LOOPS - Introduction to Java Programming - page 155

sum += currentValue;
currentValue += 0.01 ;

}

After this loop, sum is 50.50000000000003 . This loop adds the numbers from smallest to
biggest. What happens if you add numbers from biggest to smallest (i.e., 1.0 , 0.99 , 0.98 ,
. . . , 0.02 , 0.01 in this order) as follows:

double currentValue = 1.0 ;

for ( int count = 0 ; count < 100 ; count++) {

sum += currentValue;
currentValue -= 0.01 ;

}

After this loop, sum is 50.49999999999995 . Adding from biggest to smallest is less
accurate than adding from smallest to biggest. This phenomenon is an artifact of the finite-
precision arithmetic. Adding a very small number to a very big number can have no effect
if the result requires more precision than the variable can store. For example, the inaccu-
rate result of 100000000.0 + 0.000000001 is 100000000.0 . To obtain more accurate
results, carefully select the order of computation. Adding smaller numbers before bigger
numbers is one way to minimize errors.

avoiding numeric error

4.8 Case Studies
Loops are fundamental in programming. The ability to write loops is essential in
learning Java programming.

Key
Point

If you can write programs using loops, you know how to program! For this reason, this section
presents three additional examples of solving problems using loops.

4.8.1 Case Study: Finding the Greatest Common Divisor
The greatest common divisor (gcd) of the two integers 4 and 2 is 2 . The greatest common
divisor of the two integers 16 and 24 is 8 . How do you find the greatest common divisor? Let
the two input integers be n1 and n2 . You know that number 1 is a common divisor, but it may
not be the greatest common divisor. So, you can check whether k (for k , 3 , 4 , and so on)
is a common divisor for n1 and n2 , until k is greater than n1 or n2 . Store the common divisor
in a variable named gcd . Initially, gcd is 1 . Whenever a new common divisor is found, it
becomes the new gcd. When you have checked all the possible common divisors from 2 up to
n1 or n2 , the value in variable gcd is the greatest common divisor. The idea can be translated
into the following loop:

gcd

=

int gcd = 1 ; // Initial gcd is 1
int k = 2 ; // Possible gcd

while (k <= n1 && k <= n2) {
if (n1 % k == 0 && n2 % k == 0 )
gcd = k; // Update gcd
k++; // Next possible gcd

}

// After the loop, gcd is the greatest common divisor for n1 and n2

Listing 4.9 presents the program that prompts the user to enter two positive integers and
finds their greatest common divisor.