PFP Homework 1

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Automatons

Usually, a finite automaton defined as: where Q represents states, next is input alphabet, then transitions (), is starting state and finally there is a set of final states.

In our case, a finite automaton is represented by types:

type Transition = (Int, Char, Int)
type Automaton = (Int, String, [Transition], Int, [Int])

where:

  • first number N defines number of states - states will be coded by integer numbers in interval ;
  • second element is a string containing the input symbols - you can safely assume; it contains no duplicities;
  • third is a list defining the trasition function - elementary transition is a triple (q1, a, q2) representing a transition: ;
  • is a number from interval ;
  • finally, there is a list of number representing possible states that are final in defined automaton.

As examples we can use following automatons:

ex1 :: Automaton 
ex1 = (3, "ab", [(0,'a',1), (0,'b',0), (1,'a',1), (1,'b',2), (2,'a',1), (2,'b',0)], 0, [2],)

ex2 :: Automaton 
ex2 = (3, "ab", [(0,'a',1), (0,'a',0), (0,'b',0), (1,'b',2)], 0, [2])

Your task will be to create three functions:

  • First function checks if a given finite automaton is a deterministic finite automaton.
isDeterministic:: Automaton  -> Bool
ghci>isDeterministic ex1
True
ghci>isDeterministic ex2
False
  • Second function runs automaton for given string and establishes, if given input string is accepted by the given automaton. It should work for both deterministic and non-deterministic finite automatons.
isAccepting:: Automaton -> String -> Bool
ghci>isAccepting ex1 "aab"
True
ghci>isAccepting ex2 "aaa"
False
  • Finally, create a function that takes a non-deterministic automaton and produces it's deterministic equivalent (the output can be very different based on used methodology).
convert:: Automaton -> Automaton
ghci>convert ex2
(3, "ab", [(0,'a',1), (0,'b',0), (1,'a',1), (1,'b',2), (2,'a',1), (2,'b',0)], 0, [2],)