Trees (Ch. 10.2) Longin Jan Latecki Temple University based on slides by Simon Langley, Shang-Hua Teng, and William Alb

1. Trees (Ch. 10.2)Longin Jan LateckiTemple Universitybased on slides bySimon Langley, Shang-Hua Teng, and William Albritton

2. Section 10.2: Applications of Trees • Binary search trees • A simple data structure for sorted lists • Decision trees • Minimum comparisons in sorting algorithms • Prefix codes • Huffman coding

3. Basic Data Structures - Trees • Informal: a tree is a structure that looks like a real tree (up-side-down) • Formal: a tree is a connected graph with no cycles.

4. Trees - Terminology size=7 root subtree x value b e m height=2 c d a nodes leaf Every node must have its value(s) Non-leaf node has subtree(s) Non-root node has a single parent node

5. Types of Tree Binary Tree Each node has at most 2 sub-trees m-ary Trees Each node has at most m sub-trees

6. Binary Search Trees A binary search tree: • … is a binary tree. • if a node has value N, all values in its left sub-tree are less than or equal to N, and all values in its right sub-tree are greater than N.

7. Binary Search Tree Format Example: • Items are stored at individual tree nodes. • We arrange for the tree to always obey this invariant: • For every item x, • Every node in x’s left subtree is less than x. • Every node in x’s right subtree is greater than x. 7 3 12 1 5 9 15 0 2 8 11

8. This is NOT a binary search tree

9. This is a binary search tree

10. Searching a binary search tree Time per level search(t, s) { If(s == label(t)) return t; If(t is leaf) return null If(s < label(t)) search(t’s left tree, s) else search(t’s right tree, s)} O(1) O(1) h Total O(h)

11. Searching a binary search tree Time per level search( t, s ) { while(t != null) { if(s == label(t)) return t; if(s < label(t) t = leftSubTree(t); else t = rightSubTree(t); } return null; O(1) O(1) h Total O(h)

12. Here’s another function that does the same (we search for label s): TreeSearch(t, s) while (t != NULL and s != label[t]) if (s < label[t]) t = left[t]; else t = right[t]; return t;

13. Insertion in a binary search tree:we need to search before we insert Insert 6 Insert 11 6 11 6 11 6 6 11 always insert to a leaf ? Time complexity O(height_of_tree) n = size of the tree O(log n) if it is balanced

14. Insertion insertInOrder(t, s) { if(t is an empty tree) // insert here return a new tree node with value s else if( s < label(t)) t.left = insertInOrder(t.left, s ) else t.right = insertInOrder(t.right, s) return t }

15. Recursive Binary Tree Insert procedure insert(T: binary tree, x: item)v := root[T]if v = null then begin root[T] := x; return “Done” endelse if v = xreturn “Already present”else if x < vthenreturn insert(leftSubtree[T], x)else{must be x > v}return insert(rightSubtree[T], x)

16. Comparison –Insertion in an ordered list insertInOrder(list, s) { loop1: search from beginning of list, look for an item >= s loop2: shift remaining list to its right, start from the end of list insert s } Insert 6 6 6 6 6 8 9 6 2 3 4 5 7 6 7 8 9 Time complexity? O(n) n = size of the list

17. Try it!! • Build binary search trees for the following input sequences • 7, 4, 2, 6, 1, 3, 5, 7 • 7, 1, 2, 3, 4, 5, 6, 7 • 7, 4, 2, 1, 7, 3, 6, 5 • 1, 2, 3, 4, 5, 6, 7, 8 • 8, 7, 6, 5, 4, 3, 2, 1

18. Decision Trees • A decision tree represents a decision-making process. • Each possible “decision point” or situation is represented by a node. • Each possible choice that could be made at that decision point is represented by an edge to a child node. • In the extended decision trees used in decision analysis, we also include nodes that represent random events and their outcomes.

19. Coin-Weighing Problem • Imagine you have 8 coins, oneof which is a lighter counterfeit, and a free-beam balance. • No scale of weight markings is required for this problem! • How many weighings are needed to guarantee that the counterfeit coin will be found? ?

20. As a Decision-Tree Problem • In each situation, we pick two disjoint and equal-size subsets of coins to put on the scale. A given sequence ofweighings thus yieldsa decision tree withbranching factor 3. The balance then“decides” whether to tip left, tip right, or stay balanced.

21. Applying the Tree Height Theorem • The decision tree must have at least 8 leaf nodes, since there are 8 possible outcomes. • In terms of which coin is the counterfeit one. • Recall the tree-height theorem, h≥logm. • Thus the decision tree must have heighth≥ log38 = 1.893… = 2. • Let’s see if we solve the problem with only 2 weightings…

22. General Solution Strategy • The problem is an example of searching for 1 unique particular item, from among a list of n otherwise identical items. • Somewhat analogous to the adage of “searching for a needle in haystack.” • Armed with our balance, we can attack the problem using a divide-and-conquer strategy, like what’s done in binary search. • We want to narrow down the set of possible locations where the desired item (coin) could be found down from n to just 1, in a logarithmic fashion. • Each weighing has 3 possible outcomes. • Thus, we should use it to partition the search space into 3 pieces that are as close to equal-sized as possible. • This strategy will lead to the minimum possible worst-case number of weighings required.

23. Coin Balancing Decision Tree • Here’s what the tree looks like in our case: 123 vs 456 left: 123 balanced:78 right: 456 4 vs. 5 1 vs. 2 7 vs. 8 L:1 L:4 L:7 R:2 B:3 R:5 B:6 R:8

24. General Balance Strategy • On each step, putn/3of the n coins to be searched on each side of the scale. • If the scale tips to the left, then: • The lightweight fake is in the right set ofn/3≈ n/3 coins. • If the scale tips to the right, then: • The lightweight fake is in the left set of n/3≈ n/3coins. • If the scale stays balanced, then: • The fake is in the remaining set ofn− 2n/3 ≈ n/3coins that were not weighed! You can prove that this strategy always leads to a balanced 3-ary tree.

25. Data Compression • Suppose we have 3GB character data file that we wish to include in an email. • Suppose file only contains 26 letters {a,…,z}. • Suppose each letter a in {a,…,z} occurs with frequency fa. • Suppose we encode each letter by a binary code • If we use a fixed length code, we need 5 bits for each character • The resulting message length is • Can we do better?

26. Data Compression: A Smaller Example • Suppose the file only has 6 letters {a,b,c,d,e,f} with frequencies • Fixed length 3G=3000000000 bits • Variable length Fixed length Variable length

27. How to decode? • At first it is not obvious how decoding will happen, but this is possible if we use prefix codes

28. Prefix Codes • No encoding of a character can be the prefix of the longer encoding of another character: • We could not encode t as 01 and x as 01101 since 01 is a prefix of 01101 • By using a binary tree representation we generate prefix codes with letters as leaves

29. Decoding prefix codes • Follow the tree until it reaches to a leaf, and then repeat! • A message can be decoded uniquely!

30. Prefix codes allow easy decoding Decode: 11111011100 s 1011100 sa 11100 san 0 sane

31. Some Properties • Prefix codes allow easy decoding • An optimal code must be a full binary tree (a tree where every internal node has two children) • For C leaves there are C-1 internal nodes • The number of bits to encode a file is where f(c) is the freq of c, lengthT(c) is the tree depth of c, which corresponds to the code length of c

32. Optimal Prefix Coding Problem • Given is a set of n letters (c1,…, cn) with frequencies (f1,…, fn). • Construct a full binary tree T to define a prefix code that minimizes the average code length

33. Greedy Algorithms • Many optimization problems can be solved using a greedy approach • The basic principle is that local optimal decisions may be used to build an optimal solution • But the greedy approach may not always lead to an optimal solution overall for all problems • The key is knowing which problems will work with this approach and which will not • We study • The problem of generating Huffman codes

34. Greedy algorithms • A greedy algorithm always makes the choice that looks best at the moment • My everyday examples: • Driving in Los Angeles, NY, or Boston for that matter • Playing cards • Invest on stocks • Choose a university • The hope: a locally optimal choice will lead to a globally optimal solution • For some problems, it works • Greedy algorithms tend to be easier to code

35. David Huffman’s idea • A Term paper at MIT • Build the tree (code) bottom-up in a greedy fashion Each tree has a weight in its root and symbols as its leaves. We start with a forest of one vertex trees representing the input symbols. We recursively merge two trees whose sum of weights is minimal until we have only one tree.

36. The Huffman Coding algorithm- History • In 1951, David Huffman and his MIT information theory classmates given the choice of a term paper or a final exam • Huffman hit upon the idea of using a frequency-sorted binary tree and quickly proved this method the most efficient. • In doing so, the student outdid his professor, who had worked with information theory inventor Claude Shannon to develop a similar code. • Huffman built the tree from the bottom up instead of from the top down

37. Huffman Coding Algorithm • Take the two least probable symbols in the alphabet • Combine these two symbols into a single symbol, and repeat.

38. 1.0 0 0.55 1 1 0 0.45 0.3 0 1 0 1 Example • Ax={ a , b , c , d , e } • Px={0.25, 0.25, 0.2, 0.15, 0.15} a 0.25 c 0.2 d 0.15 b 0.25 e 0.15 00 10 11 010 011

39. Building the Encoding Tree

40. Building the Encoding Tree

41. Building the Encoding Tree Building the Encoding Tree

42. Building the Encoding Tree Building the Encoding Tree

43. Building the Encoding Tree Building the Encoding Tree