GENERALIZED FUZZY TOPOGRAPHIC TOPOLOGICAL MAPPING
SITI SUHANA BINTI JAMAIAN
UNIVERSITI TEKNOLOGI MALAYSIA
GENERALIZED FUZZY TOPOGRAPHIC TOPOLOGICAL MAPPING
SITI SUHANA BINTI JAMAIAN
A thesis submitted in fulfilment of the
requirements for the award of the degree of
Master of Science (Mathematics)
Faculty of Science
Universiti Teknologi Malaysia
MARCH 2009
iv
ACKNOWLEDGEMENT
First of all, I would like to express my sincere appreciation to my research
supervisor, PM. Dr. Tahir Bin Ahmad for his encouragement, guidance and critics.
His wide knowledge and his logical way of thinking have been of great value for me.
Without his continuous support and interest, this thesis would not have been the
same as presented here.
I am also very thankful to my family for giving me great supports.
Especially, I would like to give my special thanks to my fiancé, Abd Fathul
Hakim who has always supported me to complete this work.
My sincere appreciation also extends to my co-supervisor PM. Dr. Jamalludin
Bin Talib, lecturers, all my friends and others who have provided assistance at
various occasions. Their views and tips are useful indeed.
For all their contribution, support, guidance and patience, only God can repay
them.
v
Abstract
Fuzzy Topographic Topological Mapping (FTTM) is a model for solving
neuromagnetic inverse problem. FTTM consists of four topological spaces which are
connected by three homeomorphisms. FTTM 1 and FTTM 2 were designed to
present 3-D view of an unbounded single current and bounded multicurrent sources,
respectively. It has been showed that FTTM 1 and FTTM 2 are homeomorphic and
this homeomorphism will generate another 14 FTTM. There is a conjecture stated
that if there exist n numbers of FTTM, then they will generate another n 4  n new
FTTM. In this thesis, the conjecture is proven by using geometrical features of
FTTM. In the process, several definitions such as sequence of FTTM, sequence of
polygon, sequence of cube with combination of two, three and four FTTM are
developed. Some geometrical and algebraic properties of sequences of FTTM are
identified and proven. A new conjecture is also proposed in this thesis which states
that the number of generating Fkn if there are k components and n models of Fk is
nk  n . Surprisingly, the nonzero sequence of cube with combination of two, three
and four FTTM appeared in Pascal’s Triangle.
vi
Abstrak
Pemetaan Topologi Topografi Kabur (FTTM) adalah satu model untuk
menyelesaikan masalah songsangan neuromagnetik. FTTM terdiri daripada empat
ruang topologi yang mana dihubungkan dengan tiga homeomorfisma. FTTM 1 dan
FTTM 2 direka bentuk bagi menunjukkan pandangan tiga dimensi sumber arus
tunggal tidak terbatas dan sumber arus berbilang terbatas setiap satunya. Ia telah
ditunjukkan bahawa FTTM 1 dan FTTM 2 adalah homeomorfik dan homeomorfisma
ini akan membentuk 14 FTTM yang lain. Terdapat satu konjektur yang menyatakan
bahawa jika wujud sebanyak n FTTM, maka bilangan FTTM yang baru yang
dijanakan adalah n 4  n . Dalam tesis ini, konjektur itu telah dibuktikan dengan
menggunakan ciri-ciri geometri FTTM. Dalam proses itu, beberapa takrifan seperti
jujukan FTTM, jujukan poligon dan jujukan kubus dengan gabungan dua, tiga dan
empat FTTM telah dibangunkan. Ciri-ciri geometri dan aljabar jujukan FTTM telah
dikenalpasti serta dibuktikan. Satu konjektur baru juga telah dicadangkan di dalam
tesis ini yang menyatakan bahawa bilangan Fkn yang akan dijana jika terdapat
sebanyak k komponen dan n model Fk ialah nk  n . Menariknya, jujukan kubus yang
tidak kosong dengan gabungan dua, tiga dan empat FTTM wujud dalam Segitiga
Pascal.
vii
TABLE OF CONTENTS
CHAPTER
TITLE
PAGE
THESIS STATUS DECLARATION
SUPERVISOR’S DECLARATION
1
TITLE PAGE
i
DECLARATION
ii
DEDICATION
iii
ACKNOWLEDGEMENT
iv
ABSTRACT
v
ABSTRAK
vi
TABLE OF CONTENTS
vii
LIST OF TABLES
x
LIST OF FIGURES
xi
LIST OF SYMBOLS
xiii
INTRODUCTION
1.1 Background of Research
1
1.2 Statement of Problem
2
1.3 Objectives of Research
4
1.4 Scope of Research
4
1.5 Importance of Research
5
1.6 Framework of Research
5
viii
2
LITERATURE REVIEW
2.1 Introduction
8
2.2 Fuzzy Topographic Topological Mapping
8
(FTTM)
2.2.1 FTTM 1 and FTTM 2
11
2.2.2 Characteristics of FTTM
13
2.2.3 Generalized FTTM
15
2.3 Fibonacci Cube
17
2.4 Pascal’s Triangle
19
2.4.1 Patterns in Pascal’s Triangle
2.5 Conclusion
3
22
27
A CONSTRUCTIVE PROOF FOR
GENERATING FTTM
3.1 Introduction
28
3.2 Axiomatic System
28
3.3 Proving Methods
30
3.3.1 Exhaustion
31
3.3.2 Direct
32
3.3.3 Contradiction
33
3.3.4 Contrapositive
34
3.3.5 Mathematical Induction
34
3.3.6 Constructive
35
3.4 Conclusion
4
36
A SEQUENCE OF FTTM
4.1 Introduction
37
4.2 Geometrical Features of FTTM
37
4.3 Formal Definitions
45
4.4 A Sequence of FTTM
47
ix
5
4.5 A Sequence of Cubes in FTTMn
49
4.6 “Extended Cubes” in FTTMn
57
4.7 Conclusion
76
A SEQUENCE OF FTTM IN RELATION TO
PASCAL’S TRIANGLE
5.1 Introduction
77
5.2 The Relation of A Sequence of FTTM to
78
Pascal’s Triangle
5.3 The Proof of Li Yun’s Conjecture
80
5.4 Further Results
83
5.5 Coefficients of FTTM and F5
95
5.6 Relating Extension Results of FTTM to
100
Pascal’s Triangle
6
5.7 A Sequence of Polygon
103
5.8 Conclusion
107
CONCLUSION
6.1 Summary
108
6.2 Suggestions for Further Research
109
REFERENCES
110
x
LIST OF TABLES
TABLE NO.
TITLE
PAGE
2.1
Polygonal Number
26
4.1
Geometrical features of combined FTTM
40
4.2
Geometrical features of sequences of FTTM1 until
44
FTTM20
4.3
Cube with two terms FTTM in FTTMn
49
4.4
Samples of sequence of FTTM2/n
56
4.5
Cube with three terms FTTM in FTTMn
57
4.6
Samples of sequence of FTTM3/n
68
4.7
Cube with four terms FTTM in FTTMn
69
4.8
Samples of sequence of FTTM4/n
74
4.9
A Comparison between generating FTTMn and
76
conjecture made by Liau Li Yun (2006)
5.1
Two terms of F2 in F2n
86
5.2
Two terms of F3 in F3n
88
5.3
Three terms of F3 in F3n
89
5.4
Two, three, four and five terms of F5 in F5n
92
5.5
A Sequence of Polygon
105
xi
LIST OF FIGURES
FIGURE NO.
TITLE
PAGE
1.1
Model of FTTM
2
1.2
Homeomorphisms between FTTM 1 and FTTM 2
3
1.3
Framework of research
7
2.1
Human Brain
9
2.2
MEG System
10
2.3
FTTM Version 1
11
2.4
FTTM Version 2
12
2.5
Homeomorphism from S 2 to E 2
13
2.6
Equivalence of components of FTTM Version 1
13
2.7
Homeomorphism between corresponding
14
components of FTTM 1 and FTTM 2
2.8
FTTMn
16
2.9
Extended Fibonacci Cubes EFC1s
18
2.10
Modified EFC1(5)
18
2.11
Reconstruct EFC1(5)
18
2.12
Chinese Pascal’s Triangle
20
2.13
The original Pascal’s Triangle
20
2.14
Present-day Pascal’s Triangle
21
2.15
Hockey stick in Pascal’s Triangle
23
2.16
Fibonacci’s Sequence in Pascal’s Triangle
24
2.17
Triangular numbers in Pascal’s Triangle
24
2.18
Square number in Pascal’s Triangle
25
xii
4.1
A Sequence of FTTMn
47
4.2
FTTM2/4
55
4.3
FTTM3/4
67
4.4
FTTM4/5
73
5.1
The Nonzero Sequence of FTTM2/n, FTTM3/n and
78
FTTM4/n
5.2
An Alternative Version of Pascal’s Triangle
79
5.3
FTTMn
82
5.4
A Sequence of F2
83
5.5
A Sequence of F3
84
5.6
A Sequence of F5
84
5.7
A Sequence of Fk
85
5.8
Relating Extension Result to Pascal’s Triangle
101
5.9
The Triangle of Coefficients
103
xiii
LIST OF SYMBOLS
|

-
Homeomorphism
≤
-
Less than or equal
≥
-
Greater than or equal
>
-
Greater than
||
-
Concatenation operation

-
Implies
~P
-
Not P

-
Element of
-
Number of elements in a set
|
CHAPTER 1
INTRODUCTION
1.1
Background of Research
Fuzzy Topographic Topological Mapping (FTTM) was developed by Fuzzy
Research Group (FRG) at UTM in 1999. FTTM was built to solve neuromagnetic
inverse problem i.e. to determine the location of epileptic foci in epilepsy disorder
patient (Tahir et al., 2000). The model which consists of topological and fuzzy
structures is composed into three mathematical algorithms (Fauziah, Z., 2002).
FTTM have four components which are magnetic contour plane (MC), base magnetic
plane (BM), fuzzy magnetic field (FM) and topographic magnetic field (TM) as
shown in Figure 1.1. FTTM Version 1 was developed to present a 3-D view of an
unbounded single current source (Fauziah, Z., 2002 & Liau L. Y., 2001) while
FTTM Version 2 was developed to present 3-D view of a bounded multi current
source (Wan Eny Zarina et al., 2002).
2
MC
TM
BM
FM
Figure 1.1 Model of FTTM
1.2
Statement of Problem
FTTM Version 1 consists of three algorithms, which link between four
components. The four components are MC1, BM1, FM1 and TM1. Besides that, FTTM
Version 2 also consists of three algorithms, which link between four components.
The four components are MC2, BM2, FM2 and TM2. FTTM Version 1 as well as
FTTM Version 2 is specially designed to have equivalent topological structures
between its components (Liau L. Y., 2006).
In other words, there are
homeomorphisms between each component of FTTM Version 1 and FTTM Version
2 (see Figure 1.2).
Using the fact that FTTM 1 and FTTM 2 are homeomorphic componentwise, there
are at least another 14 elements of FTTM that can be identified easily.
3
FTTM
FTTM1 1
TM1


FM1



MC1

BM1

TM2




MC2
BM2

FM2
FTTM 2
Figure 1.2 Homeomorphisms between FTTM 1 and FTTM 2
Li Yun (2006) first noticed that if there were two elements of FTTM that are
homeomorphic
to
each
other
componentwise,
it
would
generate
more
homeomorphisms. The numbers of generating new elements of FTTM are
 2   2   2   2  
             2  14 elements.
 1   1   1   1  
Consequently, Liau Li Yun (2006) has proposed a conjecture; if there exist n
elements of FTTM, then the numbers of new elements are n 4  n .
4
1.3
Objectives of Research
The aims of this research are as follows;
1.4
a)
to study geometrical features of FTTM.
b)
to prove the conjecture (the number of generating FTTM).
c)
to find new features on generating FTTM.
Scope of Research
This research will focus on the goal to prove the conjecture, namely the
number of generating FTTM. In order to achieve this goal, the basic concepts of
FTTM and how Li Yun produced this conjecture will be studied. Basic concepts of
FTTM will be discussed.
The process of proving the conjecture includes understanding the geometrical
features of FTTM. Thus, number theory and discrete mathematics are the two areas
of mathematics that can provide some of the tools to accomplish the task. Methods of
proving, sequences including Fibonacci number and Pascal’s Triangle will also be
covered in this work.
5
1.5
Importance of Research
As mentioned earlier in this chapter, FTTM is a technique to determine the
location of epileptic foci in epilepsy disorder patient. By proving the conjecture,
many FTTM can be generated. In other words, other versions of FTTM besides
FTTM Version 1 and FTTM Version 2 can be developed in solving the
neuromagnetic inverse problem to determine the location of epileptic foci in epilepsy
disorder patient by proving the conjecture.
1.6
Framework of Research
This thesis consists of six chapters. The first chapter discusses the
background of the research, statement of problem, objectives of research, scope of
research, importance of the research and framework of research.
Chapter 2 consists the literature review of this work. It presents the concept
of FTTM and its generalization. This chapter also discusses extension of Fibonacci
numbers and Pascal’s Triangle.
Chapter 3 consists the methodology of this work. Some methods of proving
will be presented and the reason why these methods failed will be explained. Finally,
the suitable method to prove the conjecture will be revealed.
Chapter 4 exposes the geometrical features of FTTM. Some definitions will
be presented. This chapter consists the actual proof of the conjecture.
6
Chapter 5 consists the extension results. It reveals the relation between
generating FTTM and Pascal’s Triangle. This chapter also contains new theorems,
corollaries and conjecture.
Chapter 6 consists the summary and some recommendations for future works.
The framework of this research can be summarized in Figure 1.3.
GENERALIZED FUZZY TOPOGRAPHIC TOPOLOGICAL MAPPING
CHAP 1 (INTRODUCTION)
CHAP 2 (LITERATURE REVIEW)
Fibonacci
sequence
FTTM
CHAP 3 (A Constructive Proof for
Generating FTTM)
Constructive Proof
CHAP 4 (A Sequence of FTTM)
Fibonacci
Cube
Pascal’s
Triangle
CHAP 5 (Sequence of FTTM In Relation to
Pascal’s Triangle)
Geometrical features of
FTTM
Results in Pascal’s Triangle
Characteristics of sequence
New theorems, corollaries
and conjecture
Sequence of polygon
Sequence of cubes and
its extension
CHAP 6 (CONCLUSION)
7
Figure 1.3 Framework of research
CHAPTER 2
LITERATURE REVIEW
2.1
Introduction
In this chapter, literature reviews that are relevant to our research are
presented. Firstly, we will discuss on Fuzzy Topographic Topological Mapping
(FTTM) and its two versions, namely FTTM 1 and FTTM 2. Then, follow by
discussion on the homeomorphism between FTTM 1 and FTTM 2 and its
generalization. This chapter will also cover the concept of Fibonacci cubes and
Pascal’s Triangle.
2.2
Fuzzy Topographic Topological Mapping (FTTM)
The human brain is known as the most complex organized structure
(Hämäläinen et al., 1993). The brain consists of four main lobes namely frontal,
parietal, occipital and temporal (see Figure 2.1). The outermost layer of the brain is
called the cerebral cortex. The cerebral cortex has a total surface area about 2500cm2.
9
Neurons are the information-processing units in which a neuron consists of the cell
body, the dendrites and the axon. There are at least 1010 neurons in the cerebral
cortex. When information is being processed, small currents flow in neural system
and produce a weak magnetic field (Hämäläinen et al., 1993).
Figure 2.1 Human Brain (resource from
http://www.brainhealthandpuzzles.com/diagram_of_brain.html)
Different parts of the brain produce different patterns of magnetic field.
Besides, diseased brains can also produce abnormal magnetic fields such as stronger
currents are produced in the brain during epileptic seizure. The small area of a brain
tissue that triggers the epileptic seizure is called epileptic foci. For the successful
outcome of the surgery, it is necessary to locate accurately the epileptic foci in the
cortical region and to remove it without serious side effects (Hämäläinen et al.,
1993). Both invasive and noninvasive methods of locating the epileptic foci have
been used in the past.
Magnetoencephalography (MEG) is one of the noninvasive neuroimaging
techniques used to identify epileptic foci (see Figure 2.2). In short, MEG is the study
of magnetic fields generated by currents in the neurons (Uutela, K., 2001). A MEG
system consists of the SQUID (superconducting quantum interference device)
detectors coupled with flux transformers. SQUIDs are the only detectors capable of
10
resolving the weak magnetic fields and at the same time handling the large dynamic
range of the environment noise.
The recorded magnetic fields help in determining where the electrical
currents originate and the strength of currents. Estimating the cerebral current
sources underlying a measured distribution of the magnetic field is called
neuromagnetic inverse problem (Hämäläinen et al., 1993). There is only a method
available for solving this problem, namely Bayesian that needs priori information
(data base model) and time consuming (Tarantola, A. and Valette, B., 1982). Since
Bayesian needs priori information (data base model) and time consuming, there is a
great need to develop an alternative method to solve the neuromagnetic inverse
problem. As a result, FTTM is a new model for solving neuromagnetic inverse
problem. FTTM does not need priori information and less time consuming (Tahir et
al., 2005).
Figure 2.2 MEG System (resource from www.cranius.wordpress.comtagmri.com)
11
2.2.1 FTTM 1 and FTTM 2
FTTM Version 1 has been developed to present a 3-D view of an unbounded
single current source (Fauziah, Z., 2002 & Liau L. Y., 2001) in one angle
observation (upper of a head model). It consists of three algorithms, which link
between four components of the model as shown in Figure 2.3.
MC
TM
Algorithm 1
Algorithm 3
FM
BM
Algorithm 2
Figure 2.3 FTTM Version 1
The four components are Magnetic Contour Plane (MC), Base Magnetic
Plane (BM), Fuzzy Magnetic Field (FM) and Topographic Magnetic Field (TM). MC
is actually a magnetic field on a plane above a current source with z = 0. The plane is
lowered down to BM, which is a plane of the current source with z = -h. Then the
entire BM is fuzzified into a fuzzy environment (FM), where all the magnetic field
readings are fuzzified. Finally, a three dimensional presentation of FM is plotted on
BM. The final process is defuzzification of the fuzzified data to obtain a 3-D view of
the current source (TM).
Using magnetic field readings generated by MATLAB programming and
FTTM Version 1, Fauziah, Z. (2002) solved neuromagnetic inverse problem of MEG
for an unbounded single current source. An algorithm was written to determine
location, the direction and the magnitude of an unbounded single current source.
12
On the other hand, FTTM Version 2 has been developed to present 3-D view
of a bounded multi current source (Wan Eny Zarina et al., 2002) in 4 angles of
observation (upper, left, right and back of a head model).
It consists of three
algorithms, which link between four components of the model as shown in Figure
2.4.
MI
TMI
Algorithm 1
Algorithm 3
FMI
BMI
Algorithm 2
Figure 2.4 FTTM Version 2
The four components are Magnetic Image Plane (MI), Base Magnetic Image
Plane (BMI), Fuzzy Magnetic Image Field (FMI) and Topographic Magnetic Image
Field (TMI). MI is a plane above a current source with z = 0 containing all grey scale
readings (0DN-255DN) of magnetic field. The plane is lowered down to BMI, which
is a plane of the current source with z = -h. Then the entire base BMI is fuzzified into
a fuzzy environment (FMI), where all the gray scale readings are fuzzified. Finally,
a three dimensional presentation of FMI is plotted on BMI. The final process is
defuzzification of the fuzzified data to obtain a 3-D view of the current source (TMI).
Using magnetic field readings generated by MATLAB programming and
FTTM Version 2, Wan Eny Zarina et al., (2002) solved neuromagnetic inverse
problem for a bounded multi current sources. Wan Eny Zarina et al., (2002) also
used Fuzzy C-means to identify the number of current sources using data obtained
from simulation and experiments.
13
2.2.2 Characteristics of FTTM
FTTM Version 1 as well as FTTM Version 2 is specially designed to have
equivalent topological structures between its components.
In other words, a
homeomorphism between each component of FTTM Version 1 as well as FTTM
Version 2 exists. Initially, Liau L. Y. (2001) proved the equivalence between a unit
sphere (denoted by S 2 ) and an ellipsoid with x  1 , y  1 and z  2 (denoted by
E 2 ) (see Figure 2.5) and using the same idea, she finally proved the equivalence of
components of FTTM Version 1 (Figure 2.6).

E2


S2
C

C
Figure 2.5 Homeomorphism from S 2 to E 2
TM
BM

FM



MC
Figure 2.6 Equivalence of components of FTTM Version 1
14
Tahir et al. (2005) have shown that the components of FTTM Version 1 are
homeomorphic. On the other hand, Liau, L. Y. (2006) proved that the components of
FTTM Version 2 were homeomorphic. Liau, L. Y. (2006) also documented in her
PhD thesis that corresponding components of FTTM Version 1 and 2 are
homeomorphic (see Figure 2.7). In other words, Liau, L. Y. (2006) established the
homeomorphism between MC and MI, the homeomorphism between BM and BMI,
the homeomorphism between FM and FMI, and the homeomorphism between TM
and TMI.
FTTM
1 1
FTTM
Version

TM

FM



MC

BM
BMI

TMI




MI

FMI
FTTM Version 2
Figure 2.7 Homeomorphism between corresponding
components of FTTM 1 and FTTM 2
15
2.2.3 Generalized FTTM
Finally, Li Yun (2006) summarized FTTM as a set consisting of models with
four components and three algorithms which link between the four components
where the four components possess their own characteristics respectively and they
are homeomorphic. Generally, FTTM can be represented as follows:
FTTM = {(M, B, F, T): M  B  F  T}
(2.1)
(M', B', F', T'')  FTTM means that M', B', F', T' satisfy the conditions of M, B, F
and T respectively and M'  B', B'  F', and F'  T''.
FTTM Version 1 and FTTM Version 2 are examples of elements of FTTM. The fact
that FTTM Version 1 and FTTM Version 2 are homeomorphic componentwise (see
Figure 2.7); there are at least another 14 elements of FTTM that can be identified
easily. These 14 elements of FTTM are as follows:
(MC, BM, FM, TMI), (MC, BM, FMI, TM), (MC, BMI, FM, TM), (MI, BM, FM, TM),
(MC, BM, FMI, TMI), (MC, BMI, FMI, TM), (MI, BMI, FM, TM), (MI, BM, FM,
TMI),
(MI, BM, FMI, TM), (MC, BMI, FM, TMI), (MC, BMI, FMI, TMI), (MI, BM, FMI,
TMI), (MI, BMI, FM, TMI), and (MI, BMI, FMI, TM).
In other words, the number of new elements of FTTM that can be developed
from these two elements of FTTM is
 2   2   2   2  
             2  14 elements
 1   1   1   1  
with 2 represents two versions of FTTM; i.e. FTTM 1 and FTTM 2. Meanwhile, 1
represents each component of FTTM (MC, BM, FM, TM). The minus 2 means, the
elements of FTTM 1 (MC, BM, FM, TM) and FTTM 2 (MI, BMI, FMI, TMI) are not
included in the list of generating new elements of FTTM.
16
In general, Li Yun has proposed a conjecture as following;
If there exist n elements of FTTM illustrated as,
MC1
TM1
BM1
MC2
BM2
MC3
FM1
TM2
FM2
TM3
BM3
FM3
MCn
TMn
BMn
FMn
Figure 2.8 FTTMn
then the numbers of new elements are
 n!
 n   n   n   n  
n!
n!
n! 



n
             n = 
1
1
1
1
1!
n

1
!
1!
n

1
!
1!
n

1
!
1!
n

1
!




















=  n  n  n  n  n
= n 4  n elements.
(2.2)
Here, it is emphasized that the first components of FTTM will be referred as
MC1, BM1, FM1, TM1 and the second FTTM as MC2, BM2, FM2, TM2 and so on.
17
2.3
Fibonacci Cube
The motivation for developing sequence of FTTM that will be discussed in
detail in Chapter 4 comes from several papers which discussed on Fibonacci Cubes.
The selected papers include Wen-Jing Hsu (1993) and Jie Wu (1997). Wen-Jing Hsu
provided the definition of Fibonacci Cube based on the Fibonacci Code. Meanwhile,
Jie Wu proposed an Extended of Fibonacci Cube (EFC1) with an even number of
nodes. It is defined based on the same sequence F(i)=F(i-1)+F(i-2) as the regular
Fibonacci sequence. However, their initial conditions are different. Definition 2.1
defined the first Extended of Fibonacci Cube in the series. The symbol ║ denotes a
concatenation operation. For example, 01║{0,1}={010,011} and 01║{}= 01.
Definition 2.1: Extended of Fibonacci Cube (Jie Wu, 1997)
Assume EFC1(n) = (V1 (n),E1(n)), EFC1(n-1) = (V1(n-1),E1(n-1)) and EFC1(n-2) =
(V1(n-2),E1(n-2)). Then, V1(n) = 0║V1(n-1)  10║V1(n-2). Two nodes in EFC1(n) are
connected by an edge in E1(n) if and only if their labels differ in exactly one bit
position. As initial condition for recursion, V1 (3) = {0,1} and V1(4) = {00, 10, 11,
01}.
The important idea of these papers was that two nodes were connected by an
edge and their labels differ in exactly one bit (see Figure 2.9). The term bit here
means binary digits which are 0 and 1. Figure 2.9 shows examples of EFC1 of size
n, where n=3, 4, 5, 6 respectively. An EFC1 of size n consists of one EFC1 of size n1 and one EFC1 of size n-2. Figure 2.9 (a) shows a line with differ bits while Figure
2.9 (b) shows a square if the edges are drawn as a straight line. The bits also differ
for each vertex. Figure 2.9 (c) and (d) elaborate the idea of bit. Figure 2.9(c) can be
represented as non-perfect cube (see Figure 2.10 after modified a little bit to the
original figure. If Figure 2.9 is added by five more edges and two more vertices, then
it may set a perfect cube (see Figure 2.11). But, this representation has different bits
for each vertex.
18
0
1
00
01
10
11
(b)
(a)
000
001
010
011
100
101
(c)
0000
0001
0010
0011
0100
0101
1000
1001
1010
1011
(d)
Figure 2.9 Extended Fibonacci Cubes EFC1s:
(a) EFC1(3), (b) EFC1(4), (c) EFC1(5) and (d) EFC1(6)
010
011
000
001
000
001
010
111
101
Figure 2.10 Modified EFC1(5)
Figure 2.10 Modified EFC (5)
100
101
110
111
Figure 2.11 Reconstruct EFC1(5)
19
The idea of reconstructing EFC1(5) is really important in this research. This is
due to the fact that the first four vertices (000, 001, 010, 111) are similar to the
components of FTTM Version 1. The second four vertices (100, 101, 110, 111) can
be assumed as the components of FTTM Version 2. The extended version; i.e,
sequence of FTTM will use differ bit for each component. The concept of sequence
of FTTM is needed in order to study the geometrical feature of FTTM which will be
discussed in Chapter 4 later.
2.4
Pascal’s Triangle
Pascal’s arithmetic triangle or The Traité du Triangle Aritmétique, now
generally known as Pascal’s Triangle originally was developed by the ancient
Chinese. Figure 2.12 illustrate Pascal’s Triangle as depicted in 1303 by Chu ShihChien (Howard Eves, 1976; John Stillwell, 2002; David M. Burton, 2007). In reality,
Pascal’s Triangle is the product of a much earlier Eastern culture. Besides Chu ShihChien, others name that used the triangle in their work includes Chia Hsien (1050)
from China, Omar Khayyam (circa 1050-1130) from Arab and Al-Tusi (circa 12001275) from Persian (David M. Burton, 2007).
Even though Pascal’s Triangle was not originally developed by Blaise Pascal,
but he was the first person to discover the importance of all the patterns in Pascal’s
Triangle. There is a slight difference in Blaise Pascal’s representation of the triangle
in 1654 as shown in Figure 2.13 (David M. Burton, 2007) and present-day
representation of Pascal’s Triangle as presented in Figure 2.14 (John Stillwell, 2002).
20
Figure 2.13 The original Pascal’s Triangle (resource from
http://pages.csam.montclair.edu/~kazimir/history.html)
Figure 2.12 Chinese Pascal’s Triangle (resource from
20
http://www.mathsisfun.com/images/pascals-triangle-chinese.gif)
21
1
1
1
1
1
1
1
1
1
8
3
5
7
6
15
1
4
10
20
35
56
1
3
10
21
28
2
4
6
1
5
15
35
70
1
1
6
21
56
1
7
28
1
8
1
Figure 2.14 Present-day Pascal’s Triangle
The entries of Pascal’s Triangle can be obtained by adding the entries at the
left and right above it (V.A Uspenskii, 1974). Blaise Pascal developed the following
relations involving the numbers of the Pascal’s Triangle (Howard Eves, 1976);
a) Any element (not in the first row or the first column) of the Pascal’s Triangle
is equal to the sum of the two entries above it; to the left and to the right.
b) Any given element of the Pascal’s Triangle, decreased by 1, is equal to the
sum of all the elements above the row and to the left of the column containing the
given element.
c) The mth element in the nth row is (m+n-2)!/(m-1)!(n-1)!, where, by
definition, 0!=1.
d) The element in the mth row and nth column is equal to the element in the mth
column and nth row.
e) The sum of the elements along any diagonal is twice the sum of the elements
along the preceding diagonal.
f) The sum of the elements along the nth diagonal is 2n-1.
22
2.4.1 Patterns in Pascal’s Triangle
Pascal’s Triangle is also known as triangle of binomial coefficients.
Moreover, there are so many fascinating patterns in Pascal’s Triangle. Some of the
patterns are as follows (resource from http://britton.disted.camosun.bc.ca/pascal/
pascal.html);
a) The sum of the rows
The sum of the numbers in any row is equal to 2 to the nth power or 2n,
when n is the number of the row.
b) Prime numbers
If the first element in a row is a prime number (remember, the 0th element
of every row is 1), all the numbers in that row (excluding the 1's) are divisible
by it.
c) Hockey stick pattern
The diagonal of numbers of any length starting with any of the 1’s
bordering the sides of the triangle and ending on any number inside the
triangle is equal to the number below the last number of the diagonal, which
is not on the diagonal (see Figure 2.15).
23
Figure 2.15 Hockey stick in Pascal’s Triangle
d) Magic 11’s
If a row is made into a single number by using each element as a digit of
the number (carrying over when an element itself has more than one digit),
the number is equal to 11 to the nth power or 11n when n is the number of the
row.
e) Fibonacci’s Sequence
The sum of the numbers in the consecutive rows as shown in Figure 2.15
is the first numbers of the Fibonacci sequence. The sequence can also be
formed in a more direct way, very similar to the method used to form the
triangle, by adding two consecutive numbers in the sequence to produce the
next number. Figure 2.16 shows the details.
24
Figure 2.16 Fibonacci’s Sequence in Pascal’s Triangle
f) Triangular numbers
A triangular number is a number obtained by adding all positive integers less
than or equal to a given positive integer n. The triangular numbers can be found in
the diagonal starting at row 3 as shown in Figure 2.17.
Figure 2.17 Triangular numbers in Pascal’s Triangle
25
g) Square numbers
A square numbers are the squares of natural numbers, such as 1, 4, 9, 16,
25, etc. The square number is the sum of the two numbers in any circled area
in Figure 2.18. (The colors are different only to distinguish between the
separate "rubber bands"). The nth Square Number is equal to the nth
Triangular Number plus the (n-1)th Triangular Number
Figure 2.18 Square numbers in Pascal’s Triangle
Triangular and square numbers are two examples of polygonal number.
Polygonal numbers are the number of vertices in a figure formed by a certain
polygon. The first number in any polygonal numbers is always 1, or a point. The
second number is equal to the number of vertices of the polygon (see Table 2.1).
26
Table 2.1 Polygonal Number
Type 1st
Triangular
3rd
4 th
5th
6 th
1
3
6
10
15
21
Value
1
Pentagonal
4
9
16
25
36
Value
1
Hexagonal
5
12
22
35
51
6
15
28
45
66
Value
Square
Value
1
26
2nd
27
Some patterns in Pascal’s Triangle that are related to generating FTTM and
also the new pattern obtained from Pascal’s Triangle resulting from this research will
be discussed in Chapter 5.
2.5
Conclusion
In this chapter, we discussed on FTTM, Fibonacci Cubes and Pascal’s
Triangle. These mathematical concepts are important in this research. In the
following chapter, we will present some methods of proving and the reasons why
some of these methods fail to proof the conjecture.
CHAPTER 3
A CONSTRUCTIVE PROOF FOR GENERATING FTTM
3.1
Introduction
This chapter introduces axiomatic system. A thorough understanding of
axiomatic system is important in order to develop definitions and theorems. This
chapter discusses some method of proving the conjecture and the reason why most of
the other methods fail. These methods of proving include proofs by exhaustion,
direct, contradiction, contrapositive, mathematical induction and constructive. The
most suitable method in proving the conjecture is finally found.
3.2
Axiomatic System
In mathematics, there are two types of reasoning. They are inductive and
deductive reasonings (Philip, J. D. and Reuben, H., 1981). Inductive reasoning will be
used when the “truth” of a statement is established by making a generalization based
upon a limited number of observations or experiments. Inductive reasoning is the
29
basis for the scientific method. Deductive reasoning is another type of reasoning
when a conclusion derived from other statements accepted as a truth. The truth of a
new statement must be deduced from the truth of old statements. In other words, it is
a method from hypothesis which leads to a conclusion (Philip, J. D. and Reuben, H.,
1981).
Axiomatic system is a collection of undefined terms, axioms and proving
procedures for deriving theorems in the system. Axiomatic system has four
components. There are undefined terms, axioms, definitions and theorems. All these
are essential in order to prove a statement (Herbert Meschkowski, 1968).
Undefined terms are basic words which can be constructed for an axiomatic
system. It forms the basic for the vocabulary of an axiomatic system. In this system,
some terms are left undefined to prevent circular definitions (Nancy Rodgers, 1941). To
understand a definition, someone must understand the definition of each word in it.
Axiom is a self-evident truth which does not require proof and is assumed to
be true. Axioms form the foundation for what is true in the system. Axioms give the
basic properties of the undefined terms. They are the basic assumptions that we make
about the undefined terms (Nancy Rodgers, 1941). There are two rules for axioms. The
first is the axiom must be accepted (Herbert Meschkowski, 1968) since axioms are the
idealizations of observed phenomena in the “real” world or the concepts in
elementary mathematics (Burnett Meyer, 1974). The second is the axioms must be
perfectly self-evident concepts (Herbert Meschkowski, 1968) that require no proves and
explanations.
A definition is an abbreviation of a particular concept. When a mathematical
concept occurs repeatly, definition will be made (Daniel Solow, 1982). Definition is
the use of a word which explained by somebody and for somebody for various
purposes. The role of definition is to remove the ambiguity, change the meaning and
30
isolate the new concepts. Mathematical definition must be absolutely clear (Edward,
R. S., 2006).
Theorems are mathematical statements which can be derived from axioms
and need proofs. If a statement is proven then only it can be called as theorem
(Edward, R. S., 2006). Theorems can be derived from the axioms, previous theorems,
and definitions using proving procedure. Less important theorems are called
propositions. A less important theorem that is helpful in the proof of other
results is called a lemma. A corollary is a theorem that can be established directly
from a theorem that has been proven. A conjecture is a statement that is being
proposed to be a true statement, usually based on some partial evidence or the
intuition of an expert. Nothing is a theorem until it is proven; otherwise it is just a
conjecture.
3.3
Proving methods
A proof in mathematics is a convincing argument that some mathematical
statement, proposition or mathematical formula is true (Ted Sundstrom, 2007). It
consists of a set of assumptions that are combined accordingly to logical rules in
order to establish a valid conclusion. This validation can take one of two forms. In a
direct proof, a given conclusion can be shown to be true. In an indirect proof, a given
conclusion can be shown not to be false and therefore, assumed to be true. Proving
methods that are not direct include prove by contradiction, prove by exhaustion,
prove by infinite descent and prove by mathematical induction.
In this research, the main objective is to prove a conjecture proposed by Liau
Li Yun (2006). Some methods of proving used include proof by exhaustion, direct,
contradiction, contrapositive, mathematical induction and constructive.
31
3.3.1 Exhaustion
The first method to prove partially or for specific case of the conjecture is
proving by exhaustion. For examples, for FTTM 1, FTTM 2 and FTTM 3 are as
follows;
FTTM 1
0  14  1 .
FTTM 2
14  24  2 .
The elements are;
 MC1 , BM 1 , FM 1 , TM 2  ,  MC1 , BM1 , FM 2 , TM 1  ,  MC1 , BM 2 , FM 1 , TM1  ,
 MC2 , BM1 , FM 1 , TM 1  ,  MC1 , BM1 , FM 2 , TM 2  ,  MC1 , BM 2 , FM 1 , TM 2  ,
 MC2 , BM1 , FM 1 , TM 2  ,  MC1 , BM 2 , FM 2 , TM 1  ,  MC2 , BM 1 , FM 2 , TM1  ,
 MC2 , BM 2 , FM1 , TM 1  ,  MC1 , BM 2 , FM 2 , TM 2  ,  MC2 , BM 2 , FM 1 , TM 2  ,
 MC2 , BM 2 , FM 2 , TM 1  ,  MC2 , BM1 , FM 2 , TM 2  .
FTTM 3
78  34  3 .
The elements are;
 MC1 , BM 1 , FM 1 , TM 2  ,  MC1 , BM1 , FM 2 , TM 1  ,  MC1 , BM 2 , FM 1 , TM1  ,
 MC2 , BM1 , FM 1 , TM 1  ,  MC1 , BM1 , FM 2 , TM 2  ,  MC1 , BM 2 , FM 1 , TM 2  ,
 MC2 , BM1 , FM 1 , TM 2  ,  MC1 , BM 2 , FM 2 , TM 1  ,  MC2 , BM 1 , FM 2 , TM1  ,
 MC2 , BM 2 , FM1 , TM 1  ,  MC1 , BM 2 , FM 2 , TM 2  ,  MC2 , BM 2 , FM 1 , TM 2  ,
 MC2 , BM 2 , FM 2 , TM 1  ,  MC2 , BM1 , FM 2 , TM 2  ,
32
 MC1 , BM 1 , FM1 , TM 3  ,  MC1 , BM1 , FM 3 , TM 1  ,  MC1 , BM 3 , FM 1 , TM1  ,
 MC3 , BM1 , FM 1 , TM 1  ,  MC1 , BM1 , FM 3 , TM 3  ,  MC1 , BM 3 , FM1 , TM 3  ,
 MC3 , BM1 , FM 1 , TM 3  ,  MC1 , BM 3 , FM 3 , TM 1  ,  MC3 , BM1 , FM 3 , TM 1  ,
 MC3 , BM 3 , FM 1 , TM 1  ,  MC1 , BM 3 , FM 3 , TM 3  ,  MC3 , BM 3 , FM 1 , TM 3  ,
 MC3 , BM 3 , FM 3 , TM 1  ,  MC3 , BM 1 , FM 3 , TM 3  ,
 MC2 , BM 2 , FM 2 , TM 3  ,  MC2 , BM 2 , FM 3 , TM 2  ,  MC2 , BM 3 , FM 2 , TM 2  ,
 MC3 , BM 2 , FM 2 , TM 2  ,  MC2 , BM 2 , FM 3 , TM 3  ,  MC2 , BM 3 , FM 2 , TM 3  ,
 MC3 , BM 2 , FM 2 , TM 3  ,  MC2 , BM 3 , FM 3 , TM 2  ,  MC3 , BM 2 , FM 3 , TM 2  ,
 MC3 , BM 3 , FM 2 , TM 2  ,  MC2 , BM 3 , FM 3 , TM 3  ,  MC3 , BM 3 , FM 3 , TM 2  ,
 MC3 , BM 3 , FM 2 , TM 3  ,  MC3 , BM 2 , FM 3 , TM 3  .
 MC1 , BM 1 , FM 2 , TM 3  ,  MC2 , BM 1, FM 2 , TM 3  ,  MC3 , BM1 , FM 2 , TM 3  ,  MC1 , BM 2 , FM 2 , TM 3  ,
 MC1 , BM 3 , FM 2 , TM 3  ,  MC1, BM1 , FM 3 , TM 2  ,  MC2 , BM1 , FM 3 , TM 2  ,  MC3 , BM 1, FM 3 , TM 2  ,
 MC1 , BM 2 , FM 3 , TM 2  ,  MC1, BM 3 , FM 3 , TM 2  ,  MC1 , BM 2 , FM 3 , TM 1  ,  MC2 , BM 2 , FM 3 , TM 1  ,
 MC3 , BM 2 , FM 3 , TM 1  ,  MC1 , BM 2 , FM 3 , TM 3  ,  MC1 , BM 3 , FM 2 , TM1  ,  MC2 , BM 3 , FM 2 , TM 1  ,
 MC3 , BM 3 , FM 2 , TM1  ,  MC1, BM 3 , FM 2 , TM 2  ,  MC2 , BM 3 , FM 1, TM 1  ,  MC2 , BM 3 , FM 3 , TM 1  ,
 MC2 , BM 3 , FM1 , TM 2  ,  MC2 , BM 3 , FM1 , TM 3  ,  MC3 , BM 2 , FM 1, TM 1  ,  MC3 , BM 2 , FM 2 , TM 1  ,
 MC3 , BM 2 , FM1 , TM 2  ,  MC3 , BM 2 , FM1 , TM 3  ,  MC2 , BM1 , FM 1, TM 3  ,  MC2 , BM 2 , FM 1, TM 3  ,
 MC2 , BM 1 , FM 3 , TM 3  ,  MC3 , BM1 , FM 1, TM 2  ,  MC3 , BM 3 , FM1 , TM 2  ,  MC3 , BM 1 , FM 2 , TM 2  ,
 MC1 , BM 2 , FM 1 , TM 3  ,  MC1, BM 3 , FM 1, TM 2  ,  MC2 , BM 1 , FM 3 , TM 1  ,  MC3 , BM 1, FM 2 , TM 1  .
This method is ridiculous in proving the generality of the conjecture since if
FTTM 10, therefore, we have to produce a list of 9990 new elements of FTTM! This
method is not practical and decisive enough!
3.3.2 Direct
In mathematics and logic, a direct proof is a way of showing the truth or false
of a given statement by a straightforward combination of established facts without
making any further assumptions (Edward, R. S., 2006). In order to directly prove a
33
conditional statement of the form "If p, then q" ( p  q ), it is only necessary to
consider situations where the statement p is true, then q must be true. Before getting
started, it is important to know the meaning of every word in implication so that the
implication can simply tested to be a truth (Ulrich, D. and Pamela, G., 2003). This is the
idea as how to use direct proof.
Correspond to the problem; it is required to identify first what is p or
hypothesis and conclusion q. Therefore, the hypothesis is if there exist n elements of
FTTM and the conclusion is the numbers of generating new element are n 4  n . By
identifying the hypothesis, it is clear that there are many FTTM is represented by
geometrical object. This is because the model of FTTM is in the form of geometrical
object; i.e. p. On the other hand, q is an algebraic expression. That is explain this
method cannot be considered after all.
3.3.3 Contradiction
The proof by contradiction is also known as reductio ad absurdum, which is
in Latin means "reduce it to something absurd". A proof by contradiction establishes
the truth of a given proposition by the supposition that it is false (James L. Hein,
2003). That is, the supposition that p is false followed necessarily by the conclusion q
from not p, where q is false, which implies that p is true. The difficulty of this
method is that we do not know what kind of the contradiction is going to be (Daniel
Solow, 1982).
Correspond to the case; the same p and q in the direct proof is used. So, the
statement will be assumed p is not true but q is true. However, the statement is not
suitable in this problem. The statement will become p is true but q not. Furthermore,
34
there is no evidence or fact that can promise p become false. Thus, this method must
not be considered.
3.3.4 Contrapositive
Contrapositive is a quite powerful method in proving. It allows to attack a
proof backwards. Instead of going from the assumptions and trying to derive the
result, start by assuming the result is false and show that this violates one of the
assumptions. This makes use of the logical law of contrapositive (James L. Hein,
2003); if p then q ( p  q ) is equivalent to if not q then not p (
q
p ). This
method is the same as direct proof. The difference is needed to show p and q is false.
However, this method also cannot be used to proof the conjecture because p is
geometrical object while q is algebraic expression.
3.3.5 Mathematical Induction
Mathematical induction is a way to prove statements for all positive integers.
These statements usually include equations, but may be compounded. In its simplest
form, a statement may be given with a variable integer n; prove of the validity of the
statement for all values of n can be obtained through mathematical induction. There
are two steps in mathematical induction (Kenneth H. Rosen, 2000); the basis and the
inductive steps.
Let P(n) be a property of positive integers such that:
(1) Base Case: P(1) is true, and
(2) Inductive Step: if P(n) is true, then P(n + 1) is true.
35
Then P(n) is true for all positive integers.
Even though mathematical induction is a powerful proving technique, it
cannot be used to prove the conjecture. This is because; the left hand side of the
equation of the conjecture is a geometrical object (Figure 2.8) while the right hand
side is an algebraic expression (Equation 2.2). The predicate or P(n) is almost
impossible to be identified because of the difference between both sides. Therefore,
this type of proof cannot be used in proving the conjecture.
3.3.6 Constructive
In mathematics (James L. Hein, 2003), a constructive proof is a method of
proving that demonstrates the existence of a mathematical object with certain
properties by creating or providing a method for creating such an object.
Furthermore, constructive method can be identified by certain key word that appears
in the statement like there is, there are, there exist, for all, for each and for every
(Daniel Solow, 1982). In addition, this method never put any condition that the
statement of the problem should be identified first.
This method is very promising in order to prove the conjecture even though
the nature of the right hand side and the left hand side of the conjecture are not same.
From our study on geometrical features of FTTM, we may construct all the
ingredients to prove the conjecture. It is the only promising method available.
36
3.4
Conclusion
In this chapter, we have discussed on an axiomatic system. The reasons why
some of proving methods fail have been discussed in this chapter. Thus, proof by
constructive is the only suitable method.
CHAPTER 4
A SEQUENCE OF FTTM
4.1
Introduction
In the literature review of this research, we have mentioned that there are
FTTM Version 1 and FTTM Version 2. There are more than these two as stated in
the conjecture proposed by Liau Li Yun in 2006. The conjecture turned out to be
difficult and almost impossible to prove directly. Therefore, the geometrical features
of FTTM must be investigated in order to create such an object to proof the
conjecture. This chapter is dedicated to expose the geometrical features of FTTM.
We will develop some important definitions that are bringing to prove the conjecture.
4.2
Geometrical Features of FTTM
As stated before, FTTM have four components which are MC, BM, FM and
TM. The model of FTTM looks like a square. When FTTM Version 1 and FTTM
Version 2 are combined, they produce a cube. Consequently, as the number of FTTM
38
increases, ways to illustrate these combinations will also increase. In the following
subsection, we will show various configurations of combined FTTM.
By FTTMn, we mean the combination of n versions of FTTM. For example,
FTTM5 means 5 models of FTTM are combined together. Table 4.1 lists some
geometrical features of FTTM1, FTTM2, FTTM3 and FTTM4. They are vertices,
edges, faces and cubes of FTTM. The vertices of FTTM mean the components of
FTTM which are MC, BM, FM and TM. Meanwhile, the homeomorphisms between
respective components of FTTM are represented as the edges. The other feature is
the face. The square is one face of FTTM. Another type of its face is a triangle. The
triangle face can only exist in combined FTTMs. However, in this study we will only
discuss the square face. The last geometrical feature of FTTM is the cube. Similar to
the triangle face, a cube face can only exist in combined FTTM. Clearly, FTTM1
does not have any cube face.
Let’s denote vFTTMn as a set of vertices of FTTM in FTTMn, eFTTMn as a
set of edges of FTTM in FTTMn, fFTTMn as a set of faces of FTTM in FTTMn and
FTTM2/n as a set of cubes of FTTM in FTTMn. Let |
| denote the number of
elements in a set. For example, the set for each features for FTTM2 (see Table 4.1)
are as follows;
vFTTM 2  MC1 , MC2 , BM 1 , BM 2 , FM 1 , FM 2 , TM 1 , TM 2 
Therefore, vFTTM 2  8 .
 MC1MC2 , MC1 BM 1 , MC1TM 1 , BM 1 BM 2 , BM 1FM 1 , FM 1FM 2 , 
eFTTM 2  

 FM 1TM 1 , MC2 BM 2 , MC2TM 2 , BM 2 FM 2 , FM 2TM 2 , TM 1TM 2 
Therefore, eFTTM 2  12 .
39
 MC1 , BM 1 , FM1 , TM 1  ,  MC1 , MC2 , BM 2 , BM1  ,  MC2 , BM 2 , FM 2 , TM 2  , 
fFTTM 2  

TM 2 , FM 2 , FM1 , TM 1  ,  MC1 , MC2 , TM 2 , TM1  ,  BM1 , BM 2 , FM 2 , FM1  
Therefore, fFTTM 2  6 .
FTTM 2/2  MC1 BM 1FM 1TM 1MC2 BM 2 FM 2TM 2 
Therefore, FTTM 2/2  1 .
Subsequently, FTTM1 has 4 vertices, 4 edges, 1 face and does not have any
cube (see Table 4.1). There is no other way to draw FTTM2 except as listed in Table
4.1. Because of that reason, FTTM2 consists of 8 vertices, 12 edges, 6 faces and 1
cube. We can illustrate FTTM3 in three different ways. The first form of FTTM3 has
12 vertices, 20 edges, 11 faces and 3 cubes. The second form has 12 vertices, 24
edges, 11 squares faces and no cube. The third form FTTM3 consists of 12 vertices,
24 edges, 15 faces and 3 cubes. We can also illustrate FTTM4 in three different ways.
The first form of FTTM4 contains 16 vertices, 28 edges, 16 faces and 6 cubes.
Whereas, the second form of FTTM4 has 16 vertices, 40 edges, 28 faces and 6 cubes.
The last form of FTTM4 consists of 16 vertices, 32 edges, 12 faces and 2 cubes.
40
Table 4.1 Geometrical features of combined FTTM
FTTMn
Geometrical Features
FTTM1
FTTM2
MC1
TM1
BM1
FM1
MC1
TM1
BM1
FM1
MC2
TM2
BM2
FM2
Vertices
Edges
Faces
Cubes
4
4
1
0
8
12
6
1
40
41
MC1
FTTM3
TM1
1)
MC2
MC3
BM2
TM3
BM3
FM3
BM1
20
11
3
12
24
11
0
FM1
TM2
FM2
MC1
TM1
2)
BM1
FM1
MC2
BM2
TM2
FM2
MC3
TM3
FM3
41
BM3
12
42
MC1
TM1
3)
BM1
MC2
TM2
BM2
FM2
MC3
TM3
BM3
FM3
MC3
BM3
TM4
BM4
FM4
3
16
28
16
6
BM1
TM2
FM1
FM2
FM3
42
MC4
BM2
TM3
15
TM1
1)
MC2
24
FM1
MC1
FTTM4
12
43
MC1
TM1
2)
BM1
MC2
16
40
28
6
16
32
12
2
FM1
TM2
BM2
MC4
TM4
BM4
FM4
FM2
MC3
TM3
BM3
FM3
MC1
TM1
3)
MC2
BM1
TM2
FM1
BM2
FM2
MC3
TM3
MC4
BM3
FM3
FM4
43
BM4
TM4
44
If we continue illustrating combined FTTM until FTTMn, we can expect
some patterns of sequence for vertices, edges, faces and cubes emerge provided the
way we present combined FTTM must be consistent. We show these patterns in
Table 4.2 for FTTM1 until FTTM20. The formal definition for sequence of
geometrical features of FTTMn which are vertices, edges, faces and cubes are
presented in the following section.
Table 4.2 Geometrical features of sequences of FTTM1 until FTTM20
FTTMn
Vertices
Edges
Faces
Cubes
FTTM1
4
4
1
0
FTTM2
8
12
6
1
FTTM3
12
20
11
3
FTTM4
16
28
16
6
FTTM5
20
36
21
10
FTTM6
24
44
26
15
FTTM7
28
52
31
21
FTTM8
32
60
36
28
FTTM9
36
68
41
36
FTTM10
40
76
46
45
FTTM11
44
84
51
55
FTTM12
48
92
56
66
FTTM13
52
100
61
78
FTTM14
56
108
66
91
FTTM15
60
116
71
105
FTTM16
64
124
76
120
FTTM17
68
132
81
136
FTTM18
72
140
86
153
FTTM19
76
148
91
171
FTTM20
80
156
96
190
45
4.3
Formal Definitions
By observing the geometrical features (see Table 4.1) of FTTM and referring
to the properties of geometrical features of FTTM in Table 4.2, the following
definitions are developed. All the definitions below are analogous to the definition of
Fibonacci sequence given in (Kenneth H. Rosen, 2000).
The sequence of vertices begins with the integers 4, 8, 12, 16, 20, 24,… and so on.
We can define formally the sequence of vertices as follows.
Definition 4.1 (A Sequence of vertices in FTTMn )
The sequence of vertices for FTTMn which is vFTTM1, vFTTM2, vFTTM3,… are
defined recursively by the equation vFTTM n  vFTTM n 1  4 for n  1 and
vFTTM 0  0 .
The sequence of edges begins with the integers 4, 12, 20, 28, 36,… and so on. We
can define formally the sequence of edges as follows.
Definition 4.2 (A Sequence of edges in FTTMn )
The sequence of edges for FTTMn which is eFTTM1, eFTTM2, eFTTM3,… are
defined recursively by the equation eFTTM n  eFTTM n 1  8 for n  1 and
eFTTM1  4 .
The sequence of faces begins with the integers 1, 6, 11, 16, 21, 26, 31, 36, 41, 46,…
and so on. We can define formally the sequence of faces as follows.
46
Definition 4.3 (A Sequence of faces in FTTMn )
The sequence of faces in FTTMn is fFTTM1, fFTTM2, fFTTM3,… are defined
recursively by the equation fFTTM n  fFTTM n 1  5 for n  1 and fFTTM1  1 .
The sequence of cubes begins with the integers 0, 1, 3, 6, 10, 15, 21, … and so on.
We can define formally the sequence of cubes as follows.
Definition 4.4 (A Sequence of cubes in FTTMn )
The sequence of cubes for FTTMn which is FTTM2/1, FTTM2/2, FTTM2/3,… are
defined recursively by the equation FTTM 2/ n  FTTM 2 / n -1  (n -1) for n  1 and
FTTM 2/ 0  0 .
These definitions are important in our subsequent works.
47
4.4
A Sequence of FTTM
There are many ways to illustrate FTTMs. However, the form that Li Yun
(2006) introduced in her conjecture is the one that we are going to adopt in our work.
We formalize at particular form by the following two definitions.
Definition 4.5 (A Sequence of FTTM)
Let FTTMi   MCi , BM i , FM i , TM i  such that MCi , BM i , FM i , TM i are topological
space
with
MCi  BMi  FMi  TMi.
The
set
of
FTTMi
is
denoted
by
FTTM  FTTMi : i  1, 2,3,..., n .
A sequence of n FTTMi of FTTM is FTTM1, FTTM2, FTTM3,…, FTTMn such that
MCi  MCi 1 , BM i  BM i 1 , FM i  FM i 1 , TM i  TM i 1 .
Definition 4.5 is illustrated in Figure 4.1.
MC1
TM1
BM1
FM1
MC2
TM2
BM2
FM2
MCn
TMn
BMn
FMn
Figure 4.1 A Sequence of FTTMn
48
We adopt the term of an ordinary sequence in our newly sequence of FTTM.
Definition 4.6 (k-FTTMn )
k-FTTMn is the k-th FTTM of a sequence of FTTMn for n  k .
From now on the sequence of FTTMn must be in increasing order, otherwise
stated. Example 4.1 is presented to illustrate these definitions where the first term of
FTTM3 is called 1-FTTM3, and then followed by 2-FTTM3 and 3-FTTM3.
Example 4.1 FTTM3
MC1
TM1
1-FTTM3
BM1
MC2
FM1
TM2
2-FTTM3
BM2
MC3
FM2
TM3
3-FTTM3
BM3
FM3
49
4.5
A Sequence of Cubes in FTTMn
A cube is a combination of two FTTM in FTTMn. For example, cubes that
can be produced from the combination of two FTTM in FTTM3 are; 1-FTTM3 with
2-FTTM3, 2-FTTM3 with 3-FTTM3, and 1-FTTM3 with 3-FTTM3.
Table 4.3
summarizes number of cubes that can be generated in FTTMn for n= 1, 2, 3, 4 and 5.
Table 4.3 Cube with two terms FTTM in FTTMn
FTTMn
Generated FTTM
FTTM1
MC1
TM1
BM1
FM1
FTTM2
MC1
MC2
BM2
BM1
Cubes
0
TM1
TM2
FM1
FM2
 MC1 , BM 1 , FM 1 , TM 2  ,  MC1 , BM1 , FM 2 , TM 1  ,  MC1 , BM 2 , FM 1 , TM1  ,
 MC2 , BM1 , FM 1 , TM 1  ,  MC1 , BM1 , FM 2 , TM 2  ,  MC1 , BM 2 , FM 1 , TM 2  ,
 MC2 , BM1 , FM 1 , TM 2  ,  MC1 , BM 2 , FM 2 , TM 1  ,  MC2 , BM 1 , FM 2 , TM1  ,
 MC2 , BM 2 , FM1 , TM 1  ,  MC1 , BM 2 , FM 2 , TM 2  ,  MC2 , BM 2 , FM 1 , TM 2  ,
 MC2 , BM 2 , FM 2 , TM 1  ,  MC2 , BM1 , FM 2 , TM 2  ,
14 new elements alltogether.
1
50
FTTM3
3
MC1
TM1
BM1
MC2
FM1
TM2
BM2
FM2
MC3
TM3
BM3
FM3
 MC1 , BM 1 , FM 1 , TM 2  ,  MC1 , BM1 , FM 2 , TM 1  ,  MC1 , BM 2 , FM 1 , TM1  ,
 MC2 , BM1 , FM 1 , TM 1  ,  MC1 , BM1 , FM 2 , TM 2  ,  MC1 , BM 2 , FM 1 , TM 2  ,
 MC2 , BM1 , FM 1 , TM 2  ,  MC1 , BM 2 , FM 2 , TM 1  ,  MC2 , BM 1 , FM 2 , TM1  ,
 MC2 , BM 2 , FM1 , TM 1  ,  MC1 , BM 2 , FM 2 , TM 2  ,  MC2 , BM 2 , FM 1 , TM 2  ,
 MC2 , BM 2 , FM 2 , TM 1  ,  MC2 , BM1 , FM 2 , TM 2  ,
 MC1 , BM 1 , FM1 , TM 3  ,  MC1 , BM1 , FM 3 , TM 1  ,  MC1 , BM 3 , FM 1 , TM1  ,
 MC3 , BM1 , FM 1 , TM 1  ,  MC1 , BM1 , FM 3 , TM 3  ,  MC1 , BM 3 , FM1 , TM 3  ,
 MC3 , BM1 , FM 1 , TM 3  ,  MC1 , BM 3 , FM 3 , TM 1  ,  MC3 , BM1 , FM 3 , TM 1  ,
 MC3 , BM 3 , FM 1 , TM 1  ,  MC1 , BM 3 , FM 3 , TM 3  ,  MC3 , BM 3 , FM 1 , TM 3  ,
 MC3 , BM 3 , FM 3 , TM 1  ,  MC3 , BM 1 , FM 3 , TM 3  ,
 MC2 , BM 2 , FM 2 , TM 3  ,  MC2 , BM 2 , FM 3 , TM 2  ,  MC2 , BM 3 , FM 2 , TM 2  ,
 MC3 , BM 2 , FM 2 , TM 2  ,  MC2 , BM 2 , FM 3 , TM 3  ,  MC2 , BM 3 , FM 2 , TM 3  ,
 MC3 , BM 2 , FM 2 , TM 3  ,  MC2 , BM 3 , FM 3 , TM 2  ,  MC3 , BM 2 , FM 3 , TM 2  ,
 MC3 , BM 3 , FM 2 , TM 2  ,  MC2 , BM 3 , FM 3 , TM 3  ,  MC3 , BM 3 , FM 3 , TM 2  ,
 MC3 , BM 3 , FM 2 , TM 3  ,  MC3 , BM 2 , FM 3 , TM 3  ,
42 new elements alltogether.
51
FTTM4
MC1
TM1
BM1
MC2
BM2
MC3
FM1
TM2
FM2
TM3
BM3
FM3
MC4
TM4
BM4
FM4
 MC1 , BM 1 , FM 1 , TM 2  ,  MC1 , BM1 , FM 2 , TM 1  ,  MC1 , BM 2 , FM 1 , TM1  ,
 MC2 , BM1 , FM 1 , TM 1  ,  MC1 , BM1 , FM 2 , TM 2  ,  MC1 , BM 2 , FM 1 , TM 2  ,
 MC2 , BM1 , FM 1 , TM 2  ,  MC1 , BM 2 , FM 2 , TM 1  ,  MC2 , BM 1 , FM 2 , TM1  ,
 MC2 , BM 2 , FM1 , TM 1  ,  MC1 , BM 2 , FM 2 , TM 2  ,  MC2 , BM 2 , FM 1 , TM 2  ,
 MC2 , BM 2 , FM 2 , TM 1  ,  MC2 , BM1 , FM 2 , TM 2  ,
 MC1 , BM 1 , FM1 , TM 3  ,  MC1 , BM1 , FM 3 , TM 1  ,  MC1 , BM 3 , FM 1 , TM1  ,
 MC3 , BM1 , FM 1 , TM 1  ,  MC1 , BM1 , FM 3 , TM 3  ,  MC1 , BM 3 , FM1 , TM 3  ,
 MC3 , BM1 , FM 1 , TM 3  ,  MC1 , BM 3 , FM 3 , TM 1  ,  MC3 , BM1 , FM 3 , TM 1  ,
 MC3 , BM 3 , FM 1 , TM 1  ,  MC1 , BM 3 , FM 3 , TM 3  ,  MC3 , BM 3 , FM 1 , TM 3  ,
 MC3 , BM 3 , FM 3 , TM 1  ,  MC3 , BM 1 , FM 3 , TM 3  ,
 MC1 , BM 1 , FM 1 , TM 4  ,  MC1 , BM1 , FM 4 , TM 1  ,  MC1 , BM 4 , FM 1 , TM1  ,
 MC4 , BM1 , FM 1 , TM 1  ,  MC1 , BM1 , FM 4 , TM 4  ,  MC1 , BM 4 , FM 1 , TM 4  ,
 MC4 , BM1 , FM 1 , TM 4  ,  MC1 , BM 4 , FM 4 , TM 1  ,  MC4 , BM 1 , FM 4 , TM1  ,
 MC4 , BM 4 , FM1 , TM 1  ,  MC1 , BM 4 , FM 4 , TM 4  ,  MC4 , BM 4 , FM 1 , TM 4  ,
 MC4 , BM 4 , FM 4 , TM 1  ,  MC4 , BM1 , FM 4 , TM 4  ,
 MC2 , BM 2 , FM 2 , TM 3  ,  MC2 , BM 2 , FM 3 , TM 2  ,  MC2 , BM 3 , FM 2 , TM 2  ,
 MC3 , BM 2 , FM 2 , TM 2  ,  MC2 , BM 2 , FM 3 , TM 3  ,  MC2 , BM 3 , FM 2 , TM 3  ,
 MC3 , BM 2 , FM 2 , TM 3  ,  MC2 , BM 3 , FM 3 , TM 2  ,  MC3 , BM 2 , FM 3 , TM 2  ,
 MC3 , BM 3 , FM 2 , TM 2  ,  MC2 , BM 3 , FM 3 , TM 3  ,  MC3 , BM 3 , FM 3 , TM 2  ,
 MC3 , BM 3 , FM 2 , TM 3  ,  MC3 , BM 2 , FM 3 , TM 3  ,
6
52
 MC2 , BM 2 , FM 2 , TM 4  ,  MC2 , BM 2 , FM 4 , TM 2  ,  MC2 , BM 4 , FM 2 , TM 2  ,
 MC4 , BM 2 , FM 2 , TM 2  ,  MC2 , BM 2 , FM 4 , TM 4  ,  MC2 , BM 4 , FM 2 , TM 4  ,
 MC4 , BM 2 , FM 2 , TM 4  ,  MC2 , BM 4 , FM 4 , TM 2  ,  MC4 , BM 2 , FM 4 , TM 2  ,
 MC4 , BM 4 , FM 2 , TM 2  ,  MC2 , BM 4 , FM 4 , TM 4  ,  MC4 , BM 4 , FM 4 , TM 2  ,
 MC4 , BM 4 , FM 2 , TM 4  ,  MC4 , BM 2 , FM 4 , TM 4  ,
 MC3 , BM 3 , FM 3 , TM 4  ,  MC3 , BM 3 , FM 4 , TM 3  ,  MC3 , BM 4 , FM 3 , TM 3  ,
 MC4 , BM 3 , FM 3 , TM 3  ,  MC3 , BM 3 , FM 4 , TM 4  ,  MC3 , BM 4 , FM 3 , TM 4  ,
 MC4 , BM 3 , FM 3 , TM 4  ,  MC3 , BM 4 , FM 4 , TM 3  ,  MC4 , BM 3 , FM 4 , TM 3  ,
 MC4 , BM 4 , FM 3 , TM 3  ,  MC3 , BM 4 , FM 4 , TM 4  ,  MC4 , BM 4 , FM 4 , TM 3  ,
 MC4 , BM 4 , FM 3 , TM 4  ,  MC4 , BM 3 , FM 4 , TM 4  ,
84 new elements alltogether.
FTTM5
MC1
MC2
MC3
MC4
MC5
BM5
BM4
BM3
BM2
TM1
FM1
BM1
TM2
FM2
TM3
FM3
TM4
FM4
TM5
FM5
 MC1 , BM 1 , FM 1 , TM 2  ,  MC1 , BM1 , FM 2 , TM 1  ,  MC1 , BM 2 , FM 1 , TM1  ,
 MC2 , BM1 , FM 1 , TM 1  ,  MC1 , BM1 , FM 2 , TM 2  ,  MC1 , BM 2 , FM 1 , TM 2  ,
 MC2 , BM1 , FM 1 , TM 2  ,  MC1 , BM 2 , FM 2 , TM 1  ,  MC2 , BM 1 , FM 2 , TM1  ,
 MC2 , BM 2 , FM1 , TM 1  ,  MC1 , BM 2 , FM 2 , TM 2  ,  MC2 , BM 2 , FM 1 , TM 2  ,
 MC2 , BM 2 , FM 2 , TM 1  ,  MC2 , BM1 , FM 2 , TM 2  ,
10
53
 MC1 , BM 1 , FM1 , TM 3  ,  MC1 , BM1 , FM 3 , TM 1  ,  MC1 , BM 3 , FM 1 , TM1  ,
 MC3 , BM1 , FM 1 , TM 1  ,  MC1 , BM1 , FM 3 , TM 3  ,  MC1 , BM 3 , FM1 , TM 3  ,
 MC3 , BM1 , FM 1 , TM 3  ,  MC1 , BM 3 , FM 3 , TM 1  ,  MC3 , BM1 , FM 3 , TM 1  ,
 MC3 , BM 3 , FM 1 , TM 1  ,  MC1 , BM 3 , FM 3 , TM 3  ,  MC3 , BM 3 , FM 1 , TM 3  ,
 MC3 , BM 3 , FM 3 , TM 1  ,  MC3 , BM 1 , FM 3 , TM 3  ,
 MC1 , BM 1 , FM 1 , TM 4  ,  MC1 , BM1 , FM 4 , TM 1  ,  MC1 , BM 4 , FM 1 , TM1  ,
 MC4 , BM1 , FM 1 , TM 1  ,  MC1 , BM1 , FM 4 , TM 4  ,  MC1 , BM 4 , FM 1 , TM 4  ,
 MC4 , BM1 , FM 1 , TM 4  ,  MC1 , BM 4 , FM 4 , TM 1  ,  MC4 , BM 1 , FM 4 , TM1  ,
 MC4 , BM 4 , FM1 , TM 1  ,  MC1 , BM 4 , FM 4 , TM 4  ,  MC4 , BM 4 , FM 1 , TM 4  ,
 MC4 , BM 4 , FM 4 , TM 1  ,  MC4 , BM1 , FM 4 , TM 4  ,
 MC2 , BM 2 , FM 2 , TM 3  ,  MC2 , BM 2 , FM 3 , TM 2  ,  MC2 , BM 3 , FM 2 , TM 2  ,
 MC3 , BM 2 , FM 2 , TM 2  ,  MC2 , BM 2 , FM 3 , TM 3  ,  MC2 , BM 3 , FM 2 , TM 3  ,
 MC3 , BM 2 , FM 2 , TM 3  ,  MC2 , BM 3 , FM 3 , TM 2  ,  MC3 , BM 2 , FM 3 , TM 2  ,
 MC3 , BM 3 , FM 2 , TM 2  ,  MC2 , BM 3 , FM 3 , TM 3  ,  MC3 , BM 3 , FM 3 , TM 2  ,
 MC3 , BM 3 , FM 2 , TM 3  ,  MC3 , BM 2 , FM 3 , TM 3  ,
 MC2 , BM 2 , FM 2 , TM 4  ,  MC2 , BM 2 , FM 4 , TM 2  ,  MC2 , BM 4 , FM 2 , TM 2  ,
 MC4 , BM 2 , FM 2 , TM 2  ,  MC2 , BM 2 , FM 4 , TM 4  ,  MC2 , BM 4 , FM 2 , TM 4  ,
 MC4 , BM 2 , FM 2 , TM 4  ,  MC2 , BM 4 , FM 4 , TM 2  ,  MC4 , BM 2 , FM 4 , TM 2  ,
 MC4 , BM 4 , FM 2 , TM 2  ,  MC2 , BM 4 , FM 4 , TM 4  ,  MC4 , BM 4 , FM 4 , TM 2  ,
 MC4 , BM 4 , FM 2 , TM 4  ,  MC4 , BM 2 , FM 4 , TM 4  ,
 MC3 , BM 3 , FM 3 , TM 4  ,  MC3 , BM 3 , FM 4 , TM 3  ,  MC3 , BM 4 , FM 3 , TM 3  ,
 MC4 , BM 3 , FM 3 , TM 3  ,  MC3 , BM 3 , FM 4 , TM 4  ,  MC3 , BM 4 , FM 3 , TM 4  ,
 MC4 , BM 3 , FM 3 , TM 4  ,  MC3 , BM 4 , FM 4 , TM 3  ,  MC4 , BM 3 , FM 4 , TM 3  ,
 MC4 , BM 4 , FM 3 , TM 3  ,  MC3 , BM 4 , FM 4 , TM 4  ,  MC4 , BM 4 , FM 4 , TM 3  ,
 MC4 , BM 4 , FM 3 , TM 4  ,  MC4 , BM 3 , FM 4 , TM 4  ,
 MC1 , BM 1 , FM1 , TM 5  ,  MC1 , BM 1 , FM 5 , TM 1  ,  MC1 , BM 5 , FM1 , TM 1  ,
 MC5 , BM 1 , FM 1 , TM 1  ,  MC1 , BM 1 , FM 5 , TM 5  ,  MC1 , BM 5 , FM1 , TM 5  ,
 MC5 , BM 1 , FM 1 , TM 5  ,  MC1 , BM 5 , FM 5 , TM 1  ,  MC5 , BM 1 , FM 5 , TM 1  ,
 MC5 , BM 5 , FM 1 , TM 1  ,  MC1 , BM 5 , FM 5 , TM 5  ,  MC5 , BM 5 , FM 5 , TM 1  ,
 MC5 , BM 5 , FM 1 , TM 5  ,  MC5 , BM 1 , FM 5 , TM 5  ,
54
 MC2 , BM 2 , FM 2 , TM 5  ,  MC2 , BM 2 , FM 5 , TM 2  ,  MC2 , BM 5 , FM 2 , TM 2  ,
 MC5 , BM 2 , FM 2 , TM 2  ,  MC2 , BM 2 , FM 5 , TM 5  ,  MC2 , BM 5 , FM 2 , TM 5  ,
 MC5 , BM 2 , FM 2 , TM 5  ,  MC2 , BM 5 , FM 5 , TM 2  ,  MC5 , BM 2 , FM 5 , TM 2  ,
 MC5 , BM 5 , FM 2 , TM 2  ,  MC2 , BM 5 , FM 5 , TM 5  ,  MC5 , BM 5 , FM 5 , TM 2  ,
 MC5 , BM 5 , FM 2 , TM 5  ,  MC5 , BM 2 , FM 5 , TM 5  ,
 MC3 , BM 3 , FM 3 , TM 5  ,  MC3 , BM 3 , FM 5 , TM 3  ,  MC3 , BM 5 , FM 3 , TM 3  ,
 MC5 , BM 3 , FM 3 , TM 3  ,  MC3 , BM 3 , FM 5 , TM 5  ,  MC3 , BM 5 , FM 3 , TM 5  ,
 MC5 , BM 3 , FM 3 , TM 5  ,  MC3 , BM 5 , FM 5 , TM 3  ,  MC5 , BM 3 , FM 5 , TM 3  ,
 MC5 , BM 5 , FM 3 , TM 3  ,  MC3 , BM 5 , FM 5 , TM 5  ,  MC5 , BM 5 , FM 5 , TM 3  ,
 MC5 , BM 5 , FM 3 , TM 5  ,  MC5 , BM 3 , FM 5 , TM 5  ,
 MC4 , BM 4 , FM 4 , TM 5  ,  MC4 , BM 4 , FM 5 , TM 4  ,  MC4 , BM 5 , FM 4 , TM 4  ,
 MC5 , BM 4 , FM 4 , TM 4  ,  MC4 , BM 4 , FM 5 , TM 5  ,  MC4 , BM 5 , FM 4 , TM 5  ,
 MC5 , BM 4 , FM 4 , TM 5  ,  MC4 , BM 5 , FM 5 , TM 4  ,  MC5 , BM 4 , FM 5 , TM 4  ,
 MC5 , BM 5 , FM 4 , TM 4  ,  MC4 , BM 5 , FM 5 , TM 5  ,  MC5 , BM 5 , FM 5 , TM 4  ,
 MC5 , BM 5 , FM 4 , TM 5  ,  MC5 , BM 4 , FM 5 , TM 5  ,
140 new elements alltogether.
A combination of two FTTM will produce another 14 new elements.
Therefore, Definition 4.4 (sequence of cubes in FTTMn) can be rewritten as follows
and the definition is illustrated in Example 4.2.
Definition 4.7 (A Sequence of FTTM2/n)
FTTM2/n means the number of cubes produced by the combination of any two terms
FTTM in FTTMn with FTTM2/1 = 0. Hence, FTTM2/2 = 1, FTTM2/3 = 3, FTTM2/4 =
6 and in general;
FTTM2/n = FTTM2/n-1+ (n-1) for all n  1 .
55
Example 4.2 FTTM2/4
We can see that;
MC1
TM1
FTTM2/4= FTTM2/3 + 3
FTTM2/3= FTTM2/2 +2
BM1
MC2
FM1
TM2
FTTM2/2= FTTM2/1 +1
1
BM2
MC3
FM2
TM3
Therefore,
2
FTTM2/2= 0+1=1
BM3
MC4
4
FM3
TM4
FTTM2/3=1+2=3
3
FTTM2/4=3+3=6
BM4
5
6
FM4
Figure 4.2 FTTM2/4
That means, FTTM2/4 have six cubes as shown in Figure 4.2. They are;
1
1-FTTM4 and 2-FTTM4
4
1-FTTM4 and 3-FTTM4
2
2-FTTM4 and 3-FTTM4
5
2-FTTM4 and 4-FTTM4
3
3-FTTM4 and 4-FTTM4
6
1-FTTM4 and 4-FTTM4
Each cube will generate another 14 new elements of FTTM. For example cube
1
is the combination of 1-FTTM4 and 2-FTTM4.
 MC1, BM1, FM1,TM2  ,  MC1, BM1, FM2 ,TM1  ,  MC1, BM2 , FM1,TM1  ,  MC2 , BM1, FM1,TM1  ,
 MC1, BM1, FM2 ,TM2  ,  MC1, BM2 , FM1,TM2  ,  MC2 , BM1, FM1,TM2  ,  MC1, BM2 , FM2 ,TM1  ,
 MC2 , BM1, FM2 ,TM1  ,  MC2 , BM2 , FM1,TM1  ,  MC1, BM2 , FM2 ,TM2  ,  MC2 , BM2 , FM1,TM2  ,
 MC2 , BM2 , FM2 ,TM1  ,  MC2 , BM1, FM2 ,TM2  ,
56
Definition 4.7 introduces the sequence of FTTM2/n. The initial value for the
sequence is 0 which mean there is no cube in FTTM1. It follows by 1, 3, 6, 10 as
shown in Table 4.4. Table 4.4 reveals the number of elements that can be generated
from the combination of two FTTM in FTTMn for n=1 to10.
Table 4.4 Samples of sequence of FTTM2/n
FTTMn
FTTM2/n
14 FTTM2/n
FTTM1
0
0
FTTM2
1
14
FTTM3
3
42
FTTM4
6
84
FTTM5
10
140
FTTM6
15
210
FTTM7
21
294
FTTM8
28
392
FTTM9
36
504
FTTM10
45
630
57
4.6
“Extended Cubes” in FTTMn
Cubes can also be produced from the combination of three FTTM in FTTMn.
Table 4.5 demonstrates the characteristics of cubes in FTTMn if n= 1, 2, 3, 4 and 5.
The number of cubes that can be obtained from the combination of three FTTM in
FTTMn is different to cubes generated from two versions FTTM. Thus, a
combination of three terms of a sequence of FTTM will generate new sequence of
cubes as given in Table 4.5.
Table 4.5 Cube with three terms FTTM in FTTMn
FTTMn
Generated FTTM
FTTM1
MC1
TM1
BM1
FM1
Cubes
0
FTTM2
0
MC1
MC2
BM2
BM1
TM1
TM2
FM2
FM1
58
FTTM3
MC1
BM1
MC2
1
TM1
FM1
TM2
BM2
MC3
TM3
BM3
FM3
FM2
 MC1 , BM 1 , FM 2 , TM 3  ,  MC2 , BM1 , FM 2 , TM 3  ,  MC3 , BM 1 , FM 2 , TM 3  ,
 MC1 , BM 2 , FM 2 , TM 3  ,  MC1 , BM 2 , FM 2 , TM 3  ,  MC1 , BM 1 , FM 3 , TM 2  ,
 MC2 , BM1 , FM 3 , TM 2  ,  MC3 , BM 1 , FM 3 , TM 2  ,  MC1 , BM 2 , FM 3 , TM 2  ,
 MC1 , BM 3 , FM 3 , TM 2  ,  MC1 , BM 2 , FM 3 , TM1  ,  MC2 , BM 2 , FM 3 , TM 1  ,
 MC3 , BM 2 , FM 3 , TM 1  ,  MC1 , BM 2 , FM 3 , TM 3  ,  MC1 , BM 3 , FM 2 , TM 1  ,
 MC2 , BM 3 , FM 2 , TM1  ,  MC3 , BM 3 , FM 2 , TM 1  ,  MC1 , BM 3 , FM 2 , TM 2  ,
 MC2 , BM 3 , FM1 , TM 1  ,  MC2 , BM 3 , FM 3 , TM1  ,  MC2 , BM 3 , FM 1 , TM 2  ,
 MC2 , BM 3 , FM1 , TM 3  ,  MC3 , BM 2 , FM 1 , TM1  ,  MC3 , BM 2 , FM 2 , TM 1  ,
 MC3 , BM 2 , FM1 , TM 2  ,  MC3 , BM 2 , FM 1 , TM 3  ,  MC2 , BM 1 , FM 1 , TM 3  ,
 MC2 , BM 2 , FM1 , TM 3  ,  MC2 , BM 1 , FM 3 , TM 3  ,  MC3 , BM 1 , FM 1 , TM 2  ,
 MC3 , BM 3 , FM 1 , TM 2  ,  MC3 , BM 1 , FM 2 , TM 2  ,  MC1 , BM 2 , FM 1 , TM 3  ,
 MC1 , BM 3 , FM 1 , TM 2  ,  MC2 , BM1 , FM 3 , TM 1  ,  MC3 , BM1 , FM 2 , TM 1  ,
36 new elements alltogether.
FTTM4
MC1
TM1
BM1
MC2
BM2
MC3
BM3
MC4
TM4
BM4
FM4
FM1
TM2
TM3
FM3
FM2
4
59
 MC1 , BM 1 , FM 2 , TM 3  ,  MC2 , BM1 , FM 2 , TM 3  ,  MC3 , BM 1 , FM 2 , TM 3  ,
 MC1 , BM 2 , FM 2 , TM 3  ,  MC1 , BM 2 , FM 2 , TM 3  ,  MC1 , BM 1 , FM 3 , TM 2  ,
 MC2 , BM1 , FM 3 , TM 2  ,  MC3 , BM 1 , FM 3 , TM 2  ,  MC1 , BM 2 , FM 3 , TM 2  ,
 MC1 , BM 3 , FM 3 , TM 2  ,  MC1 , BM 2 , FM 3 , TM1  ,  MC2 , BM 2 , FM 3 , TM 1  ,
 MC3 , BM 2 , FM 3 , TM 1  ,  MC1 , BM 2 , FM 3 , TM 3  ,  MC1 , BM 3 , FM 2 , TM 1  ,
 MC2 , BM 3 , FM 2 , TM1  ,  MC3 , BM 3 , FM 2 , TM 1  ,  MC1 , BM 3 , FM 2 , TM 2  ,
 MC2 , BM 3 , FM1 , TM 1  ,  MC2 , BM 3 , FM 3 , TM1  ,  MC2 , BM 3 , FM 1 , TM 2  ,
 MC2 , BM 3 , FM1 , TM 3  ,  MC3 , BM 2 , FM 1 , TM1  ,  MC3 , BM 2 , FM 2 , TM 1  ,
 MC3 , BM 2 , FM1 , TM 2  ,  MC3 , BM 2 , FM 1 , TM 3  ,  MC2 , BM 1 , FM 1 , TM 3  ,
 MC2 , BM 2 , FM1 , TM 3  ,  MC2 , BM 1 , FM 3 , TM 3  ,  MC3 , BM 1 , FM 1 , TM 2  ,
 MC3 , BM 3 , FM 1 , TM 2  ,  MC3 , BM 1 , FM 2 , TM 2  ,  MC1 , BM 2 , FM 1 , TM 3  ,
 MC1 , BM 3 , FM 1 , TM 2  ,  MC2 , BM1 , FM 3 , TM 1  ,  MC3 , BM1 , FM 2 , TM 1  ,
 MC1 , BM 1 , FM 2 , TM 4  ,  MC2 , BM 1 , FM 2 , TM 4  ,  MC4 , BM 1 , FM 2 , TM 4  ,
 MC1 , BM 2 , FM 2 , TM 4  ,  MC1 , BM 4 , FM 2 , TM 4  ,  MC1 , BM 1 , FM 4 , TM 2  ,
 MC2 , BM1 , FM 4 , TM 2  ,  MC4 , BM1 , FM 4 , TM 2  ,  MC1 , BM 2 , FM 4 , TM 2  ,
 MC1 , BM 4 , FM 4 , TM 2  ,  MC1 , BM 2 , FM 4 , TM 1  ,  MC2 , BM 2 , FM 4 , TM 1  ,
 MC4 , BM 2 , FM 4 , TM 1  ,  MC1 , BM 2 , FM 4 , TM 4  ,  MC1 , BM 4 , FM 2 , TM 1  ,
 MC2 , BM 4 , FM 2 , TM 1  ,  MC4 , BM 4 , FM 2 , TM 1  ,  MC1 , BM 4 , FM 2 , TM 2  ,
 MC2 , BM 4 , FM1 , TM 1  ,  MC2 , BM 4 , FM 4 , TM 1  ,  MC2 , BM 4 , FM 1 , TM 2  ,
 MC2 , BM 4 , FM1 , TM 4  ,  MC4 , BM 2 , FM1 , TM 1  ,  MC4 , BM 2 , FM 2 , TM 1  ,
 MC4 , BM 2 , FM1 , TM 2  ,  MC4 , BM 2 , FM1 , TM 4  ,  MC2 , BM1 , FM 1 , TM 4  ,
 MC2 , BM 2 , FM1 , TM 4  ,  MC2 , BM1 , FM 4 , TM 4  ,  MC4 , BM1 , FM 1 , TM 2  ,
 MC4 , BM 4 , FM1 , TM 2  ,  MC4 , BM1 , FM 2 , TM 2  ,  MC1 , BM 2 , FM 1 , TM 4  ,
 MC1 , BM 4 , FM 1 , TM 2  ,  MC2 , BM 1 , FM 4 , TM 1  ,  MC4 , BM 1 , FM 2 , TM 1  ,
60
 MC1 , BM 1 , FM 3 , TM 4  ,  MC3 , BM 1 , FM 3 , TM 4  ,  MC4 , BM 1 , FM 3 , TM 4  ,
 MC1 , BM 3 , FM 3 , TM 4  ,  MC1 , BM 4 , FM 3 , TM 4  ,  MC1 , BM 1 , FM 4 , TM 3  ,
 MC3 , BM 1 , FM 4 , TM 3  ,  MC4 , BM 1 , FM 4 , TM 3  ,  MC1 , BM 3 , FM 4 , TM 3  ,
 MC1 , BM 4 , FM 4 , TM 3  ,  MC1 , BM 3 , FM 4 , TM1  ,  MC3 , BM 3 , FM 4 , TM 1  ,
 MC4 , BM 3 , FM 4 , TM1  ,  MC1 , BM 3 , FM 4 , TM 4  ,  MC1 , BM 4 , FM 3 , TM 1  ,
 MC3 , BM 4 , FM 3 , TM 1  ,  MC4 , BM 4 , FM 3 , TM 1  ,  MC1 , BM 4 , FM 3 , TM 3  ,
 MC3 , BM 4 , FM 1 , TM 1  ,  MC3 , BM 4 , FM 4 , TM1  ,  MC3 , BM 4 , FM 1 , TM 3  ,
 MC3 , BM 4 , FM 1 , TM 4  ,  MC4 , BM 3 , FM 1 , TM1  ,  MC4 , BM 3 , FM 3 , TM 1  ,
 MC4 , BM 3 , FM 1 , TM 3  ,  MC4 , BM 3 , FM 1 , TM 4  ,  MC3 , BM1 , FM 1 , TM 4  ,
 MC3 , BM 3 , FM 1 , TM 4  ,  MC3 , BM 1 , FM 4 , TM 4  ,  MC4 , BM 1 , FM 1 , TM 3  ,
 MC4 , BM 4 , FM1 , TM 3  ,  MC4 , BM 1 , FM 3 , TM 3  ,  MC1 , BM 3 , FM 1 , TM 4  ,
 MC1 , BM 4 , FM 1 , TM 3  ,  MC3 , BM 1 , FM 4 , TM 1  ,  MC4 , BM1 , FM 3 , TM 1  ,
 MC2 , BM 2 , FM 3 , TM 4  ,  MC3 , BM 2 , FM 3 , TM 4  ,  MC4 , BM 2 , FM 3 , TM 4  ,
 MC2 , BM 3 , FM 3 , TM 4  ,  MC2 , BM 4 , FM 3 , TM 4  ,  MC2 , BM 2 , FM 4 , TM 3  ,
 MC3 , BM 2 , FM 4 , TM 3  ,  MC4 , BM 2 , FM 4 , TM 3  ,  MC2 , BM 3 , FM 4 , TM 3  ,
 MC2 , BM 4 , FM 4 , TM 3  ,  MC2 , BM 3 , FM 4 , TM 4  ,  MC2 , BM 3 , FM 4 , TM 2  ,
 MC3 , BM 3 , FM 4 , TM 2  ,  MC4 , BM 3 , FM 4 , TM 2  ,  MC2 , BM 4 , FM 3 , TM 2  ,
 MC3 , BM 4 , FM 3 , TM 2  ,  MC4 , BM 4 , FM 3 , TM 2  ,  MC2 , BM 4 , FM 3 , TM 3  ,
 MC3 , BM 4 , FM 2 , TM 2  ,  MC3 , BM 4 , FM 4 , TM 2  ,  MC3 , BM 4 , FM 2 , TM 3  ,
 MC3 , BM 4 , FM 2 , TM 4  ,  MC4 , BM 3 , FM 2 , TM 2  ,  MC4 , BM 3 , FM 3 , TM 2  ,
 MC4 , BM 3 , FM 2 , TM 3  ,  MC4 , BM 3 , FM 2 , TM 4  ,  MC3 , BM 2 , FM 2 , TM 4  ,
 MC3 , BM 3 , FM 2 , TM 4  ,  MC3 , BM 2 , FM 4 , TM 4  ,  MC4 , BM 2 , FM 2 , TM 3  ,
 MC4 , BM 4 , FM 2 , TM 3  ,  MC4 , BM 2 , FM 3 , TM 3  ,  MC2 , BM 3 , FM 2 , TM 4  ,
 MC2 , BM 4 , FM 2 , TM 3  ,  MC3 , BM 2 , FM 4 , TM 2  ,  MC4 , BM 2 , FM 3 , TM 2  ,
144 new elements altogether.
61
FTTM5
10
MC1
TM1
FM1
BM1
MC2
TM2
BM2
MC3
BM3
MC4
FM2
TM3
FM3
TM4
BM4
FM4
MC5
TM5
BM5
FM5
 MC1 , BM 1 , FM 2 , TM 3  ,  MC2 , BM1 , FM 2 , TM 3  ,  MC3 , BM 1 , FM 2 , TM 3  ,
 MC1 , BM 2 , FM 2 , TM 3  ,  MC1 , BM 3 , FM 2 , TM 3  ,  MC1 , BM 1 , FM 3 , TM 2  ,
 MC2 , BM1 , FM 3 , TM 2  ,  MC3 , BM 1 , FM 3 , TM 2  ,  MC1 , BM 2 , FM 3 , TM 2  ,
 MC1 , BM 3 , FM 3 , TM 2  ,  MC1 , BM 2 , FM 3 , TM1  ,  MC2 , BM 2 , FM 3 , TM 1  ,
 MC3 , BM 2 , FM 3 , TM 1  ,  MC1 , BM 2 , FM 3 , TM 3  ,  MC1 , BM 3 , FM 2 , TM 1  ,
 MC2 , BM 3 , FM 2 , TM1  ,  MC3 , BM 3 , FM 2 , TM 1  ,  MC1 , BM 3 , FM 2 , TM 2  ,
 MC2 , BM 3 , FM1 , TM 1  ,  MC2 , BM 3 , FM 3 , TM1  ,  MC2 , BM 3 , FM 1 , TM 2  ,
 MC2 , BM 3 , FM1 , TM 3  ,  MC3 , BM 2 , FM 1 , TM1  ,  MC3 , BM 2 , FM 2 , TM 1  ,
 MC3 , BM 2 , FM1 , TM 2  ,  MC3 , BM 2 , FM 1 , TM 3  ,  MC2 , BM 1 , FM 1 , TM 3  ,
 MC2 , BM 2 , FM1 , TM 3  ,  MC2 , BM 1 , FM 3 , TM 3  ,  MC3 , BM 1 , FM 1 , TM 2  ,
 MC3 , BM 3 , FM 1 , TM 2  ,  MC3 , BM 1 , FM 2 , TM 2  ,  MC1 , BM 2 , FM 1 , TM 3  ,
 MC1 , BM 3 , FM 1 , TM 2  ,  MC2 , BM1 , FM 3 , TM 1  ,  MC3 , BM1 , FM 2 , TM 1  ,
62
 MC1 , BM 1 , FM 2 , TM 4  ,  MC2 , BM 1 , FM 2 , TM 4  ,  MC4 , BM 1 , FM 2 , TM 4  ,
 MC1 , BM 2 , FM 2 , TM 4  ,  MC1 , BM 4 , FM 2 , TM 4  ,  MC1 , BM 1 , FM 4 , TM 2  ,
 MC2 , BM1 , FM 4 , TM 2  ,  MC4 , BM1 , FM 4 , TM 2  ,  MC1 , BM 2 , FM 4 , TM 2  ,
 MC1 , BM 4 , FM 4 , TM 2  ,  MC1 , BM 2 , FM 4 , TM 1  ,  MC2 , BM 2 , FM 4 , TM 1  ,
 MC4 , BM 2 , FM 4 , TM 1  ,  MC1 , BM 2 , FM 4 , TM 4  ,  MC1 , BM 4 , FM 2 , TM 1  ,
 MC2 , BM 4 , FM 2 , TM 1  ,  MC4 , BM 4 , FM 2 , TM 1  ,  MC1 , BM 4 , FM 2 , TM 2  ,
 MC2 , BM 4 , FM1 , TM 1  ,  MC2 , BM 4 , FM 4 , TM 1  ,  MC2 , BM 4 , FM 1 , TM 2  ,
 MC2 , BM 4 , FM1 , TM 4  ,  MC4 , BM 2 , FM1 , TM 1  ,  MC4 , BM 2 , FM 2 , TM 1  ,
 MC4 , BM 2 , FM1 , TM 2  ,  MC4 , BM 2 , FM1 , TM 4  ,  MC2 , BM1 , FM 1 , TM 4  ,
 MC2 , BM 2 , FM1 , TM 4  ,  MC2 , BM1 , FM 4 , TM 4  ,  MC4 , BM1 , FM 1 , TM 2  ,
 MC4 , BM 4 , FM1 , TM 2  ,  MC4 , BM1 , FM 2 , TM 2  ,  MC1 , BM 2 , FM 1 , TM 4  ,
 MC1 , BM 4 , FM 1 , TM 2  ,  MC2 , BM 1 , FM 4 , TM 1  ,  MC4 , BM 1 , FM 2 , TM 1  ,
 MC1 , BM 1 , FM 2 , TM 5  ,  MC2 , BM 1 , FM 2 , TM 5  ,  MC5 , BM1 , FM 2 , TM 5  ,
 MC1 , BM 2 , FM 2 , TM 5  ,  MC1 , BM 5 , FM 2 , TM 5  ,  MC1 , BM1 , FM 5 , TM 2  ,
 MC2 , BM1 , FM 5 , TM 2  ,  MC5 , BM 1 , FM 5 , TM 2  ,  MC1 , BM 2 , FM 5 , TM 2  ,
 MC1 , BM 5 , FM 5 , TM 2  ,  MC1 , BM 2 , FM 5 , TM 1  ,  MC2 , BM 2 , FM 5 , TM 1  ,
 MC5 , BM 2 , FM 5 , TM 1  ,  MC1 , BM 2 , FM 5 , TM 5  ,  MC1 , BM 5 , FM 2 , TM 1  ,
 MC2 , BM 5 , FM 2 , TM 1  ,  MC5 , BM 5 , FM 2 , TM1  ,  MC1 , BM 5 , FM 2 , TM 2  ,
 MC2 , BM 5 , FM 1 , TM 1  ,  MC2 , BM 5 , FM 5 , TM 1  ,  MC2 , BM 5 , FM 1 , TM 2  ,
 MC2 , BM 5 , FM 1 , TM 5  ,  MC5 , BM 2 , FM1 , TM 1  ,  MC5 , BM 2 , FM 2 , TM 1  ,
 MC5 , BM 2 , FM 1 , TM 2  ,  MC5 , BM 2 , FM 1 , TM 5  ,  MC2 , BM 1 , FM 1 , TM 5  ,
 MC2 , BM 2 , FM1 , TM 5  ,  MC2 , BM 1 , FM 5 , TM 5  ,  MC5 , BM1 , FM 1 , TM 2  ,
 MC5 , BM 5 , FM 1 , TM 2  ,  MC5 , BM 1 , FM 2 , TM 2  ,  MC1 , BM 2 , FM 1 , TM 5  ,
 MC1 , BM 5 , FM 1 , TM 2  ,  MC2 , BM 1 , FM 5 , TM 1  ,  MC5 , BM 1 , FM 2 , TM1  ,
63
 MC1 , BM 1 , FM 3 , TM 4  ,  MC3 , BM 1 , FM 3 , TM 4  ,  MC4 , BM 1 , FM 3 , TM 4  ,
 MC1 , BM 3 , FM 3 , TM 4  ,  MC1 , BM 4 , FM 3 , TM 4  ,  MC1 , BM 1 , FM 4 , TM 3  ,
 MC3 , BM 1 , FM 4 , TM 3  ,  MC4 , BM 1 , FM 4 , TM 3  ,  MC1 , BM 3 , FM 4 , TM 3  ,
 MC1 , BM 4 , FM 4 , TM 3  ,  MC1 , BM 3 , FM 4 , TM1  ,  MC3 , BM 3 , FM 4 , TM 1  ,
 MC4 , BM 3 , FM 4 , TM1  ,  MC1 , BM 3 , FM 4 , TM 4  ,  MC1 , BM 4 , FM 3 , TM 1  ,
 MC3 , BM 4 , FM 3 , TM 1  ,  MC4 , BM 4 , FM 3 , TM 1  ,  MC1 , BM 4 , FM 3 , TM 3  ,
 MC3 , BM 4 , FM 1 , TM 1  ,  MC3 , BM 4 , FM 4 , TM1  ,  MC3 , BM 4 , FM 1 , TM 3  ,
 MC3 , BM 4 , FM 1 , TM 4  ,  MC4 , BM 3 , FM 1 , TM1  ,  MC4 , BM 3 , FM 3 , TM 1  ,
 MC4 , BM 3 , FM 1 , TM 3  ,  MC4 , BM 3 , FM 1 , TM 4  ,  MC3 , BM1 , FM 1 , TM 4  ,
 MC3 , BM 3 , FM 1 , TM 4  ,  MC3 , BM 1 , FM 4 , TM 4  ,  MC4 , BM 1 , FM 1 , TM 3  ,
 MC4 , BM 4 , FM1 , TM 3  ,  MC4 , BM 1 , FM 3 , TM 3  ,  MC1 , BM 3 , FM 1 , TM 4  ,
 MC1 , BM 4 , FM 1 , TM 3  ,  MC3 , BM 1 , FM 4 , TM 1  ,  MC4 , BM1 , FM 3 , TM 1  ,
 MC1 , BM 1 , FM 3 , TM 5  ,  MC3 , BM 1 , FM 3 , TM 5  ,  MC5 , BM 1 , FM 3 , TM 5  ,
 MC1 , BM 3 , FM 3 , TM 5  ,  MC1 , BM 5 , FM 3 , TM 5  ,  MC1 , BM 1 , FM 5 , TM 3  ,
 MC1 , BM 5 , FM 5 , TM 3  ,  MC5 , BM1 , FM 5 , TM 3  ,  MC1 , BM 3 , FM 5 , TM 3  ,
 MC1 , BM 5 , FM 5 , TM 3  ,  MC1 , BM 3 , FM 5 , TM 1  ,  MC3 , BM 3 , FM 5 , TM 1  ,
 MC5 , BM 3 , FM 5 , TM 1  ,  MC1 , BM 3 , FM 5 , TM 5  ,  MC1 , BM 5 , FM 3 , TM1  ,
 MC3 , BM 5 , FM 3 , TM 1  ,  MC5 , BM 5 , FM 3 , TM 1  ,  MC1 , BM 5 , FM 3 , TM 3  ,
 MC3 , BM 5 , FM 1 , TM 1  ,  MC3 , BM 5 , FM 5 , TM 1  ,  MC3 , BM 5 , FM 1 , TM 3  ,
 MC3 , BM 5 , FM 1 , TM 5  ,  MC5 , BM 3 , FM1 , TM 1  ,  MC5 , BM 3 , FM 3 , TM 1  ,
 MC5 , BM 3 , FM 1 , TM 3  ,  MC5 , BM 3 , FM 1 , TM 5  ,  MC3 , BM 1 , FM1 , TM 5  ,
 MC3 , BM 3 , FM 1 , TM 5  ,  MC3 , BM 3 , FM1 , TM 5  ,  MC5 , BM 1 , FM1 , TM 3  ,
 MC5 , BM 5 , FM 1 , TM 3  ,  MC5 , BM1 , FM 3 , TM 3  ,  MC1 , BM 3 , FM1 , TM 5  ,
 MC1 , BM 5 , FM 1 , TM 3  ,  MC3 , BM 1 , FM 5 , TM 1  ,  MC5 , BM 1 , FM 3 , TM1  ,
64
 MC1 , BM 1 , FM 4 , TM 5  ,  MC4 , BM 1 , FM 4 , TM 5  ,  MC5 , BM1 , FM 4 , TM 5  ,
 MC1 , BM 4 , FM 4 , TM 5  ,  MC1 , BM 5 , FM 4 , TM 5  ,  MC1 , BM1 , FM 5 , TM 4  ,
 MC4 , BM1 , FM 5 , TM 4  ,  MC5 , BM 1 , FM 5 , TM 4  ,  MC1 , BM 4 , FM 5 , TM 4  ,
 MC1 , BM 5 , FM 5 , TM 4  ,  MC1 , BM 4 , FM 5 , TM 1  ,  MC4 , BM 4 , FM 5 , TM 1  ,
 MC5 , BM 4 , FM 5 , TM 1  ,  MC1 , BM 4 , FM 5 , TM 5  ,  MC1 , BM 5 , FM 4 , TM 1  ,
 MC4 , BM 5 , FM 4 , TM 1  ,  MC5 , BM 5 , FM 4 , TM1  ,  MC1 , BM 5 , FM 4 , TM 4  ,
 MC4 , BM 5 , FM 1 , TM 1  ,  MC4 , BM 5 , FM 5 , TM 1  ,  MC4 , BM 5 , FM 1 , TM 4  ,
 MC4 , BM 5 , FM 1 , TM 5  ,  MC5 , BM 4 , FM1 , TM 1  ,  MC5 , BM 4 , FM 4 , TM 1  ,
 MC5 , BM 4 , FM 1 , TM 4  ,  MC5 , BM 4 , FM 1 , TM 5  ,  MC4 , BM 1 , FM 1 , TM 5  ,
 MC4 , BM 4 , FM1 , TM 5  ,  MC4 , BM 1 , FM 5 , TM 5  ,  MC5 , BM1 , FM 1 , TM 4  ,
 MC5 , BM 5 , FM 1 , TM 4  ,  MC5 , BM 1 , FM 4 , TM 4  ,  MC1 , BM 4 , FM 1 , TM 5  ,
 MC1 , BM 5 , FM 1 , TM 4  ,  MC4 , BM 1 , FM 5 , TM 1  ,  MC5 , BM 1 , FM 4 , TM1  ,
 MC2 , BM 2 , FM 3 , TM 4  ,  MC3 , BM 2 , FM 3 , TM 4  ,  MC4 , BM 2 , FM 3 , TM 4  ,
 MC2 , BM 3 , FM 3 , TM 4  ,  MC2 , BM 4 , FM 3 , TM 4  ,  MC2 , BM 2 , FM 4 , TM 3  ,
 MC3 , BM 2 , FM 4 , TM 3  ,  MC4 , BM 2 , FM 4 , TM 3  ,  MC2 , BM 3 , FM 4 , TM 3  ,
 MC2 , BM 4 , FM 4 , TM 3  ,  MC2 , BM 3 , FM 4 , TM 4  ,  MC2 , BM 3 , FM 4 , TM 2  ,
 MC3 , BM 3 , FM 4 , TM 2  ,  MC4 , BM 3 , FM 4 , TM 2  ,  MC2 , BM 4 , FM 3 , TM 2  ,
 MC3 , BM 4 , FM 3 , TM 2  ,  MC4 , BM 4 , FM 3 , TM 2  ,  MC2 , BM 4 , FM 3 , TM 3  ,
 MC3 , BM 4 , FM 2 , TM 2  ,  MC3 , BM 4 , FM 4 , TM 2  ,  MC3 , BM 4 , FM 2 , TM 3  ,
 MC3 , BM 4 , FM 2 , TM 4  ,  MC4 , BM 3 , FM 2 , TM 2  ,  MC4 , BM 3 , FM 3 , TM 2  ,
 MC4 , BM 3 , FM 2 , TM 3  ,  MC4 , BM 3 , FM 2 , TM 4  ,  MC3 , BM 2 , FM 2 , TM 4  ,
 MC3 , BM 3 , FM 2 , TM 4  ,  MC3 , BM 2 , FM 4 , TM 4  ,  MC4 , BM 2 , FM 2 , TM 3  ,
 MC4 , BM 4 , FM 2 , TM 3  ,  MC4 , BM 2 , FM 3 , TM 3  ,  MC2 , BM 3 , FM 2 , TM 4  ,
 MC2 , BM 4 , FM 2 , TM 3  ,  MC3 , BM 2 , FM 4 , TM 2  ,  MC4 , BM 2 , FM 3 , TM 2  ,
65
 MC2 , BM 2 , FM 3 , TM 5  ,  MC3 , BM 2 , FM 3 , TM 5  ,  MC5 , BM 2 , FM 3 , TM 5  ,
 MC2 , BM 3 , FM 3 , TM 5  ,  MC2 , BM 5 , FM 3 , TM 5  ,  MC2 , BM 2 , FM 5 , TM 3  ,
 MC3 , BM 2 , FM 5 , TM 3  ,  MC5 , BM 2 , FM 5 , TM 3  ,  MC2 , BM 3 , FM 5 , TM 3  ,
 MC2 , BM 5 , FM 5 , TM 3  ,  MC2 , BM 3 , FM 5 , TM 5  ,  MC2 , BM 3 , FM 5 , TM 2  ,
 MC3 , BM 3 , FM 5 , TM 2  ,  MC5 , BM 3 , FM 5 , TM 2  ,  MC2 , BM 5 , FM 3 , TM 2  ,
 MC3 , BM 5 , FM 3 , TM 2  ,  MC5 , BM 5 , FM 3 , TM 2  ,  MC2 , BM 5 , FM 3 , TM 3  ,
 MC3 , BM 5 , FM 2 , TM 2  ,  MC3 , BM 5 , FM 5 , TM 2  ,  MC3 , BM 5 , FM 2 , TM 3  ,
 MC3 , BM 5 , FM 2 , TM 5  ,  MC5 , BM 3 , FM 2 , TM 2  ,  MC5 , BM 3 , FM 3 , TM 2  ,
 MC5 , BM 3 , FM 2 , TM 3  ,  MC5 , BM 3 , FM 2 , TM 5  ,  MC3 , BM 2 , FM 2 , TM 5  ,
 MC3 , BM 3 , FM 2 , TM 5  ,  MC3 , BM 2 , FM 5 , TM 5  ,  MC5 , BM 2 , FM 2 , TM 3  ,
 MC5 , BM 5 , FM 2 , TM 3  ,  MC5 , BM 2 , FM 3 , TM 3  ,  MC2 , BM 3 , FM 2 , TM 5  ,
 MC2 , BM 5 , FM 2 , TM 3  ,  MC3 , BM 2 , FM 5 , TM 2  ,  MC5 , BM 2 , FM 3 , TM 2  ,
 MC2 , BM 2 , FM 4 , TM 5  ,  MC4 , BM 2 , FM 4 , TM 5  ,  MC5 , BM 2 , FM 4 , TM 5  ,
 MC2 , BM 4 , FM 4 , TM 5  ,  MC2 , BM 5 , FM 4 , TM 5  ,  MC2 , BM 2 , FM 5 , TM 4  ,
 MC4 , BM 2 , FM 5 , TM 4  ,  MC5 , BM 2 , FM 5 , TM 4  ,  MC2 , BM 4 , FM 5 , TM 4  ,
 MC2 , BM 5 , FM 5 , TM 4  ,  MC2 , BM 4 , FM 5 , TM 5  ,  MC2 , BM 4 , FM 5 , TM 2  ,
 MC4 , BM 4 , FM 5 , TM 2  ,  MC5 , BM 4 , FM 5 , TM 2  ,  MC2 , BM 5 , FM 4 , TM 2  ,
 MC4 , BM 5 , FM 4 , TM 2  ,  MC5 , BM 5 , FM 4 , TM 2  ,  MC2 , BM 5 , FM 4 , TM 4  ,
 MC4 , BM 5 , FM 2 , TM 2  ,  MC4 , BM 5 , FM 5 , TM 2  ,  MC4 , BM 5 , FM 2 , TM 4  ,
 MC4 , BM 5 , FM 2 , TM 5  ,  MC5 , BM 4 , FM 2 , TM 2  ,  MC5 , BM 4 , FM 4 , TM 2  ,
 MC5 , BM 4 , FM 2 , TM 4  ,  MC5 , BM 4 , FM 2 , TM 5  ,  MC4 , BM 2 , FM 2 , TM 5  ,
 MC4 , BM 4 , FM 2 , TM 5  ,  MC4 , BM 2 , FM 5 , TM 5  ,  MC5 , BM 2 , FM 2 , TM 4  ,
 MC5 , BM 5 , FM 2 , TM 4  ,  MC5 , BM 2 , FM 4 , TM 4  ,  MC2 , BM 4 , FM 2 , TM 5  ,
 MC2 , BM 5 , FM 2 , TM 4  ,  MC4 , BM 2 , FM 4 , TM 5  ,  MC5 , BM 2 , FM 4 , TM 5  ,
66
 MC3 , BM 3 , FM 4 , TM 5  ,  MC4 , BM 3 , FM 4 , TM 5  ,  MC5 , BM 3 , FM 4 , TM 5  ,
 MC3 , BM 4 , FM 4 , TM 5  ,  MC3 , BM 5 , FM 4 , TM 5  ,  MC3 , BM 3 , FM 5 , TM 4  ,
 MC4 , BM 3 , FM 5 , TM 4  ,  MC5 , BM 3 , FM 5 , TM 4  ,  MC5 , BM 3 , FM 5 , TM 4  ,
 MC3 , BM 5 , FM 5 , TM 4  ,  MC3 , BM 4 , FM 5 , TM 5  ,  MC3 , BM 4 , FM 5 , TM 3  ,
 MC4 , BM 4 , FM 5 , TM 3  ,  MC5 , BM 4 , FM 5 , TM 3  ,  MC3 , BM 5 , FM 4 , TM 3  ,
 MC4 , BM 5 , FM 4 , TM 3  ,  MC5 , BM 5 , FM 4 , TM 3  ,  MC3 , BM 5 , FM 4 , TM 4  ,
 MC4 , BM 5 , FM 3 , TM 3  ,  MC4 , BM 5 , FM 5 , TM 3  ,  MC4 , BM 5 , FM 3 , TM 4  ,
 MC4 , BM 5 , FM 3 , TM 5  ,  MC5 , BM 4 , FM 3 , TM 3  ,  MC5 , BM 4 , FM 4 , TM 3  ,
 MC5 , BM 4 , FM 3 , TM 4  ,  MC5 , BM 4 , FM 3 , TM 5  ,  MC4 , BM 3 , FM 3 , TM 5  ,
 MC4 , BM 4 , FM 3 , TM 5  ,  MC4 , BM 3 , FM 5 , TM 5  ,  MC5 , BM 3 , FM 3 , TM 4  ,
 MC5 , BM 5 , FM 3 , TM 4  ,  MC5 , BM 3 , FM 4 , TM 4  ,  MC3 , BM 4 , FM 3 , TM 5  ,
 MC3 , BM 5 , FM 3 , TM 4  ,  MC4 , BM 3 , FM 5 , TM 3  ,  MC5 , BM 3 , FM 4 , TM 3  ,
360 new elements alltogether.
Definition 4.8 is developed to illustrate the sequence of FTTM3/n followed by
Example 4.3.
Definition 4.8 (A Sequence of FTTM3/n)
FTTM3/n means the number of cubes produced by the combination of any three terms
FTTM in FTTMn with FTTM3/1= 0, FTTM3/2= 0. Hence, FTTM3/3 = 1, FTTM3/4 = 4,
FTTM3/5 = 10 and in general;
FTTM3/n = FTTM3/n-1+ FTTM2/n-1 for all n  1 .
67
Example 4.3 FTTM3/4
We can see that;
FTTM3/4= FTTM3/3 + FTTM2/3
FTTM3/3= FTTM3/2 + FTTM2/2
Therefore,
FTTM3/3= 0+1=1
FTTM3/4=1+3=4
MC1
TM1
BM1
MC2
BM2
MC3
FM2
TM3
BM3
FM3
MC4
TM4
BM4
FM4
FM1
TM2
2
1
3
4
Figure 4.3 FTTM3/4
That means, FTTM3/4 have four cubes as shown in Figure 4.3. They are;
1
1-FTTM4, 2-FTTM4 and 3-FTTM4
2
2-FTTM4, 3-FTTM4 and 4-FTTM4
3
1-FTTM4, 2-FTTM4 and 4-FTTM4
4
1-FTTM4, 3-FTTM4 and 4-FTTM4
68
Each cube will generate another 36 elements of FTTM. For example cube
1
is the
combination of 1-FTTM4, 2-FTTM4 and 3-FTTM4.
 MC1, BM1, FM2 ,TM3  ,  MC2 , BM1, FM2 ,TM3  ,  MC3 , BM1, FM2 , TM3  ,  MC1, BM2 , FM2 , TM3  ,
 MC1, BM3, FM2 ,TM3  ,  MC1, BM1, FM3, TM2  ,  MC2 , BM1, FM3 , TM2  ,  MC3 , BM1, FM3 , TM2  ,
 MC1, BM2 , FM3 ,TM2  ,  MC1, BM3 , FM3,TM2  ,  MC1, BM2 , FM3, TM1  ,  MC2 , BM2 , FM3 , TM1  ,
 MC3 , BM2 , FM3 ,TM1  ,  MC1, BM2 , FM3,TM3  ,  MC1, BM3, FM2 ,TM1  ,  MC2 , BM3 , FM2 , TM1  ,
 MC3 , BM3 , FM2 ,TM1  ,  MC1, BM3 , FM2 ,TM2  ,  MC2 , BM3 , FM1, TM1  ,  MC2 , BM3, FM3 , TM1  ,
 MC2 , BM3, FM1,TM2  ,  MC2 , BM3 , FM1,TM3  ,  MC3 , BM2 , FM1, TM1  ,  MC3 , BM2 , FM2 , TM1  ,
 MC3 , BM2 , FM1,TM2  ,  MC3, BM2 , FM1,TM3  ,  MC2 , BM1, FM1,TM3  ,  MC2 , BM2 , FM1, TM3  ,
 MC2 , BM1, FM3 ,TM3  ,  MC3 , BM1, FM1, TM2  ,  MC3 , BM3, FM1, TM2  ,  MC3, BM1, FM2 , TM2  ,
 MC1, BM2 , FM1,TM3  ,  MC1, BM3 , FM1,TM2  ,  MC2 , BM1, FM3, TM1  ,  MC3 , BM1, FM2 , TM1  ,
Table 4.6 summarizes the number of elements that can be generated from the
combination of three versions of FTTM in FTTMn.
Table 4.6 Samples of sequence of FTTM3/n
FTTMn
FTTM3/n
36 FTTM3/n
FTTM1
0
0
FTTM2
0
0
FTTM3
1
36
FTTM4
4
144
FTTM5
10
360
FTTM6
20
720
FTTM7
35
1260
FTTM8
56
2016
FTTM9
84
3024
FTTM10
120
4320
69
Cubes can also be generated from a combination of four FTTM. Table 4.7
demonstrates the number of generated cubes in FTTMn for n= 1, 2, 3, 4 and 5.
Table 4.7 Cube with four terms FTTM in FTTMn
FTTMn
Generated FTTM
FTTM1
MC1
TM1
BM1
FM1
Cubes
0
FTTM2
0
MC2
MC1
TM1
BM1
FM1
BM2
TM2
FM2
FTTM3
MC1
TM1
BM1
MC2
FM1
TM2
BM2
FM2
MC3
TM3
BM3
FM3
0
70
FTTM4
MC1
BM1
MC2
BM2
MC3
BM3
MC4
TM4
BM4
FM4
1
TM1
FM1
TM2
FM2
TM3
FM3
 MC1 , BM 2 , FM 3 , TM 4  ,  MC1 , BM 2 , FM 4 , TM 3  ,  MC1 , BM 3 , FM 2 , TM 4  ,
 MC1 , BM 3 , FM 4 , TM 2  ,  MC1 , BM 4 , FM 2 , TM 3  ,  MC1 , BM 4 , FM 3 , TM 2  ,
 MC2 , BM1 , FM 3 , TM 4  ,  MC2 , BM 1 , FM 4 , TM 3  ,  MC2 , BM 3 , FM 1 , TM 4  ,
 MC2 , BM 3 , FM 4 , TM1  ,  MC2 , BM 4 , FM 1 , TM 3  ,  MC2 , BM 4 , FM 3 , TM 1  ,
 MC3 , BM 1 , FM 2 , TM 4  ,  MC3 , BM 1 , FM 4 , TM 2  ,  MC3 , BM 2 , FM 1 , TM 4  ,
 MC3 , BM 2 , FM 4 , TM1  ,  MC3 , BM 4 , FM 2 , TM 1  ,  MC3 , BM 4 , FM 1 , TM 2  ,
 MC4 , BM1 , FM 2 , TM 3  ,  MC4 , BM 1 , FM 3 , TM 2  ,  MC4 , BM 2 , FM 1 , TM 3  ,
 MC4 , BM 2 , FM 3 , TM1  ,  MC4 , BM 3 , FM 1 , TM 2  ,  MC4 , BM 3 , FM 2 , TM 1  ,
24 new elements alltogether.
FTTM5
MC1
BM1
MC2
BM2
MC3
MC4
MC5
BM5
BM4
FM3
TM4
FM4
TM5
FM5
FM1
TM2
FM2
TM3
BM3
TM1
5
71
 MC1 , BM 2 , FM 3 , TM 4  ,  MC1 , BM 2 , FM 4 , TM 3  ,  MC1 , BM 3 , FM 2 , TM 4  ,
 MC1 , BM 3 , FM 4 , TM 2  ,  MC1 , BM 4 , FM 2 , TM 3  ,  MC1 , BM 4 , FM 3 , TM 2  ,
 MC2 , BM1 , FM 3 , TM 4  ,  MC2 , BM 1 , FM 4 , TM 3  ,  MC2 , BM 3 , FM 1 , TM 4  ,
 MC2 , BM 3 , FM 4 , TM1  ,  MC2 , BM 4 , FM 1 , TM 3  ,  MC2 , BM 4 , FM 3 , TM 1  ,
 MC3 , BM 1 , FM 2 , TM 4  ,  MC3 , BM 1 , FM 4 , TM 2  ,  MC3 , BM 2 , FM 1 , TM 4  ,
 MC3 , BM 2 , FM 4 , TM1  ,  MC3 , BM 4 , FM 2 , TM 1  ,  MC3 , BM 4 , FM 1 , TM 2  ,
 MC4 , BM1 , FM 2 , TM 3  ,  MC4 , BM 1 , FM 3 , TM 2  ,  MC4 , BM 2 , FM 1 , TM 3  ,
 MC4 , BM 2 , FM 3 , TM1  ,  MC4 , BM 3 , FM 1 , TM 2  ,  MC4 , BM 3 , FM 2 , TM 1  ,
 MC1 , BM 2 , FM 3 , TM 5  ,  MC1 , BM 2 , FM 5 , TM 3  ,  MC1 , BM 3 , FM 2 , TM 5  ,
 MC1 , BM 3 , FM 5 , TM 2  ,  MC1 , BM 5 , FM 2 , TM 3  ,  MC1 , BM 5 , FM 3 , TM 2  ,
 MC2 , BM1 , FM 3 , TM 5  ,  MC2 , BM1 , FM 5 , TM 3  ,  MC2 , BM 3 , FM 1 , TM 5  ,
 MC2 , BM 3 , FM 5 , TM 1  ,  MC2 , BM 5 , FM 1 , TM 3  ,  MC2 , BM 5 , FM 3 , TM 1  ,
 MC3 , BM 1 , FM 2 , TM 5  ,  MC3 , BM 1 , FM 5 , TM 2  ,  MC3 , BM 2 , FM 1 , TM 5  ,
 MC3 , BM 2 , FM 5 , TM 1  ,  MC3 , BM 5 , FM 2 , TM 1  ,  MC3 , BM 5 , FM 1 , TM 2  ,
 MC5 , BM 1 , FM 2 , TM 3  ,  MC5 , BM 1 , FM 3 , TM 2  ,  MC5 , BM 2 , FM 1 , TM 3  ,
 MC5 , BM 2 , FM 3 , TM 1  ,  MC5 , BM 3 , FM 1 , TM 2  ,  MC5 , BM 3 , FM 2 , TM 1  ,
 MC1 , BM 2 , FM 4 , TM 5  ,  MC1 , BM 2 , FM 5 , TM 4  ,  MC1 , BM 4 , FM 2 , TM 5  ,
 MC1 , BM 4 , FM 5 , TM 2  ,  MC1 , BM 5 , FM 2 , TM 4  ,  MC1 , BM 5 , FM 4 , TM 2  ,
 MC2 , BM1 , FM 4 , TM 5  ,  MC2 , BM 1 , FM 5 , TM 4  ,  MC2 , BM 4 , FM 1 , TM 5  ,
 MC2 , BM 4 , FM 5 , TM 1  ,  MC2 , BM 5 , FM 1 , TM 4  ,  MC2 , BM 5 , FM 4 , TM1  ,
 MC4 , BM1 , FM 2 , TM 5  ,  MC4 , BM 1 , FM 5 , TM 2  ,  MC4 , BM 2 , FM 1 , TM 5  ,
 MC4 , BM 2 , FM 5 , TM 1  ,  MC4 , BM 5 , FM 2 , TM 1  ,  MC4 , BM 5 , FM 1 , TM 2  ,
 MC5 , BM 1 , FM 2 , TM 4  ,  MC5 , BM 1 , FM 4 , TM 2  ,  MC5 , BM 2 , FM 1 , TM 4  ,
 MC5 , BM 2 , FM 4 , TM 1  ,  MC5 , BM 4 , FM 1 , TM 2  ,  MC5 , BM 4 , FM 2 , TM1  ,
72
 MC1 , BM 3 , FM 4 , TM 5  ,  MC1 , BM 3 , FM 5 , TM 4  ,  MC1 , BM 4 , FM 3 , TM 5  ,
 MC1 , BM 4 , FM 5 , TM 3  ,  MC1 , BM 5 , FM 3 , TM 4  ,  MC1 , BM 5 , FM 4 , TM 3  ,
 MC3 , BM 1 , FM 4 , TM 5  ,  MC3 , BM 1 , FM 5 , TM 4  ,  MC3 , BM 4 , FM 1 , TM 5  ,
 MC3 , BM 4 , FM 5 , TM 1  ,  MC3 , BM 5 , FM 1 , TM 4  ,  MC3 , BM 5 , FM 4 , TM 1  ,
 MC4 , BM1 , FM 3 , TM 5  ,  MC4 , BM1 , FM 5 , TM 3  ,  MC4 , BM 3 , FM 1 , TM 5  ,
 MC4 , BM 3 , FM 5 , TM 1  ,  MC4 , BM 5 , FM 3 , TM 1  ,  MC4 , BM 5 , FM 1 , TM 3  ,
 MC5 , BM 1 , FM 3 , TM 4  ,  MC5 , BM 1 , FM 4 , TM 3  ,  MC5 , BM 3 , FM 1 , TM 4  ,
 MC5 , BM 3 , FM 4 , TM 1  ,  MC5 , BM 4 , FM 1 , TM 3  ,  MC5 , BM 4 , FM 3 , TM 1  ,
 MC2 , BM 3 , FM 4 , TM 5  ,  MC2 , BM 3 , FM 5 , TM 4  ,  MC2 , BM 4 , FM 3 , TM 5  ,
 MC2 , BM 4 , FM 5 , TM 3  ,  MC2 , BM 5 , FM 3 , TM 4  ,  MC2 , BM 5 , FM 4 , TM 3  ,
 MC3 , BM 2 , FM 4 , TM 5  ,  MC3 , BM 2 , FM 5 , TM 4  ,  MC3 , BM 4 , FM 2 , TM 5  ,
 MC3 , BM 4 , FM 5 , TM 2  ,  MC3 , BM 5 , FM 2 , TM 4  ,  MC3 , BM 5 , FM 4 , TM 2  ,
 MC4 , BM 2 , FM 3 , TM 5  ,  MC4 , BM 2 , FM 5 , TM 3  ,  MC4 , BM 3 , FM 2 , TM 5  ,
 MC4 , BM 3 , FM 5 , TM 2  ,  MC4 , BM 5 , FM 3 , TM 2  ,  MC4 , BM 5 , FM 2 , TM 3  ,
 MC5 , BM 2 , FM 3 , TM 4  ,  MC5 , BM 2 , FM 4 , TM 3  ,  MC5 , BM 3 , FM 2 , TM 4  ,
 MC5 , BM 3 , FM 4 , TM 2  ,  MC5 , BM 4 , FM 2 , TM 3  ,  MC5 , BM 4 , FM 3 , TM 2  ,
120 new elements alltogether.
We formalize the above generated cube by the following definition.
Definition 4.9 (A Sequence of FTTM4/n)
FTTM4/n means the number of cubes produced by the combination of any four terms
FTTM in FTTMn with FTTM4/1 = FTTM4/2 = FTTM4/3= 0. Hence, FTTM4/4 = 1,
FTTM4/5 = 5, FTTM4/6 = 15 and in general;
FTTM4/n = FTTM4/n-1 + FTTM3/n-1 for all n  1 .
73
Example 4.5 FTTM4/5
From Definition 4.9, it is shows;
FTTM4/5= FTTM4/4 + FTTM3/4
FTTM4/4= FTTM4/3 + FTTM3/3
Therefore,
FTTM4/4= 0+1=1
FTTM4/5=1+4=5
MC1
TM1
BM1
MC
BM2
MC3
FM2
TM3
BM3
MC4
FM1
TM2
1
FM3
TM4
2
BM4
MC5
BM5
3
4
FM4
TM5
FM5
Figure 4.4 FTTM4/5
This means, FTTM4/5 has five cubes as shown in Figure 4.4. They are;
1
1-FTTM5, 2-FTTM5, 3-FTTM5 and 4-FTTM5
2
1-FTTM5, 2-FTTM5, 3-FTTM5 and 5-FTTM5
3
1-FTTM5, 2-FTTM5, 4-FTTM5 and 5-FTTM5
4
1-FTTM5, 3-FTTM5, 4-FTTM5 and 5-FTTM5
5
2-FTTM5, 3-FTTM5, 4-FTTM5 and 5-FTTM5
5
74
For example cube
1
which is the combination of 1-FTTM4, 2-FTTM4 and 3-
FTTM4 will produce the following new elements of FTTM.
 MC1, BM2 , FM3, TM4  ,  MC1, BM2 , FM4 ,TM3  ,  MC1, BM3, FM2 ,TM4  ,  MC1, BM3, FM4 , TM2  ,
 MC1, BM4 , FM2 ,TM3  ,  MC1, BM4 , FM3 ,TM2  ,  MC2 , BM1, FM3 ,TM4  ,  MC2 , BM1, FM4 , TM3  ,
 MC2 , BM3, FM1, TM4  ,  MC2 , BM3 , FM4 ,TM1  ,  MC2 , BM4 , FM1,TM3  ,  MC2 , BM4 , FM3 , TM1  ,
 MC3 , BM1, FM2 , TM4  ,  MC3 , BM1, FM4 ,TM2  ,  MC3, BM2 , FM1,TM4  ,  MC3, BM2 , FM4 ,TM1  ,
 MC3 , BM4 , FM2 ,TM1  ,  MC3 , BM4 , FM1,TM2  ,  MC4 , BM1, FM2 ,TM3  ,  MC4 , BM1, FM3, TM2  ,
 MC4 , BM2 , FM1,TM3  ,  MC4 , BM2 , FM3 ,TM1  ,  MC4 , BM3 , FM1,TM2  ,  MC4 , BM3, FM2 ,TM1  ,
Table 4.8 summarizes the number of elements that can be generated from the
combination of four FTTM in FTTMn for n=1 to10.
Table 4.8 Samples of sequence of FTTM4/n
FTTMn
FTTM4/n
24FTTM4/n
FTTM1
0
0
FTTM2
0
0
FTTM3
0
0
FTTM4
1
24
FTTM5
5
120
FTTM6
15
360
FTTM7
35
840
FTTM8
70
1680
FTTM9
126
3024
FTTM10
210
5040
75
We have developed three different forms of cubes for FTTM which can be
defined as combination of two, three and four of FTTM in FTTMn. If we continue
producing cubes from combination of five versions FTTM or more, one of the
generated elements may  MC 1 , BM 2 , FM 3 , TM 4 , k5  with k5 the new component of
FTTM. This contradicts to the fact that there exist only four components of FTTM.
So, it is impossible to continue developing cubes from the combination of five and
more terms of FTTM.
The number of generating FTTM in a sequence of FTTMn is the summation
of three version cubes which are FTTM2/n, FTTM3/n and FTTM4/n. The coefficients
for each version are 14, 36 and 24 respectively. These coefficients represent the
number of new elements of FTTM. We can summarize our results as an equation
given as follows;
Generating FTTM n  14FTTM 2/ n  36FTTM3/ n  24FTTM 4 / n
with FTTM 2/1  FTTM 3/1  FTTM 4 /1  0 .
for n  1
(4.1)
Table 4.9 shows our generating FTTMn when compare to conjecture made by
Liau Li Yun (2006) for FTTM1 until FTTM10. The highlighted column clearly shows
that Equation (4.1) is equal to Li Yun conjecture. Next we need to redefine FTTM2/n,
FTTM3/n and FTTM4/n to become algebraic an expressions so that generating FTTMn
can be used to prove the conjecture.
76
Table 4.9 A Comparison between generating FTTMn and conjecture made by Liau
Li Yun (2006)
FTTMn
14FTTM2/n
36FTTM3/n
24FTTM4/n
14 FTTM2/n +
36 FTTM3/n +
24FTTM4/n
FTTM1
14(0)
36(0)
24(0)
0
0
FTTM2
14(1)
36(0)
24(0)
14
14
FTTM3
14(3)
36(1)
24(0)
78
78
FTTM4
14(6)
36(4)
24(1)
252
252
FTTM5
14(10)
36(10)
24(5)
620
620
FTTM6
14(15)
36(20)
24(15)
1290
1290
FTTM7
14(21)
36(35)
24(35)
2394
2394
FTTM8
14(28)
36(56)
24(70)
4088
4088
FTTM9
14(36)
36(84)
24(126)
6552
6552
FTTM10
14(45)
36(120)
24(210)
9990
9990
4.7
n4-n
Conclusion
This chapter exposed the geometrical features of FTTM. Besides that, various
definitions have been developed based on the characteristics of geometrical features
of FTTM. The definition of sequence of cubes in FTTM for two FTTM can be
extended to three and four FTTM. We will prove Li Yun’s 2006 conjecture in the
following chapter.
CHAPTER 5
A SEQUENCE OF FTTM IN RELATION TO PASCAL’S TRIANGLE
5.1
Introduction
In Chapter 4, we have developed some tools in order to prove the conjecture
suggested by Li Yun (2006). The tools included several definitions such as a
sequence of FTTM, a sequence of vertices, a sequence of edges, a sequence of faces
and a sequence of cubes in FTTM, a sequence of FTTM2/n, a sequence of FTTM3/n
and a sequence of FTTM4/n. Interestingly, the nonzero sequence of FTTM2/n,
FTTM3/n and FTTM4/n appear in Pascal’s Triangle. Therefore, this chapter will reveal
these interesting results. Besides that, this chapter will present the actual proof of the
conjecture as well as some theorems and corollaries. We will also present some new
conjectures of our own.
78
5.2
The Relation of A Sequence of FTTM to Pascal’s Triangle
We have found that the nonzero sequence of FTTM2/n (see Definition 4.7) is
presented in the third main diagonal (highlighted by the red color) of Pascal’s
Triangle, the nonzero sequence of FTTM3/n (see Definition 4.8) is presented in the
fourth main diagonal (highlighted by the green color) of Pascal’s Triangle and the
nonzero sequence of FTTM4/n (see Definition 4.9) is presented in the fifth main
diagonal (highlighted by the blue color) of Pascal’s Triangle as shown in Figure 5.1.
 0 


 0 
 1 


 0 
 2 


 0 
 3 


 0 
 4 


 0 
 5 


 0 
 6 


 0 
 7 


 0 
 8 


 0 
 6 


 1 
 7 


 1 
 8 


 1 
 5 


 1 
 8 


 2 
 8 


 3 
 4 


 4 
 5 


 4 
 6 


 4 
 7 


 4 
 8 


 4 
 3 


 3 
 4 


 3 
 5 


 3 
 6 


 3 
 7 


 3 
 2 


 2 
 3 


 2 
 4 


 2 
 5 


 2 
 6 


 2 
 7 


 2 
 2 


 1 
 3 


 1 
 4 


 1 
 1 


 1 
 6 


 5 
 7 


 5 
 8 


 5 
 5 


 5 
 6 


 6 
 7 


 6 
 8 


 6 
 7 


 7 
 8 


 7 
 8 


 8 
Figure 5.1 The Nonzero Sequence of FTTM2/n, FTTM3/n and FTTM4/n
Each binomial coefficient in Figure 5.1 can be replaced by its numerical value in
order to get another version of Pascal’s Triangle as illustrated in Figure 5.2.
79
n=0
1
n=1
1
n=2
1
n=3
1
n=4
1
n=5
1
n=6
1
n=7
n=8
1
1
8
3
5
7
4
20
FTTM4/n=FTTM4/n-1+ FTTM3/n-1
1
5
15
35
70
FTTM3/n=FTTM3/n-1+ FTTM2/n-1
1
10
35
56
1
6
15
FTTM2/n=FTTM2/n-1+ (n-1)
3
10
21
28
2
4
6
1
1
6
21
56
1
7
28
1
8
1
Figure 5.2 An Alternative Version of Pascal’s Triangle
Figure 5.2 shows n=0 is in the first row of Pascal’s Triangle, n=1 is in the
second row of Pascal’s Triangle, n=2 as third row of Pascal’s Triangle and so on.
While the third, fourth and fifth of the main diagonal of Pascal’s Triangle represent
the formulae of sequence of FTTM2/n, FTTM3/n and FTTM4/n respectively. By
observing the Pascal’s Triangle in Figure 5.2, we can deduce the expression for
FTTM2/n. For example, if we ask for FTTM2/5 then we need to add FTTM2/ (5-1) and
(5-1) together. FTTM2/4 can be found easily from the Pascal’s Triangle with upper
and right of FTTM2/5 and 4 is upper and left of FTTM2/5. This formula can be
obtained by adding FTTM2/4 and 4 together (see Figure 5.2). It is the same concept as
to define the entries of Pascal’s Triangle where the entries are the sum of the entries
at the left and right above it. We can use the same concept to obtain FTTM3/n and
FTTM4/n.
80
5.3
The Proof of Li Yun’s Conjecture
The problem in proving the conjecture is that the statement is not totally
algebraic statements. By revealing the geometrical features of FTTM and defining
their characteristics, we have produced Equation (4.1). Since the third, fourth and
fifth of the main diagonal of Pascal’s Triangle represent the formulae of sequence of
FTTM2/n, FTTM3/n and FTTM4/n respectively; therefore we can rewrite Equation
(4.1) as follows;
 n
n
n
FTTM n  14    36    24   ,
 2
 3
4
n  4.
 5.1
By using equation (5.1), we can prove the conjecture analytically.
n
We know FTTM2/n is equal to   .
 2
Therefore,
n
n!
FTTM 2    
n
 2   2  !  n   2   !

n  n  1 n  2  !
2! n  2  !

n  n  1
.
2!
n
Similarly, FTTM3/n is equal to   .
 3
 5.2 
81
Therefore,
 n
n!
FTTM 3    
n
 3   3 !  n   3  !

n  n  1 n  2  n  3 !
3! n  3 !

n  n  1 n  2 
.
3!
 5.3
n
Finally, FTTM4/n is equal to   .
 4
Therefore,
n
n!
FTTM 4    
n
 4   4  !  n   4   !

n  n  1 n  2  n  3 n  4  !
4! n  4  !

n  n  1 n  2  n  3
.
4!
 5.4 
By replacing Equation (5.2), (5.3) and (5.4) to Equation (4.1), we have;
 n  n  1 
 n  n  1 n  2  
 n  n  1 n  2  n  3 
FTTM n  14 
  36 
  24 

3!
4!
 2! 




 7 n  n  1  6n  n  1 n  2   n  n  1 n  2  n  3
 n  n  1  7  6  n  2    n  2  n  3  
 n  n  1  7  6n  12  n 2  3n  2n  6 
  n 2  n   n 2  n  1
 n 4  n3  n 2  n3  n 2  n
 n 4  n.
82
By showing Equation (5.1) is equal to n4  n i.e. the number of generating new
elements, therefore the conjecture made by Liau Li Yun in 2006 is finally proven.
We can write the conjecture as a theorem now.
Theorem 5.1
If there exist n elements of FTTM; i.e. FTTMn as illustrated below,
MC1
TM1
BM1
MC2
BM2
MC3
MCn
BMn
FM1
TM2
FM2
TM3
BM3
TMn
FM3
FMn
Figure 5.3 FTTMn
then the numbers of new generating elements of FTTM is given as
n
n
n
FTTM n  14    36    24    n 4  n
 2
 3
4
for n  4.
(5.5)
83
5.4
Further Results
The fact that FTTM which has four components and the number of
generating FTTM is n4  n , the result can be extended to F2n where the number of
generating F2 is n2  n . Similarly, F3n which consist of three components, then the
number of generating F3 is n3  n . F5n which consist of five components, then the
number of generating F5 is n5  n . Using the ideas of set of FTTMs (refer Equation
2.1) and a sequence of FTTM (see Definition 4.5), we can define these sequences as
follows:
Definition 5.1 (A Sequence of F2)
Let F2   Ai , Bi   F2i : Ai  Bi  be the set of all two topological spaces that are
homeomorphic. A sequence of F2n is a sequence of F21, F22, F23,…, F2n such that
Ai  Ai 1 and Bi  Bi 1 .
This definition is illustrated as below;
A1
A2
A3
An
B1
B2
B3
Bn
Figure 5.4 A Sequence of F2
84
Definition 5.2 (A Sequence of F3)
Let F3   Ai , Bi , Ci   F3i : Ai  Bi  Ci  be the set of all three topological spaces
that are homeomorphic. A sequence of F3n is a sequence of F31, F32, F33,…, F3n
such that Ai  Ai 1 , Bi  Bi 1 and Ci  Ci 1 .
This definition is illustrated as below;
A1
A2
A3
An
B1
C2
B2
B3
Bn
C1
C3
Cn
Figure 5.5 A Sequence of F3
Definition 5.3 (A Sequence of F5)
Let F5   Ai , Bi , Ci , Di , Ei   F5i : Ai  Bi  Ci  Di  Ei  be the set of all five
topological spaces that are homeomorphic. A sequence of F5n is a sequence of F51,
F52, F53,…, F5n such that Ai  Ai 1 , Bi  Bi 1 , Ci  Ci 1 , Di  Di 1 and Ei  Ei 1 .
This definition is illustrated as below;
A1
A2
B2
A3
An
B3
Bn
En
Cn
E3
C3
B1
E2
C2
E1
C1
D2
D3
Dn
Figure 5.6 A Sequence of F5
D1
85
Definition 5.4 (A Sequence of Fk)
Let Fk   Ai , Bi , Ci , Di ..., ki   Fki : Ai  Bi  Ci  Di  ...  ki  be the set of all k
topological spaces that are homeomorphic. A finite sequence of Fk is a sequence of
Fk1, Fk2, Fk3,…, Fkn such that Ai  Ai 1 , Bi  Bi 1 , Ci  Ci 1 , Di  Di 1 and ki  ki 1 .
This definition is illustrated as below;
A1
A2
B2
A3
An
Bn
B3
kn
Cn
k3
C3
B1
K2
C2
k1
C1
D1
D2
D3
Dn
Figure 5.7 A Sequence of Fk
From Definition 5.1, we can observe characteristics of F2n for n = 1, 2, 3, 4
and 5 as shown in Table 5.1. These characteristics include the geometrical features of
F2n, the number and the generated new elements of F2 under combination of two
terms F2 in F21, F22, F23, F24 and F25.
86
Table 5.1 Two terms of F2 in F2n
F2n
Generated F2
A1
F21
F22/n
B1
A1
0
B1
F22
1
A2
B2
 A1 , B2  ,  A2 , B1  ,
2 new elements of F2.
A1
F23
B1
A2
3
B2
A3
B3
 A1 , B2  ,  A2 , B1  ,  A1 , B3  ,  A3 , B1  ,  A2 , B3  ,  A3 , B2  ,
6 new elements of F2.
A1
B1
F24
6
A2
A3
A4
B2
B3
B4
 A1 , B2  ,  A2 , B1  ,  A1 , B3  ,  A3 , B1  ,  A1 , B4  ,  A4 , B1  ,
 A2 , B3  ,  A3 , B2  ,  A3 , B4  ,  A4 , B3  ,  A2 , B4  ,  A4 , B2  ,
12 new elements of F2.
87
A1
B1
F25
10
A2
A3
A4
A5
B2
B3
B4
B5
 A1 , B2  ,  A2 , B1  ,  A1 , B3  ,  A3 , B1  ,  A1 , B4  ,  A4 , B1  ,  A1 , B5  ,
 A5 , B1  ,  A2 , B3  ,  A3 , B2  ,  A2 , B4  ,  A4 , B2  ,  A2 , B5  ,  A5 , B2  ,
 A3 , B4  ,  A4 , B3  ,  A3 , B5  ,  A5 , B3  ,  A4 , B5  ,  A5 , B4  ,
20 new elements of F2.
In short the combination of two F2 will produce another two new elements.
In other words, we can generate another two F2 from that combination. Therefore,
Corollary 5.1 can be stated as follows:
Corollary 5.1
n
If there exist n elements of F2 i.e. F2n  2   , n  2 , then the numbers of new
 2
generating elements are n 2  n elements.
The Proof of Corollary 5.1
By replacing Equation 5.2 to F2n, we get
 n  n  1 
F2n  2 

 2! 
 n 2  n.
88
Form Definition 5.2, we can observe characteristics of F3n for n = 1, 2, 3, 4
and 5 as shown in Table 5.2. These characteristics include the geometrical features of
F3n, the number and the generated new elements of F3 under combination of two
terms F3 in F31, F32, F33, F34 and F35.
Table 5.2 Two terms of F3 in F3n
F3n
Generated F3
F32/n
A1
F31
0
C1
B1
A1
F32
1
A2
B1
C1
B2
C2
 A1 , B1 , C2  ,  A1 , B2 , C1  ,  A2 , B1 , C1  ,
 A1 , B2 , C2  ,  A2 , B1 , C2  ,  A2 , B2 , C1  ,
6 new elements of F3.
A1
F33
3
A2
A3
B1
B3
C1
C2
B2
C3
 A1 , B1 , C2  ,  A1 , B2 , C1  ,  A2 , B1 , C1  ,  A1 , B2 , C2  ,  A2 , B1 , C2  ,
 A2 , B2 , C1  ,  A1 , B1 , C3  ,  A1 , B3 , C1  ,  A3 , B1 , C1  ,  A1 , B3 , C3  ,
 A3 , B1 , C3  ,  A3 , B3 , C1  ,  A2 , B2 , C3  ,  A2 , B3 , C2  ,  A3 , B2 , C2  ,
 A2 , B3 , C3  ,  A3 , B2 , C3  ,  A3 , B3 , C2  ,
18 new elements of F3.
89
A1
F34
6
A2
A3
A4
B3
B4
B1
B2
C1
C2
C3
C4
 A1 , B1 , C2  ,  A1 , B2 , C1  ,  A2 , B1 , C1  ,  A1 , B2 , C2  ,  A2 , B1 , C2  ,
 A2 , B2 , C1  ,  A1 , B1 , C3  ,  A1 , B3 , C1  ,  A3 , B1 , C1  ,  A1 , B3 , C3  ,
 A3 , B1 , C3  ,  A3 , B3 , C1  ,  A1 , B1 , C4  ,  A1 , B4 , C1  ,  A4 , B1 , C1  ,
 A1 , B4 , C4  ,  A4 , B1 , C4  ,  A4 , B4 , C1  ,  A2 , B2 , C3  ,  A2 , B3 , C2  ,
 A3 , B2 , C2  ,  A2 , B3 , C3  ,  A3 , B2 , C3  ,  A3 , B3 , C2  ,  A2 , B2 , C4  ,
 A2 , B4 , C2  ,  A4 , B2 , C2  ,  A2 , B4 , C4  ,  A4 , B2 , C4  ,  A4 , B4 , C2  ,
 A3 , B3 , C4  ,  A3 , B4 , C3  ,  A4 , B3 , C3  ,  A3 , B4 , C4  ,  A4 , B3 , C4  ,
 A4 , B4 , C3  ,
36 new elements of F3.
We realize that the combination of two terms F3 will produce another six new
elements. In other words, we can generate another six F3 from that combination.
Sequence of F3n also contains combination of three different versions of F3. Table
5.4 demonstrates the properties of F3 in F3n for n = 1, 2, 3 and 4.
Table 5.3 Three terms of F3 in F3n
F3n
Generated F3
A1
F31
B1
F33/n
0
C1
90
A1
F32
0
A2
B1
C1
B2
C2
F33
1
A1
A2
A3
C1
B1
C2
B2
B3
C3
 A1 , B2 , C3  ,  A1 , B3 , C2  ,  A2 , B3 , C1  ,
 A2 , B1 , C3  ,  A3 , B1 , C2  ,  A3 , B2 , C1  ,
18 new elements of F3.
A1
F34
4
A2
A3
A4
B2
B3
B4
B1
C1
C2
C3
C4
 A1 , B2 , C3  ,  A1 , B3 , C2  ,  A2 , B3 , C1  ,  A2 , B1 , C3  ,  A3 , B1 , C2  ,
 A3 , B2 , C1  ,  A1 , B2 , C4  ,  A1 , B4 , C2  ,  A2 , B4 , C1  ,  A2 , B1 , C4  ,
 A4 , B1 , C2  ,  A4 , B2 , C1  ,  A1 , B3 , C4  ,  A1 , B4 , C3  ,  A3 , B4 , C1  ,
 A3 , B1 , C4  ,  A4 , B1 , C3  ,  A4 , B3 , C1  ,  A2 , B3 , C4  ,  A2 , B4 , C3  ,
 A3 , B4 , C2  ,  A3 , B2 , C4  ,  A4 , B2 , C3  ,  A4 , B3 , C2  ,
24 new elements of F3.
91
We can observe that the combination of three F3 will produce another six
new elements. In other words, we can generate another six F3 from that combination.
Hence, the coefficients for both combinations are the same. Therefore, Corollary 5.2
can be stated as follows:
Corollary 5.2
n n
If there exist n elements of F3 i.e. F3n  6    6   , n  3 , then the numbers of
 2  3
generating new elements are n3  n .
The Proof of Corollary 5.2
By replacing Equation 5.2 and 5.3 to F3n, we have
 n  n  1 
 n  n  1 n  2  
F3n  6 
 6

3!
 2! 


 3  n2  n    n2  n   n  2
 3n 2  3n  n3  2n 2  n 2  2n
 n3  n.
Form Definition 5.3, we can observe characteristics of F5n for n = 1, 2, 3, 4
and 5 as shown in Table 5.4. These characteristics include the geometrical features of
F5n and the number of combination of two, three, four and five terms F5 in F51, F52,
F53, F54 and F55.
92
Table 5.4 Two, three, four and five terms of F5 in F5n
F5n
Generated F5
A1
F51
B1
A1
A2
B1
B2
E2
C2
0
0
0
1
0
0
0
3
1
0
0
6
4
1
0
D2
A2
B1
B2
E3
E2
C2
E1
C2
D1
D2
D3
A1
F54
A2
A4
B3
B4
E4
D4
E3
C3
B1
B2
A3
C4
0
D1
A1
C3
F55/n
E1
F53
B3
F54/n
D1
F52
A3
F53/n
E1
C1
C2
F52/n
D3
E2
C2
D2
E1
C1
D1
93
A1
F55
A2
A3
A4
A5
B4
B5
C5
B3
E4
C3
E5
C4
B2
E2
B1
E2
C1
C2
10
10
5
1
E1
D1
D2
D3
D4
D5
The coefficients of each combination of terms F5 are presented in the next
subchapter. Therefore, Corollary 5.3 can be stated as follows:
Corollary 5.3
n
n
n
n
If there exist n elements of F5 i.e., F5n  30    150    240    120   , n  5 ,
 2
 3
4
5
then the numbers of generating new elements are n5  n elements.
The Proof of Corollary 5.3
We know,
n
n!
5 
   5  ! n   5   !

n  n  1 n  2  n  3 n  4  n  5 !
5! n  5  !

n  n  1 n  2  n  3 n  4 
.
5!
(5.6)
94
By replacing Equation (5.2), (5.3), (5.4) and (5.6) to F5n, we have;
 n  n  1 
 n  n  1 n  2  
 n  n  1 n  2  n  3 
F5n  30 
  150 
  240 

3!
4!
 2! 




 n  n  1 n  2  n  3 n  4  
 120 

5!


 15n  n  1  25n  n  1 n  2   10n  n  1 n  2  n  3  n  n  1 n  2  n  3 n  4 
 n  n  1 15  25  n  2   10  n  2  n  3   n  2  n  3 n  4  
 n  n  1 15  25n  50  10n2  30n  20n  60  n3  5n 2  6n  4n 2  20n  24 
 n 2  n  n3  n 2  n  1
 n5  n 4  n3  n 2  n 4  n3  n2  n
 n5  n.
Furthermore, we can also have the following conjecture.
If every nonzero sequence of FTTM2/n, FTTM3/n and FTTM4/n appear in the
third, fourth and fifth main diagonal of Pascal’s Triangle respectively, therefore for
every nonzero sequence of Fk2 / n , Fk3/ n , Fk4 / n ,..., Fkl / n they also obey the third, fourth,
fifth, until (l+1)th main diagonal of Pascal’s Triangle with k represents the number of
components. As a result, we can write a conjecture formally as follows.
Conjecture 5.1
n
n
 n
n
n
n
Fkn  C1    C2    C3    C4    C5    ...  C p  
 2
3
 4
5
6
k
 n k  n,
nk
with k the number of component and C1, C2, C3, C4, C5, …, Cp are the coefficients for
each combination.
95
5.5
Coefficients of FTTM and F5
n n
n
The coefficients of   ,   and   in FTTM are 14, 36 and 24
 2  3
 4
n
n
respectively, while the coefficient of   in F2 is 2 and the coefficients of   and
 2
 2
n
 3  in F3 both are 6. Clearly, the number of components increases as the coefficients
 
increase. The coefficients in Theorem 5.1, Corollary 5.1 and 5.2 were obtained by
listing all the generated elements from each combination. It will consume longer time
if we adopt the procedure. Now, we will present an easier method to define the
coefficients of FTTM and F5. We start with FTTM.
Coefficients for FTTM
 n
n
n
FTTM n  14    36    24   , n  4 .
 2
3
 4
n
The   or FTTM2/n was defined as a cube with the combination of two terms
 2
FTTM in FTTMn. So, the generated elements should have two different versions of
FTTM such as (MC4, BM3, FM3, TM4) in FTTM4. From Li Yun’s conjecture, the
n
number of generating FTTM is n4  n . Therefore, the coefficient of   will be;
 2
24  2  14.
n
The   or FTTM3/n was defined as a cube with the combination of three terms
 3
FTTM in FTTMn. So, the generated elements should have three different versions of
FTTM such as (MC2, BM2, FM3, TM4) in FTTM4. From Li Yun’s conjecture, the
number of generating FTTM is n4  n . If there is FTTM3, the generated numbers of
FTTM are
34  3  78 elements.
(5.7)
96
From 78 elements in Equation 5.7, there are elements with combination of two terms
FTTM in FTTM3 which are
14  3  42 elements.
(5.8)
In FTTM3, there are 3 cubes with combination of two terms FTTM. Each cube will
generate 14 new elements. That explains the 14 times 3 in Equation 5.8.
Hence, the difference of the generated number of FTTM will be the coefficient of
n
 3  , i.e,
 
78  42  36 .
n
The   or FTTM4/n was defined as a cube with the combination of four terms
 4
FTTM in FTTMn. So, the generated elements should have four different versions of
FTTM such as (MC1, BM2, FM3, TM4) in FTTM4. From conjecture made by Li Yun,
the number of generating FTTM is n4  n . If there is FTTM4, therefore the generated
numbers of FTTM are
44  4  252 elements.
(5.9)
From 252 elements in Equation 5.9, there are elements with combination of two and
three terms of FTTM in FTTM3 which are
14  6    36  4   228 elements.
(5.10)
In FTTM3, there are 6 cubes with combination of two terms FTTM and 4 cubes with
combination of 3 terms in FTTM. Each cube will generate 14 and 36 new elements
respectively. That explains the 14 should times 6 and 36 should times 4 as in
Equation 5.10.
97
Hence, the difference of the generated number of FTTM will be the coefficient of
n
 4  , i.e,
 
252  228  24 .
Now, we are going to determine the coefficients of F5.
Coefficient for F5
Let the components of F5 be A, B, C, D and E. Therefore, if there exist n elements of
F5, then
n
n
n
n
F5n  30    150    240    120   , n  5 .
 2
 3
4
5
n
The   or F52/n is defined as a cube with the combination of two terms F5 in F5n.
 2
So, the generated elements should have two different versions of F5 such as (A2, B5,
C5, D2, E5) in F55. Correspond to subsection extension of components, the number of
n
generating F5 is n5  n . Therefore, the coefficient of   will be;
 2
25  2  30 .
n
The   or F53/n is defined as a cube with the combination of three terms F5 in F5n.
 3
So, the generated elements should have three different versions of F5 such as (A4, B2,
C2, D5, E5) in F55. Correspond to subsection extension of components, the number of
generating F5 is n5  n . If there is F53, therefore the generated numbers of F5 are
35  3  240 elements.
(5.11)
98
From 240 elements in Equation 5.11, there are elements with combination of two
terms F5 in F53 which are
30  3  90 elements.
(5.12)
In F53, there are 3 cubes with combination of two terms F5. Each cube will generate
30 new elements. That explains the 30 times 3 as in Equation 5.12.
n
Hence, the difference of the generated number of F5 will be the coefficient of   ,
 3
i.e,
240  90  150 .
n
The   or F54/n is defined as a cube with the combination of four terms F5 in F5n.
 4
So, the generated elements should have four different versions of F5 such as (A5, B2,
C3, D4, E5) in F55. Correspond to subsection extension of components, the number of
generating F5 can be n5  n . If there is F54, therefore the generated numbers of F5
are
45  4  1020 elements.
(5.13)
From 1020 elements in Equation 5.13, there are elements with combination of two
and three terms F5 in F54 which are
 30  6   150  4   780 elements.
(5.14)
In F54, there are 6 cubes with combination of two terms F5 and 4 cubes with
combination of 3 terms F5. Each cube will generate 30 and 150 new elements
respectively. That explains the 30 should times 6 and 150 should times 4 as in
Equation 5.14.
99
n
Hence, the difference of the generated number of F5 will be the coefficient of   ,
 4
i.e,
1020  780  240 .
n
The   or F55/n is defined as a cube with the combination of five terms F5 in F5n.
5
So, the generated elements should have five different versions of F5 such as (A1, B2,
C3, D4, E5) in F55. Correspond to subsection extension of components, the number of
generating F5 is n5  n . If there is F55, therefore the generated numbers of F5 is
55  5  3120 elements.
(5.15)
From 3120 elements in Equation 5.15, there are elements with combination of two,
three and four terms F5 in F55 which is
 30 10   150 10    240 15   3000
elements.
(5.16)
In F55, there are 10 cubes with combination of two terms F5, 10 cubes with
combination of 3 terms F5 and 5 terms with combination of 4 terms F5. Each cube
will generate 30, 150 and 240 new elements respectively. That explains the 30 should
times 6, 150 should times 4 and 240 should times 5 as in Equation 5.16.
n
Hence, the difference of the generated number of F5 will be the coefficient of   ,
5
i.e,
3120  3000  120 .
100
5.6
Relating Extension Results of FTTM to Pascal’s Triangle
By extending the number of components and its coefficients, we produced
Corollaries 5.1, 5.2, 5.3 and 5.4. Using the same procedure in defining FTTMn in
Chapter 4 we now have the following corollary:
Corollary 5.5
 n
 n
 n
 n
n
F6n  62    540    1560    1800    720   , n  6.
 2
 3
 4
 5
6
All the corollaries can be related to Pascal’s Triangle as shown in Figure 5.8.
101
n=0
CF3 n
CF4n
CF5n
CF6n
2
6
14
30
62
Fk2/n
6
36
150
540
Fk3/n
24
240
1560
Fk4/n
120
1800
Fk5/n
720
Fk6/n
1
n=1
1
n=2
1
n=3
1
n=4
1
n=5
1
n=6
1
n=7
1
n=8
1
n=9
n=10
CF2n
1
1
8
9
10
15
70
1
6
21
56
126
252
1
5
35
126
210
4
20
56
1
10
35
84
120
6
15
28
1
3
10
21
36
45
3
5
7
2
4
6
1
7
28
84
210
1
1
8
36
120
1
9
45
1
10
1
Figure 5.8 Relating Extension Result to Pascal’s Triangle
101
102
The big triangle in Figure 5.8 is Pascal’s Triangle while triangle on the right
is the triangle of coefficient for F2, F3 and so on until F6. If we observe the triangle
coefficient vertically, then the entries represent coefficients for F2n until F6n
respectively. On the other hand if we observe the triangle horizontally, then the
entries will represent coefficients for Fk2/n, Fk3/n until Fk6/n respectively. Meanwhile,
each row of Pascal’s Triangle is marked by n=0, n=1 and so forth. We can find the
number of generating Fk with k the number of components in Figure 5.8. For
example the coefficients of F67 is given below;
 7
 7
7
7
7
F 67  62    540    1560    1800    720   .
 2
 3
 4
 5
 6
In order to find the number of combination of two terms F6 in F67, we can just
follow through the row of Fk2/n until we hit the entry for n=7 in Pascal’s Triangle
7
which is   . We can determine the other combinations with the same manner.
 2
We can reconstruct the right triangle in Figure 5.8 to be a triangle as in Figure
5.9. There are two new patterns that can be observed from the new triangle. The
n
outer most left diagonal represents the sequence for coefficient of Fk2/n or   . The
 2
sequence is the summation of entries in each row of Pascal’s Triangle except 1
starting from row 2. The second pattern comes from the outer most right diagonal
where the first entry of the diagonal is 2 followed by 6, 24 and so on. 6 is resulted
from the product of the number of component in F3n with the previous entry of the
diagonal which is 2. 24 is resulted from the product of the number of component in
FTTMn with the previous entry of the diagonal which is 6.
103
2
6
6
14
30
62
36
150
540
F2n
F3n
24
240
1560
FTTMn
120
1800
F5n
720
F6n
Figure 5.9 The Triangle of Coefficients
5.7
A Sequence of Polygon
In Chapter 2, we described the patterns of Pascal’s Triangle. One of the
patterns is triangular numbers. Triangular numbers are examples of polygonal
numbers. Polygonal number is defined by the number of vertices in figure formed by
certain polygon (resource from http://britton.disted.camosun.bc.ca/pascal/pascal.
html). We can adopt the idea of polygonal numbers into our work. For example we
can represent F2 with a line with two vertices, similarly F3 is a triangular with three
vertices and F4 is a square with four vertices.
104
Correspond to the definition of a sequence of FTTM and k-th FTTM in
n
FTTMn,   can be defined as a combination of two terms F2 in a sequence of
 2
n
segment or F2n,   as a combination of three terms F3 in a sequence of triangle or
 3
n
F3n,   as a combination of four terms F4 in a sequence of square or F4n and so on
 4
until we produce a hexagon as shown in Table 5.5. A sequence for each combination
of term is also presented in Table 5.5. The different color of the polygon represent
the different term. The first term is marked with dark purple. As the number of
components is increasing, the brightness of color will decrease. If we observe the
first type of polygon which is only lines then the number of combination of two
terms sequence are 0, 1, 3, 6, 10, 15, and 21. We can use the same procedure for
other polygons to determine respective number of combinations and can be
n n n n
n
summarized as   ,   ,   ,   and   for a sequence of line, triangular,
 2  3  4  5
6
square, pentagonal and hexagonal.
appear in Pascal’s Triangle.
Interestingly, all the nonzero combinations
Table 5.5 A Sequence of Polygon
n
Type
1
2
3
4
5
6
0
1
3
6
10
15
0
0
1
4
10
20
0
0
0
1
5
15
Line
F2n
n
2
 
Triangular
F3n
n
3
 
Square
F4n
n
4
 
105
106
Pentagonal
F5n
n
5
 
0
0
0
0
1
6
0
0
0
0
0
1
Hexagonal
F6n
n
6
 
106
107
5.8
Conclusion
We have proved the conjecture made by Li Yun (2006) in this chapter. Some
of the results in this research can be related directly to Pascal’s Triangle. We have
also produced new conjecture which leads to a sequence of polygon and its relation
to Pascal’s Triangle.
CHAPTER 6
CONCLUSION
6.1
Summary
The aim of this research is to prove the conjecture proposed by Liau Li Yun
in 2006. In the beginning of this thesis, we briefly discussed on FTTM. We
highlighted that the left hand side of the conjecture is a ‘geometrical object’ in nature
while the right hand side is an algebraic expression. We also discussed some
methods of proving in Chapter 3 where proving by construction is the most suitable
method that can be employed to prove the conjecture. Therefore, some geometrical
features of FTTM have been investigated in order to prove the conjecture. Such
characteristics of geometrical features that have been produced are a sequence of
vertices, a sequence of edges, a sequence of faces and a sequence of cubes. These
characteristics were used to prove the conjecture in Chapter 5. Interestingly, these
characteristics of geometrical features and their relations appear in Pascal’s Triangle
and discussed in Chapter 5. On route we have established some theorems, corollaries
and conjecture as well as the establishment of relation between a sequence of
polygon to Pascal’s Triangle.
109
6.2
Suggestions for Further Research
This research can be extended by following the suggestions:
1)
Proof the conjecture proposed in Section 5.4 of this thesis.
2)
Find the other patterns of the entries of the triangle of coefficients that
were discussed in Chapter 5.
3)
Explore the concept of a sequence of FTTM and a combination of
terms to produce cubes using the established sequence of polygon in
this research.
Finally, this thesis realizes that the sequence of FTTM, in particular their
coefficients appear in Pascal’s Triangle. By exploring these phenomena further, we
discovered that the sequence of polygon also appears in Pascal’s Triangle and more
questions arise need to be answered. Thus, we must continue study and do research
as John Morton Finney said,
“I never stop studying. There are always lots to learn. When you stop
learning, that's about the end of you.”
110
REFERENCES
Burnett
Meyer
(1974).
An
Introduction
to
Axiomatic
Systems.
Boston,
Massachusetts: Prindle, Webber & Schmidt.
Daniel Solow (1982). How to Read and Do Proofs. New York: John Wiley & Son,
Inc., 23, 34.
David M. Burton (2007). The History of Mathematics: An Introduction. (6th ed.) New
York: The McGraw-Hill Componies, Inc., 456-459,
Edward, R. S. (2006). Mathematics: A Discrete Introduction. (2nd ed.) United States:
Thomson Brooks/Cole, 109.
Fauziah, Z. (2002). Algorithma Penyelesaian Masalah Songsang Arus Tunggal Tak
Terbatas MEG. Universiti Teknologi Malaysia: Tesis Sarjana.
Fuzzy Research Group (2006). Fuzzy Topographic Topological Mapping.
[Brochure]. Universiti Teknologi Malaysia.
Hämäläinen, M., Hari, R., Ilmoniemi, R. J., Knuutila, J. and Lounasmaa, O. V.
(1993). Magnetoencephalography-Theory, Instrumentation, and Applications to
Noninvasive Studies of the Working Human Brain. Rev. Mod. Phys. 65(2): 413497.
Howard Eves (1976). An Introduction to The History of Mathematics. United States:
Library of Congress Cataloging in Publication Data, 176.
Herbert Meschkowski (1968). Introduction to Modern Mathematics. London:
Harrap.
James, L. H. (2003). Discrete Mathematics. (2nd ed.) United States: Jones and
Bartlett Publishers, 253-259.
John Stillwell (2002). Mathematics and Its History. (2nd ed.) United States: Springer
Science + Business Media, LLC, 193-194.
Jie Wu (1997). Extended Fibonacci Cubes. IEEE Transactions on Parallel and
Distributed Systems, Vol. 8, No. 12.
Kenneth, H. R. (2000). Elementary Number Theory and its Applications. (4th ed.)
United States: AT & T Laboratories, 25.
111
Liau, L. Y. (2001). Homeomorfisma S2 Antara E2 Melalui Struktur Permukaan
Riemann serta Deduksi Teknik Pembuktiannya bagi Homeomorfisma Pemetaan
Topologi Topografi Kabur (FTTM). Universiti Teknologi Malaysia: Tesis
Sarjana.
Liau, L. Y. (2006). Group-Like Algebraic Structures of Fuzzy Topographic
Topological Mapping for solving Neuromagnetic Inverse Problem. Universiti
Teknologi Malaysia: PhD Thesis.
Nancy Rodgers (1941). Learning to Reason. New York: John Wiley & Sons, Inc.
Philip, J. D. and Reuben, H. (1981). The Mathematical Experience. Birkhauser
Boston: The Harvester Press Limited.
Tahir, A., Rashdi Shah, A., Fauziah, Z. and Liau. L. Y. (2000). Development of
Detection Model for Neuromagnetic Fields. Proceeding of BIOMED. September
27-28. Kuala Lumpur: Universiti Malaya.
Tahir, A., Rashdi Shah, A., Fauziah, Z., Liau. L. Y. and Wan Eny Zarina, W. A. R.
(2005). Homeomorphism of Fuzzy Topographic Topological Mapping (FTTM).
Matematika. 21(1):35-42.
Tarantola, A. and Valette, B. (1982). Inverse Problem: Quest for Information. J.
Geophys. 50: 159-170.
Ted Sundstrom (2007). Mathematical Reasoning: writing and proof. (2nd ed.) United
States: Prentice Hall, 79.
Ulrich, D. and Pamela, G. (2003). Reading, Writing and Proving: a look closer at
mathematics. United States: Springer-Verlag New York, Inc., 53.
Uutela, K. (2001). Estimating Neural Currents from Neuromagnetic Measurements.
Helsinki University of Technology: PhD. Dissertation.
V.A Uspenskii (1974). Pascal’s Triangle. The University of Chicago Press.
Wan Eny Zarina, W. A. R., Tahir, A. and Rashidi Shah, A. (2002). Simulating The
Neuronal Current Sources In The Brain. Proceeding BIOMED. September 27-28.
Kuala Lumpur: Universiti Malaya.
Wen-Jing, H. (1993). Fibonacci Cubes – A New Interconnection Topology. IEEE
Transactions on Parallel and Distributed Systems, Vol. 4, No. 1.
112
http://www.brainhealthandpuzzles.com/diagram_of_brain.html, Labeled Diagram of
the Human Brain (2008).
www.cranius.wordpress.comtagmri.com, MEG System (2008).
http://www.mathsisfun.com/images/pascals-triangle-chinese.gif, Chinese Pascal’s
Triangle (2008).
http://pages.csam.montclair.edu/~kazimir/patterns.html, Pascal’s Triangle (2007).
http://britton.disted.camosun.bc.ca/pascal/pascal.html, Pascal’s Triangle and Its
Patterns (2008).