Lecture 3. Recurrences / Heapsort
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1 Lecture 3. Recurrences / Heapsort T. H. Cormen, C. E. Leiserson and R. L. Rivest Introduction to Algorithms, 3rd Edition, MIT Press, 2009 Sungkyunkwan University Hyunseung Choo choo@skku.edu Copyright Networking Laboratory
2 Overview for Recurrences Define what a recurrence is Discuss three methods of solving recurrences Substitution method Recursion-tree method Master method Examples of each method Algorithms Networking Laboratory 2/82
3 Definition A recurrence is an equation or inequality that describes a function in terms of its value on smaller inputs. Example from MERGE-SORT T(n) = (1) if n=1 2T(n/2) + (n) if n>1 Algorithms Networking Laboratory 3/82
4 Technicalities Normally, independent variables only assume integral values Example from MERGE-SORT revisited T(n) = (1) if n=1 T( n/2 ) + T( n/2 ) + (n) if n>1 For simplicity, ignore floors and ceilings often insignificant Algorithms Networking Laboratory 4/82
5 Technicalities Boundary conditions (small n) are also glossed over T(n) = 2T(n/2) + (n) Value of T(n) assumed to be small constant for small n Algorithms Networking Laboratory 5/82
6 Substitution Method Involves two steps: Drawback Guess the form of the solution Use mathematical induction to find the constants and show the solution works Applied only in cases where it is easy to guess at solution Useful in estimating bounds on true solution even if latter is unidentified Algorithms Networking Laboratory 6/82
7 Substitution Method Example: T(n) = 2T( n/2 ) + n Guess: T(n) = O(n lg n) Prove by induction: T(n) cn lg n for suitable c>0. Algorithms Networking Laboratory 7/82
8 Inductive Proof We ll not worry about the basis case for the moment we ll choose this as needed clearly we have: T(1) = (1) cn lg n Inductive hypothesis: For values of n < k the inequality holds, i.e., T(n) cn lg n We need to show that this holds for n = k as well. Algorithms Networking Laboratory 8/82
9 Inductive Proof In particular, for n = k/2, the inductive hypothesis should hold, i.e., T( k/2 ) c k/2 lg k/2 The recurrence gives us: T(k) = 2T( k/2 ) + k Substituting the inequality above yields: T(k) 2[c k/2 lg k/2 ] + k Algorithms Networking Laboratory 9/82
10 Inductive Proof Because of the non-decreasing nature of the functions involved, we can drop the floors and obtain: T(k) 2[c (k/2) lg (k/2)] + k Which simplifies to: T(k) ck (lg k lg 2) + k Or, since lg 2 = 1, we have: T(k) ck lg k ck + k = ck lg k + (1 c)k So if c 1, T(k) ck lg k Q.E.D. Algorithms Networking Laboratory 10/82
11 Practice Problems Use the substitution method to show that T ( n) 7T ( n / 3) n T (1) 1 2 T( n) O( n 2 ) Algorithms Networking Laboratory 11/82
12 Recursion-Tree Method Straightforward technique of coming up with a good guess Can help the Substitution Method Recursion tree: visual representation of recursive call hierarchy where each node represents the cost of a single subproblem Algorithms Networking Laboratory 12/82
13 Recursion-Tree Method T(n) = 3T( n/4 ) + (n 2 ) Algorithms Networking Laboratory 13/82
14 Recursion-Tree Method T(n) = 3T( n/4 ) + (n 2 ) Algorithms Networking Laboratory 14/82
15 Recursion-Tree Method T(n) = 3T( n/4 ) + (n 2 ) Algorithms Networking Laboratory 15/82
16 Recursion-Tree Method T(n) = 3T( n/4 ) + (n 2 ) Algorithms Networking Laboratory 16/82
17 Recursion-Tree Method Gathering all the costs together: log T(n) = 4 n 1 (3/16) i cn 2 + (n log 43 ) i=0 T(n) i=0 (3/16) i cn 2 + o(n) T(n) (1/(1 3/16))cn 2 + o(n) T(n) (16/13)cn 2 + o(n) T(n) = O(n 2 ) Algorithms Networking Laboratory 17/82
18 Recursion-Tree Method T(n) = T(n/3) + T(2n/3) + O(n) Algorithms Networking Laboratory 18/82
19 Recursion-Tree Method An overestimate of the total cost: Counter-indications: log T(n) = 3/2 n 1 cn + (n log 3/22 ) i=0 T(n) = O(n lg n) + (n lg n) Notwithstanding this, use as guess : T(n) = O(n lg n) Algorithms Networking Laboratory 19/82
20 Substitution Method Recurrence: T(n) = T(n/3) + T(2n/3) + cn Guess: T(n) = O(n lg n) Prove by induction: T(n) dn lg n for suitable d>0 (we already use c) Algorithms Networking Laboratory 20/82
21 Inductive Proof Again, we ll not worry about the basis case Inductive hypothesis: For values of n < k the inequality holds, i.e., T(n) dn lg n We need to show that this holds for n = k as well. In particular, for n = k/3, and n = 2k/3, the inductive hypothesis should hold Algorithms Networking Laboratory 21/82
22 Inductive Proof That is T(k/3) d k/3 lg k/3 T(2k/3) d 2k/3 lg 2k/3 The recurrence gives us: T(k) = T(k/3) + T(2k/3) + ck Substituting the inequalities above yields: T(k) [d (k/3) lg (k/3)] + [d (2k/3) lg (2k/3)] + ck Algorithms Networking Laboratory 22/82
23 Inductive Proof Expanding, we get T(k) [d (k/3) lg k d (k/3) lg 3] + [d (2k/3) lg k d (2k/3) lg(3/2)] + ck Rearranging, we get: T(k) dk lg k d[(k/3) lg 3 + (2k/3) lg(3/2)] + ck T(k) dk lg k dk[lg 3 2/3] + ck When d c/(lg3 (2/3)), we should have the desired: T(k) dk lg k Algorithms Networking Laboratory 23/82
24 Practice Problems Use the recursion tree method to show that T( n) T( n/ 4) T( n/ 2) n 2 T( n) ( n 2 ) Algorithms Networking Laboratory 24/82
25 Master Method Provides a cookbook method for solving recurrences Recurrence must be of the form: T(n) = at(n/b) + f(n) where a 1 and b>1 are constants and asymptotically positive function. f(n) is an Algorithms Networking Laboratory 25/82
26 Master Method Theorem 4.1: Given the recurrence previously defined, we have: 1. If f(n) = O(n logba ) for some constant >0, then T(n) = (n logba ) 2. If f(n) = (n logba ), then T(n) = (n logba lg n) Algorithms Networking Laboratory 26/82
27 Master Method 3. If f(n) = (n logba+ ) for some constant >0, and if af(n/b) cf(n) for some constant c<1 and all sufficiently large n, then T(n) = (f(n)) Algorithms Networking Laboratory 27/82
28 Example Estimate bounds on the following recurrence: Use the recursion tree method to arrive at a guess then verify using induction Point out which case in the Master Method this falls in Algorithms Networking Laboratory 28/82
29 Recursion Tree Recurrence produces the following tree: Algorithms Networking Laboratory 29/82
30 Cost Summation Collecting the level-by-level costs: A geometric series with base less than one; converges to a finite sum, hence, T(n) = (n 2 ) Algorithms Networking Laboratory 30/82
31 Exact Calculation If an exact solution is preferred: Using the formula for a partial geometric series: Algorithms Networking Laboratory 31/82
32 Exact Calculation Solving further: Algorithms Networking Laboratory 32/82
33 Master Theorem (Simplified) Algorithms Networking Laboratory 33/82
34 Practice Problems Use Master theorem to find asymptotic bound. a. T( n) 4T ( n/ 2) n b. T( n) 4T ( n/ 2) n 2 c. T( n) 4T ( n/ 2) n 3 Algorithms Networking Laboratory 34/82
35 Introduction for Heapsort Heapsort Running time: O(n lg n) Like merge sort Sorts in place: only a constant number of array elements are stored outside the input array at any time Heap Like insertion sort A data structure used by Heapsort to manage information during the execution of the algorithm Can be used as an efficient priority queue Algorithms Networking Laboratory 35/82
36 Perfect Binary Tree For binary tree with height h All nodes at levels h 1 or less have 2 children (full) h = 0 h = 1 h = 2 h = 3 Algorithms Networking Laboratory 36/82
37 Complete Binary Trees For binary tree with height h All nodes at levels h 2 or less have 2 children (full) All leaves on level h are as far left as possible h = 0 h = 2 h = 1 Algorithms Networking Laboratory 37/82
38 Complete Binary Trees h = 3 Algorithms Networking Laboratory 38/82
39 Heaps Two key properties Complete binary tree Value at node Smaller than or equal to values in subtrees Greater than or equal to values in subtrees Example max-heap Y X Z X X Y Z Algorithms Networking Laboratory 39/82
40 Heap and Non-heap Examples Min-heaps Non-heaps Algorithms Networking Laboratory 40/82
41 Binary Heap An array object that can be viewed as a nearly complete binary tree Each tree node corresponds to an array element that stores the value in the tree node The tree is completely filled on all levels except possibly the lowest, which is filled from the left up to a point A has two attributes length[a]: number of elements in the array heap-size[a]: number of elements in the heap stored within A heap-size[a] length[a] max-heap and min-heap Algorithms Networking Laboratory 41/82
42 A Max-heap Algorithms Networking Laboratory 42/82
43 Length and Heap-Size Length = 10 Heap-Size = 7 Algorithms Networking Laboratory 43/82
44 Heap Computation Given the index i of a node, the indices of its parent, left child, and right child can be computed simply: PARENT ( i) : return LEFT( i) : return 2i RIGHT( i) : return 2i 1 i / 2 Algorithms Networking Laboratory 44/82
45 0 1 2 Heap Computation parent(i) = floor(i/2) left-child(i) = 2i right-child(i)= 2i Algorithms Networking Laboratory 45/82
46 Heap Property Heap property The property that the values in the node must satisfy Max-heap property, for every node i other than the root A[PARENT(i)] A[i] The value of a node is at most the value of its parent The largest element in a max-heap is stored at the root The subtree rooted at a node contains values on larger than that contained at the node itself Algorithms Networking Laboratory 46/82
47 Heap Height The height of a node in a heap The number of edges on the longest simple downward path from the node to a leaf The height of a heap is the height of its root The height of a heap of n elements is (lg n) Exercise on page 129 Algorithms Networking Laboratory 47/82
48 Heap Procedures MAX-HEAPIFY Maintains the max-heap property O(lg n) BUILD-MAX-HEAP Produces a max-heap from an unordered input array O(n) HEAPSORT Sorts an array in place O(n lg n) Algorithms Networking Laboratory 48/82
49 Maintaining the Heap Property MAX-HEAPIFY Inputs: an array A and an index i into the array Assume the binary tree rooted at LEFT(i) and RIGHT(i) are max-heaps, but A[i] may be smaller than its children violate the max-heap property MAX-HEAPIFY let the value at A[i] floats down in the max-heap Algorithms Networking Laboratory 49/82
50 Example of MAX-HEAPIFY 16 4 < Algorithms Networking Laboratory 50/82
51 Example of MAX-HEAPIFY < Algorithms Networking Laboratory 51/82
52 Example of MAX-HEAPIFY Algorithms Networking Laboratory 52/82
53 MAX-HEAPIFY Extract the indices of LEFT and RIGHT children of i Choose the largest of A[i], A[l], A[r] Float down A[i] recursively Algorithms Networking Laboratory 53/82
54 MAX-HEAPIFY Interactive version Recursive version Algorithms Networking Laboratory 54/82
55 Running Time of MAX-HEAPIFY (1) to find out the largest among A[i], A[LEFT(i)], and A[RIGHT(i)] Plus the time to run MAX-HEAPIFY on a subtree rooted at one of the children of node i The children s subtrees each have size at most 2n/3 (why?) the worst case occurs when the last row of the tree is exactly half full T(n) T(2n/3) + (1) By case 2 of the master theorem T(n) = O(lg n) 7/11 = 0.63 Algorithms Networking Laboratory 55/82
56 Heapify Example Algorithms Networking Laboratory 56/82
57 Building a Max-Heap Observation: A[( n/2 +1)..n] are all leaves of the tree Exercise on page 130 Each is a 1-element heap to begin with Upper bound on the running time O(lg n) for each call to MAX-HEAPIFY, and call n times O(n lg n) Not tight Algorithms Networking Laboratory 57/82
58 Building a Max-Heap Algorithms Networking Laboratory 58/82
59 Loop Invariant At the start of each iteration of the for loop of lines 2-3, each node i+1, i+2,.., n is the root of a max-heap Initialization: Prior to the first iteration of the loop, i = n/2. Each node n/2 +1, n/2 +2,.., n is a leaf and the root of a trivial max-heap. Maintenance: Observe that the children of node i are numbered higher than i. By the loop invariant, therefore, they are both roots of max-heaps. This is precisely the condition required for the call MAX-HEAPIFY(A, i) to make node i a max-heap root. Moreover, the MAX-HEAPIFY call preserves the property that nodes i+1, i+2,, n are all roots of max-heaps. Decrementing i in the for loop update reestablishes the loop invariant for the next iteration. Termination: At termination, i=0. By the loop invariant, each node 1, 2,, n is the root of a max-heap. In particular, node 1 is. Algorithms Networking Laboratory 59/82
60 Cost for Build-MAX-HEAP Heap-properties of an n-element heap Height = lg n At most n/2 h+1 nodes of any height h Exercise on page 135 lg n lg n n h h O( h) O( n ) O( n ) O( ) h 0 h h 2 h 0 2 h 0 n 1 h 2 Ignore the constant ½ h 0 h h (1 1 ) x k kx 2 k 0 ( 1 x) (for x < 1) Algorithms Networking Laboratory 60/82
61 Heapsort Using BUILD-MAX-HEAP to build a max-heap on the input array A[1..n], where n=length[a] Put the maximum element, A[1], to A[n] Then discard node n from the heap by decrementing heap-size(a) A[2..n-1] remain max-heaps, but A[1] may violate call MAX-HEAPIFY(A, 1) to restore the max-heap property for A[1..n-1] Repeat the above process from n down to 2 Cost: O(n lg n) BUILD-MAX-HEAP: O(n) Each of the n-1 calls to MAX-HEAPIFY takes time O(lg n) Algorithms Networking Laboratory 61/82
62 Example: Heapsort Algorithms Networking Laboratory 62/82
63 Example: Heapsort (2) Algorithms Networking Laboratory 63/82
64 Example: Heapsort (3) Algorithms Networking Laboratory 64/82
65 Example: Heapsort (4) Algorithms Networking Laboratory 65/82
66 Example: Heapsort (5) Algorithms Networking Laboratory 66/82
67 Example: Heapsort (6) Algorithms Networking Laboratory 67/82
68 Example: Heapsort (7) Algorithms Networking Laboratory 68/82
69 Example: Heapsort (8) Algorithms Networking Laboratory 69/82
70 Heapsort Algorithm Algorithms Networking Laboratory 70/82
71 Heapsort Algorithm Algorithms Networking Laboratory 71/82
72 Heap & Heap Sort Algorithm Video Content An illustration of Heap and Heap Sort. Algorithms Networking Laboratory 72/82
73 Heap & Heap Sort Algorithm Algorithms Networking Laboratory 73/82
74 Priority Queues We can implement the priority queue ADT with a heap. The operations are: Max(S) returns the maximum element Extract-Max(S) remove and return the maximum element Insert(x,S) insert element x into S Increase-Key(S,x,k) increase x s value to k Algorithms Networking Laboratory 74/82
![Extract-Max (1) Heap-Maximum(A) return A[1] O(lg n) Heap-Extract-Max(A) 1. if heapsize[a] < 1 2. then error heap underflow 3.](/docs-images/73/68800090/images/75-1.jpg "max A[1] 4. A[1] A[heapsize[A]] 5. heapsize[a] heapsize[a] 1 6. Max-Heapify(A,1) 7.")
75 Extract-Max (1) Heap-Maximum(A) return A[1] O(lg n) Heap-Extract-Max(A) 1. if heapsize[a] < 1 2. then error heap underflow 3. max A[1] 4. A[1] A[heapsize[A]] 5. heapsize[a] heapsize[a] 1 6. Max-Heapify(A,1) 7. return max Algorithms Networking Laboratory 75/82
76 Increase-Key O(lg n) Heap-Increase-key(A, i, key) 1. if key < A[i] 2. then error new key smaller than existing one 3. A[i] key 4. while i > 1 and A[parent(i)] < A[i] 5. do Exchange(A[i], parent(a[i])) 6. i parent(i) Algorithms Networking Laboratory 76/82
77 Example: increase key (1) increase 4 to 15 Algorithms Networking Laboratory 77/82
78 Example: increase key (2) Algorithms Networking Laboratory 78/82
79 Example: increase key (3) Algorithms Networking Laboratory 79/82
80 Example: increase key (4) Algorithms Networking Laboratory 80/82
81 Insert-Max O(lg n) Heap-Insert-Max(A, key) 1. heapsize[a] heapsize[a] A[heapsize[A]] - 3. Heap-Increase-Key(A, heapsize[a], key) Algorithms Networking Laboratory 81/82
82 Practice Problems Show the resulting heap after insert 20 into the following heap Algorithms Networking Laboratory 82/82
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