Gauss' method lends itself nicely to computerization.
The code below illustrates.
It operates on an matrix `a`,
pivoting with the first row, then
with the second row, etc.

```
for (pivot_row = 1; pivot_row <= n - 1; pivot_row++) {
for (row_below = pivot_row + 1; row_below <= n; row_below++) {
multiplier = a[row_below, pivot_row] / a[pivot_row, pivot_row];
for (col = pivot_row; col <= n; col++) {
a[row_below, col] -= multiplier * a[pivot_row, col];
}
}
}
```

(This code is in the C language.
Here is a brief translation.
The loop construct
`for (pivot_row = 1; pivot_row <= n - 1; pivot_row++) { ... }`
sets `pivot_row` to 1 and then iterates while
`pivot_row` is less than or equal to , each time through
incrementing `pivot_row` by one with the
"`++`" operation.
The other non-obvious construct is that the "`-=`" in the innermost
loop amounts to the
`a[row_below, col] =- multiplier * a[pivot_row, col] + a[row_below, col]}` operation.)

While this code provides a quick take on how Gauss' method can be
mechanized, it is not ready to use.
It is naive in many ways.
The most glaring way is that it assumes that
a nonzero number is always found in
the `pivot_row, pivot_row` position for use as
the pivot entry.
To make it practical,
one way in which this code needs to be reworked is to cover
the case where finding a zero in that location leads to a row swap, or to the
conclusion that the matrix is singular.

Adding some `if` statements to cover those cases is not hard,
but we will instead consider some more subtle ways in which the code is naive.
There are pitfalls arising from the computer's reliance on finite-precision
floating point arithmetic.

For example, we have seen above that we must handle as a separate case a system that is singular. But systems that are nearly singular also require care. Consider this one.

By eye we get the solution and .
But a computer has more trouble.
A computer that represents real numbers to eight significant places (as is
common, usually called **single precision**) will represent the second
equation internally as , losing the
digits in the ninth place.
Instead of reporting the correct solution, this computer will report something
that is not even close— this computer thinks that the system is singular
because the two equations are represented internally as equal.

For some intuition about how the computer could come up with something that far off, we can graph the system.

At the scale of this graph, the two lines cannot be resolved apart.
This system is nearly singular in the sense that
the two lines are nearly the same line.
Near-singularity gives this system the property that a small change in the
system can cause a large change in its solution; for instance, changing the
to changes the intersection point
from to .
This system changes radically depending on a ninth digit, which explains why
the eight-place computer has trouble.
A problem that is very sensitive to inaccuracy or uncertainties in
the input values is **ill-conditioned**.

The above example gives one way in which a system can be difficult to solve on a computer. It has the advantage that the picture of nearly-equal lines gives a memorable insight into one way that numerical difficulties can arise. Unfortunately this insight isn't very useful when we wish to solve some large system. We cannot, typically, hope to understand the geometry of an arbitrary large system. In addition, there are ways that a computer's results may be unreliable other than that the angle between some of the linear surfaces is quite small.

For an example, consider the system below, from (Hamming 1971).

The second equation gives , so and thus both variables have values that are just less than . A computer using two digits represents the system internally in this way (we will do this example in two-digit floating point arithmetic, but a similar one with eight digits is easy to invent).

The computer's row reduction step produces a second equation , which the computer rounds to two places as . Then the computer decides from the second equation that and from the first equation that . This value is fairly good, but the is quite bad. Thus, another cause of unreliable output is a mixture of floating point arithmetic and a reliance on pivots that are small.

An experienced programmer may respond that we should go to
**double precision**
where sixteen significant digits are retained.
This will indeed solve many problems.
However, there are some difficulties with it as a general approach.
For one thing, double precision takes longer than single
precision (on a '486 chip, multiplication takes eleven ticks in single
precision but fourteen in double precision (Microsoft 1993)) and has twice
the memory requirements.
So attempting to do all calculations in
double precision is just not practical.^{[citation needed]}
And besides, the above systems can obviously be tweaked to give the
same trouble in the seventeenth digit, so double precision
won't fix all problems.
What we need is a strategy to minimize the numerical
trouble arising from solving systems on a computer,
and some guidance as to how far the reported
solutions can be trusted.

Mathematicians have made a careful study of how to get the most
reliable results.
A basic improvement on the naive code above
is to not simply take the entry
in the *pivot_row*, *pivot_row*
position for the pivot,
but rather to look at all of the entries in the *pivot_row*
column below the *pivot_row* row,
and take the one that is most likely to give reliable results
(e.g., take one that is not too small).
This strategy is **partial pivoting**.
For example, to solve the troublesome system () above,
we start by looking at both equations for a best first pivot,
and taking the in
the second equation as more likely to give good results.
Then, the pivot step of gives a first equation of
, which the computer will represent as
, leading to the conclusion that
and, after back-substitution, ,
both of which are close to right.
The code from above can be adapted to this purpose.

```
for (pivot_row = 1; pivot_row <= n - 1; pivot_row++) {
/* Find the largest pivot in this column (in row max). */
max = pivot_row;
for (row_below = pivot_row + 1; pivot_row <= n; row_below++) {
if (abs(a[row_below, pivot_row]) > abs(a[max, row_below]))
max = row_below;
}
/* Swap rows to move that pivot entry up. */
for (col = pivot_row; col <= n; col++) {
temp = a[pivot_row, col];
a[pivot_row, col] = a[max, col];
a[max, col] = temp;
}
/* Proceed as before. */
for (row_below = pivot_row + 1; row_below <= n; row_below++) {
multiplier = a[row_below, pivot_row] / a[pivot_row, pivot_row];
for (col = pivot_row; col <= n; col++) {
a[row_below, col] -= multiplier * a[pivot_row, col];
}
}
}
```

A full analysis of the best way to implement Gauss' method
is outside the scope of the book (see (Wilkinson 1965)),
but the method recommended by most experts
is a variation on the code above that first finds the best pivot
among the candidates, and then scales it to a number that is less
likely to give trouble.
This is
**scaled partial pivoting**.

In addition to returning a result that is likely to be reliable,
most well-done code will return a number, called the
**condition number**
that describes the factor by which uncertainties in the input numbers
could be magnified to become inaccuracies in the results returned
(see (Rice 1993)).

The lesson of this discussion is that just because Gauss' method always works in theory, and just because computer code correctly implements that method, and just because the answer appears on green-bar paper, doesn't mean that the answer is reliable. In practice, always use a package where experts have worked hard to counter what can go wrong.

## ExercisesEdit

- Problem 1

Using two decimal places, add and .

- Problem 2

This intersect-the-lines problem contrasts with the example discussed above.

Illustrate that in this system some small change in the numbers will produce only a small change in the solution by changing the constant in the bottom equation to and solving. Compare it to the solution of the unchanged system.

- Problem 3

Solve this system by hand (Rice 1993).

- Solve it accurately, by hand.
- Solve it by rounding at each step to four significant digits.

- Problem 4

Rounding inside the computer often has an effect on the result. Assume that your machine has eight significant digits.

- Show that the machine will compute as unequal to . Thus, computer arithmetic is not associative.
- Compare the computer's version of and . Is twice the first equation the same as the second?

- Problem 5

Ill-conditioning is not only dependent on the matrix of coefficients. This example (Hamming 1971) shows that it can arise from an interaction between the left and right sides of the system. Let be a small real.

- Solve the system by hand. Notice that the 's divide out only because there is an exact cancelation of the integer parts on the right side as well as on the left.
- Solve the system by hand, rounding to two decimal places, and with .

## ReferencesEdit

- Hamming, Richard W. (1971),
*Introduction to Applied Numerical Analysis*, Hemisphere Publishing. - Rice, John R. (1993),
*Numerical Methods, Software, and Analysis*, Academic Press. - Wilkinson, J. H. (1965),
*The Algebraic Eigenvalue Problem*, Oxford University Press. - Microsoft (1993),
*Microsoft Programmers Reference*, Microsoft Press.