How to construct DFA that L accept : w contain '110' and doesn't contain '010'?

Question

L = { w belongs {0,1}* | w contain '110' and doesn't contain '010'}

I need to construct DFA that receives L.

How can I draw a DFA that will do the both of conditions?

hints will be a great help.

Answer 1

A finite state machine (or automaton) consists of:

• A finite set of symbols, called the "alphabet".
• A finite set of states. There is no information in a state other than its label.
• A transition table, which maps each pair <state, symbol> to a new state.
• The starting state: the name of the state which the machine is in before it reads the first symbol.
• The accepting state set: A subset of the set of states which define a successful match.

The machine starts in the starting state, and then reads each symbol in turn. When it reads a symbol, it uses the transition table to decide which is the next state. When it reaches the end, it announces success if the current state is in the accepting set; otherwise, it announces failure.

In a deterministic finite-state automaton (DFA), the transition table is single-valued and complete; that is, every entry is filled in with exactly one state. (Constructing a DFA usually involves adding a "sink state", which is non-accepting and has a self-transition on every symbol. This state is used to handle inputs which cannot be at the beginning of an accepted input.)

If you have a DFA which recognises a language L, you can construct a DFA for the complement of L by simply replacing the set of accepting states with the set of non-accepting states. So if you have a DFA which recognises any input containing 010, you can construct a DFA which recognises any input which does not contain 010 by using just changing the accepting state list.

If you have two DFAs which recognise the languages L ₁ and L ₂ , you can construct a new DFA which recognises only those strings in both L ₁ and L ₂ -- that is, the intersection L ₁ ∩ L ₂ -- using the Cartesian product of the two DFAs. In the new machine:

• the alphabet is the intersection of the original alphabets. (Usually, all the alphabets are the same in this operation. I just note this for completeness.)
• the states of the new machine are the Cartesian product of the states of the original machines. In other words, the new states are all the pairs <q _i , r _j > where q _i is a state of the first machine and r _j is a state of the second machine.
• The transition table is constructed using both original transition tables. If there is a transition in the first machine on symbol α from q _i to q _i' and a transition in the second machine on the same symbol α from r _j to r _j' , then in the combined machine there is a transition on α from <q _i , r _j > to <q _i' , r _j' >.
• The starting state of the combined machine is the pair of starting states of the original machines.
• The accepting states of the combined machine are all pairs <q _i , r _j > where q _i is an accepting state of the first machine and r _j is an accepting state of the second machine.

So if you can construct DFAs for:

• The language of strings which contain 110
• The language of strings which contain 010 then you can use the above procedures to construct a DFA for the contain 110 and don't contain 010 by using the above two procedures.