Were given sets on your ass to construct a non deterministic finance state atomic recognize each of these sets. So the first set were given Is the sec containing Lambda in this string zero. But as zero be the start state, then machine should accept the empty string Lambda. So as zero should be a final state, the input is zero. Then we'll move on to another final. States and Syria should be accepted. Called his final state s one. And from this, we've obtained in non deterministic financier Tom Adan that will recognize strings Lambda and zero, but no other strings in part B. We're giving this set zero 11 now, but s zero be a start state since the empty string Lambda should not be recognized. It follows that s zero should a non final state. Now, if the first input is a zero, Emma wants to move from s zero to final State s one because zero should be accepted by the machine. Now, if instead, first input is a one, then we'll move from s zero to a non final state s to don't remain s zero because this would imply that any string with just the sequence of ones should not be accepted while 11 should be accepted. Now suppose that the next input is a one. In other words, we have a one at S two and we'll move from s to to a final state which we can just use the same final state from before s one because 11 should be except And so the resulting non deterministic finance state Thomason recognises only strings zero and 11 in part C. We're getting set 011 000 In a sense will be adapting the previous machine to make this machine. So once again, well, let s see. Rabia starts date And since lamb dish not be accepted, we have the S zero is again a non final state. Now it's the input is zero and we'll move from s zero to a final state s one, since the string zero should be accepted Now, if the input is zero, see at s one and we want to move from s one to a non final state s two since 00 should not be accepted. And finally, if the input is a zero that as to and we'll move to a final state as three. Since 000 should be accepted. Now, we're not moving back to the final state s one because we don't want any sequins of triples of zeros beaks accepted. We just want 000 to be accepted. Suppose instead that the input initially is a one, and we'll move from as zero to a non final state. This will be a new non final state s for because the string one should not be accepted. And if the next input is a one at this state s four and we want to move from S four to a final state and here we can reuse one of the previous final states s three.