Instruction stringlengths 45 106 | input_code stringlengths 1 13.7k | output_code stringlengths 1 13.7k |
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Produce a language-to-language conversion: from Factor to Java, same semantics. | 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 .
| 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);
}
}
|
Produce a functionally identical Python code for the snippet given in 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 .
| >>> 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
>>>
|
Ensure the translated Python code behaves exactly like the original Factor snippet. | 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 .
| >>> 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
>>>
|
Translate the given Factor code snippet into VB without altering its behavior. | 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 .
| 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
|
Write a version of this Factor function in VB with identical behavior. | 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 .
| 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
|
Translate this program into Go but keep the logic exactly as in 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 .
| 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}}))
}
|
Please provide an equivalent version of this Factor code in Go. | 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 .
| 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}}))
}
|
Write the same algorithm in C# as shown in this Fortran implementation. | DOUBLE PRECISION FUNCTION AREA(N,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
DOUBLE COMPLEX P(N)
DOUBLE COMPLEX PP,PC
DOUBLE COMPLEX W
DOUBLE PRECISION A
INTEGER I
IF (N.LT.3) STOP "Area: at least three points are needed
W = (P(1) + P(N/3) + P(2*N/3))/3
W = SUM(P(1:N) - W)/N + W
A = 0
PC = P(N) - W
DO I = 1,N
PP = PC
PC = P(I) - W
A = (DIMAG(PC) + DIMAG(PP))*(DBLE(PC) - DBLE(PP)) + A
END DO
AREA = A/2
END FUNCTION AREA
DOUBLE PRECISION FUNCTION AREASL(N,P)
INTEGER N
DOUBLE COMPLEX P(N)
DOUBLE PRECISION A
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
END FUNCTION AREASL
INTEGER ENUFF
DOUBLE PRECISION AREA,AREASL
DOUBLE PRECISION A1,A2
PARAMETER (ENUFF = 5)
DOUBLE COMPLEX POINT(ENUFF)
DATA POINT/(3D0,4D0),(5D0,11D0),(12D0,8D0),(9D0,5D0),(5D0,6D0)/
WRITE (6,*) POINT
A1 = AREA(5,POINT)
A2 = AREASL(5,POINT)
WRITE (6,*) "A=",A1,A2
END
| 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);
}
}
}
|
Translate this program into C# but keep the logic exactly as in Fortran. | DOUBLE PRECISION FUNCTION AREA(N,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
DOUBLE COMPLEX P(N)
DOUBLE COMPLEX PP,PC
DOUBLE COMPLEX W
DOUBLE PRECISION A
INTEGER I
IF (N.LT.3) STOP "Area: at least three points are needed
W = (P(1) + P(N/3) + P(2*N/3))/3
W = SUM(P(1:N) - W)/N + W
A = 0
PC = P(N) - W
DO I = 1,N
PP = PC
PC = P(I) - W
A = (DIMAG(PC) + DIMAG(PP))*(DBLE(PC) - DBLE(PP)) + A
END DO
AREA = A/2
END FUNCTION AREA
DOUBLE PRECISION FUNCTION AREASL(N,P)
INTEGER N
DOUBLE COMPLEX P(N)
DOUBLE PRECISION A
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
END FUNCTION AREASL
INTEGER ENUFF
DOUBLE PRECISION AREA,AREASL
DOUBLE PRECISION A1,A2
PARAMETER (ENUFF = 5)
DOUBLE COMPLEX POINT(ENUFF)
DATA POINT/(3D0,4D0),(5D0,11D0),(12D0,8D0),(9D0,5D0),(5D0,6D0)/
WRITE (6,*) POINT
A1 = AREA(5,POINT)
A2 = AREASL(5,POINT)
WRITE (6,*) "A=",A1,A2
END
| 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);
}
}
}
|
Change the following Fortran code into C++ without altering its purpose. | DOUBLE PRECISION FUNCTION AREA(N,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
DOUBLE COMPLEX P(N)
DOUBLE COMPLEX PP,PC
DOUBLE COMPLEX W
DOUBLE PRECISION A
INTEGER I
IF (N.LT.3) STOP "Area: at least three points are needed
W = (P(1) + P(N/3) + P(2*N/3))/3
W = SUM(P(1:N) - W)/N + W
A = 0
PC = P(N) - W
DO I = 1,N
PP = PC
PC = P(I) - W
A = (DIMAG(PC) + DIMAG(PP))*(DBLE(PC) - DBLE(PP)) + A
END DO
AREA = A/2
END FUNCTION AREA
DOUBLE PRECISION FUNCTION AREASL(N,P)
INTEGER N
DOUBLE COMPLEX P(N)
DOUBLE PRECISION A
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
END FUNCTION AREASL
INTEGER ENUFF
DOUBLE PRECISION AREA,AREASL
DOUBLE PRECISION A1,A2
PARAMETER (ENUFF = 5)
DOUBLE COMPLEX POINT(ENUFF)
DATA POINT/(3D0,4D0),(5D0,11D0),(12D0,8D0),(9D0,5D0),(5D0,6D0)/
WRITE (6,*) POINT
A1 = AREA(5,POINT)
A2 = AREASL(5,POINT)
WRITE (6,*) "A=",A1,A2
END
| #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;
}
|
Change the following Fortran code into C++ without altering its purpose. | DOUBLE PRECISION FUNCTION AREA(N,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
DOUBLE COMPLEX P(N)
DOUBLE COMPLEX PP,PC
DOUBLE COMPLEX W
DOUBLE PRECISION A
INTEGER I
IF (N.LT.3) STOP "Area: at least three points are needed
W = (P(1) + P(N/3) + P(2*N/3))/3
W = SUM(P(1:N) - W)/N + W
A = 0
PC = P(N) - W
DO I = 1,N
PP = PC
PC = P(I) - W
A = (DIMAG(PC) + DIMAG(PP))*(DBLE(PC) - DBLE(PP)) + A
END DO
AREA = A/2
END FUNCTION AREA
DOUBLE PRECISION FUNCTION AREASL(N,P)
INTEGER N
DOUBLE COMPLEX P(N)
DOUBLE PRECISION A
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
END FUNCTION AREASL
INTEGER ENUFF
DOUBLE PRECISION AREA,AREASL
DOUBLE PRECISION A1,A2
PARAMETER (ENUFF = 5)
DOUBLE COMPLEX POINT(ENUFF)
DATA POINT/(3D0,4D0),(5D0,11D0),(12D0,8D0),(9D0,5D0),(5D0,6D0)/
WRITE (6,*) POINT
A1 = AREA(5,POINT)
A2 = AREASL(5,POINT)
WRITE (6,*) "A=",A1,A2
END
| #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;
}
|
Write the same code in C as shown below in Fortran. | DOUBLE PRECISION FUNCTION AREA(N,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
DOUBLE COMPLEX P(N)
DOUBLE COMPLEX PP,PC
DOUBLE COMPLEX W
DOUBLE PRECISION A
INTEGER I
IF (N.LT.3) STOP "Area: at least three points are needed
W = (P(1) + P(N/3) + P(2*N/3))/3
W = SUM(P(1:N) - W)/N + W
A = 0
PC = P(N) - W
DO I = 1,N
PP = PC
PC = P(I) - W
A = (DIMAG(PC) + DIMAG(PP))*(DBLE(PC) - DBLE(PP)) + A
END DO
AREA = A/2
END FUNCTION AREA
DOUBLE PRECISION FUNCTION AREASL(N,P)
INTEGER N
DOUBLE COMPLEX P(N)
DOUBLE PRECISION A
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
END FUNCTION AREASL
INTEGER ENUFF
DOUBLE PRECISION AREA,AREASL
DOUBLE PRECISION A1,A2
PARAMETER (ENUFF = 5)
DOUBLE COMPLEX POINT(ENUFF)
DATA POINT/(3D0,4D0),(5D0,11D0),(12D0,8D0),(9D0,5D0),(5D0,6D0)/
WRITE (6,*) POINT
A1 = AREA(5,POINT)
A2 = AREASL(5,POINT)
WRITE (6,*) "A=",A1,A2
END
| #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;
}
|
Write the same code in C as shown below in Fortran. | DOUBLE PRECISION FUNCTION AREA(N,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
DOUBLE COMPLEX P(N)
DOUBLE COMPLEX PP,PC
DOUBLE COMPLEX W
DOUBLE PRECISION A
INTEGER I
IF (N.LT.3) STOP "Area: at least three points are needed
W = (P(1) + P(N/3) + P(2*N/3))/3
W = SUM(P(1:N) - W)/N + W
A = 0
PC = P(N) - W
DO I = 1,N
PP = PC
PC = P(I) - W
A = (DIMAG(PC) + DIMAG(PP))*(DBLE(PC) - DBLE(PP)) + A
END DO
AREA = A/2
END FUNCTION AREA
DOUBLE PRECISION FUNCTION AREASL(N,P)
INTEGER N
DOUBLE COMPLEX P(N)
DOUBLE PRECISION A
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
END FUNCTION AREASL
INTEGER ENUFF
DOUBLE PRECISION AREA,AREASL
DOUBLE PRECISION A1,A2
PARAMETER (ENUFF = 5)
DOUBLE COMPLEX POINT(ENUFF)
DATA POINT/(3D0,4D0),(5D0,11D0),(12D0,8D0),(9D0,5D0),(5D0,6D0)/
WRITE (6,*) POINT
A1 = AREA(5,POINT)
A2 = AREASL(5,POINT)
WRITE (6,*) "A=",A1,A2
END
| #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;
}
|
Change the programming language of this snippet from Fortran to Java without modifying what it does. | DOUBLE PRECISION FUNCTION AREA(N,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
DOUBLE COMPLEX P(N)
DOUBLE COMPLEX PP,PC
DOUBLE COMPLEX W
DOUBLE PRECISION A
INTEGER I
IF (N.LT.3) STOP "Area: at least three points are needed
W = (P(1) + P(N/3) + P(2*N/3))/3
W = SUM(P(1:N) - W)/N + W
A = 0
PC = P(N) - W
DO I = 1,N
PP = PC
PC = P(I) - W
A = (DIMAG(PC) + DIMAG(PP))*(DBLE(PC) - DBLE(PP)) + A
END DO
AREA = A/2
END FUNCTION AREA
DOUBLE PRECISION FUNCTION AREASL(N,P)
INTEGER N
DOUBLE COMPLEX P(N)
DOUBLE PRECISION A
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
END FUNCTION AREASL
INTEGER ENUFF
DOUBLE PRECISION AREA,AREASL
DOUBLE PRECISION A1,A2
PARAMETER (ENUFF = 5)
DOUBLE COMPLEX POINT(ENUFF)
DATA POINT/(3D0,4D0),(5D0,11D0),(12D0,8D0),(9D0,5D0),(5D0,6D0)/
WRITE (6,*) POINT
A1 = AREA(5,POINT)
A2 = AREASL(5,POINT)
WRITE (6,*) "A=",A1,A2
END
| 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);
}
}
|
Produce a functionally identical Java code for the snippet given in Fortran. | DOUBLE PRECISION FUNCTION AREA(N,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
DOUBLE COMPLEX P(N)
DOUBLE COMPLEX PP,PC
DOUBLE COMPLEX W
DOUBLE PRECISION A
INTEGER I
IF (N.LT.3) STOP "Area: at least three points are needed
W = (P(1) + P(N/3) + P(2*N/3))/3
W = SUM(P(1:N) - W)/N + W
A = 0
PC = P(N) - W
DO I = 1,N
PP = PC
PC = P(I) - W
A = (DIMAG(PC) + DIMAG(PP))*(DBLE(PC) - DBLE(PP)) + A
END DO
AREA = A/2
END FUNCTION AREA
DOUBLE PRECISION FUNCTION AREASL(N,P)
INTEGER N
DOUBLE COMPLEX P(N)
DOUBLE PRECISION A
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
END FUNCTION AREASL
INTEGER ENUFF
DOUBLE PRECISION AREA,AREASL
DOUBLE PRECISION A1,A2
PARAMETER (ENUFF = 5)
DOUBLE COMPLEX POINT(ENUFF)
DATA POINT/(3D0,4D0),(5D0,11D0),(12D0,8D0),(9D0,5D0),(5D0,6D0)/
WRITE (6,*) POINT
A1 = AREA(5,POINT)
A2 = AREASL(5,POINT)
WRITE (6,*) "A=",A1,A2
END
| 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);
}
}
|
Write the same code in Python as shown below in Fortran. | DOUBLE PRECISION FUNCTION AREA(N,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
DOUBLE COMPLEX P(N)
DOUBLE COMPLEX PP,PC
DOUBLE COMPLEX W
DOUBLE PRECISION A
INTEGER I
IF (N.LT.3) STOP "Area: at least three points are needed
W = (P(1) + P(N/3) + P(2*N/3))/3
W = SUM(P(1:N) - W)/N + W
A = 0
PC = P(N) - W
DO I = 1,N
PP = PC
PC = P(I) - W
A = (DIMAG(PC) + DIMAG(PP))*(DBLE(PC) - DBLE(PP)) + A
END DO
AREA = A/2
END FUNCTION AREA
DOUBLE PRECISION FUNCTION AREASL(N,P)
INTEGER N
DOUBLE COMPLEX P(N)
DOUBLE PRECISION A
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
END FUNCTION AREASL
INTEGER ENUFF
DOUBLE PRECISION AREA,AREASL
DOUBLE PRECISION A1,A2
PARAMETER (ENUFF = 5)
DOUBLE COMPLEX POINT(ENUFF)
DATA POINT/(3D0,4D0),(5D0,11D0),(12D0,8D0),(9D0,5D0),(5D0,6D0)/
WRITE (6,*) POINT
A1 = AREA(5,POINT)
A2 = AREASL(5,POINT)
WRITE (6,*) "A=",A1,A2
END
| >>> 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
>>>
|
Convert this Fortran block to Python, preserving its control flow and logic. | DOUBLE PRECISION FUNCTION AREA(N,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
DOUBLE COMPLEX P(N)
DOUBLE COMPLEX PP,PC
DOUBLE COMPLEX W
DOUBLE PRECISION A
INTEGER I
IF (N.LT.3) STOP "Area: at least three points are needed
W = (P(1) + P(N/3) + P(2*N/3))/3
W = SUM(P(1:N) - W)/N + W
A = 0
PC = P(N) - W
DO I = 1,N
PP = PC
PC = P(I) - W
A = (DIMAG(PC) + DIMAG(PP))*(DBLE(PC) - DBLE(PP)) + A
END DO
AREA = A/2
END FUNCTION AREA
DOUBLE PRECISION FUNCTION AREASL(N,P)
INTEGER N
DOUBLE COMPLEX P(N)
DOUBLE PRECISION A
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
END FUNCTION AREASL
INTEGER ENUFF
DOUBLE PRECISION AREA,AREASL
DOUBLE PRECISION A1,A2
PARAMETER (ENUFF = 5)
DOUBLE COMPLEX POINT(ENUFF)
DATA POINT/(3D0,4D0),(5D0,11D0),(12D0,8D0),(9D0,5D0),(5D0,6D0)/
WRITE (6,*) POINT
A1 = AREA(5,POINT)
A2 = AREASL(5,POINT)
WRITE (6,*) "A=",A1,A2
END
| >>> 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
>>>
|
Maintain the same structure and functionality when rewriting this code in VB. | DOUBLE PRECISION FUNCTION AREA(N,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
DOUBLE COMPLEX P(N)
DOUBLE COMPLEX PP,PC
DOUBLE COMPLEX W
DOUBLE PRECISION A
INTEGER I
IF (N.LT.3) STOP "Area: at least three points are needed
W = (P(1) + P(N/3) + P(2*N/3))/3
W = SUM(P(1:N) - W)/N + W
A = 0
PC = P(N) - W
DO I = 1,N
PP = PC
PC = P(I) - W
A = (DIMAG(PC) + DIMAG(PP))*(DBLE(PC) - DBLE(PP)) + A
END DO
AREA = A/2
END FUNCTION AREA
DOUBLE PRECISION FUNCTION AREASL(N,P)
INTEGER N
DOUBLE COMPLEX P(N)
DOUBLE PRECISION A
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
END FUNCTION AREASL
INTEGER ENUFF
DOUBLE PRECISION AREA,AREASL
DOUBLE PRECISION A1,A2
PARAMETER (ENUFF = 5)
DOUBLE COMPLEX POINT(ENUFF)
DATA POINT/(3D0,4D0),(5D0,11D0),(12D0,8D0),(9D0,5D0),(5D0,6D0)/
WRITE (6,*) POINT
A1 = AREA(5,POINT)
A2 = AREASL(5,POINT)
WRITE (6,*) "A=",A1,A2
END
| 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
|
Write the same algorithm in VB as shown in this Fortran implementation. | DOUBLE PRECISION FUNCTION AREA(N,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
DOUBLE COMPLEX P(N)
DOUBLE COMPLEX PP,PC
DOUBLE COMPLEX W
DOUBLE PRECISION A
INTEGER I
IF (N.LT.3) STOP "Area: at least three points are needed
W = (P(1) + P(N/3) + P(2*N/3))/3
W = SUM(P(1:N) - W)/N + W
A = 0
PC = P(N) - W
DO I = 1,N
PP = PC
PC = P(I) - W
A = (DIMAG(PC) + DIMAG(PP))*(DBLE(PC) - DBLE(PP)) + A
END DO
AREA = A/2
END FUNCTION AREA
DOUBLE PRECISION FUNCTION AREASL(N,P)
INTEGER N
DOUBLE COMPLEX P(N)
DOUBLE PRECISION A
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
END FUNCTION AREASL
INTEGER ENUFF
DOUBLE PRECISION AREA,AREASL
DOUBLE PRECISION A1,A2
PARAMETER (ENUFF = 5)
DOUBLE COMPLEX POINT(ENUFF)
DATA POINT/(3D0,4D0),(5D0,11D0),(12D0,8D0),(9D0,5D0),(5D0,6D0)/
WRITE (6,*) POINT
A1 = AREA(5,POINT)
A2 = AREASL(5,POINT)
WRITE (6,*) "A=",A1,A2
END
| 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
|
Keep all operations the same but rewrite the snippet in C. | import Data.Bifunctor (bimap)
shoelace :: [(Double, Double)] -> Double
shoelace =
let calcSums ((x, y), (a, b)) = bimap (x * b +) (a * y +)
in (/ 2)
. abs
. uncurry (-)
. foldr calcSums (0, 0)
. (<*>) zip (tail . cycle)
main :: IO ()
main =
print $
shoelace [(3, 4), (5, 11), (12, 8), (9, 5), (5, 6)]
| #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;
}
|
Ensure the translated C code behaves exactly like the original Haskell snippet. | import Data.Bifunctor (bimap)
shoelace :: [(Double, Double)] -> Double
shoelace =
let calcSums ((x, y), (a, b)) = bimap (x * b +) (a * y +)
in (/ 2)
. abs
. uncurry (-)
. foldr calcSums (0, 0)
. (<*>) zip (tail . cycle)
main :: IO ()
main =
print $
shoelace [(3, 4), (5, 11), (12, 8), (9, 5), (5, 6)]
| #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;
}
|
Ensure the translated C# code behaves exactly like the original Haskell snippet. | import Data.Bifunctor (bimap)
shoelace :: [(Double, Double)] -> Double
shoelace =
let calcSums ((x, y), (a, b)) = bimap (x * b +) (a * y +)
in (/ 2)
. abs
. uncurry (-)
. foldr calcSums (0, 0)
. (<*>) zip (tail . cycle)
main :: IO ()
main =
print $
shoelace [(3, 4), (5, 11), (12, 8), (9, 5), (5, 6)]
| 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);
}
}
}
|
Convert this Haskell snippet to C# and keep its semantics consistent. | import Data.Bifunctor (bimap)
shoelace :: [(Double, Double)] -> Double
shoelace =
let calcSums ((x, y), (a, b)) = bimap (x * b +) (a * y +)
in (/ 2)
. abs
. uncurry (-)
. foldr calcSums (0, 0)
. (<*>) zip (tail . cycle)
main :: IO ()
main =
print $
shoelace [(3, 4), (5, 11), (12, 8), (9, 5), (5, 6)]
| 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);
}
}
}
|
Change the programming language of this snippet from Haskell to C++ without modifying what it does. | import Data.Bifunctor (bimap)
shoelace :: [(Double, Double)] -> Double
shoelace =
let calcSums ((x, y), (a, b)) = bimap (x * b +) (a * y +)
in (/ 2)
. abs
. uncurry (-)
. foldr calcSums (0, 0)
. (<*>) zip (tail . cycle)
main :: IO ()
main =
print $
shoelace [(3, 4), (5, 11), (12, 8), (9, 5), (5, 6)]
| #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;
}
|
Generate an equivalent C++ version of this Haskell code. | import Data.Bifunctor (bimap)
shoelace :: [(Double, Double)] -> Double
shoelace =
let calcSums ((x, y), (a, b)) = bimap (x * b +) (a * y +)
in (/ 2)
. abs
. uncurry (-)
. foldr calcSums (0, 0)
. (<*>) zip (tail . cycle)
main :: IO ()
main =
print $
shoelace [(3, 4), (5, 11), (12, 8), (9, 5), (5, 6)]
| #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;
}
|
Generate an equivalent Java version of this Haskell code. | import Data.Bifunctor (bimap)
shoelace :: [(Double, Double)] -> Double
shoelace =
let calcSums ((x, y), (a, b)) = bimap (x * b +) (a * y +)
in (/ 2)
. abs
. uncurry (-)
. foldr calcSums (0, 0)
. (<*>) zip (tail . cycle)
main :: IO ()
main =
print $
shoelace [(3, 4), (5, 11), (12, 8), (9, 5), (5, 6)]
| 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);
}
}
|
Write a version of this Haskell function in Java with identical behavior. | import Data.Bifunctor (bimap)
shoelace :: [(Double, Double)] -> Double
shoelace =
let calcSums ((x, y), (a, b)) = bimap (x * b +) (a * y +)
in (/ 2)
. abs
. uncurry (-)
. foldr calcSums (0, 0)
. (<*>) zip (tail . cycle)
main :: IO ()
main =
print $
shoelace [(3, 4), (5, 11), (12, 8), (9, 5), (5, 6)]
| 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);
}
}
|
Preserve the algorithm and functionality while converting the code from Haskell to Python. | import Data.Bifunctor (bimap)
shoelace :: [(Double, Double)] -> Double
shoelace =
let calcSums ((x, y), (a, b)) = bimap (x * b +) (a * y +)
in (/ 2)
. abs
. uncurry (-)
. foldr calcSums (0, 0)
. (<*>) zip (tail . cycle)
main :: IO ()
main =
print $
shoelace [(3, 4), (5, 11), (12, 8), (9, 5), (5, 6)]
| >>> 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
>>>
|
Port the provided Haskell code into Python while preserving the original functionality. | import Data.Bifunctor (bimap)
shoelace :: [(Double, Double)] -> Double
shoelace =
let calcSums ((x, y), (a, b)) = bimap (x * b +) (a * y +)
in (/ 2)
. abs
. uncurry (-)
. foldr calcSums (0, 0)
. (<*>) zip (tail . cycle)
main :: IO ()
main =
print $
shoelace [(3, 4), (5, 11), (12, 8), (9, 5), (5, 6)]
| >>> 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
>>>
|
Rewrite this program in VB while keeping its functionality equivalent to the Haskell version. | import Data.Bifunctor (bimap)
shoelace :: [(Double, Double)] -> Double
shoelace =
let calcSums ((x, y), (a, b)) = bimap (x * b +) (a * y +)
in (/ 2)
. abs
. uncurry (-)
. foldr calcSums (0, 0)
. (<*>) zip (tail . cycle)
main :: IO ()
main =
print $
shoelace [(3, 4), (5, 11), (12, 8), (9, 5), (5, 6)]
| 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
|
Convert this Haskell snippet to VB and keep its semantics consistent. | import Data.Bifunctor (bimap)
shoelace :: [(Double, Double)] -> Double
shoelace =
let calcSums ((x, y), (a, b)) = bimap (x * b +) (a * y +)
in (/ 2)
. abs
. uncurry (-)
. foldr calcSums (0, 0)
. (<*>) zip (tail . cycle)
main :: IO ()
main =
print $
shoelace [(3, 4), (5, 11), (12, 8), (9, 5), (5, 6)]
| 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
|
Write the same code in Go as shown below in Haskell. | import Data.Bifunctor (bimap)
shoelace :: [(Double, Double)] -> Double
shoelace =
let calcSums ((x, y), (a, b)) = bimap (x * b +) (a * y +)
in (/ 2)
. abs
. uncurry (-)
. foldr calcSums (0, 0)
. (<*>) zip (tail . cycle)
main :: IO ()
main =
print $
shoelace [(3, 4), (5, 11), (12, 8), (9, 5), (5, 6)]
| 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}}))
}
|
Generate an equivalent Go version of this Haskell code. | import Data.Bifunctor (bimap)
shoelace :: [(Double, Double)] -> Double
shoelace =
let calcSums ((x, y), (a, b)) = bimap (x * b +) (a * y +)
in (/ 2)
. abs
. uncurry (-)
. foldr calcSums (0, 0)
. (<*>) zip (tail . cycle)
main :: IO ()
main =
print $
shoelace [(3, 4), (5, 11), (12, 8), (9, 5), (5, 6)]
| 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}}))
}
|
Translate this program into C but keep the logic exactly as in J. | shoelace=:verb define
0.5*|+/((* 1&|.)/ - (* _1&|.)/)|:y
)
| #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;
}
|
Transform the following J implementation into C, maintaining the same output and logic. | shoelace=:verb define
0.5*|+/((* 1&|.)/ - (* _1&|.)/)|:y
)
| #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;
}
|
Keep all operations the same but rewrite the snippet in C#. | shoelace=:verb define
0.5*|+/((* 1&|.)/ - (* _1&|.)/)|:y
)
| 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);
}
}
}
|
Generate a C# translation of this J snippet without changing its computational steps. | shoelace=:verb define
0.5*|+/((* 1&|.)/ - (* _1&|.)/)|:y
)
| 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);
}
}
}
|
Ensure the translated C++ code behaves exactly like the original J snippet. | shoelace=:verb define
0.5*|+/((* 1&|.)/ - (* _1&|.)/)|:y
)
| #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;
}
|
Write the same algorithm in C++ as shown in this J implementation. | shoelace=:verb define
0.5*|+/((* 1&|.)/ - (* _1&|.)/)|:y
)
| #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;
}
|
Port the following code from J to Java with equivalent syntax and logic. | shoelace=:verb define
0.5*|+/((* 1&|.)/ - (* _1&|.)/)|:y
)
| 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);
}
}
|
Preserve the algorithm and functionality while converting the code from J to Java. | shoelace=:verb define
0.5*|+/((* 1&|.)/ - (* _1&|.)/)|:y
)
| 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);
}
}
|
Keep all operations the same but rewrite the snippet in Python. | shoelace=:verb define
0.5*|+/((* 1&|.)/ - (* _1&|.)/)|:y
)
| >>> 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
>>>
|
Produce a language-to-language conversion: from J to VB, same semantics. | shoelace=:verb define
0.5*|+/((* 1&|.)/ - (* _1&|.)/)|:y
)
| 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
|
Convert this J snippet to VB and keep its semantics consistent. | shoelace=:verb define
0.5*|+/((* 1&|.)/ - (* _1&|.)/)|:y
)
| 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
|
Convert the following code from J to Go, ensuring the logic remains intact. | shoelace=:verb define
0.5*|+/((* 1&|.)/ - (* _1&|.)/)|:y
)
| 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}}))
}
|
Convert this J block to Go, preserving its control flow and logic. | shoelace=:verb define
0.5*|+/((* 1&|.)/ - (* _1&|.)/)|:y
)
| 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}}))
}
|
Transform the following Julia implementation into C, maintaining the same output and logic. | """
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)
| #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;
}
|
Ensure the translated C code behaves exactly like the original Julia snippet. | """
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)
| #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;
}
|
Write the same algorithm in C# as shown in this Julia implementation. | """
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)
| 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);
}
}
}
|
Keep all operations the same but rewrite the snippet in C#. | """
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)
| 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);
}
}
}
|
Write the same algorithm in C++ as shown in this Julia implementation. | """
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)
| #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;
}
|
Produce a functionally identical C++ code for the snippet given in 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)
| #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;
}
|
Translate this program into Java but keep the logic exactly as in 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)
| 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);
}
}
|
Rewrite this program in Java while keeping its functionality equivalent to the Julia version. | """
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)
| 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);
}
}
|
Please provide an equivalent version of this Julia code in 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)
| >>> 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
>>>
|
Ensure the translated Python code behaves exactly like the original Julia snippet. | """
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)
| >>> 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
>>>
|
Produce a functionally identical VB code for the snippet given in 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)
| 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
|
Translate the given Julia code snippet into VB without altering its behavior. | """
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)
| 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
|
Port the provided Julia code into Go while preserving the original functionality. | """
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)
| 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}}))
}
|
Write a version of this Julia function in Go with identical behavior. | """
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)
| 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}}))
}
|
Produce a functionally identical C code for the snippet given in 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
| #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;
}
|
Convert this Lua block to C, preserving its control flow and logic. | 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
| #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;
}
|
Write the same algorithm in C# as shown in this Lua implementation. | 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 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);
}
}
}
|
Write the same algorithm in C# as shown in this Lua implementation. | 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 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);
}
}
}
|
Preserve the algorithm and functionality while converting the code from Lua to C++. | 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
| #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;
}
|
Rewrite the snippet below in C++ so it works the same as the original Lua code. | 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
| #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;
}
|
Produce a language-to-language conversion: from Lua to Java, same semantics. | 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
| 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);
}
}
|
Generate a Java translation of this Lua snippet without changing its computational steps. | 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
| 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);
}
}
|
Write the same algorithm in Python as shown in this Lua implementation. | 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
| >>> 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
>>>
|
Transform the following Lua implementation into Python, maintaining the same output and logic. | 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
| >>> 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
>>>
|
Port the following code from Lua to VB with equivalent syntax and logic. | 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
| 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
|
Please provide an equivalent version of this Lua code in VB. | 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
| 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
|
Can you help me rewrite this code in Go instead of Lua, keeping it the same logically? | 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
| 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}}))
}
|
Port the following code from Lua to Go with equivalent syntax and logic. | 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
| 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}}))
}
|
Keep all operations the same but rewrite the snippet in C. | Area[Polygon[{{3, 4}, {5, 11}, {12, 8}, {9, 5}, {5, 6}}]]
| #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;
}
|
Preserve the algorithm and functionality while converting the code from Mathematica to C. | Area[Polygon[{{3, 4}, {5, 11}, {12, 8}, {9, 5}, {5, 6}}]]
| #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;
}
|
Transform the following Mathematica implementation into C#, maintaining the same output and logic. | Area[Polygon[{{3, 4}, {5, 11}, {12, 8}, {9, 5}, {5, 6}}]]
| 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);
}
}
}
|
Please provide an equivalent version of this Mathematica code in C#. | Area[Polygon[{{3, 4}, {5, 11}, {12, 8}, {9, 5}, {5, 6}}]]
| 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);
}
}
}
|
Translate the given Mathematica code snippet into C++ without altering its behavior. | Area[Polygon[{{3, 4}, {5, 11}, {12, 8}, {9, 5}, {5, 6}}]]
| #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;
}
|
Change the programming language of this snippet from Mathematica to C++ without modifying what it does. | Area[Polygon[{{3, 4}, {5, 11}, {12, 8}, {9, 5}, {5, 6}}]]
| #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;
}
|
Port the following code from Mathematica to Java with equivalent syntax and logic. | Area[Polygon[{{3, 4}, {5, 11}, {12, 8}, {9, 5}, {5, 6}}]]
| 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);
}
}
|
Produce a functionally identical Java code for the snippet given in Mathematica. | Area[Polygon[{{3, 4}, {5, 11}, {12, 8}, {9, 5}, {5, 6}}]]
| 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);
}
}
|
Port the following code from Mathematica to Python with equivalent syntax and logic. | Area[Polygon[{{3, 4}, {5, 11}, {12, 8}, {9, 5}, {5, 6}}]]
| >>> 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
>>>
|
Write the same algorithm in Python as shown in this Mathematica implementation. | Area[Polygon[{{3, 4}, {5, 11}, {12, 8}, {9, 5}, {5, 6}}]]
| >>> 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
>>>
|
Translate the given Mathematica code snippet into VB without altering its behavior. | Area[Polygon[{{3, 4}, {5, 11}, {12, 8}, {9, 5}, {5, 6}}]]
| 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
|
Keep all operations the same but rewrite the snippet in VB. | Area[Polygon[{{3, 4}, {5, 11}, {12, 8}, {9, 5}, {5, 6}}]]
| 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
|
Produce a language-to-language conversion: from Mathematica to Go, same semantics. | Area[Polygon[{{3, 4}, {5, 11}, {12, 8}, {9, 5}, {5, 6}}]]
| 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}}))
}
|
Write a version of this Mathematica function in Go with identical behavior. | Area[Polygon[{{3, 4}, {5, 11}, {12, 8}, {9, 5}, {5, 6}}]]
| 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}}))
}
|
Change the programming language of this snippet from Nim to C without modifying what it does. | 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)
| #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;
}
|
Convert this Nim snippet to C and keep its semantics consistent. | 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)
| #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;
}
|
Can you help me rewrite this code in C# instead of Nim, keeping it the same logically? | 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)
| 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);
}
}
}
|
Write the same code in C# as shown below in 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)
| 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);
}
}
}
|
Rewrite this program in C++ while keeping its functionality equivalent to the Nim version. | 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)
| #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;
}
|
Transform the following Nim implementation into C++, maintaining the same output and logic. | 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)
| #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;
}
|
Rewrite the snippet below in Java so it works the same as the original Nim code. | 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)
| 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);
}
}
|
Translate this program into Java but keep the logic exactly as in 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)
| 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);
}
}
|
Rewrite the snippet below in Python so it works the same as the original Nim code. | 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)
| >>> 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
>>>
|
Translate this program into Python but keep the logic exactly as in 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)
| >>> 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
>>>
|
Convert this Nim snippet to VB and keep its semantics consistent. | 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)
| 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
|
Produce a functionally identical Go code for the snippet given in 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)
| 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}}))
}
|
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