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GENERATING COMBINATORIAL OBJECTS –
A NEW PERSPECTIVE
by
Alexander Chizoma Nwala
B.Sc. December 2011, Elizabeth City State University, Elizabeth City, North Carolina
A Thesis Submitted to the Faculty of
Old Dominion University in Partial Fulfillment of the
Requirements for the Degree of
MASTERS OF SCIENCE
COMPUTER SCIENCE
OLD DOMINION UNIVERSITY
May 2014
Approved by:
________________________
Stephan Olariu (Director)
________________________
Michelle C. Weigle (Member)
________________________
Hussein A. Wahab (Member)
ABSTRACT
GENERATING COMBINATORIAL OBJECTS – A NEW PERSPECTIVE
Alexander Chizoma Nwala
Old Dominion University, 2014
Director: Dr. Stephan Olariu
Combinatorics is the science of “possibilities.” This definition, while not formal is
a fair statement because all too often, in order to gain insight into the solution of many
counting problems, we explore the possibilities. In some cases we seek to know how many
options, while in other cases we seek to enumerate or list the options. Irrespective of the
scenario, combinatorics plays a vital role today. In many instances such as exploring the
options for choosing a new password for a combination lock, we employ combinatorics. In
considering the possible license plate permutations for a state, or to see if we have enough
IP addresses or telephone numbers, we employ combinatorics. In fact, our DNA in the
nucleus of our cells are combinatorial objects consisting of permutations with repetitions
of nucleobases represented as (G - Guanine, A - Adenine, T - Thymine, C - Cytosine). So
it is also fair to say combinatorics is part of our life. Combinatorics (the branch of
mathematics that deals with the arrangement of objects), offers many different ways of
counting such as combinations – the focus of this research effort. Combinations refer to a
special way of arranging objects when order is not considered. Combinations also belong
to the class of fundamental combinatorial objects, which is no surprise why we already
have many algorithms which generate combinations. However, these algorithms generate
combinations directly, meaning they are bound to have only one way of arranging their
tuples. In contrast, this research effort offers a new way of generating combinations, an
indirect way. The indirect way offers a means by which the generator can define the order
of tuples. This new way depends on two combinatorial objects: a parent combinatorial
method, M1 (an original non-recursive constant amortized time algorithm for generating
permutations with repetitions allowed), and an indexing combinatorial method (Algorithm
f) which generates a pattern that can be translated to indices corresponding to the location
of combinations in M1.
iv
Copyright, 2014, by Alexander Chizoma Nwala, All Rights Reserved.
v
This thesis is dedicated to Comfort
vi
ACKNOWLEDGMENTS
I acknowledge the exceptional guidance, patience and instruction of my thesis
supervisor Dr. Stephan Olariu, as well as the essential input and support of Dr. Michelle
C. Weigle and Dr. Hussein Abdel-Wahab of my thesis committee.
I remain grateful to my cousin Dr. Kingsley Nwala, without whom I will not be
where I am. I remain grateful to Dr. Ellis Lawrence and Ajay Gupta for exposing me to
research, Dr. Jamiiru Luttamaguzi for his unyielding guidance and Mr. Antonio Rook for
guiding me when I was a very young student.
Finally and most importantly, I remain grateful to God, and I acknowledge the
uncompromising support of my Father, Alexander E. Nwala, and the rest of my family.
vii
TABLE OF CONTENTS
Page
LIST OF FIGURES ........................................................................................................... ix
Chapter
1. INTRODUCTION ...........................................................................................................1
1.1 MOTIVATION ..................................................................................................2
2. BACKGROUND .............................................................................................................3
2.1 PERMUTATIONS .............................................................................................3
2.2 COMBINATIONS .............................................................................................5
2.3 REPRESENTING COMBINATORIAL OBJECTS .........................................7
2.4 PROPERTIES OF COMBINATORIAL ALGORITHMS ..............................12
2.5 MORE ON PLACING BALLS INTO BINS ..................................................14
2.6 COST METRIC ...............................................................................................17
3. STATE OF THE ART. ..................................................................................................18
3.1 ALGORITHM 1...............................................................................................18
3.2 ALGORITHM 2...............................................................................................19
3.3 ALGORITHM 3...............................................................................................20
3.4 ALGORITHM 4...............................................................................................21
3.5 ALGORITHM 5...............................................................................................22
4. STUDY OF THE DERIVATION OF M1 .....................................................................26
4.1 PROPERTIES OF M1 .....................................................................................31
5. STUDY TO FIND COMBINATIONS PATTERN IN M1 ...........................................33
6. TRANSLATION OF PATTERN TO ID-VARIANT 2 SOLUTION ...........................48
6.1 TRANSLATION METHOD 1 ........................................................................48
6.2 TRANSLATION METHOD 2 ........................................................................51
6.3 SUBSTITUTION INTO M1 ............................................................................53
7. BIJECTION, COST ANALYSIS AND GENERATIONG COMBINATIONS WITH
F f ..............................................................................................................................56
7.1 BIJECTION BETWEEN ID-VARIANT 2 AND COMBINATIONS ............56
7.2 COST ANALYSIS...........................................................................................56
7.3 GENERATING COMBINATIONS WITH f...................................................59
8. APPLICATIONS ...........................................................................................................60
8.1 ALGORITHM f FOR STEM CELL RESEARCH ..........................................60
viii
8.2 M1 FOR IMPLEMENTING LICENSE PLATE SERIAL FORMAT ............61
9. CONCLUSION ..............................................................................................................65
REFERENCES ..................................................................................................................67
VITA ..................................................................................................................................69
ix
LIST OF FIGURES
Figure Page
1. Permutations of Three Colored Balls .................................................................... 3
2. Four-Digit Password Choices ............................................................................... 4
3. Permutations (Left) and Combinations (Right) Over the Same Set ..................... 5
4. Specific Configuration When n = 3 and k = 3 ..................................................... 6
5. Combinations When Repetition Is Allowed In SB Notation: n = 4, k = 3 ........... 7
6. Different Ways to Represent Combinatorial Objects ........................................... 9
7. Translation of a Combinatorial Object ................................................................. 10
8. All possible Paths in a 3 × 3 Grid ......................................................................... 11
9. (73) Combinations in Lex, CoLex, Gray code and Cool-lex Order ...................... 13
10. Placing Distinguishable Balls into Distinguishable Bins ..................................... 14
11. Placing Distinguishable Balls into Distinguishable Bins (Variant 2) ................... 15
12. Algorithm 1, Generating All Combinations in Lex Order .................................... 18
13. Algorithm 2, Generating All Combinations in Lex Order .................................... 19
14. Algorithm 3, Generating Combinations in Colex Order ....................................... 20
15. Algorithm 4, Generating Combinations in Gray Code Order ............................... 21
16. Revolving Door Combinations in Gray Code Order ............................................ 22
17. Combinations; (63) ................................................................................................ 23
18. Direct Method of Generating Combinations......................................................... 25
19. Indirect Method of Generating Combinations ...................................................... 25
20. Study of the Derivation of M1 .............................................................................. 27
x
21. Non-Recursive Method for Generating Permutations (Repetitions Allowed) – M1
............................................................................................................................... 30
22. SB and M1 Pattern for Placing Five Balls into Two Bins in Bar Position Notation
(BPN) .................................................................................................................... 34
23. M1 Pattern for Placing Five Balls into Three Bins in BPN .................................. 34
24. M1 Pattern for Placing Five Balls into Four Bins in BPN .................................... 35
25. M1 Pattern for Placing Five Balls into Five Bins in BPN .................................... 36
26. Algorithm f ............................................................................................................ 44
27. Translated Pattern for n = 5, y = 2 ....................................................................... 49
28. Translation of Pattern When n = 5, y = 4 ............................................................. 51
29. Substitution into M1 Resulting from Translating p When n = 5 and k = 4 ......... 54
30. f Output for (73) in Four Different Orders ............................................................ 59
31. Prefixes of License Plate Serial Format ................................................................ 62
32. Suffixes of License Plate Serial Format................................................................ 63
33. List of License Plate Serials .................................................................................. 64
1
CHAPTER 1
INTRODUCTION
Since we live in a world of finite resources, the science of counting is very crucial
in our daily lives. Especially because the solution to many problems take the shape of
counting. For example, given the Virginia license plate format: 𝐴𝐵𝐶 − 1234, the
multiplication rule of counting tells us that there are 175,760,000 (263 × 104) license
plate choices without restriction to special characters or profanities. By means of
generating permutations with repetitions allowed, we can see that the license plate choices
begin with 𝐴𝐴𝐴 − 0000 and end with 𝑍𝑍𝑍 − 9999. The same multiplication rule tells us
that there are 4,294,967,296 (232) possible address spaces in IPv4. When we want to
know the different possible three committee members from a group of six faculty, we
employ the basic combinations rule to find out that there are 20 such combinations. In the
foregoing instances, knowing the cardinality may be a sufficient solution, but in some other
cases such as enumerating all the two-digit permutations, (as may be the case when the last
two digits of an important phone number is missing), knowing by virtue of the
multiplication rule that we must call 100 people before we find the right person is not
enough. Instead we must enumerate all such two-digit permutations starting with 00 and
ending with 99. Another case wherein enumeration is useful is the case of enumerating all
four-letter permutations, to discover the password of a simple combination lock. In this
case, knowing we have to exhaust 10,000 possibilities is not sufficient, instead we must
know the possibilities, and prefer a systematic means of generating these possibilities.
2
These counting methods discussed, belong to the field of mathematics called
combinatorics. Combinatorics deals with the arrangement of objects.
1.1 MOTIVATION
Combinatorics and its elementary elements such as permutations and combinations are
ubiquitous and crucial in mathematics. From the gamblers of the seventeenth century who
sought to understand the sample space wherein their fate was bound, to brute force attack
algorithms, combinatorics is inescapable. And any contribution to this very crucial branch
of mathematics is bound to be useful. Consequently, the goal of this research effort is to
contribute a new perspective of generating combinatorial objects, by focusing on
combinations: So far, algorithms which generate combinations do so directly, thus binding
the order of their tuple to a specific order. This research effort proposes a new way of
looking at combinations – as objects belonging to a larger set of combinatorial objects.
Therefore, combinations can be harvested from the parent combinatorial object. And if the
means to harvest combinations is based on indexing the position of combinations in the
parent combinatorial object, then the resulting combinations generation algorithm is not
bound to a specific tuple order.
3
CHAPTER 2
BACKGROUND
The fundamental science of counting is as old as mathematics. In many sciences,
counting remains the most basic form of keeping track of objects. Enumeration, which is
the counting of objects with specific properties and constraints is an important branch of
combinatorics – the branch of mathematics that deals with the arrangement of objects.
Combinatorics offers many different ways of counting such as permutations and
combinations which offer solutions to many counting problems. Permutations and
combinations also belong the class of elementary combinatorial objects.
2.1 PERMUTATIONS
A permutation refers to a specific order from the ordered arrangement of objects.
Given the finite set of all possible ways of arranging some arbitrary set of objects, an item
of the set is called a permutation. For example, given three balls (red, green, and blue), let
us consider the different ways the set of three balls can be arranged:
1
2
3
4
5
6
Fig.1 Permutations of Three Colored Balls
4
From Fig. 1, it is seen that there are six ways of arranging or permuting three different
objects. Thus, we can say that there are six elements in the set resulting from the three-
permutations of a set of six unique elements or 𝑃(6,3) = 6.
𝐺𝑖𝑣𝑒𝑛 𝑘, 1 ≤ 𝑘 ≤ 𝑛, 𝑃(𝑛, 𝑘) = 𝑛(𝑛 − 1)(𝑛 − 2)(𝑛 − 3)… (𝑛 − 𝑘 + 1) (1)
𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑠 𝑡ℎ𝑒 𝑐𝑎𝑟𝑑𝑖𝑛𝑎𝑙𝑖𝑡𝑦 𝑜𝑓 𝑘 𝑝𝑒𝑟𝑚𝑢𝑡𝑎𝑡𝑖𝑜𝑛𝑠 𝑜𝑓 𝑎 𝑠𝑒𝑡 𝑜𝑓 𝑛 𝑢𝑛𝑖𝑞𝑢𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠
𝐺𝑖𝑣𝑒𝑛 𝑘, 0 ≤ 𝑘 ≤ 𝑛, 𝑎𝑛𝑑 0! = 1,
𝑃(𝑛, 𝑘) =𝑛!
(𝑛−𝑘)! (2)
Permutations with repetition
Some problems arise where in the permutation of k objects, n objects are always
available. This means repetitive use of an object is allowed. For example, consider a four-
digit password combination lock, let us consider all the possible four-digit password
candidates. Given that each of the four positions (p1 – p4) can be occupied by integers from
0-9:
p1 p2 p3 p4
1 0 0 0 0
2 0 0 0 1
3 0 0 0 2
4 0 0 0 3
5 0 0 0 4
6 0 0 0 5
. . . . .
. . . . .
. . . . .
10000 9 9 9 9
Fig.2 Four-Digit Password Choices
5
From Fig. 2 it can be seen that there are 10,000 possible four-digit password candidates.
𝐺𝑖𝑣𝑒𝑛 𝑘 ≤ 𝑛, 𝑇ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 𝑛𝑘 𝑤𝑎𝑦𝑠 𝑜𝑓 (3)
𝑝𝑒𝑟𝑚𝑢𝑡𝑖𝑛𝑔 𝑘 𝑢𝑛𝑖𝑞𝑢𝑒 𝑜𝑏𝑗𝑒𝑐𝑡𝑠 𝑓𝑟𝑜𝑚 𝑎 𝑠𝑒𝑡 𝑜𝑓 𝑠𝑖𝑧𝑒 𝑛, 𝑤ℎ𝑒𝑛 𝑟𝑒𝑝𝑒𝑡𝑖𝑡𝑖𝑜𝑛 𝑖𝑠 𝑎𝑙𝑙𝑜𝑤𝑒𝑑
2.2 COMBINATIONS
When permutations are discussed, the order of the objects of interest is significant.
This means given the strings ABC and CAB from the three-permutation of the set of
alphabets, ABC is not the same as CAB, because order is prime. However, considering the
strings ABC and CAB under the context of a combination, they are indistinguishable. This
is because a combination is a special kind permutation when order is not of importance.
For this reason, combinations are commonly referred to as unordered selections.
Reconsidering the permutations example (Fig. 1), in the context of combinations, there is
only one possible combination resulting from the selection of three objects from a three-
set (red, green, and blue) collection. This means all six possible permutations are the same
combination:
1
2
3
4
5
6
Fig.3 Permutations (Left) and Combinations (Right) Over the Same Set
As seen from Fig. 3, combinations do not take into account order unlike permutations.
1
=
6
𝐺𝑖𝑣𝑒𝑛 𝑘, 0 ≤ 𝑘 ≤ 𝑛, 𝐶(𝑛, 𝑘) 𝑜𝑟 (𝑛𝑘) 𝑖𝑠 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑎𝑠 𝑓𝑜𝑙𝑙𝑜𝑤𝑠:
𝐶(𝑛, 𝑘) =𝑛!
𝑘!(𝑛−𝑘!) (4)
𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑠 𝑡ℎ𝑒 𝑐𝑎𝑟𝑑𝑖𝑛𝑎𝑙𝑖𝑡𝑦 𝑜𝑓 𝑘 − 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛𝑠 𝑜𝑓 𝑛 𝑑𝑖𝑠𝑖𝑛𝑐𝑡 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠
𝐺𝑖𝑣𝑒𝑛 𝑘 ≤ 𝑛, (5)
𝐶(𝑛, 𝑘) = 𝐶(𝑛, 𝑛 − 𝑘) Combinations with repetition
Combinations can be considered with or without repetitions just as permutations.
However, combinations with repetitions are more intricate. Before considering
combinations with repetitions, the following problem and its respective notation is
presented.
Placing indistinguishable balls into distinguishable bins (ID) – Variant 1
How can n (n kinds) of indistinguishable balls be placed into k distinguishable bins
- for example, how can three (n = 3) kinds (red, green, and yellow) of balls be placed into
three (k = 3) blue bins. Fig. 4 below is a specific case where three red balls are placed
within the first bin, one yellow ball is placed within the second bin, and one green within
the last bin.
Fig.4 Specific Configuration When n = 3 and k = 3
7
Stars and Bars notation (SB)
Stars and Bars [1], a notation made popular by William Feller can be used to
represent the problem of placing indistinguishable balls into distinguishable bins. Stars
represent the balls and bars - the bins. Similarly, the problem of placing n indistinguishable
balls into k distinguishable bins can be represented with n stars and k-1 bars. Fig. 4 in stars
and bars notation is:
**|*|*
The combinations problem of selecting k items from a set of n when repetition is allowed,
can be seen in terms of placing n indistinguishable balls into k distinguishable bins.
𝑇ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 𝐶(𝑛 + 𝑘 − 1, 𝑘) = 𝐶(𝑛 + 𝑘 − 1, 𝑛 − 1) 𝑤𝑎𝑦𝑠 (6)
𝑜𝑓 𝑠𝑒𝑙𝑒𝑐𝑡𝑖𝑛𝑔 𝑘 𝑖𝑡𝑒𝑚𝑠 𝑓𝑟𝑜𝑚 𝑎 𝑠𝑒𝑡 𝑜𝑓 𝑛 𝑑𝑖𝑠𝑡𝑖𝑛𝑐𝑡 𝑖𝑡𝑒𝑚𝑠 𝑤𝑖𝑡ℎ 𝑟𝑒𝑝𝑒𝑡𝑖𝑡𝑖𝑜𝑛 𝑎𝑙𝑙𝑜𝑤𝑒𝑑
||****
|*|***
|**|**
|***|*
|****|
*||***
*|*|**
*|**|*
*|***|
**||**
**|*|*
**|**|
***||*
***|*|
****||
Fig.5 Combinations When Repetition Is Allowed In SB Notation: n = 4, k = 3
2.3 REPRESENTING COMBINATORIAL OBJECTS
8
There are many ways of representing combinatorial objects such as stars and bars
notation, bit-string notation and subset notation. For example, given a set {1,2,3,4,5,6}, a
combination {1,3,6} - in subset notation, can be represented in binary string format as the
string below indicating the items which are “switched on”:
101001
In many instances, for readability, zeros are replaced with dots:
1.1..1
Fig. 6 outlines the various ways combinatorial objects such as combinations can be
represented.
9
i BPN Bitstring Subset 1 2 3 4 5 6 7
1 1 1 1 111....
1,2,3
2 1 1 2 11.1...
1,2,4
3 1 1 3 11..1..
1,2,5
4 1 1 4 11...1.
1,2,6
5 1 1 5 11....1
1,2,7
6 1 2 2 1.11...
1,3,4
7 1 2 3 1.1.1..
1,3,5
8 1 2 4 1.1..1.
1,3,6
9 1 2 5 1.1...1
1,3,7
10 1 3 3 1..11..
1,4,5
11 1 3 4 1..1.1.
1,4,6
12 1 3 5 1..1..1
1,4,7
13 1 4 4 1...11.
1,5,6
14 1 4 5 1...1.1
1,5,7
15 1 5 5 1....11
1,6,7
16 2 2 2 .111...
2,3,4
17 2 2 3 .11.1..
2,3,5
18 2 2 4 .11..1.
2,3,6
19 2 2 5 .11...1
2,3,7
20 2 3 3 .1.11..
2,4,5
21 2 3 4 .1.1.1.
2,4,6
22 2 3 5 .1.1..1
2,4,7
23 2 4 4 .1..11.
2,5,6
24 2 4 5 .1..1.1
2,5,7
25 2 5 5 .1...11
2,6,7
26 3 3 3 ..111..
3,4,5
27 3 3 4 ..11.1.
3,4,6
28 3 3 5 ..11..1
3,4,7
29 3 4 4 ..1.11.
3,5,6
30 3 4 5 ..1.1.1
3,5,7
31 3 5 5 ..1..11
3,6,7
32 4 4 4 ...111.
4,5,6
33 4 4 5 ...11.1
4,5,7
34 4 5 5 ...1.11
4,6,7
35 5 5 5 ....111
5,6,7
Fig.6 Different Ways to Represent Combinatorial Objects
10
Bar Position Notation (BPN), a notation coined in this research effort, is a
derivative of the stars and bars notation which simple takes into account the position of
bars (or 1s). For example, given the following string:
****
BPN takes into account the possible places a bar could reside:
1*2*3*4*5
Consequently, “|*|*|***” is represented as “1,2,3” in BPN.
From Fig. 6, the subset describes the combinatorial objects resulting from C(7,3). The BPN
and bit-string notations describe the combinatorial objects resulting from placing four
indistinguishable balls into four distinguishable bins. The character “1” represents the
boundary of the bins while the dots represent the balls. For this reason, the following string
”...11.1” means there are three balls in the first bin, zero balls in the second, one in the
third, and zero in the last (Fig. 7)
Finally, combinatorial objects such as combinations can be represented as paths in
an n by k grid. This is because C(n+k, k) represents all paths of length n + k consisting of
n vertical moves and k horizontal moves on the edge of an n by k grid. For example, Fig.
8 enumerates 20 (C(6, 3)) paths when n = 3 and k = 3.
...11.1 =
Fig.7 Translation of a Combinatorial Object
11
000111 001011 001101 001110
010011 010101 010110 011001
110001 110010 110100 111000
011010 011100 100011 100101
100110 101001 101010 101100
Fig.8 All Possible Paths in a 3 × 3 Grid
12
2.4 Properties of Combinatorial Algorithms
Algorithms which generate all tuples are typically of a recursive nature. On the
other hand, algorithms which given an input tuple and some other state information of some
variable(s), generate the next tuple, are typically of an iterative nature.
An Iterative algorithm which generates the next combination without a loop is
called loopless. And if the algorithm can begin from any point once given the input state
variables, the algorithm is said to be memoryless.
Many combinatorial algorithms generate combinatorial objects in different orders
such as Lexicographic or Lex, Colexicographic or CoLex, and Gray code orders.
Informally lexicographic order represents the order of tuples as seen from a
dictionary.
𝐺𝑖𝑣𝑒𝑛 𝑡𝑤𝑜 𝑠𝑒𝑡𝑠 𝐴 𝑎𝑛𝑑 𝐵, 𝑡ℎ𝑒 𝑙𝑒𝑥𝑖𝑐𝑜𝑔𝑟𝑎𝑝ℎ𝑖𝑐 𝑜𝑟𝑑𝑒𝑟 𝑜𝑣𝑒𝑟 𝑡ℎ𝑒 𝑐𝑎𝑟𝑡𝑒𝑠𝑖𝑎𝑛 𝑝𝑟𝑜𝑑𝑢𝑐𝑡: {𝑎, 𝑏} ≤ {𝑎′, 𝑏′} 𝑖𝑓𝑓 𝑎 < 𝑎′𝑜𝑟 𝑎 = 𝑎′𝑎𝑛𝑑 𝑏 ≤ 𝑏′ (7)
Informally, colexicographic order represents the order derived by applying the
following operations to all tuples:
a) Reflect tuples
b) Order lexicographically
c) Reflect tuples again
𝐴 𝑠𝑡𝑟𝑖𝑛𝑔 𝑎1𝑎2…𝑎𝑛 < 𝑏1𝑏2…𝑏𝑚 𝑖𝑠 𝑖𝑛 𝑅𝑒𝐿𝑒𝑥 𝑜𝑟𝑑𝑒𝑟 𝑖𝑓 𝑏1𝑏2…𝑏𝑚 < 𝑎1𝑎2…𝑎𝑛 (8)
𝐴 𝑠𝑡𝑟𝑖𝑛𝑔 𝑎1𝑎2…𝑎𝑛 < 𝑏1𝑏2…𝑏𝑚 𝑖𝑠 𝑖𝑛 𝐶𝑜𝐿𝑒𝑥 𝑜𝑟𝑑𝑒𝑟 𝑖𝑓 𝑎𝑛…𝑎2𝑎1 < 𝑏𝑚…𝑏2𝑏1 (9)
Gray code or minimal change order represents the order of tuples wherein
successive tuples differ by a constant number of bits. There are various possible gray
code orders such as the Cool-lex order [9].
13
Lex Colex Gray Code Cool-lex
1 ....111 111.... 111.... 111....
2 ...1.11 11.1... 1.11... .111...
3 ...11.1 1.11... .111... 1.11...
4 ...111. .111... 11.1... 11.1...
5 ..1..11 11..1.. 1..11.. .11.1..
6 ..1.1.1 1.1.1.. .1.11.. 1.1.1..
7 ..1.11. .11.1.. ..111.. .1.11..
8 ..11..1 1..11.. 1.1.1.. ..111..
9 ..11.1. .1.11.. .11.1.. 1..11..
10 ..111.. ..111.. 11..1.. 11..1..
11 .1...11 11...1. 1...11. .11..1.
12 .1..1.1 1.1..1. .1..11. 1.1..1.
13 .1..11. .11..1. ..1.11. .1.1.1.
14 .1.1..1 1..1.1. ...111. ..11.1.
15 .1.1.1. .1.1.1. 1..1.1. 1..1.1.
16 .1.11.. ..11.1. .1.1.1. .1..11.
17 .11...1 1...11. ..11.1. ..1.11.
18 .11..1. .1..11. 1.1..1. ...111.
19 .11.1.. ..1.11. .11..1. 1...11.
20 .111... ...111. 11...1. 11...1.
21 1....11 11....1 1....11 .11...1
22 1...1.1 1.1...1 .1...11 1.1...1
23 1...11. .11...1 ..1..11 .1.1..1
24 1..1..1 1..1..1 ...1.11 ..11..1
25 1..1.1. .1.1..1 ....111 1..1..1
26 1..11.. ..11..1 1...1.1 .1..1.1
27 1.1...1 1...1.1 .1..1.1 ..1.1.1
28 1.1..1. .1..1.1 ..1.1.1 ...11.1
29 1.1.1.. ..1.1.1 ...11.1 1...1.1
30 1.11... ...11.1 1..1..1 .1...11
31 11....1 1....11 .1.1..1 ..1..11
32 11...1. .1...11 ..11..1 ...1.11
33 11..1.. ..1..11 1.1...1 ....111
34 11.1... ...1.11 .11...1 1....11
35 111.... ....111 11....1 11....1
Fig.9 (73) Combinations in Lex, CoLex, Gray code and Cool-lex Order
14
2.4 MORE ON PLACING BALLS INTO BINS
The solutions to many problems present themselves as one of the different
variations of enumerating the different ways balls can be placed into bins. It must be noted
that the order in which the balls are placed within the bins does not matter. However, it
must be properly defined whether the balls and the bins are distinguishable or
indistinguishable.
Placing distinguishable balls into distinguishable bins (DD)
As seen from Fig. 10, the balls are distinguishable (red, green, and blue) as well as
the bins (B1, B2, B3, and B4). Consider a problem wherein we must count the total number
of ways four cards from a standard deck of 52 cards are to be distributed to three players.
It is clear that the cards can be seen as distinguishable balls since they are unique and the
players can be seen as distinguishable bins since they are also unique. To solve this
problem, consider that
1. 𝑇ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑝𝑙𝑎𝑦𝑒𝑟 𝑐𝑎𝑛 𝑏𝑒 𝑑𝑒𝑎𝑙𝑡 4 𝑐𝑎𝑟𝑑𝑠 𝑖𝑛 𝐶(52, 4) 𝑤𝑎𝑦𝑠 2. 𝑇ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑝𝑙𝑎𝑦𝑒𝑟 𝑐𝑎𝑛 𝑏𝑒 𝑑𝑒𝑎𝑙𝑡 4 𝑐𝑎𝑟𝑑𝑠 𝑖𝑛 𝐶(52 − 4, 4) = 𝐶(48, 4) 𝑤𝑎𝑦𝑠 3. 𝑇ℎ𝑒 𝑡ℎ𝑖𝑟𝑑 𝑝𝑙𝑎𝑦𝑒𝑟 𝑐𝑎𝑛 𝑏𝑒 𝑑𝑒𝑎𝑙𝑡 4 𝑐𝑎𝑟𝑑𝑠 𝑖𝑛 𝐶(48 − 4, 4) = 𝐶(44, 4) 𝑤𝑎𝑦𝑠 4. 𝑇ℎ𝑒 𝑓𝑜𝑢𝑟𝑡ℎ 𝑝𝑙𝑎𝑦𝑒𝑟 𝑐𝑎𝑛 𝑏𝑒 𝑑𝑒𝑎𝑙𝑡 4 𝑐𝑎𝑟𝑑𝑠 𝑖𝑛 𝐶(44 − 4, 4) = 𝐶(40, 4) 𝑤𝑎𝑦𝑠 5. 𝐵𝑎𝑠𝑒𝑑 𝑜𝑛 𝑡ℎ𝑒 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑟𝑢𝑙𝑒 𝑜𝑓 𝑐𝑜𝑢𝑛𝑡𝑖𝑛𝑔 𝑡ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 𝑎 𝑡𝑜𝑡𝑎𝑙 𝑜𝑓
6. 𝐶(52, 4) × 𝐶(48, 4) × 𝐶(44, 4) × 𝐶(40, 4) =52!
48! 4!×
48!
44! 4!×
44!
40! 4!×
40!
36! 4!
7. 𝐶(52, 4) × 𝐶(48, 4) × 𝐶(44, 4) × 𝐶(40, 4) =52!
4! 4! 4! 4!36!
B4 B1 B2 B3
Fig.10 Placing Distinguishable Balls into Distinguishable Bins
15
𝑇ℎ𝑒𝑟𝑒 𝑎𝑟𝑒𝑛!
𝑛1!𝑛2!𝑛3…𝑛𝑘! 𝑤𝑎𝑦𝑠 𝑜𝑓 𝑝𝑙𝑎𝑐𝑖𝑛𝑔 𝑛 𝑑𝑖𝑠𝑡𝑖𝑛𝑔𝑢𝑖𝑠ℎ𝑎𝑏𝑙𝑒 𝑏𝑎𝑙𝑙𝑠 𝑖𝑛𝑡𝑜 𝑘 (10)
𝑑𝑖𝑠𝑖𝑛𝑔𝑢𝑖𝑠ℎ𝑎𝑏𝑙𝑒 𝑏𝑖𝑛𝑠
Placing indistinguishable balls into distinguishable bins (ID) – Variant 2
Another variant of placing indistinguishable balls into distinguishable bins takes
into account the number of indistinguishable balls. For example, Fig. 11 describes the case
of placing four (n=4) balls into three (k=3) bins,
As seen in Fig. 11, n means n indistinguishable instances of balls (as opposed to n kinds
of balls in the Variant 1 problem).
𝑇ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 (𝑛 + 𝑘 − 1𝑘 − 1
) 𝑤𝑎𝑦𝑠 𝑜𝑓 𝑝𝑙𝑎𝑐𝑖𝑛𝑔 𝑛 𝑖𝑛𝑑𝑖𝑠𝑡𝑖𝑛𝑔𝑢𝑖𝑠ℎ𝑎𝑏𝑙𝑒 𝑏𝑎𝑙𝑙𝑠 (11)
𝑖𝑛𝑡𝑜 𝑘 𝑑𝑖𝑠𝑖𝑛𝑔𝑢𝑖𝑠ℎ𝑎𝑏𝑙𝑒 𝑏𝑖𝑛𝑠
Placing distinguishable balls into indistinguishable bins (DI)
The problem of counting the placement of distinguishable balls into
indistinguishable bins is more difficult than the already visited counterparts. In fact, there
is no closed form formula to count the compartments. However, Stirling numbers of the
second kind gives us a summation expression that equals the number of ways n
distinguishable balls can be placed into k indistinguishable bins such that no bin is empty.
It must be noted that in this case, unlike the previous visited cases (DD and ID), each bin
Fig.11 Placing Distinguishable Balls into Distinguishable Bins (Variant 2)
16
can contain any number of balls. For example, let us enumerate all the possible ways of
placing four different balls (1, 2, 3, 4) into three different bins:
1. {{1,2,3,4}}
2. {{1,2,3}, {4}}
3. {{1,2,4}, {3}}
4. {{1,3,4}, {2}}
5. {{2,3,4}, {1}}
6. {{1,2}, {3,4}}
7. {{1,3}, {2,4}}
8. {{1,4}, {2,3}}
9. {{1,2}, {3}, {4}}
10. {{1,3}, {2}, {4}}
11. {{1,4}, {2}, {3}}
12. {{2,3}, {1}, {4}}
13. {{2,4}},{1}, {3}}
14. {{3,4}, {1}, {2}}
𝑇ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑜𝑓 𝑝𝑙𝑎𝑐𝑖𝑛𝑔 𝑛 𝑑𝑖𝑠𝑡𝑖𝑛𝑔𝑢𝑖𝑠ℎ𝑎𝑏𝑙𝑒 𝑏𝑎𝑙𝑙𝑠 𝑖𝑛𝑡𝑜 𝑘 𝑖𝑛𝑑𝑖𝑠𝑡𝑖𝑛𝑔𝑢𝑖𝑠ℎ𝑎𝑏𝑙𝑒 𝑏𝑖𝑛𝑠:
∑ 𝑆(𝑛, 𝑗) = ∑1
𝑗!
𝑘𝑗=1 ∑ (−1)𝑖 (
𝑗𝑖) (𝑗 − 𝑖)𝑛
𝑗−1𝑖=0
𝑘𝑗=1 (12)
From (12), it can be seen that 𝑆(4,1) + 𝑆(4,2) + 𝑆(4,3) = 14
Placing indistinguishable balls into indistinguishable bins (II)
Consider the following problem: Express n as the sum of at most k positive integers
in a decreasing order. An expression:
∑ 𝑥𝑖 = 𝑛𝑗𝑖=1 , where 𝑥1 ≥ 𝑥2 ≥ 𝑥3 ≥ ⋯ ≥ 𝑥𝑗 = 𝑛,
where 𝑥𝑖 is a positive integer, is called a partition of n into j integers. Consequently, the
number of partitions of n integers into at most k positive integers is equal to the number of
ways of placing n indistinguishable balls into k indistinguishable bins. As is the case of
placing distinguishable balls into indistinguishable bins, unfortunately, there is no closed
form formula for counting the partitions. Given n = 6, the following are the partitions:
6 = 6
5 + 1 = 6
4 + 2 = 6
4 + 1 + 1 = 6
3 + 3 = 6
3 + 2 + 1 = 6
17
3 + 1 + 1 + 1 = 6
2 + 2 + 2 = 6
2 + 2 + 1 + 1 = 6
2.5 COST METRIC
Given a set of algorithms that achieve the same goal, most often, the deciding factor
which places an algorithm on top of the list is the cost of the algorithm based on some cost
measure. It is fair to say that one of the most desirable properties for an algorithm which
generates combinatorial objects is the Constant Amortized Time (CAT) property.
CAT
The CAT property is based on the idea that the computational cost of an algorithm is
proportional to the size of the output of the algorithm. This means that the time to process
and output of the combinatorial objects is not considered. For example, given a
combinatorial algorithm L with input x and output of X objects (each of size x), then L has
the CAT property since its running time is O(X).
18
CHAPTER 3
STATE OF THE ART
Combinations belong to the class of elementary combinatorial objects, so it comes
as no surprise that there are many algorithms for generating combinations. In spite of this,
a lot of these algorithms exhibit very similar properties such as the order of their tuples,
recursion (recursive or iterative), loop (loop or loopless), and memory property. In this
chapter, five algorithms are presented.
3.1 ALGORITHM 1
Consider Algorithm 1 [2] which generates all k-Combinations 𝑐𝑘…𝑐2𝑐1 of n
numbers:,0, 1, 2, … , 𝑛 − 1 in Lex order.
𝐺𝑖𝑣𝑒𝑛 𝑛 ≥ 𝑘 ≥ 0, 𝑎𝑛𝑑 𝑠𝑒𝑛𝑡𝑖𝑛𝑒𝑙 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 𝑐𝑘+1 𝑎𝑛𝑑 𝑐𝑘+2 𝒑𝒓𝒐𝒄𝒆𝒅𝒖𝒓𝒆 𝐿() 𝐼𝑛𝑖𝑡𝑖𝑎𝑙𝑖𝑧𝑒(); 𝒑𝒓𝒐𝒄𝒆𝒅𝒖𝒓𝒆 𝐼𝑛𝑖𝑡𝑖𝑎𝑙𝑖𝑧𝑒()
𝒘𝒉𝒊𝒍𝒆 1 = 1 𝒇𝒐𝒓 𝑗 = 1 𝒕𝒐 𝑘 𝒅𝒐
𝑉𝑖𝑠𝑖𝑡 𝑐𝑘 …𝑐2𝑐1; 𝑐𝑗 = 𝑗 − 1;
𝑗 = 1; 𝒆𝒏𝒅𝒇𝒐𝒓
𝒘𝒉𝒊𝒍𝒆 𝑐𝑗 + 1 = 𝑐𝑗+1 𝑐𝑘+1 = 𝑛;
𝑐𝑗 = 𝑗 − 1; 𝑐𝑘+2 = 0;
𝑗 = 𝑗 + 1; 𝒆𝒏𝒅
𝒆𝒏𝒅𝒘𝒉𝒊𝒍𝒆 𝒊𝒇 𝑗 > 𝑘 𝒕𝒉𝒆𝒏 𝑻𝒆𝒓𝒎𝒊𝒏𝒂𝒕𝒆; 𝒆𝒏𝒅𝒊𝒇
𝑐𝑗 = 𝑐𝑗 + 1;
𝒆𝒏𝒅𝒘𝒉𝒊𝒍𝒆 𝒆𝒏𝒅
Fig.12 Algorithm 1, Generating All Combinations in Lex Order
19
In addition, Algorithm 1 does not have the CAT property. Another algorithm which
generates combinations in Lex order is Algorithm 2 [8]. Algorithm 2 is based upon the
following recurrence relation (13):
(𝑛𝑘) =
{
1 ; 𝑛 = 0
(𝑛 − 10
) ; 𝑘 = 0
(𝑛 − 1𝑛 − 1
) ; 𝑘 = 𝑛
(𝑛 − 1𝑘
) + (𝑛 − 1𝑘 − 1
) ; 0 < 𝑘 < 𝑛
(13)
3.2 ALGORITHM 2
𝐺𝑖𝑣𝑒𝑛, 𝑗: ℕ,𝑚: ℕ, 𝑎𝑛𝑑 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑐𝑎𝑙𝑙 𝑡𝑜 𝑝𝑟𝑜𝑐𝑒𝑑𝑢𝑟𝑒 𝐶𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛(1, 0) 𝒑𝒓𝒐𝒄𝒆𝒅𝒖𝒓𝒆 𝐶𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛(𝑗,𝑚) 𝒊𝒇 𝑗 > 𝑛 𝒕𝒉𝒆𝒏 𝑉𝑖𝑠𝑖𝑡 𝑏1𝑏2𝑏3…𝑏𝑛; 𝒆𝒍𝒔𝒆 𝒊𝒇 𝑘 − 𝑚 < 𝑛 − 𝑗 + 1 𝒕𝒉𝒆𝒏 𝑏𝒋 = 0;
𝐶𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛(𝑗 + 1,𝑚); 𝒆𝒏𝒅𝒊𝒇 𝒊𝒇 𝑚 < 𝑘 𝒕𝒉𝒆𝒏 𝑏𝒋 = 1;
𝐶𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛(𝑗 + 1, 𝑚 + 1); 𝒆𝒏𝒅𝒊𝒇 𝒆𝒏𝒅𝒊𝒇 𝒆𝒏𝒅
Algorithm 2 just like algorithm 1 does not have the CAT property. However, Algorithm 2
is recursive – hence the simplicity. Also note that this algorithm generates combinations in
binary string format.
Fig.13 Algorithm 2, Generating All Combinations in Lex Order
20
The algorithms explored so far generate combinations in Lex order. Consequently,
let us consider a recursive algorithm [3] which generates combinations in CoLex order.
Note that Algorithm 3 generates combinations in binary string format. Algorithm 3 is based
on the following recurrence relation (14):
𝑡(𝑛, 𝑘) = {
1 ; 𝑛 = 01 + 𝑡(𝑛 − 1, 0) ; 𝑘 = 0
1 + 𝑡(𝑛 − 1, 𝑛 − 1) ; 𝑘 = 𝑛
1 + 𝑡(𝑛 − 1, 𝑘) + 𝑡(𝑛 − 1, 𝑘 − 1) ; 0 < 𝑘 < 𝑛
3.3 ALGORITHM 3
Similar to the already explored algorithms, Algorithm 3 does not have the CAT property
since
𝑡(𝑛,𝑘)+1
(𝑛𝑘)
=(𝑛+2)(𝑛+1)
𝑘+1(𝑛−𝑘+1) (15)
𝒑𝒓𝒐𝒄𝒆𝒅𝒖𝒓𝒆 𝐶𝑜𝑙𝑒𝑥( 𝑛: ℕ, 𝑘: ℕ )
𝒊𝒇 𝑛 = 0 𝒕𝒉𝒆𝒏 𝑉𝑖𝑠𝑖𝑡 𝑏1𝑏2𝑏3…𝑏𝑛; 𝒆𝒍𝒔𝒆 𝒊𝒇 𝑘 < 𝑛 𝒕𝒉𝒆𝒏
𝑏𝑛 = 0; 𝐶𝑜𝑙𝑒𝑥( 𝑛 − 1, 𝑘 ); 𝒆𝒏𝒅𝒊𝒇 𝒊𝒇 𝑘 > 0 𝒕𝒉𝒆𝒏
𝑏𝑛 = 1; 𝐶𝑜𝑙𝑒𝑥( 𝑛 − 1, 𝑘 − 1 ); 𝒆𝒏𝒅𝒊𝒇 𝒆𝒏𝒅𝒊𝒇 𝒆𝒏𝒅
(14)
Fig.14 Algorithm 3, Generating Combinations in Colex Order
21
The algorithms considered so far do not have the CAT property. Consequently,
Algorithm 4 [4] below has this property. Also note Algorithm 4 generates combinations in
minimal change or gray code order (successive binary strings differ by a constant number
of bits).
3.4 ALGORITHM 4
𝐺𝑖𝑣𝑒𝑛, 𝑛: ℕ, 𝑘: ℕ, 𝑎𝑛𝑑 𝑔1𝑔2𝑔3…𝑔𝑛
𝒑𝒓𝒐𝒄𝒆𝒅𝒖𝒓𝒆 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒(𝑛, 𝑘) 𝒑𝒓𝒐𝒄𝒆𝒅𝒖𝒓𝒆 𝐼𝑛𝑖𝑡𝑖𝑎𝑙𝑖𝑧𝑒()
𝐼𝑛𝑖𝑡𝑖𝑎𝑙𝑖𝑧𝑒(); 𝒇𝒐𝒓 𝑖 = 1 𝒕𝒐 𝑘 𝒅𝒐
𝑉𝑖𝑠𝑖𝑡 𝑔1𝑔2𝑔3…𝑔𝑛; 𝑔𝑖= 1;
𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒1(𝑛, 𝑘); 𝒆𝒏𝒅𝒇𝒐𝒓
𝒆𝒏𝒅 𝒇𝒐𝒓 𝑖 = 𝑘 + 1 𝒕𝒐 𝑛 𝒅𝒐
𝒑𝒓𝒐𝒄𝒆𝒅𝒖𝒓𝒆 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒1(𝑛, 𝑘) 𝑔𝑖= 0;
𝒊𝒇 𝑘 > 0 𝒂𝒏𝒅 𝑘 < 𝑛 𝒕𝒉𝒆𝒏 𝒆𝒏𝒅𝒇𝒐𝒓
𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒1(𝑛 − 1, 𝑘); 𝒆𝒏𝒅
𝒊𝒇 𝑘 = 1 𝒕𝒉𝒆𝒏 𝒑𝒓𝒐𝒄𝒆𝒅𝒖𝒓𝒆 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒2(𝑛, 𝑘)
𝑔𝑛= 1; 𝒊𝒇 𝑘 > 0 𝒂𝒏𝒅 𝑘 < 𝑛 𝒕𝒉𝒆𝒏
𝑔𝑛−1
= 0; 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒1(𝑛 − 1, 𝑘 − 1);
𝑉𝑖𝑠𝑖𝑡 𝑔1𝑔2𝑔3…𝑔𝑛; 𝒊𝒇 𝑘 = 1 𝒕𝒉𝒆𝒏
𝒆𝒍𝒔𝒆 𝑔𝑛−1
= 1;
𝑔𝑛= 1; 𝑔
𝑛= 0;
𝑔𝑘−1
= 0; 𝑉𝑖𝑠𝑖𝑡 𝑔1𝑔2𝑔3…𝑔𝑛;
𝑉𝑖𝑠𝑖𝑡 𝑔1𝑔2𝑔3…𝑔𝑛; 𝒆𝒍𝒔𝒆
𝒆𝒏𝒅𝒊𝒇 𝑔𝑘−1
= 1;
𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒2(𝑛 − 1, 𝑘 − 1); 𝑔
𝑛= 0;
𝒆𝒏𝒅𝒊𝒇 𝑉𝑖𝑠𝑖𝑡 𝑔1𝑔2𝑔3…𝑔𝑛; 𝒆𝒏𝒅 𝒆𝒏𝒅𝒊𝒇
𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒2(𝑛 − 1, 𝑘);
𝒆𝒏𝒅𝒊𝒇
𝒆𝒏𝒅
Fig.15 Algorithm 4, Generating Combinations in Gray Code Order
22
3.5 ALGORITHM 5
Finally, let us consider the CAT “Revolving – door combinations” [5], which
generates all k-Combinations in Lexicographic order based on the following sequence:
(𝑐𝑘, −𝑐𝑘−1, 𝑐𝑘−2, … , (−1)𝑘−1𝑐1) (16)
𝐺𝑖𝑣𝑒𝑛 𝑛 ≥ 𝑘 ≥ 1, 𝑐1𝑐2𝑐3…𝑐𝑘 , 𝑠𝑒𝑛𝑡𝑖𝑛𝑒𝑙 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑐𝑘+1, 𝑎𝑛𝑑 𝑔𝑙𝑜𝑏𝑎𝑙 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑗. 𝒑𝒓𝒐𝒄𝒆𝒅𝒖𝒓𝒆 𝑅()
𝐼𝑛𝑖𝑡𝑖𝑎𝑙𝑖𝑧𝑒(); 𝒇𝒖𝒏𝒕𝒊𝒐𝒏 𝑅1()
𝒓𝒆𝒑𝒆𝒂𝒕 𝒊𝒇 𝑗 = 𝑘 𝒕𝒉𝒆𝒏 𝑉𝑖𝑠𝑖𝑡 𝑐𝑘 …𝑐2𝑐1; 𝒓𝒆𝒕𝒖𝒓𝒏 𝑓𝑎𝑙𝑠𝑒; 𝒖𝒏𝒕𝒊𝒍 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒() = 𝑓𝑎𝑙𝑠𝑒; 𝒆𝒏𝒅𝒊𝒇 𝒆𝒏𝒅 𝒊𝒇 𝑐𝑗 > 𝑗 𝒕𝒉𝒆𝒏
𝒑𝒓𝒐𝒄𝒆𝒅𝒖𝒓𝒆 𝐼𝑛𝑖𝑡𝑖𝑎𝑙𝑖𝑧𝑒() 𝑐𝑗 = 𝑐𝑗−1;
𝒇𝒐𝒓 𝑖 = 1 𝑡𝑜 𝑘 𝒅𝒐 𝑐𝑗−1 = 𝑗 − 1;
𝑐𝑖 = 𝑖; 𝒓𝒆𝒕𝒖𝒓𝒏 𝑡𝑟𝑢𝑒; 𝒆𝒏𝒅𝒇𝒐𝒓 𝒆𝒏𝒅𝒊𝒇
𝑐𝑘+1 = 𝑛; 𝑗 = 𝑗 + 1; 𝒆𝒏𝒅 𝒓𝒆𝒕𝒖𝒓𝒏 𝑅2(); 𝒇𝒖𝒏𝒕𝒊𝒐𝒏 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒() 𝒆𝒏𝒅 𝑗 = 1; 𝒇𝒖𝒏𝒕𝒊𝒐𝒏 𝑅2() 𝒊𝒇 𝑘 𝑖𝑠 𝑜𝑑𝑑 𝒕𝒉𝒆𝒏 𝒊𝒇 𝑗 = 𝑘 𝒕𝒉𝒆𝒏 𝒊𝒇 𝑐1 + 1 < 𝑐1 𝒕𝒉𝒆𝒏 𝒓𝒆𝒕𝒖𝒓𝒏 𝑓𝑎𝑙𝑠𝑒;
𝑐1 = 𝑐1 + 1; 𝒆𝒏𝒅𝒊𝒇 𝒓𝒆𝒕𝒖𝒓𝒏 𝑡𝑟𝑢𝑒; 𝒊𝒇 𝑐𝑗 + 1 < 𝑐𝑗+1 𝒕𝒉𝒆𝒏
𝒆𝒍𝒔𝒆 𝑐𝑗−1 = 𝑐𝑗;
𝒓𝒆𝒕𝒖𝒓𝒏 𝑅1(); 𝑐𝑗 = 𝑐𝑗 + 1;
𝒆𝒏𝒅𝒊𝒇 𝒓𝒆𝒕𝒖𝒓𝒏 𝑡𝑟𝑢𝑒; 𝒆𝒍𝒔𝒆 𝒆𝒏𝒅𝒊𝒇 𝒊𝒇 𝑐1 > 0 𝒕𝒉𝒆𝒏 𝑗 = 𝑗 + 1;
𝑐1 = 𝑐1 − 1; 𝒓𝒆𝒕𝒖𝒓𝒏 𝑅1(); 𝒓𝒆𝒕𝒖𝒓𝒏 𝑡𝑟𝑢𝑒; 𝒆𝒏𝒅 𝒆𝒍𝒔𝒆
𝒓𝒆𝒕𝒖𝒓𝒏 𝑅2(); 𝒆𝒏𝒅𝒊𝒇 𝒆𝒏𝒅𝒊𝒇 𝒆𝒏𝒅
Fig.16 Revolving Door Combinations in Gray Code Order
23
Algorithm 1.0 2.0 3.0 4.0 5.0
1 2 3 ...111 111... 111... 2 1 0
1 2 4 ..1.11 11.1.. 1.11.. 3 2 0
1 3 4 ..11.1 1.11.. .111.. 3 2 1
2 3 4 ..111. .111.. 11.1.. 3 1 0
1 2 5 .1..11 11..1. 1..11. 4 3 0
1 3 5 .1.1.1 1.1.1. .1.11. 4 3 1
2 3 5 .1.11. .11.1. ..111. 4 3 2
1 4 5 .11..1 1..11. 1.1.1. 4 2 0
2 4 5 .11.1. .1.11. .11.1. 4 2 1
3 4 5 .111.. ..111. 11..1. 4 1 0
1 2 6 1...11 11...1 1...11 5 4 0
1 3 6 1..1.1 1.1..1 .1..11 5 4 1
2 3 6 1..11. .11..1 ..1.11 5 4 2
1 4 6 1.1..1 1..1.1 ...111 5 4 3
2 4 6 1.1.1. .1.1.1 1..1.1 5 3 0
3 4 6 1.11.. ..11.1 .1.1.1 5 3 1
1 5 6 11...1 1...11 ..11.1 5 3 2
2 5 6 11..1. .1..11 1.1..1 5 2 0
3 5 6 11.1.. ..1.11 .11..1 5 2 1
4 5 6 111... ...111 11...1 5 1 0
Fig.17 Combinations;(63)
There are many more algorithm which generate combinations such as Eades-
McKay algorithm [6] which generates combinations in gray code order, and a combinations
generating algorithm based on prefix shifts [9] which generates combinations in Cool-lex
order and [7]. However, these algorithms share a lot of properties in common such as the
order of their tuples (Lex, CoLex, Gray code, etc), as well as efficiency (CAT property).
So far the algorithms we have explored and many other combinations generating
algorithms are tied to a specific tuple order. This is because these algorithms generate
combinations directly. This direct generation of combinations need not be the case: the
(63)
24
following work outlined in the subsequent chapters explores a different method of
generating combinations, an indirect method. The consequence of this is a flexible means
of generating combinations wherein the generator defines the order of tuples. There are
two main objects which support generating combinations indirectly, namely the parent
combinatorial method, as well as the indexing combinatorial method. Since the parent
combinatorial object contains a superset of combinatorial objects including combinations,
all which is required, is a means to generate the indices corresponding to the location of
combinations. This is where the indexing method comes into play. This research effort
defines two objects required to generate combinations indirectly as follows:
Parent Combinatorial Method 1 (M1): is a non-recursive CAT algorithm for generating
permutations with repetitions allowed. M1 serves as the parent method since it contains
different kinds of combinatorial objects including combinations. Refer to chapter 2, to
explore the location of combinations in M1.
Indexing Combinatorial Method (f): this method generates a pattern which can be
translated to indices corresponding to the location of combinations in M1.
25
As outlined by Fig. 19, the parent combinatorial method (M1) contains other combinatorial
objects besides combinations. Hence, the indexing combinatorial method (f) is needed to
extract combinations from the parent method. This means that the indexing combinatorial
method has knowledge of the locations of combinations in the parent combinatorial
method.
(𝑛
𝑘) Combinations
Algorithm to generate
Combinations
…
combinations object 1
Other combinatorial objects
combinations object 2
M1
(𝑛
𝑘) Combinations f
Fig.18 Direct Method of Generating Combinations
Fig.19 Indirect Method of Generating Combinations
26
CHAPTER 4
STUDY OF THE DERIVATION OF M1
Let us consider the following study leading to the derivation of M1 – a CAT non-
recursive method for generating permutations with repetitions allowed. The goal is to find
a pattern and translate the language of the pattern into a series of mathematical formulae.
To do this, let us consider sections from six-permutations with repetitions allowed for a set
𝑈 = {𝐴, 𝐵, 𝐶}. As seen in Fig. 20, all columns from 1-6 count from the smallest alphabet
to the largest alphabet but do this in different ways. Fig. 20 is color coded to emphasize the
trends in the pattern. For instance column 1 reaches the largest alphabet quickly because it
does not repeat (no consecutive alphabets are the same). When column 1 reaches the largest
alphabet, it starts the counting process all over again. Column 2 counts but repeats the
counting alphabet, so it reaches the largest alphabet less quick when compared to column
1. When column 2 reaches the largest alphabet, it restarts the counting process. Columns
3, 4, and 5 behave in a similar way, the difference being the number of consecutive
alphabets. The last column counts with repetition but does not restart the count process.
The trend remains that column 1 is the fastest to reach the largest alphabet, column 2 is the
second fastest and so on. Also column 1 does not have consecutive characters of the same
type, column 2 counts with 3 consecutive characters of the same type, column 3 – 9, column
4 – 27, column 5 – 81, column 6 – 243.
27
Column 6 5 4 3 2 1 6 5 4 3 2 1 6 5 4 3 2 1
1 A A A A A A 38 A A B B A B 75 A A C C A C
2 A A A A A B 39 A A B B A C 76 A A C C B A
3 A A A A A C 40 A A B B B A 77 A A C C B B
4 A A A A B A 41 A A B B B B 78 A A C C B C
5 A A A A B B 42 A A B B B C 79 A A C C C A
6 A A A A B C 43 A A B B C A 80 A A C C C B
7 A A A A C A 44 A A B B C B 81 A A C C C C
8 A A A A C B 45 A A B B C C 82 A B A A A A
9 A A A A C C 46 A A B C A A 83 A B A A A B
10 A A A B A A 47 A A B C A B 84 A B A A A C
11 A A A B A B 48 A A B C A C 85 A B A A B A
12 A A A B A C 49 A A B C B A 86 A B A A B B
13 A A A B B A 50 A A B C B B 87 A B A A B C
14 A A A B B B 51 A A B C B C 88 A B A A C A
15 A A A B B C 52 A A B C C A 89 A B A A C B
16 A A A B C A 53 A A B C C B 90 A B A A C C
17 A A A B C B 54 A A B C C C 91 A B A B A A
18 A A A B C C 55 A A C A A A 92 A B A B A B
19 A A A C A A 56 A A C A A B 93 A B A B A C
20 A A A C A B 57 A A C A A C 94 A B A B B A
21 A A A C A C 58 A A C A B A 95 A B A B B B
22 A A A C B A 59 A A C A B B 96 A B A B B C
23 A A A C B B 60 A A C A B C 97 A B A B C A
24 A A A C B C 61 A A C A C A 98 A B A B C B
25 A A A C C A 62 A A C A C B 99 A B A B C C
26 A A A C C B 63 A A C A C C 100 A B A C A A
27 A A A C C C 64 A A C B A A 101 A B A C A B
28 A A B A A A 65 A A C B A B 102 A B A C A C
29 A A B A A B 66 A A C B A C 103 A B A C B A
30 A A B A A C 67 A A C B B A 104 A B A C B B
31 A A B A B A 68 A A C B B B 105 A B A C B C
32 A A B A B B 69 A A C B B C 106 A B A C C A
33 A A B A B C 70 A A C B C A 107 A B A C C B
34 A A B A C A 71 A A C B C B 108 A B A C C C
35 A A B A C B 72 A A C B C C 109 A B B A A A
36 A A B A C C 73 A A C C A A 110 A B B A A B
37 A A B B A A 74 A A C C A B 111 A B B A A C
Fig.20 Study of the Derivation of M1
28
Fig.20 Continued
Column 6 5 4 3 2 1 6 5 4 3 2 1 6 5 4 3 2 1
370 B B B C A A 411 B C A A B C 452 B C B C A B
371 B B B C A B 412 B C A A C A 453 B C B C A C
372 B B B C A C 413 B C A A C B 454 B C B C B A
373 B B B C B A 414 B C A A C C 455 B C B C B B
374 B B B C B B 415 B C A B A A 456 B C B C B C
375 B B B C B C 416 B C A B A B 457 B C B C C A
376 B B B C C A 417 B C A B A C 458 B C B C C B
377 B B B C C B 418 B C A B B A 459 B C B C C C
378 B B B C C C 419 B C A B B B 460 B C C A A A
379 B B C A A A 420 B C A B B C 461 B C C A A B
380 B B C A A B 421 B C A B C A 462 B C C A A C
381 B B C A A C 422 B C A B C B 463 B C C A B A
382 B B C A B A 423 B C A B C C 464 B C C A B B
383 B B C A B B 424 B C A C A A 465 B C C A B C
384 B B C A B C 425 B C A C A B 466 B C C A C A
385 B B C A C A 426 B C A C A C 467 B C C A C B
386 B B C A C B 427 B C A C B A 468 B C C A C C
387 B B C A C C 428 B C A C B B 469 B C C B A A
388 B B C B A A 429 B C A C B C 470 B C C B A B
389 B B C B A B 430 B C A C C A 471 B C C B A C
390 B B C B A C 431 B C A C C B 472 B C C B B A
391 B B C B B A 432 B C A C C C 473 B C C B B B
392 B B C B B B 433 B C B A A A 474 B C C B B C
393 B B C B B C 434 B C B A A B 475 B C C B C A
394 B B C B C A 435 B C B A A C 476 B C C B C B
395 B B C B C B 436 B C B A B A 477 B C C B C C
396 B B C B C C 437 B C B A B B 478 B C C C A A
397 B B C C A A 438 B C B A B C 479 B C C C A B
398 B B C C A B 439 B C B A C A 480 B C C C A C
399 B B C C A C 440 B C B A C B 481 B C C C B A
400 B B C C B A 441 B C B A C C 482 B C C C B B
401 B B C C B B 442 B C B B A A 483 B C C C B C
402 B B C C B C 443 B C B B A B 484 B C C C C A
403 B B C C C A 444 B C B B A C 485 B C C C C B
404 B B C C C B 445 B C B B B A 486 B C C C C C
405 B B C C C C 446 B C B B B B 487 C A A A A A
406 B C A A A A 447 B C B B B C 488 C A A A A B
29
Fig.20 Continued
Column 6 5 4 3 2 1 6 5 4 3 2 1 6 5 4 3 2 1
620 C B B C C B 661 C C A B B A 702 C C B C C C
621 C B B C C C 662 C C A B B B 703 C C C A A A
622 C B C A A A 663 C C A B B C 704 C C C A A B
623 C B C A A B 664 C C A B C A 705 C C C A A C
624 C B C A A C 665 C C A B C B 706 C C C A B A
625 C B C A B A 666 C C A B C C 707 C C C A B B
626 C B C A B B 667 C C A C A A 708 C C C A B C
627 C B C A B C 668 C C A C A B 709 C C C A C A
628 C B C A C A 669 C C A C A C 710 C C C A C B
629 C B C A C B 670 C C A C B A 711 C C C A C C
630 C B C A C C 671 C C A C B B 712 C C C B A A
631 C B C B A A 672 C C A C B C 713 C C C B A B
632 C B C B A B 673 C C A C C A 714 C C C B A C
633 C B C B A C 674 C C A C C B 715 C C C B B A
634 C B C B B A 675 C C A C C C 716 C C C B B B
635 C B C B B B 676 C C B A A A 717 C C C B B C
636 C B C B B C 677 C C B A A B 718 C C C B C A
637 C B C B C A 678 C C B A A C 719 C C C B C B
638 C B C B C B 679 C C B A B A 720 C C C B C C
639 C B C B C C 680 C C B A B B 721 C C C C A A
640 C B C C A A 681 C C B A B C 722 C C C C A B
641 C B C C A B 682 C C B A C A 723 C C C C A C
642 C B C C A C 683 C C B A C B 724 C C C C B A
643 C B C C B A 684 C C B A C C 725 C C C C B B
644 C B C C B B 685 C C B B A A 726 C C C C B C
645 C B C C B C 686 C C B B A B 727 C C C C C A
646 C B C C C A 687 C C B B A C 728 C C C C C B
647 C B C C C B 688 C C B B B A 729 C C C C C C
648 C B C C C C 689 C C B B B B
649 C C A A A A 690 C C B B B C
650 C C A A A B 691 C C B B C A
651 C C A A A C 692 C C B B C B
652 C C A A B A 693 C C B B C C
653 C C A A B B 694 C C B C A A
654 C C A A B C 695 C C B C A B
655 C C A A C A 696 C C B C A C
656 C C A A C B 697 C C B C B A
30
Following extensive study of the pattern from multiple permutations from different sets led
to the derivation of M1.
Given M1, the following is a usage example: Generate six-permutations (repetition
allowed) from the set 𝑈 = {𝐴, 𝐵, 𝐶} 𝑜𝑟 𝑈 = {1, 2, 3} for i = 729. Note that M1 disregards
fractions.
𝐼𝑛 𝑜𝑟𝑑𝑒𝑟 𝑡𝑜 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒 𝑡ℎ𝑒 𝑟 − 𝑝𝑒𝑟𝑚𝑢𝑡𝑎𝑡𝑖𝑜𝑛𝑠 𝑜𝑓 𝑎 𝑠𝑒𝑡 𝑤𝑖𝑡ℎ 𝑟𝑒𝑝𝑒𝑡𝑖𝑡𝑖𝑜𝑛 𝑎𝑙𝑙𝑜𝑤𝑒𝑑, 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑚𝑒𝑡ℎ𝑜𝑑 𝑀1 𝑖𝑠 𝑑𝑒𝑓𝑖𝑛𝑒𝑑:
𝐺𝑖𝑣𝑒𝑛 𝑎 𝑠𝑒𝑡 𝑈 𝑤𝑖𝑡ℎ 𝑓𝑖𝑟𝑠𝑡 𝑡𝑒𝑟𝑚 𝐴, |𝑈| = 𝑛, 1 ≤ 𝑖 ≤ 𝑛𝑟
𝑔1(𝑖) 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑠 𝑡ℎ𝑒 𝑙𝑎𝑠𝑡 𝑡𝑒𝑟𝑚
𝑔2(𝑖) 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑠 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑡𝑜 𝑙𝑎𝑠𝑡 𝑡𝑒𝑟𝑚
… 𝑔𝑗(𝑖) 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑠 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑡𝑒𝑟𝑚
𝑓(𝑖) = {
𝑖
𝑛− 1, 𝑖 𝑚𝑜𝑑 𝑛 = 0
𝑖
𝑛, 𝑖 𝑚𝑜𝑑 𝑛 ≠ 0
𝑔1(𝑖) = 𝐴 + 𝑖 − 𝑛𝑓(𝑖) − 1
𝑔2(𝑖) = {𝐴 + 𝑓(𝑖), 𝑓(𝑖) < 𝑛
𝐴 + (𝑓(𝑖) 𝑚𝑜𝑑 𝑛), 𝑓(𝑖) ≥ 𝑛
𝑔3(𝑖) =
{
𝐴 , 𝑓(𝑖) < 𝑛
𝐴 +𝑓(𝑖)
𝑛 , 𝑛 ≤ 𝑓(𝑖) < 𝑛2
𝐴 + (𝑓(𝑖)
𝑛− 𝑛)𝑚𝑜𝑑 𝑛, 𝑓(𝑖) ≥ 𝑛2
𝑔𝑗(𝑖) = {
𝐴, 𝑓(𝑖) < 𝑛𝑗−2
𝐴 + ((𝑓(𝑖)
𝑛𝑗−2)𝑚𝑜𝑑 𝑛) , 𝑓(𝑖) ≥ 𝑛𝑗−2
Fig.21 Non-Recursive Method for Generating Permutations (Repetitions Allowed) – M1
31
Solution: 𝑖 = 729, 𝑛 = 3, 𝐴 = 1
𝑓(729) =729
3− 1 = 242
𝑔1(729) = 1 + 729 − 3(242) − 1 = 3 𝑜𝑟 𝐶
𝑔2(729) = 1 + (242 𝑚𝑜𝑑 3) = 3 𝑜𝑟 𝐶
𝑔3(729) = 1 + ((242
3− 3) 𝑚𝑜𝑑 3) = 3 𝑜𝑟 𝐶
𝑔4(729) = 1 + ((242
34−2) 𝑚𝑜𝑑 3) = 3 𝑜𝑟 𝐶
𝑔5(729) = 1 + ((242
35−2) 𝑚𝑜𝑑 3) = 3 𝑜𝑟 𝐶
𝑔6(729) = 1 + ((242
36−2) 𝑚𝑜𝑑 3) = 3 𝑜𝑟 𝐶
4.1 PROPERTIES OF M1
M1 is non-recursive, loopless (finding the ith r-permutation does not involve a
loop), memoryless (can be started from anywhere given) and is of the CAT family because
given the ith permutation, the computation required to compute the (i+1)th permutation is
constant – involving a fix number of operations, without any looping. Another interesting
feature of M1 is the fact that in the listing of all the tuples of the r-permutations (repetition
allowed) of a set, M1 is capable of listing tuples in multiple order (e.g lex, reverse lex, etc).
Consequently, any indexing algorithm which generates integers (1 ≤ 𝑟 ≤ 𝑛𝑟) in a given
order O corresponding to the r-permutations of a set can be used to generate combinatorial
objects from M1 in order O. For example, if we generate integers (1 ≤ 𝑟 ≤ 𝑛𝑟) from
smallest to largest and index M1, the combinatorial objects resulting from such operation
will be in lexicographic order. If the integers are generated from largest to smallest and
32
substituted into M1 in this order, the resulting combinatorial objects will be in reverse
lexicographic order.
33
CHAPTER 5
STUDY TO FIND COMBINATIONS PATTERN IN M1
Following the derivation of a highly efficient non-recursive method for generating
permutation with repetitions allowed (M1) – the parent combinatorial method, we should
now focus of how we can extract combinations from M1. To do this, we start from the
problem of placing n indistinguishable balls (of the same kind), into k distinguishable bins
(ID-Variant 2). This approach is justified by the equivalence relation in chapter 4 which
outlines a bijection relationship between ID-Variant 2 and the combinations problem.
Premise for solution
Consider the following:
1. SB – Enumeration of Stars and Bars for the balls and bins problem
(ID-Variant 2)
2. M1 (given) – an iterative method of generating permutations with
repetitions allowed.
3. A mirror image upper triangular matrix - an upper-matrix which has
a diagonal that runs from the top right to the bottom left (unlike the
conventional top left to bottom right diagonal of an upper triangular
matrix).
The problem of placing balls into bins corresponds to indexing M1. Therefore, the
following study seeks to investigate the pattern generated by indexing M1 as it corresponds
to placing five balls into one, and then two, three, four, and finally five bins.
Case SB(5,2), M1(6,1):
34
SB M1
|***** 1
*|**** 2
**|*** 3
***|** 4
****|* 5
*****| 6 Fig.22 SB and M1 Pattern for Placing Five Balls into Two Bins in Bar Position Notation (BPN)
Case SB(5,3), M1(6,2):
1,1 2,2 3,3 4,4 5,5 6,6
1,2 2,3 3,4 4,5 5,6
1,3 2,4 3,5 4,6 6,1
1,4 2,5 3,6 5,1 6,2
1,5 2,6 4,1 5,2 6,3
1,6 3,1 4,2 5,3 6,4
2,1 3,2 4,3 5,4 6,5 Fig.23 M1 Pattern for Placing Five Balls into Three Bins in BPN
Case SB(5,4), M1(6,3):
35
1,1,1 2,2,2 3,3,3 4,4,4 5,5,5 6,6,6
1,1,2 2,2,3 3,3,4 4,4,5 5,5,6
1,1,3 2,2,4 3,3,5 4,4,6 5,6,1
1,1,4 2,2,5 3,3,6 4,5,1 5,6,2
1,1,5 2,2,6 3,4,1 4,5,2 5,6,3
1,1,6 2,3,1 3,4,2 4,5,3 5,6,4
1,2,1 2,3,2 3,4,3 4,5,4 5,6,5
1,2,2 2,3,3 3,4,4 4,5,5 5,6,6
1,2,3 2,3,4 3,4,5 4,5,6 6,1,1
1,2,4 2,3,5 3,4,6 4,6,1 6,1,2
1,2,5 2,3,6 3,5,1 4,6,2 6,1,3
1,2,6 2,4,1 3,5,2 4,6,3 6,1,4
1,3,1 2,4,2 3,5,3 4,6,4 6,1,5
1,3,2 2,4,3 3,5,4 4,6,5 6,1,6
1,3,3 2,4,4 3,5,5 4,6,6 6,2,1
1,3,4 2,4,5 3,5,6 5,1,1 6,2,2
1,3,5 2,4,6 3,6,1 5,1,2 6,2,3
1,3,6 2,5,1 3,6,2 5,1,3 6,2,4
1,4,1 2,5,2 3,6,3 5,1,4 6,2,5
1,4,2 2,5,3 3,6,4 5,1,5 6,2,6
1,4,3 2,5,4 3,6,5 5,1,6 6,3,1
1,4,4 2,5,5 3,6,6 5,2,1 6,3,2
1,4,5 2,5,6 4,1,1 5,2,2 6,3,3
1,4,6 2,6,1 4,1,2 5,2,3 6,3,4
1,5,1 2,6,2 4,1,3 5,2,4 6,3,5
1,5,2 2,6,3 4,1,4 5,2,5 6,3,6
1,5,3 2,6,4 4,1,5 5,2,6 6,4,1
1,5,4 2,6,5 4,1,6 5,3,1 6,4,2
1,5,5 2,6,6 4,2,1 5,3,2 6,4,3
1,5,6 3,1,1 4,2,2 5,3,3 6,4,4
1,6,1 3,1,2 4,2,3 5,3,4 6,4,5
1,6,2 3,1,3 4,2,4 5,3,5 6,4,6
1,6,3 3,1,4 4,2,5 5,3,6 6,5,1
1,6,4 3,1,5 4,2,6 5,4,1 6,5,2
1,6,5 3,1,6 4,3,1 5,4,2 6,5,3
1,6,6 3,2,1 4,3,2 5,4,3 6,5,4
2,1,1 3,2,2 4,3,3 5,4,4 6,5,5
2,1,2 3,2,3 4,3,4 5,4,5 6,5,6
2,1,3 3,2,4 4,3,5 5,4,6 6,6,1 Fig.24 M1 Pattern for Placing Five Balls into Four Bins in BPN
36
Fig.24 Continued
Case SB(5,5), M1(6,4):
1,1,1,1 2,2,2,2 3,3,3,3 4,4,4,4 5,5,5,5 6,6,6,6
1,1,1,2 2,2,2,3 3,3,3,4 4,4,4,5 5,5,5,6
1,1,1,3 2,2,2,4 3,3,3,5 4,4,4,6 5,5,6,1
1,1,1,4 2,2,2,5 3,3,3,6 4,4,5,1 5,5,6,2
1,1,1,5 2,2,2,6 3,3,4,1 4,4,5,2 5,5,6,3
1,1,1,6 2,2,3,1 3,3,4,2 4,4,5,3 5,5,6,4
1,1,2,1 2,2,3,2 3,3,4,3 4,4,5,4 5,5,6,5
1,1,2,2 2,2,3,3 3,3,4,4 4,4,5,5 5,5,6,6
1,1,2,3 2,2,3,4 3,3,4,5 4,4,5,6 5,6,1,1
1,1,2,4 2,2,3,5 3,3,4,6 4,4,6,1 5,6,1,2
1,1,2,5 2,2,3,6 3,3,5,1 4,4,6,2 5,6,1,3
1,1,2,6 2,2,4,1 3,3,5,2 4,4,6,3 5,6,1,4
1,1,3,1 2,2,4,2 3,3,5,3 4,4,6,4 5,6,1,5
1,1,3,2 2,2,4,3 3,3,5,4 4,4,6,5 5,6,1,6
1,1,3,3 2,2,4,4 3,3,5,5 4,4,6,6 5,6,2,1
1,1,3,4 2,2,4,5 3,3,5,6 4,5,1,1 5,6,2,2
1,1,3,5 2,2,4,6 3,3,6,1 4,5,1,2 5,6,2,3
1,1,3,6 2,2,5,1 3,3,6,2 4,5,1,3 5,6,2,4
1,1,4,1 2,2,5,2 3,3,6,3 4,5,1,4 5,6,2,5
1,1,4,2 2,2,5,3 3,3,6,4 4,5,1,5 5,6,2,6
1,1,4,3 2,2,5,4 3,3,6,5 4,5,1,6 5,6,3,1
1,1,4,4 2,2,5,5 3,3,6,6 4,5,2,1 5,6,3,2
1,1,4,5 2,2,5,6 3,4,1,1 4,5,2,2 5,6,3,3
1,1,4,6 2,2,6,1 3,4,1,2 4,5,2,3 5,6,3,4
1,1,5,1 2,2,6,2 3,4,1,3 4,5,2,4 5,6,3,5
1,1,5,2 2,2,6,3 3,4,1,4 4,5,2,5 5,6,3,6
1,1,5,3 2,2,6,4 3,4,1,5 4,5,2,6 5,6,4,1
1,1,5,4 2,2,6,5 3,4,1,6 4,5,3,1 5,6,4,2
1,1,5,5 2,2,6,6 3,4,2,1 4,5,3,2 5,6,4,3
1,1,5,6 2,3,1,1 3,4,2,2 4,5,3,3 5,6,4,4 Fig.25 M1 Pattern for Placing Five Balls into Five Bins in BPN
2,1,4 3,2,5 4,3,6 5,5,1 6,6,2
2,1,5 3,2,6 4,4,1 5,5,2 6,6,3
2,1,6 3,3,1 4,4,2 5,5,3 6,6,4
2,2,1 3,3,2 4,4,3 5,5,4 6,6,5
37
Fig.25 Continued
1,1,6,1 2,3,1,2 3,4,2,3 4,5,3,4 5,6,4,5
1,1,6,2 2,3,1,3 3,4,2,4 4,5,3,5 5,6,4,6
1,1,6,3 2,3,1,4 3,4,2,5 4,5,3,6 5,6,5,1
1,1,6,4 2,3,1,5 3,4,2,6 4,5,4,1 5,6,5,2
1,1,6,5 2,3,1,6 3,4,3,1 4,5,4,2 5,6,5,3
1,1,6,6 2,3,2,1 3,4,3,2 4,5,4,3 5,6,5,4
1,2,1,1 2,3,2,2 3,4,3,3 4,5,4,4 5,6,5,5
1,2,1,2 2,3,2,3 3,4,3,4 4,5,4,5 5,6,5,6
1,2,1,3 2,3,2,4 3,4,3,5 4,5,4,6 5,6,6,1
1,2,1,4 2,3,2,5 3,4,3,6 4,5,5,1 5,6,6,2
1,2,1,5 2,3,2,6 3,4,4,1 4,5,5,2 5,6,6,3
1,2,1,6 2,3,3,1 3,4,4,2 4,5,5,3 5,6,6,4
1,2,2,1 2,3,3,2 3,4,4,3 4,5,5,4 5,6,6,5
1,2,2,2 2,3,3,3 3,4,4,4 4,5,5,5 5,6,6,6
1,2,2,3 2,3,3,4 3,4,4,5 4,5,5,6 6,1,1,1
1,2,2,4 2,3,3,5 3,4,4,6 4,5,6,1 6,1,1,2
1,2,2,5 2,3,3,6 3,4,5,1 4,5,6,2 6,1,1,3
1,2,2,6 2,3,4,1 3,4,5,2 4,5,6,3 6,1,1,4
1,2,3,1 2,3,4,2 3,4,5,3 4,5,6,4 6,1,1,5
1,2,3,2 2,3,4,3 3,4,5,4 4,5,6,5 6,1,1,6
1,2,3,3 2,3,4,4 3,4,5,5 4,5,6,6 6,1,2,1
1,2,3,4 2,3,4,5 3,4,5,6 4,6,1,1 6,1,2,2
1,2,3,5 2,3,4,6 3,4,6,1 4,6,1,2 6,1,2,3
1,2,3,6 2,3,5,1 3,4,6,2 4,6,1,3 6,1,2,4
1,2,4,1 2,3,5,2 3,4,6,3 4,6,1,4 6,1,2,5
1,2,4,2 2,3,5,3 3,4,6,4 4,6,1,5 6,1,2,6
1,2,4,3 2,3,5,4 3,4,6,5 4,6,1,6 6,1,3,1
1,2,4,4 2,3,5,5 3,4,6,6 4,6,2,1 6,1,3,2
1,2,4,5 2,3,5,6 3,5,1,1 4,6,2,2 6,1,3,3
1,2,4,6 2,3,6,1 3,5,1,2 4,6,2,3 6,1,3,4
1,2,5,1 2,3,6,2 3,5,1,3 4,6,2,4 6,1,3,5
1,2,5,2 2,3,6,3 3,5,1,4 4,6,2,5 6,1,3,6
1,2,5,3 2,3,6,4 3,5,1,5 4,6,2,6 6,1,4,1
1,2,5,4 2,3,6,5 3,5,1,6 4,6,3,1 6,1,4,2
1,2,5,5 2,3,6,6 3,5,2,1 4,6,3,2 6,1,4,3
1,2,5,6 2,4,1,1 3,5,2,2 4,6,3,3 6,1,4,4
1,2,6,1 2,4,1,2 3,5,2,3 4,6,3,4 6,1,4,5
1,2,6,2 2,4,1,3 3,5,2,4 4,6,3,5 6,1,4,6
1,2,6,3 2,4,1,4 3,5,2,5 4,6,3,6 6,1,5,1
1,2,6,4 2,4,1,5 3,5,2,6 4,6,4,1 6,1,5,2
38
Fig.25 Continued
1,2,6,5 2,4,1,6 3,5,3,1 4,6,4,2 6,1,5,3
1,2,6,6 2,4,2,1 3,5,3,2 4,6,4,3 6,1,5,4
1,3,1,1 2,4,2,2 3,5,3,3 4,6,4,4 6,1,5,5
1,3,1,2 2,4,2,3 3,5,3,4 4,6,4,5 6,1,5,6
1,3,1,3 2,4,2,4 3,5,3,5 4,6,4,6 6,1,6,1
1,3,1,4 2,4,2,5 3,5,3,6 4,6,5,1 6,1,6,2
1,3,1,5 2,4,2,6 3,5,4,1 4,6,5,2 6,1,6,3
1,3,1,6 2,4,3,1 3,5,4,2 4,6,5,3 6,1,6,4
1,3,2,1 2,4,3,2 3,5,4,3 4,6,5,4 6,1,6,5
1,3,2,2 2,4,3,3 3,5,4,4 4,6,5,5 6,1,6,6
1,3,2,3 2,4,3,4 3,5,4,5 4,6,5,6 6,2,1,1
1,3,2,4 2,4,3,5 3,5,4,6 4,6,6,1 6,2,1,2
1,3,2,5 2,4,3,6 3,5,5,1 4,6,6,2 6,2,1,3
1,3,2,6 2,4,4,1 3,5,5,2 4,6,6,3 6,2,1,4
1,3,3,1 2,4,4,2 3,5,5,3 4,6,6,4 6,2,1,5
1,3,3,2 2,4,4,3 3,5,5,4 4,6,6,5 6,2,1,6
1,3,3,3 2,4,4,4 3,5,5,5 4,6,6,6 6,2,2,1
1,3,3,4 2,4,4,5 3,5,5,6 5,1,1,1 6,2,2,2
1,3,3,5 2,4,4,6 3,5,6,1 5,1,1,2 6,2,2,3
1,3,3,6 2,4,5,1 3,5,6,2 5,1,1,3 6,2,2,4
1,3,4,1 2,4,5,2 3,5,6,3 5,1,1,4 6,2,2,5
1,3,4,2 2,4,5,3 3,5,6,4 5,1,1,5 6,2,2,6
1,3,4,3 2,4,5,4 3,5,6,5 5,1,1,6 6,2,3,1
1,3,4,4 2,4,5,5 3,5,6,6 5,1,2,1 6,2,3,2
1,3,4,5 2,4,5,6 3,6,1,1 5,1,2,2 6,2,3,3
1,3,4,6 2,4,6,1 3,6,1,2 5,1,2,3 6,2,3,4
1,3,5,1 2,4,6,2 3,6,1,3 5,1,2,4 6,2,3,5
1,3,5,2 2,4,6,3 3,6,1,4 5,1,2,5 6,2,3,6
1,3,5,3 2,4,6,4 3,6,1,5 5,1,2,6 6,2,4,1
1,3,5,4 2,4,6,5 3,6,1,6 5,1,3,1 6,2,4,2
1,3,5,5 2,4,6,6 3,6,2,1 5,1,3,2 6,2,4,3
1,3,5,6 2,5,1,1 3,6,2,2 5,1,3,3 6,2,4,4
1,3,6,1 2,5,1,2 3,6,2,3 5,1,3,4 6,2,4,5
1,3,6,2 2,5,1,3 3,6,2,4 5,1,3,5 6,2,4,6
1,3,6,3 2,5,1,4 3,6,2,5 5,1,3,6 6,2,5,1
1,3,6,4 2,5,1,5 3,6,2,6 5,1,4,1 6,2,5,2
1,3,6,5 2,5,1,6 3,6,3,1 5,1,4,2 6,2,5,3
1,3,6,6 2,5,2,1 3,6,3,2 5,1,4,3 6,2,5,4
1,4,1,1 2,5,2,2 3,6,3,3 5,1,4,4 6,2,5,5
1,4,1,2 2,5,2,3 3,6,3,4 5,1,4,5 6,2,5,6
39
Fig.25 Continued
1,4,1,3 2,5,2,4 3,6,3,5 5,1,4,6 6,2,6,1
1,4,1,4 2,5,2,5 3,6,3,6 5,1,5,1 6,2,6,2
1,4,1,5 2,5,2,6 3,6,4,1 5,1,5,2 6,2,6,3
1,4,1,6 2,5,3,1 3,6,4,2 5,1,5,3 6,2,6,4
1,4,2,1 2,5,3,2 3,6,4,3 5,1,5,4 6,2,6,5
1,4,2,2 2,5,3,3 3,6,4,4 5,1,5,5 6,2,6,6
1,4,2,3 2,5,3,4 3,6,4,5 5,1,5,6 6,3,1,1
1,4,2,4 2,5,3,5 3,6,4,6 5,1,6,1 6,3,1,2
1,4,2,5 2,5,3,6 3,6,5,1 5,1,6,2 6,3,1,3
1,4,2,6 2,5,4,1 3,6,5,2 5,1,6,3 6,3,1,4
1,4,3,1 2,5,4,2 3,6,5,3 5,1,6,4 6,3,1,5
1,4,3,2 2,5,4,3 3,6,5,4 5,1,6,5 6,3,1,6
1,4,3,3 2,5,4,4 3,6,5,5 5,1,6,6 6,3,2,1
1,4,3,4 2,5,4,5 3,6,5,6 5,2,1,1 6,3,2,2
1,4,3,5 2,5,4,6 3,6,6,1 5,2,1,2 6,3,2,3
1,4,3,6 2,5,5,1 3,6,6,2 5,2,1,3 6,3,2,4
1,4,4,1 2,5,5,2 3,6,6,3 5,2,1,4 6,3,2,5
1,4,4,2 2,5,5,3 3,6,6,4 5,2,1,5 6,3,2,6
1,4,4,3 2,5,5,4 3,6,6,5 5,2,1,6 6,3,3,1
1,4,4,4 2,5,5,5 3,6,6,6 5,2,2,1 6,3,3,2
1,4,4,5 2,5,5,6 4,1,1,1 5,2,2,2 6,3,3,3
1,4,4,6 2,5,6,1 4,1,1,2 5,2,2,3 6,3,3,4
1,4,5,1 2,5,6,2 4,1,1,3 5,2,2,4 6,3,3,5
1,4,5,2 2,5,6,3 4,1,1,4 5,2,2,5 6,3,3,6
1,4,5,3 2,5,6,4 4,1,1,5 5,2,2,6 6,3,4,1
1,4,5,4 2,5,6,5 4,1,1,6 5,2,3,1 6,3,4,2
1,4,5,5 2,5,6,6 4,1,2,1 5,2,3,2 6,3,4,3
1,4,5,6 2,6,1,1 4,1,2,2 5,2,3,3 6,3,4,4
1,4,6,1 2,6,1,2 4,1,2,3 5,2,3,4 6,3,4,5
1,4,6,2 2,6,1,3 4,1,2,4 5,2,3,5 6,3,4,6
1,4,6,3 2,6,1,4 4,1,2,5 5,2,3,6 6,3,5,1
1,4,6,4 2,6,1,5 4,1,2,6 5,2,4,1 6,3,5,2
1,4,6,5 2,6,1,6 4,1,3,1 5,2,4,2 6,3,5,3
1,4,6,6 2,6,2,1 4,1,3,2 5,2,4,3 6,3,5,4
1,5,1,1 2,6,2,2 4,1,3,3 5,2,4,4 6,3,5,5
1,5,1,2 2,6,2,3 4,1,3,4 5,2,4,5 6,3,5,6
1,5,1,3 2,6,2,4 4,1,3,5 5,2,4,6 6,3,6,1
1,5,1,4 2,6,2,5 4,1,3,6 5,2,5,1 6,3,6,2
1,5,1,5 2,6,2,6 4,1,4,1 5,2,5,2 6,3,6,3
1,5,1,6 2,6,3,1 4,1,4,2 5,2,5,3 6,3,6,4
40
Fig.25 Continued
1,5,2,1 2,6,3,2 4,1,4,3 5,2,5,4 6,3,6,5
1,5,2,2 2,6,3,3 4,1,4,4 5,2,5,5 6,3,6,6
1,5,2,3 2,6,3,4 4,1,4,5 5,2,5,6 6,4,1,1
1,5,2,4 2,6,3,5 4,1,4,6 5,2,6,1 6,4,1,2
1,5,2,5 2,6,3,6 4,1,5,1 5,2,6,2 6,4,1,3
1,5,2,6 2,6,4,1 4,1,5,2 5,2,6,3 6,4,1,4
1,5,3,1 2,6,4,2 4,1,5,3 5,2,6,4 6,4,1,5
1,5,3,2 2,6,4,3 4,1,5,4 5,2,6,5 6,4,1,6
1,5,3,3 2,6,4,4 4,1,5,5 5,2,6,6 6,4,2,1
1,5,3,4 2,6,4,5 4,1,5,6 5,3,1,1 6,4,2,2
1,5,3,5 2,6,4,6 4,1,6,1 5,3,1,2 6,4,2,3
1,5,3,6 2,6,5,1 4,1,6,2 5,3,1,3 6,4,2,4
1,5,4,1 2,6,5,2 4,1,6,3 5,3,1,4 6,4,2,5
1,5,4,2 2,6,5,3 4,1,6,4 5,3,1,5 6,4,2,6
1,5,4,3 2,6,5,4 4,1,6,5 5,3,1,6 6,4,3,1
1,5,4,4 2,6,5,5 4,1,6,6 5,3,2,1 6,4,3,2
1,5,4,5 2,6,5,6 4,2,1,1 5,3,2,2 6,4,3,3
1,5,4,6 2,6,6,1 4,2,1,2 5,3,2,3 6,4,3,4
1,5,5,1 2,6,6,2 4,2,1,3 5,3,2,4 6,4,3,5
1,5,5,2 2,6,6,3 4,2,1,4 5,3,2,5 6,4,3,6
1,5,5,3 2,6,6,4 4,2,1,5 5,3,2,6 6,4,4,1
1,5,5,4 2,6,6,5 4,2,1,6 5,3,3,1 6,4,4,2
1,5,5,5 2,6,6,6 4,2,2,1 5,3,3,2 6,4,4,3
1,5,5,6 3,1,1,1 4,2,2,2 5,3,3,3 6,4,4,4
1,5,6,1 3,1,1,2 4,2,2,3 5,3,3,4 6,4,4,5
1,5,6,2 3,1,1,3 4,2,2,4 5,3,3,5 6,4,4,6
1,5,6,3 3,1,1,4 4,2,2,5 5,3,3,6 6,4,5,1
1,5,6,4 3,1,1,5 4,2,2,6 5,3,4,1 6,4,5,2
1,5,6,5 3,1,1,6 4,2,3,1 5,3,4,2 6,4,5,3
1,5,6,6 3,1,2,1 4,2,3,2 5,3,4,3 6,4,5,4
1,6,1,1 3,1,2,2 4,2,3,3 5,3,4,4 6,4,5,5
1,6,1,2 3,1,2,3 4,2,3,4 5,3,4,5 6,4,5,6
1,6,1,3 3,1,2,4 4,2,3,5 5,3,4,6 6,4,6,1
1,6,1,4 3,1,2,5 4,2,3,6 5,3,5,1 6,4,6,2
1,6,1,5 3,1,2,6 4,2,4,1 5,3,5,2 6,4,6,3
1,6,1,6 3,1,3,1 4,2,4,2 5,3,5,3 6,4,6,4
1,6,2,1 3,1,3,2 4,2,4,3 5,3,5,4 6,4,6,5
1,6,2,2 3,1,3,3 4,2,4,4 5,3,5,5 6,4,6,6
1,6,2,3 3,1,3,4 4,2,4,5 5,3,5,6 6,5,1,1
1,6,2,4 3,1,3,5 4,2,4,6 5,3,6,1 6,5,1,2
41
Fig.25 Continued
1,6,2,5 3,1,3,6 4,2,5,1 5,3,6,2 6,5,1,3
1,6,2,6 3,1,4,1 4,2,5,2 5,3,6,3 6,5,1,4
1,6,3,1 3,1,4,2 4,2,5,3 5,3,6,4 6,5,1,5
1,6,3,2 3,1,4,3 4,2,5,4 5,3,6,5 6,5,1,6
1,6,3,3 3,1,4,4 4,2,5,5 5,3,6,6 6,5,2,1
1,6,3,4 3,1,4,5 4,2,5,6 5,4,1,1 6,5,2,2
1,6,3,5 3,1,4,6 4,2,6,1 5,4,1,2 6,5,2,3
1,6,3,6 3,1,5,1 4,2,6,2 5,4,1,3 6,5,2,4
1,6,4,1 3,1,5,2 4,2,6,3 5,4,1,4 6,5,2,5
1,6,4,2 3,1,5,3 4,2,6,4 5,4,1,5 6,5,2,6
1,6,4,3 3,1,5,4 4,2,6,5 5,4,1,6 6,5,3,1
1,6,4,4 3,1,5,5 4,2,6,6 5,4,2,1 6,5,3,2
1,6,4,5 3,1,5,6 4,3,1,1 5,4,2,2 6,5,3,3
1,6,4,6 3,1,6,1 4,3,1,2 5,4,2,3 6,5,3,4
1,6,5,1 3,1,6,2 4,3,1,3 5,4,2,4 6,5,3,5
1,6,5,2 3,1,6,3 4,3,1,4 5,4,2,5 6,5,3,6
1,6,5,3 3,1,6,4 4,3,1,5 5,4,2,6 6,5,4,1
1,6,5,4 3,1,6,5 4,3,1,6 5,4,3,1 6,5,4,2
1,6,5,5 3,1,6,6 4,3,2,1 5,4,3,2 6,5,4,3
1,6,5,6 3,2,1,1 4,3,2,2 5,4,3,3 6,5,4,4
1,6,6,1 3,2,1,2 4,3,2,3 5,4,3,4 6,5,4,5
1,6,6,2 3,2,1,3 4,3,2,4 5,4,3,5 6,5,4,6
1,6,6,3 3,2,1,4 4,3,2,5 5,4,3,6 6,5,5,1
1,6,6,4 3,2,1,5 4,3,2,6 5,4,4,1 6,5,5,2
1,6,6,5 3,2,1,6 4,3,3,1 5,4,4,2 6,5,5,3
1,6,6,6 3,2,2,1 4,3,3,2 5,4,4,3 6,5,5,4
2,1,1,1 3,2,2,2 4,3,3,3 5,4,4,4 6,5,5,5
2,1,1,2 3,2,2,3 4,3,3,4 5,4,4,5 6,5,5,6
2,1,1,3 3,2,2,4 4,3,3,5 5,4,4,6 6,5,6,1
2,1,1,4 3,2,2,5 4,3,3,6 5,4,5,1 6,5,6,2
2,1,1,5 3,2,2,6 4,3,4,1 5,4,5,2 6,5,6,3
2,1,1,6 3,2,3,1 4,3,4,2 5,4,5,3 6,5,6,4
2,1,2,1 3,2,3,2 4,3,4,3 5,4,5,4 6,5,6,5
2,1,2,2 3,2,3,3 4,3,4,4 5,4,5,5 6,5,6,6
2,1,2,3 3,2,3,4 4,3,4,5 5,4,5,6 6,6,1,1
2,1,2,4 3,2,3,5 4,3,4,6 5,4,6,1 6,6,1,2
2,1,2,5 3,2,3,6 4,3,5,1 5,4,6,2 6,6,1,3
2,1,2,6 3,2,4,1 4,3,5,2 5,4,6,3 6,6,1,4
2,1,3,1 3,2,4,2 4,3,5,3 5,4,6,4 6,6,1,5
2,1,3,2 3,2,4,3 4,3,5,4 5,4,6,5 6,6,1,6
42
Fig.25 Continued
2,1,3,3 3,2,4,4 4,3,5,5 5,4,6,6 6,6,2,1
2,1,3,4 3,2,4,5 4,3,5,6 5,5,1,1 6,6,2,2
2,1,3,5 3,2,4,6 4,3,6,1 5,5,1,2 6,6,2,3
2,1,3,6 3,2,5,1 4,3,6,2 5,5,1,3 6,6,2,4
2,1,4,1 3,2,5,2 4,3,6,3 5,5,1,4 6,6,2,5
2,1,4,2 3,2,5,3 4,3,6,4 5,5,1,5 6,6,2,6
2,1,4,3 3,2,5,4 4,3,6,5 5,5,1,6 6,6,3,1
2,1,4,4 3,2,5,5 4,3,6,6 5,5,2,1 6,6,3,2
2,1,4,5 3,2,5,6 4,4,1,1 5,5,2,2 6,6,3,3
2,1,4,6 3,2,6,1 4,4,1,2 5,5,2,3 6,6,3,4
2,1,5,1 3,2,6,2 4,4,1,3 5,5,2,4 6,6,3,5
2,1,5,2 3,2,6,3 4,4,1,4 5,5,2,5 6,6,3,6
2,1,5,3 3,2,6,4 4,4,1,5 5,5,2,6 6,6,4,1
2,1,5,4 3,2,6,5 4,4,1,6 5,5,3,1 6,6,4,2
2,1,5,5 3,2,6,6 4,4,2,1 5,5,3,2 6,6,4,3
2,1,5,6 3,3,1,1 4,4,2,2 5,5,3,3 6,6,4,4
2,1,6,1 3,3,1,2 4,4,2,3 5,5,3,4 6,6,4,5
2,1,6,2 3,3,1,3 4,4,2,4 5,5,3,5 6,6,4,6
2,1,6,3 3,3,1,4 4,4,2,5 5,5,3,6 6,6,5,1
2,1,6,4 3,3,1,5 4,4,2,6 5,5,4,1 6,6,5,2
2,1,6,5 3,3,1,6 4,4,3,1 5,5,4,2 6,6,5,3
2,1,6,6 3,3,2,1 4,4,3,2 5,5,4,3 6,6,5,4
2,2,1,1 3,3,2,2 4,4,3,3 5,5,4,4 6,6,5,5
2,2,1,2 3,3,2,3 4,4,3,4 5,5,4,5 6,6,5,6
2,2,1,3 3,3,2,4 4,4,3,5 5,5,4,6 6,6,6,1
2,2,1,4 3,3,2,5 4,4,3,6 5,5,5,1 6,6,6,2
2,2,1,5 3,3,2,6 4,4,4,1 5,5,5,2 6,6,6,3
2,2,1,6 3,3,3,1 4,4,4,2 5,5,5,3 6,6,6,4
2,2,2,1 3,3,3,2 4,4,4,3 5,5,5,4 6,6,6,5
As seen from the series of studies (Fig. 22-25), M1 repeats SB. This is because M1
“respects” order unlike SB. For example, the string “1, 1, 1, 2” and “1, 1, 2, 1” are
equivalent since they mean that three bars are in the first position and one bar is in the
second position. As a result of this repetitive way of stating the same configuration, a
method which seeks to extract SB from M1 must eliminate the repetition. Similarly, this
43
method could simply index M1 in a regular manner according to the pattern seen in Fig.
25 which depends on the number of bins. From the series of studies, it is clear that with the
exception of case SB(5,2), M1(6,1), the manner to index M1 follows a “count down shape.”
For example, let us define an integer x as the cardinality of a mirror image of an upper-
triangular matrix, and b as an empty i by i matrix. Consider the following counting
operations due to counting the size of mirror image upper-triangular matrices where i = 6.
Case SB(5,3), M1(6,2):
6 (means there is only one mirror image upper-triangular matrix of size six in this case)
Case SB(5,4), M1(6,3):
6 5 4 3 2 1
Case SB(5,5), M1(6,4):
From our study, we can also see an interesting and well know kind of pattern – a
fractal. This is clear because the pattern is self-similar and the “count down shape” remains
the same. Based on this observation, we could conjecture that case SB(5,6), M1(6,5) is
made up of the following mirror image upper-triangular matrices:
Case SB(5,6), M1(6,5):
6 5 4 3 2 1
b 5 4 3 2 1
b b 4 3 2 1
b b b 3 2 1
b b b b 2 1
b b b b b 1
6 5 4 3 2 1 b b b b b b b b b b b b
b 5 4 3 2 1 b 5 4 3 2 1 b b b b b b
b b 4 3 2 1 b b 4 3 2 1 b b 4 3 2 1
b b b 3 2 1 b b b 3 2 1 b b b 3 2 1
b b b b 2 1 b b b b 2 1 b b b b 2 1
b b b b b 1 b b b b b 1 b b b b b 1
b b b b b b b b b b b b b b b b b b
44
As seen from Case SB(5,3), M1(6,2) to Case SB(5,6), M1(6,5), the pattern generated by
indexing M1 with entries that correspond to the position of the bars in SB is a recursive
one which can be described by the following fractal rules (Fig. 26):
Given the problem of placing n indistinguishable balls into k distinguishable bins (ID-
Variant 2), the function, fractal f(n,k) is defined as follows:
𝐿𝑒𝑡 𝑎 𝑙𝑖𝑠𝑡 𝑏𝑒 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑎𝑠 𝑎 𝑠𝑒𝑡 𝑤ℎ𝑖𝑐ℎ 𝑝𝑒𝑟𝑚𝑖𝑡𝑠 𝑟𝑒𝑝𝑒𝑡𝑖𝑡𝑖𝑜𝑛. 𝐿𝑒𝑡 𝑎 𝑙𝑖𝑠𝑡 𝑜𝑓 𝑙𝑖𝑠𝑡𝑠, 𝐴 𝑏𝑒 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑎𝑠 𝑓𝑜𝑙𝑙𝑜𝑤𝑠: 𝐴 = {{𝑥}, {𝑥 − 1}, {𝑥 − 2}, … {2}, {1}} 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝑓(𝑛: ℕ, 𝑘: ℕ ) 𝑥 = 𝑛 + 1; 𝐺𝑖𝑣𝑒𝑛 𝑎 𝑠𝑒𝑡 𝑆, 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝑏(S)
𝑦 = 𝑘 − 1; 𝒓𝒆𝒕𝒖𝒓𝒏 ⋃ {𝑏}|𝑆|𝑖=1
𝒊𝒇 𝑦 = 1 𝒕𝒉𝒆𝒏 𝒆𝒏𝒅
𝑝 = {1, 2, 3, … , 𝑥}; 𝒆𝒏𝒅𝒊𝒇 𝒊𝒇 𝑦 = 2 𝒕𝒉𝒆𝒏
𝑝 = {𝑥}; 𝒆𝒏𝒅𝒊𝒇 𝒊𝒇 𝑦 = 3 𝒕𝒉𝒆𝒏
𝑝 = {𝑥, 𝑥 − 1, 𝑥 − 2,… , 1}; 𝒆𝒏𝒅𝒊𝒇 𝒊𝒇 𝑦 > 3 𝒕𝒉𝒆𝒏 𝒇𝒐𝒓 𝑖 = 1 𝒕𝒐 𝑦 − 3 𝒅𝒐
{𝑥} = {𝐴1, 𝐴2, 𝐴3, … , 𝐴𝑥}
{𝑥 − 1} = {𝑏(𝐴𝑥)𝑥, 𝐴2, 𝐴3, … , 𝐴𝑥}
{𝑥 − 2} = {𝑏(𝐴𝑥)𝑥, 𝑏(𝐴𝑥)𝑥−1, 𝐴3, … , 𝐴𝑥} … {1} = {𝑏(𝐴𝑥)𝑥, 𝑏(𝐴𝑥)𝑥−1, 𝑏(𝐴𝑥)𝑥−2, … , 𝐴𝑥}
𝐴 = {{𝑥}, {𝑥 − 1}, {𝑥 − 2}, … , {2}, {1}} 𝒆𝒏𝒅𝒇𝒐𝒓 𝒆𝒏𝒅𝒊𝒇 𝑝 = 𝐴1 ∪ 𝐴2 ∪ 𝐴3 ∪, … ,∪ 𝐴𝑥
𝒆𝒏𝒅
b b b b b b b b b b b b b b b b b b
b b b b b b b b b b b b b b b b b b
b b b 3 2 1 b b b b b b b b b b b b
b b b b 2 1 b b b b 2 1 b b b b b b
b b b b b 1 b b b b b 1 b b b b b 1
Fig.26 Algorithm f
45
Usage of f
Example 1: Generate pattern using f when n = 5, k = 2
𝑥 = 6, 𝑦 = 1
𝑝 = {1, 2, 3, 4, 5, 6}
Example 2: Generate pattern using f when n = 5, k = 3
𝑥 = 6, 𝑦 = 2
𝑝 = {6}
Example 3: Generate pattern using f when n = 5, k = 4
𝑥 = 6, 𝑦 = 3
𝑝 = {6, 5, 4, 3, 2, 1}
Example 4: Generate pattern using f when n = 5, k = 5
𝑥 = 6, 𝑦 = 4
𝐴 = {{6}, {5}, {4}, {3}, {2}, {1}} 𝑰𝒕𝒆𝒓𝒂𝒕𝒊𝒐𝒏 𝟏:
{6} = {{6}, {5}, {4}, {3}, {2}, {1}} {5} = {{𝑏}, {5}, {4}, {3}, {2}, {1}} {4} = {{𝑏}, {𝑏}, {4}, {3}, {2}, {1}} {3} = {{𝑏}, {𝑏}, {𝑏}, {3}, {2}, {1}} {2} = {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {2}, {1}} {1} = {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {1}}
𝐴 = { {{6}, {5}, {4}, {3}, {2}, {1}},
{{𝑏}, {5}, {4}, {3}, {2}, {1}},
{{𝑏}, {𝑏}, {4}, {3}, {2}, {1}},
{{𝑏}, {𝑏}, {𝑏}, {3}, {2}, {1}},
{{𝑏}, {𝑏}, {𝑏}, {𝑏}, {2}, {1}},
{{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {1}}
{ 𝑝 = 𝐴
Example 5: Generate pattern using f when n = 5, y = 6
46
𝑥 = 6, 𝑦 = 5
𝐴 = {{6}, {5}, {4}, {3}, {2}, {1}} 𝑰𝒕𝒆𝒓𝒂𝒕𝒊𝒐𝒏 𝟏:
{6} = {{6}, {5}, {4}, {3}, {2}, {1}} 𝐴 = { {{6}, {5}, {4}, {3}, {2}, {1}},
{5} = {{𝑏}, {5}, {4}, {3}, {2}, {1}} {{𝑏}, {5}, {4}, {3}, {2}, {1}},
{4} = {{𝑏}, {𝑏}, {4}, {3}, {2}, {1}} {{𝑏}, {𝑏}, {4}, {3}, {2}, {1}},
{3} = {{𝑏}, {𝑏}, {𝑏}, {3}, {2}, {1}} {{𝑏}, {𝑏}, {𝑏}, {3}, {2}, {1}},
{2} = {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {2}, {1}} {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {2}, {1}},
{1} = {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {1}} {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {1}} }
𝑰𝒕𝒆𝒓𝒂𝒕𝒊𝒐𝒏 𝟐: {6} = {{6}, {5}, {4}, {3}, {2}, {1}} {3} = {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}} {{𝑏}, {5}, {4}, {3}, {2}, {1}} {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}} {{𝑏}, {𝑏}, {4}, {3}, {2}, {1}} {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}} {{𝑏}, {𝑏}, {𝑏}, {3}, {2}, {1}} {{𝑏}, {𝑏}, {𝑏}, {3}, {2}, {1}} {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {2}, {1}} {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {2}, {1}} {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {1}} {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {1}}
{5} = {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}} {2} = {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}} {{𝑏}, {5}, {4}, {3}, {2}, {1}} {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}} {{𝑏}, {𝑏}, {4}, {3}, {2}, {1}} {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}} {{𝑏}, {𝑏}, {𝑏}, {3}, {2}, {1}} {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}} {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {2}, {1}} {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {2}, {1}} {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {1}} {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {1}}
{4} = {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}} {1} = {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}} {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}} {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}} {{𝑏}, {𝑏}, {4}, {3}, {2}, {1}} {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}} {{𝑏}, {𝑏}, {𝑏}, {3}, {2}, {1}} {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}} {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {2}, {1}} {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}} {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {1}} {{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {1}}
47
𝐴 = { {{6}, {5}, {4}, {3}, {2}, {1}}
{{𝑏}, {5}, {4}, {3}, {2}, {1}}
{{𝑏}, {𝑏}, {4}, {3}, {2}, {1}}
{{𝑏}, {𝑏}, {𝑏}, {3}, {2}, {1}}
{{𝑏}, {𝑏}, {𝑏}, {𝑏}, {2}, {1}}
{{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {1}},
{{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}}
{{𝑏}, {5}, {4}, {3}, {2}, {1}}
{{𝑏}, {𝑏}, {4}, {3}, {2}, {1}}
{{𝑏}, {𝑏}, {𝑏}, {3}, {2}, {1}}
{{𝑏}, {𝑏}, {𝑏}, {𝑏}, {2}, {1}}
{{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {1}},
{{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}}
{{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}}
{{𝑏}, {𝑏}, {4}, {3}, {2}, {1}}
{{𝑏}, {𝑏}, {𝑏}, {3}, {2}, {1}}
{{𝑏}, {𝑏}, {𝑏}, {𝑏}, {2}, {1}}
{{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {1}},
{{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}}
{{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}}
{{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}}
{{𝑏}, {𝑏}, {𝑏}, {3}, {2}, {1}}
{{𝑏}, {𝑏}, {𝑏}, {𝑏}, {2}, {1}}
{{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {1}},
{{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}}
{{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}}
{{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}}
{{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}}
{{𝑏}, {𝑏}, {𝑏}, {𝑏}, {2}, {1}}
{{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {1}},
{{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}}
{{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}}
{{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}}
{{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}}
{{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}}
{{𝑏}, {𝑏}, {𝑏}, {𝑏}, {𝑏}, {1}}
{
𝑝 = 𝐴
48
CHAPTER 6
TRANSLATION OF PATTERN TO ID-VARIANT 2 SOLUTION
From the usage of f examples in Chapter 5, the set of patterns p look nothing like
an enumerative solution to the problem of placing n indistinguishable balls into k
distinguishable bins (ID-Variant 2). This is because p is only the container which holds
objects that will be translated to integer values. The integer values are subsequently
substituted into M1 to derive the position of the bars post translation. This chapter presents
two ways of translating p. The first way is straightforward but expensive because every
blank object (b) is visited. The second method is an optimization of the first method. Let
us consider the following translation examples using translation method 1 to translate the
set of patterns generated in chapter 5 usage of f examples.
6.1 TRANSLATION METHOD 1
Given the set of patterns p for a given x, an element pi represents a “marked” pi by
pi upper triangular matrix within an x by x upper triangular matrix. The following examples
illustrate the translation.
Example 6: Translate pattern generated by f when n = 5, k = 2, 𝑝 = {1, 2, 3, 4, 5, 6}
Since y = 1, p requires no translation, thus can be substituted directly into M1 to derive the
position of the bars corresponding with placing five balls into two bins.
Example 7: Translate pattern generated by f when n = 5, k = 3, 𝑝 = {6}
Since x = 6 and p1 = 6, 6 × 6 of the 6 × 6 upper triangular matrix is marked (blue) yielding:
49
Following translation, the set (Fig. 27) below can be substituted into M1 to get the position
of the bars corresponding to enumerating the configurations of placing five
indistinguishable balls into three distinguishable bins.
1 2 3 4 5 6 8 9 10 11 12 15 16 17 18 22 23 24 29 30 36
Fig.27 Translated Pattern for n = 5, y = 2
Example 8: Translate pattern generated by f when n = 5, k = 4, 𝑝 = {6, 5, 4, 3, 2, 1}
Since x = 6 and p1 = 6, 6 × 6 of the 6 × 6 upper triangular matrix is marked (blue) yielding:
p2 = 5, 5 × 5 of the 6 × 6 upper triangular matrix is marked (blue) yielding:
p3 = 4, 4 × 4 of the 6 × 6 upper triangular matrix is marked (blue) yielding:
b
b b
b b b
b b b b
b b b b b
1 2 3 4 5 6
7 8 9 10 11 12
13 14 15 16 17 18
19 20 21 22 23 24
25 26 27 28 29 30
31 32 33 43 35 36
b
b b
b b b
b b b b
b b b b b
1 2 3 4 5 6
7 8 9 10 11 12
13 14 15 16 17 18
19 20 21 22 23 24
25 26 27 28 29 30
31 32 33 43 35 36
b b b b b b
b
b b
b b b
b b b b
b b b b b
37 38 39 40 41 42
43 44 45 46 47 48
49 50 51 52 53 54
55 56 57 58 59 60
61 62 63 64 65 66
67 68 69 70 71 72
=
=
=
50
p4 = 3, 3 × 3 of the 6 × 6 upper triangular matrix is marked (blue) yielding:
p5 = 2, 2 × 2 of the 6 × 6 upper triangular matrix is marked (blue) yielding:
p6= 1, 1 × 1 of the 6 × 6 upper triangular matrix is marked (blue) yielding:
b b b b b b
b b b b b b
b b
b b b
b b b b
b b b b b
73 74 75 76 77 78
79 80 81 82 83 84
85 86 87 88 89 90
91 92 93 94 95 96
97 98 99 100 101 102
103 104 105 106 107 108
b b b b b b
b b b b b b
b b b b b b
b b b
b b b b
b b b b b
109 110 111 112 113 114
115 116 117 118 119 120
121 122 123 124 125 126
127 128 129 130 131 132
133 134 135 136 137 138
139 140 141 142 143 144
b b b b b b
b b b b b b
b b b b b b
b b b b b b
b b b b
b b b b b
145 146 147 148 149 150
151 152 153 154 155 156
157 158 159 160 161 162
163 164 165 166 167 168
169 170 171 172 173 174
175 176 177 178 179 180
b b b b b b
b b b b b b
b b b b b b
b b b b b b
b b b b b b
b b b b b
181 182 183 184 185 186
187 188 189 190 191 192
193 194 195 196 197 198
199 200 201 202 203 204
205 206 207 208 209 210
211 212 213 214 215 216
=
=
=
=
51
Following translation, the following set (Fig. 28) will be substituted into M1 to get the
position of the bars corresponding to enumerating the configurations of placing five
indistinguishable balls into four distinguishable bins.
1 2 3 4 5 6
8 9 10 11 12 15
16 17 18 22 23 24
29 30 36 44 45 46
47 48 51 52 53 54
58 59 60 65 66 72
87 88 89 90 94 95
96 101 102 108 130 131
132 137 138 144 173 174
180 216
Fig.28 Translation of Pattern When n = 5, k = 4
6.2 TRANSLATION METHOD 2
While translation method 1 is straightforward – involving a simple counting
operation, it is highly wasteful since it takes all the blanks into account. For this reason, it
visits more objects than needed. Translation method 2 solves this problem based on the
insight that all that is required to generate integer values, are the entry in the pattern set and
its position in the set. The following is the definition of translation method 2:
Given the set of patterns p for a given x, an element pi represents a two-dimensional array
A with pi rows and columns of length 𝑝𝑖, 𝑝𝑖 − 1, 𝑝𝑖 − 2,… , 1
For each pi , consider an offset parameter for 𝑖 ≠ 1
𝛿 = (𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑚𝑎𝑡𝑟𝑖𝑥 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 × 𝑠𝑞𝑢𝑎𝑟𝑒 𝑚𝑎𝑡𝑟𝑖𝑥) + 𝑟𝑜𝑤𝑠 𝑜𝑓 𝑙𝑒𝑛𝑔𝑡ℎ 𝑥 + 𝑟𝑜𝑤𝑠 𝑜𝑓 𝑙𝑒𝑛𝑔𝑡ℎ 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝑥
𝛿 = (𝑖 − 1)𝑥2 + (𝑥 − 𝑝𝑖)𝑥 + (𝑥 − 𝑝𝑖)
For each pi , consider a set of offsets F, |𝐹| = 𝑝𝑖, 𝑝𝑖 ≠ 𝑥, 𝑝𝑖: ℤ, 𝑖 ≠ 1
52
𝐹1 = 𝛿, 𝐹2 = 𝛿 + (𝑥 + 1), 𝐹3 = 𝛿 + 2(𝑥 + 1),…, 𝐹𝑝𝑖 = 𝛿 + (𝑝𝑖 − 1)(𝑥 + 1)
The translation rules are defined as follows:
𝒊𝒇 𝑝𝑖 = 𝑥 𝒕𝒉𝒆𝒏
A:
𝒆𝒍𝒔𝒆
A:
𝒆𝒏𝒅𝒊𝒇
Example 9: Translate pattern generated by f when n = 5, y = 4, 𝑝 = {6, 5, 4, 3, 2, 1} using
translation method 2.
𝑝1 = 6, 𝑖 = 1, 𝑥 = 6 A:
1 2 3 … 𝑥
2 + 𝑥 3 + 𝑥 … 𝑥 + 𝑥
3 + 2𝑥 … 𝑥 + 2𝑥
… 𝑥2
𝐹1 + 1 𝐹1 + 2 𝐹1 + 3 … 𝐹1 + 𝑝𝑖
𝐹2 + 1 𝐹2 + 2 … 𝐹2 + (𝑝𝑖 − 1)
𝐹3 + 1 … 𝐹3 + (𝑝𝑖 − 2)
…
𝐹𝑝𝑖 + 1
1 2 3 4 5 6
2 + 6 3 + 6 4 + 6 5 + 6 6 + 6
3 + 12 4 + 12 5 + 12 6 + 12
4 + 18 5 + 18 6 + 18
5 + 24 6 + 24
62
1 2 3 4 5 6
8 9 10 11 12
=
53
𝑝2 = 5, 𝑖 = 2, 𝑥 = 6
𝛿 = 43 |𝐹| = 5
𝐹1 = 42 𝐹2 = 50
𝐹3 = 57
𝐹4 = 64
𝐹5 = 71 A:
𝑝3, 𝑝4, 𝑝5, 𝑎𝑛𝑑 𝑝6 all follow the same path.
6.3 SUBSTITUTION INTO M1
After the pattern has been translated, the integer values can then be substituted into
M1 to derive the position of the bars as it corresponds to placing n indistinguishable balls
into k distinguishable bins. For example Fig. 29 shows the substitution of the integer values
into M1 resulting from translating the pattern p when n = 5 and k = 4.
15 16 17 18
22 23 24
29 30
36
43 + 1 43 + 2 43 + 3 43 + 4 43 + 5
50 + 1 50 + 2 50 + 3 50 + 4
57 + 1 57 + 2 57 + 3
64 + 1 64 + 2
71 + 1
44 45 46 47 48
51 52 53 54
58 59 60
65 66
72
=
54
i g3(i) g2(i) g1(i) SB - NOTATION
1 1 1 1 |||*****
2 1 1 2 ||*|****
3 1 1 3 ||**|***
4 1 1 4 ||***|**
5 1 1 5 ||****|*
6 1 1 6 ||*****|
8 1 2 2 |*||****
9 1 2 3 |*|*|***
10 1 2 4 |*|**|**
11 1 2 5 |*|***|*
12 1 2 6 |*|****|
15 1 3 3 |**||***
16 1 3 4 |**|*|**
17 1 3 5 |**|**|*
18 1 3 6 |**|***|
22 1 4 4 |***||**
23 1 4 5 |***|*|*
24 1 4 6 |***|**|
29 1 5 5 |****||*
30 1 5 6 |****|*|
36 1 6 6 |*****||
44 2 2 2 *|||****
45 2 2 3 *||*|***
46 2 2 4 *||**|**
47 2 2 5 *||***|*
48 2 2 6 *||****|
51 2 3 3 *|*||***
52 2 3 4 *|*|*|**
53 2 3 5 *|*|**|*
54 2 3 6 *|*|***|
58 2 4 4 *|**||**
59 2 4 5 *|**|*|*
60 2 4 6 *|**|**|
65 2 5 5 *|***||*
66 2 5 6 *|***|*|
72 2 6 6 *|****||
87 3 3 3 **|||*** Fig.29 Substitution into M1 Resulting from Translating p When n = 5 and k = 4
55
Fig.29 Continued
88 3 3 4 **||*|**
89 3 3 5 **||**|*
90 3 3 6 **||***|
94 3 4 4 **|*||**
95 3 4 5 **|*|*|*
96 3 4 6 **|*|**|
101 3 5 5 **|**||*
102 3 5 6 **|**|*|
108 3 6 6 **|***||
130 4 4 4 ***|||**
131 4 4 5 ***||*|*
132 4 4 6 ***||**|
137 4 5 5 ***|*||*
138 4 5 6 ***|*|*|
144 4 6 6 ***|**||
173 5 5 5 ****|||*
174 5 5 6 ****||*|
180 5 6 6 ****|*||
216 6 6 6 *****|||
56
CHAPTER 7
BIJECTION, COST ANALYSIS AND GENERATING COMBINATIONS WITH f
So far, we have focused on the solution of the combinations problem from the
perspective of placing n indistinguishable balls into k distinguishable bins (ID-Variant 2).
This chapter presents a justification of this approach by showing that the combinations
problem can be reduced to ID-Variant 2 due to (17) below:
𝐶𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛𝑠(𝑁, 𝑘) = 𝑆𝐵(𝑁 − (𝑘 + 1) + 1, 𝑘 + 1) (17)
7.1 BIJECTION BETWEEN ID-VARIANT 2 AND COMBINATIONS
Given SB(n,k) - the problem of placing n indistinguishable balls into k
distinguishable bins (also known as ID-Variant 2 in this text), and the combinations
problem, C(n,k). Since the combinations problem is solved due to a reduction (17) to
SB(n,k), show that equation (17) holds.
Solution
𝐺𝑖𝑣𝑒𝑛 𝑡ℎ𝑒 𝐶𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛𝑠 𝑝𝑟𝑜𝑏𝑙𝑒𝑚 𝐶(𝑁, 𝑘) = (𝑁𝑘) 𝑖𝑡𝑒𝑚𝑠 (18)
𝐺𝑖𝑣𝑒𝑛 𝑆𝐵(𝑁, 𝑘) = (𝑁 + 𝑘 − 1𝑘 − 1
) 𝑖𝑡𝑒𝑚𝑠 (19)
𝐹𝑟𝑜𝑚 (17), 𝑙𝑒𝑡 𝑋 = 𝑁 − (𝑘 + 1) + 1 𝑎𝑛𝑑 𝑦 = 𝑘 + 1
𝐹𝑟𝑜𝑚 (19), 𝑆𝐵(𝑋, 𝑦) = (𝑋 + 𝑦 − 1𝑦 − 1
) = (𝑁 − (𝑘 + 1) + 1 + (𝑘 + 1) − 1
(𝑘 + 1) − 1) (20)
= (𝑁𝑘)
7.2 COST ANALYSIS
57
The cost of generating combinations C, is defined as follow:
𝐶 = 𝑐𝑜𝑠𝑡 𝑜𝑓 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑛𝑔 𝑝𝑎𝑡𝑡𝑒𝑟𝑛 𝑃𝑔 +
𝑐𝑜𝑠𝑡 𝑜𝑓 𝑡𝑟𝑎𝑣𝑒𝑟𝑠𝑖𝑛𝑔 𝑝𝑎𝑡𝑡𝑒𝑟𝑛 𝑃𝑣 +
𝑐𝑜𝑠𝑡 𝑜𝑓 𝑡𝑟𝑎𝑠𝑙𝑎𝑡𝑖𝑛𝑔 𝑝𝑎𝑡𝑡𝑒𝑟𝑛 𝑃𝑡 +
𝑐𝑜𝑠𝑡 𝑜𝑓 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑝𝑎𝑡𝑡𝑒𝑟𝑛 𝑖𝑛𝑡𝑜 𝑀1 𝑃𝑠 = 𝑃𝑔 + 𝑃𝑣 + 𝑃𝑡 + 𝑃𝑠 (21)
Given that f has four cases (case y = 1 to y > 3), there are four costs C1, C2, C3, and C4 for
each case, with C4 as the largest cost. Consequently, let us examine the costs associated
with generating combinations for each case:
Given (𝑛1), since the first case does not need translation or traversing,
𝐶1 = 𝑃𝑔 + 𝑃𝑣 + 𝑃𝑡 + 𝑃𝑠 = 𝑃𝑔 + 0 + 0 + 𝑃𝑠
Since 𝑃𝑔 = 𝑃𝑠 = 𝑛
𝐶1 = 𝑛 + 0 + 0 + 𝑛 = 2𝑛
Amortized cost: 2𝑛
(𝑛1)= 2
Given (𝑛2), since 𝑝 = {𝑥}
𝐶2 = 𝑃𝑔 + 𝑃𝑣 + 𝑃𝑡 + 𝑃𝑠 = 1 + 1 + (𝑛2) + (
𝑛2) = 2 + 2 (
𝑛2)
Note that 𝑃𝑣 depends on the size of the pattern and 𝑃𝑡 depends on the (𝑛𝑘)
Amortized cost: 2
(𝑛2)+2(𝑛2)
(𝑛2)
, since 𝑛 ≥ 2, by upper bounding, 2
(22)+ 2 = 4
Given (𝑛3), since 𝑝 = {𝑥, 𝑥 − 1, 𝑥 − 2,… ,1}
𝐶3 = 𝑃𝑔 + 𝑃𝑣 + 𝑃𝑡 + 𝑃𝑠 = (𝑛 + 𝑘 − 1) + (𝑛 + 𝑘 − 1) + (𝑛3) + (
𝑛3)
𝐶3 = (𝑛 + 3 − 1) + (𝑛 + 3 − 1) + 2 (𝑛3)
𝐶3 = 2(𝑛 − 2) + 2 (𝑛3)
58
Amortized cost: 2(𝑛−2)
(𝑛3)+2(𝑛3)
(𝑛3)= 2(𝑛 − 2) ×
3!(𝑛−3)
𝑛(𝑛−1)(𝑛−2)(𝑛−3)!
Since 𝑛 ≥ 3, by upper bounding, 2 ×3!
3(2)+ 2 = 4
Given (𝑛𝑘) , 𝐶4 = 𝑃𝑔 + 𝑃𝑣 + 𝑃𝑡 + 𝑃𝑠 = 𝑃𝑔 + 𝑃𝑣 + (
𝑛𝑘) + (
𝑛𝑘)
On close examination of case four (y > 3) of f, it becomes clear that the most
straightforward implementation of the block inside the for-loop involves a count-down (x,
x-1, x-2, …, 1) which contains a counting up operation (1, 2, 3, …, x). Consequently,
𝐶4 = 𝑃𝑔 + 𝑃𝑣 + 𝑃𝑡 + 𝑃𝑠 = (𝑘 − 4)(𝑛 + 1)𝑛+1 + (𝑛 + 1)𝑘−4 + (
𝑛𝑘) + (
𝑛𝑘)
As a result of C4 , f is not CAT for the last case. The waste is clear because of the blanks
taken into consideration and the aggregating operations. Also, another problem with f is
the fact that it is not memoryless; it continuously “grows” the pattern, so this could lead to
memory problems. However, f is a naïve method developed with the intention of
correctness, not efficiency. And algorithm f verifies another way of generating
combinations – an indirect method.
59
7.3 GENERATING COMBINATIONS WITH f
7C3 Lex Relex Arbitrary1 Arbitrary2
....111 111.... ...111. 11.1...
...1.11 11.1... 1.1..1. ..111..
...11.1 11..1.. ..1..11 .11..1.
...111. 11...1. 1..11.. .1...11
..1..11 11....1 1...1.1 1....11
..1.1.1 1.11... .1...11 ..11.1.
..1.11. 1.1.1.. 11...1. 1.1..1.
..11..1 1.1..1. .1..11. 1..11..
..11.1. 1.1...1 1..1.1. 1..1.1.
..111.. 1..11.. ..11..1 .1..1.1
.1...11 1..1.1. .1.11.. .11...1
.1..1.1 1..1..1 .11..1. 1.11...
.1..11. 1...11. 11.1... ..1.1.1
.1.1..1 1...1.1 ..1.1.1 .1..11.
.1.1.1. 1....11 ..1.11. ...111.
.1.11.. .111... ...11.1 1.1...1
.11...1 .11.1.. .111... 1...1.1
.11..1. .11..1. .1.1.1. 11..1..
.11.1.. .11...1 1..1..1 .1.11..
.111... .1.11.. 1...11. ..1.11.
1....11 .1.1.1. ...1.11 .1.1.1.
1...1.1 .1.1..1 ....111 ....111
1...11. .1..11. 11....1 ..11..1
1..1..1 .1..1.1 111.... .1.1..1
1..1.1. .1...11 .11.1.. ...1.11
1..11.. ..111.. 1.1...1 111....
1.1...1 ..11.1. ..111.. ...11.1
1.1..1. ..11..1 11..1.. 1..1..1
1.1.1.. ..1.11. .11...1 1...11.
1.11... ..1.1.1 .1..1.1 11....1
11....1 ..1..11 1....11 11...1.
11...1. ...111. 1.1.1.. 1.1.1..
11..1.. ...11.1 ..11.1. .111...
11.1... ...1.11 .1.1..1 ..1..11
111.... ....111 1.11... .11.1..
Fig.30 f Output for (73) in Four Different Orders
60
CHAPTER 8
APPLICATIONS
8.1 ALGORITHM f FOR STEM CELL RESEARCH
Stem cells are cells with the peculiar property of possessing the ability to divide
into different cell types. In addition to this regenerative property [10], stem cells also divide
to replenish the supply of other cells which may be injured or worn out as a result of normal
wear and tear [15]. There are two possible outcomes of progeny cells from stem cell
division; the new cells could remain stem cells like their parent or the new cells could
become specialized cells such as muscle, brain, and red blood cells.
The main properties which distinguish stem cells from other kinds of cells are their
ability to renew through cell division even after long periods of dormancy. This property
of stem cells is the driving force for regenerative medicine [13], since it offers the
possibility of treating diseases such as type I diabetes [12], heart disease and other blood
disorders [11].
As a result of present research, it has been shown that there exist three different kinds of
stem cells, namely: embryonic stem cells, somatic or adult stem cells, and Induced
Pluripotent stem cells (IPSCs) [14].
Given the novelty of stem cell research, there are many questions surrounding stem
cells. However, a fundamental question which stands out is: What is, or what are the
regulatory factors that control stem cell proliferation and self-renewal?
The answer to these questions could have serious implications in stem cell research; as it
could enable scientist to grow embryonic and adult stem cells more efficiently in the
61
laboratory. Consequently, answers to this question could advance cell-based therapies or
drug screening [16].
Given the substantial implications surrounding the benefits of stem cells as it relates
to regenerative medicine, many studies have been conducted to understand the dynamics
of stem cells. When stem cells divides, the surrounding population decreases by one stem
cell (type 0 division) or remains constant (type 1 division), or increases by one (type 2
division). In the quest to understand the dynamics of stem cells, one such school of thought
considers stem cells as indistinguishable balls (since stem cells are identical) and the three
types of divisions as distinguishable bins. Consequently, due to (11) we can count the
configurations resulting from placing n stem cells into k division types. However, knowing
the cardinality of the sample space is insufficient for modelling. This calls for a systematic
means of enumerating the configurations. Since stem cells are modelled stochastically,
Algorithm f could be employed to generate the configurations resulting from placing n balls
into k bins in ransom tuple order.
8.2 M1 FOR IMPLEMENTING LICENSE PLATE SERIAL FORMAT
M1 could be easily adapted to generate license plate serials with a specific format.
For example, consider the following North Carolina/Virginia license plate serial format:
ABC – 1234. This representation means the first three symbols are alphabetic characters
(A-Z) and the last 4 symbols are numeric characters (0-9). Consequently, this serial format
can be implemented with two instances of M1 methods for the two different sets (alphabets
and numbers).
Firstly, let us consider generating the list of prefixes consisting of the alphabets.
Given the set 𝑈 = {𝐴, 𝐵, 𝐶}, 𝑔3 generates the first character, 𝑔2 generates the second
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character and 𝑔1 generates that last character as seen in Fig. 31:
g3 g2 g1
1 A A A
2 A A B
3 A A C
4 A A D
5 A A E
6 A A F
7 A A G
8 A A H
9 A A I
10 A A J
11 A A K
12 A A L
13 A A M
14 A A N
15 A A O
16 A A P
17 A A Q
18 A A R
19 A A S
20 A A T
21 A A U
22 A A V
23 A A W
24 A A X
25 A A Y
26 A A Z
. . .
17576 Z Z Z
Fig.31 Prefixes of License Plate Serial Format
Secondly, let us consider generating the list of suffixes consisting of numbers.
Given the set 𝑈 = {0,1,2, 3, … ,9}, 𝑔4 generates the first character, 𝑔3 generates the second,
𝑔2 – third, 𝑔1 – last character as seen in the figure Fig. 32:
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g4 g3 g2 g1
1 0 0 0 0
2 0 0 0 1
3 0 0 0 2
4 0 0 0 3
5 0 0 0 4
6 0 0 0 5
7 0 0 0 6
8 0 0 0 7
9 0 0 0 8
10 0 0 0 9
11 0 0 1 0
12 0 0 1 1
13 0 0 1 2
14 0 0 1 3
15 0 0 1 4
16 0 0 1 5
17 0 0 1 6
18 0 0 1 7
19 0 0 1 8
20 0 0 1 9
21 0 0 2 0
22 0 0 2 1
23 0 0 2 2
24 0 0 2 3
25 0 0 2 4
26 0 0 2 5
. . . .
10000 9 9 9 9
Fig.32 Suffixes of License Plate Serial Format
Finally, we can get the resultant license plate serial format list by finding the
cartesian product of the list of prefixes and the list of suffixes as outlined Fig. 33:
64
1 A A A 0 0 0 0
2 A A A 0 0 0 1
3 A A A 0 0 0 2
4 A A A 0 0 0 3
5 A A A 0 0 0 4
6 A A A 0 0 0 5
7 A A A 0 0 0 6
8 A A A 0 0 0 7
9 A A A 0 0 0 8
10 A A A 0 0 0 9
11 A A A 0 0 1 0
12 A A A 0 0 1 1
13 A A A 0 0 1 2
14 A A A 0 0 1 3
15 A A A 0 0 1 4
16 A A A 0 0 1 5
17 A A A 0 0 1 6
18 A A A 0 0 1 7
19 A A A 0 0 1 8
20 A A A 0 0 1 9
21 A A A 0 0 2 0
22 A A A 0 0 2 1
23 A A A 0 0 2 2
24 A A A 0 0 2 3
25 A A A 0 0 2 4
26 A A A 0 0 2 5
. . . . . . . .
10000 A A A 9 9 9 9
10001 B B B 0 0 0 0
10002 B B B 0 0 0 1
10003 B B B 0 0 0 2
10004 B B B 0 0 0 3
10005 B B B 0 0 0 4
10006 B B B 0 0 0 5
10007 B B B 0 0 0 6
10008 B B B 0 0 0 7
10009 B B B 0 0 0 8
10010 B B B 0 0 0 9
. . . . . . . .
175760000 Z Z Z 9 9 9 9
Fig.33 List of License Plate Serials
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CHAPTER 9
CONCLUSION
Combinatorics remains a crucial area of mathematics which shows up in many
forms. And combinatorial solutions take the form of many counting methods such as
enumerating permutations and combinations. This research effort contributes a new
perspective of looking at and generating combinatorial objects – as objects belonging to a
parent combinatorial objects, and as a result can be harvested by indexing. This results in
a flexible means of arranging tuples. This idea is presented through the perspective of
generating combinations, but could be applied to other combinatorial objects such as
permutations and multisets. The solution definition comes in the form of two original
methods M1 – the parent combinatorial method and Algorithm f – the indexing
combinatorial method.
Through an empirical study, it is shown that M1, (a non-recursive method for
generating permutations with repetitions allowed) contains the position of bars
corresponding to the placement of n indistinguishable balls into k indistinguishable bins
(ID-Variant 2). The study also shows that the position of bars take the shape of a “count-
down” fractal. Thus leading to the development of Algorithm f.
Algorithm f, a naïve counting fractal concluded the study by generating a pattern
corresponding to the position of bars, which subsequently are translated to integer values
that are substituted to yield combinations. Even though, f proved to be wasteful, f is
sufficient to verify the indirect means of generating combinations. This was the ultimate
goal (correctness) not efficiency. However, a more efficient version of f could be possible.
66
Notwithstanding, given the importance of the field of combinatorics, the results
from this research effort offer a new perspective, a perspective which could open new ways
of studying and generating combinatorial object.
67
REFERENCES
[1] W. Feller, in An Introduction to Probability Theory and Its Applications, New
York, John Wiley & Sons, 1964.
[2] D. Knuth, The Art of Programming, Generation all Combinations, Addison-Wiley,
2005, p. 4.
[3] F. Ruskey, "Combinatorial Generation," in Combinatorial Generation, 2003, p. 60.
[4] F. Ruskey, "Combinatorial Generation," in Combinatorial Generation, 2003, p.
128.
[5] D. Knuth, in The Art of Programming, Generation all Combinations, Addison-
Wiley, 2005, p. 9.
[6] P. Eades and B. McKay, An Algorithm for generating subsets of fixed size with a
strong minimal change property, vol. 19, 1984, pp. 131-133.
[7] J. Gersting, Mathematical Structures for Computer Science, 7, Ed., Freeman, 2014,
p. 285.
[8] F. Ruskey, 1 10 2003. [Online]. Available:
http://www.1stworks.com/ref/RuskeyCombGen.pdf.
[9] F. Ruskey and W. Aaron, "The Coolest Way to Generate Combinations," Discrete
Mathematics, vol. 309, no. 17, 2009.
[10] "National Institute of Health," 28 April 2002. [Online]. Available:
http://stemcells.nih.gov/info/basics/pages/basics1.aspx.
[11] "Learn.Genetics," 2014. [Online]. Available:
http://learn.genetics.utah.edu/content/stemcells/sctoday/.
[12] "California Institute for Regenerative Medicine," 2013. [Online]. Available:
http://www.cirm.ca.gov/our-progress/type-1-diabetes-fact-sheet.
[13] "National Institute of Health," 9 December 2011. [Online]. Available:
http://stemcells.nih.gov/info/scireport/Pages/2006report.aspx.
[14] "New York Stem Cell Foundation," [Online]. Available:
http://www.nyscf.org/about-stem-cells/stem-cells-101.
68
[15] "University of Rochester Medical Center," 2009. [Online]. Available:
http://www.urmc.rochester.edu/MediaLibraries/URMCMedia/life-sciences-
learning-center/documents/STUDENTStemCells10-8-09.pdf.
[16] "California Institute for Regenerative Medicine," 2013. [Online]. Available:
http://www.cirm.ca.gov/our-progress/stem-cells-therapies.
SUPPLEMENTAL SOURCES CONSULTED
[17] A. Nijenhuis and H. Wilf, Combinatorial algorithms for computers and calculators,
2 2 ed., New York: Academic Press, 1978.
[18] “National Academy of Sciences,” 2014. [Online]. Available:
http://www.nasonline.org/.
[19] “National Academy of Science,” 2014.[Online]. Available:
http://nas-sites.org/stemcells/
[20] K. Rosen, in Discrete Mathematics and Its Applications, 7 ed., New York:
McGMcGraw-Hill, 2012.
[21] T. Feil, J. Krone, in Essential Discrete Mathematics for Computer Science, New
Jerse New Jersey: Prentice Hall, 2002.
[22] G. Martin, in Counting: The Art of Enumerative Combinatorics, New York:
Springer-Verlag, 2001.
69
VITA
Alexander Chizoma Nwala
Department of Computer Science
Old Dominion University
Norfolk, VA 23529
EDUCATION
M.S., 2014, Computer Science, Old Dominion University
B.S., 2011, Computer Science, Elizabeth City State University, Elizabeth City,
North Carolina
CONFERENCES/PUBLICATION & PRESENTATIONS/SUBMITTED PAPERS
Teaching Information Security with Workflow Technology – A Case Study
Approach, submitted to Journal of Information Systems Education
Persistence: Identical Object Disambiguation, Tracking and Tagging Algorithm
PathOrigin: Robust Noise Tolerant Path Comparison Algorithm
Department of Homeland Security University Network Summit – Poster Presenter
(2011), Washington DC, Presented “Non-Recursive Mathematical Implementation
of the Probability Tree Diagram: Part I,” Proceeding published in Catastrophes &
Complex Systems: Transportation, Student Day Abstract Book March 29, 2011
Department of Homeland Security University Network Summit (2008), Anaheim
California
WORKING/INTERNSHIP EXPERIENCE
Research Assistant, Old Dominion University (May 2012 – Present)
North Carolina Department of Transportation - Summer Intern (2009/2010/2011)
Department of General Education ECSU - Tutor (Spring 2009 – Fall 2011)
Typset using Word.
