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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*...
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); } } ...
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*...
>>> 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(*point...
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*...
>>> 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(*point...
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*...
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) /...
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*...
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) /...
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*...
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...
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*...
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...
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 ...
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]....
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 ...
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]....
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 ...
#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 ...
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 ...
#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 ...
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 ...
#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(poin...
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 ...
#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(poin...
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 ...
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); } } ...
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 ...
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); } } ...
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 ...
>>> 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(*point...
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 ...
>>> 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(*point...
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 ...
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) /...
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 ...
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) /...
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...
#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(poin...
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...
#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(poin...
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...
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]....
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...
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]....
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...
#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 ...
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...
#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 ...
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...
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); } } ...
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...
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); } } ...
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...
>>> 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(*point...
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...
>>> 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(*point...
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...
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) /...
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...
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) /...
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...
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...
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...
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...
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(poin...
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(poin...
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]....
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]....
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 ...
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 ...
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); } } ...
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); } } ...
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(*point...
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) /...
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) /...
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...
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...
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(poin...
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(poin...
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]....
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]....
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 ...
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 ...
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); } } ...
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); } } ...
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(*point...
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(*point...
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) /...
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) /...
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...
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...
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(poin...
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(poin...
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]....
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]....
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 ...
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 ...
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); } } ...
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); } } ...
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(*point...
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(*point...
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) /...
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) /...
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...
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...
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(poin...
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(poin...
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]....
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]....
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 ...
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 ...
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); } } ...
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); } } ...
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(*point...
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(*point...
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) /...
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) /...
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...
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...
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 =...
#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(poin...
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 =...
#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(poin...
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 =...
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]....
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 =...
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]....
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 =...
#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 ...
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 =...
#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 ...
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 =...
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); } } ...
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 =...
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); } } ...
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 =...
>>> 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(*point...
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 =...
>>> 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(*point...
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 =...
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) /...
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 =...
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...