Content uploaded by Supanut Chaidee

Author content

All content in this area was uploaded by Supanut Chaidee on Jan 28, 2014

Content may be subject to copyright.

Content uploaded by Supanut Chaidee

Author content

All content in this area was uploaded by Supanut Chaidee on Jan 27, 2014

Content may be subject to copyright.

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.

200

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).

Proceedings of AMM2013

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).

202

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.

Proceedings of AMM2013

203

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.

204

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

.

Proceedings of AMM2013

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.

206

Proceedings of AMM2013