Sonoma State University
 
Algorithm Analysis
Instructor: Henry M. Walker

Lecturer, Sonoma State University
Professor Emeritus of Computer Science and Mathematics, Grinnell College

Although CS 415 has been well developed for several years, last year the CS faculty made a significant, long-term curricular change regarding SSU's Upper Division GE Area B Requirement.

Consequences of Data Representation on Programming

Data representation has important consequences for designing and writing accurate and reliable programs. Some major issues involve:


For example, conditional statements may yield unexpected results, loops may not terminate as planned, and computations may accumulate numerical errors.

Note: Most of the floating-point examples in this reading involve float type, since float numbers store only 7-8 digits of accuracy, and the corresponding examples utilize relatively modest-sized numbers. The same issues arise for double numbers. However, since double numbers store 16 or so digits of accuracy, examples typically involve relatively long sequences of digits and thus may seem somewhat unintuitive.


Integer overflow

The Reading on the Binary Representation of Integers observed that most programming languages, including C, allocate a specified number of bits for the storage of an integer.

Whatever the number of bits allocated, integer storage has two essential properties:

Processing that attempts to compute and/or store an integer outside the limited range yields a condition called overflow. Hardware sometimes raises error conditions that can be investigated when overflow occurs — but sometimes processing continues with numbers that may have little meaning.

Examples:

Two examples may highlight some of the difficulties that can arise from integer overflow.

Example 1: Computing an Average

Consider the following program average-2-ints.c that is designed to read two integers and compute their average:

/* program to average two numbers */

#include <stdio.h>

int main ()
{
  int num1, num2;
  printf ("Enter two integers:  ");
  scanf ("%d%d", &num1, &num2);
  int avg = (num1 + num2) / 2;
  printf ("The average of the numbers is %d\n", avg);
  
  return 0;
}

This program was run twice on a machine utilizing 32-bit integers, yielding the following two results:

Enter two integers:  3 5
The average of the numbers is 4

Enter two integers:  2000000000 2000000000
The average of the numbers is -147483648

The first run produces the expected value: 4 is indeed the average of 3 and 5.

However, the second run reports that the average of two positive numbers is negative. The issue is that the sum 2000000000 + 2000000000 is 4000000000 — a number larger than the 2,147,483,647 maximum for signed integers on a 32-bit machine. In this case, 4000000000 is interpreted as a negative number, and this number divided by 2 is still negative.


Caution: When applications utilize integer data, programmers must anticipate the potential size of the integers. Integer overflow can have devastating consequences.

Example 2: Comair

In December 2004, Comair, an airline with a hub in Cincinnati, Ohio, utilized a database system that included a table for schedule changes. Data recorded included both flights and flight crews. In the evolution of the software system, each change in the table was numbered sequentially starting at 1 and using 16-bit signed integers. After the new year, the airline had planned to upgrade the system to a 32-bit numbering system, but this change in software was deferred until after the busy holiday season.

Unfortunately, the Midwestern United States experienced many storms throughout December 2004, leading to many schedule changes for both flights and crews for the approximately 1100 flights that Comair flew daily.

The underlying software system kept up with these changes until the number of changes reached 32768 — just beyond the limit for 16-bit signed integers. At that point, the computer system crashed, and Comair had to cancel all of its flights on December 25, 2004. News reports indicated that some 30,000 travelers in 118 cities were affected.


Floating-point overflow, underflow, and round-off error

Just as integer values may be too large or too small to be stored in the space allocated for an int, real numbers may be outside the range that can be stored by a float or double. Recall from the Reading on the Binary Representation of Floating-Point Numbers that real numbers are stored in a scientific notation, including a mantissa and an exponent.

For float numbers, exponents of two normally must be between -126 and 127, although some adjustments are allowed for positive numbers smaller than 2-126. When the scientific notation yields 2128 or larger, then the number cannot be stored in a float and floating-point overflow occurs. A similar statement applies to underflow.

A different type of issue arises for floating-point numbers, because only 23 bits (single precision) or 52 bit (double precision) are available to store a mantissa. Whenever a mantissa requires additional storage, the number cannot be stored exactly, and an error, called roundoff error, arises.

Perhaps surprisingly, inherent round-off error has a direct impact on the properties of addition, as illustrated by a short computation.

Rather than computing with binary numbers and storing 23 or 52 bits of accuracy, consider the analogous problem of storing decimal numbers with 4 digits of accuracy. When considering 1.000 + 0.0007, the result should be 1.0007. However, only 4 digits of accuracy can be stored, so a choice is needed.

Although either approach is possible, the addition algorithm should be consistent.

Now consider the sum 1.000 + 0.0007 + 0.0007. The specific computation depends upon where parentheses are inserted. Additional may arise, depending upon whether truncation or rounding is used to maintain 4-digits of storage.

addition with associativity and truncation/rounding

In summary:

Conclusion: Regardless of whether truncation or rounding is used, left associativity produces a different result than right associativity. That is, addition within the computer is not associative, so many familiar properties of numbers no longer hold!

Example 3: One tenth (decimal)

The Reading on the Binary Representation of Floating-Point Numbers determined that the decimal, floating-point number 1/10 has an infinite binary representation: 0.000110011001100110011... . Since this number would require infinite storage, the decimal number 1/10 cannot be stored exactly within either single-precision or double-precision storage. Thus, the actual storage of this number contains some error (small, but non-zero0).

Example 4: The Vancouver Stock Exchange Index

Although the roundoff error for an individual number may be small, such errors can accumulate when a computer repeats the same process thousands, millions, or billions of times. The following example is taken from The Limits of Computer by Henry M. Walker, Jones and Bartlett Publishers, pp. 100-101, and is used with permission of the copyright holder.

Between 1981 and 1983, the Vancouver Stock Exchange Index was computed with each trade of stock. The computation used four decimal places in the computation, but the result was then truncated to three places of accuracy. This computation was then repeated with each of the approximately 3000 trades that occurred daily. As a result of the truncation of the calculation to three decimal places, the index lost about 1 point a day, or about 20 points per month. Thus, to correct this consistent numerical error, the index was recomputed and corrected over the weekend of November 26, 1983. Specifically, when trading concluding on November 25, the Vancouver Stock Exchange Index was quoted as 524.811; the corrected figure used to open Monday morning was 1098.892. The correction for 22 months of compounded error caused the Index to increase by 574.081 over a weekend without any changes in stock prices. In this example, computations began with correct data, but subsequent processing introduced a significant error in the result.

Floating-point comparisons in if statements

Since real numbers are stored with the potential for round-off error, the actual number stored may be slightly smaller or slightly larger than the intended number. Thus, comparisons with real numbers should take small errors into account.

Example 5 illustrates some difficulties that can arise.

When testing real numbers, these examples illustrate that some adjustment may be needed to allow for round-off error. For example, a test

  if (result == one)

might be adjusted as follows:

  double allowed_error = 0.01;
  if ((result > one - allowed_error) && (result < one + allowed_error))

Example 5:

Consider the following program one-tenth.c that adds the decimal number 0.1 (one tenth) to itself:

/* Program to compute 0.1 + 0.1 + ... 0.1 (ten times) */

#include <stdio.h>

int main ()
{
  double one = 1.0;
  double ten = 10.0;
  double one_tenth = one / ten;
  double result = one_tenth + one_tenth + one_tenth
                + one_tenth + one_tenth + one_tenth
                + one_tenth + one_tenth + one_tenth
                + one_tenth;                
  printf ("result is %20.16lf\n", result);
  if (result == one)
    printf ("adding one_tenth ten times IS one\n");
  else
    printf ("adding one_tenth ten times IS NOT one\n");
    
  return 0;
}

This program adds one_tenth 10 times and produces the following output:

result is   0.9999999999999999
adding one_tenth ten times IS NOT one

Altogether, the storage 1.0/10.0 (decimal) includes a small round-off error. Adding the number 10 times yields a number quite close to 1.0, but not exactly 1.0.


Floating-point comparisons in loops

The same issues of round-off errors arise in loops as in comparisons. For loops, such errors may impact the number of times the body of a loop is executed. Example 6 illustrates one type of difficulty.

A somewhat more subtle difficulty arises if the loop in Example 6 is changed, with the intention that the values to be added are 0.0, 0.1, 0.2, ..., 0.9, 1.0. One candidate for the loop condition might be

   while (val <= end)

However, a careful reading of the output from Example 6 shows that the computer would produce the following output within the loop.

val =      0.000000000000000;  sum =      0.000000000000000
val =      0.100000001490116;  sum =      0.100000001490116
val =      0.200000002980232;  sum =      0.300000011920929
val =      0.300000011920929;  sum =      0.600000023841858
val =      0.400000005960464;  sum =      1.000000000000000
val =      0.500000000000000;  sum =      1.500000000000000
val =      0.600000023841858;  sum =      2.099999904632568
val =      0.700000047683716;  sum =      2.799999952316284
val =      0.800000071525574;  sum =      3.599999904632568
val =      0.900000095367432;  sum =      4.500000000000000

Although the intention was for the computer to execute the loop with val having the value 1.0, the next value of val would be 1.000000119209290, a value larger than 1.0, and the loop terminates one iteration early.

As illustrated above, one way to address this problem involves including an allowed-error term:

  double allowed_error = 0.01;
  while (val <= end + allowed_error)

However, this approach would be adding 0.1 to val several times, and we know the value 0.1 is not stored exactly. Thus, each iteration of the loop would be including additional round-off error to our results. To avoid this problem, one might use an integer variable to control the loop.

  int i;
  for (i = 0; i <= 10; i++)
     {
        val = i / 10.0;
        ...

In this approach, the int variable is stored exactly, and the loop guarantees that the loop will be executed the proper number of times. Further, the value of val is recomputed from exact numbers with each iteration. Storage of the resulting real number in val may involve round-off error. However, the round-off error in val will not accumulate from iteration to iteration, since the computation for val begins fresh each time.

Example 6: Loop until equality condition

Consider the task of start at 0.0 and adding 0.1, 0.2, 0.3, 0.4, ... until the number being added is 1.0. The program float-loop.c performs this task using float numbers throughout.

/* Program to study a loop with floats */
#include <stdio.h>

int main ()
{
  float inc = 1.0/10.0;
  float val = 0.0;
  float end = 1.0;
  float sum = 0.0;

  printf ("program to loop with floats from %22.15f to %22.15f increment %22.15f\n",
          val, end, inc);

  while (val != end)
    {
      /* add to sum and print */
      sum += val;
      printf ("val = %22.15f;  sum = %22.15f\n", val, sum);

      /* increment val for loop test and next iteration */
      val += inc;
    }

  printf ("loop terminated with val = %22.15f;  sum = %22.15f\n", val, sum);

}

When this program is run, the first part of the output begins

program to loop with floats from      0.000000000000000 to    1.000000000000000 increment      0.100000001490116
val =      0.000000000000000;  sum =      0.000000000000000
val =      0.100000001490116;  sum =      0.100000001490116
val =      0.200000002980232;  sum =      0.300000011920929
val =      0.300000011920929;  sum =      0.600000023841858
val =      0.400000005960464;  sum =      1.000000000000000
val =      0.500000000000000;  sum =      1.500000000000000
val =      0.600000023841858;  sum =      2.099999904632568
val =      0.700000047683716;  sum =      2.799999952316284
val =      0.800000071525574;  sum =      3.599999904632568
val =      0.900000095367432;  sum =      4.500000000000000
val =      1.000000119209290;  sum =      5.500000000000000
val =      1.100000143051147;  sum =      6.600000381469727
val =      1.200000166893005;  sum =      7.800000667572021
val =      1.300000190734863;  sum =      9.100000381469727
val =      1.400000214576721;  sum =     10.500000953674316
val =      1.500000238418579;  sum =     12.000000953674316
val =      1.600000262260437;  sum =     13.600001335144043
val =      1.700000286102295;  sum =     15.300001144409180
val =      1.800000309944153;  sum =     17.100002288818359
val =      1.900000333786011;  sum =     19.000001907348633

As this output indicates, the value being added never is exactly 1.00. The value being added does reach 1.000000119209290, a value very close to 1.00, but not exactly the same. As a result, this program continues in an apparent infinite loop.

Interestingly, although the numbers are slightly different, the same issue arises if all variables have type double rather than type float.


Error accumulation

Since the number of bits available to store real numbers is limited, the storage of any floating-point number is subject to some round-off error. Although the amount of error for an individual number may be quite small, these numbers may be used in loops that continue for thousands, millions, billions or more iterations. As a result, round off errors can accumulate.

Although such accumulated error may have little significance for some applications and computations, the level of inaccuracy may matter significantly in other circumstances. Although a full analysis of accuracy is well beyond the scope of this course, three notes seem appropriate.

  1. Whenever floating-point numbers are involved in computations, questions should be asked regarding accuracy. How accurate are the results, and how do you know?

  2. The initial discussion of round-off error (near Example 4) considered the sum 1.000 + 0.0007 + 0.0007.

    • In this example, using the large number (1.000) first had the effect of overwhelming the small numbers (0.0007). More generally, adding very small floating-point numbers to large ones may not be noticeable — the addition may be lost in bits that cannot be stored.
    • Adding small numbers first may allow the sums to accumulate (e.g., 0.0007 + 0.0007 = 0.0014), and the result may be large enough to impact the addition of large numbers.

    Overall, a program should add small numbers together before adding the sum to larger numbers; adding numbers in ascending order can be a helpful strategy.

  3. The mathematical subject, called numerical analysis, explores the accumulation of numerical error in substantial depth. When an application requires considerable accuracy, investigation of techniques from numerical analysis may be particularly important.



created 3 June 2016 by Henry M. Walker
expanded and edited 4 June 2016 by Henry M. Walker
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For more information, please contact Henry M. Walker at walker@cs.grinnell.edu.