# Van der Corput sequence

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 ${\displaystyle 2^{0}}$; the digit in the next column to the left has a multiplier of ${\displaystyle 2^{1}}$; and so on.

So in the following table:

  0.
1.
10.
11.
...

the binary number "10" is ${\displaystyle 1\times 2^{1}+0\times 2^{0}}$.

You can also have binary digits to the right of the “point”, just as in the decimal number system. In that case, the digit in the place immediately to the right of the point has a weight of ${\displaystyle 2^{-1}}$, or ${\displaystyle 1/2}$. The weight for the second column to the right of the point is ${\displaystyle 2^{-2}}$ or ${\displaystyle 1/4}$. 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 ${\displaystyle 0\times 2^{-1}+1\times 2^{-2}}$ or ${\displaystyle 1/4}$.

Distribution of 2500 points each: Van der Corput (top) vs pseudorandom
Members of the sequence lie within the interval ${\displaystyle 0\leq x<1}$. 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 ${\displaystyle 1\times 2^{3}+0\times 2^{2}+1\times 2^{1}+1\times 2^{0}}$.
Reflected this would become .1101 or ${\displaystyle 1\times 2^{-1}+1\times 2^{-2}+0\times 2^{-3}+1\times 2^{-4}}$

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

## 360 Assembly

Translation of: BBC BASIC

The program uses two ASSIST macros (XDECO,XPRNT) to keep the code as short as possible.

*        Van der Corput sequence   31/01/2017VDCS     CSECT         USING  VDCS,R13           base register         B      72(R15)            skip savearea         DC     17F'0'             savearea         STM    R14,R12,12(R13)    prolog         ST     R13,4(R15)         " <-         ST     R15,8(R13)         " ->         LR     R13,R15            " addressability         ZAP    B,=P'2'            b=2  (base)         ZAP    M,=P'-1'           m=-1         SR     R6,R6              i=0LOOPI    CH     R6,=H'10'          do i=0 to 10         BH     ELOOPI         AP     M,=P'1'            w=m+1         ZAP    V,=P'0'            v=0         ZAP    S,=P'1'            s=1         ZAP    N,M                n=mWHILE    CP     N,=P'0'            do while n<>0         BE     EWHILE         MP     S,B                s=s*b         ZAP    PL16,N             n         DP     PL16,B             n/b         ZAP    W,PL16+8(8)        w=n mod b          MP     W,=P'100000'       *100000         ZAP    PL16,W             w         DP     PL16,S             w/s         ZAP    W,PL16(8)          w=w/s         AP     V,W                v=v+(n mod b)*100000/s         ZAP    PL16,N             n         DP     PL16,B             n/b         ZAP    N,PL16(8)          n=n/b         B      WHILEEWHILE   XDECO  R6,XDEC            edit i         MVC    PG+0(3),XDEC+9     output i         MVC    PG+3(3),=C' 0.'         UNPK   Z,V                unpack v         OI     Z+L'Z-1,X'F0'      edit v         MVC    PG+6(5),Z+11       output v  (v/100000)         XPRNT  PG,L'PG            print buffer         LA     R6,1(R6)           i=i+1         B      LOOPIELOOPI   L      R13,4(0,R13)       epilog          LM     R14,R12,12(R13)    " restore         XR     R15,R15            " rc=0         BR     R14                exitB        DS     PL8M        DS     PL8V        DS     PL8S        DS     PL8N        DS     PL8W        DS     PL8                packed Z        DS     ZL16               zonedPL16     DS     PL16               packed maxPG       DC     CL80' '            bufferXDEC     DS     CL12               work area for xdeco         YREGS         END    VDCS
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
10 0.31250


## 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


## 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

## AutoHotkey

Works with: AutoHotkey_L
SetFormat, FloatFast, 0.5for 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

## AWK

 # syntax: GAWK -f VAN_DER_CORPUT_SEQUENCE.AWK# converted from BBC BASICBEGIN {    printf("base")    for (i=0; i<=9; i++) {      printf(" %7d",i)    }    printf("\n")    for (base=2; base<=5; base++) {      printf("%-4s",base)      for (i=0; i<=9; i++) {        printf(" %7.5f",vdc(i,base))      }      printf("\n")    }    exit(0)}function vdc(n,b,  s,v) {    s = 1    while (n) {      s *= b      v += (n % b) / s      n /= b      n = int(n)    }    return(v)} 

Output:

base       0       1       2       3       4       5       6       7       8       9
2    0.00000 0.50000 0.25000 0.75000 0.12500 0.62500 0.37500 0.87500 0.06250 0.56250
3    0.00000 0.33333 0.66667 0.11111 0.44444 0.77778 0.22222 0.55556 0.88889 0.03704
4    0.00000 0.25000 0.50000 0.75000 0.06250 0.31250 0.56250 0.81250 0.12500 0.37500
5    0.00000 0.20000 0.40000 0.60000 0.80000 0.04000 0.24000 0.44000 0.64000 0.84000


      @% = &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  ## 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 = 10for (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 ## 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 ## 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  ## 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... ## 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 | ## 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 ## 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] ## 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] ## Elixir Works with: Elixir version 1.1 defmodule Van_der_corput do def sequence( n, base \\ 2 ) do "0." <> (Integer.to_string(n, base) |> String.reverse ) end def float( n, base \\ 2 ) do Integer.digits(n, base) |> Enum.reduce(0, fn i,acc -> (i + acc) / base end) end def fraction( n, base \\ 2 ) do str = Integer.to_string(n, base) |> String.reverse denominator = Enum.reduce(1..String.length(str), 1, fn _,acc -> acc*base end) reduction( String.to_integer(str, base), denominator ) end defp reduction( 0, _ ), do: "0" defp reduction( numerator, denominator ) do gcd = gcd( numerator, denominator ) "#{ div(numerator, gcd) }/#{ div(denominator, gcd) }" end defp gcd( a, 0 ), do: a defp gcd( a, b ), do: gcd( b, rem(a, b) )end funs = [ {"Float(Base):", &Van_der_corput.sequence/2}, {"Float(Decimal):", &Van_der_corput.float/2 }, {"Fraction:", &Van_der_corput.fraction/2} ]Enum.each(funs, fn {title, fun} -> IO.puts title Enum.each(2..5, fn base -> IO.puts " Base #{ base }: #{ Enum.map_join(0..9, ", ", &fun.(&1, base)) }" end)end) Output: Float(Base): 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 Float(Decimal): Base 2: 0.0, 0.5, 0.25, 0.75, 0.125, 0.625, 0.375, 0.875, 0.0625, 0.5625 Base 3: 0.0, 0.3333333333333333, 0.6666666666666666, 0.1111111111111111, 0.4444444444444444, 0.7777777777777778, 0.2222222222222222, 0.5555555555555555, 0.8888888888888888, 0.037037037037037035 Base 4: 0.0, 0.25, 0.5, 0.75, 0.0625, 0.3125, 0.5625, 0.8125, 0.125, 0.375 Base 5: 0.0, 0.2, 0.4, 0.6, 0.8, 0.04, 0.24, 0.44000000000000006, 0.64, 0.8400000000000001 Fraction: 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  ## 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  ## ERRE PROGRAM VAN_DER_CORPUT !! for rosettacode.org! PROCEDURE VDC(N%,B%->RES) LOCAL V,S% S%=1 WHILE N%>0 DO S%*=B% V+=(N% MOD B%)/S% N%=N% DIV B% END WHILE RES=VEND PROCEDURE BEGIN FOR BASE%=2 TO 5 DO PRINT("Base";STR$(BASE%);":")        FOR NUMBER%=0 TO 9 DO          VDC(NUMBER%,BASE%->RES)          WRITE("#.##### ";RES;)        END FOR        PRINT      END FOREND PROGRAM
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


## 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 vdcend 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



## 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]

## Factor

Works with: Factor version 0.98
USING: formatting fry io kernel math math.functions math.parsermath.ranges sequences ;IN: rosetta-code.van-der-corput : vdc ( n base -- x )    [ >base string>digits <reversed> ]    [ nip '[ 1 + neg _ swap ^ * ] ] 2bi map-index sum ; : vdc-demo ( -- )    2 5 [a,b] [        dup "Base %d: " printf 10 <iota>        [ swap vdc "%-5u " printf ] with each nl    ] each ; MAIN: vdc-demo
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


## 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

## Fortran

This is straightforward once one remembers that the obvious scheme for extracting digits from a number produces them from the low-order end to the high-order end. This reversal is normally annoying, but here a "reflection" is desired. The source is old-style, except for using F90's ability to have a function (or subroutine) name appear on its END statement with this checked by the compiler. Because the MODULE protocol introduced by F90 is not bothered with, the type of the function has to be declared in all routines invoking it if the default type based on the form of the name does not suffice. Single precision suffices, but the F90 compiler moans that the type of the function itself has not been explicitly declared. Ah well.
      FUNCTION VDC(N,BASE)	!Calculates a Van der Corput number...Converts 1234 in decimal to 4321 in V, and P = 10000.       INTEGER N	!For this integer,       INTEGER BASE	!In this base.       INTEGER I	!A copy of N that can be damaged.       INTEGER P	!Successive powers of BASE.       INTEGER V	!Accumulates digits.        P = 1		! = BASE**0        V = 0		!Start with no digits, as if N = 0.        I = N		!Here we go.        DO WHILE (I .NE. 0)	!While something remains,          V = V*BASE + MOD(I,BASE)	!Extract its low-order digit.          I = I/BASE			!Reduce it by a power.          P = P*BASE			!And track the power.        END DO			!Thus extract the digits in reverse order: right-to-left.        VDC = V/FLOAT(P)	!The power is one above the highest digit.      END FUNCTION VDC	!Numerology is weird.       PROGRAM POKE      INTEGER FIRST,LAST	!Might as well document some constants.      PARAMETER (FIRST = 0,LAST = 9)	!Thus, the first ten values.      INTEGER I,BASE		!Steppers.      REAL VDC			!Stop the compiler moaning about undeclared items.       WRITE (6,1) FIRST,LAST,(I, I = FIRST,LAST)	!Announce.    1 FORMAT ("Calculates values ",I0," to ",I0," of the ",     1 "Van der Corput sequence, in various bases."/     2 "Base",666I9)       DO BASE = 2,13	!A selection of bases.        WRITE (6,2) BASE,(VDC(I,BASE), I = FIRST,LAST)	!Show the specified span.    2   FORMAT (I4,666F9.6)	!Aligns with FORMAT 1.      END DO		!On to the next base.       END

Output: six-digit precision is about the most that single precision offers.

Calculates values 0 to 9 of the Van der Corput sequence, in various bases.
Base        0        1        2        3        4        5        6        7        8        9
2 0.000000 0.500000 0.250000 0.750000 0.125000 0.625000 0.375000 0.875000 0.062500 0.562500
3 0.000000 0.333333 0.666667 0.111111 0.444444 0.777778 0.222222 0.555556 0.888889 0.037037
4 0.000000 0.250000 0.500000 0.750000 0.062500 0.312500 0.562500 0.812500 0.125000 0.375000
5 0.000000 0.200000 0.400000 0.600000 0.800000 0.040000 0.240000 0.440000 0.640000 0.840000
6 0.000000 0.166667 0.333333 0.500000 0.666667 0.833333 0.027778 0.194444 0.361111 0.527778
7 0.000000 0.142857 0.285714 0.428571 0.571429 0.714286 0.857143 0.020408 0.163265 0.306122
8 0.000000 0.125000 0.250000 0.375000 0.500000 0.625000 0.750000 0.875000 0.015625 0.140625
9 0.000000 0.111111 0.222222 0.333333 0.444444 0.555556 0.666667 0.777778 0.888889 0.012346
10 0.000000 0.100000 0.200000 0.300000 0.400000 0.500000 0.600000 0.700000 0.800000 0.900000
11 0.000000 0.090909 0.181818 0.272727 0.363636 0.454545 0.545455 0.636364 0.727273 0.818182
12 0.000000 0.083333 0.166667 0.250000 0.333333 0.416667 0.500000 0.583333 0.666667 0.750000
13 0.000000 0.076923 0.153846 0.230769 0.307692 0.384615 0.461538 0.538462 0.615385 0.692308


## FreeBASIC

' version 03-12-2016' compile with: fbc -s console Function num_base(number As ULongInt, _base_ As UInteger) As String     If _base_ > 9 Then        Print "base not handled by function"        Sleep 5000        Return ""    End If     Dim As ULongInt n    Dim As String ans     While number <> 0        n = number Mod _base_        ans = Str(n) + ans        number = number \ _base_    Wend     If ans = "" Then ans = "0"     Return "." + ans End Function ' ------=< MAIN >=------ Dim As ULong k, lFor k = 2 To 5    Print "Base = "; k    For l = 0 To 12        Print left(num_base(l, k) + "      ",6);    Next    Print : printNext ' empty keyboard bufferWhile Inkey <> "" : WendPrint : Print "hit any key to end program"SleepEnd
Output:
Base = 2
.0    .1    .10   .11   .100  .101  .110  .111  .1000 .1001 .1010 .1011 .1100

Base = 3
.0    .1    .2    .10   .11   .12   .20   .21   .22   .100  .101  .102  .110

Base = 4
.0    .1    .2    .3    .10   .11   .12   .13   .20   .21   .22   .23   .30

Base = 5
.0    .1    .2    .3    .4    .10   .11   .12   .13   .14   .20   .21   .22

## 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


## 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.Listimport Data.Ratioimport System.Environmentimport Text.Printf -- A wrapper type for Rationals to make them look nicer when we print them.newtype Rat = Rat Rationalinstance 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] -> IntegerdigitsToNum 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 -> Ratvdc 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


## 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 xend procedure round(n,d)    places := 10 ^ d    return real(integer(n*places + 0.5)) / placesend

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 resultend

It produces the same output as shown above.

## J

Solution:

vdc=: ([ %~ %@[ #. #.inv)"0 _

Examples:

   2 vdc i.10                NB. 1st 10 nums of Van der Corput sequence in base 20 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 nums0 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 50 1r2 1r4 3r4  1r8  5r8  3r8   7r8  1r16  9r160 1r3 2r3 1r9  4r9  7r9  2r9   5r9   8r9  1r270 1r4 1r2 3r4 1r16 5r16 9r16 13r16   1r8   3r80 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 ("base 2", etc.). 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.

## 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

## jq

Works with: jq version 1.4

The neat thing about the following implementation of vdc(base) is that it shows how the task can be accomplished in two separate steps without the need to construct an intermediate array.

# vdc(base) converts an input decimal integer to a decimal number based on the van der# Corput sequence using base 'base', e.g. (4 | vdc(2)) is 0.125.#def vdc(base):   # The helper function converts a stream of residuals to a decimal,  # e.g. if base is 2, then decimalize( (0,0,1) ) yields 0.125  def decimalize(stream):    reduce stream as $d # state: [accumulator, power] ( [0, 1/base]; .[1] as$power | [ .[0] + ($d *$power), $power / base] ) | .[0]; if . == 0 then 0 else decimalize(recurse( if . == 0 then empty else ./base | floor end ) % base) end ; Example: def round(n): (if . < 0 then -1 else 1 end) as$s  | $s*10*.*n | if (floor%10)>4 then (.+5) else . end | ./10 | floor/n | .*$s; range(2;6) | . as $base | "Base \(.): \( [ range(0;11) | vdc($base)|round(1000) ] )"
Output:
 $jq -n -f -c -r van_der_corput_sequence.jqBase 2: [0,0.5,0.25,0.75,0.125,0.625,0.375,0.875,0.063,0.563,0.313]Base 3: [0,0.333,0.667,0.111,0.444,0.778,0.222,0.556,0.889,0.037,0.37]Base 4: [0,0.25,0.5,0.75,0.063,0.313,0.563,0.813,0.125,0.375,0.625]Base 5: [0,0.2,0.4,0.6,0.8,0.04,0.24,0.44,0.64,0.84,0.08] ## Julia vandercorput(num::Integer, base::Integer) = sum(d * Float64(base) ^ -ex for (ex, d) in enumerate(digits(num, base))) for base in 2:9 @printf("%10s %i:", "Base", base) for num in 0:9 @printf("%7.3f", vandercorput(num, base)) end println(" [...]")end Output:  Base 2: 0.000 0.500 0.250 0.750 0.125 0.625 0.375 0.875 0.063 0.563... Base 3: 0.000 0.333 0.667 0.111 0.444 0.778 0.222 0.556 0.889 0.037... Base 4: 0.000 0.250 0.500 0.750 0.063 0.313 0.563 0.813 0.125 0.375... Base 5: 0.000 0.200 0.400 0.600 0.800 0.040 0.240 0.440 0.640 0.840... Base 6: 0.000 0.167 0.333 0.500 0.667 0.833 0.028 0.194 0.361 0.528... Base 7: 0.000 0.143 0.286 0.429 0.571 0.714 0.857 0.020 0.163 0.306... Base 8: 0.000 0.125 0.250 0.375 0.500 0.625 0.750 0.875 0.016 0.141... Base 9: 0.000 0.111 0.222 0.333 0.444 0.556 0.667 0.778 0.889 0.012...  ## Kotlin Translation of: C // version 1.1.2 data class Rational(val num: Int, val denom: Int) fun vdc(n: Int, base: Int): Rational { var p = 0 var q = 1 var nn = n while (nn != 0) { p = p * base + nn % base q *= base nn /= base } val num = p val denom = q while (p != 0) { nn = p p = q % p q = nn } return Rational(num / q, denom / q)} fun main(args: Array<String>) { for (b in 2..5) { print("base$b:")        for (i in 0..9) {            val(num, denom) = vdc(i, b)            if (num != 0) print("  $num/$denom")            else print("  0")        }        println()    }}
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


## 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 mend

## 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}

## 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

## 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$/* 3 2 """ """ """ """ """ """ """(%o126) "2" "1" + "2" "0" + "2" "1" + "1" """ """ """ """ """ """ """ - 4 - 3 - 2 - 1 """ """ """ """ """ """ """ """(%o129) "1" "2" + "0" "2" + "1" "2" + "1" "2" """ """ """ """ """ """ """ """ */ ## Modula-2 MODULE Sequence;FROM FormatString IMPORT FormatString;FROM Terminal IMPORT WriteString,WriteLn,ReadChar; PROCEDURE vc(n,base : INTEGER; VAR num,denom : INTEGER);VAR p,q : INTEGER;BEGIN p := 0; q := 1; WHILE n#0 DO p := p * base + (n MOD base); q := q * base; n := n DIV base END; num := p; denom := q; WHILE p#0 DO n := p; p := q MOD p; q := n END; num := num DIV q; denom := denom DIV qEND vc; VAR buf : ARRAY[0..31] OF CHAR; d,n,i,b : INTEGER;BEGIN FOR b:=2 TO 5 DO FormatString("base %i:", buf, b); WriteString(buf); FOR i:=0 TO 9 DO vc(i,b,n,d); IF n#0 THEN FormatString(" %i/%i", buf, n, d); WriteString(buf) ELSE WriteString(" 0") END END; WriteLn END; ReadCharEND Sequence. ## PARI/GP VdC(n)=n=binary(n);sum(i=1,#n,if(n[i],1.>>(#n+1-i)));VdC(n)=sum(i=1,#binary(n),if(bittest(n,i-1),1.>>i)); \\ Alternate approachvector(10,n,VdC(n)) Output: [0.500000000, 0.250000000, 0.750000000, 0.125000000, 0.625000000, 0.375000000, 0.875000000, 0.0625000000, 0.562500000, 0.312500000] ## Pascal Tested with Free Pascal Program VanDerCorput;{$IFDEF FPC}  {$MODE DELPHI}{$ELSE}  {$APPTYPE CONSOLE}{$ENDIF} type  tvdrCallback = procedure (nom,denom: NativeInt); { Base=2function rev2(n,Pot:NativeUint):NativeUint;var  r : Nativeint;begin  r := 0;  while Pot > 0 do  Begin    r := r shl 1 OR (n AND 1);    n := n shr 1;    dec(Pot);  end;  rev2 := r;end;} function reverse(n,base,Pot:NativeUint):NativeUint;var  r,c : Nativeint;begin  r := 0;//No need to test n> 0 in this special case, n starting in upper half  while Pot > 0 do  Begin    c := n div base;    r := n+(r-c)*base;    n := c;    dec(Pot);  end;  reverse := r;end; procedure VanDerCorput(base,count:NativeUint;f:tvdrCallback);//calculates count nominater and denominater of Van der Corput sequence// to base var Pot, denom,nom, i : NativeUint;Begin  denom := 1;  Pot := 0;  while count > 0 do  Begin    IF Pot = 0 then      f(0,1);    //start in upper half    i := denom;    inc(Pot);    denom := denom *base;     repeat      nom := reverse(i,base,Pot);      IF count > 0 then        f(nom,denom)      else        break;      inc(i);      dec(count);    until i >= denom;  end;end; procedure vdrOutPut(nom,denom: NativeInt);Begin  write(nom,'/',denom,'  ');end; var i : NativeUint;Begin  For i := 2 to 5 do  Begin    write(' Base ',i:2,' :');    VanDerCorput(i,9,@vdrOutPut);    writeln;  end;end. 
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  2/4  3/4  1/16  5/16  9/16  13/16  2/16  6/16
Base  5 :0/1  1/5  2/5  3/5  4/5  1/25  6/25  11/25  16/25  21/25

## 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";}

## Perl 6

Works with: rakudo version 2016.08

First a cheap implementation in base 2, using string operations.

constant VdC = map { :2("0." ~ .base(2).flip) }, ^Inf;.say for VdC[^16];

Here is a more elaborate version using the polymod built-in integer method:

sub VdC($base = 2) { map { [+]$_ && .polymod($base xx *) Z/ [\*]$base xx *    }, ^Inf} .say for VdC[^10];
Output:
0
0.5
0.25
0.75
0.125
0.625
0.375
0.875
0.0625
0.5625

Here is 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 = $base, {$_  *  $base } ... *; [+] do for (flat @values Z @denoms) ->$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.

## Phix

Not entirely sure what to print, so decided to print in three different ways.
It struck me straightaway that the VdC of say 123 is 321/1000, which seems trivial in any base or desired format.

enum BASE, FRAC, DECIMALconstant DESC = {"Base","Fraction","Decimal"} function vdc(integer n, atom base, integer flag)object res = ""atom num = 0, denom = 1, digit, g    while n do        denom *= base        digit = remainder(n,base)        n = floor(n/base)        if flag=BASE then            res &= digit+'0'        else            num = num*base+digit        end if    end while    if flag=FRAC then        g = gcd(num,denom)        return {num/g,denom/g}    elsif flag=DECIMAL then        return num/denom    end if    return {iff(length(res)=0?"0":"0."&res)}end function procedure show_vdc(integer flag, string fmt)object v    for i=2 to 5 do        printf(1,"%s %d: ",{DESC[flag],i})        for j=0 to 9 do            v = vdc(j,i,flag)            if flag=FRAC and v[1]=0 then                printf(1,"0 ")            else                printf(1,fmt,v)            end if        end for        puts(1,"\n")    end forend procedure show_vdc(BASE,"%s ")show_vdc(FRAC,"%d/%d ")show_vdc(DECIMAL,"%g ")
Output:
Base 2: 0 0.1 0.01 0.11 0.001 0.101 0.011 0.111 0.0001 0.1001
Base 3: 0 0.1 0.2 0.01 0.11 0.21 0.02 0.12 0.22 0.001
Base 4: 0 0.1 0.2 0.3 0.01 0.11 0.21 0.31 0.02 0.12
Base 5: 0 0.1 0.2 0.3 0.4 0.01 0.11 0.21 0.31 0.41
Fraction 2: 0 1/2 1/4 3/4 1/8 5/8 3/8 7/8 1/16 9/16
Fraction 3: 0 1/3 2/3 1/9 4/9 7/9 2/9 5/9 8/9 1/27
Fraction 4: 0 1/4 1/2 3/4 1/16 5/16 9/16 13/16 1/8 3/8
Fraction 5: 0 1/5 2/5 3/5 4/5 1/25 6/25 11/25 16/25 21/25
Decimal 2: 0 0.5 0.25 0.75 0.125 0.625 0.375 0.875 0.0625 0.5625
Decimal 3: 0 0.333333 0.666667 0.111111 0.444444 0.777778 0.222222 0.555556 0.888889 0.037037
Decimal 4: 0 0.25 0.5 0.75 0.0625 0.3125 0.5625 0.8125 0.125 0.375
Decimal 5: 0 0.2 0.4 0.6 0.8 0.04 0.24 0.44 0.64 0.84


## 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

## 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


## Prolog

% vdc( N, Base, Out )% Out = the Van der Corput representation of N in given Basevdc( 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 newlineswrite_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 lineprint_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 .


## PureBasic

Procedure.d nBase(n.i,b.i)  Define r.d,s.i=1    While n    s*b    r+(Mod(n,b)/s)    n=Int(n/b)  Wend    ProcedureReturn r    EndProcedure Define.i b,cOpenConsole("van der Corput - Sequence")For b=2 To 5  Print("Base "+Str(b)+": ")  For c=0 To 9        Print(StrD(nBase(c,b),5)+~"\t")  Next  PrintN("")NextInput()
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

## 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]>>> 

### 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]

### 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]

## 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 |# 

## REXX

### binary version

This REXX version only handles binary (base 2).

Virtually any integer (including negative) is allowed and is accurate (no rounding).

A range of integers (for output) is also supported.

/*REXX program converts an integer (or a range)  ──►  a Van der Corput number in base 2.*/numeric digits 1000                              /*handle almost anything the user wants*/parse arg a b .                                  /*obtain the optional arguments from CL*/if a==''  then parse value  0  10   with   a  b  /*Not specified?  Then use the defaults*/if b==''  then b=a                               /*assume a  range  for a single number.*/       do j=a  to b                               /*traipse through the range of numbers.*/      _=VdC( abs(j) )                            /*convert absolute value of an integer.*/      leading=substr('-',   2 + sign(j) )        /*if needed,  elide the leading sign.  */      say leading || _                           /*show number, with leading minus sign?*/      end   /*j*/exit                                             /*stick a fork in it,  we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/VdC: procedure;    y=x2b( d2x( arg(1) ) )  + 0   /*convert to  hexadecimal, then binary.*/     if y==0  then return 0                      /*handle the special case of zero.     */              else return '.'reverse(y)          /*heavy lifting is performed 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

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 program converts an  integer  (or a range)  ──►  a Van der Corput number,        *//*─────────────── in base 2,  or optionally, any other base up to and including base 90.*/numeric digits 1000                              /*handle almost anything the user wants*/parse arg a b r .                                /*obtain optional arguments from the CL*/if a=='' | a=="," then parse value 0 10 with a b /*Not specified?  Then use the defaults*/if b=='' | b=="," then b=a                       /* "      "         "   "   "      "   */if r=='' | r=="," then r=2                       /* "      "         "   "   "      "   */z=                                               /*a placeholder for a list of numbers. */                do j=a  to b                     /*traipse through the range of integers*/                _=VdC( abs(j), abs(r) )          /*convert the ABSolute value of integer*/                _=substr('-',  2 + sign(j) )_    /*if needed, keep the leading  -  sign.*/                if r>0  then say _               /*if positive base, then just show it. */                        else z=z _               /*     ··· else append (build) a list. */                end   /*j*/ if z\==''  then say strip(z)                     /*if a list is wanted, then display it.*/exit                                             /*stick a fork in it,  we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/base: procedure; parse arg x, toB, inB           /*get a number,  toBase,  and  inBase. */  /*╔══════════════════════════════════════════════════════════════════════════════════╗    ║ Input to this function:    x       (X   is required  and it must be an integer). ║    ║                          toBase    the base to convert   X   to    (default=10). ║    ║                          inBase    the base  X  is expressed in    (default=10). ║    ║                                                                                  ║    ║                                    toBase & inBase  have a limit of:   2 ──► 90  ║    ╚══════════════════════════════════════════════════════════════════════════════════╝*/      @abc= 'abcdefghijklmnopqrstuvwxyz'         /*the lowercase Latin alphabet letters.*/      @[email protected];          upper @abcU           /*go whole hog & extend with uppercase.*/      @@@= 0123456789 || @abc || @abcU           /*prefix them with the decimal digits. */      @@@= @@@'<>[]{}()[email protected]#$%^&*_+-=|\/;:' /*add some special characters as well, */ /*──those chars should all be viewable.*/ numeric digits 1000 /*what the hey, support bigun' numbers.*/ maxB=length(@@@) /*maximum base (radix) supported here. */ if toB=='' then toB=10 /*if omitted, then assume default (10)*/ if inB=='' then inB=10 /* " " " " " " */ #=0 /* [↓] convert base inB X ──► base 10*/ do j=1 for length(x) /*process each "numeral" in the string.*/ _=substr(x, j, 1) /*pick off a "digit" (numeral) from X.*/ v=pos(_, @@@) /*get the value of this "digit"/numeral*/ if v==0 | v>inB then call erd /*is it an illegal "digit" (numeral) ? */ #=# * inB + v - 1 /*construct new number, digit by digit.*/ end /*j*/ y= /* [↓] convert base 10 # ──► base toB.*/ do while #>=toB /*deconstruct the new number (#). */ y=substr(@@@, # // toB + 1, 1)y /* construct the output number, ··· */ #=# % toB /* ··· and also whittle down #. */ end /*while*/ return substr(@@@, # + 1, 1)y /*return a constructed "numeric" string*//*──────────────────────────────────────────────────────────────────────────────────────*/erd: say 'the character ' v " isn't a legal numeral for base " inB'.'; exit 13/*──────────────────────────────────────────────────────────────────────────────────────*/VdC: return '.'reverse(base(arg(1), arg(2))) /*convert the #, reverse the #, append.*/ (A negative base indicates to show numbers as a list.) output when using the input of: 0 30 -2 .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  output when using the input of: 1 30 -3 .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  output when using the input of: 1 30 -4 .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  output when using the input of: 1 30 -5 .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  output when using the input of: 55582777 55582804 -80 .V[Is1 .W[Is1 .X[Is1 .Y[Is1 .Z[Is1 .<[Is1 .>[Is1 .[[Is1 .][Is1 .{[Is1 .}[Is1 .([Is1 .)[Is1 .?[Is1 .~[Is1 .![Is1 [email protected][Is1 .#[Is1 .$[Is1 .%[Is1 .^[Is1 .&[Is1 .*[Is1 .0]Is1 .1]Is1 .2]Is1 .3]Is1 .4]Is1


## Ring

 decimals(4)for base = 2 to 5    see "base " + string(base) + " : "    for number = 0 to 9        see "" + corput(number, base) + " "    next    see nl next func corput n, b     vdc = 0     denom = 1     while n            denom *= b           rem = n % b           n = floor(n/b)           vdc += rem / denom     end     return vdc

Output:

base 2 : 0 0.5000 0.2500 0.7500 0.1250 0.6250 0.3750 0.8750 0.0625 0.5625
base 3 : 0 0.3333 0.6667 0.1111 0.4444 0.7778 0.2222 0.5556 0.8889 0.0370
base 4 : 0 0.2500 0.5000 0.7500 0.0625 0.3125 0.5625 0.8125 0.1250 0.3750
base 5 : 0 0.2000 0.4000 0.6000 0.8000 0.0400 0.2400 0.4400 0.6400 0.8400


## 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


## Scala

object VanDerCorput extends App {    def compute(n: Int, base: Int = 2) =        Iterator.from(0).            scanLeft(1)((a, _) => a * base).            map(b => (n - 1) / b -> b).            takeWhile(_._1 != 0).            foldLeft(0d)((a, b) => a + (b._1 % base).toDouble / b._2 / base)     val n = scala.io.StdIn.readInt    val b = scala.io.StdIn.readInt    (1 to n).foreach(x => println(compute(x, b)))}
Output:
n: 30
base: 2
0.0
0.5
0.25
0.75
0.125
0.625
0.375
0.875
0.0625
0.5625
0.3125
0.8125
0.1875
0.6875
0.4375
0.9375
0.03125
0.53125
0.28125
0.78125
0.15625
0.65625
0.40625
0.90625
0.09375
0.59375
0.34375
0.84375
0.21875
0.71875

## 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  ## Sidef Translation of: Perl func vdc(value, base=2) { while (value[-1] > 0) { value.append(value[-1] / base -> int) } var (x, sum) = (1, 0) value.each { |i| sum += ((i % base) / (x *= base)) } return sum} for base in (2..5) { var seq = 10.of {|i| vdc([i], base) } "base %d: %s\n".printf(base, seq.map{|n| "%.4f" % n}.join(', '))} Output: base 2: 0.0000, 0.5000, 0.2500, 0.7500, 0.1250, 0.6250, 0.3750, 0.8750, 0.0625, 0.5625 base 3: 0.0000, 0.3333, 0.6667, 0.1111, 0.4444, 0.7778, 0.2222, 0.5556, 0.8889, 0.0370 base 4: 0.0000, 0.2500, 0.5000, 0.7500, 0.0625, 0.3125, 0.5625, 0.8125, 0.1250, 0.3750 base 5: 0.0000, 0.2000, 0.4000, 0.6000, 0.8000, 0.0400, 0.2400, 0.4400, 0.6400, 0.8400 ## Swift Translation of: C func vanDerCorput(n: Int, base: Int, num: inout Int, denom: inout Int) { var n = n, p = 0, q = 1 while n != 0 { p = p * base + (n % base) q *= base n /= base } num = p denom = q while p != 0 { n = p p = q % p q = n } num /= q denom /= q} var num = 0var denom = 0 for base in 2...5 { print("base \(base): 0 ", terminator: "") for n in 1..<10 { vanDerCorput(n: n, base: base, num: &num, denom: &denom) print("\(num)/\(denom) ", terminator: "") } print()} 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 ## Stata Stata has builtin functions in Mata to compute Halton sequences, which are generalizations of the Van der Corput sequence. See halton in Stata help, and two articles in the Stata Journal: Scrambled Halton sequences in Mata by Stanislav Kolenikov and Generating Halton sequences using Mata by David M. Drukker and Richard Gates. mata// 5th term of Van der Corput sequencehalton(1,1,5) .625 // the first 10 terms of Van der Corput sequencehalton(10,1) 1 +---------+ 1 | .5 | 2 | .25 | 3 | .75 | 4 | .125 | 5 | .625 | 6 | .375 | 7 | .875 | 8 | .0625 | 9 | .5625 | 10 | .3125 | +---------+ // the first 10 terms of Van der Corput sequence in base 3ghalton(10,3,0) 1 +---------------+ 1 | .3333333333 | 2 | .6666666667 | 3 | .1111111111 | 4 | .4444444444 | 5 | .7777777778 | 6 | .2222222222 | 7 | .5555555556 | 8 | .8888888889 | 9 | .037037037 | 10 | .3703703704 | +---------------+ end Reproduce the plot in the task description: clearmatast_addobs(2500)st_addvar("double","x")st_addvar("double","y")st_addvar("double","z")k=1::2500st_store(k,1,k)st_store(k,2,0.5*runiform(2500,1))st_store(k,3,0.5:+0.5*halton(2500,1))endtwoway scatter y x, msize(tiny) color(blue) /// || scatter z x, msize(tiny) color(green) legend(off) xtitle("") /// title(Distribution: Van der Corput (top) vs pseudorandom) /// ylabel(, angle(0) format(%3.1f)) ## 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 : $result}]} Note that the above procedure will produce terms of the Van der Corput sequence by default. # Print the first 10 terms of the Van der Corput sequencefor {set i 1} {$i <= 10} {incr i} {    puts "vanDerCorput($i) = [digitReverse$i]"} # In other basesforeach 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


## VBA

Translation of: Phix
Base only.
Private Function vdc(ByVal n As Integer, BASE As Variant) As Variant    Dim res As String    Dim digit As Integer, g As Integer, denom As Integer    denom = 1    Do While n        denom = denom * BASE        digit = n Mod BASE        n = n \ BASE        res = res & CStr(digit) '+ "0"    Loop    vdc = IIf(Len(res) = 0, "0", "0." & res)End Function Public Sub show_vdc()    Dim v As Variant, j As Integer    For i = 2 To 5        Debug.Print "Base "; i; ": ";        For j = 0 To 9            v = vdc(j, i)            Debug.Print v; " ";        Next j        Debug.Print    Next iEnd Sub
Output:
Base  2 : 0 0.1 0.01 0.11 0.001 0.101 0.011 0.111 0.0001 0.1001
Base  3 : 0 0.1 0.2 0.01 0.11 0.21 0.02 0.12 0.22 0.001
Base  4 : 0 0.1 0.2 0.3 0.01 0.11 0.21 0.31 0.02 0.12
Base  5 : 0 0.1 0.2 0.3 0.4 0.01 0.11 0.21 0.31 0.41 

## VBScript

'http://rosettacode.org/wiki/Van_der_Corput_sequence'Van der Corput Sequence fucntion call = VanVanDerCorput(number,base) Base2 = "0" : Base3 = "0" : Base4 = "0" : Base5 = "0"Base6 = "0" : Base7 = "0" : Base8 = "0" : Base9 = "0" l = 1h = 1Do Until l = 9	'Set h to the value of l after each function call	'as it sets it to 0 - see lines 37 to 40.	Base2 = Base2 & ", " & VanDerCorput(h,2) : h = l	Base3 = Base3 & ", " & VanDerCorput(h,3) : h = l	Base4 = Base4 & ", " & VanDerCorput(h,4) : h = l	Base5 = Base5 & ", " & VanDerCorput(h,5) : h = l	Base6 = Base6 & ", " & VanDerCorput(h,6) : h = l	l = l + 1Loop WScript.Echo "Base 2: " & Base2WScript.Echo "Base 3: " & Base3WScript.Echo "Base 4: " & Base4WScript.Echo "Base 5: " & Base5WScript.Echo "Base 6: " & Base6 'Van der Corput SequenceFunction VanDerCorput(n,b)	k = RevString(Dec2BaseN(n,b))	For i = 1 To Len(k)		VanDerCorput = VanDerCorput + (CLng(Mid(k,i,1)) * b^-i)	NextEnd Function 'Decimal to Base N ConversionFunction Dec2BaseN(q,c)	Dec2BaseN = ""	Do Until q = 0		Dec2BaseN = CStr(q Mod c) & Dec2BaseN		q = Int(q / c)	LoopEnd Function 'Reverse StringFunction RevString(s)	For j = Len(s) To 1 Step -1		RevString = RevString & Mid(s,j,1)	NextEnd Function
Output:
Base 2: 0, 0.5, 0.5, 0.25, 0.75, 0.125, 0.625, 0.375, 0.875
Base 3: 0, 0.333333333333333, 0.666666666666667, 0.111111111111111, 0.444444444444444, 0.777777777777778, 0.222222222222222, 0.555555555555556, 0.888888888888889
Base 4: 0, 0.25, 0.5, 0.75, 0.0625, 0.3125, 0.5625, 0.8125, 0.125
Base 5: 0, 0.2, 0.4, 0.6, 0.8, 0.04, 0.24, 0.44, 0.64
Base 6: 0, 0.166666666666667, 0.333333333333333, 0.5, 0.666666666666667, 0.833333333333333, 2.77777777777778E-02, 0.194444444444444, 0.361111111111111


## Visual Basic .NET

Translation of: C
Module Module1     Function ToBase(n As Integer, b As Integer) As String        Dim result = ""        If b < 2 Or b > 16 Then            Throw New ArgumentException("The base is out of range")        End If         Do            Dim remainder = n Mod b            result = "0123456789ABCDEF"(remainder) + result            n = n \ b        Loop While n > 0         Return result    End Function     Sub Main()        For b = 2 To 5            Console.WriteLine("Base = {0}", b)            For i = 0 To 12                Dim s = "." + ToBase(i, b)                Console.Write("{0,6} ", s)            Next            Console.WriteLine()            Console.WriteLine()        Next    End Sub End Module
Output:
Base = 2
.0     .1    .10    .11   .100   .101   .110   .111  .1000  .1001  .1010  .1011  .1100

Base = 3
.0     .1     .2    .10    .11    .12    .20    .21    .22   .100   .101   .102   .110

Base = 4
.0     .1     .2     .3    .10    .11    .12    .13    .20    .21    .22    .23    .30

Base = 5
.0     .1     .2     .3     .4    .10    .11    .12    .13    .14    .20    .21    .22

## XPL0

include c:\cxpl\codes;  \intrinsic 'code' declarations func real VdC(N);       \Return Nth term of van der Corput sequence in base 2int  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


## 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