1 / 68

COSC 3101A - Design and Analysis of Algorithms 3

COSC 3101A - Design and Analysis of Algorithms 3. Recurrences Master’s Method Heapsort and Priority Queue. Many of the slides are taken from Monica Nicolescu’s slides, Univ. Nevada, Reno, monica@cs.unr.edu. Fibonacci numbers. Leonardo Pisano

ivria
Télécharger la présentation

COSC 3101A - Design and Analysis of Algorithms 3

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. COSC 3101A - Design and Analysis of Algorithms3 Recurrences Master’s Method Heapsort and Priority Queue Many of the slides are taken from Monica Nicolescu’s slides, Univ. Nevada, Reno, monica@cs.unr.edu

  2. Fibonacci numbers Leonardo Pisano Born: 1170 in (probably) Pisa (now in Italy)Died: 1250 in (possibly) Pisa (now in Italy) • F(n) = F(n-1) + F(n-2) • F(1) = 0, F(2) = 1 • F(3) = 1, F(4) = 2, F(5) = 3, and so on 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ... “A certain man put a pair of rabbits in a place surrounded on all sides by a wall. How many pairs of rabbits can be produced from that pair in a year if it is supposed that every month each pair begets a new pair which from the second month on becomes productive? “ COSC3101A

  3. Review: Recursive Algorithms • A recursive algorithm is an algorithm which solves a problem by repeatedly reducing the problem to a smaller version of itself .. until it reaches a form for which a solution exists. • Compared with iterative algorithms: • More memory • More computation • Simpler, natural way of thinking about the problem COSC3101A

  4. Review: Recursive Example (1) N! (N factorial) is defined for positive values as: N! = 0 if N = 0 N! = N * (N-1) * (N-2) * (N-3) * … * 1 if N > 0 For example: 5! = 5*4*3*2*1 = 120 3! = 3*2*1 = 6 The definition of N! can be restated as: N! = N * (N-1)! Where 0! = 1 This is the same N!, but it is defined by reducing the problem to a smaller version of the original (hence, recursively).  COSC3101A

  5. Review: Recursive Example (2) int factorial(int num){ if (num == 0) return 1; else return num * factorial(num – 1); } Assume that factorial is called with the argument 4:  Call to factorial(4) will return  4 * value returned by factorial(3)  Call to factorial(3) will return  3 * value returned by factorial(2) Call to factorial(2) will return  2 * value returned by factorial(1) Call to factorial(1) will return  1 * value returned by factorial(0)   Call to factorial(0) returns  1 COSC3101A

  6. Review: Recursive Example (3) The call to factorial(0) has returned, so now factorial(1) can finish: Call to factorial(1) returns  1 * value returned by factorial(0) 1 => 1 The call to factorial(1) has returned, so now factorial(2) can finish: Call to factorial(2) returns  2 * value returned by factorial(1) 1 => 2 The call to factorial(2) has returned, so now factorial(3) can finish: Call to factorial(3) returns  3 * value returned by factorial(2) 2 => 6 The call to factorial(3) has returned, so now factorial(4) can finish: Call to factorial(4) returns  4 * value returned by factorial(3) 6=>24 recursion results in a large number of ‘activation records’ (one per method call) being placed on the system stack COSC3101A

  7. Recurrences Def.: Recurrence = an equation or inequality that describes a function in terms of its value on smaller inputs, and one or more base cases E.g.: Fibonacci numbers: • Recurrence: F(n) = F(n-1) + F(n-2) • Boundary conditions: F(1) = 0, F(2) = 1 • Compute: F(3) = 1, F(4) = 2, F(5) = 3, and so on In many cases, the running time of an algorithm is expressed as a recurrence! COSC3101A

  8. Recurrences and Running Time • Recurrences arise when an algorithm contains recursive calls to itself • What is the actual running time of the algorithm? • Need to solve the recurrence • Find an explicit formula of the expression (the generic term of the sequence) COSC3101A

  9. Typical Recurrences Running Time • T(n) = T(n-1) + n Θ(n2) • Recursive algorithm that loops through the input to eliminate one item • T(n) = T(n/2) + c Θ(lgn) • Recursive algorithm that halves the input in one step • T(n) = T(n/2) + n Θ(n) • Recursive algorithm that halves the input but must examine every item in the input • T(n) = 2T(n/2) + 1 Θ(n) • Recursive algorithm that splits the input into 2 halves and does a constant amount of other work COSC3101A

  10. Recurrences - Intuition • For a recurrence of the type: • It takes f(n) to make the processing for the problem of size n • The algorithm divides the problem into a subproblems, each of size n/b • T(n) = number of subproblems * Running time(n/b) + processing of the problem of size n COSC3101A

  11. Methods for Solving Recurrences • Iteration Method • Substitution Method • Recursion Tree Method • Master Method COSC3101A

  12. Iteration Method • Expand (iterate) the recurrence • Express The function as a summation of terms dependent only on n and the initial condition COSC3101A

  13. Iteration Method – Example(1) T(n) = c + T(n/2) T(n) = c + T(n/2) T(n/2) = c + T(n/4) = c + c + T(n/4) T(n/4) = c + T(n/8) = c + c + c + T(n/8) Assume n = 2k T(n) = c + c + … + c + T(1) = clgn + T(1) = Θ(lgn) k times COSC3101A

  14. Iteration Method – Example(2) T(n) = n + 2T(n/2) T(n) = n + 2T(n/2) T(n/2) = n/2 + 2T(n/4) = n + 2(n/2 + 2T(n/4)) = n + n + 4T(n/4) = n + n + 4(n/4 + 2T(n/8)) = n + n + n + 8T(n/8) … = in + 2iT(n/2i) = kn + 2kT(1) = nlgn + nT(1) = Θ(nlgn) Assume: n = 2k COSC3101A

  15. Substitution method (1) • Guess a solution • Experiences, creativity • iteration method, recursion-tree method • Use induction to prove that the solution works COSC3101A

  16. Substitution method (2) • Guess a solution • T(n) = O(g(n)) • Induction goal: apply the definition of the asymptotic notation • T(n) ≤ c g(n), for some c > 0 and n ≥ n0 • Induction hypothesis: T(k) ≤ c g(k) for all k ≤ n • Prove the induction goal • Use the induction hypothesis to find some values of the constants c and n0 for which the induction goal holds COSC3101A

  17. Substitution Method – Example 1-1 T(n) = T(n-1) + n • Guess: T(n) = O(n2) • Induction goal: T(n) ≤ c n2, for some c and n ≥ n0 • Induction hypothesis: T(k) ≤ ck2 for all k ≤ n • Proof of induction goal: T(n) = T(n-1) + n ≤ c (n-1)2 + n = cn2 – (2cn – c - n) ≤ cn2 if: 2cn – c – n ≥ 0  c ≥ n/(2n-1)  c ≥ 1/(2 – 1/n) • For n ≥ 1  2 – 1/n ≥ 1 any c ≥ 1 will work COSC3101A

  18. Substitution Method – Example 1-2 T(n) = T(n-1) + n • Boundary conditions: • Base case: n0 = 1  T(1) = 1 has to verify condition: T(1) ≤ c (1)2 1 ≤ c OK! • We can similarly prove that T(n) = (n2)<your practice> • And therefore: T(n) = (n2) COSC3101A

  19. Substitution Method – Example 2-1 T(n) = 2T(n/2) + n • Guess: T(n) = O(nlgn) • Induction goal: T(n) ≤ cn lgn, for some c and n ≥ n0 • Induction hypothesis: T(n/2) ≤ cn lg(n/2) • Proof of induction goal: T(n) = T(n/2) + n ≤ 2c (n/2)lg(n/2) + n = cn lgn – cn + n ≤ cn lgn if: - cn + n ≤ 0  c ≥ 1 COSC3101A

  20. Substitution Method – Example 2-2 T(n) = 2T(n/2) + n • Boundary conditions: • Base case: n0 = 1  T(1) = 1 has to verify condition: T(1) ≤ cn0lgn0 1 ≤ c * 1 * lg1 = 0 – contradiction • Choose n0 = 2  T(2) = 4 has to verify condition: T(2) ≤ c * 2 * lg2  4 ≤ 2c  choose c = 2 • We can similarly prove that T(n) = (nlgn)<your practice> • And therefore: T(n) = (nlgn) COSC3101A

  21. Changing variables T(n) = 2T( ) + lgn • Rename: m = lgn n = 2m T (2m) = 2T(2m/2) + m • Rename: S(m) = T(2m) S(m) = 2S(m/2) + m  S(m) = O(mlgm) (demonstrated before) T(n) = T(2m) = S(m) = O(mlgm)=O(lgnlglgn) Idea: transform the recurrence to one that you have seen before COSC3101A

  22. Recursion-tree method Convert the recurrence into a tree: • Each node represents the cost incurred at various levels of recursion • Sum up the costs of all levels Used to “guess” a solution for the recurrence COSC3101A

  23. W(n) = 2W(n/2) + n2 Subproblem size at level i is: n/2i Subproblem size hits 1 when 1 = n/2i  i = lgn Cost of the problem at level i = n2/2iNo. of nodes at level i = 2i Total cost:  W(n) = O(n2) Recursion-tree Example 1 COSC3101A

  24. Recursion-tree Example 2 E.g.:T(n) = 3T(n/4) + cn2 • Subproblem size at level i is: n/4i • Subproblem size hits 1 when 1 = n/4i  i = log4n • Cost of a node at level i = c(n/4i)2 • Number of nodes at level i = 3i last level has 3log4n = nlog43nodes • Total cost:  T(n) = O(n2) COSC3101A

  25. regularity condition Master method • “Cookbook” for solving recurrences of the form: where, a ≥ 1, b > 1, and f(n) > 0 Case 1: if f(n) = O(nlogba-) for some  > 0, then: T(n) = (nt) Case 2: if f(n) = (nlogba), then: T(n) = (nlogba lgn) Case 3: if f(n) = (nlogba+) for some  > 0, and if af(n/b) ≤ cf(n) for some c < 1 and all sufficiently large n, then: T(n) = (f(n)) COSC3101A

  26. Why nlogba? • Case 1: • If f(n) is dominated by nlogba: • T(n) = (nlogbn) • Case 3: • If f(n) dominates nlogba: • T(n) = (f(n)) • Case 2: • If f(n) = (nlogba): • T(n) = (nlogba logn) • Assume n = bk k = logbn • At the end of iteration i = k: COSC3101A

  27. Examples (1) T(n) = 2T(n/2) + n a = 2, b = 2, log22 = 1 Comparenlog22withf(n) = n  f(n) = (n)  Case 2  T(n) = (nlgn) COSC3101A

  28. Examples (2) T(n) = 2T(n/2) + n2 a = 2, b = 2, log22 = 1 Comparenwithf(n) = n2  f(n) = (n1+) Case 3 verify regularity cond. a f(n/b) ≤ c f(n)  2 n2/4 ≤ c n2  c = ½ is a solution (c<1)  T(n) = (n2) COSC3101A

  29. Examples (3) T(n) = 2T(n/2) + a = 2, b = 2, log22 = 1 Comparen withf(n) = n1/2  f(n) = O(n1-) Case 1  T(n) = (n) COSC3101A

  30. Examples (4) T(n) = 3T(n/4) + nlgn a = 3, b = 4, log43 = 0.793 Comparen0.793 withf(n) = nlgn f(n) = (nlog43+)Case 3 Check regularity condition: 3(n/4)lg(n/4) ≤ (3/4)nlgn = c f(n), c=3/4 T(n) = (nlgn) COSC3101A

  31. Examples (5) T(n) = 2T(n/2) + nlgn a = 2, b = 2, log22 = 1 • Compare n with f(n) = nlgn • seems like case 3 should apply • f(n) must be polynomially larger by a factor of n • In this case it is only larger by a factor of lgn COSC3101A

  32. A Job Scheduling Application • Job scheduling • The key is the priority of the jobs in the queue • The job with the highest priority needs to be executed next • Operations • Insert, remove maximum • Data structures • Priority queues • Ordered array/list, unordered array/list COSC3101A

  33. Example COSC3101A

  34. PQ Implementations & Cost Worst-case asymptotic costs for a PQ with N items Insert Remove max ordered array N 1 ordered list N 1 unordered array 1 N unordered list 1 N Can we implement both operations efficiently? COSC3101A

  35. Background on Trees • Def:Binary tree = structure composed of a finite set of nodes that either: • Contains no nodes, or • Is composed of three disjoint sets of nodes: a root node, a left subtree and a right subtree root 4 Right subtree Left subtree 1 3 2 16 9 10 14 8 COSC3101A

  36. 4 4 1 3 1 3 2 16 9 10 2 16 9 10 14 8 7 12 Complete binary tree Full binary tree Special Types of Trees • Def:Full binary tree = a binary tree in which each node is either a leaf or has degree exactly 2. • Def:Complete binary tree = a binary tree in which all leaves have the same depth and all internal nodes have degree 2. COSC3101A

  37. The Heap Data Structure • Def:A heap is a nearly complete binary tree with the following two properties: • Structural property: all levels are full, except possibly the last one, which is filled from left to right • Order (heap) property: for any node x Parent(x) ≥ x 8 It doesn’t matter that 4 in level 1 is smaller than 5 in level 2 7 4 5 2 Heap COSC3101A

  38. Definitions • Height of a node = the number of edges on a longest simple path from the node down to a leaf • Depth of a node = the length of a path from the root to the node • Height of tree = height of root node = lgn, for a heap of n elements Height of root = 3 4 1 3 Height of (2)= 1 Depth of (10)= 2 2 16 9 10 14 8 COSC3101A

  39. Array Representation of Heaps • A heap can be stored as an array A. • Root of tree is A[1] • Left child of A[i] = A[2i] • Right child of A[i] = A[2i + 1] • Parent of A[i] = A[ i/2 ] • Heapsize[A] ≤ length[A] • The elements in the subarray A[(n/2+1) .. n] are leaves • The root is the maximum element of the heap A heap is a binary tree that is filled in order COSC3101A

  40. Heap Types • Max-heaps (largest element at root), have the max-heap property: • for all nodes i, excluding the root: A[PARENT(i)] ≥ A[i] • Min-heaps (smallest element at root), have the min-heap property: • for all nodes i, excluding the root: A[PARENT(i)] ≤ A[i] COSC3101A

  41. Operations on Heaps • Maintain the max-heap property • MAX-HEAPIFY • Create a max-heap from an unordered array • BUILD-MAX-HEAP • Sort an array in place • HEAPSORT • Priority queue operations COSC3101A

  42. Operations on Priority Queues • Max-priority queues support the following operations: • INSERT(S, x): inserts element x into set S • EXTRACT-MAX(S): removes and returns element of S with largest key • MAXIMUM(S): returns element of S with largest key • INCREASE-KEY(S, x, k): increases value of element x’s key to k (Assume k ≥ x’s current key value) COSC3101A

  43. Maintaining the Heap Property • Suppose a node is smaller than a child • Left and Right subtrees of i are max-heaps • Invariant: • the heap condition is violated only at that node • To eliminate the violation: • Exchange with larger child • Move down the tree • Continue until node is not smaller than children COSC3101A

  44. Maintaining the Heap Property Alg:MAX-HEAPIFY(A, i, n) • l ← LEFT(i) • r ← RIGHT(i) • ifl ≤ n and A[l] > A[i] • thenlargest ←l • elselargest ←i • ifr ≤ n and A[r] > A[largest] • thenlargest ←r • iflargest  i • then exchange A[i] ↔ A[largest] • MAX-HEAPIFY(A, largest, n) • Assumptions: • Left and Right subtrees of i are max-heaps • A[i] may be smaller than its children COSC3101A

  45. A[2] violates the heap property A[4] violates the heap property Heap property restored Example MAX-HEAPIFY(A, 2, 10) A[2]  A[4] A[4]  A[9] COSC3101A

  46. MAX-HEAPIFY Running Time • Intuitively: • A heap is an almost complete binary tree  must process O(lgn) levels, with constant work at each level • Running time of MAX-HEAPIFY is O(lgn) • Can be written in terms of the height of the heap, as being O(h) • Since the height of the heap is lgn COSC3101A

  47. Alg:BUILD-MAX-HEAP(A) n = length[A] fori ← n/2downto1 do MAX-HEAPIFY(A, i, n) Convert an array A[1 … n] into a max-heap (n = length[A]) The elements in the subarray A[(n/2+1) .. n] are leaves Apply MAX-HEAPIFY on elements from 1 to n/2 1 4 2 3 1 3 4 5 6 7 2 16 9 10 8 9 10 14 8 7 Building a Heap A: COSC3101A

  48. 1 1 1 1 4 16 4 4 2 2 3 3 2 3 2 3 14 1 10 3 1 3 1 3 4 4 5 5 6 6 7 7 4 5 6 7 4 5 6 7 8 2 16 7 9 9 10 3 2 16 9 10 14 16 9 10 8 8 9 9 10 10 8 9 10 8 9 10 14 2 8 4 1 7 14 8 7 2 8 7 1 1 4 4 2 3 2 3 1 10 16 10 4 5 6 7 4 5 6 7 14 16 9 3 14 7 9 3 8 9 10 8 9 10 2 8 7 2 8 1 Example: A i = 5 i = 4 i = 3 i = 2 i = 1 COSC3101A

  49. 1 4 2 3 1 3 4 5 6 7 2 16 9 10 8 9 10 14 8 7 Correctness of BUILD-MAX-HEAP • Loop invariant: • At the start of each iteration of the for loop, each node i + 1, i + 2,…, n is the root of a max-heap • Initialization: • i = n/2: Nodes n/2+1, n/2+2, …, n are leaves  they are the root of trivial max-heaps COSC3101A

  50. 1 4 2 3 1 3 4 5 6 7 2 16 9 10 8 9 10 14 8 7 Correctness of BUILD-MAX-HEAP • Maintenance: • MAX-HEAPIFY makes node i a max-heap root and preserves the property that nodes i + 1, i + 2, …, n are roots of max-heaps • Decrementing i in the for loop reestablishes the loop invariant • Termination: • i = 0  each node 1, 2, …, n is the root of a max-heap (by the loop invariant) COSC3101A

More Related