2 votes 2 votes Abbas2131 asked Nov 12, 2017 Abbas2131 837 views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply Red_devil commented Nov 12, 2017 i edited by Red_devil Nov 12, 2017 reply Follow Share the answer is DCFL.because L1-L2 will contain al string which has atleast one b so L1-L2=(a^nb^+c^n) which is DCFL 0 votes 0 votes Abbas2131 commented Nov 12, 2017 reply Follow Share So the ans is ? 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes L1= {epsilon,ac,abc,aabcc,b,bb,bbb.....} L2={epsilon, ab,aabb,aaabbb..............} as we can see only epsilon is common L1-L2 ={ ac,abc,aabcc,b,bb,bbb......} ={a^n b^m c^n | m>=0,n>0 or m>0,n>=0} which is DCFL Akash Mittal answered Nov 12, 2017 Akash Mittal comment Share Follow See all 3 Comments See all 3 3 Comments reply Abbas2131 commented Nov 12, 2017 reply Follow Share @akash If we take L1-L2=L1 intersection L2' So L2'={abb,ba,aab,b,a,bb,...} (anything expct L2) L1={abc,abbc,b,bb,aabcc,...} Now common between them {b,bb,bbb...} wiz. b+ So wouldnt that be Regular ? 0 votes 0 votes Akash Mittal commented Nov 12, 2017 reply Follow Share @abbas2131 the above lang. also contain b* if we put n=0,m>0 0 votes 0 votes Abbas2131 commented Nov 12, 2017 reply Follow Share But its about intersection og L1 and L2' So why re we taking any string with c in it ffom L1, because L2' wont have it. 0 votes 0 votes Please log in or register to add a comment.