1. More Trees Discrete Structures (CS 173)
03/14/13
Magritte
Discrete Structures (CS 173)
Derek Hoiem, University of Illinois

2. Last class: trees and CFGs
Trees are a special graph with root and no cycles, with many uses
Sorting, clustering, finding similar values
Decision tree: machine learning, modeling choices
Parse trees: representing hierarchical structures
Context free grammars: generate parse trees
Proofs on trees: split at root, use inductive hypothesis on subtrees headed by the root’s children

3. Tree terminology
Nodes: root, internal, leaf, level, tree height Relations: parent/child/sibling, ancestor/descendant
root
level = 0
parent, child
level = 1
internal
subtree
level = 2
level = 3
leaves

4. This lecture: more trees
Recursion trees for illustrating computation in recursive functions
Another tree proof

5. Useful formulas 𝑚 𝑎 𝑏 = 𝑚 𝑎𝑏 𝑚 𝑎+𝑏 = 𝑚 𝑎 𝑚 𝑏 2 log 2 𝑛 =𝑛
𝑘=0 𝑛 𝑟 𝑘 = 1− 𝑟 𝑛+1 1−𝑟 𝑘=𝑚 𝑛 𝑟 𝑘 = 𝑟 𝑚 − 𝑟 𝑛+1 1−𝑟 𝑘=𝑖 𝑛 𝑘 = 𝑛(𝑛+1) 2
𝑘=0 𝑛 2 𝑘 = 𝑘=1 𝑛 2 𝑘 =
𝑘=0 𝑛 2 −2𝑘 = 𝑘=0 𝑛 2 𝑘+2 =
𝑚 𝑎 𝑏 = 𝑚 𝑎𝑏
𝑚 𝑎+𝑏 = 𝑚 𝑎 𝑚 𝑏
2 log 2 𝑛 =𝑛
log 𝑎 (𝑏) = log 2 (𝑏)/ log 2 𝑎
2 log 4 (𝑛)+2 =

6. Recursion trees 𝑇 1 =𝑐 𝑇 𝑛 =2𝑇 𝑛/2 +𝑛
cost of input
cost of subproblem
split in 2
How many levels before base case?
Sum of values in each level?
How many leaf nodes?
Total cost = sum of leaf costs + sum of internal costs:

7. Recursion trees
𝑇 1 =𝑐 𝑇 𝑛 =2𝑇 𝑛−1 +𝑑

8. Recursion trees
A 1 =𝑐 A 𝑛 =4𝐴 𝑛/2 +𝑛

9. Tree induction proof
If 𝑇 is a binary tree with root 𝑟, then its rank is (a) 0 if 𝑟 has no children (b) 1+𝑞 if 𝑟 has two children, both with rank 𝑞 (c) otherwise, the maximum rank of any of the children

10. Tree induction proof
If 𝑇 is a binary tree with root 𝑟, then its rank is (a) 0 if 𝑟 has no children (b) 1+𝑞 if 𝑟 has two children, both with rank 𝑞 (c) otherwise, the maximum rank of any of the children Claim: A tree with rank 𝑞 has at least 2 𝑞 leaves.

11. Have a great break!