Lecture 7, Feb 23

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Lecture 7, Feb 23
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CSC373Algorithm Design and AnalysisAnnouncements
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Divide and Conquer
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Evaluating T(n) aT(n/b)nc logkn
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Sorting Problem Specification

• Precondition The input is a list of n values
  with the same value possibly repeated.
• Post condition The output is a list consisting
  of the same n values in non-decreasing order.

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Recursive Sorts

• Given list of objects to be sorted
• Split the list into two sublists.
• Recursively sort the two sublists.
• Combine the two sorted sublists into one entirely
  sorted list.

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Four Recursive Sorts
Size of Sublists
n/2,n/2
n-1,1
Minimal effort splitting Lots of effort
recombining Lots of effort splittingMinimal
effort recombining
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Merge Sort
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Merge Sort
Split Set into Two
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Merge Sort
Merge two sorted lists into one
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Merge Sort Sort
Time T(n)
2T(n/2) Q(n)
Q(n log(n))
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Quick Sort
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Quick Sort
Partition set into two using randomly chosen
pivot
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Quick Sort
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Quick Sort
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Quick Sort
Let pivot be the first element in the list?
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88
98
30
31
62
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25
79
52
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Quick Sort
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If the list is already sorted, then the slit is
worst case unbalanced.
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Quick Sort
Best Time
Worst Time
Expected Time
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Quick Sort
T(n) 2T(n/2) Q(n) Q(n log(n))
Best Time
Worst Time
Q(n2)
Expected Time
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Quick Sort
T(n) 2T(n/2) Q(n) Q(n log(n))
Best Time
T(n) T(0) T(n-1) Q(n)
Worst Time
Q(n2)
Expected Time
T(n) T(1/3n) T(2/3n) Q(n)
Q(n log(n))
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Quick Sort
Expected Time
T(n) T(1/3n) T(2/3n) Q(n)
Q(n log(n))
Top work Q(n)
Q(n)
2nd level
Q(n)
3rd level
levels
Q(log(n))
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Four Recursive Sorts
Size of Sublists
n/2,n/2
n-1,1
Minimal effort splitting Lots of effort
recombining Lots of effort splittingMinimal
effort recombining
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Kth Element Problem Specification

• Precondition The input is a list of n values
  and an integer k.
• Post condition The kth smallest in the list.
  (or the first k smallest)

k3
Output 25
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Kth Element Problem Specification
Partition set into two using randomly chosen
pivot
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Kth Element Problem Specification
88
98
52
62
79
r5 elements on left
k3
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Kth Element Problem Specification
88
98
52
62
79
r5 elements on left
k8
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Kth Element Problem Specification
T(n)
Best Time
1 T(n/2) Q(n)
Q(n)
Worst Time
n0 1
Expected Time
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Kth Element Problem Specification
T(n)
Best Time
1 T(n/2) Q(n)
Q(n)
T(n) 1 T(n-1) Q(n)
Worst Time
Q(n2)
Expected Time
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Kth Element Problem Specification
T(n)
Best Time
1 T(n/2) Q(n)
Q(n)
T(n) 1 T(n-1) Q(n)
Worst Time
Q(n2)
T(n) 1 T(2/3n) Q(n)
Expected Time
Q(n)
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Counting inversions

• Given a list of numbers. a1, a2, , an.
• Goal count the number of pairs (i,j) such that
  iltj but aigtaj.
• The naïve algorithm works in O(n2).
• Example input 1 4 3 2
• output 3

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Counting inversions - idea

• Divide the list into two halves
• a1, a2, , am and am1, a2, , an.
• Sort and compute the number of inversions in each
  half.
• Merge the two halves together, counting the
  inversions between elements from the two halves.
• We would like to do the merging in linear time.

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Merge-and-Count

• We count an inversion (a,b), when we insert b
  into the merged result C.
• When we insert from A, add nothing to the number
  of the inversions.

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Merge-and-Count

• When we insert bj from B, every element remaining
  in A is bigger than and contributes 1 to the
  number of inversions.
• The entire counting is done in linear time.

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The algorithm
O(n)
2 T(n/2)
O(n)
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The running time

• T(n) 2 T(n/2) O(n)
• log a/log b 1
• c 1
• T(n) O(n log n)

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Finding the Closest Pair of Points

• Given a set of points on the plane.
• The goal is to find the pair that is closest
  together.
• The naïve algorithm works in time O(n2) (checking
  all the pairs).
• We would like to find an algorithm that works in
  time O(n log n).

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Designing the algorithm

• Denote the points by Pp1, p2, , pn.
• pi(xi,yi).
• For simplicity assume that all the coordinates
  are different.
• For P, keep the orderings Px and Py of points in
  increasing x and y coordinate order,
  respectively.

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Divide

• Find L such that half of the points are to the
  left of L (Q), and hald are to the right (R).
• Done in O(log n) time.

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Conquer

• We recursively determine the closest pair of
  points (q0,q1) in Q, and (r0,r1) in R.

• Let ?min(d(q0,q1), d(r0,r1)).
• What if there are q in Q and r in R with d(q,r)lt
  ??

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Dealing with points near L

• So need to restrict attention to a narrow band
  near L.
• Denote this set of points by S.
• By a single pass through Py, we can order our set
  S and to get Sy in O(n) time.

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Key observation

• We will only need to look at a constant number of
  potential candidates for each point in S.
• The number 15 can be reduced.

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Proof ok Key observation

• No two points are in the same box.
• q and r with d(q,r)lt ? cannot be separated by
  more than 3 layers of boxes.
• If q is in the bottom row, r must be in one of
  the squares on the picture.

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Summary of the algorithm
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Summary of the algorithm 2
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Running time of the algorithm
O(1)
2 T(n/2)
O(n)
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Running time of the algorithm 2
O(n)
O(1)
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Running time 3

• The running time is given by
• T(n) 2 T(n/2) O(n).
• log a/log b 1
• c 1
• T(n) O(n log n)