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MathematicsforComputerScience Computablestringsin{0,1} MIT 6.042J/18.062J An infinite string s in {0,1} is computable iff some Noncomputable procedure computes its digits . Sets (Procedure applied to argument n returns nth digit

  1. Mathematics for Computer Science Computable strings in {0,1} ω MIT 6.042J/18.062J An infinite string s in {0,1} ω is computable iff some Noncomputable procedure computes its digits . Sets (Procedure applied to argument n returns nth digit of s.) Albert R Meyer, March 4, 2015 Albert R Meyer, March 4, 2015 halting.1 halting.2 {ASCII}* is countable N oncomputable strings in {0,1} ω Only countably many finite So only countably many computable infinite binary strings. (List them in ASCII strings. But {0,1} ω is uncountable, so order of length.) there must be noncomputable Procedures can be expressed in strings in {0,1} ω ASCII, so only —in fact, uncountably many! countably many procedures. Albert R Meyer, March 4, 2015 Albert R Meyer, March 4, 2015 halting.3 halting.4 1

  2. The Halting Problem The Halting Problem There is no test procedure for String procedure P takes a String halting of arbitrary procedures. argument: P("no") returns 2 The Halting Problem P("albert") returns "meyer" is not decidable P("&&%99!!") causes an error by computational procedures P("what now?") runs forever. Albert R Meyer, March 4, 2013 halting.6 Albert R Meyer, March 4, 2015 halting.5 The Halting Problem The Halting Problem Suppose there was a Let s be an ASCII string procedure Q that decided defining P s . HALTS: Say s HALTS iff Q(s) returns "yes” if s HALTS P s (s) returns something. returns "no” otherwise Albert R Meyer, March 4, 2013 Albert R Meyer, March 4, 2013 halting.7 halting.8 2

  3. The Halting Problem The Halting Problem Modify Q to Q': So Q'(s) returns "yes” s HALTS iff if Q(s) returns "no" Q'(s) returns nothing Q'(s) returns nothing if Q(s) returns "yes” Albert R Meyer, March 4, 2013 Albert R Meyer, March 4, 2013 halting.9 halting.10 The Halting Problem The Halting Problem CONTRADICTION: Let t be the text for Q’ t HALTS iff NOT ( t HALTS) So by def of HALTS: There can't be such a Q: t HALTS iff Q'( t ) returns it is impossible to write a and by def of Q’: procedure that decides Q'( t ) returns iff NOT ( t HALTS) whether strings HALT Albert R Meyer, March 4, 2013 Albert R Meyer, March 4, 2013 halting.11 halting.12 3

  4. The Type-checking Problem The Type-checking Problem There is no string procedure that Use C to get a HALTS Tester H: type-checks perfectly, because: to compute H(s), construct a Suppose C was a type-checking new program text, s’, that procedure: for program text s acts like a slightly modified C(s) returns “yes” if s would cause interpreter for s. Namely: a run-time type error returns “no” otherwise. Albert R Meyer, March 4, 2013 Albert R Meyer, March 4, 2013 halting.13 halting.14 The Type-checking Problem The Type-checking Problem • s’ skips any command that Then compute C(s’) and would cause s to make a return the same value. run-time type error. So s HALTS • s’ purposely makes a type- iff s’ makes run-time type error error when it finds that s iff C(s’) = “yes” HALTS. iff H(s) = “yes” Albert R Meyer, March 4, 2013 Albert R Meyer, March 4, 2013 halting.15 halting.16 4

  5. No run-time properties The Type-checking Problem are decidable H solves the Halting The same reasoning shows that there is no perfect Problem, a contradiction. checker for essentially any So C must not error check property of procedure correctly. outcomes. Albert R Meyer, March 4, 2013 Albert R Meyer, March 4, 2013 halting.18 halting.19 5

  6. MIT OpenCourseWare http s ://ocw.mit.edu 6.042J / 18.062J Mathematics for Computer Science Spring 20 15 For information about citing these materials or our Terms of Use, visit: http s ://ocw.mit.edu/terms.

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