Decidability

1
Decidability
2

• Why study un-solvability?
• When a problem is algorithmically unsolvable, we
  realize that the problem must be simplified or
  altered before we can find an algorithm solution
• A glimpse of the unsolvable can stimulate your
  imagination and help you gain an important
  perspective on computation

3

• Decidable problems concerning Regular languages
• ADFAltB,wgt B is a DFA that accepts w
• Thm
• ADFA is a decidable language

4

• Proof
• MOn input ltB,wgt, where B is DFA
• and w is a string
• 1. Simulate B on input w
• 2. If the simulation ends in an accept state,
  accept.
• If it ends in a non-accepting state, reject.

5

• ANFA ltB,wgt B is an NFA that accepts w
• Thm
• ANFA is a decidable language.

6

• Proof
• N On input ltB,wgt where B is an NFA,
• and w is a string
• 1. Convert NFA B to an equivalent DFA C using
  the
• procedure in Thm. 1.19
• 2. Run TM M from the previous Thm on input ltC,wgt
• 3. If M accepts, accept otherwise reject.

7

• AREXltR,wgt R is a regular expression that
  generates string w
• Thm
• AREX is a decidable language.

8

• Proof
• P On input ltR,wgt where R is a regular
  expression and w is a string.
• 1. Convert R to an equivalent DFA A by the
• procedure given in Thm 1.28
• 2. Run TM M on input ltA,wgt
• 3. If Maccepts , accept if M rejects, reject.
  

9

• EDFA ltAgt A is a DFA and L(A)Ø
• Thm
• EDFA is a decidable language

10

• Proof
• T On input ltAgt where A is a DFA
• 1. Mark the start state of A.
• 2. Repeat until no new state get marked
• 3. Mark any state that has a transition
• coming into it from any state that is
• already marked.
• 4. If no accept state is marked, accept
• otherwise reject.

11

• EQDFAltA,Bgt A and B are DFAs and
• L(A)L(B)
• Thm
• EQDFA is a decidable language.

12

• Proof
• L(C) Ø iff L(A)L(B)
• L(C) is regular, since regular languages are
  closed under complementation, union, and
  intersection.
• F On input ltA,Bgt, where A and B are DFAs
• 1. Construct DFA C as described
• 2. Run the TM in the previous Thm on input ltCgt.
• 3. If accepts, accept otherwise reject.

13
Decidable problems concerning Context-free
languages

• ACFGltG,wgt G is a CFG that generates
• string w
• Thm
• ACFG is a decidable language

14

• Proof
• S On input ltG,wgt, where G is a CFG
• and w is a string
• 1. Convert G to Chomsky normal form.
• 2. List all derivations with 2n-1 steps,
• where n is the length of w.
• 3. If any of these derivations generate w,
  accept
• otherwise reject.

15

• ECFGltGgt G is a CFG and L(G)Ø
• Thm
• ECFG is a decidable language.

16

• Proof
• R On input ltGgt, where G is a CFG
• 1. Mark all terminal symbols in G
• 2. Repeat until no new variables get marked
• 3. Mark all variables A with
• A?U1 Uk
• marked
• 4. If the start symbol is not marked, ACCEPT
• else reject.

17

• EQCFGltG,Hgt G and H are CFLs and L(G)L(H)
• Not decidable
• Note that CFL is NOT closed under
  complementation or intersection. So the same
  technique for EQDFA DOES NOT work
• Thm
• Every CFL is decidable.

18

• Proof
• Let G be a CFG for A and design a TM MG that
  decides A
• MG On input w
• 1. Run TM S on input ltG,wgt
• 2. If this machine accepts, ACCEPTS
• otherwise REJECTS.

Defined before
19
Halting Problem

• ATMltM,wgt M is a TM and M accepts w
• Thm
• ATM is un-decidable .

20

• (conti.)
• UOn input ltM,wgt, where M is a TM
• and w is a string
• 1. Simulate M on input w.
• 2. If M ever enters its accept state, accept
• if M ever enters reject state, reject.
  
• U recognizes ATM but NOT decides it!

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• Diagonalization Method (Georg Cantor in 1873)
• A and B are the same size if there is a
• 1-1, onto function f A?B

f is 1-1 f(a) ? f(b) , if a? b
f
A
B
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• Eg
• N1,2,3, E2,4,6,
• f(n)2n

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• Def
• A is countable if either it is finite
• or it has the same size as N .

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• Eg

25

• Thm
• R is uncountable

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• Proof
• By contradiction.
• Suppose there is a 1-1 onto f between N and R.
• Thus x?f(n) for any n ?

x0.4201? The i-th digit of x is not equal to
the i-th of f(i)
27

• ATMltM,wgt M is a TM and M accepts w
• Thm
• ATM is un-decidable.

28

• Proof
• Assume ATM is decidable.
• Suppose H decides ATM and
• H(ltM,wgt) accept if M accepts w
• reject if M does not accept w
• Define
• D On input ltMgt, where M is a TM
• 1. Run H on input lt M,ltMgt gt.
• 2. Output the opposite of what H outputs
• i.e. if H accepts, REJECT,
• and if H rejects, ACCEPT.

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• (Proof conti.)
• D(ltMgt) accept if M does not accept ltMgt
• reject if M accepts ltMgt
• D(ltDgt) accept if D does not accept ltDgt
• reject if D accepts ltDgt
• ?
• Thus, ATM is un-decidable.

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• 1.
• (i, j) is accept if Mi accepts ltMjgt

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• 2.
• (i, j) is the value of H on input ltMi ,ltMjgtgt.

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• 3.

Contradiction!
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• Turing-unrecognizable language
• A language is co-Turing-recognizable if it is
  the Complement of a Turing-recognizable language.
• Thm
• A language is decidable iff it is both
  Turing-recognizable and co-Turing-recognizable.

34

• Proof
• ?
• If A is decidable, then both A and are
• Turing-recognizable.

Yes
Yes
Yes
MA
x
MA
x
No
No
No
35

• (Proof conti.)
• ?
• If both A and are Turing-recognizable,
• let M1 recognize A and M2 recognize .
• M decides A.

M
M1
Yes
Yes
x
M2
No
Yes
36

• Cor
• is not Turing-recognizable.