Closest-pair problem: Difference between revisions
(→{{header|Smalltalk}}: corrected according to perl and "new" pseudocode / updated "times" of the the d&c, not of the bruteforce approach) |
m (deleted an unuseful instruction) |
||
| Line 32: | Line 32: | ||
S ← { p ∈ xP : |x<sub>M</sub> - p<sub>x</sub>| < d<sub>min</sub> } |
S ← { p ∈ xP : |x<sub>M</sub> - p<sub>x</sub>| < d<sub>min</sub> } |
||
yP ← S ordered by the y coordinate, in ascending order |
yP ← S ordered by the y coordinate, in ascending order |
||
closest ← ∞ |
|||
nP ← number of points in yP |
nP ← number of points in yP |
||
closest ← d<sub>min</sub> |
closest ← d<sub>min</sub> |
||
Revision as of 16:20, 11 May 2009
You are encouraged to solve this task according to the task description, using any language you may know.
| This page uses content from Wikipedia. The original article was at Closest-pair problem. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance) |
The aim of this task is to provide a function to find the closest two points among a set of given points in two dimensions, i.e. to solve the Closest pair of points problem in the planar case.
The straightforward solution is a O(n2) algorithm (which we can call brute-force algorithm); the pseudocode (using indexes) could be simply:
bruteForceClosestPair of P(1), P(2), ... P(N)
if N < 2 then
return ∞
else
minDistance ← |P(1) - P(2)|
foreach i ∈ [1, N-1]
foreach j ∈ [i+1, N]
if |P(i) - P(j)| < minDistance then
minDistance ← |P(i) - P(j)|
endif
return minDistance
endif'
A better algorithm is based on the recursive divide&conquer approach, as explained also at Wikipedia, which is O(n log n); a pseudocode could be:
closestPair of P(1), P(2), ... P(N)
if N ≤ 3 then
return closest points of P using brute-force algorithm
else
xP ← P ordered by the x coordinate, in ascending order
PL ← points of xP from 1 to ⌈N/2⌉
PR ← points of xP from ⌈N/2⌉+1 to N
dL ← closestPair of PL
dR ← closestPair of PR
dmin ← min{dL, dR}
xM ← xP(⌈N/2⌉)
S ← { p ∈ xP : |xM - px| < dmin }
yP ← S ordered by the y coordinate, in ascending order
nP ← number of points in yP
closest ← dmin
for i from 1 to nP - 1
k ← i + 1
while k ≤ nP and yP(k)y - yP(i)y < dmin
closest ← min{|yP(k) - yP(i)|, closest}
k ← k + 1
endwhile
return closest
endif
References
Perl
<lang perl>#! /usr/bin/perl use strict; use List::Util qw(min); 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( scalar(@arr) > 0 ) {
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 $ra = shift;
my @arr = @$ra;
my $N = @arr;
return closest_pair_simple($ra) if ( scalar(@arr) <= 3 );
my $inf = 1e600;
my @xP = sort { ${$a}[0] <=> ${$b}[0] } @arr;
my $midx = ceil($N/2)-1;
my @PL = @xP[0 .. $midx];
my @PR = @xP[$midx+1 .. $N-1];
my ($al, $bl, $dL) = closest_pair(\@PL);
my ($ar, $br, $dR) = closest_pair(\@PR);
my ($m1, $m2, $dmin) = ($al, $bl, $dL);
($m1, $m2, $dmin) = ($ar, $br, $dR) if ( $dR < $dL );
my $xm = ${$xP[$midx]}[0];
my @S = ();
foreach my $p (@xP) {
push @S, $p if ( abs($xm - ${$p}[0]) < $dmin );
}
my @yP = sort { ${$a}[1] <=> ${$b}[1] } @S;
if ( scalar(@yP) > 0 ) {
my ( $w1, $w2, $closest ) = ($m1, $m2, $dmin); foreach my $i (0 .. ($#yP - 1)) {
my $k = $i + 1; while ( ($k <= $#yP) && ( (${$yP[$k]}[1] - ${$yP[$i]}[1]) < $dmin) ) { my $d = dist($yP[$k], $yP[$i]); ($w1, $w2, $closest) = ($yP[$k], $yP[$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";
exit 0;</lang>
Time for the brute-force algorithm gave 40.63user 0.12system 0:41.06elapsed, while the divide&conqueer algorithm gave 0.38user 0.00system 0:00.38elapsed with 5000 points.
Smalltalk
These class methods return a three elements array, where the first two items are the points, while the third is the distance between them (which having the two points, can be computed; but it was easier to keep it already computed in the third position of the array).
<lang smalltalk>Object subclass: ClosestPair [
ClosestPair class >> raiseInvalid: arg [
SystemExceptions.InvalidArgument
signalOn: arg
reason: 'specify at least two points'
]
ClosestPair class >> bruteForce: pointsList [ |dist from to points|
(pointsList size < 2) ifTrue: [ ^ FloatD infinity ].
points := pointsList asOrderedCollection.
from := points at: 1. to := points at: 2.
dist := from dist: to.
[ points isEmpty ]
whileFalse: [ |p0|
p0 := points removeFirst.
points do: [ :p |
((p0 dist: p) <= dist)
ifTrue: [ from := p0. to := p. dist := p0 dist: p. ]
]
].
^ { from. to. from dist: to }
]
ClosestPair class >> orderByX: points [ ^ points asSortedCollection: [:a :b| (a x) < (b x) ] ] ClosestPair class >> orderByY: points [ ^ points asSortedCollection: [:a :b| (a y) < (b y) ] ]
ClosestPair class >> splitLeft: pointsList [ ^ pointsList copyFrom: 1 to: ((pointsList size / 2) ceiling) ] ClosestPair class >> splitRight: pointsList [ |s| ^ pointsList copyFrom: (((pointsList size / 2) ceiling) + 1) to: (pointsList size). ]
ClosestPair class >> minBetween: a and: b [
(a at: 3) < (b at: 3)
ifTrue: [ ^a ]
ifFalse: [ ^b ]
]
ClosestPair class >> recursiveDAndC: pointsList [
|orderedByX pR pL minL minR minDist middleVLine joiningStrip tDist nP|
(pointsList size <= 3)
ifTrue: [ ^ self bruteForce: pointsList ].
orderedByX := self orderByX: pointsList.
pR := self splitLeft: orderedByX.
pL := self splitRight: orderedByX.
minR := self recursiveDAndC: pR.
minL := self recursiveDAndC: pL.
minDist := self minBetween: minR and: minL.
middleVLine := (orderedByX at: ((orderedByX size / 2) ceiling)) x.
joiningStrip := self
orderByY: (pointsList
select: [ :p |
((middleVLine - (p x)) abs) < (minDist at: 3)
]
).
tDist := minDist.
nP := joiningStrip size.
1 to: (nP - 1) do: [ :i | |k|
k := i + 1.
[ (k <= nP)
& ( ((joiningStrip at: (k min: nP)) y - (joiningStrip at: i) y) < (minDist at: 3) ) ]
whileTrue: [
((joiningStrip at: i) dist: (joiningStrip at: k)) < (tDist at: 3)
ifTrue: [ tDist := { joiningStrip at: i. joiningStrip at: k.
(joiningStrip at: i) dist: (joiningStrip at: k) } ].
k := k + 1.
]
].
^ tDist
]
].</lang>
Testing
<lang smalltalk>|coll cp ran| "Let's use the same seed to be sure of the results..." ran := Random seed: 100.
coll := (1 to: 10000 collect: [ :a |
Point x: (ran next)*10 y: (ran next)*10 ]).
cp := ClosestPair bruteForce: coll. ((cp at: 3) asScaledDecimal: 7) displayNl.
"or"
cp := ClosestPair recursiveDAndC: coll. ((cp at: 3) asScaledDecimal: 7) displayNl.</lang>
The brute-force approach with 10000 points, run with the time tool, gave
224.21user 1.31system 3:46.84elapsed 99%CPU
while the recursive divide&conquer algorithm gave
2.37user 0.01system 0:02.56elapsed 93%CPU
(Of course these results must be considered relative and taken cum grano salis; time counts also the time taken to initialize the Smalltalk environment, and I've taken no special measures to avoid the system load falsifying the results)
Tcl
Each point is represented as a list of two floating-point numbers, the first being the x coordinate, and the second being the y. <lang Tcl># A trivial helper... proc distance {p1 p2} {
lassign $p1 x1 y1
lassign $p2 x2 y2
expr {
hypot($x1-$x2, $y1-$y2)
}
}
proc bruteForceClosestPair {points} {
if {[package provide Tcl] >= "8.5"} {
set min Inf
} else {
set min [distance [lindex $points 0] [lindex $points 1]]
}
for {set i 0} {$i < [llength $points]-1} {incr i} {
for {set j $i} {[incr j] < [llength $points]} {} {
set dist [distance [lindex $points $i] [lindex $points $j]]
if {$dist < $min} {
set min $dist
}
}
}
return $min
}
- Some testing code:
set points {} for {set n 0} {$n < 10000} {incr n} {
lappend points [list [expr {10*rand()}] [expr {10*rand()}]]
} set t [time {
set d [bruteForceClosestPair $points]
}] puts "closest was $d, found in [lindex $t 0] microseconds"</lang>Example output
closest was 0.0006785546466922088, found in 94942709 microseconds
Note that the lindex and llength commands are both O(1).