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The adjacency matrix for the complete graph on n vertices is an n-by-n boolean matrix with false value at each entry on the diagonal and true value at every other entry. The incidence matrices Mn for the complete graphs on n vertices are as follows: Its has n rows and n(n 1)/2 columns (see Theorem 15.2 on page 286). If n = 2, it is the 2-by-1 matrix containing true in both entries. If n > 2, it is the matrix A concatenated horizontally with the matrix obtained from Mn 1 by placing one row of all false values on top of it. The four matrices are shown in Figure 15.53. a. b. c. d. e. f. a. b. c. d. e. f. The digraph is shown in Figure 15.54. Yes, this is a digraph: It has at least one one-way edge. No, the digraph is not strongly connected: There is no path from C to D. Yes, the digraph is weakly connected: Its embedded (undirected) graph is connected. No, the digraph is not acyclic: It contains the cycle AFEDA. Its adjacency matrix is shown in Figure 15.55 on page 315. The digraph G1 is shown in Figure 15.56 on page 315. Yes it is a digraph: Its adjacency matrix is not symmetric. No, this is not a simple digraph because it has a loop. Yes, this digraph is strongly connected. Yes, this digraph is weakly connected. No, the digraph is not acyclic: It contains the cycle ADB.

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CHAP. 12]

which we want to be less than or equal to 8::

cosh-'(I I d ) C O S ~ - I (QJR,)

25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79

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; }

- cosh-'( l / d ) COS~-~(I,X)

find Ha(s)H,(-s).

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public AVLTree attach(Object object) { if (root == null) { // tree is empty root = object; return this; } if (comp.compare(object,root) < 0) { // insert into left subtree if (left == null) { left = new AVLTree(object,comp); ++size; --balance; } else { int lb = left.balance; left = left.attach(object); if (left.balance != lb && left.balance != 0) { --balance; } } if (balance < -1) { if (left.balance > 0) { left = left.rotateLeft(); } return rotateRight(); } } else { // insert into right subtree if (right == null) { right = new AVLTree(object,comp); ++size; ++balance; } else { int rb = right.balance; right = right.attach(object); if (right.balance != rb && right.balance != 0) { ++balance; } } if (balance > 1) { if (right.balance < 0) { right = right.rotateRight(); } return rotateLeft(); } } return this; } private AVLTree rotateRight() // see Problem 12.5 on page 240

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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.

= 0. I ,

private AVLTree rotateLeft() { AVLTree x = this, y = right, z = y.left; x.right = z; y.left = x; int xb = x.balance, yb = y.balance; if (yb < 1) { --x.balance; y.balance = ( xb>0 yb-1 : xb+yb-2 );

where TN(x)is an Nth-order Chebyshev polynomial that is defined recursively as follows,

Theorem. Every subtree of an AVL tree is an AVL tree. Proof: The proof that every subtree of a binary search tree is a binary search tree is given in Problem 12.6. If a S is a subtree of an AVL tree T, then every node is S is also in T. Therefore, the balance number at every node in S is 1, 0, or 1.

with To(x) = 1 and Tl(x) = x. Therefore, to find the third-order Chebyshev polynomial, we first find T2(.u) as follows, T2(x)= 2xTl(x) - To(x) = 2x2 - I and then we have T3(x) = 2xT2(x) - Tl(x) = 4x3 - 2x - x = x(4x2 - 3) Thus, the denominator polynomial in I Hu(jR)12is

CHAP. 12]

we make the substitution R = s / j in 1H,(jC2)l2as follows:

HEAPS A heap is a complete binary tree whose elements have keys that satisfy the following heap property: the keys along any path from root to leaf are descending (i.e., nonincreasing). EXAMPLE 13.1 A Heap

398 9.28

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

Show that the bilinear transformation maps the jQ-axis in the s-plane onto the unit circle, lzl = 1, and maps the left-half s-plane. Re(s) < 0 insidc the unit circle, l z ( < 1.

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