# DFA Minimization

Posted By on April 29, 2016

## DFA Minimization using Myphill-Nerode Theorem

### Algorithm

 Input: DFA Output: Minimized DFA Step 1 Draw a table for all pairs of states (Qi, Qj) not necessarily connected directly [All are unmarked initially] Step 2 Consider every state pair (Qi, Qj) in the DFA where Qi ∈ F and Qj ∉ F or vice versa and mark them. [Here F is the set of final states]. Step 3 Repeat this step until we cannot mark anymore states − If there is an unmarked pair (Qi, Qj), mark it if the pair {δ(Qi, A), δ (Qi, A)} is marked for some input alphabet. Step 4 Combine all the unmarked pair (Qi, Qj) and make them a single state in the reduced DFA.

### Example

Let us use above algorithm to minimize the DFA shown below.

Step 1 − We draw a table for all pair of states.

 a b c d e f a b c d e f

Step 2 − We mark the state pairs −

 a b c d e f a b c ✔ ✔ d ✔ ✔ e ✔ ✔ f ✔ ✔ ✔

Step 3 − We will try to mark the state pairs, with green colored check mark, transitively. If we input 1 to state ‘a’ and ‘f’, it will go to state ‘c’ and ‘f’ respectively. (c, f) is already marked, hence we will mark pair (a, f). Now, we input 1 to state ‘b’ and ‘f’; it will go to state ‘d’ and ‘f’ respectively. (d, f) is already marked, hence we will mark pair (b, f).

 a b c d e f a b c ✔ ✔ d ✔ ✔ e ✔ ✔ f ✔ ✔ ✔ ✔ ✔

After step 3, we have got state combinations {a, b} {c, d} {c, e} {d, e} that are unmarked.

We can recombine {c, d} {c, e} {d, e} into {c, d, e}

Hence we got two combined states as − {a, b} and {c, d, e}

So the final minimized DFA will contain three states {f}, {a, b} and {c, d, e}

## DFA Minimization using Equivalence Theorem

If X and Y are two states in a DFA, we can combine these two states into {X, Y} if they are not distinguishable. Two states are distinguishable, if there is at least one string S, such that one of δ (X, S) and δ (Y, S) is accepting and another is not accepting. Hence, a DFA is minimal if and only if all the states are distinguishable.

### Algorithm

 Step 1 All the states Q are divided in two partitions − final statesand non-final states and are denoted by P0. All the states in a partition are 0th equivalent. Take a counter k and initialize it with 0. Step 2 Increment k by 1. For each partition in Pk, divide the states in Pk into two partitions if they are k-distinguishable. Two states within this partition X and Y are k-distinguishable if there is an input S such that δ(X, S) and δ(Y, S) are (k-1)-distinguishable. Step 3 If Pk ≠ Pk-1, repeat Step 2, otherwise go to Step 4. Step 4 Combine kth equivalent sets and make them the new states of the reduced DFA.

### Example

Let us consider the following DFA −

q δ(q,0) δ(q,1)
a b c
b a d
c e f
d e f
e e f
f f f

Let us apply above algorithm to the above DFA −

• P0 = {(c,d,e), (a,b,f)}
• P1 = {(c,d,e), (a,b),(f)}
• P2 = {(c,d,e), (a,b),(f)}

Hence, P1 = P2.

There are three states in the reduced DFA. The reduced DFA is as follows −

The State table of DFA is as follows −

Q δ(q,0) δ(q,1)
(a, b) (a, b) (c,d,e)
(c,d,e) (c,d,e) (f)
(f) (f) (f)

Its graphical representation would be as follows −

#### Posted by Akash Kurup

Founder and C.E.O, World4Engineers Educationist and Entrepreneur by passion. Orator and blogger by hobby

Website: http://world4engineers.com