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This page uses content from Wikipedia. The original article was at Benford's_law. The list of authors can be seen in the page history. As with Rosetta Code, the text of Wikipedia is available under the GNU FDL. (See links for details on variance)

Benford's law, also called the first-digit law, refers to the frequency distribution of digits in many (but not all) real-life sources of data. In this distribution, the number 1 occurs as the first digit about 30% of the time, while larger numbers occur in that position less frequently: 9 as the first digit less than 5% of the time. This distribution of first digits is the same as the widths of gridlines on a logarithmic scale. Benford's law also concerns the expected distribution for digits beyond the first, which approach a uniform distribution.

This result has been found to apply to a wide variety of data sets, including electricity bills, street addresses, stock prices, population numbers, death rates, lengths of rivers, physical and mathematical constants, and processes described by power laws (which are very common in nature). It tends to be most accurate when values are distributed across multiple orders of magnitude.

A set of numbers is said to satisfy Benford's law if the leading digit $d$ ( $d\in \{1,\ldots ,9\}$ ) occurs with probability

P(d)=\log _{10}(d+1)-\log _{10}(d)=\log _{10}\left(1+{\frac {1}{d}}\right)

For this task, write (a) routine(s) to calculate the distribution of first significant (non-zero) digits in a collection of numbers, then display the actual vs. expected distribution in the way most convenient for your language (table / graph / histogram / whatever).

Use the first 1000 numbers from the Fibonacci sequence as your data set. No need to show how the Fibonacci numbers are obtained. You can generate them or load them from a file; whichever is easiest. Display your actual vs expected distribution.

For extra credit: Show the distribution for one other set of numbers from a page on Wikipedia. State which Wikipedia page it can be obtained from and what the set enumerates. Again, no need to display the actual list of numbers or the code to load them.

See also:

numberphile.com.
A starting page on Wolfram Mathworld is Benfords Law .

Aime

<lang aime>text sum(text a, text b) {

   data d;
   integer e, f, n, r;

   e = length(a);
   f = length(b);

   r = 0;

   n = min(e, f);
   while (n) {
       n -= 1;
       e -= 1;
       f -= 1;
       r += character(a, e) - '0';
       r += character(b, f) - '0';
       b_insert(d, 0, r % 10 + '0');
       r /= 10;
   }

   if (f) {
       e = f;
       a = b;
   }

   while (e) {
       e -= 1;
       r += character(a, e) - '0';
       b_insert(d, 0, r % 10 + '0');
       r /= 10;
   }

   if (r) {
       b_insert(d, 0, r + '0');
   }

   return b_string(d);

}

text fibs(list l, integer n) {

   integer c, i;
   text a, b, w;

   l_r_integer(l, 1, 1);

   a = "0";
   b = "1";
   i = 1;
   while (i < n) {
       w = sum(a, b);
       a = b;
       b = w;
       c = character(w, 0) - '0';
       l_r_integer(l, c, 1 + l_q_integer(l, c));
       i += 1;
   }

   return w;

}

integer main(void) {

   integer i, n;
   list f;
   real m;

   n = 1000;

   i = 10;
   while (i) {
       i -= 1;
       lb_p_integer(f, 0);
   }

   fibs(f, n);

   m = 100r / n;

   o_text("\t\texpected\t   found\n");
   i = 0;
   while (i < 9) {
       i += 1;
       o_winteger(8, i);
       o_wpreal(16, 3, 3, 100 * log10(1 + 1r / i));
       o_wpreal(16, 3, 3, l_q_integer(f, i) * m);
       o_text("\n");
   }

   return 0;

}</lang>

Output:

		expected	   found
       1          30.102          30.1  
       2          17.609          17.7  
       3          12.493          12.5  
       4           9.691           9.600
       5           7.918           8    
       6           6.694           6.7  
       7           5.799           5.6  
       8           5.115           5.300
       9           4.575           4.5

AutoHotkey

Works with: AutoHotkey_L

(AutoHotkey1.1+)

<lang AutoHotkey>SetBatchLines, -1 fib := NStepSequence(1, 1, 2, 1000) Out := "Digit`tExpected`tObserved`tDeviation`n" n := [] for k, v in fib d := SubStr(v, 1, 1) , n[d] := n[d] ? n[d] + 1 : 1 for k, v in n Exppected := 100 * Log(1+ (1 / k)) , Observed := (v / fib.MaxIndex()) * 100 , Out .= k "`t" Exppected "`t" Observed "`t" Abs(Exppected - Observed) "`n" MsgBox, % Out

NStepSequence(v1, v2, n, k) { a := [v1, v2] Loop, % k - 2 { a[j := A_Index + 2] := 0 Loop, % j < n + 2 ? j - 1 : n a[j] := BigAdd(a[j - A_Index], a[j]) } return, a }

BigAdd(a, b) { if (StrLen(b) > StrLen(a)) t := a, a := b, b := t LenA := StrLen(a) + 1, LenB := StrLen(B) + 1, Carry := 0 Loop, % LenB - 1 Sum := SubStr(a, LenA - A_Index, 1) + SubStr(B, LenB - A_Index, 1) + Carry , Carry := Sum // 10 , Result := Mod(Sum, 10) . Result Loop, % I := LenA - LenB { if (!Carry) { Result := SubStr(a, 1, I) . Result break } Sum := SubStr(a, I, 1) + Carry , Carry := Sum // 10 , Result := Mod(Sum, 10) . Result , I-- } return, (Carry ? Carry : "") . Result }</lang>NStepSequence() is available here. Output:

Digit	Expected	Observed	Deviation
1	30.103000	30.100000	0.003000
2	17.609126	17.700000	0.090874
3	12.493874	12.500000	0.006126
4	9.691001	9.600000	0.091001
5	7.918125	8.000000	0.081875
6	6.694679	6.700000	0.005321
7	5.799195	5.600000	0.199195
8	5.115252	5.300000	0.184748
9	4.575749	4.500000	0.075749

AWK

syntax: GAWK -f BENFORDS_LAW.AWK

BEGIN {

   n = 1000
   for (i=1; i<=n; i++) {
     arr[substr(fibonacci(i),1,1)]++
   }
   print("digit expected observed deviation")
   for (i=1; i<=9; i++) {
     expected = log10(i+1) - log10(i)
     actual = arr[i] / n
     deviation = expected - actual
     printf("%5d %8.4f %8.4f %9.4f\n",i,expected*100,actual*100,abs(deviation*100))
   }
   exit(0)

} function fibonacci(n, a,b,c,i) {

   a = 0
   b = 1
   for (i=1; i<=n; i++) {
     c = a + b
     a = b
     b = c
   }
   return(c)

} function abs(x) { if (x >= 0) { return x } else { return -x } } function log10(x) { return log(x)/log(10) } </lang>

output:

digit expected observed deviation
    1  30.1030  30.0000    0.1030
    2  17.6091  17.7000    0.0909
    3  12.4939  12.5000    0.0061
    4   9.6910   9.6000    0.0910
    5   7.9181   8.0000    0.0819
    6   6.6947   6.7000    0.0053
    7   5.7992   5.7000    0.0992
    8   5.1153   5.3000    0.1847
    9   4.5757   4.5000    0.0757

C

<lang c>#include <stdio.h>

include <stdlib.h>
include <math.h>

float *benford_distribution(void) {

   static float prob[9];
   for (int i = 1; i < 10; i++)
       prob[i - 1] = log10f(1 + 1.0 / i);

   return prob;

}

float *get_actual_distribution(char *fn) {

   FILE *input = fopen(fn, "r");
   if (!input)
   {
       perror("Can't open file");
       exit(EXIT_FAILURE);
   }

   int tally[9] = { 0 };
   char c;
   int total = 0;
   while ((c = getc(input)) != EOF)
   {
       /* get the first nonzero digit on the current line */
       while (c < '1' || c > '9')
           c = getc(input);

       tally[c - '1']++;
       total++;

       /* discard rest of line */
       while ((c = getc(input)) != '\n' && c != EOF)
           ;
   }
   fclose(input);
   
   static float freq[9];
   for (int i = 0; i < 9; i++)
       freq[i] = tally[i] / (float) total;

   return freq;

}

int main(int argc, char **argv) {

   if (argc != 2)
   {
       printf("Usage: benford <file>\n");
       return EXIT_FAILURE;
   }

   float *actual = get_actual_distribution(argv[1]);
   float *expected = benford_distribution();

   puts("digit\tactual\texpected");
   for (int i = 0; i < 9; i++)
       printf("%d\t%.3f\t%.3f\n", i + 1, actual[i], expected[i]);

   return EXIT_SUCCESS;

}</lang>

Output:

Use with a file which should contain a number on each line.

$ ./benford fib1000.txt
digit   actual  expected
1       0.301   0.301
2       0.177   0.176
3       0.125   0.125
4       0.096   0.097
5       0.080   0.079
6       0.067   0.067
7       0.056   0.058
8       0.053   0.051
9       0.045   0.046

C++

<lang cpp>//to cope with the big numbers , I used the Class Library for Numbers( CLN ) //if used prepackaged you can compile writing "g++ -std=c++11 -lcln yourprogram.cpp -o yourprogram"

include <cln/integer.h>
include <cln/integer_io.h>
include <iostream>
include <algorithm>
include <vector>
include <iomanip>
include <sstream>
include <string>
include <cstdlib>
include <cmath>
include <map>

using namespace cln ;

class NextNum { public :

  NextNum ( cl_I & a , cl_I & b ) : first( a ) , second ( b ) { }
  cl_I operator( )( ) {
     cl_I result = first + second ;
     first = second ;
     second = result ;
     return result ;
  }

private :

  cl_I first ;
  cl_I second ;

} ;

void findFrequencies( const std::vector<cl_I> & fibos , std::map<int , int> &numberfrequencies ) {

  for ( cl_I bignumber : fibos ) {
     std::ostringstream os ;
     fprintdecimal ( os , bignumber ) ;//from header file cln/integer_io.h
     int firstdigit = std::atoi( os.str( ).substr( 0 , 1 ).c_str( )) ;
     auto result = numberfrequencies.insert( std::make_pair( firstdigit , 1 ) ) ;
     if ( ! result.second )

numberfrequencies[ firstdigit ]++ ;

}

int main( ) {

  std::vector<cl_I> fibonaccis( 1000 ) ;
  fibonaccis[ 0 ] = 0 ;
  fibonaccis[ 1 ] = 1 ;
  cl_I a = 0 ;
  cl_I b = 1 ;
  //since a and b are passed as references to the generator's constructor
  //they are constantly changed !
  std::generate_n( fibonaccis.begin( ) + 2 , 998 , NextNum( a , b ) ) ;
  std::cout << std::endl ;
  std::map<int , int> frequencies ;
  findFrequencies( fibonaccis , frequencies ) ;
  std::cout << "                found                    expected\n" ;
  for ( int i = 1 ; i < 10 ; i++ ) {
     double found = static_cast<double>( frequencies[ i ] ) / 1000 ;
     double expected = std::log10( 1 + 1 / static_cast<double>( i )) ;
     std::cout << i << " :" << std::setw( 16 ) << std::right << found * 100 << " %" ;
     std::cout.precision( 3 ) ;
     std::cout << std::setw( 26 ) << std::right << expected * 100 << " %\n" ;
  }
  return 0 ;

} </lang>

Output:

                found                    expected
1 :            30.1 %                      30.1 %
2 :            17.7 %                      17.6 %
3 :            12.5 %                      12.5 %
4 :             9.5 %                      9.69 %
5 :               8 %                      7.92 %
6 :             6.7 %                      6.69 %
7 :             5.6 %                       5.8 %
8 :             5.3 %                      5.12 %
9 :             4.5 %                      4.58 %

CoffeeScript

<lang coffeescript>fibgen = () ->

   a = 1; b = 0
   return () ->
       ([a, b] = [b, a+b])[1]

leading = (x) -> x.toString().charCodeAt(0) - 0x30

f = fibgen()

benford = (0 for i in [1..9]) benford[leading(f()) - 1] += 1 for i in [1..1000]

log10 = (x) -> Math.log(x) * Math.LOG10E

actual = benford.map (x) -> x * 0.001 expected = (log10(1 + 1/x) for x in [1..9])

console.log "Leading digital distribution of the first 1,000 Fibonacci numbers" console.log "Digit\tActual\tExpected" for i in [1..9]

   console.log i + "\t" + actual[i - 1].toFixed(3) + '\t' + expected[i - 1].toFixed(3)</lang>

Output:

Leading digital distribution of the first 1,000 Fibonacci numbers
Digit   Actual  Expected
1       0.301   0.301
2       0.177   0.176
3       0.125   0.125
4       0.096   0.097
5       0.080   0.079
6       0.067   0.067
7       0.056   0.058
8       0.053   0.051
9       0.045   0.046

Common Lisp

<lang lisp>(defun calculate-distribution (numbers)

 "Return the frequency distribution of the most significant nonzero 
  digits in the given list of numbers. The first element of the list 
  is the frequency for digit 1, the second for digit 2, and so on."
 
 (defun nonzero-digit-p (c)
   "Check whether the character is a nonzero digit"
   (and (digit-char-p c) (char/= c #\0)))

 (defun first-digit (n)
   "Return the most significant nonzero digit of the number or NIL if
    there is none."
   (let* ((s (write-to-string n))
          (c (find-if #'nonzero-digit-p s)))
     (when c
       (digit-char-p c))))

 (let ((tally (make-array 9 :element-type 'integer :initial-element 0)))
   (loop for n in numbers 
         for digit = (first-digit n)
         when digit 
         do (incf (aref tally (1- digit))))
   (loop with total = (length numbers)
         for digit-count across tally
         collect (/ digit-count total))))

(defun calculate-benford-distribution ()

 "Return the frequency distribution according to Benford's law.
  The first element of the list is the probability for digit 1, the second 
  element the probability for digit 2, and so on."
 (loop for i from 1 to 9
       collect (log (1+ (/ i)) 10)))

(defun benford (numbers)

 "Print a table of the actual and expected distributions for the given
  list of numbers."
 (let ((actual-distribution (calculate-distribution numbers))
       (expected-distribution (calculate-benford-distribution)))
   (write-line "digit actual expected")
   (format T "~:{~3D~9,3F~8,3F~%~}" 
           (map 'list #'list '(1 2 3 4 5 6 7 8 9)
                             actual-distribution
                             expected-distribution))))</lang>

; *fib1000* is a list containing the first 1000 numbers in the Fibonnaci sequence
> (benford *fib1000*)
digit actual expected
  1    0.301   0.301
  2    0.177   0.176
  3    0.125   0.125
  4    0.096   0.097
  5    0.080   0.079
  6    0.067   0.067
  7    0.056   0.058
  8    0.053   0.051
  9    0.045   0.046

D

Translation of: Scala

<lang d>import std.stdio, std.range, std.math, std.conv, std.bigint;

double[2][9] benford(R)(R seq) if (isForwardRange!R && !isInfinite!R) {

   typeof(return) freqs = 0;
   uint seqLen = 0;
   foreach (d; seq)
       if (d != 0) {
           freqs[d.text[0] - '1'][1]++;
           seqLen++;
       }

   foreach (immutable i, ref p; freqs)
       p = [log10(1.0 + 1.0 / (i + 1)), p[1] / seqLen];
   return freqs;

}

void main() {

   auto fibs = recurrence!q{a[n - 1] + a[n - 2]}(1.BigInt, 1.BigInt);

   writefln("%9s %9s %9s", "Actual", "Expected", "Deviation");
   foreach (immutable i, immutable p; fibs.take(1000).benford)
       writefln("%d: %5.2f%% | %5.2f%% | %5.4f%%",
                i+1, p[1] * 100, p[0] * 100, abs(p[1] - p[0]) * 100);

}</lang>

Output:

   Actual  Expected Deviation
1: 30.10% | 30.10% | 0.0030%
2: 17.70% | 17.61% | 0.0908%
3: 12.50% | 12.49% | 0.0061%
4:  9.60% |  9.69% | 0.0910%
5:  8.00% |  7.92% | 0.0818%
6:  6.70% |  6.69% | 0.0053%
7:  5.60% |  5.80% | 0.1992%
8:  5.30% |  5.12% | 0.1847%
9:  4.50% |  4.58% | 0.0757%

Alternative Version

The output is the same. <lang d>import std.stdio, std.range, std.math, std.conv, std.bigint,

      std.algorithm, std.array;

auto benford(R)(R seq) if (isForwardRange!R && !isInfinite!R) {

   return seq.filter!q{a != 0}.map!q{a.text[0]-'1'}.array.sort().group;

}

void main() {

   auto fibs = recurrence!q{a[n - 1] + a[n - 2]}(1.BigInt, 1.BigInt);
   auto expected = iota(1, 10).map!(d => log10(1.0 + 1.0 / d));

   enum N = 1_000;
   writefln("%9s %9s %9s", "Actual", "Expected", "Deviation");
   foreach (immutable i, immutable f; fibs.take(N).benford)
       writefln("%d: %5.2f%% | %5.2f%% | %5.4f%%", i + 1,
                f * 100.0 / N, expected[i] * 100,
                abs((f / double(N)) - expected[i]) * 100);

}</lang>

Erlang

<lang Erlang> -module( benfords_law ). -export( [actual_distribution/1, distribution/1, task/0] ).

actual_distribution( Ns ) -> lists:foldl( fun first_digit_count/2, dict:new(), Ns ).

distribution( N ) -> math:log10( 1 + (1 / N) ).

task() -> Total = 1000, Fibonaccis = fib( Total ), Actual_dict = actual_distribution( Fibonaccis ), Keys = lists:sort( dict:fetch_keys( Actual_dict) ), io:fwrite( "Digit Actual Benfords expected~n" ), [io:fwrite("~p ~p ~p~n", [X, dict:fetch(X, Actual_dict) / Total, distribution(X)]) || X <- Keys].

fib( N ) -> fib( N, 0, 1, [] ). fib( 0, Current, _, Acc ) -> lists:reverse( [Current | Acc] ); fib( N, Current, Next, Acc ) -> fib( N-1, Next, Current+Next, [Current | Acc] ).

first_digit_count( 0, Dict ) -> Dict; first_digit_count( N, Dict ) -> [Key | _] = erlang:integer_to_list( N ), dict:update_counter( Key - 48, 1, Dict ). </lang>

Output:

7> benfords_law:task().
Digit   Actual  Benfords expected
1       0.301   0.3010299956639812
2       0.177   0.17609125905568124
3       0.125   0.12493873660829993
4       0.096   0.09691001300805642
5       0.08    0.07918124604762482
6       0.067   0.06694678963061322
7       0.056   0.05799194697768673
8       0.053   0.05115252244738129
9       0.045   0.04575749056067514

Forth

<lang forth>: 3drop drop 2drop ;

f2drop fdrop fdrop ;

int-array create cells allot does> swap cells + ;

1st-fib 0e 1e ;

next-fib ftuck f+ ;

1st-digit ( fp -- n )

   pad 6 represent 3drop pad c@ [char] 0 - ;

10 int-array counts

tally

   0 counts 10 cells erase
   1st-fib
   1000 0 DO
       1 fdup 1st-digit counts +!
       next-fib
   LOOP f2drop ;

benford ( d -- fp )

   s>f 1/f 1e f+ flog ;

tab 9 emit ;

heading ( -- )

   cr ." Leading digital distribution of the first 1,000 Fibonacci numbers:"
   cr ." Digit" tab ." Actual" tab ." Expected" ;

.fixed ( n -- ) \ print count as decimal fraction

   s>d <# # # # [char] . hold #s #> type space ;

report ( -- )

   precision  3 set-precision
   heading
   10 1 DO
       cr i 3 .r
       tab i counts @ .fixed
       tab i benford f.
   LOOP
   set-precision ;

compute-benford tally report ;</lang>

Output:

Gforth 0.7.2, Copyright (C) 1995-2008 Free Software Foundation, Inc.
Gforth comes with ABSOLUTELY NO WARRANTY; for details type `license'
Type `bye' to exit
compute-benford 
Leading digital distribution of the first 1,000 Fibonacci numbers:
Digit	Actual	Expected
  1	0.301 	0.301 
  2	0.177 	0.176 
  3	0.125 	0.125 
  4	0.096 	0.0969 
  5	0.080 	0.0792 
  6	0.067 	0.0669 
  7	0.056 	0.058 
  8	0.053 	0.0512 
  9	0.045 	0.0458  ok

Fortran

FORTRAN 90. Compilation and output of this program using emacs compile command and a fairly obvious Makefile entry: <lang fortran>-*- mode: compilation; default-directory: "/tmp/" -*- Compilation started at Sat May 18 01:13:00

a=./f && make $a && $a f95 -Wall -ffree-form f.F -o f

 0.301030010      0.176091254      0.124938756       9.69100147E-02   7.91812614E-02   6.69467747E-02   5.79919666E-02   5.11525236E-02   4.57575098E-02 THE LAW
 0.300999999      0.177000001      0.125000000       9.60000008E-02   7.99999982E-02   6.70000017E-02   5.70000000E-02   5.29999994E-02   4.50000018E-02 LEADING FIBONACCI DIGIT

Compilation finished at Sat May 18 01:13:00</lang>

<lang fortran>subroutine fibber(a,b,c,d)

 ! compute most significant digits, Fibonacci like.
 implicit none
 integer (kind=8), intent(in) :: a,b
 integer (kind=8), intent(out) :: c,d
 d = a + b
 if (15 .lt. log10(float(d))) then
   c = b/10
   d = d/10
 else
   c = b
 endif

end subroutine fibber

integer function leadingDigit(a)

 implicit none
 integer (kind=8), intent(in) :: a
 integer (kind=8) :: b
 b = a
 do while (9 .lt. b)
   b = b/10
 end do
 leadingDigit = transfer(b,leadingDigit)

end function leadingDigit

real function benfordsLaw(a)

 implicit none
 integer, intent(in) :: a
 benfordsLaw = log10(1.0 + 1.0 / a)

end function benfordsLaw

program benford

 implicit none

 interface

   subroutine fibber(a,b,c,d)
     implicit none
     integer (kind=8), intent(in) :: a,b
     integer (kind=8), intent(out) :: c,d
   end subroutine fibber

   integer function leadingDigit(a)
     implicit none
     integer (kind=8), intent(in) :: a
   end function leadingDigit

   real function benfordsLaw(a)
     implicit none
     integer, intent(in) :: a
   end function benfordsLaw

 end interface

 integer (kind=8) :: a, b, c, d
 integer :: i, count(10)
 data count/10*0/
 a = 1
 b = 1
 do i = 1, 1001
   count(leadingDigit(a)) = count(leadingDigit(a)) + 1
   call fibber(a,b,c,d)
   a = c
   b = d
 end do
 write(6,*) (benfordsLaw(i),i=1,9),'THE LAW'
 write(6,*) (count(i)/1000.0 ,i=1,9),'LEADING FIBONACCI DIGIT'

end program benford</lang>

Go

<lang go>package main

import (

   "fmt"
   "math"

)

func Fib1000() []float64 {

   a, b, r := 0., 1., [1000]float64{}
   for i := range r {
       r[i], a, b = b, b, b+a
   }
   return r[:]

}

func main() {

   show(Fib1000(), "First 1000 Fibonacci numbers")

}

func show(c []float64, title string) {

   var f [9]int
   for _, v := range c {
       f[fmt.Sprintf("%g", v)[0]-'1']++
   }
   fmt.Println(title)
   fmt.Println("Digit  Observed  Predicted")
   for i, n := range f {
       fmt.Printf("  %d  %9.3f  %8.3f\n", i+1, float64(n)/float64(len(c)),
           math.Log10(1+1/float64(i+1)))
   }

}</lang>

Output:

First 1000 Fibonacci numbers
Digit  Observed  Predicted
  1      0.301     0.301
  2      0.177     0.176
  3      0.125     0.125
  4      0.096     0.097
  5      0.080     0.079
  6      0.067     0.067
  7      0.056     0.058
  8      0.053     0.051
  9      0.045     0.046

Haskell

<lang haskell>import qualified Data.Map as M import Data.Char (digitToInt)

fstdigit :: Integer -> Int fstdigit = digitToInt . head . show

n = 1000::Int fibs = 1:1:zipWith (+) fibs (tail fibs) fibdata = map fstdigit $ take n fibs freqs = M.fromListWith (+) $ zip fibdata (repeat 1)

tab :: [(Int, Double, Double)] tab = [(d,

      (fromIntegral (M.findWithDefault 0 d freqs) /(fromIntegral n) ),
       logBase 10.0 $ 1 + 1/(fromIntegral d) ) | d<-[1..9]]

main = print tab</lang>

Output:

[(1,0.301,0.301029995663981),
(2,0.177,0.176091259055681),
(3,0.125,0.1249387366083),
(4,0.096,0.0969100130080564),
(5,0.08,0.0791812460476248),
(6,0.067,0.0669467896306132),
(7,0.056,0.0579919469776867),
(8,0.053,0.0511525224473813),
(9,0.045,0.0457574905606751)]

Icon and Unicon

The following solution works in both languages.

<lang unicon>global counts, total

procedure main()

  counts := table(0)
  total := 0.0
  every benlaw(fib(1 to 1000))

  every i := 1 to 9 do 
     write(i,": ",right(100*counts[string(i)]/total,9)," ",100*P(i))

end

procedure benlaw(n)

  if counts[n ? (tab(upto('123456789')),move(1))] +:= 1 then total +:= 1

end

procedure P(d)

  return log(1+1.0/d, 10)

end

procedure fib(n) # From Fibonacci Sequence task

   return fibMat(n)[1]

end

procedure fibMat(n)

   if n <= 0 then return [0,0]
   if n  = 1 then return [1,0]
   fp := fibMat(n/2)
   c := fp[1]*fp[1] + fp[2]*fp[2]
   d := fp[1]*(fp[1]+2*fp[2])
   if n%2 = 1 then return [c+d, d]
   else return [d, c]

end</lang>

Sample run:

->benlaw
1:      30.1 30.10299956639811
2:      17.7 17.60912590556812
3:      12.5 12.49387366082999
4:       9.6 9.69100130080564
5:       8.0 7.918124604762481
6:       6.7 6.694678963061322
7:       5.6 5.799194697768673
8:       5.3 5.115252244738128
9:       4.5 4.575749056067514
->

J

We show the correlation coefficient of Benford's law with the leading digits of the first 1000 Fibonacci numbers is almost unity. <lang J>log10 =: 10&^. benford =: log10@:(1+%) assert '0.30 0.18 0.12 0.10 0.08 0.07 0.06 0.05 0.05' -: 5j2 ": benford >: i. 9

append_next_fib =: , +/@:(_2&{.) assert 5 8 13 -: append_next_fib 5 8

leading_digits =: {.@":&> assert '581' -: leading_digits 5 8 13x

count =: #/.~ /: ~. assert 2 1 3 4 -: count 'XCXBAXACXC' NB. 2 A's, 1 B, 3 C's, and some X's.

normalize =: % +/ assert 1r3 2r3 -: normalize 1 2x

FIB =: append_next_fib ^: (1000-#) 1 1 LDF =: leading_digits FIB

TALLY_BY_KEY =: count LDF assert 9 -: # TALLY_BY_KEY NB. If all of [1-9] are present then we know what the digits are.

mean =: +/ % # center=: - mean mp =: $:~ :(+/ .*) num =: mp&:center den =: %:@:(*&:(+/@:(*:@:center))) r =: num % den NB. r is the LibreOffice correl function assert '_0.982' -: 6j3 ": 1 2 3 r 6 5 3 NB. confirmed using LibreOffice correl function

assert '0.9999' -: 6j4 ": (normalize TALLY_BY_KEY) r benford >: i.9

assert '0.9999' -: 6j4 ": TALLY_BY_KEY r benford >: i.9 NB. Of course we don't need normalization</lang>

Java

<lang Java>import java.math.BigInteger;

public class Benford {

   private static interface NumberGenerator {
       BigInteger[] getNumbers();
   }

   private static class FibonacciGenerator implements NumberGenerator {
       public BigInteger[] getNumbers() {
           final BigInteger[] fib = new BigInteger[ 1000 ];
           fib[ 0 ] = fib[ 1 ] = BigInteger.ONE;
           for ( int i = 2; i < fib.length; i++ )
               fib[ i ] = fib[ i - 2 ].add( fib[ i - 1 ] );
           return fib;
       }
   }

   private final int[] firstDigits = new int[ 9 ];
   private final int   count;

   private Benford( final NumberGenerator ng ) {
       final BigInteger[] numbers = ng.getNumbers();
       count = numbers.length;
       for ( final BigInteger number : numbers )
           firstDigits[ Integer.valueOf( number.toString().substring( 0, 1 ) ) - 1 ]++;
   }

   public String toString() {
       final StringBuilder result = new StringBuilder();
       for ( int i = 0; i < firstDigits.length; i++ )
           result.append( i + 1 )
               .append( '\t' ).append( firstDigits[ i ] / ( double )count )
               .append( '\t' ).append( Math.log10( 1 + 1d / ( i + 1 ) ) )
               .append( '\n' );
       return result.toString();
   }

   public static void main( final String[] args ) {
       System.out.println( new Benford( new FibonacciGenerator() ) );
   }

}</lang> The output is:

1	0.301	0.3010299956639812
2	0.177	0.17609125905568124
3	0.125	0.12493873660829993
4	0.096	0.09691001300805642
5	0.08	0.07918124604762482
6	0.067	0.06694678963061322
7	0.056	0.05799194697768673
8	0.053	0.05115252244738129
9	0.045	0.04575749056067514

To use other number sequences, implement a suitable NumberGenerator, construct a Benford instance with it and print it.

Julia

<lang Julia>fib(n) = ([one(n) one(n) ; one(n) zero(n)]^n)[1,2]

ben(l) = [count(x->x==i, map(n->string(n)[1],l)) for i='1':'9']./length(l)

benford(l) = [[1:9] ben(l) log10(1+1/[1:9])]</lang>

Output:

julia> benford([fib(big(n)) for n = 1:1000])
9x3 Array{Float64,2}:
 1.0  0.301  0.30103  
 2.0  0.177  0.176091 
 3.0  0.125  0.124939 
 4.0  0.096  0.09691  
 5.0  0.08   0.0791812
 6.0  0.067  0.0669468
 7.0  0.056  0.0579919
 8.0  0.053  0.0511525
 9.0  0.045  0.0457575

Liberty BASIC

Using function from http://rosettacode.org/wiki/Fibonacci_sequence#Liberty_BASIC <lang lb> dim bin(9)

N=1000 for i = 0 to N-1

   num$ = str$(fiboI(i))
   d=val(left$(num$,1))
   'print num$, d
   bin(d)=bin(d)+1

next print

print "Digit", "Actual freq", "Expected freq" for i = 1 to 9

   print i, bin(i)/N, using("#.###", P(i))

   P = log10(d+1)-log10(d)

end function

function log10(x)

   log10 = log(x)/log(10)

end function

function fiboI(n)

   a = 0
   b = 1
   for i = 1 to n
       temp = a + b
       a = b
       b = temp
   next i
   fiboI = a

end function </lang>

Output:

Digit         Actual freq   Expected freq
1             0.301         0.301
2             0.177         0.176
3             0.125         0.125
4             0.095         0.097
5             0.08          0.079
6             0.067         0.067
7             0.056         0.058
8             0.053         0.051
9             0.045         0.046

Mathematica

<lang mathematica>fibdata = Array[First@IntegerDigits@Fibonacci@# &, 1000]; Table[{d, N@Count[fibdata, d]/Length@fibdata, Log10[1. + 1/d]}, {d, 1,

   9}] // Grid</lang>

Output:

1	0.301	0.30103
2	0.177	0.176091
3	0.125	0.124939
4	0.096	0.09691
5	0.08	0.0791812
6	0.067	0.0669468
7	0.056	0.0579919
8	0.053	0.0511525
9	0.045	0.0457575

NetRexx

<lang NetRexx>/* NetRexx */ options replace format comments java crossref symbols nobinary

runSample(arg) return

-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ method brenfordDeveation(nlist = Rexx[]) public static

 observed = 0
 loop n_ over nlist
   d1 = n_.left(1)
   if d1 = 0 then iterate n_
   observed[d1] = observed[d1] + 1
   end n_
 say ' '.right(4) 'Observed'.right(11) 'Expected'.right(11) 'Deviation'.right(11)
 loop n_ = 1 to 9
   actual = (observed[n_] / (nlist.length - 1))
   expect = Rexx(Math.log10(1 + 1 / n_))
   deviat = expect - actual
   say n_.right(3)':' (actual * 100).format(3, 6)'%' (expect * 100).format(3, 6)'%' (deviat * 100).abs().format(3, 6)'%'
   end n_
 return

-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ method fibonacciList(size = 1000) public static returns Rexx[]

 fibs = Rexx[size + 1]
 fibs[0] = 0
 fibs[1] = 1
 loop n_ = 2 to size
   fibs[n_] = fibs[n_ - 1] + fibs[n_ - 2]
   end n_
 return fibs

-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ method runSample(arg) private static

 parse arg n_ .
 if n_ =  then n_ = 1000
 fibList = fibonacciList(n_)
 say 'Fibonacci sequence to' n_
 brenfordDeveation(fibList)
 return

</lang>

Output:

Fibonacci sequence to 1000
        Observed    Expected   Deviation
  1:  30.100000%  30.103000%   0.003000%
  2:  17.700000%  17.609126%   0.090874%
  3:  12.500000%  12.493874%   0.006127%
  4:   9.600000%   9.691001%   0.091001%
  5:   8.000000%   7.918125%   0.081875%
  6:   6.700000%   6.694679%   0.005321%
  7:   5.600000%   5.799195%   0.199195%
  8:   5.300000%   5.115252%   0.184748%
  9:   4.500000%   4.575749%   0.075749%

PARI/GP

<lang parigp>distribution(v)={ my(t=vector(9,n,sum(i=1,#v,v[i]==n))); print("Digit\tActual\tExpected"); for(i=1,9,print(i, "\t", t[i], "\t", round(#v*(log(i+1)-log(i))/log(10)))) }; dist(f)=distribution(vector(1000,n,digits(f(n))[1])); lucas(n)=fibonacci(n-1)+fibonacci(n+1); dist(fibonacci) dist(lucas)</lang>

Output:

Digit   Actual  Expected
1       301     301
2       177     176
3       125     125
4       96      97
5       80      79
6       67      67
7       56      58
8       53      51
9       45      46

Digit   Actual  Expected
1       301     301
2       174     176
3       127     125
4       97      97
5       79      79
6       66      67
7       59      58
8       51      51
9       46      46

Perl

<lang Perl>#!/usr/bin/perl use strict ; use warnings ; use POSIX qw( log10 ) ;

my @fibonacci = ( 0 , 1 ) ; while ( @fibonacci != 1000 ) {

  push @fibonacci , $fibonacci[ -1 ] + $fibonacci[ -2 ] ;

} my @actuals ; my @expected ; for my $i( 1..9 ) {

  my $sum = 0 ;
  map { $sum++ if $_ =~ /\A$i/ } @fibonacci ;
  push @actuals , $sum / 1000  ;
  push @expected , log10( 1 + 1/$i ) ;

} print " Observed Expected\n" ; for my $i( 1..9 ) {

  print "$i : " ;
  my $result = sprintf ( "%.2f" , 100 * $actuals[ $i - 1 ] ) ;
  printf "%11s %%" , $result ;
  $result = sprintf ( "%.2f" , 100 * $expected[ $i - 1 ] ) ;
  printf "%15s %%\n" , $result ;

}</lang>

Output:

 
         Observed         Expected
1 :       30.10 %          30.10 %
2 :       17.70 %          17.61 %
3 :       12.50 %          12.49 %
4 :        9.50 %           9.69 %
5 :        8.00 %           7.92 %
6 :        6.70 %           6.69 %
7 :        5.60 %           5.80 %
8 :        5.30 %           5.12 %
9 :        4.50 %           4.58 %

Perl 6

<lang perl6>sub benford(@a) { bag +« @a».comb: /<( <[ 1..9 ]> )> <[ , . \d ]>*/ }

sub show(%distribution) {

   printf "%9s %9s  %s\n", <Actual Expected Deviation>;
   for 1 .. 9 -> $digit {
       my $actual = %distribution{$digit} * 100 / [+] %distribution.values;
       my $expected = (1 + 1 / $digit).log(10) * 100;
       printf "%d: %5.2f%% | %5.2f%% | %.2f%%\n",
         $digit, $actual, $expected, abs($expected - $actual);
   }

}

multi MAIN($file) { show benford $file.IO.lines } multi MAIN() { show benford ( 1, 1, 2, *+* ... * )[^1000] }</lang>

Output: First 1000 Fibonaccis

   Actual  Expected  Deviation
1: 30.10% | 30.10% | 0.00%
2: 17.70% | 17.61% | 0.09%
3: 12.50% | 12.49% | 0.01%
4:  9.60% |  9.69% | 0.09%
5:  8.00% |  7.92% | 0.08%
6:  6.70% |  6.69% | 0.01%
7:  5.60% |  5.80% | 0.20%
8:  5.30% |  5.12% | 0.18%
9:  4.50% |  4.58% | 0.08%

Extra credit: Square Kilometers of land under cultivation, by country / territory. First column from Wikipedia: Land use statistics by country.

   Actual  Expected  Deviation
1: 33.33% | 30.10% | 3.23%
2: 18.31% | 17.61% | 0.70%
3: 13.15% | 12.49% | 0.65%
4:  8.45% |  9.69% | 1.24%
5:  9.39% |  7.92% | 1.47%
6:  5.63% |  6.69% | 1.06%
7:  4.69% |  5.80% | 1.10%
8:  5.16% |  5.12% | 0.05%
9:  1.88% |  4.58% | 2.70%

PL/I

<lang PL/I> (fofl, size, subrg): Benford: procedure options(main); /* 20 October 2013 */

  declare sc(1000) char(1), f(1000) float (16);
  declare d fixed (1);

  call Fibonacci(f);
  call digits(sc, f);

  put skip list ('digit  expected     obtained');
  do d= 1 upthru 9;
     put skip edit (d, log10(1 + 1/d), tally(sc, trim(d))/1000)
        (f(3), 2 f(13,8) );
  end;

Fibonacci: procedure (f);

  declare f(*) float (16);
  declare i fixed binary;

  f(1), f(2) = 1;
  do i = 3 to 1000;
     f(i) = f(i-1) + f(i-2);
  end;

end Fibonacci;

digits: procedure (sc, f);

  declare sc(*) char(1), f(*) float (16);
  sc = substr(trim(f), 1, 1);

end digits;

tally: procedure (sc, d) returns (fixed binary);

  declare sc(*) char(1), d char(1);
  declare (i, t) fixed binary;
  t = 0;
  do i = 1 to 1000;
     if sc(i) = d then t = t + 1;
  end;
  return (t);

end tally; end Benford; </lang> Results:

digit  expected     obtained 
  1   0.30103000   0.30099487
  2   0.17609126   0.17698669
  3   0.12493874   0.12500000
  4   0.09691001   0.09599304
  5   0.07918125   0.07998657
  6   0.06694679   0.06698608
  7   0.05799195   0.05599976
  8   0.05115252   0.05299377
  9   0.04575749   0.04499817

PL/pgSQL

<lang SQL> WITH recursive constant(val) AS ( select 1000. ) , fib(a,b) AS ( SELECT CAST(0 AS numeric), CAST(1 AS numeric) UNION ALL SELECT b,a+b FROM fib ) , benford(first_digit, probability_real, probability_theoretical) AS ( SELECT *, CAST(log(1. + 1./CAST(first_digit AS INT)) AS NUMERIC(5,4)) probability_theoretical FROM ( SELECT first_digit, CAST(COUNT(1)/(select val from constant) AS NUMERIC(5,4)) probability_real FROM ( SELECT SUBSTRING(CAST(a AS VARCHAR(100)),1,1) first_digit FROM fib WHERE SUBSTRING(CAST(a AS VARCHAR(100)),1,1) <> '0' LIMIT (select val from constant) ) t GROUP BY first_digit ) f ORDER BY first_digit ASC ) select * from benford cross join

    (select cast(corr(probability_theoretical,probability_real) as numeric(5,4)) correlation
     from benford) c

</lang>

Prolog

Works with: SWI Prolog version 6.2.6 by Jan Wielemaker, University of Amsterdam

Note: SWI Prolog implements arbitrary precision integer arithmetic through use of the GNU MP library <lang Prolog>%_________________________________________________________________ % Does the Fibonacci sequence follow Benford's law? %~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ % Fibonacci sequence generator fib(C, [P,S], C, N) :- N is P + S. fib(C, [P,S], Cv, V) :- succ(C, Cn), N is P + S, !, fib(Cn, [S,N], Cv, V).

fib(0, 0). fib(1, 1). fib(C, N) :- fib(2, [0,1], C, N). % Generate from 3rd sequence on

% The benford law calculated benford(D, Val) :- Val is log10(1+1/D).

% Retrieves the first characters of the first 1000 fibonacci numbers % (excluding zero) firstchar(V) :- fib(C,N), N =\= 0, atom_chars(N, [Ch|_]), number_chars(V, [Ch]), (C>999-> !; true).

% Increment the n'th list item (1 based), result -> third argument. incNth(1, [Dh|Dt], [Ch|Dt]) :- !, succ(Dh, Ch). incNth(H, [Dh|Dt], [Dh|Ct]) :- succ(Hn, H), !, incNth(Hn, Dt, Ct).

% Calculate the frequency of the all the list items freq([], D, D). freq([H|T], D, C) :- incNth(H, D, L), !, freq(T, L, C).

% Output the results writeHdr :- format('~t~w~15| - ~t~w\n', ['Benford', 'Measured']). writeData(Benford, Freq) :- format('~t~2f%~15| - ~t~2f%\n', [Benford*100, Freq*100]).

go :- % main goal findall(B, (between(1,9,N), benford(N,B)), Benford), findall(C, firstchar(C), Fc), freq(Fc, Freq), writeHdr, maplist(writeData, Benford, Freq).</lang>

Output:

?- go.
        Benford - Measured
         30.10% - 30.10%
         17.61% - 17.70%
         12.49% - 12.50%
          9.69% - 9.60%
          7.92% - 8.00%
          6.69% - 6.70%
          5.80% - 5.60%
          5.12% - 5.30%
          4.58% - 4.50%

Python

Works with Python 3.X & 2.7 <lang python>from __future__ import division from itertools import islice, count from collections import Counter from math import log10 from random import randint

expected = [log10(1+1/d) for d in range(1,10)]

def fib():

   a,b = 1,1
   while True:
       yield a
       a,b = b,a+b

powers of 3 as a test sequence

def power_of_threes():

   return (3**k for k in count(0))

def heads(s):

   for a in s: yield int(str(a)[0])

def show_dist(title, s):

   c = Counter(s)
   size = sum(c.values())
   res = [c[d]/size for d in range(1,10)]

   print("\n%s Benfords deviation" % title)
   for r, e in zip(res, expected):
       print("%5.1f%% %5.1f%%  %5.1f%%" % (r*100., e*100., abs(r - e)*100.))

def rand1000():

   while True: yield randint(1,9999)

if __name__ == '__main__':

   show_dist("fibbed", islice(heads(fib()), 1000))
   show_dist("threes", islice(heads(power_of_threes()), 1000))

   # just to show that not all kind-of-random sets behave like that
   show_dist("random", islice(heads(rand1000()), 10000))</lang>

Output:

fibbed Benfords deviation
 30.1%  30.1%    0.0%
 17.7%  17.6%    0.1%
 12.5%  12.5%    0.0%
  9.6%   9.7%    0.1%
  8.0%   7.9%    0.1%
  6.7%   6.7%    0.0%
  5.6%   5.8%    0.2%
  5.3%   5.1%    0.2%
  4.5%   4.6%    0.1%

threes Benfords deviation
 30.0%  30.1%    0.1%
 17.7%  17.6%    0.1%
 12.3%  12.5%    0.2%
  9.8%   9.7%    0.1%
  7.9%   7.9%    0.0%
  6.6%   6.7%    0.1%
  5.9%   5.8%    0.1%
  5.2%   5.1%    0.1%
  4.6%   4.6%    0.0%

random Benfords deviation
 11.2%  30.1%   18.9%
 10.9%  17.6%    6.7%
 11.6%  12.5%    0.9%
 11.1%   9.7%    1.4%
 11.6%   7.9%    3.7%
 11.4%   6.7%    4.7%
 10.3%   5.8%    4.5%
 11.0%   5.1%    5.9%
 10.9%   4.6%    6.3%

R

<lang R> pbenford <- function(d){

 return(log10(1+(1/d)))

}

get_lead_digit <- function(number){

 return(as.numeric(substr(number,1,1)))

}

fib_iter <- function(n){

 first <- 1
 second <- 0
 for(i in 1:n){
   sum <- first + second
   first <- second
   second <- sum
 }
 return(sum)

}

fib_sequence <- mapply(fib_iter,c(1:1000)) lead_digits <- mapply(get_lead_digit,fib_sequence)

observed_frequencies <- table(lead_digits)/1000 expected_frequencies <- mapply(pbenford,c(1:9))

data <- data.frame(observed_frequencies,expected_frequencies) colnames(data) <- c("digit","obs.frequency","exp.frequency") dev_percentage <- abs((data$obs.frequency-data$exp.frequency)*100) data <- data.frame(data,dev_percentage)

print(data) </lang>

Output:

digit obs.frequency exp.frequency dev_percentage
    1         0.301       0.30103       0.003000
    2         0.177       0.17609       0.090874
    3         0.125       0.12494       0.006126
    4         0.096       0.09691       0.091001
    5         0.080       0.07918       0.081875
    6         0.067       0.06695       0.005321
    7         0.056       0.05799       0.199195
    8         0.053       0.05115       0.184748
    9         0.045       0.04576       0.075749

Racket

<lang Racket>#lang racket

(define (log10 n) (/ (log n) (log 10)))

(define (first-digit n)

 (quotient n (expt 10 (inexact->exact (floor (log10 n))))))

(define N 10000)

(define fibs

 (let loop ([n N] [a 0] [b 1])
   (if (zero? n) '() (cons b (loop (sub1 n) b (+ a b))))))

(define v (make-vector 10 0)) (for ([n fibs])

 (define f (first-digit n))
 (vector-set! v f (add1 (vector-ref v f))))

(printf "N OBS EXP\n") (define (pct n) (~r (* n 100.0) #:precision 1 #:min-width 4)) (for ([i (in-range 1 10)])

 (printf "~a: ~a% ~a%\n" i
         (pct (/ (vector-ref v i) N))
         (pct (log10 (+ 1 (/ i))))))

Output


N OBS EXP
1: 30.1% 30.1%
2: 17.6% 17.6%
3: 12.5% 12.5%
4: 9.7% 9.7%
5: 7.9% 7.9%
6: 6.7% 6.7%
7: 5.8% 5.8%
8: 5.1% 5.1%
9: 4.6% 4.6%</lang>

REXX

The REXX language practically hasn't any high math functions, so the E, LN, and LOG functions were included herein. Note that the E and LN10 functions return a limited amount of accuracy, and they should be greater than 50 digits (in this case).

Note that prime numbers don't lend themselves to Benford's law. Apologies to those people's eyes looking at the prime number generator. Uf-ta!

For the extra credit stuff, I choose to generate the primes and factorials rather than find a web-page with them listed, as each list is very easy to generate. <lang rexx>/*REXX program demonstrates some common trig functions (30 digits shown)*/ numeric digits 50 /*use only 50 digits for LN, LOG.*/ parse arg N .; if N== then N=1000 /*allow sample size specification*/

              /*══════════════apply Benford's law to Fibonacci numbers.*/

@.=1; do j=3 to N; jm1=j-1; jm2=j-2; @.j=@.jm2+@.jm1; end /*j*/ call show_results "Benford's law applied to" N 'Fibonacci numbers'

              /*══════════════apply Benford's law to prime numbers.    */

p=0; do j=2 until p==N; if \isPrime(j) then iterate; p=p+1; @.p=j;end call show_results "Benford's law applied to" N 'prime numbers'

              /*══════════════apply Benford's law to factorials.       */
        do j=1  for N;      @.j=!(j);  end  /*j*/

call show_results "Benford's law applied to" N 'factorial products' exit /*stick a fork in it, we're done.*/ /*──────────────────────────────────SHOW_RESULTS subroutine─────────────*/ show_results: w1=max(length('observed'),length(N-2)) ; say pad=' '; w2=max(length('expected' ),length(N )) say pad 'digit' pad center('observed',w1) pad center('expected',w2) say pad '─────' pad center(,w1,'─') pad center(,w2,'─') pad arg(1) !.=0; do j=1 for N; _=left(@.j,1); !._=!._+1; end /*get 1st digits.*/

       do k=1  for 9                  /*show results for Fibonacci nums*/
       say pad center(k,5) pad center(format(!.k/N,,length(N-2)),w1),
                           pad center(format(log(1+1/k),,length(N)+2),w2)
       end   /*k*/

return /*──────────────────────────────────one─line subroutines───────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────*/ !: procedure; parse arg x; !=1; do j=2 to x; !=!*j; end /*j*/; return ! e: return 2.7182818284590452353602874713526624977572470936999595749669676277240766303535 isprime: procedure; parse arg x; if wordpos(x,'2 3 5 7')\==0 then return 1; if x//2==0 then return 0; if x//3==0 then return 0; do j=5 by 6 until j*j>x; if x//j==0 then return 0; if x//(j+2)==0 then return 0; end; return 1 ln10:return 2.30258509299404568401799145468436420760110148862877297603332790096757260967735248023599720508959829834196778404228624863340952546508280675666628736909878168948290720832555468084379989482623319852839350530896538 ln:procedure expose $.;parse arg x,f;if x==10 then do;_=ln10();xx=format(_);if xx\==_ then return xx;end;call e;ig=x>1.5;is=1-2*(ig\==1);ii=0;xx=x;return .ln_comp() .ln_comp:do while ig&xx>1.5|\ig&xx<.5;_=e();do k=-1;iz=xx*_**-is;if k>=0&(ig&iz<1|\ig&iz>.5) then leave;_=_*_;izz=iz;end;xx=izz;ii=ii+is*2**k;end;x=x*e()**-ii-1;z=0;_=-1;p=z;do k=1;_=-_*x;z=z+_/k;if z=p then leave;p=z;end;return z+ii log:return ln(arg(1))/ln(10)</lang> output when using the default input:

    digit     observed      expected
    ─────     ────────      ────────     Benford's law applied to 1000 Fibonacci numbers
      1         0.301       0.301030
      2         0.177       0.176091
      3         0.125       0.124939
      4         0.096       0.096910
      5         0.080       0.079181
      6         0.067       0.066947
      7         0.056       0.057992
      8         0.053       0.051153
      9         0.045       0.045757

    digit     observed      expected
    ─────     ────────      ────────     Benford's law applied to 1000 prime numbers
      1         0.160       0.301030
      2         0.146       0.176091
      3         0.139       0.124939
      4         0.139       0.096910
      5         0.131       0.079181
      6         0.135       0.066947
      7         0.118       0.057992
      8         0.017       0.051153
      9         0.015       0.045757

    digit     observed      expected
    ─────     ────────      ────────     Benford's law applied to 1000 factorial products
      1         0.293       0.301030
      2         0.176       0.176091
      3         0.124       0.124939
      4         0.102       0.096910
      5         0.069       0.079181
      6         0.087       0.066947
      7         0.051       0.057992
      8         0.051       0.051153
      9         0.047       0.045757

output when using 100,000 primes:

    digit     observed      expected
    ─────     ────────      ────────     Benford's law applied to 100000 prime numbers
      1        0.31087      .3010300
      2        0.09142      .1760912
      3        0.08960      .1249387
      4        0.08747      .0969100
      5        0.08615      .0791812
      6        0.08458      .0669467
      7        0.08435      .0579919
      8        0.08326      .0511525
      9        0.08230      .0457574

Ruby

Translation of: Python

<lang ruby>EXPECTED = (1..9).map{|d| Math.log10(1+1.0/d)}

def fib(n)

 a,b = 0,1
 n.times.map{ret, a, b = a, b, a+b; ret}

end

powers of 3 as a test sequence

def power_of_threes(n)

 n.times.map{|k| 3**k}

end

def heads(s)

 s.map{|a| a.to_s[0].to_i}

end

def show_dist(title, s)

 s = heads(s)
 c = Array.new(10, 0)
 s.each{|x| c[x] += 1}
 size = s.size.to_f
 res = (1..9).map{|d| c[d]/size}
 puts "\n    %s Benfords deviation" % title
 res.zip(EXPECTED).each.with_index(1) do |(r, e), i|
   puts "%2d: %5.1f%%  %5.1f%%  %5.1f%%" % [i, r*100, e*100, (r - e).abs*100]
 end

end

def random(n)

 n.times.map{rand(1..n)}

end

show_dist("fibbed", fib(1000)) show_dist("threes", power_of_threes(1000))

just to show that not all kind-of-random sets behave like that

show_dist("random", random(10000))</lang>

Output:

    fibbed Benfords deviation
 1:  30.1%   30.1%    0.0%
 2:  17.7%   17.6%    0.1%
 3:  12.5%   12.5%    0.0%
 4:   9.5%    9.7%    0.2%
 5:   8.0%    7.9%    0.1%
 6:   6.7%    6.7%    0.0%
 7:   5.6%    5.8%    0.2%
 8:   5.3%    5.1%    0.2%
 9:   4.5%    4.6%    0.1%

    threes Benfords deviation
 1:  30.0%   30.1%    0.1%
 2:  17.7%   17.6%    0.1%
 3:  12.3%   12.5%    0.2%
 4:   9.8%    9.7%    0.1%
 5:   7.9%    7.9%    0.0%
 6:   6.6%    6.7%    0.1%
 7:   5.9%    5.8%    0.1%
 8:   5.2%    5.1%    0.1%
 9:   4.6%    4.6%    0.0%

    random Benfords deviation
 1:  10.9%   30.1%   19.2%
 2:  10.9%   17.6%    6.7%
 3:  11.7%   12.5%    0.8%
 4:  10.8%    9.7%    1.1%
 5:  11.2%    7.9%    3.3%
 6:  11.9%    6.7%    5.2%
 7:  10.7%    5.8%    4.9%
 8:  11.1%    5.1%    6.0%
 9:  10.8%    4.6%    6.2%

Run BASIC

<lang runbasic> N = 1000 for i = 0 to N - 1

   n$	= str$(fibonacci(i))
   j	= val(left$(n$,1))
   actual(j) = actual(j) +1

next print

html "

"

for i = 1 to 9

html ""

Digit	Actual	Expected
";i;"	";using("##.###",actual(i)/10);"	";using("##.###", frequency(i)*100);"

"

end

function frequency(n)

   frequency = log10(n+1) - log10(n)

end function

function log10(n)

   log10 = log(n) / log(10)

end function

function fibonacci(n)

   b = 1
   for i = 1 to n
       temp		= fibonacci + b
       fibonacci	= b
       b		= temp
   next i

end function </lang>

Digit	Actual	Expected
1	30.100	30.103
2	17.700	17.609
3	12.500	12.494
4	9.500	9.691
5	8.000	7.918
6	6.700	6.695
7	5.600	5.799
8	5.300	5.115
9	4.500	4.576

Scala

<lang scala>// Fibonacci Sequence (begining with 1,1): 1 1 2 3 5 8 13 21 34 55 ... val fibs : Stream[BigInt] = { def series(i:BigInt,j:BigInt):Stream[BigInt] = i #:: series(j, i+j); series(1,0).tail.tail }

/**

* Given a numeric sequence, return the distribution of the most-signicant-digit 
* as expected by Benford's Law and then by actual distribution.
*/

def benford[N:Numeric]( data:Seq[N] ) : Map[Int,(Double,Double)] = {

 import scala.math._
 
 val maxSize = 10000000  // An arbitrary size to avoid problems with endless streams
 
 val size = (data.take(maxSize)).size.toDouble
 
 val distribution = data.take(maxSize).groupBy(_.toString.head.toString.toInt).map{ case (d,l) => (d -> l.size) }
 
 (for( i <- (1 to 9) ) yield { (i -> (log10(1D + 1D / i), (distribution(i) / size))) }).toMap

}

{

 println( "Fibonacci Sequence (size=1000): 1 1 2 3 5 8 13 21 34 55 ...\n" )
 println( "%9s %9s %9s".format( "Actual", "Expected", "Deviation" ) )

 benford( fibs.take(1000) ).toList.sorted foreach { 
   case (k, v) => println( "%d: %5.2f%% | %5.2f%% | %5.4f%%".format(k,v._2*100,v._1*100,math.abs(v._2-v._1)*100) ) 
 }

}</lang>

Output:

Fibonacci Sequence (size=1000): 1 1 2 3 5 8 13 21 34 55 ...

   Actual  Expected Deviation
1: 30.10% | 30.10% | 0.0030%
2: 17.70% | 17.61% | 0.0909%
3: 12.50% | 12.49% | 0.0061%
4:  9.60% |  9.69% | 0.0910%
5:  8.00% |  7.92% | 0.0819%
6:  6.70% |  6.69% | 0.0053%
7:  5.60% |  5.80% | 0.1992%
8:  5.30% |  5.12% | 0.1847%
9:  4.50% |  4.58% | 0.0757%

Sidef

<lang ruby>var fibonacci = [0, 1] ; {

   fibonacci.append(fibonacci[-1] + $fibonacci[-2]);

} * (1000 - fibonacci.len);

var (actuals, expected) = ([], []);

{ |i|

   var num = 0;
   fibonacci.each { |j| j.digit(-1) == i && (num++)};
   actuals.append(num / 1000);
   expected.append(1 + (1/i) -> log10);

} * 9;

"%17s%17s\n".printf("Observed","Expected"); { |i|

   "%d : %11s %%%15s %%\n".printf(
           i, "%.2f".sprintf(100 *  actuals[i - 1]),
              "%.2f".sprintf(100 * expected[i - 1]),
   );

} * 9;</lang>

Output:

         Observed         Expected
1 :       30.10 %          30.10 %
2 :       17.70 %          17.61 %
3 :       12.50 %          12.49 %
4 :        9.50 %           9.69 %
5 :        8.00 %           7.92 %
6 :        6.70 %           6.69 %
7 :        5.60 %           5.80 %
8 :        5.30 %           5.12 %
9 :        4.50 %           4.58 %

SQL

If we load some numbers into a table, we can do the sums without too much difficulty. I tried to make this as database-neutral as possible, but I only had Oracle handy to test it on.

The query is the same for any number sequence you care to put in the benford table.

<lang SQL>-- Create table create table benford (num integer);

-- Seed table insert into benford (num) values (1); insert into benford (num) values (1); insert into benford (num) values (2);

-- Populate table insert into benford (num)

 select
   ult + penult
 from
   (select max(num) as ult from benford),
   (select max(num) as penult from benford where num not in (select max(num) from benford))

-- Repeat as many times as desired -- in Oracle SQL*Plus, press "Slash, Enter" a lot of times -- or wrap this in a loop, but that will require something db-specific...

-- Do sums select

 digit,
 count(digit) / numbers as actual,
 log(10, 1 + 1 / digit) as expected

from

 (
   select
     floor(num/power(10,length(num)-1)) as digit
   from
     benford
 ),
 (
   select
     count(*) as numbers
   from
     benford
 )

group by digit, numbers order by digit;

-- Tidy up drop table benford;</lang>

Output:

I only loaded the first 100 Fibonacci numbers before my fingers were sore from repeating the data load. 8~)

     DIGIT     ACTUAL   EXPECTED
---------- ---------- ----------
         1         .3 .301029996
         2        .18 .176091259
         3        .13 .124938737
         4        .09 .096910013
         5        .08 .079181246
         6        .06  .06694679
         7        .05 .057991947
         8        .07 .051152522
         9        .04 .045757491

9 rows selected.

Tcl

<lang tcl>proc benfordTest {numbers} {

   # Count the leading digits (RE matches first digit in each number,
   # even if negative)
   set accum {1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9 0}
   foreach n $numbers {

if {[regexp {[1-9]} $n digit]} { dict incr accum $digit }

   # Print the report
   puts " digit | measured | theory"
   puts "-------+----------+--------"
   dict for {digit count} $accum {

puts [format "%6d | %7.2f%% | %5.2f%%" $digit \ [expr {$count * 100.0 / [llength $numbers]}] \ [expr {log(1+1./$digit)/log(10)*100.0}]]

}</lang> Demonstrating with Fibonacci numbers: <lang tcl>proc fibs n {

   for {set a 1;set b [set i 0]} {$i < $n} {incr i} {

lappend result [set b [expr {$a + [set a $b]}]]

   }
   return $result

} benfordTest [fibs 1000]</lang>

Output:

 digit | measured | theory
-------+----------+--------
     1 |   30.10% | 30.10%
     2 |   17.70% | 17.61%
     3 |   12.50% | 12.49%
     4 |    9.60% |  9.69%
     5 |    8.00% |  7.92%
     6 |    6.70% |  6.69%
     7 |    5.60% |  5.80%
     8 |    5.30% |  5.12%
     9 |    4.50% |  4.58%

@@ Line 955: / Line 955: @@
 ben(l) = [count(x->x==i, map(n->string(n)[1],l)) for i='1':'9']./length(l)
-benford(l) = collect(zip(1:9, ben(l), log10(1.+1./[1:9])))</lang>
+benford(l) = [[1:9] ben(l) log10(1+1/[1:9])]</lang>
 {{Out}}
 <pre>julia> benford([fib(big(n)) for n = 1:1000])
--element Array{(Int64,Float64,Float64),1}:
+x3 Array{Float64,2}:
- (1,0.301,0.3010299956639812)
+.0  0.301  0.30103
- (2,0.177,0.17609125905568124)
+.0  0.177  0.176091
- (3,0.125,0.12493873660829993)
+.0  0.125  0.124939
- (4,0.096,0.09691001300805642)
+.0  0.096  0.09691
- (5,0.08,0.07918124604762482)
+.0  0.08   0.0791812
- (6,0.067,0.06694678963061322)
+.0  0.067  0.0669468
- (7,0.056,0.05799194697768673)
+.0  0.056  0.0579919
- (8,0.053,0.05115252244738129)
+.0  0.053  0.0511525
- (9,0.045,0.04575749056067514)
+.0  0.045  0.0457575</pre>
-</pre>
 =={{header|Liberty BASIC}}==