Calculating Polynomials

1. Calculating Polynomials • We will use a generic polynomial form of:where the coefficient values are known constants • The value of x will be the input and the result is the value of the polynomial using this x value

2. Standard Evaluation Algorithm result = a[0] + a[1]*x xPower = x for i = 2 to n do xPower = xPower * x result = result + a[i]*xPower end for return result

3. Analysis • Before the loop, there is • One multiplication • One addition • The for loop is done N-1 times • There are two multiplications in the loop • There is one addition in the loop • There are a total of • 2N – 1 multiplications • N additions

4. Horner’s Method • Based on the factorization of a polynomial • The generic equation is factored as: • For example, the equation: • would be factored as:

5. Horner’s Method Algorithm result = a[n] for i = n - 1 down to 0 do result = result * x result = result + a[i] end for return result

6. Analysis • The for loop is done N times • There is one multiplication in the loop • There is one addition in the loop • There are a total of • N multiplications • N additions • Saves N – 1 multiplications over the standard algorithm

7. Preprocessed Coefficients • Uses a factorization of a polynomial based on polynomials of about half the original polynomial’s degree • For example, where the standard algorithm would do 255 multiplications to calculate x256, we could square x and then square the result seven more times to get the same result

8. Preprocessed Coefficients • Used with polynomials that are monic (an=1) and have a highest power that is one less than a power of 2 • If the polynomial has highest power that is 2k– 1, we can factor it as:where j = 2k-1

9. Preprocessed Coefficients • If we choose b so that it is aj-1 – 1, q(x) and r(x) will both be monic, and so the process can be recursively applied to them as well

10. Preprocessed Coefficients Example • For the example equation of:because the highest degree is 3 = 22 –1, j would be 21 = 2, and b would have a value of a1 – 1 = 6 • This makes one factor x2 + 6, and we divide p(x) by this polynomial to find q(x) and r(x)

11. Preprocessed Coefficients Example • The division is:which gives

12. Analysis • We analyze preprocessed coefficients by developing recurrence relations for the number of multiplications and additions • In the factorization, we break the polynomial up into two smaller polynomials and do one additional multiplication and two additional additions

13. Analysis • Because we required that N = 2k– 1, we get: M(1) = 0 A(1) = 0M(k) = 2M(k–1) + 1 A(k) = 2A(k–1) + 2 • Solving these equations gives about N/2 multiplications and (3N – 1)/2 additions • We need to include the k – 1 (or lg N) multiplications needed to calculate the series of values x2, x4, x8, …

14. Polynomial Algorithm Comparison • For the example polynomial: • Standard algorithm: • 5 multiplications and 3 additions • Horner’s method • 3 multiplications and 3 additions • Preprocessed coefficients • 2 multiplications and 4 additions

15. Polynomial Algorithm Comparison • In general, for a polynomial of degree N: • Standard algorithm: • 2N – 1 multiplications and N additions • Horner’s method • N multiplications and N additions • Preprocessed coefficients • N/2 + lg N multiplications and (3N – 1)/2 additions

16. Linear Equations • A system of linear equations is a set of N equations in N unknowns: a11x1 + a12x2 + a13x3 + … + a1NxN = b1 a21x1 + a22x2 + aa3x3 + … + a2NxN = b2  aNx1 + aN2x2 + aN3x3 + … + aNNxN = bN

17. Linear Equations • When these equations are used, the coefficients (a values) are constants and the results (b values) are typically input values • We want to determine the x values that satisfy these equations and produce the indicated results

18. Linear Equation Example • An example set of linear equations with three unknowns is:2x1 – 4x2 + 6x3 = 146x1 – 6x2 + 6x3 = 244x1 + 2x2 + 2x3 = 18

19. Solving Linear Equations • One method to determine a solution would be to substitute one equation into another • For example, we solve the first equation for x1 and then substitute this into the rest of the equations • This substitution reduces the number of unknowns and equations

20. Solving Linear Equations • If we repeat this, we get to one unknown and one equation and can then back up to get the values of the rest • If there are many equations, this process can be time consuming, prone to error, and is not easily computerized

21. Gauss-Jordan Method • This method is based on the previous idea • We store the equation constants in a matrix with N rows and N+1 columns • For the example, we would get:

22. Gauss-Jordan Method • We perform operations on the rows until we eventually have the identity matrix in the first N columns and then the unknown values will be in the final column:

23. Gauss-Jordan Method • On each pass, we pick a new row and divide it by the first element that is not zero • We then subtract multiples of this row from all of the others to create all zeros in a column except in this row • When we have done this N times, each row will have one value of 1 and the last column will have the unknown values

24. Example • Consider the example again: • We begin by dividing the first row by 2, and then subtract 6 times it from the second row and 4 times it from the third row

25. Example • Now, we divide the second row by 6, and then subtract -2 times it from the first row and 10 times it from the third row

26. Example • Now, we divide the third row by 10, and then subtract -1 times it from the first row and -2 times it from the second row

27. Example • This gives the result of: • And we have that x1 = 3, x2 = 1, and x3 = 2