1. 1 Foundations of Software Design Fall 2002 Marti Hearst Lecture 29: Computability, Turing Machines, Can Computers Think?

2. 2 Computability Is there anything a computer cannot compute? Linked to the notion of what is an algorithm.

3. 3 Alan Turing An amazing scientist –Helped solved the Enigma Machine (WWII) –Advances in Probability Theory Invented the theory behind computers –Turing Machine –Turing Test

4. 4 Turing Machines Anything that can be computed by a finite set of rules can be computed by a Turing machine. Turing Equivalence –We saw that finite automata are less powerful than TMs Can’t compute Thus not Turing-equivalent –Modern computers and programming languages are all Turing-equivalent

5. 5 Adapted from slide by Costas Busch, http://www.cs.rpi.edu/courses/fall01/modcomp/ Regular Languages Context-Free Languages Languages accepted by Turing Machines

6. 6 Adapted from slide by Costas Busch, http://www.cs.rpi.edu/courses/fall01/modcomp/ A Turing Machine...... Tape Read-Write head Control Unit

7. 7 Adapted from slide by Costas Busch, http://www.cs.rpi.edu/courses/fall01/modcomp/ The Tape...... Read-Write head No boundaries -- infinite length The head moves Left or Right

8. 8 Adapted from slide by Costas Busch, http://www.cs.rpi.edu/courses/fall01/modcomp/...... Read-Write head The head at each time step: 1. Reads a symbol 2. Writes a symbol 3. Moves Left or Right

9. 9 Adapted from slide by Costas Busch, http://www.cs.rpi.edu/courses/fall01/modcomp/...... Example: Time 0...... Time 1 1. Reads 2. Writes 3. Moves Left

10. 10 Adapted from slide by Costas Busch, http://www.cs.rpi.edu/courses/fall01/modcomp/ The Input String...... Blank symbol head Head starts at the leftmost position of the input string Input string

11. 11 Adapted from slide by Costas Busch, http://www.cs.rpi.edu/courses/fall01/modcomp/ States & Transitions Read Write Move Left Move Right

12. 12 Adapted from slide by Costas Busch, http://www.cs.rpi.edu/courses/fall01/modcomp/ Example:...... Time 1 current state

13. 13 Adapted from slide by Costas Busch, http://www.cs.rpi.edu/courses/fall01/modcomp/...... Time 1...... Time 2

14. 14 Adapted from slide by Costas Busch, http://www.cs.rpi.edu/courses/fall01/modcomp/ Determinism Allowed Not Allowed Turing Machines are deterministic

15. 15 Adapted from slide by Costas Busch, http://www.cs.rpi.edu/courses/fall01/modcomp/ Halting The machine halts if there are no possible transitions to follow

16. 16 Adapted from slide by Costas Busch, http://www.cs.rpi.edu/courses/fall01/modcomp/ Example:...... No possible transition HALT!!!

17. 17 Adapted from slide by Costas Busch, http://www.cs.rpi.edu/courses/fall01/modcomp/ Final States Allowed Not Allowed Final states have no outgoing transitions In a final state the machine halts

18. 18 Adapted from slide by Costas Busch, http://www.cs.rpi.edu/courses/fall01/modcomp/ Acceptance Accept Input If machine halts in a final state Reject Input If machine halts in a non-final state or If machine enters an infinite loop

19. 19 Adapted from slide by Costas Busch, http://www.cs.rpi.edu/courses/fall01/modcomp/ Infinite Loop Example A Turing machine for language

20. 20 Adapted from slide by Costas Busch, http://www.cs.rpi.edu/courses/fall01/modcomp/ Time 0

21. 21 Adapted from slide by Costas Busch, http://www.cs.rpi.edu/courses/fall01/modcomp/ Time 1

22. 22 Adapted from slide by Costas Busch, http://www.cs.rpi.edu/courses/fall01/modcomp/ Time 2

23. 23 Adapted from slide by Costas Busch, http://www.cs.rpi.edu/courses/fall01/modcomp/ Time 2 Time 3 Time 4 Time 5... Infinite Loop

24. 24 Adapted from slides by Lee Wee Sun, http://www.comp.nus.edu.sg/~cs1305/2000/ Church-Turing Thesis The Church-Turing thesis says that Turing machine algorithms are the same as our intuitive notion of algorithms. –Most people think the Church Turing thesis is correct. –It means, among other things that You can simulate any computer program that runs on any hardware in any language with any other program on any other hardware using any other language. Tasks that humans agree is an algorithm can always be executed on a computer.

25. 25 Adapted from slides by Lee Wee Sun, http://www.comp.nus.edu.sg/~cs1305/2000/ The Halting Problem An example of something that is not computable. Created by Turing in 1936 to define a problem which no algorithmic procedure can solve. Can we write a program that will take in a user's program and inputs and decide whether –it will eventually stop, or –it will run infinitely in some infinite loop ?

26. 26 Adapted from slides by Lee Wee Sun, http://www.comp.nus.edu.sg/~cs1305/2000/ Proof (by contradiction) Assume that it is possible to write a program to solve the Halting Problem. Denote this program by HaltAnswerer(prog,inputs). HaltAnswerer(prog,inputs) will –return yes if prog will halt on inputs and –no otherwise. A program is just a string of characters –E.g. your Java program is just a long string of characters An input can also be considered as just a string of characters So HaltAnswerer is effectively just working on two strings

27. 27 Adapted from slides by Lee Wee Sun, http://www.comp.nus.edu.sg/~cs1305/2000/ Proof (cont.) We can now write another program Loopy(prog) that uses HaltAnswerer The program Loopy(prog) does the following: [1] If HaltAnswerer(prog,prog) returns yes, Loopy will go into an infinite loop. [2] If HaltAnswerer(prog,prog) returns no, Loopy will halt.

28. 28 Adapted from slides by Lee Wee Sun, http://www.comp.nus.edu.sg/~cs1305/2000/ Proof (cont.) [1] If HaltAnswerer(prog,prog) returns yes, Loopy will go into an infinite loop. [2] If HaltAnswerer(prog,prog) returns no, Loopy will halt. Consider what happens when we run Loopy(Loopy). –If Loopy loops infinitely, HaltAnswerer(Loopy,Loopy) return no which by [2] above means Loopy will halt. –If Loopy halts, HaltAnswerer(Loopy,Loopy) will return yes which by [1] above means Loopy will loop infinitely. Conclusion: Our assumption that it is possible to write a program to solve the Halting Problem has resulted in a contradiction.

29. 29 Diagonalization Not Diagon Alley A proof by contradiction technique –Uses the notion of Infinity The name comes from a proof that shows you can’t ever list all numbers. –Assume you list all possible binary numbers –Diagonalization shows you can always construct a new number that is not yet in the list Have to assume numbers can have infinite length –Construct the new number by choosing the opposite of the number on the diagonal.

30. 30 Diagonalization B100000… B200101… B310110… B401111… B511000… … D11001…1…

31. 31 The Universal Turing Machine There are an infinite number of Turing Machines There are an infinite number of calculations that can be done with a finite set of rules. However, we can define a Universal Turing Machine which can simulate all possible TMs –Comes from the definition of TMs –You convert the description of the TM and its input into two tapes, and use these as the input to the UTM

32. 32 The Halting Problem Halting Problem: –There is no procedure for telling whether an arbitrary TM will halt on a given input. Use Diagonalization to show this. –Again, proof by contradiction –Assume there is a rule for deciding if a TM will halt. Construct a table as follows: –List all Turing machines down the side –List the possible inputs across the top –In position (j,i) put the result of executing Turing machine j on input i If it halts, output H If it doesn’t halt, output ?

33. 33 Diagonalization 12345… T1HH?HH… T2??HHH… T3HHHHH… T4HHH?H… T5HHHHH… … D?H?H?…

34. 34 Diagonalization on the Halting Problem Now define a new TM called D that will halt for all inputs. It outputs –H if TMi(i) does not halt –? if TMi(i) does halt We already said that the assumption is that we can always decide if a TM halts. Also, we said this table lists all possible TMs. –So D must be in the table. But this means that we are saying that D outputs halt if it doesn’t halt! –To see this, give D as input to itself. This is a contradiction. Hence the premise does not hold: We cannot determine if an arbitrary program will halt.

35. 35 “Going Meta” This proof based in part on Gödel’s Theorem If you are interested in these kinds of questions (and other things related to “going meta” like compiler compilers), see –Gödel, Escher, Bach: An Eternal Golden Braid, by Douglas Hofstadter.

36. 36 The Turing Test An observer –Interacts with a keyboard and monitor –Has to distinguish which of two respondents is a computer and which is human. There is a contest with a $100,000 prize! –For the first computer whose responses are indistinguishable from a human's. –The Loebner Prize http://www.loebner.net/Prizef/loebner-prize.htmlhttp://www.loebner.net/Prizef/loebner-prize.html It’s actually pretty easy to fool people over the short term. Chat room ’bots work quite well.

37. 37 What is Intelligence? Do androids dream … ? What would it take for a computer’s thoughts to be indistinguishable from a human’s? THIS is the deepest question of CS.