Distribution of 0 digits in factorial series: Difference between revisions
Distribution of 0 digits in factorial series (view source)
Revision as of 12:08, 8 February 2024
, 3 months agoAdd C# implementation
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{{
Large Factorials and the Distribution of '0' in base 10 digits.
Line 41:
permanently falls below 0.16. This task took many hours in the Python example, though I wonder if there is a faster
algorithm out there.
=={{header|11l}}==
{{trans|Python}}
<syntaxhighlight lang="11l">F facpropzeros(n, verbose = 1B)
V proportions = [0.0] * n
V (fac, psum) = (BigInt(1), 0.0)
L(i) 0 .< n
fac *= i + 1
V d = String(fac)
psum += sum(d.map(x -> Int(x == ‘0’))) / Float(d.len)
proportions[i] = psum / (i + 1)
I verbose
print(‘The mean proportion of 0 in factorials from 1 to #. is #..’.format(n, psum / n))
R proportions
L(n) [100, 1000, 10000]
facpropzeros(n)</syntaxhighlight>
{{out}}
<pre>
The mean proportion of 0 in factorials from 1 to 100 is 0.246753186.
The mean proportion of 0 in factorials from 1 to 1000 is 0.203544551.
The mean proportion of 0 in factorials from 1 to 10000 is 0.173003848.
</pre>
=== Base 1000 version ===
<syntaxhighlight lang="11l">F zinit()
V zc = [0] * 999
L(x) 1..9
zc[x - 1] = 2
zc[10 * x - 1] = 2
zc[100 * x - 1] = 2
L(y) (10.<100).step(10)
zc[y + x - 1] = 1
zc[10 * y + x - 1] = 1
zc[10 * (y + x) - 1] = 1
R zc
F meanfactorialdigits()
V zc = zinit()
V rfs = [1]
V (total, trail, first) = (0.0, 1, 0)
L(f) 2 .< 50000
V (carry, d999, zeroes) = (0, 0, (trail - 1) * 3)
V (j, l) = (trail, rfs.len)
L j <= l | carry != 0
I j <= l
carry = rfs[j - 1] * f + carry
d999 = carry % 1000
I j <= l
rfs[j - 1] = d999
E
rfs.append(d999)
zeroes += I d999 == 0 {3} E zc[d999 - 1]
carry I/= 1000
j++
L rfs[trail - 1] == 0
trail++
d999 = rfs.last
d999 = I d999 >= 100 {0} E I d999 < 10 {2} E 1
zeroes -= d999
V digits = rfs.len * 3 - d999
total += Float(zeroes) / digits
V ratio = total / f
I f C [100, 1000, 10000]
print(‘The mean proportion of zero digits in factorials to #. is #.’.format(f, ratio))
I ratio >= 0.16
first = 0
E I first == 0
first = f
print(‘The mean proportion dips permanently below 0.16 at ’first‘.’)
meanfactorialdigits()</syntaxhighlight>
{{out}}
<pre>
The mean proportion of zero digits in factorials to 100 is 0.246753186
The mean proportion of zero digits in factorials to 1000 is 0.203544551
The mean proportion of zero digits in factorials to 10000 is 0.173003848
The mean proportion dips permanently below 0.16 at 47332.
</pre>
=={{header|Arturo}}==
<syntaxhighlight lang="arturo">su: 0.0
f: 1
lim: 100
loop 1..10000 'n [
'f * n
str: to :string f
'su + (enumerate str 'c -> c = `0`) // size str
if n = lim [
print [n ":" su // n]
'lim * 10
]
]</syntaxhighlight>
{{out}}
<pre>100 : 0.2467531861674322
1000 : 0.2035445511031646
10000 : 0.1730038482418671</pre>
=={{header|C#}}==
{{trans|Java}}
<syntaxhighlight lang="C#">
using System;
using System.Collections.Generic;
using System.Numerics;
public class DistributionInFactorials
{
public static void Main(string[] args)
{
List<int> limits = new List<int> { 100, 1_000, 10_000 };
foreach (int limit in limits)
{
MeanFactorialDigits(limit);
}
}
private static void MeanFactorialDigits(int limit)
{
BigInteger factorial = BigInteger.One;
double proportionSum = 0.0;
double proportionMean = 0.0;
for (int n = 1; n <= limit; n++)
{
factorial = factorial * n;
string factorialString = factorial.ToString();
int digitCount = factorialString.Length;
long zeroCount = factorialString.Split('0').Length - 1;
proportionSum += (double)zeroCount / digitCount;
proportionMean = proportionSum / n;
}
string result = string.Format("{0:F8}", proportionMean);
Console.WriteLine("Mean proportion of zero digits in factorials from 1 to " + limit + " is " + result);
}
}
</syntaxhighlight>
{{out}}
<pre>
Mean proportion of zero digits in factorials from 1 to 100 is 0.24675319
Mean proportion of zero digits in factorials from 1 to 1000 is 0.20354455
Mean proportion of zero digits in factorials from 1 to 10000 is 0.17300385
</pre>
=={{header|C++}}==
{{trans|Phix}}
<syntaxhighlight lang="cpp">#include <array>
#include <chrono>
#include <iomanip>
#include <iostream>
#include <vector>
auto init_zc() {
std::array<int, 1000> zc;
zc.fill(0);
zc[0] = 3;
for (int x = 1; x <= 9; ++x) {
zc[x] = 2;
zc[10 * x] = 2;
zc[100 * x] = 2;
for (int y = 10; y <= 90; y += 10) {
zc[y + x] = 1;
zc[10 * y + x] = 1;
zc[10 * (y + x)] = 1;
}
}
return zc;
}
template <typename clock_type>
auto elapsed(const std::chrono::time_point<clock_type>& t0) {
auto t1 = clock_type::now();
auto duration =
std::chrono::duration_cast<std::chrono::milliseconds>(t1 - t0);
return duration.count();
}
int main() {
auto zc = init_zc();
auto t0 = std::chrono::high_resolution_clock::now();
int trail = 1, first = 0;
double total = 0;
std::vector<int> rfs{1};
std::cout << std::fixed << std::setprecision(10);
for (int f = 2; f <= 50000; ++f) {
int carry = 0, d999, zeroes = (trail - 1) * 3, len = rfs.size();
for (int j = trail - 1; j < len || carry != 0; ++j) {
if (j < len)
carry += rfs[j] * f;
d999 = carry % 1000;
if (j < len)
rfs[j] = d999;
else
rfs.push_back(d999);
zeroes += zc[d999];
carry /= 1000;
}
while (rfs[trail - 1] == 0)
++trail;
d999 = rfs.back();
d999 = d999 < 100 ? (d999 < 10 ? 2 : 1) : 0;
zeroes -= d999;
int digits = rfs.size() * 3 - d999;
total += double(zeroes) / digits;
double ratio = total / f;
if (ratio >= 0.16)
first = 0;
else if (first == 0)
first = f;
if (f == 100 || f == 1000 || f == 10000) {
std::cout << "Mean proportion of zero digits in factorials to " << f
<< " is " << ratio << ". (" << elapsed(t0) << "ms)\n";
}
}
std::cout << "The mean proportion dips permanently below 0.16 at " << first
<< ". (" << elapsed(t0) << "ms)\n";
}</syntaxhighlight>
{{out}}
<pre>
Mean proportion of zero digits in factorials to 100 is 0.2467531862. (0ms)
Mean proportion of zero digits in factorials to 1000 is 0.2035445511. (1ms)
Mean proportion of zero digits in factorials to 10000 is 0.1730038482. (152ms)
The mean proportion dips permanently below 0.16 at 47332. (4598ms)
</pre>
=={{header|Go}}==
Line 47 ⟶ 285:
{{libheader|Go-rcu}}
Timings here are 2.8 seconds for the basic task and 182.5 seconds for the stretch goal.
<
import (
Line 87 ⟶ 325:
fmt.Println(" (stays below 0.16 after this)")
fmt.Printf("%6s = %12.10f\n", "50,000", sum / 50000)
}</
{{out}}
Line 102 ⟶ 340:
{{trans|Phix}}
Much quicker than before with 10,000 now being reached in 0.35 seconds and the stretch goal in about 5.5 seconds.
<
import (
Line 189 ⟶ 427:
fmt.Println(" (stays below 0.16 after this)")
fmt.Printf("%6s = %12.10f\n", "50,000", total/50000)
}</
{{out}}
Line 196 ⟶ 434:
</pre>
=={{header|Java}}==
<syntaxhighlight lang="java">
import java.math.BigInteger;
import java.util.List;
public final class DistributionInFactorials {
public static void main(String[] aArgs) {
List<Integer> limits = List.of( 100, 1_000, 10_000 );
for ( Integer limit : limits ) {
meanFactorialDigits(limit);
}
}
private static void meanFactorialDigits(Integer aLimit) {
BigInteger factorial = BigInteger.ONE;
double proportionSum = 0.0;
double proportionMean = 0.0;
for ( int n = 1; n <= aLimit; n++ ) {
factorial = factorial.multiply(BigInteger.valueOf(n));
String factorialString = factorial.toString();
int digitCount = factorialString.length();
long zeroCount = factorialString.chars().filter( ch -> ch == '0' ).count();
proportionSum += (double) zeroCount / digitCount;
proportionMean = proportionSum / n;
}
String result = String.format("%.8f", proportionMean);
System.out.println("Mean proportion of zero digits in factorials from 1 to " + aLimit + " is " + result);
}
}
</syntaxhighlight>
{{ out }}
<pre>
Mean proportion of zero digits in factorials from 1 to 100 is 0.24675319
Mean proportion of zero digits in factorials from 1 to 1000 is 0.20354455
Mean proportion of zero digits in factorials from 1 to 10000 is 0.17300385
</pre>
=={{header|jq}}==
'''Works with jq'''
The precision of jq's integer arithmetic is not up to this task, so in the following we borrow from the "BigInt" library and use a string representation of integers.
Unfortunately, although gojq (the Go implementation of jq) does support unbounded-precision integer arithmetic, it is unsuited for the task because of memory management issues.
'''From BigInt.jq'''
<syntaxhighlight lang="jq">
# multiply two decimal strings, which may be signed (+ or -)
def long_multiply(num1; num2):
def stripsign:
.[0:1] as $a
| if $a == "-" then [ -1, .[1:]]
elif $a == "+" then [ 1, .[1:]]
else [1, .]
end;
def adjustsign(sign):
if sign == 1 then . else "-" + . end;
# mult/2 assumes neither argument has a sign
def mult(num1;num2):
(num1 | explode | map(.-48) | reverse) as $a1
| (num2 | explode | map(.-48) | reverse) as $a2
| reduce range(0; num1|length) as $i1
([]; # result
reduce range(0; num2|length) as $i2
(.;
($i1 + $i2) as $ix
| ( $a1[$i1] * $a2[$i2] + (if $ix >= length then 0 else .[$ix] end) ) as $r
| if $r > 9 # carrying
then
.[$ix + 1] = ($r / 10 | floor) + (if $ix + 1 >= length then 0 else .[$ix + 1] end )
| .[$ix] = $r - ( $r / 10 | floor ) * 10
else
.[$ix] = $r
end
)
)
| reverse | map(.+48) | implode;
(num1|stripsign) as $a1
| (num2|stripsign) as $a2
| if $a1[1] == "0" or $a2[1] == "0" then "0"
elif $a1[1] == "1" then $a2[1]|adjustsign( $a1[0] * $a2[0] )
elif $a2[1] == "1" then $a1[1]|adjustsign( $a1[0] * $a2[0] )
else mult($a1[1]; $a2[1]) | adjustsign( $a1[0] * $a2[0] )
end;
</syntaxhighlight>
'''The task'''
<syntaxhighlight lang="jq">
def count(s): reduce s as $x (0; .+1);
def meanfactorialdigits:
def digits: tostring | explode;
def nzeros: count( .[] | select(. == 48) ); # "0" is 48
. as $N
| 0.16 as $goal
| label $out
| reduce range( 1; 1+$N ) as $i ( {factorial: "1", proportionsum: 0.0, first: null };
.factorial = long_multiply(.factorial; $i|tostring)
| (.factorial|digits) as $d
| .proportionsum += ($d | (nzeros / length))
| (.proportionsum / $i) as $propmean
| if .first
then if $propmean > $goal then .first = null else . end
elif $propmean <= $goal then .first = $i
else .
end)
| "Mean proportion of zero digits in factorials to \($N) is \(.proportionsum/$N);" +
(if .first then " mean <= \($goal) from N=\(.first) on." else " goal (\($goal)) unmet." end);
# The task:
100, 1000, 10000 | meanfactorialdigits</syntaxhighlight>
{{out}}
<pre>
Mean proportion of zero digits in factorials to 100 is 0.24675318616743216; goal (0.16) unmet.
Mean proportion of zero digits in factorials to 1000 is 0.20354455110316458; goal (0.16) unmet.
Mean proportion of zero digits in factorials to 10000 is 0.17300384824186707; goal (0.16) unmet.
</pre>
=={{header|Julia}}==
<
factoril, proportionsum = big"1", 0.0
for i in 1:N
Line 218 ⟶ 581:
@time meanfactorialdigits(50000, 0.16)
</
<pre>
Mean proportion of zero digits in factorials to 100 is 0.24675318616743216
Line 228 ⟶ 591:
</pre>
=== Base 1000 version ===
{{trans|Pascal, Phix}}
<syntaxhighlight lang="julia">function init_zc()
zc = zeros(Int, 999)
for x in 1:9
zc[x] = 2 # 00x
zc[10*x] = 2 # 0x0
zc[100*x] = 2 # x00
for y in 10:10:90
zc[y+x] = 1 # 0yx
zc[10*y+x] = 1 # y0x
zc[10*(y+x)] = 1 # yx0
end
end
return zc
end
function meanfactorialzeros(N = 50000, verbose = true)
zc = init_zc()
rfs = [1]
total, trail, first, firstratio = 0.0, 1, 0, 0.0
for f in 2:N
carry, d999, zeroes = 0, 0, (trail - 1) * 3
j, l = trail, length(rfs)
while j <= l || carry != 0
if j <= l
carry = (rfs[j]) * f + carry
end
d999 = carry % 1000
if j <= l
rfs[j] = d999
else
push!(rfs, d999)
end
zeroes += (d999 == 0) ? 3 : zc[d999]
carry ÷= 1000
j += 1
end
while rfs[trail] == 0
trail += 1
end
# d999 = quick correction for length and zeroes:
d999 = rfs[end]
d999 = d999 < 100 ? d999 < 10 ? 2 : 1 : 0
zeroes -= d999
digits = length(rfs) * 3 - d999
total += zeroes / digits
ratio = total / f
if ratio >= 0.16
first = 0
firstratio = 0.0
elseif first == 0
first = f
firstratio = ratio
end
if f in [100, 1000, 10000]
verbose && println("Mean proportion of zero digits in factorials to $f is $ratio")
end
end
verbose && println("The mean proportion dips permanently below 0.16 at $first.")
end
meanfactorialzeros(100, false)
@time meanfactorialzeros()
</syntaxhighlight>{{out}}
<pre>
Mean proportion of zero digits in factorials to 100 is 0.24675318616743216
Mean proportion of zero digits in factorials to 1000 is 0.20354455110316458
Mean proportion of zero digits in factorials to 10000 is 0.17300384824186707
The mean proportion dips permanently below 0.16 at 47332.
4.638323 seconds (50.08 k allocations: 7.352 MiB)
</pre>
=={{header|Mathematica}}/{{header|Wolfram Language}}==
<syntaxhighlight lang="mathematica">ClearAll[ZeroDigitsFractionFactorial]
ZeroDigitsFractionFactorial[n_Integer] := Module[{m},
m = IntegerDigits[n!];
Count[m, 0]/Length[m]
]
ZeroDigitsFractionFactorial /@ Range[6] // Mean // N
ZeroDigitsFractionFactorial /@ Range[25] // Mean // N
ZeroDigitsFractionFactorial /@ Range[100] // Mean // N
ZeroDigitsFractionFactorial /@ Range[1000] // Mean // N
ZeroDigitsFractionFactorial /@ Range[10000] // Mean // N
fracs = ParallelMap[ZeroDigitsFractionFactorial, Range[50000], Method -> ("ItemsPerEvaluation" -> 100)];
means = Accumulate[N@fracs]/Range[Length[fracs]];
len = LengthWhile[Reverse@means, # < 0.16 &];
50000 - len + 1</syntaxhighlight>
{{out}}
<pre>0.111111
0.267873
0.246753
0.203545
0.173004
47332</pre>
=={{header|Nim}}==
===Task===
{{libheader|bignum}}
<
import bignum
Line 247 ⟶ 707:
lim *= 10
echo()
echo getMonoTime() - t0</
{{out}}
Line 260 ⟶ 720:
At each step, we eliminate the trailing zeroes to reduce the length of the number and save some time. But this is not much, about 8%.
<
import bignum
Line 281 ⟶ 741:
echo "Permanently below 0.16 at n = ", first
echo "Execution time: ", getMonoTime() - t0</
{{out}}
Line 290 ⟶ 750:
The most time consuming is converting to string and search for zeros.<BR>
Therefor I do not convert to string.I divide the base in sections of 3 digits with counting zeros in a lookup table.
<
{$IFDEF FPC} {$MODE DELPHI} {$Optimization ON,ALL} {$ENDIF}
uses
Line 478 ⟶ 938:
inc(i);
writeln('First ratio < 0.16 ', i:8,SumOfRatio[i]/i:20:17);
end.</
{{out}}
<pre> 100 0.246753186167432
Line 486 ⟶ 946:
First ratio < 0.16 47332 0.15999999579985665
Real time: 4.898 s CPU share: 99.55 % // 2.67s on 2200G freepascal 3.2.2</pre>
=={{header|Perl}}==
{{libheader|ntheory}}
<syntaxhighlight lang="perl">use strict;
use warnings;
use ntheory qw/factorial/;
for my $n (100, 1000, 10000) {
my($sum,$f) = 0;
$f = factorial $_ and $sum += ($f =~ tr/0//) / length $f for 1..$n;
printf "%5d: %.5f\n", $n, $sum/$n;
}</syntaxhighlight>
{{out}}
<pre> 100: 0.24675
1000: 0.20354
10000: 0.17300</pre>
=={{header|Phix}}==
Using "string math" to create reversed factorials, for slightly easier skipping of "trailing" zeroes,
but converted to base 1000 and with the zero counting idea from Pascal, which sped it up threefold.
<!--<
<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>
<span style="color: #004080;">sequence</span> <span style="color: #000000;">rfs</span> <span style="color: #0000FF;">=</span> <span style="color: #0000FF;">{</span><span style="color: #000000;">1</span><span style="color: #0000FF;">}</span> <span style="color: #000080;font-style:italic;">-- reverse factorial(1) in base 1000</span>
Line 555 ⟶ 1,029:
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"The mean proportion dips permanently below 0.16 at %d. (%s)\n"</span><span style="color: #0000FF;">,{</span><span style="color: #000000;">first</span><span style="color: #0000FF;">,</span><span style="color: #000000;">e</span><span style="color: #0000FF;">})</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<!--</
{{out}}
<pre>
Line 566 ⟶ 1,040:
=== trailing zeroes only ===
Should you only be interested in the ratio of trailing zeroes, you can do that much faster:
<!--<
<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>
<span style="color: #004080;">atom</span> <span style="color: #000000;">t0</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">time</span><span style="color: #0000FF;">(),</span>
Line 594 ⟶ 1,068:
<span style="color: #004080;">string</span> <span style="color: #000000;">e</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">elapsed</span><span style="color: #0000FF;">(</span><span style="color: #7060A8;">time</span><span style="color: #0000FF;">()-</span><span style="color: #000000;">t0</span><span style="color: #0000FF;">)</span>
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"The mean proportion dips permanently below 0.07 at %d. (%s)\n"</span><span style="color: #0000FF;">,{</span><span style="color: #000000;">first</span><span style="color: #0000FF;">,</span><span style="color: #000000;">e</span><span style="color: #0000FF;">})</span>
<!--</
{{out}}
<pre>
Line 602 ⟶ 1,076:
The mean proportion dips permanently below 0.07 at 31549. (0.1s)
</pre>
=={{header|Python}}==
<
proportions = [0.0] * N
fac, psum = 1, 0.0
Line 626 ⟶ 1,099:
print("The mean proportion dips permanently below 0.16 at {}.".format(n + 2))
</
<pre>
The mean proportion of 0 in factorials from 1 to 100 is 0.24675318616743216.
Line 634 ⟶ 1,107:
</pre>
The means can be plotted, showing a jump from 0 to over 0.25, followed by a slowly dropping curve:
<
plt.plot([i+1 for i in range(len(props))], props)
</syntaxhighlight>
=== Base 1000 version ===
{{trans|Go via Phix via Pascal}}
<syntaxhighlight lang="python">def zinit():
zc = [0] * 999
for x in range(1, 10):
zc[x - 1] = 2 # 00x
zc[10 * x - 1] = 2 # 0x0
zc[100 * x - 1] = 2 # x00
for y in range(10, 100, 10):
zc[y + x - 1] = 1 # 0yx
zc[10 * y + x - 1] = 1 # y0x
zc[10 * (y + x) - 1] = 1 # yx0
return zc
def meanfactorialdigits():
zc = zinit()
rfs = [1]
total, trail, first = 0.0, 1, 0
for f in range(2, 50000):
carry, d999, zeroes = 0, 0, (trail - 1) * 3
j, l = trail, len(rfs)
while j <= l or carry != 0:
if j <= l:
carry = rfs[j-1] * f + carry
d999 = carry % 1000
if j <= l:
rfs[j-1] = d999
else:
rfs.append(d999)
zeroes += 3 if d999 == 0 else zc[d999-1]
carry //= 1000
j += 1
while rfs[trail-1] == 0:
trail += 1
# d999 is a quick correction for length and zeros
d999 = rfs[-1]
d999 = 0 if d999 >= 100 else 2 if d999 < 10 else 1
zeroes -= d999
digits = len(rfs) * 3 - d999
total += zeroes / digits
ratio = total / f
if f in [100, 1000, 10000]:
print("The mean proportion of zero digits in factorials to {} is {}".format(f, ratio))
if ratio >= 0.16:
first = 0
elif first == 0:
first = f
print("The mean proportion dips permanently below 0.16 at {}.".format(first))
import time
TIME0 = time.perf_counter()
meanfactorialdigits()
print("\nTotal time:", time.perf_counter() - TIME0, "seconds.")
</syntaxhighlight>{{out}}
<pre>
The mean proportion of zero digits in factorials to 100 is 0.24675318616743216
The mean proportion of zero digits in factorials to 1000 is 0.20354455110316458
The mean proportion of zero digits in factorials to 10000 is 0.17300384824186707
The mean proportion dips permanently below 0.16 at 47332.
Total time: 648.3583232999999 seconds.
</pre>
=={{header|Raku}}==
Works, but depressingly slow for 10000.
<syntaxhighlight lang="raku"
sink 10000!; # prime the iterator to allow multithreading
sub zs ($n) { ( constant zero-share = (^Inf).race(:32batch).map: { (.!.comb.Bag){'0'} / .!.chars } )[$n+1] }
.say for (
Line 650 ⟶ 1,194:
,1000
,10000
).map: -> \n { "{n}: {([+] (^n).map: *.&zs) / n}" }</
{{out}}
<pre>100: 0.
1000: 0.
10000: 0.
</pre>
=={{header|REXX}}==
<
parse arg $ /*obtain optional arguments from the CL*/
if $='' | $="," then $= 100 1000 10000 /*not specified? Then use the default.*/
Line 688 ⟶ 1,231:
do k=1 for z; != ! * k; y= y + countstr(0, !) / length(!)
end /*k*/
return y/z</
{{out|output|text= when using the default inputs:}}
<pre>
Line 697 ⟶ 1,240:
10,000 │ 0.1730038482418660531800366428930706156810278809057883361518852958446868172...
───────────┴────────────────────────────────────────────────────────────────────────────────
</pre>
=={{header|Rust}}==
{{trans|Phix}}
<syntaxhighlight lang="rust">fn init_zc() -> Vec<usize> {
let mut zc = vec![0; 1000];
zc[0] = 3;
for x in 1..=9 {
zc[x] = 2;
zc[10 * x] = 2;
zc[100 * x] = 2;
let mut y = 10;
while y <= 90 {
zc[y + x] = 1;
zc[10 * y + x] = 1;
zc[10 * (y + x)] = 1;
y += 10;
}
}
zc
}
fn main() {
use std::time::Instant;
let zc = init_zc();
let t0 = Instant::now();
let mut trail = 1;
let mut first = 0;
let mut total: f64 = 0.0;
let mut rfs = vec![1];
for f in 2..=50000 {
let mut carry = 0;
let mut d999: usize;
let mut zeroes = (trail - 1) * 3;
let len = rfs.len();
let mut j = trail - 1;
while j < len || carry != 0 {
if j < len {
carry += rfs[j] * f;
}
d999 = carry % 1000;
if j < len {
rfs[j] = d999;
} else {
rfs.push(d999);
}
zeroes += zc[d999];
carry /= 1000;
j += 1;
}
while rfs[trail - 1] == 0 {
trail += 1;
}
d999 = rfs[rfs.len() - 1];
d999 = if d999 < 100 {
if d999 < 10 {
2
} else {
1
}
} else {
0
};
zeroes -= d999;
let digits = rfs.len() * 3 - d999;
total += (zeroes as f64) / (digits as f64);
let ratio = total / (f as f64);
if ratio >= 0.16 {
first = 0;
} else if first == 0 {
first = f;
}
if f == 100 || f == 1000 || f == 10000 {
let duration = t0.elapsed();
println!(
"Mean proportion of zero digits in factorials to {} is {:.10}. ({}ms)",
f,
ratio,
duration.as_millis()
);
}
}
let duration = t0.elapsed();
println!(
"The mean proportion dips permanently below 0.16 at {}. ({}ms)",
first,
duration.as_millis()
);
}</syntaxhighlight>
{{out}}
<pre>
Mean proportion of zero digits in factorials to 100 is 0.2467531862. (0ms)
Mean proportion of zero digits in factorials to 1000 is 0.2035445511. (1ms)
Mean proportion of zero digits in factorials to 10000 is 0.1730038482. (149ms)
The mean proportion dips permanently below 0.16 at 47332. (4485ms)
</pre>
=={{header|Ruby}}==
<syntaxhighlight lang="ruby">[100, 1000, 10_000].each do |n|
v = 1
total_proportion = (1..n).sum do |k|
v *= k
digits = v.digits
Rational(digits.count(0), digits.size)
end
puts "The mean proportion of 0 in factorials from 1 to #{n} is #{(total_proportion/n).to_f}."
end</syntaxhighlight>
{{out}}
<pre>The mean proportion of 0 in factorials from 1 to 100 is 0.24675318616743222.
The mean proportion of 0 in factorials from 1 to 1000 is 0.20354455110316463.
The mean proportion of 0 in factorials from 1 to 10000 is 0.17300384824186604.
</pre>
=={{header|Sidef}}==
<syntaxhighlight lang="ruby">func mean_factorial_digits(n, d = 0) {
var v = 1
var total = 0.float
for k in (1..n) {
v *= k
total += v.digits.count(d)/v.len
}
total / n
}
say mean_factorial_digits(100)
say mean_factorial_digits(1000)
say mean_factorial_digits(10000)</syntaxhighlight>
{{out}}
<pre>
0.246753186167432217778415887197352699112940703327
0.203544551103164635640043803171145530298574116789
0.173003848241866053180036642893070615681027880906
</pre>
Line 704 ⟶ 1,381:
{{libheader|Wren-fmt}}
Very slow indeed, 10.75 minutes to reach N = 10,000.
<
import "./fmt" for Fmt
var fact = BigInt.one
Line 719 ⟶ 1,396:
Fmt.print("$,6d = $12.10f", n, sum / n)
}
}</
{{out}}
Line 732 ⟶ 1,409:
{{trans|Phix}}
Around 60 times faster than before with 10,000 now being reached in about 10.5 seconds. Even the stretch goal is now viable and comes in at 5 minutes 41 seconds.
<
var rfs = [1] // reverse factorial(1) in base 1000
Line 797 ⟶ 1,474:
Fmt.write("$,6d = $12.10f", first, firstRatio)
System.print(" (stays below 0.16 after this)")
Fmt.print("$,6d = $12.10f", 50000, total/50000)</
{{out}}
| |||