162 views which of the statement is/are correct

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If L={a+b+} then LR={b+a+}

So L contains strings {ab,aab,aaab,abbb...} and LR={ba,baa,bba,...}

No strings common.

If L={a+} then LR={a+}

So L contains strings {a,aa,aaa...} and LR contains {a,aa,aaa...}

All strings common.

So S1 is false.

If L1={a} and L2={a*}

Then L1L2={aa*} and L2L1={a*a}  Both are same hence commutative in this case also..

S2 is false..

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Okay done :)

If L={a+b+} then LR={b+a+}

So L contains strings {ab,aab,aaab,abbb...} and LR={ba,baa,bba,...}

No strings common.

If L={a+} then LR={a+}

So L contains strings {a,aa,aaa...} and LR contains {a,aa,aaa...}

All strings common.

So S1 is false.

--------------------------------------------------------

If L1={a} and L2={a*}

Then L1L2={aa*} and L2L1={a*a}  Both are same hence commutative in this case also..

S2 is false..

selected

## Related questions

1
176 views
Min number of states in equivalent DFA ______________ will it be 4 or 5 ??
$P_{1}:$ {$<M>|M$ is a TM that accepts atleast $2$ strings of different length} $P_{2}:$ {$<M>|M$ is a TM and there exists an input whose length less than $100,$ on which $M$ halts } The number of problem which is $RE$ but not $REC$ _____________
The pushdown automata $M=\left \{ \left ( q_{0},q_{1},q_{2} \right ),\left ( a,b \right ) ,\left ( 0,1 \right ),\partial ,q_{0},0,\left \{ q_{0} \right \}\right \}$ $\partial \left ( q_{0},a,0 \right )=\left ( q_{1},10 \right )$ ... think here $q_{2}$ is accepting state , not $q_{0}$ As last transition going from $q_{2}$ to $q_{0}$ and not $q_{0}$ to $q_{2}$ Am I right?