algorithms

Gaussian elimination
Under our algebraic definition, merely writing down the coordinates of a vertex involves
solving a system of linear equations. How is this done?
We are given a system of n linear equations in n unknowns, say n = 4 and
x 1 − 2x 3 = 2
x 2 + x 3 = 3
x 1 + x 2 − x 4 = 4
x 2 + 3x 3 + x 4 = 5
The high school method for solving such systems is to repeatedly apply the following rule:
if we add a multiple of one equation to another equation, the overall system of equations
remains equivalent. For example, adding −1 times the first equation to the third one, we get
the equivalent system
x 1 − 2x 3 = 2
x 2 + x 3 = 3
x 2 + 2x 3 − x 4 = 2
x 2 + 3x 3 + x 4 = 5
This transformation is clever in the following sense: it eliminates the variable x 1 from the
third equation, leaving just one equation with x 1 . In other words, ignoring the first equation,
we have a system of three equations in three unknowns: we decreased n by 1! We can solve
this smaller system to get x 2 , x 3 , x 4 , and then plug these into the first equation to get x 1 .
This suggests an algorithm—once more due to Gauss.
procedure gauss(E, X)
Input: A system E = {e 1 , . . . , e n } of equations in n unknowns X = {x 1 , . . . , x n }:
e 1 : a 11 x 1 + a 12 x 2 + · · · + a 1n x n = b 1 ; · · · ; e n : a n1 x 1 + a n2 x 2 + · · · + a nn x n = b n
Output: A solution of the system, if one exists
if all coefficients a i1 are zero:
halt with message ‘‘either infeasible or not linearly independent’’
if n = 1: return b 1 /a 11
choose the coefficient a p1 of largest magnitude, and swap equations e 1 , e p
for i = 2 to n:
e i = e i − (a i1 /a 11 ) · e 1
(x 2 , . . . , x n ) = gauss(E − {e 1 }, X − {x 1 })
x 1 = (b 1 − ∑ j>1 a 1jx j )/a 11
return (x 1 , . . . , x n )
(When choosing the equation to swap into first place, we pick the one with largest |a p1 | for
reasons of numerical accuracy; after all, we will be dividing by a p1 .)
Gaussian elimination uses O(n 2 ) arithmetic operations to reduce the problem size from
n to n − 1, and thus uses O(n 3 ) operations overall. To show that this is also a good estimate
of the total running time, we need to argue that the numbers involved remain polynomially
bounded—for instance, that the solution (x 1 , . . . , x n ) does not require too much more
precision to write down than the original coefficients a ij and b i . Do you see why this is true?
221