Transcript Document

KOLAM DESIGNS BASED ON
FIBONACCI NUMBERS
S. Naranan
30 January 2008
Copyright: Prof. S. Naranan, Chennai, India
FIBONACCI SERIES

Definition: Start with two numbers 0 and 1
F(0) = 0 F(1) = 1
F(n) = F(n-1) + F (n-2) n >1
The series 0 1 1 2 3 5 8 13 21 34 55 ...
is called the Fibonacci Series
 History: Leonardo of Pisa (1170 – 1250)
Liber Abaci (1202).
Crusader for the
“Hindu-Arabic” number system in Europe
Decimal notation and symbol zero. (slide)
 Origin: Fibonacci Series is the solution to
the “Rabbit Problem”:
There is a pair of rabbits to start with. Each
pair gives birth to a new pair once a month
starting 2 months after birth. Rabbits don’t
die. How many rabbits are there after 12 m?
The Fibonacci Series gives the number of pairs of
rabbits month by month.
SANSKRIT PROSODY

Sanskrit verse has words of two kinds: (1) one
syllable (S) and two syllables (L).

In how many ways can a cadence of n syllables
be created? (Acharya Hemachandra, 1150)
n
1
2
3
4
5
# of cadences
S
1= F(2)
SS L
2 =F(3)
SSS LS SL
3 =F(4)
SSSS LSS SLS SSL LL
5 =F(5)
SSSSS LSSS SLSS SSLS LLS
SSSL LSL SLL
8 =F(6)
Ans: F (n+1).
To find the number of cadences of length n, add
an S to all cadences of length n - 1 and L to all
cadences of length n – 2.
F(n) = F(n – 1) + F (n – 2)
Hemachandra’s work precedes Fibonacci’s by 50
years.
 Equivalently: in how many ways can n be
made up as sums of 1’s and 2’s, treating the
order as important? Ans. F(n+1). (Knuth)
GOLDEN RATIO

Fibonacci Series: 0 1 1 2 3 5 8 13 21 ....
Define ratio R(n) = F(n+1)/F(n) n = 1,2,3....
1/1=1 2/1=2 3/2=1.5 5/3=1.667 8/5=1.6
 For large n, R(n) tends to 1.618034 ....
called the Golden Ratio (φ). The ratios are
alternately more and less than φ. (slide)
φ = (1 + 51/2)/2
 For large n, the series tends to a geometric
progression with common ratio φ. To
calculate φ
If (a b c) are three consecutive FN’s
b/a = φ = c/b = (a+b)/b = 1+ (1/φ)
φ2 – φ – 1 = 0
Solutions: φ = (1 + 51/2)/2 or (1 – 51/2)/2.
 Euclid: |------A-------|---B-----|
If A/B = (A+B)/A, then A/B = φ.
FIBONACCI SERIES
[F (n)] = 1 1 2 3 5 8 13 21 34 55 …………….
Golden Ratio =
phi =
1.61803399
n
F (n) F (n+1)
R = F(n+1)/F(n)
R - phi
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
1
1
2
3
5
8
13
21
34
55
89
144
233
377
610
987
1597
2584
4181
6765
1
2
1.5
1.666666667
1.6
1.625
1.615384615
1.619047619
1.617647059
1.618181818
1.617977528
1.618055556
1.618025751
1.618037135
1.618032787
1.618034448
1.618033813
1.618034056
1.618033963
1.618033999
-0.61803399
0.38196601
-0.11803399
0.04863268
-0.01803399
0.00696601
-0.00264937
0.00101363
-0.00038693
0.00014783
-0.00005646
0.00002157
-0.00000824
0.00000315
-0.00000120
0.00000046
-0.00000018
0.00000007
-0.00000003
0.00000001
1
2
3
5
8
13
21
34
55
89
144
233
377
610
987
1597
2584
4181
6765
10946
GOLDEN RATIO (contd.)

Formula for F(n)
1/2
n
n
F(n) = (1/5 )[φ – (1-φ) ]
1/2
where φ = 1.618034... = (1 + 5 )/2
1/2
.... (1-φ) =-0.618304... = (1 - 5 )/2.
 For large n
n
n
F(n) = φ /5
1/2
1/2
F(n) = φ /5 (rounded to nearest integer)
n 1/2
F(n) = integral part of [φ /5 + ½]
a non-recursive formula for F(n).
CONTINUED FRACTION EXPANSION OF φ

Continued fraction (CF): e.g.
28/11 = 2 + 6/11 = 2 + [1/(11/6)]
11/6 = 1 + 5/6 = 1 + [1/(6/5)]
6/5 = 1 + 1/5
CF of 28/11 = (2 1 1 5)
(2 1 1 5 are the coefficients of CF)
CF of ratio p/q is a unique representation.

CF of the ratio F(n+1)/F(n)
e.g 13/8 = (1 1 1 1 1 1)
CF of F(n+1)/F(n) = (1 1 1 1 ...........1)
in which there are n coefs all 1’s.
This is also apparent from
φ = 1+ (1/φ) = 1+[1/(1+(1/φ))] .....
FIBONACCI NUMBERS AND THE G.C.D.
ALGORITHM

Given an arbitrary pair of integers (p, q) p > q
What is the maximum number of steps in the
computation of gcd (p, q) ?
 Solution: The number of steps is maximum when
p, q are consecutive Fibonacci Numbers F(n),
F(n+1).
 Lame’s Theorem:
# of steps in gcd (p, q) < 5 (# of decimal digits in q)
gcd (13,8) = gcd [F(7), F(6)]. # of steps is 4.

Above result follows from:
(a) # of steps in gcd [F(n+1), F(n)] is n-2
(b) F(n) = φn/51/2.

Applications of FNs in Computer Science:

sorting of data, information retrieval

generation of random numbers

methods of approximation ........
GOLDEN RATIO IN NATURE

Plant World: helices in structure of stalks,
stems, tendrils, seeds, flowers, cones and leaves.
# of turns made along a helical path as you
move from one leaf to the leaf directly above it
tends to be F(n): 2 3 5 8... (Phyllotaxy, Kepler)
 Spiral pattern of arrangement sunflower
seeds:
o Two sets of logarithmic spirals,
clockwise and anti-clockwise. # of
spirals in the sets F(n+1), F(n). e.g. 55,
34 (common) 144, 89
and 233, 144 ! (slide)
 Spiral shapes of sea-shells (slide)
 Flower petals: 5 are the most common.
FIBONACCI SPIRAL
13
5
5
8
1
1
3
1
3
2
2
8
FIBONACCI NUMBERS (MISC)

Fibonacci primes:
2 3 5 13 89 233 1597 286559 .......
Are there infinite Fibonacci primes?
Unsolved problem.
 If F(n) is prime, then n is prime (only
exception is F(4) = 3.
 Converse is not true:
If n is prime then F(n) is not always prime
(e.g. F(19) = 4181 = 37 x13).
If the converse were true, it is easily seen that
the number of Fibonacci primes is infinite.

If n is composite, then F(n) is composite.

Every positive integer is a sum of FNs in a
unique way:
n = F(k1) + F(k2) + F(k3) ..... + F(kn)
in which no two consecutive FNs appear.
e.g. 28 = 21 + 5 + 2
φ IN GEOMETRY

The Golden ratio (φ) occurs in pentagons,
decagons and 3-dimension Platonic solids.
1/2
cos 36 = (1+5 )/4 = φ/2
 If (a b c) are three consecutive FNs
2
b –ca=±1
2
n
F(n+1) F(n-1) – F(n) = (-1)
 “Vanishing trick” based on the above
identity. (slides)
FIBONACCI NUMBERS
THE VANISHING TRICK (5,8,13)
8
13
D
A
B
5
B
A
8
C
13 x 5 = 65
13
D
C
D
8 x 8 = 64
5
B
A
C
8 x 8 = 13 x 5 - 1
13 x 5 = 65
Missing Unit Square is the parallelogram (angle 1.3 deg )
FIBONACCI VANISHING TRICK (3,5,8)
8
5
D
A
B
3
B
A
5
C
8 x 3 = 24
D
C
8
5 x 5 = 25
D
3
B
A
5x5=8x3+1
C
8 x 3 = 24
Extra Unit Square is the parallelogram (angle 3.4 deg)
VARIANTS OF FIBONACCI RECURSION

Generalized Fibonacci Series:
G(1)= , G(2) = , G(n) = G(n-1) + G(n-2) n >2
 If (,) = (1,1) then G(n) becomes F(n).
 If (,)= (2,1) then G(n) is L(n), the Lucas
Series: 2 1 3 4 7 11 18 29 47 ......
 Formula for G(n):
G(n) = (1/51/2) [ φn –  (1-φ)n]

Ratio of successive terms approaches φ as n
increases as in Fibonacci Series.

“Tribonacci” Series:
F(1) =1, F(2) = 1, F(3) = 2
F(n) = F(n-1)+F(n-2)+F(n-3) n > 3
1 1 2 4 7 13 24 44 81 ......
Ratio of successive terms approaches 1.83792...
Generalization of recursion to m sums. As m
increases the ratio of successive terms tends to 2.
VARIANTS (contd)

“Rabbit” problem:
Original: Each pair gives birth to a new pair
once a month starting two months after birth.
F(n) = F(n-1) + F(n-2)
1 1 2 3 5 8 13 21 34 ....
 If each pair starts reproducing after three
months
F(n) = F(n-1) + F(n-3)
1 1 2 3 4 6 9 13 19 ....
 If each pair starts reproducing after one
month
F(n) = 2 F(n-1) = 2n-1

Random Fibonacci Recursion.
P(n) = P(n-1) + P(n-2) if coin flip is Heads
P(n) = |P(n-1) - P(n-2)| if coin flip is Tails
Ratio of consecutive terms approaches 1.1319...
(Vishwanath’s constant)
KOLAM DESIGNS

Decorative geometrical patterns on a regular
grid of dots. A line drawing of curves and loops
around the dots (Pulli Kolam)

They adorn entrances to households, places of
worship. Designs are simple to large ones of
bewildering complexity.

South Indian Folk Art is 1000+ years old.
Nurtured by generations of women – housewives
and housemaids - in rural and urban areas.

Broad rules with few constraints allow intricate,
complex and creative designs.
Some Small Popular Kolams
KOLAMS BASED ON FIBONACCI
NUMBERS

Fibonacci Numbers: 0 1 1 2 3 5 8 13 ...

Fibonacci Recursion: F(n) = F(n-1) + F(n-2)

Fibonacci Kolams:
Square
3x3, 5x5, 8x8, 13x13 .....
Rectangular
2x3, 3x5, 5x8, 8x13 ....
They can be created using a modular approach
starting from small kolams (2x2, 2x3) using the
Fibonacci Recursion.
 To create square and rectangular kolams of
arbitrary size, we can use Generalized
Fibonacci Series.
GROUND RULES

Square and Rectangular grids

Four-fold symmetry (rotational) only for square
grids

No empty unit cells

Single loop.

In Folk Art of Kolams, generally
o four-fold symmetry for square kolams is
mandatory
o but empty unit cells and multiple loops are
allowed
o however, single loops are special and more
difficult to achieve. (Anthadhi Kolam)
no loose ends ! Closed loops capture evil spirits from
entering homes !
BASIC EQUATIONS
 Fibonacci Numbers:
0 1 1 2 3 5 8 13 21 34 55 89 ....
Let Q (a b c d) be a quartet of consecutive
Fibonacci numbers. e.g. Q (2 3 5 8)

Identities relating a b c d
b c = b2 + a b
(1)
d 2 = a2 + 4 b c
(2)

Proof: (1) c = b+a, multiplying by b
b c = b2 + a b
(2) d – a = b+c-a = b+b = 2b
d + a = c+b+a= c+c = 2c
Multiplying the two eqs.
d 2 – a2 = 4 b c
d 2 = a2 + 4 b c

Geometrical interpretation
(1) Big rectangle = Square + Small Rectangle
(2) Big square = Small square + 4 Rectangles
BASIC EQUATIONS (contd)
 Unique
feature
of
the
Fibonacci
Construction of a square using a smaller
square in the center and four cyclically
placed rectangles is that four-fold symmetry
is built into it. The basic rectangle module
is rotated by 90, 180 and 270 degrees.
 In standard notation basic equations (1) (2)
are
F(n-2) F(n-1) = F(n-2)2 + F(n-2) F(n-3) n > 2
F(n)2 = F(n-3)2 + 4 F(n-2) F (n-1) n > 3

Some examples: (a b c d)
(0 1 1 2)
22 = 02 + 4 (1 x 1)
(1 1 2 3)
32 = 12 + 4 (1 x 2)
(1 2 3 5)
52 = 12 + 4 (2 x 3)
(2 3 5 8)
82 = 22 + 4 (3 x 5)
(3 5 8 13)
132 = 32 + 4 (5 x 8)
(5 8 13 21)
21 2 = 52 + 4 (8 x 13)
Square Fibonacci Kolam
c
a
b
b
c
a
c
c
b
b
c
5 x 5 (1 2 3 5)
3 x 3 (1 1 2 3) and 3 x 5
2 x 2 (0 1 1 2) and 8 x 8 (2 3 5 8)
Process for Construction
n
Fn-1. Fn-2
Fn x Fn
Fn-1. Fn-2
Fn x Fn
0x0
0
1x1
1
3
1x1
1x1
4
1x2
2
1x2
2x2
5
2x3
3x5
5x8
8 x 13
13 x 21
21 x 34
BLOCK I
RECT KOLAMS
5x5
5
8x8
6
13 x 13
7
21 x 21
8
34 x 34
9
13 x 21
21 x 21
10
4
8 x 13
13 x 13
9
3x3
5x8
8x8
8
3
3x5
5x5
7
2x2
2x3
3x3
6
n
21 x 34
BLOCK II
MODULES
BLOCK III
SQUARE KOLAMS
5x8
8 x 13
13 x 13 : ( 3 5 8 13)
21 x 21 : ( 5 8 13 21)
FIBONACCI KOLAMS OF ARBITRARY SIZE
 Square and Rectangular Kolams of any
desired size (n x n, m x n) can be created
using Generalized Fibonacci Numbers
(GFN).
 Generalized Fibonacci Series:
Given G(1) = α, G(2) = β
n >2
G(n) = G(n-1)+G(n-2)
(α β) = (1 1) gives Fibonacci Series
(α β) = (2 1) gives Lucas Series:
2 1 3 4 7 11 18 29 ........
Basic equations for (a b c d) 4 consecutive GFNs

are the same as for the Fibonacci Series:
d2= a2 + 4 b c
b c = b2 + a b
Square Fibonacci Kolam
c
a
b
b
c
a
c
c
b
b
c
Lucas Kolam (4 x 4)
(2 1 3 4)
(0 2 2 4)
Lucas Kolams ( 7 x 7, 9 x 9)
(1 3 4 7)
(1 4 5 9)
G F Kolams (6 x 6 and 9 x 9)
(0 3 3 6)
(0 5 5 10)
G F Kolam (5 x 5 and 6 x 6)
(4 1 5 6)
(3 1 4 5)
RECIPE FOR GF KOLAM n x n

Choose an appropriate quartet Q (a b c d)

Set d = n

Let c be approximately 0.6 to 0.7 times d
(rounded to nearest integer)

Since d and c are fixed, b = d – c and a = c – b.

For example: for a 17 x 17 square GF Kolam
d = 17 c = 11 b = 6 a = 5
Q (5 6 11 17): 172 = 52 + 4 (6 x 11)
17 x 17 kolam has a 5 x 5 kolam in the
centre enveloped by 4 rectangles 6 x 11
cyclically placed. (slide)
 There are in all 7 sets of 4 splices (28) to
knit together the 5 constituents to produce a
single loop 17 x 17 kolam. With 6 sets of 4
splices (24) we get five loops (slide)
Table 1: Summary of Q = (a b c d) for
Generalized Fibonacci Square Kolams of size n
n
a
b
2m +1
(m = 1,2..)
4t
(t = 1,2…)
4t+2
(t = 1,2…)
1+2k
m-k
2k
2t -k
2k
c
d
k
m+k+ 1 2m +1 0,1,2,3.
2t +k
2t +1-k 2t+1+ k
4t
1,3,5…
4t +2
0,2,4,6..
Table 2: Choices of Q = (a b c d )
and kopt for some n
n
kopt
a
b
c
d=n
14
15
16
17
18
19
20
21
22
23
24
2
1
1
2
2
2
3
2
2
2
3
4
3
2
5
4
5
6
5
4
5
6
5
6
7
6
7
7
7
8
9
9
9
9
9
9
11
11
12
13
13
13
14
15
14
15
16
17
18
19
20
21
22
23
24
17 x 17 (5 6 11 17) – 5 Loops
17 x 17 (5 6 11 17) – Single Loop
Composition of GFK Rectangles
12
12
7
7
7
7
12
12
2
2
12 x 19
5
1
5
1
2
2
2
5
2
5
Fig 6a. COMPOSITION
OF 12 X 19 RECTANGLE
Fig 6a. COMPOSITION OF 12 X 19 RECTANGLE
7
7
7 x 18
4
7
7
4
4
7
4
7
7
7
3
3
1
3
2
2
5
Composition of GFK Rectangles
12
7
7
12
2
5
1
2
2
5
Fig 6a. COMPOSITION OF 12 X 19 RECTANGLE
7
7
4
4
7
7
3
1
3
2
2
5
Fig 6b. COMPOSITION OF 7 X 18 RECTANGLE
FIGURE 6
Fibonacci Kolam ( 2 x 2, Single Loop)
FIGURE 7(a)
FIGURE 7(a)
FIGURE 7(a)
FIGURE 7(b)
FIGURE 7(b)
Fibonacci Kolam ( 2 x 2, Multiple Loop)
FIGURE 8(a)
FIGURE 8(a)
FIGURE 8(b)
Fibonacci Rectangular Kolams ( 2 x 3)
E
H
R
G
U
S
5 x 5 (1 2 3 5)
TABLE 4. GROUP TABLE:
SYMMETRY OPERATORS OF RECTANGLES
(X
Y)
(Y -X)
I
R(90)
I
R(90) R(180) R(-90)
M(x)
M(y)
M(45) M(-45)
I
R(90) R(180) R(-90)
M(x)
M(y)
M(45) M(-45)
R(90) R(180) R(-90)
R(90)
M(-45) M(45)
M(y)
M(x)
M(y)
(-X -Y) R(180)
R(180) R(-90)
(-Y -X)
R(-90)
R(-90)
I
(X -Y)
M(x)
M(x)
M(-45)
M(y)
M(45)
(-X Y)
M(y)
M(y)
M(45)
M(x)
M(-45) R(180)
(Y
M(45)
M(45)
M(y)
M(-45)
M(x)
R(-90)
(-Y -X) M(-45)
M(-45)
M(x)
M(45)
M(y)
R(90) R(-90) R(180)
X)
I
I
R(90) R(180) M(45) M(-45)
I
M(-45) M(45)
M(x)
R(180) R(-90)
I
R(90)
M(x)
M(y)
R(90)
R(90) R(-90)
I
R(180)
I
TABLE 3. SYMMETRY PROPERTIES OF
2 X 3 RECTANGULAR KOLAMS
KOLAM
K
SYMMETRY
OPERATION
I
(X Y)
R(90)
(Y -X)
R(180)
=Mo
(-X -Y)
R(-90)
(-Y X)
Mx
(X -Y)
My
(-X Y)
M(45)
=MyR(90)
(Y X)
M(-45)
=MxR(90)
(-Y -X)
Diamond Grid Kolam D (13)
FIGURE 10: STRUCTURE OF THE DIAMOND GRID
CONCLUSION AND SUMMARY

Kolams are decorative designs in South Indian
Art drawn around a grid of points.

Describe a general scheme for kolam designs
based on the numbers in the Fibonacci Series.

Squares (32 52 82 132 212) and rectangles (2x3,
3x5, 5x8, 8x13) are presented.

Modular approach permits extension to larger
kolams and computer-aided design.
This
enhances the level of creativity of the art.

The algorithmic procedure for kolam design is
generalized for kolams of arbitrary size using
Generalized Fibonacci Series.

Procedure can be generalized to grids other than
square and rectangular.

Challenging mathematical problem: enumeration of kolams of a given size