Let L be a customary language. Then, at that point, there exists a steady 'c' to such an extent that for each string w in L −
|w| ≥ c
We can break w into three strings, w = xyz, to such an extent that −
|y| > 0
|xy| ≤ c
For all k ≥ 0, the string xykz is additionally in L.
Utilizations of Siphoning Lemma
Siphoning Lemma is to be applied to show that specific dialects are not standard. It ought to never be utilized to show a language is normal.
Assuming L is customary, it fulfills Siphoning Lemma.
On the off chance that L doesn't fulfill Siphoning Lemma, it is non-customary.
Technique to demonstrate that a language L isn't standard
From the beginning, we need to accept that L is normal.
In this way, the siphoning lemma ought to hold for L.
Utilize the siphoning lemma to get an inconsistency −
Select w to such an extent that |w| ≥ c
Select y with the end goal that |y| ≥ 1
Select x with the end goal that |xy| ≤ c
Appoint the excess string to z.
Select k with the end goal that the subsequent string isn't in L.
Thus L isn't customary.
Issue
Demonstrate that L = {aibi | I ≥ 0} isn't customary.
Arrangement −
From the start, we expect that L is standard and n is the quantity of states.
Let w = anbn. Subsequently |w| = 2n ≥ n.
By siphoning lemma, let w = xyz, where |xy| ≤ n.
Let x = ap, y = aq, and z = arbn, where p + q + r = n, p ≠ 0, q ≠ 0, r ≠ 0. In this manner |y| ≠ 0.
Let k = 2. Then, at that point, xy2z = apa2qarbn.
Number of as = (p + 2q + r) = (p + q + r) + q = n + q
Thus, xy2z = an+q bn. Since q ≠ 0, xy2z isn't of the structure anbn.
Hence, xy2z isn't in L. Subsequently L isn't ordinary.
If (Q, ∑, δ, q0, F) be a DFA that accepts a language L, then the complement of the DFA can be obtained by swapping its accepting states with its non-accepting states and vice versa.
We will take an example and elaborate this below −
This DFA accepts the language
L = {a, aa, aaa , ............. }
over the alphabet
∑ = {a, b}
So, RE = a+.
Now we will swap its accepting states with its non-accepting states and vice versa and will get the following −
This DFA accepts the language
Ľ = {ε, b, ab ,bb,ba, ............... }
over the alphabet
∑ = {a, b}
Note − If we want to complement an NFA, we have to first convert it to DFA and then have to swap states as in the previous method.
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