Demonstrating that ( L { 1^{n^2} mid n geq 0 } ) is Not a Context-Free Language Using the Pumping Lemma

Introduction: In the realm of formal language theory, understanding whether a given language is context-free or not is crucial. One of the most powerful tools in this context is the Pumping Lemma for context-free languages. In this article, we will demonstrate through a detailed proof that the language ( L { 1^{n^2} mid n geq 0 } ) is not context-free using the Pumping Lemma.

The Pumping Lemma for Context-Free Languages

The Pumping Lemma for context-free languages states that for any context-free language ( L ), there exists a constant ( p ) (the pumping length) such that any string ( s ) in ( L ) of length at least ( p ) can be divided into five parts ( s uvwxy ) satisfying the following conditions:

( |vwx| leq p ) (the length of the middle part is at most ( p )) ( |vx| geq 1 ) (at least one of ( v ) or ( x ) is non-empty) For all ( i geq 0 ), the string ( uv^iwx^iy ) is also in ( L ).

Proof by Contradiction

Assumption: We start by assuming ( L { 1^{n^2} mid n geq 0 } ) is a context-free language. According to the Pumping Lemma, there exists a constant ( p ). Let's choose a string ( s 1^{p^2} ). Clearly, ( s in L ) since ( p^2 ) is a perfect square, and ( |s| p^2 geq p ).

Decomposition of the String

According to the Pumping Lemma, we can split ( s ) into five parts, ( s uvwxy ), such that:

( |vwx| leq p ) (i.e., the length of the middle part is at most ( p )) ( |vx| geq 1 ) (at least one of ( v ) or ( x ) is non-empty)

Since ( vwx leq p ), the segments ( v ) and ( x ) can only contain 1s and their total length must be less than or equal to ( p ).

Pumping the String

Consider the string ( s uv^2wx^2y ). The pumping will increase the number of 1s in ( s ) because we are repeating ( v ) and ( x ).

Length of the Pumped String

Let ( v k ) and ( x m ). The length of the new string ( s ) becomes:

[ s u v w x y v w x u v x y v x p^2 k m p^2 k m ]

Therefore, the length of ( s ) is ( p^2 k m ).

Checking if ( s in L )

For ( s ) to be in ( L ), the length ( p^2 k m ) must also be a perfect square. Let ( n ) such that ( s n^2 ). Thus, we need to find ( n ) such that:

[ n^2 p^2 k m ]

Rearranging gives:

[ n^2 - p^2 k m ]

This can be factored as:

[ (n - p)(n p) k m ]

Since ( k m ) is a fixed positive integer as ( |vx| geq 1 ), ( n - p ) and ( n p ) must both be integers that multiply to give ( k m ).

Contradiction

For sufficiently large ( p ), ( n - p ) and ( n p ) would need to differ by ( 2p ) and thus cannot both be factors of ( k m ) in a way that maintains ( n ) as an integer. This results in a contradiction as ( k m ) being a fixed integer cannot satisfy this condition for all ( p ).

Conclusion

Since we reached a contradiction, we conclude that our initial assumption that ( L ) is context-free must be false. Therefore, ( L { 1^{n^2} mid n geq 0 } ) is not a context-free language.