1. Big-O & Recursion

2. Recursion
What is the Big-O of each?

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

4. Recursion
What calls get made?

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

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

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

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

9. 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)

10. Proofs

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

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

13. 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

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

15. 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

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

17. 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

18. 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

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

20. 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

21. 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 𝑙𝑜𝑔 2 𝑛
𝑇 𝑛 =1+ 𝑙𝑜𝑔 2 𝑛 ≈ 𝑙𝑜𝑔 2 𝑛

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

23. 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 = 4T(n/4) + 2n = 4(2T(n/8) + (n/4) + 1) + 2n = 8T(n/8) + 3n + 3
= … = 2kT(n/2k) + kn + k
Level 1
Level 2
Level 3
Level k

24. 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

25. 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) = 4T(n - 2) = 4(2T(n - 3) + 1) = 8T(n - 3) + 7
= … = 2kT(n - k) + (2k – 1)
Level 1
Level 2
Level 3
Level k

26. 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 𝑛−1 −1 = 2 𝑛− 𝑛−1 −1 =2 2 𝑛−1 −1 ≈ 2 𝑛−1 ≈ 2 𝑛