Van der Corput sequence

From Rosetta Code
Jump to: navigation, search
Task
Van der Corput sequence
You are encouraged to solve this task according to the task description, using any language you may know.

When counting integers in binary, if you put a (binary) point to the right of the count then the column immediately to the left denotes a digit with a multiplier of 20; the next column to the lefts digit has a multiplier of 21 and so on.

So in the following table:

  0.
  1.
 10.
 11.
 ...

The binary number "10" is 1 \times 2^1 + 0 \times 2^0.

You can have binary digits to the right of the “point” just as in the decimal number system too. in this case, the digit in the place immediately to the right of the point has a weight of 2 − 1, or 1 / 2. The weight for the second column to the right of the point is 2 − 2 or 1 / 4. And so on.

If you take the integer binary count of the first table, and reflect the digits about the binary point, you end up with the van der Corput sequence of numbers in base 2.

  .0
  .1
  .01
  .11
  ...

The third member of the sequence: binary 0.01 is therefore 0 \times 2^{-1} + 1 \times 2^{-2} or 1 / 4.


Distribution of 2500 points each: Van der Corput (top) vs pseudorandom
Members of the sequence lie within the interval 0 \leq x < 1. Points within the sequence tend to be evenly distributed which is a useful trait to have for Monte Carlo simulations.

This sequence is also a superset of the numbers representable by the "fraction" field of an old IEEE floating point standard. In that standard, the "fraction" field represented the fractional part of a binary number beginning with "1." e.g. 1.101001101.

Hint

A hint at a way to generate members of the sequence is to modify a routine used to change the base of an integer:

>>> def base10change(n, base):
digits = []
while n:
n,remainder = divmod(n, base)
digits.insert(0, remainder)
return digits
 
>>> base10change(11, 2)
[1, 0, 1, 1]

the above showing that 11 in decimal is 1\times 2^3 + 0\times 2^2 + 1\times 2^1 + 1\times 2^0.
Reflected this would become .1101 or 1\times 2^{-1} + 1\times 2^{-2} + 0\times 2^{-3} + 1\times 2^{-4}

Task Description

  • Create a function/method/routine that given n, generates the n'th term of the van der Corput sequence in base 2.
  • Use the function to compute and display the first ten members of the sequence. (The first member of the sequence is for n=0).
  • As a stretch goal/extra credit, compute and show members of the sequence for bases other than 2.

See also

Contents

[edit] ActionScript

This implementation uses logarithms to computes the nth term of the sequence at any base. Numbers in the output are rounded to 6 decimal places to hide any floating point inaccuracies.

 
package {
 
import flash.display.Sprite;
import flash.events.Event;
 
public class VanDerCorput extends Sprite {
 
public function VanDerCorput():void {
if (stage) init();
else addEventListener(Event.ADDED_TO_STAGE, init);
}
 
private function init(e:Event = null):void {
 
removeEventListener(Event.ADDED_TO_STAGE, init);
 
var base2:Vector.<Number> = new Vector.<Number>(10, true);
var base3:Vector.<Number> = new Vector.<Number>(10, true);
var base4:Vector.<Number> = new Vector.<Number>(10, true);
var base5:Vector.<Number> = new Vector.<Number>(10, true);
var base6:Vector.<Number> = new Vector.<Number>(10, true);
var base7:Vector.<Number> = new Vector.<Number>(10, true);
var base8:Vector.<Number> = new Vector.<Number>(10, true);
 
var i:uint;
 
for ( i = 0; i < 10; i++ ) {
base2[i] = Math.round( _getTerm(i, 2) * 1000000 ) / 1000000;
base3[i] = Math.round( _getTerm(i, 3) * 1000000 ) / 1000000;
base4[i] = Math.round( _getTerm(i, 4) * 1000000 ) / 1000000;
base5[i] = Math.round( _getTerm(i, 5) * 1000000 ) / 1000000;
base6[i] = Math.round( _getTerm(i, 6) * 1000000 ) / 1000000;
base7[i] = Math.round( _getTerm(i, 7) * 1000000 ) / 1000000;
base8[i] = Math.round( _getTerm(i, 8) * 1000000 ) / 1000000;
}
 
trace("Base 2: " + base2.join(', '));
trace("Base 3: " + base3.join(', '));
trace("Base 4: " + base4.join(', '));
trace("Base 5: " + base5.join(', '));
trace("Base 6: " + base6.join(', '));
trace("Base 7: " + base7.join(', '));
trace("Base 8: " + base8.join(', '));
 
}
 
private function _getTerm(n:uint, base:uint = 2):Number {
 
var r:Number = 0, p:uint, digit:uint;
var baseLog:Number = Math.log(base);
 
while ( n > 0 ) {
p = Math.pow( base, uint(Math.log(n) / baseLog) );
 
digit = n / p;
n %= p;
r += digit / (p * base);
}
 
return r;
 
}
 
}
 
}
 
Output:
Base 2: 0, 0.5, 0.25, 0.75, 0.125, 0.625, 0.375, 0.875, 0.0625, 0.5625
Base 3: 0, 0.333333, 0.666667, 0.111111, 0.444444, 0.777778, 0.222222, 0.555556, 0.888889, 0.037037
Base 4: 0, 0.25, 0.5, 0.75, 0.0625, 0.3125, 0.5625, 0.8125, 0.125, 0.375
Base 5: 0, 0.2, 0.4, 0.6, 0.8, 0.04, 0.24, 0.44, 0.64, 0.84
Base 6: 0, 0.166667, 0.333333, 0.5, 0.666667, 0.833333, 0.027778, 0.194444, 0.361111, 0.527778
Base 7: 0, 0.142857, 0.285714, 0.428571, 0.571429, 0.714286, 0.857143, 0.020408, 0.163265, 0.306122
Base 8: 0, 0.125, 0.25, 0.375, 0.5, 0.625, 0.75, 0.875, 0.015625, 0.140625

[edit] Ada

with Ada.Text_IO;
 
procedure Main is
package Float_IO is new Ada.Text_IO.Float_IO (Float);
function Van_Der_Corput (N : Natural; Base : Positive := 2) return Float is
Value  : Natural  := N;
Result  : Float  := 0.0;
Exponent : Positive := 1;
begin
while Value > 0 loop
Result  := Result +
Float (Value mod Base) / Float (Base ** Exponent);
Value  := Value / Base;
Exponent := Exponent + 1;
end loop;
return Result;
end Van_Der_Corput;
begin
for Base in 2 .. 5 loop
Ada.Text_IO.Put ("Base" & Integer'Image (Base) & ":");
for N in 1 .. 10 loop
Ada.Text_IO.Put (' ');
Float_IO.Put (Item => Van_Der_Corput (N, Base), Exp => 0);
end loop;
Ada.Text_IO.New_Line;
end loop;
end Main;
Output:
Base 2:  0.50000  0.25000  0.75000  0.12500  0.62500  0.37500  0.87500  0.06250  0.56250  0.31250
Base 3:  0.33333  0.66667  0.11111  0.44444  0.77778  0.22222  0.55556  0.88889  0.03704  0.37037
Base 4:  0.25000  0.50000  0.75000  0.06250  0.31250  0.56250  0.81250  0.12500  0.37500  0.62500
Base 5:  0.20000  0.40000  0.60000  0.80000  0.04000  0.24000  0.44000  0.64000  0.84000  0.08000

[edit] AutoHotkey

Works with: AutoHotkey_L
SetFormat, FloatFast, 0.5
for i, v in [2, 3, 4, 5, 6] {
seq .= "Base " v ": "
Loop, 10
seq .= VanDerCorput(A_Index - 1, v) (A_Index = 10 ? "`n" : ", ")
}
MsgBox, % seq
 
VanDerCorput(n, b, r=0) {
while n
r += Mod(n, b) * b ** -A_Index, n := n // b
return, r
}
Output:
Base 2: 0, 0.50000, 0.25000, 0.75000, 0.12500, 0.62500, 0.37500, 0.87500, 0.06250, 0.56250
Base 3: 0, 0.33333, 0.66667, 0.11111, 0.44444, 0.77778, 0.22222, 0.55555, 0.88889, 0.03704
Base 4: 0, 0.25000, 0.50000, 0.75000, 0.06250, 0.31250, 0.56250, 0.81250, 0.12500, 0.37500
Base 5: 0, 0.20000, 0.40000, 0.60000, 0.80000, 0.04000, 0.24000, 0.44000, 0.64000, 0.84000
Base 6: 0, 0.16667, 0.33333, 0.50000, 0.66667, 0.83333, 0.02778, 0.19445, 0.36111, 0.52778

[edit] BBC BASIC

      @% = &20509
FOR base% = 2 TO 5
PRINT "Base " ; STR$(base%) ":"
FOR number% = 0 TO 9
PRINT FNvdc(number%, base%);
NEXT
PRINT
NEXT
END
 
DEF FNvdc(n%, b%)
LOCAL v, s%
s% = 1
WHILE n%
s% *= b%
v += (n% MOD b%) / s%
n% DIV= b%
ENDWHILE
= v
Output:
Base 2:
  0.00000  0.50000  0.25000  0.75000  0.12500  0.62500  0.37500  0.87500  0.06250  0.56250
Base 3:
  0.00000  0.33333  0.66667  0.11111  0.44444  0.77778  0.22222  0.55556  0.88889  0.03704
Base 4:
  0.00000  0.25000  0.50000  0.75000  0.06250  0.31250  0.56250  0.81250  0.12500  0.37500
Base 5:
  0.00000  0.20000  0.40000  0.60000  0.80000  0.04000  0.24000  0.44000  0.64000  0.84000

[edit] bc

This solution hardcodes the literal 10 because numeric literals in bc can use any base from 2 to 16. This solution only works with integer bases from 2 to 16.

/*
* Return the _n_th term of the van der Corput sequence.
* Uses the current _ibase_.
*/
define v(n) {
auto c, r, s
 
s = scale
scale = 0 /* to use integer division */
 
/*
* c = count digits of n
* r = reverse the digits of n
*/
for (0; n != 0; n /= 10) {
c += 1
r = (10 * r) + (n % 10)
}
 
/* move radix point to left of digits */
scale = length(r) + 6
r /= 10 ^ c
 
scale = s
return r
}
 
t = 10
for (b = 2; b <= 4; b++) {
"base "; b
obase = b
for (i = 0; i < 10; i++) {
ibase = b
" "; v(i)
ibase = t
}
obase = t
}
quit

Some of the calculations are not exact, because bc performs calculations using base 10. So the program prints a result like .202222221 (base 3) when the exact result would be .21 (base 3).

Output:
base 2
  0.00000000000000
  .10000000000000
  .01000000000000
  .11000000000000
  .00100000000000
  .10100000000000
  .01100000000000
  .11100000000000
  .00010000000000
  .10010000000000
base 3
  0.000000000
  .022222222
  .122222221
  .002222222
  .102222222
  .202222221
  .012222222
  .112222221
  .212222221
  .000222222
base 4
  0.0000000
  .1000000
  .2000000
  .3000000
  .0100000
  .1100000
  .2100000
  .310000000
  .0200000
  .1200000

[edit] C

#include <stdio.h>
 
void vc(int n, int base, int *num, int *denom)
{
int p = 0, q = 1;
 
while (n) {
p = p * base + (n % base);
q *= base;
n /= base;
}
 
*num = p;
*denom = q;
 
while (p) { n = p; p = q % p; q = n; }
*num /= q;
*denom /= q;
}
 
int main()
{
int d, n, i, b;
for (b = 2; b < 6; b++) {
printf("base %d:", b);
for (i = 0; i < 10; i++) {
vc(i, b, &n, &d);
if (n) printf("  %d/%d", n, d);
else printf(" 0");
}
printf("\n");
}
 
return 0;
}
Output:
base 2:  0  1/2  1/4  3/4  1/8  5/8  3/8  7/8  1/16  9/16
base 3:  0  1/3  2/3  1/9  4/9  7/9  2/9  5/9  8/9  1/27
base 4:  0  1/4  1/2  3/4  1/16  5/16  9/16  13/16  1/8  3/8
base 5:  0  1/5  2/5  3/5  4/5  1/25  6/25  11/25  16/25  21/25

[edit] C++

Translation of: Perl 6
#include <cmath>
#include <iostream>
 
double vdc(int n, double base = 2)
{
double vdc = 0, denom = 1;
while (n)
{
vdc += fmod(n, base) / (denom *= base);
n /= base; // note: conversion from 'double' to 'int'
}
return vdc;
}
 
int main()
{
for (double base = 2; base < 6; ++base)
{
std::cout << "Base " << base << "\n";
for (int n = 0; n < 10; ++n)
{
std::cout << vdc(n, base) << " ";
}
std::cout << "\n\n";
}
}
Output:
Base 2
0 0.5 0.25 0.75 0.125 0.625 0.375 0.875 0.0625 0.5625 

Base 3
0 0.333333 0.666667 0.111111 0.444444 0.777778 0.222222 0.555556 0.888889 0.037037 

Base 4
0 0.25 0.5 0.75 0.0625 0.3125 0.5625 0.8125 0.125 0.375 

Base 5
0 0.2 0.4 0.6 0.8 0.04 0.24 0.44 0.64 0.84 

[edit] C#

This is based on the C version.
It uses LINQ and enumeration over a collection to package the sequence and make it easy to use. Note that the iterator returns a generic Tuple whose items are the numerator and denominator for the item.

 
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
 
namespace VanDerCorput
{
/// <summary>
/// Computes the Van der Corput sequence for any number base.
/// The numbers in the sequence vary from zero to one, including zero but excluding one.
/// The sequence possesses low discrepancy.
/// Here are the first ten terms for bases 2 to 5:
///
/// base 2: 0 1/2 1/4 3/4 1/8 5/8 3/8 7/8 1/16 9/16
/// base 3: 0 1/3 2/3 1/9 4/9 7/9 2/9 5/9 8/9 1/27
/// base 4: 0 1/4 1/2 3/4 1/16 5/16 9/16 13/16 1/8 3/8
/// base 5: 0 1/5 2/5 3/5 4/5 1/25 6/25 11/25 16/25 21/25
/// </summary>
/// <see cref="http://rosettacode.org/wiki/Van_der_Corput_sequence"/>
public class VanDerCorputSequence: IEnumerable<Tuple<long,long>>
{
/// <summary>
/// Number base for the sequence, which must bwe two or more.
/// </summary>
public int Base { get; private set; }
 
/// <summary>
/// Maximum number of terms to be returned by iterator.
/// </summary>
public long Count { get; private set; }
 
/// <summary>
/// Construct a sequence for the given base.
/// </summary>
/// <param name="iBase">Number base for the sequence.</param>
/// <param name="count">Maximum number of items to be returned by the iterator.</param>
public VanDerCorputSequence(int iBase, long count = long.MaxValue) {
if (iBase < 2)
throw new ArgumentOutOfRangeException("iBase", "must be two or greater, not the given value of " + iBase);
Base = iBase;
Count = count;
}
 
/// <summary>
/// Compute nth term in the Van der Corput sequence for the base specified in the constructor.
/// </summary>
/// <param name="n">The position in the sequence, which may be zero or any positive number.</param>
/// This number is always an integral power of the base.</param>
/// <returns>The Van der Corput sequence value expressed as a Tuple containing a numerator and a denominator.</returns>
public Tuple<long,long> Compute(long n)
{
long p = 0, q = 1;
long numerator, denominator;
while (n != 0)
{
p = p * Base + (n % Base);
q *= Base;
n /= Base;
}
numerator = p;
denominator = q;
while (p != 0)
{
n = p;
p = q % p;
q = n;
}
numerator /= q;
denominator /= q;
return new Tuple<long,long>(numerator, denominator);
}
 
/// <summary>
/// Compute nth term in the Van der Corput sequence for the given base.
/// </summary>
/// <param name="iBase">Base to use for the sequence.</param>
/// <param name="n">The position in the sequence, which may be zero or any positive number.</param>
/// <returns>The Van der Corput sequence value expressed as a Tuple containing a numerator and a denominator.</returns>
public static Tuple<long, long> Compute(int iBase, long n)
{
var seq = new VanDerCorputSequence(iBase);
return seq.Compute(n);
}
 
/// <summary>
/// Iterate over the Van Der Corput sequence.
/// The first value in the sequence is always zero, regardless of the base.
/// </summary>
/// <returns>A tuple whose items are the Van der Corput value given as a numerator and denominator.</returns>
public IEnumerator<Tuple<long, long>> GetEnumerator()
{
long iSequenceIndex = 0L;
while (iSequenceIndex < Count)
{
yield return Compute(iSequenceIndex);
iSequenceIndex++;
}
}
 
System.Collections.IEnumerator System.Collections.IEnumerable.GetEnumerator()
{
return GetEnumerator();
}
}
 
class Program
{
static void Main(string[] args)
{
TestBasesTwoThroughFive();
 
Console.WriteLine("Type return to continue...");
Console.ReadLine();
}
 
static void TestBasesTwoThroughFive()
{
foreach (var seq in Enumerable.Range(2, 5).Select(x => new VanDerCorputSequence(x, 10))) // Just the first 10 elements of the each sequence
{
Console.Write("base " + seq.Base + ":");
foreach(var vc in seq)
Console.Write(" " + vc.Item1 + "/" + vc.Item2);
Console.WriteLine();
}
}
}
}
 
 
Output:
base 2: 0/1 1/2 1/4 3/4 1/8 5/8 3/8 7/8 1/16 9/16
base 3: 0/1 1/3 2/3 1/9 4/9 7/9 2/9 5/9 8/9 1/27
base 4: 0/1 1/4 1/2 3/4 1/16 5/16 9/16 13/16 1/8 3/8
base 5: 0/1 1/5 2/5 3/5 4/5 1/25 6/25 11/25 16/25 21/25
base 6: 0/1 1/6 1/3 1/2 2/3 5/6 1/36 7/36 13/36 19/36
Type return to continue...

[edit] Clojure

(defn van-der-corput
"Get the nth element of the van der Corput sequence."
([n]
;; Default base = 2
(van-der-corput n 2))
([n base]
(let [s (/ 1 base)] ;; A multiplicand to shift to the right of the decimal.
;; We essentially want to reverse the digits of n and put them after the
;; decimal point. So, we repeatedly pull off the lowest digit of n, scale
;; it to the right of the decimal point, and accumulate that.
(loop [sum 0
n n
scale s]
(if (zero? n)
sum ;; Base case: no digits left, so we're done.
(recur (+ sum (* (rem n base) scale)) ;; Accumulate the least digit
(quot n base) ;; Drop a digit of n
(* scale s))))))) ;; Move farther past the decimal
 
(clojure.pprint/print-table
(cons :base (range 10)) ;; column headings
(for [base (range 2 6)] ;; rows
(into {:base base}
(for [n (range 10)] ;; table entries
[n (van-der-corput n base)]))))
Output:
| :base | 0 |   1 |   2 |   3 |    4 |    5 |    6 |     7 |     8 |     9 |
|-------+---+-----+-----+-----+------+------+------+-------+-------+-------|
|     2 | 0 | 1/2 | 1/4 | 3/4 |  1/8 |  5/8 |  3/8 |   7/8 |  1/16 |  9/16 |
|     3 | 0 | 1/3 | 2/3 | 1/9 |  4/9 |  7/9 |  2/9 |   5/9 |   8/9 |  1/27 |
|     4 | 0 | 1/4 | 1/2 | 3/4 | 1/16 | 5/16 | 9/16 | 13/16 |   1/8 |   3/8 |
|     5 | 0 | 1/5 | 2/5 | 3/5 |  4/5 | 1/25 | 6/25 | 11/25 | 16/25 | 21/25 |


[edit] Common Lisp

(defun van-der-Corput (n base)
(loop for d = 1 then (* d base) while (<= d n)
finally
(return (/ (parse-integer
(reverse (write-to-string n :base base))
:radix base)
d))))
 
(loop for base from 2 to 5 do
(format t "Base ~a: ~{~6a~^~}~%" base
(loop for i to 10 collect (van-der-Corput i base))))
Output:
Base 2: 0     1/2   1/4   3/4   1/8   5/8   3/8   7/8   1/16  9/16  5/16  
Base 3: 0     1/3   2/3   1/9   4/9   7/9   2/9   5/9   8/9   1/27  10/27 
Base 4: 0     1/4   1/2   3/4   1/16  5/16  9/16  13/16 1/8   3/8   5/8   
Base 5: 0     1/5   2/5   3/5   4/5   1/25  6/25  11/25 16/25 21/25 2/25

[edit] D

double vdc(int n, in double base=2.0) pure nothrow @safe @nogc {
double vdc = 0.0, denom = 1.0;
while (n) {
denom *= base;
vdc += (n % base) / denom;
n /= base;
}
return vdc;
}
 
void main() {
import std.stdio, std.algorithm, std.range;
 
foreach (immutable b; 2 .. 6)
writeln("\nBase ", b, ": ", 10.iota.map!(n => vdc(n, b)));
}
Output:
Base 2: [0, 0.5, 0.25, 0.75, 0.125, 0.625, 0.375, 0.875, 0.0625, 0.5625]

Base 3: [0, 0.333333, 0.666667, 0.111111, 0.444444, 0.777778, 0.222222, 0.555556, 0.888889, 0.037037]

Base 4: [0, 0.25, 0.5, 0.75, 0.0625, 0.3125, 0.5625, 0.8125, 0.125, 0.375]

Base 5: [0, 0.2, 0.4, 0.6, 0.8, 0.04, 0.24, 0.44, 0.64, 0.84]

[edit] Ela

open random number list
 
vdc bs n = vdc' 0.0 1.0 n
where vdc' v d n
| n > 0 = vdc' v' d' n'
| else = v
where
d' = d * bs
rem = n % bs
n' = truncate (n / bs)
v' = v + rem / d'

Test (with base 2.0, using non-strict map function on infinite list):

take 10 <| map' (vdc 2.0) [1..]
Output:
[0.5,0.25,0.75,0.125,0.625,0.375,0.875,0.0625,0.5625,0.3125]

[edit] Erlang

I liked the bc output-in-same-base, but think this is the way it should look.

 
-module( van_der_corput ).
 
-export( [sequence/1, sequence/2, task/0] ).
 
sequence( N ) -> sequence( N, 2 ).
 
sequence( 0, _Base ) -> 0.0;
sequence( N, Base ) -> erlang:list_to_float( "0." ++ lists:flatten([erlang:integer_to_list(X) || X <- sequence_loop(N, Base)]) ).
 
task() -> [task(X) || X <- lists:seq(2, 5)].
 
 
 
sequence_loop( 0, _Base ) -> [];
sequence_loop( N, Base ) ->
New_n = N div Base,
Digit = N rem Base,
[Digit | sequence_loop( New_n, Base )].
 
task( Base ) ->
io:fwrite( "Base ~p:", [Base] ),
[io:fwrite( " ~p", [sequence(X, Base)] ) || X <- lists:seq(0, 9)],
io:fwrite( "~n" ).
 
Output:
34> van_der_corput:task().
Base 2: 0.0 0.1 0.01 0.11 0.001 0.101 0.011 0.111 0.0001 0.1001
Base 3: 0.0 0.1 0.2 0.01 0.11 0.21 0.02 0.12 0.22 0.001
Base 4: 0.0 0.1 0.2 0.3 0.01 0.11 0.21 0.31 0.02 0.12
Base 5: 0.0 0.1 0.2 0.3 0.4 0.01 0.11 0.21 0.31 0.41

[edit] Euphoria

Translation of: D
function vdc(integer n, atom base)
atom vdc, denom, rem
vdc = 0
denom = 1
while n do
denom *= base
rem = remainder(n,base)
n = floor(n/base)
vdc += rem / denom
end while
return vdc
end function
 
for i = 2 to 5 do
printf(1,"Base %d\n",i)
for j = 0 to 9 do
printf(1,"%g ",vdc(j,i))
end for
puts(1,"\n\n")
end for
Output:
Base 2
0 0.5 0.25 0.75 0.125 0.625 0.375 0.875 0.0625 0.5625

Base 3
0 0.333333 0.666667 0.111111 0.444444 0.777778 0.222222 0.555556 0.888889 0.037037

Base 4
0 0.25 0.5 0.75 0.0625 0.3125 0.5625 0.8125 0.125 0.375

Base 5
0 0.2 0.4 0.6 0.8 0.04 0.24 0.44 0.64 0.84

[edit] F#

open System
 
let vdc n b =
let rec loop n denom acc =
if n > 0l then
let m, remainder = Math.DivRem(n, b)
loop m (denom * b) (acc + (float remainder) / (float (denom * b)))
else acc
loop n 1 0.0
 
 
[<EntryPoint>]
let main argv =
printfn "%A" [ for n in 0 .. 9 -> (vdc n 2) ]
printfn "%A" [ for n in 0 .. 9 -> (vdc n 5) ]
0
Output:
[0.0; 0.5; 0.25; 0.75; 0.125; 0.625; 0.375; 0.875; 0.0625; 0.5625]
[0.0; 0.2; 0.4; 0.6; 0.8; 0.04; 0.24; 0.44; 0.64; 0.84]

[edit] Forth

: fvdc ( base n -- f )
0e 1e ( F: vdc denominator )
begin dup while
over s>d d>f f*
over /mod ( base rem n )
swap s>d d>f fover f/
frot f+ fswap
repeat 2drop fdrop ;
 
: test 10 0 do 2 i fvdc cr f. loop ;
Output:
test
0.
0.5
0.25
0.75
0.125
0.625
0.375
0.875
0.0625
0.5625  ok

[edit] Go

package main
 
import "fmt"
 
func v2(n uint) (r float64) {
p := .5
for n > 0 {
if n&1 == 1 {
r += p
}
p *= .5
n >>= 1
}
return
}
 
func newV(base uint) func(uint) float64 {
invb := 1 / float64(base)
return func(n uint) (r float64) {
p := invb
for n > 0 {
r += p * float64(n%base)
p *= invb
n /= base
}
return
}
}
 
func main() {
fmt.Println("Base 2:")
for i := uint(0); i < 10; i++ {
fmt.Println(i, v2(i))
}
fmt.Println("Base 3:")
v3 := newV(3)
for i := uint(0); i < 10; i++ {
fmt.Println(i, v3(i))
}
}
Output:
Base 2:
0 0
1 0.5
2 0.25
3 0.75
4 0.125
5 0.625
6 0.375
7 0.875
8 0.0625
9 0.5625
Base 3:
0 0
1 0.3333333333333333
2 0.6666666666666666
3 0.1111111111111111
4 0.4444444444444444
5 0.7777777777777777
6 0.2222222222222222
7 0.5555555555555556
8 0.8888888888888888
9 0.037037037037037035

[edit] Haskell

The function vdc returns the nth exact, arbitrary precision van der Corput number for any base ≥ 2 and any n. (A reasonable value is returned for negative values of n.)

import Data.List
import Data.Ratio
import System.Environment
import Text.Printf
 
-- A wrapper type for Rationals to make them look nicer when we print them.
newtype Rat = Rat Rational
instance Show Rat where
show (Rat n) = show (numerator n) ++ "/" ++ show (denominator n)
 
-- Convert a list of base b digits to its corresponding number. We assume the
-- digits are valid base b numbers and that their order is from least to most
-- significant.
digitsToNum :: Integer -> [Integer] -> Integer
digitsToNum b = foldr1 (\d acc -> b * acc + d)
 
-- Convert a number to the list of its base b digits. The order will be from
-- least to most significant.
numToDigits :: Integer -> Integer -> [Integer]
numToDigits _ 0 = [0]
numToDigits b n = unfoldr step n
where step 0 = Nothing
step m = let (q,r) = m `quotRem` b in Just (r,q)
 
-- Return the n'th element in the base b van der Corput sequence. The base
-- must be ≥ 2.
vdc :: Integer -> Integer -> Rat
vdc b n | b < 2 = error "vdc: base must be ≥ 2"
| otherwise = let ds = reverse $ numToDigits b n
in Rat (digitsToNum b ds % b ^ length ds)
 
-- Print the base followed by a sequence of van der Corput numbers.
printVdc :: (Integer,[Rat]) -> IO ()
printVdc (b,ns) = putStrLn $ printf "Base %d:" b
++ concatMap (printf " %5s" . show) ns
 
-- To print the n'th van der Corput numbers for n in [2,3,4,5] call the program
-- with no arguments. Otherwise, passing the base b, first n, next n and
-- maximum n will print the base b numbers for n in [firstN, nextN, ..., maxN].
main :: IO ()
main = do
args <- getArgs
let (bases, nums) = case args of
[b, f, s, m] -> ([read b], [read f, read s..read m])
_ -> ([2,3,4,5], [0..9])
mapM_ printVdc [(b,rs) | b <- bases, let rs = map (vdc b) nums]
Output:
for small bases:
$ ./vandercorput 
Base 2:   0/1   1/2   1/4   3/4   1/8   5/8   3/8   7/8  1/16  9/16
Base 3:   0/1   1/3   2/3   1/9   4/9   7/9   2/9   5/9   8/9  1/27
Base 4:   0/1   1/4   1/2   3/4  1/16  5/16  9/16 13/16   1/8   3/8
Base 5:   0/1   1/5   2/5   3/5   4/5  1/25  6/25 11/25 16/25 21/25
Output:
for a larger base. (Base 123 for n ∈ [50, 100, …, 300].)
$ ./vandercorput 123 50 100 300
Base 123: 50/123 100/123 3322/15129 9472/15129 494/15129 6644/15129

[edit] Icon and Unicon

The following solution works in both Icon and Unicon:

procedure main(A)
base := integer(get(A)) | 2
every writes(round(vdc(0 to 9,base),10)," ")
write()
end
 
procedure vdc(n, base)
e := 1.0
x := 0.0
while x +:= 1(((0 < n) % base) / (e *:= base), n /:= base)
return x
end
 
procedure round(n,d)
places := 10 ^ d
return real(integer(n*places + 0.5)) / places
end

and a sample run is:

->vdc
0.0 0.5 0.25 0.75 0.125 0.625 0.375 0.875 0.0625 0.5625 
->vdc 3
0.0 0.3333333333 0.6666666667 0.1111111111 0.4444444444 0.7777777778 0.2222222222 0.5555555556 0.8888888889 0.037037037 
->vdc 5
0.0 0.2 0.4 0.6 0.8 0.04 0.24 0.44 0.64 0.84 
->vdc 123
0.0 0.0081300813 0.0162601626 0.0243902439 0.0325203252 0.0406504065 0.0487804878 0.0569105691 0.0650406504 0.07317073170000001 
->

An alternate, Unicon-specific implementation of vdc patterned after the functional Perl 6 solution is:

procedure vdc(n, base)
s1 := create |((0 < 1(.n, n /:= base)) % base)
s2 := create 2(e := 1.0, |(e *:= base))
every (result := 0) +:= |s1() / s2()
return result
end

It produces the same output as shown above.

[edit] J

Solution:

vdc=: ([ %~ %@[ #. #.inv)"0 _

Examples:

   2 vdc i.10                NB. 1st 10 nums of Van der Corput sequence in base 2
0 0.5 0.25 0.75 0.125 0.625 0.375 0.875 0.0625 0.5625
2x vdc i.10 NB. as above but using rational nums
0 1r2 1r4 3r4 1r8 5r8 3r8 7r8 1r16 9r16
2 3 4 5x vdc i.10 NB. 1st 10 nums of Van der Corput sequence in bases 2 3 4 5
0 1r2 1r4 3r4 1r8 5r8 3r8 7r8 1r16 9r16
0 1r3 2r3 1r9 4r9 7r9 2r9 5r9 8r9 1r27
0 1r4 1r2 3r4 1r16 5r16 9r16 13r16 1r8 3r8
0 1r5 2r5 3r5 4r5 1r25 6r25 11r25 16r25 21r25

In other words: use the left argument as the "base" to structure the sequence numbers into digits. Then use the reciprocal of the left argument as the "base" to re-represent this sequence and divide that result by the left argument to get the Van der Corput sequence number.

[edit] Java

Translation of: Perl 6

Using (denom *= 2) as the denominator is not a recommended way of doing things since it is not clear when the multiplication and assignment happen. Comparing this to the "++" operator, it looks like it should do the doubling and assignment second. Comparing it to the "++" operator used as a preincrement operator, it looks like it should do the doubling and assignment first. Comparing it to the behavior of parentheses, it looks like it should do the doubling and assignment first. Luckily for us, it works the same in Java as in Perl 6 (doubling and assignment first). It was kept the Perl 6 way to help with the comparison. Normally, we would initialize denom to 2 (since that is the denominator of the leftmost digit), use it alone in the vdc sum, and then double it after.

public class VanDerCorput{
public static double vdc(int n){
double vdc = 0;
int denom = 1;
while(n != 0){
vdc += n % 2.0 / (denom *= 2);
n /= 2;
}
return vdc;
}
 
public static void main(String[] args){
for(int i = 0; i <= 10; i++){
System.out.println(vdc(i));
}
}
}
Output:
0.0
0.5
0.25
0.75
0.125
0.625
0.375
0.875
0.0625
0.5625
0.3125

[edit] Lua

function vdc(n, base)
local digits = {}
while n ~= 0 do
local m = math.floor(n / base)
table.insert(digits, n - m * base)
n = m
end
m = 0
for p, d in pairs(digits) do
m = m + math.pow(base, -p) * d
end
return m
end

[edit] Mathematica

VanDerCorput[n_,base_:2]:=Table[
FromDigits[{Reverse[IntegerDigits[k,base]],0},base],
{k,n}]


VanDerCorput[10,2]
->{1/2,1/4,3/4,1/8,5/8,3/8,7/8,1/16,9/16,5/16}

VanDerCorput[10,3]
->{1/3, 2/3, 1/9, 4/9, 7/9, 2/9, 5/9, 8/9, 1/27, 10/27}

VanDerCorput[10,4]
->{1/4, 1/2, 3/4, 1/16, 5/16, 9/16, 13/16, 1/8, 3/8, 5/8}

VanDerCorput[10,5]
->{1/5, 2/5, 3/5, 4/5, 1/25, 6/25, 11/25, 16/25, 21/25, 2/25}

[edit] MATLAB / Octave

    function x = corput (n)
b = dec2bin(1:n)-'0'; % generate sequence of binary numbers from 1 to n
l = size(b,2); % get number of binary digits
w = (1:l)-l-1; % 2.^w are the weights
x = b * ( 2.^w'); % matrix times vector multiplication for
end;
Output:
 corput(10)
 ans =

   0.500000
   0.250000
   0.750000
   0.125000
   0.625000
   0.375000
   0.875000
   0.062500
   0.562500
   0.312500

[edit] Maxima

Define two helper functions

/* convert a decimal integer to a list of digits in base `base' */
dec2digits(d, base):= block([digits: []],
while (d>0) do block([newdi: mod(d, base)],
digits: cons(newdi, digits),
d: round( (d - newdi) / base)),
digits)$
 
dec2digits(123, 10);
/* [1, 2, 3] */
dec2digits( 8, 2);
/* [1, 0, 0, 0] */
/* convert a list of digits in base `base' to a decimal integer */
digits2dec(l, base):= block([s: 0, po: 1],
for di in reverse(l) do (s: di*po + s, po: po*base),
s)$
 
digits2dec([1, 2, 3], 10);
/* 123 */
digits2dec([1, 0, 0, 0], 2);
/* 8 */

The main function

vdc(n, base):= makelist(
digits2dec(
dec2digits(k, base),
1/base) / base,
k, n);
 
vdc(10, 2);
/*
1 1 3 1 5 3 7 1 9 5
(%o123) [-, -, -, -, -, -, -, --, --, --]
2 4 4 8 8 8 8 16 16 16
*/
 
vdc(10, 5);
/*
1 2 3 4 1 6 11 16 21 2
(%o124) [-, -, -, -, --, --, --, --, --, --]
5 5 5 5 25 25 25 25 25 25
*/

digits2dec can by used with symbols to produce the same example as in the task description

 
/* 11 in decimal is */
digits: digits2dec([box(1), box(0), box(1), box(1)], box(2));
aux: expand(digits2dec(digits, 1/base) / base)$
simp: false$
/* reflected this would become ... */
subst(box(2), base, aux);
simp: true$
 
/*
 
3 2
""" """ """ """ """ """ """
(%o126) "2" "1" + "2" "0" + "2" "1" + "1"
""" """ """ """ """ """ """
 
- 4 - 3 - 2 - 1
""" """ """ """ """ """ """ """
(%o129) "1" "2" + "0" "2" + "1" "2" + "1" "2"
""" """ """ """ """ """ """ """
 
*/

[edit] PARI/GP

VdC(n)=n=binary(n);sum(i=1,#n,if(n[i],1.>>(#n+1-i)));
VdC(n)=sum(i=1,#binary(n),if(bittest(n,i-1),1.>>i)); \\ Alternate approach
vector(10,n,VdC(n))
Output:
[0.500000000, 0.250000000, 0.750000000, 0.125000000, 0.625000000, 0.375000000, 0.875000000, 0.0625000000, 0.562500000, 0.312500000]

[edit] Perl

Translation of: Perl6
sub vdc {
my @value = shift;
my $base = shift // 2;
use integer;
push @value, $value[-1] / $base while $value[-1] > 0;
my ($x, $sum) = (1, 0);
no integer;
$sum += ($_ % $base) / ($x *= $base) for @value;
return $sum;
}
 
for my $base ( 2 .. 5 ) {
print "base $base: ", join ' ', map { vdc($_, $base) } 0 .. 10;
print "\n";
}

[edit] Perl 6

First we present a fairly standard imperative version in which we mutate three variables in parallel:

sub vdc($num, $base = 2) {
my $n = $num;
my $vdc = 0;
my $denom = 1;
while $n {
$vdc += $n mod $base / ($denom *= $base);
$n div= $base;
}
$vdc;
}
 
for 2..5 -> $b {
say "Base $b";
say (vdc($_,$b) for ^10).perl;
say '';
}
Output:
Base 2
(0, 1/2, 1/4, 3/4, 1/8, 5/8, 3/8, 7/8, 1/16, 9/16)

Base 3
(0, 1/3, 2/3, 1/9, 4/9, 7/9, 2/9, 5/9, 8/9, 1/27)

Base 4
(0, 1/4, 1/2, 3/4, 1/16, 5/16, 9/16, 13/16, 1/8, 3/8)

Base 5
(0, 1/5, 2/5, 3/5, 4/5, 1/25, 6/25, 11/25, 16/25, 21/25)

Here is a functional version that produces the same output:

sub vdc($value, $base = 2) {
my @values := $value, { $_ div $base } ... 0;
my @denoms := $base, { $_ * $base } ... *;
[+] do for @values Z @denoms -> $v, $d {
$v mod $base / $d;
}
}

We first define two sequences, one finite, one infinite. When we zip those sequences together, the finite sequence terminates the loop (which, since a Perl 6 loop returns all its values, is merely another way of writing a map). We then sum with [+], a reduction of the + operator. (We could have in-lined the sequences or used a traditional map operator, but this way seems more readable than the typical FP solution.) The do is necessary to introduce a statement where a term is expected, since Perl 6 distinguishes "sentences" from "noun phrases" as a natural language might.

[edit] PicoLisp

(scl 6)
 
(de vdc (N B)
(default B 2)
(let (R 0 A 1.0)
(until (=0 N)
(inc 'R (* (setq A (/ A B)) (% N B)))
(setq N (/ N B)) )
R ) )
 
(for B (2 3 4)
(prinl "Base: " B)
(for N (range 0 9)
(prinl N ": " (round (vdc N B) 4)) ) )
Output:
Base: 2
0: 0.0000
1: 0.5000
2: 0.2500
3: 0.7500
4: 0.1250
5: 0.6250
6: 0.3750
7: 0.8750
8: 0.0625
9: 0.5625
Base: 3
0: 0.0000
1: 0.3333
2: 0.6667
3: 0.1111
4: 0.4444
5: 0.7778
6: 0.2222
7: 0.5556
8: 0.8889
9: 0.0370
Base: 4
0: 0.0000
1: 0.2500
2: 0.5000
3: 0.7500
4: 0.0625
5: 0.3125
6: 0.5625
7: 0.8125
8: 0.1250
9: 0.3750

[edit] PL/I

 
vdcb: procedure (an) returns (bit (31)); /* 6 July 2012 */
declare an fixed binary (31);
declare (n, i) fixed binary (31);
declare v bit (31) varying;
 
n = an; v = ''b;
do i = 1 by 1 while (n > 0);
if iand(n, 1) = 1 then v = v || '1'b; else v = v || '0'b;
n = isrl(n, 1);
end;
return (v);
end vdcb;
 
declare i fixed binary (31);
 
do i = 0 to 10;
put skip list ('0.' || vdcb(i));
end;
 
Output:
0.0000000000000000000000000000000 
0.1000000000000000000000000000000 
0.0100000000000000000000000000000 
0.1100000000000000000000000000000 
0.0010000000000000000000000000000 
0.1010000000000000000000000000000 
0.0110000000000000000000000000000 
0.1110000000000000000000000000000 
0.0001000000000000000000000000000 
0.1001000000000000000000000000000 
0.0101000000000000000000000000000 

[edit] Prolog

% vdc( N, Base, Out )
% Out = the Van der Corput representation of N in given Base
vdc( 0, _, [] ).
vdc( N, Base, Out ) :-
Nr is mod(N, Base),
Nq is N // Base,
vdc( Nq, Base, Tmp ),
Out = [Nr|Tmp].
 
% Writes every element of a list to stdout; no newlines
write_list( [] ).
write_list( [H|T] ) :-
write( H ),
write_list( T ).
 
% Writes the Nth Van der Corput item.
print_vdc( N, Base ) :-
vdc( N, Base, Lst ),
write('0.'),
write_list( Lst ).
print_vdc( N ) :-
print_vdc( N, 2 ).
 
% Prints the first N+1 elements of the Van der Corput
% sequence, each to its own line
print_some( 0, _ ) :-
write( '0.0' ).
print_some( N, Base ) :-
M is N - 1,
print_some( M, Base ),
nl,
print_vdc( N, Base ).
print_some( N ) :-
print_some( N, 2 ).
 
test :-
writeln('First 10 members in base 2:'),
print_some( 9 ),
nl,
write('7th member in base 4 (stretch goal) => '),
print_vdc( 7, 4 ).
 
Output:
(result of test):
First 10 members in base 2:
0.0
0.1
0.01
0.11
0.001
0.101
0.011
0.111
0.0001
0.1001
7th member in base 4 (stretch goal) => 0.31
true .

[edit] Python

(Python3.x)

The multi-base sequence generator

def vdc(n, base=2):
vdc, denom = 0,1
while n:
denom *= base
n, remainder = divmod(n, base)
vdc += remainder / denom
return vdc

Sample output

Base 2 and then 3:

>>> [vdc(i) for i in range(10)]
[0, 0.5, 0.25, 0.75, 0.125, 0.625, 0.375, 0.875, 0.0625, 0.5625]
>>> [vdc(i, 3) for i in range(10)]
[0, 0.3333333333333333, 0.6666666666666666, 0.1111111111111111, 0.4444444444444444, 0.7777777777777777, 0.2222222222222222, 0.5555555555555556, 0.8888888888888888, 0.037037037037037035]
>>>

[edit] As fractions

We can get the output as rational numbers if we use the fraction module (and change its string representation to look like a fraction):

>>> from fractions import Fraction
>>> Fraction.__repr__ = lambda x: '%i/%i' % (x.numerator, x.denominator)
>>> [vdc(i, base=Fraction(2)) for i in range(10)]
[0, 1/2, 1/4, 3/4, 1/8, 5/8, 3/8, 7/8, 1/16, 9/16]

[edit] Stretch goal

Sequences for different bases:

>>> for b in range(3,6):
print('\nBase', b)
print([vdc(i, base=Fraction(b)) for i in range(10)])
 
Base 3
[0, 1/3, 2/3, 1/9, 4/9, 7/9, 2/9, 5/9, 8/9, 1/27]
 
Base 4
[0, 1/4, 1/2, 3/4, 1/16, 5/16, 9/16, 13/16, 1/8, 3/8]
 
Base 5
[0, 1/5, 2/5, 3/5, 4/5, 1/25, 6/25, 11/25, 16/25, 21/25]

[edit] Racket

Following the suggestion.

#lang racket
(define (van-der-Corput n base)
(if (zero? n)
0
(let-values ([(q r) (quotient/remainder n base)])
(/ (+ r (van-der-Corput q base))
base))))

By digits, extracted arithmetically.

#lang racket
(define (digit-length n base)
(if (< n base) 1 (add1 (digit-length (quotient n base) base))))
(define (digit n i base)
(remainder (quotient n (expt base i)) base))
(define (van-der-Corput n base)
(for/sum ([i (digit-length n base)]) (/ (digit n i base) (expt base (+ i 1)))))

Output.

(for ([base (in-range 2 (add1 5))])
(printf "Base ~a: " base)
(for ([n (in-range 0 10)])
(printf "~a " (van-der-Corput n base)))
(newline))
 
#| Base 2: 0 1/2 1/4 3/4 1/8 5/8 3/8 7/8 1/16 9/16
Base 3: 0 1/3 2/3 1/9 4/9 7/9 2/9 5/9 8/9 1/27
Base 4: 0 1/4 1/2 3/4 1/16 5/16 9/16 13/16 1/8 3/8
Base 5: 0 1/5 2/5 3/5 4/5 1/25 6/25 11/25 16/25 21/25 |#

[edit] REXX

[edit] binary version

This version only handles binary (base 2).
Virtually any integer (including negative) is allowed and is accurate (no rounding).
A range of integers is also supported.

/*REXX pgm converts an integer (or a range)──►van der Corput # in base 2*/
numeric digits 1000 /*handle anything the user wants.*/
parse arg a b . /*obtain the number(s) [maybe]. */
if a=='' then do; a=0; b=10; end /*if none specified, use defaults*/
if b=='' then b=a /*assume a "range" of a single #.*/
 
do j=a to b /*traipse through the range. */
_=vdC(abs(j)) /*convert ABS value of integer.*/
leading=substr('-',2+sign(j)) /*if needed, elide leading sign.*/
say leading || _ /*show number (with leading -  ?)*/
end /*j*/
exit /*stick a fork in it, we're done.*/
/*──────────────────────────────────VDC [van der Corput] subroutine─────*/
vdC: procedure; y=x2b(d2x(arg(1)))+0 /*convert to hex, then binary. */
if y==0 then return 0 /*handle special case of zero. */
else return '.'reverse(y) /*heavy lifting by REXX*/

output when using the default input of:   0   10

0
.1
.01
.11
.001
.101
.011
.111
.0001
.1001
.0101

===any radix up to 90=== (and This version handles what the first version does, plus any radix up to (and including) base 90.
It can also support a list (enabled when the base is negative).

/*REXX pgm converts an  integer  (or a range) ──► van der Corput number */
/*in base 2, or optionally, any other base up to and including base 90.*/
numeric digits 1000 /*handle anything the user wants.*/
parse arg a b r . /*obtain the number(s) [maybe]. */
if a=='' then do; a=0; b=10; end /*if none specified, use defaults*/
if b=='' then b=a /*assume a "range" of a single #.*/
if r=='' then r=2 /*assume a radix (base) of 2. */
z= /*placeholder for a list of nums.*/
 
do j=a to b /*traipse through the range. */
_=vdC(abs(j), abs(r)) /*convert ABS value of integer.*/
_=substr('-', 2+sign(j))_ /*if needed, keep leading - sign.*/
if r>0 then say _ /*if positive base, just show it.*/
else z=z _ /* ··· else build a list· */
end /*j*/
 
if z\=='' then say strip(z) /*if list wanted, then show it. */
exit /*stick a fork in it, we're done.*/
/*──────────────────────────────────BASE subroutine (up to base 90)─────*/
base: procedure; parse arg x,toB,inB /*get a number, toBase, inBase */
/*┌────────────────────────────────────────────────────────────────────┐
┌─┘ Input to this subroutine (all must be positive whole numbers): └─┐
│ │
│ x (is required). │
│ toBase the base to convert X to. │
│ inBase the base X is expressed in. │
│ │
│ toBase or inBase can be omitted which causes the default of │
└─┐ 10 to be used. The limits of both are: 2 ──► 90. ┌─┘
└────────────────────────────────────────────────────────────────────┘*/

@abc='abcdefghijklmnopqrstuvwxyz' /*Latin lowercase alphabet chars.*/
@abcU=@abc; upper @abcU /*go whole hog and extend chars. */
@@@=0123456789 || @abc || @abcU /*prefix 'em with numeric digits.*/
@@@=@@@'<>[]{}()?~!@#$%^&*_+-=|\/;:~' /*add some special chars as well,*/
/*spec. chars should be viewable.*/
numeric digits 1000 /*what the hey, support biggies. */
maxB=length(@@@) /*max base (radix) supported here*/
parse arg x,toB,inB /*get a number, toBase, inBase */
if toB=='' then toB=10 /*if skipped, assume default (10)*/
if inB=='' then inB=10 /* " " " " " */
/*══════════════════════════════════convert X from base inB ──► base 10.*/
#=0; do j=1 for length(x)
_=substr(x,j,1) /*pick off a "digit" from X. */
v=pos(_,@@@) /*get the value of this "digits".*/
if v==0 | v>inB then call erd x,j,inB /*illegal "digit" ? */
#=#*inB + v - 1 /*construct new num, dig by dig. */
end /*j*/
/*══════════════════════════════════convert # from base 10 ──► base toB.*/
y=; do while #>=toB /*deconstruct the new number (#).*/
y=substr(@@@,(#//toB)+1,1)y /* construct the output number. */
#=# % toB /*··· and whittle # down also. */
end /*while*/
 
return substr(@@@,#+1,1)y
/*──────────────────────────────────VDC [van der Corput] subroutine─────*/
vdC: return '.'reverse(base(arg(1),arg(2))) /*convert, reverse, append.*/

output when using the multiple inputs of (where a negative base indicates to show numbers as a list):

0 30 -2
1 30 -3
1 30 -4
1 30 -5
55582777 55582804 -80

(All outputs are a single line list.)

.0 .1 .01 .11 .001 .101 .011 .111 .0001 .1001 .0101 .1101 .0011 .1011 .0111 .1111 .00001 .10001 .01001 .11001 .00101 .10101 .01101 .11101 .00011 .10011 .01011 .11011 .00111 .10111 .01111
.1 .2 .01 .11 .21 .02 .12 .22 .001 .101 .201 .011 .111 .211 .021 .121 .221 .002 .102 .202 .012 .112 .212 .022 .122 .222 .0001 .1001 .2001 .0101
.1 .2 .3 .01 .11 .21 .31 .02 .12 .22 .32 .03 .13 .23 .33 .001 .101 .201 .301 .011 .111 .211 .311 .021 .121 .221 .321 .031 .131 .231
.1 .2 .3 .4 .01 .11 .21 .31 .41 .02 .12 .22 .32 .42 .03 .13 .23 .33 .43 .04 .14 .24 .34 .44 .001 .101 .201 .301 .401 .011
.V[Is1 .W[Is1 .X[Is1 .Y[Is1 .Z[Is1 .<[Is1 .>[Is1 .[[Is1 .][Is1 .{[Is1 .}[Is1 .([Is1 .)[Is1 .?[Is1 .~[Is1 .![Is1 .@[Is1 .#[Is1 .$[Is1 .%[Is1 .^[Is1 .&[Is1 .*[Is1 .0]Is1 .1]Is1 .2]Is1 .3]Is1 .4]Is1

[edit] Ruby

The multi-base sequence generator

def vdc(n, base=2)
str = n.to_s(base).reverse
str.to_i(base).quo(base ** str.length)
end
 
(2..5).each do |base|
puts "Base #{base}: " + Array.new(10){|i| vdc(i,base)}.join(", ")
end

Sample output

Base 2: 0/1, 1/2, 1/4, 3/4, 1/8, 5/8, 3/8, 7/8, 1/16, 9/16
Base 3: 0/1, 1/3, 2/3, 1/9, 4/9, 7/9, 2/9, 5/9, 8/9, 1/27
Base 4: 0/1, 1/4, 1/2, 3/4, 1/16, 5/16, 9/16, 13/16, 1/8, 3/8
Base 5: 0/1, 1/5, 2/5, 3/5, 4/5, 1/25, 6/25, 11/25, 16/25, 21/25

[edit] Seed7

Translation of: D
$ include "seed7_05.s7i";
include "float.s7i";
 
const func float: vdc (in var integer: number, in integer: base) is func
result
var float: vdc is 0.0;
local
var integer: denom is 1;
var integer: remainder is 0;
begin
while number <> 0 do
denom *:= base;
remainder := number rem base;
number := number div base;
vdc +:= flt(remainder) / flt(denom);
end while;
end func;
 
const proc: main is func
local
var integer: base is 0;
var integer: number is 0;
begin
for base range 2 to 5 do
writeln;
writeln("Base " <& base);
for number range 0 to 9 do
write(vdc(number, base) digits 6 <& " ");
end for;
writeln;
end for;
end func;
Output:

Base 2
0.000000 0.500000 0.250000 0.750000 0.125000 0.625000 0.375000 0.875000 0.062500 0.562500 

Base 3
0.000000 0.333333 0.666667 0.111111 0.444444 0.777778 0.222222 0.555556 0.888889 0.037037 

Base 4
0.000000 0.250000 0.500000 0.750000 0.062500 0.312500 0.562500 0.812500 0.125000 0.375000 

Base 5
0.000000 0.200000 0.400000 0.600000 0.800000 0.040000 0.240000 0.440000 0.640000 0.840000 

[edit] Tcl

The core of this is code to handle digit reversing. Note that this also tackles negative numbers (by preserving the sign independently).

proc digitReverse {n {base 2}} {
set n [expr {[set neg [expr {$n < 0}]] ? -$n : $n}]
set result 0.0
set bit [expr {1.0 / $base}]
for {} {$n > 0} {set n [expr {$n / $base}]} {
set result [expr {$result + $bit * ($n % $base)}]
set bit [expr {$bit / $base}]
}
return [expr {$neg ? -$result : $result}]
}

Note that the above procedure will produce terms of the Van der Corput sequence by default.

# Print the first 10 terms of the Van der Corput sequence
for {set i 1} {$i <= 10} {incr i} {
puts "vanDerCorput($i) = [digitReverse $i]"
}
 
# In other bases
foreach base {3 4 5} {
set seq {}
for {set i 1} {$i <= 10} {incr i} {
lappend seq [format %.5f [digitReverse $i $base]]
}
puts "${base}: [join $seq {, }]"
}
Output:
vanDerCorput(1) = 0.5
vanDerCorput(2) = 0.25
vanDerCorput(3) = 0.75
vanDerCorput(4) = 0.125
vanDerCorput(5) = 0.625
vanDerCorput(6) = 0.375
vanDerCorput(7) = 0.875
vanDerCorput(8) = 0.0625
vanDerCorput(9) = 0.5625
vanDerCorput(10) = 0.3125
3: 0.33333, 0.66667, 0.11111, 0.44444, 0.77778, 0.22222, 0.55556, 0.88889, 0.03704, 0.37037
4: 0.25000, 0.50000, 0.75000, 0.06250, 0.31250, 0.56250, 0.81250, 0.12500, 0.37500, 0.62500
5: 0.20000, 0.40000, 0.60000, 0.80000, 0.04000, 0.24000, 0.44000, 0.64000, 0.84000, 0.08000

[edit] XPL0

include c:\cxpl\codes;  \intrinsic 'code' declarations
 
func real VdC(N); \Return Nth term of van der Corput sequence in base 2
int N;
real V, U;
[V:= 0.0; U:= 0.5;
repeat N:= N/2;
if rem(0) then V:= V+U;
U:= U/2.0;
until N=0;
return V;
];
 
int N;
for N:= 0 to 10-1 do
[IntOut(0, N); RlOut(0, VdC(N)); CrLf(0)]
Output:
0    0.00000
1    0.50000
2    0.25000
3    0.75000
4    0.12500
5    0.62500
6    0.37500
7    0.87500
8    0.06250
9    0.56250

[edit] zkl

Translation of: Python
fcn vdc(n,base=2){
vdc:=0.0; denom:=1;
while(n){ reg remainder;
denom *= base;
n, remainder = n.divr(base);
vdc += (remainder.toFloat() / denom);
}
vdc
}
Translation of: Ruby
fcn vdc(n,base=2){
str:=n.toString(base).reverse();
str.toInt(base).toFloat()/(base.toFloat().pow(str.len()))
}
Output:
[0..10].apply(vdcR).println("base 2");
L(0,0.5,0.25,0.75,0.125,0.625,0.375,0.875,0.0625,0.5625,0.3125)base 2

[0..10].apply(vdc.fp1(3)).println("base 3");
L(0,0.333333,0.666667,0.111111,0.444444,0.777778,0.222222,0.555556,0.888889,0.037037,0.37037)base 3
Personal tools
Namespaces

Variants
Actions
Community
Explore
Misc
Toolbox