minimum lines, maximum points There are $P$ points in the 2-dimensional plane. Through each point, we draw two orthogonal lines: one horizontal (parallel to x axis), one vertical (parallel to y axis). Obviously, some of these lines conicide. How can we arrange the points such as the total number of lines is minimized?
I started with solving the dual problem (is that the correct name?): suppose we are allowed exactly $L$ lines, what is the maximum number of points we can have? 
Suppose the number of different x values is $X$, and the number of different y values is $Y$. Then the number of lines is $L=X+Y$, and the maximum number of points is $P=X*Y$. So we have the following optimization problem:
$maximize(X*Y)$
$such-that: X+Y=L$
AFAIK, the solution to this problem is to take X and Y as close as possible, e.g., if $L$ is even, take $X=Y=L/2$, and then $P=(L/2)^2$. 
However, I don't know how to go from this to the original problem. For example, if we have $P=14$ points, then we can arrange them in a 7-by-2 grid and have $L=9$ lines, but we can also arrange them in a 4-by-4 grid with 2 points missing, then we will have only $L=8$ lines. So, how can I find the solution to the original minimization problem:
$minimize(X+Y)$
$such-that: X*Y>=P$
?
 A: You're correct. We will place the points one at a time, in a zigzag greedy fashion, so that no lattice points are wasted. Redefine $n$ to be the number of points. Now for each $i\in \{1,...,n\}$, let $P_i$ be the $i$th point to be placed. Notice that since any integer $i$ lies between two consecutive perfect squares of the form $k^2$ and $k^2 + 2k + 1$ (where $k=\lfloor \sqrt{i} \rfloor$), we can uniquely express $i$ in the form $i=k^2+r$, where $r\in \{0,1,...,2k\}$. We now map each point to the Cartesian plane as follows:
$$ P_i =
\begin{cases}
(k,k)   & \text{if } r = 0 \\
(k+1,r) & \text{if } r \in \{1,...,k\} \\
(r-k,k+1) & \text{if } r \in \{k+1,...,2k\} \\
\end{cases}
$$
For example, since $14=3^2+5$ and $5\in \{4,5,6\}$, we would place the $14$th point $P_{14}$ at $(5-3,3+1)=(2,4)$.
Note that if $L(n)$ is the total number of drawn lines after placing down the $n$ points $P_1,P_2,...,P_n$, and if we express $n$ in the form $n=k^2+r$ where $r\in \{0,1,...,2k\}$ as above, then:
$$ L(n) =
\begin{cases}
2k   & \text{if } r = 0 \\
2k+1 & \text{if } r \in \{1,...,k\} \\
2k+2 & \text{if } r \in \{k+1,...,2k\} \\
\end{cases}
$$
