied under partial correctness it returns yes We claim that we can construct an algorithm to decide the Halting-No-Input Problem Let be a program Algorithm is given by Run algorithm on program true false and return the negated answer (continued on next slide) 15/31

18 Partial Correctness Problem Problem: Given a Hoare triple, does satisfy the triple under partial correctness? Theorem: The Partial Correctness Problem is undecidable Proof: (Continued) Now, by construction, halts and returns yes if and only if program halts This is true since the Hoare triple true false is never satisfied and so by the definition of partial correctness, this triple satisfies partial correctness if and only if does not halt Thus algorithm will return no if and only if program halts This is the negation of what we want our program do so flip the answer Thus if algorithm exists, then the Halting-No-Input Problem is decidable This is a contradiction Thus the Partial Correctness problem is undecidable 16/31

19 Another Undecidable Problem Provability of a formula in Predicate Logic: Given a formula, does have a proof? (Or, equivalently, is valid?) No algorithm exists to solve provability: Theorem Provability is undecidable 17/31

20 Provability In Predicate Logic Is Undecidable Outline of proof: Suppose that some algorithm decides provability 1 Devise an algorithm to solve the following problem: Given a program and input, produce a formula such that has a proof iff halts on input 2 Combine algorithm with the above algorithm to get an algorithm that decides the Halting Problem: On input, give the formula as input to But no algorithm decides the Halting problem Thus no algorithm decides provability 18/31

21 The Technique: Diagonalization The technique used in the proof of the undecidability of the halting problem is called diagonalization It was originally devised by Georg Cantor (in 1873) for a different purpose Cantor was concerned with the problem of measuring the sizes of infinite sets Are some infinite sets larger than others? Example The set of even natural numbers is the same size (!!) as the set of all natural numbers Example The set of all infinite sequences over is larger than the set of natural numbers (Idea of proof: (A) Assume that each sequence has a number (B) find a sequence that is different from every numbered sequence) Diagonalization Arguments 19/31

22 Countable sets A set is countable if there is a one-to-one correspondence between and the set of natural numbers (N) How do we prove that a set is uncountable? Countability 20/31

23 Cantor s diagonal argument Consider an infinite sequence, where each element is an infinite sequence of 1s or 0s ( is a binary string of infinite length) Countability 21/31

24 Cantor s diagonal argument There is a sequence such that if, then = 0, and otherwise Define Then is not any of the sequences on the list Countability 22/31

25 Cantor s diagonal argument By construction, is not contained in the countable sequence Let be a set consisting of all infinite sequences of 0s and 1s By definition, must contain and Since is not in, the set cannot coincide with Therefore, is uncountable; it cannot be placed in one-to-one correspondence with N Countability 23/31

26 Cantor s diagonal argument Therefore, the set of even integers is smaller than the set of all infinite sequences over Countability 24/31

27 R is Uncountable We will show that a subset of R, per example, is uncountable Let be the following countable sequence of elements from : Countability 25/31

28 R is Uncountable It is possible to build a new number in such a way that if, where is one digit natural number, then is an one-digit natural number different than Countability 26/31

29 R is Uncountable By the construction, is not contained in the countable set Therefore the countable set cannot coincide with We obtain that is uncountable Countability 27/31

30 R is Uncountable (long proof) We build an one-to-one correspondence between the set and a subset of R Let function, where is a string in For example, Observe that, and Hence, is not a bijection Countability 28/31

31 Cantor s diagonal argument (long proof) To produce a bijection from to the interval R: From remove the numbers having two binary expansions and form From, remove the strings appearing after the binary point in the binary expansions of 0, 1, and the numbers in sequence and form from to is defined by: If is the th string in sequence, let be the th number in sequence ; otherwise, let Countability 29/31

32 Cantor s diagonal argument (long proof) To build a bijection from to R, we use, a bijection from to R The linear function provides a bijection from to The composite function provides a bijection from to R The function is a bijection from to R Countability 30/31

33 Cantor s diagonal argument Using diagonalization, one can show that (for example): Q N Z N N The set of all functions from N to N is uncountable Countability 31/31