Lecture 5 Divide and Conquer: closest pair of points and numbers multiplication

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Lecture 5Divide and Conquer closest pair of
points andnumbers multiplication

• COMP 523 Advanced Algorithmic Techniques
• Lecturer Dariusz Kowalski

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Overview

• Previous lecture
• Sorting algorithms in (expected) time O(n log n)
• Merge Sort
• Quick Sort and randomisation
• This lecture
• Another kind of merging closest pair of points
• Beyond ?(n log n) numbers multiplication

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Problem

• Closest pair of points
• Input n points on the plane
• Output two points having the closest distance
• Remarks
• Exhaustive search algorithm gives time O(n2)
• Similar approach gives ALL pairs of closest points

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Solution

• Preprocessing
• Sort points according to the first coordinate
  (obtain horizontal list H) and according the
  second coordinate (obtain vertical list V)
• Main Algorithm
• Partition list H input into halves in linear time
  (draw vertical line L containing medium point)
• Solve the problem for each half of the input
  separately, obtaining two pairs of closest
  points let ? be the smallest distance from the
  obtained ones
• Find if there is a pair which has distance
  smaller than ? - if yes then find the smallest
  distance pair

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Main difficulty

• Find if there is a pair which has distance
  smaller than ? -
• if yes then find the smallest distance pair
• How to do it in linear time?

?
?
?
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Closest pair in the strip

• 1. Find a sublist P of V containing all points
  with a distance at most ? from the line L - in
  linear time
• 2. For each element in P check its distance to
  the next 8 elements and remember the closest pair
  ever found

?
?
?
L
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Why to check only 8 points ahead?

• 1. Partition the strip into squares of length ?/2
  as shown in the picture.
• 2. Each square contains at most 1 point - by
  definition of ?.
• 3. If there are at least 2 squares between points
  then they can not be the closest points.
• 4. There are at most 8 squares to check.

?
?/2
?/2
?/2
?/2
?/2
?/2
L
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Time complexity

• Preprocessing sorting in time O(n log n)
• Main Algorithm recursion in time O(n log n)
• T(n) ? 2 T(n/2) 9.5n
• T(2) 1

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Beyond ?(n log n) - integers multiplication

• Input two integers x and y, each consisting of n
  bits
• Output multiply these integers
• Naïve Algorithm
• Multiply each bit i of x by y, add (i-1) zeroes
  at the end
• Add the numbers
• Time ?(n2) of bit operations

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Divide and Conquer approach

• Let x x1 2n/2 x2 and y y1 2n/2 y2.
• Let p x1 x2 and q y1 y2. Then
• x y (x1 2n/2 x2) (y1 2n/2 y2)
• x1 y1 2n/2 (x2 y1 x1 y2) 2n x2 y2
• x1 y1 2n/2 (pq - x1 y1- x2 y2) 2n x2 y2
• Algorithm
• Compute p and q in linear time
• Compute recursively x1 y1 , x2 y2 , pq
• Perform x1 y1 2n/2 (pq - x1 y1- x2 y2) 2n x2
  y2 in linear time

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Time complexity

• T(n) ? 3T(n/2) c n
• T(1) ? c
• Solution T(n) ? d nlog 3, for some constant d ?
  3c
• Proof For n 1 straightforward. General case
• T(n) ? 3T(n/2) c n ? 9T(n/4) 3c(n/2) c n
• ? c n (1 3/2 (3/2)2 (3/2)log n)
• c n 2(3/2)log n 1 ? 3c n nlog 3 - 1
• 3c nlog 3 O(n1.59)

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Conclusions

• Divide and conquer - closest points algorithm in
  time O(n log n)
• Beyond ?(n log n) O(nlog 3) time algorithm for
  integer multiplication

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Textbook and Exercises

• Read Chapter 5 from Section 5.4
• How to add two n-bits integers in linear time?
• Solved Exercise 1 from the textbook, chapter 5
  Divide and Conquer
• Exercises 1, 6, 7 from the textbook, chapter 5
  Divide and Conquer