# Benford's law

Benford's law
You are encouraged to solve this task according to the task description, using any language you may know.
 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 ${\displaystyle d}$  (${\displaystyle d\in \{1,\ldots ,9\}}$) occurs with probability

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

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.

## 8th

: n:log10e  1 10 ln /  ;

with: n

: n:log10  \ n -- n
ln log10e * ;

: benford  \ x -- x
1 swap / 1+ log10 ;

: fibs \ xt n
swap >r
0.0 1.0 rot
( dup r@ w:exec tuck + ) swap times
2drop rdrop ;

var counts

: init
a:new ( 0 a:push ) 9 times counts ! ;

: leading \ n -- n
"%g" s:strfmt
0 s:@ '0 - nip ;

: bump-digit \ n --
1 swap
counts @ swap 1- ' + a:op! drop ;

: count-fibs \ --
( leading bump-digit ) 1000 fibs ;

counts @  ( 0.001 * ) a:map  counts ! ;

: spaces \ n --
' space swap times ;

: .fw \ s n --
>r s:len r> rot . swap - spaces ;

"Digit" 8 .fw "Expected" 10 .fw "Actual" 10 .fw cr ;

: .digit \ n --
>s 8 .fw ;

: .actual \ n --
"%.3f" s:strfmt 10 .fw ;

: .expected \ n --
"%.4f" s:strfmt 10 .fw ;

: report \ --
counts @
( swap 1+ dup benford swap
.digit .expected .actual cr )
a:each drop ;

: benford-test

;with

benford-test
bye
Output:
Digit   Expected  Actual
1       0.3010    0.301
2       0.1761    0.177
3       0.1249    0.125
4       0.0969    0.096
5       0.0792    0.080
6       0.0669    0.067
7       0.0580    0.056
8       0.0512    0.053
9       0.0458    0.045


## 11l

Translation of: D
F get_fibs()
V a = 1.0
V b = 1.0
[Float] r
L 1000
r [+]= a
(a, b) = (b, a + b)
R r

F benford(seq)
V freqs = [(0.0, 0.0)] * 9
V seq_len = 0
L(d) seq
I d != 0
freqs[String(d)[0].code - ‘1’.code][1]++
seq_len++

L(&f) freqs
f = (log10(1.0 + 1.0 / (L.index + 1)), f[1] / seq_len)
R freqs

print(‘#9 #9 #9’.format(‘Actual’, ‘Expected’, ‘Deviation’))

L(p) benford(get_fibs())
print(‘#.: #2.2% | #2.2% | #.4%’.format(L.index + 1, p[1] * 100, p[0] * 100, abs(p[1] - p[0]) * 100))
Output:
   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%


## ABC

HOW TO RETURN fibonacci.numbers n:
PUT 1, 1 IN a, b
PUT {} IN fibo
FOR i IN {1..n}:
INSERT a IN fibo
PUT b, a+b IN a, b
RETURN fibo

HOW TO RETURN digit.distribution nums:
PUT {} IN digits
FOR i IN {1..9}: PUT i IN digits["i"]
PUT {} IN dist
FOR i IN {1..9}: PUT 0 IN dist[i]
FOR n IN nums:
PUT digits["n"|1] IN digit
PUT dist[digit] + 1 IN dist[digit]
FOR i IN {1..9}:
PUT dist[i] / #nums IN dist[i]
RETURN dist

PUT digit.distribution fibonacci.numbers 1000 IN observations

WRITE "Digit"<<6, "Expected">>10, "Observed">>10/
FOR d IN {1..9}:
WRITE d<<6, ((10 log (1 + 1/d))>>10)|10, observations[d]>>10/
Output:
Digit   Expected  Observed
1     0.30102999     0.301
2     0.17609125     0.177
3     0.12493873     0.125
4     0.09691001     0.096
5     0.07918124      0.08
6     0.06694678     0.067
7     0.05799194     0.056
8     0.05115252     0.053
9     0.04575749     0.045

The program reads the Fibonacci-Numbers from the standard input. Each input line is supposed to hold N, followed by Fib(N).

with Ada.Text_IO, Ada.Numerics.Generic_Elementary_Functions;

procedure Benford is

subtype Nonzero_Digit is Natural range 1 .. 9;
function First_Digit(S: String) return Nonzero_Digit is
(if S(S'First) in '1' .. '9'
then Nonzero_Digit'Value(S(S'First .. S'First))
else First_Digit(S(S'First+1 .. S'Last)));

procedure Print(D: Nonzero_Digit; Counted, Sum: Natural) is
Actual: constant Float := Float(Counted) / Float(Sum);
Expected: constant Float := Math.Log(1.0 + 1.0 / Float(D), Base => 10.0);
Deviation: constant Float := abs(Expected-Actual);
begin
N_IO.Put(D, 5);
N_IO.Put(Counted, 14);
F_IO.Put(Float(Sum)*Expected, Fore => 16, Aft => 1, Exp => 0);
F_IO.Put(100.0*Actual, Fore => 9, Aft => 2, Exp => 0);
F_IO.Put(100.0*Expected, Fore => 11, Aft => 2, Exp => 0);
F_IO.Put(100.0*Deviation, Fore => 13, Aft => 2, Exp => 0);
end Print;

Cnt: array(Nonzero_Digit) of Natural := (1 .. 9 => 0);
D: Nonzero_Digit;
Sum: Natural := 0;
Counter: Positive;

begin
-- each line in the input file holds Counter, followed by Fib(Counter)
N_IO.Get(Counter);
-- Counter and skip it, we just don't need it
-- read the rest of the line and extract the first digit
Cnt(D) := Cnt(D)+1;
Sum := Sum + 1;
end loop;
& "   Expected[%]   Difference[%]");
for I in Nonzero_Digit loop
Print(I, Cnt(I), Sum);
end loop;
end Benford;

Output:
>./benford < fibo.txt
Digit  Found[total]   Expected[total]    Found[%]   Expected[%]   Difference[%]
1           301             301.0       30.10         30.10            0.00
2           177             176.1       17.70         17.61            0.09
3           125             124.9       12.50         12.49            0.01
4            96              96.9        9.60          9.69            0.09
5            80              79.2        8.00          7.92            0.08
6            67              66.9        6.70          6.69            0.01
7            56              58.0        5.60          5.80            0.20
8            53              51.2        5.30          5.12            0.18
9            45              45.8        4.50          4.58            0.08

### Extra Credit

Input is the list of primes below 100,000 from [1]. Since each line in that file holds prime and only a prime, but no ongoing counter, we must slightly modify the program by commenting out a single line:

      -- N_IO.Get(Counter);


We can also edit out the declaration of the variable "Counter" ...or live with a compiler warning about never reading or assigning that variable.

Output:

As it turns out, the distribution of the first digits of primes is almost flat and does not seem follow Benford's law:

>./benford < primes-to-100k.txt
Digit  Found[total]   Expected[total]    Found[%]   Expected[%]   Difference[%]
1          1193            2887.5       12.44         30.10           17.67
2          1129            1689.1       11.77         17.61            5.84
3          1097            1198.4       11.44         12.49            1.06
4          1069             929.6       11.14          9.69            1.45
5          1055             759.5       11.00          7.92            3.08
6          1013             642.2       10.56          6.69            3.87
7          1027             556.3       10.71          5.80            4.91
8          1003             490.7       10.46          5.12            5.34
9          1006             438.9       10.49          4.58            5.91

## Aime

text
sum(text a, text b)
{
data d;
integer e, f, n, r;

e = ~a;
f = ~b;

r = 0;

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

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

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

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

d;
}

text
fibs(list l, integer n)
{
integer c, i;
text a, b, w;

l[1] = 1;

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

w;
}

integer
main(void)
{
integer i, n;
list f;
real m;

n = 1000;

f.pn_integer(0, 10, 0);

fibs(f, n);

m = 100r / n;

o_text("\t\texpected\t   found\n");
i = 0;
while (i < 9) {
i += 1;
o_form("%8d/p3d3w16//p3d3w16/\n", i, 100 * log10(1 + 1r / i), f[i] * m);
}

0;
}
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  

## ALGOL 68

Works with: ALGOL 68G version Any - tested with release 2.8.3.win32

Uses Algol 68G's LONG LONG INT which has programmer specifiable precision.

BEGIN
# set the number of digits for LONG LONG INT values #
PR precision 256 PR
# returns the probability of the first digit of each non-zero number in s #
PROC digit probability = ( []LONG LONG INT s )[]REAL:
BEGIN
[ 1 : 9 ]REAL result;
# count the number of times each digit is the first #
[ 1 : 9 ]INT  count := ( 0, 0, 0, 0, 0, 0, 0, 0, 0 );
FOR i FROM LWB s TO UPB s DO
LONG LONG INT v := ABS s[ i ];
IF v /= 0 THEN
WHILE v > 9 DO v OVERAB 10 OD;
count[ SHORTEN SHORTEN v ] +:= 1
FI
OD;
# calculate the probability of each digit #
INT number of elements = ( UPB s + 1 ) - LWB s;
FOR i TO 9 DO
result[ i ] := IF number of elements = 0 THEN 0 ELSE count[ i ] / number of elements FI
OD;
result
END # digit probability # ;
# outputs the digit probabilities of some numbers and those expected by Benford's law #
PROC compare to benford = ( []REAL actual )VOID:
FOR i TO 9 DO
print( ( "Benford: ", fixed( log( 1 + ( 1 / i ) ), -7, 3 )
, " actual: ", fixed( actual[ i ], -7, 3 )
, newline
)
)
OD # compare to benford # ;
# generate 1000 fibonacci numbers #
[ 0 : 1000 ]LONG LONG INT fn;
fn[ 0 ] := 0;
fn[ 1 ] := 1;
FOR i FROM 2 TO UPB fn DO fn[ i ] := fn[ i - 1 ] + fn[ i - 2 ] OD;
# get the probabilities of each first digit of the fibonacci numbers and #
# compare to the probabilities expected by Benford's law #
compare to benford( digit probability( fn ) )
END
Output:
Benford:   0.301 actual:   0.301
Benford:   0.176 actual:   0.177
Benford:   0.125 actual:   0.125
Benford:   0.097 actual:   0.096
Benford:   0.079 actual:   0.080
Benford:   0.067 actual:   0.067
Benford:   0.058 actual:   0.056
Benford:   0.051 actual:   0.053
Benford:   0.046 actual:   0.045


## APL

Works with: Dyalog APL
task←{
benf ← ≢÷⍨(⍳9)(+/∘.=)(⍎⊃∘⍕)¨

fibs ← (⊢,(+/¯2↑⊢))⍣998⊢1 1

exp ← 10⍟1+÷⍳9
obs ← benf fibs

⎕←'Expected  Actual'⍪5⍕exp,[1.5]obs
}

Output:
Expected  Actual
0.30103 0.30100
0.17609 0.17700
0.12494 0.12500
0.09691 0.09600
0.07918 0.08000
0.06695 0.06700
0.05799 0.05600
0.05115 0.05300
0.04576 0.04500

## Arturo

fib: [a b]: <= [0 1]
do.times:998 -> [a b]: @[b 'fib ++ <= a+b]

leading: fib | map 'x -> first ~"|x|"
| tally

print "digit    actual    expected"
loop 1..9 'd ->
print [
log 1 + 1//d 10
]

Output:
digit    actual    expected
1        0.301     0.3010299956639811
2        0.177     0.1760912590556812
3        0.125     0.1249387366082999
4        0.095     0.09691001300805641
5        0.08      0.0791812460476248
6        0.067     0.06694678963061322
7        0.056     0.05799194697768673
8        0.053     0.05115252244738128
9        0.045     0.04575749056067514

## AutoHotkey

Works with: AutoHotkey_L

(AutoHotkey1.1+)

SetBatchLines, -1
fib := NStepSequence(1, 1, 2, 1000)
Out := "DigittExpectedtObservedtDeviationn"
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
}

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
}


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) }

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


## BCPL

BCPL doesn't do floating point well, so I use integer math to compute the most significant digits of the Fibonacci sequence and use a table that has the values of log10(d + 1/d)

GET "libhdr"

MANIFEST {
fib_sz = 1_000_000_000 // keep 9 significant digits for computing F(n)
}

LET msd(n) = VALOF {
UNTIL n < 10 DO
n /:= 10
RESULTIS n
}

LET fibonacci(n, tally) BE {
LET a, b, c, e = 0, 1, ?, 0
FOR i = 1 TO n {
TEST e = 0
THEN tally[msd(b)] +:= 1
ELSE tally[b / (fib_sz / 10)] +:= 1
c := a + b
IF c > fib_sz {
a := a / 10 - (a MOD 10 >= 5) // subtract, since condition evalutes to
b := b / 10 - (b MOD 10 >= 5) // eithr 0 or -1.
c := a + b
e +:= 1 // keep track of exponent, just 'cuz
}
a, b := b, c
}
}

LET start() = VALOF {
// expected value of benford: log10(d + 1/d)
LET expected = TABLE 0, 301, 176, 125, 97, 79, 67, 58, 51, 46
LET actual = VEC 9
FOR i = 0 TO 9 DO actual!i := 0
fibonacci(1000, actual)
writes("*nLeading digital distribution of the first 1,000 Fibonacci numbers*n")
writes("Digit*tActual*tExpected*n")
FOR i = 1 TO 9
writef("%i *t%0.3d *t%0.3d *n", i, actual!i, expected!i)
RESULTIS 0
}
Output:
BCPL 32-bit Cintcode System (13 Jan 2020)
0.000>
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


## BASIC256

n = 1000
dim actual(n) fill 0

for nr = 1 to n
num$= string(fibonacci(nr)) j = int(left(num$,1))
actual[j] += 1
next

print "First 1000 Fibonacci numbers"
print "Digit  ", "Actual freq  ", "Expected freq"
for i = 1 to 9
freq = frequency(i)*100
print "  "; ljust(i,4), rjust(actual[i]/10,5), rjust(freq,5)
next
end

function frequency(n)
return (log10(n+1) - log10(n))
end function

function fibonacci(f)
f = int(f)
a = 0 : b = 1 : c = 0 : n = 0

while n < f
a = b
b = c
c = a + b
n += 1
end while

return c
end function

Output:
First 1000 Fibonacci numbers
Digit         Actual freq   Expected freq
1            30.1         30.1029995664
2            17.7         17.6091259056
3            12.5         12.4938736608
4             9.6         9.69100130081
5             8.0         7.91812460476
6             6.7         6.694679
7             5.6         5.79919469777
8             5.3         5.11525224474
9             4.5         4.57574905607

## 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;
}

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++ //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 ; }  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 %  ## Chipmunk Basic Translation of: Yabasic Works with: Chipmunk Basic version 3.6.4 100 cls 110 n = 1000 120 dim actual(n) 130 for nr = 1 to n 140 num$ = str$(fibonacci(nr)) 150 j = val(left$(num$,1)) 160 actual(j) = actual(j)+1 170 next 180 print "First 1000 Fibonacci numbers" 190 print "Digit Actual freq Expected freq" 200 for i = 1 to 9 210 freq = frequency(i)*100 220 print format$(i,"###");
230   print using "        ##.###";actual(i)/10;
240   print using "           ##.###";freq
250 next
260 end
270 sub frequency(n)
280   frequency = (log10(n+1)-log10(n))
290 end sub
300 sub log10(n)
310   log10 = log(n)/log(10)
320 end sub
330 sub fibonacci(n)
335 rem https://rosettacode.org/wiki/Fibonacci_sequence#Chipmunk_Basic
340   n1 = 0
350   n2 = 1
360   for k = 1 to abs(n)
370     sum = n1+n2
380     n1 = n2
390     n2 = sum
400   next k
410   if n < 0 then
420     fibonacci = n1*((-1)^((-n)+1))
430   else
440     fibonacci = n1
450   endif
460 end sub


## Clojure

(ns example
(:gen-class))

(defn abs [x]
(if (> x 0)
x
(- x)))

(defn calc-benford-stats [digits]
" Frequencies of digits in data "
(let [y (frequencies digits)
tot (reduce + (vals y))]
[y tot]))

(defn show-benford-stats [v]
" Prints in percent the actual, Benford expected, and difference"
(let [fd (map (comp first str) v)]        ; first digit of each record
(doseq [q (range 1 10)
:let [[y tot] (calc-benford-stats fd)
d (first (str q))         ; reference digit
f (/ (get y d 0) tot 0.01)  ; percent of occurence of digit
p (* (Math/log10 (/ (inc q) q)) 100)  ; Benford expected percent
e (abs (- f p))]]                     ; error (difference)
(println (format "%3d %10.2f %10.2f %10.2f"
q
f
p
e)))))

; Generate fibonacci results
(def fib (lazy-cat [0N 1N] (map + fib (rest fib))))

;(def fib-digits (map (comp first str) (take 10000 fib)))
(def fib-digits (take 10000 fib))
(def header "         found-%    expected-%  diff")

(println "Fibonacci Results")
(show-benford-stats fib-digits)
;
; Universal Constants from Physics (using first column of data)
(println "Universal Constants from Physics")
(let [
data-parser (fn [s]
(let [x (re-find #"\s{10}-?[0|/\.]*([1-9])" s)]
(if (not (nil? x))    ; Skips records without number
(second x)
x)))

input (slurp "http://physics.nist.gov/cuu/Constants/Table/allascii.txt")

(data-parser line))
z (filter identity y)]
(show-benford-stats z))

; Sunspots
(println "Sunspots average count per month since 1749")
(let [
data-parser (fn [s]
(nth (re-find #"(.+?\s){3}([1-9])" s) 2))

; Sunspot data loaded from file (saved from ;https://solarscience.msfc.nasa.gov/greenwch/SN_m_tot_V2.0.txt")
; (note: attempting to load directly from url causes https Trust issues, so saved to file after loading to Browser)
input (slurp "SN_m_tot_V2.0.txt")
(data-parser line))]

(show-benford-stats y))

Output:
Fibonacci Results
found-%    expected-%  diff
1      30.11      30.10       0.01
2      17.62      17.61       0.01
3      12.49      12.49       0.00
4       9.68       9.69       0.01
5       7.92       7.92       0.00
6       6.68       6.69       0.01
7       5.80       5.80       0.00
8       5.13       5.12       0.01
9       4.56       4.58       0.02
Universal Constants from Physics
found-%    expected-%  diff
1      34.34      30.10       4.23
2      18.67      17.61       1.07
3       9.04      12.49       3.46
4       8.43       9.69       1.26
5       8.43       7.92       0.52
6       7.23       6.69       0.53
7       3.31       5.80       2.49
8       5.12       5.12       0.01
9       5.42       4.58       0.85
Sunspots average count per month since 1749
found-%    expected-%  diff
1      37.44      30.10       7.34
2      16.28      17.61       1.33
3       7.16      12.49       5.34
4       6.88       9.69       2.81
5       6.35       7.92       1.57
6       6.04       6.69       0.66
7       7.25       5.80       1.45
8       5.57       5.12       0.46
9       5.76       4.58       1.18


## 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)

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

(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))))

; *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

## Crystal

Translation of: Ruby
require "big"

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

def fib(n)
a, b = 0.to_big_i, 1.to_big_i
(0...n).map { ret, a, b = a, b, a + b; ret }
end

# powers of 3 as a test sequence
def power_of_threes(n)
(0...n).map { |k| 3.to_big_i ** k }
end

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

def show_dist(title, s)
c = Array.new(10, 0)
s.each{ |x| c[x] += 1 }
siz = s.size
res = (1..9).map{ |d| c[d] / siz }
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)
(0...n).map { |i| 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))

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:  11.2%   30.1%   18.9%
2:  10.8%   17.6%    6.8%
3:  10.8%   12.5%    1.7%
4:  11.0%    9.7%    1.3%
5:  11.1%    7.9%    3.1%
6:  10.8%    6.7%    4.1%
7:  10.8%    5.8%    5.0%
8:  11.2%    5.1%    6.1%
9:  12.3%    4.6%    7.7%


## D

Translation of: Scala
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);
}

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.

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);
}


See Pascal.

## EasyLang

func$add a$ b$. for i to higher len a$ len b$a = number substr a$ i 1
b = number substr b$i 1 r = a + b + c c = r div 10 r$ &= r mod 10
.
if c > 0
r$&= c . return r$
.
#
len fibdist[] 9
proc mkfibdist . .
# generate 1000 fibonacci numbers as
# (reversed) strings, because 53 bit
# integers are too small
#
n = 1000
prev$= 0 val$ = 1
fibdist[1] = 1
for i = 2 to n
h$= add prev$ val$prev$ = val$val$ = h$ind = number substr val$ len val$1 fibdist[ind] += 1 . for i to len fibdist[] fibdist[i] = fibdist[i] / n . . mkfibdist # len benfdist[] 9 proc mkbenfdist . . for i to 9 benfdist[i] = log10 (1 + 1.0 / i) . . mkbenfdist # numfmt 3 0 print "Actual Expected" for i to 9 print fibdist[i] & " " & benfdist[i] . Output: Actual Expected 0.301 0.301 0.177 0.176 0.125 0.125 0.096 0.097 0.080 0.079 0.067 0.067 0.056 0.058 0.053 0.051 0.045 0.046  ## Elixir defmodule Benfords_law do def distribution(n), do: :math.log10( 1 + (1 / n) ) def task(total \\ 1000) do IO.puts "Digit Actual Benfords expected" fib(total) |> Enum.group_by(fn i -> hd(to_char_list(i)) end) |> Enum.map(fn {key,list} -> {key - ?0, length(list)} end) |> Enum.sort |> Enum.each(fn {x,len} -> IO.puts "#{x} #{len / total} #{distribution(x)}" end) end defp fib(n) do # suppresses zero Stream.unfold({1,1}, fn {a,b} -> {a,{b,a+b}} end) |> Enum.take(n) end end Benfords_law.task  Output: 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  ## 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 ).  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  ## F# For Fibonacci code, see https://rosettacode.org/wiki/Fibonacci_sequence#F.89 open System let fibonacci = Seq.unfold (fun (x, y) -> Some(x, (y, x + y))) (0I,1I) let fibFirstNumbers nth = fibonacci |> Seq.take nth |> Seq.map (fun n -> n.ToString().[0] |> string |> Int32.Parse) let fibFirstNumbersFrequency nth = let firstNumbers = fibFirstNumbers nth |> Seq.toList let counts = firstNumbers |> List.countBy id |> List.sort |> List.filter (fun (k, _) -> k <> 0) let total = firstNumbers |> List.length |> float counts |> List.map (fun (_, v) -> float v/total) let benfordLaw d = log10(1.0 + (1.0/float d)) let benfordLawFigures = [1..9] |> List.map benfordLaw let run () = printfn "Frequency of the first digit (1 through 9) in the Fibonacci sequence:" fibFirstNumbersFrequency 1000 |> List.iter (fun f -> printf$"{f:N5} ")
printfn "\nBenford's law for 1 through 9:"
benfordLawFigures |> List.iter (fun f -> printf $"{f:N5} ")  Output: Frequency of the first digit (1 through 9) in the Fibonacci sequence: 0.30100 0.17700 0.12500 0.09500 0.08000 0.06700 0.05600 0.05300 0.04500 Benford's law for 1 through 9: 0.30103 0.17609 0.12494 0.09691 0.07918 0.06695 0.05799 0.05115 0.04576  ## Factor USING: assocs compiler.tree.propagation.call-effect formatting kernel math math.functions math.statistics math.text.utils sequences ; IN: rosetta-code.benfords-law : expected ( n -- x ) recip 1 + log10 ; : next-fib ( vec -- vec' ) [ last2 ] keep [ + ] dip [ push ] keep ; : data ( -- seq ) V{ 1 1 } clone 998 [ next-fib ] times ; : 1st-digit ( n -- m ) 1 digit-groups last ; : leading ( -- seq ) data [ 1st-digit ] map ; : .header ( -- ) "Digit" "Expected" "Actual" "%-10s%-10s%-10s\n" printf ; : digit-report ( digit digit-count -- digit expected actual ) dupd [ expected ] dip 1000 /f ; : .digit-report ( digit digit-count -- ) digit-report "%-10d%-10.4f%-10.4f\n" printf ; : main ( -- ) .header leading histogram [ .digit-report ] assoc-each ; MAIN: main  Output: Digit Expected Actual 1 0.3010 0.3010 2 0.1761 0.1770 3 0.1249 0.1250 4 0.0969 0.0960 5 0.0792 0.0800 6 0.0669 0.0670 7 0.0580 0.0560 8 0.0512 0.0530 9 0.0458 0.0450  ## 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 ;  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: -*- 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  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  ## FreeBASIC Library: GMP ' version 27-10-2016 ' compile with: fbc -s console #Define max 1000 ' total number of Fibonacci numbers #Define max_sieve 15485863 ' should give 1,000,000 #Include Once "gmp.bi" ' uses the GMP libary Dim As ZString Ptr z_str Dim As ULong n, d ReDim As ULong digit(1 To 9) Dim As Double expect, found Dim As mpz_ptr fib1, fib2 fib1 = Allocate(Len(__mpz_struct)) : Mpz_init_set_ui(fib1, 0) fib2 = Allocate(Len(__mpz_struct)) : Mpz_init_set_ui(fib2, 1) digit(1) = 1 ' fib2 For n = 2 To max Swap fib1, fib2 ' fib1 = 1, fib2 = 0 mpz_add(fib2, fib1, fib2) ' fib1 = 1, fib2 = 1 (fib1 + fib2) z_str = mpz_get_str(0, 10, fib2) d = Val(Left(*z_str, 1)) ' strip the 1 digit on the left off digit(d) = digit(d) +1 Next mpz_clear(fib1) : DeAllocate(fib1) mpz_clear(fib2) : DeAllocate(fib2) Print Print "First 1000 Fibonacci numbers" Print "nr: total found expected difference" For d = 1 To 9 n = digit(d) found = n / 10 expect = (Log(1 + 1 / d) / Log(10)) * 100 Print Using " ## ##### ###.## % ###.## % ##.### %"; _ d; n ; found; expect; expect - found Next ReDim digit(1 To 9) ReDim As UByte sieve(max_sieve) 'For d = 4 To max_sieve Step 2 ' sieve(d) = 1 'Next Print : Print "start sieve" For d = 3 To sqr(max_sieve) If sieve(d) = 0 Then For n = d * d To max_sieve Step d * 2 sieve(n) = 1 Next End If Next digit(2) = 1 ' 2 Print "start collecting first digits" For n = 3 To max_sieve Step 2 If sieve(n) = 0 Then d = Val(Left(Trim(Str(n)), 1)) digit(d) = digit(d) +1 End If Next Dim As ulong total For n = 1 To 9 total = total + digit(n) Next Print Print "First";total; " primes" Print "nr: total found expected difference" For d = 1 To 9 n = digit(d) found = n / total * 100 expect = (Log(1 + 1 / d) / Log(10)) * 100 Print Using " ## ######## ###.## % ###.## % ###.### %"; _ d; n ; found; expect; expect - found Next ' empty keyboard buffer While InKey <> "" : Wend Print : Print "hit any key to end program" Sleep End  Output: First 1000 Fibonacci numbers nr: total found expected difference 1 301 30.10 % 30.10 % 0.003 % 2 177 17.70 % 17.61 % -0.091 % 3 125 12.50 % 12.49 % -0.006 % 4 96 9.60 % 9.69 % 0.091 % 5 80 8.00 % 7.92 % -0.082 % 6 67 6.70 % 6.69 % -0.005 % 7 56 5.60 % 5.80 % 0.199 % 8 53 5.30 % 5.12 % -0.185 % 9 45 4.50 % 4.58 % 0.076 % start sieve start collecting first digits First1000000 primes nr: total found expected difference 1 415441 41.54 % 30.10 % -11.441 % 2 77025 7.70 % 17.61 % 9.907 % 3 75290 7.53 % 12.49 % 4.965 % 4 74114 7.41 % 9.69 % 2.280 % 5 72951 7.30 % 7.92 % 0.623 % 6 72257 7.23 % 6.69 % -0.531 % 7 71564 7.16 % 5.80 % -1.357 % 8 71038 7.10 % 5.12 % -1.989 % 9 70320 7.03 % 4.58 % -2.456 % ## Fōrmulæ Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for storage and transfer purposes more than visualization and edition. Programs in Fōrmulæ are created/edited online in its website. In this page you can see and run the program(s) related to this task and their results. You can also change either the programs or the parameters they are called with, for experimentation, but remember that these programs were created with the main purpose of showing a clear solution of the task, and they generally lack any kind of validation. Solution: The following function calculates the distribution of the first digit of a list of integer, positive numbers: Example: Benford distribution. Benford distribution is, by definition (using a precision of 10 digits): Comparison chart. The following function creates a chart, in order to compare the distribution of the first digis in a givel list or numbers, and the Benford distribution: Testing. Testing with the previous list of numbers: Case 1. Use the first 1,000 numbers from the Fibonacci sequence as your data set Case 2. The sequence of the fisrt 1,000 natural numbers does not follow the Benford distribution: Case 3. The following example is for the list of the fist 1,000 factorial numbers. ## FutureBasic Short t, i, j, k, m Double a(9), z Double phi, psi CFStringRef s print @"Benford:" for i = 1 to 9 a(i) = log10( 1 + 1 / i ) print fn StringWithFormat( @"%.3f ", a(i) ), next // Fibonacci according to DeMoivre and Binet for t = 1 to 9 : a(t) = 0 : next // Clean array phi = ( 1 + sqr(5) ) / 2 psi = ( 1 - sqr(5) ) / 2 for i = 1 to 1000 z = ( phi^i - psi^i ) / sqr( 5 ) s = fn StringWithFormat( @"%e", z) // Get first digit t = fn StringIntegerValue( left( s, 1 ) ) a(t) = a(t) + 1 next print @"\n\nFibonacci:" for i = 1 to 9 print fn StringWithFormat( @"%.3f ", a(i) / 1000 ), next // Multiplication tables for t = 1 to 9 : a(t) = 0 : next // Clean array for i = 1 to 10 for j = 1 to 10 for k = 1 to 10 for m = 1 to 10 z = i * j * k * m s = fn StringWithFormat( @"%e", z ) t = fn StringIntegerValue( left( s, 1 ) ) a(t) = a(t) + 1 next next next next print @"\n\nMultiplication:" for i = 1 to 9 print fn StringWithFormat( @"%.3f ", a(i) / 1e4 ), next // Factorials according to DeMoivre and Stirling for t = 1 to 9 : a(t) = 0 : next // Clean array for i = 10 to 110 z = sqr(2 * pi * i ) * (i / exp(1) )^i s = fn StringWithFormat( @"%e", z ) t = fn StringIntegerValue( left( s, 1 ) ) a(t) = a(t) + 1 next print @"\n\nFactorials:" for i = 1 to 9 print fn StringWithFormat( @"%.2f ", a(i) / 100 ), next handleevents } Output: Benford: 0.301 0.176 0.125 0.097 0.079 0.067 0.058 0.051 0.046 Fibonacci: 0.301 0.177 0.125 0.096 0.080 0.067 0.056 0.053 0.045 Multiplication: 0.302 0.181 0.124 0.103 0.071 0.061 0.055 0.055 0.047 Factorials: 0.35 0.16 0.12 0.06 0.06 0.06 0.02 0.10 0.08  ## 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))) } }  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  ## Groovy Solution: Uses Fibonacci sequence analytic formula Translation of: Java def tallyFirstDigits = { size, generator -> def population = (0..<size).collect { generator(it) } def firstDigits = [0]*10 population.each { number -> firstDigits[(number as String)[0] as int] ++ } firstDigits }  Test: def digitCounts = tallyFirstDigits(1000, aFib) println "d actual predicted" (1..<10).each { printf ("%d %10.6f %10.6f\n", it, digitCounts[it]/1000, Math.log10(1.0 + 1.0/it)) }  Output: d actual predicted 1 0.301000 0.301030 2 0.177000 0.176091 3 0.125000 0.124939 4 0.095000 0.096910 5 0.080000 0.079181 6 0.067000 0.066947 7 0.056000 0.057992 8 0.053000 0.051153 9 0.045000 0.045757 ## GW-BASIC  This example is in need of improvement: Does not show actual vs expected values for Fibonaccis as task specifies. 10 DEF FNBENFORD(N)=LOG(1+1/N)/LOG(10) 20 CLS 30 PRINT "One digit Benford's Law" 40 FOR i = 1 TO 9 50 PRINT i,:PRINT USING "##.######";FNBENFORD(i) 60 NEXT i 70 END Output:  1 0.301030 2 0.176091 3 0.124939 4 0.096910 5 0.079181 6 0.066947 7 0.057992 8 0.051153 9 0.045758  ## 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

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.

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


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

Solution

benford=: 10&^.@(1 + %)                  NB. expected frequency of first digit y
Digits=: '123456789'
firstSigDigits=: {.@(-. -.&Digits)@":"0  NB. extract first significant digit from numbers

freq=: (] % +/)@:<:@(#/.~)@,             NB. calc frequency of values (x) in y


Required Example

   First1000Fib=: (, +/@:(_2&{.)) ^: (1000-#) 1 1
NB. Expected vs Actual frequencies for Digits 1-9
Digits ((] ,. benford)@"."0@[ ,. (freq firstSigDigits)) First1000Fib
1   0.30103 0.301
2  0.176091 0.177
3  0.124939 0.125
4   0.09691 0.096
5 0.0791812  0.08
6 0.0669468 0.067
7 0.0579919 0.056
8 0.0511525 0.053
9 0.0457575 0.045


Alternatively

We show the correlation coefficient of Benford's law with the leading digits of the first 1000 Fibonacci numbers is almost unity.

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

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

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  ## Java import java.math.BigInteger; import java.util.Locale; public class BenfordsLaw { private static BigInteger[] generateFibonacci(int n) { BigInteger[] fib = new BigInteger[n]; fib[0] = BigInteger.ONE; fib[1] = BigInteger.ONE; for (int i = 2; i < fib.length; i++) { fib[i] = fib[i - 2].add(fib[i - 1]); } return fib; } public static void main(String[] args) { BigInteger[] numbers = generateFibonacci(1000); int[] firstDigits = new int[10]; for (BigInteger number : numbers) { firstDigits[Integer.valueOf(number.toString().substring(0, 1))]++; } for (int i = 1; i < firstDigits.length; i++) { System.out.printf(Locale.ROOT, "%d %10.6f %10.6f%n", i, (double) firstDigits[i] / numbers.length, Math.log10(1.0 + 1.0 / i)); } } }  The output is: 1 0.301000 0.301030 2 0.177000 0.176091 3 0.125000 0.124939 4 0.096000 0.096910 5 0.080000 0.079181 6 0.067000 0.066947 7 0.056000 0.057992 8 0.053000 0.051153 9 0.045000 0.045757 To use other number sequences, implement a suitable NumberGenerator, construct a Benford instance with it and print it. ## JavaScript const fibseries = n => [...Array(n)] .reduce( (fib, _, i) => i < 2 ? ( fib ) : fib.concat(fib[i - 1] + fib[i - 2]), [1, 1] ); const benford = array => [1, 2, 3, 4, 5, 6, 7, 8, 9] .map(val => [val, array .reduce( (sum, item) => sum + ( ${item} [0] === ${val} ), 0 ) / array.length, Math.log10(1 + 1 / val) ]); console.log(benford(fibseries(1000)))  Output: 0: (3) [1, 0.301, 0.3010299956639812] 1: (3) [2, 0.177, 0.17609125905568124] 2: (3) [3, 0.125, 0.12493873660829992] 3: (3) [4, 0.096, 0.09691001300805642] 4: (3) [5, 0.08, 0.07918124604762482] 5: (3) [6, 0.067, 0.06694678963061322] 6: (3) [7, 0.056, 0.05799194697768673] 7: (3) [8, 0.053, 0.05115252244738129] 8: (3) [9, 0.045, 0.04575749056067514] ## jq Works with: jq version 1.4 This implementation shows the observed and expected number of occurrences together with the χ² statistic. For the sake of being self-contained, the following includes a generator for Fibonacci numbers, and a prime number generator that is inefficient but brief and can generate numbers within an arbitrary range. # Generate the first n Fibonacci numbers: 1, 1, ... # Numerical accuracy is insufficient beyond about 1450. def fibonacci(n): # input: [f(i-2), f(i-1), countdown] def fib: (.[0] + .[1]) as$sum
| if .[2] <= 0 then empty
elif .[2] == 1 then $sum else$sum, ([ .[1], $sum, .[2] - 1 ] | fib) end; [1, 0, n] | fib ; # is_prime is tailored to work with jq 1.4 def is_prime: if . == 2 then true else 2 < . and . % 2 == 1 and . as$in
| (($in + 1) | sqrt) as$m
| (((($m - 1) / 2) | floor) + 1) as$max
| reduce range(1; $max) as$i
(true; if . then ($in % ((2 *$i) + 1)) > 0 else false end)
end ;

# primes in [m,n)
def primes(m;n):
range(m;n) | select(is_prime);

def runs:
reduce .[] as $item ( []; if . == [] then [ [$item, 1] ]
else  .[length-1] as $last | if$last[0] == $item then (.[0:length-1] + [ [$item, $last[1] + 1] ] ) else . + [[$item, 1]]
end
end ) ;

# Inefficient but brief:
def histogram: sort | runs;

def benford_probability:
tonumber
| if . > 0 then ((1 + (1 /.)) | log) / (10|log)
else 0
end ;

# benford takes a stream and produces an array of [ "d", observed, expected ]
def benford(stream):
[stream | tostring | .[0:1] ] | histogram as $histogram | reduce ($histogram | .[] | .[0]) as $digit ([]; . + [$digit, ($digit|benford_probability)] ) | map(select(type == "number")) as$probabilities
| ([ $histogram | .[] | .[1] ] | add) as$total
| reduce range(0; $histogram|length) as$i
([]; . + ([$histogram[$i] + [$total *$probabilities[$i]] ] ) ) ; # given an array of [value, observed, expected] values, # produce the χ² statistic def chiSquared: reduce .[] as$triple
(0;
if $triple[2] == 0 then . else . + ($triple[1] as $o |$triple[2] as $e | ($o - $e) | (.*.)/$e)
end) ;

# truncate n places after the decimal point;
# return a string since it can readily be converted back to a number
def precision(n):
tostring as $s |$s | index(".")
| if . then $s[0:.+n+1] else$s end ;

# Right-justify but do not truncate
def rjustify(n):
length as $length | if n <=$length then . else " " * (n-length) + . end; # Attempt to align decimals so integer part is in a field of width n def align(n): index(".") asix
| if n < $ix then . elif$ix then (.[0:$ix]|rjustify(n)) +.[$ix:]
else rjustify(n)
end ;

# given an array of [value, observed, expected] values,
# produce rows of the form: value observed expected
def print_rows(prec):
.[] | map( precision(prec)|align(5) + "  ") | add ;

benford(stream) as $array | heading, " Digit Observed Expected", ($array | print_rows(2) ),
"",
" χ² = \( $array | chiSquared | precision(4))", "" ; def task: report("First 100 fibonacci numbers:"; fibonacci( 100) ), report("First 1000 fibonacci numbers:"; fibonacci(1000) ), report("Primes less than 1000:"; primes(2;1000)), report("Primes between 1000 and 10000:"; primes(1000;10000)), report("Primes less than 100000:"; primes(2;100000)) ; task Output: First 100 fibonacci numbers: Digit Observed Expected 1 30 30.10 2 18 17.60 3 13 12.49 4 9 9.69 5 8 7.91 6 6 6.69 7 5 5.79 8 7 5.11 9 4 4.57 χ² = 1.0287 First 1000 fibonacci numbers: Digit Observed Expected 1 301 301.02 2 177 176.09 3 125 124.93 4 96 96.91 5 80 79.18 6 67 66.94 7 56 57.99 8 53 51.15 9 45 45.75 χ² = 0.1694 Primes less than 1000: Digit Observed Expected 1 25 50.57 2 19 29.58 3 19 20.98 4 20 16.28 5 17 13.30 6 18 11.24 7 18 9.74 8 17 8.59 9 15 7.68 χ² = 45.0162 Primes between 1000 and 10000: Digit Observed Expected 1 135 319.39 2 127 186.83 3 120 132.55 4 119 102.82 5 114 84.01 6 117 71.03 7 107 61.52 8 110 54.27 9 112 48.54 χ² = 343.5583 Primes less than 100000: Digit Observed Expected 1 1193 2887.47 2 1129 1689.06 3 1097 1198.41 4 1069 929.56 5 1055 759.50 6 1013 642.15 7 1027 556.25 8 1003 490.65 9 1006 438.90 χ² = 3204.8072 ## Julia # Benford's Law P(d) = log10(1+1/d) function benford(numbers) firstdigit(n) = last(digits(n)) counts = zeros(9) foreach(n -> counts[firstdigit(n)] += 1, numbers) counts ./ sum(counts) end struct Fibonacci end Base.iterate(::Fibonacci, (a, b) = big.((0, 1))) = b, (b, a + b) sample = Iterators.take(Fibonacci(), 1000) observed = benford(sample) .* 100 expected = P.(1:9) .* 100 table = Real[1:9 observed expected] using Plots plot([observed expected]; title = "Benford's Law", seriestype = [:bar :line], linewidth = [0 5], xticks = 1:9, xlabel = "first digit", ylabel = "frequency %", label = ["1000 Fibonacci numbers" "P(d)=log10(1+1/d)"]) using Printf println("Benford's Law\nFrequency of first digit\nin 1000 Fibonacci numbers") println("digit observed expected") foreach(i -> @printf("%3d%9.2f%%%9.2f%%\n", table[i,:]...), 1:9)  Output: Benford's Law Frequency of first digit in 1000 Fibonacci numbers digit observed expected 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% ## Kotlin import java.math.BigInteger interface NumberGenerator { val numbers: Array<BigInteger> } class Benford(ng: NumberGenerator) { override fun toString() = str private val firstDigits = IntArray(9) private val count = ng.numbers.size.toDouble() private val str: String init { for (n in ng.numbers) { firstDigits[n.toString().substring(0, 1).toInt() - 1]++ } str = with(StringBuilder()) { for (i in firstDigits.indices) { append(i + 1).append('\t').append(firstDigits[i] / count) append('\t').append(Math.log10(1 + 1.0 / (i + 1))).append('\n') } toString() } } } object FibonacciGenerator : NumberGenerator { override val numbers: Array<BigInteger> by lazy { val fib = Array<BigInteger>(1000, { BigInteger.ONE }) for (i in 2 until fib.size) fib[i] = fib[i - 2].add(fib[i - 1]) fib } } fun main(a: Array<String>) = println(Benford(FibonacciGenerator))  ## Liberty BASIC Using function from http://rosettacode.org/wiki/Fibonacci_sequence#Liberty_BASIC 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))
next

function P(d)
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
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


## Lua

actual = {}
expected = {}
for i = 1, 9 do
actual[i] = 0
expected[i] = math.log10(1 + 1 / i)
end

n = 0
file = io.open("fibs1000.txt", "r")
for line in file:lines() do
digit = string.byte(line, 1) - 48
actual[digit] = actual[digit] + 1
n = n + 1
end
file:close()

print("digit   actual  expected")
for i = 1, 9 do
print(i, actual[i] / n, expected[i])
end

Output:
digit   actual  expected
1       0.301   0.30102999566398
2       0.177   0.17609125905568
3       0.125   0.1249387366083
4       0.096   0.096910013008056
5       0.08    0.079181246047625
6       0.067   0.066946789630613
7       0.056   0.057991946977687
8       0.053   0.051152522447381
9       0.045   0.045757490560675

## Mathematica / Wolfram Language

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

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

## MATLAB

Translation of: Julia
benfords_law();

function benfords_law
% Benford's Law
P = @(d) log10(1 + 1./d);

% Benford function
function counts = benford(numbers)
firstdigit = @(n) floor(mod(n / 10^floor(log10(n)), 10));
counts = zeros(1, 9);
for i = 1:length(numbers)
digit = firstdigit(numbers(i));
if digit ~= 0
counts(digit) = counts(digit) + 1;
end
end
counts = counts ./ sum(counts);
end

% Generate Fibonacci numbers
function fibNums = fibonacci(n)
fibNums = zeros(1, n);
a = 0;
b = 1;
for i = 1:n
c = b;
b = a + b;
a = c;
fibNums(i) = b;
end
end

% Sample
sample = fibonacci(1000);

% Observed and expected frequencies
observed = benford(sample) * 100;
expected = arrayfun(P, 1:9) * 100;

% Table
mytable = [1:9; observed; expected]';

% Plotting
bar(1:9, observed);
hold on;
plot(1:9, expected, 'LineWidth', 2);
hold off;
title("Benford's Law");
xlabel("First Digit");
ylabel("Frequency %");
legend("1000 Fibonacci Numbers", "P(d) = log10(1 + 1/d)");
xticks(1:9);

% Displaying the results
fprintf("Benford's Law\nFrequency of first digit\nin 1000 Fibonacci numbers\n");
disp(table(mytable(:,1),mytable(:,2),mytable(:,3),'VariableNames',{'digit', 'observed(%)', 'expected(%)'}))
end

Output:
Benford's Law
Frequency of first digit
in 1000 Fibonacci numbers
digit    observed(%)    expected(%)
_____    ___________    ___________

1           30          30.103
2         17.7          17.609
3         12.5          12.494
4          9.6           9.691
5            8          7.9181
6          6.7          6.6947
7          5.7          5.7992
8          5.3          5.1153
9          4.5          4.5757


## 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

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%


## Nim

import math
import strformat

type

# Non zero digit range.
Digit = range[1..9]

# Count array used to compute a distribution.
Count = array[Digit, Natural]

# Array to store frequencies.
Distribution = array[Digit, float]

####################################################################################################
# Fibonacci numbers generation.

import bignum

proc fib(count: int): seq[Int] =
## Build a sequence of "count auccessive Finonacci numbers.
result.setLen(count)
result[0..1] = @[newInt(1), newInt(1)]
for i in 2..<count:
result[i] = result[i-1] + result[i-2]

####################################################################################################
# Benford distribution.

proc benfordDistrib(): Distribution =
## Compute Benford distribution.

for d in 1..9:
result[d] = log10(1 + 1 / d)

const BenfordDist = benfordDistrib()

#---------------------------------------------------------------------------------------------------

template firstDigit(n: Int): Digit =
# Return the first (non null) digit of "n".
ord(($n)[0]) - ord('0') #--------------------------------------------------------------------------------------------------- proc actualDistrib(s: openArray[Int]): Distribution = ## Compute actual distribution of first digit. ## Null values are ignored. var counts: Count for val in s: if not val.isZero(): inc counts[val.firstDigit()] let total = sum(counts) for d in 1..9: result[d] = counts[d] / total #--------------------------------------------------------------------------------------------------- proc display(distrib: Distribution) = ## Display an actual distribution versus the Benford reference distribution. echo "d actual expected" echo "---------------------" for d in 1..9: echo fmt"{d} {distrib[d]:6.4f} {BenfordDist[d]:6.4f}" #——————————————————————————————————————————————————————————————————————————————————————————————————— let fibSeq = fib(1000) let distrib = actualDistrib(fibSeq) echo "Fibonacci numbers first digit distribution:\n" distrib.display()  Output: Fibonacci numbers first digit distribution: d actual expected --------------------- 1 0.3010 0.3010 2 0.1770 0.1761 3 0.1250 0.1249 4 0.0960 0.0969 5 0.0800 0.0792 6 0.0670 0.0669 7 0.0560 0.0580 8 0.0530 0.0512 9 0.0450 0.0458 ## Oberon-2 Works with: oo2c version 2 MODULE BenfordLaw; IMPORT LRealStr, LRealMath, Out := NPCT:Console; VAR r: ARRAY 1000 OF LONGREAL; d: ARRAY 10 OF LONGINT; a: LONGREAL; i: LONGINT; PROCEDURE Fibb(VAR r: ARRAY OF LONGREAL); VAR i: LONGINT; BEGIN r[0] := 1.0;r[1] := 1.0; FOR i := 2 TO LEN(r) - 1 DO r[i] := r[i - 2] + r[i - 1] END END Fibb; PROCEDURE Dist(r [NO_COPY]: ARRAY OF LONGREAL; VAR d: ARRAY OF LONGINT); VAR i: LONGINT; str: ARRAY 256 OF CHAR; BEGIN FOR i := 0 TO LEN(r) - 1 DO LRealStr.RealToStr(r[i],str); INC(d[ORD(str[0]) - ORD('0')]) END END Dist; BEGIN Fibb(r); Dist(r,d); Out.String("First 1000 fibonacci numbers: ");Out.Ln; Out.String(" digit ");Out.String(" observed ");Out.String(" predicted ");Out.Ln; FOR i := 1 TO LEN(d) - 1 DO a := LRealMath.ln(1.0 + 1.0 / i ) / LRealMath.ln(10); Out.Int(i,5);Out.LongRealFix(d[i] / 1000.0,9,3);Out.LongRealFix(a,10,3);Out.Ln END END BenfordLaw.  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  ## OCaml For the Fibonacci sequence, we use the function from https://rosettacode.org/wiki/Fibonacci_sequence#Arbitrary_Precision Note the remark about the compilation of the program there. open Num let fib = let rec fib_aux f0 f1 = function | 0 -> f0 | 1 -> f1 | n -> fib_aux f1 (f1 +/ f0) (n - 1) in fib_aux (num_of_int 0) (num_of_int 1) ;; let create_fibo_string = function n -> string_of_num (fib n) ;; let rec range i j = if i > j then [] else i :: (range (i + 1) j) let n_max = 1000 ;; let numbers = range 1 n_max in let get_first_digit = function s -> Char.escaped (String.get s 0) in let first_digits = List.map get_first_digit (List.map create_fibo_string numbers) in let data = Array.create 9 0 in let fill_data vec = function n -> vec.(n - 1) <- vec.(n - 1) + 1 in List.iter (fill_data data) (List.map int_of_string first_digits) ; Printf.printf "\nFrequency of the first digits in the Fibonacci sequence:\n" ; Array.iter (Printf.printf "%f ") (Array.map (fun x -> (float x) /. float (n_max)) data) ; let xvalues = range 1 9 in let benfords_law = function x -> log10 (1.0 +. 1.0 /. float (x)) in Printf.printf "\nPrediction of Benford's law:\n " ; List.iter (Printf.printf "%f ") (List.map benfords_law xvalues) ; Printf.printf "\n" ;;  Output: Frequency of the first digits in the Fibonacci sequence: 0.301000 0.177000 0.125000 0.096000 0.080000 0.067000 0.056000 0.053000 0.045000 Prediction of Benford's law: 0.301030 0.176091 0.124939 0.096910 0.079181 0.066947 0.057992 0.051153 0.045757  ## PARI/GP 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) 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 ## Pascal program fibFirstdigit; {$IFDEF FPC}{$MODE Delphi}{$ELSE}{$APPTYPE CONSOLE}{$ENDIF}
uses
sysutils;
type
tDigitCount = array[0..9] of LongInt;
var
s: Ansistring;
dgtCnt,
expectedCnt : tDigitCount;

procedure GetFirstDigitFibonacci(var dgtCnt:tDigitCount;n:LongInt=1000);
//summing up only the first 9 digits
//n = 1000 -> difference to first 9 digits complete fib < 100 == 2 digits
var
Begin
for a in dgtCnt do dgtCnt[a] := 0;
a := 0;b := 1;
while n > 0 do
Begin
c := a+b;
//overflow? round and divide by base 10
IF c < a then
Begin a := (a+5) div 10;b := (b+5) div 10;c := a+b;end;
a := b;b := c;
s := IntToStr(a);inc(dgtCnt[Ord(s[1])-Ord('0')]);
dec(n);
end;
end;

procedure InitExpected(var dgtCnt:tDigitCount;n:LongInt=1000);
var
i: integer;
begin
for i := 1 to 9  do
dgtCnt[i] := trunc(n*ln(1 + 1 / i)/ln(10));
end;

var
reldiff: double;
i,cnt: integer;
begin
cnt := 1000;
InitExpected(expectedCnt,cnt);
GetFirstDigitFibonacci(dgtCnt,cnt);
writeln('Digit  count  expected  rel diff');
For i := 1 to 9 do
Begin
reldiff := 100*(expectedCnt[i]-dgtCnt[i])/expectedCnt[i];
writeln(i:5,dgtCnt[i]:7,expectedCnt[i]:10,reldiff:10:5,' %');
end;
end.

Digit  Count  Expected  rel Diff
1    301       301   0.00000 %
2    177       176  -0.56818 %
3    125       124  -0.80645 %
4     96        96   0.00000 %
5     80        79  -1.26582 %
6     67        66  -1.51515 %
7     56        57   1.75439 %
8     53        51  -3.92157 %
9     45        45   0.00000 %

## 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 ; }  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 %  ## Phix Translation of: Go with javascript_semantics function benford(sequence s, string title) sequence f = repeat(0,9) for i=1 to length(s) do integer fdx = sprint(s[i])[1]-'0' f[fdx] += 1 end for sequence res = {title,"Digit Observed% Predicted%"} for i=1 to length(f) do atom o = f[i]/length(s)*100, e = log10(1+1/i)*100 res = append(res,sprintf(" %d %9.3f %8.3f", {i, o, e})) end for return res end function function fib(integer lim) atom a=0, b=1 sequence res = repeat(0,lim) for i=1 to lim do {res[i], a, b} = {b, b, b+a} end for return res end function sequence res = {benford(fib(1000),"First 1000 Fibonacci numbers"), benford(get_primes(-10000),"First 10000 Prime numbers"), benford(sq_power(3,tagset(500)),"First 500 powers of three")} papply(true,printf,{1,{"%-40s%-40s%-40s\n"},columnize(res)})  Output: First 1000 Fibonacci numbers First 10000 Prime numbers First 500 powers of three Digit Observed% Predicted% Digit Observed% Predicted% Digit Observed% Predicted% 1 30.100 30.103 1 16.010 30.103 1 30.000 30.103 2 17.700 17.609 2 11.290 17.609 2 17.600 17.609 3 12.500 12.494 3 10.970 12.494 3 12.400 12.494 4 9.600 9.691 4 10.690 9.691 4 9.800 9.691 5 8.000 7.918 5 10.550 7.918 5 8.000 7.918 6 6.700 6.695 6 10.130 6.695 6 6.600 6.695 7 5.600 5.799 7 10.270 5.799 7 5.800 5.799 8 5.300 5.115 8 10.030 5.115 8 5.200 5.115 9 4.500 4.576 9 10.060 4.576 9 4.600 4.576  ## Picat go => N = 1000, Fib = [fib(I) : I in 1..N], check_benford(Fib), nl. % Check a list of numbers for Benford's law check_benford(L) => Len = L.len, println(len=Len), Count = [F[1].to_integer() : Num in L, F=Num.to_string()].occurrences(), P = new_map([I=D/Len : I=D in Count]), println("Benford (percent):"), foreach(D in 1..9) B = benford(D)*100, PI = P.get(D,0)*100, Diff = abs(PI - B), printf("%d: count=%5d observed: %0.2f%% Benford: %0.2f%% diff=%0.3f\n", D,Count.get(D,0),PI,B,Diff) end, nl. benford(D) = log10(1+1/D). % create an occurrences map of a list occurrences(List) = Map => Map = new_map(), foreach(E in List) Map.put(E, cond(Map.has_key(E),Map.get(E)+1,1)) end. Output: Benford (percent): 1: count= 301 observed: 30.10% Benford: 30.10% diff=0.003 2: count= 177 observed: 17.70% Benford: 17.61% diff=0.091 3: count= 125 observed: 12.50% Benford: 12.49% diff=0.006 4: count= 96 observed: 9.60% Benford: 9.69% diff=0.091 5: count= 80 observed: 8.00% Benford: 7.92% diff=0.082 6: count= 67 observed: 6.70% Benford: 6.69% diff=0.005 7: count= 56 observed: 5.60% Benford: 5.80% diff=0.199 8: count= 53 observed: 5.30% Benford: 5.12% diff=0.185 9: count= 45 observed: 4.50% Benford: 4.58% diff=0.076  Extra credit: Using data from https://en.wikipedia.org/wiki/Land_use_statistics_by_country "Cultivated land (km^2)" Output: Benford (percent): 1: count= 72 observed: 33.64% Benford: 30.10% diff=3.542 2: count= 38 observed: 17.76% Benford: 17.61% diff=0.148 3: count= 29 observed: 13.55% Benford: 12.49% diff=1.058 4: count= 19 observed: 8.88% Benford: 9.69% diff=0.812 5: count= 20 observed: 9.35% Benford: 7.92% diff=1.428 6: count= 12 observed: 5.61% Benford: 6.69% diff=1.087 7: count= 9 observed: 4.21% Benford: 5.80% diff=1.594 8: count= 11 observed: 5.14% Benford: 5.12% diff=0.025 9: count= 4 observed: 1.87% Benford: 4.58% diff=2.707 ## PicoLisp Picolisp does not support floating point math, but it can call libc math functions and convert the results to a fixed point number for e.g. natural logarithm. (scl 4) (load "@lib/misc.l") (load "@lib/math.l") (setq LOG10E 0.4343) (de fibo (N) (let (A 0 B 1 C NIL) (make (link B) (do (dec N) (setq C (+ A B) A B B C) (link B))))) (setq Actual (let ( Digits (sort (mapcar '((N) (format (car (chop N)))) (fibo 1000))) Count 0 ) (make (for (Ds Digits Ds (cdr Ds)) (inc 'Count) (when (n== (car Ds) (cadr Ds)) (link Count) (setq Count 0)))))) (setq Expected (mapcar '((D) (*/ (log (+ 1. (/ 1. D))) LOG10E 1.)) (range 1 9))) (prinl "Digit\tActual\tExpected") (let (As Actual Bs Expected) (for D 9 (prinl D "\t" (format (pop 'As) 3) "\t" (round (pop 'Bs) 3)))) (bye) Output: 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  ## 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; 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 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  ## PowerShell The sample file was not found. I selected another that contained the first two-thousand in the Fibonacci sequence, so there is a small amount of extra filtering. $url  = "https://oeis.org/A000045/b000045.txt"
$file = "$env:TEMP\FibonacciNumbers.txt"
(New-Object System.Net.WebClient).DownloadFile($url,$file)

$benford = Get-Content -Path$file |
Select-Object -Skip 1 -First 1000 |
ForEach-Object {(($_ -split " ")[1].ToString().ToCharArray())[0]} | Group-Object | Select-Object -Property @{Name="Digit" ; Expression={[int]($_.Name)}},
Count,
@{Name="Actual"  ; Expression={$_.Count/1000}}, @{Name="Expected"; Expression={[double]("{0:f5}" -f [Math]::Log10(1 + 1 /$_.Name))}}

$benford | Sort-Object -Property Digit | Format-Table -AutoSize Remove-Item -Path$file -Force -ErrorAction SilentlyContinue

Output:
Digit Count Actual Expected
----- ----- ------ --------
1   301  0.301  0.30103
2   177  0.177  0.17609
3   125  0.125  0.12494
4    96  0.096  0.09691
5    80   0.08  0.07918
6    67  0.067  0.06695
7    56  0.056  0.05799
8    53  0.053  0.05115
9    45  0.045  0.04576


## 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

%_________________________________________________________________
% 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).

freq([H|T], Freq) :-
length([H|T], Len), min_list([H|T], Min), max_list([H|T], Max),
findall(0, between(Min,Max,_), In),
freq([H|T], In, F),	  % Frequency stored in F
findall(N, (member(V, F), N is V/Len), Freq). % Normalise F->Freq

% 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).

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%

## PureBasic

#MAX_N=1000
NewMap d1.i()
Dim fi.s(#MAX_N)
fi(0)="0" : fi(1)="1"
Declare.s Sigma(sx.s,sy.s)

For I=2 To #MAX_N
fi(I)=Sigma(fi(I-2),fi(I-1))
Next

For I=1 To #MAX_N
d1(Left(fi(I),1))+1
Next

Procedure.s Sigma(sx.s, sy.s)
Define i.i, v1.i, v2.i, r.i
Define s.s, sa.s
sy=ReverseString(sy) : s=ReverseString(sx)
For i=1 To Len(s)*Bool(Len(s)>Len(sy))+Len(sy)*Bool(Len(sy)>=Len(s))
v1=Val(Mid(s,i,1))
v2=Val(Mid(sy,i,1))
r+v1+v2
sa+Str(r%10)
r/10
Next i
If r : sa+Str(r%10) : EndIf
ProcedureReturn ReverseString(sa)
EndProcedure

OpenConsole("Benford's law: Fibonacci sequence 1.."+Str(#MAX_N))

Print(~"Dig.\t\tCnt."+~"\t\tExp.\t\tDif.\n\n")
ForEach d1()
Print(RSet(MapKey(d1()),4," ")+~"\t:\t"+RSet(Str(d1()),3," ")+~"\t\t")
ex=Int(#MAX_N*Log(1+1/Val(MapKey(d1())))/Log(10))
PrintN(RSet(Str(ex),3," ")+~"\t\t"+RSet(StrF((ex-d1())*100/ex,5),8," ")+" %")
Next

PrintN(~"\nPress Enter...")
Input()

Output:
Dig.            Cnt.            Exp.            Dif.

1    :       301             301              0.00000 %
2    :       177             176             -0.56818 %
3    :       125             124             -0.80645 %
4    :        96              96              0.00000 %
5    :        80              79             -1.26582 %
6    :        67              66             -1.51515 %
7    :        56              57              1.75439 %
8    :        53              51             -3.92157 %
9    :        45              45              0.00000 %

Press Enter...

## Python

Works with Python 3.X & 2.7

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))

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__':

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

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

pbenford <- function(d){
return(log10(1+(1/d)))
}

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))

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)

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

(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%


## Raku

(formerly Perl 6)

Works with: rakudo version 2016-10-24
sub benford(@a) { bag +« @a».substr(0,1) }

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] }


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%

## REXX

The REXX language (for the most part) hasn't any high math functions, so the   e,   ln,   and log   functions were included herein.

For the extra credit stuff, it was chosen to generate Fibonacci and factorials rather than find a web─page with them listed,   as each list is very easy to generate.

/*REXX pgm demonstrates Benford's law applied to 2 common functions (30 dec. digs used).*/
numeric digits length( e() )  -  length(.)       /*width of (e)  for LN & LOG precision.*/
parse arg N .;  if N=='' | N==","  then N= 1000  /*allow sample size to be specified.   */
pad= "   "                                       /*W1, W2: # digs past the decimal point*/
w1= max(2 + length('observed'), length(N-2) )    /*for aligning output for a number.    */
w2= max(2 + length('expected'), length(N  ) )    /* "      "    frequency distributions.*/
LN10= ln(10)                                     /*calculate the  ln(10)   {used by LOG}*/
call coef                                        /*generate nine frequency coefficients.*/
call fib                                         /*generate   N   Fibonacci numbers.    */
call show "Benford's law applied to"      N      'Fibonacci numbers'
call fact                                        /*generate   N   factorials.           */
call show "Benford's law applied to"      N      'factorial products'
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
coef:       do j=1  for 9; #.j=pad center(format(log(1+1/j),,length(N)+2),w2); end; return
fact: @.=1; do j=2  for N-1;       a= j-1;                   @.j= @.a * j;     end; return
fib:  @.=1; do j=3  for N-2;       a= j-1;       b= a-1;     @.j= @.a + @.b;   end; return
e:    return 2.71828182845904523536028747135266249775724709369995957496696762772407663035
log:  return   ln( arg(1) )    /   LN10
/*──────────────────────────────────────────────────────────────────────────────────────*/
ln: procedure; parse arg x; e= e();  _= e;  ig= (x>1.5);  is= 1 - 2 * (ig\=1); i= 0;  s= x
do while ig&s>1.5  |  \ig&s<.5             /*nitty─gritty part of  LN  calculation*/
do k=-1; iz=s*_**-is; if k>=0&(ig&iz<1|\ig&iz>.5)  then leave; _=_*_; izz=iz;  end
s=izz;  i= i + is* 2**k; end  /*while*/;    x= x * e** - i - 1;  z= 0;  _= -1;  p= z
do k=1;  _= -_ * x; z= z + _/k; if z=p  then leave;  p= z;  end /*k*/;  return z+i
/*──────────────────────────────────────────────────────────────────────────────────────*/
!.=0;     do j=1  for N;   _= left(@.j, 1);     !._= !._ + 1  /*get the 1st digit.*/
end   /*j*/
end   /*k*/
return

output   when using the default (1000 numbers)   for the 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 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 the following for the input:     10000
     digit       observed       expected
───────     ──────────     ──────────     Benford's law applied to 10000 Fibonacci numbers
1          0.3011       0.3010300
2          0.1762       0.1760913
3          0.1250       0.1249387
4          0.0968       0.0969100
5          0.0792       0.0791812
6          0.0668       0.0669468
7          0.0580       0.0579919
8          0.0513       0.0511525
9          0.0456       0.0457575

digit       observed       expected
───────     ──────────     ──────────     Benford's law applied to 10000 factorial products
1          0.2956       0.3010300
2          0.1789       0.1760913
3          0.1276       0.1249387
4          0.0963       0.0969100
5          0.0794       0.0791812
6          0.0715       0.0669468
7          0.0571       0.0579919
8          0.0510       0.0511525
9          0.0426       0.0457575


## Ring

# Project : Benford's law

decimals(3)
n= 1000
actual = list(n)
for x = 1 to len(actual)
actual[x] = 0
next

for nr = 1 to n
n1 = string(fibonacci(nr))
j = number(left(n1,1))
actual[j] = actual[j] + 1
next

see "Digit   " + "Actual   " + "Expected" + nl
for m = 1 to 9
fr = frequency(m)*100
see "" + m + "   " + (actual[m]/10) + "   " + fr + nl
next

func frequency(n)
freq = log10(n+1) - log10(n)
return freq

func log10(n)
log1 = log(n) / log(10)
return log1

func fibonacci(y)
if y = 0 return 0 ok
if y = 1 return 1 ok
if y > 1 return fibonacci(y-1) + fibonacci(y-2) ok

Output:

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


## RPL

Here we use an interesting RPL feature that allows to pass an arithmetic expression or a function name as an argument. Thus, the code below can check the 'benfordance' of various series without needing to edit it. The resulting array being a matrix, some additional operations allow to compute the maximum difference at digit level between the tested function and Benford's law, as a distance indicator.

Works with: Halcyon Calc version 4.2.7
≪ → sn min max
≪ { 9 2 } 0 CON
min max FOR j
{ 0 1 } 1 j sn EVAL
IF DUP THEN
MANT FLOOR PUT
DUP2 GET 1 + PUT
ELSE
3 DROPN
END
NEXT
max min - 1 + /
1 9 FOR j
{ 0 2 } 1 j PUT
j INV 1 + LOG PUT
NEXT
DUP [ 1 0 ] *
OVER [ 0 1 ] * -
RNRM
≫
≫
'BENFD' STO

'→FIB' 1 100 BENFD
≪ LOG ≫ 1 10000 BENFD


Output:

4: [[ 0.301 0.301029995664 ]
[ 0.177 0.176091259056 ]
[ 0.125 0.124938736608 ]
[ 0.096 9.69100130081E-02 ]
[ 0.08 7.91812460476E-02 ]
[ 6.7E-02 6.69467896306E-02 ]
[ 0.056 5.79919469777E-02 ]
[ 0.053 5.11525224474E-02 ]
[ 0.045 4.57574905607E-02 ]]
3: 1.8847477552E-02
2: [[ 9E-04 0.301029995664 ]
[ 9E-03 0.176091259056 ]
[ 0.09001 0.124938736608 ]
[ 0.90001 9.69100130081E-02 ]
[ 0.00001 7.91812460476E-02 ]
[ 0.00002 6.69467896306E-02 ]
[ 0.00001 5.79919469777E-02 ]
[ 0.00001 5.11525224474E-02 ]
[ 0.00002 4.57574905607E-02 ]]
1: 0.775161263392


## Ruby

Translation of: Python
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

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

def show_dist(title, 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))

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

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 "<table border=1><TR bgcolor=wheat><TD>Digit<td>Actual<td>Expected</td><tr>"
for i = 1 to 9
html "<tr align=right><td>";i;"</td><td>";using("##.###",actual(i)/10);"</td><td>";using("##.###", frequency(i)*100);"</td></tr>"
next
html "</table>"
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
 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

## Rust

Works with: rustc version 1.12 stable

This solution uses the num create for arbitrary-precision integers and the num_traits create for the zero and one implementations. It computes the Fibonacci numbers from scratch via the fib function.

extern crate num_traits;
extern crate num;

use num::bigint::{BigInt, ToBigInt};
use num_traits::{Zero, One};
use std::collections::HashMap;

// Return a vector of all fibonacci results from fib(1) to fib(n)
fn fib(n: usize) -> Vec<BigInt> {
let mut result = Vec::with_capacity(n);
let mut a = BigInt::zero();
let mut b = BigInt::one();

result.push(b.clone());

for i in 1..n {
let t = b.clone();
b = a+b;
a = t;
result.push(b.clone());
}

result
}

// Return the first digit of a BigInt
fn first_digit(x: &BigInt) -> u8 {
let zero = BigInt::zero();
assert!(x > &zero);

// parse the first digit of the stringified integer
*&s[..1].parse::<u8>().unwrap()
}

fn main() {
const N: usize = 1000;
let mut counter: HashMap<u8, u32> = HashMap::new();
for x in fib(N) {
let d = first_digit(&x);
*counter.entry(d).or_insert(0) += 1;
}

println!("{:>13}    {:>10}", "real", "predicted");
for y in 1..10 {
println!("{}: {:10.3} v. {:10.3}", y, *counter.get(&y).unwrap_or(&0) as f32 / N as f32,
(1.0 + 1.0 / (y as f32)).log10());
}

}

Output:
         real     predicted
1:      0.301 v.      0.301
2:      0.177 v.      0.176
3:      0.125 v.      0.125
4:      0.096 v.      0.097
5:      0.080 v.      0.079
6:      0.067 v.      0.067
7:      0.056 v.      0.058
8:      0.053 v.      0.051
9:      0.045 v.      0.046


## 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) )
}
}

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%


## Scheme

Works with: Chez Scheme
; Compute the probability of leading digit d (an integer [1,9]) according to Benford's law.

(define benford-probability
(lambda (d)
(- (log (1+ d) 10) (log d 10))))

; Determine the distribution of the leading digit of the sequence provided by the given
; generator.  Return as a vector of 10 elements -- the 0th element will always be 0.

(lambda (seqgen count)
(let ((digcounts (make-vector 10 0)))
(do ((index 0 (1+ index)))
((>= index count))
(let* ((value (seqgen))
(string (format "~a" value))
(vector-map (lambda (digcnt) (/ digcnt count)) digcounts))))

; Create a Fibonacci sequence generator.

(define make-fibgen
(lambda ()
(let ((fn-2 0) (fn-1 1))
(lambda ()
(let ((fn fn-1))
(set! fn-1 (+ fn-2 fn-1))
(set! fn-2 fn)
fn)))))

; Create a sequence generator that returns elements of the given vector.

(define make-vecgen
(lambda (vector)
(let ((index 0))
(lambda ()
(set! index (1+ index))
(vector-ref vector (1- index))))))

; Read all the values in the given file into a list.

(lambda (filenm)
(call-with-input-file filenm
(lambda (ip)
(if (eof-object? value)
'()

; Display a table of digit, Benford's law, sequence distribution, and difference.

(define display-table
(lambda (seqnam seqgen count)
(printf "~%~3@a ~11@a ~11@a ~11@a~%" "dig" "Benford's" seqnam "difference")
(do ((digit 1 (1+ digit)))
((> digit 9))
(let* ((fraction (vector-ref dist digit))
(benford (benford-probability digit))
(diff (- fraction benford)))
(printf "~3d ~11,5f ~11,5f ~11,5f~%" digit benford fraction diff))))))

; Emit tables of various sequence distributions.

(display-table "Fib/1000" (make-fibgen) 1000)
(display-table "Rnd/1T/1M" (lambda () (1+ (random 1000000000000))) 1000000)
(display-table "Craters/D" (make-vecgen craters) (vector-length craters)))

Output:

The first one thousand Fibonnaci numbers.

dig   Benford's    Fib/1000  difference
1     0.30103     0.30100    -0.00003
2     0.17609     0.17700     0.00091
3     0.12494     0.12500     0.00006
4     0.09691     0.09600    -0.00091
5     0.07918     0.08000     0.00082
6     0.06695     0.06700     0.00005
7     0.05799     0.05600    -0.00199
8     0.05115     0.05300     0.00185
9     0.04576     0.04500    -0.00076
Output:

One million pseudo-random numbers from one to one trillion.

dig   Benford's   Rnd/1T/1M  difference
1     0.30103     0.11055    -0.19048
2     0.17609     0.11100    -0.06509
3     0.12494     0.11126    -0.01368
4     0.09691     0.11129     0.01438
5     0.07918     0.11141     0.03223
6     0.06695     0.11088     0.04394
7     0.05799     0.11120     0.05321
8     0.05115     0.11112     0.05997
9     0.04576     0.11128     0.06553
Output:

Diameters of 1,595 Lunar craters in meters.
From the Gazetteer of Planetary Nomenclature [2], referred to by the Wikipedia Planetary nomenclature page [3].

dig   Benford's   Craters/D  difference
1     0.30103     0.23950    -0.06153
2     0.17609     0.10658    -0.06951
3     0.12494     0.11912    -0.00582
4     0.09691     0.12414     0.02723
5     0.07918     0.11034     0.03116
6     0.06695     0.10596     0.03901
7     0.05799     0.06959     0.01160
8     0.05115     0.06646     0.01531
9     0.04576     0.05831     0.01255

## SETL

program benfords_law;
fibos := fibo_list(1000);

expected := [log(1 + 1/d)/log 10 : d in [1..9]];
actual := benford(fibos);

print('d   Expected     Actual');
loop for d in [1..9] do
print(d, '  ', fixed(expected(d), 7, 5), '  ', fixed(actual(d), 7, 5));
end loop;

proc benford(list);
dist := [];
loop for n in list do
dist(val(str n)(1)) +:= 1;
end loop;
return [d / #list : d in dist];
end proc;

proc fibo_list(n);
a := 1;
b := 1;
fibs := [];
loop while n>0 do
fibs with:= a;
[a, b] := [b, a+b];
n -:= 1;
end loop;
return fibs;
end proc;
end program;
Output:
d   Expected     Actual
1    0.30103    0.30100
2    0.17609    0.17700
3    0.12494    0.12500
4    0.09691    0.09600
5    0.07918    0.08000
6    0.06695    0.06700
7    0.05799    0.05600
8    0.05115    0.05300
9    0.04576    0.04500

## Sidef

var (actuals, expected) = ([], [])
var fibonacci = 1000.of {|i| fib(i).digit(-1) }

for i in (1..9) {
var num = fibonacci.count_by {|j| j == i }
actuals.append(num / 1000)
expected.append(1 + (1/i) -> log10)
}

"%17s%17s\n".printf("Observed","Expected")

for i in (1..9) {
"%d : %11s %%%15s %%\n".printf(
i, "%.2f".sprintf(100 *  actuals[i - 1]),
"%.2f".sprintf(100 * expected[i - 1]),
)
}

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.

-- 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;

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.

## Stata

clear
set obs 1000
scalar phi=(1+sqrt(5))/2
gen fib=(phi^_n-(-1/phi)^_n)/sqrt(5)
gen k=real(substr(string(fib),1,1))
hist k, discrete                      // show a histogram
qui tabulate k, matcell(f)            // compute frequencies

mata
f=st_matrix("f")
p=log10(1:+1:/(1::9))*sum(f)
// print observed vs predicted probabilities
f,p
1             2
+-----------------------------+
1 |          297   301.0299957  |
2 |          178   176.0912591  |
3 |          127   124.9387366  |
4 |           96   96.91001301  |
5 |           80   79.18124605  |
6 |           67   66.94678963  |
7 |           57   57.99194698  |
8 |           53   51.15252245  |
9 |           45   45.75749056  |
+-----------------------------+


Assuming the data are random, one can also do a goodness of fit chi-square test:

// chi-square statistic
chisq=sum((f-p):^2:/p)
chisq
.2219340262
// p-value
chi2tail(8,chisq)
.9999942179
end


The p-value is very close to 1, showing that the observed distribution is very close to the Benford law.

The fit is not as good with the sequence (2+sqrt(2))^n:

clear
set obs 500
scalar s=2+sqrt(2)
gen a=s^_n
gen k=real(substr(string(a),1,1))
hist k, discrete
qui tabulate k, matcell(f)

mata
f=st_matrix("f")
p=log10(1:+1:/(1::9))*sum(f)
f,p
1             2
+-----------------------------+
1 |          134   150.5149978  |
2 |           99   88.04562953  |
3 |           68    62.4693683  |
4 |           34    48.4550065  |
5 |           33   39.59062302  |
6 |           33   33.47339482  |
7 |           33   28.99597349  |
8 |           33   25.57626122  |
9 |           33   22.87874528  |
+-----------------------------+

chisq=sum((f-p):^2:/p)
chisq
16.26588528

chi2tail(8,chisq)
.0387287805
end


Now the p-value is less than the usual 5% risk, and one would reject the hypothesis that the data follow the Benford law.

## Swift

import Foundation

/* Reads from a file and returns the content as a String */

var ret:String = ""

let path = Foundation.URL(string: "file://"+file)

do {
ret = try String(contentsOf: path!, encoding: String.Encoding.utf8)
}
catch {
exit(-1)
}

return ret
}

/* Calculates the probability following Benford's law */
func benford(digit z:Int) -> Double {

if z<=0 || z>9 {
perror("Argument must be between 1 and 9.")
return 0
}

return log10(Double(1)+Double(1)/Double(z))
}

// get CLI input
if CommandLine.arguments.count < 2 {
print("Usage: Benford [FILE]")
exit(-1)
}

let pathToFile = CommandLine.arguments[1]

// Read from given file and parse into lines
let lines = content.components(separatedBy: "\n")

var digitCount:UInt64 = 0
var countDigit:[UInt64] = [0,0,0,0,0,0,0,0,0]

// check digits line by line
for line in lines {
if line == "" {
continue
}
let charLine = Array(line.characters)
switch(charLine[0]){
case "1":
countDigit[0] += 1
digitCount += 1
break
case "2":
countDigit[1] += 1
digitCount += 1
break
case "3":
countDigit[2] += 1
digitCount += 1
break
case "4":
countDigit[3] += 1
digitCount += 1
break
case "5":
countDigit[4] += 1
digitCount += 1
break
case "6":
countDigit[5] += 1
digitCount += 1
break
case "7":
countDigit[6] += 1
digitCount += 1
break
case "8":
countDigit[7] += 1
digitCount += 1
break
case "9":
countDigit[8] += 1
digitCount += 1
break
default:
break
}

}

// print result
print("Digit\tBenford [%]\tObserved [%]\tDeviation")
print("~~~~~\t~~~~~~~~~~~~\t~~~~~~~~~~~~\t~~~~~~~~~")
for i in 0..<9 {
let temp:Double = Double(countDigit[i])/Double(digitCount)
let ben = benford(digit: i+1)
print(String(format: "%d\t%.2f\t\t%.2f\t\t%.4f", i+1,ben*100,temp*100,ben-temp))
}

Output:
$./Benford Usage: Benford [FILE]$ ./Benford Fibonacci.txt
Digit	Benford [%]	Observed [%]	Deviation
~~~~~	~~~~~~~~~~~~	~~~~~~~~~~~~	~~~~~~~~~
1	30.10		30.10		0.0000
2	17.61		17.70		-0.0009
3	12.49		12.50		-0.0001
4	9.69		9.60		0.0009
5	7.92		8.00		-0.0008
6	6.69		6.70		-0.0001
7	5.80		5.60		0.0020
8	5.12		5.30		-0.0018
9	4.58		4.50		0.0008

## 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}]]
}
}


Demonstrating with Fibonacci numbers:

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 $result } benfordTest [fibs 1000]  Output:  digit | measured | theory -------+----------+-------- 1 | 30.10% | 30.10% 2 | 17.70% | 17.61% 3 | 12.50% | 12.49% 4 | 9.60% | 9.69% 5 | 8.00% | 7.92% 6 | 6.70% | 6.69% 7 | 5.60% | 5.80% 8 | 5.30% | 5.12% 9 | 4.50% | 4.58%  ## True BASIC Translation of: Chipmunk Basic FUNCTION log10(n) LET log10 = LOG(n)/LOG(10) END FUNCTION FUNCTION frequency(n) LET frequency = (log10(n+1)-log10(n)) END FUNCTION FUNCTION fibonacci(n) !https://rosettacode.org/wiki/Fibonacci_sequence#True_BASIC LET n1 = 0 LET n2 = 1 FOR k = 1 TO ABS(n) LET sum = n1+n2 LET n1 = n2 LET n2 = sum NEXT k IF n < 0 THEN LET fibonacci = n1*((-1)^((-n)+1)) ELSE LET fibonacci = n1 END FUNCTION CLEAR LET n = 1000 DIM actual(0) MAT REDIM actual(n) FOR nr = 1 TO n LET num$ = STR$(fibonacci(nr)) LET j = VAL((num$)[1:1])
LET actual(j) = actual(j)+1
NEXT nr
PRINT "First 1000 Fibonacci numbers"
PRINT "Digit   Actual freq      Expected freq"
FOR i = 1 TO 9
LET freq = frequency(i)*100
PRINT  USING "###": i;
PRINT  USING "        ##.###": actual(i)/10;
PRINT  USING "           ##.###": freq
NEXT i
END


## Uiua

This example is a great application of Benford's law, as the discrepancy between the expected and actual series highlights how the larger Fibonacci numbers exceed Uiua's ability to store large numbers to the required accuracy.

F ← |1 memo(⨬(+∩F-1.-1|⋅1)<2.)
Lim ← 1000
&p$"Expected: _"⁅×Limₙ10+1÷:1↘1⇡10 &p$"Got:      _"⊕⧻⊛.≡(⊢°⋕F)⇡Lim
Output:
Expected: [301 176 125 97 79 67 58 51 46]
Got:      [301 177 125 80 53 67 45 96 56]


## VBA (Visual Basic for Application)

Sub BenfordLaw()

Dim BenResult(1 To 9) As Long

BENref = "30,1%|17,6%|12,5%|9,7%|7,9%|6,7%|5,8%|5,1%|4,6%"

For Each c In Selection.Cells
If InStr(1, "-0123456789", Left(c, 1)) > 0 Then
For i = 1 To 9
If CInt(Left(Abs(c), 1)) = i Then BenResult(i) = BenResult(i) + 1: Exit For
Next
End If
Next
Total= Application.Sum(BenResult)
biggest= Len(CStr(BenResult(1)))

txt = "#   |   Values    |     Real     |  Expected   " & vbCrLf
For i = 1 To 9
If BenResult(i) > 0 Then
txt = txt & "#" & i & " | " & vbTab
txt = txt & String((biggest - Len(CStr(BenResult(i)))) * 2, " ") & Format(BenResult(i), "0") & " | " & vbTab
txt = txt & String((Len(CStr(Format(BenResult(1) / Total, "##0.0%"))) - Len(CStr(Format(BenResult(i) / Total, "##0.0%")))) * 2, " ") & Format(BenResult(i) / Total, "##0.0%") & " | " & vbTab
txt = txt & Format(Split(BENref, "|")(i - 1), "     ##0.0%") & vbCrLf
End If
Next

MsgBox txt, vbOKOnly, "Finish"

End Sub

}


## Visual FoxPro

#DEFINE CTAB CHR(9)
#DEFINE COMMA ","
#DEFINE CRLF CHR(13) + CHR(10)
LOCAL i As Integer, n As Integer, n1 As Integer, rho As Double, c As String
n = 1000
LOCAL ARRAY a[n,2], res[1]
CLOSE DATABASES ALL
CREATE CURSOR fibo(dig C(1))
INDEX ON dig TAG dig COLLATE "Machine"
SET ORDER TO 0
*!* Populate the cursor with the leading digit of the first 1000 Fibonacci numbers
a[1,1] = "1"
a[1,2] = 1
a[2,1] = "1"
a[2,2] = 1
FOR i = 3 TO n
a[i,2] = a[i-2,2] + a[i-1,2]
a[i,1] = LEFT(TRANSFORM(a[i,2]), 1)
ENDFOR
APPEND FROM ARRAY a FIELDS dig
CREATE CURSOR results (digit I, count I, prob B(6), expected B(6))
INSERT INTO results ;
SELECT dig, COUNT(1), COUNT(1)/n, Pr(VAL(dig)) FROM fibo GROUP BY dig ORDER BY dig
n1 = RECCOUNT()
*!* Correlation coefficient
SELECT (n1*SUM(prob*expected) - SUM(prob)*SUM(expected))/;
(SQRT(n1*SUM(prob*prob) - SUM(prob)*SUM(prob))*SQRT(n1*SUM(expected*expected) - SUM(expected)*SUM(expected))) ;
FROM results INTO ARRAY res
rho = CAST(res[1] As B(6))
SET SAFETY OFF
COPY TO benford.txt TYPE CSV
c = FILETOSTR("benford.txt")
*!* Replace commas with tabs
c = STRTRAN(c, COMMA, CTAB) + CRLF + "Correlation Coefficient: " + TRANSFORM(rho)
STRTOFILE(c, "benford.txt", 0)
SET SAFETY ON

FUNCTION Pr(d As Integer) As Double
RETURN LOG10(1 + 1/d)
ENDFUNC

Output:
digit	count	prob	expected
1		301	0.301000	0.301000
2		177	0.177000	0.176100
3		125	0.125000	0.124900
4		 96	0.096000	0.096900
5		 80	0.080000	0.079200
6		 67	0.067000	0.066900
7		 56	0.056000	0.058000
8		 53	0.053000	0.051200
9		 45	0.045000	0.045800

Correlation Coefficient: 0.999908


## V (Vlang)

Translation of: Go
Library: Wren-fmt
import math

fn fib1000() []f64 {
mut a, mut b, mut r := 0.0, 1.0, []f64{len:1000}
for i in 0..r.len {
r[i], a, b = b, b, b+a
}
return r
}

fn main() {
show(fib1000(), "First 1000 Fibonacci numbers")
}

fn show(c []f64, title string) {
mut f := [9]int{}
for v in c {
f["$v"[0]-'1'[0]]++ } println(title) println("Digit Observed Predicted") for i, n in f { println("${i+1}  ${f64(n)/f64(c.len):9.3f}${math.log10(1+1/f64(i+1)):8.3f}")
}
}
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


## Wren

Translation of: Go
Library: Wren-fmt
import "./fmt" for Fmt

var fib1000 = Fn.new {
var a = 0
var b = 1
var r = List.filled(1000, 0)
for (i in 0...r.count) {
var oa = a
var ob = b
r[i] = ob
a = ob
b = ob + oa
}
return r
}

var LN10 = 2.3025850929940457

var log10 = Fn.new { |x| x.log / LN10 }

var show = Fn.new { |c, title|
var f = List.filled(9, 0)
for (v in c) {
var t = "%(v)".bytes[0] - 49
f[t] = f[t] + 1
}
System.print(title)
System.print("Digit  Observed  Predicted")
for (i in 0...f.count) {
var n = f[i]
var obs = Fmt.f(9, n/c.count, 3)
var t = log10.call(1/(i + 1) + 1)
var pred = Fmt.f(8, t, 3)
System.print("  %(i+1) %(obs)  %(pred)")
}
}

show.call(fib1000.call(), "First 1000 Fibonacci numbers:")

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


## XPL0

The program is run like this: benford < fibdig.txt

Luckly, there's no need to show how the first digits of the Fibonacci sequence were obtained.

int  D, N, Counts(10);
real A, E;
[for D:= 0 to 9 do Counts(D):= 0;
for N:= 1 to 1000 do
[D:= ChIn(1) & $0F; \ASCII to binary Counts(D):= Counts(D)+1; ]; Text(0, "Digit Actual Expected Difference^m^j"); for D:= 1 to 9 do [IntOut(0, D); ChOut(0, ^ ); A:= float(Counts(D))/1000.; RlOut(0, A); E:= Log(1. + 1./float(D)); RlOut(0, E); RlOut(0, E-A); CrLf(0); ]; ] Output: Digit Actual Expected Difference 1 0.30100 0.30103 0.00003 2 0.17700 0.17609 -0.00091 3 0.12500 0.12494 -0.00006 4 0.09600 0.09691 0.00091 5 0.08000 0.07918 -0.00082 6 0.06700 0.06695 -0.00005 7 0.05600 0.05799 0.00199 8 0.05300 0.05115 -0.00185 9 0.04500 0.04576 0.00076  ## Yabasic Using function from https://www.rosettacode.org/wiki/Fibonacci_sequence#Yabasic n = 1000 dim actual(n) for nr = 1 to n num$ = str$(fibonacci(nr)) j = val(left$(num$,1)) actual(j) = actual(j) + 1 next print "First 1000 Fibonacci numbers" print "Digit ", "Actual freq ", "Expected freq" for i = 1 to 9 freq = frequency(i)*100 print i using("###"), " ", (actual(i)/10) using("##.###"), " ", freq using("##.###") next end sub frequency(n) return (log10(n+1) - log10(n)) end sub sub log10(n) return log(n) / log(10) end sub sub fibonacci(n) local n1, n2, k, sum n1 = 0 n2 = 1 for k = 1 to abs(n) sum = n1 + n2 n1 = n2 n2 = sum next k if n < 0 then return n1 * ((-1) ^ ((-n) + 1)) else return n1 end if end sub  Output: First 1000 Fibonacci numbers Digit Actual freq Expected freq 1 30.100 30.103 2 17.700 17.609 3 12.500 12.494 4 9.600 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 ## zkl Translation of: Go show( // use list (fib(1)...fib(1000)) --> (1..4.34666e+208) (0).pump(1000,List,fcn(ab){ab.append(ab.sum(0.0)).pop(0)}.fp(L(1,1))), "First 1000 Fibonacci numbers"); fcn show(data,title){ f:=(0).pump(9,List,Ref.fp(0)); // (Ref(0),Ref(0)... foreach v in (data){ // eg 1.49707e+207 ("g" format) --> "1" (first digit) f[v.toString()[0].toInt()-1].inc(); } println(title); println("Digit Observed Predicted"); foreach i,n in ([1..].zip(f)){ // -->(1,Ref)...(9,Ref) println(" %d %9.3f %8.3f".fmt(i,n.value.toFloat()/data.len(), (1.0+1.0/i).log10())) } } Translation of: CoffeeScript var BN=Import("zklBigNum"); fcn fibgen(a,b) { return(a,self.fcn.fp(b,a+b)) } //-->L(fib,fcn) benford := [0..9].pump(List,Ref.fp(0)).copy(); //L(Ref(0),...) const N=1000; [1..N].reduce('wrap(fiber,_){ n,f:=fiber; benford[n.toString()[0]].inc(); // first digit of fib f() // next (fib,fcn) pair },fibgen(BN(1),BN(1))); // de-ref Refs ie convert to int to float, divide by N actual := benford.apply(T("value","toFloat",'/(N))); expected := [1..9].apply(fcn(x){(1.0 + 1.0/x).log10()}); println("Leading digital distribution of the first 1,000 Fibonacci numbers"); println("Digit\tActual\tExpected"); foreach i in ([1..9]){ println("%d\t%.3f\t%.3f".fmt(i,actual[i], expected[i-1])); } 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  ## ZX Spectrum Basic Translation of: Liberty BASIC 10 RANDOMIZE 20 DIM b(9) 30 LET n=100 40 FOR i=1 TO n 50 GO SUB 1000 60 LET n$=STR$fiboI 70 LET d=VAL n$(1)
80 LET b(d)=b(d)+1
90 NEXT i
100 PRINT "Digit";TAB 6;"Actual freq";TAB 18;"Expected freq"
110 FOR i=1 TO 9
120 LET pdi=(LN (i+1)/LN 10)-(LN i/LN 10)
130 PRINT i;TAB 6;b(i)/n;TAB 18;pdi
140 NEXT i
150 STOP
1000 REM Fibonacci
1010 LET fiboI=0: LET b=1
1020 FOR j=1 TO i
1030 LET temp=fiboI+b
1040 LET fiboI=b
1050 LET b=temp
1060 NEXT j
1070 RETURN


The results obtained are adjusted fairly well, except for the number 8. This occurs with Sinclair BASIC, Sam BASIC and SpecBAS fits.