Updates HW#1 has been delayed until next MONDAY. There were two errors in the assignment Merge sort runs in Θ(n log n). Insertion sort runs in Θ(n2). Consider a modification to Merge sort where n/k sublists of length k are sorted using Insertion sort and then merged by Merge sort.

Updates A) Show each n/k sublist can be sorted by Insertion Sort in Θ(n/k) worst case time. B) Show that the sublists can be merged in Θ(n lg(n/k)) worst case time. C) Skip. D) How should k be chosen in practice? (hint: philosophical)

Lecture 3: Solving Recurrences

Recurrences Recurrence: An equation that describes a function in terms of its value on smaller inputs. Example (MergeSort): T(n) =Θ(n) if n = 1, 2T(n/2) + Θ(n) if n>1

Recurrences We will study three techniques: 1.The substitution method 2.The recursion-tree method 3.The master method Sometimes more than one works well, sometimes none work well.

The Substitution Method The most general method: 1.Guess the form of the solution. 2.Verify using induction. 3.Solve for the constants.

Substitution Method: Example 1 Example: T(n) = 4T(n/2) + 100n Guess: T(n) = O(n 3 )

Substitution Method: Example 1 Example: T(n) = 4T(n/2) + 100n Proof: 1.Assume that T(1) = Θ(1). 2.Guess T(n) = O(n 3 ). We want to prove T(n) 0. 3.Verify by substituting into the recurrence. We assume this holds for T(n/2) 0, that is solving for constants, when c > 200 and n > 1. Always verify the base case(s) as well. Here Θ(1) < cn 3 for sufficiently large c.

A Tighter Bound? Example: T(n) = 4T(n/2) + 100n Guess: T(n) = O(n 2 )

A Tighter Bound? Example: T(n) = 4T(n/2) + 100n Proof: 1.Assume that T(1) = Θ(1). 2.Guess T(n) = O(n 2 ). We want to prove T(n) 0. 3.Verify by substituting into the recurrence. We assume this holds for T(n/2) 0. No valid assignment of c and n > 0 can be made! We have made a bad guess. We MUST prove the EXACT form of the inductive hypothesis (our guess).

Subtleties.. Example: T(n) = 4T(n/2) + 100n Guess T(n) = O(n 2 -n).

Subtleties.. Example: T(n) = 4T(n/2) + 100n Proof: 1.Assume that T(1) = Θ(1). 2.Guess T(n) = O(n 2 -n). We want to prove T(n) 0 and c 2 >0. 3.Verify by substituting into the recurrence. We assume this holds for T(n/2) 0, that is, c 2 >100.

Recursion Tree Method The Recursion Tree Method doesn’t necc. build a proof, but a good guess. You might use this method with the substitution method. The idea is to draw the tree and do the “accounting” the way we did with MergeSort last time. This can be unreliable, but it provides good intuition.

Example of recursion tree Solve T(n) = T(n/4) + T(n/2) + n 2 :

Example of recursion tree T(n)T(n) Solve T(n) = T(n/4) + T(n/2) + n 2 :

Example of recursion tree T(n/4) T(n/2) n2n2 Solve T(n) = T(n/4) + T(n/2) + n 2 :

Example of recursion tree Solve T(n) = T(n/4) + T(n/2) + n 2 : n2n2 (n/4) 2 (n/2) 2 T(n/16)T(n/8) T(n/4)

Example of recursion tree (n/16) 2 (n/8) 2 (n/4) 2 (n/2) 2  (1) … Solve T(n) = T(n/4) + T(n/2) + n 2 : n2n2

Example of recursion tree Solve T(n) = T(n/4) + T(n/2) + n 2 : (n/16) 2 (n/8) 2 (n/4) 2 (n/2) 2  (1) … n2n2

Example of recursion tree Solve T(n) = T(n/4) + T(n/2) + n 2 : (n/16) 2 (n/8) 2 (n/4) 2 (n/2) 2  (1) … n2n2

Example of recursion tree Solve T(n) = T(n/4) + T(n/2) + n 2 : (n/16) 2 (n/8) 2 (n/4) 2  (1) … n2n2 (n/2) 2 …

Example of recursion tree Solve T(n) = T(n/4) + T(n/2) + n 2 : (n/16) 2 (n/8) 2 (n/4) 2  (1) … … Total = =  (n 2 ) n2n2 (n/2) 2 geometric series

The Master Method The Master Theorem: Let a > 1 and b > 1 be constants. Let f(n) be a function and T(n) be defined on the non-negative integers as, T(n) = a T(n/b) + f(n). T(n) can be asymptotically bounded as follows: 1.If f(n) = O(n logba-ε ), then T(n) = Θ(n logba ) 2.If f(n) = Θ(n logba ), then T(n) = Θ(n logba lg n) 3.If f(n) = Ω(n logba+ ε ) and if af(n/b)<cf(n) for c < 1 and all sufficiently large n, then T(n) = Θ(f(n)).