Big-O & Recursion

Recursion What is the Big-O of each?

Recursion What is the Big-O of each? Need Number of recursive calls Work done at each recursive call

Recursion What calls get made?

Recursion What calls get made? 16, 8, 4, 2, 1 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0

Recursion Number of recursive calls 16, 8, 4, 2, 1 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0 O(logn) O(n)

Recursion BigO Assume print is O(1) All work is O(1) = logn calls * O(1) work per call = O(logn)

Recursion BigO All work is O(1) = n calls * O(1) work per call = O(n)

Recursion BigO Work is is O(1) + O(n) = O(n) O(n) recursive calls = O(n) calls * O(n) work per call = O(n2) foo(array, size) if(size == 0) return for(int i = 0; i < size; i++) cout << array[i] foo(array, size - 1) Recursive calls for size = 10: 10, 9, 8, 7, … , 0 O(n)

Proofs

Recurrence Relationships Recurrence Relationship: Recursively defined math function: Used to: Formally establish recursive BigO Tackle complex recursion (multiple calls)

Recursion What is the Big-O? T(n) = T(n - 1) + 1 T(0) = 1

Recursion What is the Big-O? T(n) = T(n - 1) + 1 Level 1 = … = T(n - k) + k Level 1 Level 2 Level 3 Level k

Recursion What is the Big-O? Know: T(n) = T(n - k) + k T(0) = 1 So solve n – k = 0 … k = n

Recursion What is the Big-O? Given: T(n) = T(n - k) + k k = n Plug in k: 𝑇 𝑛 =𝑇 𝑛 −𝑛 + 𝑛 𝑇 𝑛 =𝑇 0 + 𝑛 𝑇 𝑛 =1 + 𝑛 𝑇 𝑛 ≈ 𝑛

Recursion What is the Big-O? T(n) = T(n/2) + 1 T(1) = 1

Recursion What is the Big-O? T(n) = T(n/2) + 1 Level 1 = … = T(n/2k) + k Level 1 Level 2 Level 3 Level k

Recursion What is the Big-O? T(n) = T(n/2k) + k Level k So what is k? How many levels are there? Level k

Recursion What is the Big-O? T(n) = T(n/2k) + k Know T(1) = 1 Need to solve n/2k = 1

Recursion What is the Big-O? T(n) = T(n/2k) + k Know T(1) = 1 Need to solve n/2k = 1 n = 2k log2(n) = k

Recursion What is the Big-O? Given: T(n) = T(n/2k) + k k = log2(n) Plug in k: 𝑇 𝑛 =𝑇 𝑛 2 𝑙𝑜𝑔 2 𝑛 + 𝑙𝑜𝑔 2 𝑛 𝑇 𝑛 =𝑇 𝑛 𝑛 + 𝑙𝑜𝑔 2 𝑛 𝑇 𝑛 =𝑇 1 + 𝑙𝑜𝑔 2 𝑛 𝑇 𝑛 =1+ 𝑙𝑜𝑔 2 𝑛 ≈ 𝑙𝑜𝑔 2 𝑛

Common Recurrences Big O's to recognize: Our focus is intuitive Math in 231/232 & Analysis of Algorithms

Derivations if you want them… T(n) = 2T(n/2) + cn + c T(n) = 2T(n/2) + n + 1 = 2(2T(n/4) + (n/2) + 1) + n + 1 = 4T(n/4) + 2n + 2 = 4T(n/4) + 2n + 2 = 4(2T(n/8) + (n/4) + 1) + 2n + 2 = 8T(n/8) + 3n + 3 = … = 2kT(n/2k) + kn + k Level 1 Level 2 Level 3 Level k

Derivations if you want them… T(n) = 2T(n/2) + cn + c Level k T(n) = 2k T(n/2k) + kn + k T(1) = 1, solve 1 = n/2k k = log2n Substitute: 𝑇 𝑛 = 2 𝑙𝑜𝑔 2 𝑛 𝑇 𝑛 2 𝑙𝑜𝑔 2 𝑛 + 𝑙𝑜𝑔 2 𝑛 𝑛+ 𝑙𝑜𝑔 2 𝑛 =𝑛𝑇 𝑛 𝑛 + 𝑛 𝑙𝑜𝑔 2 𝑛+ 𝑙𝑜𝑔 2 𝑛 =𝑛𝑇 1 + 𝑛 𝑙𝑜𝑔 2 𝑛+ 𝑙𝑜𝑔 2 𝑛 =𝑛+ 𝑛 𝑙𝑜𝑔 2 𝑛+ 𝑙𝑜𝑔 2 𝑛 ≈𝑛 𝑙𝑜𝑔 2 𝑛

Derivations if you want them… T(n) = 2T(n - 1) + c T(n) = 2T(n - 1) + 1 = 2(2T(n - 2) + 1) + 1 = 4T(n - 2) + 3 = 4T(n - 2) + 3 = 4(2T(n - 3) + 1) + 3 = 8T(n - 3) + 7 = … = 2kT(n - k) + (2k – 1) Level 1 Level 2 Level 3 Level k

Derivations if you want them… T(n) = 2T(n - 1) + c Level k T(n) = 2kT(n - k) + (2k – 1) T(1) = 1, solve 1 = n - k k = n - 1 Substitute: 𝑇 𝑛 = 2 𝑛−1 𝑇 𝑛 −(𝑛 −1) +( 2 𝑛−1 −1) = 2 𝑛−1 𝑇 1 + 2 𝑛−1 −1 = 2 𝑛−1 + 2 𝑛−1 −1 =2 2 𝑛−1 −1 ≈ 2 𝑛−1 ≈ 2 𝑛