How to resolve the algorithm Closest-pair problem step by step in the Swift programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Closest-pair problem step by step in the Swift programming language
Table of Contents
Problem Statement
Provide a function to find the closest two points among a set of given points in two dimensions, i.e. to solve the Closest pair of points problem in the planar case. The straightforward solution is a O(n2) algorithm (which we can call brute-force algorithm); the pseudo-code (using indexes) could be simply: A better algorithm is based on the recursive divide&conquer approach, as explained also at Wikipedia's Closest pair of points problem, which is O(n log n); a pseudo-code could be:
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Closest-pair problem step by step in the Swift programming language
Source code in the swift programming language
import Foundation
struct Point {
var x: Double
var y: Double
func distance(to p: Point) -> Double {
let x = pow(p.x - self.x, 2)
let y = pow(p.y - self.y, 2)
return (x + y).squareRoot()
}
}
extension Collection where Element == Point {
func closestPair() -> (Point, Point)? {
let (xP, xY) = (sorted(by: { $0.x < $1.x }), sorted(by: { $0.y < $1.y }))
return Self.closestPair(xP, xY)?.1
}
static func closestPair(_ xP: [Element], _ yP: [Element]) -> (Double, (Point, Point))? {
guard xP.count > 3 else { return xP.closestPairBruteForce() }
let half = xP.count / 2
let xl = Array(xP[..<half])
let xr = Array(xP[half...])
let xm = xl.last!.x
let (yl, yr) = yP.reduce(into: ([Element](), [Element]()), {cur, el in
if el.x > xm {
cur.1.append(el)
} else {
cur.0.append(el)
}
})
guard let (distanceL, pairL) = closestPair(xl, yl) else { return nil }
guard let (distanceR, pairR) = closestPair(xr, yr) else { return nil }
let (dMin, pairMin) = distanceL > distanceR ? (distanceR, pairR) : (distanceL, pairL)
let ys = yP.filter({ abs(xm - $0.x) < dMin })
var (closest, pairClosest) = (dMin, pairMin)
for i in 0..<ys.count {
let p1 = ys[i]
for k in i+1..<ys.count {
let p2 = ys[k]
guard abs(p2.y - p1.y) < dMin else { break }
let distance = abs(p1.distance(to: p2))
if distance < closest {
(closest, pairClosest) = (distance, (p1, p2))
}
}
}
return (closest, pairClosest)
}
func closestPairBruteForce() -> (Double, (Point, Point))? {
guard count >= 2 else { return nil }
var closestPoints = (self.first!, self[index(after: startIndex)])
var minDistance = abs(closestPoints.0.distance(to: closestPoints.1))
guard count != 2 else { return (minDistance, closestPoints) }
for i in 0..<count {
for j in i+1..<count {
let (iIndex, jIndex) = (index(startIndex, offsetBy: i), index(startIndex, offsetBy: j))
let (p1, p2) = (self[iIndex], self[jIndex])
let distance = abs(p1.distance(to: p2))
if distance < minDistance {
minDistance = distance
closestPoints = (p1, p2)
}
}
}
return (minDistance, closestPoints)
}
}
var points = [Point]()
for _ in 0..<10_000 {
points.append(Point(
x: .random(in: -10.0...10.0),
y: .random(in: -10.0...10.0)
))
}
print(points.closestPair()!)
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