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ADVANCED ALGORITHMS

Polynomials and the FFT(UNIT-3)

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1. Polynomials : A polynomial in the variable x over an algebraic field F represents a function A(x) as a formal sum :

A(x) = aj xj j = 0 to n-1

Here, we call the values a0, a1, a2, a3,….,an-1 as the coefficients of the polynomial.A polynomial has degree k, if its highest non-zero coefficient is ak.

Any integer strictly greater than the degree of a polynomial ‘k’, is degree-bound ‘n’. i.e., n > k.

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The representation of a Polynomial :

Ex-1 : A(x) = 𝑥3 − 2 − 1 𝑥Here, A(x) has degree 3.

A(x) has degree bounds 4,5,6,..A(x) has coefficients (-1,-2,0,1)

Ex-2 : 𝑥4 + 𝑥2 − 1

Here, A(x) has degree 4. A(x) has degree bounds 5,6,7,.. A(x) has coefficients (-1,0,1,0,1)

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Coefficient Representation : (CR)

A CR of a polynomial of degree bound ‘n’ is a vector of coefficients

a =

Adding Polynomials : Let a = (a0,a1,a2,… , an-1)

b = (b0,b1,b2,… , bn-1)

Sum is the coefficient vector C, wherec = (c0,c1,c2,… , cn-1)

Here, cj = aj + bj j = 0,1,2,..,n-1.

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Multiplying Polynomials :Multiplying of two degree-bound ‘n’ polynomials A(x) and B(x) :

C(x) =

Ex-3 : Add & Multiply the following two polynomials :A(x) = x2 - 10 x + 9B(x) = x2 + 4 x - 5

A(x) + B(x) = 2x2 - 6 x + 4

A(x) X B(x) = x4 – 6 x3 – 36 x2 + 86 x - 45

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2. Horner’s Rule :Given the Polynomial :p(x) = ai xi

= a0 + a1 x + a2 x2 + … + an xn

i = 0 to n

Here, i = 0,1,2,..,n-1,n ai is a real number.

The above polynomial can be written in the form :p(x) = a0 + x (a1 + x (a2 + x (a3 + …. +

x (an-1 + an x )……)) Let x = x0

Now, assume bn := an

bn-1 := an-1 + bn xo ………….. b0 := a0 + b1 xo

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Here let us substitute b values into expression of p(x0).

So, p(x0) = a0 + x0 (a1 + x0 (a2 + x0 (a3 + …. + x0 (an-1 + bn x0 )……))

= a0 + x0 (a1 + x0 (a2 + x0 (a3 + …. + x0 bn-1 )……))

= ………… = a0 + x0 (a1 + x0 (a2 + x0 b3 ))

= a0 + x0 (a1 + x0 b2)

= a0 + x0 b1 = b0

Hence p(x0) = b0

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Ex-4 : Evaluate 2x3 - 6x2 + 2x - 1 for x = 3

Synthetic Division : (Horner’s Method)

x0 | x3 x2 x1 x0

---------------------------------------- 3 | 2 - 6 2 - 1

| 6 0 6-----------------------------------------

2 0 2 5

i.e., The Value of p(x0) is 5.

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Ex-5 : Evaluate x3 - 6x2 + 11 x - 6 for x = 2

Synthetic Division : (Horner’s Method)

x0 | x3 x2 x1 x0

---------------------------------------- 2 | 1 - 6 11 - 6

| 2 - 8 6-----------------------------------------

1 - 4 3 0

i.e., The Value of p(x0) is 0.

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3. Point-Value Representation : A Point-Value Representation of a polynomial A(x) of degree-bound n is a set of n point-value pairs :

{(x0 ,y0), (x1 ,y1), (x2 ,y2),…, (xn-1 ,yn-1),}

such that all of the xk are distinct and yk = A(xk) for k = 0,1,2,…,n-1.

Ex-6 : A(x) = x3 - 2 x + 1Here, xk : 0, 1, 2, 3

A(xk ) : 1, 0, 5, 22 { (0,1), (1,0), (2,5), (3,22) }

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Interpolation of PVR : The inverse of evaluation is Interpolation. - determines coefficient form of polynomial from

point value representation.

Theorem : (Uniqueness of an interpolating polynomial)

“For any set {(x0 ,y0), (x1 ,y1), (x2 ,y2),…, (xn-1 ,yn-1),}

of n point-value pairs all the xk values are distinct,

there is a unique polynomial A(x) of degree-bound n such that yk = A(xk) for k = 0, 1, 2, …, n-1.”

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The above theorem can be written as :

The matrix on the left is denoted V(x0 , x1 , x2 , …, xn-1). It is known as : Vander-monde Matrix.

i.e., a = V -1 y.

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Lagrange’s Formula :

A(x) = yk Z, wherek = 0 to n-1

Z = j k (x – xj) / j k (xk – xj)

Ex-7 : Consider the PVR :[ (0,1), (1,0), (2,5), (3,22) ]

Here, 1.[(x-1)(x-2)(x-3)] / [(0-1) (0-2) (0-3)]= (x3 - 6x2 + 11 x - 6) / - 6= (-x3 + 6x2 - 11 x + 6) / 6 ….(1)

And 0.[(x-0)(x-2)(x-3)] / [(1-0) (1-2) (1-3)]= 0 ….(2)

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And 5.[(x-0)(x-1)(x-3)] / [(2-0) (2-1) (2-3)]= 5. (x3 - 4x2 + 3 x ) / (-2)= (-15 x3 + 60 x2 - 45 x ) / 6 …..(3)

And 22.[(x-0)(x-1)(x-2)] / [(3-0) (3-1) (3-2)]= 22 (x3 - 3 x2 + 2 x ) / 6= (22 x3 - 66 x2 + 44 x ) / 6 ……(4)

(1) + (2) + (3) + (4) : (6 x3 + 0 x2 - 12 x + 6) / 6= x3 - 2 x + 1 …Answer

The above is the polynomial for the given PVR.

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Adding Polynomials using PVR : Let C(x) = A(x) + B(x)

A : [(x0 ,y0), (x1 ,y1), (x2 ,y2),…, (xn-1 ,yn-1)]

B : [(x0 ,z0), (x1 ,z1), (x2 ,z2),…, (xn-1 ,zn-1)]

C : [ (x0 , y0+ z0), (x1 , y1+ z1), ….,(xn-1 , yn-1+ zn-1) ]

Ex-8 : Find the value of C(x) in PVR, whereA(x) = x3 - 2 x + 1B(x) = x3 + x2 + 1

Let x = (0,1,2,3)

Then A : [(0,1), (1,0), (2, 5), (3,22) ] B : [ (0,1), (1,3), (2, 13), (3, 37) ]

Hence C : [(0,2), (1, 3), (2, 18), (3, 59) ]

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Multiplying Polynomials using PVR :Let C(x) = A(x) . B(x) A : [(x0 ,y0), (x1 ,y1), (x2 ,y2),…, (x2n-1 ,y2n-1)]

B : [(x0 ,z0), (x1 ,z1), (x2 ,z2),…, (x2n-1 ,z2n-1)]

C : [ (x0 , y0 z0), (x1 , y1 z1),….,(x2n-1 , y2n-1+ z2n-1) ]

Ex-9 : Find the value of C(x) in PVR, whereA(x) = x3 - 2 x + 1 ; B(x) = x3 + x2 + 1

Let x = (-3,-2,-1,0,1,2,3)

A : [(-3,20), (-2,-3), (-1,2),(0,1), (1,0), (2, 5), (3,22) ]B : [ (-3,-17),(-2,-3), (-1, 1),(0,1), (1,3), (2, 13), (3,17)]C : [(-3,340),(-2,9),(-1,2),(0,1), (1, 0), (2, 65), (3,814) ]

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Ex-8 Contd… There are FOUR cordinates :

C : [(0,2), (1, 3), (2, 18), (3, 59) ]

Basing on First & Second Coordinates : - 2 x3 + 12 x2 - 22 x + 12 …..(1) 9 x3 - 45 x2 + 54 x …..(2)

Basing on Third & Fourth Coordinates :- 54 x3 + 216 x2 - 162 x …..(1) 59 x3 - 177 x2 + 118 x …..(2)

The denominator is 6. Finally 2 x3 + x2 - 2 x + 2

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4. Complex Roots of Unity : A complex nth root of unity (1) is a complex number such that n = 1. There are exactly n complex nth roots of unity :

e2 i k / n for k = 0, 1, 2, …., n-1 where eiu = cos (u) + i sin (u)

Using e2 i k / n = cos(2 k / n) + i sin(2 k / n)

we can check that it is a root.

(e2 i k / n)n = e2 i k

= cos(2 k) + i sin (2k) = 1 + 0 = 1

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Degrees : 00 450 900 1800

SIN 0 1/ 2 1 0COS 1 1/ 2 0 -1

Principal nth Root of Unity : The Value n = e 2 ∏ i /n

is called the ‘principal nth root of unity’.

All of the other complex nth roots of unity are powers of n .

The n complex nth roots of unity are : n

0, n1, n

2,….., nn-1

Note : Here, nn = n

0 = 1

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Ex-10 : Find the 8th complex roots of unity.

The two roots of unity are : +1, - 1The complex 4th roots of unity are :

1, - 1, i, - i where (-1) = iIn general, the n complex roots of unity are equally spaced around the circle of unit radius centered at the origin of the complex plane.

Note : nj . n

k = nj+k = n

(j + k) mod n

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Visualizing 8 Complex 8th Roots of Unity :

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First Root : 80

= 88

= 1Here, k = 0, n = 8

So, 80 = e 2 ∏ i k / n = e 0 = 1

Second Root : Here, k = 1, n = 88

1 = e 2 ∏ i k / n = e 2 ∏ i / 8 = e ∏ i / 4

by using the formula eiu = cos (u) + i sin (u)

So, 81

= cos (/4) + i sin (/4)

= 1 / 2 + i / 2

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Third Root : Here, k = 2, n = 88

2 = e 2 ∏ i k / n = e 2 ∏ i 2 / 8 = e ∏ i / 2

So, 82

= cos (/2) + i sin (/2) = i

Fourth Root : Here, k = 3, n = 88

3 = e 2 ∏ i k / n = e 2 ∏ i 3 / 8 = e ∏ i 3 / 4

So, 83

= cos (3/4) + i sin (3/4)= cos ( - /4) + i sin ( - /4)= - cos /4 + i sin /4)

[since cos ( - x) = - cos x, sin ( - x) = sin x]

= - 1 / 2 + i /2

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Fifth Root : Here, k = 4, n = 88

4 = e 2 ∏ i k / n = e 2 ∏ i 4 / 8 = e ∏ i

So, 84

= cos () + i sin () = - 1

Sixth Root : Here, k = 5, n = 88

5 = e 2 ∏ i k / n = e 2 ∏ i 5 / 8 = e ∏ i 5 / 4

So, 85

= cos (5/4) + i sin (5/4)= cos ( + /4) + i sin ( + /4)= - cos /4 + i (- sin /4)

[since cos ( + x) = - cos x, sin ( + x) = - sin x]

= - 1 / 2 - i /2

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Seventh Root : Here, k = 6, n = 88

6 = e 2 ∏ i k / n = e 2 ∏ i 6 / 8 = e 3∏ i / 2

So, 86

= cos ( + /2 ) + i sin ( + /2) = - cos (/2) + i (- sin (/2) )

= - 0 + i ( - 1) = - iEighth Root : Here, k = 7, n = 8

87 = e 2 ∏ i k / n = e 2 ∏ i 7 / 8 = e ∏ i 7 / 4

So, 87 = cos (7/4) + i sin (7/4)

= cos ( + 3/4) + i sin ( + 3/4)= - cos 3/4 + i (- sin 3/4)

[since cos ( + x) = - cos x, sin ( + x) = - sin x]= cos /4 + i (- sin /4) = 1 / 2 - i /2