The shortest distance between 3 points.

1. Fermat Point
Gurpreet Sidhu
April 1, 2012
1 A Modern Scenario
Chancellor of the Exchequer, George Osbourne, of Great Britain announced in his 2012 budget that he
will be creating ten super-connected broadband cities [22]. These are London, Edinburgh, Belfast, Cardiff,
Birmingham, Bradford, Bristol, Leeds, Manchester and Newcastle. Mr Osborne said that he wanted the UK to
become “Europe’s technology centre”. Concerned about the costs of the project, he tells his researchers to find
the most cost effective way of doing this.
A short while later, one of the researchers comes back after spending a few hours on Google, with a program
called “GeoSteiner” [21] and claims that it can find the true shortest network between all ten cities. After
contacting the writers of the program, Professor Martin Zachariasen1 of the University of Copenhagen confirms
that the researcher is correct and that the algorithm within the GeoSteiner software can find the shortest “street
network” between up to 3000 points. He affirms the researchers claim and points out that, between the ten
cities up to 37.5 km of infrastructure can be saved compared to a simple straight line road network. He also
sends the output from the program as can be seen in Figure 1. The researcher points out that this is a 3.61%
saving and at a conservative estimate could save the treasury £2.8 million (£75,000 per km).
Figure 1: 10 cities in the UK, connected on the left by straight road segments only, connected on the right with
three new nodes N1 , N2 and N3 , what are these nodes?...
1 Professor Martin Zachariasen is the Head of Department of Computer Science, I am grateful for his personal help with the
GeoSteiner program.
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2. 2 Introduction
The modern scenario presents us with an opportunity to explore exactly what has gone into solving this type
of problem. Some keywords that were mentioned were ‘street network’2 , ‘algorithm’3 and the reader may have
noticed some new ‘nodes’ in Figure 1. These nodes are the key to solving this particular type of problem.
Originally called Fermat points they are now more commonly known as Steiner points. I will present a brief
history of the Fermat point and its inherent special properties, extending it to the Steiner tree problem and
finally revisiting and summing with our super-connected ten cities example.
In geometry the Fermat point of a triangle, also called the Torricelli point, is a point such that the total
distance from the three vertices of the triangle to the ‘point’ is the minimum possible. It is so named because
this problem was first raised by Pierre Fermat (1601 - 1665) early in the 17th Century in a private letter to
Torricelli at the end of his celebrated essay on maxima and minima. Torricelli’s solution was published by his
pupil Viviani in 1659 [16] [15].
The Fermat point gives a solution to the Steiner tree problem for three points. We will discuss the Steiner
tree problem in detail later on in this article, as an extension, for problems involving more than three points
and consequently n points.
Remark 1. It should be noted that in mathematical literature from 1941 onwards the Fermat point becomes the
Steiner point ever since the publication of What is Mathematics? by Courant and Robbins [14]. In which they
present the problem posed by Fermat as a special case of a general problem that Jakob Steiner had worked on in
his studies of combinatorial optimization [11]. In fact, the authors not only fail to refer to Fermat but they also
fail to note the keen interest of other scientists such as Gauß or Jarnik and K¨ssler for the general problem [17].
o
Figure 2: The Fermat Point ‘P’
2A set of lines and points that represent a system connecting each point to all others by some route.
3A set of instructions for solving a problem, usually with the requirement that the procedure terminate at some point.
2

3. 3 History of the Fermat Point
Torricelli (1608 - 1647)
Evangelista Torricelli’s solution formulated around 1640 was a simple geometric one involving circles. If you
construct circles through the equilateral triangles generated by the sides of the given triangle, in our case
ABC, you will find that the circles intersect at the required point. This came to be known as the Torricelli
point, yielding the minimal path between points A, B and C. It can be noted that only two such circles are
needed.
Figure 3: Torricelli’s Circles Solution
Cavalieri (1598 - 1647)
Bonaventura Francesco Cavalieri in his final year of life published Excercitationes Geometricae and noted an
important property of the Torricelli point. He found that the angles created by the segments joining the point
to the given triangles vertices were equal.
∠AP B = ∠BP C = ∠CP A
Figure 4: Cavalieri
Simpson (1710 - 1761)
Thomas Simpson found an alternative solution in 1750. He used line segments connecting the new vertices
A , B and C of the equilateral triangles to the opposite vertices of the original given triangle. These lines are
referred to as Simpson Lines.
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4. Figure 5: Simpson Lines
Heinen
In 1834 Franz Heinen noted that the Simpson lines are all of equal length, and each Simpson line is the sum
of the distances of the original points to the Fermat point.
AP + BP + CP = AA = BB = CC
He also made the observation that if one of the vertices of the given triangle was equal to 120◦ then that vertex
itself became the Fermat point. At this point we could examine what happens when an angle is greater than
120◦ but we take a look at this later in our complementary problem, section 5, to the simple Fermat problem. It
should also be noted that it was Franz Heinen that solved the original problem posed by Fermat in its entirety.
Courant and Robbins
Richard Courant and Herbert Robbins published their book What is Mathematics? in 1941. They presented
the problem as the Steiner Tree Problem after Jakob Steiner’s work on his studies of combinatorial optimisation
and offered a rigorous proof to its solution [14]. To move forward, our solutions would be better explained if we
now move away from simple geometry with triangles to a more direct method of describing and solving for the
problem of the Fermat point. In order to proceed we first need to know some basic definitions.
Figure 6: A Tree and A Minimal Spanning Tree
Definition 1. Tree. A tree is a connected graph that has no cycles [18].
Definition 2. Spanning Tree. Let G be a connected graph. A spanning tree in G is a subgraph of G that includes
every vertex of G and is also a tree [18].
Definition 3. Minimal Spanning Tree. A minimal spanning tree for a graph G is the spanning tree for G that
has the shortest length [18].
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5. Figure 7: 3 Points In A Plane
It can be seen from Figure 6 that the minimal spanning tree of three points in the plane, is simply the points
connected using the two shortest edges. Figure 7 shows three points in a plane forming a triangle in which all
angles are less than 120◦ . In this case we can construct the minimal spanning tree as seen in A1 B1 C1 and
construct the shortest network using the Fermat point as seen in A2 B2 C2 . Following on from the observation
made by Heinen we can note that if ∠BAC = 120◦ or above then A1 B1 C1 is both the minimal spanning tree
and the shortest network with P1 at the same point as A1 . The minimal spanning tree is of importance as we
go on further to consider more than three points in the plane, from it we can ascertain the shortest distance
between points using only the points themselves and straight lines between them. However, using Fermat points,
we can create a ‘street network’ using the new Fermat points to connect to the original points in the plane. This
new network as we have found in A2 B2 C2 is shorter in length than the minimal spanning tree of A1 B1 C1 .
We shall revisit this later in Section 6 as there is a significant relationship between these lengths.
4 Geometric Proof
Courant and Robbins offered a rigorous proof in 1941, but a much more simple and elegant proof can be
found in Introduction to Geometry by H S M Coxeter in 1961[4]. If we first consider any point P inside the
ABC and join it to points A, B, C and rotate the new formed inner ABP 60◦ anticlockwise to form
BC P so that ABC and P BP are both equilateral triangles as in Figure 8 then
AP + BP + CP = C P + P P + P C
which is a broken line connecting C’ to C. Such a path is minimal when this line is straight
∠BP C = 180◦ − ∠BP P = 120◦
and
∠AP B = ∠C P B = 180◦ − ∠P P B = 120◦
Figure 8: Geometric Proof
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6. 5 A Complementary Problem
We have already noted that when an angle of a triangle is equal to 120◦ then that point itself becomes the
Fermat point and the minimal spanning tree is the minimum distance between the three points of that given
triangle. Now we examine the case when an angle is greater than 120◦ in which we find a new point P , outside
of the area enclosed by the triangle as in Figure 9. This new point P subtends an angle ∠AP B = 120◦ , and can
easily be geometrically produced by using a hybrid of the earlier techniques used to find the Fermat point. This
method was first used by Melzak [1] in 1961 and was part of an algorithm to find a Steiner tree accommodating
n points, which we shall discuss in detail. We simply construct a circumscribed circle through the projected
equilateral triangle of the largest side opposite the angle greater than 120◦ . In our case the side AB. Now the
point P can easily be found on the circle by constructing a Simpson line extending through the new vertex of
the equilateral triangle and point C [14]. Bern and Graham [8] call this the replacement point, it is used in the
construction of Steiner trees, as it can replace two points without changing the length of the overall solution.
Figure 9: A Complementary Problem
Certainly P does not solve our minimum distances problem but it does have some relation to it. Our original
problem was to minimise AP + BP + CP . In this complementary case we have found the solution to minimising
AP + BP − CP .
6 The Steiner Tree Problem
Courant and Robbins, however mistakenly, propelled “The Steiner Tree Problem” to stardom, with no mention
of Torricelli, Simpson or Heinen who had each contributed to the original problem posed by Fermat. As is the
way in mathematics, Jakob Steiner’s contribution is not well known or understood but his general problem is
now on a par with the travelling sales man problem in combinatorial optimisation and his name will be forever
assigned with the general Fermat problem.
Figure 10: Street Networks
Suppose we want to find the shortest network between more than three points in the euclidean plane. It
makes sense to move away from three fixed points and generalise to the case when we have n given points, A1 ,
A2 , A3 , ... , An . An example may be that we want to connect four towns by constructing streets. Now your first
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7. impressions may be to construct a simple minimal spanning tree or you may even think to make two diagonals
which intersect forming cross roads connecting all four points, as illustrated in Figure 10.
Definition 4. Steiner Minimal Tree. The shortest network interconnecting a set A of n points that may contain
Steiner points S which are not originally in the set A [8]. All leaves must be regular points. Any two edges meet
at an angle of atleast 120◦ . Every Steiner point has degree of exactly three [3].
Definition 5. Topology. A connection matrix, or any equivalent description, specifying which pairs of points
from the set A (A1 , A2 , ... , An ) and any Steiner points (S1 , S2 , ... , Sn−2 ) have a connecting line. Thus the
topology specifies the connections but not the positions of S1 , S2 , ... , Sn−2 [3].
Definition 6. Full Steiner Tree/Topology. A topology with s = n − 2 will be called a full Steiner topology. The
corresponding relatively minimal tree will be called a full Steiner tree [3].
Definition 7. Steiner Minimal Length. The shortest Steiner minimal tree network is the Steiner minimal
length [14].
We are now at a cross road, as we can follow the route of a singular point which finds the minimum distance
from the given n points. Or we can try to find a Steiner minimal tree which only allows Steiner points to be
established in addition to the given n points. As you can see in Figure 10 even with a simple example of four
points, the singular point case in the middle is well defeated by the Steiner tree which has two Steiner points S1
and S2 . The total lengths in each case can be seen below each figure and the minimal spanning tree length is
better reduced by the Steiner tree. This leads us on to an interesting point raised earlier about the relationship
between the minimal spanning tree length and the Steiner tree length, the Steiner Ratio conjecture which we
will discuss shortly.
Remark 2. It should be noted that the Steiner tree shown in Figure 10 is only one of a possible twenty seven
topologies, each with a different Steiner length [3]. It may not be the case that the one depicted is the Steiner
minimal length. Topologies can have zero Steiner points or a maximum of n − 2 Steiner points. A formula for
finding the number of full Steiner topologies was given by Winter and Zachariasen [12] in 1996;
(2n − 4)!
F ull Steiner T opologies = (1)
2n−2 (n − 2)!
Melzak’s Algorithm: 1961
Z. A. Melzak is accredited with being the first person to develop an algorithm for the Steiner problem [1]. His
method uses brute force to find its way through all possible topologies of Steiner trees from n points, so that you
may then conclude from all possibilities the shortest Steiner tree. His method is very simple, it only requires
a compass and ruler construction as in Figure 9. The algorithm firstly separates the set of n given points into
every possible subset of given points. Then the algorithm creates a number of possible Steiner trees for each
subset by using Melzaks hybrid construction which makes use of all cases of Fermat point constructions, namely
when a point subtends an angle of less than 120◦ , exactly 120◦ or in the case of greater than 120◦ replacing two
points with a single replacement point, remember this does not alter the distance. A worked example of this
method is shown in Figure 11.
The problem with this method however is that as soon as the number of points is large for example n > 10
it becomes very difficult, in the case of n = 10 you must consider 512 subsets of the original points, for each
subset if a replacement point is needed that too can be on either side of the line segment you are constructing
against. After a pair of points in the subset is replaced by one of the two possible replacement points, each
subsequent step of the algorithm replaces either two given points, a given point and a replacement point or two
replacement points with another replacement point until the subset is reduced to three points [8]. Considering
our modern day scenario of connecting ten cities (n = 10), of the 512 subsets, the two point subsets do not
require much work, but each of the 45 eight point subsets have two million replacement sequences and more
than 18,000 ways to recombine the subsets into trees [8]. Once the Steiner point is found for the reduced three
point problem, the algorithm starts again by considering all possibilities of the new reduced problem which has
one less point in its set, because two points have been replaced by one.
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8. Figure 11: Melzak’s Algorithm. The seven point problem (A, B, C, D, E, F, G) has been reduced to a three
point problem (A, B, C) and a five point problem (C, D, E, F, G) by separating at the point C. To construct a
possible Steiner tree, possible replacement points have been constructed using the hybrid method (W, X, Y, Z).
Replacement X reduces the five points down to four points (C, D, E, X). Replacements W and Y reduce down
to three points (D, W, X) and (C, E, Y ). The line W Y intersects the two circles circumscribed around the
equilateral triangles DW X and CEY producing the two Steiner points S2 and S3 . The intersection of line XS2
with circle F GX locates another Stenier point S1 . S4 is easily found by the line segment from replacement point
Z to point C intersecting with the circle ABZ. Since the best partitioning and pairing cannot be determined
in advance, all possibilities must be considered to find the shortest tree.
Number of Steiner Points
The reader may have noticed that we have already cited that the maximum number of Steiner points from a
set of n points is n − 2. We now present a proof by Gilbert and Pollak in 1968 for why this is the case. It is
clear that every tree has one more point than it has lines. If we consider any tree constructed with or without
Steiner points S on a set A of given points (A1 , A2 , ... , An ), we can deduce the number of lines to be n + S − 1.
We can count the number lines forwards and backwards since each line has two ends, resulting in 2(n + S − 1)
incident lines. Another way to count these lines in both directions is to sum the number of vertices Ai with
k incident lines and add to it the number of Steiner points each with three incident lines. Equating these two
counting methods together we have;
k
3S + knk = 2(n + S − 1)
i=1
= 2((n1 + n2 + ... + nk ) + S − 1)
k
=2 nk + 2S − 2
i=1
k k
S=2 nk − knk − 2
i=1 i=1
inputting some values f or k we can simplif y to
S = n1 − 2 − n3 − 2n4 − 3n5 ...
S ≤n−2
S =n−2 (2)
if and only when each point Ai has only one edge, which as we know already is when we have a F ull Steiner T ree.
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9. Steiner Ratio Conjecture
Looking back at our Fermat problem when n = 3 we simplify the case of the triangle further still by making
it an equilateral triangle with each side having one unit length.
Conjecture 1. In 1968, Edgar Gilbert and Henry Pollak [3] conjectured that no matter how the set of n points
are originally located, the Steiner length never falls short of the minimal spanning length by more than 13.34%.
Equivalently [13], √
Steiner M inimal Length 3
ρ= ≥ = 0.866... (3)
M inimal Spanning Length 2
Figure 12: Steiner Ratio
The Steiner ratio conjecture is important as it used to replace a hard problem by an easy one when it comes
to trying to find the shortest street network for a set of n points an example is given under the next section.
Presently, lets revisit the four points in Figure 10. We found that one of the Steiner topologies out of a possible
twenty seven √ yielded a Steiner length of 7.95, now we can conjecture that at most we could be hopeful for is
ρ = SM L = 23 giving a Steiner minimal length of 7.205. We have saved 4.44% compared with the minimal
8.32
spanning tree in this example. The conjecture was verified by Pollak in 1978 for n = 4 [5], by Du et al in 1985
for n = 5 [6] and by Rubinstein and Thomas in 1991 for n = 6 [10].
For the general case of n points, proofs of the lower bound of ρ have crept up from 0.5 as reported in 1968 [3]
by E. F. Moore over the years to 0.824 in 1985 by Chung and Graham [7] until in 1990 Du and Hwang published
an article claiming that the Steiner ratio conjecture first proposed by Gilbert and Pollak is true [9]. However a
recent article published by Ivanov and Tuzhilin in 2011 [19] highlights that some of the conjectures used in the
proof by Du and Hwang are incorrect and so the Steiner ratio conjecture is still open.
Steiner Tree Algorithms
In order to compute all possible Steiner trees, an algorithm has to be developed. Over the years since Melzak,
mathematicians have found better and faster ways by developing techniques and theorems which exclude many
of the permutations arising with Melzak’s technique. This paper will not explore them. For the interested
reader, some commonly known are the ‘pruning’ and ‘wedge’ property techniques. The development of the
computer and the number of operations a modern CPU can do per second has elevated the number of n points
for which a minimal Steiner tree can be computed. As with our modern day scenario, the GeoSteiner software
program can on any local machine, calculate the minimal Steiner tree for up to 3000 points in a few minutes,
a number which was unimaginable only 50 years ago. An algorithm is assessed by the time it takes to reach a
result, 50 years ago it could take up to 30 minutes for n = 30.
There are two types, ones which can be solved in polynomial time, and those which cannot are exponential
time algorithms. Equation 1 for calculating the number of full Steiner topologies is super exponential. To
have an idea of how fast this function grows, here are some data points. f (2) = 1, f (4) = 3, f (8) = 10, 395,
f (10) = 2, 027, 025 and f (12) = 654, 729, 075. Due to the fact that finding all possible topologies for a n
point problem requires a super exponential equation, any algorithm based on this is exponential, and thus is a
NP-Hard problem4 .
4 Non-Deterministic polynomial-time hard problems are a class of problems, which are yet to be proven that they can be solved
in polynomial time.
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10. If a problem is NP-Hard as is the Steiner tree problem, it does not mean that we are at a dead end, because
there are other ways to tackle the problem for large numbers of n. It may be hard for the reader to imagine a
problem similar to our ten cities problem which needed more than 3000 points. Yet the Steiner tree problem
and permutations5 of it are applied in many other areas of interest. Such as the design of electronic circuit
boards, where the shortest network is required between electronic components such as transistors etc not only
to keep costs down but also to decrease the resistance on electrons reaching one point from another. Hence
reducing heat and making the circuit more efficient. Another example where n can easily be imagined in very
large numbers is the interaction of weak forces between multiquarks at the atomic scale.
For these types of problems we can make use of equation 3 where we know that we can never achieve a saving
of more than 13.34% compared to that of the minimal spanning tree. At this point the reader may note that the
surely for large n the minimal spanning tree would be just as hard to find, but there is a procedure called the
‘Greedy Algorithm’6 which makes this process very easy. In which case ‘heuristic’7 algorithms for the Steiner
tree problem are used to find a tree such that an acceptable amount of saving is achieved for the problem in
question.
7 Conclusion
Figure 13: A Novel Idea
Exploring the Fermat point and consequently the Steiner tree problem has been intensely enjoyable on my part.
The problem is so simple to explain yet still eludes mathematicians in its complexity. There is a vast amount of
material on the subject which the reader can immerse themselves in. I hope the reader can understand to some
degree how the GeoSteiner software algorithm is working, and that they are assured that George Osbourne will
find no other shorter distance than that of the minimal Steiner tree between the ten cities.
During my research, exploring the history and modern techniques of finding Steiner points, I came upon my
own original idea which with little effort seems to work for n = 3 and n = 4. Suppose you can construct a
tessellating grid of equilateral triangles, you will find that the grid itself will contain bigger equilateral triangles
made up of four smaller ones. Now take three or four points which do not all lie on a straight line, and do not
subtend more than 120◦ between any of their adjacent points. Then by simply superimposing our points on
the grid we can use equation 2 to find a full Steiner tree for our original points when each of the points touches
an arc on the equilateral grid. This is illustrated in Figure 13. This procedure can be carried out by inspection
or by a computer algorithm. I do not know if it will work in every case or if it will work for more than four
points. I guess my idea is just another line of thought and if it merits any further investigation, I leave this to
the reader.
5 RectilinearSteiner tree problem where only horizontal and vertical line segments are allowed. Steiner trees on spherical surfaces.
6 JosephKruskal 1956, the Greedy Algorithm finds a subset of the edges forming a tree where the total weight of all the edges
is minimised.
7 Heuristic techniques make use of experience, simplifying the problem using ‘pruning’ as well as ’wedging’ and other techniques.
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