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