The correct grammar for the language L = 0^x 1^y | x=0 , y=1 is:

S- XY | Ꜫ , X-0X | Ꜫ , Y-1 Y | 1

This is because the language L consists of strings of 0's followed by 1's. The number of 0's (x) can be zero or more, and the number of 1's (y) must be one or more.

The grammar S- XY | Ꜫ , X-0X | Ꜫ , Y-1 Y | 1 correctly generates this language.

Here's how:

- S is the start symbol. It can produce the string XY or the empty string Ꜫ. The empty string is included because x can be zero, meaning a string of 0's is not required.
- X generates the 0's in the string. It can produce another 0 and another X (0X), or it can stop producing 0's (Ꜫ).
- Y generates the 1's in the string. It can produce another 1 and another Y (1Y), or it can stop producing 1's after at least one 1 has been produced (1). This is why the production Y-1 Y | 1 is used instead of Y-1 Y | Ꜫ. The latter would allow for zero 1's, which is not allowed in the language L.

Question

The correct grammar for the language L = 0^x 1^y | x>=0 , y>=1 is:

S-> XY | Ꜫ , X->0X | Ꜫ , Y->1 Y | 1

This is because the language L consists of strings of 0's followed by 1's. The number of 0's (x) can be zero or more, and the number of 1's (y) must be one or more.

The grammar S-> XY | Ꜫ , X->0X | Ꜫ , Y->1 Y | 1 correctly generates this language.

Here's how:

- S is the start symbol. It can produce the string XY or the empty string Ꜫ. The empty string is included because x can be zero, meaning a string of 0's is not required.
- X generates the 0's in the string. It can produce another 0 and another X (0X), or it can stop producing 0's (Ꜫ).
- Y generates the 1's in the string. It can produce another 1 and another Y (1Y), or it can stop producing 1's after at least one 1 has been produced (1). This is why the production Y->1 Y | 1 is used instead of Y->1 Y | Ꜫ. The latter would allow for zero 1's, which is not allowed in the language L.

Knowee AI · Accepted Answer

The correct grammar for the language L = 0^x 1^y | x>=0 , y>=1 is:

S-> XY | Ꜫ , X->0X | Ꜫ , Y->1 Y | 1

This is because the language L consists of strings of 0's followed by 1's. The number of 0's (x) can be zero or more, and the number of 1's (y) must be one or more.

The grammar S-> XY | Ꜫ , X->0X | Ꜫ , Y->1 Y | 1 correctly generates this language.

Here's how:

- S is the start symbol. It can produce the string XY or the empty string Ꜫ. The empty string is included because x can be zero, meaning a string of 0's is not required.
- X generates the 0's in the string. It can produce another 0 and another X (0X), or it can stop producing 0's (Ꜫ).
- Y generates the 1's in the string. It can produce another 1 and another Y (1Y), or it can stop producing 1's after at least one 1 has been produced (1). This is why the production Y->1 Y | 1 is used instead of Y->1 Y | Ꜫ. The latter would allow for zero 1's, which is not allowed in the language L.

For the given language L = 0 x 1 y | x>=0 , y>=1 , choose the grammar for the language. Options : S-> XY | Ꜫ , X->0X | Ꜫ , Y->1 Y | 1 S-> XY , X->0X | 0 , Y->1 Y | Ꜫ S-> XY , X->0X | Ꜫ , Y->1 Y | 1 None of above mentioned

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