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12.1 To insert a new record with key 16 into the tree shown in Figure 12.16, the initial search would lead to the first leaf node. Since that is a five-way search tree, that first leaf node has overflowed, causing it to be split into two leaf nodes and moving its middle key 19 up to its parent node, as shown in Figure 12.17. But now that parent node has overflowed. So it also gets split, moving its middle key up to its parent node, as shown in Figure 12.18. Two other ordering of the seven keys in Example 12.5 on page 235 that will produce the same BST: a. 44, 22, 33, 77, 55, 99, 88 b. 44, 22, 77, 33, 55, 99, 88 An array of objects could be sorted by inserting their objects into a binary search tree and then using an inorder traversal to copy them back into the array. The BST property guarantees that the inorder traversal will visit the elements in order. If an AVL tree is used, then each insertion runs in O(lgn) time, so building the tree with n elements will require O(n lgn) time. The subsequent inorder traversal also has O(n lgn) complexity, so the entire algorithm sorts the array in O(n lgn) time. All except a are binary search trees.

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(a) The system y(n) = r2(n)u(n) is rnernoryless (i.e.. the response of the system at time n depends only on the

public static String factor(int n) // precondition: n > 1 // returns the prime factorization of n; // example: factor(4840) returns "2*2*2*5*11*11"

input at time n and on no other values of the input). Therefore, this system is causal. when (b) The system y(n) = x(ln1) is an example of a noncausal system. This may be seen by looking at the outpu~ n < 0. In particular, note that y(- I) = s ( l ) . Therefore. the output of the system at time 11 = -1 depends on the value of the input at a future time.

private AVLTree rotateRight() { AVLTree x = this, y = left, z = y.left; x.left = z; y.left = x; int xb = x.balance; int yb = y.balance;

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2.24 Christian Goldbach (1690 1764) conjectured in 1742 that every even number greater than 2 is the sum of two primes. Write a program that tests the Goldbach conjecture for all even numbers less than 100. Use the Primes class from Problem 2.22. Your first 10 lines of output should look like this:

(c) For this system, in order to compute the output y(n) at time n all we need to know is the value of the input x ( n ) at times n, n - 3, and n - 10. Therefore. this system must be causal.

(d) This system is noncausal, which may be seen by evaluating v(n) for 11 < 0. For example,

CHAP. 12]

10 12 14 16 18 20 22

Because y(- I) depends on the value of .r(2), which occurs after time n = - I , this system is noncausal

if (yb > 1) { ++x.balance; y.balance = ( xb<0 yb+1 : xb+yb+2 ); } else if (yb > xb) { x.balance += yb-1; ++y.balance; } else { y.balance = xb+2; } return y; }

The cascade structure is derived by factoring the numerator and denominator polynomials of H(z):

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= = = = = = =

This factorization corresponds to a cascade of first-order filters, each having one pole and one zero. In general. the coefficients n and Bk will be complex. However, if h ( n ) is real. the roots of H(z) will occur in complex k

Theorem. Every subtree of a binary search tree is a binary search tree. Proof: Let T be a binary search tree, and let S be a subtree of T. Let x be any element in S, and let L and R be the left and right subtrees of x in S. Then, since S is a subtree of T, x is also an element of T, and L and R are the left and right subtrees of x in T. Therefore, y x z for every y L and every z R because T has the BST property. Thus, S also has the BST property.

conjugate pairs, and these complex conjugate factors may be combined to form second-order factors with real coefficients:

2.25 Pierre de Fermat (1601 1665) conjectured that there are infinitely many prime numbers of p 2 the form n = 2 +1 for some integer p. These numbers are called Fermat primes. For exam1 2 ple, 5 is a Fermat prime because it is a prime number and it has the form 2 +1. Write a program that finds all the Fermat primes that are in the range of the int type. Use the Primes class from Problem 2.22 and the Math.pow() method. Your first 5 lines of output should look like this:

A sixth-order 1IR filter implemented as a cascade of three second-order systems in direct form I1 is shown in Fig. 8-9.

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