1.1 Roundoff Errors
Consider the common fractions
,
,
,
,
, and
. In the decimal, or base 10, number system, we can represent
,
, and
exactly as 0.5, 0.25, and 0.2, respectively. In the decimal system, we can represent any fraction whose denominator divides evenly into a power of 10 exactly as a decimal fraction. But the denominator 3 doesn't divide evenly, and so
repeats infinitely: 0.3333 . . . . Neither does the denominator 6, and
is 0.16666 . . . . Neither does the denominator 7, and
is 0.142857142857 . . ., where the last
group
of six digits repeats infinitely.
Like most modern computers, the Java virtual machine uses the binary, or base 2, number system. How well can we represent these fractions in base 2? Program 1-1 prints and sums some of their values. See Listing 1-1.
Listing 1-1 Fractions.
package numbercruncher.program1_1;
/**
* PROGRAM 1-1: Fractions
*
* Print and sum the values of the fractions 1/2, 1/3, 1/4, and 1/5
* to look for any roundoff errors.
*/
public class Fractions
{
private static final float HALF = 1/2f;
private static final float THIRD = 1/3f;
private static final float QUARTER = 1/4f;
private static final float FIFTH = 1/5f;
private static final float SIXTH = 1/6f;
private static final float SEVENTH = 1/7f;
private static final int FACTOR = 840;
public static void main(String args[])
{
System.out.println("1/2 = " + HALF);
System.out.println("1/3 = " + THIRD);
System.out.println("1/4 = " + QUARTER);
System.out.println("1/5 = " + FIFTH);
System.out.println("1/6 = " + SIXTH);
System.out.println("1/7 = " + SEVENTH);
float sum = 0;
System.out.println();
for (int i = 0; i < FACTOR; ++i) sum += HALF;
System.out.println("1/2 summed " + FACTOR + " times = " + sum +
" (should be " + FACTOR/2 + ")");
sum = 0;
for (int i = 0; i < FACTOR; ++i) sum += THIRD;
System.out.println("1/3 summed " + FACTOR + " times = " + sum +
" (should be " + FACTOR/3 + ")");
sum = 0;
for (int i = 0; i < FACTOR; ++i) sum += QUARTER;
System.out.println("1/4 summed " + FACTOR + " times = " + sum +
" (should be " + FACTOR/4 + ")");
sum = 0;
for (int i = 0; i < FACTOR; ++i) sum += FIFTH;
System.out.println("1/5 summed " + FACTOR + " times = " + sum +
" (should be " + FACTOR/5 + ")");
sum = 0;
for (int i = 0; i < FACTOR; ++i) sum += SIXTH;
System.out.println("1/6 summed " + FACTOR + " times = " + sum +
" (should be " + FACTOR/6 + ")");
sum = 0;
for (int i = 0; i < FACTOR; ++i) sum += SEVENTH;
System.out.println("1/7 summed " + FACTOR + " times = " + sum +
" (should be " + FACTOR/7 + ")");
}
}
Output:
1/2 = 0.5
1/3 = 0.33333334
1/4 = 0.25
1/5 = 0.2
1/6 = 0.16666667
1/7 = 0.14285715
1/2 summed 840 times = 420.0 (should be 420)
1/3 summed 840 times = 279.99915 (should be 280)
1/4 summed 840 times = 210.0 (should be 210)
1/5 summed 840 times = 167.99858 (should be 168)
1/6 summed 840 times = 139.99957 (should be 140)
1/7 summed 840 times = 120.001114 (should be 120)
We appended an
f
to some of the numeric literals in the program to make them single-precision
float
numbers. That way,
1/2f
uses floating-point instead of integer arithmetic.
The first set of the program's output lines doesn't look too bad;
,
, and
appear to be fine. There's a small roundoff error for
C it's a bit odd that the rightmost digit got rounded
up,
but the error is quite small. There are similarly small roundoff errors for
and
.
The second set of the output lines shows what happens when we let even small errors accumulate. Although there was a rounding
up
error in
, we now see that it
accumulated
a rounding
down
error! Evidently, there was initially a tiny hidden error in
that accumulated a rounding down error.
also accumulated a rounding down error, and
accumulated a rounding up error.
and
apparently had no errors, but then, of course, they are exact powers of 2:
and
.
