Instruction stringlengths 45 106 | input_code stringlengths 1 13.7k | output_code stringlengths 1 13.7k |
|---|---|---|
Translate the given F# code snippet into VB without altering its behavior. | 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... | 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;... |
Preserve the algorithm and functionality while converting the code from F# to Go. | 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... | 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... |
Keep all operations the same but rewrite the snippet in C. | 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 ^ .
| #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... |
Translate the given Factor code snippet into C# without altering its behavior. | 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 ^ .
| 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] )... |
Rewrite the snippet below in C++ so it works the same as the original Factor code. | 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 ^ .
| 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 Factor code into Java without altering its purpose. | 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 ^ .
| 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;
... |
Maintain the same structure and functionality when rewriting this code in Python. | 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 ^ .
| 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 Factor snippet without changing its computational steps. | 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 ^ .
| 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 functionally identical Go code for the snippet given in Factor. | 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 ^ .
| 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 Forth to C, 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.
| #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 the following code from Forth to C#, ensuring the logic remains intact. | : 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.
| 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] )... |
Rewrite the snippet below in C++ so it works the same as the original Forth code. | : 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.
| 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);
... |
Can you help me rewrite this code in Java instead of Forth, keeping it the same logically? | : 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.
| 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;
... |
Maintain the same structure and functionality when rewriting this code in Python. | : 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.
| 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 Forth, keeping it the same logically? | : 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.
| 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 Forth snippet to Go and keep its semantics consistent. | : 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.
| 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 Fortran snippet without changing its computational steps. | 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... | 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] )... |
Transform the following Fortran implementation into C++, maintaining the same output and logic. | 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... | 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);
... |
Port the following code from Fortran to C with equivalent syntax and logic. | 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... | #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... |
Maintain the same structure and functionality when rewriting this code in Go. | 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... | 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 code in Java as shown below in Fortran. | 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... | 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 Fortran snippet to Python and keep its semantics consistent. | 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... | 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 the following code from Fortran to PHP, ensuring the logic remains intact. | 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];
}
|
Write a version of this Groovy function in C with identical behavior. | 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 ... | #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 Groovy to C#. | 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 ... | 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 following code from Groovy to C++ with equivalent syntax and logic. | 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 ... | 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);
... |
Ensure the translated Java code behaves exactly like the original Groovy snippet. | 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 ... | 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 Groovy block to Python, preserving its control flow and logic. | 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 ... | 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 Groovy implementation. | 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 ... | 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;... |
Translate the given Groovy code snippet into Go without altering its behavior. | 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 ... | 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... |
Ensure the translated C code behaves exactly like the original Haskell snippet. | n `nthRoot` x = fst $ until (uncurry(==)) (\(_,x0) -> (x0,((n-1)*x0+x/x0**(n-1))/n)) (x,x/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... |
Can you help me rewrite this code in C# instead of Haskell, keeping it the same logically? | n `nthRoot` x = fst $ until (uncurry(==)) (\(_,x0) -> (x0,((n-1)*x0+x/x0**(n-1))/n)) (x,x/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] )... |
Rewrite the snippet below in C++ so it works the same as the original Haskell code. | n `nthRoot` x = fst $ until (uncurry(==)) (\(_,x0) -> (x0,((n-1)*x0+x/x0**(n-1))/n)) (x,x/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);
... |
Preserve the algorithm and functionality while converting the code from Haskell to Java. | n `nthRoot` x = fst $ until (uncurry(==)) (\(_,x0) -> (x0,((n-1)*x0+x/x0**(n-1))/n)) (x,x/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;
... |
Write the same algorithm in Python as shown in this Haskell implementation. | n `nthRoot` x = fst $ until (uncurry(==)) (\(_,x0) -> (x0,((n-1)*x0+x/x0**(n-1))/n)) (x,x/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
|
Produce a language-to-language conversion: from Haskell to VB, same semantics. | n `nthRoot` x = fst $ until (uncurry(==)) (\(_,x0) -> (x0,((n-1)*x0+x/x0**(n-1))/n)) (x,x/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;... |
Preserve the algorithm and functionality while converting the code from Haskell to Go. | n `nthRoot` x = fst $ until (uncurry(==)) (\(_,x0) -> (x0,((n-1)*x0+x/x0**(n-1))/n)) (x,x/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... |
Translate this program into C but keep the logic exactly as in Icon. | 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,... | #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 a version of this Icon function in C# with identical behavior. | 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,... | 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] )... |
Ensure the translated C++ code behaves exactly like the original Icon snippet. | 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,... | 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);
... |
Preserve the algorithm and functionality while converting the code from Icon to Java. | 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,... | 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;
... |
Generate a Python translation of this Icon snippet without changing its computational steps. | 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,... | 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
|
Ensure the translated VB code behaves exactly like the original Icon snippet. | 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,... | 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. | 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,... | 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 J to C without modifying what it does. | '`N X NP' =. (0 { [)`(1 { [)`(2 { [)
iter =. N %~ (NP * ]) + X % ] ^ NP
nth_root =: (, , _1+[) iter^:_ f. ]
10 nth_root 7131.5^10
7131.5
| #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... |
Write the same code in C# as shown below in J. | '`N X NP' =. (0 { [)`(1 { [)`(2 { [)
iter =. N %~ (NP * ]) + X % ] ^ NP
nth_root =: (, , _1+[) iter^:_ f. ]
10 nth_root 7131.5^10
7131.5
| 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] )... |
Port the following code from J to C++ with equivalent syntax and logic. | '`N X NP' =. (0 { [)`(1 { [)`(2 { [)
iter =. N %~ (NP * ]) + X % ] ^ NP
nth_root =: (, , _1+[) iter^:_ f. ]
10 nth_root 7131.5^10
7131.5
| 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);
... |
Write the same algorithm in Java 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
| 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 this program in Python while keeping its functionality equivalent to the J version. | '`N X NP' =. (0 { [)`(1 { [)`(2 { [)
iter =. N %~ (NP * ]) + X % ] ^ NP
nth_root =: (, , _1+[) iter^:_ f. ]
10 nth_root 7131.5^10
7131.5
| 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
|
Change the programming language of this snippet from J to VB without modifying what it does. | '`N X NP' =. (0 { [)`(1 { [)`(2 { [)
iter =. N %~ (NP * ]) + X % ] ^ NP
nth_root =: (, , _1+[) iter^:_ f. ]
10 nth_root 7131.5^10
7131.5
| 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;... |
Port the provided J code into Go while preserving the original functionality. | '`N X NP' =. (0 { [)`(1 { [)`(2 { [)
iter =. N %~ (NP * ]) + X % ] ^ NP
nth_root =: (, , _1+[) iter^:_ f. ]
10 nth_root 7131.5^10
7131.5
| 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... |
Change the programming language of this snippet from Julia to C without modifying what it does. | 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 =... | #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... |
Translate this program into C# but keep the logic exactly as in Julia. | 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 =... | 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 Julia code into C++ while preserving the original functionality. | 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 =... | 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 the snippet below in Java so it works the same as the original Julia code. | 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 =... | 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;
... |
Generate an equivalent Python version of this Julia code. | 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 =... | 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 Julia snippet without changing its computational steps. | 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 =... | 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 provided Julia code into Go while preserving the original functionality. | 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 =... | 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... |
Convert this Lua block to C, preserving its control flow and logic. | function nroot(root, num)
return num^(1/root)
end
| #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... |
Produce a language-to-language conversion: from Lua to C#, same semantics. | function nroot(root, num)
return num^(1/root)
end
| 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] )... |
Maintain the same structure and functionality when rewriting this code in C++. | function nroot(root, num)
return num^(1/root)
end
| 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);
... |
Maintain the same structure and functionality when rewriting this code in Java. | function nroot(root, num)
return num^(1/root)
end
| 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 the following code from Lua to Python, ensuring the logic remains intact. | function nroot(root, num)
return num^(1/root)
end
| 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. | function nroot(root, num)
return num^(1/root)
end
| 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 Lua to Go with equivalent syntax and logic. | function nroot(root, num)
return num^(1/root)
end
| 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... |
Convert this Mathematica snippet to C and keep its semantics consistent. | Root[A,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... |
Produce a functionally identical C# code for the snippet given in Mathematica. | Root[A,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] )... |
Port the following code from Mathematica to C++ with equivalent syntax and logic. | Root[A,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);
... |
Generate a Java translation of this Mathematica snippet without changing its computational steps. | Root[A,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 Mathematica snippet to Python and keep its semantics consistent. | Root[A,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
|
Rewrite the snippet below in VB so it works the same as the original Mathematica code. | Root[A,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;... |
Change the programming language of this snippet from Mathematica to Go without modifying what it does. | Root[A,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... |
Port the provided MATLAB code into C while preserving the original functionality. | 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
| #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... |
Can you help me rewrite this code in C# instead of MATLAB, keeping it the same logically? | 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
| 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++. | 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
| 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);
... |
Preserve the algorithm and functionality while converting the code from MATLAB to Java. | 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
| 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 following code from MATLAB to Python 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
| 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 following code from MATLAB to VB 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
| 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 MATLAB block to Go, preserving its control flow 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
| 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... |
Maintain the same structure and functionality when rewriting this code in C. | 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)
| #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 Nim to C#. | 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)
| 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] )... |
Write a version of this Nim function in C++ with identical behavior. | 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)
| 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);
... |
Write a version of this Nim function in Java with identical behavior. | 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)
| 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;
... |
Preserve the algorithm and functionality while converting the code from Nim to Python. | 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)
| 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
|
Please provide an equivalent version of this Nim code in VB. | 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)
| 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 following Nim code into Go without altering its purpose. | 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)
| 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 a version of this OCaml function in C with identical behavior. | 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" ... | #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 OCaml code into C# without altering its purpose. | 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" ... | 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] )... |
Rewrite this program in C++ while keeping its functionality equivalent to the OCaml version. | 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" ... | 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);
... |
Write the same code in Java as shown below 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" ... | 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;
... |
Produce a functionally identical Python code for the snippet given 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" ... | 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
|
Ensure the translated VB code behaves exactly like the original OCaml snippet. | 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" ... | 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 OCaml to Go without modifying what it does. | 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" ... | 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... |
Rewrite the snippet below in C so it works the same as the original Perl code. | use strict;
sub nthroot ($$)
{
my ( $n, $A ) = @_;
my $x0 = $A / $n;
my $m = $n - 1.0;
while(1) {
my $x1 = ($m * $x0 + $A / ($x0 ** $m)) / $n;
return $x1 if abs($x1 - $x0) < abs($x0 * 1e-9);
$x0 = $x1;
}
}
| #include <stdio.h>
#include <float.h>
double pow_ (double x, int e) {
int i;
double r = 1;
for (i = 0; i < e; i++) {
r *= x;
}
return r;
}
double root (int n, double x) {
double d, r = 1;
if (!x) {
return 0;
}
if (n < 1 || (x < 0 && !(n&1))) {
return 0.0 / 0... |
Change the following Perl code into C# without altering its purpose. | use strict;
sub nthroot ($$)
{
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;
}
}
| 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 an equivalent C++ version of this Perl code. | use strict;
sub nthroot ($$)
{
my ( $n, $A ) = @_;
my $x0 = $A / $n;
my $m = $n - 1.0;
while(1) {
my $x1 = ($m * $x0 + $A / ($x0 ** $m)) / $n;
return $x1 if abs($x1 - $x0) < abs($x0 * 1e-9);
$x0 = $x1;
}
}
| double NthRoot(double m_nValue, double index, double guess, double pc)
{
double result = guess;
double result_next;
do
{
result_next = (1.0/index)*((index-1.0)*result+(m_nValue)/(pow(result,(index-1.0))));
result = result_next;
pc--;
}while(pc>1);
... |
Produce a functionally identical Java code for the snippet given in Perl. | use strict;
sub nthroot ($$)
{
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;
}
}
| 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;
... |
Generate a Python translation of this Perl snippet without changing its computational steps. | use strict;
sub nthroot ($$)
{
my ( $n, $A ) = @_;
my $x0 = $A / $n;
my $m = $n - 1.0;
while(1) {
my $x1 = ($m * $x0 + $A / ($x0 ** $m)) / $n;
return $x1 if abs($x1 - $x0) < abs($x0 * 1e-9);
$x0 = $x1;
}
}
| from decimal import Decimal, getcontext
def nthroot (n, A, precision):
getcontext().prec = precision
n = Decimal(n)
x_0 = A / n
x_1 = 1
while True:
x_0, x_1 = x_1, (1 / n)*((n - 1)*x_0 + (A / (x_0 ** (n - 1))))
if x_0 == x_1:
return x_1
|
Port the following code from Perl to VB with equivalent syntax and logic. | use strict;
sub nthroot ($$)
{
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;
}
}
| 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 Perl snippet. | 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;
}
}
| 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... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.