and the


Evaluation by Horner's Rule

Given a polynomial of degree n,

p(x) = anxn + an-1xn-1 + ... + a1x1 + a0
one might suspect that n+(n-1)+(n-2)+...+1 = n(n+1)/2 multiplications would be needed to evaluate p(x) for a given x. However Horner's Rule shows that it can be rewritten so that only n multiplications are needed:
p(x) = (((anx + an-1)x + an-2)x + ... a1)x + a0

Incidentally, this is exactly the way that integer constants are evaluated from strings of characters (digits):

12345 = 1*104 + 2*103 + 3*102 + 4*101 + 5*100
      = (((1*10 + 2)*10 + 3*10 + 4)*10 + 5

- just think of the digit values as the coefficients and the `base' of the number system as x.

Horner's rule also pops up for calculating the "rolling hash value"in Rabin's [string searching] algorithm.

-- 1999 L.A.

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© L. Allison   http://www.allisons.org/ll/   (or as otherwise indicated),
Faculty of Information Technology (Clayton), Monash University, Australia 3800 (6/'05 was School of Computer Science and Software Engineering, Fac. Info. Tech., Monash University,
was Department of Computer Science, Fac. Comp. & Info. Tech., '89 was Department of Computer Science, Fac. Sci., '68-'71 was Department of Information Science, Fac. Sci.)
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