Let me define a **degree $n$ colored arrangement of circles** on $S^2$ to be a collection $\mathcal{C}$ of $n$ disjoint, smoothly embedded circles $C\_1,\dotsc, C\_n\subset S^2$ together with a continuous map (temperature)


$$ T: S^2 \setminus (C\_1\cup\cdots \cup C\_n)\to \lbrace 1,-1\rbrace $$ 


such that for any circle $C\_k$, the temperatures on opposite sides of $C\_k$ are different. In other words, if $\gamma:(-\varepsilon,\varepsilon)\to S^2$, $t\mapsto \gamma(t)$, is a short smooth path that intersects exactly one circle $C\_k$, and it does so transversally at $t=0$, then 


$$\lim\_{t\searrow 0} \; T\bigl(\; \gamma(t)\;\bigr)= - \lim\_{t\nearrow 0}\; T\bigl(\;\gamma(t)\;\bigr). $$


Think of the components of $T^{-1}(-1)$ as *icy regions* and of the components of $T^{-1}(1)$ as *liquid regions*. We will refer to $T$ as a *coloring* of the *arrangement of circles* $\mathcal{C}$.


Two arrangements of circles $\mathcal{C}$ and $\mathcal{C}'$ are called **equivalent** if there exists an orientation preserving diffeomorphism of $S^2$ that maps one arrangement to the other. We denote by $\mathcal{A}\_n$ the set of equivalence classes of arrangements on $n$ disjoint circles on $S^2$.


Two degree $n$ colored arrangements $(\mathcal{C}, T)$ and $(\mathcal{C}', T')$ are called **equivalent** if there exists an orientation preserving diffeomorphism $\Psi: S^2\to S^2$ mapping the circles in $\mathcal{C}$ to the circles in $\mathcal{C}'$ and such that $T'\circ \Psi= T$.


Two colorings of a given arrangement $\mathcal{C}$ of $n$ disjoint circles on $S^2$ are called **equivalent** if the colored arrangements $(\mathcal{C}, T)$ and $(\mathcal{C}, T')$ are equivalent. We get in this fashion a map


$$ K\_n : \mathcal{A}\_n\to\mathbb{Z}$$


that associates to an arrangement $\mathcal{C}$ the nummber of inequivalent colorings of $\mathcal{C}$. Here are some questions I find interesting.


$\mathbf{Q\_0}$ Find the cardinality of $\mathcal{A}\_n$ ans its large $n$ asymptotics.


$\mathbf{Q\_1}$ Find the cardinality of $\mathcal{S}\_n$ and its large $n$ asymptotics. 


$\mathbf{Q\_1^\*}$ Investigate the map $K\_n$.


$\mathbf{Q\_2}$ We declare an colored arrangement $(\mathcal{C}, T)$ to be **selfdual** if it is equivalent to $(\mathcal{C}, -T)$. Find the number of selfdual colored arrangements of degree $n$ and its large $n$ asymptotics.


$\mathbf{Q\_3}$ Equip $\mathcal{S}\_n$ with the uniform probability density and denote by $X\_n:\mathcal{S}\_n\to \mathbb{R}$ the random variable o that associates to a colored arrangement $(\mathcal{C}, T)$ the Euler characteristic of the liquid region $T^{-1}(1)$. Denote by $\mu\_n$ the probability distribution of $X\_n$. Investigate the large $n$ behavior of $\mu\_n$.


$\mathbf{Q\_4}$ Can you think of a good notion of a random arrangement of $n$ disjoint circles on $S^2$?



Colored arrangements of circles on the two sphere
Consider the graph obtained by assigning to each region a vertex, and two vertices being joined by an edge iff the two regions are neighbors. It is easy to see that this is a tree, with the empty circles as terminal vertices: once you crossed a circle, the only way to return to the same region is by the same edge.


The tree isomorphism corresponds to the equivalence of arrangements of circles.


So the cardinality of $\mathcal A\_n$ is the same as for the [unlabeled trees](http://en.wikipedia.org/wiki/Tree_graph#Unlabeled_trees). In the same wiki reference is mentioned he asymptotic estimate given by Otter (1948). For the cardinality of $\mathcal A\_n$ is no known formula.


For the $\mathcal S\_n$, there are only two ways to color a tree so that each edge connects two vertices of opposite colors. They may be not distinct (*i.e.* may be **selfdual**), since it is possible to have a tree automorphism which switches the colors. The asymptotics is similar to the uncolored case, because there are only one or two distinct colorings of the same tree.


The notion of [randomness for trees](http://en.wikipedia.org/wiki/Random_tree) depends on the purpose.


The answers are partial, mainly because the answers to the corresponding questions for the trees are partial known, but I hope that the connection with trees may help.


