I'm currently using van Benthem's "Modal logic for open minds", ed. 2010.
In page 16 (and later in exercises), he considers a model whose relations are shown by directed graphs (the so called process graphs) and then, ranging modal operators only over $\top$ and $\bot$, he makes formulas that are true only at a considered world. Now then the world is defined by the unique formula: World $w$ will be the only world satisfying a formula like $\phi$.
Now the problem is that the process is not explained, and I'm afraid if there is an algorithm or something that is missing from my view. [Indeed, all I got is that for a dead-lock (of course if there is only one), $\Box\bot$ can always be a choice!]
Please clarify me on this,
Thanks.
Well, I am afraid, I cannot give you the exact algorithm for solving such problems, but I could try indicate the reasoning which helps me. The aim of such exercises is to show that for distinct states in a model there is a modal formula that distinguishes them (otherwise, they are modally indistinguishable, i.e. they are the same from the modal point of view). So, each world has something "special". Let us take exercise 1a. We start with the state which requires a formula of minimal modal depth. This state is $3$, and the unique formula for the state is $\square \bot$ (since $\bot$ holds in no state and there are no transitions from $3$). Then, we take state $2$. There is only one transition, and we can say, that $(M,2) \models \square \square \bot \wedge \Diamond \square \bot$ . We need conjunction, because $\square \square \bot$ is true in $3$, and $\Diamond \square \bot$ is true in $4$. However, $\square \square \bot \wedge \Diamond \square \bot$ is true only in $2$. So, now you may see, that we work our way "backwards", from a shallow modal depth to a deeper one. For $1$ we have $(M,1) \models \Diamond (\square \square \bot \wedge \Diamond \square \bot)$. We do not use $\square$ here, since $3$ makes all the squares vacuously true. And for the state $4$ we have $(M,4) \models \Diamond (\Diamond (\square \square \bot \wedge \Diamond \square \bot))$. In the exercise 1b the difficulty is due to the fact, that states $2$ and $4$ are indistinguishable (i.e. bisimilar), they agree on transitions and valuation. Hope, this helps.