Complexity

• The number of non-input sequitur operations applied < 2n where n is the input length.

• Amortized Complexity Argument

  • Let s = the sum of the right hand sides of all the production rules. Let r = the number of rules.
  • We evaluate s - r/2.
  • Initially s - r/2 = 1/2 because s = 1 and r = 1.
  • s - r/2 > 0 at all times because each rule has at least 1 symbol on the right hand side.
  • s - r/2 increases by 1 for every input operation.
  • s - r/2 decreases by at least 1/2 for each non-input sequitur rule applied.

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