Closest-pair problem

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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:

bruteForceClosestPair of P(1), P(2), ... P(N) if N < 2 then return ∞ else minDistance ← |P(1) - P(2)| minPoints ← { P(1), P(2) } foreach i ∈ [1, N-1] foreach j ∈ [i+1, N] if |P(i) - P(j)| < minDistance then minDistance ← |P(i) - P(j)| minPoints ← { P(i), P(j) } endif endfor endfor return minDistance, minPoints endif

A better algorithm is based on the recursive divide and conquer approach, as explained also at Wikipedia's Closest pair of points problem, which is O(nlog(n)) a pseudo-code could be:

closestPair of (xP, yP) where xP is P(1) .. P(N) sorted by x coordinate, and yP is P(1) .. P(N) sorted by y coordinate (ascending order) if N ≤ 3 then return closest points of xP using brute-force algorithm else xL ← points of xP from 1 to ⌈N/2⌉ xR ← points of xP from ⌈N/2⌉+1 to N xm ← xP(⌈N/2⌉)x yL ← { p ∈ yP : px ≤ xm } yR ← { p ∈ yP : px > xm } (dL, pairL) ← closestPair of (xL, yL) (dR, pairR) ← closestPair of (xR, yR) (dmin, pairMin) ← (dR, pairR) if dL < dR then (dmin, pairMin) ← (dL, pairL) endif yS ← { p ∈ yP : |xm - px| < dmin } nS ← number of points in yS (closest, closestPair) ← (dmin, pairMin) for i from 1 to nS - 1 k ← i + 1 while k ≤ nS and yS(k)y - yS(i)y < dmin if |yS(k) - yS(i)| < closest then (closest, closestPair) ← (|yS(k) - yS(i)|, {yS(k), yS(i)}) endif k ← k + 1 endwhile endfor return closest, closestPair endif

For the input, expect the argument to be an array of objects (points) with x and y members set to numbers. For the output, return an object containing the key:value pairs for distance and pair (the pair of two closest points).

References and further readings:

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