1 votes 1 votes Consider the languages $L_{1}= \phi$ and $L_{2}=\{1\}$. Which one of the following represents $L_{1}^{\ast}\cup L_{2}^{\ast} L_{1}^{\ast}$? $\{\in \}$ $\{\in,1\}$ $\phi$ $1^{\ast}$ Theory of Computation ugcnetcse-jan2017-paper3 theory-of-computation regular-language + – go_editor asked Mar 24, 2020 • recategorized May 24, 2020 go_editor 1.6k views answer comment Share Follow See 1 comment See all 1 1 comment reply akash.dinkar12 commented Oct 9, 2017 reply Follow Share L 1* ∪ L 2* L 1* φ* ∪ (1)*.φ* * > . > ∪ (priority) ε ∪ (1)*.ε ε ∪ (1)* 1* 1 votes 1 votes Please log in or register to add a comment.
2 votes 2 votes option 4 L1 = ϕ, and L2 = {1} L1 * is also ϕ L2* will be 1* so L1* U L2* L1* = 1* Debasmita Bhoumik answered Jan 28, 2017 Debasmita Bhoumik comment Share Follow See 1 comment See all 1 1 comment reply Tanuj Guha Thakurta commented Feb 16, 2019 reply Follow Share L1* is not ϕ but {ε} 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes $L_1^* = \phi ^* = \epsilon$ $L_2^*L_1^* = \{1\}^* . \epsilon = 1^*$ $L_1^*\ \cup L_2^*L_1^* = \epsilon \ \cup 1^* = 1^*$ Option (D) Rishabh Gupta 2 answered Nov 7, 2017 Rishabh Gupta 2 comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes ∅*=∈ ∅.anything =∅ ∈ union ∅ =∈ A is answer Smriti012 answered Jan 31, 2017 Smriti012 comment Share Follow See 1 comment See all 1 1 comment reply akash_chauhan commented Jul 18, 2022 reply Follow Share As, L1* = epsilon therefore, L2*. L1* = 1* . epsilon ( not 1*. phi ) which results in 1* i. e Option D. ( not option A) 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes 4 is answer because l1 is empty so no use of union and l2 is singleton so it will be 4 abhi1997 answered Feb 2, 2017 abhi1997 comment Share Follow See all 0 reply Please log in or register to add a comment.