数据结构与算法分析Java语言版课后答案


Weiss Java Solutions pages p. i (front) Windfall Software, PCA ZzTEX 12.2 Publisher Greg Tobin Executive Editor Michael Hirsch Editorial Assistant Lindsey Triebel Marketing Manager Michelle Brown Marketing Assistant Dana Lopreato Digital Asset Manager Marianne Groth Composition Windfall Software, using ZzTEX Access the latest information about Addison-Wesley titles from our World Wide Web site: http://www.aw- bc.com/computing Many of the designations used by manufacturers and sellers to distinguish their products are claimed as trademarks. Where those designations appear in this book, and Addison-Wesley was aware of a trademark claim, the designations have been printed in initial caps or all caps. Copyright © 2007 by Pearson Education, Inc. For information on obtaining permission for use of material in this work, please submit a written request to Pearson Education, Inc., Rights and Contract Department, 75 Arlington Street, Suite 300, Boston, MA 02116 or fax your request to (617) 848-7047. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or any other media embodiments now known or hereafter to become known, without the prior written permission of the publisher. Printed in the United States of America. 12345678910—PDF— 9 08 07 06 Weiss Java Solutions pages p. ii (front) Windfall Software, PCA ZzTEX 12.2 CONTENTS Preface v Chapter 1 Introduction 1 Chapter 2 Algorithm Analysis 3 Chapter 3 Lists, Stacks, and Queues 7 Chapter 4 Trees 29 Chapter 5 Hashing 47 Chapter 6 Priority Queues (Heaps) 53 Chapter 7 Sorting 61 Chapter 8 The Disjoint Set Class 67 Chapter 9 Graph Algorithms 71 Chapter 10 Algorithm Design Techniques 85 Chapter 11 Amortized Analysis 95 Chapter 12 Advanced Data Structures and Implementation 99 iii Weiss Java Solutions pages p. iii (front) Windfall Software, PCA ZzTEX 12.2 Weiss Java Solutions pages p. iv (front) Windfall Software, PCA ZzTEX 12.2 PREFACE Included in this manual are answers to many of the exercises in the textbook Data Structures and Algorithm Analysis in Java, second edition, published by Addison-Wesley. These answers reflect the state of the book in the first printing of the second edition. Specifically omitted are general programming questions and any question whose solution is pointed to by a reference at the end of the chapter. Solutions vary in degree of completeness; generally, minor details are left to the reader. For clarity, the few code segments that are present are meant to be pseudo-Java rather than completely perfect code. Errors can be reported to weiss@fiu.edu. v Weiss Java Solutions pages p. v (front) Windfall Software, PCA ZzTEX 12.2 Weiss Java Solutions pages p. vi (front) Windfall Software, PCA ZzTEX 12.2 CHAPTER 1 Introduction 1.4 The general way to do this is to write a procedure with heading void processFile( String fileName ); which opens fileName, does whatever processing is needed, and then closes it. If a line of the form #include SomeFile is detected, then the call processFile( SomeFile ); is made recursively. Self-referential includes can be detected by keeping a list of files for which a call to processFile has not yet terminated, and checking this list before making a new call to processFile. 1.5 public static int ones( int n ) { if(n<2) return n; returnn%2+ones(n/2); } 1.7 (a) The proof is by induction. The theorem is clearly true for 0 < X ≤ 1, since it is true for X = 1, and for X < 1, log X is negative. It is also easy to see that the theorem holds for 1 < X ≤ 2, since it is true for X = 2, and for X < 2, log X is at most 1. Suppose the theorem is true for p < X ≤ 2p (where p is a positive integer), and consider any 2p < Y ≤ 4p (p ≥ 1). Then log Y = 1+ log(Y/2)< 1 + Y/2 < Y/2 + Y/2 ≤ Y, where the first inequality follows by the inductive hypothesis. (b) Let 2X = A. Then AB = (2X)B = 2XB. Thus log AB = XB. Since X = log A, the theorem is proved. 1.8 (a) The sum is 4/3 and follows directly from the formula. (b) S = 1 4 + 2 42 + 3 43 + ....4S = 1 + 2 4 + 3 42 + .... Subtracting the first equation from the second gives 3S = 1 + 1 4 + 2 42 + .... By part (a), 3S = 4/3soS = 4/9. (c) S = 1 4 + 4 42 + 9 43 + ....4S = 1 + 4 4 + 9 42 + 16 43 + .... Subtracting the first equation from the second gives 3S = 1 + 3 4 + 5 42 + 7 43 + .... Rewriting, we get 3S = 2 ∞ i=0 i 4i + ∞ i=0 1 4i . Thus 3S = 2(4/9) + 4/3 = 20/9. Thus S = 20/27. (d) Let SN = ∞ i=0 iN 4i . Follow the same method as in parts (a)–(c) to obtain a formula for SN in terms of SN−1, SN−2,...,S0 and solve the recurrence. Solving the recurrence is very difficult. 1.9 N i=N/2 1 i = N i=1 1 i − N/2−1 i=1 1 i ≈ ln N − ln N/2 ≈ ln 2. 1.10 24 = 16 ≡ 1 (mod 5). (24)25 ≡ 125 (mod 5). Thus 2100 ≡ 1 (mod 5). 1 Weiss Java Solutions pages p. 1 (chap01) Windfall Software, PCA ZzTEX 12.2 2 Chapter 1 Introduction 1.11 (a) Proof is by induction. The statement is clearly true for N = 1 and N = 2. Assume true for N = 1,2,...,k. Then k+1 i=1 Fi = k i=1 Fi + Fk+1. By the induction hypothesis, the value of the sum on the right is Fk+2 − 2 + Fk+1 = Fk+3 − 2, where the latter equality follows from the definition of the Fibonacci numbers. This proves the claim for N = k + 1, and hence for all N. (b) As in the text, the proof is by induction. Observe that φ + 1= φ2. This implies that φ−1 + φ−2 = 1. For N = 1 and N = 2, the statement is true. Assume the claim is true for N = 1,2,...,k. Fk+1 = Fk + Fk−1 by the definition, and we can use the inductive hypothesis on the right-hand side, obtaining Fk+1 <φk + φk−1 <φ−1φk+1 + φ−2φk+1 Fk+1 <(φ−1 + φ−2)φk+1 <φk+1 and proving the theorem. (c) See any of the advanced math references at the end of the chapter. The derivation involves the use of generating functions. 1.12 (a) N i=1 (2i − 1) = 2 N i=1 i − N i=1 1 = N(N + 1) − N = N2. (b) The easiest way to prove this is by induction. The case N = 1 is trivial. Otherwise, N+1 i=1 i3 = (N + 1)3 + N i=1 i3 = (N + 1)3 + N2(N + 1)2 4 = (N + 1)2  N2 4 + (N + 1)  = (N + 1)2  N2 + 4N + 4 4  = (N + 1)2(N + 2)2 22 =  (N + 1)(N + 2) 2 2 =  N+1 i=1 i 2 Weiss Java Solutions pages p. 2 (chap01) Windfall Software, PCA ZzTEX 12.2 CHAPTER 2 Algorithm Analysis 2.1 2/N,37, √ N, N, N log log N, N log N, N log(N2), N log2 N, N1.5, N2, N2 log N, N3,2N/2,2N. N log N and N log(N2) grow at the same rate. 2.2 (a) True. (b) False. A counterexample is T1(N) = 2N, T2(N) = N, and f(N) = N. (c) False. A counterexample is T1(N) = N2, T2(N) = N, and f(N) = N2. (d) False. The same counterexample as in part (c) applies. 2.3 We claim that N log N is the slower growing function. Tosee this, suppose otherwise. Then, N/ √ log N would grow slower than log N. Taking logs of both sides, we find that, under this assumption, /  log N log N grows slower than log log N. But the first expression simplifies to   log N.If L = log N, then we are claiming that  √ L grows slower than log L, or equivalently, that 2L grows slower than log2 L. But we know that log2 L = o(L), so the original assumption is false, proving the claim. 2.4 Clearly, logk1 N = o(logk2 N) if k1 < k2, so we need to worry only about positive integers. The claim is clearly true for k = 0 and k = 1. Suppose it is true for k < i. Then, by L ’ Hˆopital’s rule, limN→∞ logi N N = limN→∞ ilogi−1 N N The second limit is zero by the inductive hypothesis, proving the claim. 2.5 Let f(N) = 1 when N is even, and N when N is odd. Likewise, let g(N) = 1 when N is odd, and N when N is even. Then the ratio f(N)/g(N) oscillates between 0 and inf. 2.6 (a) 22N (b) O(log log D) 2.7 For all these programs, the following analysis will agree with a simulation: (I) The running time is O(N). (II) The running time is O(N2). (III) The running time is O(N3). (IV) The running time is O(N2). (V) j can be as large as i2, which could be as large as N2. k can be as large as j, which is N2. The running time is thus proportional to N.N2. N2, which is O(N5). (VI) The if statement is executed at most N3 times, by previous arguments, but it is true only O(N2) times (because it is true exactly i times for each i). Thus the innermost loop is only executed O(N2) 3 Weiss Java Solutions pages p. 3 (chap02) Windfall Software, PCA ZzTEX 12.2 4 Chapter 2 Algorithm Analysis times. Each time through, it takes O(j2) = O(N2) time, for a total of O(N4). This is an example where multiplying loop sizes can occasionally give an overestimate. 2.8 (a) It should be clear that all algorithms generate only legal permutations. The first two algorithms have tests to guarantee no duplicates; the third algorithm works by shuffling an array that initially has no duplicates, so none can occur. It is also clear that the first two algorithms are completely random, and that each permutation is equally likely. The third algorithm, due to R. Floyd, is not as obvious; the correctness can be proved by induction. See J. Bentley, “Programming Pearls,” Communications of the ACM 30 (1987), 754–757. Note that if the second line of algorithm 3 is replaced with the statement swapReferences( a[i], a[ randInt( 0, n-1)]); then not all permutations are equally likely. To see this, notice that for N = 3, there are 27 equally likely ways of performing the three swaps, depending on the three random integers. Since there are only 6 permutations, and 6 does not evenly divide 27, each permutation cannot possibly be equally represented. (b) For the first algorithm, the time to decide if a random number to be placed in a[i] has not been used earlier is O(i). The expected number of random numbers that need to be tried is N/(N − i). This is obtained as follows: i of the N numbers would be duplicates. Thus the probability of success is (N − i)/N. Thus the expected number of independent trials is N/(N − i). The time bound is thus N−1 i=0 Ni N − i < N−1 i=0 N2 N − i < N2 N−1 i=0 1 N − i < N2 N j=1 1 j = O(N2 log N) The second algorithm saves a factor of i for each random number, and thus reduces the time bound to O(N log N) on average. The third algorithm is clearly linear. (c,d) The running times should agree with the preceding analysis if the machine has enough memory. If not, the third algorithm will not seem linear because of a drastic increase for large N. (e) The worst-case running time of algorithms I and II cannot be bounded because there is always a finite probability that the program will not terminate by some given time T. The algorithm does, however, terminate with probability 1. The worst-case running time of the third algorithm is linear— its running time does not depend on the sequence of random numbers. 2.9 Algorithm 1 at 10,000 is about 38 minutes and at 100,000 is about 26 days. Algorithms 1–4 at 1 million are approximately: 72 years, 4 hours, 0.7 seconds, and 0.03 seconds respectively. These calculations assume a machine with enough memory to hold the entire array. 2.10 (a) O(N) (b) O(N2) (c) The answer depends on how many digits past the decimal point are computed. Each digit costs O(N). 2.11 (a) Five times as long, or 2.5 ms. (b) Slightly more than five times as long. (c) 25 times as long, or 12.5 ms. (d) 125 times as long, or 62.5 ms. 2.12 (a) 12000 times as large a problem, or input size 1,200,000. (b) input size of approximately 425,000. Weiss Java Solutions pages p. 4 (chap02) Windfall Software, PCA ZzTEX 12.2 Solutions 5 (c) √ 12000 times as large a problem, or input size 10,954. (d) 120001/3 times as large a problem, or input size 2,289. 2.13 (a) O(N2). (b) O(N log N). 2.15 Use a variation of binary search to get an O(log N) solution (assuming the array is preread). 2.20 (a) Test to see if N is an odd number (or 2) and is not divisible by 3, 5, 7, . . . , √ N. (b) O( √ N), assuming that all divisions count for one unit of time. (c) B = O(log N). (d) O(2B/2). (e) If a 20-bit number can be tested in time T, then a 40-bit number would require about T2 time. (f) B is the better measure because it more accurately represents the size of the input. 2.21 The running time is proportional to N times the sum of the reciprocals of the primes less than N. This is O(N log log N). See Knuth, Volume 2. 2.22 Compute X2, X4, X8, X10, X20, X40, X60, and X62. 2.23 Maintain an array that can be filled in a for loop. The array will contain X, X2, X4,uptoX2log N . The binary representation of N (which can be obtained by testing even or odd and then dividing by 2, until all bits are examined) can be used to multiply the appropriate entries of the array. 2.24 For N = 0orN = 1, the number of multiplies is zero. If b(N) is the number of ones in the binary representation of N, then if N > 1, the number of multiplies used is log N+b(N) − 1 2.25 (a) A. (b) B. (c) The information given is not sufficient to determine an answer. We have only worst-case bounds. (d) Yes. 2.26 (a) Recursion is unnecessary if there are two or fewer elements. (b) One way to do this is to note that if the first N − 1elements have a majority, then the last element cannot change this. Otherwise, the last element could be a majority. Thus if N is odd, ignore the last element. Run the algorithm as before. If no majority element emerges, then return the Nth element as a candidate. (c) The running time is O(N), and satisfies T(N) = T(N/2) + O(N). (d) One copy of the original needs to be saved. After this, the B array, and indeed the recursion, can be avoided by placing each Bi in the A array. The difference is that the original recursive strategy implies that O(log N) arrays are used; this guarantees only two copies. 2.27 Start from the top-right corner. With a comparison, either a match is found, we go left, or we go down. Therefore, the number of comparisons is linear. 2.28 (a,c) Find the two largest numbers in the array. (b,d) Similar solutions; (b) is described here. The maximum difference is at least zero (i ≡ j), so that can be the initial value of the answer to beat. At any point in the algorithm, we have the current value j, and the current low point i.Ifa[j]− a[i] is larger than the current best, update the best difference. Weiss Java Solutions pages p. 5 (chap02) Windfall Software, PCA ZzTEX 12.2 6 Chapter 2 Algorithm Analysis If a[j] is less than a[i], reset the current low point to i. Start with i at index 0, j at index 0. j just scans the array, so the running time is O(N). 2.29 Otherwise, we could perform operations in parallel by cleverly encoding several integers into one. For instance, if A = 001, B = 101, C = 111, D = 100, we could add A and B at the same time as C and D by adding 00A00C + 00B00D. We could extend this to add N pairs of numbers at once in unit cost. 2.31 No. If low = 1, high = 2, then mid = 1, and the recursive call does not make progress. 2.33 No. As in Exercise 2.31, no progress is made. 2.34 See my textbook Data Structures and Problem Solving using Java for an explanation. Weiss Java Solutions pages p. 6 (chap02) Windfall Software, PCA ZzTEX 12.2 CHAPTER 3 Lists, Stacks, and Queues 3.1 public static void printLots(List L, List P) { Iterator iterL = L.iterator(); Iterator iterP = P.iterator(); AnyType itemL=null; Integer itemP=0; int start = 0; while ( iterL.hasNext() && iterP.hasNext() ) { itemP = iterP.next(); System.out.println("Looking for position " + itemP); while ( start < itemP && iterL.hasNext() ) { start++; itemL = iterL.next(); } System.out.println( itemL ); } } 3.2 (a) For singly linked lists: // beforeP is the cell before the two adjacent cells that are to be swapped. // Error checks are omitted for clarity. public static void swapWithNext( Node beforep ) { Node p, afterp; p = beforep.next; afterp = p.next; // Both p and afterp assumed not null. p.next = afterp.next; beforep.next = afterp; 7 Weiss Java Solutions pages p. 7 (chap03) Windfall Software, PCA ZzTEX 12.2 8 Chapter 3 Lists, Stacks, and Queues afterp.next = p; } (b) For doubly linked lists: // p and afterp are cells to be switched. Error checks as before. public static void swapWithNext( Node p ) { Node beforep, afterp; beforep = p.prev; afterp = p.next; p.next = afterp.next; beforep.next = afterp; afterp.next = p; p.next.prev = p; p.prev = afterp; afterp.prev = beforep; } 3.3 public boolean contains( AnyType x ) { Node p = beginMarker.next; while( p != endMarker && !(p.data.equals(x) )) { p = p.next; } return (p != endMarker); } 3.4 public static > void intersection(List L1, List L2, List Intersect) { ListIterator iterL1 = L1.listIterator(); ListIterator iterL2 = L2.listIterator(); AnyType itemL1=null, itemL2=null; // get first item in each list if ( iterL1.hasNext() && iterL2.hasNext() ) { itemL1 = iterL1.next(); itemL2 = iterL2.next(); } Weiss Java Solutions pages p. 8 (chap03) Windfall Software, PCA ZzTEX 12.2 Solutions 9 while ( itemL1 != null && itemL2 != null ) { int compareResult = itemL1.compareTo(itemL2); if ( compareResult == 0 ) { Intersect.add(itemL1); itemL1 = iterL1.hasNext()?iterL1.next():null; itemL2 = iterL2.hasNext()?iterL2.next():null; } else if ( compareResult<0) { itemL1 = iterL1.hasNext()?iterL1.next():null; } else { itemL2 = iterL2.hasNext()?iterL2.next():null; } } } 3.5 public static > void union(List L1, List L2, List Result) { ListIterator iterL1 = L1.listIterator(); ListIterator iterL2 = L2.listIterator(); AnyType itemL1=null, itemL2=null; // get first item in each list if ( iterL1.hasNext() && iterL2.hasNext() ) { itemL1 = iterL1.next(); itemL2 = iterL2.next(); } while ( itemL1 != null && itemL2 != null ) { int compareResult = itemL1.compareTo(itemL2); if ( compareResult == 0 ) { Result.add(itemL1); itemL1 = iterL1.hasNext()?iterL1.next():null; itemL2 = iterL2.hasNext()?iterL2.next():null; } Weiss Java Solutions pages p. 9 (chap03) Windfall Software, PCA ZzTEX 12.2 10 Chapter 3 Lists, Stacks, and Queues else if ( compareResult<0) { Result.add(itemL1); itemL1 = iterL1.hasNext()?iterL1.next():null; } else { Result.add(itemL2); itemL2 = iterL2.hasNext()?iterL2.next():null; } } } 3.6 A basic algorithm is to iterate through the list, removing every Mth item. This can be improved by two observations. The first is that an item M distance away is the same as an item that is only M mod N away. This is useful when M is large enough to cause iteration through the list multiple times. The second is that an item M distance away in the forward direction is the same as an item (M − N) away in the backwards direction. This improvement is useful when M is more than N/2 (half the list). The solution shown below uses these two observations. Note that the list size changes as items are removed. The worst case running time is O(N min(M, N)), though with the improvements given the algorithm might be significantly faster. If M = 1, the algorithm is linear. public static void pass(int m, int n) { int i, j, mPrime, numLeft; ArrayList L = new ArrayList(); for (i=1; i<=n; i++) L.add(i); ListIterator iter = L.listIterator(); Integer item=0; numLeft = n; mPrime=m%n; for (i=0; i items ) { Iterator iter = items.iterator(); while ( iter.hasNext() ) { add(iter.next()); } } This runs in O(N) time, where N is the size of the items collection. Weiss Java Solutions pages p. 11 (chap03) Windfall Software, PCA ZzTEX 12.2 12 Chapter 3 Lists, Stacks, and Queues 3.10 public void removeAll( Iterable items ) { AnyType item, element; Iterator iterItems = items.iterator(); while ( iterItems.hasNext()) { item = iterItems.next(); Iterator iterList = iterator(); while ( iterList.hasNext()) { element = iterList.next(); if ( element.equals(item) ) iterList.remove(); } } } This runs in O(MN), where M is the size of the items, and N is the size of the list. 3.11 import java.util.*; public class SingleList { SingleList() { init(); } boolean add( Object x ) { if (contains(x)) return false; else { Node p = new Node(x); p.next = head.next; head.next = p; theSize++; } return true; } boolean remove(Object x) { if (!contains(x)) return false; else { Node p = head.next; Weiss Java Solutions pages p. 12 (chap03) Windfall Software, PCA ZzTEX 12.2 Solutions 13 Node trailer = head; while (!p.data.equals(x)) { trailer = p; p = p.next; } trailer.next = p.next; theSize--; } return true; } int size() { return theSize; } void print() { Node p = head.next; while (p != null) { System.out.print(p.data+""); p = p.next; } System.out.println(); } boolean contains( Object x ) { Node p = head.next; while (p != null) { if (x.equals(p.data)) return true; else p = p.next; } return false; } void init() { theSize = 0; head = new Node(); head.next = null; } private Node head; private int theSize; Weiss Java Solutions pages p. 13 (chap03) Windfall Software, PCA ZzTEX 12.2 14 Chapter 3 Lists, Stacks, and Queues private class Node { Node() { this(null, null); } Node(Object d) { this(d, null); } Node(Object d, Node n) { data = d; next = n; } Object data; Node next; } } 3.12 import java.util.*; public class SingleListSorted { SingleListSorted() { init(); } boolean add( Comparable x ) { if (contains(x)) return false; else { Node p = head.next; Node trailer = head; while (p != null && p.data.compareTo(x) < 0) { trailer = p; p = p.next; } trailer.next = new Node(x); trailer.next.next = p; theSize++; } return true; } Weiss Java Solutions pages p. 14 (chap03) Windfall Software, PCA ZzTEX 12.2 Solutions 15 boolean remove(Comparable x) { if (!contains(x)) return false; else { Node p = head.next; Node trailer = head; while (!p.data.equals(x)) { trailer = p; p = p.next; } trailer.next = p.next; theSize--; } return true; } int size() { return theSize; } void print() { Node p = head.next; while (p != null) { System.out.print(p.data+""); p = p.next; } System.out.println(); } boolean contains( Comparable x ) { Node p = head.next; while (p != null && p.data.compareTo(x) <= 0) { if (x.equals(p.data)) return true; else p = p.next; } return false; } void init() { Weiss Java Solutions pages p. 15 (chap03) Windfall Software, PCA ZzTEX 12.2 16 Chapter 3 Lists, Stacks, and Queues theSize = 0; head = new Node(); head.next = null; } private Node head; private int theSize; private class Node { Node() { this(null, null); } Node(Comparable d) { this(d, null); } Node(Comparable d, Node n) { data = d; next = n; } Comparable data; Node next; } } 3.13 public java.util.Iterator iterator() { return new ArrayListIterator( ); } public java.util.ListIterator listIterator() { return new ArrayListIterator( ); } private class ArrayListIterator implements java.util.ListIterator { private int current = 0; boolean backwards = false; public boolean hasNext() { return current < size(); } public AnyType next() { if ( !hasNext() ) throw new java.util.NoSuchElementException(); backwards = false; return theItems[ current++ ]; } Weiss Java Solutions pages p. 16 (chap03) Windfall Software, PCA ZzTEX 12.2 Solutions 17 public boolean hasPrevious() { return current > 0; } public AnyType previous() { if ( !hasPrevious() ) throw new java.util.NoSuchElementException(); backwards = true; return theItems[ --current ]; } public void add( AnyType x ) { MyArrayList.this.add( current++, x ); } public void remove() { if (backwards) MyArrayList.this.remove( current-- ); else MyArrayList.this.remove( --current ); } public void set( AnyType newVal ) { MyArrayList.this.set( current, newVal ); } public int nextIndex() { throw new java.lang.UnsupportedOperationException(); } public int previousIndex() { throw new java.lang.UnsupportedOperationException(); } } 3.14 public java.util.Iterator iterator() { return new LinkedListIterator( ); } public java.util.ListIterator listIterator() { return new LinkedListIterator( ); } private class LinkedListIterator implements java.util.ListIterator { Weiss Java Solutions pages p. 17 (chap03) Windfall Software, PCA ZzTEX 12.2 18 Chapter 3 Lists, Stacks, and Queues private Node current = beginMarker.next; private int expectedModCount = modCount; private boolean okToRemove = false; public boolean hasNext() { return current != endMarker; } public AnyType next() { if( modCount != expectedModCount ) throw new java.util.ConcurrentModificationException( ); if( !hasNext( ) ) throw new java.util.NoSuchElementException( ); AnyType nextItem = current.data; current = current.next; okToRemove = true; return nextItem; } public boolean hasPrevious() { return current.prev != beginMarker; } public AnyType previous() { if ( modCount != expectedModCount ) throw new java.util.ConcurrentModificationException(); if ( !hasPrevious() ) throw new java.util.NoSuchElementException(); current = current.prev; AnyType previousItem = current.data; okToRemove = true; return previousItem; } public void add( AnyType x ) { if ( modCount != expectedModCount ) throw new java.util.ConcurrentModificationException(); MyLinkedList.this.addBefore( current.next, x ); } public void remove() { Weiss Java Solutions pages p. 18 (chap03) Windfall Software, PCA ZzTEX 12.2 Solutions 19 if( modCount != expectedModCount ) throw new java.util.ConcurrentModificationException( ); if( !okToRemove ) throw new IllegalStateException( ); MyLinkedList.this.remove( current.prev ); okToRemove = false; } public void set( AnyType newVal ) { if ( modCount != expectedModCount ) throw new java.util.ConcurrentModificationException(); MyLinkedList.this.set( current.next, newVal ); } public int nextIndex() { throw new java.lang.UnsupportedOperationException(); } public int previousIndex() { throw new java.lang.UnsupportedOperationException(); } } // and change MyLinkedList as follows: public AnyType set( int idx, AnyType newVal ) { return set( getNode( idx ), newVal ); } private AnyType set( Node p, AnyType newVal ) { AnyType oldVal = p.data; p.data = newVal; return oldVal; } 3.16 Iterator reverseIterator() { return new ArrayListReverseIterator( ); } private class ArrayListReverseIterator implements java.util.Iterator Weiss Java Solutions pages p. 19 (chap03) Windfall Software, PCA ZzTEX 12.2 20 Chapter 3 Lists, Stacks, and Queues { private int current = size()-1; public boolean hasNext() { return current >= 0; } public AnyType next() { if ( !hasNext() ) throw new java.util.NoSuchElementException(); return theItems[ current-- ]; } public void remove() { MyArrayList.this.remove( --current ); } } 3.18 public void addFirst( AnyType x ) { addBefore( beginMarker.next, x ); } public void addLast( AnyType x ) { addBefore( endMarker, x ); } public void removeFirst( ) { remove( beginMarker.next ); } public void removeLast( ) { remove( endMarker.prev ); } public AnyType getFirst( ) { return get( 0 ); } public AnyType getLast( ) { return get( size()-1);} 3.19 Without head or tail nodes the operations of inserting and deleting from the end becomes an O(N) operation where the N is the number of elements in the list. The algorithm must walk down the list before inserting at the end. Without the head node insert needs a special case to account for when something is inserted before the first node. 3.20 (a) The advantages are that it is simpler to code, and there is a possible saving if deleted keys are subsequently reinserted (in the same place). The disadvantage is that it uses more space, because each cell needs an extra bit (which is typically a byte), and unused cells are not freed. 3.22 The following function evaluates a postfix expression, using +, −, ∗, /, and ^ ending in =. It requires spaces between all operators and =. Weiss Java Solutions pages p. 20 (chap03) Windfall Software, PCA ZzTEX 12.2 Solutions 21 static double evalPostFix() { Stack s = new Stack(); String token; Double a, b, result=0.0; boolean isNumber; Scanner sc = new Scanner(System.in); token = sc.next(); while (token.charAt(0) != ’=’) { try { isNumber = true; result = Double.parseDouble(token); } catch (Exception e) { isNumber = false; } if (isNumber) s.push(result); else { switch (token.charAt(0)) { case ’+’: a = s.pop(); b = s.pop(); s.push(a+b); break; case ’-’: a = s.pop(); b = s.pop(); s.push(a-b); break; case ’*’: a = s.pop(); b = s.pop(); s.push(a*b); break; case ’/’: a = s.pop(); b = s.pop(); s.push(a/b); break; case ’^’: a = s.pop(); b = s.pop(); s.push(Math.exp(a*Math.log(b))); break; } } token = sc.next(); } return s.peek(); } Weiss Java Solutions pages p. 21 (chap03) Windfall Software, PCA ZzTEX 12.2 22 Chapter 3 Lists, Stacks, and Queues 3.23 (a, b) This function will read in from standard input an infix expression of single lower case characters and the operators +, −, /, ∗, ^, and ( ), and outputs a postfix expression. static void InFixToPostFix() { Stack s = new Stack(); String expression; Character token; int i=0; Scanner sc = new Scanner(System.in); expression = sc.next(); while ((token = expression.charAt(i++)) != ’=’) { if (token >= ’a’ && token <= ’z’) System.out.print(token+""); else { switch (token) { case ’)’: while ( !s.empty() && s.peek() != ’(’ ) { System.out.print(s.pop() + " "); } s.pop(); break; case ’(’: s.push(token); break; case ’^’: while ( !s.empty() && !(s.peek() == ’^’ || s.peek() == ’(’) ) { System.out.print(s.pop()); } s.push(token); break; case ’*’: case ’/’: while ( !s.empty() && s.peek() != ’+’ && s.peek() != ’-’ && s.peek() != ’(’ ) { System.out.print(s.pop()); } s.push(token); break; case ’+’: case ’-’: while ( !s.empty() && s.peek() != ’(’ ) { System.out.print(s.pop() + " "); } s.push(token); break; } } } while (!s.empty()) Weiss Java Solutions pages p. 22 (chap03) Windfall Software, PCA ZzTEX 12.2 Solutions 23 { System.out.print(s.pop()); } System.out.println(); } 3.24 Two stacks can be implemented in an array by having one grow from the low end of the array up, and the other from the high end down. 3.25 (a) Let E be our extended stack. We will implement E with two stacks. One stack, which we’ll call S, is used to keep track of the push and pop operations, and the other M, keeps track of the minimum. To implement E.push (x), we perform S.push (x).Ifx is smaller than or equal to the top element in stack M, then we also perform M.push (x). To implement E.pop( ) we perform S.pop().Ifx is equal to the top element in stack M, then we also M.pop().E.findMin( ) is performed by examining the top of M. All these operations are clearly O(1). (b) This result follows from a theorem in Chapter 7 that shows that sorting must take (N log N) time. O(N) operations in the repertoire, including deleteMin, would be sufficient to sort. 3.26 Three stacks can be implemented by having one grow from the bottom up, another from the top down and a third somewhere in the middle growing in some (arbitrary) direction. If the third stack collides with either of the other two, it needs to be moved. A reasonable strategy is to move it so that its center (at the time of the move) is halfway between the tops of the other two stacks. 3.27 Stack space will not run out because only 49 calls will be stacked. However, the running time is exponential, as shown in Chapter 2, and thus the routine will not terminate in a reasonable amount of time. 3.28 Since the LinkedList class supports adding and removing from the first and end of the list, the Deque class shown below simply wraps these operations. public class Deque { Deque() { L = new LinkedList(); } void push(AnyType x) { L.addFirst(x); } AnyType pop() { return L.removeFirst(); } void inject(AnyType x) { L.addLast(x); } AnyType eject() { return L.removeLast(); } LinkedList L; } 3.29 Reversal of a linked list can be done recursively using a stack, but this requires O(N) extra space. The following solution is similar to strategies employed in garbage collection algorithms. At the top of the while loop, the list from the start to previousPos is already reversed, whereas the rest of the list, Weiss Java Solutions pages p. 23 (chap03) Windfall Software, PCA ZzTEX 12.2 24 Chapter 3 Lists, Stacks, and Queues from currentPos to the end is normal. This algorithm uses only constant extra space. This solution reverses the list which can then be printed in reverse order. void reverseList() { Node currentPos, nextPos, previousPos; previousPos = null; currentPos = head.next; // skip header node nextPos = currentPos.next; while( nextPos != null) { currentPos.next = previousPos; previousPos = currentPos; currentPos = nextPos; nextPos = nextPos.next; } currentPos.next = previousPos; head.next = currentPos; } 3.30 (c) This follows well-known statistical theorems. See Sleator and Tarjan’s paper in Chapter 11 for references. 3.31 public class SingleStack { SingleStack() { head = null; } void push(AnyType x) { Node p = new Node(x, head); head = p; } AnyType top() { return head.data; } void pop() { head = head.next; } Weiss Java Solutions pages p. 24 (chap03) Windfall Software, PCA ZzTEX 12.2 Solutions 25 private class Node { Node() { this(null, null); } Node(AnyType x) { this(x, null); } Node(AnyType x, Node p) { data = x; next = p; } AnyType data; Node next; } private Node head; } 3.32 public class SingleQueue { SingleQueue() { front = null; rear = null; } void enqueue(AnyType x) { Node p = new Node(x, null); if (rear != null) rear = rear.next = p; else front = rear = p; } AnyType dequeue() { AnyType temp = front.data; Node p = front; if (front.next == null) // only 1 node front = rear = null; else front = front.next; return temp; } private class Node Weiss Java Solutions pages p. 25 (chap03) Windfall Software, PCA ZzTEX 12.2 26 Chapter 3 Lists, Stacks, and Queues { Node() { this(null, null); } Node(AnyType x) { this(x, null); } Node(AnyType x, Node p) { data = x; next = p; } AnyType data; Node next; } private Node front, rear; } 3.33 import java.util.*; public class SingleQueueArray { SingleQueueArray() { this(101); } // note: actually holds one less than given size SingleQueueArray(int s) { maxSize = s; front = 0; rear = 0; elements = new ArrayList(maxSize); } void enqueue(AnyType x) { if ( !full() ) { if (elements.size() < maxSize) // add elements until size is reached elements.add(x); else elements.set(rear, x); // after size is reached, use set rear = (rear + 1) % maxSize; } } Weiss Java Solutions pages p. 26 (chap03) Windfall Software, PCA ZzTEX 12.2 Solutions 27 AnyType dequeue() { AnyType temp=null; if ( !empty() ) { temp = elements.get(front); front = (front+1) % maxSize; } return temp; } boolean empty() { return front == rear; } boolean full() { return (rear + 1) % maxSize == front; } private int front, rear; private int maxSize; private ArrayList elements; } 3.34 (b) Use two iterators p and q, both initially at the start of the list. Advance p one step at a time, and q two steps at a time. If q reaches the end there is no cycle; otherwise, p and q will eventually catch up to each other in the middle of the cycle. 3.35 (a) Does not work in constant time for insertions at the end. (b) Because of the circularity, we can access the front item in constant time, so this works. 3.36 Copy the value of the item in the next node (that is, the node that follows the referenced node) into the current node (that is, the node being referenced). Then do a deletion of the next node. 3.37 (a) Add a copy of the node in position p after position p; then change the value stored in position p to x. (b) Set p.data = p.next.data and set p.next = p.next.next. Note that the tail node guarantees that there is always a next node. Weiss Java Solutions pages p. 27 (chap03) Windfall Software, PCA ZzTEX 12.2 Weiss Java Solutions pages p. 28 (chap04) Windfall Software, PCA ZzTEX 12.2 CHAPTER 4 Trees 4.1 (a) A. (b) G, H, I, L, M, and K. 4.2 For node B: (a) A. (b) D and E. (c) C. (d) 1. (e) 3. 4.3 4. 4.4 There are N nodes. Each node has two links, so there are 2N links. Each node but the root has one incoming link from its parent, which accounts for N − 1 links. The rest are null. 4.5 Proof is by induction. The theorem is trivially true for h = 0. Assume true for h = 1,2,...,k.Atree of height k + 1 can have two subtrees of height at most k. These can have at most 2k+1 − 1 nodes each by the induction hypothesis. These 2k+2 − 2 nodes plus the root prove the theorem for height k + 1 and hence for all heights. 4.6 This can be shown by induction. Alternatively, let N = number of nodes, F = number of full nodes, L = number of leaves, and H = number of half nodes (nodes with one child). Clearly, N = F + H + L. Further, 2F + H = N − 1 (see Exercise 4.4). Subtracting yields L − F = 1. 4.7 This can be shown by induction. In a tree with no nodes, the sum is zero, and in a one-node tree, the root is a leaf at depth zero, so the claim is true. Suppose the theorem is true for all trees with at most k nodes. Consider any tree with k + 1 nodes. Such a tree consists of an i node left subtree and a k − i node right subtree. By the inductive hypothesis, the sum for the left subtree leaves is at most one with respect to the left tree root. Because all leaves are one deeper with respect to the original tree than with respect to the subtree, the sum is at most 1/2 with respect to the root. Similar logic implies that the sum for leaves in the right subtree is at most 1/2, proving the theorem. The equality is true if and only if there are no nodes with one child. If there is a node with one child, the equality cannot be true because adding the second child would increase the sum to higher than 1. If no nodes have one child, then we can find and remove two sibling leaves, creating a new tree. It is easy to see that this new tree has the same sum as the old. Applying this step repeatedly, we arrive at a single node, whose sum is 1. Thus the original tree had sum 1. 4.8 (a) -**ab+cde. (b) ((a*b)*(c+d))-e. (c) ab*cd+*e-. 29 Weiss Java Solutions pages p. 29 (chap04) Windfall Software, PCA ZzTEX 12.2 30 Chapter 4 Trees 4.9 4.10 import java.io.*; public class FileList { public void list(File f) { list(f, 0); } public void list(File f, int depth) { printName(f, depth); if (f.isDirectory()) { File [] files = f.listFiles(); for (File i : files) list(i, depth+1); } } void printName(File f, int depth) { String name = f.getName(); for (int i=0; i> { private static class BinaryNode { BinaryNode( AnyType theElement ) { this( theElement, null, null, null ); } BinaryNode( AnyType theElement, BinaryNode lt, BinaryNode rt, BinaryNode pt ) { element = theElement; left = lt; right = rt; parent = pt; } AnyType element; BinaryNode left; BinaryNode right; BinaryNode parent; } public java.util.Iterator iterator() { return new MyTreeSetIterator( ); } private class MyTreeSetIterator implements java.util.Iterator { private BinaryNode current = findMin(root); private BinaryNode previous; private int expectedModCount = modCount; private boolean okToRemove = false; private boolean atEnd = false; public boolean hasNext() { return !atEnd; } public AnyType next() { if( modCount != expectedModCount ) throw new java.util.ConcurrentModificationException( ); if( !hasNext( ) ) throw new java.util.NoSuchElementException( ); AnyType nextItem = current.element; Weiss Java Solutions pages p. 31 (chap04) Windfall Software, PCA ZzTEX 12.2 32 Chapter 4 Trees previous = current; // if there is a right child, next node is min in right subtree if (current.right != null) { current = findMin(current.right); } else { // else, find ancestor that it is left of BinaryNode child = current; current = current.parent; while ( current != null && current.left != child) { child = current; current = current.parent; } if (current == null) atEnd = true; } okToRemove = true; return nextItem; } public void remove() { if( modCount != expectedModCount ) throw new java.util.ConcurrentModificationException( ); if( !okToRemove ) throw new IllegalStateException( ); MyTreeSet.this.remove( previous.element ); okToRemove = false; } } public MyTreeSet() { root = null; } public void makeEmpty() { modCount++; root = null; } Weiss Java Solutions pages p. 32 (chap04) Windfall Software, PCA ZzTEX 12.2 Solutions 33 public boolean isEmpty() { return root == null; } public boolean contains( AnyType x ) { return contains( x, root ); } public AnyType findMin() throws UnderflowException { if ( isEmpty() ) throw new UnderflowException(); else return findMin( root ).element; } public AnyType findMax() throws UnderflowException { if ( isEmpty() ) throw new UnderflowException(); else return findMax( root ).element; } public void insert( AnyType x ) { root = insert( x, root, null ); } public void remove( AnyType x ) { root = remove( x, root ); } public void printTree() { if ( isEmpty() ) System.out.println( "Empty tree" ); else printTree( root ); } private void printTree( BinaryNode t ) { if(t!=null ) { printTree( t.left ); System.out.println( t.element ); printTree( t.right ); } } private boolean contains( AnyType x, BinaryNode t ) { Weiss Java Solutions pages p. 33 (chap04) Windfall Software, PCA ZzTEX 12.2 34 Chapter 4 Trees if(t==null ) return false; int compareResult = x.compareTo( t.element ); if ( compareResult < 0) return contains( x, t.left ); else if ( compareResult > 0) return contains( x, t.right ); else return true; // match } private BinaryNode findMin( BinaryNode t ) { if(t==null ) return null; else if ( t.left == null ) return t; return findMin( t.left ); } private BinaryNode findMax( BinaryNode t ) { if(t==null ) return null; else if ( t.right == null ) return t; return findMax( t.right ); } private BinaryNode insert( AnyType x, BinaryNode t, BinaryNode pt ) { if(t==null ) { modCount++; return new BinaryNode( x, null, null, pt); } int compareResult = x.compareTo( t.element ); if ( compareResult < 0) t.left = insert( x, t.left, t ); else if ( compareResult > 0) t.right = insert( x, t.right, t ); else Weiss Java Solutions pages p. 34 (chap04) Windfall Software, PCA ZzTEX 12.2 Solutions 35 ; // duplicate return t; } private BinaryNode remove( AnyType x, BinaryNode t ) { if(t==null ) return t; // not found int compareResult = x.compareTo( t.element ); if ( compareResult < 0) t.left = remove( x, t.left ); else if ( compareResult > 0) t.right = remove( x, t.right ); else if ( t.left != null && t.right != null ) // two children { t.element = findMin( t.right ).element; t.right = remove( t.element, t.right ); } else { modCount++; BinaryNode oneChild; oneChild = ( t.left != null ) ? t.left : t.right; oneChild.parent = t.parent; // update parent link t = oneChild; } return t; } private BinaryNode root; int modCount = 0; } 4.13 This solution does not use header and tail nodes. import java.util.*; class UnderflowException extends Exception { }; public class MyTreeSet2> { private static class BinaryNode { BinaryNode( AnyType theElement ) Weiss Java Solutions pages p. 35 (chap04) Windfall Software, PCA ZzTEX 12.2 36 Chapter 4 Trees { this( theElement, null, null, null, null ); } BinaryNode( AnyType theElement, BinaryNode lt, BinaryNode rt, BinaryNode nt, BinaryNode pv ) { element = theElement; left = lt; right = rt; next = nt; prev = pv; } AnyType element; BinaryNode left; BinaryNode right; BinaryNode next; BinaryNode prev; } public java.util.Iterator iterator() { return new MyTreeSet2Iterator( ); } private class MyTreeSet2Iterator implements java.util.Iterator { private BinaryNode current = findMin(root); private BinaryNode previous; private int expectedModCount = modCount; private boolean okToRemove = false; private boolean atEnd = false; public boolean hasNext() { return !atEnd; } public AnyType next() { if( modCount != expectedModCount ) throw new java.util.ConcurrentModificationException( ); if( !hasNext( ) ) throw new java.util.NoSuchElementException( ); AnyType nextItem = current.element; previous = current; current = current.next; if (current == null) atEnd = true; okToRemove = true; return nextItem; } Weiss Java Solutions pages p. 36 (chap04) Windfall Software, PCA ZzTEX 12.2 Solutions 37 public void remove() { if( modCount != expectedModCount ) throw new java.util.ConcurrentModificationException( ); if( !okToRemove ) throw new IllegalStateException( ); MyTreeSet2.this.remove( previous.element ); okToRemove = false; } } public MyTreeSet2() { root = null; } public void makeEmpty() { modCount++; root = null; } public boolean isEmpty() { return root == null; } public boolean contains( AnyType x ) { return contains( x, root ); } public AnyType findMin() throws UnderflowException { if ( isEmpty() ) throw new UnderflowException(); else return findMin( root ).element; } public AnyType findMax() throws UnderflowException { if ( isEmpty() ) throw new UnderflowException(); else return findMax( root ).element; } public void insert( AnyType x ) { root = insert( x, root, null, null ); } public void remove( AnyType x ) { root = remove( x, root ); } Weiss Java Solutions pages p. 37 (chap04) Windfall Software, PCA ZzTEX 12.2 38 Chapter 4 Trees public void printTree() { if ( isEmpty() ) System.out.println( "Empty tree" ); else printTree( root ); } private void printTree( BinaryNode t ) { if(t!=null ) { printTree( t.left ); System.out.println( t.element ); printTree( t.right ); } } private boolean contains( AnyType x, BinaryNode t ) { if(t==null ) return false; int compareResult = x.compareTo( t.element ); if ( compareResult < 0) return contains( x, t.left ); else if ( compareResult > 0) return contains( x, t.right ); else return true; // match } private BinaryNode findMin( BinaryNode t ) { if(t==null ) return null; else if ( t.left == null ) return t; return findMin( t.left ); } private BinaryNode findMax( BinaryNode t ) { if(t==null ) return null; else if ( t.right == null ) Weiss Java Solutions pages p. 38 (chap04) Windfall Software, PCA ZzTEX 12.2 Solutions 39 return t; return findMax( t.right ); } private BinaryNode insert( AnyType x, BinaryNode t, BinaryNode nt, BinaryNode pv ) { if(t==null ) { modCount++; BinaryNode newNode = new BinaryNode( x, null, null, nt, pv); if (nt != null) { nt.prev = newNode; } if (pv != null) { pv.next = newNode; } return newNode; } int compareResult = x.compareTo( t.element ); if ( compareResult < 0) t.left = insert( x, t.left, t, pv ); else if ( compareResult > 0) { t.right = insert( x, t.right, nt, t ); } else ; // duplicate return t; } private BinaryNode remove( AnyType x, BinaryNode t ) { if(t==null ) return t; // not found int compareResult = x.compareTo( t.element ); if ( compareResult < 0) t.left = remove( x, t.left ); else if ( compareResult > 0) t.right = remove( x, t.right ); else if ( t.left != null && t.right != null ) // two children Weiss Java Solutions pages p. 39 (chap04) Windfall Software, PCA ZzTEX 12.2 40 Chapter 4 Trees { t.element = findMin( t.right ).element; t.right = remove( t.element, t.right ); } else { modCount++; t.prev.next = t.next; // update next and prev links t.next.prev = t.prev; t = ( t.left != null ) ? t.left : t.right; } return t; } private BinaryNode root; int modCount = 0; } 4.14 (a) Keep a bit array B.Ifi is in the tree, then B[i] is true; otherwise, it is false. Repeatedly generate random integers until an unused one is found. If there are N elements already in the tree, then M − N are not, and the probability of finding one of these is (M − N)/M. Thus the expected number of trials is M/(M − N) = α/(α − 1). (b) To find an element that is in the tree, repeatedly generate random integers until an already-used integer is found. The probability of finding one is N/M, so the expected number of trials is M/N = α. (c) The total cost for one insert and one delete is alpha/(α − 1) + α = 1 + α + 1/(α − 1). Setting alpha = 2 minimizes this cost. 4.18 (a) N(0) = 1, N(1) = 2, N(h) = N(h − 1) + N(h − 2) + 1. (b) The heights are one less than the Fibonacci numbers. 4.19 4.20 It is easy to verify by hand that the claim is true for 1≤ k ≤ 3. Suppose it is true for k = 1, 2, 3, . . . h. Then after the first 2h − 1 insertions, 2h−1 is at the root, and the right subtree is a balanced tree containing 2h−1 + 1 through 2h − 1. Each of the next 2h−1 insertions, namely, 2h through 2h + 2h−1 − 1, insert a new maximum and get placed in the right subtree, eventually forming a perfectly balanced right subtree of height h − 1. This follows by the induction hypothesis because the right subtree may be viewed as being formed from the successive insertion of 2h−1 + 1 through 2h + 2h−1 − 1. The next insertion forces an imbalance at the root, and thus a single rotation. It is easy Weiss Java Solutions pages p. 40 (chap04) Windfall Software, PCA ZzTEX 12.2 Solutions 41 to check that this brings 2h to the root and creates a perfectly balanced left subtree of height h − 1. The new key is attached to a perfectly balanced right subtree of height h − 2 as the last node in the right path. Thus the right subtree is exactly as if the nodes 2h + 1 through 2h + 2h−1 were inserted in order. By the inductive hypothesis, the subsequent successive insertion of 2h + 2h−1 + 1 through 2h+1 − 1 will create a perfectly balanced right subtree of height h − 1. Thus after the last insertion, both the left and the right subtrees are perfectly balanced, and of the same height, so the entire tree of 2h+1 − 1 nodes is perfectly balanced (and has height h). 4.21 The two remaining functions are mirror images of the text procedures. Just switch right and left everywhere. 4.24 After applying the standard binary search tree deletion algorithm, nodes on the deletion path need to have their balance changed, and rotations may need to be performed. Unlike insertion, more than one node may need rotation. 4.25 (a) O(log log N). (b) The minimum AVL tree of height 127 (8-byte ints range from −128 to 127). This is a huge tree. 4.26 static AvlNode doubleRotateWithLeft( AvlNode k3 ) { AvlNode k1, k2; k1 = k3.left; k2 = k1.right; k1.right = k2.left; k3.left = k2.right; k2.left = k1; k2.right = k3; k1.height = max( height(k1.left), height(k1.right))+1; k3.height = max( height(k3.left), height(k3.right))+1; k2.height = max( k1. height, k3.height)+1; return k3; } 4.27 After accessing 3, Weiss Java Solutions pages p. 41 (chap04) Windfall Software, PCA ZzTEX 12.2 42 Chapter 4 Trees After accessing 9, After accessing 1, After accessing 5, 4.28 Weiss Java Solutions pages p. 42 (chap04) Windfall Software, PCA ZzTEX 12.2 Solutions 43 4.29 (a) An easy proof by induction. 4.31 (a–c) All of these routines take linear time. static int countNodes( Node t ) { if (t == null) return 0; return 1 + countNodes(t.left) + countNodes(t.right); } static int countLeaves( Node t ) { if (t == null) return 0; else if ( t.left == null && t.right == null) return 1; return countLeaves(t.left) + countLeaves(t.right); } static int countFull( Node t ) { if (t == null) return 0; int tIsFull = ( t.left != null && t.right != null)?1:0; return tIsFull + countFull(t.left) + countFull(t.right); } 4.32 Have the recursive routine return a triple that consists of a boolean (whether the tree is a BST) and the minimum and maximum items. Then a tree is a BST if it is empty or both subtrees are (recursively) BSTs, and the value in the node lies between the maximum of the left subtree and the minimum of the right subtrees. Coding details are omitted. 4.33 static Node removeLeaves( Node t ) { if (t == null || ( t.left == null && t.right == null )) return null; t.left = removeLeaves( t.left ); t.right = removeLeaves( t.right ); return t; } 4.34 We assume the existence of a method randInt(lower, upper) that generates a uniform random integer in the appropriate closed interval. static Node makeRandomTree( int lower, int upper ) { Node t = null; int randomValue; Weiss Java Solutions pages p. 43 (chap04) Windfall Software, PCA ZzTEX 12.2 44 Chapter 4 Trees if ( lower <= upper ) t = new Node( randomValue = randInt( lower, upper ), makeRandomTree( lower, randomValue-1), makeRandomTree( randomValue + 1, upper ) ); return t; } static Node makeRandomTree( int n ) { return makeRandomTree( 1, n ); } 4.35 // LastNode[0] is the address containing the last value that was assigned to a node. // This is a standard trick to get a call by reference. static Node genTree( int height, int [] lastNode ) { Node t = null; if ( height >= 0 ) { t = new Node(); t.left = genTree( height - 1, lastNode[0] ); t.element = ++lastNode[0]; t.right = genTree( height - 2, lastNode[0] ); } return t; } static Node minAvlTree( int h) { int lastNodeAssigned[]={0}; return genTree( h, lastNodeAssigned ); } 4.36 There are two obvious ways of solving this problem. One way mimics Exercise 4.34 by replacing randInt(lower,upper) with (lower+upper) / 2. This requires computing 2h+1 − 1, which is not that difficult. The other mimics the previous exercise by noting that the heights of the subtrees are both h − 1. 4.37 This is known as one-dimensional range searching. The time is O(K) to perform the inorder traversal, if a significant number of nodes are found, and also proportional to the depth of the tree, if we get to some leaves (for instance, if no nodes are found). Since the average depth is O(log N), this gives an O(K + log N) average bound. static void printRange( Comparable lower, Comparable upper, BinaryNode t ) { if( t != null ) { if( lower.compareTo( t.element ) <= 0 ) printRange( lower, upper, t.left ); if( lower.compareTo( t.element ) <= 0 && t.element.compareTo( upper ) <= 0 ) Weiss Java Solutions pages p. 44 (chap04) Windfall Software, PCA ZzTEX 12.2 Solutions 45 System.out.println( t.element ); if( t.element.compareTo( upper ) <= 0 ) printRange( lower, upper, t.right ); } } 4.38 This exercise and Exercise 4.39 are likely programming assignments, so we do not provide code here. 4.44 4.46 The function shown here is clearly a linear time routine because in the worst case it does a traversal on both t1 and t2. static boolean similar( Node t1, Node t2 ) { if( t1 == NULL || t2 == NULL ) return t1 == NULL && t2 == NULL; return similar( t1.left, t2.left ) && similar( t1.right, t2.right ); } 4.48 The easiest solution is to compute, in linear time, the inorder numbers of the nodes in both trees. If the inorder number of the root of T2 is x, then find x in T1 and rotate it to the root. Recursively apply this strategy to the left and right subtrees of T1 (by looking at the values in the root of T2’s left and right subtrees). If dN is the depth of x, then the running time satisfies T(N) = T(i) + T(N − i − 1) + dN, where i is the size of the left subtree. In the worst case, dN is always O(N), and i is always 0, so the worst-case running time is quadratic. Under the plausible assumption that all values of i are equally likely, then even if dN is always O(N), the average value of T(N) is O(N log N). This is a common recurrence that was already formulated in the chapter and is solved in Chapter 7. Under the more reasonable assumption that dN is typically logarithmic, then the running time is O(N). 4.49 Add a data member to each node indicating the size of the tree it roots. This allows computation of its inorder traversal number. 4.50 (a) You need an extra bit for each thread. (c) You can do tree traversals somewhat easier and without recursion. The disadvantage is that it reeks of old-style hacking. 4.51 (a) Both values are 0. (b) The root contributes (N − 2)/N full nodes on average, because the root is full as long as it does not contain the largest or smallest item. The remainder of the equation is the expected contribution of the subtrees. (d) The average number of leaves is (N + 1)/3. Weiss Java Solutions pages p. 45 (chap04) Windfall Software, PCA ZzTEX 12.2 Weiss Java Solutions pages p. 46 (chap05) Windfall Software, PCA ZzTEX 12.2 CHAPTER 5 Hashing 5.1 (a) On the assumption that we add collisions to the end of the list (which is the easier way if a hash table is being built by hand), the separate chaining hash table that results is shown here. (b) 47 Weiss Java Solutions pages p. 47 (chap05) Windfall Software, PCA ZzTEX 12.2 48 Chapter 5 Hashing (c) (d) 1989 cannot be inserted into the table because hash2(1989) = 6, and the alternative locations 5, 1, 7, and 3 are already taken. The table at this point is as follows: 5.2 When rehashing, we choose a table size that is roughly twice as large and prime. In our case, the appropriate new table size is 19, with hash function h(x) = x (mod 19). (a) Scanning down the separate chaining hash table, the new locations are 4371 in list 1, 1323 in list 12, 6173 in list 17, 4344 in list 12, 4199 in list 0, 9679 in list 8, and 1989 in list 13. (b) The new locations are 9679 in bucket 8, 4371 in bucket 1, 1989 in bucket 13, 1323 in bucket 12, 6173 in bucket 17, 4344 in bucket 14 (because both 12 and 13 are already occupied), and 4199 in bucket 0. (c) The new locations are 9679 in bucket 8, 4371 in bucket 1, 1989 in bucket 13, 1323 in bucket 12, 6173 in bucket 17, 4344 in bucket 16 (because both 12 and 13 are already occupied), and 4199 in bucket 0. (d) The new locations are 9679 in bucket 8, 4371 in bucket 1, 1989 in bucket 13, 1323 in bucket 12, 6173 in bucket 17, 4344 in bucket 15 because 12 is already occupied, and 4199 in bucket 0. 5.4 We must be careful not to rehash too often. Let p be the threshold (fraction of table size) at which we rehash to a smaller table. Then if the new table has size N, it contains 2pN elements. This table will require rehashing after either 2N − 2pN insertions or pN deletions. Balancing these costs suggests that a good choice is p = 2/3. For instance, suppose we have a table of size 300. If we rehash at 200 elements, then the new table size is N = 150, and we can do either 100 insertions or 100 deletions until a new rehash is required. Weiss Java Solutions pages p. 48 (chap05) Windfall Software, PCA ZzTEX 12.2 Solutions 49 If we know that insertions are more frequent than deletions, then we might choose p to be somewhat larger. If p is too close to 1.0, however, then a sequence of a small number of deletions followed by insertions can cause frequent rehashing. In the worst case, if p = 1.0, then alternating deletions and insertions both require rehashing. 5.5 No; this does not take deletions into account. 5.6 (b) If the number of deleted cells is small, then we spend extra time looking for inactive cells that are not likely to be found. If the number of deleted cells is large, then we may get improvement. 5.7 In a good library implementation, the length method should be inlined. 5.8 Separate chaining hashing requires the use of links, which costs some memory, and the standard method of implementing calls on memory allocation routines, which typically are expensive. Linear probing is easily implemented, but performance degrades severely as the load factor increases because of primary clustering. Quadratic probing is only slightly more difficult to implement and gives good performance in practice. An insertion can fail if the table is half empty, but this is not likely. Even if it were, such an insertion would be so expensive that it wouldn’t matter and would almost certainly point up a weakness in the hash function. Double hashing eliminates primary and secondary clustering, but the computation of a second hash function can be costly. Gonnet and Baeza-Yates [8] compare several hashing strategies; their results suggest that quadratic probing is the fastest method. 5.10 The old values would remain valid if the hashed values were less than the old table size. 5.11 Sorting the MN records and eliminating duplicates would require O(MN log MN) time using a standard sorting algorithm. If terms are merged by using a hash function, then the merging time is constant per term for a total of O(MN). If the output polynomial is small and has only O(M + N) terms, then it is easy to sort it in O((M + N) log(M + N)) time, which is less than O(MN). Thus the total is O(MN). This bound is better because the model is less restrictive: Hashing is performing operations on the keys rather than just comparison between the keys. A similar bound can be obtained by using bucket sort instead of a standard sorting algorithm. Operations such as hashing are much more expensive than comparisons in practice, so this bound might not be an improvement. On the other hand, if the output polynomial is expected to have only O(M + N) terms, then using a hash table saves a huge amount of space, since under these conditions, the hash table needs only O(M + N) space. Another method of implementing these operations is to use a search tree instead of a hash table; a balanced tree is required because elements are inserted in the tree with too much order. A splay tree might be particularly well suited for this type of a problem because it does well with sequential accesses. Comparing the different ways of solving the problem is a good programming assignment. 5.12 To each hash table slot, we can add an extra data member that we’ll call whereOnStack, and we can keep an extra stack. When an insertion is first performed into a slot, we push the address (or number) of the slot onto the stack and set the whereOnStack data member to refer to the top of the stack. When we access a hash table slot, we check that whereOnStack refers to a valid part of the stack and that the entry in the (middle of the) stack that is referred to by the whereOnStack data member has that hash table slot as an address. 5.16 (a) The cost of an unsuccessful search equals the cost of an insertion. (b) I(λ) = 1 λ λ 0  1 1 − x − x − ln(1 − x)  dx = 1 − ln(1 − λ) − λ/2 Weiss Java Solutions pages p. 49 (chap05) Windfall Software, PCA ZzTEX 12.2 50 Chapter 5 Hashing 5.17 public class Map { public Map() { items = new QuadraticProbingHashTable>(); } public void put( KeyType key, ValueType val ) { Entry e = new Entry(key, val); items.insert( e ); } public ValueType get( KeyType key ) { ValueType v = (ValueType) new Object(); Entry e = new Entry(key, v); e = items.find(e); return e.val; } public boolean isEmpty() { return items.isEmpty(); } public void makeEmpty() { items.makeEmpty(); } private QuadraticProbingHashTable> items; private static class Entry { Entry( KeyType k, ValueType v ) { key=k; val=v; } public int hashCode() { return key.hashCode(); } //public boolean equals( Entry rhs ) public boolean equals( Object rhs ) { return rhs instanceof Entry && key.equals( ((Entry)rhs).key ); } Weiss Java Solutions pages p. 50 (chap05) Windfall Software, PCA ZzTEX 12.2 Solutions 51 private KeyType key; private ValueType val; } } The QuadraticProbingHashTable.java file needs these additional methods (not given in the text) for the above solution to work: public boolean isEmpty( ) { return currentSize == 0; } public AnyType find( AnyType x ) { int currentPos = findPos( x ); return isActive( currentPos ) ? array[ currentPos ].element : null; } 5.19 Weiss Java Solutions pages p. 51 (chap05) Windfall Software, PCA ZzTEX 12.2 Weiss Java Solutions pages p. 52 (chap06) Windfall Software, PCA ZzTEX 12.2 CHAPTER 6 Priority Queues (Heaps) 6.1 Yes. When an element is inserted, we compare it to the current minimum and change the minimum if the new element is smaller. deleteMin operations are expensive in this scheme. 6.2 6.3 The result of three deleteMins, starting with both of the heaps in Exercise 6.2, is as follows: 6.4 (a) 4N (b) O(N2) (c) O(N4.1) (d) O(2N) 6.5 public void insert( AnyType x ) { if ( currentSize == array.length-1) enlargeArray( array.length*2+1); // Percolate up int hole = ++currentSize; 53 Weiss Java Solutions pages p. 53 (chap06) Windfall Software, PCA ZzTEX 12.2 54 Chapter 6 Priority Queues (Heaps) for(;hole>1&&x.compareTo( array[ hole/2])<0;hole /= 2) array[ hole ] = array[ hole/2 ]; array[0]=array[ hole]=x; } 6.6 225. To see this, start with i = 1 and position at the root. Follow the path toward the last node, doubling i when taking a left child, and doubling i and adding one when taking a right child. 6.7 (a) We show that H(N), which is the sum of the heights of nodes in a complete binary tree of N nodes, is N − b(N), where b(N) is the number of ones in the binary representation of N. Observe that for N = 0 and N = 1, the claim is true. Assume that it is true for values of k up to and including N − 1. Suppose the left and right subtrees have L and R nodes, respectively. Since the root has height log N, we have H(N) =log N+H(L) + H(R) =log N+L − b(L) + R − b(R) = N − 1 + (log N−b(L) − b(R)) The second line follows from the inductive hypothesis, and the third follows because L + R = N − 1. Now the last node in the tree is in either the left subtree or the right subtree. If it is in the left subtree, then the right subtree is a perfect tree, and b(R) =log N−1. Further, the binary representation of N and L are identical, with the exception that the leading 10 in N becomes 1 in L. (For instance, if N =37=100101, L = 10101.) It is clear that the second digit of N must be zero if the last node is in the left subtree. Thus in this case, b(L) = b(N), and H(N) = N − b(N) If the last node is in the right subtree, then b(L) =log N. The binary representation of R is identical to N, except that the leading 1 is not present. (For instance, if N =27=101011, L = 01011.) Thus b(R) = b(N) − 1, and again H(N) = N − b(N) (b) Run a single-elimination tournament among eight elements. This requires seven comparisons and generates ordering information indicated by the binomial tree shown here. The eighth comparison is between b and c.Ifc is less than b, then b is made a child of c. Otherwise, both c and d are made children of b. (c) A recursive strategy is used. Assume that N = 2k. A binomial tree is built for the N elements as in part (b). The largest subtree of the root is then recursively converted into a binary heap of 2k−1 elements. The last element in the heap (which is the only one on an extra level) is then inserted into the binomial queue consisting of the remaining binomial trees, thus forming another binomial tree of 2k−1 elements. At that point, the root has a subtree that is a heap of 2k−1 − 1 elements and another subtree that is a binomial tree of 2k−1 elements. Recursively convert that subtree into a heap; now Weiss Java Solutions pages p. 54 (chap06) Windfall Software, PCA ZzTEX 12.2 Solutions 55 the whole structure is a binary heap. The running time for N = 2k satisfies T(N) = 2T(N/2) + log N. The base case is T(8) = 8. 6.9 Let D1, D2,...,Dk be random variables representing the depth of the smallest, second smallest, and kth smallest elements, respectively.We are interested in calculating E(Dk). In what follows, we assume that the heap size N is one less than a power of two (that is, the bottom level is completely filled) but sufficiently large so that terms bounded by O(1/N) are negligible. Without loss of generality, we may assume that the kth smallest element is in the left subheap of the root. Let pj,k be the probability that this element is the jth smallest element in the subheap. Lemma. For k > 1, E(Dk) = k−1 j=1 pj,k(E(Dj) + 1). Proof. An element that is at depth d in the left subheap is at depth d + 1 in the entire subheap. Since E(Dj + 1) = E(Dj) + 1, the theorem follows. Since by assumption, the bottom level of the heap is full, each of second, third,...,k − 1th smallest elements are in the left subheap with probability of 0.5. (Technically, the probability should be half − 1/(N − 1) of being in the right subheap and half + 1/(N − 1) of being in the left, since we have already placed the kth smallest in the right. Recall that we have assumed that terms of size O(1/N) can be ignored.) Thus pj,k = pk−j,k = 1 2k−2 k − 2 j − 1 Theorem. E(Dk) ≤ log k. Proof. The proof is by induction. The theorem clearly holds for k = 1 and k = 2. We then show that it holds for arbitrary k > 2 on the assumption that it holds for all smaller k. Now, by the inductive hypothesis, for any 1 ≤ j ≤ k − 1, E(Dj) + E(Dk−j) ≤ log j + log k − j Since f(x) = log x is convex for x > 0, log j + log k − j ≤ 2 log(k/2) Thus E(Dj) + E(Dk−j) ≤ log(k/2) + log(k/2) Furthermore, since pj,k = pk−j,k, pj,kE(Dj) + pk−j,kE(Dk−j) ≤ pj,k log(k/2) + pk−j,k log(k/2) From the lemma, E(Dk) = k−1 j=1 pj,k(E(Dj) + 1) = 1 + k−1 j=1 pj,kE(Dj) Weiss Java Solutions pages p. 55 (chap06) Windfall Software, PCA ZzTEX 12.2 56 Chapter 6 Priority Queues (Heaps) Thus E(Dk) ≤ 1 + k−1 j=1 pj,k log(k/2) ≤ 1 + log(k/2) k−1 j=1 pj,k ≤ 1 + log(k/2) ≤ log k completing the proof. It can also be shown that asymptotically, E(Dk) ≈ log(k − 1) − 0.273548. 6.10 (a) Perform a preorder traversal of the heap. (b) Works for leftist and skew heaps. The running time is O(Kd) for d-heaps. 6.12 Simulations show that the linear time algorithm is the faster, not only on worst-case inputs, but also on random data. 6.13 (a) If the heap is organized as a (min) heap, then starting at the hole at the root, find a path down to a leaf by taking the minimum child. The requires roughly log N comparisons. To find the correct place where to move the hole, perform a binary search on the log N elements. This takes O(log log N) comparisons. (b) Find a path of minimum children, stopping after log N − log log N levels. At this point, it is easy to determine if the hole should be placed above or below the stopping point. If it goes below, then continue finding the path, but perform the binary search on only the last log log N elements on the path, for a total of log N + log log log N comparisons. Otherwise, perform a binary search on the first log N − log log N elements. The binary search takes at most log log N comparisons, and the path finding took only log N − log log N, so the total in this case is log N. So the worst case is the first case. (c) The bound can be improved to log N + log ∗N + O(1), where log ∗N is the inverse Ackerman function (see Chapter 8). This bound can be found in reference [17]. 6.14 The parent is at position (i + d − 2)/d. The children are in positions (i − 1)d + 2,...,id + 1. 6.15 (a) O((M + dN) logd N). (b) O((M + N) log N). (c) O(M + N2). (d) d = max(2, M/N). (See the related discussion at the end of Section 11.4.) 6.16 Starting from the second most signficant digit in i, and going toward the least significant digit, branch left for 0s, and right for 1s. 6.17 (a) Place negative infinity as a root with the two heaps as subtrees. Then do a deleteMin. (b) Place negative infinity as a root with the larger heap as the left subheap, and the smaller heap as the right subheap. Then do a deleteMin. (c) SKETCH: Split the larger subheap into smaller heaps as follows: on the left-most path, remove two subheaps of height r − 1, then one of height r, r + 1, and so one, until l − 2. Then merge the trees, going smaller to higher, using the results of parts (a) and (b), with the extra nodes on the left path substituting for the insertion of infinity, and subsequent deleteMin. Weiss Java Solutions pages p. 56 (chap06) Windfall Software, PCA ZzTEX 12.2 Solutions 57 6.19 6.20 6.21 This theorem is true, and the proof is very much along the same lines as Exercise 4.20. 6.22 If elements are inserted in decreasing order, a leftist heap consisting of a chain of left children is formed. This is the best because the right path length is minimized. 6.23 (a) If a decreaseKey is performed on a node that is very deep (very left), the time to percolate up would be prohibitive. Thus the obvious solution doesn’t work. However, we can still do the operation efficiently by a combination of remove and insert.Toremove an arbitrary node x in the heap, replace x by the merge of its left and right subheaps. This might create an imbalance for nodes on the path from x’s parent to the root that would need to be fixed by a child swap. However, it is easy to show that at most log N nodes can be affected, preserving the time bound. This is discussed in Chapter 11. 6.24 Lazy deletion in leftist heaps is discussed in the paper by Cheriton and Tarjan [10]. The general idea is that if the root is marked deleted, then a preorder traversal of the heap is formed, and the frontier of marked nodes is removed, leaving a collection of heaps. These can be merged two at a time by placing all the heaps on a queue, removing two, merging them, and placing the result at the end of the queue, terminating when only one heap remains. 6.25 (a) The standard way to do this is to divide the work into passes. A new pass begins when the first element reappears in a heap that is dequeued. The first pass takes roughly 2 ∗ 1 ∗ (N/2) time units because there are N/2 merges of trees with one node each on the right path. The next pass takes 2 ∗ 2 ∗ (N/4) time units because of the roughly N/4 merges of trees with no more than two nodes on the right path. The third pass takes 2 ∗ 3 ∗ (N/8) time units, and so on. The sum converges to 4N. (b) It generates heaps that are more leftist. Weiss Java Solutions pages p. 57 (chap06) Windfall Software, PCA ZzTEX 12.2 58 Chapter 6 Priority Queues (Heaps) 6.26 6.27 6.28 This claim is also true, and the proof is similar in spirit to Exercise 4.20 or 6.21. 6.29 Yes. All the single operation estimates in Exercise 6.25 become amortized instead of worst-case, but by the definition of amortized analysis, the sum of these estimates is a worst-case bound for the sequence. 6.30 Clearly the claim is true for k = 1. Suppose it is true for all values i = 1,2,...,k.ABk+1 tree is formed by attaching a Bk tree to the root of a Bk tree. Thus by induction, it contains a B0 through Bk−1 tree, as well as the newly attached Bk tree, proving the claim. 6.31 Proof is by induction. Clearly the claim is true for k = 1. Assume true for all values i = 1,2,...,k. A Bk+1 tree is formed by attaching a Bk tree to the original Bk tree. The original thus had k d nodes at depth d. The attached tree had k d − 1 nodes at depth d − 1, which are now at depth d. Adding these two terms and using a well-known formula establishes the theorem. Weiss Java Solutions pages p. 58 (chap06) Windfall Software, PCA ZzTEX 12.2 Solutions 59 6.32 6.33 This is established in Chapter 11. 6.38 Don’t keep the key values in the heap, but keep only the difference between the value of the key in a node and the value of the parent’s key. 6.39 O(N + k log N) is a better bound than O(N log k). The first bound is O(N) if k = O(N/ log N). The second bound is more than this as soon as k grows faster than a constant. For the other values (N/ log N) = k = o(N), the first bound is better. When k = (N), the bounds are identical. Weiss Java Solutions pages p. 59 (chap06) Windfall Software, PCA ZzTEX 12.2 Weiss Java Solutions pages p. 60 (chap07) Windfall Software, PCA ZzTEX 12.2 CHAPTER 7 Sorting 7.1 7.2 O(N), because the while loop terminates immediately. Of course, accidentally changing the test to include equalities raises the running time to quadratic for this type of input. 7.3 The inversion that existed between a[i] and a[i + k] is removed. This shows at least one inversion is removed. For each of the k − 1 elements a[i + 1], a[i + 2],...,a[i + k − 1], at most two inversions can be removed by the exchange. This gives a maximum of 2(k − 1) + 1 = 2k − 1. 7.4 7.5 (a) (N2). The 2-sort removes at most only three inversions at a time; hence the algorithm is (N2). The 2-sort is two insertion sorts of size N/2, so the cost of that pass is O(N2). The 1-sort is also O(N2), so the total is O(N2). 7.6 Part (a) is an extension of the theorem proved in the text. Part (b) is fairly complicated; see reference [12]. 7.7 See reference [12]. 7.8 Use the input specified in the hint. If the number of inversions is shown to be (N2), then the bound follows, since no increments are removed until an ht/2 sort. If we consider the pattern formed hk through h2k−1, where k = t/2 + 1, we find that it has length N = hk(hk + 1) − 1, and the number of inversions is roughly hk 4/24, which is (N2). 7.9 (a) O(N log N). No exchanges, but each pass takes O(N). (b) O(N log N). It is easy to show that after an hk sort, no element is farther than hk from its rightful position. Thus if the increments satisfy hk+1 ≤ chk for a constant c, which implies O(log N) increments, then the bound is O(N log N). 61 Weiss Java Solutions pages p. 61 (chap07) Windfall Software, PCA ZzTEX 12.2 62 Chapter 7 Sorting 7.10 (a) No, because it is still possible for consecutive increments to share a common factor. An example is the sequence 1, 3, 9, 21, 45, ht+1 = 2ht + 3. (b) Yes, because consecutive increments are relatively prime. The running time becomes O(N3/2). 7.11 The input is read in as 142, 543, 123, 65, 453, 879, 572, 434, 111, 242, 811, 102 The result of the heapify is 879, 811, 572, 434, 543, 123, 142, 65, 111, 242, 453, 102 879 is removed from the heap and placed at the end. We’ll place it in italics to signal that it is not part of the heap. 102 is placed in the hole and bubbled down, obtaining 811, 543, 572, 434, 453, 123, 142, 65, 111, 242, 102, 879 Continuing the process, we obtain 572, 543, 142, 434, 453, 123, 102, 65, 111, 242, 811, 879 543, 453, 142, 434, 242, 123, 102, 65, 111, 572, 811, 879 453, 434, 142, 111, 242, 123, 102, 65, 543, 572, 811, 879 434, 242, 142, 111, 65, 123, 102, 453, 543, 572, 811, 879 242, 111, 142, 102, 65, 123, 434, 453, 543, 572, 811, 879 142, 111, 123, 102, 65, 242, 434, 453, 543, 572, 811, 879 123, 111, 65, 102, 142, 242, 434, 453, 543, 572, 811, 879 111, 102, 65, 123, 142, 242, 434, 453, 543, 572, 811, 879 102, 65, 111, 123, 142, 242, 434, 453, 543, 572, 811, 879 65, 102, 111, 123, 142, 242, 434, 453, 543, 572, 811, 879 7.13 Heapsort uses at least (roughly) N log N comparisons on any input, so there are no particularly good inputs. This bound is tight; see the paper by Schaeffer and Sedgewick [18]. This result applies for almost all variations of heapsort, which have different rearrangement strategies. See Y. Ding and M. A. Weiss, “Best Case Lower Bounds for Heapsort,” Computing 49 (1992). 7.14 If the root is stored in position low, then the left child of node i is stored at position 2i + 1 − low. This requires a small change to the heapsort code. 7.15 First the sequence {3, 1, 4, 1} is sorted. To do this, the sequence {3, 1} is sorted. This involves sorting {3} and {1}, which are base cases, and merging the result to obtain {1, 3}. The sequence {4, 1} is likewise sorted into {1, 4}. Then these two sequences are merged to obtain {1, 1, 3, 4}. The second half is sorted similarly, eventually obtaining {2, 5, 6, 9}. The merged result is then easily computed as {1, 1, 2, 3, 4, 5, 6, 9}. 7.16 private static > void mergeSort( AnyType [] a, AnyType [] tmpArray, int left, int right ) { int n = a.length; for ( int subListSize = 1; subListSize < n; subListSize *= 2 ) { int part1Start = 0; while ( part1Start + subListSize K. The sort is O(N log N); the scan is O(N). 7.50 (b) After sorting the items do a scan using an outer loop that considers all positions p, and an inner loop that searches for two additional items that sum to K − a[p]. The two loops give a quadratic algorithm. Weiss Java Solutions pages p. 66 (chap07) Windfall Software, PCA ZzTEX 12.2 CHAPTER 8 The Disjoint Set Class 8.1 We assume that unions operated on the roots of the trees containing the arguments. Also, in case of ties, the second tree is made a child of the first. Arbitrary union and union by height give the same answer (shown as the first tree) for this problem. Union by size gives the second tree. 8.2 In both cases, have nodes 16 and 0 point directly to the root. 8.4 Claim: A tree of height h has at least 2h nodes. The proof is by induction. A tree of height 0 clearly has at least 1 node, and a tree of height 1 clearly has at least 2. Let T be the tree of height h with fewest nodes. Thus at the time of T’s last union, it must have been a tree of height h − 1, since otherwise T would have been smaller at that time than it is now and still would have been of height h, which is impossible by assumption of T’s minimality. Since T’s height was updated, it must have been as a result of a union with another tree of height h − 1. By the induction hypothesis, we know that at the time of the union, T had at least 2h−1 nodes, as did the tree attached to it, for a total of 2h nodes, proving the claim. Thus an N-node tree has depth at most log N. 8.5 All answers are O(M) because in all cases alpha(M, N) = 1. 8.6 It is easy to see that this invariant is maintained in each disjoint set as the algorithm proceeds. 8.7 Run the normal algorithm; never remove the prespecified wall. Let s1 and s2 be squares on the opposite side of the wall. Do not remove any wall that results in s1 being connected to the ending point or s2 being connected to the starting point, and continue until there are only two disjoint sets remaining. 67 Weiss Java Solutions pages p. 67 (chap08) Windfall Software, PCA ZzTEX 12.2 68 Chapter 8 The Disjoint Set Class 8.8 (a) When we perform a union we push onto a stack the two roots and the old values of their parents. To implement a deunion, we only have to pop the stack and restore the values. This strategy works fine in the absence of path compression. (b) If path compression is implemented, the strategy described in part (a) does not work, because path compression moves elements out of subtrees. For instance, the sequence union(1,2), union(3,4), union(1,3), find(4), deunion(1,3) will leave 4 in set 1 if path compression is implemented. 8.9 We assume that the tree is implemented with links instead of a simple array. Thus find will return a reference instead of an actual set name. We will keep an array to map set numbers to their tree nodes. union and find are implemented in the standard manner. To perform remove(x), first perform a find(x) with path compression. Then mark the node containing x as vacant. Create a new one-node tree with x and have it pointed to by the appropriate array entry. The time to perform a remove is the same as the time to perform a find, except that there potentially could be a large number of vacant nodes. To take care of this, after N removes are performed, perform a find on every node, with path compression. If a find(x) returns a vacant root, then place x in the root node, and make the old node containing x vacant. The results of Exercise 8.11 guarantee that this will take linear time, which can be charged to the N removes. At this point, all vacant nodes (indeed all nonroot nodes) are children of a root, and vacant nodes can be disposed (if an array of references to them has been kept). This also guarantees that there are never more than 2N nodes in the forest and preserves the Mα(M, N) asymptotic time bound. 8.11 Suppose there are u union and f find operations. Each union costs constant time, for a total of u.A find costs one unit per vertex visited. We charge, as in the text, under the following slightly modified rules: (A) the vertex is a root or child of the root (B) otherwise Essentially, all vertices are in one rank group. During any find, there can be at most two rule (A) charges, for a total of 2f. Each vertex can be charged at most once under rule (B) because after path compression it will be a child of the root. The number of vertices that are not roots or children of roots is clearly bounded by u, independent of the unioning strategy, because each union changes exactly one vertex from root to nonroot status, and this bounds the number of type (B) nodes. Thus the total rule (B) charges are at most u. Adding all charges gives a bound of 2f + 2u, which is linear in the number of operations. 8.13 For each vertex v, let the pseudorank Rv be defined as log Sv, where Sv is the number of descendents (including itself) of v in the final tree, after all union operations are performed, ignoring path compression. Although the pseudorank is not maintained by the algorithm, it is not hard to show that the pseudorank satisfies the same properties as the ranks do in union-by-rank. Clearly, a vertex with pseudorank Rv has at least 2Rv descendents (by its definition), and the number of vertices of pseudorank R is at most N/2R. The union-by-size rule ensures that the parent of a node has twice as many descendents as the node, so the pseudoranks monotonically increase on the path toward the root if there is no path compression. The argument in Lemma 8.3 tells us that path compression does not destroy this property. If we partition the vertices by pseudoranks and assign the charges in the same manner as in the text proof for union-by-rank, the same steps follow, and the identical bound is obtained. 8.14 This is most conveniently implemented without recursion and is faster because, even if full path compression is implemented nonrecursively,it requires two passes up the tree. This requires only one. Weiss Java Solutions pages p. 68 (chap08) Windfall Software, PCA ZzTEX 12.2 Solutions 69 We leave the coding to the reader since comparing the various union and find strategies is a reasonable programming project. The worst-case running time remains the same because the properties of the ranks are unchanged. Instead of charging one unit to each vertex on the path to the root, we can charge two units to alternating vertices (namely, the vertices whose parents are altered by path halving). These vertices get parents of higher rank, as before, and the same kind of analysis bounds the total charges. Weiss Java Solutions pages p. 69 (chap08) Windfall Software, PCA ZzTEX 12.2 Weiss Java Solutions pages p. 70 (chap09) Windfall Software, PCA ZzTEX 12.2 CHAPTER 9 Graph Algorithms 9.1 The following ordering is arrived at by using a queue and assumes that vertices appear on an adjacency list alphabetically. The topological order that results is then s, G, D, H, A, B, E, I, F,C, t 9.2 Assuming the same adjacency list, the topological order produced when a stack is used is s, G, H, D, A, E, I, F,B, C, t Because a topological sort processes vertices in the same manner as a breadth-first search, it tends to produce a more natural ordering. 9.4 The idea is the same as in Exercise 5.12. 9.5 (a) (Unweighted paths) A → B, A → C, A → B → G, A → B → E, A → C → D, A → B → E → F. (b) (Weighted paths) A → C, A → B, A → B → G, A → B → G → E, A → B → G → E → F, A → B → G → E → D. 9.6 We’ll assume that Dijkstra’salgorithm is implemented with a priority queue of vertices that uses the de- creaseKey operation. Dijkstra’salgorithm uses |E| decreaseKey operations, which cost O(logd |V|) each, and |V| deleteMin operations, which cost O(d logd |V|) each. The running time is thus O(|E| logd |V|+ |V|d logd |V|). The cost of the decreaseKey operations balances the insert operations when d =|E|/|V|. For a sparse graph, this might give a value of d that is less than 2; we can’t allow this, so d is chosen to be max(2, |E|/|V|). This gives a running time of O(|E| log2+|E|/|V| |V|), which is a slight theoretical improvement. Moret and Shapiro report (indirectly) that d-heaps do not improve the running time in practice. 9.7 (a) The graph shown here is an example. Dijkstra’s algorithm gives a path from A to C of cost 2, when the path from A to B to C has cost 1. (b) We define a pass of the algorithm as follows: Pass 0 consists of marking the start vertex as known and placing its adjacent vertices on the queue. For j > 0, pass j consists of marking as known all vertices on the queue at the end of pass j − 1. Each pass requires linear time, since during a pass, a vertex is placed on the queue at most once. It is easy to show by induction that if there is a shortest 71 Weiss Java Solutions pages p. 71 (chap09) Windfall Software, PCA ZzTEX 12.2 72 Chapter 9 Graph Algorithms path from s to v containing k edges, then dv will equal the length of this path by the beginning of pass k. Thus there are at most |V| passes, giving an O(|E||V|) bound. 9.8 See the comments for Exercise 9.19. 9.10 (a) Use an array count such that for any vertex u, count[u] is the number of distinct paths from s to u known so far. When a vertex v is marked as known, its adjacency list is traversed. Let w be a vertex on the adjacency list. If dv + cv,w = dw, then increment count[w] by count[v] because all shortest paths from s to v with last edge (v, w) give a shortest path to w. If dv + cv,w < dw, then pw and dw get updated. All previously known shortest paths to w are now invalid, but all shortest paths to v now lead to shortest paths for w, so set count[w] to equal count[v]. Note: Zero-cost edges mess up this algorithm. (b) Use an array numEdges such that for any vertex u, numEdges[u]is the shortest number of edges on a path of distance du from s to u known so far. Thus numEdges is used as a tiebreaker when selecting the vertex to mark. As before, v is the vertex marked known, and w is adjacent to v. If dv + cv,w = dw, then change pw to v and numEdges[w] to numEdges[v]+1 if numEdges[v]+1 < numEdges[w]. If dv + cv,w < dw, then update pw and dw, and set numEdges[w] to numEdges[v]+1. 9.11 (This solution is not unique.) First, send four units of flow along the path s, G, H, I, t. This gives the following residual graph: Next, send three units of flow along s, D, E, F,t. The residual graph that results is as follows: Now two units of flow are sent along the path s, G, D, A, B, C, t, yielding the following residual graph: Weiss Java Solutions pages p. 72 (chap09) Windfall Software, PCA ZzTEX 12.2 Solutions 73 One unit of flow is then sent along s, D, A, E, C, t: Finally, one unit of flow can go along the path s, A, E, C, t: The preceding residual graph has no path from s to t. Thus the algorithm terminates. The final flow graph, which carries 11 units, is as follows: This flow is not unique. For instance, two units of the flow that goes from G to D to A to E could gobyGtoHtoE. 9.12 Let T be the tree with root r, and children r1, r2,...,rk, which are the roots of T1, T2,...,Tk, which have maximum incoming flow of c1, c2,...,ck, respectively. By the problem statement, we may take the maximum incoming flow of r to be infinity.The recursive pseudo-method findMaxFlow( T, incomingCap ) finds the value of the maximum flow in T (finding the actual flow is a matter of bookkeeping); the flow is guaranteed not to exceed incomingCap. Weiss Java Solutions pages p. 73 (chap09) Windfall Software, PCA ZzTEX 12.2 74 Chapter 9 Graph Algorithms If T is a leaf, then findMaxFlow returns incomingCap since we have assumed a sink of infinite capacity. Otherwise, a standard postorder traversal can be used to compute the maximum flow in linear time. // FlowType is an int or double static FlowType findMaxFlow( Tree T, FlowType incomingCap ) { FlowType childFlow, totalFlow; if( T.hasNoSubtrees( ) ) return incomingCap; else { totalFlow = 0; for( each subtree $T_i$ of T ) { childFlow = findMaxFlow( $T_i$, min( incomingCap, $c_i$ ) ); totalFlow += childFlow; incomingCap -= childFlow; } return totalFlow; } } 9.13 (a) Assume that the graph is connected and undirected. If it is not connected, then apply the algorithm to the connected components. Initially, mark all vertices as unknown. Pick any vertex v, color it red, and perform a depth-first search. When a node is first encountered, color it blue if the DFS has just come from a red node, and red otherwise. If at any point, the depth-first search encounters an edge between two identical colors, then the graph is not bipartite; otherwise, it is. A breadth-first search (that is, using a queue) also works. This problem, which is essentially two-coloring a graph, is clearly solvable in linear time. This contrasts with three-coloring, which is NP-complete. (b) Construct an undirected graph with a vertex for each instructor, a vertex for each course, and an edge between (v, w) if instructor v is qualified to teach course w. Such a graph is bipartite; a matching of M edges means that M courses can be covered simultaneously. (c) Give each edge in the bipartite graph a weight of 1, and direct the edge from the instructor to the course. Add a vertex s with edges of weight 1 from s to all instructor vertices. Add a vertex t with edges of weight 1 from all course vertices to t. The maximum flow is equal to the maximum matching. (d) The running time is O(|E||V|half ) because this is the special case of the network flow problem mentioned in the text. All edges have unit cost, and every vertex (except s and t) has either an indegree or outdegree of 1. 9.14 This is a slight modification of Dijkstra’s algorithm. Let fi be the best flow from s to i at any point in the algorithm. Initially, fi = 0 for all vertices, except s: fs = inf. At each stage, we select v such that fv is maximum among all unknown vertices. Then for each w adjacent to v, the cost of the flow to w using v as an intermediate is min(fv, cv,w). If this value is higher than the current value of fw, then fw and pw are updated. Weiss Java Solutions pages p. 74 (chap09) Windfall Software, PCA ZzTEX 12.2 Solutions 75 9.15 One possible minimum spanning tree is shown here. This solution is not unique. 9.16 Both work correctly. The proof makes no use of the fact that an edge must be nonnegative. 9.17 The proof of this fact can be found in any good graph theory book. A more general theorem follows: Theorem. Let G = (V, E) be an undirected, unweighted graph, and let A be the adjacency matrix for G (which contains either 1s or 0s). Let D be the matrix such that D[v][v] is equal to the degree of v; all nondiagonal matrices are 0. Then the number of spanning trees of G is equal to the determinant of A + D. 9.19 The obvious solution using elementary methods is to bucket sort the edge weights in linear time. Then the running time of Kruskal’s algorithm is dominated by the union/find operations and is O(|E|α(|E|, |V|)). The Van-Emde Boas priority queues (see Chapter 6 references) give an immediate O(|E| log log |V|) running time for Dijkstra’s algorithm, but this isn’t even as good as a Fibonacci heap implementation. More sophisticated priority queue methods that combine these ideas have been proposed, including M. L. Fredman and D. E. Willard, “Trans-dichotomous Algorithms for Minimum Spanning Trees and Shortest Paths,” Proceedings of the Thirty-first Annual IEEE Symposium on the Foundations of Computer Science (1990), 719–725. The paper presents a linear-time minimum spanning tree algorithm and an O(|E|+|V| log |V|/ log log |V|) implementation of Dijkstra’s algorithm if the edge costs are suitably small. Various improvements have been discovered since. 9.20 Since the minimum spanning tree algorithm works for negative edge costs, an obvious solution is to replace all the edge costs by their negatives and use the minimum spanning tree algorithm. Alternatively, change the logic so that < is replaced by >, min by max, and vice versa. 9.21 We start the depth-first search at A and visit adjacent vertices alphabetically. The articulation points are C, E, and F. C is an articulation point because low[B]≥ num[C]; E is an articulation point because low[H]≥ num[E]; and F is an articulation point because low[G]≥ num[F]. The depth-first spanning tree is shown in the following Figure. Weiss Java Solutions pages p. 75 (chap09) Windfall Software, PCA ZzTEX 12.2 76 Chapter 9 Graph Algorithms 9.22 The only difficult part is showing that if some nonroot vertex a is an articulation point, then there is no back edge between any proper descendent of a and a proper ancestor of a in the depth-first spanning tree. We prove this by a contradiction. Let u and v be two vertices such that every path from u to v goes through a. At least one of u and v is a proper descendent of a, since otherwise there is a path from u to v that avoids a. Assume without loss of generality that u is a proper descendent of a. Let c be the child of a that contains u as a descendent. If there is no back edge between a descendent of c and a proper ancestor of a, then the theorem is true immediately, so suppose for the sake of contradiction that there is a back edge (s, t). Then either v is a proper descendent of a or it isn’t. In the second case, by taking a path from u to s to t to v, we can avoid a, which is a contradiction. In the first case, clearly v cannot be a descendent of c, so let c be the child of a that contains v as a descendent. By a similar argument as before, the only possibility is that there is a back edge (s, t) between a descendent of c and a proper ancestor of a. Then there is a path from u to s to t to t to s to v; this path avoids a, which is also a contradiction. 9.23 (a) Do a depth-first search and count the number of back edges. (b) This is the feedback edge set problem. See reference [1] or [20]. 9.24 Let (v, w) be a cross edge. Since at the time w is examined it is already marked, and w is not a descendent of v (else it would be a forward edge), processing for w is already complete when processing for v commences. Thus under the convention that trees (and subtrees) are placed left to right, the cross edge goes from right to left. 9.25 Suppose the vertices are numbered in preorder and postorder. If (v, w) is a tree edge, then v must have a smaller preorder number than w. It is easy to see that the converse is true. If (v, w) is a cross edge, then v must have both a larger preorder and postorder number than w. The converse is shown as follows: Because v has a larger preorder number, w cannot be a descendent Weiss Java Solutions pages p. 76 (chap09) Windfall Software, PCA ZzTEX 12.2 Solutions 77 of v; because it has a larger postorder number, v cannot be a descendent of w; thus they must be in different trees. Otherwise, v has a larger preorder number but is not a cross edge. To test if (v, w) is a back edge, keep a stack of vertices that are active in the depth-first search call (that is, a stack of vertices on the path from the current root). By keeping a bit array indicating presence on the stack, we can easily decide if (v, w) is a back edge or a forward edge. 9.26 The first depth-first spanning tree is Gr, with the order in which to perform the second depth-first search, is shown next. The strongly connected components are F and all other vertices. 9.28 This is the algorithm mentioned in the references. 9.29 Because an edge (v, w) is implicitly processed, it is placed on a stack. If v is determined to be an articulation point because low[w]≥ num[v], then the stack is popped until edge (v, w) is removed: The set of popped edges is a biconnected component. An edge (v, w) is not placed on the stack if the edge (w, v) was already processed as a back edge. 9.30 Let (u, v) be an edge of the breadth-first spanning tree. (u, v) are connected, thus they must be in the same tree. Let the root of the tree be r; if the shortest path from r to u is du, then u is at level du; likewise, v is at level dv.If(u, v) were a back edge, then du > dv, and v is visited before u. But if there were an edge between u and v, and v is visited first, then there would be a tree edge (v, u), and not a back edge (u, v). Likewise, if (u, v) were a forward edge, then there is some w, distinct from u and v, on the path from u to v; this contradicts the fact that dv = dw + 1. Thus only tree edges and cross edges are possible. Weiss Java Solutions pages p. 77 (chap09) Windfall Software, PCA ZzTEX 12.2 78 Chapter 9 Graph Algorithms 9.31 Perform a depth-first search. The return from each recursive call implies the edge traversal in the opposite direction. The time is clearly linear. 9.33 If there is an Euler circuit, then it consists of entering and exiting nodes; the number of entrances clearly must equal the number of exits. If the graph is not strongly connected, there cannot be a cycle connecting all the vertices. To prove the converse, an algorithm similar in spirit to the undirected version can be used. 9.34 Neither of the proposed algorithms works. For example, as shown, a depth-first search of a bicon- nected graph that follows A, B, C, D is forced back to A, where it is stranded. 9.35 These are classic graph theory results. Consult any graph theory text for a solution to this exercise. 9.36 All the algorithms work without modification for multigraphs. 9.37 Obviously, G must be connected. If each edge of G can be converted to a directed edge and produce a strongly connected graph G, then G is convertible. Then, if the removal of a single edge disconnects G, G is not convertible, since this would also disconnect G. This is easy to test by checking to see if there are any single-edge biconnected components. Otherwise, perform a depth-first search on G and direct each tree edge away from the root and each back edge toward the root. The resulting graph is strongly connected because, for any vertex v, we can get to a higher level than v by taking some (possibly 0) tree edges and a back edge. We can apply this until we eventually get to the root, and then follow tree edges down to any other vertex. 9.38 (b) Define a graph where each stick is represented by a vertex. If stick Si is above Sj and thus must be removed first, then place an edge from Si to Sj. A legal pick-up ordering is given by a topological sort; if the graph has a cycle, then the sticks cannot be picked up. 9.39 Use a depth-first search, marking colors when a new vertex is visited, starting at the root, and returning false if a color clash is detected along a backedge. 9.40 Use a greedy algorithm: At each step, choose the vertex with the highest connectivity to vertices not already chosen. Proof that only 1/4 of the edges are not touched can be done with a straightforward calculation of edges that remain after each step. With E edges and V vertices, the total connectivity is 2E at the start, so at most (V − 2)/V of the edges remain after the first step. Then at most (V − 3)/(V − 1) of those edges remain after the second step, and so on. Multiply it out, and you get about 1/4 of the edges remaining. 9.41 If no vertex has indegree 0, we can find a cycle by tracing backwards through vertices with positive indegree; since every vertex on the trace back has a positive indegree, we eventually reach a vertex twice, and the cycle has been found. 9.42 The basic idea is that we examine cell A[s, t]: If it is 0, then t is not a sink; otherwise s is not a sink. Each examination can eliminate one vertex; after |V|−1 queries, we have one candidate that can be checked in 2|V|−1 additional tests. To do the first step, after the test A[s, t] is performed, replace the eliminated vertex with the next vertex in the the sequence of vertices. Weiss Java Solutions pages p. 78 (chap09) Windfall Software, PCA ZzTEX 12.2 Solutions 79 9.43 Perform a postorder traversal, computing node sizes; the first node that has a size greater than N/2 vertices is the node to remove. 9.44 This is essentially the algorithm described in Section 9.3.4 for critical paths. 9.45 These are all recursive depth first searches. 9.46 This is a single source unweighted shortest path problem. 9.47 This is a weighted shortest path problem: Adjacent squares have a cost of 1 if there is no wall, and P + 1 if there is a wall. 9.48 (a) Use the same ideas from the shortest path algorithm, but place unmarked adjacent squares at the front of a double-ended queue, and unmarked adjacent wall-separate squares at the end. Note that when dequeues are performed from the front, a stale square (in other words, one that was marked after it entered the deque) may emerge and would be ignored. (b) Use Exercise 9.47, with a penalty equal to the number of squares in the maze. 9.49 The following implementation does not use the map class. The use of ArrayList instead will speed up the algorithm since access now takes O(1) instead of O(log N), if the list of words can fit in main memory. One could use maps to also solve the problem. Using maps with keys and data both as strings allows the maps to store adjacent words in the word ladder directly. The Chapter 4 code can be modified for this problem by just adding words that differ in length by one then checking to see if the cost should be 1 or p when Dijkstra’s algorithm is applied. import java.util.*; import java.io.*; public class WordLadder { static final int INFINITY = 99999; private static class Vertex { Vertex() { adj = new ArrayList(); weight = new ArrayList(); } public ArrayList adj; // Adjacency list public ArrayList weight; // Weight list public boolean known; public int dist; public String name; public int path; }; /** * Print shortest path to v after dijkstra has run. Weiss Java Solutions pages p. 79 (chap09) Windfall Software, PCA ZzTEX 12.2 80 Chapter 9 Graph Algorithms * Assume that the path exists. */ static void printPath( int vIndex, ArrayList V ) { if( vIndex >= 0 && V.get(vIndex).path > -1 ) { printPath( V.get(vIndex).path, V ); System.out.println(" to "); } System.out.println(V.get(vIndex).name); } static void dijkstra( int sIndex, int tIndex, ArrayList Vertices ) { int smallestDist; int smallestVertex; Vertex v, s, t; int n = Vertices.size(); Vertices.get(sIndex).dist = 0; for(;;) { smallestDist = INFINITY; smallestVertex = -1; for( int i=0; i readWords( Scanner in ) { String oneLine = new String(); ArrayList v = new ArrayList(); while ( in.hasNext() ) { oneLine = in.next(); Vertex w = new Vertex(); w.name = oneLine; w.known = false; w.path = -1; w.dist = INFINITY; v.add(w); } return v; } static int oneCharOff( String word1, String word2, int p) { String big, small, shrink; int cost; if( Math.abs ((int)(word1.length() - word2.length()))>1) return 0; else if (word1.length() == word2.length()) { int diffs = 0; for( inti=0;i1) return 0; if (diffs == 1) return 1; } if (word2.length() > word1.length ()) { Weiss Java Solutions pages p. 81 (chap09) Windfall Software, PCA ZzTEX 12.2 82 Chapter 9 Graph Algorithms big = word2; small = word1; } else { big = word1; small = word2; } for(inti=0;i words, int p) { int cost; for(inti=0;i 0) { words.get(i).adj.add(j); words.get(i).weight.add(cost); words.get(j).adj.add(i); words.get(j).weight.add(cost); } } } public static void main(String args[]) { int p=0; int w1Index, w2Index; String w1 = new String(); String w2 = new String(); Scanner input = new Scanner(System.in); Scanner dict; Weiss Java Solutions pages p. 82 (chap09) Windfall Software, PCA ZzTEX 12.2 Solutions 83 try { dict = new Scanner(new File("dict.txt")); } catch (Exception e) { System.out.println("Error opening dictionary file"); return; } System.out.println("What is the cost of single char deletions: "); p = input.nextInt(); ArrayList words = readWords( dict ); do { System.out.println("Enter two words in the dictionary: "); w1 = input.next(); w2 = input.next(); // Find their indices (here is where a map would be superior // However all other accesses are now in O(1) time for (w1Index = 0; w1Index < words.size() && (words.get(w1Index).name).compareTo(w1) != 0 ; w1Index++); for (w2Index = 0; w2Index < words.size() && (words.get(w2Index).name).compareTo(w2) != 0 ; w2Index++); } while (w1Index >= words.size() || w2Index >= words.size()); fillAdjacencies(words, p); // make the adjacency list dijkstra( w1Index, w2Index, words ); // use dijkstra’s algorithm System.out.println(); printPath(w2Index, words); // print the result System.out.println(); } } 9.50 Each team is a vertex; if X has defeated Y, draw an edge from X to Y. 9.51 Each currency is a vertex; draw an edge of cost log C between vertices to represent a currency exchange rate, C. A negative cycle represents an arbitrage play. 9.52 Each course is a vertex; if X is a prerequisite for Y, draw an edge from X to Y Find the longest path in the acyclic graph. 9.53 Each actor is a vertex; an edge connects two vertices if the actors have a shared movie role. The graph is not drawn explicitly, but edges can be deduced as needed by examining cast lists. Weiss Java Solutions pages p. 83 (chap09) Windfall Software, PCA ZzTEX 12.2 84 Chapter 9 Graph Algorithms 9.54 Given an instance of clique, form the graph G that is the complement graph of G: (v, w) is an edge in G if and only if it is not an edge in G. Then G has a vertex cover of at most |V|−K if G has a clique of size at least K. (The vertices that form the vertex cover are exactly those not in the clique.) The details of the proof are left to the reader. 9.55 A proof can be found in Garey and Johnson [20]. 9.56 Clearly, the baseball card collector problem (BCCP) is in NP, because it is easy to check if K packets contain all the cards. To show it is NP-complete, we reduce vertex cover to it. Let G = (V, E), and let K be an instance of vertex cover. For each vertex v, place all edges adjacent to v in packet Pv. The K packets will contain all edges (baseball cards) if G can be covered by K vertices. Weiss Java Solutions pages p. 84 (chap09) Windfall Software, PCA ZzTEX 12.2 CHAPTER 10 Algorithm Design Techniques 10.1 First, we show that if N evenly divides P, then each of j(i−1)P+1 through jiP must be placed as the ith job on some processor. Suppose otherwise. Then in the supposed optimal ordering, we must be able to find some jobs jx and jy such that jx is the tth job on some processor and jy is the t + 1th job on some processor but tx > ty. Let jz be the job immediately following jx. If we swap jy and jz, it is easy to check that the mean processing time is unchanged and thus still optimal. But now jy follows jx, which is impossible because we know that the jobs on any processor must be in sorted order from the results of the one processor case. Let je1, je2,...,jeM be the extra jobs if N does not evenly divide P. It is easy to see that the processing time for these jobs depends only on how quickly they can be scheduled and that they must be the last scheduled job on some processor. It is easy to see that the first M processors must have jobs j(i−1)P+1 through jiP+M; we leave the details to the reader. 10.3 10.4 One method is to generate code that can be evaluated by a stack machine. The two operations are push (the one node tree corresponding to) a symbol onto a stack and combine, which pops two trees off the stack, merges them, and pushes the result back on. For the example in the text, the stack instructions are push(s), push(nl), combine, push(t), combine, push(a), combine, push(e), combine, push(i), push (sp), combine, combine. By encoding a combine with a 0 and a push with a 1 followed by the symbol, the total extra space is 2N − 1 bits if all the symbols are of equal length. Generating the stack machine code can be done with a simple recursive procedure and is left to the reader. 10.6 Maintain two queues, Q1 and Q2. Q1 will store single-node trees in sorted order, and Q2 will store multinode trees in sorted order. Place the initial single-node trees on Q1, enqueueing the smallest weight tree first. Initially, Q2 is empty.Examine the first two entries of each of Q1 and Q2, and dequeue the two smallest. (This requires an easily implemented extension to the ADT.) Merge the tree and place the result at the end of Q2. Continue this step until Q1 is empty and only one tree is left in Q2. 85 Weiss Java Solutions pages p. 85 (chap10) Windfall Software, PCA ZzTEX 12.2 86 Chapter 10 Algorithm Design Techniques 10.9 Toimplement first fit, we keep track of bins bi, which have more room than any of the lower numbered bins. A theoretically easy way to do this is to maintain a splay tree ordered by empty space. To insert w, we find the smallest of these bins, which has at least w empty space; after w is added to the bin, if the resulting amount of empty space is less than the inorder predecessor in the tree, the entry can be removed; otherwise, a decreaseKey is performed. To implement best fit, we need to keep track of the amount of empty space in each bin. As before, a splay tree can keep track of this. To insert an item of size w, perform an insert of w. If there is a bin that can fit the item exactly, the insert will detect it and splay it to the root; the item can be added and the root deleted. Otherwise, the insert has placed w at the root (which eventually needs to be removed). We find the minimum element M in the right subtree, which brings M to the right subtree’s root, attach the left subtree to M, and delete w. We then perform an easily implemented decreaseKey on M to reflect the fact that the bin is less empty. 10.10 Next fit: 12 bins (.42, .25, .27), (.07, .72), (.86, .09), (.44, .50), (.68), (.73), (.31), (.78, .17), (.79), (.37), (.73, .23), (.30). First fit: 10 bins (.42, .25, .27), (.07, .72, .09), (.86), (.44, .50), (.68, .31), (.73, .17), (.78), (.79), (.37, .23, .30), (.73). Best fit: 10 bins (.42, .25, .27), (.07, .72, .09), (.86), (.44, .50), (.68, .31), (.73, .23), (.78, .17), (.79), (.37, .30 ), (.73). First fit decreasing: 10 bins (.86, .09), (.79, .17), (.78, .07), (.73, .27), (.73, .25), (.72, .23), (.68, .31), (.50, .44), (.42, .37), (.30). Best fit decreasing: 10 bins (.86, .09), (.79, .17), (.78), (.73, .27), (.73, .25), (.72, .23), (.68, .31), (.50, .44), (.42, .37, .07), (.30). Note that use of 10 bins is optimal. 10.12 We prove the second case, leaving the first and third (which give the same results as Theorem 10.6) to the reader. Observe that logp N = logp (bm) = mp logp b Working this through, Equation (10.9) becomes T(N) = T(bm) = am m i=0 bk a i ip logp b If a = bk, then T(N) = am logp b m i=0 ip = O(ammp+1 logp b) Since m = log N/ log b, and am = Nk, and b is a constant, we obtain T(N) = O(Nk logp+1 N) 10.13 The easiest way to prove this is by an induction argument. 10.14 Divide the unit square into N − 1 square grids each with side 1/ √ N − 1. Since there are N points, some grid must contain two points. Thus the shortest distance is conservatively given by at most 2/(N − 1). 10.15 The results of the previous exercise imply that the width of the strip is O(1/ √ N). Because the width of the strip is O(1/ √ N), and thus covers only O(1/ √ N) of the area of the square, we expect a similar fraction of the points to fall in the strip. Thus only O(N/ √ N) points are expected in the strip; this is O( √ N). Weiss Java Solutions pages p. 86 (chap10) Windfall Software, PCA ZzTEX 12.2 Solutions 87 10.17 The recurrence works out to T(N) = T(2N/3) + T(N/3) + O(N) This is not linear, because the sum is not less than one. The running time is O(N log N). 10.18 The recurrence for median-of-median-of-seven partitioning is T(N) = T(5N/7) + T(N/7) + O(N) If all we are concerned about is the linearity, then median-of-median-of-seven can be used. 10.21 We derive the values of s and δ, following the style in the original paper [17]. Let Rt,X be the rank of element t in some sample X. If a sample S of elements is chosen randomly from S, and |S|=s, |S|=N, then we’ve already seen that E(Rt,S) = N + 1 s + 1 Rt,S where E means expected value. For instance, if t is the third largest in a sample of 5 elements, then in a group of 19 elements it is expected to be the tenth largest. We can also calculate the variance: V(Rt,S) =  (Rt,S)(s − Rt,S + 1)(N + 1)(N − s) (s + 1)2(s + 2) = O(N/ √ s) We choose v1 and v2 so that E(Rv1,S) + 2dV(Rv1,S) ≈ k ≈ E(Rv2,S) − 2dV(Rv2,S) where d indicates how many variances we allow. (The larger d is, the less likely the element we are looking for will not be in S.) The probability that k is not between v1 and v2 is 2 ∞ d erf(x)dx = O(e−d2/d) If d = log1/2 N, then this probability is o(1/N), specifically O(1/(N log N)). This means that the expected work in this case is O(log−1 N) because O(N) work is performed with very small probability. These mean and variance equations imply Rv1,S ≥ k (s + 1) (N + 1) − d √ s and Rv2,S ≤ k (s + 1) (N + 1) + d √ s This gives equation (A): δ = d √ s = √ s log1/2 N (A) If we first pivot around v2, the cost is N comparisons. If we now partition elements in S that are less than v2 around v1, the cost is Rv2,S, which has expected value k + δ N+1 s+1 . Thus the total cost of partitioning is N + k + δ N+1 s+1 . The cost of the selections to find v1 and v2 in the sample S is O(s). Thus the total expected number of comparisons is N + k + O(s) + O(Nδ/s) The low order term is minimized when s = Nδ/s (B) Weiss Java Solutions pages p. 87 (chap10) Windfall Software, PCA ZzTEX 12.2 88 Chapter 10 Algorithm Design Techniques Combining Equations (A) and (B), we see that s2 = Nδ = √ sN log1/2 N (C) s3/2 = N log1/2 N (D) s = N2/3 log1/3 N (E) δ = N1/3 log2/3 N (F) 10.22 First, we calculate 12*43. In this case, XL = 1, XR = 2, YL = 4, YR = 3, D1 =−1, D2 =−1, XLYL = 4, XRYR = 6, D1D2 = 1, D3 = 11, and the result is 516. Next, we calculate 34*21. In this case, XL = 3, XR = 4, YL = 2, YR = 1, D1 =−1, D2 =−1, XLYL = 6, XRYR = 4, D1D2 = 1, D3 = 11, and the result is 714. Third, we calculate 22*22. Here, XL = 2, XR = 2, YL = 2, YR = 2, D1 = 0, D2 = 0, XLYL = 4, XRYR = 4, D1D2 = 0, D3 = 8, and the result is 484. Finally, we calculate 1234*4321. XL = 12, XR = 34, YL = 43, YR = 21, D1 =−22, D2 =−2. By previous calculations, XLYL = 516, XRYR = 714, and D1D2 = 484. Thus D3 = 1714, and the result is 714 + 171400 + 5160000 = 5332114. 10.23 The multiplication evaluates to (ac − bd) + (bc + ad)i. Compute ac, bd, and (a − b)(d − c) + ac + bd. 10.24 The algebra is easy to verify. The problem with this method is that if X and Y are positive N bit numbers, their sum might be an N + 1 bit number. This causes complications. 10.26 Matrix multiplication is not commutative, so the algorithm couldn’t be used recursively on matrices if commutativity was used. 10.27 If the algorithm doesn’t use commutativity (which turns out to be true), then a divide and conquer algorithm gives a running time of O(Nlog70 143640) = O(N2.795). 10.28 1150 scalar multiplications are used if the order of evaluation is ((A1A2)(((A3A4)A5)A6)) 10.29 (a) Let the chain be a 1 × 1 matrix, a 1 × A matrix, and an A × B matrix. Multiplication by using the first two matrices first makes the cost of the chain A + AB. The alternative method gives a cost of AB + B,soifA > B, then the algorithm fails. Thus a counterexample is multiplying a 1× 1 matrix bya1× 3 matrix by a 3 × 2 matrix. (b, c) A counterexample is multiplying a 1 × 1 matrix by a 1 × 2 matrix by a 2 × 3 matrix. 10.31 The optimal binary search tree is the same one that would be obtained by a greedy strategy: I is at the root and has children and and it; a and or are leaves; the total cost is 2.14. 10.33 This theorem is from F.Yao’s paper, reference [60]. 10.34 A recursive procedure is clearly called for; if there is an intermediate vertex, stopOver on the path from s to t, then we want to print out the path from s to stopOver and then stopOver to t. We don’t want to print out stopOver twice, however, so the method does not print out the first or last vertex on the path and reserves that for the driver. // Print the path between s and t, except do not print // the first or last vertex. Print a trailing " to " only. static void printPath1( int [][] path, int s, int t ) { int stopOver = path[ s ][ t ]; if( s != t && stopOver != -1 ) { Weiss Java Solutions pages p. 88 (chap10) Windfall Software, PCA ZzTEX 12.2 Solutions 89 printPath1( path, s, stopOver ); System.out.print( "" + stopOver+"to"); printPath1( path, stopOver, t ); } } // Assume the existence of a path of length at least 1 static void printPath( int [][] path, int s, int t ) { System.out.print( ""+s+"to"); printPath1( path, s, t ); System.out.println( t ); } 10.38 If the modulus is a power of two, then the least significant bit of the “random” number oscillates. Thus flip will always return heads and tails alternately, and the level chosen for a skip list insertion will always be one or two. Consequently, the performance of the skip list will be (N) per operation. 10.39 (a) 25 ≡ 32 mod 341, 210 ≡ 1mod 341. Since 322 ≡ 1mod 341, this proves that 341 is not prime. We can also immediately conclude that 2340 ≡ 1 mod 341 by raising the last equation to the 34th power. The exponentiation would continue as follows: 220 ≡ 1mod 341, 221 ≡ 2 mod 341, 242 ≡ 4 mod 341, 284 ≡ 16 mod 341, 285 ≡ 32 mod 341, 2170 ≡ 1 mod 341, and 2340 ≡ 1 mod 341. (b) If A = 2, then although 2560 ≡ 1 mod 561, 2280 ≡ 1 mod 561proves that 561 is not prime. If A = 3, then 3561 ≡ 375 mod 561, which proves that 561 is not prime. A = 4 obviously doesn’t fool the al- gorithm, since 4140 ≡ 1 mod 561. A = 5 fools the algorithm: 51 ≡ 5 mod 561, 52 ≡ 25 mod 561, 54 ≡ 64 mod 561, 58 ≡ 169 mod 561, 516 ≡ 511mod 561, 517 = 311mod 561, 534 ≡ 229 mod 561, 535 ≡ 23 mod 561, 570 ≡ 529 mod 561, 5140 ≡ 463 mod 561, 5280 ≡ 67 mod 561, 5560 ≡ 1 mod 561. 10.41 The two point sets are 0, 4, 6, 9, 16, 17 and 0, 5, 7, 13, 16, 17. 10.42 To find all point sets, we backtrack even if found == true, and print out information when line 2 is executed. In the case where there are some duplicate distances, it is possible that several symmetries will be printed. 10.43 10.46 We place circles in order. Suppose we are trying to place circle j of radius rj. If some circle i of radius ri is centered at xi, then j is tangent to i if it is placed at xi + 2√rirsj. To see this, notice that the line connecting the centers has length ri + rj, and the difference in y-coordinates of the centers is |rj − ri|. The difference in x-coordinates follows from the Pythagorean theorem. Weiss Java Solutions pages p. 89 (chap10) Windfall Software, PCA ZzTEX 12.2 90 Chapter 10 Algorithm Design Techniques To place circle j, we compute where it would be placed if it were tangent to each of the first j − 1 circles, selecting the maximum value. If this value is less than rj, then we place circle j at xj. The running time is O(N2). 10.47 Construct a minimum spanning tree T of G, pick any vertex in the graph, and then find a path in T that goes through every edge exactly once in each direction. (This is done by a depth-first search; see Exercise 9.31.) This path has twice the cost of the minimum spanning tree, but it is not a simple cycle. Make it a simple cycle, without increasing the total cost, by bypassing a vertex when it is seen a second time (except that if the start vertex is seen, close the cycle) and going to the next unseen vertex on the path, possibly bypassing several vertices in the process. The cost of this direct route cannot be larger than original because of the triangle inequality. If there were a tour of cost K, then by removing one edge on the tour, we would have a minimum spanning tree of cost less than K (assuming that edge weights are positive). Thus the minimum spanning tree is a lower bound on the optimal traveling salesman tour. This implies that the algorithm is within a factor of 2 of optimal. 10.48 If there are two players, then the problem is easy, so assume k > 1. If the players are numbered 1 through N, then divide them into two groups: 1 through N/2 and N/2 + 1 though N.Ontheith day, for 1≤ i ≤ N/2, player p in the second group plays players ((p + i) mod N/2) + 1 in the first group. Thus after N/2 days, everyone in group 1 has played everyone in group 2. In the last N/2 − 1 days, recursively conduct round-robin tournaments for the two groups of players. 10.49 Divide the players into two groups of size N/2 and N/2, respectively, and recursively arrange the players in any order. Then merge the two lists (declare that px > py if x has defeated y, and py > px if y has defeated x; exactly one is possible) in linear time as is done in mergesort. 10.51 Divide and conquer algorithms (among others) can be used for both problems, but neither is trivial to implement. See the computational geometry references for more information. 10.52 (a) Use dynamic programming. Let Sk = the best setting of words wk, wk+1,...wN, Uk = the ugliness of this setting, and lk = for this setting, (a link to) the word that starts the second line. To compute Sk−1, try putting wk−1, wk,...,wM all on the first line for k < M and M i=k−1 wi < L. Compute the ugliness of each of these possibilities by, for each M, computing the ugliness of setting the first line and adding Um+1. Let M be the value of M that yields the minimum ugliness. Then Uk−1 = this value, and lk−1 = M + 1. Compute values of U and l starting with the last word and working back to the first. The minimum ugliness of the paragraph is U1; the actual setting can be found by starting at l1 and following the links in l since this will yield the first word on each line. Weiss Java Solutions pages p. 90 (chap10) Windfall Software, PCA ZzTEX 12.2 Solutions 91 (b) The running time is quadratic in the case where the number of words that can fit on a line is consistently (N). The space is linear to keep the arrays U and l. If the line length is restricted to some constant, then the running time is linear because only O(1) words can go on a line. (c) Put as many words on a line as can fit. This clearly minimizes the number of lines, and hence the ugliness, as can be shown by a simple calculation. 10.53 An obvious O(N2) solution to construct a graph with vertices 1, 2,...,N and place an edge (v, w) in G iff av < aw. This graph must be acyclic, thus its longest path can be found in time linear in the number of edges; the whole computation thus takes O(N2) time. Let BEST(k) be the increasing subsequence of exactly k elements that has the minimum last element. Let t be the length of the maximum increasing subsequence. We show how to update BEST(k) as we scan the input array. Let LAST(k) be the last element in BEST(k). It is easy to show that if i < j, LAST(i)0&&j>0) { if (M[i][j] == 1 + M[i-1][j-1] && w1.charAt(i-1) == w2.charAt(j-1) ) // there was a match { common = w1.charAt(i-1)+common; i--; j--; } else if (M[i-1][j] > M[i][j-1]) i--; else j--; } return common; } public static void main(String args[]) { String w1 = new String("dynamic"); String w2 = new String("programming"); String w3 = lcs(w1, w2); System.out.println(w3); } } 10.56 (a) A dynamic programming solution resolves part (a). Let FITS(i, s) be 1 if a subset of the first i items sums to exactly s; FITS(i,0) is always 1. Then FITS(x, t) is 1 if either FITS(x − 1, t − ax) or FITS(x − 1, t) is 1, and 0 otherwise. (b) This doesn’t show that P = NP, because the size of the problem is a function of N and log K. Only log K bits are needed to represent K; thus an O(NK) solution is exponential in the input size. 10.57 (a) Let the minimum number of coins required to give x cents in change be COIN(x); COIN(0) = 0. Then COIN(x) is one more than the minimum value of COIN(x − ci), giving a dynamic programming solution. Weiss Java Solutions pages p. 92 (chap10) Windfall Software, PCA ZzTEX 12.2 Solutions 93 (b) Let WAYS(x, i) be the number of ways to make x cents in change without using the first i coin types. If there are N types of coins, then WAYS(x, N) = 0ifx! = 0, and WAYS(0, i) = 1. Then WAYS(x, i − 1) is equal to the sum of WAYS(x − pci, i), for integer values of p no larger than x/ci (but including 0). 10.59 (a) Place eight queens randomly on the board, making sure that no two are on the same row or column. This is done by generating a random permutation of 1..8. There are only 5040 such permutations, and 92 of these give a solution. 10.60 (a) Since the knight leaves every square once, it makes B2 moves. If the squares are alternately colored black and white like a checkerboard, then a knight always moves to a different colored square. If B is odd, then so is B2, which means that at the end of the tour the knight will be on a different colored square than at the start of the tour. Thus the knight cannot be at the original square. 10.61 (a) If the graph has a cycle, then the recursion does not always make progress toward a base case, and thus an infinite loop will result. (b) If the graph is acyclic, the recursive call makes progress, and the algorithm terminates. This could be proved formally by induction. (c) This algorithm is exponential. 10.62 (a) A simple dynamic programming algorithm, in which we scan in row order suffices. Each square computes the maximum sized square for which it could be in the lower right corner. That information is easily obtained from the square that is above it and the square that is to its left. (b) See the April 1998 issue of Dr.Dobb’s Journal for an article by David Vandevoorde on the maximal rectangle problem. 10.67 The final score is 20-16; black wins. Weiss Java Solutions pages p. 93 (chap10) Windfall Software, PCA ZzTEX 12.2 Weiss Java Solutions pages p. 94 (chap11) Windfall Software, PCA ZzTEX 12.2 CHAPTER 11 Amortized Analysis 11.1 When the number of trees after the insertions is more than the number before. 11.2 Although each insertion takes roughly log N, and each deleteMin takes 2 log N actual time, our accounting system is charging these particular operations as 2 for the insertion and 3 log N − 2 for the deleteMin. The total time is still the same; this is an accounting gimmick. If the number of insertions and deleteMins are roughly equivalent, then it really is just a gimmick and not very meaningful; the bound has more significance if, for instance, there are N insertions and O(N/ log N) deleteMins (in which case, the total time is linear). 11.3 Insert the sequence N, N + 1, N − 1, N + 2, N − 2, N + 3,...,1,2N into an initially empty skew heap. The right path has N nodes, so any operation could take (N) time. 11.5 We implement decreaseKey(x) as follows: If lowering the value of x creates a heap order violation, then cut x from its parent, which creates a new skew heap H1 with the new value of x as a root, and also makes the old skew heap (H) smaller. This operation might also increase the potential of H, but only by at most log N. We now merge H and H1. The total amortized time of the merge is O(log N), so the total time of the decreaseKey operation is O(log N). 11.8 For the zig-zig case, the actual cost is 2, and the potential change is Rf (X) + Rf (P) + Rf (G) − Ri(X) − Ri(P) − Ri(G). This gives an amortized time bound of ATzig−zig = 2 + Rf (X) + Rf (P) + Rf (G) − Ri(X) − Ri(P) − Ri(G) Since Rf (X) = Ri(G), this reduces to = 2 + Rf (P) + Rf (G) − Ri(X) − Ri(P) Also, Rf (X)>Rf (P) and Ri(X)
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