Benford's law: Difference between revisions
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c = Array.new(10, 0) |
c = Array.new(10, 0) |
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s.each{ |x| c[x] += 1 } |
s.each{ |x| c[x] += 1 } |
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siz = s.size |
siz = s.size |
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res = (1..9).map{ |d| c[d] / siz } |
res = (1..9).map{ |d| c[d] / siz } |
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puts "\n %s Benfords deviation" % title |
puts "\n %s Benfords deviation" % title |
Revision as of 17:02, 13 March 2020
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 () occurs with probability
For this task, write (a) routine(s) to calculate the distribution of first significant (non-zero) digits in a collection of numbers, then display the actual vs. expected distribution in the way most convenient for your language (table / graph / histogram / whatever).
Use the first 1000 numbers from the Fibonacci sequence as your data set. No need to show how the Fibonacci numbers are obtained.
You can generate them or load them from a file; whichever is easiest.
Display your actual vs expected distribution.
For extra credit: Show the distribution for one other set of numbers from a page on Wikipedia. State which Wikipedia page it can be obtained from and what the set enumerates. Again, no need to display the actual list of numbers or the code to load them.
- See also:
- numberphile.com.
- A starting page on Wolfram Mathworld is Benfords Law .
11l
<lang 11l>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))</lang>
- 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%
8th
<lang 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 ;
- adjust \ --
counts @ ( 0.001 * ) a:map counts ! ;
- spaces \ n --
' space swap times ;
- .fw \ s n --
>r s:len r> rot . swap - spaces ;
- .header \ --
"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 \ --
.header counts @ ( swap 1+ dup benford swap .digit .expected .actual cr ) a:each drop ;
- benford-test
init count-fibs adjust report ;
- with
benford-test bye </lang>
- 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
Ada
The program reads the Fibonacci-Numbers from the standard input. Each input line is supposed to hold N, followed by Fib(N).
<lang Ada>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))); package N_IO is new Ada.Text_IO.Integer_IO(Natural); procedure Print(D: Nonzero_Digit; Counted, Sum: Natural) is package Math is new Ada.Numerics.Generic_Elementary_Functions(Float); package F_IO is new Ada.Text_IO.Float_IO(Float); 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
while not Ada.Text_IO.End_Of_File loop -- 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 D := First_Digit(Ada.Text_IO.Get_Line); -- read the rest of the line and extract the first digit Cnt(D) := Cnt(D)+1; Sum := Sum + 1; end loop; Ada.Text_IO.Put_Line(" Digit Found[total] Expected[total] Found[%]" & " Expected[%] Difference[%]"); for I in Nonzero_Digit loop Print(I, Cnt(I), Sum); Ada.Text_IO.New_Line; end loop;
end Benford;</lang>
- 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:
<lang Ada> -- N_IO.Get(Counter);</lang>
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
<lang 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;
}</lang>
- Output:
expected found 1 30.102 30.1 2 17.609 17.7 3 12.493 12.5 4 9.691 9.600 5 7.918 8 6 6.694 6.7 7 5.799 5.6 8 5.115 5.300 9 4.575 4.5
ALGOL 68
Uses Algol 68G's LONG LONG INT which has programmer specifiable precision. <lang algol68>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</lang>
- 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
AutoHotkey
(AutoHotkey1.1+)
<lang AutoHotkey>SetBatchLines, -1 fib := NStepSequence(1, 1, 2, 1000) Out := "Digit`tExpected`tObserved`tDeviation`n" n := [] for k, v in fib d := SubStr(v, 1, 1) , n[d] := n[d] ? n[d] + 1 : 1 for k, v in n Exppected := 100 * Log(1+ (1 / k)) , Observed := (v / fib.MaxIndex()) * 100 , Out .= k "`t" Exppected "`t" Observed "`t" Abs(Exppected - Observed) "`n" MsgBox, % Out
NStepSequence(v1, v2, n, k) { a := [v1, v2] Loop, % k - 2 { a[j := A_Index + 2] := 0 Loop, % j < n + 2 ? j - 1 : n a[j] := BigAdd(a[j - A_Index], a[j]) } return, a }
BigAdd(a, b) { if (StrLen(b) > StrLen(a)) t := a, a := b, b := t LenA := StrLen(a) + 1, LenB := StrLen(B) + 1, Carry := 0 Loop, % LenB - 1 Sum := SubStr(a, LenA - A_Index, 1) + SubStr(B, LenB - A_Index, 1) + Carry , Carry := Sum // 10 , Result := Mod(Sum, 10) . Result Loop, % I := LenA - LenB { if (!Carry) { Result := SubStr(a, 1, I) . Result break } Sum := SubStr(a, I, 1) + Carry , Carry := Sum // 10 , Result := Mod(Sum, 10) . Result , I-- } return, (Carry ? Carry : "") . Result }</lang>NStepSequence() is available here. Output:
Digit Expected Observed Deviation 1 30.103000 30.100000 0.003000 2 17.609126 17.700000 0.090874 3 12.493874 12.500000 0.006126 4 9.691001 9.600000 0.091001 5 7.918125 8.000000 0.081875 6 6.694679 6.700000 0.005321 7 5.799195 5.600000 0.199195 8 5.115252 5.300000 0.184748 9 4.575749 4.500000 0.075749
AWK
<lang AWK>
- syntax: GAWK -f BENFORDS_LAW.AWK
BEGIN {
n = 1000 for (i=1; i<=n; i++) { arr[substr(fibonacci(i),1,1)]++ } print("digit expected observed deviation") for (i=1; i<=9; i++) { expected = log10(i+1) - log10(i) actual = arr[i] / n deviation = expected - actual printf("%5d %8.4f %8.4f %9.4f\n",i,expected*100,actual*100,abs(deviation*100)) } exit(0)
} function fibonacci(n, a,b,c,i) {
a = 0 b = 1 for (i=1; i<=n; i++) { c = a + b a = b b = c } return(c)
} function abs(x) { if (x >= 0) { return x } else { return -x } } function log10(x) { return log(x)/log(10) } </lang>
- Output:
digit expected observed deviation 1 30.1030 30.0000 0.1030 2 17.6091 17.7000 0.0909 3 12.4939 12.5000 0.0061 4 9.6910 9.6000 0.0910 5 7.9181 8.0000 0.0819 6 6.6947 6.7000 0.0053 7 5.7992 5.7000 0.0992 8 5.1153 5.3000 0.1847 9 4.5757 4.5000 0.0757
C
<lang c>#include <stdio.h>
- include <stdlib.h>
- include <math.h>
float *benford_distribution(void) {
static float prob[9]; for (int i = 1; i < 10; i++) prob[i - 1] = log10f(1 + 1.0 / i);
return prob;
}
float *get_actual_distribution(char *fn) {
FILE *input = fopen(fn, "r"); if (!input) { perror("Can't open file"); exit(EXIT_FAILURE); }
int tally[9] = { 0 }; char c; int total = 0; while ((c = getc(input)) != EOF) { /* get the first nonzero digit on the current line */ while (c < '1' || c > '9') c = getc(input);
tally[c - '1']++; total++;
/* discard rest of line */ while ((c = getc(input)) != '\n' && c != EOF) ; } fclose(input); static float freq[9]; for (int i = 0; i < 9; i++) freq[i] = tally[i] / (float) total;
return freq;
}
int main(int argc, char **argv) {
if (argc != 2) { printf("Usage: benford <file>\n"); return EXIT_FAILURE; }
float *actual = get_actual_distribution(argv[1]); float *expected = benford_distribution();
puts("digit\tactual\texpected"); for (int i = 0; i < 9; i++) printf("%d\t%.3f\t%.3f\n", i + 1, actual[i], expected[i]);
return EXIT_SUCCESS;
}</lang>
- Output:
Use with a file which should contain a number on each line.
$ ./benford fib1000.txt digit actual expected 1 0.301 0.301 2 0.177 0.176 3 0.125 0.125 4 0.096 0.097 5 0.080 0.079 6 0.067 0.067 7 0.056 0.058 8 0.053 0.051 9 0.045 0.046
C++
<lang cpp>//to cope with the big numbers , I used the Class Library for Numbers( CLN ) //if used prepackaged you can compile writing "g++ -std=c++11 -lcln yourprogram.cpp -o yourprogram"
- include <cln/integer.h>
- include <cln/integer_io.h>
- include <iostream>
- include <algorithm>
- include <vector>
- include <iomanip>
- include <sstream>
- include <string>
- include <cstdlib>
- include <cmath>
- include <map>
using namespace cln ;
class NextNum { public :
NextNum ( cl_I & a , cl_I & b ) : first( a ) , second ( b ) { } cl_I operator( )( ) { cl_I result = first + second ; first = second ; second = result ; return result ; }
private :
cl_I first ; cl_I second ;
} ;
void findFrequencies( const std::vector<cl_I> & fibos , std::map<int , int> &numberfrequencies ) {
for ( cl_I bignumber : fibos ) { std::ostringstream os ; fprintdecimal ( os , bignumber ) ;//from header file cln/integer_io.h int firstdigit = std::atoi( os.str( ).substr( 0 , 1 ).c_str( )) ; auto result = numberfrequencies.insert( std::make_pair( firstdigit , 1 ) ) ; if ( ! result.second )
numberfrequencies[ firstdigit ]++ ;
}
}
int main( ) {
std::vector<cl_I> fibonaccis( 1000 ) ; fibonaccis[ 0 ] = 0 ; fibonaccis[ 1 ] = 1 ; cl_I a = 0 ; cl_I b = 1 ; //since a and b are passed as references to the generator's constructor //they are constantly changed ! std::generate_n( fibonaccis.begin( ) + 2 , 998 , NextNum( a , b ) ) ; std::cout << std::endl ; std::map<int , int> frequencies ; findFrequencies( fibonaccis , frequencies ) ; std::cout << " found expected\n" ; for ( int i = 1 ; i < 10 ; i++ ) { double found = static_cast<double>( frequencies[ i ] ) / 1000 ; double expected = std::log10( 1 + 1 / static_cast<double>( i )) ; std::cout << i << " :" << std::setw( 16 ) << std::right << found * 100 << " %" ; std::cout.precision( 3 ) ; std::cout << std::setw( 26 ) << std::right << expected * 100 << " %\n" ; } return 0 ;
} </lang>
- Output:
found expected 1 : 30.1 % 30.1 % 2 : 17.7 % 17.6 % 3 : 12.5 % 12.5 % 4 : 9.5 % 9.69 % 5 : 8 % 7.92 % 6 : 6.7 % 6.69 % 7 : 5.6 % 5.8 % 8 : 5.3 % 5.12 % 9 : 4.5 % 4.58 %
Clojure
<lang lisp>(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") (println header) (show-benford-stats fib-digits)
- Universal Constants from Physics (using first column of data)
(println "Universal Constants from Physics") (println header) (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")
y (for [line (line-seq (java.io.BufferedReader. (java.io.StringReader. input)))] (data-parser line)) z (filter identity y)] (show-benford-stats z))
- Sunspots
(println "Sunspots average count per month since 1749") (println header) (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") y (for [line (line-seq (java.io.BufferedReader. (java.io.StringReader. input)))] (data-parser line))]
(show-benford-stats y))
</lang>
- 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
<lang coffeescript>fibgen = () ->
a = 1; b = 0 return () -> ([a, b] = [b, a+b])[1]
leading = (x) -> x.toString().charCodeAt(0) - 0x30
f = fibgen()
benford = (0 for i in [1..9]) benford[leading(f()) - 1] += 1 for i in [1..1000]
log10 = (x) -> Math.log(x) * Math.LOG10E
actual = benford.map (x) -> x * 0.001 expected = (log10(1 + 1/x) for x in [1..9])
console.log "Leading digital distribution of the first 1,000 Fibonacci numbers" console.log "Digit\tActual\tExpected" for i in [1..9]
console.log i + "\t" + actual[i - 1].toFixed(3) + '\t' + expected[i - 1].toFixed(3)</lang>
- Output:
Leading digital distribution of the first 1,000 Fibonacci numbers Digit Actual Expected 1 0.301 0.301 2 0.177 0.176 3 0.125 0.125 4 0.096 0.097 5 0.080 0.079 6 0.067 0.067 7 0.056 0.058 8 0.053 0.051 9 0.045 0.046
Common Lisp
<lang lisp>(defun calculate-distribution (numbers)
"Return the frequency distribution of the most significant nonzero digits in the given list of numbers. The first element of the list is the frequency for digit 1, the second for digit 2, and so on." (defun nonzero-digit-p (c) "Check whether the character is a nonzero digit" (and (digit-char-p c) (char/= c #\0)))
(defun first-digit (n) "Return the most significant nonzero digit of the number or NIL if there is none." (let* ((s (write-to-string n)) (c (find-if #'nonzero-digit-p s))) (when c (digit-char-p c))))
(let ((tally (make-array 9 :element-type 'integer :initial-element 0))) (loop for n in numbers for digit = (first-digit n) when digit do (incf (aref tally (1- digit)))) (loop with total = (length numbers) for digit-count across tally collect (/ digit-count total))))
(defun calculate-benford-distribution ()
"Return the frequency distribution according to Benford's law. The first element of the list is the probability for digit 1, the second element the probability for digit 2, and so on." (loop for i from 1 to 9 collect (log (1+ (/ i)) 10)))
(defun benford (numbers)
"Print a table of the actual and expected distributions for the given list of numbers." (let ((actual-distribution (calculate-distribution numbers)) (expected-distribution (calculate-benford-distribution))) (write-line "digit actual expected") (format T "~:{~3D~9,3F~8,3F~%~}" (map 'list #'list '(1 2 3 4 5 6 7 8 9) actual-distribution expected-distribution))))</lang>
; *fib1000* is a list containing the first 1000 numbers in the Fibonnaci sequence > (benford *fib1000*) digit actual expected 1 0.301 0.301 2 0.177 0.176 3 0.125 0.125 4 0.096 0.097 5 0.080 0.079 6 0.067 0.067 7 0.056 0.058 8 0.053 0.051 9 0.045 0.046
Crystal
<lang 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
def heads(s)
s.map { |a| a.to_s[0].to_i }
end
def show_dist(title, s)
s = heads(s) c = Array.new(10, 0) s.each{ |x| c[x] += 1 } 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))</lang>
- Output:
fibbed Benfords deviation 1: 30.1% 30.1% 0.0% 2: 17.7% 17.6% 0.1% 3: 12.5% 12.5% 0.0% 4: 9.5% 9.7% 0.2% 5: 8.0% 7.9% 0.1% 6: 6.7% 6.7% 0.0% 7: 5.6% 5.8% 0.2% 8: 5.3% 5.1% 0.2% 9: 4.5% 4.6% 0.1% threes Benfords deviation 1: 30.0% 30.1% 0.1% 2: 17.7% 17.6% 0.1% 3: 12.3% 12.5% 0.2% 4: 9.8% 9.7% 0.1% 5: 7.9% 7.9% 0.0% 6: 6.6% 6.7% 0.1% 7: 5.9% 5.8% 0.1% 8: 5.2% 5.1% 0.1% 9: 4.6% 4.6% 0.0% random Benfords deviation 1: 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
<lang d>import std.stdio, std.range, std.math, std.conv, std.bigint;
double[2][9] benford(R)(R seq) if (isForwardRange!R && !isInfinite!R) {
typeof(return) freqs = 0; uint seqLen = 0; foreach (d; seq) if (d != 0) { freqs[d.text[0] - '1'][1]++; seqLen++; }
foreach (immutable i, ref p; freqs) p = [log10(1.0 + 1.0 / (i + 1)), p[1] / seqLen]; return freqs;
}
void main() {
auto fibs = recurrence!q{a[n - 1] + a[n - 2]}(1.BigInt, 1.BigInt);
writefln("%9s %9s %9s", "Actual", "Expected", "Deviation"); foreach (immutable i, immutable p; fibs.take(1000).benford) writefln("%d: %5.2f%% | %5.2f%% | %5.4f%%", i+1, p[1] * 100, p[0] * 100, abs(p[1] - p[0]) * 100);
}</lang>
- Output:
Actual Expected Deviation 1: 30.10% | 30.10% | 0.0030% 2: 17.70% | 17.61% | 0.0908% 3: 12.50% | 12.49% | 0.0061% 4: 9.60% | 9.69% | 0.0910% 5: 8.00% | 7.92% | 0.0818% 6: 6.70% | 6.69% | 0.0053% 7: 5.60% | 5.80% | 0.1992% 8: 5.30% | 5.12% | 0.1847% 9: 4.50% | 4.58% | 0.0757%
Alternative Version
The output is the same. <lang d>import std.stdio, std.range, std.math, std.conv, std.bigint,
std.algorithm, std.array;
auto benford(R)(R seq) if (isForwardRange!R && !isInfinite!R) {
return seq.filter!q{a != 0}.map!q{a.text[0]-'1'}.array.sort().group;
}
void main() {
auto fibs = recurrence!q{a[n - 1] + a[n - 2]}(1.BigInt, 1.BigInt); auto expected = iota(1, 10).map!(d => log10(1.0 + 1.0 / d));
enum N = 1_000; writefln("%9s %9s %9s", "Actual", "Expected", "Deviation"); foreach (immutable i, immutable f; fibs.take(N).benford) writefln("%d: %5.2f%% | %5.2f%% | %5.4f%%", i + 1, f * 100.0 / N, expected[i] * 100, abs((f / double(N)) - expected[i]) * 100);
}</lang>
Elixir
<lang 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</lang>
- 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
<lang Erlang> -module( benfords_law ). -export( [actual_distribution/1, distribution/1, task/0] ).
actual_distribution( Ns ) -> lists:foldl( fun first_digit_count/2, dict:new(), Ns ).
distribution( N ) -> math:log10( 1 + (1 / N) ).
task() -> Total = 1000, Fibonaccis = fib( Total ), Actual_dict = actual_distribution( Fibonaccis ), Keys = lists:sort( dict:fetch_keys( Actual_dict) ), io:fwrite( "Digit Actual Benfords expected~n" ), [io:fwrite("~p ~p ~p~n", [X, dict:fetch(X, Actual_dict) / Total, distribution(X)]) || X <- Keys].
fib( N ) -> fib( N, 0, 1, [] ). fib( 0, Current, _, Acc ) -> lists:reverse( [Current | Acc] ); fib( N, Current, Next, Acc ) -> fib( N-1, Next, Current+Next, [Current | Acc] ).
first_digit_count( 0, Dict ) -> Dict; first_digit_count( N, Dict ) -> [Key | _] = erlang:integer_to_list( N ), dict:update_counter( Key - 48, 1, Dict ). </lang>
- Output:
7> benfords_law:task(). Digit Actual Benfords expected 1 0.301 0.3010299956639812 2 0.177 0.17609125905568124 3 0.125 0.12493873660829993 4 0.096 0.09691001300805642 5 0.08 0.07918124604762482 6 0.067 0.06694678963061322 7 0.056 0.05799194697768673 8 0.053 0.05115252244738129 9 0.045 0.04575749056067514
Factor
<lang 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</lang>
- 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
<lang forth>: 3drop drop 2drop ;
- f2drop fdrop fdrop ;
- int-array create cells allot does> swap cells + ;
- 1st-fib 0e 1e ;
- next-fib ftuck f+ ;
- 1st-digit ( fp -- n )
pad 6 represent 3drop pad c@ [char] 0 - ;
10 int-array counts
- tally
0 counts 10 cells erase 1st-fib 1000 0 DO 1 fdup 1st-digit counts +! next-fib LOOP f2drop ;
- benford ( d -- fp )
s>f 1/f 1e f+ flog ;
- tab 9 emit ;
- heading ( -- )
cr ." Leading digital distribution of the first 1,000 Fibonacci numbers:" cr ." Digit" tab ." Actual" tab ." Expected" ;
- .fixed ( n -- ) \ print count as decimal fraction
s>d <# # # # [char] . hold #s #> type space ;
- report ( -- )
precision 3 set-precision heading 10 1 DO cr i 3 .r tab i counts @ .fixed tab i benford f. LOOP set-precision ;
- compute-benford tally report ;</lang>
- Output:
Gforth 0.7.2, Copyright (C) 1995-2008 Free Software Foundation, Inc. Gforth comes with ABSOLUTELY NO WARRANTY; for details type `license' Type `bye' to exit compute-benford Leading digital distribution of the first 1,000 Fibonacci numbers: Digit Actual Expected 1 0.301 0.301 2 0.177 0.176 3 0.125 0.125 4 0.096 0.0969 5 0.080 0.0792 6 0.067 0.0669 7 0.056 0.058 8 0.053 0.0512 9 0.045 0.0458 ok
Fortran
FORTRAN 90. Compilation and output of this program using emacs compile command and a fairly obvious Makefile entry: <lang fortran>-*- mode: compilation; default-directory: "/tmp/" -*- Compilation started at Sat May 18 01:13:00
a=./f && make $a && $a f95 -Wall -ffree-form f.F -o f
0.301030010 0.176091254 0.124938756 9.69100147E-02 7.91812614E-02 6.69467747E-02 5.79919666E-02 5.11525236E-02 4.57575098E-02 THE LAW 0.300999999 0.177000001 0.125000000 9.60000008E-02 7.99999982E-02 6.70000017E-02 5.70000000E-02 5.29999994E-02 4.50000018E-02 LEADING FIBONACCI DIGIT
Compilation finished at Sat May 18 01:13:00</lang>
<lang fortran>subroutine fibber(a,b,c,d)
! compute most significant digits, Fibonacci like. implicit none integer (kind=8), intent(in) :: a,b integer (kind=8), intent(out) :: c,d d = a + b if (15 .lt. log10(float(d))) then c = b/10 d = d/10 else c = b endif
end subroutine fibber
integer function leadingDigit(a)
implicit none integer (kind=8), intent(in) :: a integer (kind=8) :: b b = a do while (9 .lt. b) b = b/10 end do leadingDigit = transfer(b,leadingDigit)
end function leadingDigit
real function benfordsLaw(a)
implicit none integer, intent(in) :: a benfordsLaw = log10(1.0 + 1.0 / a)
end function benfordsLaw
program benford
implicit none
interface
subroutine fibber(a,b,c,d) implicit none integer (kind=8), intent(in) :: a,b integer (kind=8), intent(out) :: c,d end subroutine fibber
integer function leadingDigit(a) implicit none integer (kind=8), intent(in) :: a end function leadingDigit
real function benfordsLaw(a) implicit none integer, intent(in) :: a end function benfordsLaw
end interface
integer (kind=8) :: a, b, c, d integer :: i, count(10) data count/10*0/ a = 1 b = 1 do i = 1, 1001 count(leadingDigit(a)) = count(leadingDigit(a)) + 1 call fibber(a,b,c,d) a = c b = d end do write(6,*) (benfordsLaw(i),i=1,9),'THE LAW' write(6,*) (count(i)/1000.0 ,i=1,9),'LEADING FIBONACCI DIGIT'
end program benford</lang>
FreeBASIC
<lang freebasic>' 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</lang>
- 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 %
Go
<lang go>package main
import (
"fmt" "math"
)
func Fib1000() []float64 {
a, b, r := 0., 1., [1000]float64{} for i := range r { r[i], a, b = b, b, b+a } return r[:]
}
func main() {
show(Fib1000(), "First 1000 Fibonacci numbers")
}
func show(c []float64, title string) {
var f [9]int for _, v := range c { f[fmt.Sprintf("%g", v)[0]-'1']++ } fmt.Println(title) fmt.Println("Digit Observed Predicted") for i, n := range f { fmt.Printf(" %d %9.3f %8.3f\n", i+1, float64(n)/float64(len(c)), math.Log10(1+1/float64(i+1))) }
}</lang>
- Output:
First 1000 Fibonacci numbers Digit Observed Predicted 1 0.301 0.301 2 0.177 0.176 3 0.125 0.125 4 0.096 0.097 5 0.080 0.079 6 0.067 0.067 7 0.056 0.058 8 0.053 0.051 9 0.045 0.046
Groovy
Solution:
Uses Fibonacci sequence analytic formula
<lang groovy>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
}</lang>
Test: <lang groovy>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))
}</lang>
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
Haskell
<lang haskell>import qualified Data.Map as M import Data.Char (digitToInt)
fstdigit :: Integer -> Int fstdigit = digitToInt . head . show
n = 1000 :: Int
fibs = 1 : 1 : zipWith (+) fibs (tail fibs)
fibdata = map fstdigit $ take n fibs
freqs = M.fromListWith (+) $ zip fibdata (repeat 1)
tab :: [(Int, Double, Double)] tab =
[ ( d , fromIntegral (M.findWithDefault 0 d freqs) / fromIntegral n , logBase 10.0 $ 1 + 1 / fromIntegral d) | d <- [1 .. 9] ]
main = print tab</lang>
- Output:
[(1,0.301,0.301029995663981), (2,0.177,0.176091259055681), (3,0.125,0.1249387366083), (4,0.096,0.0969100130080564), (5,0.08,0.0791812460476248), (6,0.067,0.0669467896306132), (7,0.056,0.0579919469776867), (8,0.053,0.0511525224473813), (9,0.045,0.0457574905606751)]
Icon and Unicon
The following solution works in both languages.
<lang unicon>global counts, total
procedure main()
counts := table(0) total := 0.0 every benlaw(fib(1 to 1000))
every i := 1 to 9 do write(i,": ",right(100*counts[string(i)]/total,9)," ",100*P(i))
end
procedure benlaw(n)
if counts[n ? (tab(upto('123456789')),move(1))] +:= 1 then total +:= 1
end
procedure P(d)
return log(1+1.0/d, 10)
end
procedure fib(n) # From Fibonacci Sequence task
return fibMat(n)[1]
end
procedure fibMat(n)
if n <= 0 then return [0,0] if n = 1 then return [1,0] fp := fibMat(n/2) c := fp[1]*fp[1] + fp[2]*fp[2] d := fp[1]*(fp[1]+2*fp[2]) if n%2 = 1 then return [c+d, d] else return [d, c]
end</lang>
Sample run:
->benlaw 1: 30.1 30.10299956639811 2: 17.7 17.60912590556812 3: 12.5 12.49387366082999 4: 9.6 9.69100130080564 5: 8.0 7.918124604762481 6: 6.7 6.694678963061322 7: 5.6 5.799194697768673 8: 5.3 5.115252244738128 9: 4.5 4.575749056067514 ->
J
We show the correlation coefficient of Benford's law with the leading digits of the first 1000 Fibonacci numbers is almost unity. <lang J>log10 =: 10&^. benford =: log10@:(1+%) assert '0.30 0.18 0.12 0.10 0.08 0.07 0.06 0.05 0.05' -: 5j2 ": benford >: i. 9
append_next_fib =: , +/@:(_2&{.)
assert 5 8 13 -: append_next_fib 5 8
leading_digits =: {.@":&> assert '581' -: leading_digits 5 8 13x
count =: #/.~ /: ~. assert 2 1 3 4 -: count 'XCXBAXACXC' NB. 2 A's, 1 B, 3 C's, and some X's.
normalize =: % +/ assert 1r3 2r3 -: normalize 1 2x
FIB =: append_next_fib ^: (1000-#) 1 1 LDF =: leading_digits FIB
TALLY_BY_KEY =: count LDF
assert 9 -: # TALLY_BY_KEY NB. If all of [1-9] are present then we know what the digits are.
mean =: +/ % # center=: - mean mp =: $:~ :(+/ .*) num =: mp&:center den =: %:@:(*&:(+/@:(*:@:center))) r =: num % den NB. r is the LibreOffice correl function assert '_0.982' -: 6j3 ": 1 2 3 r 6 5 3 NB. confirmed using LibreOffice correl function
assert '0.9999' -: 6j4 ": (normalize TALLY_BY_KEY) r benford >: i.9
assert '0.9999' -: 6j4 ": TALLY_BY_KEY r benford >: i.9 NB. Of course we don't need normalization</lang>
Java
<lang Java>import java.math.BigInteger; import java.util.Locale;
public class BenfordsLaw {
private static BigInteger[] generateFibonacci(int n) { BigInteger[] fib = new BigInteger[n]; fib[0] = BigInteger.ONE; fib[1] = BigInteger.ONE; for (int i = 2; i < fib.length; i++) { fib[i] = fib[i - 2].add(fib[i - 1]); } return fib; }
public static void main(String[] args) { BigInteger[] numbers = generateFibonacci(1000);
int[] firstDigits = new int[10]; for (BigInteger number : numbers) { firstDigits[Integer.valueOf(number.toString().substring(0, 1))]++; }
for (int i = 1; i < firstDigits.length; i++) { System.out.printf(Locale.ROOT, "%d %10.6f %10.6f%n", i, (double) firstDigits[i] / numbers.length, Math.log10(1.0 + 1.0 / i)); } }
}</lang> The output is:
1 0.301000 0.301030 2 0.177000 0.176091 3 0.125000 0.124939 4 0.096000 0.096910 5 0.080000 0.079181 6 0.067000 0.066947 7 0.056000 0.057992 8 0.053000 0.051153 9 0.045000 0.045757
To use other number sequences, implement a suitable NumberGenerator, construct a Benford instance with it and print it.
JavaScript
<lang javascript>const fibseries = n => [...Array(n)]
.reduce( (fib, _, i) => i < 2 ? ( fib ) : fib.concat(fib[i - 1] + fib[i - 2]), [1, 1] );
const benford = array => [1, 2, 3, 4, 5, 6, 7, 8, 9]
.map(val => [val, array .reduce( (sum, item) => sum + ( `${item}` [0] === `${val}` ), 0 ) / array.length, Math.log10(1 + 1 / val) ]);
console.log(benford(fibseries(1000)))</lang>
- 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
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.<lang jq># 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(".") as $ix | 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 ;
def report(heading; stream):
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</lang>
- 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
<lang Julia>fib(n) = ([one(n) one(n) ; one(n) zero(n)]^n)[1,2]
ben(l) = [count(x->x==i, map(n->string(n)[1],l)) for i='1':'9']./length(l)
benford(l) = [Number[1:9;] ben(l) log10(1.+1./[1:9;])]</lang>
- Output:
julia> benford([fib(big(n)) for n = 1:1000]) 9x3 Array{Number,2}: 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
Kotlin
<lang scala>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))</lang>
Liberty BASIC
Using function from http://rosettacode.org/wiki/Fibonacci_sequence#Liberty_BASIC <lang lb> dim bin(9)
N=1000 for i = 0 to N-1
num$ = str$(fiboI(i)) d=val(left$(num$,1)) 'print num$, d bin(d)=bin(d)+1
next print
print "Digit", "Actual freq", "Expected freq" for i = 1 to 9
print i, bin(i)/N, using("#.###", P(i))
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 </lang>
- Output:
Digit Actual freq Expected freq 1 0.301 0.301 2 0.177 0.176 3 0.125 0.125 4 0.095 0.097 5 0.08 0.079 6 0.067 0.067 7 0.056 0.058 8 0.053 0.051 9 0.045 0.046
Lua
<lang 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</lang>
- 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
<lang mathematica>fibdata = Array[First@IntegerDigits@Fibonacci@# &, 1000]; Table[{d, N@Count[fibdata, d]/Length@fibdata, Log10[1. + 1/d]}, {d, 1,
9}] // Grid</lang>
- Output:
1 0.301 0.30103 2 0.177 0.176091 3 0.125 0.124939 4 0.096 0.09691 5 0.08 0.0791812 6 0.067 0.0669468 7 0.056 0.0579919 8 0.053 0.0511525 9 0.045 0.0457575
NetRexx
<lang NetRexx>/* NetRexx */ options replace format comments java crossref symbols nobinary
runSample(arg) return
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ method brenfordDeveation(nlist = Rexx[]) public static
observed = 0 loop n_ over nlist d1 = n_.left(1) if d1 = 0 then iterate n_ observed[d1] = observed[d1] + 1 end n_ say ' '.right(4) 'Observed'.right(11) 'Expected'.right(11) 'Deviation'.right(11) loop n_ = 1 to 9 actual = (observed[n_] / (nlist.length - 1)) expect = Rexx(Math.log10(1 + 1 / n_)) deviat = expect - actual say n_.right(3)':' (actual * 100).format(3, 6)'%' (expect * 100).format(3, 6)'%' (deviat * 100).abs().format(3, 6)'%' end n_ return
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ method fibonacciList(size = 1000) public static returns Rexx[]
fibs = Rexx[size + 1] fibs[0] = 0 fibs[1] = 1 loop n_ = 2 to size fibs[n_] = fibs[n_ - 1] + fibs[n_ - 2] end n_ return fibs
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ method runSample(arg) private static
parse arg n_ . if n_ = then n_ = 1000 fibList = fibonacciList(n_) say 'Fibonacci sequence to' n_ brenfordDeveation(fibList) return
</lang>
- Output:
Fibonacci sequence to 1000 Observed Expected Deviation 1: 30.100000% 30.103000% 0.003000% 2: 17.700000% 17.609126% 0.090874% 3: 12.500000% 12.493874% 0.006127% 4: 9.600000% 9.691001% 0.091001% 5: 8.000000% 7.918125% 0.081875% 6: 6.700000% 6.694679% 0.005321% 7: 5.600000% 5.799195% 0.199195% 8: 5.300000% 5.115252% 0.184748% 9: 4.500000% 4.575749% 0.075749%
Oberon-2
<lang oberon2> 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. </lang>
- Output:
First 1000 fibonacci numbers: digit observed predicted 1 0.301 0.301 2 0.177 0.176 3 0.125 0.125 4 0.096 0.097 5 0.080 0.079 6 0.067 0.067 7 0.056 0.058 8 0.053 0.051 9 0.045 0.046
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.
<lang ocaml>
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" ;;
</lang>
- 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
<lang parigp>distribution(v)={ my(t=vector(9,n,sum(i=1,#v,v[i]==n))); print("Digit\tActual\tExpected"); for(i=1,9,print(i, "\t", t[i], "\t", round(#v*(log(i+1)-log(i))/log(10)))) }; dist(f)=distribution(vector(1000,n,digits(f(n))[1])); lucas(n)=fibonacci(n-1)+fibonacci(n+1); dist(fibonacci) dist(lucas)</lang>
- Output:
Digit Actual Expected 1 301 301 2 177 176 3 125 125 4 96 97 5 80 79 6 67 67 7 56 58 8 53 51 9 45 46 Digit Actual Expected 1 301 301 2 174 176 3 127 125 4 97 97 5 79 79 6 66 67 7 59 58 8 51 51 9 46 46
Pascal
<lang 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
a,b,c : LongWord;//about 9.6 decimals
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.</lang>
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
<lang Perl>#!/usr/bin/perl use strict ; use warnings ; use POSIX qw( log10 ) ;
my @fibonacci = ( 0 , 1 ) ; while ( @fibonacci != 1000 ) {
push @fibonacci , $fibonacci[ -1 ] + $fibonacci[ -2 ] ;
} my @actuals ; my @expected ; for my $i( 1..9 ) {
my $sum = 0 ; map { $sum++ if $_ =~ /\A$i/ } @fibonacci ; push @actuals , $sum / 1000 ; push @expected , log10( 1 + 1/$i ) ;
} print " Observed Expected\n" ; for my $i( 1..9 ) {
print "$i : " ; my $result = sprintf ( "%.2f" , 100 * $actuals[ $i - 1 ] ) ; printf "%11s %%" , $result ; $result = sprintf ( "%.2f" , 100 * $expected[ $i - 1 ] ) ; printf "%15s %%\n" , $result ;
}</lang>
- Output:
Observed Expected 1 : 30.10 % 30.10 % 2 : 17.70 % 17.61 % 3 : 12.50 % 12.49 % 4 : 9.50 % 9.69 % 5 : 8.00 % 7.92 % 6 : 6.70 % 6.69 % 7 : 5.60 % 5.80 % 8 : 5.30 % 5.12 % 9 : 4.50 % 4.58 %
Perl 6
<lang perl6>sub benford(@a) { bag +« @a».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] }</lang>
Output: First 1000 Fibonaccis
Actual Expected Deviation 1: 30.10% | 30.10% | 0.00% 2: 17.70% | 17.61% | 0.09% 3: 12.50% | 12.49% | 0.01% 4: 9.60% | 9.69% | 0.09% 5: 8.00% | 7.92% | 0.08% 6: 6.70% | 6.69% | 0.01% 7: 5.60% | 5.80% | 0.20% 8: 5.30% | 5.12% | 0.18% 9: 4.50% | 4.58% | 0.08%
Extra credit: Square Kilometers of land under cultivation, by country / territory. First column from Wikipedia: Land use statistics by country.
Actual Expected Deviation 1: 33.33% | 30.10% | 3.23% 2: 18.31% | 17.61% | 0.70% 3: 13.15% | 12.49% | 0.65% 4: 8.45% | 9.69% | 1.24% 5: 9.39% | 7.92% | 1.47% 6: 5.63% | 6.69% | 1.06% 7: 4.69% | 5.80% | 1.10% 8: 5.16% | 5.12% | 0.05% 9: 1.88% | 4.58% | 2.70%
Phix
<lang Phix>procedure main(sequence s, string title) sequence f = repeat(0,9)
for i=1 to length(s) do f[sprint(s[i])[1]-'0'] += 1 end for puts(1,title) puts(1,"Digit Observed% Predicted%\n") for i=1 to length(f) do printf(1," %d %9.3f %8.3f\n", {i, f[i]/length(s)*100, log10(1+1/i)*100}) end for
end procedure main(fib(1000),"First 1000 Fibonacci numbers\n") main(primes(10000),"First 10000 Prime numbers\n") main(threes(500),"First 500 powers of three\n")</lang> Supporting staff: <lang Phix>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
function primes(integer lim) integer n = 1, k, p sequence res = {2}
while length(res)<lim do k = 3 p = 1 n += 2 while k*k<=n and p do p = floor(n/k)*k!=n k += 2 end while if p then res = append(res,n) end if end while return res
end function
function threes(integer lim) sequence res = repeat(0,lim)
for i=1 to lim do res[i] = power(3,i) end for return res
end function
constant INVLN10 = 0.43429_44819_03251_82765 function log10(object x1)
return log(x1) * INVLN10
end function</lang>
- Output:
(put into columns by hand)
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
PL/I
<lang PL/I> (fofl, size, subrg): Benford: procedure options(main); /* 20 October 2013 */
declare sc(1000) char(1), f(1000) float (16); declare d fixed (1);
call Fibonacci(f); call digits(sc, f);
put skip list ('digit expected obtained'); do d= 1 upthru 9; put skip edit (d, log10(1 + 1/d), tally(sc, trim(d))/1000) (f(3), 2 f(13,8) ); end;
Fibonacci: procedure (f);
declare f(*) float (16); declare i fixed binary;
f(1), f(2) = 1; do i = 3 to 1000; f(i) = f(i-1) + f(i-2); end;
end Fibonacci;
digits: procedure (sc, f);
declare sc(*) char(1), f(*) float (16); sc = substr(trim(f), 1, 1);
end digits;
tally: procedure (sc, d) returns (fixed binary);
declare sc(*) char(1), d char(1); declare (i, t) fixed binary; t = 0; do i = 1 to 1000; if sc(i) = d then t = t + 1; end; return (t);
end tally; end Benford; </lang> Results:
digit expected obtained 1 0.30103000 0.30099487 2 0.17609126 0.17698669 3 0.12493874 0.12500000 4 0.09691001 0.09599304 5 0.07918125 0.07998657 6 0.06694679 0.06698608 7 0.05799195 0.05599976 8 0.05115252 0.05299377 9 0.04575749 0.04499817
PL/pgSQL
<lang SQL> WITH recursive constant(val) AS ( select 1000. ) , fib(a,b) AS ( SELECT CAST(0 AS numeric), CAST(1 AS numeric) UNION ALL SELECT b,a+b FROM fib ) , benford(first_digit, probability_real, probability_theoretical) AS ( SELECT *, CAST(log(1. + 1./CAST(first_digit AS INT)) AS NUMERIC(5,4)) probability_theoretical FROM ( SELECT first_digit, CAST(COUNT(1)/(select val from constant) AS NUMERIC(5,4)) probability_real FROM ( SELECT SUBSTRING(CAST(a AS VARCHAR(100)),1,1) first_digit FROM fib WHERE SUBSTRING(CAST(a AS VARCHAR(100)),1,1) <> '0' LIMIT (select val from constant) ) t GROUP BY first_digit ) f ORDER BY first_digit ASC ) select * from benford cross join
(select cast(corr(probability_theoretical,probability_real) as numeric(5,4)) correlation from benford) c
</lang>
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. <lang PowerShell> $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 </lang>
- 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
Note: SWI Prolog implements arbitrary precision integer arithmetic through use of the GNU MP library <lang Prolog>%_________________________________________________________________ % Does the Fibonacci sequence follow Benford's law? %~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ % Fibonacci sequence generator fib(C, [P,S], C, N) :- N is P + S. fib(C, [P,S], Cv, V) :- succ(C, Cn), N is P + S, !, fib(Cn, [S,N], Cv, V).
fib(0, 0). fib(1, 1). fib(C, N) :- fib(2, [0,1], C, N). % Generate from 3rd sequence on
% The benford law calculated benford(D, Val) :- Val is log10(1+1/D).
% Retrieves the first characters of the first 1000 fibonacci numbers % (excluding zero) firstchar(V) :- fib(C,N), N =\= 0, atom_chars(N, [Ch|_]), number_chars(V, [Ch]), (C>999-> !; true).
% Increment the n'th list item (1 based), result -> third argument. incNth(1, [Dh|Dt], [Ch|Dt]) :- !, succ(Dh, Ch). incNth(H, [Dh|Dt], [Dh|Ct]) :- succ(Hn, H), !, incNth(Hn, Dt, Ct).
% Calculate the frequency of the all the list items freq([], D, D). freq([H|T], D, C) :- incNth(H, D, L), !, freq(T, L, C).
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).</lang>
- Output:
?- go. Benford - Measured 30.10% - 30.10% 17.61% - 17.70% 12.49% - 12.50% 9.69% - 9.60% 7.92% - 8.00% 6.69% - 6.70% 5.80% - 5.60% 5.12% - 5.30% 4.58% - 4.50%
PureBasic
<lang 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()</lang>
- 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 <lang python>from __future__ import division from itertools import islice, count from collections import Counter from math import log10 from random import randint
expected = [log10(1+1/d) for d in range(1,10)]
def fib():
a,b = 1,1 while True: yield a a,b = b,a+b
- powers of 3 as a test sequence
def power_of_threes():
return (3**k for k in count(0))
def heads(s):
for a in s: yield int(str(a)[0])
def show_dist(title, s):
c = Counter(s) size = sum(c.values()) res = [c[d]/size for d in range(1,10)]
print("\n%s Benfords deviation" % title) for r, e in zip(res, expected): print("%5.1f%% %5.1f%% %5.1f%%" % (r*100., e*100., abs(r - e)*100.))
def rand1000():
while True: yield randint(1,9999)
if __name__ == '__main__':
show_dist("fibbed", islice(heads(fib()), 1000)) show_dist("threes", islice(heads(power_of_threes()), 1000))
# just to show that not all kind-of-random sets behave like that show_dist("random", islice(heads(rand1000()), 10000))</lang>
- Output:
fibbed Benfords deviation 30.1% 30.1% 0.0% 17.7% 17.6% 0.1% 12.5% 12.5% 0.0% 9.6% 9.7% 0.1% 8.0% 7.9% 0.1% 6.7% 6.7% 0.0% 5.6% 5.8% 0.2% 5.3% 5.1% 0.2% 4.5% 4.6% 0.1% threes Benfords deviation 30.0% 30.1% 0.1% 17.7% 17.6% 0.1% 12.3% 12.5% 0.2% 9.8% 9.7% 0.1% 7.9% 7.9% 0.0% 6.6% 6.7% 0.1% 5.9% 5.8% 0.1% 5.2% 5.1% 0.1% 4.6% 4.6% 0.0% random Benfords deviation 11.2% 30.1% 18.9% 10.9% 17.6% 6.7% 11.6% 12.5% 0.9% 11.1% 9.7% 1.4% 11.6% 7.9% 3.7% 11.4% 6.7% 4.7% 10.3% 5.8% 4.5% 11.0% 5.1% 5.9% 10.9% 4.6% 6.3%
R
<lang R> pbenford <- function(d){
return(log10(1+(1/d)))
}
get_lead_digit <- function(number){
return(as.numeric(substr(number,1,1)))
}
fib_iter <- function(n){
first <- 1 second <- 0 for(i in 1:n){ sum <- first + second first <- second second <- sum } return(sum)
}
fib_sequence <- mapply(fib_iter,c(1:1000)) lead_digits <- mapply(get_lead_digit,fib_sequence)
observed_frequencies <- table(lead_digits)/1000 expected_frequencies <- mapply(pbenford,c(1:9))
data <- data.frame(observed_frequencies,expected_frequencies) colnames(data) <- c("digit","obs.frequency","exp.frequency") dev_percentage <- abs((data$obs.frequency-data$exp.frequency)*100) data <- data.frame(data,dev_percentage)
print(data) </lang>
- Output:
digit obs.frequency exp.frequency dev_percentage 1 0.301 0.30103 0.003000 2 0.177 0.17609 0.090874 3 0.125 0.12494 0.006126 4 0.096 0.09691 0.091001 5 0.080 0.07918 0.081875 6 0.067 0.06695 0.005321 7 0.056 0.05799 0.199195 8 0.053 0.05115 0.184748 9 0.045 0.04576 0.075749
Racket
<lang Racket>#lang racket
(define (log10 n) (/ (log n) (log 10)))
(define (first-digit n)
(quotient n (expt 10 (inexact->exact (floor (log10 n))))))
(define N 10000)
(define fibs
(let loop ([n N] [a 0] [b 1]) (if (zero? n) '() (cons b (loop (sub1 n) b (+ a b))))))
(define v (make-vector 10 0)) (for ([n fibs])
(define f (first-digit n)) (vector-set! v f (add1 (vector-ref v f))))
(printf "N OBS EXP\n") (define (pct n) (~r (* n 100.0) #:precision 1 #:min-width 4)) (for ([i (in-range 1 10)])
(printf "~a: ~a% ~a%\n" i (pct (/ (vector-ref v i) N)) (pct (log10 (+ 1 (/ i))))))
- Output
- N OBS EXP
- 1
- 30.1% 30.1%
- 2
- 17.6% 17.6%
- 3
- 12.5% 12.5%
- 4
- 9.7% 9.7%
- 5
- 7.9% 7.9%
- 6
- 6.7% 6.7%
- 7
- 5.8% 5.8%
- 8
- 5.1% 5.1%
- 9
- 4.6% 4.6%</lang>
REXX
The REXX language (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. <lang rexx>/*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
/*──────────────────────────────────────────────────────────────────────────────────────*/ show: say; say pad ' digit ' pad center("observed",w1) pad center('expected',w2)
say pad '───────' pad center("", w1, '─') pad center("",w2,'─') pad arg(1) !.=0; do j=1 for N; _= left(@.j, 1); !._= !._ + 1 /*get the 1st digit.*/ end /*j*/ do f=1 for 9; say pad center(f,7) pad center(format(!.f/N,,length(N-2)),w1) #.f end /*k*/ return</lang>
- 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
=={{header|Ring}}== <lang 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 </lang> Output: <pre> 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
Ruby
<lang ruby>EXPECTED = (1..9).map{|d| Math.log10(1+1.0/d)}
def fib(n)
a,b = 0,1 n.times.map{ret, a, b = a, b, a+b; ret}
end
- powers of 3 as a test sequence
def power_of_threes(n)
n.times.map{|k| 3**k}
end
def heads(s)
s.map{|a| a.to_s[0].to_i}
end
def show_dist(title, s)
s = heads(s) c = Array.new(10, 0) s.each{|x| c[x] += 1} size = s.size.to_f res = (1..9).map{|d| c[d]/size} puts "\n %s Benfords deviation" % title res.zip(EXPECTED).each.with_index(1) do |(r, e), i| puts "%2d: %5.1f%% %5.1f%% %5.1f%%" % [i, r*100, e*100, (r - e).abs*100] end
end
def random(n)
n.times.map{rand(1..n)}
end
show_dist("fibbed", fib(1000)) show_dist("threes", power_of_threes(1000))
- just to show that not all kind-of-random sets behave like that
show_dist("random", random(10000))</lang>
- Output:
fibbed Benfords deviation 1: 30.1% 30.1% 0.0% 2: 17.7% 17.6% 0.1% 3: 12.5% 12.5% 0.0% 4: 9.5% 9.7% 0.2% 5: 8.0% 7.9% 0.1% 6: 6.7% 6.7% 0.0% 7: 5.6% 5.8% 0.2% 8: 5.3% 5.1% 0.2% 9: 4.5% 4.6% 0.1% threes Benfords deviation 1: 30.0% 30.1% 0.1% 2: 17.7% 17.6% 0.1% 3: 12.3% 12.5% 0.2% 4: 9.8% 9.7% 0.1% 5: 7.9% 7.9% 0.0% 6: 6.6% 6.7% 0.1% 7: 5.9% 5.8% 0.1% 8: 5.2% 5.1% 0.1% 9: 4.6% 4.6% 0.0% random Benfords deviation 1: 10.9% 30.1% 19.2% 2: 10.9% 17.6% 6.7% 3: 11.7% 12.5% 0.8% 4: 10.8% 9.7% 1.1% 5: 11.2% 7.9% 3.3% 6: 11.9% 6.7% 5.2% 7: 10.7% 5.8% 4.9% 8: 11.1% 5.1% 6.0% 9: 10.8% 4.6% 6.2%
Run BASIC
<lang runbasic> N = 1000 for i = 0 to N - 1
n$ = str$(fibonacci(i)) j = val(left$(n$,1)) actual(j) = actual(j) +1
next print
html "
"for i = 1 to 9
html ""next
html "Digit | Actual | Expected |
";i;" | ";using("##.###",actual(i)/10);" | ";using("##.###", frequency(i)*100);" |
"
end
function frequency(n)
frequency = log10(n+1) - log10(n)
end function
function log10(n)
log10 = log(n) / log(10)
end function
function fibonacci(n)
b = 1 for i = 1 to n temp = fibonacci + b fibonacci = b b = temp next i
end function </lang>
Digit | Actual | Expected |
1 | 30.100 | 30.103 |
2 | 17.700 | 17.609 |
3 | 12.500 | 12.494 |
4 | 9.500 | 9.691 |
5 | 8.000 | 7.918 |
6 | 6.700 | 6.695 |
7 | 5.600 | 5.799 |
8 | 5.300 | 5.115 |
9 | 4.500 | 4.576 |
Rust
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.
<lang rust> 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);
let s = x.to_str_radix(10);
// 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()); }
} </lang>
- 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
<lang scala>// Fibonacci Sequence (begining with 1,1): 1 1 2 3 5 8 13 21 34 55 ... val fibs : Stream[BigInt] = { def series(i:BigInt,j:BigInt):Stream[BigInt] = i #:: series(j, i+j); series(1,0).tail.tail }
/**
* Given a numeric sequence, return the distribution of the most-signicant-digit * as expected by Benford's Law and then by actual distribution. */
def benford[N:Numeric]( data:Seq[N] ) : Map[Int,(Double,Double)] = {
import scala.math._ val maxSize = 10000000 // An arbitrary size to avoid problems with endless streams val size = (data.take(maxSize)).size.toDouble val distribution = data.take(maxSize).groupBy(_.toString.head.toString.toInt).map{ case (d,l) => (d -> l.size) } (for( i <- (1 to 9) ) yield { (i -> (log10(1D + 1D / i), (distribution(i) / size))) }).toMap
}
{
println( "Fibonacci Sequence (size=1000): 1 1 2 3 5 8 13 21 34 55 ...\n" ) println( "%9s %9s %9s".format( "Actual", "Expected", "Deviation" ) )
benford( fibs.take(1000) ).toList.sorted foreach { case (k, v) => println( "%d: %5.2f%% | %5.2f%% | %5.4f%%".format(k,v._2*100,v._1*100,math.abs(v._2-v._1)*100) ) }
}</lang>
- Output:
Fibonacci Sequence (size=1000): 1 1 2 3 5 8 13 21 34 55 ... Actual Expected Deviation 1: 30.10% | 30.10% | 0.0030% 2: 17.70% | 17.61% | 0.0909% 3: 12.50% | 12.49% | 0.0061% 4: 9.60% | 9.69% | 0.0910% 5: 8.00% | 7.92% | 0.0819% 6: 6.70% | 6.69% | 0.0053% 7: 5.60% | 5.80% | 0.1992% 8: 5.30% | 5.12% | 0.1847% 9: 4.50% | 4.58% | 0.0757%
Sidef
<lang ruby>var (actuals, expected) = ([], []) var fibonacci = 1000.of {|i| fib(i).digit(0) }
for i (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 (1..9) {
"%d : %11s %%%15s %%\n".printf( i, "%.2f".sprintf(100 * actuals[i - 1]), "%.2f".sprintf(100 * expected[i - 1]), )
}</lang>
- Output:
Observed Expected 1 : 30.10 % 30.10 % 2 : 17.70 % 17.61 % 3 : 12.50 % 12.49 % 4 : 9.50 % 9.69 % 5 : 8.00 % 7.92 % 6 : 6.70 % 6.69 % 7 : 5.60 % 5.80 % 8 : 5.30 % 5.12 % 9 : 4.50 % 4.58 %
SQL
If we load some numbers into a table, we can do the sums without too much difficulty. I tried to make this as database-neutral as possible, but I only had Oracle handy to test it on.
The query is the same for any number sequence you care to put in the benford table.
<lang SQL>-- Create table create table benford (num integer);
-- Seed table insert into benford (num) values (1); insert into benford (num) values (1); insert into benford (num) values (2);
-- Populate table insert into benford (num)
select ult + penult from (select max(num) as ult from benford), (select max(num) as penult from benford where num not in (select max(num) from benford))
-- Repeat as many times as desired -- in Oracle SQL*Plus, press "Slash, Enter" a lot of times -- or wrap this in a loop, but that will require something db-specific...
-- Do sums select
digit, count(digit) / numbers as actual, log(10, 1 + 1 / digit) as expected
from
( select floor(num/power(10,length(num)-1)) as digit from benford ), ( select count(*) as numbers from benford )
group by digit, numbers order by digit;
-- Tidy up drop table benford;</lang>
- Output:
I only loaded the first 100 Fibonacci numbers before my fingers were sore from repeating the data load. 8~)
DIGIT ACTUAL EXPECTED ---------- ---------- ---------- 1 .3 .301029996 2 .18 .176091259 3 .13 .124938737 4 .09 .096910013 5 .08 .079181246 6 .06 .06694679 7 .05 .057991947 8 .07 .051152522 9 .04 .045757491 9 rows selected.
Stata
<lang 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 | +-----------------------------+</lang>
Assuming the data are random, one can also do a goodness of fit chi-square test:
<lang stata>// chi-square statistic chisq=sum((f-p):^2:/p) chisq
.2219340262
// p-value chi2tail(8,chisq)
.9999942179
end</lang>
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:
<lang stata>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</lang>
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
<lang Swift>import Foundation
/* Reads from a file and returns the content as a String */ func readFromFile(fileName file:String) -> String{
var ret:String = "" let path = Foundation.URL(string: "file://"+file) do { ret = try String(contentsOf: path!, encoding: String.Encoding.utf8) } catch { print("Could not read from file!") 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 content = readFromFile(fileName: pathToFile) 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))
}</lang>
- 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
<lang tcl>proc benfordTest {numbers} {
# Count the leading digits (RE matches first digit in each number, # even if negative) set accum {1 0 2 0 3 0 4 0 5 0 6 0 7 0 8 0 9 0} foreach n $numbers {
if {[regexp {[1-9]} $n digit]} { dict incr accum $digit }
}
# Print the report puts " digit | measured | theory" puts "-------+----------+--------" dict for {digit count} $accum {
puts [format "%6d | %7.2f%% | %5.2f%%" $digit \ [expr {$count * 100.0 / [llength $numbers]}] \ [expr {log(1+1./$digit)/log(10)*100.0}]]
}
}</lang> Demonstrating with Fibonacci numbers: <lang tcl>proc fibs n {
for {set a 1;set b [set i 0]} {$i < $n} {incr i} {
lappend result [set b [expr {$a + [set a $b]}]]
} return $result
} benfordTest [fibs 1000]</lang>
- Output:
digit | measured | theory -------+----------+-------- 1 | 30.10% | 30.10% 2 | 17.70% | 17.61% 3 | 12.50% | 12.49% 4 | 9.60% | 9.69% 5 | 8.00% | 7.92% 6 | 6.70% | 6.69% 7 | 5.60% | 5.80% 8 | 5.30% | 5.12% 9 | 4.50% | 4.58%
Visual FoxPro
<lang vfp>
- 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 </lang>
- 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
zkl
<lang zkl>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())) }
}</lang>
<lang zkl>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])); }</lang>
- Output:
First 1000 Fibonacci numbers Digit Observed Predicted 1 0.301 0.301 2 0.177 0.176 3 0.125 0.125 4 0.096 0.097 5 0.080 0.079 6 0.067 0.067 7 0.056 0.058 8 0.053 0.051 9 0.045 0.046
ZX Spectrum Basic
<lang zxbasic>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 </lang> The results obtained are adjusted fairly well, except for the number 8. This occurs with Sinclair BASIC, Sam BASIC and SpecBAS fits.
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