DFA over alphabet {a,b}

Question

Let M be a DFA over the alphabet ∑ = {a,b} with exactly 2 states. Suppose further that M accepts a finite number 'n' of distinct words. what is the maximum value of 'n' .

A) 1

B) 2

C) 3

D) 4

truly speaking i didn't get the question very clear.

<answer given is A>

Answer 1

M accepts only finite number of words. So, we should have a dead state and one out of two states must be a dead state and this must be reachable from the start state. Further there shouldn't be a loop in the path from the start state to any final state. So, the only option is for the transition on both a and b from start state to go to the dead state. This would make the DFA accept only "empty" string by making the start state final. So, I suppose n = 0.

Comments

Sir € will not considered as a word??

If we say a language L={€,a,b} then size of language set is 3. So ans should be 1. Isn't it?

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but question asks for no. of distinct words- I dont count empty string as a word. Since, no other option matches 1 is the best one to choose.

Answer 2

a machine will give you finite no. of output . means 

eg . a machine contains 2 length string   ∑ = {a,b} .......  so the machine will accept finite number of elements(  aa, ab ba bb ).

but in case of infinite length line  a machine starts with "a" ... then (a,ab,abb,aab,ababababba, .........so on  ) .

and if a dfa contains any cycle then it will produce infinitely many sting , so for finite number of elements the dfa must not contain any cycle . and it has only 2 states . so according to the condition if you draw this machine you will find out that 1st state must be final state and the 2nd one is used as dead state and this machine will only accept £ .  If you considered £ as word then answer will be 1 otherwise 0 . peace 

Comments

sir can you please explain further why second state will be dead state and why not final or non-final state?