Length of shortest hamiltonian path in a circle Let's say I have a circle of radius $r$.
I will place $N$ points inside this circle, and then find the shortest hamiltonian path going through all these points.

Of course, I know that this shortest hamiltonain path will be shorter than $2 r\cdot N$
But is there a way to have a better upper-bound ?
Are there research papers about this kind of problem ?
Edit
The problem can be stated as follow:
let $D = \{z \in \mathbb{C}, |z| \leq 1\}$
$$u_n = \sup_{(p_i)\in D^n}(\min_{\sigma \in S_n}\left(\sum_{k=1}^n|p_{\sigma(i+1)}-p_{\sigma(i)}|\right))$$
Find the best possible bound for $u_n$
 A: Summary: For any choice of $N$ points, we can get a Hamiltonian cycle of length at most $8r\sqrt N$. There exist configurations for which we can't do better than about $r \sqrt{\pi N}$.

For a bad configuration, chop the circle up into about $\pi k^2$ square tiles of side length $\frac rk$, and put a point in the center of each tile. There is a distance of at least $\frac rk$ between any two points, so any Hamiltonian cycle (which consists of about $\pi k^2$ line segments between the points) must have length at least about $\pi k^2 \cdot \frac rk = \pi kr$. Here, $N \approx \pi k^2$, so $k \approx \sqrt{\frac N \pi}$, and the length $\pi kr$ is about $r\sqrt{\pi N}$.

To find a Hamiltonian cycle of length at most $8r \sqrt N$ given $N$ arbitrary points, put the circle into a $2r \times 2r$ square. Chop the square up into $4^k$ square tiles of side length $\frac{2r}{2^k}$, where $k$ is chosen sufficiently large that no two points are inside the same square tile. Then, draw an order-$k$ Moore curve through the centers of the square tiles:

(The image is from Wikipedia.)
Our plan will be to visit the $N$ points in the order that the Moore curve visits the tiles that they're in.
To prove that this does well, we'll need to talk about Hilbert curves, since a Moore curve is made up of four of those. The important thing to know about these is that a Hilbert curve of order $k$ takes $4^k - 1$ steps inside a $2^k \times 2^k$ grid, and both the first $4^{k-1}-1$ and the last $4^{k-1}-1$ steps are copies of a Hilbert curve of order $k-1$.
Lemma 1. Suppose a Hilbert curve of some order is drawn through the centers of square tiles of side length $\ell$. If $P$ is a point in the $1^{\text{st}}$ square tile, and $Q$ is a point in the $n^{\text{th}}$ tile, then the length $PQ$ is at most $2\ell\sqrt{2n}$.
Proof. Choose $j$ such that $4^{j-1} \le n < 4^j$: the least number such that $P$ and $Q$ are both inside an order-$j$ Hilbert curve. That means that they're both contained inside a $2^j \times 2^j$ grid, so the length $PQ$ is at most $2^j \sqrt 2 \cdot \ell$. We get the bound above by bounding $2^{j-1} \le \sqrt n$ (from $4^{j-1} \le n$). $\square$
Lemma 2. Now suppose $P$ and $Q$ are (respectively) points in the $p^{\text{th}}$ tile and the $q^{\text{th}}$ tile visited by a Moore curve, where the tiles still have side length $\ell$, and $p<q$. Then the length $PQ$ is at most $4\ell\sqrt{q-p}$.
Proof. In between $p$ and $q$, there is a step $r$ where we go from the end of a Hilbert curve containing the $p^{\text{th}}$ tile to the start of a Hilbert curve containing the $q^{\text{th}}$ tile. Let $R$ be a point on the boundary between those tiles. By Lemma 1 applied twice, $PR \le 2\ell\sqrt{2(r-p)}$ and $RQ \le 2\ell\sqrt{2(q-r)}$. By the triangle inequality, $PQ \le PR + RQ$, and because $x \mapsto \sqrt x$ is concave, the sum is maximized when $r = \frac{p+q}{2}$. In that case, we get $PR + RQ \le 2\ell\sqrt{2 \cdot \frac{q-p}{2}} + 2\ell\sqrt{2 \cdot \frac{q-p}{2}} = 4\ell\sqrt{q-p}$. $\square$
We are ready to find the length of our Hamiltonian cycle. Suppose our points are $P_1, P_2, \dots, P_N$, numbered in the order that the Moore curve visits them. For $1 \le i \le N-1$, let $d_i$ be the the number of steps it takes our Moore curve to go from $P_i$ to $P_{i+1}$; let $d_N$ be the number of steps it takes our Moore curve to go from $P_N$ back to $P_1$, keeping in mind that the Moore curve is closed and symmetric so a form of Lemma 2 still applies. Then the total length $P_1P_2 + \dots + P_N P_1$ is at most $4\ell\sqrt{d_1} + \dots + 4\ell\sqrt{d_N}$, where $\ell = \frac{2r}{2^k}$. By concavity, this sum is maximized when $d_1 = \dots = d_N = \frac{4^k}{N}$, in which case each $\sqrt{d_i}$ is $\frac{2^k}{\sqrt N}$, and their sum is $2^k \sqrt N$. Multiplying by $4\ell$, we get $8r \sqrt N$, as promised.
