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 # of cadences 1 S 1= F(2) 2 SS L 2 =F(3) 3 SSS LS SL 3 =F(4) 4 SSSS LSS SLS SSL LL 5 =F(5) 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 - phi1 1 1 1 -0.618033992 1 2 2 0.381966013 2 3 1.5 -0.118033994 3 5 1.666666667 0.048632685 5 8 1.6 -0.018033996 8 13 1.625 0.006966017 13 21 1.615384615 -0.002649378 21 34 1.619047619 0.001013639 34 55 1.617647059 -0.00038693

10 55 89 1.618181818 0.0001478311 89 144 1.617977528 -0.0000564612 144 233 1.618055556 0.0000215713 233 377 1.618025751 -0.0000082414 377 610 1.618037135 0.0000031515 610 987 1.618032787 -0.0000012016 987 1597 1.618034448 0.0000004617 1597 2584 1.618033813 -0.0000001818 2584 4181 1.618034056 0.0000000719 4181 6765 1.618033963 -0.0000000320 6765 10946 1.618033999 0.00000001

GOLDEN RATIO (contd.) Formula for F(n)

F(n) = (1/51/2)[φn – (1-φ)n]

where φ = 1.618034... = (1 + 51/2)/2 .... (1-φ) =-0.618304... = (1 - 51/2)/2.

For large n F(n) = φn/51/2

F(n) = φn/51/2 (rounded to nearest integer) F(n) = integral part of [φn/51/2 + 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

8

8

5

5

3

32

2

1 1

1

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.

cos 36 = (1+51/2)/4 = φ/2

If (a b c) are three consecutive FNs b2 – c a = ± 1

F(n+1) F(n-1) – F(n)2 = (-1)n

“Vanishing trick” based on the above identity. (slides)

FIBONACCI NUMBERSTHE VANISHING TRICK (5,8,13)

13 x 5 = 65

8 x 8 = 64

8 x 8 = 13 x 5 - 113 x 5 = 65

Missing Unit Square is the parallelogram (angle 1.3 deg)

AB

CD

A

BD

C

13

5

8

8

A

BD

C

13

5

FIBONACCI VANISHING TRICK (3,5,8)

8 x 3 = 24

5 x 5 = 25

5 x 5 = 8 x 3 + 1

8 x 3 = 24

Extra Unit Square is the parallelogram (angle 3.4 deg)

A B

C

D

A

C

D

B

A

C

D

B

5

5

3

8

8

3

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) d2 = 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.

d2 – a2 = 4 b c d2 = 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) 212 = 52 + 4 (8 x 13)

FIGURE 1

c

a c b

b

c

b c

b

ca

Square Fibonacci Kolam

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)

n Fn-1. Fn-2 Fn x Fn Fn-1. Fn-2 Fn x Fn n

0 x 0 0

1 x 1 13 1 x 1

1 x 1 24 1 x 2 1 x 2

2 x 2 2 x 2 35 2 x 3 2 x 3

3 x 3 3 x 3 46 3 x 5 3 x 5

5 x 5 5 x 5 57 5 x 8 5 x 8

8 x 8 8 x 8 68 8 x 13 8 x 13

13 x 13 13 x 13 79 13 x 21 13 x 21

21 x 21 21 x 21 810 21 x 34 21 x 34

34 x 34 9

BLOCK I BLOCK II BLOCK III RECT KOLAMS MODULES SQUARE KOLAMS

FIGURE 6. RECURSIVE PROCEDURE FOR CONSTRUCTION OF SQUARE AND RECTANGULAR FIBONACCI KOLAMS

Process for Construction

5 x 8

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 G(n) = G(n-1)+G(n-2) 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: b c = b2 + a b d2= a2 + 4 b c

FIGURE 1

c

a c b

b

c

b c

b

ca

Square Fibonacci Kolam

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)

(3 1 4 5)

(4 1 5 6)

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 c d k

2m +1 1+2k m-k m+k+ 1 2m +1 0,1,2,3.(m = 1,2..)

4t 2k 2t -k 2t +k 4t 1,3,5…(t = 1,2…)

4t+2 2k 2t +1-k 2t+1+ k 4t +2 0,2,4,6..(t = 1,2…)

Table 2: Choices of Q = (a b c d ) and kopt for some n

n kopt a b c d = n

14 2 4 5 9 1415 1 3 6 9 1516 1 2 7 9 1617 2 5 6 11 1718 2 4 7 11 1819 2 5 7 12 1920 3 6 7 13 2021 2 5 8 13 2122 2 4 9 13 2223 2 5 9 14 2324 3 6 9 15 24

17 x 17 (5 6 11 17) – 5 Loops

17 x 17 (5 6 11 17) – Single Loop

Fig 6a. COMPOSITION OF 12 X 19 RECTANGLE

Fig 6b. COMPOSITION OF 7 X 18 RECTANGLE

FIGURE 6

7

7 7

74

4

3

3

1 2 52

Composition of GFK Rectangles

7 x 18

12 x 19

Composition of GFK Rectangles

Fig 6a. COMPOSITION OF 12 X 19 RECTANGLE

Fig 6b. COMPOSITION OF 7 X 18 RECTANGLE

FIGURE 6

12

12

7

7

5

5 2

2

2

7

7 7

7 4

4

3

3

1 2 5 2

1

FIGURE 7(a)

FIGURE 7(b)

FIGURE 7(a)

FIGURE 7(b)

FIGURE 7(a)

FIGURE 7(b)

Fibonacci Kolam ( 2 x 2, Single Loop)

FIGURE 8(a)

FIGURE 8(b)

FIGURE 8(a)

FIGURE 8(b)

Fibonacci Kolam ( 2 x 2, Multiple Loop)

Fibonacci Rectangular Kolams ( 2 x 3)

E R G

H U S

5 x 5 (1 2 3 5)

I R(90) R(180) R(-90) M(x) M(y) M(45) M(-45)

(X Y) I I R(90) R(180) R(-90) M(x) M(y) M(45) M(-45)

(Y -X) R(90) R(90) R(180) R(-90) I M(-45) M(45) M(y) M(x)

(-X -Y) R(180) R(180) R(-90) I R(90) M(y) M(x) M(-45) M(45)

(-Y -X) R(-90) R(-90) I R(90) R(180) M(45) M(-45) M(x) M(y)

(X -Y) M(x) M(x) M(-45) M(y) M(45) I R(180) R(-90) R(90)

(-X Y) M(y) M(y) M(45) M(x) M(-45) R(180) I R(90) R(-90)

(Y X) M(45) M(45) M(y) M(-45) M(x) R(-90) R(90) I R(180)

(-Y -X) M(-45) M(-45) M(x) M(45) M(y) R(90) R(-90) R(180) I

TABLE 4. GROUP TABLE:

SYMMETRY OPERATORS OF RECTANGLES

TABLE 3. SYMMETRY PROPERTIES OF

2 X 3 RECTANGULAR KOLAMS

KOLAMK

SYMMETRYOPERATION

I (X Y)

R(90) (Y -X)

R(180) (-X -Y)=Mo

R(-90) (-Y X)

Mx (X -Y)

My (-X Y)

M(45) (Y X)=MyR(90)

M(-45) (-Y -X)=MxR(90)

FIGURE 10: STRUCTURE OF THE DIAMOND GRID

Diamond Grid Kolam D (13)

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: enumera-tion of kolams of a given size