# A deterministic FA for $0^*1^*$ is required

A deterministic finite automaton without $$\epsilon$$ steps for the language $$0^*1^*$$ is required. Any nice picture ? I have created a NFA for this language which has 2 states $$Q_1,Q_2$$, both are accepting, $$Q_1$$ is initial, there is a move under $$0$$ from $$Q_1$$ to $$Q_1$$, a move under $$1$$ from $$Q_1$$ to $$Q_2$$ and finally a move from $$Q_2$$ to $$Q_2$$ under $$1$$.

• An empty string or any string containing only $$0$$s is accepted, so any $$0^*$$ is accepted.
• Zero or more $$0$$s followed by one or more $$1$$s is accepted, so any $$0^*11^*$$ is accepted.
• Zero or more $$0$$s followed by one or more $$1$$s followed by $$0$$ is rejected, so any $$0^*11^*0$$ is rejected.
• Any extension to a rejected string is also rejected (i.e. there is no route back from a rejecting state to an accepting state), so any $$0^*11^*0(0+1)^*$$ is rejected.