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FORMAL LANGUAGES, AUTOMATA AND COMPUTABILITY

15-453. FORMAL LANGUAGES, AUTOMATA AND COMPUTABILITY. A TM = { (M,w) | M is a TM that accepts string w }. HALT TM = { (M,w) | M is a TM that halts on string w }. E TM = { M | M is a TM and L(M) =  }. REG TM = { M | M is a TM and L(M) is regular}.

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FORMAL LANGUAGES, AUTOMATA AND COMPUTABILITY

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  1. 15-453 FORMAL LANGUAGES, AUTOMATA AND COMPUTABILITY

  2. ATM = { (M,w) | M is a TM that accepts string w } HALTTM = { (M,w) | M is a TM that halts on string w } ETM = { M | M is a TM and L(M) =  } REGTM = { M | M is a TM and L(M) is regular} EQTM = {( M, N) | M, N are TMs and L(M) =L(N)} ALLPDA = { P | P is a PDA and L(P) = Σ* } ALL UNDECIDABLE Use Reductions to Prove Which are SEMI-DECIDABLE?

  3. Let f : Σ*  Σ* be a computable function such that w  A  f(w)  B Σ* Σ* B A f f Say: A is mapping reducible to B; Write: A m B Also, A mB, why?

  4. Theorem: If A m B and B is (semi) decidable, then A is (semi) decidable Let M decide B and let f be a reduction from A to B Proof: We build a machine N that decides A as follows: On input w: 1. Compute f(w) 2. Run M on f(w)

  5. ATM = { (M,w) | M is a TM that accepts string w } HALTTM = { (M,w) | M is a TM that halts on string w } CLAIM: ATMmHALTTM CONSTRUCT f : Σ*  Σ* f: (M,w)  (M’, w) where M’( w) = M(w) if M(w) accepts Loops otherwise So, (M, w ) ATM  (M’, w) HALTTM So HALTTMis NOT DECIDABLE, but it is SEMI-DECIDABLE (Why?)

  6. ATM = { (M,w) | M is a TM that accepts string w } ETM = { M | M is a TM and L(M) =  } CLAIM: ATMm ETM ATMmETM CONSTRUCT f : Σ*  Σ* f: (M,w)  Mwwhere Mw (s) = M(w) if s = w Loops otherwise So, M(w) accepts  L (Mw)   So, (M, w ) ATM  Mw  ETM So ETM is NOT DECIDABLE, but it is SEMI-DECIDABLE (why?) Is ETM SEMI-DECIDABLE?

  7. ATM = { (M,w) | M is a TM that accepts string w } REGTM = { M | M is a TM and L(M) is regular} CLAIM: ATMmREGTM SoREGTMis UNDECIDABLE CONSTRUCT f : Σ*  Σ* f: (M,w)  M’wwhere M’w (s) = accept if s = 0n1n M(w) otherwise So, L (M’w) = Σ* if M(w) accepts {0n1n} if not So, (M, w ) ATM  M’w  REGTM Is REG SEMI-DECIDABLE?

  8. ATM = { (M,w) | M is a TM that accepts string w } REGTM = { M | M is a TM and L(M) is regular} CLAIM: ATMmREGTM So, REG NOTSEMI-DECIDABLE CONSTRUCT f : Σ*  Σ* f: (M,w)  M”wwhere M”w (s) = accept if s = 0n1n and M(w) accepts Loop otherwise So, L (M’w) = {0n1n}if M(w) accepts  if not So, (M, w ) ATM  M”w  REGTM So, REG NOTSEMI-DECIDABLE

  9. ETM = { M | M is a TM and L(M) =  } EQTM = {( M, N) | M, N are TMs and L(M) =L(N)} CLAIM: ETMmEQTM SoEQTMis UNDECIDABLE CONSTRUCT f : Σ*  Σ* f: M  (M, M  ) where M  (s) = Loops So, M E TM  (M, M  )  EQTM NO, since, Is EQTMSEMI-DECIDABLE? ATMmETMmEQTM

  10. ATM = { (M,w) | M is a TM that accepts string w } ALLPDA = { P | P is a PDA and L(P) = Σ* } CLAIM: ATMmALLPDA ATMmALLPDA CONSTRUCT f : Σ*  Σ* f: (M,w)  Pwwhere Pw (s) = accept iff s is NOT an accepting computation of M(w) So, (M, w ) ATM  Pw  ALLPDA So, (M, w ) ATM  Pw  ALLPDA

  11. COMPUTATION HISTORIES An accepting computation history is a sequence of configurations C1,C2,…,Ck, where 1. C1 is the start configuration, 2. Ck is an accepting configuration, 3. Each Ci follows from Ci-1 An rejecting computation history is a sequence of configurations C1,C2,…,Ck, where 1. C1 is the start configuration, 2. Ck is a rejecting configuration, 3. Each Ci follows from Ci-1 M accepts w if and only if there exists an accepting computation history that starts with C1=q0w

  12. P will recognize all strings (read as sequences of configurations) that: 1. Do not start with C1 2. Do not end with an accepting configuration 3. Where some Ci does not properly yield Ci+1 ε,ε → ε ε,ε → ε ε,ε → ε Non-deterministic checks for 1, 2, and 3.

  13. qreject x → x, L 2n 0 → 0, L { 0 | n ≥ 0 } q2  → , R  → , L x → x, R x → x, R q0 q1 q3 0 → , R 0 → x, R x → x, R 0 → 0, R  → , R 0 → x, R  → , R qaccept q4 x → x, R  → , R

  14. x → x, L 2n 0 → 0, L { 0 | n ≥ 0 } q2  → , R  → , L x → x, R x → x, R q0 q1 q3 0 → , R 0 → x, R x → x, R 0 → 0, R  → , R qreject 0 → x, R  → , R qaccept q4 x → x, R  → , R q00000 q1000 xq300 x0q40 x0xq3 x0q2x xq20x q2x0x q2x0x

  15. P recognizes all strings except: #C1# C2R#C3 #C4R#C5 #C6R#….# Ck If k is odd, putCk on stack and see if Ck+1R follows properly: For example, If =uaqibv and (qi,b) = (qj,c,R), then Ck properly yields Ck+1 Ck+1 = uacqjv

  16. P recognizes all strings except: #C1# C2R#C3 #C4R#C5 #C6R#….# Ck If k is odd, putCk on stack and see if Ck+1R follows properly: If =uaqibv and (qi,b) = (qj,c,L), then Ck properly yields Ck+1 Ck+1 = uqjacv

  17. P recognizes all strings except: #C1# C2R#C3 #C4R#C5 #C6R#….# Ck If k is even, putCkRon stack and see if Ck+1 follows properly.

  18. q00000 q1000 xq300 x0q40 ODD x0xq3 x0q2x xq20x q2x0x : #q00000#000q1#xq300#0q40x #x0xq3# ... #

  19. q00000 q1000 xq300 x0q40 ODD x0xq3 x0q2x xq20x q2x0x : #q00000#000q1#xq300#0q40x #x0xq3# ... #

  20. q00000 q1000 xq300 x0q40 ODD x0xq3 x0q2x xq20x q2x0x : #q00000#000q1#xq300#0q40x #x0xq3# ... #

  21. q00000 q1000 xq300 x0q40 EVEN x0xq3 x0q2x xq20x q2x0x : #q00000#000q1#xq300#0q40x #x0xq3# ... #

  22. q00000 q1000 xq300 x0q40 EVEN x0xq3 x0q2x xq20x q2x0x : #q00000#000q1#xq300#0q40x #x0xq3# ... #

  23. THE POST CORRESPONDENCE PROBLEM

  24. ba ba b b a a bcb ab ab a a a THE PCP GAME

  25. aaa aaa a a a c a aa aa aa a c a a

  26. abc abc ca ca b b a a a ab ab ab ca ca a c a c

  27. abc acc ca ab ca a

  28. GENERAL RULE #1 If every top string is longer than the corresponding bottom one, there can’t be a match

  29. aab acc caa b b c aa b b a a a

  30. GENERAL RULE #2 If there is a domino with the same string on the top and on the bottom, there is a match

  31. POST CORRESPONDENCE PROBLEM Given a collection of dominos, is there a match? PCP = { P | P is a set of dominos with a match } PCP is undecidable!

  32. THE FPCP GAME … is just like the PCP game except that a match has to start with the first domino

  33. aaa aaa a a a c a aa aa aa a c a a FPCP

  34. ba b b a bcb ab a a FPCP

  35. Theorem:FPCP is undecidable Proof: Assume machine C decides FPCP We will show how to use C to decide ATM

  36. aba caa a … bb d c Given (M,w) we will construct a set of dominos P where a match is an accepting computation history for M on w P = C P has a match?

  37. qreject x → x, L 2n 0 → 0, L { 0 | n ≥ 0 } q2  → , R  → , L x → x, R x → x, R q0 q1 q3 0 → , R 0 → x, R x → x, R 0 → 0, R  → , R 0 → x, R  → , R qaccept q4 x → x, R  → , R

  38. x → x, L 2n 0 → 0, L { 0 | n ≥ 0 } q2  → , R  → , L x → x, R x → x, R q0 q1 q3 0 → , R 0 → x, R x → x, R 0 → 0, R  → , R qreject 0 → x, R  → , R qaccept q4 x → x, R  → , R q00000 q1000 xq300 x0q40 x0xq3 x0q2x xq20x q2x0x q2x0x #q00000#q1000#xq300#x0q40#x0xq3# ... # :

  39. Given (M,w), we will construct an instance P of FPCP in 7 steps

  40. # #q0w1w2…wn# STEP 1 Put into P START

  41. cqa qa pcb bp STEP 2 If (q,a) = (p,b,R) then add STEP 3 for all c  Γ If (q,a) = (p,b,L) then add RULES

  42. qreject x → x, L 2n 0 → 0, L { 0 | n ≥ 0 } q2  → , R  → , L x → x, R x → x, R q0 q1 q3 0 → , R 0 → x, R x → x, R 0 → 0, R  → , R 0 → x, R  → , R qaccept q4 x → x, R  → , R

  43. x → x, L 2n 0 → 0, L { 0 | n ≥ 0 } q2  → , R #  → , L x → x, R x → x, R #q00000# q0 q1 q3 0 → , R 0 → x, R x → x, R 0 → 0, R  → , R qreject 0 → x, R  → , R xq20 q20 xq3 0q3 q3 0q20 q10 q00 qaccept q4 q1 xq3 q2 q20 q2x q20 q2x0 q200 x → x, R  → , R …

  44. a # # # a # STEP 4 for all a  Γ add STEP 5 add CONTINUE

  45. qacca aqacc a # # qacc qacc # # a STEP 4 for all a  Γ add STEP 5 add STEP 6 add for all a  Γ

  46. # # # q00 q00 q10 q10 q1 q1x q0x q0 #q00000# #q00000# #q00000# xq3 xq3 qa q1 q1 xq1 xqr qr q20 xq20 0q3 0q20 q3x q2 q30 q40 q4 q4x q2x0 q1 q20 q20 xq3 q200 0q4 xq3 qr xq4 #  0 x 0 0 0  0 0 # q3x xq3 q3 0q3x xq3x # q2x q2xx   # 0 # 0 x 0 0 0 0 q2 # q20x q2x xqacc qacc qacc qaccx 0qacc qacc0 0 0 x x x x

  47. qacc## # STEP 7 add END

  48. # q00 q10  qacc0 qacc 0qacc qacc## qacc q0 #q00# 0qrej  0q1 qrej qacc qacc qacc qacc # # # q1 0 qreject qacc 0 # # q00 # 0 q1 # # # 0 0qacc qacc # #q00# 0q1 # 0 qacc # # # 0 qacc qacc 0 → 0, R  → , R q0 q1 qaccept 0→ 0, R  → , R

  49. Given (M,w), we can construct an instance of FPCP that has a match if and only if M accepts w

  50.  t1 tk t2 t1 t1 t2 tk  b1 b2 b1 bk b1 b2 bk Can convert an instance of FPCP into one of PCP: Let u = u1u2…un,define: u =  u1  u2  u3  …  un u = u1  u2  u3  …  un u =  u1  u2  u3  …  un FPCP: … PCP: …

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