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Maintain the same structure and functionality when rewriting this code in C.
$epsilon=1E-10 function power($x,$e){ $ret=1 for($i=1;$i -le $e;$i++){ $ret*=$x } return $ret } function root($y,$n){ if (0+$n -lt 0){$tmp=-$n} else {$tmp=$n} $ans=1 do{ $d = ($y/(power $ans ($tmp-1)) - $ans)/$tmp $ans+=$d } while ($d -lt -$epsilon -or $d -gt $epsilon) if (0+$n -lt 0...
#include <stdio.h> #include <float.h> double pow_ (double x, int e) { int i; double r = 1; for (i = 0; i < e; i++) { r *= x; } return r; } double root (int n, double x) { double d, r = 1; if (!x) { return 0; } if (n < 1 || (x < 0 && !(n&1))) { return 0.0 / 0...
Convert this PowerShell block to C#, preserving its control flow and logic.
$epsilon=1E-10 function power($x,$e){ $ret=1 for($i=1;$i -le $e;$i++){ $ret*=$x } return $ret } function root($y,$n){ if (0+$n -lt 0){$tmp=-$n} else {$tmp=$n} $ans=1 do{ $d = ($y/(power $ans ($tmp-1)) - $ans)/$tmp $ans+=$d } while ($d -lt -$epsilon -or $d -gt $epsilon) if (0+$n -lt 0...
static void Main(string[] args) { Console.WriteLine(NthRoot(81,2,.001)); Console.WriteLine(NthRoot(1000,3,.001)); Console.ReadLine(); } public static double NthRoot(double A,int n, double p) { double _n= (double) n; double[] x = new double[2]; x[0] = A; x[1] = A/_n; while(Math.Abs(x[0] -x[1] )...
Preserve the algorithm and functionality while converting the code from PowerShell to C++.
$epsilon=1E-10 function power($x,$e){ $ret=1 for($i=1;$i -le $e;$i++){ $ret*=$x } return $ret } function root($y,$n){ if (0+$n -lt 0){$tmp=-$n} else {$tmp=$n} $ans=1 do{ $d = ($y/(power $ans ($tmp-1)) - $ans)/$tmp $ans+=$d } while ($d -lt -$epsilon -or $d -gt $epsilon) if (0+$n -lt 0...
double NthRoot(double m_nValue, double index, double guess, double pc) { double result = guess; double result_next; do { result_next = (1.0/index)*((index-1.0)*result+(m_nValue)/(pow(result,(index-1.0)))); result = result_next; pc--; }while(pc>1); ...
Convert this PowerShell snippet to Java and keep its semantics consistent.
$epsilon=1E-10 function power($x,$e){ $ret=1 for($i=1;$i -le $e;$i++){ $ret*=$x } return $ret } function root($y,$n){ if (0+$n -lt 0){$tmp=-$n} else {$tmp=$n} $ans=1 do{ $d = ($y/(power $ans ($tmp-1)) - $ans)/$tmp $ans+=$d } while ($d -lt -$epsilon -or $d -gt $epsilon) if (0+$n -lt 0...
public static double nthroot(int n, double A) { return nthroot(n, A, .001); } public static double nthroot(int n, double A, double p) { if(A < 0) { System.err.println("A < 0"); return -1; } else if(A == 0) { return 0; } double x_prev = A; double x = A / n; while(Math.abs(x - x_prev) > p) { x_prev = x; ...
Convert this PowerShell block to Python, preserving its control flow and logic.
$epsilon=1E-10 function power($x,$e){ $ret=1 for($i=1;$i -le $e;$i++){ $ret*=$x } return $ret } function root($y,$n){ if (0+$n -lt 0){$tmp=-$n} else {$tmp=$n} $ans=1 do{ $d = ($y/(power $ans ($tmp-1)) - $ans)/$tmp $ans+=$d } while ($d -lt -$epsilon -or $d -gt $epsilon) if (0+$n -lt 0...
from decimal import Decimal, getcontext def nthroot (n, A, precision): getcontext().prec = precision n = Decimal(n) x_0 = A / n x_1 = 1 while True: x_0, x_1 = x_1, (1 / n)*((n - 1)*x_0 + (A / (x_0 ** (n - 1)))) if x_0 == x_1: return x_1
Rewrite the snippet below in VB so it works the same as the original PowerShell code.
$epsilon=1E-10 function power($x,$e){ $ret=1 for($i=1;$i -le $e;$i++){ $ret*=$x } return $ret } function root($y,$n){ if (0+$n -lt 0){$tmp=-$n} else {$tmp=$n} $ans=1 do{ $d = ($y/(power $ans ($tmp-1)) - $ans)/$tmp $ans+=$d } while ($d -lt -$epsilon -or $d -gt $epsilon) if (0+$n -lt 0...
Private Function nth_root(y As Double, n As Double) Dim eps As Double: eps = 0.00000000000001 Dim x As Variant: x = 1 Do While True d = (y / x ^ (n - 1) - x) / n x = x + d e = eps * x If d > -e And d < e Then Exit Do End If Loop Debug.Print y; n;...
Maintain the same structure and functionality when rewriting this code in Go.
$epsilon=1E-10 function power($x,$e){ $ret=1 for($i=1;$i -le $e;$i++){ $ret*=$x } return $ret } function root($y,$n){ if (0+$n -lt 0){$tmp=-$n} else {$tmp=$n} $ans=1 do{ $d = ($y/(power $ans ($tmp-1)) - $ans)/$tmp $ans+=$d } while ($d -lt -$epsilon -or $d -gt $epsilon) if (0+$n -lt 0...
func root(a float64, n int) float64 { n1 := n - 1 n1f, rn := float64(n1), 1/float64(n) x, x0 := 1., 0. for { potx, t2 := 1/x, a for b := n1; b > 0; b >>= 1 { if b&1 == 1 { t2 *= potx } potx *= potx } x0, x = x, rn*(n1f*x...
Produce a language-to-language conversion: from R to C, same semantics.
nthroot <- function(A, n, tol=sqrt(.Machine$double.eps)) { ifelse(A < 1, x0 <- A * n, x0 <- A / n) repeat { x1 <- ((n-1)*x0 + A / x0^(n-1))/n if(abs(x1 - x0) > tol) x0 <- x1 else break } x1 } nthroot(7131.5^10, 10) nthroot(7, 0.5)
#include <stdio.h> #include <float.h> double pow_ (double x, int e) { int i; double r = 1; for (i = 0; i < e; i++) { r *= x; } return r; } double root (int n, double x) { double d, r = 1; if (!x) { return 0; } if (n < 1 || (x < 0 && !(n&1))) { return 0.0 / 0...
Generate a C# translation of this R snippet without changing its computational steps.
nthroot <- function(A, n, tol=sqrt(.Machine$double.eps)) { ifelse(A < 1, x0 <- A * n, x0 <- A / n) repeat { x1 <- ((n-1)*x0 + A / x0^(n-1))/n if(abs(x1 - x0) > tol) x0 <- x1 else break } x1 } nthroot(7131.5^10, 10) nthroot(7, 0.5)
static void Main(string[] args) { Console.WriteLine(NthRoot(81,2,.001)); Console.WriteLine(NthRoot(1000,3,.001)); Console.ReadLine(); } public static double NthRoot(double A,int n, double p) { double _n= (double) n; double[] x = new double[2]; x[0] = A; x[1] = A/_n; while(Math.Abs(x[0] -x[1] )...
Preserve the algorithm and functionality while converting the code from R to C++.
nthroot <- function(A, n, tol=sqrt(.Machine$double.eps)) { ifelse(A < 1, x0 <- A * n, x0 <- A / n) repeat { x1 <- ((n-1)*x0 + A / x0^(n-1))/n if(abs(x1 - x0) > tol) x0 <- x1 else break } x1 } nthroot(7131.5^10, 10) nthroot(7, 0.5)
double NthRoot(double m_nValue, double index, double guess, double pc) { double result = guess; double result_next; do { result_next = (1.0/index)*((index-1.0)*result+(m_nValue)/(pow(result,(index-1.0)))); result = result_next; pc--; }while(pc>1); ...
Produce a functionally identical Java code for the snippet given in R.
nthroot <- function(A, n, tol=sqrt(.Machine$double.eps)) { ifelse(A < 1, x0 <- A * n, x0 <- A / n) repeat { x1 <- ((n-1)*x0 + A / x0^(n-1))/n if(abs(x1 - x0) > tol) x0 <- x1 else break } x1 } nthroot(7131.5^10, 10) nthroot(7, 0.5)
public static double nthroot(int n, double A) { return nthroot(n, A, .001); } public static double nthroot(int n, double A, double p) { if(A < 0) { System.err.println("A < 0"); return -1; } else if(A == 0) { return 0; } double x_prev = A; double x = A / n; while(Math.abs(x - x_prev) > p) { x_prev = x; ...
Rewrite the snippet below in Python so it works the same as the original R code.
nthroot <- function(A, n, tol=sqrt(.Machine$double.eps)) { ifelse(A < 1, x0 <- A * n, x0 <- A / n) repeat { x1 <- ((n-1)*x0 + A / x0^(n-1))/n if(abs(x1 - x0) > tol) x0 <- x1 else break } x1 } nthroot(7131.5^10, 10) nthroot(7, 0.5)
from decimal import Decimal, getcontext def nthroot (n, A, precision): getcontext().prec = precision n = Decimal(n) x_0 = A / n x_1 = 1 while True: x_0, x_1 = x_1, (1 / n)*((n - 1)*x_0 + (A / (x_0 ** (n - 1)))) if x_0 == x_1: return x_1
Convert this R block to VB, preserving its control flow and logic.
nthroot <- function(A, n, tol=sqrt(.Machine$double.eps)) { ifelse(A < 1, x0 <- A * n, x0 <- A / n) repeat { x1 <- ((n-1)*x0 + A / x0^(n-1))/n if(abs(x1 - x0) > tol) x0 <- x1 else break } x1 } nthroot(7131.5^10, 10) nthroot(7, 0.5)
Private Function nth_root(y As Double, n As Double) Dim eps As Double: eps = 0.00000000000001 Dim x As Variant: x = 1 Do While True d = (y / x ^ (n - 1) - x) / n x = x + d e = eps * x If d > -e And d < e Then Exit Do End If Loop Debug.Print y; n;...
Generate an equivalent Go version of this R code.
nthroot <- function(A, n, tol=sqrt(.Machine$double.eps)) { ifelse(A < 1, x0 <- A * n, x0 <- A / n) repeat { x1 <- ((n-1)*x0 + A / x0^(n-1))/n if(abs(x1 - x0) > tol) x0 <- x1 else break } x1 } nthroot(7131.5^10, 10) nthroot(7, 0.5)
func root(a float64, n int) float64 { n1 := n - 1 n1f, rn := float64(n1), 1/float64(n) x, x0 := 1., 0. for { potx, t2 := 1/x, a for b := n1; b > 0; b >>= 1 { if b&1 == 1 { t2 *= potx } potx *= potx } x0, x = x, rn*(n1f*x...
Port the provided Racket code into C while preserving the original functionality.
#lang racket (define (nth-root number root (tolerance 0.001)) (define (acceptable? next current) (< (abs (- next current)) tolerance)) (define (improve current) (/ (+ (* (- root 1) current) (/ number (expt current (- root 1)))) root)) (define (loop current) (define next-guess (improve current))...
#include <stdio.h> #include <float.h> double pow_ (double x, int e) { int i; double r = 1; for (i = 0; i < e; i++) { r *= x; } return r; } double root (int n, double x) { double d, r = 1; if (!x) { return 0; } if (n < 1 || (x < 0 && !(n&1))) { return 0.0 / 0...
Change the following Racket code into C# without altering its purpose.
#lang racket (define (nth-root number root (tolerance 0.001)) (define (acceptable? next current) (< (abs (- next current)) tolerance)) (define (improve current) (/ (+ (* (- root 1) current) (/ number (expt current (- root 1)))) root)) (define (loop current) (define next-guess (improve current))...
static void Main(string[] args) { Console.WriteLine(NthRoot(81,2,.001)); Console.WriteLine(NthRoot(1000,3,.001)); Console.ReadLine(); } public static double NthRoot(double A,int n, double p) { double _n= (double) n; double[] x = new double[2]; x[0] = A; x[1] = A/_n; while(Math.Abs(x[0] -x[1] )...
Convert the following code from Racket to C++, ensuring the logic remains intact.
#lang racket (define (nth-root number root (tolerance 0.001)) (define (acceptable? next current) (< (abs (- next current)) tolerance)) (define (improve current) (/ (+ (* (- root 1) current) (/ number (expt current (- root 1)))) root)) (define (loop current) (define next-guess (improve current))...
double NthRoot(double m_nValue, double index, double guess, double pc) { double result = guess; double result_next; do { result_next = (1.0/index)*((index-1.0)*result+(m_nValue)/(pow(result,(index-1.0)))); result = result_next; pc--; }while(pc>1); ...
Generate a Java translation of this Racket snippet without changing its computational steps.
#lang racket (define (nth-root number root (tolerance 0.001)) (define (acceptable? next current) (< (abs (- next current)) tolerance)) (define (improve current) (/ (+ (* (- root 1) current) (/ number (expt current (- root 1)))) root)) (define (loop current) (define next-guess (improve current))...
public static double nthroot(int n, double A) { return nthroot(n, A, .001); } public static double nthroot(int n, double A, double p) { if(A < 0) { System.err.println("A < 0"); return -1; } else if(A == 0) { return 0; } double x_prev = A; double x = A / n; while(Math.abs(x - x_prev) > p) { x_prev = x; ...
Port the provided Racket code into Python while preserving the original functionality.
#lang racket (define (nth-root number root (tolerance 0.001)) (define (acceptable? next current) (< (abs (- next current)) tolerance)) (define (improve current) (/ (+ (* (- root 1) current) (/ number (expt current (- root 1)))) root)) (define (loop current) (define next-guess (improve current))...
from decimal import Decimal, getcontext def nthroot (n, A, precision): getcontext().prec = precision n = Decimal(n) x_0 = A / n x_1 = 1 while True: x_0, x_1 = x_1, (1 / n)*((n - 1)*x_0 + (A / (x_0 ** (n - 1)))) if x_0 == x_1: return x_1
Port the provided Racket code into VB while preserving the original functionality.
#lang racket (define (nth-root number root (tolerance 0.001)) (define (acceptable? next current) (< (abs (- next current)) tolerance)) (define (improve current) (/ (+ (* (- root 1) current) (/ number (expt current (- root 1)))) root)) (define (loop current) (define next-guess (improve current))...
Private Function nth_root(y As Double, n As Double) Dim eps As Double: eps = 0.00000000000001 Dim x As Variant: x = 1 Do While True d = (y / x ^ (n - 1) - x) / n x = x + d e = eps * x If d > -e And d < e Then Exit Do End If Loop Debug.Print y; n;...
Produce a language-to-language conversion: from Racket to Go, same semantics.
#lang racket (define (nth-root number root (tolerance 0.001)) (define (acceptable? next current) (< (abs (- next current)) tolerance)) (define (improve current) (/ (+ (* (- root 1) current) (/ number (expt current (- root 1)))) root)) (define (loop current) (define next-guess (improve current))...
func root(a float64, n int) float64 { n1 := n - 1 n1f, rn := float64(n1), 1/float64(n) x, x0 := 1., 0. for { potx, t2 := 1/x, a for b := n1; b > 0; b >>= 1 { if b&1 == 1 { t2 *= potx } potx *= potx } x0, x = x, rn*(n1f*x...
Port the provided COBOL code into C while preserving the original functionality.
IDENTIFICATION DIVISION. PROGRAM-ID. Nth-Root. AUTHOR. Bill Gunshannon. INSTALLATION. DATE-WRITTEN. 4 Feb 2020. ************************************************************ ** Program Abstract: ** Compute the Nth Root of a positive real number. ** ...
#include <stdio.h> #include <float.h> double pow_ (double x, int e) { int i; double r = 1; for (i = 0; i < e; i++) { r *= x; } return r; } double root (int n, double x) { double d, r = 1; if (!x) { return 0; } if (n < 1 || (x < 0 && !(n&1))) { return 0.0 / 0...
Preserve the algorithm and functionality while converting the code from COBOL to C#.
IDENTIFICATION DIVISION. PROGRAM-ID. Nth-Root. AUTHOR. Bill Gunshannon. INSTALLATION. DATE-WRITTEN. 4 Feb 2020. ************************************************************ ** Program Abstract: ** Compute the Nth Root of a positive real number. ** ...
static void Main(string[] args) { Console.WriteLine(NthRoot(81,2,.001)); Console.WriteLine(NthRoot(1000,3,.001)); Console.ReadLine(); } public static double NthRoot(double A,int n, double p) { double _n= (double) n; double[] x = new double[2]; x[0] = A; x[1] = A/_n; while(Math.Abs(x[0] -x[1] )...
Please provide an equivalent version of this COBOL code in C++.
IDENTIFICATION DIVISION. PROGRAM-ID. Nth-Root. AUTHOR. Bill Gunshannon. INSTALLATION. DATE-WRITTEN. 4 Feb 2020. ************************************************************ ** Program Abstract: ** Compute the Nth Root of a positive real number. ** ...
double NthRoot(double m_nValue, double index, double guess, double pc) { double result = guess; double result_next; do { result_next = (1.0/index)*((index-1.0)*result+(m_nValue)/(pow(result,(index-1.0)))); result = result_next; pc--; }while(pc>1); ...
Rewrite this program in Java while keeping its functionality equivalent to the COBOL version.
IDENTIFICATION DIVISION. PROGRAM-ID. Nth-Root. AUTHOR. Bill Gunshannon. INSTALLATION. DATE-WRITTEN. 4 Feb 2020. ************************************************************ ** Program Abstract: ** Compute the Nth Root of a positive real number. ** ...
public static double nthroot(int n, double A) { return nthroot(n, A, .001); } public static double nthroot(int n, double A, double p) { if(A < 0) { System.err.println("A < 0"); return -1; } else if(A == 0) { return 0; } double x_prev = A; double x = A / n; while(Math.abs(x - x_prev) > p) { x_prev = x; ...
Can you help me rewrite this code in Python instead of COBOL, keeping it the same logically?
IDENTIFICATION DIVISION. PROGRAM-ID. Nth-Root. AUTHOR. Bill Gunshannon. INSTALLATION. DATE-WRITTEN. 4 Feb 2020. ************************************************************ ** Program Abstract: ** Compute the Nth Root of a positive real number. ** ...
from decimal import Decimal, getcontext def nthroot (n, A, precision): getcontext().prec = precision n = Decimal(n) x_0 = A / n x_1 = 1 while True: x_0, x_1 = x_1, (1 / n)*((n - 1)*x_0 + (A / (x_0 ** (n - 1)))) if x_0 == x_1: return x_1
Generate a VB translation of this COBOL snippet without changing its computational steps.
IDENTIFICATION DIVISION. PROGRAM-ID. Nth-Root. AUTHOR. Bill Gunshannon. INSTALLATION. DATE-WRITTEN. 4 Feb 2020. ************************************************************ ** Program Abstract: ** Compute the Nth Root of a positive real number. ** ...
Private Function nth_root(y As Double, n As Double) Dim eps As Double: eps = 0.00000000000001 Dim x As Variant: x = 1 Do While True d = (y / x ^ (n - 1) - x) / n x = x + d e = eps * x If d > -e And d < e Then Exit Do End If Loop Debug.Print y; n;...
Rewrite the snippet below in Go so it works the same as the original COBOL code.
IDENTIFICATION DIVISION. PROGRAM-ID. Nth-Root. AUTHOR. Bill Gunshannon. INSTALLATION. DATE-WRITTEN. 4 Feb 2020. ************************************************************ ** Program Abstract: ** Compute the Nth Root of a positive real number. ** ...
func root(a float64, n int) float64 { n1 := n - 1 n1f, rn := float64(n1), 1/float64(n) x, x0 := 1., 0. for { potx, t2 := 1/x, a for b := n1; b > 0; b >>= 1 { if b&1 == 1 { t2 *= potx } potx *= potx } x0, x = x, rn*(n1f*x...
Change the programming language of this snippet from REXX to C without modifying what it does.
class nth_root method main(args=String[]) static if args.length < 2 then do say "at least 2 arguments expected" exit end x = args[0] root = args[1] if args.length > 2 then digs = args[2] if root=='' then root=2 if digs = null, digs = '' then digs=20 numeric digits digs ...
#include <stdio.h> #include <float.h> double pow_ (double x, int e) { int i; double r = 1; for (i = 0; i < e; i++) { r *= x; } return r; } double root (int n, double x) { double d, r = 1; if (!x) { return 0; } if (n < 1 || (x < 0 && !(n&1))) { return 0.0 / 0...
Write the same algorithm in C# as shown in this REXX implementation.
class nth_root method main(args=String[]) static if args.length < 2 then do say "at least 2 arguments expected" exit end x = args[0] root = args[1] if args.length > 2 then digs = args[2] if root=='' then root=2 if digs = null, digs = '' then digs=20 numeric digits digs ...
static void Main(string[] args) { Console.WriteLine(NthRoot(81,2,.001)); Console.WriteLine(NthRoot(1000,3,.001)); Console.ReadLine(); } public static double NthRoot(double A,int n, double p) { double _n= (double) n; double[] x = new double[2]; x[0] = A; x[1] = A/_n; while(Math.Abs(x[0] -x[1] )...
Port the provided REXX code into C++ while preserving the original functionality.
class nth_root method main(args=String[]) static if args.length < 2 then do say "at least 2 arguments expected" exit end x = args[0] root = args[1] if args.length > 2 then digs = args[2] if root=='' then root=2 if digs = null, digs = '' then digs=20 numeric digits digs ...
double NthRoot(double m_nValue, double index, double guess, double pc) { double result = guess; double result_next; do { result_next = (1.0/index)*((index-1.0)*result+(m_nValue)/(pow(result,(index-1.0)))); result = result_next; pc--; }while(pc>1); ...
Please provide an equivalent version of this REXX code in Java.
class nth_root method main(args=String[]) static if args.length < 2 then do say "at least 2 arguments expected" exit end x = args[0] root = args[1] if args.length > 2 then digs = args[2] if root=='' then root=2 if digs = null, digs = '' then digs=20 numeric digits digs ...
public static double nthroot(int n, double A) { return nthroot(n, A, .001); } public static double nthroot(int n, double A, double p) { if(A < 0) { System.err.println("A < 0"); return -1; } else if(A == 0) { return 0; } double x_prev = A; double x = A / n; while(Math.abs(x - x_prev) > p) { x_prev = x; ...
Keep all operations the same but rewrite the snippet in Python.
class nth_root method main(args=String[]) static if args.length < 2 then do say "at least 2 arguments expected" exit end x = args[0] root = args[1] if args.length > 2 then digs = args[2] if root=='' then root=2 if digs = null, digs = '' then digs=20 numeric digits digs ...
from decimal import Decimal, getcontext def nthroot (n, A, precision): getcontext().prec = precision n = Decimal(n) x_0 = A / n x_1 = 1 while True: x_0, x_1 = x_1, (1 / n)*((n - 1)*x_0 + (A / (x_0 ** (n - 1)))) if x_0 == x_1: return x_1
Port the provided REXX code into VB while preserving the original functionality.
class nth_root method main(args=String[]) static if args.length < 2 then do say "at least 2 arguments expected" exit end x = args[0] root = args[1] if args.length > 2 then digs = args[2] if root=='' then root=2 if digs = null, digs = '' then digs=20 numeric digits digs ...
Private Function nth_root(y As Double, n As Double) Dim eps As Double: eps = 0.00000000000001 Dim x As Variant: x = 1 Do While True d = (y / x ^ (n - 1) - x) / n x = x + d e = eps * x If d > -e And d < e Then Exit Do End If Loop Debug.Print y; n;...
Port the following code from REXX to Go with equivalent syntax and logic.
class nth_root method main(args=String[]) static if args.length < 2 then do say "at least 2 arguments expected" exit end x = args[0] root = args[1] if args.length > 2 then digs = args[2] if root=='' then root=2 if digs = null, digs = '' then digs=20 numeric digits digs ...
func root(a float64, n int) float64 { n1 := n - 1 n1f, rn := float64(n1), 1/float64(n) x, x0 := 1., 0. for { potx, t2 := 1/x, a for b := n1; b > 0; b >>= 1 { if b&1 == 1 { t2 *= potx } potx *= potx } x0, x = x, rn*(n1f*x...
Change the programming language of this snippet from Ruby to C without modifying what it does.
def nthroot(n, a, precision = 1e-5) x = Float(a) begin prev = x x = ((n - 1) * prev + a / (prev ** (n - 1))) / n end while (prev - x).abs > precision x end p nthroot(5,34)
#include <stdio.h> #include <float.h> double pow_ (double x, int e) { int i; double r = 1; for (i = 0; i < e; i++) { r *= x; } return r; } double root (int n, double x) { double d, r = 1; if (!x) { return 0; } if (n < 1 || (x < 0 && !(n&1))) { return 0.0 / 0...
Rewrite the snippet below in C# so it works the same as the original Ruby code.
def nthroot(n, a, precision = 1e-5) x = Float(a) begin prev = x x = ((n - 1) * prev + a / (prev ** (n - 1))) / n end while (prev - x).abs > precision x end p nthroot(5,34)
static void Main(string[] args) { Console.WriteLine(NthRoot(81,2,.001)); Console.WriteLine(NthRoot(1000,3,.001)); Console.ReadLine(); } public static double NthRoot(double A,int n, double p) { double _n= (double) n; double[] x = new double[2]; x[0] = A; x[1] = A/_n; while(Math.Abs(x[0] -x[1] )...
Change the programming language of this snippet from Ruby to C++ without modifying what it does.
def nthroot(n, a, precision = 1e-5) x = Float(a) begin prev = x x = ((n - 1) * prev + a / (prev ** (n - 1))) / n end while (prev - x).abs > precision x end p nthroot(5,34)
double NthRoot(double m_nValue, double index, double guess, double pc) { double result = guess; double result_next; do { result_next = (1.0/index)*((index-1.0)*result+(m_nValue)/(pow(result,(index-1.0)))); result = result_next; pc--; }while(pc>1); ...
Change the following Ruby code into Java without altering its purpose.
def nthroot(n, a, precision = 1e-5) x = Float(a) begin prev = x x = ((n - 1) * prev + a / (prev ** (n - 1))) / n end while (prev - x).abs > precision x end p nthroot(5,34)
public static double nthroot(int n, double A) { return nthroot(n, A, .001); } public static double nthroot(int n, double A, double p) { if(A < 0) { System.err.println("A < 0"); return -1; } else if(A == 0) { return 0; } double x_prev = A; double x = A / n; while(Math.abs(x - x_prev) > p) { x_prev = x; ...
Convert this Ruby snippet to Python and keep its semantics consistent.
def nthroot(n, a, precision = 1e-5) x = Float(a) begin prev = x x = ((n - 1) * prev + a / (prev ** (n - 1))) / n end while (prev - x).abs > precision x end p nthroot(5,34)
from decimal import Decimal, getcontext def nthroot (n, A, precision): getcontext().prec = precision n = Decimal(n) x_0 = A / n x_1 = 1 while True: x_0, x_1 = x_1, (1 / n)*((n - 1)*x_0 + (A / (x_0 ** (n - 1)))) if x_0 == x_1: return x_1
Write the same algorithm in VB as shown in this Ruby implementation.
def nthroot(n, a, precision = 1e-5) x = Float(a) begin prev = x x = ((n - 1) * prev + a / (prev ** (n - 1))) / n end while (prev - x).abs > precision x end p nthroot(5,34)
Private Function nth_root(y As Double, n As Double) Dim eps As Double: eps = 0.00000000000001 Dim x As Variant: x = 1 Do While True d = (y / x ^ (n - 1) - x) / n x = x + d e = eps * x If d > -e And d < e Then Exit Do End If Loop Debug.Print y; n;...
Rewrite the snippet below in Go so it works the same as the original Ruby code.
def nthroot(n, a, precision = 1e-5) x = Float(a) begin prev = x x = ((n - 1) * prev + a / (prev ** (n - 1))) / n end while (prev - x).abs > precision x end p nthroot(5,34)
func root(a float64, n int) float64 { n1 := n - 1 n1f, rn := float64(n1), 1/float64(n) x, x0 := 1., 0. for { potx, t2 := 1/x, a for b := n1; b > 0; b >>= 1 { if b&1 == 1 { t2 *= potx } potx *= potx } x0, x = x, rn*(n1f*x...
Generate a C translation of this Scala snippet without changing its computational steps.
fun nthRoot(x: Double, n: Int): Double { if (n < 2) throw IllegalArgumentException("n must be more than 1") if (x <= 0.0) throw IllegalArgumentException("x must be positive") val np = n - 1 fun iter(g: Double) = (np * g + x / Math.pow(g, np.toDouble())) / n var g1 = x var g2 = iter(g1) whi...
#include <stdio.h> #include <float.h> double pow_ (double x, int e) { int i; double r = 1; for (i = 0; i < e; i++) { r *= x; } return r; } double root (int n, double x) { double d, r = 1; if (!x) { return 0; } if (n < 1 || (x < 0 && !(n&1))) { return 0.0 / 0...
Write the same code in C# as shown below in Scala.
fun nthRoot(x: Double, n: Int): Double { if (n < 2) throw IllegalArgumentException("n must be more than 1") if (x <= 0.0) throw IllegalArgumentException("x must be positive") val np = n - 1 fun iter(g: Double) = (np * g + x / Math.pow(g, np.toDouble())) / n var g1 = x var g2 = iter(g1) whi...
static void Main(string[] args) { Console.WriteLine(NthRoot(81,2,.001)); Console.WriteLine(NthRoot(1000,3,.001)); Console.ReadLine(); } public static double NthRoot(double A,int n, double p) { double _n= (double) n; double[] x = new double[2]; x[0] = A; x[1] = A/_n; while(Math.Abs(x[0] -x[1] )...
Keep all operations the same but rewrite the snippet in C++.
fun nthRoot(x: Double, n: Int): Double { if (n < 2) throw IllegalArgumentException("n must be more than 1") if (x <= 0.0) throw IllegalArgumentException("x must be positive") val np = n - 1 fun iter(g: Double) = (np * g + x / Math.pow(g, np.toDouble())) / n var g1 = x var g2 = iter(g1) whi...
double NthRoot(double m_nValue, double index, double guess, double pc) { double result = guess; double result_next; do { result_next = (1.0/index)*((index-1.0)*result+(m_nValue)/(pow(result,(index-1.0)))); result = result_next; pc--; }while(pc>1); ...
Change the programming language of this snippet from Scala to Java without modifying what it does.
fun nthRoot(x: Double, n: Int): Double { if (n < 2) throw IllegalArgumentException("n must be more than 1") if (x <= 0.0) throw IllegalArgumentException("x must be positive") val np = n - 1 fun iter(g: Double) = (np * g + x / Math.pow(g, np.toDouble())) / n var g1 = x var g2 = iter(g1) whi...
public static double nthroot(int n, double A) { return nthroot(n, A, .001); } public static double nthroot(int n, double A, double p) { if(A < 0) { System.err.println("A < 0"); return -1; } else if(A == 0) { return 0; } double x_prev = A; double x = A / n; while(Math.abs(x - x_prev) > p) { x_prev = x; ...
Change the programming language of this snippet from Scala to Python without modifying what it does.
fun nthRoot(x: Double, n: Int): Double { if (n < 2) throw IllegalArgumentException("n must be more than 1") if (x <= 0.0) throw IllegalArgumentException("x must be positive") val np = n - 1 fun iter(g: Double) = (np * g + x / Math.pow(g, np.toDouble())) / n var g1 = x var g2 = iter(g1) whi...
from decimal import Decimal, getcontext def nthroot (n, A, precision): getcontext().prec = precision n = Decimal(n) x_0 = A / n x_1 = 1 while True: x_0, x_1 = x_1, (1 / n)*((n - 1)*x_0 + (A / (x_0 ** (n - 1)))) if x_0 == x_1: return x_1
Maintain the same structure and functionality when rewriting this code in VB.
fun nthRoot(x: Double, n: Int): Double { if (n < 2) throw IllegalArgumentException("n must be more than 1") if (x <= 0.0) throw IllegalArgumentException("x must be positive") val np = n - 1 fun iter(g: Double) = (np * g + x / Math.pow(g, np.toDouble())) / n var g1 = x var g2 = iter(g1) whi...
Private Function nth_root(y As Double, n As Double) Dim eps As Double: eps = 0.00000000000001 Dim x As Variant: x = 1 Do While True d = (y / x ^ (n - 1) - x) / n x = x + d e = eps * x If d > -e And d < e Then Exit Do End If Loop Debug.Print y; n;...
Ensure the translated Go code behaves exactly like the original Scala snippet.
fun nthRoot(x: Double, n: Int): Double { if (n < 2) throw IllegalArgumentException("n must be more than 1") if (x <= 0.0) throw IllegalArgumentException("x must be positive") val np = n - 1 fun iter(g: Double) = (np * g + x / Math.pow(g, np.toDouble())) / n var g1 = x var g2 = iter(g1) whi...
func root(a float64, n int) float64 { n1 := n - 1 n1f, rn := float64(n1), 1/float64(n) x, x0 := 1., 0. for { potx, t2 := 1/x, a for b := n1; b > 0; b >>= 1 { if b&1 == 1 { t2 *= potx } potx *= potx } x0, x = x, rn*(n1f*x...
Produce a functionally identical C code for the snippet given in Swift.
extension FloatingPoint where Self: ExpressibleByFloatLiteral { @inlinable public func power(_ e: Int) -> Self { var res = Self(1) for _ in 0..<e { res *= self } return res } @inlinable public func root(n: Int, epsilon: Self = 2.220446049250313e-16) -> Self { var d = Self(0) v...
#include <stdio.h> #include <float.h> double pow_ (double x, int e) { int i; double r = 1; for (i = 0; i < e; i++) { r *= x; } return r; } double root (int n, double x) { double d, r = 1; if (!x) { return 0; } if (n < 1 || (x < 0 && !(n&1))) { return 0.0 / 0...
Convert this Swift block to C#, preserving its control flow and logic.
extension FloatingPoint where Self: ExpressibleByFloatLiteral { @inlinable public func power(_ e: Int) -> Self { var res = Self(1) for _ in 0..<e { res *= self } return res } @inlinable public func root(n: Int, epsilon: Self = 2.220446049250313e-16) -> Self { var d = Self(0) v...
static void Main(string[] args) { Console.WriteLine(NthRoot(81,2,.001)); Console.WriteLine(NthRoot(1000,3,.001)); Console.ReadLine(); } public static double NthRoot(double A,int n, double p) { double _n= (double) n; double[] x = new double[2]; x[0] = A; x[1] = A/_n; while(Math.Abs(x[0] -x[1] )...
Produce a functionally identical C++ code for the snippet given in Swift.
extension FloatingPoint where Self: ExpressibleByFloatLiteral { @inlinable public func power(_ e: Int) -> Self { var res = Self(1) for _ in 0..<e { res *= self } return res } @inlinable public func root(n: Int, epsilon: Self = 2.220446049250313e-16) -> Self { var d = Self(0) v...
double NthRoot(double m_nValue, double index, double guess, double pc) { double result = guess; double result_next; do { result_next = (1.0/index)*((index-1.0)*result+(m_nValue)/(pow(result,(index-1.0)))); result = result_next; pc--; }while(pc>1); ...
Transform the following Swift implementation into Java, maintaining the same output and logic.
extension FloatingPoint where Self: ExpressibleByFloatLiteral { @inlinable public func power(_ e: Int) -> Self { var res = Self(1) for _ in 0..<e { res *= self } return res } @inlinable public func root(n: Int, epsilon: Self = 2.220446049250313e-16) -> Self { var d = Self(0) v...
public static double nthroot(int n, double A) { return nthroot(n, A, .001); } public static double nthroot(int n, double A, double p) { if(A < 0) { System.err.println("A < 0"); return -1; } else if(A == 0) { return 0; } double x_prev = A; double x = A / n; while(Math.abs(x - x_prev) > p) { x_prev = x; ...
Convert this Swift block to Python, preserving its control flow and logic.
extension FloatingPoint where Self: ExpressibleByFloatLiteral { @inlinable public func power(_ e: Int) -> Self { var res = Self(1) for _ in 0..<e { res *= self } return res } @inlinable public func root(n: Int, epsilon: Self = 2.220446049250313e-16) -> Self { var d = Self(0) v...
from decimal import Decimal, getcontext def nthroot (n, A, precision): getcontext().prec = precision n = Decimal(n) x_0 = A / n x_1 = 1 while True: x_0, x_1 = x_1, (1 / n)*((n - 1)*x_0 + (A / (x_0 ** (n - 1)))) if x_0 == x_1: return x_1
Produce a language-to-language conversion: from Swift to VB, same semantics.
extension FloatingPoint where Self: ExpressibleByFloatLiteral { @inlinable public func power(_ e: Int) -> Self { var res = Self(1) for _ in 0..<e { res *= self } return res } @inlinable public func root(n: Int, epsilon: Self = 2.220446049250313e-16) -> Self { var d = Self(0) v...
Private Function nth_root(y As Double, n As Double) Dim eps As Double: eps = 0.00000000000001 Dim x As Variant: x = 1 Do While True d = (y / x ^ (n - 1) - x) / n x = x + d e = eps * x If d > -e And d < e Then Exit Do End If Loop Debug.Print y; n;...
Change the programming language of this snippet from Swift to Go without modifying what it does.
extension FloatingPoint where Self: ExpressibleByFloatLiteral { @inlinable public func power(_ e: Int) -> Self { var res = Self(1) for _ in 0..<e { res *= self } return res } @inlinable public func root(n: Int, epsilon: Self = 2.220446049250313e-16) -> Self { var d = Self(0) v...
func root(a float64, n int) float64 { n1 := n - 1 n1f, rn := float64(n1), 1/float64(n) x, x0 := 1., 0. for { potx, t2 := 1/x, a for b := n1; b > 0; b >>= 1 { if b&1 == 1 { t2 *= potx } potx *= potx } x0, x = x, rn*(n1f*x...
Write the same algorithm in C as shown in this Tcl implementation.
proc nthroot {n A} { expr {pow($A, 1.0/$n)} }
#include <stdio.h> #include <float.h> double pow_ (double x, int e) { int i; double r = 1; for (i = 0; i < e; i++) { r *= x; } return r; } double root (int n, double x) { double d, r = 1; if (!x) { return 0; } if (n < 1 || (x < 0 && !(n&1))) { return 0.0 / 0...
Port the provided Tcl code into C# while preserving the original functionality.
proc nthroot {n A} { expr {pow($A, 1.0/$n)} }
static void Main(string[] args) { Console.WriteLine(NthRoot(81,2,.001)); Console.WriteLine(NthRoot(1000,3,.001)); Console.ReadLine(); } public static double NthRoot(double A,int n, double p) { double _n= (double) n; double[] x = new double[2]; x[0] = A; x[1] = A/_n; while(Math.Abs(x[0] -x[1] )...
Generate a C++ translation of this Tcl snippet without changing its computational steps.
proc nthroot {n A} { expr {pow($A, 1.0/$n)} }
double NthRoot(double m_nValue, double index, double guess, double pc) { double result = guess; double result_next; do { result_next = (1.0/index)*((index-1.0)*result+(m_nValue)/(pow(result,(index-1.0)))); result = result_next; pc--; }while(pc>1); ...
Convert this Tcl block to Java, preserving its control flow and logic.
proc nthroot {n A} { expr {pow($A, 1.0/$n)} }
public static double nthroot(int n, double A) { return nthroot(n, A, .001); } public static double nthroot(int n, double A, double p) { if(A < 0) { System.err.println("A < 0"); return -1; } else if(A == 0) { return 0; } double x_prev = A; double x = A / n; while(Math.abs(x - x_prev) > p) { x_prev = x; ...
Convert this Tcl snippet to Python and keep its semantics consistent.
proc nthroot {n A} { expr {pow($A, 1.0/$n)} }
from decimal import Decimal, getcontext def nthroot (n, A, precision): getcontext().prec = precision n = Decimal(n) x_0 = A / n x_1 = 1 while True: x_0, x_1 = x_1, (1 / n)*((n - 1)*x_0 + (A / (x_0 ** (n - 1)))) if x_0 == x_1: return x_1
Can you help me rewrite this code in VB instead of Tcl, keeping it the same logically?
proc nthroot {n A} { expr {pow($A, 1.0/$n)} }
Private Function nth_root(y As Double, n As Double) Dim eps As Double: eps = 0.00000000000001 Dim x As Variant: x = 1 Do While True d = (y / x ^ (n - 1) - x) / n x = x + d e = eps * x If d > -e And d < e Then Exit Do End If Loop Debug.Print y; n;...
Convert this Tcl block to Go, preserving its control flow and logic.
proc nthroot {n A} { expr {pow($A, 1.0/$n)} }
func root(a float64, n int) float64 { n1 := n - 1 n1f, rn := float64(n1), 1/float64(n) x, x0 := 1., 0. for { potx, t2 := 1/x, a for b := n1; b > 0; b >>= 1 { if b&1 == 1 { t2 *= potx } potx *= potx } x0, x = x, rn*(n1f*x...
Preserve the algorithm and functionality while converting the code from Rust to PHP.
fn nthRoot(n: f64, A: f64) -> f64 { let p = 1e-9_f64 ; let mut x0 = A / n ; loop { let mut x1 = ( (n-1.0) * x0 + A / f64::powf(x0, n-1.0) ) / n; if (x1-x0).abs() < (x0*p).abs() { return x1 }; x0 = x1 } } fn main() { println!("{}", nthRoot(3. , 8. )); }
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Produce a functionally identical PHP code for the snippet given in Ada.
with Ada.Text_IO; use Ada.Text_IO; procedure Test_Nth_Root is generic type Real is digits <>; function Nth_Root (Value : Real; N : Positive) return Real; function Nth_Root (Value : Real; N : Positive) return Real is type Index is mod 2; X : array (Index) of Real := (Value, Value); ...
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Write the same code in PHP as shown below in Arturo.
nthRoot: function [a,n][ N: to :floating n result: a x: a / N while [0.000000000000001 < abs result-x][ x: result result: (1//n) * add (n-1)*x a/pow x n-1 ] return result ] print nthRoot 34.0 5 print nthRoot 42.0 10 print nthRoot 5.0 2
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Produce a language-to-language conversion: from AutoHotKey to PHP, same semantics.
p := 0.000001 MsgBox, % nthRoot( 10, 7131.5**10, p) "`n" . nthRoot( 5, 34.0 , p) "`n" . nthRoot( 2, 2 , p) "`n" . nthRoot(0.5, 7 , p) "`n" nthRoot(n, A, p) {  x1 := A x2 := A / n While Abs(x1 - x2) > p { x1 := x2 x2 := ((n-1)*x2+A/x2**(n-1)...
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Port the provided AWK code into PHP while preserving the original functionality.
BEGIN { print nthroot(8,3) print nthroot(16,2) print nthroot(16,4) print nthroot(125,3) print nthroot(3,3) print nthroot(3,2) } function nthroot(y,n) { eps = 1e-15; x = 1; do { d = ( y / ( x^(n-1) ) - x ) / n ; x += d; e = eps*x; } while ( d < -e || d > e ) return ...
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Maintain the same structure and functionality when rewriting this code in PHP.
*FLOAT 64 @% = &D0D PRINT "Cube root of 5 is "; FNroot(3, 5, 0) PRINT "125th root of 5643 is "; FNroot(125, 5643, 0) END DEF FNroot(n%, a, d) LOCAL x0, x1 : x0 = a / n% : REPEAT x1 = ((n% - 1)*x0 + a/x0^(n%-1)) / n% SWAP x0, x1 UNTIL ABS (x0 ...
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Maintain the same structure and functionality when rewriting this code in PHP.
(ns test-project-intellij.core (:gen-class)) (defn abs [x] " Absolute value" (if (< x 0) (- x) x)) (defn power [x n] " x to power n, where n = 0, 1, 2, ... " (apply * (repeat n x))) (defn calc-delta [A x n] " nth rooth algorithm delta calculation " (/ (- (/ A (power x (- n 1))) x) n)) (defn nth-root ...
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Rewrite the snippet below in PHP so it works the same as the original Common_Lisp code.
(defun nth-root (n a &optional (epsilon .0001) (guess (1- n))) (assert (and (> n 1) (> a 0))) (flet ((next (x) (/ (+ (* (1- n) x) (/ a (expt x (1- n)))) n))) (do* ((xi guess xi+1) (xi+1 (next xi) (next xi))) ((< (abs (- xi+1 xi)) epsilon) xi+1))))
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Translate the given D code snippet into PHP without altering its behavior.
import std.stdio, std.math; real nthroot(in int n, in real A, in real p=0.001) pure nothrow { real[2] x = [A, A / n]; while (abs(x[1] - x[0]) > p) x = [x[1], ((n - 1) * x[1] + A / (x[1] ^^ (n-1))) / n]; return x[1]; } void main() { writeln(nthroot(10, 7131.5 ^^ 10)); writeln(nthroot(6, 64)...
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Keep all operations the same but rewrite the snippet in PHP.
USES Math; function NthRoot(A, Precision: Double; n: Integer): Double; var x_p, X: Double; begin x_p := Sqrt(A); while Abs(A - Power(x_p, n)) > Precision do begin x := (1/n) * (((n-1) * x_p) + (A/(Power(x_p, n - 1)))); x_p := x; end; Result := x_p; end;
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Write the same algorithm in PHP as shown in this Elixir implementation.
defmodule RC do def nth_root(n, x, precision \\ 1.0e-5) do f = fn(prev) -> ((n - 1) * prev + x / :math.pow(prev, (n-1))) / n end fixed_point(f, x, precision, f.(x)) end defp fixed_point(_, guess, tolerance, next) when abs(guess - next) < tolerance, do: next defp fixed_point(f, _, tolerance, next), do...
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Produce a functionally identical PHP code for the snippet given in Erlang.
fixed_point(F, Guess, Tolerance) -> fixed_point(F, Guess, Tolerance, F(Guess)). fixed_point(_, Guess, Tolerance, Next) when abs(Guess - Next) < Tolerance -> Next; fixed_point(F, _, Tolerance, Next) -> fixed_point(F, Next, Tolerance, F(Next)).
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Produce a functionally identical PHP code for the snippet given in F#.
let nthroot n A = let rec f x = let m = n - 1. let x' = (m * x + A/x**m) / n match abs(x' - x) with | t when t < abs(x * 1e-9) -> x' | _ -> f x' f (A / double n) [<EntryPoint>] let main args = if args.Length <> 2 then eprintfn "usage: nthroot n A" exi...
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Transform the following Factor implementation into PHP, maintaining the same output and logic.
USING: kernel locals math math.functions prettyprint ; :: th-root ( a n -- a^1/n ) a [ a over n 1 - ^ /f over n 1 - * + n /f swap over 1e-5 ~ not ] loop ; 34 5 th-root . 34 5 recip ^ .
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Produce a language-to-language conversion: from Forth to PHP, same semantics.
: th-root { F: a F: n -- a^1/n } a begin a fover n 1e f- f** f/ fover n 1e f- f* f+ n f/ fswap fover 1e-5 f~ until ; 34e 5e th-root f. 34e 5e 1/f f** f.
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Write the same algorithm in PHP as shown in this Fortran implementation.
program NthRootTest implicit none print *, nthroot(10, 7131.5**10) print *, nthroot(5, 34.0) contains function nthroot(n, A, p) real :: nthroot integer, intent(in) :: n real, intent(in) :: A real, intent(in), optional :: p real :: rp, x(2) if ( A < 0 ) then s...
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Translate this program into PHP but keep the logic exactly as in Groovy.
import static Constants.tolerance import static java.math.RoundingMode.HALF_UP def root(double base, double n) { double xOld = 1 double xNew = 0 while (true) { xNew = ((n - 1) * xOld + base/(xOld)**(n - 1))/n if ((xNew - xOld).abs() < tolerance) { break } xOld = xNew } (xNew as ...
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Write a version of this Haskell function in PHP with identical behavior.
n `nthRoot` x = fst $ until (uncurry(==)) (\(_,x0) -> (x0,((n-1)*x0+x/x0**(n-1))/n)) (x,x/n)
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Transform the following Icon implementation into PHP, maintaining the same output and logic.
procedure main() showroot(125,3) showroot(27,3) showroot(1024,10) showroot(39.0625,4) showroot(7131.5^10,10) end procedure showroot(a,n) printf("%i-th root of %i = %i\n",n,a,root(a,n)) end procedure root(a,n,p) if n < 0 | type(n) !== "integer" then runerr(101,n) if a < 0 then runerr(205,...
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Write the same algorithm in PHP as shown in this J implementation.
'`N X NP' =. (0 { [)`(1 { [)`(2 { [) iter =. N %~ (NP * ]) + X % ] ^ NP nth_root =: (, , _1+[) iter^:_ f. ] 10 nth_root 7131.5^10 7131.5
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Change the following Julia code into PHP without altering its purpose.
function nthroot(n::Integer, r::Real) r < 0 || n == 0 && throw(DomainError()) n < 0 && return 1 / nthroot(-n, r) r > 0 || return 0 x = r / n prevdx = r while true y = x ^ (n - 1) dx = (r - y * x) / (n * y) abs(dx) ≥ abs(prevdx) && return x x += dx prevdx =...
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Keep all operations the same but rewrite the snippet in PHP.
function nroot(root, num) return num^(1/root) end
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Keep all operations the same but rewrite the snippet in PHP.
Root[A,n]
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Port the following code from MATLAB to PHP with equivalent syntax and logic.
function answer = nthRoot(number,root) format long answer = number / root; guess = number; while not(guess == answer) guess = answer; answer = (1/root)*( ((root - 1)*guess) + ( number/(guess^(root - 1)) ) ); end end
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Preserve the algorithm and functionality while converting the code from Nim to PHP.
import math proc nthRoot(a: float; n: int): float = var n = float(n) result = a var x = a / n while abs(result-x) > 1e-15: x = result result = (1/n) * (((n-1)*x) + (a / pow(x, n-1))) echo nthRoot(34.0, 5) echo nthRoot(42.0, 10) echo nthRoot(5.0, 2)
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Translate this program into PHP but keep the logic exactly as in OCaml.
let nthroot ~n ~a ?(tol=0.001) () = let nf = float n in let nf1 = nf -. 1.0 in let rec iter x = let x' = (nf1 *. x +. a /. (x ** nf1)) /. nf in if tol > abs_float (x -. x') then x' else iter x' in iter 1.0 ;; let () = Printf.printf "%g\n" (nthroot 10 (7131.5 ** 10.0) ()); Printf.printf "%g\n" ...
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Please provide an equivalent version of this Perl code in PHP.
use strict; sub nthroot ($$) { my ( $n, $A ) = @_; my $x0 = $A / $n; my $m = $n - 1.0; while(1) { my $x1 = ($m * $x0 + $A / ($x0 ** $m)) / $n; return $x1 if abs($x1 - $x0) < abs($x0 * 1e-9); $x0 = $x1; } }
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Change the following PowerShell code into PHP without altering its purpose.
$epsilon=1E-10 function power($x,$e){ $ret=1 for($i=1;$i -le $e;$i++){ $ret*=$x } return $ret } function root($y,$n){ if (0+$n -lt 0){$tmp=-$n} else {$tmp=$n} $ans=1 do{ $d = ($y/(power $ans ($tmp-1)) - $ans)/$tmp $ans+=$d } while ($d -lt -$epsilon -or $d -gt $epsilon) if (0+$n -lt 0...
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Convert this R block to PHP, preserving its control flow and logic.
nthroot <- function(A, n, tol=sqrt(.Machine$double.eps)) { ifelse(A < 1, x0 <- A * n, x0 <- A / n) repeat { x1 <- ((n-1)*x0 + A / x0^(n-1))/n if(abs(x1 - x0) > tol) x0 <- x1 else break } x1 } nthroot(7131.5^10, 10) nthroot(7, 0.5)
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Preserve the algorithm and functionality while converting the code from Racket to PHP.
#lang racket (define (nth-root number root (tolerance 0.001)) (define (acceptable? next current) (< (abs (- next current)) tolerance)) (define (improve current) (/ (+ (* (- root 1) current) (/ number (expt current (- root 1)))) root)) (define (loop current) (define next-guess (improve current))...
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Convert this COBOL snippet to PHP and keep its semantics consistent.
IDENTIFICATION DIVISION. PROGRAM-ID. Nth-Root. AUTHOR. Bill Gunshannon. INSTALLATION. DATE-WRITTEN. 4 Feb 2020. ************************************************************ ** Program Abstract: ** Compute the Nth Root of a positive real number. ** ...
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Can you help me rewrite this code in PHP instead of REXX, keeping it the same logically?
class nth_root method main(args=String[]) static if args.length < 2 then do say "at least 2 arguments expected" exit end x = args[0] root = args[1] if args.length > 2 then digs = args[2] if root=='' then root=2 if digs = null, digs = '' then digs=20 numeric digits digs ...
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Rewrite this program in PHP while keeping its functionality equivalent to the Ruby version.
def nthroot(n, a, precision = 1e-5) x = Float(a) begin prev = x x = ((n - 1) * prev + a / (prev ** (n - 1))) / n end while (prev - x).abs > precision x end p nthroot(5,34)
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Keep all operations the same but rewrite the snippet in PHP.
fun nthRoot(x: Double, n: Int): Double { if (n < 2) throw IllegalArgumentException("n must be more than 1") if (x <= 0.0) throw IllegalArgumentException("x must be positive") val np = n - 1 fun iter(g: Double) = (np * g + x / Math.pow(g, np.toDouble())) / n var g1 = x var g2 = iter(g1) whi...
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Convert the following code from Swift to PHP, ensuring the logic remains intact.
extension FloatingPoint where Self: ExpressibleByFloatLiteral { @inlinable public func power(_ e: Int) -> Self { var res = Self(1) for _ in 0..<e { res *= self } return res } @inlinable public func root(n: Int, epsilon: Self = 2.220446049250313e-16) -> Self { var d = Self(0) v...
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Maintain the same structure and functionality when rewriting this code in PHP.
proc nthroot {n A} { expr {pow($A, 1.0/$n)} }
function nthroot($number, $root, $p = P) { $x[0] = $number; $x[1] = $number/$root; while(abs($x[1]-$x[0]) > $p) { $x[0] = $x[1]; $x[1] = (($root-1)*$x[1] + $number/pow($x[1], $root-1))/$root; } return $x[1]; }
Convert this C block to Rust, preserving its control flow and logic.
#include <stdio.h> #include <float.h> double pow_ (double x, int e) { int i; double r = 1; for (i = 0; i < e; i++) { r *= x; } return r; } double root (int n, double x) { double d, r = 1; if (!x) { return 0; } if (n < 1 || (x < 0 && !(n&1))) { return 0.0 / 0...
fn nthRoot(n: f64, A: f64) -> f64 { let p = 1e-9_f64 ; let mut x0 = A / n ; loop { let mut x1 = ( (n-1.0) * x0 + A / f64::powf(x0, n-1.0) ) / n; if (x1-x0).abs() < (x0*p).abs() { return x1 }; x0 = x1 } } fn main() { println!("{}", nthRoot(3. , 8. )); }