Recurrences

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Information about Recurrences
Education

Published on February 17, 2014

Author: sandpoonia

Source: slideshare.net

Description

Substitution, Iteration , Master Theorem

Algorithms Sandeep Kumar Poonia Head Of Dept. CS/IT B.E., M.Tech., UGC-NET LM-IAENG, LM-IACSIT,LM-CSTA, LM-AIRCC, LM-SCIEI, AM-UACEE

Algorithms Merge Sort Solving Recurrences The Master Theorem Introduction to heapsort Quicksort Sandeep Kumar Poonia

Merge Sort MergeSort(A, left, right) { if (left < right) { mid = floor((left + right) / 2); MergeSort(A, left, mid); MergeSort(A, mid+1, right); Merge(A, left, mid, right); } } // Merge() takes two sorted subarrays of A and // merges them into a single sorted subarray of A // (how long should this take?) Sandeep Kumar Poonia

Merge Sort: Example  Show MergeSort() running on the array A = {10, 5, 7, 6, 1, 4, 8, 3, 2, 9}; Sandeep Kumar Poonia

Analysis of Merge Sort Statement Effort MergeSort(A, left, right) { if (left < right) { mid = floor((left + right) / 2); MergeSort(A, left, mid); MergeSort(A, mid+1, right); Merge(A, left, mid, right); } }  So T(n) = (1) when n = 1, and 2T(n/2) + (n) when n > 1  So what (more succinctly) is T(n)? Sandeep Kumar Poonia T(n) (1) (1) T(n/2) T(n/2) (n)

Recurrences  The expression: c n 1    T ( n)   n 2T    cn n  1  2  is a recurrence.  Recurrence: an equation that describes a function in terms of its value on smaller functions Sandeep Kumar Poonia

Recurrence Examples 0 n0  s ( n)   c  s(n  1) n  0 0 n0  s ( n)   n  s(n  1) n  0 c n 1    T ( n)   n 2T    c n  1  2    c n 1  T ( n)    n aT    cn n  1  b Sandeep Kumar Poonia

Solving Recurrences Substitution method  Iteration method  Master method  Sandeep Kumar Poonia

Solving Recurrences  The substitution method    A.k.a. the “making a good guess method” Guess the form of the answer, then use induction to find the constants and show that solution works Examples: = 2T(n/2) + (n)  T(n) = (n lg n)  T(n) = 2T(n/2) + n  ???  T(n) Sandeep Kumar Poonia

Solving Recurrences  The substitution method    A.k.a. the “making a good guess method” Guess the form of the answer, then use induction to find the constants and show that solution works Examples: = 2T(n/2) + (n)  T(n) = (n lg n)  T(n) = 2T(n/2) + n  T(n) = (n lg n)  T(n) = 2T(n/2 )+ 17) + n  ???  T(n) Sandeep Kumar Poonia

Solving Recurrences  The substitution method    A.k.a. the “making a good guess method” Guess the form of the answer, then use induction to find the constants and show that solution works Examples: = 2T(n/2) + (n)  T(n) = (n lg n)  T(n) = 2T(n/2) + n  T(n) = (n lg n)  T(n) = 2T(n/2+ 17) + n  (n lg n)  T(n) Sandeep Kumar Poonia

Solving Recurrences  Another option is the “iteration method”     Expand the recurrence Work some algebra to express as a summation Evaluate the summation We will show several examples Sandeep Kumar Poonia

0 n0  s ( n)   c  s(n  1) n  0  s(n) = c + s(n-1) c + c + s(n-2) 2c + s(n-2) 2c + c + s(n-3) 3c + s(n-3) … kc + s(n-k) = ck + s(n-k) Sandeep Kumar Poonia

0 n0  s ( n)   c  s(n  1) n  0  So far for n >= k we have   s(n) = ck + s(n-k) What if k = n?  s(n) = cn + s(0) = cn Sandeep Kumar Poonia

0 n0  s ( n)   c  s(n  1) n  0  So far for n >= k we have   s(n) = ck + s(n-k) What if k = n?  s(n) = cn + s(0) = cn 0 n0  s ( n)   c  s(n  1) n  0  So  Thus in general  s(n) = cn Sandeep Kumar Poonia

0 n0  s ( n)   n  s(n  1) n  0  = = = = = = s(n) n + s(n-1) n + n-1 + s(n-2) n + n-1 + n-2 + s(n-3) n + n-1 + n-2 + n-3 + s(n-4) … n + n-1 + n-2 + n-3 + … + n-(k-1) + s(n-k) Sandeep Kumar Poonia

0 n0  s ( n)   n  s(n  1) n  0  = = = = = = s(n) n + s(n-1) n + n-1 + s(n-2) n + n-1 + n-2 + s(n-3) n + n-1 + n-2 + n-3 + s(n-4) … n + n-1 + n-2 + n-3 + … + n-(k-1) + s(n-k) n = i i  n  k 1 Sandeep Kumar Poonia  s(n  k )

0 n0  s ( n)   n  s(n  1) n  0  So far for n >= k we have n i i  n  k 1 Sandeep Kumar Poonia  s(n  k )

0 n0  s ( n)   n  s(n  1) n  0  So far for n >= k we have n i  s(n  k ) i  n  k 1  What if k = n? Sandeep Kumar Poonia

0 n0  s ( n)   n  s(n  1) n  0  So far for n >= k we have n i  s(n  k ) i  n  k 1  What if k = n? n 1  i  s(0)   i  0  n 2 i 1 i 1 n Sandeep Kumar Poonia n

0 n0  s ( n)   n  s(n  1) n  0  So far for n >= k we have n i  s(n  k ) i  n  k 1  What if k = n? n 1  i  s(0)   i  0  n 2 i 1 i 1 n  Thus in general n 1 s ( n)  n 2 Sandeep Kumar Poonia n

c n 1   n T (n)  2T    c n 1  2   T(n) = 2T(n/2) + c 2(2T(n/2/2) + c) + c 22T(n/22) + 2c + c 22(2T(n/22/2) + c) + 3c 23T(n/23) + 4c + 3c 23T(n/23) + 7c 23(2T(n/23/2) + c) + 7c 24T(n/24) + 15c … 2kT(n/2k) + (2k - 1)c Sandeep Kumar Poonia

c n 1   n T (n)  2T    c n 1  2   So far for n > 2k we have   T(n) = 2kT(n/2k) + (2k - 1)c What if k = lg n?  T(n) = 2lg n T(n/2lg n) + (2lg n - 1)c = n T(n/n) + (n - 1)c = n T(1) + (n-1)c = nc + (n-1)c = (2n - 1)c Sandeep Kumar Poonia

The Master Theorem  Given: a divide and conquer algorithm    An algorithm that divides the problem of size n into a subproblems, each of size n/b Let the cost of each stage (i.e., the work to divide the problem + combine solved subproblems) be described by the function f(n) Then, the Master Theorem gives us a cookbook for the algorithm’s running time: Sandeep Kumar Poonia

The Master Theorem  if T(n) = aT(n/b) + f(n) then    n logb a     logb a T (n)   n log n    f (n)       Sandeep Kumar Poonia    f (n)  O n logb a      0 logb a f ( n)   n   c 1   f (n)   n logb a  AND  af (n / b)  cf (n) for large n       

Using The Master Method  T(n) = 9T(n/3) + n    a=9, b=3, f(n) = n nlog a = nlog 9 = (n2) Since f(n) = O(nlog 9 - ), where =1, case 1 applies: b 3 3    T (n)   nlogb a when f (n)  O nlogb a   Thus the solution is T(n) = (n2) Sandeep Kumar Poonia 

Sorting Revisited  So far we’ve talked about two algorithms to sort an array of numbers    What is the advantage of merge sort? What is the advantage of insertion sort? Next on the agenda: Heapsort  Combines advantages of both previous algorithms Sandeep Kumar Poonia

Heaps  A heap can be seen as a complete binary tree: 16 14 10 8 2   7 4 9 1 What makes a binary tree complete? Is the example above complete? Sandeep Kumar Poonia 3

Heaps  A heap can be seen as a complete binary tree: 16 14 10 8 2  7 4 1 9 1 1 3 1 1 The book calls them “nearly complete” binary trees; can think of unfilled slots as null pointers Sandeep Kumar Poonia 1

Heaps  In practice, heaps are usually implemented as arrays: 16 14 A = 16 14 10 8 7 9 3 2 4 8 1 = 2 Sandeep Kumar Poonia 10 7 4 1 9 3

Heaps  To represent a complete binary tree as an array:      The root node is A[1] Node i is A[i] The parent of node i is A[i/2] (note: integer divide) The left child of node i is A[2i] The right child of node i is A[2i + 1] 16 14 A = 16 14 10 8 7 9 3 2 4 8 1 = 2 Sandeep Kumar Poonia 10 7 4 1 9 3

Referencing Heap Elements  So… Parent(i) { return i/2; } Left(i) { return 2*i; } right(i) { return 2*i + 1; } An aside: How would you implement this most efficiently?  Another aside: Really?  Sandeep Kumar Poonia

The Heap Property  Heaps also satisfy the heap property: A[Parent(i)]  A[i] for all nodes i > 1    In other words, the value of a node is at most the value of its parent Where is the largest element in a heap stored? Definitions:   The height of a node in the tree = the number of edges on the longest downward path to a leaf The height of a tree = the height of its root Sandeep Kumar Poonia

Heap Height What is the height of an n-element heap? Why?  This is nice: basic heap operations take at most time proportional to the height of the heap  Sandeep Kumar Poonia

Heap Operations: Heapify()  Heapify(): maintain the heap property     Given: a node i in the heap with children l and r Given: two subtrees rooted at l and r, assumed to be heaps Problem: The subtree rooted at i may violate the heap property (How?) Action: let the value of the parent node “float down” so subtree at i satisfies the heap property  What do you suppose will be the basic operation between i, l, and r? Sandeep Kumar Poonia

Heap Operations: Heapify() Heapify(A, i) { l = Left(i); r = Right(i); if (l <= heap_size(A) && A[l] > A[i]) largest = l; else largest = i; if (r <= heap_size(A) && A[r] > A[largest]) largest = r; if (largest != i) Swap(A, i, largest); Heapify(A, largest); } Sandeep Kumar Poonia

Heapify() Example 16 4 10 14 2 7 8 3 1 A = 16 4 10 14 7 Sandeep Kumar Poonia 9 9 3 2 8 1

Heapify() Example 16 4 10 14 2 7 8 3 1 A = 16 4 10 14 7 Sandeep Kumar Poonia 9 9 3 2 8 1

Heapify() Example 16 4 10 14 2 7 8 3 1 A = 16 4 10 14 7 Sandeep Kumar Poonia 9 9 3 2 8 1

Heapify() Example 16 14 10 4 2 7 8 3 1 A = 16 14 10 4 Sandeep Kumar Poonia 9 7 9 3 2 8 1

Heapify() Example 16 14 10 4 2 7 8 3 1 A = 16 14 10 4 Sandeep Kumar Poonia 9 7 9 3 2 8 1

Heapify() Example 16 14 10 4 2 7 8 3 1 A = 16 14 10 4 Sandeep Kumar Poonia 9 7 9 3 2 8 1

Heapify() Example 16 14 10 8 2 7 4 3 1 A = 16 14 10 8 Sandeep Kumar Poonia 9 7 9 3 2 4 1

Heapify() Example 16 14 10 8 2 7 4 3 1 A = 16 14 10 8 Sandeep Kumar Poonia 9 7 9 3 2 4 1

Heapify() Example 16 14 10 8 2 7 4 3 1 A = 16 14 10 8 Sandeep Kumar Poonia 9 7 9 3 2 4 1

Analyzing Heapify(): Informal Aside from the recursive call, what is the running time of Heapify()?  How many times can Heapify() recursively call itself?  What is the worst-case running time of Heapify() on a heap of size n?  Sandeep Kumar Poonia

Analyzing Heapify(): Formal Fixing up relationships between i, l, and r takes (1) time  If the heap at i has n elements, how many elements can the subtrees at l or r have?   Draw it Answer: 2n/3 (worst case: bottom row 1/2 full)  So time taken by Heapify() is given by  T(n)  T(2n/3) + (1) Sandeep Kumar Poonia

Analyzing Heapify(): Formal So we have T(n)  T(2n/3) + (1)  By case 2 of the Master Theorem, T(n) = O(lg n)  Thus, Heapify() takes linear time  Sandeep Kumar Poonia

Heap Operations: BuildHeap()  We can build a heap in a bottom-up manner by running Heapify() on successive subarrays   Fact: for array of length n, all elements in range A[n/2 + 1 .. n] are heaps (Why?) So:  Walk backwards through the array from n/2 to 1, calling Heapify() on each node.  Order of processing guarantees that the children of node i are heaps when i is processed Sandeep Kumar Poonia

BuildHeap() // given an unsorted array A, make A a heap BuildHeap(A) { heap_size(A) = length(A); for (i = length[A]/2 downto 1) Heapify(A, i); } Sandeep Kumar Poonia

BuildHeap() Example  Work through example A = {4, 1, 3, 2, 16, 9, 10, 14, 8, 7} 4 1 3 2 14 Sandeep Kumar Poonia 16 8 7 9 10

Analyzing BuildHeap()  Each call to Heapify() takes O(lg n) time There are O(n) such calls (specifically, n/2)  Thus the running time is O(n lg n)     Is this a correct asymptotic upper bound? Is this an asymptotically tight bound? A tighter bound is O(n)  How can this be? Is there a flaw in the above reasoning? Sandeep Kumar Poonia

Analyzing BuildHeap(): Tight  To Heapify() a subtree takes O(h) time where h is the height of the subtree   h = O(lg m), m = # nodes in subtree The height of most subtrees is small Fact: an n-element heap has at most n/2h+1 nodes of height h  CLR 7.3 uses this fact to prove that BuildHeap() takes O(n) time  Sandeep Kumar Poonia

Heapsort  Given BuildHeap(), an in-place sorting algorithm is easily constructed:   Maximum element is at A[1] Discard by swapping with element at A[n]  Decrement heap_size[A]  A[n] now contains correct value  Restore heap property at A[1] by calling Heapify()  Repeat, always swapping A[1] for A[heap_size(A)] Sandeep Kumar Poonia

Heapsort Heapsort(A) { BuildHeap(A); for (i = length(A) downto 2) { Swap(A[1], A[i]); heap_size(A) -= 1; Heapify(A, 1); } } Sandeep Kumar Poonia

Analyzing Heapsort The call to BuildHeap() takes O(n) time  Each of the n - 1 calls to Heapify() takes O(lg n) time  Thus the total time taken by HeapSort() = O(n) + (n - 1) O(lg n) = O(n) + O(n lg n) = O(n lg n)  Sandeep Kumar Poonia

Priority Queues Heapsort is a nice algorithm, but in practice Quicksort (coming up) usually wins  But the heap data structure is incredibly useful for implementing priority queues     A data structure for maintaining a set S of elements, each with an associated value or key Supports the operations Insert(), Maximum(), and ExtractMax() What might a priority queue be useful for? Sandeep Kumar Poonia

Priority Queue Operations Insert(S, x) inserts the element x into set S  Maximum(S) returns the element of S with the maximum key  ExtractMax(S) removes and returns the element of S with the maximum key  How could we implement these operations using a heap?  Sandeep Kumar Poonia

Tying It Into The Real World  And now, a real-world example… Sandeep Kumar Poonia

Tying It Into The “Real World”  And now, a real-world example…combat billiards       Sort of like pool... Except you’re trying to kill the other players… And the table is the size of a polo field… And the balls are the size of Suburbans... And instead of a cue you drive a vehicle with a ram on it Figure 1: boring traditional pool Problem: how do you simulate the physics? Sandeep Kumar Poonia

Combat Billiards: Simulating The Physics  Simplifying assumptions:  G-rated version: No players  Just  n balls bouncing around No spin, no friction  Easy to calculate the positions of the balls at time Tn from time Tn-1 if there are no collisions in between  Simple elastic collisions Sandeep Kumar Poonia

Simulating The Physics  Assume we know how to compute when two moving spheres will intersect   Given the state of the system, we can calculate when the next collision will occur for each ball At each collision Ci:  Advance the system to the time Ti of the collision  Recompute the next collision for the ball(s) involved  Find the next overall collision Ci+1 and repeat  How should we keep track of all these collisions and when they occur? Sandeep Kumar Poonia

Implementing Priority Queues HeapInsert(A, key) // what’s running time? { heap_size[A] ++; i = heap_size[A]; while (i > 1 AND A[Parent(i)] < key) { A[i] = A[Parent(i)]; i = Parent(i); } A[i] = key; } Sandeep Kumar Poonia

Implementing Priority Queues HeapMaximum(A) { // This one is really tricky: return A[i]; } Sandeep Kumar Poonia

Implementing Priority Queues HeapExtractMax(A) { if (heap_size[A] < 1) { error; } max = A[1]; A[1] = A[heap_size[A]] heap_size[A] --; Heapify(A, 1); return max; } Sandeep Kumar Poonia

Back To Combat Billiards      Extract the next collision Ci from the queue Advance the system to the time Ti of the collision Recompute the next collision(s) for the ball(s) involved Insert collision(s) into the queue, using the time of occurrence as the key Find the next overall collision Ci+1 and repeat Sandeep Kumar Poonia

Using A Priority Queue For Event Simulation More natural to use Minimum() and ExtractMin()  What if a player hits a ball?    Need to code up a Delete() operation How? What will the running time be? Sandeep Kumar Poonia

Quicksort Sorts in place  Sorts O(n lg n) in the average case  Sorts O(n2) in the worst case  So why would people use it instead of merge sort?  Sandeep Kumar Poonia

Quicksort  Another divide-and-conquer algorithm  The array A[p..r] is partitioned into two nonempty subarrays A[p..q] and A[q+1..r]  Invariant: All elements in A[p..q] are less than all elements in A[q+1..r]   The subarrays are recursively sorted by calls to quicksort Unlike merge sort, no combining step: two subarrays form an already-sorted array Sandeep Kumar Poonia

Quicksort Code Quicksort(A, p, r) { if (p < r) { q = Partition(A, p, r); Quicksort(A, p, q); Quicksort(A, q+1, r); } } Sandeep Kumar Poonia

Partition  Clearly, all the action takes place in the partition() function   Rearranges the subarray in place End result:  Two subarrays  All values in first subarray  all values in second   Returns the index of the “pivot” element separating the two subarrays How do you suppose we implement this function? Sandeep Kumar Poonia

Partition In Words  Partition(A, p, r):   Select an element to act as the “pivot” (which?) Grow two regions, A[p..i] and A[j..r]  All elements in A[p..i] <= pivot  All elements in A[j..r] >= pivot      Increment i until A[i] >= pivot Decrement j until A[j] <= pivot Swap A[i] and A[j] Repeat until i >= j Return j Sandeep Kumar Poonia

Partition Code Partition(A, p, r) x = A[p]; Illustrate on i = p - 1; A = {5, 3, 2, 6, 4, 1, 3, 7}; j = r + 1; while (TRUE) repeat j--; until A[j] <= x; What is the running time of repeat partition()? i++; until A[i] >= x; if (i < j) Swap(A, i, j); else return j; Sandeep Kumar Poonia

Review: Analyzing Quicksort  What will be the worst case for the algorithm?   What will be the best case for the algorithm?   Partition is balanced Which is more likely?   Partition is always unbalanced The latter, by far, except... Will any particular input elicit the worst case?  Yes: Already-sorted input Sandeep Kumar Poonia

Review: Analyzing Quicksort  In the worst case: T(1) = (1) T(n) = T(n - 1) + (n)  Works out to T(n) = (n2) Sandeep Kumar Poonia

Review: Analyzing Quicksort  In the best case: T(n) = 2T(n/2) + (n)  What does this work out to? T(n) = (n lg n) Sandeep Kumar Poonia

Review: Analyzing Quicksort (Average Case)  Intuitively, a real-life run of quicksort will produce a mix of “bad” and “good” splits    Randomly distributed among the recursion tree Pretend for intuition that they alternate between best-case (n/2 : n/2) and worst-case (n-1 : 1) What happens if we bad-split root node, then good-split the resulting size (n-1) node? Sandeep Kumar Poonia

Review: Analyzing Quicksort (Average Case)  Intuitively, a real-life run of quicksort will produce a mix of “bad” and “good” splits    Randomly distributed among the recursion tree Pretend for intuition that they alternate between best-case (n/2 : n/2) and worst-case (n-1 : 1) What happens if we bad-split root node, then good-split the resulting size (n-1) node?  We end up with three subarrays, size 1, (n-1)/2, (n-1)/2  Combined cost of splits = n + n -1 = 2n -1 = O(n)  No worse than if we had good-split the root node! Sandeep Kumar Poonia

Review: Analyzing Quicksort (Average Case) Intuitively, the O(n) cost of a bad split (or 2 or 3 bad splits) can be absorbed into the O(n) cost of each good split  Thus running time of alternating bad and good splits is still O(n lg n), with slightly higher constants  How can we be more rigorous?  Sandeep Kumar Poonia

Analyzing Quicksort: Average Case  For simplicity, assume:   All inputs distinct (no repeats) Slightly different partition() procedure  partition around a random element, which is not included in subarrays  all splits (0:n-1, 1:n-2, 2:n-3, … , n-1:0) equally likely What is the probability of a particular split happening?  Answer: 1/n  Sandeep Kumar Poonia

Analyzing Quicksort: Average Case So partition generates splits (0:n-1, 1:n-2, 2:n-3, … , n-2:1, n-1:0) each with probability 1/n  If T(n) is the expected running time,  1 n1 T n    T k   T n  1  k   n  n k 0 What is each term under the summation for?  What is the (n) term for?  Sandeep Kumar Poonia

Analyzing Quicksort: Average Case  So… 1 n 1 T n    T k   T n  1  k   n  n k 0 2 n 1   T k   n  n k 0 Sandeep Kumar Poonia Write it on the board

Analyzing Quicksort: Average Case  We can solve this recurrence using the dreaded substitution method     Guess the answer Assume that the inductive hypothesis holds Substitute it in for some value < n Prove that it follows for n Sandeep Kumar Poonia

Analyzing Quicksort: Average Case  We can solve this recurrence using the dreaded substitution method  Guess the answer  What’s    the answer? Assume that the inductive hypothesis holds Substitute it in for some value < n Prove that it follows for n Sandeep Kumar Poonia

Analyzing Quicksort: Average Case  We can solve this recurrence using the dreaded substitution method  Guess the answer  T(n)    = O(n lg n) Assume that the inductive hypothesis holds Substitute it in for some value < n Prove that it follows for n Sandeep Kumar Poonia

Analyzing Quicksort: Average Case  We can solve this recurrence using the dreaded substitution method  Guess the answer  T(n)  = O(n lg n) Assume that the inductive hypothesis holds  What’s   the inductive hypothesis? Substitute it in for some value < n Prove that it follows for n Sandeep Kumar Poonia

Analyzing Quicksort: Average Case  We can solve this recurrence using the dreaded substitution method  Guess the answer  T(n)  Assume that the inductive hypothesis holds  T(n)   = O(n lg n)  an lg n + b for some constants a and b Substitute it in for some value < n Prove that it follows for n Sandeep Kumar Poonia

Analyzing Quicksort: Average Case  We can solve this recurrence using the dreaded substitution method  Guess the answer  T(n)  Assume that the inductive hypothesis holds  T(n)  = O(n lg n)  an lg n + b for some constants a and b Substitute it in for some value < n  What  value? Prove that it follows for n Sandeep Kumar Poonia

Analyzing Quicksort: Average Case  We can solve this recurrence using the dreaded substitution method  Guess the answer  T(n)  Assume that the inductive hypothesis holds  T(n)   an lg n + b for some constants a and b Substitute it in for some value < n  The  = O(n lg n) value k in the recurrence Prove that it follows for n Sandeep Kumar Poonia

Analyzing Quicksort: Average Case  We can solve this recurrence using the dreaded substitution method  Guess the answer  T(n)  Assume that the inductive hypothesis holds  T(n)   an lg n + b for some constants a and b Substitute it in for some value < n  The  = O(n lg n) value k in the recurrence Prove that it follows for n  Grind Sandeep Kumar Poonia through it…

Analyzing Quicksort: Average Case 2 n 1 T n    T k   n  n k 0 2 n 1   ak lg k  b   n  n k 0 The recurrence to be solved Plug in inductive hypothesis What are we doing here? 2  n 1  What are we doing here?  b   ak lg k  b   n  Expand out the k=0 case n  k 1  2 n 1 2b   ak lg k  b    n  n k 1 n 2b/n is just a constant, What are we doing here? so fold it into (n) 2 n 1   ak lg k  b   n  n k 1 Note: leaving the same recurrence as the book Sandeep Kumar Poonia

Analyzing Quicksort: Average Case 2 n 1 T n    ak lg k  b   n  n k 1 2 n 1 2 n 1   ak lg k   b  n  n k 1 n k 1 The recurrence to be solved Distribute the summation What are we doing here? 2a n 1 2b the summation:  k lg k  (n  1)  n  Evaluateare we doing here? What  b+b+…+b = b (n-1) n k 1 n 2a n 1   k lg k  2b  n  n k 1 This summation gets its own set of slides later Sandeep Kumar Poonia Since n-1<n,we doing here? What are 2b(n-1)/n < 2b

Analyzing Quicksort: Average Case 2a n 1 T n    k lg k  2b  n  n k 1     The recurrence to be solved 2a  1 2 1 2 We’ll prove this  n lg n  n   2b  n  What the hell? later n 2 8  a an lg n  n  2b  n  Distribute the (2a/n) term What are we doing here? 4 a  Remember, our goal is to get  an lg n  b   n   b  n  What are we doing here? T(n)  an lg n + b 4   Pick a large enough that an lg n  b How did we do this? an/4 dominates (n)+b Sandeep Kumar Poonia

Analyzing Quicksort: Average Case  So T(n)  an lg n + b for certain a and b     Thus the induction holds Thus T(n) = O(n lg n) Thus quicksort runs in O(n lg n) time on average (phew!) Oh yeah, the summation… Sandeep Kumar Poonia

Tightly Bounding The Key Summation n 1 n 2 1 n 1 k 1 k 1 k  n 2  n 2 1 n 1 k 1 k  n 2   k lg k   k lg k   k lg k   k lg k   k lg n Sandeep Kumar Poonia n 1 k 1  n 2 1 k  n 2   k lg k  lg n  k Split the summation for a What are we doing here? tighter bound The lg k in the second term What are we doing here? is bounded by lg n Move the lg n outside the What are we doing here? summation

Tightly Bounding The Key Summation n 1  k lg k  k 1 n 2 1  k lg k  lg n k 1 n 1 k The summation bound so far k  n 2   n 1 k 1  n 2 1 k  n 2   k lgn 2  lg n  k n 2 1  k lg n  1  lg n k 1 n 1 k n 1 k 1 Sandeep Kumar Poonia lg n/2 = lg n we doing here? What are - 1 k  n 2  n 2 1 k  n 2   lg n  1 The lg k in the first term is What are we doing here? bounded by lg n/2  k  lg n  k Move (lg n - 1) outside the What are we doing here? summation

Tightly Bounding The Key Summation n 2 1 n 1  k lg k  lg n  1 k 1  lg n  k  lg n k 1 n 2 1 k  k 1 k  k  lg n k 1 n 1 k Distribute the (lg n - 1) What are we doing here? k  n 2  n 2 1 k 1 The summation bound so far k  n 2  n 2 1 n 1 k 1  lg n k  k  n  1(n)   lg n  2   Sandeep Kumar Poonia n 1 The summations overlap in What are we doing here? range; combine them n 2 1 k k 1 The Guassian series here? What are we doing

Tightly Bounding The Key Summation  n  1(n)   k lg k   2  lg n   k   k 1 k 1 n 2 1 1  nn  1lg n   k 2 k 1 n 2 1 n 1 The summation bound so far Rearrange first term, place What are we doing here? upper bound on second 1 1  n  n   nn  1lg n     1 2 2  2  2  1 2 1 2 n  n lg n  n lg n  n  2 8 4  Sandeep Kumar Poonia  X Guassian series What are we doing? Multiply it What are we doing? all out

Tightly Bounding The Key Summation  n 1  1 2 1 2 n  k lg k  2 n lg n  n lg n  8 n  4 k 1 1 2 1 2  n lg n  n when n  2 2 8 Done!!! Sandeep Kumar Poonia

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