Distribution of 0 digits in factorial series
(Redirected from Distribution of 0 Digits in factorial series)
You are encouraged to solve this task according to the task description, using any language you may know.
Large Factorials and the Distribution of '0' in base 10 digits.
- About the task
We can see that some features of factorial numbers (the series of numbers 1!, 2!, 3!, ...) come about because such numbers are the product of a series of counting numbers, and so those products have predictable factors. For example, all factorials above 1! are even numbers, since they have 2 as a factor. Similarly, all factorials from 5! up end in a 0, because they have 5 and 2 as factors, and thus have 10 as a factor. In fact, the factorial integers add another 0 at the end of the factorial for every step of 5 upward: 5! = 120, 10! = 3628800, 15! = 1307674368000, 16! = 20922789888000 and so on.
Because factorial numbers, which quickly become quite large, continue to have another terminal 0 on the right hand side of the number for every factor of 5 added to the factorial product, one might think that the proportion of zeros in a base 10 factorial number might be close to 1/5. However, though the factorial products add another terminating 0 every factor of 5 multiplied into the product, as the numbers become quite large, the number of digits in the factorial product expands exponentially, and so the number above the terminating zeros tends toward 10% of each digit from 0 to 1 as the factorial becomes larger. Thus, as the factorials become larger, the proportion of 0 digits in the factorial products shifts slowly from around 1/5 toward 1/10, since the number of terminating zeros in n! increases only in proportion to n, whereas the number of digits of n! in base 10 increases exponentially.
- The task
Create a function to calculate the mean of the proportions of 0 digits out of the total digits found in each factorial product from 1! to N!. This proportion of 0 digits in base 10 should be calculated using the number as printed as a base 10 integer.
Example: for 1 to 6 we have 1!, 2!, 3!, 4!, 5!, 6!, or (1, 2, 6, 24, 120, 720), so we need the mean of (0/1, 0/1, 0/1, 0/2, 1/3, 1/3) = (2/3) (totals of each proportion) / 6 (= N), or 0.1111111...
Example: for 1 to 25 the mean of the proportions of 0 digits in the factorial products series of N! with N from 1 to 25 is 0.26787.
Do this task for 1 to N where N is in (100, 1000, and 10000), so, compute the mean of the proportion of 0 digits for each product in the series of each of the factorials from 1 to 100, 1 to 1000, and 1 to 10000.
- Stretch task
Find the N in 10000 < N < 50000 where the mean of the proportions of 0 digits in the factorial products from 1 to N 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.
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)- Output:
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.
Base 1000 version
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()- Output:
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.
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
]
]
- Output:
100 : 0.2467531861674322 1000 : 0.2035445511031646 10000 : 0.1730038482418671
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);
}
}
- Output:
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
C++
#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";
}
- Output:
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)
Go
Brute force
Timings here are 2.8 seconds for the basic task and 182.5 seconds for the stretch goal.
package main
import (
"fmt"
big "github.com/ncw/gmp"
"rcu"
)
func main() {
fact := big.NewInt(1)
sum := 0.0
first := int64(0)
firstRatio := 0.0
fmt.Println("The mean proportion of zero digits in factorials up to the following are:")
for n := int64(1); n <= 50000; n++ {
fact.Mul(fact, big.NewInt(n))
bytes := []byte(fact.String())
digits := len(bytes)
zeros := 0
for _, b := range bytes {
if b == '0' {
zeros++
}
}
sum += float64(zeros)/float64(digits)
ratio := sum / float64(n)
if n == 100 || n == 1000 || n == 10000 {
fmt.Printf("%6s = %12.10f\n", rcu.Commatize(int(n)), ratio)
}
if first > 0 && ratio >= 0.16 {
first = 0
firstRatio = 0.0
} else if first == 0 && ratio < 0.16 {
first = n
firstRatio = ratio
}
}
fmt.Printf("%6s = %12.10f", rcu.Commatize(int(first)), firstRatio)
fmt.Println(" (stays below 0.16 after this)")
fmt.Printf("%6s = %12.10f\n", "50,000", sum / 50000)
}
- Output:
The mean proportion of zero digits in factorials up to the following are: 100 = 0.2467531862 1,000 = 0.2035445511 10,000 = 0.1730038482 47,332 = 0.1599999958 (stays below 0.16 after this) 50,000 = 0.1596200546
'String math' and base 1000
Much quicker than before with 10,000 now being reached in 0.35 seconds and the stretch goal in about 5.5 seconds.
package main
import (
"fmt"
"rcu"
)
var rfs = []int{1} // reverse factorial(1) in base 1000
var zc = make([]int, 999)
func init() {
for x := 1; x <= 9; x++ {
zc[x-1] = 2 // 00x
zc[10*x-1] = 2 // 0x0
zc[100*x-1] = 2 // x00
var y = 10
for y <= 90 {
zc[y+x-1] = 1 // 0yx
zc[10*y+x-1] = 1 // y0x
zc[10*(y+x)-1] = 1 // yx0
y += 10
}
}
}
func main() {
total := 0.0
trail := 1
first := 0
firstRatio := 0.0
fmt.Println("The mean proportion of zero digits in factorials up to the following are:")
for f := 2; f <= 10000; f++ {
carry := 0
d999 := 0
zeros := (trail - 1) * 3
j := trail
l := len(rfs)
for j <= l || carry != 0 {
if j <= l {
carry = rfs[j-1]*f + carry
}
d999 = carry % 1000
if j <= l {
rfs[j-1] = d999
} else {
rfs = append(rfs, d999)
}
if d999 == 0 {
zeros += 3
} else {
zeros += zc[d999-1]
}
carry /= 1000
j++
}
for rfs[trail-1] == 0 {
trail++
}
// d999 = quick correction for length and zeros
d999 = rfs[len(rfs)-1]
if d999 < 100 {
if d999 < 10 {
d999 = 2
} else {
d999 = 1
}
} else {
d999 = 0
}
zeros -= d999
digits := len(rfs)*3 - d999
total += float64(zeros) / float64(digits)
ratio := total / float64(f)
if ratio >= 0.16 {
first = 0
firstRatio = 0.0
} else if first == 0 {
first = f
firstRatio = ratio
}
if f == 100 || f == 1000 || f == 10000 {
fmt.Printf("%6s = %12.10f\n", rcu.Commatize(f), ratio)
}
}
fmt.Printf("%6s = %12.10f", rcu.Commatize(first), firstRatio)
fmt.Println(" (stays below 0.16 after this)")
fmt.Printf("%6s = %12.10f\n", "50,000", total/50000)
}
- Output:
Same as 'brute force' version.
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);
}
}
- Output:
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
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
# 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;The task
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- Output:
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.
Julia
function meanfactorialdigits(N, goal = 0.0)
factoril, proportionsum = big"1", 0.0
for i in 1:N
factoril *= i
d = digits(factoril)
zero_proportion_in_fac = count(x -> x == 0, d) / length(d)
proportionsum += zero_proportion_in_fac
propmean = proportionsum / i
if i > 15 && propmean <= goal
println("The mean proportion dips permanently below $goal at $i.")
break
end
if i == N
println("Mean proportion of zero digits in factorials to $N is ", propmean)
end
end
end
@time foreach(meanfactorialdigits, [100, 1000, 10000])
@time meanfactorialdigits(50000, 0.16)
- Output:
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 3.030182 seconds (297.84 k allocations: 1.669 GiB, 0.83% gc time, 0.28% compilation time) The mean proportion dips permanently below 0.16 at 47332. 179.157788 seconds (3.65 M allocations: 59.696 GiB, 1.11% gc time)
Base 1000 version
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()
- Output:
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)
Mathematica /Wolfram Language
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
- Output:
0.111111 0.267873 0.246753 0.203545 0.173004 47332
Nim
Task
import strutils, std/monotimes
import bignum
let t0 = getMonoTime()
var sum = 0.0
var f = newInt(1)
var lim = 100
for n in 1..10_000:
f *= n
let str = $f
sum += str.count('0') / str.len
if n == lim:
echo n, ":\t", sum / float(n)
lim *= 10
echo()
echo getMonoTime() - t0
- Output:
100: 0.2467531861674322 1000: 0.2035445511031646 10000: 0.1730038482418671 (seconds: 2, nanosecond: 857794404)
Stretch task
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 strutils, std/monotimes
import bignum
let t0 = getMonoTime()
var sum = 0.0
var first = 0
var f = newInt(1)
var count0 = 0
for n in 1..<50_000:
f *= n
while f mod 10 == 0: # Reduce the length of "f".
f = f div 10
inc count0
let str = $f
sum += (str.count('0') + count0) / (str.len + count0)
if sum / float(n) < 0.16:
if first == 0: first = n
else:
first = 0
echo "Permanently below 0.16 at n = ", first
echo "Execution time: ", getMonoTime() - t0
- Output:
Permanently below 0.16 at n = 47332 Execution time: (seconds: 190, nanosecond: 215845101)
Pascal
Doing the calculation in Base 1,000,000,000 like in Primorial_numbers#alternative.
The most time consuming is converting to string and search for zeros.
Therefor I do not convert to string.I divide the base in sections of 3 digits with counting zeros in a lookup table.
program Factorial;
{$IFDEF FPC} {$MODE DELPHI} {$Optimization ON,ALL} {$ENDIF}
uses
sysutils;
type
tMul = array of LongWord;
tpMul = pLongWord;
const
LongWordDec = 1000*1000*1000;
LIMIT = 50000;
var
CountOfZero : array[0..999] of byte;
SumOfRatio :array[0..LIMIT] of extended;
procedure OutMul(pMul:tpMul;Lmt :NativeInt);
// for testing
Begin
write(pMul[lmt]);
For lmt := lmt-1 downto 0 do
write(Format('%.9d',[pMul[lmt]]));
writeln;
end;
procedure InitCoZ;
//Init Lookup table for 3 digits
var
x,y : integer;
begin
fillchar(CountOfZero,SizeOf(CountOfZero),#0);
CountOfZero[0] := 3; //000
For x := 1 to 9 do
Begin
CountOfZero[x] := 2; //00x
CountOfZero[10*x] := 2; //0x0
CountOfZero[100*x] := 2; //x00
y := 10;
repeat
CountOfZero[y+x] := 1; //0yx
CountOfZero[10*y+x] := 1; //y0x
CountOfZero[10*(y+x)] := 1; //yx0
inc(y,10)
until y > 100;
end;
end;
function getFactorialDecDigits(n:NativeInt):NativeInt;
var
res: extended;
Begin
result := -1;
IF (n > 0) AND (n <= 1000*1000) then
Begin
res := 0;
repeat res := res+ln(n); dec(n); until n < 2;
result := trunc(res/ln(10))+1;
end;
end;
function CntZero(pMul:tpMul;Lmt :NativeInt):NativeUint;
//count zeros in Base 1,000,000,000 number
var
q,r : LongWord;
i : NativeInt;
begin
result := 0;
For i := Lmt-1 downto 0 do
Begin
q := pMul[i];
r := q DIV 1000;
result +=CountOfZero[q-1000*r];//q-1000*r == q mod 1000
q := r;
r := q DIV 1000;
result +=CountOfZero[q-1000*r];
q := r;
r := q DIV 1000;
result +=CountOfZero[q-1000*r];
end;
//special case first digits no leading '0'
q := pMul[lmt];
while q >= 1000 do
begin
r := q DIV 1000;
result +=CountOfZero[q-1000*r];
q := r;
end;
while q > 0 do
begin
r := q DIV 10;
result += Ord( q-10*r= 0);
q := r;
end;
end;
function GetCoD(pMul:tpMul;Lmt :NativeInt):NativeUint;
//count of decimal digits
var
i : longWord;
begin
result := 9*Lmt;
i := pMul[Lmt];
while i > 1000 do
begin
i := i DIV 1000;
inc(result,3);
end;
while i > 0 do
begin
i := i DIV 10;
inc(result);
end;
end;
procedure DoChecks(pMul:tpMul;Lmt,i :NativeInt);
//(extended(1.0)* makes TIO.RUN faster // only using FPU?
Begin
SumOfRatio[i] := SumOfRatio[i-1] + (extended(1.0)*CntZero(pMul,Lmt))/GetCoD(pMul,Lmt);
end;
function MulByI(pMul:tpMul;UL,i :NativeInt):NativeInt;
var
prod : Uint64;
j : nativeInt;
carry : LongWord;
begin
result := UL;
carry := 0;
For j := 0 to result do
Begin
prod := i*pMul[0]+Carry;
Carry := prod Div LongWordDec;
pMul[0] := Prod - LongWordDec*Carry;
inc(pMul);
end;
IF Carry <> 0 then
Begin
inc(result);
pMul[0]:= Carry;
End;
end;
procedure getFactorialExact(n:NativeInt);
var
MulArr : tMul;
pMul : tpMul;
i,ul : NativeInt;
begin
i := getFactorialDecDigits(n) DIV 9 +10;
Setlength(MulArr,i);
pMul := @MulArr[0];
Ul := 0;
pMul[Ul]:= 1;
i := 1;
repeat
UL := MulByI(pMul,UL,i);
//Now do what you like to do with i!
DoChecks(pMul,UL,i);
inc(i);
until i> n;
end;
procedure Out_(i: integer);
begin
if i > LIMIT then
EXIT;
writeln(i:8,SumOfRatio[i]/i:18:15);
end;
var
i : integer;
Begin
InitCoZ;
SumOfRatio[0]:= 0;
getFactorialExact(LIMIT);
Out_(100);
Out_(1000);
Out_(10000);
Out_(50000);
i := limit;
while i >0 do
Begin
if SumOfRatio[i]/i >0.16 then
break;
dec(i);
end;
inc(i);
writeln('First ratio < 0.16 ', i:8,SumOfRatio[i]/i:20:17);
end.
- Output:
100 0.246753186167432
1000 0.203544551103165
10000 0.173003848241866
50000 0.159620054602269
First ratio < 0.16 47332 0.15999999579985665
Real time: 4.898 s CPU share: 99.55 % // 2.67s on 2200G freepascal 3.2.2
PascalABC.NET
##
var fact := 1bi;
var sum := 0.0;
println('The mean proportion of zero digits in factorials up to the following are:');
for var n := 1 to 10_000 do
begin
fact *= n;
var factstr := fact.toString;
var zeros := factstr.count(x -> x = '0');
sum := sum + zeros / factstr.length;
if n in |100, 1000, 10000| then
writeln(n:5, sum / n:20);
end;
- Output:
The mean proportion of zero digits in factorials up to the following are: 100 0.246753186167432 1000 0.203544551103165 10000 0.173003848241867
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;
}
- Output:
100: 0.24675 1000: 0.20354 10000: 0.17300
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.
with javascript_semantics sequence rfs = {1} -- reverse factorial(1) in base 1000 function init_zc() sequence zc = repeat(0,999) for x=1 to 9 do zc[x] = 2 -- 00x zc[10*x] = 2 -- 0x0 zc[100*x] = 2 -- x00 for y=10 to 90 by 10 do zc[y+x] = 1 -- 0yx zc[10*y+x] = 1 -- y0x zc[10*(y+x)] = 1 -- yx0 end for end for return zc end function constant zc = init_zc() atom t0 = time(), total = 0 integer trail = 1, first = 0 for f=2 to iff(platform()=JS?10000:50000) do integer carry = 0, d999, zeroes = (trail-1)*3, j = trail, l = length(rfs) while j<=l or carry do if j<=l then carry = (rfs[j])*f+carry end if d999 = remainder(carry,1000) if j<=l then rfs[j] = d999 else rfs &= d999 end if zeroes += iff(d999=0?3:zc[d999]) carry = floor(carry/1000) j += 1 end while while rfs[trail]=0 do trail += 1 end while -- d999 := quick correction for length and zeroes: d999 = rfs[$] d999 = iff(d999<100?iff(d999<10?2:1):0) zeroes -= d999 integer digits = length(rfs)*3-d999 total += zeroes/digits atom ratio = total/f if ratio>=0.16 then first = 0 elsif first=0 then first = f end if if find(f,{100,1000,10000}) then string e = elapsed(time()-t0) printf(1,"Mean proportion of zero digits in factorials to %d is %.10f (%s)\n",{f,ratio,e}) end if end for if platform()!=JS then string e = elapsed(time()-t0) printf(1,"The mean proportion dips permanently below 0.16 at %d. (%s)\n",{first,e}) end if
- Output:
Mean proportion of zero digits in factorials to 100 is 0.2467531862 (0s) Mean proportion of zero digits in factorials to 1000 is 0.2035445511 (0.2s) Mean proportion of zero digits in factorials to 10000 is 0.1730038482 (2.3s) The mean proportion dips permanently below 0.16 at 47332. (1 minute and 2s)
(stretch goal removed under pwa/p2js since otherwise you'd get a blank screen for 2 or 3 minutes)
trailing zeroes only
Should you only be interested in the ratio of trailing zeroes, you can do that much faster:
with javascript_semantics atom t0 = time(), f10 = log10(1), total = 0 integer first = 0 for f=2 to 50000 do f10 += log10(f) integer digits = ceil(f10), zeroes = 0, v = 5 while v<=f do zeroes += floor(f/v) v *= 5 end while total += zeroes/digits atom ratio = total/f if ratio>=0.07 then first = 0 elsif first=0 then first = f end if if find(f,{100,1000,10000}) then printf(1,"Mean proportion of trailing zeroes in factorials to %d is %f\n",{f,ratio}) end if end for string e = elapsed(time()-t0) printf(1,"The mean proportion dips permanently below 0.07 at %d. (%s)\n",{first,e})
- Output:
Mean proportion of trailing zeroes in factorials to 100 is 0.170338 Mean proportion of trailing zeroes in factorials to 1000 is 0.116334 Mean proportion of trailing zeroes in factorials to 10000 is 0.081267 The mean proportion dips permanently below 0.07 at 31549. (0.1s)
Python
def facpropzeros(N, verbose = True):
proportions = [0.0] * N
fac, psum = 1, 0.0
for i in range(N):
fac *= i + 1
d = list(str(fac))
psum += sum(map(lambda x: x == '0', d)) / len(d)
proportions[i] = psum / (i + 1)
if verbose:
print("The mean proportion of 0 in factorials from 1 to {} is {}.".format(N, psum / N))
return proportions
for n in [100, 1000, 10000]:
facpropzeros(n)
props = facpropzeros(47500, False)
n = (next(i for i in reversed(range(len(props))) if props[i] > 0.16))
print("The mean proportion dips permanently below 0.16 at {}.".format(n + 2))
- Output:
The mean proportion of 0 in factorials from 1 to 100 is 0.24675318616743216. The mean proportion of 0 in factorials from 1 to 1000 is 0.20354455110316458. The mean proportion of 0 in factorials from 1 to 10000 is 0.17300384824186707. The mean proportion dips permanently below 0.16 at 47332.
The means can be plotted, showing a jump from 0 to over 0.25, followed by a slowly dropping curve:
import matplotlib.pyplot as plt
plt.plot([i+1 for i in range(len(props))], props)
Base 1000 version
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.")
- Output:
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.
Quackery
[ $ "bigrat.qky" loadfile ] now!
[ stack ] is digits ( --> s )
[ stack ] is zeroes ( --> s )
[ 0 digits put
0 zeroes put
[ 1 digits tally
10 /mod 0 =
zeroes tally
dup 0 = until ]
drop
zeroes take
digits take
reduce ] is zeroprop ( n --> n/d )
[ dup dip
[ 0 n->v
1 ]
times
[ i^ 1+ *
dup zeroprop
rot dip v+ ]
drop
rot n->v v/ ] is distribution ( n --> n/d )
100 distribution 10 point$ echo$ cr
1000 distribution 10 point$ echo$ cr
10000 distribution 10 point$ echo$ cr- Output:
0.2467531862 0.2035445511 0.1730038482
Raku
Works, but depressingly slow for 10000.
sub postfix:<!> (Int $n) { ( constant factorial = 1, 1, |[\*] 2..* )[$n] }
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 (
100
,1000
,10000
).map: -> \n { "{n}: {([+] (^n).map: *.&zs) / n}" }
- Output:
100: 0.24675318616743216 1000: 0.20354455110316458 10000: 0.17300384824186605
REXX
/*REXX program computes the mean of the proportion of "0" digits a series of factorials.*/
parse arg $ /*obtain optional arguments from the CL*/
if $='' | $="," then $= 100 1000 10000 /*not specified? Then use the default.*/
#= words($) /*the number of ranges to be used here.*/
numeric digits 100 /*increase dec. digs, but only to 100. */
big= word($, #); != 1 /*obtain the largest number in ranges. */
do i=1 for big /*calculate biggest ! using 100 digs.*/
!= ! * i /*calculate the factorial of BIG. */
end /*i*/
if pos('E', !)>0 then do /*In exponential format? Then get EXP.*/
parse var ! 'E' x /*parse the exponent from the number. */
numeric digits x+1 /*set the decimal digits to X plus 1.*/
end /* [↑] the +1 is for the dec. point.*/
title= ' mean proportion of zeros in the (decimal) factorial products for N'
say ' N │'center(title, 80) /*display the title for the output. */
say '───────────┼'center("" , 80, '─') /* " a sep " " " */
do j=1 for #; n= word($, j) /*calculate some factorial ranges. */
say center( commas(n), 11)'│' left(0dist(n), 75)... /*show results for above range.*/
end /*j*/
say '───────────┴'center("" , 80, '─') /*display a foot sep for the output. */
exit 0 /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
commas: parse arg ?; do jc=length(?)-3 to 1 by -3; ?=insert(',', ?, jc); end; return ?
/*──────────────────────────────────────────────────────────────────────────────────────*/
0dist: procedure; parse arg z; != 1; y= 0
do k=1 for z; != ! * k; y= y + countstr(0, !) / length(!)
end /*k*/
return y/z
- output when using the default inputs:
N │ mean proportion of zeros in the (decimal) factorial products for N
───────────┼────────────────────────────────────────────────────────────────────────────────
100 │ 0.2467531861674322177784158871973526991129407033266153063813195937196095976...
1,000 │ 0.2035445511031646356400438031711455302985741167890402203486699704599684047...
10,000 │ 0.1730038482418660531800366428930706156810278809057883361518852958446868172...
───────────┴────────────────────────────────────────────────────────────────────────────────
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()
);
}
- Output:
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)
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
- Output:
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.
Sidef
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)
- Output:
0.246753186167432217778415887197352699112940703327 0.203544551103164635640043803171145530298574116789 0.173003848241866053180036642893070615681027880906
Wren
Brute force
Very slow indeed, 10.75 minutes to reach N = 10,000.
import "./big" for BigInt
import "./fmt" for Fmt
var fact = BigInt.one
var sum = 0
System.print("The mean proportion of zero digits in factorials up to the following are:")
for (n in 1..10000) {
fact = fact * n
var bytes = fact.toString.bytes
var digits = bytes.count
var zeros = bytes.count { |b| b == 48 }
sum = sum + zeros / digits
if (n == 100 || n == 1000 || n == 10000) {
Fmt.print("$,6d = $12.10f", n, sum / n)
}
}
- Output:
The mean proportion of zero digits in factorials up to the following are: 100 = 0.2467531862 1,000 = 0.2035445511 10,000 = 0.1730038482
'String math' and base 1000
Around 60 times faster than before with 10,000 now being reached in about 10.5 seconds. Even the stretch goal is now viable and comes in at 5 minutes 41 seconds.
import "./fmt" for Fmt
var rfs = [1] // reverse factorial(1) in base 1000
var init = Fn.new { |zc|
for (x in 1..9) {
zc[x-1] = 2 // 00x
zc[10*x - 1] = 2 // 0x0
zc[100*x - 1] = 2 // x00
var y = 10
while (y <= 90) {
zc[y + x - 1] = 1 // 0yx
zc[10*y + x - 1] = 1 // y0x
zc[10*(y + x) - 1] = 1 // yx0
y = y + 10
}
}
}
var zc = List.filled(999, 0)
init.call(zc)
var total = 0
var trail = 1
var first = 0
var firstRatio = 0
System.print("The mean proportion of zero digits in factorials up to the following are:")
for (f in 2..50000) {
var carry = 0
var d999 = 0
var zeros = (trail-1) * 3
var j = trail
var l = rfs.count
while (j <= l || carry != 0) {
if (j <= l) carry = rfs[j-1]*f + carry
d999 = carry % 1000
if (j <= l) {
rfs[j-1] = d999
} else {
rfs.add(d999)
}
zeros = zeros + ((d999 == 0) ? 3 : zc[d999-1])
carry = (carry/1000).floor
j = j + 1
}
while (rfs[trail-1] == 0) trail = trail + 1
// d999 = quick correction for length and zeros
d999 = rfs[-1]
d999 = (d999 < 100) ? ((d999 < 10) ? 2 : 1) : 0
zeros = zeros - d999
var digits = rfs.count * 3 - d999
total = total + zeros/digits
var ratio = total / f
if (ratio >= 0.16) {
first = 0
firstRatio = 0
} else if (first == 0) {
first = f
firstRatio = ratio
}
if (f == 100 || f == 1000 || f == 10000) {
Fmt.print("$,6d = $12.10f", f, ratio)
}
}
Fmt.write("$,6d = $12.10f", first, firstRatio)
System.print(" (stays below 0.16 after this)")
Fmt.print("$,6d = $12.10f", 50000, total/50000)
- Output:
The mean proportion of zero digits in factorials up to the following are: 100 = 0.2467531862 1,000 = 0.2035445511 10,000 = 0.1730038482 47,332 = 0.1599999958 (stays below 0.16 after this) 50,000 = 0.1596200546