AutomaPy 1.1.0

This package refers to the topic of automata theory, which includes DFA, NDFA, Mealy machines, Moore machines and Finite state machine.

AutomaPy

The package contains a set of tools and algorithms for theoretical computer science, which could include automata theory as well as other topics.

Examples of How To Use (AutomaPy)

Create a program that implements a machine that accepts strings ending with '101'.

from AutomaPy import DFA



dfa = DFA()



dfa.addState("A" , {"0": "A", "1":"B"}, initial_state = True)

dfa.addState("B" , {"0": "C", "1":"B"})

dfa.addState("C" , {"0": "A", "1":"D"})

dfa.addState("D" , {"0": "C", "1":"B"}, final_state=True)



print(dfa.endingWithOneZeroOne("101"))

print(dfa.endingWithOneZeroOne("01101"))

print(dfa.endingWithOneZeroOne("011011"))

The code creates a DFA object and defines four states: A, B, C, and D. The transitions between these states are defined by dictionaries that map input symbols (0 or 1) to the next state. The initial state is set to A, and the final state is set to D.

The code then uses the endingWithOneZeroOne method of the DFA object to check if the given strings end with the pattern "101". This method takes a string as input and returns True if the DFA accepts the string (i.e., if it ends in the final state), and False otherwise.

The first call to endingWithOneZeroOne with the input string "101" returns True because the DFA transitions from state A to state B on the first "1", then from state B to state C on the "0", and finally from state C to state D on the "1", which is the final state.

The second call to endingWithOneZeroOne with the input string "01101" also returns True because the DFA follows the same transitions as in the first case, but with an additional "0" input that takes it from state A to state C.

The third call to endingWithOneZeroOne with the input string "011011" returns False because the DFA transitions from state A to state C on the first "0", then from state C to state A on the second "1", and finally from state A to state B on the third "1", which is not the final state.

Design a Program for creating machine that accepts three consecutive one.

from AutomaPy import DFA



dfa = DFA()



dfa.addState("A" , {"0": "A", "1":"B"}, initial_state = True)

dfa.addState("B" , {"0": "A", "1":"C"})

dfa.addState("C" , {"0": "A", "1":"D"})

dfa.addState("D" , {"0": "F", "1":"E"}, final_state=True)

dfa.addState("E" , {"0": "E", "1":"E"})

dfa.addState("F", {"0": "F", "1":"G"}, final_state=True)

dfa.addState("G", {"0": "F", "1":"H"}, final_state=True)

dfa.addState("H", {"0": "F", "1":"D"}, final_state=True)



print(dfa.threeConsecutiveOne("010111"))

print(dfa.threeConsecutiveOne("01011"))

print(dfa.threeConsecutiveOne("010101"))

The code creates a DFA object and defines eight states: A, B, C, D, E, F, G, and H. The transitions between these states are defined by dictionaries that map input symbols (0 or 1) to the next state. The initial state is set to A, and the final states are set to D, F, G, and H.

The code then uses the threeConsecutiveOne method of the DFA object to check if the given strings contain three consecutive ones. This method takes a string as input and returns True if the DFA accepts the string (i.e., if it ends in one of the final states), and False otherwise.

The first call to threeConsecutiveOne with the input string "010111" returns True because the DFA transitions from state A to state B on the first "0", then from state B to state C on the first "1", then from state C to state D on the second "1", and finally to one of the final states (F, G, or H) on the third "1".

The second call to threeConsecutiveOne with the input string "01011" returns False because the DFA transitions from state A to state B on the first "0", then from state B to state C on the first "1", then from state C to state D on the second "1", but it does not reach a final state because there is no third "1".

The third call to threeConsecutiveOne with the input string "010101" returns False because the DFA transitions from state A to state B on the first "0", then from state B to state A on the first "1", then from state A to state B on the second "0", then from state B to state A on the second "1", and finally from state A to state B on the third "0", which is not a valid transition according to the DFA definition.

Write a program for tokenization of given input.

from AutomaPy import DFA



dfa = DFA()



print(dfa.tokenize("This is an example of tokenization."))

Design a program for accepting decimal number divisible by 2.

from AutomaPy import DFA



dfa = DFA()



dfa.addState("A", {"0": "A", "1": "B"}, initial_state=True, final_state=True)

dfa.addState("B", {"0": "A", "1": "B"})



print(dfa.decimalNumberDivisibleByTwo("10")) # Decimal number of "10" is 2

print(dfa.decimalNumberDivisibleByTwo("110")) # Decimal number of "10" is 6

print(dfa.decimalNumberDivisibleByTwo("101")) # Decimal number of "10" is 5

Design a program for creating a machine which accepts string having equal no. of 1’s and 0’s.

from AutomaPy import DFA



dfa = DFA()



dfa.addState("A", {"0": "B", "1": "B"}, initial_state=True, final_state=True)

dfa.addState("B", {"0": "A", "1": "A"}, final_state=True)



print(dfa.equalNumberOfOneZero("10"))

print(dfa.equalNumberOfOneZero("101100"))

print(dfa.equalNumberOfOneZero("1011"))

Design a program for creating a machine which count number of 1's and 0's in a given string.

from AutomaPy import DFA



dfa.addState("A", {"0": "A", "1": "A"}, initial_state=True)



print(dfa.countNumberOfOneZero("0101"))

print(dfa.countNumberOfOneZero("01"))

print(dfa.countNumberOfOneZero("011111"))

print(dfa.countNumberOfOneZero("00000"))