Steiner Graphs worksheet

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1. Suppose there are four cities, located at the corners of a square with
sides of length 4. A network of roads is to connect all four cities. Is it
possible to construct a network of length 12? 11.5? 11(*)?
2. Given a triangle ABC, find a point T such that T A + T B + T C is
minimal.
First suppose that ABC is a triangle with largest angle at most 120 ◦ .
a1. Consider a rotation around A through the angle of 60 ◦ . Suppose
that the rotation sends C to D and T to N. Prove that DN + NT +
T B = T A + T C + T B.
a2. Prove that AT + BT + CT BD. When does the equality hold?
Now suppose that ∠A 120 ◦ .
b1. Let M ∉ ∠A. Prove that MA + MB + MC > AB + AC.
b2. Let M ∈ ∠A. Once again consider the rotation around A through
the angle of 60 ◦ . Suppose that the rotation sends C to D and T to N.
Prove that AB + AC BM + AM + CM
3. Given N points A 1 , . . . , A N in the plane, the goal is to connect them
by a network consisting of line segments of minimum total length. An
example is shown on Figure 1. There can be other points (apart from
A 1 , . . . , A N ) where two or more line segments meet. These are marked
by B i on the figure. They are called Steiner points. The initial points
A 1 , . . . , A N are called real points.
Suppose the network of minimal total length is found.
a. Prove that the angle between any two line segments incident to the
common point is greater or equal than 120 ◦ .
b. Prove that there are no more than 3 line segments incident to a
real point.
c. Prove that there are exactly 3 line segments incident to a Steiner
point, and the angles between them are equal to 120 ◦ .
Define a Steiner tree as a network with the properties (a)-(c). We’ve
just proved that a minimal network is a Steiner tree.
4. A real point A is called a deadlock vertex, if there’s only one line
segment that connects it to another point B, and B is also real. A
pair of deadlock vertices is a pair of two real points A, B, such
that each of A and B has exactly one line segment incident to it, and
these two line segments are incident to a common Steiner point.

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Figure 1: An example of Steiner tree
A
1
A2
A3
B1
A
4
A5
B2
A
6
B
3
A7
B
4
A
8

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Figure 2: Deleting deadlock vertex A
B C B C
A
(In Figure 1 A 2 , A 4 are deadlock vertices, A 7 and A 8 form a pair of
deadlock vertices.)
Prove that every Steiner tree has either a deadlock vertex or a pair of
deadlock vertices.
Hint: Consider a path in the Steiner tree consisting of the maximal
number of line segments.
5. Try to guess a minimal network for four points at the corners of a
square.
6. Now we would like to construct all the Steiner trees for N given points.
Firstly we will verify that the problem of constructing Steiner trees for
N points reduces to the analogous problem for N − 1 points.
For that we will prove that from any given Steiner tree S N for N points
one can construct some Steiner tree S N−1 for N − 1 points.
Note that S N has either a deadlock vertex or a pair of deadlock vertices.
a. In the first case just delete the deadlock vertex (see Figure 2). The
remainder is a Steiner tree for N − 1 points.
b. In the second case everything is a little bit more complicated (see
Figure 3). Let A 1 , A 2 be a pair of deadlock vertices, A 1 B, A 2 B be
the incident line segments, the Steiner point B is connected with a
point K. In order to decrease N we could delete the points A 1 , A 2 ,
but the result would consist of N − 2 real points and wouldn’t be a
Steiner tree (why?). So we would like to put an additional real point
M somewhere.
Construct the equilateral triangle ADM.
b1. Prove that M belongs to the circle circumscribed about triangle
ADM.
b2. Prove that B ∈ KM. Construct a Steiner tree for N − 1 points.
7. We will construct Steiner trees by induction.

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Figure 3: Deleting a deadlock pair A
K
K
A
1
B
A
2
M
M
a. Construct a Steiner tree for N = 2
b. Suppose we know how to construct a Steiner tree for N − 1 points.
Using the previous problem, think about an algorithm that would
allow us to construct all the Steiner trees for N points.
Hint: The first way is to delete an arbitrary point A j temporarily, to
construct a Steiner tree S for A i , i = 1, . . . , j −1, j +1, . . . , N. Connect
A j with an arbitrary vertex A k from S. Check whether the result is a
Steiner tree. Repeat for all A k , A j possible.
Using the previous problem, think about the second way.
8. Construct all the Steiner trees for a triangle.
9. Construct all the Steiner trees for a square.