# 0.3 The steiner tree problem  (Page 3/3)

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## Proof

If $q$ were such a vertex, the Steiner minimal tree would contain either a edge from $q$ to ${v}_{i}$ not containing ${v}_{j}$ , or vice versa. If the SMT contains a edge from $q$ to ${v}_{i}$ , for example, the SMT can be shortened by deleting $\left[{v}_{i},v\right]$ and adding $\left[q,{v}_{j}\right]$ , a contradiction.

Similar statements can be made to further specify where a Steiner hull may be.

## Remark

It is easy to see that a full Steiner topology with $n$ terminals has $2n-3$ edges. Let $f\left(n\right),n\ge 2$ , denote the number of full Steiner topologies with $n-2$ Steiner points. Adding a Steiner point to the middle of any of the edges and connecting it to a new terminal results in a full Steiner tree for $n+1$ terminals. Geometrically, this causes the edge to become two separate edges, connected by the new Steiner point. In order to be consistent with the fact that edges must meet at ${120}^{\circ }$ angles at Steiner points, the two new edges will rotate to meet at an angle of ${120}^{\circ }$ with each other and with the edge connecting the new Steiner point to the new terminal. Thus there are $2n-3$ ways to create a new full Steiner tree, so

$f\left(n+1\right)=\left(2n-3\right)f\left(n\right)$

which has the solution

$f\left(n\right)=\frac{\left(2n-4\right)!}{{2}^{n-2}\left(n-2\right)!}$

Let $F\left(n,k\right)$ denote the number of Steiner topologies with $n$ terminals and $k$ Steiner points such that no terminal is of degree 3. Then $F\left(n,k\right)$ can be obtained from f(k) by first selecting $k+2$ terminals and a full Steiner topology on it, and then adding the remaining $n-k-2$ terminals one at a time at interior points of some edges, creating a "kink" in each edge. The first terminal can go to one of the $2k+1$ edges, the second to one of $2k+2$ edges, and so on, until the $\left(n-k-2\right)$ th point is added to one of the $2k+\left(n-k-2\right)=n+k-2$ edges. Thus

$F\left(n,k\right)=\left(\genfrac{}{}{0pt}{}{n}{k+2}\right)f\left(k\right)\frac{\left(n+k-2\right)!}{\left(2k\right)!}$

Letting ${n}_{3}$ denote the number of terminals of degree 3

$F\left(n\right)=\sum _{k=0}^{n-2}\sum _{{n}_{3}=0}^{\frac{n-k-2}{2}}\left(\genfrac{}{}{0pt}{}{n}{{n}_{3}}\right)F\left(n-{n}_{3},k+{n}_{3}\right)\frac{\left(n+{n}_{3}\right)!}{k!}$

A table containing the values of $f\left(n\right)$ and $F\left(n\right)$ for $n=2,...,8$ is given.

 $n$ 2 3 4 5 6 7 8 $f\left(n\right)$ 1 1 3 15 105 945 10395 $F\left(n\right)$ 1 4 31 360 5625 110880 2643795

## Computational complexity

The optimization problem:

Given : A set $N$ of terminals in the Euclidean plane

Find : A Steiner tree of shortest length spanning $N$

can be recast as a decision problem:

Given : A set $N$ of terminals in the Euclidean plane and an integer $B$

Decide : Is there a Steiner tree $T$ that spans $N$ such that $|T|\le B$

The Steiner tree problem is NP-complete. This means that it is at least as hard as any problem in NP, but the solution is still verifiable in polynomial time. However, the problem cannot be solved in polynomial time. The discrete Euclidean Steiner problem, however, is not known to be NP-complete.

Given : A set $N$ of terminals with integer coordinates in the Euclidean plane and an integer $B$ .

Decide : Is there a Steiner tree $T$ that spans $N$ such that all Steiner points have integer coordinates and the discrete length of $T$ is less than or equal to $B$ , where the discrete length of each edge of $T$ is the smallest integer not less than the length of that edge.

There is a version of this problem that is in P: Given a finite set in a Banach-Minkowski plane, a minimal Steiner tree for this set can be found in polynomially bounded time with an algorithm that executes only graph theoretic functions.

## The steiner ratio

The Steiner ratio is the smallest possible ratio between the total length of a minimum Steiner tree and the total length of a minimum spanning tree. It is conjectured that, in the Euclidean Steiner problem, the Steiner ratio is $\frac{\sqrt{3}}{2}$ . This is called the Gilbert-Pollak conjecture, and was though to be proven in 1990, but some people do not accept this proof. Gilbert and Pollak also conjectured the ratio for higher dimensional spaces, but this was disproven.

It's easy to see that this is true in the four point case. Alternatively, the Steiner ratio is sometimes given as the total length of the minimum spanning tree over the length of the minimum Steiner tree. The Steiner ratio for the rectilinear Steiner tree, which is when all edges of the Steiner tree are perpendicular, is $\frac{3}{2}$ .

## Other applications

Steiner trees have many biological applications. They can be used to describe the ways proteins connect and fold. Steiner trees are also used in phylogeny. Organic chemistry diagrams often include Steiner trees. For example, CH ${}_{6}$ N ${}_{3}$ is a regular hexagon Steiner tree.

Steiner trees have many computer science applications as well, such as in network design and circuit layout.

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