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This is 1 less than the number of times n can be divided by 2, so the value returned, 1 + lg(n/2), is correct First, we prove the conjecture that the call print(left, right) will print n! distinct strings, all having the same prefix string left, where n = rightlength() If n = 1, the method prints left+right and returns; that is 1! (distinct) string Assume that when rightlength() = n 1, the call print(left,right) prints (n 1)! distinct strings all having the same left prefix string Then, when rightlength() = n, the for loop makes n calls of the form print(left+temp,ss), where temp is a distinct character and ss = ssubstring(1,n) Since the length of ssubstring(1,n) is n 1, each of those calls will print (n 1)! distinct strings all having the same left+temp prefix string..

num_list[sorted_index + 1] = newNum } end To repeat, the variable number_index keeps track of where the algorithm is in the unsorted set of numbers The algorithm starts with the second number (number_index = 2) Then the algorithm compares the number to the largest number that has been sorted so far, num_list[sorted_index] If the number is smaller than the previously sorted number, the algorithm moves the previously sorted number up one position in num_list, and checks the new number against the next largest number in the previously sorted elements of num_list Finally, the algorithm will encounter a previously sorted number which is smaller than the number being inserted, or it will find itself past the starting position of num_list At that point, the number can be inserted into the num_list The algorithm completes when all of the positions in the num_list have been sorted.

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The implementation symbol is used to show that a class implements an interface. EXAMPLE 1.7 Implementing the Comparable Interface

Change line 1 of Example 1.5 on page 6 to:

To analyze the running time of the insertion sort, we note first that the performance will be proportional to n, the number of elements to be sorted We also note that each element to be sorted must be compared one or many times with the elements already sorted In the best case, the elements will be sorted already, and each element will require only a single comparison, so the best-case performance of the insertion sort is (n) In the worst case, the elements to be sorted will be in reverse order, so that every element will require comparison with every element already sorted The second number will be compared with the first, the third with the second and first, the fourth with the third, second, and first, etc If there were four numbers in reverse order, the number of comparisons would be six.

public class Ratio implements Comparable { public int compareTo(Object object) { if (object==this) { return 0; } else if (!(object instanceof Ratio)) { throw new IllegalArgumentException("Ratio type expected"); } Ratio that = (Ratio)object; normalize(this); normalize(that); return (this.num*that.den - that.num*this.den); } private static void normalize(Ratio x) { if (x.num == 0) { // x == Ratio.ZERO x.den = 1; } else if (x.den < 0) { // change sign of num and den: x.num *= -1; x.den *= -1; } }

In general, the number of comparisons in the worst case for the insertion sort will be: n2/2 - n/2 The number of comparisons will grow as the square of the number of elements to be sorted The negative term of -n/2, and the division of n2 by the constant 2, mean that the rate of growth in number of comparisons will not be the full rate that n2 would imply However, for very large values of n, those terms other than.

and then add these two methods:

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

CHAP. 2]

The Comparable interface requires the compareTo() method, as specified at line 2. The purpose of the method is to indicate whether the implicit argument is less than, equal to, or greater than the explicit argument. The indicator is the sign of the returned integer: Negative indicates that this < object, zero indicates that this = object, and positive indicates that this > object. In the case of ratios a/b and c/d, we can tell whether a/b < c/d by cross-multiplying and checking whether ad < bc. Since that case should be indicated by returning a negative number, we can simply return ad bc. Indeed, the value of that expression will be negative, zero, or positive, when a/b < c/d, a/b = c/d, or a/b > c/d, respectively. However, that arithmetic trick works only if b > 0 and d > 0. To ensure that the denominators are not negative, we employ the static utility method defined at line 14. It ensures that the denominator of its argument is not negative.

Figure 1.7 illustrates how the implementation of an interface is represented in UML. The association symbol is a dashed arrow pointing to the interface. The diagram for the interface marks its name with the stereotype interface . Note that an interface rectangle has only two parts, while a class rectangle has three. This is because interfaces have no fields. Figure 1.7 also illustrates how to represent package membership. The Comparable interface is part of the java.lang package, as indicated by the tab above the interface symbol. POLYMORPHISM

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