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Task

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&conquer approach,   as explained also at   Wikipedia's Closest pair of points problem,   which is   O(n log 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



References and further readings

   Closest pair of points problem
   Closest Pair (McGill)
   Closest Pair (UCSB)
   Closest pair (WUStL)
   Closest pair (IUPUI) 


 
decimals(10)
x = list(10)
y = list(10)
x[1] = 0.654682
y[1] = 0.925557
x[2] = 0.409382
y[2] = 0.619391
x[3] = 0.891663
y[3] = 0.888594
x[4] = 0.716629
y[4] = 0.996200
x[5] = 0.477721
y[5] = 0.946355
x[6] = 0.925092
y[6] = 0.818220
x[7] = 0.624291
y[7] = 0.142924
x[8] = 0.211332
y[8] = 0.221507
x[9] = 0.293786
y[9] = 0.691701
x[10] = 0.839186
y[10] = 0.728260
 
min = 10000
for i = 1 to 9
    for j = i+1 to 10
        dsq = pow((x[i] - x[j]),2) + pow((y[i] - y[j]),2)
        if dsq < min min = dsq  mini = i minj = j ok
    next
next
see "closest pair is : " + mini + " and " + minj + " at distance " + sqrt(min)
 