Numerical Approach to the Optimal Partitioning of a Square Problem

Abstract

Optimal partitioning of a square problem is the search for least-diameter way to partition a unit square into n pieces. The problem has been settled for some cases. Unfortunately, the proof for each n has its own unique trick with no relation to the proof for the other cases. In this work, we introduce a new approach to this problem. We construct an algorithm to generate all of the combinatorial patterns of the effective partitions. Then, we use the numerical optimization to find the minimum diameter. The results agree with the previous works for n = 4 to 7.

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Proceedings of the 18
th
Annual Meeting in Mathematics (AMM2013)
Thaksin University, Phatthalung, Thailand, March 14-16, 2013
Numerical Approach to
the Optimal Partitioning of a Square Problem
Supanut Chaidee
1,∗
and Wacharin Wichiramala
2
Department of Mathematics and Computer Science
Faculty of Science, Chulalongkorn University
254 Phayathai Road, Pathumwan, Bangkok, THAILAND
1
schaidee@hotmail.com,
2
wacharin.w@chula.ac.th
Abstract : Optimal partitioning of a square problem is the search for least-diameter way
to partition a unit square into n pieces. The problem has been settled for some cases.
Unfortunately, the proof for each n has its own unique trick with no relation to the proof for
the other cases. In this work, we introduce a new approach to this problem. We construct
an algorithm to generate all of the combinatorial patterns of the effective partitions. Then,
we use the numerical optimization to find the minimum diameter. The results agree with
the previous works for n = 4 to 7.
Keywords : Optimal Partitioning; Square Dissecting.
2000 Mathematics Subject Classification : 65D18; 65D99; 74P20
1 Introduction
Partitioning the square into pieces of minimum diameter was introduced by Wanceles in
[5], where he settled the case of 3 pieces, see figure 1 (left), and published in the American
Mathematical Monthly journal as follows:
The dissection of a unit square into one 1/8 ×1 and two 1/2 ×7/8 rectangular pieces
that is possible to cover the unit square with three sets each having diameter d =
√
65/8.
Prove that the unit square cannot be covered by three sets of all of which have diameter
less than d.
In 1959 [4], Page solved this problem for the case of 5 pieces, see figure 1 (right), by
answering the following question:What is the minimum d such that the unit square may be
covered with five sets each having diameter d?
In 1965, Graham studied a similar problem on partitioning the equilateral triangle into
∗
Corresponding author e-mail : schaidee@hotmail.com
Copyright
c
 2013 by the AMM2013. All rights reserve.

n pieces for n up to 15 in [3]. After that, Guy and Selfridge studied the problem on the unit
square in 1973 [2].
Figure 1 : The graphical solutions are shown for n = 3 (left) and n = 5 (right).
For S denoted a closed unit square, we define the family Π of all partition π of S into
n disjoint subset S
i
such that ∪S
i
= S. We denote
d
n
= inf
π∈Π
max
1≤i≤n
sup
P,Q∈S
i
d(P,Q) (1.1)
where P,Q are points in S
i
, d(P,Q) is the Euclidian distance between them.
Guy and Selfridge found the exact minimum diameter d
n
for n up to 10 and gave
numerical bounds of the diameter for some n. Later in 1986, Jepsen worked independently
to prove the case n upto 6 in [1].
However, the proof method of the optimal diameter proposed in every literature cannot
be applied to prove in any other cases. That is, the proof of each case is independent from
the other cases. In this work, we use numerical approach to solve this problem as follows.
Since each effective partition is composed of convex polygons, we may consider such
partition as an embedded planar graph with straight edges and vertices of degree at least
3. Then, we construct an algorithm to list all of the possible combinatorial patterns of
these graphs. For each combinatorial pattern, we use the numerical optimization to get the
minimal diameter. Finally, we compare minimum diameter from those patterns to get d
n
.
In addition, the result agree with the previous works for n = 4 to 7.
2 Relationship between Effective Partition and Correspond-
ing Embedded Planar Graphs
Firstly, we give a basic definition about graph.
Definition 2.1. A graph G consists of a finite nonempty set V of objects called vertices.
The set E of 2-element subsets of V is called edges. The sets V and E are the vertex set
and edge set, respectively.
So, the graph G is the ordered pair of V and E, i.e. G = (V ,E). We denote |V | and
|E| by V and E, respectively.
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Proceedings of AMM2013

2.1 Properties of an embedding planar graph corresponding to effec-
tive partitions
As we consider polygon pieces from partitioning of a square, we scope on a diameter of a
polygon. It is well-known that a polygon P surrounded by vertices v
1
,v
2
,..., v
k
has diame-
ter diam(P) = max
i, j
d(v
i
,v
j
).
Since we partition a square into the pieces of convex polygon, we consider the graph
corresponding to the polygon. We first consider the figure 2 that shows the optimal diam-
eter for n = 8 in [2].
Figure 2 : An effective partition for the case n = 8 in [2].
A planar straight-line graph G determines in general a subdivision of the plane. Each
face F of the subdivision, together with the edges of G that are on the boundary of F,
forms a polygon in the plane.
By optimal partitioning, the partitioned piece must be a polygon which has straight
side. From the consideration on each polygon piece, we have a planar embedding which is
the graph of degree greater than 3. So, the considering planar graphs do not have loop and
multiple edges.
In process of enumerating combinatorial patterns corresponding to the partition, we
observe that each point can have degree 3 or greater. However, it is sufficient to consider
points of degree 3 by the following reason:
For each combinatorial pattern of a partitioning, we note that the optimal diameter can
be obtained by max
1≤i≤n
sup
P,Q∈S
i
d(P,Q) where S
i
is a piece of polygon. This diameter is
the represented diameter of that admissible pattern. So we find the infimum of all of the
combinatorial patterns. Nevertheless, there are some cases which have the point degree 4
or greater. If we generate graph by vertices of degree 3 in the figure 3 (left), we can join the
vertex together to the figure 3 (right) in the step of diameter optimization. That is, v
i
,v
j
are
coincident with (x
0
,y
0
). Therefore, we can consider d
n
by combinatorial pattern of degree
3. From now, the word combinatorial pattern is represented to combinatorial pattern of
degree 3.
Figure 3 : The 2 vertices of degree 3 (left) merge together to be the vertex of degree 4
(right).
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201

Let π be a combinatorial pattern and Π be a set of all of the combinatorial patterns. For a
piece S
k
of π, we have diam S
k
= max
i, j
d(v
i
,v
j
). So, the diameter of π obtains by
diam π = max
k
diam S
k
. So, the optimal diameter of Π can be observed by d
n
= inf
patternΠ
(diam π) over the partitions of degree 3.
2.2 The conditions on the square
Let S be a square on the plane, G be the graph corresponding to a partition of the square,
V be the set of all vertices and E be the set of all of the edges of graph G. We denote that
V
1
,V
2
,V
3
and V
4
are the sets of all vertices on the top, left, bottom and right sides of the
square respectively. We also specify V
0
as the set of points in the square which are not on
the side of the square. The cardinal number of V
i
for i = 0,1,2,3,4, is denoted by V
i
. We
write the 5-tuple in the form (V
0
,V
1
,V
2
,V
3
,V
4
).
Excluding the four vertices of the square, every vertex has degree 3. We note that
for each vertex in V
i
, for i = 1,2,3, 4 has exactly one remaining edge connecting to other
vertices and each vertex in V
0
can connect to 3 edges. We also determine that V
i
≥ 1 for
each i.
3 The Proposed Method
In this section, we provide the properties of combinatorial patterns. Then we use these
properties as the candidates to construct algorithm to generate all possible combinatorial
patterns. The final step is to optimize the diameter for each pattern and compare them to
obtain the minimum diameter d
n
for n = 4 to 7.
We first state a lemma concerning with the partitioning of a square.
Lemma 3.1. A subdivided partition has smaller or equal diameter.
Proof. Let P = {S
i
} and P
0
= {S
0
j
} be partitions such that each S
i
in P is a subset of some
S
0
j
in P
0
. Then diam(P) ≤ diam(P
0
).
By the previous lemma, we have d
4
=
√
2/2 and d
5
= 5
√
17/32 and also obtain that
(d
n
) is the decreasing sequence.
Lemma 3.2. For n ≥4, there is no edge from a vertex in V
1
to a vertex in V
3
nor a vertex
in V
2
to a vertex in V
4
.
Proof. Since d
4
=
√
2/2 by Lemma 3.1, we have d
n
≤
√
2/2 for each n ≥ 4 by the fact
that (d
n
) is decreasing. If there is an edge from a vertex in V
i
to the opposite side, then the
diameter will be greater than 1, contradiction.
Lemma 3.3. If there is an edge connecting the vertices of consecutive sides, then we can
find another partition with a smaller or equal diameter.
Proof. Without loss of generality, suppose that there is an edge connecting the vertex in
V
1
to the vertex in V
2
. We will construct another partition with smaller or equal diameter
as follows (see the figure 4).
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Proceedings of AMM2013

Figure 4 : The process of constructing a new partition from the old partition.
From the figure 4 (left), we first remove the edge v
1
v
2
and assign the original S
1
to be the
part of the new S
2
in the figure 4 (middle). The diameter of the new partition does not
increase since diameter of S
1
and pieces at the upper left corner are bounded below by the
distance v
3
v
4
. Next, we will make another partition by subdividing the current S
2
to be the
new S
2
and S
1
in the figure 4 (right). From Lemma 3.1, the diameter of the last partition
does not increase.
Lemma 3.4. For n ≥ 4, let v
k
be a vertex in V
0
. If v
i
∈ V
i
connects to v
k
, then there is no
vertex from the opposite side v
j
∈ V
j
that connects to v
k
.
Proof. Suppose that there is an edge v
i
v
k
and v
k
v
j
. Since the degree of v
k
is equal to 3,
the remaining edge must be on one side of v
i
v
j
. The piece on the opposite side of v
k
has
diameter at least 1, a contradiction.
We list all possible distinct combinatorial patterns by putting the vertices on sides of
a square according to the number of vertices on sides. From the previous lemmas, that is
enough to consider a combinatorial pattern with the following properties.
1. Every internal vertex has degree 3.
2. There are no edges from a vertex in V
i
to a vertex in V
j
.
3. There are no loops and multiple eges.
4. The combinatorial pattern must be a planar graph.
Next, we pair up a vertex in V
1
and V
3
to a vertex in V
0
. After that, we connect a
vertex in V
2
and V
4
to a point in V
0
by all possibilities which generate planar graph which
each vertex has degree 3. Connection of the points must satisfy the properties as above.
The figure 5 shows all of the possible distinct combinatorial types for case n = 4. Case
n = 5 is shown by the figure 6.
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Figure 5 : All of the possible distinct combinatorial patterns for n = 4
Figure 6 : All of the possible distinct combinatorial patterns for n = 5
We next pick each of combinatorial distinct pattern to optimize the diameter in the
optimizing algorithm. We first define the Euclidian distance of two points by d(p,q) =
p
(x
1
−x
2
)
2
+ (y
1
−y
2
)
2
where p = (x
1
,y
1
) and q = (x
2
,y
2
). We find the distance be-
tween two points in each face and consider max
p,q
d(p,q) where p,q are in the same face.
Therefore, the position of points are used to be variables. Since the diameter function is
a convex function and all conditions force the domain to be convex, this numerical mini-
mization is confirmed theoretically to reach minimum by very high accuracy. For each n,
we find optimal diameter of each combinatorial pattern. Then, we find the infimum among
these. The result releases d
n
for each n.
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Proceedings of AMM2013

4 Numerical Results and Conclusion
From the method and algorithm, we use Wolfram Mathematica to program and obtain the
optimal diameter in case of n = 4 to 7 in order to reassure the validity of our algorithm.
We define D
n
to be the optimal diameter obtaining from the algorithm. The following
table shows the optimal diameter of n = 4 to 7 obtained from our method and algorithm,
compared to d
n
from [2].
n The combinatorial pattern The optimizing result d
n
D
n
4 0.707107 0.707107
5 0.644236 0.644235
6 0.593997 0.593996
7 0.548584 0.548584
Table 1 : The table of the combinatorial patterns leading to the optimal partition
with d
n
and D
n
.
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205

Our programming results of D
4
to D
7
match the both results of d
4
to d
7
upto the
fifth decimal place and the same optimal partition in [2]. We can apply this method and
algorithm to the larger n. In order to get the result faster, we need more conditions on
optimal partitions to reduce unnecessary combinatorial patterns.
Acknowledgement(s) : The authors wish to thank the Development and Promotion of Sci-
ence and Technology Talents Project (DPST) by the Institute for the Promotion of Teaching
Science and Technology (IPST), Ministry of Education, Thailand for the financial support.
References
[1] C.H. Jepsen, Coloring points in the unit square, The American Mathematical
Monthly, 17 : 3 (December, 1986), 231-237.
[2] R.K. Guy and J.L. Selfridge, Optimal covering of the square, Colloquia Mathematica
Societatis Janos Bolyai, (1973), 745-799.
[3] R.L. Graham, On partitions of an equilateral triangle, Canadian Journal of Mathe-
matics, 19 : 2 (1967), 394-409.
[4] Y. Page and J.L. Selfridge, Elementary problems and solutions : E1374, The Ameri-
can Mathematical Monthly, 67 : 2 (February, 1960), 185.
[5] G.K. Wanceslas and J. Lipman, Elementary Problems and Solutions : E1311, The
American Mathematical Monthly, 65 : 10 (December, 1958), 775.
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Proceedings of AMM2013