Applied Course Test 1

Question

The number of states in a minimal DFA that accepts set of all strings beginning with 1 that, when interpreted as a binary integer is a multiple of 5 over the alphabet={0,1}. For example, strings 101, 1010 and 1111 are in the language?

Comments

6 states? (0 should be rejected right?)

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ya.. can u pls explain how?

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answer added.

Answer 1

Make DFA for divisible by 5 and exclude '0' string from that DFA  So based on this minimal DFA will have 7 states and looks like below-

Approach is-

in divisible by 5 DFA there will be 5 equivalence classes{0,1,2,3,4} and as mentioned in question strings are start with 1 so we need to reject all string with 0.

Comments

@Shubhgupta ur dfa is accepting the strings that are starting with zero.. u should add a dead state transition for zero from q0.... so the no of states will be 7 as said by you earlier.. can u plz explain what procedure did u follow for the states q1,q3,q2,q4,q5..

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@anjali007, i have previously added that dead state but it doesn't make more sense because if a string 0101 then its binary equivalent is 5 right? but i think as mentioned in question DFA accepts stating with 1 only.

Correct image added.

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@Shubhgupta and for the rest of the five states did u follow any "particular approach" or just analysed the states?

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@anjali007, as i mentioned in answer q1 represents string with 1 remainder, q2 with 2 remainder, q3 with 3 remainder and q4 with 4 remainder. and i have created qf new because we don't want to accept 0.

is it clear?

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Yup...thanks

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@Shubhgupta

Please check if this is the apporach you used

first considering : Integer of binary string as multiple of 5 ==> 5 states q0,q1,q2,q3,q4

                0        1

-->*q0      q0       q1

q1            q2       q3

q2            q4        q0

q3            q1        q2

q4             q3       q4

Now we have to ignore strings starting with one hence after modification of transitions we get dfa as in selected and

if starting with 1 condition not mentioned then it will '6' states right??

 

 

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yes approach is correct but 5 states will be q0,q1,q2,q3,q4 not q5.

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Ya tht was done in hurry

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is This Logic Right or Not?

        we know modulo 5 DFA need  minimum 5 state, in this all state work in cycle with each other state.

Here 0 is not start point means we have to reject it so 1 extra state for it and also State 0 is not doing its work so we  need 1 Extra state for it to do the job of it.

 so total    5+1+1=7