Use the pumping lemma to prove that the following languages

Use the pumping lemma to prove that the following languages are not context-free.

a. {0ⁿ 1²ⁿ 0ⁿ : n 0}

b. {w {1,2,3,4} : the number of 1s equals the number of 2s,

and the number of 3s equals the number of 4s}

Solution

Let L be a daily language. Then there exists a relentless ‘c’ specified for each string w in L

|w| c

We can break w into 3 strings, w = xyz, specified

|y| > 0
|xy| c
For all k zero, the string xykz is additionally in L.
Applications of Pumping Lemma
Pumping Lemma is to be applied to point out that sure languages don\'t seem to be regular. It ought to ne\'er be wont to show a language is regular.

If L is regular, it satisfies Pumping Lemma.

If L is non-regular, it doesn\'t satisfy Pumping Lemma.

Method to prove that a language L isn\'t regular
At first, we\'ve got to assume that L is regular.

So, the pumping lemma ought to hold for L.

Use the pumping lemma to get a contradiction

Select w specified |w| c

Select y specified |y| one

Select x specified |xy| c

Assign the remaining string to z.

Select k specified the ensuing string isn\'t in L.

Hence L isn\'t regular.

Problem
Prove that L = i 0 isn\'t regular.

Solution

At first, we tend to assume that L is regular and n is that the range of states.

Let w = anbn. so |w| = 2n n.

By pumping lemma, let w = xyz, wherever |xy| n.

Let x = ap, y = aq, and z = arbn, wherever p + letter of the alphabet + r = n.p 0, q 0, r 0. so |y| zero

Let k = 2. Then xy2z = apa2qarbn.

Number of as = (p + 2q + r) = (p + letter of the alphabet + r) + letter of the alphabet = n + letter of the alphabet

Hence, xy2z = an+q bn. Since letter of the alphabet zero, xy2z isn\'t of the shape anbn.

Thus, xy2z isn\'t in L. thence L isn\'t regular.

Complement of a DFA
If (Q, , , q0, F) be a DFA that accepts a language L, then the complement of the DFA are often obtained by swapping its acceptive states with its non-accepting states and the other way around.Let L be a daily language. Then there exists a relentless ‘c’ specified for each string w in L

|w| c

We can break w into 3 strings, w = xyz, specified

|y| > 0
|xy| c
For all k zero, the string xykz is additionally in L.
Applications of Pumping Lemma
Pumping Lemma is to be applied to point out that sure languages don\'t seem to be regular. It ought to ne\'er be wont to show a language is regular.

If L is regular, it satisfies Pumping Lemma.

If L is non-regular, it doesn\'t satisfy Pumping Lemma.

Method to prove that a language L isn\'t regular
At first, we\'ve got to assume that L is regular.

So, the pumping lemma ought to hold for L.

Use the pumping lemma to get a contradiction

Select w specified |w| c

Select y specified |y| one

Select x specified |xy| c

Assign the remaining string to z.

Select k specified the ensuing string isn\'t in L.

Hence L isn\'t regular.

Problem
Prove that L = i 0 isn\'t regular.

Solution

At first, we tend to assume that L is regular and n is that the range of states.

Let w = anbn. so |w| = 2n n.

By pumping lemma, let w = xyz, wherever |xy| n.

Let x = ap, y = aq, and z = arbn, wherever p + letter of the alphabet + r = n.p 0, q 0, r 0. so |y| zero

Let k = 2. Then xy2z = apa2qarbn.

Number of as = (p + 2q + r) = (p + letter of the alphabet + r) + letter of the alphabet = n + letter of the alphabet

Hence, xy2z = an+q bn. Since letter of the alphabet zero, xy2z isn\'t of the shape anbn.

Thus, xy2z isn\'t in L. thence L isn\'t regular.

Complement of a DFA
If (Q, , , q0, F) be a DFA that accepts a language L, then the complement of the DFA are often obtained by swapping its acceptive states with its non-accepting states and the other way around.