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According to the given language, which among the following expressions does it correspond to ?

Language $L=\{x\in\{0,1\}\mid x\text{ is of length 4 or less}\}$.

  1. $(0+1+0+1+0+1+0+1)^4$
  2. $(0+1)^4$
  3. $(01)^4$
  4. $(0+1+\varepsilon)^4$
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D is the correct answer.
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option D because of empty string .
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0

6 Answers

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3 votes

Given that,  $L=\{x\in \{0,1\}|x\; \text{is length 4 or less}\}$

$\text{Expression (E)} =$ Length $4$ or less means, length $0$ or length $(1)$ or length $(2)$ or length $(3)$ or length $4$

$E = \varepsilon + (0+1)^{1} + (0+1)^{2}+(0+1)^{3} + (0+1)^{4} = (\varepsilon+0+1)^{4}$

So, the correct answer is $(D).$

References:

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1 vote
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When length is less than or equal to n then regular expression is $(0+1+ epsilon)^n$

Why epsilon is needed? Else we will get only strings of length n.

So D is correct.
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$\text{option A, B, C does not generate minimal string of length 0 --- } \epsilon$

hence option D is correct answer
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The extended notation would be (0+1)4 but however, we may allow some or all the factors to be ε. Thus ε needs to be included in the given regular expression.

correct option -D
Answer:

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