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This means it might contain formatting issues, incorrect code, conceptual problems, or other severe issues.
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{{task}}
Given the n + 1 vertices x[0], y[0] .. x[N], y[N] of a simple polygon described in a clockwise direction, then the polygon's area can be calculated by:
abs( (sum(x[0]*y[1] + ... x[n-1]*y[n]) + x[N]*y[0]) -
(sum(x[1]*y[0] + ... x[n]*y[n-1]) + x[0]*y[N])
) / 2
(Where abs returns the absolute value)
;Task: Write a function/method/routine to use the the [[wp:Shoelace formula|Shoelace formula]] to calculate the area of the polygon described by the ordered points: (3,4), (5,11), (12,8), (9,5), and (5,6)
Show the answer here, on this page.
360 Assembly
* SHOELACE 25/02/2019
SHOELACE CSECT
USING SHOELACE,R15 base register
MVC SUPS(8),POINTS x(nt+1)=x(1); y(nt+1)=y(1)
LA R9,0 area=0
LA R7,POINTS @x(1)
LA R6,NT do i=1 to nt
LOOP L R3,0(R7) x(i)
M R2,12(R7) *y(i+1)
L R5,8(R7) x(i+1)
M R4,4(R7) *y(i)
SR R3,R5 x(i)*y(i+1)-x(i+1)*y(i)
AR R9,R3 area=area+x(i)*y(i+1)-x(i+1)*y(i)
LA R7,8(R7) @x(i++)
BCT R6,LOOP enddo
LPR R9,R9 area=abs(area)
SRA R9,1 area=area/2
XDECO R9,PG edit area
XPRNT PG,L'PG print area
BR R14 return to caller
NT EQU (SUPS-POINTS)/8 nt number of points
POINTS DC F'3',F'4',F'5',F'11',F'12',F'8',F'9',F'5',F'5',F'6'
SUPS DS 2F x(nt+1),y(nt+1)
PG DC CL12' ' buffer
REGEQU
END SHOELACE
{{out}}
30
Ada
{{works with|Ada|Ada|83}}
with Ada.Text_IO;
procedure Shoelace_Formula_For_Polygonal_Area
is
type Point is record
x, y : Float;
end record;
type Polygon is array (Positive range <>) of Point;
function Shoelace(input : in Polygon) return Float
is
sum_1 : Float := 0.0;
sum_2 : Float := 0.0;
tmp : constant Polygon := input & input(input'First);
begin
for I in tmp'First .. tmp'Last - 1 loop
sum_1 := sum_1 + tmp(I).x * tmp(I+1).y;
sum_2 := sum_2 + tmp(I+1).x * tmp(I).y;
end loop;
return abs(sum_1 - sum_2) / 2.0;
end Shoelace;
my_polygon : constant Polygon :=
((3.0, 4.0),
(5.0, 11.0),
(12.0, 8.0),
(9.0, 5.0),
(5.0, 6.0));
begin
Ada.Text_IO.Put_Line(Shoelace(my_polygon)'Img);
end Shoelace_Formula_For_Polygonal_Area;
{{out}}
3.00000E+01
ALGOL 60
Optimized version: '''begin''' '''comment''' Shoelace formula for polygonal area - Algol 60; '''real''' '''array''' x[1:33],y[1:33]; '''integer''' i,n; '''real''' a; ininteger(0,n); '''for''' i:=1 '''step''' 1 '''until''' n '''do''' '''begin''' inreal(0,x[i]); inreal(0,y[i]) '''end'''; x[i]:=x[1]; y[i]:=y[1]; a:=0; '''for''' i:=1 '''step''' 1 '''until''' n '''do''' a:=a+x[i]*y[i+1]-x[i+1]*y[i]; a:=abs(a/2.); outreal(1,a) '''end''' {{out}}
30.00
Non-optimized version: '''begin''' '''comment''' Shoelace formula for polygonal area - Algol 60; '''real''' '''array''' x[1:32],y[1:32]; '''integer''' i,j,n; '''real''' a; ininteger(0,n); '''for''' i:=1 '''step''' 1 '''until''' n '''do''' '''begin''' inreal(0,x[i]); inreal(0,y[i]) '''end'''; a:=0; '''for''' i:=1 '''step''' 1 '''until''' n '''do''' '''begin''' j:='''if''' i=n '''then''' 1 '''else''' i+1; a:=a+x[i]*y[j]-x[j]*y[i] '''end'''; a:=abs(a/2.); outreal(1,a) '''end''' {{out}}
30.00
ALGOL 68
BEGIN
# returns the area of the polygon defined by the points p using the Shoelace formula #
OP AREA = ( [,]REAL p )REAL:
BEGIN
[,]REAL points = p[ AT 1, AT 1 ]; # normalise array bounds to start at 1 #
IF 2 UPB points /= 2 THEN
# the points do not have 2 coordinates #
-1
ELSE
REAL result := 0;
INT n = 1 UPB points;
IF n > 1 THEN
# there at least two points #
[]REAL x = points[ :, 1 ];
[]REAL y = points[ :, 2 ];
FOR i TO 1 UPB points - 1 DO
result +:= x[ i ] * y[ i + 1 ];
result -:= x[ i + 1 ] * y[ i ]
OD;
result +:= x[ n ] * y[ 1 ];
result -:= x[ 1 ] * y[ n ]
FI;
( ABS result ) / 2
FI
END # AREA # ;
# test case as per the task #
print( ( fixed( AREA [,]REAL( ( 3.0, 4.0 ), ( 5.0, 11.0 ), ( 12.0, 8.0 ), ( 9.0, 5.0 ), ( 5.0, 6.0 ) ), -6, 2 ), newline ) )
END
{{out}}
30.00
C
Reads the points from a file whose name is supplied via the command line, prints out usage if invoked incorrectly.
#include<stdlib.h>
#include<stdio.h>
#include<math.h>
typedef struct{
double x,y;
}point;
double shoelace(char* inputFile){
int i,numPoints;
double leftSum = 0,rightSum = 0;
point* pointSet;
FILE* fp = fopen(inputFile,"r");
fscanf(fp,"%d",&numPoints);
pointSet = (point*)malloc((numPoints + 1)*sizeof(point));
for(i=0;i<numPoints;i++){
fscanf(fp,"%lf %lf",&pointSet[i].x,&pointSet[i].y);
}
fclose(fp);
pointSet[numPoints] = pointSet[0];
for(i=0;i<numPoints;i++){
leftSum += pointSet[i].x*pointSet[i+1].y;
rightSum += pointSet[i+1].x*pointSet[i].y;
}
free(pointSet);
return 0.5*fabs(leftSum - rightSum);
}
int main(int argC,char* argV[])
{
if(argC==1)
printf("\nUsage : %s <full path of polygon vertices file>",argV[0]);
else
printf("The polygon area is %lf square units.",shoelace(argV[1]));
return 0;
}
Input file, first line specifies number of points followed by the ordered vertices set with one vertex on each line.
5
3 4
5 11
12 8
9 5
5 6
Invocation and output :
C:\rosettaCode>shoelace.exe polyData.txt
The polygon area is 30.000000 square units.
C++
{{trans|D}}
#include <iostream>
#include <tuple>
#include <vector>
using namespace std;
double shoelace(vector<pair<double, double>> points) {
double leftSum = 0.0;
double rightSum = 0.0;
for (int i = 0; i < points.size(); ++i) {
int j = (i + 1) % points.size();
leftSum += points[i].first * points[j].second;
rightSum += points[j].first * points[i].second;
}
return 0.5 * abs(leftSum - rightSum);
}
void main() {
vector<pair<double, double>> points = {
make_pair( 3, 4),
make_pair( 5, 11),
make_pair(12, 8),
make_pair( 9, 5),
make_pair( 5, 6),
};
auto ans = shoelace(points);
cout << ans << endl;
}
{{out}}
30
C#
{{trans|Java}}
using System;
using System.Collections.Generic;
namespace ShoelaceFormula {
using Point = Tuple<double, double>;
class Program {
static double ShoelaceArea(List<Point> v) {
int n = v.Count;
double a = 0.0;
for (int i = 0; i < n - 1; i++) {
a += v[i].Item1 * v[i + 1].Item2 - v[i + 1].Item1 * v[i].Item2;
}
return Math.Abs(a + v[n - 1].Item1 * v[0].Item2 - v[0].Item1 * v[n - 1].Item2) / 2.0;
}
static void Main(string[] args) {
List<Point> v = new List<Point>() {
new Point(3,4),
new Point(5,11),
new Point(12,8),
new Point(9,5),
new Point(5,6),
};
double area = ShoelaceArea(v);
Console.WriteLine("Given a polygon with vertices [{0}],", string.Join(", ", v));
Console.WriteLine("its area is {0}.", area);
}
}
}
{{out}}
Given a polygon with vertices [(3, 4), (5, 11), (12, 8), (9, 5), (5, 6)],
its area is 30.
D
import std.stdio;
Point[] pnts = [{3,4}, {5,11}, {12,8}, {9,5}, {5,6}];
void main() {
auto ans = shoelace(pnts);
writeln(ans);
}
struct Point {
real x, y;
}
real shoelace(Point[] pnts) {
real leftSum = 0, rightSum = 0;
for (int i=0; i<pnts.length; ++i) {
int j = (i+1) % pnts.length;
leftSum += pnts[i].x * pnts[j].y;
rightSum += pnts[j].x * pnts[i].y;
}
import std.math : abs;
return 0.5 * abs(leftSum - rightSum);
}
unittest {
auto ans = shoelace(pnts);
assert(ans == 30);
}
{{out}}
30
=={{header|F_Sharp|F#}}==
// Shoelace formula for area of polygon. Nigel Galloway: April 11th., 2018
let fN(n::g) = abs(List.pairwise(n::g@[n])|>List.fold(fun n ((nα,gα),(nβ,gβ))->n+(nα*gβ)-(gα*nβ)) 0.0)/2.0
printfn "%f" (fN [(3.0,4.0); (5.0,11.0); (12.0,8.0); (9.0,5.0); (5.0,6.0)])
{{out}}
30.000000
Factor
By constructing a circular from a sequence, we can index elements beyond the length of the sequence, wrapping around to the beginning. We can also change the beginning of the sequence to an arbitrary index. This allows us to use 2map to cleanly obtain a sum.
USING: circular kernel math prettyprint sequences ;
IN: rosetta-code.shoelace
CONSTANT: input { { 3 4 } { 5 11 } { 12 8 } { 9 5 } { 5 6 } }
: align-pairs ( pairs-seq -- seq1 seq2 )
<circular> dup clone [ 1 ] dip
[ change-circular-start ] keep ;
: shoelace-sum ( seq1 seq2 -- n )
[ [ first ] [ second ] bi* * ] 2map sum ;
: shoelace-area ( pairs-seq -- area )
[ align-pairs ] [ align-pairs swap ] bi
[ shoelace-sum ] 2bi@ - abs 2 / ;
input shoelace-area .
{{out}}
30
=={{header|Fōrmulæ}}==
In [http://wiki.formulae.org/Shoelace_formula_for_polygonal_area this] page you can see the solution of this task.
Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text ([http://wiki.formulae.org/Editing_F%C5%8Drmul%C3%A6_expressions more info]). Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for transportation effects more than visualization and edition.
The option to show Fōrmulæ programs and their results is showing images. Unfortunately images cannot be uploaded in Rosetta Code.
Fortran
Fortran 90
Except for the use of "END FUNCTION ''name'' instead of just END, and the convenient function SUM with array span expressions (so SUM(P) rather than a DO-loop to sum the elements of array P), both standardised with F90, this would be acceptable to F66, which introduced complex number arithmetic. Otherwise, separate X and Y arrays would be needed, but complex numbers seemed convenient seeing as (x,y) pairs are involved. But because the MODULE facility of F90 has not been used, routines invoking functions must declare the type of the function names, especially if the default types are unsuitable, as here. In function AREA, the x and y parts are dealt with together, but in AREASL they might be better as separate arrays, thus avoiding the DIMAG and DBLE functions to extract the x and y parts. Incidentally, the x and y parts can be interchanged and the calculation still works. Comparing the two resulting areas might give some indication of their accuracy.
If the MODULE protocol were used, the size of an array parameter is passed as a secret additional parameter accessible via the special function UBOUND, but otherwise it must be passed as an explicit parameter. A quirk of the compiler requires that N be declared before it appears in DOUBLE COMPLEX P(N) so as it is my practice to declare parameters in the order specified, here N comes before P. However, it is not clear whether specifying P(N) does much good (as in array index checking) as an alternative is to specify P(*) meaning merely that the array has one dimension, or even P(12345) to the same effect, with no attention to the actual numerical value. See for example [[Array_length#Fortran]]
DOUBLE PRECISION FUNCTION AREA(N,P) !Calculates the area enclosed by the polygon P.
C Uses the mid-point rule for integration. Consider the line joining (x1,y1) to (x2,y2)
C The area under that line (down to the x-axis) is the y-span midpoint (y1 + y2)/2 times the width (x2 - x1)
C This is the trapezoidal rule for a single interval, and follows from simple geometry.
C Now consider a sequence of such points heading in the +x direction: each successive interval's area is positive.
C Follow with a sequence of points heading in the -x direction, back to the first point: their areas are all negative.
C The resulting sum is the area below the +x sequence and above the -x sequence: the area of the polygon.
C The point sequence can wobble as it wishes and can meet the other side, but it must not cross itself
c as would be done in a figure 8 drawn with a crossover instead of a meeting.
C A clockwise traversal (as for an island) gives a positive area; use anti-clockwise for a lake.
INTEGER N !The number of points.
DOUBLE COMPLEX P(N) !The points.
DOUBLE COMPLEX PP,PC !Point Previous and Point Current.
DOUBLE COMPLEX W !Polygon centre. Map coordinates usually have large offsets.
DOUBLE PRECISION A !The area accumulator.
INTEGER I !A stepper.
IF (N.LT.3) STOP "Area: at least three points are needed!" !Good grief.
W = (P(1) + P(N/3) + P(2*N/3))/3 !An initial working average.
W = SUM(P(1:N) - W)/N + W !A good working average is the average itself.
A = 0 !The area enclosed by the point sequence.
PC = P(N) - W !The last point is implicitly joined to the first.
DO I = 1,N !Step through the positions.
PP = PC !Previous position.
PC = P(I) - W !Current position.
A = (DIMAG(PC) + DIMAG(PP))*(DBLE(PC) - DBLE(PP)) + A !Area integral component.
END DO !On to the next position.
AREA = A/2 !Divide by two once.
END FUNCTION AREA !The units are those of the points.
DOUBLE PRECISION FUNCTION AREASL(N,P) !Area enclosed by polygon P, by the "shoelace" method.
INTEGER N !The number of points.
DOUBLE COMPLEX P(N) !The points.
DOUBLE PRECISION A !A scratchpad.
A = SUM(DBLE(P(1:N - 1)*DIMAG(P(2:N)))) + DBLE(P(N))*DIMAG(P(1))
1 - SUM(DBLE(P(2:N)*DIMAG(P(1:N - 1)))) - DBLE(P(1))*DIMAG(P(N))
AREASL = A/2 !The midpoint formula requires a halving.
END FUNCTION AREASL !Negative for clockwise, positive for anti-clockwise.
INTEGER ENUFF
DOUBLE PRECISION AREA,AREASL !The default types are not correct.
DOUBLE PRECISION A1,A2 !Scratchpads, in case of a debugging WRITE within the functions.
PARAMETER (ENUFF = 5) !The specification.
DOUBLE COMPLEX POINT(ENUFF) !Could use X and Y arrays instead.
DATA POINT/(3D0,4D0),(5D0,11D0),(12D0,8D0),(9D0,5D0),(5D0,6D0)/ !"D" for double precision.
WRITE (6,*) POINT
A1 = AREA(5,POINT)
A2 = AREASL(5,POINT)
WRITE (6,*) "A=",A1,A2
END
Output: WRITE (6,*) means write to output unit six (standard output) with free-format (the *). Note the different sign convention.
(3.00000000000000,4.00000000000000) (5.00000000000000,11.0000000000000)
(12.0000000000000,8.00000000000000) (9.00000000000000,5.00000000000000)
(5.00000000000000,6.00000000000000)
A= 30.0000000000000 -30.0000000000000
The "shoelace" method came as a surprise to me, as I've always used what I had thought the "obvious" method. Note that function AREA makes one pass through the point data not two, and because map coordinate values often have large offsets a working average is used to reduce the loss of precision. This requires faith that SUM(P(1:N) - W) will be evaluated as written, not as SUM(P(1:N)) - N*W with even greater optimisation opportunity awaiting in cancelling further components of the expression. For example, the New Zealand metric grid has (2510000,6023150) as (Easting,Northing) or (x,y) at its central point of 41°S 173°E rather than (0,0) so seven digits of precision are used up. If anyone wants a copy of a set of point sequences for NZ (30,000 positions, 570KB) with lots of islands and lakes, even a pond in an island in a lake in the North Island...
Fortran I
In orginal FORTRAN 1957:
C SHOELACE FORMULA FOR POLYGONAL AREA
DIMENSION X(33),Y(33)
READ 101,N
DO 1 I=1,N
1 READ 102,X(I),Y(I)
X(I)=X(1)
Y(I)=Y(1)
A=0
DO 2 I=1,N
2 A=A+X(I)*Y(I+1)-X(I+1)*Y(I)
A=ABSF(A/2.)
PRINT 303,A
STOP
101 FORMAT(I2)
102 FORMAT(2F6.2)
303 FORMAT(F10.2)
{{in}}
5
3.00 4.00
5.00 11.00
12.00 8.00
9.00 5.00
5.00 6.00
{{out}}
30.00
FreeBASIC
' version 18-08-2017
' compile with: fbc -s console
Type _point_
As Double x, y
End Type
Function shoelace_formula(p() As _point_ ) As Double
Dim As UInteger i
Dim As Double sum
For i = 1 To UBound(p) -1
sum += p(i ).x * p(i +1).y
sum -= p(i +1).x * p(i ).y
Next
sum += p(i).x * p(1).y
sum -= p(1).x * p(i).y
Return Abs(sum) / 2
End Function
' ------=< MAIN >=------
Dim As _point_ p_array(1 To ...) = {(3,4), (5,11), (12,8), (9,5), (5,6)}
Print "The area of the polygon ="; shoelace_formula(p_array())
' empty keyboard buffer
While Inkey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
{{out}}
The area of the polygon = 30
Go
package main
import "fmt"
type point struct{ x, y float64 }
func shoelace(pts []point) float64 {
sum := 0.
p0 := pts[len(pts)-1]
for _, p1 := range pts {
sum += p0.y*p1.x - p0.x*p1.y
p0 = p1
}
return sum / 2
}
func main() {
fmt.Println(shoelace([]point{{3, 4}, {5, 11}, {12, 8}, {9, 5}, {5, 6}}))
}
{{out}}
30
Haskell
main :: IO ()
main = print (shoelace [(3, 4), (5, 11), (12, 8), (9, 5), (5, 6)])
-- Calculate the area of a polygon formed by the list of coordinates
-- Coordinates are of the form (x, y).
shoelace :: [(Double, Double)] -> Double
shoelace ps = 0.5 * abs (leftSum - rightSum)
where
(leftSum, rightSum) = foldr calcSums (0, 0) interlaced
calcSums ((xi, yi), (nxi, nyi)) (l, r) = (l + xi * nyi, r + nxi * yi)
interlaced = zip ps (tail (cycle ps))
{{out}}
30.0
J
Implementation:
shoelace=:verb define
0.5*|+/((* 1&|.)/ - (* _1&|.)/)|:y
)
Task example:
shoelace 3 4,5 11,12 8,9 5,:5 6
30
Exposition:
We start with our list of coordinate pairs
3 4,5 11,12 8,9 5,:5 6
3 4
5 11
12 8
9 5
5 6
But the first thing we do is transpose them so that x coordinates and y coordinates are the two items we are working with:
|:3 4,5 11,12 8,9 5,:5 6
3 5 12 9 5
4 11 8 5 6
We want to rotate the y list by one (in each direction) and multiply the x list items by the corresponding y list items. Something like this, for example:
3 5 12 9 5* 1|.4 11 8 5 6
33 40 60 54 20
Or, rephrased:
(* 1&|.)/|:3 4,5 11,12 8,9 5,:5 6
33 40 60 54 20
We'll be subtracting what we get when we rotate in the other direction, which looks like this:
((* 1&|.)/ - (* _1&|.)/)|:3 4,5 11,12 8,9 5,:5 6
15 20 _72 _18 _5
Finally, we add up that list, take the absolute value (there are contexts where signed area is interesting - for example, some graphics application - but that was not a part of this task) and divide that by 2.
Java
{{trans|Kotlin}} {{works with|Java|9}}
import java.util.List;
public class ShoelaceFormula {
private static class Point {
int x, y;
Point(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public String toString() {
return String.format("(%d, %d)", x, y);
}
}
private static double shoelaceArea(List<Point> v) {
int n = v.size();
double a = 0.0;
for (int i = 0; i < n - 1; i++) {
a += v.get(i).x * v.get(i + 1).y - v.get(i + 1).x * v.get(i).y;
}
return Math.abs(a + v.get(n - 1).x * v.get(0).y - v.get(0).x * v.get(n - 1).y) / 2.0;
}
public static void main(String[] args) {
List<Point> v = List.of(
new Point(3, 4),
new Point(5, 11),
new Point(12, 8),
new Point(9, 5),
new Point(5, 6)
);
double area = shoelaceArea(v);
System.out.printf("Given a polygon with vertices %s,%n", v);
System.out.printf("its area is %f,%n", area);
}
}
{{out}}
Given a polygon with vertices [(3, 4), (5, 11), (12, 8), (9, 5), (5, 6)],
its area is 30.000000,
Julia
{{works with|Julia|0.6}} {{trans|Python}}
"""
Assumes x,y points go around the polygon in one direction.
"""
shoelacearea(x, y) =
abs(sum(i * j for (i, j) in zip(x, append!(y[2:end], y[1]))) -
sum(i * j for (i, j) in zip(append!(x[2:end], x[1]), y))) / 2
x, y = [3, 5, 12, 9, 5], [4, 11, 8, 5, 6]
@show x y shoelacearea(x, y)
{{out}}
x = [3, 5, 12, 9, 5]
y = [4, 11, 8, 5, 6]
shoelacearea(x, y) = 30.0
Kotlin
// version 1.1.3
class Point(val x: Int, val y: Int) {
override fun toString() = "($x, $y)"
}
fun shoelaceArea(v: List<Point>): Double {
val n = v.size
var a = 0.0
for (i in 0 until n - 1) {
a += v[i].x * v[i + 1].y - v[i + 1].x * v[i].y
}
return Math.abs(a + v[n - 1].x * v[0].y - v[0].x * v[n -1].y) / 2.0
}
fun main(args: Array<String>) {
val v = listOf(
Point(3, 4), Point(5, 11), Point(12, 8), Point(9, 5), Point(5, 6)
)
val area = shoelaceArea(v)
println("Given a polygon with vertices at $v,")
println("its area is $area")
}
{{out}}
Given a polygon with vertices at [(3, 4), (5, 11), (12, 8), (9, 5), (5, 6)],
its area is 30.0
Lua
function shoeArea(ps)
local function det2(i,j)
return ps[i][1]*ps[j][2]-ps[j][1]*ps[i][2]
end
local sum = #ps>2 and det2(#ps,1) or 0
for i=1,#ps-1 do sum = sum + det2(i,i+1)end
return math.abs(0.5 * sum)
end
Using an accumulator helper inner function
function shoeArea(ps)
local function ssum(acc, p1, p2, ...)
if not p2 or not p1 then
return math.abs(0.5 * acc)
else
return ssum(acc + p1[1]*p2[2]-p1[2]*p2[1], p2, ...)
end
end
return ssum(0, ps[#ps], table.unpack(ps))
end
local p = {{3,4}, {5,11}, {12,8}, {9,5}, {5,6}}
print(shoeArea(p))-- 30
both version handle special cases of less than 3 point as 0 area result.
Mathematica
Geometry objects built-in in the Wolfram Language
Area[Polygon[{{3, 4}, {5, 11}, {12, 8}, {9, 5}, {5, 6}}]]
{{out}}
30
min
{{works with|min|0.19.3}}
((((first) map) ((last) map)) cleave) :dezip
(((first) (rest)) cleave append) :rotate
((0 <) (-1 *) when) :abs
(
=b =a a size :n 0 :i () =list
(i n <) (
a i get b i get ' prepend list append #list
i succ @i
) while list
) :rezip
(rezip (-> *) map sum) :cross-sum
(
((dezip rotate) (dezip swap rotate)) cleave
((id) (cross-sum) (id) (cross-sum)) spread
- abs 2 /
) :shoelace
((3 4) (5 11) (12 8) (9 5) (5 6)) shoelace print
{{out}}
30.0
MiniScript
shoelace = function(vertices)
sum = 0
points = vertices.len
for i in range(0,points-2)
sum = sum + vertices[i][0]*vertices[i+1][1]
end for
sum = sum + vertices[points-1][0]*vertices[0][1]
for i in range(points-1,1)
sum = sum - vertices[i][0]*vertices[i-1][1]
end for
sum = sum - vertices[0][0]*vertices[points-1][1]
return abs(sum)/2
end function
verts = [[3,4],[5,11],[12,8],[9,5],[5,6]]
print "The polygon area is " + shoelace(verts)
{{out}}
The polygon area is 30
=={{header|Modula-2}}==
MODULE ShoelaceFormula;
FROM RealStr IMPORT RealToStr;
FROM FormatString IMPORT FormatString;
FROM Terminal IMPORT WriteString,WriteLn,ReadChar;
TYPE
Point = RECORD
x,y : INTEGER;
END;
PROCEDURE PointToString(self : Point; VAR buf : ARRAY OF CHAR);
BEGIN
FormatString("(%i, %i)", buf, self.x, self.y);
END PointToString;
PROCEDURE ShoelaceArea(v : ARRAY OF Point) : REAL;
VAR
a : REAL;
i,n : INTEGER;
BEGIN
n := HIGH(v);
a := 0.0;
FOR i:=0 TO n-1 DO
a := a + FLOAT(v[i].x * v[i+1].y - v[i+1].x * v[i].y);
END;
RETURN ABS(a + FLOAT(v[n].x * v[0].y - v[0].x * v[n].y)) / 2.0;
END ShoelaceArea;
VAR
v : ARRAY[0..4] OF Point;
buf : ARRAY[0..63] OF CHAR;
area : REAL;
i : INTEGER;
BEGIN
v[0] := Point{3,4};
v[1] := Point{5,11};
v[2] := Point{12,8};
v[3] := Point{9,5};
v[4] := Point{5,6};
area := ShoelaceArea(v);
WriteString("Given a polygon with verticies ");
FOR i:=0 TO HIGH(v) DO
PointToString(v[i], buf);
WriteString(buf);
WriteString(" ");
END;
WriteLn;
RealToStr(area, buf);
WriteString("its area is ");
WriteString(buf);
WriteLn;
ReadChar;
END ShoelaceFormula.
Perl
use strict;
use warnings;
use feature 'say';
sub area_by_shoelace {
my $area;
our @p;
$#_ > 0 ? @p = @_ : (local *p = shift);
$area += $p[$_][0] * $p[($_+1)%@p][1] for 0 .. @p-1;
$area -= $p[$_][1] * $p[($_+1)%@p][0] for 0 .. @p-1;
return abs $area/2;
}
my @poly = ( [3,4], [5,11], [12,8], [9,5], [5,6] );
say area_by_shoelace( [3,4], [5,11], [12,8], [9,5], [5,6] );
say area_by_shoelace( [ [3,4], [5,11], [12,8], [9,5], [5,6] ] );
say area_by_shoelace( @poly );
say area_by_shoelace( \@poly );
{{out}}
30
30
30
30
Perl 6
Index and mod offset
{{works with|Rakudo|2017.07}}
sub area-by-shoelace(@p) {
(^@p).map({@p[$_;0] * @p[($_+1)%@p;1] - @p[$_;1] * @p[($_+1)%@p;0]}).sum.abs / 2
}
say area-by-shoelace( [ (3,4), (5,11), (12,8), (9,5), (5,6) ] );
{{out}}
30
Slice and rotation
{{works with|Rakudo|2017.07}}
sub area-by-shoelace ( @p ) {
my @x := @p».[0];
my @y := @p».[1];
my $s := ( @x Z* @y.rotate( 1) ).sum
- ( @x Z* @y.rotate(-1) ).sum;
return $s.abs / 2;
}
say area-by-shoelace( [ (3,4), (5,11), (12,8), (9,5), (5,6) ] );
{{out}}
30
Phix
enum X, Y
function shoelace(sequence s)
atom t = 0
if length(s)>2 then
s = append(s,s[1])
for i=1 to length(s)-1 do
t += s[i][X]*s[i+1][Y] - s[i+1][X]*s[i][Y]
end for
end if
return abs(t)/2
end function
constant test = {{3,4},{5,11},{12,8},{9,5},{5,6}}
?shoelace(test)
{{out}}
30
PowerBASIC
{{Trans|Visual Basic}}
#COMPILE EXE
#DIM ALL
#COMPILER PBCC 6
FUNCTION ShoelaceArea(x() AS DOUBLE, y() AS DOUBLE) AS DOUBLE
LOCAL i, j AS LONG
LOCAL Area AS DOUBLE
j = UBOUND(x())
FOR i = LBOUND(x()) TO UBOUND(x())
Area += (y(j) + y(i)) * (x(j) - x(i))
j = i
NEXT i
FUNCTION = ABS(Area) / 2
END FUNCTION
FUNCTION PBMAIN () AS LONG
REDIM x(0 TO 4) AS DOUBLE, y(0 TO 4) AS DOUBLE
ARRAY ASSIGN x() = 3, 5, 12, 9, 5
ARRAY ASSIGN y() = 4, 11, 8, 5, 6
CON.PRINT STR$(ShoelaceArea(x(), y()))
CON.WAITKEY$
END FUNCTION
{{out}}
30
Python
def area_by_shoelace(x, y):
"Assumes x,y points go around the polygon in one direction"
return abs( sum(i * j for i, j in zip(x, y[1:] + y[:1]))
-sum(i * j for i, j in zip(x[1:] + x[:1], y ))) / 2
>>> points = [(3,4), (5,11), (12,8), (9,5), (5,6)]
>>> x, y = zip(*points)
>>> area_by_shoelace(x, y)
30.0
>>>
Or, defined in terms of '''reduce''' and '''cycle''': {{Trans|Haskell}} {{Works with|Python|3.7}}
'''Polygonal area by shoelace formula'''
from itertools import (cycle, islice)
from functools import (reduce)
# shoelaceArea :: [(Float, Float)] -> Float
def shoelaceArea(xys):
'''Area of polygon with vertices
at (x, y) points in xys.
'''
def go(a, tpl):
l, r = a
(x, y), (dx, dy) = tpl
return (l + x * dy, r + y * dx)
(nl, nr) = reduce(
go,
zip(xys, tail(cycle(xys))),
(0, 0)
)
return abs(nl - nr) / 2
# TEST ----------------------------------------------------
# main :: IO()
def main():
'''Sample calculation'''
ps = [(3, 4), (5, 11), (12, 8), (9, 5), (5, 6)]
print(__doc__ + ':')
print(repr(ps) + ' -> ' + str(shoelaceArea(ps)))
# GENERIC -------------------------------------------------
# tail :: [a] -> [a]
# tail :: Gen [a] -> [a]
def tail(xs):
'''The elements following the head of a
(non-empty) list or generator stream.
'''
if isinstance(xs, list):
return xs[1:]
else:
list(islice(xs, 1)) # First item dropped.
return xs
if __name__ == '__main__':
main()
{{Out}}
Polygonal area by shoelace formula:
[(3, 4), (5, 11), (12, 8), (9, 5), (5, 6)] -> 30.0
Racket
#lang racket/base
(struct P (x y))
(define (area . Ps)
(define (A P-a P-b)
(+ (for/sum ((p_i Ps)
(p_i+1 (in-sequences (cdr Ps)
(in-value (car Ps)))))
(* (P-a p_i) (P-b p_i+1)))))
(/ (abs (- (A P-x P-y) (A P-y P-x))) 2))
(module+ main
(area (P 3 4) (P 5 11) (P 12 8) (P 9 5) (P 5 6)))
{{out}}
30
REXX
endpoints as exceptions
/*REXX program uses a Shoelace formula to calculate the area of an N-sided polygon. */
parse arg pts; $polygon = 'polygon area of ' /*get optional args from the CL.*/
if pts='' then pts= '(3,4),(5,11),(12,8),(9,5),(5,6)' /*Not specified? Use default. */
@= pts /*elide extra blanks; save pts.*/
do #=1 until @='' /*perform destructive parse on @*/
parse var @ '(' x.# "," y.# ')' "," @ /*obtain X and Y coördinates*/
end /*#*/
A= 0 /*initialize the area to zero.*/
do j=1 for #; jp=j+1; if jp># then jp=1 /*adjust for J for overflow. */
jm=j-1; if jm==0 then jm=# /* " " " " underflow. */
A=A + x.j * (y.jp - y.jm) /*compute a part of the area. */
end /*j*/
say $polygon # " points: " pts ' is ───► ' abs(A/2) /*stick a fork in it, we're done*/
{{out|output|text= when using the default input:}}
polygon area of 5 points: (3,4),(5,11),(12,8),(9,5),(5,6) is ───► 30
===endpoints as wrap-around=== This REXX version uses a different method to define the '''X0, Y0''', and '''Xn+1''', '''Yn+1''' data points (and not treat them as exceptions).
When calculating the area for many polygons (or where the number of polygon sides is large), this method would be faster.
/*REXX program uses a Shoelace formula to calculate the area of an N-sided polygon.*/
parse arg pts; $polygon = 'polygon area of ' /*get optional args from the CL.*/
if pts='' then pts= "(3,4),(5,11),(12,8),(9,5),(5,6)" /*Not specified? Use default. */
A= 0; @= space(pts, 0) /*init A; elide blanks from pts.*/
do #=1 until @=='' /*perform destructive parse on @*/
parse var @ '(' x.# "," y.# ')' "," @ /*obtain X and Y coördinates*/
end /*n*/
e= #+1; parse value y.1 y.# with y.e y.0 /*define Y.n+1 & Y.0 points.*/
do j=1 for #; jm= j - 1; jp= j + 1 /*compute J-1 & J+1 indices.*/
A=A + x.j * (y.jm - y.jp) /*compute a portion of the area.*/
end /*j*/ /* [↓] use one half of | A | */
say $polygon # " points: " pts ' is ───► ' abs(A/2) /*stick a fork in it, we're done*/
{{out|output|text= is the same as the 1st REXX version.}}
somewhat simplified
reformatted and suitable for ooRexx. (x.0 etc. not needed)
{{out}}
```txt
polygon area of 5 points: (3,4),(5,11),(12,8),(9,5),(5,6) is ---> 30
even simpler
Using the published algorithm
{{out}}
```txt
polygon area of 5 points: (3,4),(5,11),(12,8),(9,5),(5,6) is ---> 30
Ring
# Project : Shoelace formula for polygonal area
p = [[3,4], [5,11], [12,8], [9,5], [5,6]]
see "The area of the polygon = " + shoelace(p)
func shoelace(p)
sum = 0
for i = 1 to len(p) -1
sum = sum + p[i][1] * p[i +1][2]
sum = sum - p[i +1][1] * p[i][2]
next
sum = sum + p[i][1] * p[1][2]
sum = sum - p[1][1] * p[i][2]
return fabs(sum) / 2
Output:
The area of the polygon = 30
Ruby
Point = Struct.new(:x,:y) do
def shoelace(other)
x * other.y - y * other.x
end
end
class Polygon
def initialize(*coords)
@points = coords.map{|c| Point.new(*c) }
end
def area
points = @points + [@points.first]
points.each_cons(2).sum{|p1,p2| p1.shoelace(p2) }.abs.fdiv(2)
end
end
puts Polygon.new([3,4], [5,11], [12,8], [9,5], [5,6]).area # => 30.0
Scala
case class Point( x:Int,y:Int ) { override def toString = "(" + x + "," + y + ")" }
case class Polygon( pp:List[Point] ) {
require( pp.size > 2, "A Polygon must consist of more than two points" )
override def toString = "Polygon(" + pp.mkString(" ", ", ", " ") + ")"
def area = {
// Calculate using the Shoelace Formula
val xx = pp.map( p => p.x )
val yy = pp.map( p => p.y )
val overlace = xx zip yy.drop(1)++yy.take(1)
val underlace = yy zip xx.drop(1)++xx.take(1)
(overlace.map( t => t._1 * t._2 ).sum - underlace.map( t => t._1 * t._2 ).sum).abs / 2.0
}
}
// A little test...
{
val p = Polygon( List( Point(3,4), Point(5,11), Point(12,8), Point(9,5), Point(5,6) ) )
assert( p.area == 30.0 )
println( "Area of " + p + " = " + p.area )
}
{{out}}
Area of Polygon( (3,4), (5,11), (12,8), (9,5), (5,6) ) = 30.0
Sidef
{{trans|Perl 6}}
func area_by_shoelace (*p) {
var x = p.map{_[0]}
var y = p.map{_[1]}
var s = (
(x ~Z* y.rotate(+1)).sum -
(x ~Z* y.rotate(-1)).sum
)
s.abs / 2
}
say area_by_shoelace([3,4], [5,11], [12,8], [9,5], [5,6])
{{out}}
30
Swift
{{trans|Scala}}
import Foundation
struct Point {
var x: Double
var y: Double
}
extension Point: CustomStringConvertible {
var description: String {
return "Point(x: \(x), y: \(y))"
}
}
struct Polygon {
var points: [Point]
var area: Double {
let xx = points.map({ $0.x })
let yy = points.map({ $0.y })
let overlace = zip(xx, yy.dropFirst() + yy.prefix(1)).map({ $0.0 * $0.1 }).reduce(0, +)
let underlace = zip(yy, xx.dropFirst() + xx.prefix(1)).map({ $0.0 * $0.1 }).reduce(0, +)
return abs(overlace - underlace) / 2
}
init(points: [Point]) {
self.points = points
}
init(points: [(Double, Double)]) {
self.init(points: points.map({ Point(x: $0.0, y: $0.1) }))
}
}
let poly = Polygon(points: [
(3,4),
(5,11),
(12,8),
(9,5),
(5,6)
])
print("\(poly) area = \(poly.area)")
{{out}}
Polygon(points: [Point(x: 3.0, y: 4.0), Point(x: 5.0, y: 11.0), Point(x: 12.0, y: 8.0), Point(x: 9.0, y: 5.0), Point(x: 5.0, y: 6.0)]) area = 30.0
=={{header|TI-83 BASIC}}== {{works with|TI-83 BASIC|TI-84Plus 2.55MP}}
[[3,4][5,11][12,8][9,5][5,6]]->[A]
Dim([A])->N:0->A
For(I,1,N)
I+1->J:If J>N:Then:1->J:End
A+[A](I,1)*[A](J,2)-[A](J,1)*[A](I,2)->A
End
Abs(A)/2->A
{{out}}
30
VBA
{{trans|Phix}}
Option Base 1
Public Enum axes
u = 1
v
End Enum
Private Function shoelace(s As Collection) As Double
Dim t As Double
If s.Count > 2 Then
s.Add s(1)
For i = 1 To s.Count - 1
t = t + s(i)(u) * s(i + 1)(v) - s(i + 1)(u) * s(i)(v)
Next i
End If
shoelace = Abs(t) / 2
End Function
Public Sub polygonal_area()
Dim task() As Variant
task = [{3,4;5,11;12,8;9,5;5,6}]
Dim tcol As New Collection
For i = 1 To UBound(task)
tcol.Add Array(task(i, u), task(i, v))
Next i
Debug.Print shoelace(tcol)
End Sub
{{out}}
30
VBScript
' Shoelace formula for polygonal area - VBScript
Dim points, x(),y()
points = Array(3,4, 5,11, 12,8, 9,5, 5,6)
n=(UBound(points)+1)\2
Redim x(n+1),y(n+1)
j=0
For i = 1 To n
x(i)=points(j)
y(i)=points(j+1)
j=j+2
Next 'i
x(i)=points(0)
y(i)=points(1)
For i = 1 To n
area = area + x(i)*y(i+1) - x(i+1)*y(i)
Next 'i
area = Abs(area)/2
msgbox area,,"Shoelace formula"
{{out}}
30
Visual Basic
{{works with|Visual Basic|5}} {{works with|Visual Basic|6}} {{works with|VBA|Access 97}} {{works with|VBA|6.5}} {{works with|VBA|7.1}}
Option Explicit
Public Function ShoelaceArea(x() As Double, y() As Double) As Double
Dim i As Long, j As Long
Dim Area As Double
j = UBound(x())
For i = LBound(x()) To UBound(x())
Area = Area + (y(j) + y(i)) * (x(j) - x(i))
j = i
Next i
ShoelaceArea = Abs(Area) / 2
End Function
Sub Main()
Dim v As Variant
Dim n As Long, i As Long, j As Long
v = Array(3, 4, 5, 11, 12, 8, 9, 5, 5, 6)
n = (UBound(v) - LBound(v) + 1) \ 2 - 1
ReDim x(0 To n) As Double, y(0 To n) As Double
j = 0
For i = 0 To n
x(i) = v(j)
y(i) = v(j + 1)
j = j + 2
Next i
Debug.Print ShoelaceArea(x(), y())
End Sub
{{out}}
30
Visual Basic .NET
{{trans|C#}}
Option Strict On
Imports Point = System.Tuple(Of Double, Double)
Module Module1
Function ShoelaceArea(v As List(Of Point)) As Double
Dim n = v.Count
Dim a = 0.0
For i = 0 To n - 2
a += v(i).Item1 * v(i + 1).Item2 - v(i + 1).Item1 * v(i).Item2
Next
Return Math.Abs(a + v(n - 1).Item1 * v(0).Item2 - v(0).Item1 * v(n - 1).Item2) / 2.0
End Function
Sub Main()
Dim v As New List(Of Point) From {
New Point(3, 4),
New Point(5, 11),
New Point(12, 8),
New Point(9, 5),
New Point(5, 6)
}
Dim area = ShoelaceArea(v)
Console.WriteLine("Given a polygon with vertices [{0}],", String.Join(", ", v))
Console.WriteLine("its area is {0}.", area)
End Sub
End Module
{{out}}
Given a polygon with vertices [(3, 4), (5, 11), (12, 8), (9, 5), (5, 6)],
its area is 30.
zkl
By the "book":
fcn areaByShoelace(points){ // ( (x,y),(x,y)...)
xs,ys:=Utils.Helpers.listUnzip(points); // (x,x,...), (y,y,,,)
( xs.zipWith('*,ys[1,*]).sum(0) + xs[-1]*ys[0] -
xs[1,*].zipWith('*,ys).sum(0) - xs[0]*ys[-1] )
.abs().toFloat()/2;
}
or an iterative solution:
fcn areaByShoelace2(points){ // ( (x,y),(x,y)...)
xs,ys:=Utils.Helpers.listUnzip(points); // (x,x,...), (y,y,,,)
N:=points.len();
N.reduce('wrap(s,n){ s + xs[n]*ys[(n+1)%N] - xs[(n+1)%N]*ys[n] },0)
.abs().toFloat()/2;
}
points:=T(T(3,4), T(5,11), T(12,8), T(9,5), T(5,6));
areaByShoelace(points).println();
areaByShoelace2(points).println();
{{out}}
30
30