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(These are the number of leaves in the left subtree, the number of leaves in the right subtree, the number of internal nodes in the left subtree, and the number of internal nodes in the right subtree, respectively) Then l = lL + lR = 2h 1 + 2h 1 = 2 2h 1 = 2h and m = mL + mR + 1 = (2h 1 1) + (2h 1 1) + 1 = 2 2h 1 1 = 2I 1 Therefore, by the (Second) Principle of Mathematical Induction, the formulas must be true for full binary trees of any height h 0 By simply adding the formulas for m and l, we obtain the first corollary Corollary 111 The full binary tree of height h has a total of n = 2h+1 1 nodes.

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f Aki; :

(1.15)

1 2 3 4 5 6 7 8 9 10

By solving the formula n = 2h+1 1 for h, we obtain this corollary: Corollary 112 The full binary tree with n nodes has height h = lg(n+1) 1 Note that the formula in Corollary 112 is correct even in the special case where n = 0: The empty binary tree has height h = lg(n+1) 1 = lg(0+1) 1 = lg(1) 1 = 0 1 = 1..

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For the particular solution, it is necessary to find the sequence yp(n) that satisfies the difference equation for the given x(n). In general, this requires some creativity and insight. However, for many of the typical inputs that we are interested in, the solution will have the same form as the input. Table 1-2 lists the particular solution for some commonly encountered inputs. For example, if x ( n ) = a n u ( n ) , the particular solution will be of the form

provided a is not a root of the characteristic equation. The constant C is found by substituting the solution into the difference equation. Note that for x ( n ) = CS(n) the particular solution is zero. Because x ( n ) = 0 for n > 0, the unit sample only affects the initial condition of y(n). Table 1-2 The Particular Solution to an LCCDE for Several Different Inputs

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public class TestStringEquality { static public void main(String[] args) { String x = new String("ABCDE"); String y = new String("ABCDE"); System.out.println("x = " + x); System.out.println("y = " + y); System.out.println("(x == y) = " + (x == y)); System.out.println("x.equals(y) = " + x.equals(y)); } }

CHAP. 11]

y(n) - 0.25y(n - 2) = x(n) for x ( n ) = u(n) assuming initial conditions of y(- 1) = 1 and y(-2) = 0. We begin by finding the particular solution. From Table 1-2 we see that for x(n) = u(n) y,(n) = C I Substituting this solution into the difference equation we find C I -0.25CI = 1 I 4 -1 - 0.25 3 To find the homogeneous solution, we set yh(n) = zn, which gives the characteristic polynomial In order for this to hold, we must have

The next corollary applies Corollary 11.1 together with the fact that the full binary tree of height h has more nodes than any other binary tree of height h. Corollary 11.3 In any binary tree of height h, h + 1 n 2h+1 1 and lg n h where n is the number of its nodes. n 1

The output is:

cl=--

z2 - 0.25 = 0

IDENTITY, EQUALITY, AND ISOMORPHISM In a computer, two objects are identically equal if they occupy the same space in memory, so they have the same address. In other words, there really only one object, but with two different names. That meaning of equality is reflected in Java by the equality operator. If x and y are references to objects, then the condition (x == y) will be true only if x and y both refer to the same object. But the normal concept of equality in mathematics is that the two things have the same value. This distinction is handled in Java by the equals() method, defined in the Object class (see 4) and thus inherited by every class. As defined there, it has the same effect as the equals operator: x.equals(y) means x == y. But that equals() method is intended to be overridden in subclasses so that it will return true not only when the two objects are identically equal, but also when they are separate objects that are the same in whatever sense the class designer intends. For example, x.equals(y) could be defined to be true for distinct instances x and y of Point class if they have the same coordinates. EXAMPLE 11.6 Testing Equality of Strings

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