Question

Part B - Automata Construction

1. Draw a DFA which accepts the following language over the alphabet of {0,1}: the set of all strings such that the number of 0s is divisible by 2 and the number of 1s is divisible by 5. Your DFA must handle all intput strings in {0,1}*.
   Here is a methodical way to do this:
   • Figure out all the final states and label each with the shortest string it accepts,
   • work backwards from these states to the starting state, labelling each state you create with the shortest string accepted so far,
   • add the missing transitions.
   The labelling of the states with the shortest string it accepts is important in order to not get confused.
2. Draw a DFA which accepts the following language over the alphabet of {0,1}: the set of all strings such that the number of 0s is not divisible by 2 or the number of 1s is not divisible by 5. Your DFA must handle all intput strings in {0,1}*.
   (Hint: look at solution of previous question)
3. Draw the simplest possible (i.e. with fewest number of states) NFA which accepts the following language over the alphabet of {a, e, i, o, u, b, p, r, s, t}: the set of strings which start with a vowel, then two consonants, then 0 or more vowels, followed by the first vowel.
4. Assuming that you have the following three automata, A₁, A₂, A₃ recognizing the languages L₁, L₂, L₃ respectively, shown here as black boxes:
   
   1. Draw an NFA called A which recognizes the language L₂ (L₁ | L₃) L₂*
   2. What is the starting state of A?
   3. What are A's final states?

S1 F1 · F32 4 鈴

Answer 1

Here I - [0,1). Since DFA accepts the set of all strings such that the number of 0s is divisible by 2 and the number of 1s is divisible by 5, we need to İnsert states to count the number of 0's and number of 1's up to 2 and 5 respectively Let a state is represented by qy(O sis1,0s j S 4). where i shows the number of 0's and j shows the number of 1's up to the state qij. Definitely, the state goo will be the initial state Transition from the state lij is defined as follows lij.0), where k - i1) mod2 (1)-ip, where p -j+ 1) mod 5 For example. δ(qoo, 0)-q10, δ(Joo, 1)-q01-0(9.0, 0)-Joo, δ(9.1, 0)-q01 etc The state qoo will be the final state. The transition diagram of the DFA will be as follows kj' WhereK(+ 1) mod 1 IO Iu LA 03 9 In second part where number of 0s is not divisible by 2 or the number of 1s is not divisible by 5, the DFA will remain same but states except qoo will be final states

Third part:

Draw the simplest possible (i.e. with fewest number of states) NFA which accepts the following language over the alphabet of {a, e, i, o, u, b, p, r, s, t}: the set of strings which start with a vowel, then two consonants, then 0 or more vowels, followed by the first vowel.

Here the part ‘followed by first vowel’ is not clear? If it is the first vowel (from where the string is started), then DFA/ NFA is not possible as some additional memory required to remember it and we have to go for Pushdown automata.

Last part:

(b) starting state of A is S2

(c) All states of A1 and A3 are final states