I believe (although I'm not that good at math) that each child at least sees one ear. So the minimum amount should be $n\cdot n$ i think?

No, it is wrong. Child can see back of head of neigbour, so he does not see ear.

Just to be sure that I understand correctly: * Every cell contains one child. * Every child looks north, south, east, or west into an adjacent cell. * No child sees the face (front) of another child. * Every child sees either the ear or the back of another child. Is this all correct? (My guess is that the answer under these conditions should be $n+2$.)

test20 wrote: Just to be sure that I understand correctly: * Every cell contains one child. * Every child looks north, south, east, or west into an adjacent cell. * No child sees the face (front) of another child. * Every child sees either the ear or the back of another child. Is this all correct? (My guess is that the answer under these conditions should be $n+2$.) Yes, it is correct.

I'm getting $2n$ I tried using a spiral, starting from the bottom left. Each child looks forward, until the child will see an ear, then they turn to the right. Using induction, it is pretty simple to prove that this method results in $2n$. How are you getting $n+2$?

Ok, basically we are placing arrows (north, south, east, or west) upon squares of the board subject to the condition of facing "within" the board. We can make this a digraph; in particular there is a cycle, and it follows that at least two rows and columns with at least two such children. Suppose some row has no such child. It follows that every arrow is "north" or "south", and thus that every column contains such a child (by containment). This proves the minimum of $n+2$. I'm on mobile, so no construction, but I assure you that it exists Edit: this doesn't work exactly, replace "children" with corners and it should, I'll look into the details when I get home...

As promised (though perhaps a bit tardy ): (Uniquely) Associate to every child a node of a graph that is in the precise center of the square upon which she stands. For each child, construct a directed edge from her node to the node of the person whom she faces; it follows that every node has outdegree one, and furthermore by hypothesis that every edge has a unique direction. Define a "corner" of the graph to be a node among whose associated edges there are two forming a $90^{\circ}$ angle. One can readily show that for each corner, there is (at least) a child facing the one associated to said node with the desired property. It follows that there are at least as many such children as there are nodes; we will show that there must be at least $n+2$ corners, upon which point we may conclude the result. Recall that any digraph in which each vertex has outdegree at least one must necessarily contain a cycle; the one we have constructed is no exception. Because every edge of our digraph is parallel to the edges of the original square, and because no vertex directs to itself, we may additionally conclude that there are at least two columns and at least two rows containing at least two corners. Now suppose that some row has no corners. One can show that this implies that no edge is contained within said row, and thus that every column must contain some corner. The claim follows $\Box$ Now for a construction: Let the children stand on the $n^2$ lattice points of $S=\{0,1,\dots,n\}\times \{0,1,\dots,n-1\}$. For $(x,y)\in S$, let the child at $(x,y)$ face: \[\begin{cases} \text{South} & \text{if}\ x>0\text{ and }y>0\\ \text{West} & \text{if}\ x>0\text{ and }y=0\\ \text{North} & \text{if}\ x=0\text{ and }y\in\{0,1,\dots,n-2\}\\ \text{East} & \text{if}\ (x,y)=(0,n-1)\\ \end{cases} \]It may be checked that this construction works $\blacksquare$