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 Algorithm 2 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 3
Step 1 − All the states Q are divided in two partitions − final states and 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 the 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 −
Q | δ(q,0) | δ(q,1) |
---|---|---|
(a, b) | (a, b) | (c,d,e) |
(c,d,e) | (c,d,e) | (f) |
(f) | (f) | (f) |
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