CS 3240: Languages and Computation

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CS 3240 Languages and Computation

• Decidable Languages and The Halting Problem

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Decidability

• A language is decidable if some Turing machine
  decides it
• Not all languages are decidable
• How to show a language is decidable?
• Write a decider that decides it
• Accepts w iff w is in the language
• Must halt on all inputs

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DFA Acceptance Problem

• Consider the language ADFA ltB,wgt B is a DFA
  that accepts the string w
• Theorem ADFA is a decidable language
• Proof Consider the following TM, M
• On input string ltB,wgt, where B is a DFA and w is
  an input to B
• Simulate B on input w
• If simulation ends in accept state, accept.
  Otherwise, reject.

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NFA Acceptance Problem

• Consider the language
• ANFA ltB,wgt B is a NFA that accepts the
  string w
• Theorem ANFA is a decidable language
• Proof Consider the following TM, NOn input
  string ltB,wgt
• Convert B to a DFA C
• Run TM M from previous slide on ltC,wgt
• If M accepts, accept. Otherwise, reject.

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RE Acceptance Problem

• Consider the language
• AREX ltR,wgt R is an RE that generates the
  string w
• Theorem AREX is a decidable language
• Proof Consider the following TM, P
• On input string ltR,wgt
• Convert R to a DFA C using algorithm discussed in
  class and in text
• Run TM M from earlier slide on ltC,wgt
• If M accepts, accept. Otherwise, reject.

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Emptiness Testing Problem

• Consider the language
• EDFA ltAgt A is a DFA and L(A) ?
• Theorem EDFA is a decidable language
• Proof Consider the following TM, T
• On input string ltAgt, where A is a DFA
• Mark the start state
• Repeat until no new states get marked
• Mark any state that has a transition coming into
  it from any state already marked
• If no accept states are marked, accept.
  Otherwise, reject.

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DFA Equivalence Problem

• EQDFA ltA,Bgt A and B are DFAs and L(A)
  L(B)
• Theorem EQDFA is a decidable language
• Proof Consider the following language
• (L(A) ? L(B)) ? (L(A) ? L(B))

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DFA Equivalence Problem

• (L(A) ? L(B)) ? (L(A) ? L(B))

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DFA Equivalence Problem

• EQDFA ltA,Bgt A and B are DFAs and L(A)
  L(B)
• Theorem EQDFA is a decidable language
• Proof idea Consider DFA C that accepts L(C)
  (L(A) ? L(B)) ? (L(A) ? L(B))
• How do we know such a DFA exists?
• If L(C) ?, then L(A) L(B)

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TM That Decides EQDFA

• On input string ltA,Bgt, where A and B are DFAs
• Create DFA C such that
• L(C) (L(A) ? L(B)) ? (L(A) ? L(B))
• Submit C to Turing machine T that decides EDFA
• If T accepts C, accept. Otherwise, reject.

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Deciders and CFGs

• Consider the following language
• ACFG ltG,wgt G is a CFG that generates string
  w
• Is ACFG decidable?
• How can we get a TM to simulate a CFG?
• Must be certain CFG tries a finite number of
  steps!
• Solution Use Chomsky normal form

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Chomsky Normal Form Review

• All rules are of the form
• A ?BC
• A ?a
• Where A, B, and C are any variables (B and C
  cannot be the start variable)
• S ?eis the only erule, where S is the start
  variable

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How Many Steps to Generate w?

• If w 0
• 1 step
• If w n gt 0?
• 2n 1 steps

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TM Simulating ACFG

• On input ltGgt, where G is a CFG
• Convert G into Chomsky normal form
• If w 0
• If there is an S ? e rule, accept
• Otherwise, reject
• List all derivations with 2w-1 steps
• If any generate w, accept
• Otherwise, reject

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Empty CFGs

• Consider the following language
• ECFG ltGgt G is a CFG and L(G) ?
• Theorem ECFG is decidable
• Can we use the TM in ACFG to prove this?
• No. There are infinitely many possible strings
  in S
• We need to check if there is any way to get from
  the start variable to some string of terminals

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Work Backwards

• On input ltGgt, where G is a CFG
• Mark all terminals
• Repeat until no new variables are marked
• Mark any variable A if G has a rule A?U1U2Uk
  where U1, U2, , Uk are all marked
• If S is marked, reject Otherwise, accept

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What About EQCFG?

• Recall for EQDFA, we considered
• (L(A) ? L(B)) ? (L(A) ? L(B))
• Will this work for CFGs?
• No. CFGs are not closed under complementation or
  intersection
• EQCFG is not a decidable language!
• We will show this later

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Decidability of CFLs

• Theorem Every context-free language L is
  decidable
• Can we make a TM that simulates a PDA?
• No. May run into infinite loop.
• Proof idea For each w, we need to decide
  whether or not w is in L. Let G be a CFG for L.
  This problem boils down to ACFG, which we showed
  is decidable.

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Summary of Decidable Languages
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Turing Machine Acceptance Problem

• Consider the following language
• ATMltM,wgt M is a TM that accepts w
• Theorem ATM is Turing-recognizable
• Theorem ATM is undecidable
• Proof idea Construct a universal Turing machine
  recognizes, but does not decide, ATM

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The Universal Turing Machine

• U On input ltM, wgt, where M is a TM and w is a
  string
• Simulate M on input w
• If M ever enters its accept state, accept
• If M ever enters its reject state, reject

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Why Cant U Decide ATM?

• Intuitively, if M never halts on w, then U never
  halts on ltM,wgt
• Given a TM M and a string w, does M halt on input
  w?
• This is also known as the halting problem
• It is undecidable
• We will prove this more rigorously later

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Comparing Sizes of Infinite Sets

• Given two infinite sets A and B, is there a way
  to determine whether AB or if AgtB?
• Functional correspondence can show two infinite
  sets have the same number of elements
• Diagonalization can show one infinite set has
  more elements than another

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Functional Correspondence

• Let f be a function from A to B
• f is called injective (one-to-one) if
• f(a1) ? f(a2) whenever a1 ? a2
• f is called surjective (onto) if
• For every b ? B, there is some a ? A such that
  f(a) b
• f is called a correspondence if it is injective
  and surjective
• A correspondence is a way to pair elements of the
  two sets

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Example Correspondence

• Consider fZ ? P, where Z 0,1,2, and P
  positive squares
• P 1, 4, 9, 16, 25,
• f(x) (x1)2
• Is f one-to-one?
• Yes
• Is f onto?
• Yes
• Therefore Z P

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Countable Sets

• Let N 1, 2, 3, be the set of natural
  numbers
• The set A is countable if
• A is finite, or
• A N
• Some example of countable sets
• Integers
• x x ? Z and x 1 (mod 3)

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The Positive Rational Numbers

• Is Q m/n m,n ? N countable?
• Yes!
• Construct a one-to-one mapping

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The Real Numbers

• Is R (set of positive real numbers) countable?
• No!

X 4.1337 Diagonalization
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The Set of All Infinite Binary Strings

• Is the set of all infinite binary strings
  countable?
• No
• Diagonalization also works to prove this is not
  countable
• On the other hand, the set of finite length
  binary strings is countable!
• Let xb be the binary representation of x
• f(x) xb is 1-to-1 and onto function from N to
  the set of finite binary strings

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Languages in S

• Is the set of all languages in S countable?
• No
• It has the same cardinality as the set of binary
  strings
• S e, a, b, aa, ab, ba, bb, aaa, aab,
• A a, ab, aaa,
• ?A 0 1 0 0 1 0 0 1

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S vs. Languages in S

• The set S is countable
• Let S n
• Every string in S can be associated with a
  unique number, y, in base-(n1)
• E.g., if S a, b, c, we can associate the
  string acba with the value 143342241140
  121
• Let f(x) be the string associated with x
• The set of all languages in Sis the power set of
  S

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Turing Machines

• Is the set of all TMs countable?
• Yes
• Since all Turing machines can be represented by a
  string of finite length, the set of all Turing
  machines are countable
• Theorem Some languages are not
  Turing-recognizable
• Proof idea There are more languages than there
  are Turing machines

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Undecidability of ATM

• Theorem ATM is undecidable
• Proof (By contradiction) Assume ATM is
  decidable and let H be a decider for ATM
• H(ltM,wgt)

accept if M accepts w reject if M does not
accept w
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Proof (cont.)

• Consider the TM D that submits the string ltMgt as
  input to the TM M
• D On input ltMgt, where M is a TM
• Run H on input ltM,ltMgtgt
• If H accepts ltM,ltMgtgt, reject
• If H rejects ltM,ltMgtgt, accept
• Since H is a decider, it must accept or reject
• Therefore, D is a decider as well

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Proof (cont.)

• What happens if Ds input is ltDgt?
• D(ltDgt)
• D cannot exist!
• Therefore, H cannot exist i.e., ATM is
  undecidable

reject if D accepts ltDgt accept if D does not
accept ltDgt
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Undecidable languages
Turing recognizable
Co-Turing recognizable
Decidable