Shoelace formula for polygonal area

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 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.

11l

Translation of: Python

<lang 11l>F area_by_shoelace(x, y)

  R abs(sum(zip(x, y[1..] [+] y[0.<1]).map((i, j) -> i * j))
       -sum(zip(x[1..] [+] x[0.<1], y).map((i, j) -> i * j))) / 2

V points = [(3, 4), (5, 11), (12, 8), (9, 5), (5, 6)] V x = points.map(p -> p[0]) V y = points.map(p -> p[1])

print(area_by_shoelace(x, y))</lang>

Output:

Ada

Works with: Ada version 83

<lang Ada>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;</lang>

Output:

 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

Output:

     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

Output:

     30.00

ALGOL 68

<lang algol68>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 </lang>

Output:

 30.00

APL

Works with: Dyalog APL

<lang APL>shoelace ← 2÷⍨|∘(((1⊃¨⊢)+.×1⌽2⊃¨⊢)-(1⌽1⊃¨⊢)+.×2⊃¨⊢)</lang>

Output:

      shoelace (3 4) (5 11) (12 8) (9 5) (5 6)
30

C

Reads the points from a file whose name is supplied via the command line, prints out usage if invoked incorrectly. <lang C>

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; } </lang> Input file, first line specifies number of points followed by the ordered vertices set with one vertex on each line.

Invocation and output :

C:\rosettaCode>shoelace.exe polyData.txt
The polygon area is 30.000000 square units.

C#

Translation of: Java

<lang csharp>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);
       }
   }

}</lang>

Output:

Given a polygon with vertices [(3, 4), (5, 11), (12, 8), (9, 5), (5, 6)],
its area is 30.

C++

Translation of: D

<lang cpp>#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; }</lang>

Output:

Cowgol

<lang cowgol>include "cowgol.coh";

typedef Coord is uint16; # floating point types are not supported

record Point is

   x: Coord;
   y: Coord;

end record;

sub shoelace(p: [Point], length: intptr): (area: Coord) is

   var left: Coord := 0;
   var right: Coord := 0;
   
   var y0 := p.y;
   var x0 := p.x;
   
   while length > 1 loop
       var xp := p.x;
       var yp := p.y;
       p := @next p;
       left := left + xp * p.y;
       right := right + yp * p.x;
       length := length - 1;
   end loop;
   
   left := left + y0 * p.x;
   right := right + x0 * p.y;
   if left < right then
       area := right - left;
   else
       area := left - right;
   end if;
   
   area := area / 2;

end sub;

var polygon: Point[] := {{3,4},{5,11},{12,8},{9,5},{5,6}}; print_i16(shoelace(&polygon[0], @sizeof polygon)); print_nl();</lang>

Output:

<lang fsharp> // 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)])</lang>

Output:

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. <lang factor>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 .</lang>

Output:

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 <lang 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</lang>

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: <lang fortran> 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)

</lang>

Input:

 5
  3.00  4.00 
  5.00 11.00
 12.00  8.00
  9.00  5.00
  5.00  6.00

Output:

     30.00

FreeBASIC

<lang 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</lang>

Output:

The area of the polygon = 30

Fōrmulæ

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Programs in Fōrmulæ are created/edited online in its website, However they run on execution servers. By default remote servers are used, but they are limited in memory and processing power, since they are intended for demonstration and casual use. A local server can be downloaded and installed, it has no limitations (it runs in your own computer). Because of that, example programs can be fully visualized and edited, but some of them will not run if they require a moderate or heavy computation/memory resources, and no local server is being used.

In this page you can see the program(s) related to this task and their results.

Go

<lang 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}}))

}</lang>

Output:

Haskell

<lang Haskell>main :: IO () main = print (shoelace [(3, 4), (5, 11), (12, 8), (9, 5), (5, 6)])

-- The area of a polygon formed by the list of (x, y) coordinates.

shoelace :: [(Double, Double)] -> Double shoelace =

 let calcSums ((xi, yi), (nxi, nyi)) (l, r) = (l + xi * nyi, r + nxi * yi)
 in (/ 2) .
    abs . uncurry (-) . foldr calcSums (0, 0) . (<*>) zip (tail . cycle)</lang>

Output:

30.0

J

Implementation:

<lang J>shoelace=:verb define

 0.5*|+/((* 1&|.)/ - (* _1&|.)/)|:y

)</lang>

Task example:

<lang J> shoelace 3 4,5 11,12 8,9 5,:5 6 30</lang>

Exposition:

We start with our list of coordinate pairs

<lang J> 3 4,5 11,12 8,9 5,:5 6

3  4
5 11

12 8

9  5
5  6</lang>

But the first thing we do is transpose them so that x coordinates and y coordinates are the two items we are working with:

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:

Or, rephrased:

We'll be subtracting what we get when we rotate in the other direction, which looks like this:

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

Translation of: Kotlin

Works with: Java version 9

<lang Java>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);
   }

}</lang>

Output:

Given a polygon with vertices [(3, 4), (5, 11), (12, 8), (9, 5), (5, 6)],
its area is 30.000000,

JavaScript

<lang javascript>(() => {

   'use strict';

   // shoelaceArea :: [(Float, Float)] -> Float
   const shoeLaceArea = vertices => abs(
       uncurry(subtract)(
           foldl(
               a => x => map(
                   b => a[int(b)] + x[0][int(b)] * x[1][int(!b)]
               )([0, 1])
           )([0, 0])(
               ap(zip)(compose(tail, cycle))(
                   vertices
               )
           )
       )
   ) / 2;

   // ----------------------- TEST -----------------------
   const main = () => {
       const ps = [
           [3, 4],
           [5, 11],
           [12, 8],
           [9, 5],
           [5, 6]
       ];
       return unlines([
           'Polygonal area by shoelace formula:',
           JSON.stringify(ps) + ' -> ' + shoeLaceArea(ps)
       ]);
   };

   // ---------------- GENERIC FUNCTIONS -----------------

   // abs :: Num -> Num
   const abs =
       // Absolute value of a given number - without the sign.
       Math.abs;

   // ap :: (a -> b -> c) -> (a -> b) -> a -> c
   const ap = f =>
       // Applicative instance for functions.
       // f(x) applied to g(x).
       g => x => f(x)(
           g(x)
       );

   // compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
   const compose = (...fs) =>
       fs.reduce(
           (f, g) => x => f(g(x)),
           x => x
       );

   // cycle :: [a] -> Generator [a]
   function* cycle(xs) {
       const lng = xs.length;
       let i = 0;
       while (true) {
           yield(xs[i])
           i = (1 + i) % lng;
       }
   }

   // foldl :: (a -> b -> a) -> a -> [b] -> a
   const foldl = f =>
       a => xs => xs.reduce((x, y) => f(x)(y), a);

   // int :: Bool -> Int
   const int = bln =>
       bln ? (
           1
       ) : 0;

   // length :: [a] -> Int
   const length = xs =>
       // Returns Infinity over objects without finite
       // length. This enables zip and zipWith to choose
       // the shorter argument when one is non-finite,
       // like cycle, repeat etc
       (Array.isArray(xs) || 'string' === typeof xs) ? (
           xs.length
       ) : Infinity;

   // map :: (a -> b) -> [a] -> [b]
   const map = f =>
       // The list obtained by applying f
       // to each element of xs.
       // (The image of xs under f).
       xs => (
           Array.isArray(xs) ? (
               xs
           ) : xs.split()
       ).map(f);

   // subtract :: Num -> Num -> Num
   const subtract = x =>
       y => y - x;

   // tail :: [a] -> [a]
   const tail = xs =>
       // A new list consisting of all
       // items of xs except the first.
       'GeneratorFunction' !== xs.constructor.constructor.name ? (
           0 < xs.length ? xs.slice(1) : []
       ) : (take(1)(xs), xs);

   // take :: Int -> [a] -> [a]
   // take :: Int -> String -> String
   const take = n =>
       // The first n elements of a list,
       // string of characters, or stream.
       xs => 'GeneratorFunction' !== xs
       .constructor.constructor.name ? (
           xs.slice(0, n)
       ) : [].concat.apply([], Array.from({
           length: n
       }, () => {
           const x = xs.next();
           return x.done ? [] : [x.value];
       }));

   // uncurry :: (a -> b -> c) -> ((a, b) -> c)
   const uncurry = f =>
       // A function over a pair, derived
       // from a curried function.
       function() {
           const
               args = arguments,
               xy = Boolean(args.length % 2) ? (
                   args[0]
               ) : args;
           return f(xy[0])(xy[1]);
       };

   // unlines :: [String] -> String
   const unlines = xs =>
       // A single string formed by the intercalation
       // of a list of strings with the newline character.
       xs.join('\n');

   // zip :: [a] -> [b] -> [(a, b)]
   const zip = xs =>
       // Use of `take` and `length` here allows for zipping with non-finite
       // lists - i.e. generators like cycle, repeat, iterate.
       ys => {
           const
               lng = Math.min(length(xs), length(ys)),
               vs = take(lng)(ys);
           return take(lng)(xs).map(
               (x, i) => [x, vs[i]]
           );
       };

   // MAIN ---
   return main();

})();</lang>

Output:

Polygonal area by shoelace formula:
[[3,4],[5,11],[12,8],[9,5],[5,6]] -> 30

Julia

Works with: Julia version 0.6

Translation of: Python

<lang julia>""" 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)</lang>

Output:

x = [3, 5, 12, 9, 5]
y = [4, 11, 8, 5, 6]
shoelacearea(x, y) = 30.0

Kotlin

<lang scala>// 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")

}</lang>

Output:

Given a polygon with vertices at [(3, 4), (5, 11), (12, 8), (9, 5), (5, 6)],
its area is 30.0

Lambdatalk

<lang scheme> {def shoelace

{lambda {:pol}
 {abs
  {/ 
   {- 
    {+ {S.map {{lambda {:pol :i} {* {car {A.get :i :pol}}
                                    {cdr {A.get {+ :i 1} :pol}}}} :pol}
              {S.serie 0 {- {A.length :pol} 2}}}
                                 {* {car {A.get {- {A.length :pol} 1} :pol}}
                                    {cdr {A.get 0 :pol}}}}
    {+ {S.map {{lambda {:pol :i} {* {car {A.get {+ :i 1} :pol}}
                                    {cdr {A.get :i :pol}}}} :pol}
              {S.serie 0 {- {A.length :pol} 2}}} 
                                 {* {car {A.get 0 :pol}}
                                    {cdr {A.get {- {A.length :pol} 1} :pol}}}}} 2}}}}

-> shoelace

{def pol

 {A.new {cons 3 4}
        {cons 5 11}
        {cons 12 8}
        {cons 9 5}
        {cons 5 6}}}

-> pol = [(3 4),(5 11),(12 8),(9 5),(5 6)]

{shoelace {pol}} -> 30 </lang>

Lua

<lang 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</lang> Using an accumulator helper inner function <lang lua>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 </lang> both version handle special cases of less than 3 point as 0 area result.

Maple

<lang Maple> with(ArrayTools):

module Point()

option object;
local x := 0;
local y := 0;

export getX::static := proc(self::Point, $)
 return self:-x;
end proc;

export getY::static := proc(self::Point, $)
 return self:-y
end proc;

export ModuleApply::static := proc()
 Object(Point, _passed);
end proc;

export ModuleCopy::static := proc(new::Point, proto::Point, X, Y, $)
 new:-x := X;
 new:-y := Y;
end proc;

export ModulePrint::static := proc(self::Point)
 return cat("(", self:-x, ",", self:-y, ")");
end proc;

end module:

module Polygon()

option object;
local vertices := Array([Point(0,0)]);

export getVertices::static := proc(self::Polygon)
 return self:-vertices;
end proc;

export area::static := proc(self::Polygon)
 local i, N := ArrayNumElems(self:-vertices);
 local total := getX(self:-vertices[N]) * getY(self:-vertices[1]) - getX(self:-vertices[1]) * getY(self:-vertices[N]);
 total += map(`+`, seq(getX(self:-vertices[i]) * getY(self:-vertices[i+1]), i = 1..(N-1))) - map(`+`, seq(getX(self:-vertices[i+1]) * getY(self:-vertices[i]), i = 1..(N-1)));
 return abs(total / 2);
end proc;

export ModuleApply::static := proc()
 Object(Polygon, _passed);
end proc;

export ModuleCopy::Static := proc(new::Polygon, proto::Polygon, Ps, $)
 new:-vertices := Ps;
end proc;

export ModulePrint::static := proc(self::Polygon)
 return self:-vertices;
end proc;

end module:

P1 := Polygon(Array([Point(3,4), Point(5,11), Point(12,8), Point(9,5), Point(5,6)])): area(P1); </lang>

Output:

Mathematica/Wolfram Language

Geometry objects built-in in the Wolfram Language <lang Mathematica>Area[Polygon[{{3, 4}, {5, 11}, {12, 8}, {9, 5}, {5, 6}}]]</lang>

Output:

min

Works with: min version 0.19.3

<lang min>((((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</lang>

Output:

30.0

MiniScript

<lang 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) </lang>

Output:

The polygon area is 30

Modula-2

<lang modula2>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.</lang>

Nim

<lang nim>type

 Point = tuple
   x: float
   y: float

func shoelace(points: openArray[Point]): float =

 var leftSum, rightSum = 0.0
 for i in 0..<len(points):
   var j = (i + 1) mod len(points)
   leftSum  += points[i].x * points[j].y
   rightSum += points[j].x * points[i].y
 0.5 * abs(leftSum - rightSum)

var points = [(3.0, 4.0), (5.0, 11.0), (12.0, 8.0), (9.0, 5.0), (5.0, 6.0)]

echo shoelace(points)</lang>

Output:

30.0

Perl

<lang 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 );</lang>

Output:

Phix

<lang 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)</lang>

Output:

An alternative solution, which does not need the X,Y enum, and gives the same output: <lang Phix>function shoelace(sequence s)

   atom t = 0
   integer j = length(s)
   if j!=0 then
       sequence {x,y} = columnize(s)
       for i=1 to j do
           t += (y[j] + y[i]) * (x[j] - x[i])
           j = i
       end for
   end if
   return abs(t)/2

end function</lang>

PowerBASIC

Translation of: Visual Basic

<lang powerbasic>#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</lang>

Output:

Python

<lang 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 >>>

Even simpler:
In python we can take an advantage of that x[-1] refers to the last element in an array, same as x[N-1].
Introducing the index i=[0,1,2,...,N-1]; i-1=[-1,0,...,N-2]; N is the number of vertices of a polygon.
Thus x[i] is a sequence of the x-coordinate of the polygon vertices, x[i-1] is the sequence shifted by 1 index.
Note that the shift must be negative. The positive shift x[i+1] results in an error: x[N] index out of bound.

import numpy as np

x,y are arrays containing coordinates of the polygon vertices

x=np.array([3,5,12,9,5]) y=np.array([4,11,8,5,6]) i=np.arange(len(x))

Area=np.sum(x[i-1]*y[i]-x[i]*y[i-1])*0.5 # signed area, positive if the vertex sequence is counterclockwise

Area=np.abs(np.sum(x[i-1]*y[i]-x[i]*y[i-1])*0.5) # one line of code for the shoelace formula

Remember that applying the Shoelace formula
will result in a loss of precision if x,y have big offsets.
Remove the offsets first, e.g.
x=x-np.mean(x);y=y-np.mean(y)
or
x=x-x[0];y=y-y[0]
before applying the Shoelace formula.

</lang>

Or, defined in terms of reduce and cycle:

Translation of: Haskell

Works with: Python version 3.7

<lang python>Polygonal area by shoelace formula

from itertools import cycle, islice from functools import reduce from operator import sub

--------- SHOELACE FORMULA FOR POLYGONAL AREA ----------

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

   return abs(sub(*reduce(
       go,
       zip(
           xys,
           islice(cycle(xys), 1, None)
       ),
       (0, 0)
   ))) / 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)))

if __name__ == '__main__':

   main()</lang>

Output:

Polygonal area by shoelace formula:
[(3, 4), (5, 11), (12, 8), (9, 5), (5, 6)]  ->  30.0

Racket

<lang 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)))</lang>

Output:

Raku

(formerly Perl 6)

Index and mod offset

Works with: Rakudo version 2017.07

<lang perl6>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) ] );</lang>

Output:

Slice and rotation

Works with: Rakudo version 2017.07

<lang perl6>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) ] ); </lang>

Output:

REXX

wrap-around endpoints

<lang rexx>/*REXX program uses a Shoelace formula to calculate the area of an N─sided polygon.*/ parse arg $; if $= then $= "(3,4),(5,11),(12,8),(9,5),(5,6)" /*Use the default?*/ A= 0; @= space($, 0) /*init A; elide blanks from pts.*/

        do #=1  until @==;      parse var  @    '('   x.#   ","   y.#   ')'   ","   @
        end   /*#*/                                    /* [↨]  get X and Y coördinates.*/

z= #+1; y.0= y.#; y.z= y.1 /*define low & high Y end points*/

        do j=1  for #;  jm= j-1;  jp= j+1;   A= A + x.j*(y.jm - y.jp) /*portion of area*/
        end   /*j*/                                    /*stick a fork in it, we're done*/

say 'polygon area of ' # " points: " $ ' is ───► ' abs(A/2)</lang>

output when using the default input:

polygon area of  5  points:  (3,4),(5,11),(12,8),(9,5),(5,6)   is ───►  30

somewhat simplified

reformatted and suitable for ooRexx. (x.0 etc. not needed) <lang>/*REXX program uses a Shoelace formula to calculate the area of an N-sided polygon. */ parse arg pts /*obtain optional arguments from the CL*/ if pts= then pts= '(3,4),(5,11),(12,8),(9,5),(5,6)' /*Not specified? Use default. */ pts=space(pts,0); z=pts /*elide extra blanks; save pts.*/ do n=1 until z= /*perform destructive parse on z*/

 parse var z '(' x.n ',' y.n ')' ',' z                 /*obtain X and Y coördinates    */
 end

z=n+1; y.z=y.1 /* take care of end points */

      y.0=y.n

A=0 /*initialize the area to zero.*/ do j=1 for n;

 jp=j+1;
 jm=j-1;
 A=A+x.j*(y.jp-y.jm)                                   /*compute a part of the area.   */
 end

A=abs(A/2) /*obtain half of the ¦ A ¦ sum*/ say 'polygon area of' n 'points:' pts 'is --->' A</lang>

Output:

polygon area of 5 points: (3,4),(5,11),(12,8),(9,5),(5,6) is ---> 30

even simpler

Using the published algorithm <lang>/*REXX program uses a Shoelace formula to calculate the area of an N-sided polygon. */ parse arg pts /*obtain optional arguments from the CL*/ if pts= then pts= '(3,4),(5,11),(12,8),(9,5),(5,6)' /*Not specified? Use default. */ pts=space(pts,0); z=pts /*elide extra blanks; save pts.*/ do n=1 until z= /*perform destructive parse on z*/

 parse var z '(' x.n ',' y.n ')' ',' z                 /*obtain X and Y coördinates    */
 end

a=0 Do i=1 To n-1

 j=i+1
 a=a+x.i*y.j-x.j*y.i
 End

a=a+x.n*y.1-x.1*y.n a=abs(a)/2 say 'polygon area of' n 'points:' pts 'is --->' a</lang>

Output:

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

</lang> Output:

The area of the polygon = 30

Ruby

<lang 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 </lang>

Scala

<lang 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 ) } </lang>

Output:

Area of Polygon( (3,4), (5,11), (12,8), (9,5), (5,6) ) = 30.0

Sidef

Translation of: Raku

<lang ruby>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])</lang>

Output:

Swift

Translation of: Scala

<lang swift>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)")</lang>

Output:

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

TI-83 BASIC

Works with: TI-83 BASIC version TI-84Plus 2.55MP

<lang ti83b>[[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</lang>

Output:

VBA

Translation of: Phix

<lang vb>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</lang>

Output:

VBScript

<lang vb>' 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" </lang>

Output:

Visual Basic

Works with: Visual Basic version 5

Works with: Visual Basic version 6

Works with: VBA version Access 97

Works with: VBA version 6.5

Works with: VBA version 7.1

<lang vb>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</lang>

Output:

Visual Basic .NET

Translation of: C#

<lang vbnet>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</lang>

Output:

Given a polygon with vertices [(3, 4), (5, 11), (12, 8), (9, 5), (5, 6)],
its area is 30.

Wren

<lang ecmascript>var shoelace = Fn.new { |pts|

   var area = 0
   for (i in 0...pts.count-1) {
       area = area + pts[i][0]*pts[i+1][1] - pts[i+1][0]*pts[i][1]
   }
   return (area + pts[-1][0]*pts[0][1] - pts[0][0]*pts[-1][1]).abs / 2

}

var pts = [ [3, 4], [5, 11], [12, 8], [9, 5], [5, 6] ] System.print("The polygon with vertices at %(pts) has an area of %(shoelace.call(pts)).")</lang>

Output:

The polygon with vertices at [[3, 4], [5, 11], [12, 8], [9, 5], [5, 6]] has an area of 30.

XPL0

<lang XPL0>proc real Shoelace(N, X, Y); int N, X, Y; int S, I; [S:= 0; for I:= 0 to N-2 do

       S:= S + X(I)*Y(I+1) - X(I+1)*Y(I);

S:= S + X(I)*Y(0) - X(0)*Y(I); return float(abs(S)) / 2.0; ];

RlOut(0, Shoelace(5, [3, 5, 12, 9, 5], [4, 11, 8, 5, 6]))</lang>

Output:

   30.00000

zkl

By the "book": <lang zkl>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;

}</lang> or an iterative solution: <lang zkl>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;

}</lang> <lang zkl>points:=T(T(3,4), T(5,11), T(12,8), T(9,5), T(5,6)); areaByShoelace(points).println(); areaByShoelace2(points).println();</lang>

Output:

30
30