Two states ( qi, qj ) are distinguishable in partition Pk if for any input symbol a, δ ( qi, a ) and δ ( qj, a ) are in different sets in partition Pk-1. So minimal DFA will have two states. Step 1 − Draw a table for all pairs of states (Q i, Q j) not necessarily connected directly [All are unmarked initially]. P0 will have two sets of states. Minimization of DFA Suppose there is a DFA D Q, Σ, q0, δ, F > which recognizes a language L. Then the minimized DFA D Q’, Σ, q0, δ’, F’ > can be constructed for language L as: Step 1: We will divide Q (set of states) into two sets. To calculate P1, we will check whether sets of partition P0 can be partitioned or not: i) For set { q1, q2, q4 } : δ ( q1, 0 ) = δ ( q2, 0 ) = q2 and δ ( q1, 1 ) = δ ( q2, 1 ) = q5, So q1 and q2 are not distinguishable. Minimization of DFA Using Equivalence Theorem- Step-01: Eliminate all the dead states and inaccessible states from the given DFA (if any). Therefore, statement 3 is also false. ii) For set { q0, q3, q5 } : δ ( q0, 0 ) = q3 and δ ( q3, 0 ) = q0 δ ( q0, 1) = q1 and δ( q3, 1 ) = q4 Moves of q0 and q3 on input symbol 0 are q3 and q0 respectively which are in same set in partition P0. Most popular in Theory of Computation & Automata, More related articles in Theory of Computation & Automata, We use cookies to ensure you have the best browsing experience on our website. This can be a problem because then we will not know which action to invoke in the accepting state. Statement 3 says that the DFA is minimal. So, this is the final partition. v) For set { q5 }: Since we have only one state in this set, it can’t be further partitioned. If we input 1 to state ‘a’ and ‘f’, it will go to state ‘c’ and ‘f’ respectively. Now, we input 1 to state ‘b’ and ‘f’; it will go to state ‘d’ and ‘f’ respectively.

For each partition in Pk, divide the states in Pk into two partitions if they are k-distinguishable. All those non-final states which transit to itself for all input symbols in ∑ are called as dead states. The reduced DFA is as follows −. So, P1 = { { q1, q2, q4 }, { q0, q3}, { q5 } }. of states in minimized DFA will be equal to no. Similarly, δ ( q1, 0 ) = δ ( q4, 0 ) = q2 and δ ( q1, 1 ) = δ ( q4, 1 ) = q5, So q1 and q4 are not distinguishable. Input − DFA. So, q0 and q3 are not distinguishable. DFA minimization stands for converting a given DFA to its equivalent DFA with minimum number of states. So, q0 and q3 are not distinguishable. The DFA is present in the panel on the left, and a tree of states is present in the screen to the right. Step 3 − We will try to mark the state pairs, with green colored check mark, transitively.

Minimización de DFA - DFA minimization. Step 1 − All the states Q are divided in two partitions − final states and non-final states and are denoted by P0. Attention reader! One set will contain q1, q2, q4 which are final states of DFA and another set will contain remaining states.

How to find whether two states in partition Pk are distinguishable ? Since, P0 = P1, P1 is the final DFA.

Which of the following is false? 2 and 4 only C. 2 and 3 only D. 3 and 4 only. Let us apply the above algorithm to the above DFA −, There are three states in the reduced DFA. Step 3 − If Pk ≠ Pk-1, repeat Step 2, otherwise go to Step 4. So, P2 = { { q1, q2, q4 }, { q0, q3 }, { q5 } } Since, P1=P2. If X and Y are two states in a DFA, we can combine these two states into {X, Y} if they are not distinguishable. No. Output − Minimized DFA. A. So correct option is (D). Complement of L(A) is context-free. 2. q0 and q1 can be merged. Let us use Algorithm 2 to minimize the DFA shown below.

So P0 = { { q1, q2, q4 }, { q0, q3, q5 } }. A accepts all strings over { 0, 1 } of length atleast two. All the states in a partition are 0th equivalent. Ejemplo DFA. In each set of Pk-1, we will take all possible pair of states.

Similarly, δ ( q1, 0 ) = δ ( q4, 0 ) = q2 and δ ( q1, 1 ) = δ ( q4, 1 ) = q5, So q1 and q4 are not distinguishable. Example Consider the following DFA shown in figure. Minimization of DFA Suppose there is a DFA D < Q, Σ, q0, δ, F > which recognizes a language L. Then the minimized DFA D < Q’, Σ, q0, δ’, F’ > can be constructed for language L as: Step 1: We will divide Q (set of states) into two sets. Similarly, q0 and q3 are merged into one.

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