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Randomized Algorithms

Randomized Algorithms. Eduardo Laber Loana T. Nogueira. Quicksort. Objective. Quicksort. Objective Sort a list of n elements. An Idea. An Idea. Imagine if we could find an element y  S such that half the members of S are smaller than y, then we could use the following scheme. An Idea.

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Randomized Algorithms

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  1. Randomized Algorithms Eduardo Laber Loana T. Nogueira

  2. Quicksort • Objective

  3. Quicksort • Objective • Sort a list of n elements

  4. An Idea

  5. An Idea Imagine if we could find an element y  S such that half the members of S are smaller than y, then we could use the following scheme

  6. An Idea Imagine if we could find an element y  S such that half the members of S are smaller than y, then we could use the following scheme • Partition S\{y} into two sets S1 and S2

  7. An Idea Imagine if we could find an element y  S such that half the members of S are smaller than y, then we could use the following scheme • Partition S\{y} into two sets S1 and S2 • S1: elements of S that are smaller than y • S2: elements of S that are greater than y

  8. An Idea Imagine if we could find an element y  S such that half the members of S are smaller than y, then we could use the following scheme • Partition S\{y} into two sets S1 and S2 • S1: elements of S that are smaller than y • S2: elements of S that are greater than y • Recursively sort S1 and S2

  9. Suppose we know how to find y

  10. Suppose we know how to find y Time to find y: cn steps, for some constant c

  11. Suppose we know how to find y Time to find y: cn steps, for some constant c we could partition S\{y} into S1 and S2 in n-1 additional steps

  12. Suppose we know how to find y Time to find y: cn steps, for some constant c we could partition S\{y} into S1 and S2 in n-1 additional steps The total number os steps in out sorting procedure would be given by the recurrence T(n)  2T(n/2) + (c+1)n

  13. Suppose we know how to find y Time to find y: cn steps, for some constant c we could partition S\{y} into S1 and S2 in n-1 additional steps The total number os steps in out sorting procedure would be given by the recurrence  c’nlogn T(n)  2T(n/2) + (c+1)n

  14. What´s the problem with the scheme above? Quicksort

  15. What´s the problem with the scheme above? Quicksort How to find y?

  16. Deterministic Quicksort • Let y be the first element of S

  17. Deterministic Quicksort • Let y be the first element of S • Split S into two sets: S< and S>

  18. Deterministic Quicksort • Let y be the first element of S • Split S into two sets: S< and S> • S< : elements smaller than y • S> : elements greater than y

  19. Deterministic Quicksort • Let y be the first element of S • Split S into two sets: S< and S> • S< : elements smaller than y • S> : elements greater than y • Qsort ( S< ), Qsort ( S> )

  20. Performance • Worst Case: O( n2 ) • Avarage Case: O( nlogn )

  21. Performance • Worst Case: O( n2 ) (The set is already sorted) • Avarage Case: O( nlogn )

  22. Performance • Worst Case: O( n2 )(The set is already sorted) • Avarage Case: O( nlogn )

  23. An Randomzied Algorithm

  24. An Randomzied Algorithm • An algorithm that makes choice (random) during the algorithm execution

  25. Randomized Quicksort (RandQS)

  26. Randomized Quicksort (RandQS) • Choose an element y uniformly at random of S

  27. Randomized Quicksort (RandQS) • Choose an element y uniformly at random of S • Every element of S has equal probability fo being chosen

  28. Randomized Quicksort (RandQS) • Choose an element y uniformly at random of S • Every element of S has equal probability fo being chosen • By comparing each element fo S with y, determine S< and S>

  29. Randomized Quicksort (RandQS) • Choose an element y uniformly at random of S • Every element of S has equal probability fo being chosen • By comparing each element fo S with y, determine S< and S> • Recursively sort S< and S>

  30. Randomized Quicksort (RandQS) • Choose an element y uniformly at random of S • Every element of S has equal probability fo being chosen • By comparing each element fo S with y, determine S< and S> • Recursively sort S< and S> • OUTPUT: S<, followed by y and then S>

  31. Intuition • For some instance Quicksort works very bad  O( n2 )

  32. Intuition • For some instance Quicksort works very bad  O( n2 ) • Randomization produces different executions for the same input. There is no instance for which RandQS works bad in avarage

  33. Analysis

  34. Analysis • For sorting algorithms: we measure the running time of RandQS in terms of the number of comparisons it performs

  35. Analysis • For sorting algorithms: we measure the running time of RandQS in terms of the number of comparisons it performs This is the dominat cost in any reasonable implementation

  36. Analysis • For sorting algorithms: we measure the running time of RandQS in terms of the number of comparisons it performs This is the dominat cost in any reasonable implementation Our Goal: Analyse the expected number of comparisons in an execution of RandQS

  37. Analysis Si: the ith smallest element of S

  38. Analysis Si: the ith smallest element of S S1 is the smallest element of S

  39. Analysis Si: the ith smallest element of S S1 is the smallest element of S Sn is the largest element of S

  40. Analysis Si: the ith smallest element of S Define the random variable S1 is the smallest element of S Sn is the largest element of S 1, if Si and Sj are compared xik= 0, otherwise

  41. Analysis Si: the ith smallest element of S Define the random variable Dado um experimento aleatório com espaço amostral S, uma variável aleatória é uma função que associa a cada elemento amostral um número real S1 is the smallest element of S Sn is the largest element of S 1, if Si and Sj are compared xik= 0, otherwise

  42. Analysis Xij is a count of comparisons between Si and Sj: n   The total number of comparisons: Xij i = 1 j > i

  43. Analysis Xij is a count of comparisons between Si and Sj: n   The total number of comparisons: Xij i = 1 j > i We are interested in the expected number of comparisons n   E[ ] Xij i = 1 j > i

  44. Analysis Xij is a count of comparisons between Si and Sj: n   The total number of comparisons: Xij i = 1 j > i By the linearity of E[] n We are interested in the expected number of comparisons n     E[Xij] E[ ] = Xij i = 1 j > i i = 1 j > i

  45. Analysis Pij: the probability that Si and Sj are compared in an execution

  46. Analysis Pij: the probability that Si and Sj are compared in an execution Since Xij only assumes the values 0 and 1, ij ij ij ij

  47. Analysis – Binary Tree T of RandQS • Each node is labeled with a distinct element of S T y

  48. Analysis – Binary Tree T of RandQS • Each node is labeled with a distinct element of S T y S<

  49. Analysis – Binary Tree T of RandQS • Each node is labeled with a distinct element of S T y S> S<

  50. Analysis – Binary Tree T of RandQS • Each node is labeled with a distinct element of S T y S> S< The root of T is compared to the elements in the two sub-trees, but no comparisons is perfomed between an element of the left and righ sub-trees

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