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The conditional expression ( n c min n : min) evaluates to n if n < min, and it evaluates to min otherwise. So assigning that value to min is equivalent to the first line of the for loop in the example. Similarly, the assignment max = ( n > max n : min ) is equivalent to the second line in the other for loop.

Fig. 17-25

(min=max=n; n>O;) { min = (n c min n : min); max = (n > max n : min); tin >> n;

Fig. 17-26

Implement the Euclidean Algorithm for finding the greatest common divisor of two given pos\ itive integers.

(f) 8.36 (a)

Fig. 17-27

The Euclidean Algorithm transforms a pair of positive integers (m, n) into a pair (d, 0) by repeatedly dividing the larger integer by the smaller integer and replacing the larger with the remainder. When the remainder is 0, the other integer in the pair will be the greatest common divisor of the original pair (and of all the intermediate pairs). For example, if m is 532 and y2 is 112, then the Euclidean Algorithm reduces the pair (532,112) to CWO) by

Fig. 17-28

(532,112) + (112,84) + (84,28) + (28,0)

So 28 is the greatest common divisor of 532 and 112. This result can be verified from the facts that 532 = 28.19 a n d 112 = 28.8. The reason that the Euclidean Algorithm works is that each pair in the sequence has the same set of divisors, which are precisely the factors of the greatest common divisor. In the example above, that common set of divisors is { 1, 2, 4, 7, 14, 28). The reason that this set of divisors is invariant under the reduction process is that when m = n-q + r, a number is a common divisor of m and y1 if and only if it is a common divisor of y2 and r.

Fig. 17-29

main0 // begin scope of main0 { int m, n, r; tout << "Enter two positive integers: I'; tin >> m >> n; if (m c n) { int temp = m; m = n; n = temp; } // make m >= n tout << "The g.c.d. of ' CC m CC ' and ' CC n CC' ' is "; while (n > 0) { r =m%n; m = n; n = r; tout CC m CC endl;

Relations among cubic units Find the volume V of (a) A cubic foot in cubic inches (b) A cubic yard in cubic feet (c) A liter (cubic decimeter) in cubic centimeters

Elr&ex .tpfTu .pssitive. integers: .$3a 2x2. .; : ,.

(a) V e3 for a cube. Since 1 ft So 1 ft3 1728 in3. (b) V e3 for a cube. Since 1 yd So 1 yd3 27ft3. (c) V e3 again. Since 1 dm3 So 1 liter 1000 cm3. 12 in, V 3 ft, V 10 cm3, V 123 33 103 1728 27 1000

Suppose we wanted to calculate the square root of the value represented by the expression Area / 3.141593, using the library function Sqr. To do so, we could write

rhe g;c &, of 532 t&d Ia12 i s 28:

Finding volumes or cubes Find the volume V of a cube, in cubic feet, if one edge is (a) 4 in, (b) 4 ft, (c) 4 yd.

, ,

,: . . : . 1. : -: . ..

To find the volume in cubic feet, we must express the side in feet. (a) V (b) V (c) V e3 and, since 4 in e3 43 64 ft3 12 ft, V 123 1728 ft3

.,-,.

1 3 ft,

Write and test a program that reads a given number of pairs (x, y) of real numbers and then computes the least-squares regression line for the data set. Use the equation y = mx + b where