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# Closest-pair problem

(Redirected from 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 of points 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)

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
returnelse
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

## 360 Assembly

*        Closest Pair Problem      10/03/2017
CLOSEST CSECT
USING CLOSEST,R13 base register
B 72(R15) skip savearea
DC 17F'0' savearea
STM R14,R12,12(R13) save previous context
ST R13,4(R15) link backward
ST R15,8(R13) link forward
LR R13,R15 set addressability
LA R6,1 i=1
LA R7,2 j=2
BAL R14,DDCALC dd=(px(i)-px(j))^2+(py(i)-py(j))^2
BAL R14,DDSTORE ddmin=dd; ii=i; jj=j
LA R6,1 i=1
DO WHILE=(C,R6,LE,N) do i=1 to n
LA R7,1 j=1
DO WHILE=(C,R7,LE,N) do j=1 to n
BAL R14,DDCALC dd=(px(i)-px(j))^2+(py(i)-py(j))^2
IF CP,DD,GT,=P'0' THEN if dd>0 then
IF CP,DD,LT,DDMIN THEN if dd<ddmin then
BAL R14,DDSTORE ddmin=dd; ii=i; jj=j
ENDIF , endif
ENDIF , endif
LA R7,1(R7) j++
ENDDO , enddo j
LA R6,1(R6) i++
ENDDO , enddo i
ZAP WPD,DDMIN ddmin
DP WPD,=PL8'2' ddmin/2
ZAP SQRT2,WPD(8) sqrt2=ddmin/2
ZAP SQRT1,DDMIN sqrt1=ddmin
DO WHILE=(CP,SQRT1,NE,SQRT2) do while sqrt1<>sqrt2
ZAP SQRT1,SQRT2 sqrt1=sqrt2
ZAP WPD,DDMIN ddmin
DP WPD,SQRT1 /sqrt1
ZAP WP1,WPD(8) ddmin/sqrt1
AP WP1,SQRT1 +sqrt1
ZAP WPD,WP1 ~
DP WPD,=PL8'2' /2
ZAP SQRT2,WPD(8) sqrt2=(sqrt1+(ddmin/sqrt1))/2
ENDDO , enddo while
MVC PG,=CL80'the minimum distance '
ZAP WP1,SQRT2 sqrt2
BAL R14,EDITPK edit
MVC PG+21(L'WC),WC output
XPRNT PG,L'PG print buffer
XPRNT =CL22'is between the points:',22
MVC PG,PGP init buffer
L R1,II ii
SLA R1,4 *16
LA R4,PXY-16(R1) @px(ii)
MVC WP1,0(R4) px(ii)
BAL R14,EDITPK edit
MVC PG+3(L'WC),WC output
MVC WP1,8(R4) py(ii)
BAL R14,EDITPK edit
MVC PG+21(L'WC),WC output
XPRNT PG,L'PG print buffer
MVC PG,PGP init buffer
L R1,JJ jj
SLA R1,4 *16
LA R4,PXY-16(R1) @px(jj)
MVC WP1,0(R4) px(jj)
BAL R14,EDITPK edit
MVC PG+3(L'WC),WC output
MVC WP1,8(R4) py(jj)
BAL R14,EDITPK edit
MVC PG+21(L'WC),WC output
XPRNT PG,L'PG print buffer
L R13,4(0,R13) restore previous savearea pointer
LM R14,R12,12(R13) restore previous context
XR R15,R15 rc=0
BR R14 exit
DDCALC EQU * ---- dd=(px(i)-px(j))^2+(py(i)-py(j))^2
LR R1,R6 i
SLA R1,4 *16
LA R4,PXY-16(R1) @px(i)
LR R1,R7 j
SLA R1,4 *16
LA R5,PXY-16(R1) @px(j)
ZAP WP1,0(8,R4) px(i)
ZAP WP2,0(8,R5) px(j)
SP WP1,WP2 px(i)-px(j)
ZAP WPS,WP1 =
MP WP1,WPS (px(i)-px(j))*(px(i)-px(j))
ZAP WP2,8(8,R4) py(i)
ZAP WP3,8(8,R5) py(j)
SP WP2,WP3 py(i)-py(j)
ZAP WPS,WP2 =
MP WP2,WPS (py(i)-py(j))*(py(i)-py(j))
AP WP1,WP2 (px(i)-px(j))^2+(py(i)-py(j))^2
ZAP DD,WP1 dd=(px(i)-px(j))^2+(py(i)-py(j))^2
BR R14 ---- return
DDSTORE EQU * ---- ddmin=dd; ii=i; jj=j
ZAP DDMIN,DD ddmin=dd
ST R6,II ii=i
ST R7,JJ jj=j
BR R14 ---- return
EDITPK EQU * ----
EDMK WM,WP1 edit and mark
BCTR R1,0 -1
MVC 0(1,R1),WM+17 set sign
MVC WC,WM len17<-len18
BR R14 ---- return
N DC A((PGP-PXY)/16)
PXY DC PL8'0.654682',PL8'0.925557',PL8'0.409382',PL8'0.619391'
DC PL8'0.891663',PL8'0.888594',PL8'0.716629',PL8'0.996200'
DC PL8'0.477721',PL8'0.946355',PL8'0.925092',PL8'0.818220'
DC PL8'0.624291',PL8'0.142924',PL8'0.211332',PL8'0.221507'
DC PL8'0.293786',PL8'0.691701',PL8'0.839186',PL8'0.728260'
PGP DC CL80' [+xxxxxxxxx.xxxxxx,+xxxxxxxxx.xxxxxx]'
MASK DC C' ',7X'20',X'21',X'20',C'.',6X'20',C'-' CL18 15num
II DS F
JJ DS F
DD DS PL8
DDMIN DS PL8
SQRT1 DS PL8
SQRT2 DS PL8
WP1 DS PL8
WP2 DS PL8
WP3 DS PL8
WPS DS PL8
WPD DS PL16
WM DS CL18
WC DS CL17
PG DS CL80
YREGS
END CLOSEST
Output:
the minimum distance          0.077910
is between the points:
[         0.891663,         0.888594]
[         0.925092,         0.818220]

Dimension independent, but has to be defined at procedure call time (could be a parameter). Output is simple, can be formatted using Float_IO.

closest.adb: (uses brute force algorithm)

procedure Closest is
package Math is new Ada.Numerics.Generic_Elementary_Functions (Float);

Dimension : constant := 2;
type Vector is array (1 .. Dimension) of Float;
type Matrix is array (Positive range <>) of Vector;

-- calculate the distance of two points
function Distance (Left, Right : Vector) return Float is
Result : Float := 0.0;
Offset : Natural := 0;
begin
loop
Result := Result + (Left(Left'First + Offset) - Right(Right'First + Offset))**2;
Offset := Offset + 1;
exit when Offset >= Left'Length;
end loop;
return Math.Sqrt (Result);
end Distance;

-- determine the two closest points inside a cloud of vectors
function Get_Closest_Points (Cloud : Matrix) return Matrix is
Result : Matrix (1..2);
Min_Distance : Float;
begin
if Cloud'Length(1) < 2 then
raise Constraint_Error;
end if;
Result := (Cloud (Cloud'First), Cloud (Cloud'First + 1));
Min_Distance := Distance (Cloud (Cloud'First), Cloud (Cloud'First + 1));
for I in Cloud'First (1) .. Cloud'Last(1) - 1 loop
for J in I + 1 .. Cloud'Last(1) loop
if Distance (Cloud (I), Cloud (J)) < Min_Distance then
Min_Distance := Distance (Cloud (I), Cloud (J));
Result := (Cloud (I), Cloud (J));
end if;
end loop;
end loop;
return Result;
end Get_Closest_Points;

Test_Cloud : constant Matrix (1 .. 10) := ( (5.0, 9.0), (9.0, 3.0),
(2.0, 0.0), (8.0, 4.0),
(7.0, 4.0), (9.0, 10.0),
(1.0, 9.0), (8.0, 2.0),
(0.0, 10.0), (9.0, 6.0));
Closest_Points : Matrix := Get_Closest_Points (Test_Cloud);

Second_Test : constant Matrix (1 .. 10) := ( (0.654682, 0.925557), (0.409382, 0.619391),
(0.891663, 0.888594), (0.716629, 0.9962),
(0.477721, 0.946355), (0.925092, 0.81822),
(0.624291, 0.142924), (0.211332, 0.221507),
(0.293786, 0.691701), (0.839186, 0.72826));
Second_Points : Matrix := Get_Closest_Points (Second_Test);
begin
Ada.Text_IO.Put_Line ("P1: " & Float'Image (Closest_Points (1) (1)) & " " & Float'Image (Closest_Points (1) (2)));
Ada.Text_IO.Put_Line ("P2: " & Float'Image (Closest_Points (2) (1)) & " " & Float'Image (Closest_Points (2) (2)));
Ada.Text_IO.Put_Line ("Distance: " & Float'Image (Distance (Closest_Points (1), Closest_Points (2))));
Ada.Text_IO.Put_Line ("Closest Points 2:");
Ada.Text_IO.Put_Line ("P1: " & Float'Image (Second_Points (1) (1)) & " " & Float'Image (Second_Points (1) (2)));
Ada.Text_IO.Put_Line ("P2: " & Float'Image (Second_Points (2) (1)) & " " & Float'Image (Second_Points (2) (2)));
Ada.Text_IO.Put_Line ("Distance: " & Float'Image (Distance (Second_Points (1), Second_Points (2))));
end Closest;
Output:
Closest Points:
P1:  8.00000E+00  4.00000E+00
P2:  7.00000E+00  4.00000E+00
Distance:  1.00000E+00
Closest Points 2:
P1:  8.91663E-01  8.88594E-01
P2:  9.25092E-01  8.18220E-01
Distance:  7.79101E-02

## AutoHotkey

ClosestPair(points){
if (points.count() <= 3)
return bruteForceClosestPair(points)
split := xSplit(Points)
LP := split.1 ; left points
LD := ClosestPair(LP) ; recursion : left closest pair
RP := split.2 ; right points
RD := ClosestPair(RP) ; recursion : right closest pair
minD := min(LD, RD) ; minimum of LD & RD
xmin := Split.3 - minD ; strip left boundary
xmax := Split.3 + minD ; strip right boundary
S := strip(points, xmin, xmax)
if (s.count()>=2)
{
SD := ClosestPair(S) ; recursion : strip closest pair
return min(SD, minD)
}
return minD
}
;---------------------------------------------------------------
strip(points, xmin, xmax){
strip:=[]
for i, coord in points
if (coord.1 >= xmin) && (coord.1 <= xmax)
strip.push([coord.1, coord.2])
return strip
}
;---------------------------------------------------------------
bruteForceClosestPair(points){
minD := []
loop, % points.count()-1{
p1 := points.RemoveAt(1)
loop, % points.count(){
p2 := points[A_Index]
d := dist(p1, p2)
minD.push(d)
}
}
return min(minD*)
}
;---------------------------------------------------------------
dist(p1, p2){
return Sqrt((p2.1-p1.1)**2 + (p2.2-p1.2)**2)
}
;---------------------------------------------------------------
xSplit(Points){
xL := [], xR := []
p := xSort(Points)
Loop % Ceil(p.count()/2)
xL.push(p.RemoveAt(1))
while p.count()
xR.push(p.RemoveAt(1))
mid := (xL[xl.count(),1] + xR[1,1])/2
return [xL, xR, mid]
}
;---------------------------------------------------------------
xSort(Points){
S := [], Res :=[]
for i, coord in points
S[coord.1, coord.2] := true
for x, coord in S
for y, v in coord
res.push([x, y])
return res
}
;---------------------------------------------------------------
Examples:
points := [[1, 1], [12, 30], [40, 50], [5, 1], [12, 10], [3, 4], [17,25], [45,50],[51,34],[2,1],[2,2],[10,10]]
MsgBox % ClosestPair(points)
Output:
1.000000

## AWK

# syntax: GAWK -f CLOSEST-PAIR_PROBLEM.AWK
BEGIN {
x[++n] = 0.654682 ; y[n] = 0.925557
x[++n] = 0.409382 ; y[n] = 0.619391
x[++n] = 0.891663 ; y[n] = 0.888594
x[++n] = 0.716629 ; y[n] = 0.996200
x[++n] = 0.477721 ; y[n] = 0.946355
x[++n] = 0.925092 ; y[n] = 0.818220
x[++n] = 0.624291 ; y[n] = 0.142924
x[++n] = 0.211332 ; y[n] = 0.221507
x[++n] = 0.293786 ; y[n] = 0.691701
x[++n] = 0.839186 ; y[n] = 0.728260
min = 1E20
for (i=1; i<=n-1; i++) {
for (j=i+1; j<=n; j++) {
dsq = (x[i]-x[j])^2 + (y[i]-y[j])^2
if (dsq < min) {
min = dsq
mini = i
minj = j
}
}
}
printf("distance between (%.6f,%.6f) and (%.6f,%.6f) is %g\n",x[mini],y[mini],x[minj],y[minj],sqrt(min))
exit(0)
}

Output:
distance between (0.891663,0.888594) and (0.925092,0.818220) is 0.0779102

## BASIC256

Versión de fuerza bruta:

Dim x(9)
x = {0.654682, 0.409382, 0.891663, 0.716629, 0.477721, 0.925092, 0.624291, 0.211332, 0.293786, 0.839186}
Dim y(9)
y = {0.925557, 0.619391, 0.888594, 0.996200, 0.946355, 0.818220, 0.142924, 0.221507, 0.691701, 0.728260}

minDist = 1^30
For i = 0 To 8
For j = i+1 To 9
dist = (x[i] - x[j])^2 + (y[i] - y[j])^2
If dist < minDist Then minDist = dist : minDisti = i : minDistj = j
Next j
Next i
Print "El par más cercano es "; minDisti; " y "; minDistj; " a una distancia de "; Sqr(minDist)
End

Output:
El par más cercano es 2 y 5 a una distancia de 0,077910191355

## BBC BASIC

To find the closest pair it is sufficient to compare the squared-distances, it is not necessary to perform the square root for each pair!

DIM x(9), y(9)

FOR I% = 0 TO 9
NEXT

min = 1E30
FOR I% = 0 TO 8
FOR J% = I%+1 TO 9
dsq = (x(I%) - x(J%))^2 + (y(I%) - y(J%))^2
IF dsq < min min = dsq : mini% = I% : minj% = J%
NEXT
NEXT I%
PRINT "Closest pair is ";mini% " and ";minj% " at distance "; SQR(min)
END

DATA 0.654682, 0.925557
DATA 0.409382, 0.619391
DATA 0.891663, 0.888594
DATA 0.716629, 0.996200
DATA 0.477721, 0.946355
DATA 0.925092, 0.818220
DATA 0.624291, 0.142924
DATA 0.211332, 0.221507
DATA 0.293786, 0.691701
DATA 0.839186, 0.728260

Output:
Closest pair is 2 and 5 at distance 0.0779101913

## C#

We provide a small helper class for distance comparisons:

class Segment
{
public Segment(PointF p1, PointF p2)
{
P1 = p1;
P2 = p2;
}

public readonly PointF P1;
public readonly PointF P2;

public float Length()
{
return (float)Math.Sqrt(LengthSquared());
}

public float LengthSquared()
{
return (P1.X - P2.X) * (P1.X - P2.X)
+ (P1.Y - P2.Y) * (P1.Y - P2.Y);
}
}

Brute force:

Segment Closest_BruteForce(List<PointF> points)
{
int n = points.Count;
var result = Enumerable.Range( 0, n-1)
.SelectMany( i => Enumerable.Range( i+1, n-(i+1) )
.Select( j => new Segment( points[i], points[j] )))
.OrderBy( seg => seg.LengthSquared())
.First();

return result;
}

And divide-and-conquer.

public static Segment MyClosestDivide(List<PointF> points)
{
return MyClosestRec(points.OrderBy(p => p.X).ToList());
}

private static Segment MyClosestRec(List<PointF> pointsByX)
{
int count = pointsByX.Count;
if (count <= 4)
return Closest_BruteForce(pointsByX);

// left and right lists sorted by X, as order retained from full list
var leftByX = pointsByX.Take(count/2).ToList();
var leftResult = MyClosestRec(leftByX);

var rightByX = pointsByX.Skip(count/2).ToList();
var rightResult = MyClosestRec(rightByX);

var result = rightResult.Length() < leftResult.Length() ? rightResult : leftResult;

// There may be a shorter distance that crosses the divider
// Thus, extract all the points within result.Length either side
var midX = leftByX.Last().X;
var bandWidth = result.Length();
var inBandByX = pointsByX.Where(p => Math.Abs(midX - p.X) <= bandWidth);

// Sort by Y, so we can efficiently check for closer pairs
var inBandByY = inBandByX.OrderBy(p => p.Y).ToArray();

int iLast = inBandByY.Length - 1;
for (int i = 0; i < iLast; i++ )
{
var pLower = inBandByY[i];

for (int j = i + 1; j <= iLast; j++)
{
var pUpper = inBandByY[j];

// Comparing each point to successivly increasing Y values
// Thus, can terminate as soon as deltaY is greater than best result
if ((pUpper.Y - pLower.Y) >= result.Length())
break;

if (Segment.Length(pLower, pUpper) < result.Length())
result = new Segment(pLower, pUpper);
}
}

return result;
}

However, the difference in speed is still remarkable.

var randomizer = new Random(10);
var points = Enumerable.Range( 0, 10000).Select( i => new PointF( (float)randomizer.NextDouble(), (float)randomizer.NextDouble())).ToList();
Stopwatch sw = Stopwatch.StartNew();
var r1 = Closest_BruteForce(points);
sw.Stop();
Debugger.Log(1, "", string.Format("Time used (Brute force) (float): {0} ms", sw.Elapsed.TotalMilliseconds));
Stopwatch sw2 = Stopwatch.StartNew();
var result2 = Closest_Recursive(points);
sw2.Stop();
Debugger.Log(1, "", string.Format("Time used (Divide & Conquer): {0} ms",sw2.Elapsed.TotalMilliseconds));
Assert.Equal(r1.Length(), result2.Length());
Output:
Time used (Brute force) (float): 145731.8935 ms
Time used (Divide & Conquer): 1139.2111 ms

Non Linq Brute Force:

Segment Closest_BruteForce(List<PointF> points)
{
Trace.Assert(points.Count >= 2);

int count = points.Count;

// Seed the result - doesn't matter what points are used
// This just avoids having to do null checks in the main loop below
var result = new Segment(points[0], points[1]);
var bestLength = result.Length();

for (int i = 0; i < count; i++)
for (int j = i + 1; j < count; j++)
if (Segment.Length(points[i], points[j]) < bestLength)
{
result = new Segment(points[i], points[j]);
bestLength = result.Length();
}

return result;
}

Targeted Search: Much simpler than divide and conquer, and actually runs faster for the random points. Key optimization is that if the distance along the X axis is greater than the best total length you already have, you can terminate the inner loop early. However, as only sorts in the X direction, it degenerates into an N^2 algorithm if all the points have the same X.

Segment Closest(List<PointF> points)
{
Trace.Assert(points.Count >= 2);

int count = points.Count;
points.Sort((lhs, rhs) => lhs.X.CompareTo(rhs.X));

var result = new Segment(points[0], points[1]);
var bestLength = result.Length();

for (int i = 0; i < count; i++)
{
var from = points[i];

for (int j = i + 1; j < count; j++)
{
var to = points[j];

var dx = to.X - from.X;
if (dx >= bestLength)
{
break;
}

if (Segment.Length(from, to) < bestLength)
{
result = new Segment(from, to);
bestLength = result.Length();
}
}
}

return result;
}

## C++

/*
Author: Kevin Bacon
Date: 04/03/2014
*/

#include <iostream>
#include <vector>
#include <utility>
#include <cmath>
#include <random>
#include <chrono>
#include <algorithm>
#include <iterator>

typedef std::pair<double, double> point_t;
typedef std::pair<point_t, point_t> points_t;

double distance_between(const point_t& a, const point_t& b) {
return std::sqrt(std::pow(b.first - a.first, 2)
+ std::pow(b.second - a.second, 2));
}

std::pair<double, points_t> find_closest_brute(const std::vector<point_t>& points) {
if (points.size() < 2) {
return { -1, { { 0, 0 }, { 0, 0 } } };
}
auto minDistance = std::abs(distance_between(points.at(0), points.at(1)));
points_t minPoints = { points.at(0), points.at(1) };
for (auto i = std::begin(points); i != (std::end(points) - 1); ++i) {
for (auto j = i + 1; j < std::end(points); ++j) {
auto newDistance = std::abs(distance_between(*i, *j));
if (newDistance < minDistance) {
minDistance = newDistance;
minPoints.first = *i;
minPoints.second = *j;
}
}
}
return { minDistance, minPoints };
}

std::pair<double, points_t> find_closest_optimized(const std::vector<point_t>& xP,
const std::vector<point_t>& yP) {
if (xP.size() <= 3) {
return find_closest_brute(xP);
}
auto N = xP.size();
auto xL = std::vector<point_t>();
auto xR = std::vector<point_t>();
std::copy(std::begin(xP), std::begin(xP) + (N / 2), std::back_inserter(xL));
std::copy(std::begin(xP) + (N / 2), std::end(xP), std::back_inserter(xR));
auto xM = xP.at((N-1) / 2).first;
auto yL = std::vector<point_t>();
auto yR = std::vector<point_t>();
std::copy_if(std::begin(yP), std::end(yP), std::back_inserter(yL), [&xM](const point_t& p) {
return p.first <= xM;
});
std::copy_if(std::begin(yP), std::end(yP), std::back_inserter(yR), [&xM](const point_t& p) {
return p.first > xM;
});
auto p1 = find_closest_optimized(xL, yL);
auto p2 = find_closest_optimized(xR, yR);
auto minPair = (p1.first <= p2.first) ? p1 : p2;
auto yS = std::vector<point_t>();
std::copy_if(std::begin(yP), std::end(yP), std::back_inserter(yS), [&minPair, &xM](const point_t& p) {
return std::abs(xM - p.first) < minPair.first;
});
auto result = minPair;
for (auto i = std::begin(yS); i != (std::end(yS) - 1); ++i) {
for (auto k = i + 1; k != std::end(yS) &&
((k->second - i->second) < minPair.first); ++k) {
auto newDistance = std::abs(distance_between(*k, *i));
if (newDistance < result.first) {
result = { newDistance, { *k, *i } };
}
}
}
return result;
}

void print_point(const point_t& point) {
std::cout << "(" << point.first
<< ", " << point.second
<< ")";
}

int main(int argc, char * argv[]) {
std::default_random_engine re(std::chrono::system_clock::to_time_t(
std::chrono::system_clock::now()));
std::uniform_real_distribution<double> urd(-500.0, 500.0);
std::vector<point_t> points(100);
std::generate(std::begin(points), std::end(points), [&urd, &re]() {
return point_t { 1000 + urd(re), 1000 + urd(re) };
});
auto answer = find_closest_brute(points);
std::sort(std::begin(points), std::end(points), [](const point_t& a, const point_t& b) {
return a.first < b.first;
});
auto xP = points;
std::sort(std::begin(points), std::end(points), [](const point_t& a, const point_t& b) {
return a.second < b.second;
});
auto yP = points;
std::cout << "Min distance (brute): " << answer.first << " ";
std::cout << ", ";
answer = find_closest_optimized(xP, yP);
std::cout << "\nMin distance (optimized): " << answer.first << " ";
std::cout << ", ";
return 0;
}
Output:
Min distance (brute): 6.95886 (932.735, 1002.7), (939.216, 1000.17)
Min distance (optimized): 6.95886 (932.735, 1002.7), (939.216, 1000.17)

## Clojure

(defn distance [[x1 y1] [x2 y2]]
(let [dx (- x2 x1), dy (- y2 y1)]
(Math/sqrt (+ (* dx dx) (* dy dy)))))

(defn brute-force [points]
(let [n (count points)]
(when (< 1 n)
(apply min-key first
(for [i (range 0 (dec n)), :let [p1 (nth points i)],
j (range (inc i) n), :let [p2 (nth points j)]]
[(distance p1 p2) p1 p2])))))

(defn combine [yS [dmin pmin1 pmin2]]
(apply min-key first
(conj (for [[p1 p2] (partition 2 1 yS)
:let [[_ py1] p1 [_ py2] p2]
:while (< (- py1 py2) dmin)]
[(distance p1 p2) p1 p2])
[dmin pmin1 pmin2])))

(defn closest-pair
([points]
(closest-pair
(sort-by first points)
(sort-by second points)))
([xP yP]
(if (< (count xP) 4)
(brute-force xP)
(let [[xL xR] (partition-all (Math/ceil (/ (count xP) 2)) xP)
[xm _] (last xL)
{yL true yR false} (group-by (fn [[px _]] (<= px xm)) yP)
dL&pairL (closest-pair xL yL)
dR&pairR (closest-pair xR yR)
[dmin pmin1 pmin2] (min-key first dL&pairL dR&pairR)
{yS true} (group-by (fn [[px _]] (< (Math/abs (- xm px)) dmin)) yP)]
(combine yS [dmin pmin1 pmin2])))))

## Common Lisp

Points are conses whose cars are x coördinates and whose cdrs are y coördinates. This version includes the optimizations given in the McGill description of the algorithm.

(defun point-distance (p1 p2)
(destructuring-bind (x1 . y1) p1
(destructuring-bind (x2 . y2) p2
(let ((dx (- x2 x1)) (dy (- y2 y1)))
(sqrt (+ (* dx dx) (* dy dy)))))))

(defun closest-pair-bf (points)
(let ((pair (list (first points) (second points)))
(dist (point-distance (first points) (second points))))
(dolist (p1 points (values pair dist))
(dolist (p2 points)
(unless (eq p1 p2)
(let ((pdist (point-distance p1 p2)))
(when (< pdist dist)
(setf (first pair) p1
(second pair) p2
dist pdist))))))))

(defun closest-pair (points)
(labels
((cp (xp &aux (length (length xp)))
(if (<= length 3)
(multiple-value-bind (pair distance) (closest-pair-bf xp)
(values pair distance (sort xp '< :key 'cdr)))
(let* ((xr (nthcdr (1- (floor length 2)) xp))
(xm (/ (+ (caar xr) (caadr xr)) 2)))
(psetf xr (rest xr)
(rest xr) '())
(multiple-value-bind (lpair ldist yl) (cp xp)
(multiple-value-bind (rpair rdist yr) (cp xr)
(multiple-value-bind (dist pair)
(if (< ldist rdist)
(values ldist lpair)
(values rdist rpair))
(let* ((all-ys (merge 'vector yl yr '< :key 'cdr))
(ys (remove-if #'(lambda (p)
(> (abs (- (car p) xm)) dist))
all-ys))
(ns (length ys)))
(dotimes (i ns)
(do ((k (1+ i) (1+ k)))
((or (= k ns)
(> (- (cdr (aref ys k))
(cdr (aref ys i)))
dist)))
(let ((pd (point-distance (aref ys i)
(aref ys k))))
(when (< pd dist)
(setf dist pd
(first pair) (aref ys i)
(second pair) (aref ys k))))))
(values pair dist all-ys)))))))))
(multiple-value-bind (pair distance)
(cp (sort (copy-list points) '< :key 'car))
(values pair distance))))

## D

### Compact Versions

import std.stdio, std.typecons, std.math, std.algorithm,
std.random, std.traits, std.range, std.complex;

auto bruteForceClosestPair(T)(in T[] points) pure nothrow @nogc {
// return pairwise(points.length.iota, points.length.iota)
// .reduce!(min!((i, j) => abs(points[i] - points[j])));
auto minD = Unqual!(typeof(T.re)).infinity;
T minI, minJ;
foreach (immutable i, const p1; points.dropBackOne)
foreach (const p2; points[i + 1 .. \$]) {
immutable dist = abs(p1 - p2);
if (dist < minD) {
minD = dist;
minI = p1;
minJ = p2;
}
}
return tuple(minD, minI, minJ);
}

auto closestPair(T)(T[] points) pure nothrow {
static Tuple!(typeof(T.re), T, T) inner(in T[] xP, /*in*/ T[] yP)
pure nothrow {
if (xP.length <= 3)
return xP.bruteForceClosestPair;
const Pl = xP[0 .. \$ / 2];
const Pr = xP[\$ / 2 .. \$];
immutable xDiv = Pl.back.re;
auto Yr = yP.partition!(p => p.re <= xDiv);
immutable dl_pairl = inner(Pl, yP[0 .. yP.length - Yr.length]);
immutable dr_pairr = inner(Pr, Yr);
immutable dm_pairm = dl_pairl[0]<dr_pairr[0] ? dl_pairl : dr_pairr;
immutable dm = dm_pairm[0];
const nextY = yP.filter!(p => abs(p.re - xDiv) < dm).array;

if (nextY.length > 1) {
auto minD = typeof(T.re).infinity;
size_t minI, minJ;
foreach (immutable i; 0 .. nextY.length - 1)
foreach (immutable j; i + 1 .. min(i + 8, nextY.length)) {
immutable double dist = abs(nextY[i] - nextY[j]);
if (dist < minD) {
minD = dist;
minI = i;
minJ = j;
}
}
return dm <= minD ? dm_pairm :
typeof(return)(minD, nextY[minI], nextY[minJ]);
} else
return dm_pairm;
}

points.sort!q{ a.re < b.re };
const xP = points.dup;
points.sort!q{ a.im < b.im };
return inner(xP, points);
}

void main() {
alias C = complex;
auto pts = [C(5,9), C(9,3), C(2), C(8,4), C(7,4), C(9,10), C(1,9),
C(8,2), C(0,10), C(9,6)];
pts.writeln;
writeln("bruteForceClosestPair: ", pts.bruteForceClosestPair);
writeln(" closestPair: ", pts.closestPair);

rndGen.seed = 1;
Complex!double[10_000] points;
foreach (ref p; points)
p = C(uniform(0.0, 1000.0) + uniform(0.0, 1000.0));
writeln("bruteForceClosestPair: ", points.bruteForceClosestPair);
writeln(" closestPair: ", points.closestPair);
}
Output:
[5+9i, 9+3i, 2+0i, 8+4i, 7+4i, 9+10i, 1+9i, 8+2i, 0+10i, 9+6i]
bruteForceClosestPair: Tuple!(double, Complex!double, Complex!double)(1, 8+4i, 7+4i)
closestPair: Tuple!(double, Complex!double, Complex!double)(1, 7+4i, 8+4i)
bruteForceClosestPair: Tuple!(double, Complex!double, Complex!double)(1.76951e-05, 1040.2+0i, 1040.2+0i)
closestPair: Tuple!(double, Complex!double, Complex!double)(1.76951e-05, 1040.2+0i, 1040.2+0i)

About 1.87 seconds run-time for data generation and brute force version, and about 0.03 seconds for data generation and divide & conquer (10_000 points in both cases) with ldc2 compiler.

### Faster Brute-force Version

import std.stdio, std.random, std.math, std.typecons, std.complex,
std.traits;

Nullable!(Tuple!(size_t, size_t))
bfClosestPair2(T)(in Complex!T[] points) pure nothrow @nogc {
auto minD = Unqual!(typeof(points[0].re)).infinity;
if (points.length < 2)
return typeof(return)();

size_t minI, minJ;
foreach (immutable i; 0 .. points.length - 1)
foreach (immutable j; i + 1 .. points.length) {
auto dist = (points[i].re - points[j].re) ^^ 2;
if (dist < minD) {
dist += (points[i].im - points[j].im) ^^ 2;
if (dist < minD) {
minD = dist;
minI = i;
minJ = j;
}
}
}

return typeof(return)(tuple(minI, minJ));
}

void main() {
alias C = Complex!double;
auto rng = 31415.Xorshift;
C[10_000] pts;
foreach (ref p; pts)
p = C(uniform(0.0, 1000.0, rng), uniform(0.0, 1000.0, rng));

immutable ij = pts.bfClosestPair2;
if (ij.isNull)
return;
writefln("Closest pair: Distance: %f p1, p2: %f, %f",
abs(pts[ij[0]] - pts[ij[1]]), pts[ij[0]], pts[ij[1]]);
}
Output:
Closest pair: Distance: 0.019212  p1, p2: 9.74223+119.419i, 9.72306+119.418i

About 0.12 seconds run-time for brute-force version 2 (10_000 points) with with LDC2 compiler.

## Elixir

defmodule Closest_pair do
# brute-force algorithm:
def bruteForce([p0,p1|_] = points), do: bf_loop(points, {distance(p0, p1), {p0, p1}})

defp bf_loop([_], acc), do: acc
defp bf_loop([h|t], acc), do: bf_loop(t, bf_loop(h, t, acc))

defp bf_loop(_, [], acc), do: acc
defp bf_loop(p0, [p1|t], {minD, minP}) do
dist = distance(p0, p1)
if dist < minD, do: bf_loop(p0, t, {dist, {p0, p1}}),
else: bf_loop(p0, t, {minD, minP})
end

defp distance({p0x,p0y}, {p1x,p1y}) do
:math.sqrt( (p1x - p0x) * (p1x - p0x) + (p1y - p0y) * (p1y - p0y) )
end

# recursive divide&conquer approach:
def recursive(points) do
recursive(Enum.sort(points), Enum.sort_by(points, fn {_x,y} -> y end))
end

def recursive(xP, _yP) when length(xP) <= 3, do: bruteForce(xP)
def recursive(xP, yP) do
{xL, xR} = Enum.split(xP, div(length(xP), 2))
{xm, _} = hd(xR)
{yL, yR} = Enum.partition(yP, fn {x,_} -> x < xm end)
{dL, pairL} = recursive(xL, yL)
{dR, pairR} = recursive(xR, yR)
{dmin, pairMin} = if dL<dR, do: {dL, pairL}, else: {dR, pairR}
yS = Enum.filter(yP, fn {x,_} -> abs(xm - x) < dmin end)
merge(yS, {dmin, pairMin})
end

defp merge([_], acc), do: acc
defp merge([h|t], acc), do: merge(t, merge_loop(h, t, acc))

defp merge_loop(_, [], acc), do: acc
defp merge_loop(p0, [p1|_], {dmin,_}=acc) when dmin <= elem(p1,1) - elem(p0,1), do: acc
defp merge_loop(p0, [p1|t], {dmin, pair}) do
dist = distance(p0, p1)
if dist < dmin, do: merge_loop(p0, t, {dist, {p0, p1}}),
else: merge_loop(p0, t, {dmin, pair})
end
end

data = [{0.654682, 0.925557}, {0.409382, 0.619391}, {0.891663, 0.888594}, {0.716629, 0.996200},
{0.477721, 0.946355}, {0.925092, 0.818220}, {0.624291, 0.142924}, {0.211332, 0.221507},
{0.293786, 0.691701}, {0.839186, 0.728260}]

IO.inspect Closest_pair.bruteForce(data)
IO.inspect Closest_pair.recursive(data)

data2 = for _ <- 1..5000, do: {:rand.uniform, :rand.uniform}
IO.puts "\nBrute-force:"
IO.inspect :timer.tc(fn -> Closest_pair.bruteForce(data2) end)
IO.puts "Recursive divide&conquer:"
IO.inspect :timer.tc(fn -> Closest_pair.recursive(data2) end)
Output:
{0.07791019135517516, {{0.891663, 0.888594}, {0.925092, 0.81822}}}
{0.07791019135517516, {{0.891663, 0.888594}, {0.925092, 0.81822}}}

Brute-force:
{9579000,
{2.068674444452469e-4,
{{0.9397601102440695, 0.020420581980209674},
{0.9399398976079764, 0.020522908141823986}}}}
Recursive divide&conquer:
{109000,
{2.068674444452469e-4,
{{0.9397601102440695, 0.020420581980209674},
{0.9399398976079764, 0.020522908141823986}}}}

## F#

Brute force:

let closest_pairs (xys: Point []) =
let n = xys.Length
seq { for i in 0..n-2 do
for j in i+1..n-1 do
yield xys.[i], xys.[j] }
|> Seq.minBy (fun (p0, p1) -> (p1 - p0).LengthSquared)

For example:

closest_pairs
[|Point(0.0, 0.0); Point(1.0, 0.0); Point (2.0, 2.0)|]

gives:

(0,0, 1,0)

Divide And Conquer:

open System;
open System.Drawing;
open System.Diagnostics;

let Length (seg : (PointF * PointF) option) =
match seg with
| None -> System.Single.MaxValue
| Some(line) ->
let f = fst line
let t = snd line

let dx = f.X - t.X
let dy = f.Y - t.Y
sqrt (dx*dx + dy*dy)

let Shortest a b =
if Length(a) < Length(b) then
a
else
b

let rec ClosestBoundY from maxY (ptsByY : PointF list) =
match ptsByY with
| [] -> None
| hd :: tl ->
if hd.Y > maxY then
None
else
let toHd = Some(from, hd)
let bestToRest = ClosestBoundY from maxY tl
Shortest toHd bestToRest

let rec ClosestWithinRange ptsByY maxDy =
match ptsByY with
| [] -> None
| hd :: tl ->
let fromHd = ClosestBoundY hd (hd.Y + maxDy) tl
let fromRest = ClosestWithinRange tl maxDy
Shortest fromHd fromRest

// Cuts pts half way through it's length
// Order is not maintained in result lists however
let Halve pts =
let rec ShiftToFirst first second n =
match (n, second) with
| 0, _ -> (first, second) // finished the split, so return current state
| _, [] -> (first, []) // not enough items, so first takes the whole original list
| n, hd::tl -> ShiftToFirst (hd :: first) tl (n-1) // shift 1st item from second to first, then recurse with n-1

let n = (List.length pts) / 2
ShiftToFirst [] pts n

let rec ClosestPair (pts : PointF list) =
if List.length pts < 2 then
None
else
let ptsByX = pts |> List.sortBy(fun(p) -> p.X)

let (left, right) = Halve ptsByX
let leftResult = ClosestPair left
let rightResult = ClosestPair right

let bestInHalf = Shortest leftResult rightResult
let bestLength = Length bestInHalf

let divideX = List.head(right).X
let inBand = pts |> List.filter(fun(p) -> Math.Abs(p.X - divideX) < bestLength)

let byY = inBand |> List.sortBy(fun(p) -> p.Y)
let bestCross = ClosestWithinRange byY bestLength
Shortest bestInHalf bestCross

let GeneratePoints n =
let rand = new Random()
[1..n] |> List.map(fun(i) -> new PointF(float32(rand.NextDouble()), float32(rand.NextDouble())))

let timer = Stopwatch.StartNew()
let pts = GeneratePoints (50 * 1000)
let closest = ClosestPair pts
let takenMs = timer.ElapsedMilliseconds

printfn "Closest Pair '%A'. Distance %f" closest (Length closest)
printfn "Took %d [ms]" takenMs

## Fantom

(Based on the Ruby example.)

class Point
{
Float x
Float y

// create a random point
new make (Float x := Float.random * 10, Float y := Float.random * 10)
{
this.x = x
this.y = y
}

Float distance (Point p)
{
((x-p.x)*(x-p.x) + (y-p.y)*(y-p.y)).sqrt
}

override Str toStr () { "(\$x, \$y)" }
}

class Main
{
// use brute force approach
static Point[] findClosestPair1 (Point[] points)
{
if (points.size < 2) return points // list too small
Point[] closestPair := [points[0], points[1]]
Float closestDistance := points[0].distance(points[1])

(1..<points.size).each |Int i|
{
((i+1)..<points.size).each |Int j|
{
Float trydistance := points[i].distance(points[j])
if (trydistance < closestDistance)
{
closestPair = [points[i], points[j]]
closestDistance = trydistance
}
}
}

return closestPair
}

// use recursive divide-and-conquer approach
static Point[] findClosestPair2 (Point[] points)
{
if (points.size <= 3) return findClosestPair1(points)
points.sort |Point a, Point b -> Int| { a.x <=> b.x }
bestLeft := findClosestPair2 (points[0..(points.size/2)])
bestRight := findClosestPair2 (points[(points.size/2)..-1])

Float minDistance
Point[] closePoints := [,]
if (bestLeft[0].distance(bestLeft[1]) < bestRight[0].distance(bestRight[1]))
{
minDistance = bestLeft[0].distance(bestLeft[1])
closePoints = bestLeft
}
else
{
minDistance = bestRight[0].distance(bestRight[1])
closePoints = bestRight
}
yPoints := points.findAll |Point p -> Bool|
{
(points.last.x - p.x).abs < minDistance
}.sort |Point a, Point b -> Int| { a.y <=> b.y }

closestPair := [,]
closestDist := Float.posInf

for (Int i := 0; i < yPoints.size - 1; ++i)
{
for (Int j := (i+1); j < yPoints.size; ++j)
{
if ((yPoints[j].y - yPoints[i].y) >= minDistance)
{
break
}
else
{
dist := yPoints[i].distance (yPoints[j])
if (dist < closestDist)
{
closestDist = dist
closestPair = [yPoints[i], yPoints[j]]
}
}
}
}
if (closestDist < minDistance)
return closestPair
else
return closePoints
}

public static Void main (Str[] args)
{
Int numPoints := 10 // default value, in case a number not given on command line
if ((args.size > 0) && (args[0].toInt(10, false) != null))
{
numPoints = args[0].toInt(10, false)
}

Point[] points := [,]
numPoints.times { points.add (Point()) }

Int t1 := Duration.now.toMillis
echo (findClosestPair1(points.dup))
Int t2 := Duration.now.toMillis
echo ("Time taken: \${(t2-t1)}ms")
echo (findClosestPair2(points.dup))
Int t3 := Duration.now.toMillis
echo ("Time taken: \${(t3-t2)}ms")
}
}

Output:
\$ fan closestPoints 1000
[(1.4542885676006445, 8.238581003965352), (1.4528464044751888, 8.234724407229772)]
Time taken: 88ms
[(1.4528464044751888, 8.234724407229772), (1.4542885676006445, 8.238581003965352)]
Time taken: 80ms
\$ fan closestPoints 10000
[(3.454790171891945, 5.307252398266497), (3.4540208686702245, 5.308350223433488)]
Time taken: 6248ms
[(3.454790171891945, 5.307252398266497), (3.4540208686702245, 5.308350223433488)]
Time taken: 228ms

## FreeBASIC

Versión de fuerza bruta:

Dim As Integer i, j
Dim As Double minDist = 1^30
Dim As Double x(9), y(9), dist, mini, minj

Data 0.654682, 0.925557
Data 0.409382, 0.619391
Data 0.891663, 0.888594
Data 0.716629, 0.996200
Data 0.477721, 0.946355
Data 0.925092, 0.818220
Data 0.624291, 0.142924
Data 0.211332, 0.221507
Data 0.293786, 0.691701
Data 0.839186, 0.728260

For i = 0 To 9
Next i

For i = 0 To 8
For j = i+1 To 9
dist = (x(i) - x(j))^2 + (y(i) - y(j))^2
If dist < minDist Then
minDist = dist
mini = i
minj = j
End If
Next j
Next i

Print "El par más cercano es "; mini; " y "; minj; " a una distancia de "; Sqr(minDist)
End

Output:
El par más cercano es 2 y 5 a una distancia de 0.07791019135517516

## Go

Brute force

package main

import (
"fmt"
"math"
"math/rand"
"time"
)

type xy struct {
x, y float64
}

const n = 1000
const scale = 100.

func d(p1, p2 xy) float64 {
return math.Hypot(p2.x-p1.x, p2.y-p1.y)
}

func main() {
rand.Seed(time.Now().Unix())
points := make([]xy, n)
for i := range points {
points[i] = xy{rand.Float64() * scale, rand.Float64() * scale}
}
p1, p2 := closestPair(points)
fmt.Println(p1, p2)
fmt.Println("distance:", d(p1, p2))
}

func closestPair(points []xy) (p1, p2 xy) {
if len(points) < 2 {
panic("at least two points expected")
}
min := 2 * scale
for i, q1 := range points[:len(points)-1] {
for _, q2 := range points[i+1:] {
if dq := d(q1, q2); dq < min {
p1, p2 = q1, q2
min = dq
}
}
}
return
}

O(n)

// implementation following algorithm described in
// http://www.cs.umd.edu/~samir/grant/cp.pdf
package main

import (
"fmt"
"math"
"math/rand"
"time"
)

// number of points to search for closest pair
const n = 1e6

// size of bounding box for points.
// x and y will be random with uniform distribution in the range [0,scale).
const scale = 100.

// point struct
type xy struct {
x, y float64 // coordinates
key int64 // an annotation used in the algorithm
}

func d(p1, p2 xy) float64 {
return math.Hypot(p2.x-p1.x, p2.y-p1.y)
}

func main() {
rand.Seed(time.Now().Unix())
points := make([]xy, n)
for i := range points {
points[i] = xy{rand.Float64() * scale, rand.Float64() * scale, 0}
}
p1, p2 := closestPair(points)
fmt.Println(p1, p2)
fmt.Println("distance:", d(p1, p2))
}

func closestPair(s []xy) (p1, p2 xy) {
if len(s) < 2 {
panic("2 points required")
}
var dxi float64
// step 0
for s1, i := s, 1; ; i++ {
// step 1: compute min distance to a random point
// (for the case of random data, it's enough to just try
// to pick a different point)
rp := i % len(s1)
xi := s1[rp]
dxi = 2 * scale
for p, xn := range s1 {
if p != rp {
if dq := d(xi, xn); dq < dxi {
dxi = dq
}
}
}

// step 2: filter
invB := 3 / dxi // b is size of a mesh cell
mx := int64(scale*invB) + 1 // mx is number of cells along a side
// construct map as a histogram:
// key is index into mesh. value is count of points in cell
hm := map[int64]int{}
for ip, p := range s1 {
key := int64(p.x*invB)*mx + int64(p.y*invB)
s1[ip].key = key
hm[key]++
}
// construct s2 = s1 less the points without neighbors
s2 := make([]xy, 0, len(s1))
nx := []int64{-mx - 1, -mx, -mx + 1, -1, 0, 1, mx - 1, mx, mx + 1}
for i, p := range s1 {
nn := 0
for _, ofs := range nx {
nn += hm[p.key+ofs]
if nn > 1 {
s2 = append(s2, s1[i])
break
}
}
}

// step 3: done?
if len(s2) == 0 {
break
}
s1 = s2
}
// step 4: compute answer from approximation
invB := 1 / dxi
mx := int64(scale*invB) + 1
hm := map[int64][]int{}
for i, p := range s {
key := int64(p.x*invB)*mx + int64(p.y*invB)
s[i].key = key
hm[key] = append(hm[key], i)
}
nx := []int64{-mx - 1, -mx, -mx + 1, -1, 0, 1, mx - 1, mx, mx + 1}
var min = scale * 2
for ip, p := range s {
for _, ofs := range nx {
for _, iq := range hm[p.key+ofs] {
if ip != iq {
if d1 := d(p, s[iq]); d1 < min {
min = d1
p1, p2 = p, s[iq]
}
}
}
}
}
return p1, p2
}

## Groovy

Point class:

class Point {
final Number x, y
Point(Number x = 0, Number y = 0) { this.x = x; this.y = y }
Number distance(Point that) { ((this.x - that.x)**2 + (this.y - that.y)**2)**0.5 }
String toString() { "{x:\${x}, y:\${y}}" }
}

Brute force solution. Incorporates X-only and Y-only pre-checks in two places to cut down on the square root calculations:

def bruteClosest(Collection pointCol) {
assert pointCol
List l = pointCol
int n = l.size()
assert n > 1
if (n == 2) return [distance:l[0].distance(l[1]), points:[l[0],l[1]]]
def answer = [distance: Double.POSITIVE_INFINITY]
(0..<(n-1)).each { i ->
((i+1)..<n).findAll { j ->
(l[i].x - l[j].x).abs() < answer.distance &&
(l[i].y - l[j].y).abs() < answer.distance
}.each { j ->
if ((l[i].x - l[j].x).abs() < answer.distance &&
(l[i].y - l[j].y).abs() < answer.distance) {
def dist = l[i].distance(l[j])
if (dist < answer.distance) {
answer = [distance:dist, points:[l[i],l[j]]]
}
}
}
}
}

Elegant (divide-and-conquer reduction) solution. Incorporates X-only and Y-only pre-checks in two places (four if you count the inclusion of the brute force solution) to cut down on the square root calculations:

def elegantClosest(Collection pointCol) {
assert pointCol
List xList = (pointCol as List).sort { it.x }
List yList = xList.clone().sort { it.y }
reductionClosest(xList, xList)
}

def reductionClosest(List xPoints, List yPoints) {
// assert xPoints && yPoints
// assert (xPoints as Set) == (yPoints as Set)
int n = xPoints.size()
if (n < 10) return bruteClosest(xPoints)

int nMid = Math.ceil(n/2)
List xLeft = xPoints[0..<nMid]
List xRight = xPoints[nMid..<n]
Number xMid = xLeft[-1].x
List yLeft = yPoints.findAll { it.x <= xMid }
List yRight = yPoints.findAll { it.x > xMid }
if (xRight[0].x == xMid) {
yLeft = xLeft.collect{ it }.sort { it.y }
yRight = xRight.collect{ it }.sort { it.y }
}

Map aLeft = reductionClosest(xLeft, yLeft)
Map aRight = reductionClosest(xRight, yRight)
Map aMin = aRight.distance < aLeft.distance ? aRight : aLeft
List yMid = yPoints.findAll { (xMid - it.x).abs() < aMin.distance }
int nyMid = yMid.size()
if (nyMid < 2) return aMin

Map answer = aMin
(0..<(nyMid-1)).each { i ->
((i+1)..<nyMid).findAll { j ->
(yMid[j].x - yMid[i].x).abs() < aMin.distance &&
(yMid[j].y - yMid[i].y).abs() < aMin.distance &&
yMid[j].distance(yMid[i]) < aMin.distance
}.each { k ->
if ((yMid[k].x - yMid[i].x).abs() < answer.distance && (yMid[k].y - yMid[i].y).abs() < answer.distance) {
def ikDist = yMid[i].distance(yMid[k])
if ( ikDist < answer.distance) {
answer = [distance:ikDist, points:[yMid[i],yMid[k]]]
}
}
}
}
}

Benchmark/Test:

def random = new Random()

(1..4).each {
def point10 = (0..<(10**it)).collect { new Point(random.nextInt(1000001) - 500000,random.nextInt(1000001) - 500000) }

def startE = System.currentTimeMillis()
def closestE = elegantClosest(point10)
def elapsedE = System.currentTimeMillis() - startE
println """
\${10**it} POINTS
-----------------------------------------
Elegant reduction:
elapsed: \${elapsedE/1000} s
closest: \${closestE}
"""

def startB = System.currentTimeMillis()
def closestB = bruteClosest(point10)
def elapsedB = System.currentTimeMillis() - startB
println """Brute force:
elapsed: \${elapsedB/1000} s
closest: \${closestB}

Speedup ratio (B/E): \${elapsedB/elapsedE}
=========================================
"""

}

Results:

10 POINTS
-----------------------------------------
Elegant reduction:
elapsed: 0.019 s
closest: [distance:85758.5249173515, points:[{x:310073, y:-27339}, {x:382387, y:18761}]]

Brute force:
elapsed: 0.001 s
closest: [distance:85758.5249173515, points:[{x:310073, y:-27339}, {x:382387, y:18761}]]

Speedup ratio (B/E): 0.0526315789
=========================================

100 POINTS
-----------------------------------------
Elegant reduction:
elapsed: 0.019 s
closest: [distance:3166.229934796271, points:[{x:-343735, y:-244394}, {x:-341099, y:-246148}]]

Brute force:
elapsed: 0.027 s
closest: [distance:3166.229934796271, points:[{x:-343735, y:-244394}, {x:-341099, y:-246148}]]

Speedup ratio (B/E): 1.4210526316
=========================================

1000 POINTS
-----------------------------------------
Elegant reduction:
elapsed: 0.241 s
closest: [distance:374.22586762542215, points:[{x:411817, y:-83016}, {x:412038, y:-82714}]]

Brute force:
elapsed: 0.618 s
closest: [distance:374.22586762542215, points:[{x:411817, y:-83016}, {x:412038, y:-82714}]]

Speedup ratio (B/E): 2.5643153527
=========================================

10000 POINTS
-----------------------------------------
Elegant reduction:
elapsed: 1.957 s
closest: [distance:79.00632886041473, points:[{x:187928, y:-452338}, {x:187929, y:-452259}]]

Brute force:
elapsed: 51.567 s
closest: [distance:79.00632886041473, points:[{x:187928, y:-452338}, {x:187929, y:-452259}]]

Speedup ratio (B/E): 26.3500255493
=========================================

BF solution:

import Data.List (minimumBy, tails, unfoldr, foldl1') --'

import System.Random (newStdGen, randomRs)

import Control.Arrow ((&&&))

import Data.Ord (comparing)

vecLeng [[a, b], [p, q]] = sqrt \$ (a - p) ^ 2 + (b - q) ^ 2

findClosestPair =
foldl1'' ((minimumBy (comparing vecLeng) .) . (. return) . (:)) .
concatMap (\(x:xs) -> map ((x :) . return) xs) . init . tails

testCP = do
g <- newStdGen
let pts :: [[Double]]
pts = take 1000 . unfoldr (Just . splitAt 2) \$ randomRs (-1, 1) g
print . (id &&& vecLeng) . findClosestPair \$ pts

main = testCP

foldl1'' = foldl1'
Output:
*Main> testCP
([[0.8347201880148426,0.40774840545089647],[0.8348731214261784,0.4087113189531284]],9.749825850154334e-4)
(4.02 secs, 488869056 bytes)

## Icon and Unicon

This is a brute force solution. It combines reading the points with computing the closest pair seen so far.

record point(x,y)

procedure main()
minDist := 0
minPair := &null
every (points := [],p1 := readPoint()) do {
if *points == 1 then minDist := dSquared(p1,points[1])
every minDist >=:= dSquared(p1,p2 := !points) do minPair := [p1,p2]
push(points, p1)
}

if \minPair then {
write("(",minPair[1].x,",",minPair[1].y,") -> ",
"(",minPair[2].x,",",minPair[2].y,")")
}
else write("One or fewer points!")
end

procedure readPoint() # Skips lines that don't have two numbers on them
suspend !&input ? point(numeric(tab(upto(', '))), numeric((move(1),tab(0))))
end

procedure dSquared(p1,p2) # Compute the square of the distance
return (p2.x-p1.x)^2 + (p2.y-p1.y)^2 # (sufficient for closeness)
end

## IS-BASIC

100 PROGRAM "Closestp.bas"
110 NUMERIC X(1 TO 10),Y(1 TO 10)
120 FOR I=1 TO 10
140 PRINT X(I),Y(I)
150 NEXT
160 LET MN=INF
170 FOR I=1 TO 9
180 FOR J=I+1 TO 10
190 LET DSQ=(X(I)-X(J))^2+(Y(I)-Y(J))^2
200 IF DSQ<MN THEN LET MN=DSQ:LET MINI=I:LET MINJ=J
210 NEXT
220 NEXT
230 PRINT "Closest pair is (";X(MINI);",";Y(MINI);") and (";X(MINJ);",";Y(MINJ);")":PRINT "at distance";SQR(MN)
240 DATA 0.654682,0.925557
250 DATA 0.409382,0.619391
260 DATA 0.891663,0.888594
270 DATA 0.716629,0.996200
280 DATA 0.477721,0.946355
290 DATA 0.925092,0.818220
300 DATA 0.624291,0.142924
310 DATA 0.211332,0.221507
320 DATA 0.293786,0.691701
330 DATA 0.839186,0.728260

## J

Solution of the simpler (brute-force) problem:

vecl   =:  +/"1&.:*:                  NB. length of each vector
dist =: <@:[email protected]:({: -"1 }:)\ NB. calculate all distances among vectors
minpair=: ({~ > {[email protected](\$ #: [email protected],)@:= <./@;)dist NB. find one pair of the closest points
closestpairbf =: (; [email protected]:-/)@minpair NB. the pair and their distance

Examples of use:

]pts=:10 2 [email protected]\$ 0
0.654682 0.925557
0.409382 0.619391
0.891663 0.888594
0.716629 0.9962
0.477721 0.946355
0.925092 0.81822
0.624291 0.142924
0.211332 0.221507
0.293786 0.691701
0.839186 0.72826

closestpairbf pts
+-----------------+---------+
|0.891663 0.888594|0.0779104|
|0.925092 0.81822| |
+-----------------+---------+

The program also works for higher dimensional vectors:

]pts=:10 4 [email protected]\$ 0
0.559164 0.482993 0.876 0.429769
0.217911 0.729463 0.97227 0.132175
0.479206 0.169165 0.495302 0.362738
0.316673 0.797519 0.745821 0.0598321
0.662585 0.726389 0.658895 0.653457
0.965094 0.664519 0.084712 0.20671
0.840877 0.591713 0.630206 0.99119
0.221416 0.114238 0.0991282 0.174741
0.946262 0.505672 0.776017 0.307362
0.262482 0.540054 0.707342 0.465234

closestpairbf pts
+------------------------------------+--------+
|0.217911 0.729463 0.97227 0.132175|0.708555|
|0.316673 0.797519 0.745821 0.0598321| |
+------------------------------------+--------+

## Java

Both the brute-force and the divide-and-conquer methods are implemented.

Code:

import java.util.*;

public class ClosestPair
{
public static class Point
{
public final double x;
public final double y;

public Point(double x, double y)
{
this.x = x;
this.y = y;
}

public String toString()
{ return "(" + x + ", " + y + ")"; }
}

public static class Pair
{
public Point point1 = null;
public Point point2 = null;
public double distance = 0.0;

public Pair()
{ }

public Pair(Point point1, Point point2)
{
this.point1 = point1;
this.point2 = point2;
calcDistance();
}

public void update(Point point1, Point point2, double distance)
{
this.point1 = point1;
this.point2 = point2;
this.distance = distance;
}

public void calcDistance()
{ this.distance = distance(point1, point2); }

public String toString()
{ return point1 + "-" + point2 + " : " + distance; }
}

public static double distance(Point p1, Point p2)
{
double xdist = p2.x - p1.x;
double ydist = p2.y - p1.y;
return Math.hypot(xdist, ydist);
}

public static Pair bruteForce(List<? extends Point> points)
{
int numPoints = points.size();
if (numPoints < 2)
return null;
Pair pair = new Pair(points.get(0), points.get(1));
if (numPoints > 2)
{
for (int i = 0; i < numPoints - 1; i++)
{
Point point1 = points.get(i);
for (int j = i + 1; j < numPoints; j++)
{
Point point2 = points.get(j);
double distance = distance(point1, point2);
if (distance < pair.distance)
pair.update(point1, point2, distance);
}
}
}
return pair;
}

public static void sortByX(List<? extends Point> points)
{
Collections.sort(points, new Comparator<Point>() {
public int compare(Point point1, Point point2)
{
if (point1.x < point2.x)
return -1;
if (point1.x > point2.x)
return 1;
return 0;
}
}
);
}

public static void sortByY(List<? extends Point> points)
{
Collections.sort(points, new Comparator<Point>() {
public int compare(Point point1, Point point2)
{
if (point1.y < point2.y)
return -1;
if (point1.y > point2.y)
return 1;
return 0;
}
}
);
}

public static Pair divideAndConquer(List<? extends Point> points)
{
List<Point> pointsSortedByX = new ArrayList<Point>(points);
sortByX(pointsSortedByX);
List<Point> pointsSortedByY = new ArrayList<Point>(points);
sortByY(pointsSortedByY);
return divideAndConquer(pointsSortedByX, pointsSortedByY);
}

private static Pair divideAndConquer(List<? extends Point> pointsSortedByX, List<? extends Point> pointsSortedByY)
{
int numPoints = pointsSortedByX.size();
if (numPoints <= 3)
return bruteForce(pointsSortedByX);

int dividingIndex = numPoints >>> 1;
List<? extends Point> leftOfCenter = pointsSortedByX.subList(0, dividingIndex);
List<? extends Point> rightOfCenter = pointsSortedByX.subList(dividingIndex, numPoints);

List<Point> tempList = new ArrayList<Point>(leftOfCenter);
sortByY(tempList);
Pair closestPair = divideAndConquer(leftOfCenter, tempList);

tempList.clear();
sortByY(tempList);
Pair closestPairRight = divideAndConquer(rightOfCenter, tempList);

if (closestPairRight.distance < closestPair.distance)
closestPair = closestPairRight;

tempList.clear();
double shortestDistance =closestPair.distance;
double centerX = rightOfCenter.get(0).x;
for (Point point : pointsSortedByY)
if (Math.abs(centerX - point.x) < shortestDistance)

for (int i = 0; i < tempList.size() - 1; i++)
{
Point point1 = tempList.get(i);
for (int j = i + 1; j < tempList.size(); j++)
{
Point point2 = tempList.get(j);
if ((point2.y - point1.y) >= shortestDistance)
break;
double distance = distance(point1, point2);
if (distance < closestPair.distance)
{
closestPair.update(point1, point2, distance);
shortestDistance = distance;
}
}
}
return closestPair;
}

public static void main(String[] args)
{
int numPoints = (args.length == 0) ? 1000 : Integer.parseInt(args[0]);
List<Point> points = new ArrayList<Point>();
Random r = new Random();
for (int i = 0; i < numPoints; i++)
System.out.println("Generated " + numPoints + " random points");
long startTime = System.currentTimeMillis();
Pair bruteForceClosestPair = bruteForce(points);
long elapsedTime = System.currentTimeMillis() - startTime;
System.out.println("Brute force (" + elapsedTime + " ms): " + bruteForceClosestPair);
startTime = System.currentTimeMillis();
Pair dqClosestPair = divideAndConquer(points);
elapsedTime = System.currentTimeMillis() - startTime;
System.out.println("Divide and conquer (" + elapsedTime + " ms): " + dqClosestPair);
if (bruteForceClosestPair.distance != dqClosestPair.distance)
System.out.println("MISMATCH");
}
}
Output:
java ClosestPair 10000
Generated 10000 random points
Brute force (1594 ms): (0.9246533850872104, 0.098709007587097)-(0.924591196030625, 0.09862206991823985) : 1.0689077146927108E-4
Divide and conquer (250 ms): (0.924591196030625, 0.09862206991823985)-(0.9246533850872104, 0.098709007587097) : 1.0689077146927108E-4

## JavaScript

Using bruteforce algorithm, the bruteforceClosestPair method below expects an array of objects with x- and y-members set to numbers, and returns an object containing the members distance and points.

function distance(p1, p2) {
var dx = Math.abs(p1.x - p2.x);
var dy = Math.abs(p1.y - p2.y);
return Math.sqrt(dx*dx + dy*dy);
}

function bruteforceClosestPair(arr) {
if (arr.length < 2) {
return Infinity;
} else {
var minDist = distance(arr[0], arr[1]);
var minPoints = arr.slice(0, 2);

for (var i=0; i<arr.length-1; i++) {
for (var j=i+1; j<arr.length; j++) {
if (distance(arr[i], arr[j]) < minDist) {
minDist = distance(arr[i], arr[j]);
minPoints = [ arr[i], arr[j] ];
}
}
}
return {
distance: minDist,
points: minPoints
};
}
}

divide-and-conquer method:

var Point = function(x, y) {
this.x = x;
this.y = y;
};
Point.prototype.getX = function() {
return this.x;
};
Point.prototype.getY = function() {
return this.y;
};

var mergeSort = function mergeSort(points, comp) {
if(points.length < 2) return points;

var n = points.length,
i = 0,
j = 0,
leftN = Math.floor(n / 2),
rightN = leftN;

var leftPart = mergeSort( points.slice(0, leftN), comp),
rightPart = mergeSort( points.slice(rightN), comp );

var sortedPart = [];

while((i < leftPart.length) && (j < rightPart.length)) {
if(comp(leftPart[i], rightPart[j]) < 0) {
sortedPart.push(leftPart[i]);
i += 1;
}
else {
sortedPart.push(rightPart[j]);
j += 1;
}
}
while(i < leftPart.length) {
sortedPart.push(leftPart[i]);
i += 1;
}
while(j < rightPart.length) {
sortedPart.push(rightPart[j]);
j += 1;
}
return sortedPart;
};

var closestPair = function _closestPair(Px, Py) {
if(Px.length < 2) return { distance: Infinity, pair: [ new Point(0, 0), new Point(0, 0) ] };
if(Px.length < 3) {
//find euclid distance
var d = Math.sqrt( Math.pow(Math.abs(Px[1].x - Px[0].x), 2) + Math.pow(Math.abs(Px[1].y - Px[0].y), 2) );
return {
distance: d,
pair: [ Px[0], Px[1] ]
};
}

var n = Px.length,
leftN = Math.floor(n / 2),
rightN = leftN;

var Xl = Px.slice(0, leftN),
Xr = Px.slice(rightN),
Xm = Xl[leftN - 1],
Yl = [],
Yr = [];
//separate Py
for(var i = 0; i < Py.length; i += 1) {
if(Py[i].x <= Xm.x)
Yl.push(Py[i]);
else
Yr.push(Py[i]);
}

var dLeft = _closestPair(Xl, Yl),
dRight = _closestPair(Xr, Yr);

var minDelta = dLeft.distance,
closestPair = dLeft.pair;
if(dLeft.distance > dRight.distance) {
minDelta = dRight.distance;
closestPair = dRight.pair;
}

//filter points around Xm within delta (minDelta)
var closeY = [];
for(i = 0; i < Py.length; i += 1) {
if(Math.abs(Py[i].x - Xm.x) < minDelta) closeY.push(Py[i]);
}
//find min within delta. 8 steps max
for(i = 0; i < closeY.length; i += 1) {
for(var j = i + 1; j < Math.min( (i + 8), closeY.length ); j += 1) {
var d = Math.sqrt( Math.pow(Math.abs(closeY[j].x - closeY[i].x), 2) + Math.pow(Math.abs(closeY[j].y - closeY[i].y), 2) );
if(d < minDelta) {
minDelta = d;
closestPair = [ closeY[i], closeY[j] ]
}
}
}

return {
distance: minDelta,
pair: closestPair
};
};

var points = [
new Point(0.748501, 4.09624),
new Point(3.00302, 5.26164),
new Point(3.61878, 9.52232),
new Point(7.46911, 4.71611),
new Point(5.7819, 2.69367),
new Point(2.34709, 8.74782),
new Point(2.87169, 5.97774),
new Point(6.33101, 0.463131),
new Point(7.46489, 4.6268),
new Point(1.45428, 0.087596)
];

var sortX = function (a, b) { return (a.x < b.x) ? -1 : ((a.x > b.x) ? 1 : 0); }
var sortY = function (a, b) { return (a.y < b.y) ? -1 : ((a.y > b.y) ? 1 : 0); }

var Px = mergeSort(points, sortX);
var Py = mergeSort(points, sortY);

console.log(JSON.stringify(closestPair(Px, Py))) // {"distance":0.0894096443343775,"pair":[{"x":7.46489,"y":4.6268},{"x":7.46911,"y":4.71611}]}

var points2 = [new Point(37100, 13118), new Point(37134, 1963), new Point(37181, 2008), new Point(37276, 21611), new Point(37307, 9320)];

Px = mergeSort(points2, sortX);
Py = mergeSort(points2, sortY);

console.log(JSON.stringify(closestPair(Px, Py))); // {"distance":65.06919393998976,"pair":[{"x":37134,"y":1963},{"x":37181,"y":2008}]}

## jq

Works with: jq version 1.4

The solution presented here is essentially a direct translation into jq of the pseudo-code presented in the task description, but "closest_pair" is added so that any list of [x,y] points can be presented, and extra lines are added to ensure that xL and yL have the same lengths.

Infrastructure:

# This definition of "until" is included in recent versions (> 1.4) of jq
# Emit the first input that satisfied the condition
def until(cond; next):
def _until:
if cond then . else (next|_until) end;
_until;

# Euclidean 2d distance
def dist(x;y):
[x[0] - y[0], x[1] - y[1]] | map(.*.) | add | sqrt;

# P is an array of points, [x,y].
# Emit the solution in the form [dist, [P1, P2]]
def bruteForceClosestPair(P):
(P|length) as \$length
| if \$length < 2 then null
else
reduce range(0; \$length-1) as \$i
( null;
reduce range(\$i+1; \$length) as \$j
(.;
dist(P[\$i]; P[\$j]) as \$d
| if . == null or \$d < .[0] then [\$d, [ P[\$i], P[\$j] ] ] else . end ) )
end;

def closest_pair:

def abs: if . < 0 then -. else . end;
def ceil: floor as \$floor
| if . == \$floor then \$floor else \$floor + 1 end;

# xP is an array [P(1), .. P(N)] sorted by x coordinate, and
# yP is an array [P(1), .. P(N)] sorted by y coordinate (ascending order).
# if N <= 3 then return closest points of xP using the brute-force algorithm.
def closestPair(xP; yP):
if xP|length <= 3 then bruteForceClosestPair(xP)
else
((xP|length)/2|ceil) as \$N
| xP[0:\$N] as \$xL
| xP[\$N:] as \$xR
| xP[\$N-1][0] as \$xm # middle
| (yP | map(select(.[0] <= \$xm ))) as \$yL0 # might be too long
| (yP | map(select(.[0] > \$xm ))) as \$yR0 # might be too short
| (if \$yL0|length == \$N then \$yL0 else \$yL0[0:\$N] end) as \$yL
| (if \$yL0|length == \$N then \$yR0 else \$yL0[\$N:] + \$yR0 end) as \$yR
| closestPair(\$xL; \$yL) as \$pairL # [dL, pairL]
| closestPair(\$xR; \$yR) as \$pairR # [dR, pairR]
| (if \$pairL[0] < \$pairR[0] then \$pairL else \$pairR end) as \$pair # [ dmin, pairMin]
| (yP | map(select( ((\$xm - .[0])|abs) < \$pair[0]))) as \$yS
| (\$yS | length) as \$nS
| \$pair[0] as \$dmin
| reduce range(0; \$nS - 1) as \$i
( [0, \$pair]; # state: [k, [d, [P1,P2]]]
.[0] = \$i + 1
| until( .[0] as \$k | \$k >= \$nS or (\$yS[\$k][1] - \$yS[\$i][1]) >= \$dmin;
.[0] as \$k
| dist(\$yS[\$k]; \$yS[\$i]) as \$d
| if \$d < .[1][0]
then [\$k+1, [ \$d, [\$yS[\$k], \$yS[\$i]]]]
else .[0] += 1
end) )
| .[1]
end;
closestPair( sort_by(.[0]); sort_by(.[1])) ;

Example from the Mathematica section:

def data:
[[0.748501, 4.09624],
[3.00302, 5.26164],
[3.61878, 9.52232],
[7.46911, 4.71611],
[5.7819, 2.69367],
[2.34709, 8.74782],
[2.87169, 5.97774],
[6.33101, 0.463131],
[7.46489, 4.6268],
[1.45428, 0.087596] ];

data | closest_pair
Output:
\$jq -M -c -n -f closest_pair.jq
[0.0894096443343775,[[7.46489,4.6268],[7.46911,4.71611]]]

## Julia

Works with: Julia version 0.6

Brute-force algorithm:

function closestpair(P::Vector{Vector{T}}) where T <: Number
N = length(P)
if N < 2 return (Inf, ()) end
mindst = norm(P[1] - P[2])
minpts = (P[1], P[2])
for i in 1:N-1, j in i+1:N
tmpdst = norm(P[i] - P[j])
if tmpdst < mindst
mindst = tmpdst
minpts = (P[i], P[j])
end
end
return mindst, minpts
end

closestpair([[0, -0.3], [1., 1.], [1.5, 2], [2, 2], [3, 3]])

## Kotlin

// version 1.1.2

typealias Point = Pair<Double, Double>

fun distance(p1: Point, p2: Point) = Math.hypot(p1.first- p2.first, p1.second - p2.second)

fun bruteForceClosestPair(p: List<Point>): Pair<Double, Pair<Point, Point>> {
val n = p.size
if (n < 2) throw IllegalArgumentException("Must be at least two points")
var minPoints = p[0] to p[1]
var minDistance = distance(p[0], p[1])
for (i in 0 until n - 1)
for (j in i + 1 until n) {
val dist = distance(p[i], p[j])
if (dist < minDistance) {
minDistance = dist
minPoints = p[i] to p[j]
}
}
return minDistance to Pair(minPoints.first, minPoints.second)
}

fun optimizedClosestPair(xP: List<Point>, yP: List<Point>): Pair<Double, Pair<Point, Point>> {
val n = xP.size
if (n <= 3) return bruteForceClosestPair(xP)
val xL = xP.take(n / 2)
val xR = xP.drop(n / 2)
val xm = xP[n / 2 - 1].first
val yL = yP.filter { it.first <= xm }
val yR = yP.filter { it.first > xm }
val (dL, pairL) = optimizedClosestPair(xL, yL)
val (dR, pairR) = optimizedClosestPair(xR, yR)
var dmin = dR
var pairMin = pairR
if (dL < dR) {
dmin = dL
pairMin = pairL
}
val yS = yP.filter { Math.abs(xm - it.first) < dmin }
val nS = yS.size
var closest = dmin
var closestPair = pairMin
for (i in 0 until nS - 1) {
var k = i + 1
while (k < nS && (yS[k].second - yS[i].second < dmin)) {
val dist = distance(yS[k], yS[i])
if (dist < closest) {
closest = dist
closestPair = Pair(yS[k], yS[i])
}
k++
}
}
return closest to closestPair
}

fun main(args: Array<String>) {
val points = listOf(
listOf(
5.0 to 9.0, 9.0 to 3.0, 2.0 to 0.0, 8.0 to 4.0, 7.0 to 4.0,
9.0 to 10.0, 1.0 to 9.0, 8.0 to 2.0, 0.0 to 10.0, 9.0 to 6.0
),
listOf(
0.654682 to 0.925557, 0.409382 to 0.619391, 0.891663 to 0.888594,
0.716629 to 0.996200, 0.477721 to 0.946355, 0.925092 to 0.818220,
0.624291 to 0.142924, 0.211332 to 0.221507, 0.293786 to 0.691701,
0.839186 to 0.728260
)
)
for (p in points) {
val (dist, pair) = bruteForceClosestPair(p)
println("Closest pair (brute force) is \${pair.first} and \${pair.second}, distance \$dist")
val xP = p.sortedBy { it.first }
val yP = p.sortedBy { it.second }
val (dist2, pair2) = optimizedClosestPair(xP, yP)
println("Closest pair (optimized) is \${pair2.first} and \${pair2.second}, distance \$dist2\n")
}
}
Output:
Closest pair (brute force) is (8.0, 4.0) and (7.0, 4.0), distance 1.0
Closest pair (optimized)   is (7.0, 4.0) and (8.0, 4.0), distance 1.0

Closest pair (brute force) is (0.891663, 0.888594) and (0.925092, 0.81822), distance 0.07791019135517516
Closest pair (optimized)   is (0.891663, 0.888594) and (0.925092, 0.81822), distance 0.07791019135517516

## Liberty BASIC

NB array terms can not be READ directly.

N =10

dim x( N), y( N)

firstPt =0
secondPt =0

for i =1 to N
read f: x( i) =f
read f: y( i) =f
next i

minDistance =1E6

for i =1 to N -1
for j =i +1 to N
dxSq =( x( i) -x( j))^2
dySq =( y( i) -y( j))^2
D =abs( ( dxSq +dySq)^0.5)
if D <minDistance then
minDistance =D
firstPt =i
secondPt =j
end if
next j
next i

print "Distance ="; minDistance; " between ( "; x( firstPt); ", "; y( firstPt); ") and ( "; x( secondPt); ", "; y( secondPt); ")"

end

data 0.654682, 0.925557
data 0.409382, 0.619391
data 0.891663, 0.888594
data 0.716629, 0.996200
data 0.477721, 0.946355
data 0.925092, 0.818220
data 0.624291, 0.142924
data 0.211332, 0.221507
data 0.293786, 0.691701
data 0.839186, 0.72826

Distance =0.77910191e-1 between ( 0.891663, 0.888594) and ( 0.925092, 0.81822)

## Maple

ClosestPair := module()

local
ModuleApply := proc(L::list,\$)
local Lx, Ly, out;
Ly := sort(L, 'key'=(i->i[2]), 'output'='permutation');
Lx := sort(L, 'key'=(i->i[1]), 'output'='permutation');
out := Recurse(L, Lx, Ly, 1, numelems(L));
return sqrt(out[1]), out[2];
end proc; # ModuleApply

local
BruteForce := proc(L, Lx, r1:=1, r2:=numelems(L), \$)
local d, p, n, i, j;
d := infinity;
for i from r1 to r2-1 do
for j from i+1 to r2 do
n := dist( L[Lx[i]], L[Lx[j]] );
if n < d then
d := n;
p := [ L[Lx[i]], L[Lx[j]] ];
end if;
end do; # j
end do; # i
return (d, p);
end proc; # BruteForce

local dist := (p, q)->(( (p[1]-q[1])^2+(p[2]-q[2])^2 ));

local Recurse := proc(L, Lx, Ly, r1, r2)
local m, xm, rDist, rPair, lDist, lPair, minDist, minPair, S, i, j, Lyr, Lyl;

if r2-r1 <= 3 then
return BruteForce(L, Lx, r1, r2);
end if;

m := ceil((r2-r1)/2)+r1;
xm := (L[Lx[m]][1] + L[Lx[m-1]][1])/2;

(Lyr, Lyl) := selectremove( i->L[i][1] < xm, Ly);

(rDist, rPair) := thisproc(L, Lx, Lyr, r1, m-1);
(lDist, lPair) := thisproc(L, Lx, Lyl, m, r2);

if rDist < lDist then
minDist := rDist;
minPair := rPair;
else
minDist := lDist;
minPair := lPair;
end if;

S := [ seq( `if`(abs(xm - L[i][1])^2< minDist, L[i], NULL ), i in Ly ) ];

for i from 1 to nops(S)-1 do
for j from i+1 to nops(S) do
if abs( S[i][2] - S[j][2] )^2 >= minDist then
break;
elif dist(S[i], S[j]) < minDist then
minDist := dist(S[i], S[j]);
minPair := [S[i], S[j]];
end if;
end do;
end do;

return (minDist, minPair);

end proc; #Recurse

end module; #ClosestPair
Output:

> L := RandomTools:-Generate(list(list(float(range=0..1),2),512)):
> ClosestPair(L);
0.002576770304, [[0.4265584800, 0.7443097852], [0.4240649736, 0.7449595321]]

## Mathematica / Wolfram Language

nearestPair[data_] :=
Block[{pos, dist = N[Outer[EuclideanDistance, data, data, 1]]},
pos = Position[dist, Min[DeleteCases[Flatten[dist], 0.]]];
data[[pos[[1]]]]]
Output:
nearestPair[{{0.748501, 4.09624}, {3.00302, 5.26164}, {3.61878,
9.52232}, {7.46911, 4.71611}, {5.7819, 2.69367}, {2.34709,
8.74782}, {2.87169, 5.97774}, {6.33101, 0.463131}, {7.46489,
4.6268}, {1.45428, 0.087596}}]

{{7.46911, 4.71611}, {7.46489, 4.6268}}

## MATLAB

This solution is an almost direct translation of the above pseudo-code into MATLAB.

function [closest,closestpair] = closestPair(xP,yP)

N = numel(xP);

if(N <= 3)

%Brute force closestpair
if(N < 2)
closest = +Inf;
closestpair = {};
else
closest = norm(xP{1}-xP{2});
closestpair = {xP{1},xP{2}};

for i = ( 1:N-1 )
for j = ( (i+1):N )
if ( norm(xP{i} - xP{j}) < closest )
closest = norm(xP{i}-xP{j});
closestpair = {xP{i},xP{j}};
end %if
end %for
end %for
end %if (N < 2)
else

halfN = ceil(N/2);

xL = { xP{1:halfN} };
xR = { xP{halfN+1:N} };
xm = xP{halfN}(1);

%cellfun( @(p)le(p(1),xm),yP ) is the same as { p ∈ yP : px ≤ xm }
yLIndicies = cellfun( @(p)le(p(1),xm),yP );

yL = { yP{yLIndicies} };
yR = { yP{~yLIndicies} };

[dL,pairL] = closestPair(xL,yL);
[dR,pairR] = closestPair(xR,yR);

if dL < dR
dmin = dL;
pairMin = pairL;
else
dmin = dR;
pairMin = pairR;
end

%cellfun( @(p)lt(norm(xm-p(1)),dmin),yP ) is the same as
%{ p ∈ yP : |xm - px| < dmin }
yS = {yP{ cellfun( @(p)lt(norm(xm-p(1)),dmin),yP ) }};
nS = numel(yS);

closest = dmin;
closestpair = pairMin;

for i = (1:nS-1)
k = i+1;

while( (k<=nS) && (yS{k}(2)-yS{i}(2) < dmin) )

if norm(yS{k}-yS{i}) < closest
closest = norm(yS{k}-yS{i});
closestpair = {yS{k},yS{i}};
end

k = k+1;
end %while
end %for
end %if (N <= 3)
end %closestPair
Output:
[distance,pair]=closestPair({[0 -.3],[1 1],[1.5 2],[2 2],[3 3]},{[0 -.3],[1 1],[1.5 2],[2 2],[3 3]})

distance =

0.500000000000000

pair =

[1x2 double] [1x2 double] %The pair is [1.5 2] and [2 2] which is correct

## Microsoft Small Basic

' Closest Pair Problem
s="0.654682,0.925557,0.409382,0.619391,0.891663,0.888594,0.716629,0.996200,0.477721,0.946355,0.925092,0.818220,0.624291,0.142924,0.211332,0.221507,0.293786,0.691701,0.839186,0.728260,"
i=0
While s<>""
i=i+1
For j=1 To 2
k=Text.GetIndexOf(s,",")
ss=Text.GetSubText(s,1,k-1)
s=Text.GetSubTextToEnd(s,k+1)
pxy[i][j]=ss
EndFor
EndWhile
n=i
i=1
j=2
dd=Math.Power(pxy[i][1]-pxy[j][1],2)+Math.Power(pxy[i][2]-pxy[j][2],2)
ddmin=dd
ii=i
jj=j
For i=1 To n
For j=1 To n
dd=Math.Power(pxy[i][1]-pxy[j][1],2)+Math.Power(pxy[i][2]-pxy[j][2],2)
If dd>0 Then
If dd<ddmin Then
ddmin=dd
ii=i
jj=j
EndIf
EndIf
EndFor
EndFor
sqrt1=ddmin
sqrt2=ddmin/2
For i=1 To 20
If sqrt1=sqrt2 Then
Goto exitfor
EndIf
sqrt1=sqrt2
sqrt2=(sqrt1+(ddmin/sqrt1))/2
EndFor
exitfor:
TextWindow.WriteLine("the minimum distance "+sqrt2)
TextWindow.WriteLine("is between the points:")
TextWindow.WriteLine(" ["+pxy[ii][1]+","+pxy[ii][2]+"] and")
TextWindow.WriteLine(" ["+pxy[jj][1]+","+pxy[jj][2]+"]")
Output:
the minimum distance 0,0779101913551750943201426138
is between the points:
[0.891663,0.888594] and
[0.925092,0.818220]

## OCaml

type point = { x : float; y : float }

let cmpPointX (a : point) (b : point) = compare a.x b.x
let cmpPointY (a : point) (b : point) = compare a.y b.y

let distSqrd (seg : (point * point) option) =
match seg with
| None -> max_float
| Some(line) ->
let a = fst line in
let b = snd line in

let dx = a.x -. b.x in
let dy = a.y -. b.y in

dx*.dx +. dy*.dy

let dist seg =
sqrt (distSqrd seg)

let shortest l1 l2 =
if distSqrd l1 < distSqrd l2 then
l1
else
l2

let halve l =
let n = List.length l in
BatList.split_at (n/2) l

let rec closestBoundY from maxY (ptsByY : point list) =
match ptsByY with
| [] -> None
| hd :: tl ->
if hd.y > maxY then
None
else
let toHd = Some(from, hd) in
let bestToRest = closestBoundY from maxY tl in
shortest toHd bestToRest

let rec closestInRange ptsByY maxDy =
match ptsByY with
| [] -> None
| hd :: tl ->
let fromHd = closestBoundY hd (hd.y +. maxDy) tl in
let fromRest = closestInRange tl maxDy in
shortest fromHd fromRest

let rec closestPairByX (ptsByX : point list) =
if List.length ptsByX < 2 then
None
else
let (left, right) = halve ptsByX in
let leftResult = closestPairByX left in
let rightResult = closestPairByX right in

let bestInHalf = shortest leftResult rightResult in
let bestLength = dist bestInHalf in

let divideX = (List.hd right).x in
let inBand = List.filter(fun(p) -> abs_float(p.x -. divideX) < bestLength) ptsByX in

let byY = List.sort cmpPointY inBand in
let bestCross = closestInRange byY bestLength in
shortest bestInHalf bestCross

let closestPair pts =
let ptsByX = List.sort cmpPointX pts in
closestPairByX ptsByX

let parsePoint str =
let sep = Str.regexp_string "," in
let tokens = Str.split sep str in
let xStr = List.nth tokens 0 in
let yStr = List.nth tokens 1 in

let xVal = (float_of_string xStr) in
let yVal = (float_of_string yStr) in

{ x = xVal; y = yVal }

let loadPoints filename =
let ic = open_in filename in
let result = ref [] in
try
while true do
let s = input_line ic in
if s <> "" then
let p = parsePoint s in
result := p :: !result;
done;
!result
with End_of_file ->
close_in ic;
!result
;;

let start = Sys.time() in
let c = closestPair loaded in
let taken = Sys.time() -. start in
Printf.printf "Took %f [s]\n" taken;

match c with
| None -> Printf.printf "No closest pair\n"
| Some(seg) ->
let a = fst seg in
let b = snd seg in

Printf.printf "(%f, %f) (%f, %f) Dist %f\n" a.x a.y b.x b.y (dist c)

## Oz

Translation of pseudocode:

declare
fun {Distance X1#Y1 X2#Y2}
{Sqrt {Pow X2-X1 2.0} + {Pow Y2-Y1 2.0}}
end

%% brute force
fun {BFClosestPair Points=P1|P2|_}
Ps = {List.toTuple unit Points} %% for efficient random access
N = {Width Ps}
MinDist = {NewCell {Distance P1 P2}}
MinPoints = {NewCell P1#P2}
in
for I in 1..N-1 do
for J in I+1..N do
IJDist = {Distance Ps.I Ps.J}
in
if IJDist < @MinDist then
MinDist := IJDist
MinPoints := Ps.I#Ps.J
end
end
end
@MinPoints
end

%% divide and conquer
fun {ClosestPair Points}
case {ClosestPair2
{Sort Points {LessThanBy X}}
{Sort Points {LessThanBy Y}}}
of Distance#Pair then
Pair
end
end

%% XP: points sorted by X, YP: sorted by Y
%% returns a pair Distance#Pair
fun {ClosestPair2 XP YP}
N = {Length XP} = {Length YP}
in
if N =< 3 then
P = {BFClosestPair XP}
in
{Distance P.1 P.2}#P
else
XL XR
{List.takeDrop XP (N div 2) ?XL ?XR}
XM = {Nth XP (N div 2)}.X
YL YR
{List.partition YP fun {\$ P} P.X =< XM end ?YL ?YR}
DL#PairL = {ClosestPair2 XL YL}
DR#PairR = {ClosestPair2 XR YR}
DMin#PairMin = if DL < DR then DL#PairL else DR#PairR end
YSList = {Filter YP fun {\$ P} {Abs XM-P.X} < DMin end}
YS = {List.toTuple unit YSList} %% for efficient random access
NS = {Width YS}
Closest = {NewCell DMin}
ClosestPair = {NewCell PairMin}
in
for I in 1..NS-1 do
for K in I+1..NS while:YS.K.Y - YS.I.Y < DMin do
DistKI = {Distance YS.K YS.I}
in
if DistKI < @Closest then
Closest := DistKI
ClosestPair := YS.K#YS.I
end
end
end
@Closest#@ClosestPair
end
end

%% To access components when points are represented as pairs
X = 1
Y = 2

%% returns a less-than predicate that accesses feature F
fun {LessThanBy F}
fun {\$ A B}
A.F < B.F
end
end

fun {Random Min Max}
Min +
{Int.toFloat {OS.rand}} * (Max-Min)
/ {Int.toFloat {OS.randLimits _}}
end

fun {RandomPoint}
{Random 0.0 100.0}#{Random 0.0 100.0}
end

Points = {MakeList 5}
in
{ForAll Points RandomPoint}
{Show Points}
{Show {ClosestPair Points}}

## PARI/GP

closestPair(v)={
my(r=norml2(v[1]-v[2]),at=[1,2]);
for(a=1,#v-1,
for(b=a+1,#v,
if(norml2(v[a]-v[b])<r,
at=[a,b];
r=norml2(v[a]-v[b])
)
)
);
[v[at[1]],v[at[2]]]
};

## Pascal

Brute force only calc square of distance, like AWK etc... As fast as D .

program closestPoints;
{\$IFDEF FPC}
{\$MODE Delphi}
{\$ENDIF}
const
PointCnt = 10000;//31623;
type
TdblPoint = Record
ptX,
ptY : double;
end;
tPtLst = array of TdblPoint;

tMinDIstIdx = record
md1,
md2 : NativeInt;
end;

function ClosPointBruteForce(var ptl :tPtLst):tMinDIstIdx;
Var
i,j,k : NativeInt;
mindst2,dst2: double; //square of distance, no need to sqrt
p0,p1 : ^TdblPoint; //using pointer, since calc of ptl[?] takes much time
Begin
i := Low(ptl);
j := High(ptl);
result.md1 := i;result.md2 := j;
mindst2 := sqr(ptl[i].ptX-ptl[j].ptX)+sqr(ptl[i].ptY-ptl[j].ptY);
repeat
p0 := @ptl[i];
p1 := p0; inc(p1);
For k := i+1 to j do
Begin
dst2:= sqr(p0^.ptX-p1^.ptX)+sqr(p0^.ptY-p1^.ptY);
IF mindst2 > dst2 then
Begin
mindst2 := dst2;
result.md1 := i;
result.md2 := k;
end;
inc(p1);
end;
inc(i);
until i = j;
end;

var
PointLst :tPtLst;
cloPt : tMinDIstIdx;
i : NativeInt;
Begin
randomize;
setlength(PointLst,PointCnt);
For i := 0 to PointCnt-1 do
with PointLst[i] do
Begin
ptX := random;
ptY := random;
end;
cloPt:= ClosPointBruteForce(PointLst) ;
i := cloPt.md1;
Writeln('P[',i:4,']= x: ',PointLst[i].ptX:0:8,
' y: ',PointLst[i].ptY:0:8);
i := cloPt.md2;
Writeln('P[',i:4,']= x: ',PointLst[i].ptX:0:8,
' y: ',PointLst[i].ptY:0:8);
end.
Output:
PointCnt = 10000

//without randomize always same results //32-Bit P[ 324]= x: 0.26211815 y: 0.45851455 P[3391]= x: 0.26217852 y: 0.45849116 real 0m0.114s //fpc 3.1.1 32 Bit -O4 -MDelphi..cpu i4330 3.5 Ghz //64-Bit doubles the speed comp switch -O2 ..-O4 same timings P[ 324]= x: 0.26211815 y: 0.45851455 P[3391]= x: 0.26217852 y: 0.45849116 real 0m0.059s //fpc 3.1.1 64 Bit -O4 -MDelphi..cpu i4330 3.5 Ghz

//with randomize P[ 47]= x: 0.12408823 y: 0.04501338 P[9429]= x: 0.12399629 y: 0.04496700 //32-Bit

PointCnt = { 10000*sqrt(10) } 31623;-> real 0m1.112s 10x times runtime

## Perl

The divide & conquer technique is about 100x faster than the brute-force algorithm.

#! /usr/bin/perl
use strict;
use POSIX qw(ceil);

sub dist
{
my ( \$a, \$b) = @_;
return sqrt( (\$a->[0] - \$b->[0])**2 +
(\$a->[1] - \$b->[1])**2 );
}

sub closest_pair_simple
{
my \$ra = shift;
my @arr = @\$ra;
my \$inf = 1e600;
return \$inf if scalar(@arr) < 2;
my ( \$a, \$b, \$d ) = (\$arr[0], \$arr[1], dist(\$arr[0], \$arr[1]));
while( @arr ) {
my \$p = pop @arr;
foreach my \$l (@arr) {
my \$t = dist(\$p, \$l);
(\$a, \$b, \$d) = (\$p, \$l, \$t) if \$t < \$d;
}
}
return (\$a, \$b, \$d);
}

sub closest_pair
{
my \$r = shift;
my @ax = sort { \$a->[0] <=> \$b->[0] } @\$r;
my @ay = sort { \$a->[1] <=> \$b->[1] } @\$r;
return closest_pair_real(\@ax, \@ay);
}

sub closest_pair_real
{
my (\$rx, \$ry) = @_;
my @xP = @\$rx;
my @yP = @\$ry;
my \$N = @xP;
return closest_pair_simple(\$rx) if scalar(@xP) <= 3;

my \$inf = 1e600;
my \$midx = ceil(\$N/2)-1;

my @PL = @xP[0 .. \$midx];
my @PR = @xP[\$midx+1 .. \$N-1];

my \$xm = \${\$xP[\$midx]}[0];

my @yR = ();
my @yL = ();
foreach my \$p (@yP) {
if ( \${\$p}[0] <= \$xm ) {
push @yR, \$p;
} else {
push @yL, \$p;
}
}

my (\$al, \$bl, \$dL) = closest_pair_real(\@PL, \@yR);
my (\$ar, \$br, \$dR) = closest_pair_real(\@PR, \@yL);

my (\$m1, \$m2, \$dmin) = (\$al, \$bl, \$dL);
(\$m1, \$m2, \$dmin) = (\$ar, \$br, \$dR) if \$dR < \$dL;

my @yS = ();
foreach my \$p (@yP) {
push @yS, \$p if abs(\$xm - \${\$p}[0]) < \$dmin;
}

if ( @yS ) {
my ( \$w1, \$w2, \$closest ) = (\$m1, \$m2, \$dmin);
foreach my \$i (0 .. (\$#yS - 1)) {

my \$k = \$i + 1;
while ( (\$k <= \$#yS) && ( (\${\$yS[\$k]}[1] - \${\$yS[\$i]}[1]) < \$dmin) ) {
my \$d = dist(\$yS[\$k], \$yS[\$i]);
(\$w1, \$w2, \$closest) = (\$yS[\$k], \$yS[\$i], \$d) if \$d < \$closest;
\$k++;
}

}
return (\$w1, \$w2, \$closest);

} else {
return (\$m1, \$m2, \$dmin);
}
}

my @points = ();
my \$N = 5000;

foreach my \$i (1..\$N) {
push @points, [rand(20)-10.0, rand(20)-10.0];
}

my (\$a, \$b, \$d) = closest_pair_simple(\@points);
print "\$d\n";

my (\$a1, \$b1, \$d1) = closest_pair(\@points);
print "\$d1\n";

## Phix

Brute force and divide and conquer (translated from pseudocode) approaches compared

function bruteForceClosestPair(sequence s)
atom {x1,y1} = s[1], {x2,y2} = s[2], dx = x1-x2, dy = y1-y2, mind = dx*dx+dy*dy
sequence minp = s[1..2]
for i=1 to length(s)-1 do
{x1,y1} = s[i]
for j=i+1 to length(s) do
{x2,y2} = s[j]
dx = x1-x2
dx = dx*dx
if dx<mind then
dy = y1-y2
dx += dy*dy
if dx<mind then
mind = dx
minp = {s[i],s[j]}
end if
end if
end for
end for
return {sqrt(mind),minp}
end function

sequence testset = sq_rnd(repeat({1,1},10000))
atom t0 = time()
sequence points
atom d
{d,points} = bruteForceClosestPair(testset)
-- (Sorting the final point pair makes brute/dc more likely to tally. Note however
-- when >1 equidistant pairs exist, brute and dc may well return different pairs;
-- it is only a problem if they decide to return different minimum distances.)
atom {{x1,y1},{x2,y2}} = sort(points)
printf(1,"Closest pair: {%f,%f} {%f,%f}, distance=%f (%3.2fs)\n",{x1,y2,x2,y2,d,time()-t0})

t0 = time()
constant X = 1, Y = 2
sequence xP = sort(testset)

function byY(sequence p1, p2)
return compare(p1[Y],p2[Y])
end function
sequence yP = custom_sort(routine_id("byY"),testset)

function distsq(sequence p1,p2)
atom {x1,y1} = p1, {x2,y2} = p2
x1 -= x2
y1 -= y2
return x1*x1 + y1*y1
end function

function closestPair(sequence 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)
integer N, midN, k, nS
sequence xL, xR, yL, yR, pairL, pairR, pairMin, yS, cPair
atom xm, dL, dR, dmin, closest

N = length(xP)
if length(yP)!=N then ?9/0 end if -- (sanity check)
if N<=3 then
return bruteForceClosestPair(xP)
end if
midN = floor(N/2)
xL = xP[1..midN]
xR = xP[midN+1..N]
xm = xP[midN][X]
yL = {}
yR = {}
for i=1 to N do
if yP[i][X]<=xm then
yL = append(yL,yP[i])
else
yR = append(yR,yP[i])
end if
end for
{dL, pairL} = closestPair(xL, yL)
{dR, pairR} = closestPair(xR, yR)
{dmin, pairMin} = {dR, pairR}
if dL<dR then
{dmin, pairMin} = {dL, pairL}
end if
yS = {}
for i=1 to length(yP) do
if abs(xm-yP[i][X])<dmin then
yS = append(yS,yP[i])
end if
end for
nS = length(yS)
{closest, cPair} = {dmin*dmin, pairMin}
for i=1 to nS-1 do
k = i + 1
while k<=nS and (yS[k][Y]-yS[i][Y])<dmin do
d = distsq(yS[k],yS[i])
if d<closest then
{closest, cPair} = {d, {yS[k], yS[i]}}
end if
k += 1
end while
end for
return {sqrt(closest), cPair}
end function

{d,points} = closestPair(xP,yP)
{{x1,y1},{x2,y2}} = sort(points) -- (see note above)
printf(1,"Closest pair: {%f,%f} {%f,%f}, distance=%f (%3.2fs)\n",{x1,y2,x2,y2,d,time()-t0})
Output:
Closest pair: {0.0328051,0.0966250} {0.0328850,0.0966250}, distance=0.000120143 (2.37s)
Closest pair: {0.0328051,0.0966250} {0.0328850,0.0966250}, distance=0.000120143 (0.14s)

## PicoLisp

(de closestPairBF (Lst)
(let Min T
(use (Pt1 Pt2)
(for P Lst
(for Q Lst
(or
(== P Q)
(>=
(setq N
(let (A (- (car P) (car Q)) B (- (cdr P) (cdr Q)))
(+ (* A A) (* B B)) ) )
Min )
(setq Min N Pt1 P Pt2 Q) ) ) )
(list Pt1 Pt2 (sqrt Min)) ) ) )

Test:

: (scl 6)
-> 6

: (closestPairBF
(quote
(0.654682 . 0.925557)
(0.409382 . 0.619391)
(0.891663 . 0.888594)
(0.716629 . 0.996200)
(0.477721 . 0.946355)
(0.925092 . 0.818220)
(0.624291 . 0.142924)
(0.211332 . 0.221507)
(0.293786 . 0.691701)
(0.839186 . 0.728260) ) )
-> ((891663 . 888594) (925092 . 818220) 77910)

## PL/I

/* Closest Pair Problem */
closest: procedure options (main);
declare n fixed binary;

get list (n);
begin;
declare 1 P(n),
2 x float,
2 y float;
declare (i, ii, j, jj) fixed binary;
declare (distance, min_distance initial (0) ) float;

get list (P);
min_distance = sqrt( (P.x(1) - P.x(2))**2 + (P.y(1) - P.y(2))**2 );
ii = 1; jj = 2;
do i = 1 to n;
do j = 1 to n;
distance = sqrt( (P.x(i) - P.x(j))**2 + (P.y(i) - P.y(j))**2 );
if distance > 0 then
if distance < min_distance then
do;
min_distance = distance;
ii = i; jj = j;
end;
end;
end;
put skip edit ('The minimum distance ', min_distance,
' is between the points [', P.x(ii),
',', P.y(ii), '] and [', P.x(jj), ',', P.y(jj), ']' )
(a, f(6,2));
end;
end closest;

## Prolog

Brute force version, works with SWI-Prolog, tested on version 7.2.3.

% main predicate, find and print closest point
do_find_closest_points(Points) :-
points_closest(Points, points(point(X1,Y1),point(X2,Y2),Dist)),
format('Point 1 : (~p, ~p)~n', [X1,Y1]),
format('Point 1 : (~p, ~p)~n', [X2,Y2]),
format('Distance: ~p~n', [Dist]).

% Find the distance between two points
distance(point(X1,Y1), point(X2,Y2), points(point(X1,Y1),point(X2,Y2),Dist)) :-
Dx is X2 - X1,
Dy is Y2 - Y1,
Dist is sqrt(Dx * Dx + Dy * Dy).

% find the closest point that relatest to another point
point_closest(Points, Point, Closest) :-
select(Point, Points, Remaining),
maplist(distance(Point), Remaining, PointList),
foldl(closest, PointList, 0, Closest).

% find the closest point/dist pair for all points
points_closest(Points, Closest) :-
maplist(point_closest(Points), Points, ClosestPerPoint),
foldl(closest, ClosestPerPoint, 0, Closest).

% used by foldl to get the lowest point/distance combination
closest(points(P1,P2,Dist), 0, points(P1,P2,Dist)).
closest(points(_,_,Dist), points(P1,P2,Dist2), points(P1,P2,Dist2)) :-
Dist2 < Dist.
closest(points(P1,P2,Dist), points(_,_,Dist2), points(P1,P2,Dist)) :-
Dist =< Dist2.

To test, pass in a list of points.

do_find_closest_points([
point(0.654682, 0.925557),
point(0.409382, 0.619391),
point(0.891663, 0.888594),
point(0.716629, 0.996200),
point(0.477721, 0.946355),
point(0.925092, 0.818220),
point(0.624291, 0.142924),
point(0.211332, 0.221507),
point(0.293786, 0.691701),
point(0.839186, 0.728260)
]).

Output:
Point 1 : (0.925092, 0.81822)
Point 1 : (0.891663, 0.888594)
Distance: 0.07791019135517516
true ;
false.

## PureBasic

Brute force version

Procedure.d bruteForceClosestPair(Array P.coordinate(1))
Protected N=ArraySize(P()), i, j
Protected mindistance.f=Infinity(), t.d
Shared a, b
If N<2
a=0: b=0
Else
For i=0 To N-1
For j=i+1 To N
t=Pow(Pow(P(i)\x-P(j)\x,2)+Pow(P(i)\y-P(j)\y,2),0.5)
If mindistance>t
mindistance=t
a=i: b=j
EndIf
Next
Next
EndIf
ProcedureReturn mindistance
EndProcedure

Implementation can be as

Structure coordinate
x.d
y.d
EndStructure

Dim DataSet.coordinate(9)
Define i, x.d, y.d, a, b

;- Load data from datasection
Restore DataPoints
For i=0 To 9
DataSet(i)\x=x
DataSet(i)\y=y
Next i

If OpenConsole()
PrintN("Mindistance= "+StrD(bruteForceClosestPair(DataSet()),6))
PrintN("Point 1= "+StrD(DataSet(a)\x,6)+": "+StrD(DataSet(a)\y,6))
PrintN("Point 2= "+StrD(DataSet(b)\x,6)+": "+StrD(DataSet(b)\y,6))
Print(#CRLF\$+"Press ENTER to quit"): Input()
EndIf

DataSection
DataPoints:
Data.d 0.654682, 0.925557, 0.409382, 0.619391, 0.891663, 0.888594
Data.d 0.716629, 0.996200, 0.477721, 0.946355, 0.925092, 0.818220
Data.d 0.624291, 0.142924, 0.211332, 0.221507, 0.293786, 0.691701, 0.839186, 0.72826
EndDataSection
Output:
Mindistance= 0.077910
Point 1= 0.891663: 0.888594
Point 2= 0.925092: 0.818220

Press ENTER to quit

## Python

"""
Compute nearest pair of points using two algorithms

First algorithm is 'brute force' comparison of every possible pair.
Second, 'divide and conquer', is based on:
www.cs.iupui.edu/~xkzou/teaching/CS580/Divide-and-conquer-closestPair.ppt
"""

from random import randint, randrange
from operator import itemgetter, attrgetter

infinity = float('inf')

# Note the use of complex numbers to represent 2D points making distance == abs(P1-P2)

def bruteForceClosestPair(point):
numPoints = len(point)
if numPoints < 2:
return infinity, (None, None)
return min( ((abs(point[i] - point[j]), (point[i], point[j]))
for i in range(numPoints-1)
for j in range(i+1,numPoints)),
key=itemgetter(0))

def closestPair(point):
xP = sorted(point, key= attrgetter('real'))
yP = sorted(point, key= attrgetter('imag'))
return _closestPair(xP, yP)

def _closestPair(xP, yP):
numPoints = len(xP)
if numPoints <= 3:
return bruteForceClosestPair(xP)
Pl = xP[:numPoints/2]
Pr = xP[numPoints/2:]
Yl, Yr = [], []
xDivider = Pl[-1].real
for p in yP:
if p.real <= xDivider:
Yl.append(p)
else:
Yr.append(p)
dl, pairl = _closestPair(Pl, Yl)
dr, pairr = _closestPair(Pr, Yr)
dm, pairm = (dl, pairl) if dl < dr else (dr, pairr)
# Points within dm of xDivider sorted by Y coord
closeY = [p for p in yP if abs(p.real - xDivider) < dm]
numCloseY = len(closeY)
if numCloseY > 1:
# There is a proof that you only need compare a max of 7 next points
closestY = min( ((abs(closeY[i] - closeY[j]), (closeY[i], closeY[j]))
for i in range(numCloseY-1)
for j in range(i+1,min(i+8, numCloseY))),
key=itemgetter(0))
return (dm, pairm) if dm <= closestY[0] else closestY
else:
return dm, pairm

def times():
''' Time the different functions
'''

import timeit

functions = [bruteForceClosestPair, closestPair]
for f in functions:
print 'Time for', f.__name__, timeit.Timer(
'%s(pointList)' % f.__name__,
'from closestpair import %s, pointList' % f.__name__).timeit(number=1)

pointList = [randint(0,1000)+1j*randint(0,1000) for i in range(2000)]

if __name__ == '__main__':
pointList = [(5+9j), (9+3j), (2+0j), (8+4j), (7+4j), (9+10j), (1+9j), (8+2j), 10j, (9+6j)]
print pointList
print ' bruteForceClosestPair:', bruteForceClosestPair(pointList)
print ' closestPair:', closestPair(pointList)
for i in range(10):
pointList = [randrange(11)+1j*randrange(11) for i in range(10)]
print '\n', pointList
print ' bruteForceClosestPair:', bruteForceClosestPair(pointList)
print ' closestPair:', closestPair(pointList)
print '\n'
times()
times()
times()
Output:
followed by timing comparisons

(Note how the two algorithms agree on the minimum distance, but may return a different pair of points if more than one pair of points share that minimum separation):

[(5+9j), (9+3j), (2+0j), (8+4j), (7+4j), (9+10j), (1+9j), (8+2j), 10j, (9+6j)]
bruteForceClosestPair: (1.0, ((8+4j), (7+4j)))
closestPair: (1.0, ((8+4j), (7+4j)))

[(10+6j), (7+0j), (9+4j), (4+8j), (7+5j), (6+4j), (1+9j), (6+4j), (1+3j), (5+0j)]
bruteForceClosestPair: (0.0, ((6+4j), (6+4j)))
closestPair: (0.0, ((6+4j), (6+4j)))

[(4+10j), (8+5j), (10+3j), (9+7j), (2+5j), (6+7j), (6+2j), (9+6j), (3+8j), (5+1j)]
bruteForceClosestPair: (1.0, ((9+7j), (9+6j)))
closestPair: (1.0, ((9+7j), (9+6j)))

[(10+0j), (3+10j), (10+7j), (1+8j), (5+10j), (8+8j), (4+7j), (6+2j), (6+10j), (9+3j)]
bruteForceClosestPair: (1.0, ((5+10j), (6+10j)))
closestPair: (1.0, ((5+10j), (6+10j)))

[(3+7j), (5+3j), 0j, (2+9j), (2+5j), (9+6j), (5+9j), (4+3j), (3+8j), (8+7j)]
bruteForceClosestPair: (1.0, ((3+7j), (3+8j)))
closestPair: (1.0, ((4+3j), (5+3j)))

[(4+3j), (10+9j), (2+7j), (7+8j), 0j, (3+10j), (10+2j), (7+10j), (7+3j), (1+4j)]
bruteForceClosestPair: (2.0, ((7+8j), (7+10j)))
closestPair: (2.0, ((7+8j), (7+10j)))

[(9+2j), (9+8j), (6+4j), (7+0j), (10+2j), (10+0j), (2+7j), (10+7j), (9+2j), (1+5j)]
bruteForceClosestPair: (0.0, ((9+2j), (9+2j)))
closestPair: (0.0, ((9+2j), (9+2j)))

[(3+3j), (8+2j), (4+0j), (1+1j), (9+10j), (5+0j), (2+3j), 5j, (5+0j), (7+0j)]
bruteForceClosestPair: (0.0, ((5+0j), (5+0j)))
closestPair: (0.0, ((5+0j), (5+0j)))

[(1+5j), (8+3j), (8+10j), (6+8j), (10+9j), (2+0j), (2+7j), (8+7j), (8+4j), (1+2j)]
bruteForceClosestPair: (1.0, ((8+3j), (8+4j)))
closestPair: (1.0, ((8+3j), (8+4j)))

[(8+4j), (8+6j), (8+0j), 0j, (10+7j), (10+6j), 6j, (1+3j), (1+8j), (6+9j)]
bruteForceClosestPair: (1.0, ((10+7j), (10+6j)))
closestPair: (1.0, ((10+7j), (10+6j)))

[(6+8j), (10+1j), 3j, (7+9j), (4+10j), (4+7j), (5+7j), (6+10j), (4+7j), (2+4j)]
bruteForceClosestPair: (0.0, ((4+7j), (4+7j)))
closestPair: (0.0, ((4+7j), (4+7j)))

Time for bruteForceClosestPair 4.57953371169
Time for closestPair 0.122539596513
Time for bruteForceClosestPair 5.13221177552
Time for closestPair 0.124602707886
Time for bruteForceClosestPair 4.83609397284
Time for closestPair 0.119326618327
>>>

## R

Works with: R version 2.8.1+

Brute force solution as per wikipedia pseudo-code

closest_pair_brute <-function(x,y,plotxy=F) {
xy = cbind(x,y)
cp = bruteforce(xy)
cat("\n\nShortest path found = \n From:\t\t(",cp[1],',',cp[2],")\n To:\t\t(",cp[3],',',cp[4],")\n Distance:\t",cp[5],"\n\n",sep="")
if(plotxy) {
plot(x,y,pch=19,col='black',main="Closest Pair", asp=1)
points(cp[1],cp[2],pch=19,col='red')
points(cp[3],cp[4],pch=19,col='red')
}
distance <- function(p1,p2) {
x1 = (p1[1])
y1 = (p1[2])
x2 = (p2[1])
y2 = (p2[2])
sqrt((x2-x1)^2 + (y2-y1)^2)
}
bf_iter <- function(m,p,idx=NA,d=NA,n=1) {
dd = distance(p,m[n,])
if((is.na(d) || dd<=d) && p!=m[n,]){d = dd; idx=n;}
if(n == length(m[,1])) { c(m[idx,],d) }
else bf_iter(m,p,idx,d,n+1)
}
bruteforce <- function(pmatrix,n=1,pd=c(NA,NA,NA,NA,NA)) {
p = pmatrix[n,]
ppd = c(p,bf_iter(pmatrix,p))
if(ppd[5]<pd[5] || is.na(pd[5])) pd = ppd
if(n==length(pmatrix[,1])) pd
else bruteforce(pmatrix,n+1,pd)
}
}

Quicker brute force solution for R that makes use of the apply function native to R for dealing with matrices. It expects x and y to take the form of separate vectors.

closestPair<-function(x,y)
{
distancev <- function(pointsv)
{
x1 <- pointsv[1]
y1 <- pointsv[2]
x2 <- pointsv[3]
y2 <- pointsv[4]
sqrt((x1 - x2)^2 + (y1 - y2)^2)
}
pairstocompare <- t(combn(length(x),2))
pointsv <- cbind(x[pairstocompare[,1]],y[pairstocompare[,1]],x[pairstocompare[,2]],y[pairstocompare[,2]])
pairstocompare <- cbind(pairstocompare,apply(pointsv,1,distancev))
minrow <- pairstocompare[pairstocompare[,3] == min(pairstocompare[,3])]
if (!is.null(nrow(minrow))) {print("More than one point at this distance!"); minrow <- minrow[1,]}
cat("The closest pair is:\n\tPoint 1: ",x[minrow[1]],", ",y[minrow[1]],
"\n\tPoint 2: ",x[minrow[2]],", ",y[minrow[2]],
"\n\tDistance: ",minrow[3],"\n",sep="")
c(distance=minrow[3],x1.x=x[minrow[1]],y1.y=y[minrow[1]],x2.x=x[minrow[2]],y2.y=y[minrow[2]])
}

This is the quickest version, that makes use of the 'dist' function of R. It takes a two-column object of x,y-values as input, or creates such an object from seperate x and y-vectors.

closest.pairs <- function(x, y=NULL, ...){
# takes two-column object(x,y-values), or creates such an object from x and y values
if(!is.null(y)) x <- cbind(x, y)

distances <- dist(x)
min.dist <- min(distances)
point.pair <- combn(1:nrow(x), 2)[, which.min(distances)]

cat("The closest pair is:\n\t",
sprintf("Point 1: %.3f, %.3f \n\tPoint 2: %.3f, %.3f \n\tDistance: %.3f.\n",
x[point.pair[1],1], x[point.pair[1],2],
x[point.pair[2],1], x[point.pair[2],2],
min.dist),
sep="" )
c( x1=x[point.pair[1],1],y1=x[point.pair[1],2],
x2=x[point.pair[2],1],y2=x[point.pair[2],2],
distance=min.dist)
}
Example
x = (sample(-1000.00:1000.00,100))
y = (sample(-1000.00:1000.00,length(x)))
cp = closest.pairs(x,y)
#cp = closestPair(x,y)
plot(x,y,pch=19,col='black',main="Closest Pair", asp=1)
points(cp["x1.x"],cp["y1.y"],pch=19,col='red')
points(cp["x2.x"],cp["y2.y"],pch=19,col='red')
#closest_pair_brute(x,y,T)

Performance
system.time(closest_pair_brute(x,y), gcFirst = TRUE)
Shortest path found =
From: (32,-987)
To: (25,-993)
Distance: 9.219544

user system elapsed
0.35 0.02 0.37

system.time(closest.pairs(x,y), gcFirst = TRUE)
The closest pair is:
Point 1: 32.000, -987.000
Point 2: 25.000, -993.000
Distance: 9.220.

user system elapsed
0.08 0.00 0.10

system.time(closestPair(x,y), gcFirst = TRUE)
The closest pair is:
Point 1: 32, -987
Point 2: 25, -993
Distance: 9.219544

user system elapsed
0.17 0.00 0.19

Using dist function for brute force, but divide and conquer (as per pseudocode) for speed:

closest.pairs.bruteforce <- function(x, y=NULL)
{
if (!is.null(y))
{
x <- cbind(x,y)
}
d <- dist(x)
cp <- x[combn(1:nrow(x), 2)[, which.min(d)],]
list(p1=cp[1,], p2=cp[2,], d=min(d))
}

closest.pairs.dandc <- function(x, y=NULL)
{
if (!is.null(y))
{
x <- cbind(x,y)
}
if (sd(x[,"x"]) < sd(x[,"y"]))
{
x <- cbind(x=x[,"y"],y=x[,"x"])
swap <- TRUE
}
else
{
swap <- FALSE
}
xp <- x[order(x[,"x"]),]
.cpdandc.rec <- function(xp,yp)
{
n <- dim(xp)[1]
if (n <= 4)
{
closest.pairs.bruteforce(xp)
}
else
{
xl <- xp[1:floor(n/2),]
xr <- xp[(floor(n/2)+1):n,]
cpl <- .cpdandc.rec(xl)
cpr <- .cpdandc.rec(xr)
if (cpl\$d<cpr\$d) cp <- cpl else cp <- cpr
cp
}
}
cp <- .cpdandc.rec(xp)

yp <- x[order(x[,"y"]),]
xm <- xp[floor(dim(xp)[1]/2),"x"]
ys <- yp[which(abs(xm - yp[,"x"]) <= cp\$d),]
nys <- dim(ys)[1]
if (!is.null(nys) && nys > 1)
{
for (i in 1:(nys-1))
{
k <- i + 1
while (k <= nys && ys[i,"y"] - ys[k,"y"] < cp\$d)
{
d <- sqrt((ys[k,"x"]-ys[i,"x"])^2 + (ys[k,"y"]-ys[i,"y"])^2)
if (d < cp\$d) cp <- list(p1=ys[i,],p2=ys[k,],d=d)
k <- k + 1
}
}
}
if (swap)
{
list(p1=cbind(x=cp\$p1["y"],y=cp\$p1["x"]),p2=cbind(x=cp\$p2["y"],y=cp\$p2["x"]),d=cp\$d)
}
else
{
cp
}
}

# Test functions
cat("How many points?\n")
n <- scan(what=integer(),n=1)
x <- rnorm(n)
y <- rnorm(n)
tstart <- proc.time()[3]
cat("Closest pairs divide and conquer:\n")
print(cp <- closest.pairs.dandc(x,y))
cat(sprintf("That took %.2f seconds.\n",proc.time()[3] - tstart))
plot(x,y)
points(c(cp\$p1["x"],cp\$p2["x"]),c(cp\$p1["y"],cp\$p2["y"]),col="red")
tstart <- proc.time()[3]
cat("\nClosest pairs brute force:\n")
print(closest.pairs.bruteforce(x,y))
cat(sprintf("That took %.2f seconds.\n",proc.time()[3] - tstart))

Output:
How many points?
1: 500
Closest pairs divide and conquer:
\$p1
x          y
1.68807938 0.05876328

\$p2
x          y
1.68904694 0.05878173

\$d
[1] 0.0009677302

That took 0.43 seconds.

Closest pairs brute force:
\$p1
x          y
1.68807938 0.05876328

\$p2
x          y
1.68904694 0.05878173

\$d
[1] 0.0009677302

That took 6.38 seconds.

## Racket

The brute force solution using complex numbers to represent pairs.

#lang racket
(define (dist z0 z1) (magnitude (- z1 z0)))
(define (dist* zs) (apply dist zs))

(define (closest-pair zs)
(if (< (length zs) 2)
-inf.0
(first
(sort (for/list ([z0 zs])
(list z0 (argmin (λ(z) (if (= z z0) +inf.0 (dist z z0))) zs)))
< #:key dist*))))

(define result (closest-pair '(0+1i 1+2i 3+4i)))
(displayln (~a "Closest points: " result))
(displayln (~a "Distance: " (dist* result)))

The divide and conquer algorithm using a struct to represent points

#lang racket
(struct point (x y) #:transparent)

(define (closest-pair ps)
(check-type ps)
(cond [(vector? ps) (if (> (vector-length ps) 1)
(closest-pair/sorted (vector-sort ps left?)
(vector-sort ps below?))
(error 'closest-pair "2 or more points are needed" ps))]
[(sequence? ps) (closest-pair (for/vector ([x (in-sequences ps)]) x))]
[else (error 'closest-pair "closest pair only supports sequence types (excluding hash)")]))

;; accept any sequence type except hash
;; any other exclusions needed?
(define (check-type ps)
(cond [(hash? ps) (error 'closest-pair "Hash tables are not supported")]
[(sequence? ps) #t]
[else (error 'closest-pair "Only sequence types are supported")]))

;; vector -> vector -> list
(define (closest-pair/sorted Px Py)
(define L (vector-length Px))
(cond [(= L 2) (vector->list Px)]
[(= L 3) (apply min-pair (combinations (vector->list Px) 2))]
[else (let*-values ([(Qx Rx) (vector-split-at Px (floor (/ L 2)))]
; Rx-min is the left most point in Rx
[(Rx-min) (vector-ref Rx 0)]
; instead of sorting Qx, Rx by y
; - Qy are members of Py to left of Rx-min
; - Ry are the remaining members of Py
[(Qy Ry) (vector-partition Py (curryr left? Rx-min))]
[(pair1) (closest-pair/sorted Qx Qy)]
[(pair2) (closest-pair/sorted Rx Ry)]
[(delta) (min (distance^2 pair1) (distance^2 pair2))]
[(pair3) (closest-split-pair Px Py delta)])
; pair3 is null when there are no split pairs closer than delta
(min-pair pair1 pair2 pair3))]))

(define (closest-split-pair Px Py delta)
(define Lp (vector-length Px))
(define x-mid (point-x (vector-ref Px (floor (/ Lp 2)))))
(define Sy (for/vector ([p (in-vector Py)]
#:when (< (abs (- (point-x p) x-mid)) delta))
p))
(define Ls (vector-length Sy))
(define-values (_ best-pair)
(for*/fold ([new-best delta]
[new-best-pair null])
([i (in-range (sub1 Ls))]
[j (in-range (+ i 1) (min (+ i 7) Ls))]
[Sij (in-value (list (vector-ref Sy i)
(vector-ref Sy j)))]
[dij (in-value (distance^2 Sij))]
#:when (< dij new-best))
(values dij Sij)))
best-pair)

;; helper procedures

;; same as partition except for vectors
;; it's critical to maintain the relative order of elements
(define (vector-partition Py pred)
(define-values (left right)
(for/fold ([Qy null]
[Ry null])
([p (in-vector Py)])
(if (pred p)
(values (cons p Qy) Ry)
(values Qy (cons p Ry)))))
(values (list->vector (reverse left))
(list->vector (reverse right))))

; is p1 (strictly) left of p2
(define (left? p1 p2) (< (point-x p1) (point-x p2)))

; is p1 (strictly) below of p2
(define (below? p1 p2) (< (point-y p1) (point-y p2)))

;; return the pair with minimum distance
(define (min-pair . pairs)
(argmin distance^2 pairs))

;; pairs are passed around as a list of 2 points
;; distance is only for comparison so no need to use sqrt
(define (distance^2 pair)
(cond [(null? pair) +inf.0]
[else (define a (first pair))
(define b (second pair))
(+ (sqr (- (point-x b) (point-x a)))
(sqr (- (point-y b) (point-y a))))]))

; points on a quadratic curve, shuffled
(define points
(shuffle
(for/list ([ i (in-range 1000)]) (point i (* i i)))))
(match-define (list (point p1x p1y) (point p2x p2y)) (closest-pair points))
(printf "Closest points on a quadratic curve (~a,~a) (~a,~a)\n" p1x p1y p2x p2y)

Output:

Closest points: (0+1i 1+2i)
Distance: 1.4142135623730951

Closest points on a quadratic curve (0,0) (1,1)

## Raku

(formerly Perl 6)

Translation of: Perl 5

We avoid taking square roots in the slow method because the squares are just as comparable. (This doesn't always work in the fast method because of distance assumptions in the algorithm.)

sub MAIN (\$N = 5000) {
my @points = (^\$N).map: { [rand * 20 - 10, rand * 20 - 10] }

my (\$af, \$bf, \$df) = closest_pair(@points);
say "fast \$df at [\$af], [\$bf]";

my (\$as, \$bs, \$ds) = closest_pair_simple(@points);
say "slow \$ds at [\$as], [\$bs]";
}

sub dist-squared(\$a,\$b) {
(\$a[0] - \$b[0]) ** 2 +
(\$a[1] - \$b[1]) ** 2;
}

sub closest_pair_simple(@arr is copy) {
return Inf if @arr < 2;
my (\$a, \$b, \$d) = flat @arr[0,1], dist-squared(|@arr[0,1]);
while @arr {
my \$p = pop @arr;
for @arr -> \$l {
my \$t = dist-squared(\$p, \$l);
(\$a, \$b, \$d) = \$p, \$l, \$t if \$t < \$d;
}
}
return \$a, \$b, sqrt \$d;
}

sub closest_pair(@r) {
my @ax = @r.sort: { .[0] }
my @ay = @r.sort: { .[1] }
return closest_pair_real(@ax, @ay);
}

sub closest_pair_real(@rx, @ry) {
return closest_pair_simple(@rx) if @rx <= 3;

my @xP = @rx;
my @yP = @ry;
my \$N = @xP;

my \$midx = ceiling(\$N/2)-1;

my @PL = @xP[0 .. \$midx];
my @PR = @xP[\$midx+1 ..^ \$N];

my \$xm = @xP[\$midx][0];

my @yR;
my @yL;
push (\$_[0] <= \$xm ?? @yR !! @yL), \$_ for @yP;

my (\$al, \$bl, \$dL) = closest_pair_real(@PL, @yR);
my (\$ar, \$br, \$dR) = closest_pair_real(@PR, @yL);

my (\$m1, \$m2, \$dmin) = \$dR < \$dL
?? (\$ar, \$br, \$dR)
!! (\$al, \$bl, \$dL);

my @yS = @yP.grep: { abs(\$xm - .[0]) < \$dmin }

if @yS {
my (\$w1, \$w2, \$closest) = \$m1, \$m2, \$dmin;
for 0 ..^ @yS.end -> \$i {
for \$i+1 ..^ @yS -> \$k {
last unless @yS[\$k][1] - @yS[\$i][1] < \$dmin;
my \$d = sqrt dist-squared(@yS[\$k], @yS[\$i]);
(\$w1, \$w2, \$closest) = @yS[\$k], @yS[\$i], \$d if \$d < \$closest;
}

}
return \$w1, \$w2, \$closest;

} else {
return \$m1, \$m2, \$dmin;
}
}

## REXX

Programming note:   this REXX version allows two (or more) points to be identical, and will
manifest itself as a minimum distance of zero   (the variable   dd   on line 17).

/*REXX program  solves the   closest pair   of  points  problem  (in two dimensions).   */
parse arg N low high seed . /*obtain optional arguments from the CL*/
if N=='' | N=="," then N= 100 /*Not specified? Then use the default.*/
if low=='' | low=="," then low= 0 /* " " " " " " */
if high=='' | high=="," then high= 20000 /* " " " " " " */
if datatype(seed, 'W') then call random ,,seed /*seed for RANDOM (BIF) repeatability.*/
w=length(high); w=w + (w//2==0) /*W: for aligning the output columns.*/
/*╔══════════════════════╗*/ do j=1 for N /*generate N random points*/
/*║ generate N points. ║*/ @x.j= random(low, high) /* " a random X */
/*╚══════════════════════╝*/ @y.j= random(low, high) /* " " " Y */
end /*j*/ /*X & Y make the point.*/
A=1; B=2 /* [↓] MINDD is actually the squared*/
minDD= (@x.A - @x.B)**2 + (@y.A - @y.B)**2 /*distance between the first two points*/
/* [↓] use of XJ & YJ speed things up.*/
do j=1 for N-1; xj= @x.j; yj= @y.j /*find minimum distance between a ··· */
do k=j+1 to N /* ··· point and all the other points.*/
dd= (xj - @x.k)**2 + (yj - @y.k)**2 /*compute squared distance from points.*/
if dd<minDD then parse value dd j k with minDD A B
end /*k*/ /* [↑] needn't take SQRT of DD (yet).*/
end /*j*/ /* [↑] when done, A & B are the points*/
\$= 'For ' N " points, the minimum distance between the two points: "
say \$ center("x", w, '═')" " center('y', w, "═") ' is: ' sqrt(abs(minDD))/1
say left('', length(\$) - 1) "["right(@x.A, w)',' right(@y.A, w)"]"
say left('', length(\$) - 1) "["right(@x.B, w)',' right(@y.B, w)"]"
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
sqrt: procedure; parse arg x; if x=0 then return 0; d=digits(); m.=9; numeric form; h=d+6
numeric digits; parse value format(x,2,1,,0) 'E0' with g 'E' _ .; g= g *.5'e'_ % 2
do j=0 while h>9; m.j= h; h= h % 2 + 1; end /*j*/
do k=j+5 to 0 by -1; numeric digits m.k; g= (g+x/g)*.5; end /*k*/; return g
output   when using the default input of:     100
For  100  points, the minimum distance between the two points:   ══x══  ══y══   is:  219.228192
[ 7277,  1625]
[ 7483,  1700]
output   when using the input of:     200
For  200  points, the minimum distance between the two points:   ══x══  ══y══   is:  39.408121
[17604, 19166]
[17627, 19198]
output   when using the input of:     1000
For  1000  points, the minimum distance between the two points:   ══x══  ══y══   is:  5.09901951
[ 6264, 19103]
[ 6263, 19108]

## Ring

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)

Output:

closest pair is : 3 and 6 at distance 0.0779101914

## Ruby

Point = Struct.new(:x, :y)

def distance(p1, p2)
Math.hypot(p1.x - p2.x, p1.y - p2.y)
end

def closest_bruteforce(points)
mindist, minpts = Float::MAX, []
points.combination(2) do |pi,pj|
dist = distance(pi, pj)
if dist < mindist
mindist = dist
minpts = [pi, pj]
end
end
[mindist, minpts]
end

def closest_recursive(points)
return closest_bruteforce(points) if points.length <= 3
xP = points.sort_by(&:x)
mid = points.length / 2
xm = xP[mid].x
dL, pairL = closest_recursive(xP[0,mid])
dR, pairR = closest_recursive(xP[mid..-1])
dmin, dpair = dL<dR ? [dL, pairL] : [dR, pairR]
yP = xP.find_all {|p| (xm - p.x).abs < dmin}.sort_by(&:y)
closest, closestPair = dmin, dpair
0.upto(yP.length - 2) do |i|
(i+1).upto(yP.length - 1) do |k|
break if (yP[k].y - yP[i].y) >= dmin
dist = distance(yP[i], yP[k])
if dist < closest
closest = dist
closestPair = [yP[i], yP[k]]
end
end
end
[closest, closestPair]
end

points = Array.new(100) {Point.new(rand, rand)}
p ans1 = closest_bruteforce(points)
p ans2 = closest_recursive(points)
fail "bogus!" if ans1[0] != ans2[0]

require 'benchmark'

points = Array.new(10000) {Point.new(rand, rand)}
Benchmark.bm(12) do |x|
x.report("bruteforce") {ans1 = closest_bruteforce(points)}
x.report("recursive") {ans2 = closest_recursive(points)}
end

Sample output

[0.005299616045889868, [#<struct Point x=0.24805908871087445, y=0.8413503128160198>, #<struct Point x=0.24355227214243136, y=0.8385620275629906>]]
[0.005299616045889868, [#<struct Point x=0.24355227214243136, y=0.8385620275629906>, #<struct Point x=0.24805908871087445, y=0.8413503128160198>]]
user     system      total        real
bruteforce    43.446000   0.000000  43.446000 ( 43.530062)
recursive      0.187000   0.000000   0.187000 (  0.190000)

## Run BASIC

Courtesy http://dkokenge.com/rbp

n =10                              ' 10 data points input
dim x(n)
dim y(n)

pt1 = 0 ' 1st point
pt2 = 0 ' 2nd point

for i =1 to n ' read in data
next i

minDist = 1000000

for i =1 to n -1
for j =i +1 to n
distXsq =(x(i) -x(j))^2
disYsq =(y(i) -y(j))^2
d =abs((dxSq +disYsq)^0.5)
if d <minDist then
minDist =d
pt1 =i
pt2 =j
end if
next j
next i

print "Distance ="; minDist; " between ("; x(pt1); ", "; y(pt1); ") and ("; x(pt2); ", "; y(pt2); ")"

end

data 0.654682, 0.925557
data 0.409382, 0.619391
data 0.891663, 0.888594
data 0.716629, 0.996200
data 0.477721, 0.946355
data 0.925092, 0.818220
data 0.624291, 0.142924
data 0.211332, 0.221507
data 0.293786, 0.691701
data 0.839186, 0.72826

## Scala

import scala.collection.mutable.ListBuffer
import scala.util.Random

object ClosestPair {
case class Point(x: Double, y: Double){
def distance(p: Point) = math.hypot(x-p.x, y-p.y)

override def toString = "(" + x + ", " + y + ")"
}

case class Pair(point1: Point, point2: Point) {
val distance: Double = point1 distance point2

override def toString = {
point1 + "-" + point2 + " : " + distance
}
}

def sortByX(points: List[Point]) = {
points.sortBy(point => point.x)
}

def sortByY(points: List[Point]) = {
points.sortBy(point => point.y)
}

def divideAndConquer(points: List[Point]): Pair = {
val pointsSortedByX = sortByX(points)
val pointsSortedByY = sortByY(points)

divideAndConquer(pointsSortedByX, pointsSortedByY)
}

def bruteForce(points: List[Point]): Pair = {
val numPoints = points.size
if (numPoints < 2)
return null
var pair = Pair(points(0), points(1))
if (numPoints > 2) {
for (i <- 0 until numPoints - 1) {
val point1 = points(i)
for (j <- i + 1 until numPoints) {
val point2 = points(j)
val distance = point1 distance point2
if (distance < pair.distance)
pair = Pair(point1, point2)
}
}
}
return pair
}

private def divideAndConquer(pointsSortedByX: List[Point], pointsSortedByY: List[Point]): Pair = {
val numPoints = pointsSortedByX.size
if(numPoints <= 3) {
return bruteForce(pointsSortedByX)
}

val dividingIndex = numPoints >>> 1
val leftOfCenter = pointsSortedByX.slice(0, dividingIndex)
val rightOfCenter = pointsSortedByX.slice(dividingIndex, numPoints)

var tempList = leftOfCenter.map(x => x)
//println(tempList)
tempList = sortByY(tempList)
var closestPair = divideAndConquer(leftOfCenter, tempList)

tempList = rightOfCenter.map(x => x)
tempList = sortByY(tempList)

val closestPairRight = divideAndConquer(rightOfCenter, tempList)

if (closestPairRight.distance < closestPair.distance)
closestPair = closestPairRight

tempList = List[Point]()
val shortestDistance = closestPair.distance
val centerX = rightOfCenter(0).x

for (point <- pointsSortedByY) {
if (Math.abs(centerX - point.x) < shortestDistance)
tempList = tempList :+ point
}

closestPair = shortestDistanceF(tempList, shortestDistance, closestPair)
closestPair
}

private def shortestDistanceF(tempList: List[Point], shortestDistance: Double, closestPair: Pair ): Pair = {
var shortest = shortestDistance
var bestResult = closestPair
for (i <- 0 until tempList.size) {
val point1 = tempList(i)
for (j <- i + 1 until tempList.size) {
val point2 = tempList(j)
if ((point2.y - point1.y) >= shortestDistance)
return closestPair
val distance = point1 distance point2
if (distance < closestPair.distance)
{
bestResult = Pair(point1, point2)
shortest = distance
}
}
}

closestPair
}

def main(args: Array[String]) {
val numPoints = if(args.length == 0) 1000 else args(0).toInt

val points = ListBuffer[Point]()
val r = new Random()
for (i <- 0 until numPoints) {
points.+=:(new Point(r.nextDouble(), r.nextDouble()))
}
println("Generated " + numPoints + " random points")

var startTime = System.currentTimeMillis()
val bruteForceClosestPair = bruteForce(points.toList)
var elapsedTime = System.currentTimeMillis() - startTime
println("Brute force (" + elapsedTime + " ms): " + bruteForceClosestPair)

startTime = System.currentTimeMillis()
val dqClosestPair = divideAndConquer(points.toList)
elapsedTime = System.currentTimeMillis() - startTime
println("Divide and conquer (" + elapsedTime + " ms): " + dqClosestPair)
if (bruteForceClosestPair.distance != dqClosestPair.distance)
println("MISMATCH")
}
}

Output:
scala ClosestPair 1000
Generated 1000 random points
Brute force (981 ms): (0.41984960343173994, 0.4499078600557793)-(0.4198255166110827, 0.45044969701435) : 5.423720721077961E-4
Divide and conquer (52 ms): (0.4198255166110827, 0.45044969701435)-(0.41984960343173994, 0.4499078600557793) : 5.423720721077961E-4

## Seed7

This is the brute force algorithm:

const type: point is new struct
var float: x is 0.0;
var float: y is 0.0;
end struct;

const func float: distance (in point: p1, in point: p2) is
return sqrt((p1.x-p2.x)**2+(p1.y-p2.y)**2);

const func array point: closest_pair (in array point: points) is func
result
var array point: result is 0 times point.value;
local
var float: dist is 0.0;
var float: minDistance is Infinity;
var integer: i is 0;
var integer: j is 0;
var integer: savei is 0;
var integer: savej is 0;
begin
for i range 1 to pred(length(points)) do
for j range succ(i) to length(points) do
dist := distance(points[i], points[j]);
if dist < minDistance then
minDistance := dist;
savei := i;
savej := j;
end if;
end for;
end for;
if minDistance <> Infinity then
result := [] (points[savei], points[savej]);
end if;
end func;

## Sidef

Translation of: Raku
func dist_squared(a, b) {
sqr(a[0] - b[0]) + sqr(a[1] - b[1])
}

func closest_pair_simple(arr) {
arr.len < 2 && return Inf
var (a, b, d) = (arr[0, 1], dist_squared(arr[0,1]))
arr.clone!
while (arr) {
var p = arr.pop
for l in arr {
var t = dist_squared(p, l)
if (t < d) {
(a, b, d) = (p, l, t)
}
}
}
return(a, b, d.sqrt)
}

func closest_pair_real(rx, ry) {
rx.len <= 3 && return closest_pair_simple(rx)

var N = rx.len
var midx = (ceil(N/2)-1)
var (PL, PR) = rx.part(midx)

var xm = rx[midx][0]

var yR = []
var yL = []

for item in ry {
(item[0] <= xm ? yR : yL) << item
}

var (al, bl, dL) = closest_pair_real(PL, yR)
var (ar, br, dR) = closest_pair_real(PR, yL)

al == Inf && return (ar, br, dR)
ar == Inf && return (al, bl, dL)

var (m1, m2, dmin) = (dR < dL ? [ar, br, dR]...
: [al, bl, dL]...)

var yS = ry.grep { |a| abs(xm - a[0]) < dmin }

var (w1, w2, closest) = (m1, m2, dmin)
for i in (0 ..^ yS.end) {
for k in (i+1 .. yS.end) {
yS[k][1] - yS[i][1] < dmin || break
var d = dist_squared(yS[k], yS[i]).sqrt
if (d < closest) {
(w1, w2, closest) = (yS[k], yS[i], d)
}
}
}

return (w1, w2, closest)
}

func closest_pair(r) {
var ax = r.sort_by { |a| a[0] }
var ay = r.sort_by { |a| a[1] }
return closest_pair_real(ax, ay);
}

var N = 5000
var points = N.of { [1.rand*20 - 10, 1.rand*20 - 10] }
var (af, bf, df) = closest_pair(points)
say "#{df} at (#{af.join(' ')}), (#{bf.join(' ')})"

## Swift

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()!)
Output:
(Point(x: 5.279430517795172, y: 8.85108182685002), Point(x: 5.278427575530877, y: 8.851990433099456))

## 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.

package require Tcl 8.5

# retrieve the x-coordinate
proc x p {lindex \$p 0}
# retrieve the y-coordinate
proc y p {lindex \$p 1}

proc distance {p1 p2} {
expr {hypot(([x \$p1]-[x \$p2]), ([y \$p1]-[y \$p2]))}
}

proc closest_bruteforce {points} {
set n [llength \$points]
set mindist Inf
set minpts {}
for {set i 0} {\$i < \$n - 1} {incr i} {
for {set j [expr {\$i + 1}]} {\$j < \$n} {incr j} {
set p1 [lindex \$points \$i]
set p2 [lindex \$points \$j]
set dist [distance \$p1 \$p2]
if {\$dist < \$mindist} {
set mindist \$dist
set minpts [list \$p1 \$p2]
}
}
}
return [list \$mindist \$minpts]
}

proc closest_recursive {points} {
set n [llength \$points]
if {\$n <= 3} {
return [closest_bruteforce \$points]
}
set xP [lsort -real -increasing -index 0 \$points]
set mid [expr {int(ceil(\$n/2.0))}]
set PL [lrange \$xP 0 [expr {\$mid-1}]]
set PR [lrange \$xP \$mid end]
set procname [lindex [info level 0] 0]
lassign [\$procname \$PL] dL pairL
lassign [\$procname \$PR] dR pairR
if {\$dL < \$dR} {
set dmin \$dL
set dpair \$pairL
} else {
set dmin \$dR
set dpair \$pairR
}

set xM [x [lindex \$PL end]]
foreach p \$xP {
if {abs(\$xM - [x \$p]) < \$dmin} {
lappend S \$p
}
}
set yP [lsort -real -increasing -index 1 \$S]
set closest Inf
set nP [llength \$yP]
for {set i 0} {\$i <= \$nP-2} {incr i} {
set yPi [lindex \$yP \$i]
for {set k [expr {\$i+1}]; set yPk [lindex \$yP \$k]} {
\$k < \$nP-1 && ([y \$yPk]-[y \$yPi]) < \$dmin
} {incr k; set yPk [lindex \$yP \$k]} {
set dist [distance \$yPk \$yPi]
if {\$dist < \$closest} {
set closest \$dist
set closestPair [list \$yPi \$yPk]
}
}
}
expr {\$closest < \$dmin ? [list \$closest \$closestPair] : [list \$dmin \$dpair]}
}

# testing
set N 10000
for {set i 1} {\$i <= \$N} {incr i} {
lappend points [list [expr {rand()*100}] [expr {rand()*100}]]
}

# instrument the number of calls to [distance] to examine the
# efficiency of the recursive solution
trace add execution distance enter comparisons
proc comparisons args {incr ::comparisons}

puts [format "%-10s  %9s  %9s  %s" method compares time closest]
foreach method {bruteforce recursive} {
set ::comparisons 0
set time [time {set ::dist(\$method) [closest_\$method \$points]} 1]
puts [format "%-10s  %9d  %9d  %s" \$method \$::comparisons [lindex \$time 0] [lindex \$::dist(\$method) 0]]
}
Output:
method      compares      time closest
bruteforce  49995000 512967207 0.0015652738546658382
recursive      14613    488094 0.0015652738546658382

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

## Ursala

The brute force algorithm is easy. Reading from left to right, clop is defined as a function that forms the Cartesian product of its argument, and then extracts the member whose left side is a minimum with respect to the floating point comparison relation after deleting equal pairs and attaching to the left of each remaining pair the sum of the squares of the differences between corresponding coordinates.

#import flo

clop = @iiK0 fleq\$-&l+ *EZF ^\~& plus+ sqr~~+ minus~~bbI

The divide and conquer algorithm following the specification given above is a little more hairy but not much longer. The eudist library function is used to compute the distance between points.

#import std
#import flo

clop =

^(fleq-<&l,fleq-<&r); @blrNCCS ~&lrbhthPX2X+ ~&a^& fleq\$-&l+ leql/8?al\^(eudist,~&)*altK33htDSL -+
^C/~&rr ^(eudist,~&)*tK33htDSL+ @rlrlPXPlX ~| fleq^\~&lr abs+ [email protected],
^/~&ar @farlK30K31XPGbrlrjX3J ^/~&arlhh @W lesser [email protected]+-

test program:

test_data =

<
(1.547290e+00,3.313053e+00),
(5.250805e-01,-7.300260e+00),
(7.062114e-02,1.220251e-02),
(-4.473024e+00,-5.393712e+00),
(-2.563714e+00,-3.595341e+00),
(-2.132372e+00,2.358850e+00),
(2.366238e+00,-9.678425e+00),
(-1.745694e+00,3.276434e+00),
(8.066843e+00,-9.101268e+00),
(-8.256901e+00,-8.717900e+00),
(7.397744e+00,-5.366434e+00),
(2.060291e-01,2.840891e+00),
(-6.935319e+00,-5.192438e+00),
(9.690418e+00,-9.175753e+00),
(3.448993e+00,2.119052e+00),
(-7.769218e+00,4.647406e-01)>

#cast %eeWWA

example = clop test_data
Output:

The output shows the minimum distance and the two points separated by that distance. (If the brute force algorithm were used, it would have displayed the square of the distance.)

9.957310e-01: (
(-2.132372e+00,2.358850e+00),
(-1.745694e+00,3.276434e+00))

## VBA

Option Explicit

Private Type MyPoint
X As Single
Y As Single
End Type

Private Type MyPair
p1 As MyPoint
p2 As MyPoint
End Type

Sub Main()
Dim points() As MyPoint, i As Long, BF As MyPair, d As Single, Nb As Long
Dim T#
Randomize Timer
Nb = 10
Do
ReDim points(1 To Nb)
For i = 1 To Nb
points(i).X = Rnd * Nb
points(i).Y = Rnd * Nb
Next
d = 1000000000000#
T = Timer
BF = BruteForce(points, d)
Debug.Print "For " & Nb & " points, runtime : " & Timer - T & " sec."
Debug.Print "point 1 : X:" & BF.p1.X & " Y:" & BF.p1.Y
Debug.Print "point 2 : X:" & BF.p2.X & " Y:" & BF.p2.Y
Debug.Print "dist : " & d
Debug.Print "--------------------------------------------------"
Nb = Nb * 10
Loop While Nb <= 10000
End Sub

Private Function BruteForce(p() As MyPoint, mindist As Single) As MyPair
Dim i As Long, j As Long, d As Single, ClosestPair As MyPair
For i = 1 To UBound(p) - 1
For j = i + 1 To UBound(p)
d = Dist(p(i), p(j))
If d < mindist Then
mindist = d
ClosestPair.p1 = p(i)
ClosestPair.p2 = p(j)
End If
Next
Next
BruteForce = ClosestPair
End Function

Private Function Dist(p1 As MyPoint, p2 As MyPoint) As Single
Dist = Sqr((p1.X - p2.X) ^ 2 + (p1.Y - p2.Y) ^ 2)
End Function

Output:
For 10 points, runtime : 0 sec.
point 1 : X:7,199265 Y:7,690955
point 2 : X:7,16863 Y:7,681544
dist : 3,204883E-02
--------------------------------------------------
For 100 points, runtime : 0 sec.
point 1 : X:48,97898 Y:96,54872
point 2 : X:48,78981 Y:96,95755
dist : 0,4504737
--------------------------------------------------
For 1000 points, runtime : 0,44921875 sec.
point 1 : X:576,9511 Y:398,5834
point 2 : X:577,364 Y:398,3212
dist : 0,4891393
--------------------------------------------------
For 10000 points, runtime : 47,46875 sec.
point 1 : X:8982,698 Y:1154,133
point 2 : X:8984,763 Y:1152,822
dist : 2,445694
--------------------------------------------------

## Visual FoxPro

CLOSE DATABASES ALL
CREATE CURSOR pairs(id I, xcoord B(6), ycoord B(6))
INSERT INTO pairs VALUES (1, 0.654682, 0.925557)
INSERT INTO pairs VALUES (2, 0.409382, 0.619391)
INSERT INTO pairs VALUES (3, 0.891663, 0.888594)
INSERT INTO pairs VALUES (4, 0.716629, 0.996200)
INSERT INTO pairs VALUES (5, 0.477721, 0.946355)
INSERT INTO pairs VALUES (6, 0.925092, 0.818220)
INSERT INTO pairs VALUES (7, 0.624291, 0.142924)
INSERT INTO pairs VALUES (8, 0.211332, 0.221507)
INSERT INTO pairs VALUES (9, 0.293786, 0.691701)
INSERT INTO pairs VALUES (10, 0.839186, 0.728260)

SELECT p1.id As id1, p2.id As id2, ;
(p1.xcoord-p2.xcoord)^2 + (p1.ycoord-p2.ycoord)^2 As dist2 ;
FROM pairs p1 JOIN pairs p2 ON p1.id < p2.id ORDER BY 3 INTO CURSOR tmp

GO TOP
? "Closest pair is " + TRANSFORM(id1) + " and " + TRANSFORM(id2) + "."
? "Distance is " + TRANSFORM(SQRT(dist2))

Output:
Visual FoxPro uses 1 based indexing,

Closest pair is 3 and 6.
Distance is 0.077910.

## XPL0

The brute force method is simpler than the recursive solution and is perfectly adequate, even for a thousand points.

include c:\cxpl\codes;          \intrinsic 'code' declarations

proc ClosestPair(P, N); \Show closest pair of points in array P
real P; int N;
real Dist2, MinDist2;
int I, J, SI, SJ;
[MinDist2:= 1e300;
for I:= 0 to N-2 do
[for J:= I+1 to N-1 do
[Dist2:= sq(P(I,0)-P(J,0)) + sq(P(I,1)-P(J,1));
if Dist2 < MinDist2 then \squared distances are sufficient for compares
[MinDist2:= Dist2;
SI:= I; SJ:= J;
];
];
];
IntOut(0, SI); Text(0, " -- "); IntOut(0, SJ); CrLf(0);
RlOut(0, P(SI,0)); Text(0, ","); RlOut(0, P(SI,1));
Text(0, " -- ");
RlOut(0, P(SJ,0)); Text(0, ","); RlOut(0, P(SJ,1));
CrLf(0);
];

real Data;
[Format(1, 6);
Data:= [[0.654682, 0.925557], \0 test data from BASIC examples
[0.409382, 0.619391], \1
[0.891663, 0.888594], \2
[0.716629, 0.996200], \3
[0.477721, 0.946355], \4
[0.925092, 0.818220], \5
[0.624291, 0.142924], \6
[0.211332, 0.221507], \7
[0.293786, 0.691701], \8
[0.839186, 0.728260]]; \9
ClosestPair(Data, 10);
]
Output:
2 -- 5
0.891663,0.888594 -- 0.925092,0.818220

## zkl

An ugly solution in both time and space.

class Point{
fcn init(_x,_y){ var[const] x=_x, y=_y; }
fcn distance(p){ (p.x-x).hypot(p.y-y) }
fcn toString { String("Point(",x,",",y,")") }
}

// find closest two points using brute ugly force:
// find all combinations of two points, measure distance, pick smallest
fcn closestPoints(points){
pairs:=Utils.Helpers.pickNFrom(2,points);
triples:=pairs.apply(fcn([(p1,p2)]){ T(p1,p2,p1.distance(p2)) });
triples.reduce(fcn([(_,_,d1)]p1,[(_,_,d2)]p2){
if(d1 < d2) p1 else p2
});
}
points:=T( 5.0, 9.0,  9.0, 3.0,
2.0, 0.0, 8.0, 4.0,
7.0, 4.0, 9.0, 10.0,
1.0, 9.0, 8.0, 2.0,
0.0, 10.0, 9.0, 6.0 ).pump(List,Void.Read,Point);

closestPoints(points).println(); //-->L(Point(8,4),Point(7,4),1)

points:=T( 0.654682, 0.925557, 0.409382, 0.619391,
0.891663, 0.888594, 0.716629, 0.9962,
0.477721, 0.946355, 0.925092, 0.81822,
0.624291, 0.142924, 0.211332, 0.221507,
0.293786, 0.691701, 0.839186, 0.72826)
closestPoints(points).println();
Output:
L(Point(8,4),Point(7,4),1)
L(Point(0.925092,0.81822),Point(0.891663,0.888594),0.0779102)

## ZX Spectrum Basic

Translation of: BBC_BASIC
10 DIM x(10): DIM y(10)
20 FOR i=1 TO 10
40 NEXT i
50 LET min=1e30
60 FOR i=1 TO 9
70 FOR j=i+1 TO 10
80 LET p1=x(i)-x(j): LET p2=y(i)-y(j): LET dsq=p1*p1+p2*p2
90 IF dsq<min THEN LET min=dsq: LET mini=i: LET minj=j
100 NEXT j
110 NEXT i
120 PRINT "Closest pair is ";mini;" and ";minj;" at distance ";SQR min
130 STOP
140 DATA 0.654682,0.925557
150 DATA 0.409382,0.619391
160 DATA 0.891663,0.888594
170 DATA 0.716629,0.996200
180 DATA 0.477721,0.946355
190 DATA 0.925092,0.818220
200 DATA 0.624291,0.142924
210 DATA 0.211332,0.221507
220 DATA 0.293786,0.691701
230 DATA 0.839186,0.728260