Algorithms. Algorithms. Algorithms 3.1 SYMBOL TABLES. API elementary implementations ordered operations

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1 OBT DGWIK KVIN WYN lgorithms OBT DGWIK KVIN WYN Data structures 3.1 YBOL TBL lgorithms F O U T D I T I O N PI elementary implementations mart data structures and dumb code works a lot better than the other way around. ric. aymond Last updated on 3/3/16 8:23 2 ymbol tables lgorithms OBT DGWIK KVIN WYN 3.1 YBOL TBL PI elementary implementations Key-value pair abstraction. Insert a value with specified key. Given a key, search for the corresponding value. x. DN lookup. Insert domain name with specified IP address. Given domain name, find corresponding IP address. domain name IP address key value 4

2 ymbol table applications ymbol tables: context application purpose of search key value dictionary find definition word definition book index find relevant pages term list of page numbers file share find song to download name of song computer ID financial account process transactions account number transaction details web search find relevant web pages keyword list of page names compiler find properties of variables variable name type and value routing table route Internet packets destination best route lso known as: maps, dictionaries, associative arrays. Generalizes arrays. Keys need not be between 0 and N 1. Language support. xternal libraries:, VisualBasic, tandard L, bash,... Built-in libraries: Java, #, ++, cala,... Built-in to language: wk, Perl, PP, Tcl, Javacript, Python, uby, Lua. every array is an associative array every object is an associative array table is the only "primitive" data structure DN find IP address domain name IP address reverse DN find domain name IP address domain name genomics find markers DN string known positions file system find file on disk filename location on disk is_awesome = Python : True, Java : False print is_awesome[ Python ] legal Python code 5 6 Basic symbol table PI ssociative array abstraction. ssociate one value with each key. public class T<Key, Value> onventions Values are not null. java.util.ap allows null values ethod get() returns null if key not present. ethod put() overwrites old value with new value. T() create an empty symbol table void put(key key, Value val) put key-value pair into the table Value get(key key) value paired with key boolean contains(key key) is there a value paired with key? Iterable<Key> keys() all the keys in the table void delete(key key) remove key (and its value) from table boolean ismpty() is the table empty? a[key] = val; a[key] asy to implement contains(). public boolean contains(key key) return get(key)!= null; int size() number of key-value pairs in the table 7 8

3 Keys and values Value type. ny generic type. specify omparable in PI. Key type: several natural assumptions. Life is good. ssume keys are omparable, use compareto(). Life sucks ssume keys are any generic type, use equals() to test equality. ssume keys are any generic type, use equals() to test equality; use hashode() to scramble key (next Wednesday). Life is good again. Best practices. Use immutable types for symbol table keys. Immutable in Java: Integer, Double, tring, java.io.file, utable in Java: tringbuilder, java.net.ul, arrays,... quality test ll Java classes inherit a method equals(). Java requirements. For any references x, y and z: eflexive: x.equals(x) is true. ymmetric: x.equals(y) iff y.equals(x). Transitive: if x.equals(y) and y.equals(z), then x.equals(z). Non-null: x.equals(null) is false. Default implementation. (x == y) do x and y refer to the same object? ustomized implementations. Integer, Double, tring, java.io.file, User-defined implementations. ome care needed. Useful for assignment equivalence relation 9 10 Implementing equals for user-defined types Implementing equals for user-defined types eems easy. eems easy, but requires some care. typically unsafe to use equals() with inheritance (would violate symmetry) public class Date implements omparable<date> private final int month; private final int day; private final int year;... public boolean equals(date that) public final class Date implements omparable<date> private final int month; private final int day; private final int year;... public boolean equals(object y) if (y == this) return true; if (y == null) return false; if (y.getlass()!= this.getlass()) return false; must be Object. Why? xperts still debate. optimize for true object equality check for null objects must be in the same class (religion: getlass() vs. instanceof) if (this.day!= that.day ) return false; if (this.month!= that.month) return false; if (this.year!= that.year ) return false; return true; check that all significant fields are the same Date that = (Date) y; if (this.day!= that.day ) return false; if (this.month!= that.month) return false; if (this.year!= that.year ) return false; return true; cast is guaranteed to succeed check that all significant fields are the same 11 12

4 quals design "tandard" recipe for user-defined types. Optimization for reference equality. heck against null. heck that two objects are of the same type; cast. ompare each significant field: if field is a primitive type, use == if field is an object, use equals() if field is an array, apply to each entry Best practices. No need to use calculated fields that depend on other fields. ompare fields mostly likely to differ first. ake compareto() consistent with equals(). apply rule recursively can use rrays.deepquals(a, b) but not a.equals(b) x.equals(y) if and only if (x.compareto(y) == 0) but use Double.compare() with double (to deal with -0.0 and NaN) e.g., cached anhattan distance Useful for assignment Frequency counter implementation public class Frequencyounter public static void main(tring[] args) T<tring, Integer> st = new T<tring, Integer>(); while (!tdin.ismpty()) tring word = tdin.readtring(); if (!st.contains(word)) st.put(word, 1); else st.put(word, st.get(word) + 1); tring max = ""; st.put(max, 0); for (tring word : st.keys()) if (st.get(word) > st.get(max)) max = word; tdout.println(max + " " + st.get(max)); public class T<Key, Value> void Value boolean print a string with max frequency T() put(key key, Value val) get(key key) contains(key key) create T update frequency of word in T Binary search in an ordered array 3.1 YBOL TBL Data structure. aintain parallel arrays for keys and values, sorted by keys. earch. Use binary search to find key. Proposition. t most ~ lg N compares to search a sorted array of length N. lgorithms PI elementary implementations get("p") keys[] L P vals[] keys[] OBT DGWIK KVIN WYN 16

5 Binary search in an ordered array lementary symbol tables: quiz 1 Data structure. aintain parallel arrays for keys and values, sorted by keys. earch. Use binary search to find key. public Value get(key key) int lo = 0, hi = N-1; while (lo <= hi) int mid = lo + (hi - lo) / 2; int cmp = key.compareto(keys[mid]); if (cmp < 0) hi = mid - 1; else if (cmp > 0) lo = mid + 1; else if (cmp == 0) return vals[mid]; return null; no matching key kipped in class Implementing binary search was. asier than I thought. B. bout what I expected.. arder than I thought. D. uch harder than I thought.. I don't know Binary search: insert lementary T implementations: summary Data structure. aintain an ordered array of key-value pairs. Insert. Use binary search to find place to insert; shift all larger keys over. Proposition. Takes linear time in the worst case. implementation guarantee average case search insert search hit insert operations on keys put("p", 10) keys[] vals[] P sequential search (unordered array or list) N N N N equals() binary search (ordered array) log N N log N N compareto() can do with log N compares, but requires N array accesses hallenge. fficient implementations of both search and insert

6 xamples of ordered symbol table PI keys values lgorithms OBT DGWIK KVIN WYN 3.1 YBOL TBL PI elementary implementations min() get(09:00:13) floor(09:05:00) select(7) keys(09:15:00, 09:25:00) ceiling(09:30:00) max() 09:00:00 hicago 09:00:03 Phoenix 09:00:13 ouston 09:00:59 hicago 09:01:10 ouston 09:03:13 hicago 09:10:11 eattle 09:10:25 eattle 09:14:25 Phoenix 09:19:32 hicago 09:19:46 hicago 09:21:05 hicago 09:22:43 eattle 09:22:54 eattle 09:25:52 hicago 09:35:21 hicago 09:36:14 eattle 09:37:44 Phoenix size(09:15:00, 09:25:00) is 5 rank(09:10:25) is 7 22 Ordered symbol table PI ank in a sorted array public class T<Key extends omparable<key>, Value> Key min() smallest key Key max() largest key Key floor(key key) largest key less than or equal to key Key ceiling(key key) smallest key greater than or equal to key int rank(key key) number of keys less than key Key select(int k) key of rank k Problem. Given a sorted array of N distinct keys, find the number of keys strictly less than a given query key. easy modification to binary search public Value get(key key) int lo = 0, hi = N-1; while (lo <= hi) int mid = lo + (hi - lo) / 2; int cmp = key.compareto(keys[mid]); if (cmp < 0) hi = mid - 1; else if (cmp > 0) lo = mid + 1; else if (cmp == 0) return vals[mid]; return null; lo public int rank(key key) mid 23 24

7 Binary search: ordered symbol table operations summary lgorithms OBT DGWIK KVIN WYN sequential search binary search search N log N insert N N min / max N 1 floor / ceiling N log N rank N log N select N 1 lgorithms F O U T D I T I O N 3.2 BINY T BTs iteration deletion (see book) OBT DGWIK KVIN WYN order of growth of the running time for ordered symbol table operations 25 Last updated on 3/3/16 8:23 Binary search trees earch tree lgorithms OBT DGWIK KVIN WYN 3.2 BINY T BTs iteration deletion Definition. BT is a binary tree in symmetric order. binary tree is either: mpty. Two disjoint binary trees (left and right). earch tree. ach node has a key, and every node s key is: Larger than all keys in its left subtree. maller than all keys in its right subtree. a subtree left link of a left link null links Binary search tree = Binary (search tree) = a search tree that s binary also (Binary search) tree = a tree that supports binary search Q. What are the differences between a heap and a binary search tree? parent of and keys smaller than 9 root right child of root key keys larger than value associated with 28

8 BT representation in Java BT implementation (skeleton) Java definition. BT is a reference to a root Node. Node is composed of four fields: Key and a Value. reference to the left and right subtree. public class BT<Key extends omparable<key>, Value> private Node root; private class Node /* see previous slide */ root of BT smaller keys larger keys public void put(key key, Value val) /* see next slides */ private class Node private Key key; private Value val; private Node left, right; public Node(Key key, Value val) this.key = key; this.val = val; BT Node key left val right BT with smaller keys BT with larger keys Binary search tree public Value get(key key) /* see next slides */ public Iterable<Key> iterator() /* see slides in next section */ public void delete(key key) /* see textbook */ Key and Value are generic types; Key is omparable BT earch BT earch earch (get). epeat: If less, if greater, if equal, if, search miss earch (get). epeat: If less, go left; if greater, go right; if equal, return value (search hit) if null, return null (search miss) only keys are shown only keys are shown 31 32

BT search: Java implementation BT put: non-recursive implementation Get. eturn value corresponding to given key, or null if no such key. public Value get(key key) Node x = root; while (x!

9 BT search: Java implementation BT put: non-recursive implementation Get. eturn value corresponding to given key, or null if no such key. public Value get(key key) Node x = root; while (x!= null) int cmp = key.compareto(x.key); if (cmp < 0) x = x.left; else if (cmp > 0) x = x.right; else if (cmp == 0) return x.val; return null; ost. Number of compares = 1 + depth of node. kipped in class epeat: If less, if greater, if equal, if null, G put G only keys are shown BT put: non-recursive implementation epeat: If less, go left if greater, go right if equal, replace value and return if null, insert new node and return G put G only keys are shown BT put: tricky recursive Java implementation Put. ssociate value with key. public void put(key key, Value val) root = put(root, key, val); private Node put(node x, Key key, Value val) if (x == null) return new Node(key, val); int cmp = key.compareto(x.key); if (cmp < 0) x.left = put(x.left, key, val); else if (cmp > 0) x.right = put(x.right, key, val); else if (cmp == 0) x.val = val; return x; Warning: concise but tricky code; read carefully! ost. Number of compares = 1 + depth of node. kipped in class 35 36

10 BT practice Q. Draw the tree when the following keys are inserted:, L, O,, P, I, G, Q. Draw the tree when the following keys are inserted:,, G, I, L, O, P, Tree shape any BTs correspond to same set of keys. Number of compares for search/insert = 1 + depth of node. best case typical case Bottom line. Tree shape depends on order of insertion. worst case BT insertion: random order visualization T implementations: summary x. Insert keys in random order. ~ 2 ln N. implementation guarantee average case search insert search hit insert operations on keys sequential search (unordered list) N N N N equals() binary search (ordered array) log N N log N N compareto() BT N N log N log N compareto() Why not shuffle to ensure a (probabilistic) guarantee of log N? xpected node depth ~ 2 ln N

11 Binary search trees: inorder traversal lgorithms OBT DGWIK KVIN WYN 3.2 BINY T BTs iteration deletion In what order does the traverse(root) code print out the keys in the BT? private void traverse(node x) if (x == null) return; traverse(x.left); tdout.println(x.key); traverse(x.right);. B.. D.. None of the above. Practice 42 Inorder traversal Traverse left subtree. nqueue key. Traverse right subtree. public Iterable<Key> keys() Queue<Key> q = new Queue<Key>(); inorder(root, q); return q; private void inorder(node x, Queue<Key> q) if (x == null) return; inorder(x.left, q); q.enqueue(x.key); inorder(x.right, q); BT key val left right BT with smaller keys BT with larger keys smaller keys, in order key larger keys, in order all keys, in order Binary search trees: quiz 1 Given N distinct keys, what is the name of this sorting algorithm? 1. huffle the keys. 2. Insert the keys into a BT, one at a time. 3. Do an inorder traversal of the BT.. Insertion sort. B. ergesort.. Quicksort. D. None of the above.. I don't know. Trick question Property. Inorder traversal of a BT yields keys in ascending order

12 orrespondence between BTs and quicksort partitioning BTs: mathematical analysis Proposition. If N distinct keys are inserted into a BT in random order, the expected number of compares for a search/insert is ~ 2 ln N. P Pf. 1 1 correspondence with quicksort partitioning. T D O U I Y L But Worst-case height is N 1. [ when client provides keys, they may not be in random order, and we have no control over probability of worst case ] emark. orrespondence is 1 1 if array has no duplicate keys Binary search trees: preorder traversal Binary search trees: postorder traversal In what order does the traverse(root) code print out the keys in the BT? private void traverse(node x) if (x == null) return; tdout.println(x.key); traverse(x.left); traverse(x.right); In what order does the traverse(root) code print out the keys in the BT? private void traverse(node x) if (x == null) return; traverse(x.left); traverse(x.right); tdout.println(x.key);.. B.. Practice B.. Practice D. D.. None of the above.. None of the above

13 Level-order traversal of a binary tree equired order: Process root. Process children of root, from left to right. Process grandchildren of root, from left to right. Useful for assignment 3.2 BINY T T queue.enqueue(root); while (!queue.ismpty()) Node x = queue.dequeue(); if (x == null) continue; tdout.println(x.item); queue.enqueue(x.left); queue.enqueue(x.right); lgorithms OBT DGWIK KVIN WYN BTs iteration deletion level order traversal: T 49 inimum and maximum inimum. mallest key in BT. aximum. Largest key in BT. Floor and ceiling Floor. Largest key in BT query key. eiling. mallest key in BT query key. min () m max floor(g) () m ceiling(q) floor(d) Q. ow to find the min / max? Q. ow to find the floor / ceiling? 51 52

14 omputing the floor omputing the floor Floor. Largest key in BT k? ase 1. [ key in node x = k ] The floor of k is k. ase 2. [ key in node x > k ] The floor of k is in the left subtree of x. ase 3. [ key in node x < k ] The floor of k can't be in left subtree of x: it is either in the right subtree of x or it is the key in node x. finding floor() finding floor(g) is less than G so floor(g) could be on the right is greater than G so floor(g) must be on the left omputing the floor function 53 public Key floor(key key) return floor(root, key); private Key floor(node x, Key key) if (x == null) return null; int cmp = key.compareto(x.key); if (cmp == 0) return x; if (cmp < 0) return floor(x.left, key); Key t = floor(x.right, key); if (t!= null) return t; else return x.key; floor(g)in left subtree is null an out of time about here in class finding floor(g) G is less than so floor(g) must be on the left G is greater than so floor(g) could be on the right result omputing the floor function 54 ank and select BT implementation: subtree counts Q. ow to implement rank() and select() efficiently for BTs?. In each node, store the number of nodes in its subtree. 2 subtree count private class Node private Key key; private Value val; private Node left; private Node right; private int count; private Node put(node x, Key key, Value val) number of nodes in subtree public int size() return size(root); private int size(node x) if (x == null) return new Node(key, val, 1); int cmp = key.compareto(x.key); if (cmp < 0) x.left = put(x.left, key, val); else if (cmp > 0) x.right = put(x.right, key, val); else if (cmp == 0) x.val = val; x.count = 1 + size(x.left) + size(x.right); return x; if (x == null) return 0; return x.count; ok to call when x is null initialize subtree count to 1 56

15 omputing the rank ank ank. ow many keys in BT < k? ase 1. [ k < key in node ] Keys in left subtree? Key in node? 0 Keys in right subtree? 0 ase 2. [ k > key in node ] Keys in left subtree? Key in node. 1 count all Keys in right subtree? count ase 3. [ k = key in node ] Keys in left subtree? Key in node. 0 Keys in right subtree? 0 count node count ank. ow many keys in BT < k? asy recursive algorithm (3 cases!) public int rank(key key) return rank(key, root); private int rank(key key, Node x) if (x == null) return 0; int cmp = key.compareto(x.key); node count if (cmp < 0) return rank(key, x.left); else if (cmp > 0) return 1 + size(x.left) + rank(key, x.right); else if (cmp == 0) return size(x.left); BT: ordered symbol table operations summary T implementations: summary sequential search binary search BT implementation guarantee average case search insert search hit insert ordered ops? key interface search N log N h sequential search (unordered list) N N N N equals() insert N N h min / max N 1 h floor / ceiling N log N h rank N log N h h = height of BT (proportional to log N if keys inserted in random order) binary search (ordered array) log N N log N N compareto() BT N N log N log N compareto() red-black BT log N log N log N log N compareto() select N 1 h ordered iteration N log N N N order of growth of running time of ordered symbol table operations Next lecture. Guarantee logarithmic performance for all operations

3.2 BINARY SEARCH TREES. BSTs ordered operations iteration deletion. Algorithms ROBERT SEDGEWICK KEVIN WAYNE.

3.2 BINARY SEARCH TREES. BSTs ordered operations iteration deletion. Algorithms ROBERT SEDGEWICK KEVIN WAYNE. 3.2 BINY T lgorithms BTs ordered operations iteration deletion OBT DGWIK KVIN WYN http://algs4.cs.princeton.edu Binary search trees Definition. BT is a binary tree in symmetric order. binary tree is either: