A Hierarchy of Formal Languages and Automata Wen-Guey Tzeng Computer Science Department National Chiao Tung University Recursively enumerable language • A language L is recursively enumerable if there exists a Turing machine M such that L(M)=L. – For a string w in L, M(w) halts and enters a final state. – For a string w, not in L, M(w) • Either, halt and enter a non-final state • Or, run forever (non-halt) 2016 Spring 2 • Why is it called ”recursively enumerable” ? – “recursive” = “computational” = “algorithmical” • We can list (enumerate) all strings in a r.e. language. – Later ! 2016 Spring 3 Recursive language • A language L is recursive if there exists a Turing machine M such that – L(M)=L – For every string w in *, M(w) always halts. All languages 2016 Spring 4 Theorem: If L is recursive, we can enumerate all strings in L. • Assume that M accepts L and halts on every input. • Let * be w1=, w2=0, w3=1, w4=00, w5=01, … in the alphabetical order. • Algorithm for enumerating strings in L: 1. i=0; 2. i=i+1; 3. Run M(wi) and wait till halt. 4. If M(wi) enters a final state 5. then list wi 6. Goto 2. 2016 Spring 5 M(w) w1 = w2 = 0 w3 = 1 w1 = 00 . wk . 1st move 2nd Halt & move non-final 3rd Halt & X move final 4th Halt & X X move non-final . X X X . rth Halt & X X X move final . Halt & Halt & . X X X X . final non-final 2016 Spring 6 Theorem: If L is recursive, so is 퐿ത • Assume that M=(Q, , , , , q0, F) accepts L and halts on all inputs. – If wL, M halts and enters pF – If wL, M halts and enter qF • Let M’=(Q, , , , , q0, F’), where F’=Q-F. – If w퐿ത, M halts and enters p F’ – If wL, M halts and enter q F’ – Thus, L(M’)=퐿ത. • Since M’ halts on all inputs, 퐿ത is recursive. 2016 Spring 7 Theorem: If L is r.e., all strings in L can be enumerated. • Assume that M accepts L. • Let * be w1=, w2=0, w3=1, w4=00, w5=01, … in the alphabetical order. • Wrong statement 3. Run M(wi) and wait till halt. 2016 Spring 8 • Algorithm for enumerating L 1. i=0; 2. i=i+1; 3. For j=1 to i; 4. Run M(wj) for the (i-j+1)-th step. 5. If M(wj) halts and enters a final state 6. then list wj 7. Goto 2. 2016 Spring 9 M(w) w1 = w2 = 0 w3 = 1 w1 = 00 . wk . 1st move 2nd Halt & Halt & move final non-final 3rd move X X 4th Halt & X X move non-final . X X X . rth Halt & X X X move final . X forever X X forever X . 2016 Spring 10 • Why does the algorithm work? – If wk is in L and accepted by M in r steps, it will be listed when j=k and i=r+j-1, within finite steps. • Thus, every w L will be listed within finite steps and every w L won’t be listed. 2016 Spring 11 Questions ? All languages L2 L1 • Does language L1 exist? – R.E., but not recursive • Does language L2 exist? – Not recursively enumerable 2016 Spring 12 A non-r.e. language • Let ={a}. • Let M1, M2, … be a list of all TM over . i i • Let L = {a : a L(Mi)}. a aa aaa . ak . M1 o x 0 . o . M2 o o x . o . M3 x x x . x . Mk o x o . x . L o o x . x . 2016 Spring 13 i i Theorem: L = {a : a L(Mi)} is r.e. • Algorithm A: Input: ak; ; 1. List the k-th TM Mk k 2. Run Mk(a ) and wait for the result If Mk(ak) halts on a final state then A enters an accepting configuration If Mk(ak) halts on a non-final state then A enters a non-accepting configuration 2016 Spring 14 – What is A’s running result on input ak? • when akL • when akL 2016 Spring 15 i i • Let 퐿 = {a : a L(Mi)} a aa aaa . ak . M1 o x 0 . o . M2 o o x . o . M3 x x x . x . Mk o x o . x . 퐿 o o x . x . 퐿 x x o . o . 2016 Spring 16 • Theorem: 퐿 = S is not r.e. – Assume that S is r.e. and accepted by TM Mk such that L(Mk)=S for some k. – Contradiction: k k • If a S, then, by definition of S, a L(Mk)=S. k k • If a S, then, by definition of S, a L(Mk)=S. – Thus, Mk does not exist and S is not r.e. 2016 Spring 17 • Theorem: L is r.e., but not recursive. – L is r.e. (why?) – If L is recursive, so is 퐿. – But, 퐿 is not recursive. Thus, L is not recursive. All languages L2 L1 2016 Spring 18 2016 Spring 19.