1. Compsci 201 Trees, Tries, Tradeoffs
Owen Astrachan
Jeff Forbes
November 1, 2017
11/1/17
Compsci 201, Fall 2017, Tree, Tries, Tradoffs

2. Compsci 201, Fall 2017, Tree, Tries, Tradoffs
R is for … R
Programming language of choice in Stats
Random
From Monte-Carlo to [0,1)
Recursion
Base case of wonderful stuff
Refactoring
Better not different
11/1/17
Compsci 201, Fall 2017, Tree, Tries, Tradoffs

3. Compsci 201, Fall 2017, Tree, Tries, Tradoffs
Plan for the Day
Tree Review
From Theory to Practice
From Recurrences to Code
What are the main ideas in DNA LinkStrand
Why Splicing matters with low level links
Trees, Tries, Tradeoffs
11/1/17
Compsci 201, Fall 2017, Tree, Tries, Tradoffs

4. Compsci 201, Fall 2017, Tree, Tries, Tradoffs
From Links to …
What is the DNA/LinkStrand assignment about?
Why do we study linked lists
How do you work in a group?
11/1/17
Compsci 201, Fall 2017, Tree, Tries, Tradoffs

5. Basics of cutAndSplice
Find enzyme like ‘gat’
Replace with splicee like ‘gggtttaaa’
Strings and StringBuilder for creating new strings
Complexity of “hello” + “world”, or A+B
String: A + B, StringBuilder: B
11/1/17
Compsci 201, Fall 2017, Tree, Tries, Tradoffs

6. What do linked lists get us?
Faster run-time, much better use of memory
We splice in constant time? Re-use strings
11/1/17
Compsci 201, Fall 2017, Tree, Tries, Tradoffs

7. String Concatenation Examined
Runtime of stringConcat(“hello”,N)
Depends on size of ret, 5, 10, 15, … 5*N
public String stringConcat(String s, int reps) {
String ret = "";
for(int k=0; k < reps; k++) {
ret += s; }
return ret;
11/1/17
Compsci 201, Fall 2017, Tree, Tries, Tradoffs

8. StringBuilder Examined
Runtime of builderConcat(“hello”,N)
… + 5 a total of N times
public String builderConcat(String s, int reps) {
StringBuilder ret = new StringBuilder();
for(int k=0; k < reps; k++) {
ret.append(s); }
return ret.toString();
11/1/17
Compsci 201, Fall 2017, Tree, Tries, Tradoffs

9. Compsci 201, Fall 2017, Tree, Tries, Tradoffs
Theory and Practice
The JVM can sometimes optimize your code
Don’t optimize what you don’t have to …
WOTO
11/1/17
Compsci 201, Fall 2017, Tree, Tries, Tradoffs

10. Compsci 201, Fall 2017, Tree, Tries, Tradoffs
dana boyd
Dr. danah boyd is a Senior Researcher at Microsoft Research, … a Visiting Researcher at Harvard Law School, …Her work examines everyday practices involving social media, with specific attention to youth engagement, privacy, and risky behaviors. She recently co-authored Hanging Out, Messing Around, and Geeking Out: Kids Living and Learning with New Media
"From day one, Mark Zuckerberg wanted Facebook to become a social utility. He succeeded. Facebook is now a utility for many. The problem with utilities is that they get regulated."
11/1/17
Compsci 201, Fall 2017, Tree, Tries, Tradoffs

11. Trees: no data structure lovelier?

12. Trees from Bottom to Top
Trees!! Trying to get the best of a few worlds:
efficient lookup: like sorted array
efficient add: like linked list
range-queries: see java.util.NavigableSet
Reminder: hashing is really, really fast
O(1) add, search, delete independent of N
BUT! No order info, worst case can be bad

13. Binary Trees
Binary trees: Not just for searching, used in many contexts,
Game trees, collisions, …
Cladistics, genomics, quad trees, …
Search is O(log n) like sorted array
Average case. Note: worst case also be O(log n), e.g., use a balanced tree
insertion/deletion O(1), once location found

14. How do search trees work?
Change doubly-linked lists, no longer linear
Similar to binary search, everything less goes left, everything greater goes right
How do we search?
How do we insert?
Insert: “koala”
“koala”
“llama”
“tiger”
“monkey”
“jaguar”
“elephant”
“giraffe”
“pig”
“hippo”
“leopard”
“koala”
“koala”
“koala”
“koala”

15. Review: tree terminologyA B E D F C G
Binary tree is a structure:
empty
root node with left and right subtrees
Tree Terminology
parent and child: A is parent of B, E is child of B
leaf node has no children, internal node has 1 or 2 children
path is sequence of nodes (edges), N1, N2, … Nk
Ni is parent of Ni+1
depth (level) of node: length of root-to-node path
level of root is 1 (measured in nodes)
height of node: length of longest node-to-leaf path
height of tree is height of root

16. A TreeNode by any other name…
What does this look like? Doubly linked list?
public class TreeNode {
TreeNode left;
TreeNode right;
String info;
TreeNode(String s,
TreeNode llink, TreeNode rlink){
info = s;
left = llink;
right = rlink; }
“llama”
“tiger”
“giraffe”

17. Tree function: Tree height
Compute tree height (longest root-to-leaf path)
int height(Tree root) {
if (root == null) return 0;
else {
return 1 + Math.max(height(root.left),
height(root.right)); }
Find height of left subtree, height of right subtree
Use results to determine height of tree

18. Tree function: Leaf Count
Calculate Number of Leaf Nodes
int leafCount(Tree root) {
if (root == null) return 0;
if (root.left == null &&
root.right == null) return 1;
return leafCount(root.left) +
leafCount(root.right); }
Similar to height: but has two base case(s)
Use results of recursive calls to determine # leaves

19. Tree functions Analyzed
int height(Tree root) {
if (root == null) return 0;
else {
return 1 + Math.max(height(root.left),
height(root.right)); }
Let T(n) be time for height to run on n-node tree
T(n) = 2T(n/2) + O(1) - roughly balanced
T(n) = T(n-1) + T(1) + O(1)
= T(n-1) + O(1) - unbalanced

20. Good Search Trees and Bad Trees

21. Bad Trees and Good Trees
11/1/17
Compsci 201, Fall 2017, Tree, Tries, Tradoffs

22. Don’t do this at home?
Let T(n) be time for height to execute (n-node tree)
T(n) = T(n/2) + T(n/2) + O(1)
T(n) = 2 T(n/2) + 1
T(n) = 2 [2(T(n/4) + 1] + 1
T(n) = 4T(n/4)
T(n) = 8T(n/8) , eureka!
T(n) = 2kT(n/2k) + 2k why true?
T(n) = nT(1) + O(n) is O(n) if we let n=2k
Different recurrence, same solution if unbalanced

23. Compsci 201, Fall 2017, Tree, Tries, Tradoffs
Recurrence Table
Recurrence
Algorithm
Solution
T(n) = T(n/2) + O(1)
Binary search
O(log n)
T(n) = T(n-1) + O(1)
Sequential search
O(n)
T(n) = T(n-1) + O(n)
Selection sort
O(n2)
T(n) = T(n/2) + O(n)
Find median or kth
T(n) = 2T(n/2) + O(1)
Tree traversal
T(n) = 2T(n/2) + O(n)
Quick or merge sort
O(n log n)
T(n) = 2T(n-1) + O(1)
Towers of Hanoi
O(2n)
Develop recurrence, look up solution
Remember: goal is big-Oh, recurrence helps
11/1/17
Compsci 201, Fall 2017, Tree, Tries, Tradoffs

24. Balanced Trees and Complexity
A tree is height-balanced if
Left and right subtrees are height-balanced
Left and right heights differ by at most one
boolean isBalanced(Tree root){
if (root == null) return true;
return
isBalanced(root.left) && isBalanced(root.right) &&
Math.abs(height(root.left)–height(root.right)) <= 1; }

25. What is complexity? Assume trees “balanced” in analyzing complexity
Roughly half the nodes in each subtree
Leads to easier analysis
How to develop recurrence relation?
What is T(n)? Time func executes on n-node tree
What other work? Express recurrence, solve it
How to solve recurrence relation
Plug, expand, plug, expand, find pattern
Proof requires induction to verify correctness

26. Recurrence relation T(n): time for isBalanced to execute (n-node tree)
T(n) = T(n/2) + T(n/2) + O(n)
T(n) = 2 T(n/2) + n
T(n) = 2 [2(T(n/4) + n/2] + n
T(n) = 4T(n/4) + n + n = 4T(n/4) + 2n
T(n) = 8T(n/8) + 3n, eureka!
T(n) = 2kT(n/2k) + kn why true?
T(n) = nT(1) + n log(n) let n=2k, so k=log n
So, solution for T(n) = 2T(n/2) + O(n) is
O(n log n) -- base 2, but base doesn't matter

27. Printing a search tree in order
When is root printed?
After left subtree, before right subtree.
void visit(TreeNode t){
if (t != null) {
visit(t.left);
System.out.println(t.info);
visit(t.right); }
Inorder traversal
“llama”
“tiger”
“monkey”
“jaguar”
“elephant”
“giraffe”
“pig”
“hippo”
“leopard”

28. Tree traversals Inorder visits search tree in order
Visit left-subtree, process root, visit right-subtree
elephant, giraffe, jaguar, llama, monkey, tiger
Navigate following nodes/links?
Visit on passing under
Second time by node
“llama”
“tiger”
“monkey”
“jaguar”
“elephant”
“giraffe”

29. Tree traversals Preorder good for reading/writing trees
Process root, then Visit left-subtree, visit right-subtree
llama, giraffe,elephant jaguar, tiger, monkey
Navigate following nodes/links?
Visit on passing-by to left
First time by node
“llama”
“tiger”
“monkey”
“jaguar”
“elephant”
“giraffe”

30. Tree traversals Post order good for deleting tree
Visit left-subtree, visit right-subtree, process root
elephant, jaguar, giraffe, monkey, tiger, llama
Navigate following nodes/links?
Visit on passing up
Third time by node
“llama”
“tiger”
“monkey”
“jaguar”
“elephant”
“giraffe”

31. Compsci 201, Fall 2017, Tree, Tries, Tradoffs
Tree WOTO
Pride in social group? Urban dictionary?
11/1/17
Compsci 201, Fall 2017, Tree, Tries, Tradoffs

32. What does insertion look like?
Simple recursive insertion into tree (accessed by root)
root = insert("foo", root);
TreeNode insert(TreeNode t, String s) {
if (t == null) t = new Tree(s,null,null);
else if (s.compareTo(t.info) <= 0)
t.left = insert(t.left,s);
else
t.right = insert(t.right,s);
return t; }

33. Notes on tree insert and search
Ineach recursive insert call
Tree parameter in the call is either the left or right field of some node in the original tree
Will be assignment to a .left or .right field!
Idiom t = treeMethod(t,…) used
Good trees go bad, what happens and why?
Insert alpha, beta, gamma, delta, epsilon, …

34. Insert and Removal For insertion we can use iteration (see BSTSet)
Traverse left or right and “look ahead” to add
Removal is tricky, depends on number of children
Straightforward when zero or one child
Complicated when two children, find successor
See set code for complete cases
If right child, straightforward
Otherwise find node that’s left child of its parent (why?)

35. Compsci 201, Fall 2017, Linked Lists & More
Wordladder Story
Ladder from ‘white’ to ‘house’
white, while, whale, shale, …
I can do that… optimally
My brother was an English major
My ladder is 16, his is 15, how?
There's a ladder that's 14 words!
The key is ‘sough’
Guarantee optimality!
QUEUE
I heard the puzzle on the way into Duke in the mid 90's. I called my brother and said I'd solve the problem. He said he would too. He used his brain, I wrote a program. Called him an hour alter and said "got it". He said "got it". I said my code was provably optimal and had a 16-word ladder. He said he had 15. WHAT?
I added "sough" to dictionary and got a 14-word ladder 
10/20/17
Compsci 201, Fall 2017, Linked Lists & More

36. Queue for shortest path
public boolean findLadder(String[] words,
String first, String last){
Queue<String> qu = new LinkedList<>();
Set<String> set = new HashSet<>();
qu.add(first);
while (qu.size() > 0){
String current = qu.remove();
if (oneAway(current,last)) return true;
for(String s : words){
if (! set.contains(s) && oneAway(from,s)){
qu.add(s);
set.(s); } }
return false;
10/20/17
Compsci 201, Fall 2017, Linked Lists & More

37. Shortest Path reprised
How does Queue ensure we find shortest path?
Where are words one away from first?
Where are words two away from first?
Why do we need to avoid revisiting a word, when?
Why do we use a set for this? Why a HashSet?
Alternatives?
If we want the ladder, not just whether it exists
What's path from white to house? We know there is one.
All words one-away from first are on Q and are added when first is dequeued/removed first time through loop. Each of these 1-away-from-start words comes off the queue and words that are one-away from these, or 2-away from start, are put on the Q. BUT it's a queue so all 2-away words go on after the 1-away.
In general, we remove N-away words before (N+1)-away words because of FIFO structure
10/20/17
Compsci 201, Fall 2017, Linked Lists & More

38. Compsci 201, Fall 2017, Linked Lists & More
Shortest path proof
All words one away from start on queue first iteration
What is first value of current when loop entered?
All one-away words dequeued before two-away
See previous assertion, property of queues
Two-away before 3-away, …
Each 2-away word is one away from a 1-away word
So all enqueued after one-away, before three-away
Any w seen/dequeued that's n:away is:
Seen before every n+k:away word for k >=1!
Don't dwell on this, but walk through it as a semi-formal argument, akin to a hand-wavy proof
10/20/17
Compsci 201, Fall 2017, Linked Lists & More

39. Keeping track of ladder
Find w, a one-away word from current
Enqueue w if not seen
Call map.put(w,current)
Remember keys are unique!
Put word on queue once!
map.put("lot", "hot")
map.put("dot", "hot")
map.put("hat", "hot")
Remind students that we need to keep track of how words are connected to reconstruct the ladder. Keys in a map are unique, but we only put a word on the queue once! Why? Using a set. We can make this one time the time we make a word the key in a map. The value is what caused the word to be put onto the queue, see examples
10/20/17
Compsci 201, Fall 2017, Linked Lists & More

40. Reconstructing Word Ladder
Run WordLaddersFull
See map and call to map.put(word,current)
What about when returning the ladder, why is the returned ladder in reverse order?
What do we know about code when statement adding (key,value) to map runs?
Run program with "white" "house" and with "voter" "crazy" and with "voter" "smart"
10/20/17
Compsci 201, Fall 2017, Linked Lists & More