Closest-pair problem

Closest-pair problem
You are encouraged to solve this task according to the task description, using any language you may know.

This page uses content from Wikipedia. The original article was at Closest-pair problem. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)

The aim of this task is to 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 pseudocode (using indexes) could be simply:

bruteForceClosestPair of P(1), P(2), ... P(N), N > 1
minDistance ← |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)|
    endif
return minDistance

A better algorithm is based on the recursive divide&conquer approach, as explained also at Wikipedia, which is O(n log n); a pseudocode could be:

closestPair of P(1), P(2), ... P(N), N > 1
if N ≤ 3 then
  return closest points of P using brute-force algorithm
else
  xP ← P ordered by the x coordinate, in ascending order
  PL ← points of xP from 1 to ⌊N/2⌋
  PR ← points of xP from ⌊N/2⌋+1 to N
  if number of points in PL is odd then
    append xP(⌊N/2⌋+1) to PL
  endif
  if number of points in PR is odd then
    prepend xP(⌊N/2⌋) to PR
  endif
  dL ← ClosestPair of PL
  dR ← ClosestPair of PR
  dmin ← min{dL, dR}
  xM ← (xP(⌊N/2⌋+1)x - xP(⌊N/2⌋)x) / 2
  S ← { p ∈ P : |xM - px| < dmin }
  yP ← S ordered by the y coordinate, in ascending order
  closest ← ∞
  nP ← number of points in yP
  for i from 1 to nP
    for j from max{i-3, 1} to min{nP, i+3}
      if i ≠ j then
        if |yP(i) - yP(j)| < closest then
          closest ← |yP(i) - yP(j)|
        endif
      endif
    endfor
  endfor
  return min{dmin, closest}
endif

Notes

• The pseudocode works with at least two points. It could be rewritten so that a single point is treated like if we want to know the distance from it and a point at infinity (i.e. the distance is infinity)

References

• Closest pair of points problem
• Closest Pair (McGill)
• Closest Pair (UCBS)
• Closest Pair (WUStL)

Smalltalk

Works with: GNU Smalltalk

These class methods return a three elements array, where the first two items are the points, while the third is the distance between them (which having the two points, can be computed; but it was easier to keep it already computed in the third position of the array).

<lang smalltalk>Object subclass: ClosestPair [

 ClosestPair class >> raiseInvalid: arg [
     SystemExceptions.InvalidArgument
       signalOn: arg
       reason: 'specify at least two points'
 ]

 ClosestPair class >> bruteForce: pointsList [ |dist from to points|
   (pointsList size < 2) ifTrue: [ self raiseInvalid: pointsList ].
   points := pointsList asOrderedCollection.
   from := points at: 1. to := points at: 2.
   dist := from dist: to.
   [ points isEmpty ]
   whileFalse: [ |p0|
     p0 := points removeFirst.
     points do: [ :p |
       ((p0 dist: p) <= dist)
       ifTrue: [ from := p0. to := p. dist := p0 dist: p. ]
     ]
   ].
   ^ { from. to. from dist: to }
 ]

 ClosestPair class >> orderByX: points [
   ^ points asSortedCollection: [:a :b| (a x) < (b x) ]
 ]
 ClosestPair class >> orderByY: points [
   ^ points asSortedCollection: [:a :b| (a y) < (b y) ]
 ]

 ClosestPair class >> splitLeft: pointsList [ |s|
   s := pointsList size // 2.
   s odd ifTrue: [ s := s + 1 ].
   ^ pointsList copyFrom: 1 to: s
 ]
 ClosestPair class >> splitRight: pointsList [ |s|
   s := (pointsList size) - (pointsList size // 2).
   s odd ifTrue: [ s := s - 1 ].
   ^ pointsList copyFrom: s to: (pointsList size).
 ]

 ClosestPair class >> minBetween: a and: b [
   (a at: 3) < (b at: 3)
     ifTrue: [ ^a ]
     ifFalse: [ ^b ]
 ]

 ClosestPair class >> recursiveDAndC: pointsList [
   |orderedByX pR pL minL minR minDist middleVLine joiningStrip tDist|
   (pointsList size < 2) ifTrue: [ self raiseInvalid: pointsList ].
   (pointsList size <= 3)
     ifTrue: [ ^ self bruteForce: pointsList ].
   orderedByX := self orderByX: pointsList.
   pR := self splitLeft: orderedByX.
   pL := self splitRight: orderedByX.
   minR := self recursiveDAndC: pR.
   minL := self recursiveDAndC: pL.
   minDist := self minBetween: minR and: minL.

   middleVLine := (((orderedByX at: (orderedByX size)//2+1) x) +
                  ((orderedByX at: (orderedByX size)//2) x)) / 2.
   joiningStrip := self 
                     orderByY: (pointsList 
                                  select: [ :p |
                                     ((middleVLine - (p x)) abs) < (minDist at: 3) 
                                  ]
                                ).
   tDist := { nil. nil.
              FloatD infinity
            }.
   1 to: (joiningStrip size) do: [ :i |
     ((i-3) max: 1)
       to: ((i+3) min: (joiningStrip size))
       do: [ :j |
         j ~= i ifTrue: [ |t|
           t := (joiningStrip at: i) dist: (joiningStrip at: j).
           t < (tDist at:3 ) 
             ifTrue: [ 
               tDist := { joiningStrip at: i.
                          joiningStrip at: j.
                          t }. 
             ]
         ]
       ]
   ].
   ^ self minBetween: minDist and: tDist
 ]

].</lang>

Testing

<lang smalltalk>|coll cp ran| "Let's use the same seed to be sure of the results..." ran := Random seed: 100.

coll := (1 to: 10000 collect: [ :a |

         Point x: (ran next)*10 y: (ran next)*10 ]).

cp := ClosestPair bruteForce: coll. ((cp at: 3) asScaledDecimal: 7) displayNl.

"or"

cp := ClosestPair recursiveDAndC: coll. ((cp at: 3) asScaledDecimal: 7) displayNl.</lang>

The brute-force approach with 10000 points, run with the time tool, gave

224.21user 1.31system 3:46.84elapsed 99%CPU

while the recursive divide&conquer algorithm gave

4.65user 0.00system 0:05.07elapsed 91%CPU

(Of course these results must be considered relative and taken cum grano salis; time counts also the time taken to initialize the Smalltalk environment, and I've taken no special measures to avoid the system load falsifying the results)

Tcl

Each point is represented as a list of two floating-point numbers, the first being the x coordinate, and the second being the y. <lang Tcl># A trivial helper... proc distance {p1 p2} {

   lassign $p1 x1 y1
   lassign $p2 x2 y2
   expr {
       hypot($x1-$x2, $y1-$y2)
   }

}

proc bruteForceClosestPair {points} {

   if {[package provide Tcl] >= "8.5"} {
       set min Inf
   } else {
       set min [distance [lindex $points 0] [lindex $points 1]]
   }

   for {set i 0} {$i < [llength $points]-1} {incr i} {
       for {set j $i} {[incr j] < [llength $points]} {} {
           set dist [distance [lindex $points $i] [lindex $points $j]]
           if {$dist < $min} {
               set min $dist
           }
       }
   }
   return $min

}

1. Some testing code:

set points {} for {set n 0} {$n < 10000} {incr n} {

   lappend points [list [expr {10*rand()}] [expr {10*rand()}]]

} set t [time {

   set d [bruteForceClosestPair $points]

}] puts "closest was $d, found in [lindex $t 0] microseconds"</lang>Example output

closest was 0.0006785546466922088, found in 94942709 microseconds

Note that the lindex and llength commands are both O(1).