Gate2017 cs Q49

  • Option : A
  • Explanation :
    Union of context free language is also context free language.
    L1 = { an bn cm| m >= 0 and n >= 0 } and
    L2 = { am bncn| n >= 0 and m >= 0 }
    L3 = L1 ∪ L2 = { anbncm∪ ambncn| n >= 0, m >= 0 } is also context free.
    L1 says number of a’s should be equal to number of b’s and L2 says number of b’s should be equal to number of c’s. Their union says either of two conditions to be true. So it is also context free language.
    Intersection of CFG may or may not be CFG.
    L3 = L1 ∩ L2 = { anbncn| n >= 0 } need not be context free
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