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Data Divide-And-Conquer Structures Divide-And-Conquer Data Structures Taking Stock Last Time Master Theorem Practice Some Sorting Algs Beginning of Java Collections Interfaces Jeff Linderoth Department of Industrial and Systems Engineering Lehigh University IE170: Algorithms in Systems Engineering: Lecture 7 This Time Hashes Trees and Binary Search Trees More on Java Collections Interfaces lab == fun January 29, 2007 Jeff Linderoth Divide-And-Conquer Data Structures IE170:Lecture 7 Master Theorem Jeff Linderoth Divide-And-Conquer Data Structures IE170:Lecture 7 Master Theorem The Java Collections Interfaces In the remainder of the class, we will be using the Java Collections Interface: http://java.sun.com/docs/books/ tutorial/collections/TOC.html Important: Most of what I will say only works if you set the "code level" to Java 5.0 in eclipse! Preferences, Java Compiler: Set this to 5.0 The interfaces form a hierarchy: (A subset of) the Collections Interface public interface Collection<E> extends Iterable<E> { // Basic operations int size(); boolean isEmpty(); boolean contains(Object element); boolean add(E element); //optional boolean remove(Object element); //optional Iterator<E> iterator(); // Array operations Object[] toArray(); <T> T[] toArray(T[] a); } Jeff Linderoth IE170:Lecture 7 Jeff Linderoth IE170:Lecture 7 Divide-And-Conquer Data Structures Master Theorem Divide-And-Conquer Data Structures Master Theorem Set Hash? A Set is a Collection that cannot contain duplicate elements. It models the mathematical set abstraction. The Set interface contains only methods inherited from Collection and adds the restriction that duplicate elements are prohibited. Set is still an interface. There are 3 implementations of Set in Java. HashSet TreeSet LinkedHashSet No, Cheech. A hash table is a data structure in which we can "look up" (or search) for an element efficiently. The expected time to search for an element in a has table in O(1). (Worst case time in (n)). Think of a hash table as an array With a regular array, we find the element whose "key" is j in position j of the array. j = 17; val = a[j]; . This is called direct addressing and it takes O(1) on your regular ol' random access computer. This form of direct addressing works when we can afford to have an array with one position for every possible key Jeff Linderoth Divide-And-Conquer Data Structures IE170:Lecture 7 Master Theorem Jeff Linderoth Divide-And-Conquer Data Structures IE170:Lecture 7 Master Theorem Example More on Hash In a hash table the number of keys stored is small relative to the number of possible keys A hash table is an array. Given a key k, we don't use k as the index into the array rather, we have a hash function h, and we use h(k) as an index into the array. Given a "universe" of keys K. Think of K as all the words in a dictionary, for example This look great. However, what happens if h(k1 ) = h(k2 ) for k1 = k2 ? Two keys hash to the same value. The elements collide This is typically handled by chaining Instead of storing a key k (or later key value pair (k, v)) at every position in the array, we store a linked list of keys. Example: h : K {0, 1, . . . m - 1}, so that h(k) gets mapped to an integer between 0 and m - 1 for every k K We say that k hashes to h(k) Jeff Linderoth IE170:Lecture 7 Jeff Linderoth IE170:Lecture 7 Divide-And-Conquer Data Structures Master Theorem Divide-And-Conquer Data Structures Master Theorem A (Fairly) Obvious point BAD hash function. h(k) = 3. If all keys hash to the same value, then looking up a key takes (n). (Since it is just a list). We would like a hash function to be "random" in the sense that a key k is equally likely to has into any of the m slots in the hash table (array). If we have such a function, then we can show that the average n time required to search for a key is (1 + m ) When hashing keys that are not numbers, you must convert them to numbers, e.g.: beer = -142 + 24 + 53 + 52 + 181 = 42. Average Hash Search Time The number of elements to be searched is 1 more than the number of elements that appear before x in x s list. Assuming we insert items into the list at the beginning, then this is the number of elements that were inserted after x. By definition: P(h(ki ) = h(kj )) = 1 m Let Xij be indicator random variable that is equal to one if and only if h(ki ) = h(kj ) Then just compute: E 1 n n i=1 1 + n j=i+1 Xij . Jeff Linderoth Divide-And-Conquer Data Structures IE170:Lecture 7 Master Theorem Jeff Linderoth Divide-And-Conquer Data Structures IE170:Lecture 7 Master Theorem Hash Functions Modular Hash Function Let m be (roughly) the size of your hash table: h(k) = k mod m Good choice of m: A prime number not too close to an exact power of 2 Multiplicative Hash Function h(k) = m(kA mod 1) Multiply key k by A, take fractional part, and multiply by m If m = 2p this can be done very fast with bit shifting A = ( 5 - 1)/2 seems a good value Back to the Java Collections So now you know what a Java HashSet is. A LinkedHashSet is a HashSet that also keeps track of the order in which elements were inserted. (Think of laying a linked list on top of the Hash Table) A TreeSet stores its elements in a alertred-black tree. In order to understand red-black trees, we must know about binary search trees. Hash table is "good" at insert(), search(), delete(). But what if you also want to support (efficiently) minimum(), maximum() Jeff Linderoth IE170:Lecture 7 Jeff Linderoth IE170:Lecture 7 Divide-And-Conquer Data Structures Java Collections Divide-And-Conquer Data Structures Java Collections Trees A tree is a set of items organized into a hierarchical structure (think of a family tree). When organized in this way, we call the items nodes. Each node has a single designated parent and one or more children. There is a single designated node, called the root, with no parent. Any node with no children is called a leaf. Any node with children is called internal. A tree in which all nodes have 2 or fewer children is called a binary tree. Storing a list of items in a tree structure allows us to represent additional relationships among the items in the list. Jeff Linderoth Divide-And-Conquer Data Structures IE170:Lecture 7 Java Collections Binary Tree Data Structures To store a tree of keys k, or maybe (key, value) pairs: (k, v), we need a data structure supporting three basic operations left l: Points to the left child right r: Points to the right child parent p: Points to the parent This allows us to traverse tree the and perform other operations on it. The level of a node in the tree is the number of recursive calls to parent() needed to reach the root. The depth of the tree is the maximum level of any of its nodes. A balanced tree is one in which all leaves are at levels k or k - 1, where k is the depth of the tree. Jeff Linderoth Divide-And-Conquer Data Structures IE170:Lecture 7 Java Collections Data Structures for Storing Trees Array The root is stored in position 0. The children of the node in position i are stored in positions 2i + 1 and 2i + 2. This determines a unique storage location for every node in the tree and makes it easy to find a node's parent and children. Using an array, the basic operations can be performed very efficiently. If the tree is unbalanced or dynamic, a linked list may be better. Data Structures for Storing Trees Linked List In a linked list, each item is stored along with explicit pointers to its parent and children. This allows for easy addition and deletion of nodes from the tree. Jeff Linderoth IE170:Lecture 7 Jeff Linderoth IE170:Lecture 7 Divide-And-Conquer Data Structures Java Collections Divide-And-Conquer Data Structures Java Collections Binary Search Tree Binary Search Trees There are lots of binary trees that can satisfy this property. A binary search tree is a data structrue that is conceptualized as a binary tree, but has one additional property: It is obvious that the number of binary tree on n nodes bn is bn = Binary Search Tree Property If y is in the left subtree of x, then k(y) k(x) 1 n+1 2n n bn = 4n (1 + O(1/n)) n3/2 And not all of these (exponentially many) are created equal. In fact, we would like to keep our binary search trees "short", because most of the operations we would like to support are a function of the height h of the tree. Jeff Linderoth Divide-And-Conquer Data Structures IE170:Lecture 7 Java Collections Jeff Linderoth Divide-And-Conquer Data Structures IE170:Lecture 7 Java Collections Operations Short Is Beautiful search() takes O(h) minimum(), maximum() also take O(h) Slightly less obvious is that insert(), delete() also take O(h) Thus we would like to keep out binary search trees "short" (h is small). successor(x) How would I know "next biggest" element? If right subtree is not empty: minimum(r(x)) If right subtree is empty: Walk up tree until you make the first "right" move insert(x) Just walk down the tree and put it in. It will go "at the bottom" Jeff Linderoth IE170:Lecture 7 Jeff Linderoth IE170:Lecture 7 Divide-And-Conquer Data Structures Java Collections Divide-And-Conquer Data Structures Java Collections delete() red-black Trees red-black trees are simply a way to keep binary search trees short. (Or balanced) If 0 or 1 child, deletion is fairly easy If 2 children, deletion is made easier by the following fact: Binary Search Tree Property If a node has 2 children, then its successor will not have a left child its predecessor will not have a right child Balanced here means that no path on the tree is more than twice as long as another path. An implication of this is that its maximum height is 2 lg(n + 1) search(), minimum(), maximum(), all take O(lg n) It's implementation is complicated, so we won't cover it insert(): also runs in O lg(n) delete(): runs in O lg(n) (but it is more complicated to maintain the "red-black" property) Jeff Linderoth Divide-And-Conquer Data Structures IE170:Lecture 7 Java Collections Jeff Linderoth Divide-And-Conquer Data Structures IE170:Lecture 7 Java Collections Back to the Java Collections Lists red-black trees remain sorted You don't really have any control over the order in which things will appear in a HashSet If you care about that you should use a LinkedHashSet, which lays a linked list on top of the HashSet In general, Sets are not for ordered collections of items, for that, you should use a list A List is an ordered Collection (sometimes called a sequence). Lists may contain duplicate elements. In addition to the operations inherited from Collection, the List interface includes operations for the following: Positional access: manipulate elements based on their numerical position in the list Search: searches for a specified object in the list and returns its numerical position Jeff Linderoth IE170:Lecture 7 Jeff Linderoth IE170:Lecture 7 Divide-And-Conquer Data Structures Java Collections Divide-And-Conquer Data Structures Java Collections (Subset of) List Interface public interface List<E> extends Collection<E> { // Positional access E get(int index); E set(int index, E element); //optional boolean add(E element); //optional void add(int index, E element); //optional E remove(int index); //optional // Search int indexOf(Object o); int lastIndexOf(Object o); // Iteration ListIterator<E> listIterator(); ListIterator<E> listIterator(int index); } Jeff Linderoth Divide-And-Conquer Data Structures IE170:Lecture 7 Java Collections Java List Implementations Two List Implementations 1 2 ArrayList: which is usually the better-performing LinkedList: offers better performance under certain circumstances, (i.e. if lots of add/remove in the middle if the list) Jeff Linderoth Divide-And-Conquer Data Structures IE170:Lecture 7 Java Collections Java Lists have extended iterators List Stuff ListIterator<E> listIterator(): gives iterator at beginning ListIterator<E> listIterator(int index): gives iterator at specified index The index refers to the element that would be returned by an initial call to next() The cursor is always between two elements: the one that would be returned by a call to previous() the one that would be returned by a call to next() public interface ListIterator<E> extends Iterator<E> { boolean hasNext(); E next(); boolean hasPrevious(); E previous(); int nextIndex(); int previousIndex(); void remove(); //optional void set(E e); //optional void add(E e); //optional } The n + 1 valid index values correspond to the n + 1 gaps between elements, from the gap before the first element to the gap after the last one. Jeff Linderoth IE170:Lecture 7 Jeff Linderoth IE170:Lecture 7 Divide-And-Conquer Data Structures Java Collections Next Time A bit on Java Collection Map Interface Move on to Heaps (Chapter 6) We have covered chapters 1-4, 10-11, and Appendices A and B News New Homework Posted! Let's have a little quiz on 2/7 Homework is due 2/5: No late homework accepted. (I need to hand out solutions and discuss in class on 2/5). Jeff Linderoth IE170:Lecture 7
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Path: Lehigh >> IE >> 170 Spring, 2007
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Path: Lehigh >> IE >> 170 Spring, 2007
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Path: Lehigh >> IE >> 170 Spring, 2007
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Path: Lehigh >> CSE >> 450 Spring, 2008
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Path: Lehigh >> CSE >> 432 Fall, 2008
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Path: Lehigh >> CSE >> 450 Spring, 2008
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Path: Lehigh >> CSE >> 450 Spring, 2008
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Path: Lehigh >> CSE >> 450 Spring, 2008
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Path: Lehigh >> CSE >> 450 Spring, 2008
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Path: Lehigh >> CSE >> 450 Spring, 2008
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Path: Lehigh >> CSE >> 450 Spring, 2008
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Path: Lehigh >> CSE >> 450 Spring, 2008
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Path: Lehigh >> CSE >> 450 Spring, 2008
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Path: Lehigh >> CSE >> 450 Spring, 2008
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Path: Lehigh >> CSE >> 450 Spring, 2008
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Path: UC Davis >> NPB >> 114 Spring, 2008
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Path: Lehigh >> CSE >> 450 Spring, 2008
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Path: Lehigh >> CSE >> 450 Spring, 2008
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Path: Lehigh >> IE >> 426 Fall, 2006
Description: 1 4 ...
sudoku2
Path: Lehigh >> IE >> 426 Fall, 2006
Description: 1 2 2 3 ...
jeffdoku
Path: Lehigh >> IE >> 426 Fall, 2006
Description: 6 8 2 1 4 5 6 1 7 6 3 1 4 2 6 7 5 7 5 6 8 9 ...
jeffdoku
Path: Lehigh >> IE >> 426 Fall, 2006
Description: 6 8 2 1 4 5 6 1 7 6 3 1 4 2 6 7 5 7 5 6 8 9 ...
Practice%20MT%202
Path: UC Davis >> NPB >> 114 Spring, 2008
Description: NPB 114 Practice MT#2 Matching (1 pt each) a. Acinar cell b. Endocrine cell c. Parietal cell _ _ _ d. Chief cell e. Mucous cell 1. Produces an alkaline fluid to protect the stomach 2. Its product is released into the bloodstream 3. Produces pepsinog...
sudoku
Path: Lehigh >> IE >> 426 Fall, 2006
Description: 6 8 2 8 6 7 5 7 4 1 3 4 5 6 5 1 4 7 3 9 1 4 2 2 5 6 8 9 7 6 ...
sudoku
Path: Lehigh >> IE >> 426 Fall, 2006
Description: 6 8 2 8 6 7 5 7 4 1 3 4 5 6 5 1 4 7 3 9 1 4 2 2 5 6 8 9 7 6 ...
news
Path: Lehigh >> IE >> 426 Fall, 2006
Description: Scenario Mean Stdev Buy Optimal q c r 100 30 100 85 0.7 0.5 0.05 YOUR CHOICE OPTIMAL Demand Sell Salvage Profit Sell Salvage 1 121 100 0 70 85 0 2 71 71 29 51.15 71 14 3 110 100 0 70 85 0 67 67 33 48.55 67 18 59 59 41 43.35 59 26 51 51 49 38.15 51 3...
hw1-survey
Path: Lehigh >> IE >> 426 Fall, 2006
Description: Informal Homework Survey September 14, 2006 Please answer the following questions. This is an anonymous survey, but even if it wasn\'t, I wouldn\'t hold your answers against you. Difficulty On a scale of 1-10, with a 10 being \"I hate you. Why are you ...
survey
Path: Lehigh >> IE >> 426 Fall, 2006
Description: IE426 Course Survey-Quiz #0 Name: email: Background Mathematics Mathematicians are like Frenchmen: whatever you say to them they translate into their own language and forthwith it is something entirely different.\" -Johann Wolfgang von Goethe Please...
wap
Path: Lehigh >> IE >> 426 Fall, 2006
Description: IE 426 Case Study Integer Programming 1 Wireless Capacity Expansion Planning Note: This is a real consulting problem. The names have been changed to protect the innocent. Prof. Linderoth will be acting as the client. You have been contracted by a...
