CS Lunch Mary Allen Wilkes Wednesday 12:15 Kendade 307 2 Divide - - PowerPoint PPT Presentation

CS Lunch Mary Allen Wilkes Wednesday 12:15 Kendade 307

1

Divide and Conquer

Divide-and-conquer. Break up problem into several parts. Solve each part recursively. Combine solutions to sub-problems into overall solution. Most common usage. Break up problem of size n into two equal parts of size n/2. Solve two parts recursively. Combine two solutions into overall solution in linear time.

2

Closest Pair of Points

Closest pair. Given n points in the plane, find a pair with smallest Euclidean distance between them. Fundamental geometric primitive. Graphics, computer vision, geographic information systems, molecular modeling, air traffic control. Brute force. Check all pairs of points p and q with Θ(n2) comparisons.

3 Slides14 - ClosestPairPoints.key - March 20, 2019

Closest Pair of Points

1-dimensional version 4

Closest Pair of Points

1-D version. Sort points For each point, find the distance between a point and the point that follows it. Remember the smallest.

5-1

Closest Pair of Points

1-D version. Sort points For each point, find the distance between a point and the point that follows it. Remember the smallest.

Cost

5-2 Slides14 - ClosestPairPoints.key - March 20, 2019

Closest Pair of Points

1-D version. Sort points For each point, find the distance between a point and the point that follows it. Remember the smallest.

Cost O(n log n)

5-3

Closest Pair of Points

1-D version. Sort points For each point, find the distance between a point and the point that follows it. Remember the smallest.

Cost O(n log n) O(n)

5-4

Closest Pair of Points

1-D version. Sort points For each point, find the distance between a point and the point that follows it. Remember the smallest.

Cost O(n log n) O(n) Total is O(n log n)

5-5 Slides14 - ClosestPairPoints.key - March 20, 2019

Closest Pair of Points

Divide: draw vertical line L so that n/2 points on each side.

L

6-1

Closest Pair of Points

Divide: draw vertical line L so that n/2 points on each side.

L

6-2

Closest Pair of Points

Solve: find closest pair in each side recursively.

L

7-1 Slides14 - ClosestPairPoints.key - March 20, 2019

Closest Pair of Points

Solve: find closest pair in each side recursively.

12 L

7-2

Closest Pair of Points

Solve: find closest pair in each side recursively.

12 21 L

7-3

Closest Pair of Points

Combine: find closest pair with one point in each side. Return best of 3 solutions. How do we do this without comparing each point on left with each point on right???

12 21 8 L

8 Slides14 - ClosestPairPoints.key - March 20, 2019

Closest Pair of Points

Let δ be the minimum between pair on left and pair on right If there exists a pair with one point in each side and whose distance < δ, find that pair.

12 21 δ = min(12, 21) L

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Closest Pair of Points

Observation: only need to consider points within δ of line L.

12 21 δ L δ = min(12, 21)

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12 21

1 2 3 4 5 6 7

Closest Pair of Points

Sort points in 2δ-strip by their y coordinate.

L δ = min(12, 21) δ

11 Slides14 - ClosestPairPoints.key - March 20, 2019

12 21

1 2 3 4 5 6 7

Closest Pair of Points

Only need to check distances of those within 15 positions in sorted list!!!!

L δ = min(12, 21) δ

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Closest Pair of Points

Let si be the point in the 2δ-strip, with the ith smallest y-coordinate.

• Claim. If |i – j| ≥ 16, then the distance

between si and sj is at least δ. Proof: No two points lie in same ½δ-by-½δ box. Two points separated by at least 3 rows have distance ≥ 3(½δ).

δ

27 29 30 31 28 26 25

δ

½δ

3 rows

½δ ½δ

39

i j

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Closest Pair of Points

Let si be the point in the 2δ-strip, with the ith smallest y-coordinate.

• Claim. If |i – j| ≥ 16, then the distance

between si and sj is at least δ. Proof: No two points lie in same ½δ-by-½δ box. Two points separated by at least 3 rows have distance ≥ 3(½δ). If a point is within δ of point 27, it must be in one of the blue boxes. There are only 15 blue boxes!

δ

27 29 30 31 28 26 25

δ

½δ

2 rows

½δ ½δ

39

i j

14 Slides14 - ClosestPairPoints.key - March 20, 2019

Closest Pair Algorithm

Closest-Pair(p1, …, pn) { Compute separation line L such that half the points
 are on one side and half on the other side. δ1 = Closest-Pair(left half) δ2 = Closest-Pair(right half) δ = min(δ1, δ2) Delete all points further than δ from separation line L Sort remaining points by y-coordinate. Scan points in y-order and calculate distance to all points with a y value within δ. If any of these distances is less than δ, update δ. return δ. }

15-1

Closest Pair Algorithm

Closest-Pair(p1, …, pn) { Compute separation line L such that half the points
 are on one side and half on the other side. δ1 = Closest-Pair(left half) δ2 = Closest-Pair(right half) δ = min(δ1, δ2) Delete all points further than δ from separation line L Sort remaining points by y-coordinate. Scan points in y-order and calculate distance to all points with a y value within δ. If any of these distances is less than δ, update δ. return δ. }

Cost

15-2

Closest Pair Algorithm

Closest-Pair(p1, …, pn) { Compute separation line L such that half the points
 are on one side and half on the other side. δ1 = Closest-Pair(left half) δ2 = Closest-Pair(right half) δ = min(δ1, δ2) Delete all points further than δ from separation line L Sort remaining points by y-coordinate. Scan points in y-order and calculate distance to all points with a y value within δ. If any of these distances is less than δ, update δ. return δ. }

O(n log n)

Cost

15-3 Slides14 - ClosestPairPoints.key - March 20, 2019

Closest Pair Algorithm

Closest-Pair(p1, …, pn) { Compute separation line L such that half the points
 are on one side and half on the other side. δ1 = Closest-Pair(left half) δ2 = Closest-Pair(right half) δ = min(δ1, δ2) Delete all points further than δ from separation line L Sort remaining points by y-coordinate. Scan points in y-order and calculate distance to all points with a y value within δ. If any of these distances is less than δ, update δ. return δ. }

O(n log n) 2T(n / 2)

Cost

15-4

Closest Pair Algorithm

Closest-Pair(p1, …, pn) { Compute separation line L such that half the points
 are on one side and half on the other side. δ1 = Closest-Pair(left half) δ2 = Closest-Pair(right half) δ = min(δ1, δ2) Delete all points further than δ from separation line L Sort remaining points by y-coordinate. Scan points in y-order and calculate distance to all points with a y value within δ. If any of these distances is less than δ, update δ. return δ. }

O(n log n) 2T(n / 2) O(n)

Cost

15-5

Closest Pair Algorithm

Closest-Pair(p1, …, pn) { Compute separation line L such that half the points
 are on one side and half on the other side. δ1 = Closest-Pair(left half) δ2 = Closest-Pair(right half) δ = min(δ1, δ2) Delete all points further than δ from separation line L Sort remaining points by y-coordinate. Scan points in y-order and calculate distance to all points with a y value within δ. If any of these distances is less than δ, update δ. return δ. }

O(n log n) 2T(n / 2) O(n) O(n log n)

Cost

15-6 Slides14 - ClosestPairPoints.key - March 20, 2019

Closest Pair Algorithm

Closest-Pair(p1, …, pn) { Compute separation line L such that half the points
 are on one side and half on the other side. δ1 = Closest-Pair(left half) δ2 = Closest-Pair(right half) δ = min(δ1, δ2) Delete all points further than δ from separation line L Sort remaining points by y-coordinate. Scan points in y-order and calculate distance to all points with a y value within δ. If any of these distances is less than δ, update δ. return δ. }

O(n log n) 2T(n / 2) O(n) O(n log n) O(n)

Cost

15-7

Closest Pair Algorithm

Closest-Pair(p1, …, pn) { Compute separation line L such that half the points
 are on one side and half on the other side. δ1 = Closest-Pair(left half) δ2 = Closest-Pair(right half) δ = min(δ1, δ2) Delete all points further than δ from separation line L Sort remaining points by y-coordinate. Scan points in y-order and calculate distance to all points with a y value within δ. If any of these distances is less than δ, update δ. return δ. }

O(n log n) 2T(n / 2) O(n) O(n log n) O(n)

Cost T(n) ≤ 2T(n/2) + O(n log n) T(n) = O(n log2 n)

15-8 Slides14 - ClosestPairPoints.key - March 20, 2019