Emptiness

Automata

Author

Prof. Calvin

Sketch

  • CFGs
    • Acceptance
    • Emptiness
    • Equivalence
  • TMs
    • Acceptance

Acceptance

Statement of Theorem

\[ \text{Decidable}(A_{CFG} := \{\langle G, w \rangle \space | \space w \in G:\text{CFG}\}) \]

  • As with NFAs, we might have infinite loops here.
  • Remember - rules can return an empty string!
  • Can’t go forward or backward.
  • We introduce CNF - “Chomsky Normal Form”

Chomsky Normal Form (CNF)

  • Two legal forms
    • \(A \rightarrow BC \space | \space \{A,B,C\} \subset V\)
    • \(A \rightarrow a \space | \space A \in V \land a \in \Sigma\)
  • This enforces uniformity.

Original Grammar

  • Two legal forms
    • \(A \rightarrow BC \space | \space \{A,B,C\} \subset V\)
    • \(A \rightarrow a \space | \space A \in V \land a \in \Sigma\)
  • \(S \rightarrow AB \space | \space a\)
  • \(A \rightarrow \varepsilon \space | \space b\)
  • \(B \rightarrow c \space | \space AB\)

Eliminate Epsilon

  • \(S \rightarrow AB \space | \space a\)
  • \(A \rightarrow \varepsilon \space | \space b\)
  • \(B \rightarrow c \space | \space AB\)
  • \(S \rightarrow bB \space | \space a \space | \space B\)
  • \(B \rightarrow c \space | \space bB \space | \space b\)

Eliminate \(V \times V\)

  • \(S \rightarrow bB \space | \space a \space | \space B\)
  • \(B \rightarrow c \space | \space bB \space | \space b\)
  • \(S \rightarrow bB \space | \space a \space | \space c \space | \space b\)
  • \(B \rightarrow c \space | \space bB \space | \space b\)

Eliminate Long RHS

  • \(S \rightarrow bB \space | \space a \space | \space c \space | \space b\)
  • \(B \rightarrow c \space | \space bB \space | \space b\)
  • \(S \rightarrow XB \space | \space a \space | \space c \space | \space b\)
  • \(B \rightarrow c \space | \space XB \space | \space b\)
  • \(X \rightarrow b\)

Reorder

  • \(S \rightarrow XB \space | \space a \space | \space c \space | \space b\)
  • \(B \rightarrow c \space | \space XB \space | \space b\)
  • \(X \rightarrow b\)
  • \(S \rightarrow XB \space | \space a \space | \space b \space | \space c\)
  • \(B \rightarrow XB \space | \space b \space | \space c\)
  • \(X \rightarrow b\)

Lemma

  • \(w \in L(H:\text{CNF})\) takes \(2|w| - 1\) steps
  • Left as an exercise.
    • Not too bad - think about the two rule types.

Theorem 14

\[ \text{Decidable}(A_{CFG} := \{\langle G, w \rangle \space | \space w \in G:\text{CFG}\}) \]

  1. Convert to CNF
  2. Test all derivations of length \(2|w| - 1\)

Corollary

  • We note that this also captures context free languages.
  • But a wrinkle!
    • From a CFL, we needn’t necessarily know the CFG
    • We don’t need it! It suffices to exist by definition
    • We only claim something is decidable, not that we know how to decide!

Understanding check

  • Is \(A_{PDA}\) decidable.
    • Just convert.

Emptiness

Statement of Theorem

\[ \text{Decidable}(EQ_{CFG} := \{\langle G, H \rangle \space | \space L(G:\text{CFG}) = L(H:\text{CFG}) \}) \]

  • Probably not gonna fly to try every string.
  • What about closures, well…
  • This is undecidable
  • We will return to this with more robust proof techniques.

Statement of Theorem

\[ \text{Decidable}(AMBIG_{CFG} := \{\langle G\rangle \space | \space G:\text{CFG is not ambigious}\}) \]

  • This is undecidable

Understanding Check

  • Why can we \(EQ_{DFA}\) but not \(EQ_{CFG}\)?
    • We lack closures we need.

Turing Machines

Statement of Theorem

\[ \text{Decidable}(A_{TM} := \{\langle M,w \rangle \space | \space w \in M:\text{TM}\}) \]

  • This turns out not to be true.
  • However, an incrementally weaker result:

\[ \text{Recognizable}(A_{TM} := \{\langle M,w \rangle \space | \space w \in M:\text{TM}\}) \]

  • We are able to show this now.

Theorem 15

\[ \text{Recognizable}(A_{TM} := \{\langle M,w \rangle \space | \space w \in M:\text{TM}\}) \]

  • Intuition:
    • Take machine \(U\)
    • \(U\) simulates \(M\) on one of its tapes.
    • \(U\) accepts if \(M\) accepts
    • \(U\) rejects if \(M\) rejects

Problem

  • There is no guarantee that \(M\) (and \(\therefore U\)) halt!
  • How to deal with this?
    • Can we say “if \(M\) doesn’t halt?”
    • How would we know!
  • To recognize, we need only accept those accepting \(w\)
    • It doesn’t matter if \(U\) halts.

Theorem 15

\[ \text{Recognizable}(A_{TM} := \{\langle M,w \rangle \space | \space w \in M:\text{TM}\}) \]

  • Intuition:
    • Take machine \(U\)
    • \(U\) simulates \(M\) on one of its tapes.
    • \(U\) accepts if \(M\) accepts
    • \(U\) rejects if \(M\) rejects
    • Consider anything else a rejection.

\(U\)

  • \(U\) was the name Turing gave the original “universal computing machine”
    • It was the subject of many latter results in:
      • Computing
      • Information
      • Mathematics
      • Philosophy
      • Linguistics

Example of \(U\)

  • We use \(U\) to denote a machine that can compute anything.
    • It has some combination of states and symbols.
    • One such is:

a 7-state 4-symbol universal Turing machine in 1962 using 2-tag systems. (Marvin Minksy)

Use of \(U\)

  • We use the term \(U\) in our proof to refer to a machine that could do anything.
  • A Universal machine.
  • The development of \(U\) is out-of-scope for this class, but with familiar with modern devices like stored memory computers and RISC processors.

Theorem 15

\[ \text{Recognizable}(A_{TM} := \{\langle M,w \rangle \space | \space w \in M:\text{TM}\}) \]

Proof.

  • Take machine \(U\)
    1. \(U\) simulates \(M\) on one of its tapes.
    2. \(U\) accepts if \(M\) accepts

Summary

  • \(A_{DFA}, A_{NFA}, E_{DFA}, EQ_{DFA}, A_{CFG}, E_{CFG}\) are decidable.
  • \(A_{TM}\) is recognizable.

Stretch Break