Left factorials: Difference between revisions
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'''Left factorials''', <big><big>!n</big></big>, may refer to either ''subfactorials'' or to ''factorial sums'';
<br>the same notation can be confusingly seen being used for the two different definitions.
Sometimes, ''subfactorials'' (also known as ''derangements'') may use any of the notations:
Line 10:
:::::::* <big><big> <b> <span style="font-family:serif">!''n''</span> </b> </big></big>
:::::::* <big><big> <b> <span style="font-family:serif">''n''¡</span> </b> </big></big>
(It may not be visually obvious, but the last example uses an upside-down exclamation mark.)
<big><big>
::::: <math> !n = \sum_{k=0}^{n-1} k! </math>
</big></big>
:::: where
<big><big>
::::: <math>!0 = 0</math>
</big></big>
Line 26 ⟶ 27:
;Task
Display the left factorials for:
* zero through ten (inclusive)
* 20 through 110 (inclusive) by tens
<br>
Display the length (in decimal digits) of the left factorials for:
*
;Also see:
* The OEIS entry: [http://oeis.org/A003422 A003422 left factorials]
* The MathWorld entry: [http://mathworld.wolfram.com/LeftFactorial.html left factorial]
Line 44 ⟶ 45:
* [http://rosettacode.org/wiki/Permutations/Derangements permutations/derangements (subfactorials)]
<br><br>
=={{header|11l}}==
{{trans|D}}
<syntaxhighlight lang="11l">F left_fact(n)
BigInt result = 0
BigInt factorial = 1
L(i) 1 .. n
result += factorial
factorial *= i
R result
print(‘First 11 left factorials:’)
print((0..10).map(i -> left_fact(i)))
print("\n20 through 110 (inclusive) by tens:")
L(i) (20..110).step(10)
print(left_fact(i))
print("\nDigits in 1,000 through 10,000 by thousands:")
print((1000..10000).step(1000).map(i -> String(left_fact(i)).len))</syntaxhighlight>
{{out}}
<pre>
First 11 left factorials:
[0, 1, 2, 4, 10, 34, 154, 874, 5914, 46234, 409114]
20 through 110 (inclusive) by tens:
128425485935180314
9157958657951075573395300940314
20935051082417771847631371547939998232420940314
620960027832821612639424806694551108812720525606160920420940314
141074930726669571000530822087000522211656242116439949000980378746128920420940314
173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314
906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314
16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314
942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314
145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314
Digits in 1,000 through 10,000 by thousands:
[2565, 5733, 9128, 12670, 16322, 20062, 23875, 27749, 31678, 35656]
</pre>
=={{header|ALGOL 68}}==
Uses the Algol 68G LONG LONG INT type which has programmer definable precision.
{{works with|ALGOL 68G|Any - tested with release 2.8.3.win32}}
<
PR precision 36000 PR
# stores left factorials in an array #
Line 115 ⟶ 156:
OD
END
</syntaxhighlight>
{{out}}
<pre>
Line 153 ⟶ 194:
35656
</pre>
=={{header|Arturo}}==
<syntaxhighlight lang="arturo">lfactorial: function [n][
if zero? n -> return 0
fold 0..dec n [x y] -> x + factorial y
]
print "First eleven:"
0..10 | map => lfactorial
| print
print "\n20th through 110th by tens:"
r: range.step: 10 20 110
r | map => lfactorial
| loop => print
print "\nDigits in 1,000th through 10,000th by thousands:"
r: range.step: 1000 1000 10000
r | map'x -> size ~"|lfactorial x|"
| print</syntaxhighlight>
{{out}}
<pre>First eleven:
0 1 2 4 10 34 154 874 5914 46234 409114
20th through 110th by tens:
128425485935180314
9157958657951075573395300940314
20935051082417771847631371547939998232420940314
620960027832821612639424806694551108812720525606160920420940314
141074930726669571000530822087000522211656242116439949000980378746128920420940314
173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314
906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314
16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314
942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314
145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314
Digits in 1,000th through 10,000th by thousands:
2565 5733 9128 12670 16322 20062 23875 27749 31678 35656</pre>
=={{header|AWK}}==
Old posix AWK doesn't support computing with large numbers. However modern gawk can use GMP if the flag -M is used.
<syntaxhighlight lang="awk">
#!/usr/bin/gawk -Mf
function left_factorial(num) {
Line 180 ⟶ 262:
}
}
</syntaxhighlight>
{{out}}
<pre>
Line 215 ⟶ 297:
!10000 has 35656 digits
</pre>
=={{header|BBC BASIC}}==
{{works with|BBC BASIC for Windows}}
Use the 'Mapm' library.
<
Result$="0" : A$="1"
Line 228 ⟶ 311:
IF I% > 999 IF I% MOD 1000 = 0 PRINT "!";I% " has " LENFNMAPM_FormatDec(Result$,0) " digits"
NEXT
END</
{{out}}
<pre>
Line 265 ⟶ 348:
=={{header|Bracmat}}==
{{trans|D}}
<
= result factorial i
. 0:?result
Line 301 ⟶ 384:
. (=L.@(!arg:? [?L)&out$!L)
)
)</
{{out}}
<pre>First 11 left factorials:
Line 342 ⟶ 425:
=={{header|C}}==
{{libheader|GMP}}
<syntaxhighlight lang="c">
#include <stdio.h>
#include <stdlib.h>
Line 393 ⟶ 476:
return 0;
}
</syntaxhighlight>
{{out}}
<pre>
Line 428 ⟶ 511:
!10000 has 35656 digits
</pre>
=={{header|C sharp|C#}}==
<syntaxhighlight lang="csharp">
using System;
using System.Numerics;
namespace LeftFactorial
{
class Program
{
static void Main(string[] args)
{
for (int i = 0; i <= 10; i++)
{
Console.WriteLine(string.Format("!{0} = {1}", i, LeftFactorial(i)));
}
for (int j = 20; j <= 110; j += 10)
{
Console.WriteLine(string.Format("!{0} = {1}", j, LeftFactorial(j)));
}
for (int k = 1000; k <= 10000; k += 1000)
{
Console.WriteLine(string.Format("!{0} has {1} digits", k, LeftFactorial(k).ToString().Length));
}
Console.ReadKey();
}
private static BigInteger Factorial(int number)
{
BigInteger accumulator = 1;
for (int factor = 1; factor <= number; factor++)
{
accumulator *= factor;
}
return accumulator;
}
private static BigInteger LeftFactorial(int n)
{
BigInteger result = 0;
for (int i = 0; i < n; i++)
{
result += Factorial(i);
}
return result;
}
}
}
</syntaxhighlight>
{{out}}
<pre>
!0 = 0
!1 = 1
!2 = 2
!3 = 4
!4 = 10
!5 = 34
!6 = 154
!7 = 874
!8 = 5914
!9 = 46234
!10 = 409114
!20 = 128425485935180314
!30 = 9157958657951075573395300940314
!40 = 20935051082417771847631371547939998232420940314
!50 = 620960027832821612639424806694551108812720525606160920420940314
!60 = 141074930726669571000530822087000522211656242116439949000980378746128920420940314
!70 = 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314
!80 = 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314
!90 = 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314
!100 = 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314
!110 = 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314
!1000 has 2565 digits
!2000 has 5733 digits
!3000 has 9128 digits
!4000 has 12670 digits
!5000 has 16322 digits
!6000 has 20062 digits
!7000 has 23875 digits
!8000 has 27749 digits
!9000 has 31678 digits
!10000 has 35656 digits
</pre>
Faster Implementation
<syntaxhighlight lang="csharp">
using System;
using System.Numerics;
namespace LeftFactorial
{
class Program
{
static void Main(string[] args)
{
for (int i = 0; i <= 10; i++)
{
Console.WriteLine(string.Format("!{0} : {1}", i, LeftFactorial(i)));
}
for (int j = 20; j <= 110; j += 10)
{
Console.WriteLine(string.Format("!{0} : {1}", j, LeftFactorial(j)));
}
for (int k = 1000; k <= 10000; k += 1000)
{
Console.WriteLine(string.Format("!{0} : has {1} digits", k, LeftFactorial(k).ToString().Length));
}
Console.ReadKey();
}
private static BigInteger LeftFactorial(int n)
{
BigInteger result = 0;
BigInteger subResult = 1;
for (int i = 0; i < n; i++)
{
if (i == 0)
{
subResult = 1;
}
else
{
subResult *= i;
}
result += subResult;
}
return result;
}
}
}
</syntaxhighlight>
=={{header|C++}}==
<syntaxhighlight lang="cpp">
#include <vector>
#include <string>
Line 558 ⟶ 786:
}
}
</syntaxhighlight>
{{out}}
<pre>
Line 594 ⟶ 822:
</pre>
===Faster alternative===
{{libheader|GMP}}
<syntaxhighlight lang="cpp">#include <iostream>
#include <gmpxx.h>
template <typename integer>
class left_factorial_generator {
public:
integer next() {
factorial_
return
}
private:
unsigned int n_ = 1;
integer factorial_ = 1;
integer next_ = 0;
};
int main() {
left_factorial_generator<mpz_class> lf;
int i = 0;
std::cout << "Left factorials 0 through 10:\n";
for (; i <= 10; ++i)
std::cout << "!" << i << " = " << lf.next() << '\n';
std::cout << "Left factorials 20 through 110, by tens:\n";
for (; i <= 110; ++i) {
auto n = lf.next();
if (i % 10 == 0)
std::cout << "!" << i << " = " << n << '\n';
}
std::cout << "Lengths of left factorials 1000 through 10000, by thousands:\n";
for (; i <= 10000; ++i) {
auto n = lf.next();
if (i % 1000 == 0)
std::cout << "length of !" << i << " = " << n.get_str().size() << '\n';
}
return 0;
}</syntaxhighlight>
{{out}}
<pre>
Left factorials 0 through 10:
!0 = 0
!1 = 1
Line 662 ⟶ 877:
!9 = 46234
!10 = 409114
Left factorials 20 through 110, by tens:
!20 = 128425485935180314
!30 = 9157958657951075573395300940314
Line 672 ⟶ 888:
!100 = 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314
!110 = 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314
Lengths of left factorials 1000 through 10000, by thousands:
length of !1000 = 2565
length of !2000 = 5733
length of !3000 = 9128
length of !4000 = 12670
length of !5000 = 16322
length of !6000 = 20062
length of !7000 = 23875
length of !8000 = 27749
length of !9000 = 31678
length of !10000 = 35656
</pre>
=={{header|Clojure}}==
<
(:gen-class))
Line 759 ⟶ 921:
(doseq [n (range 1000 10001 1000)]
(println (format "!%-5d has %5d digits" n (count (str (biginteger (left-factorial n)))))))
</syntaxhighlight>
{{Output}}
<pre>
Line 796 ⟶ 958:
=={{header|Common Lisp}}==
<
(defun fact (n)
(reduce #'* (loop for i from 1 to n collect i)))
Line 809 ⟶ 971:
(format t "1000 -> 10000 by 1000~&")
(format t "~{~a digits~&~}" (loop for i from 1000 upto 10000 by 1000 collect (length (format nil "~a" (left-fac i)))))
</syntaxhighlight>
{{out}}
<pre>
Line 839 ⟶ 1,001:
=={{header|D}}==
<
BigInt leftFact(in uint n) pure nothrow /*@safe*/ {
Line 856 ⟶ 1,018:
writefln("\nDigits in 1,000 through 10,000 by thousands:\n%s",
iota(1_000, 10_001, 1_000).map!(i => i.leftFact.text.length));
}</
{{out}}
<pre>First 11 left factorials:
Line 878 ⟶ 1,040:
=={{header|EchoLisp}}==
We use the 'bigint' library and memoization : (remember 'function).
<
(lib 'bigint)
(define (!n n)
Line 884 ⟶ 1,046:
(+ (!n (1- n)) (factorial (1- n)))))
(remember '!n)
</syntaxhighlight>
Output:
<
(for ((n 11)) (printf "!n(%d) = %d" n (!n n)))
(for ((n (in-range 20 120 10))) (printf "!n(%d) = %d" n (!n n)))
Line 925 ⟶ 1,087:
Digits of !n(9000) = 31678
Digits of !n(10000) = 35656
</syntaxhighlight>
=={{header|Elixir}}==
<
def calc(0), do: 0
def calc(n) do
Line 947 ⟶ 1,109:
digits = LeftFactorial.calc(i) |> to_char_list |> length
IO.puts "!#{i} has #{digits} digits"
end)</
{{out}}
Line 986 ⟶ 1,148:
=={{header|F_Sharp|F#}}==
===The Functıon===
<
// Generate Sequence of Left Factorials: Nigel Galloway, March 5th., 2019.
let LF=Seq.unfold(fun (Σ,n,g)->Some(Σ,(Σ+n,n*g,g+1I))) (0I,1I,1I)
</syntaxhighlight>
===The Tasks===
;Display LF 0..10
<
LF |> Seq.take 11|>Seq.iter(printfn "%A")
</syntaxhighlight>
{{out}}
<pre>
Line 1,010 ⟶ 1,172:
</pre>
;Display LF 20..110 in steps of 10
<
LF |> Seq.skip 20 |> Seq.take 91 |> Seq.iteri(fun n g->if n%10=0 then printfn "%A" g)
</syntaxhighlight>
{{out}}
<pre>
Line 1,027 ⟶ 1,189:
</pre>
;Display the length (in decimal digits) of LF 1000 .. 10000 in steps of 1000
<
LF |> Seq.skip 1000 |> Seq.take 9001 |> Seq.iteri(fun n g->if n%1000=0 then printfn "%d" (string g).Length)
</syntaxhighlight>
{{out}}
<pre>
Line 1,043 ⟶ 1,205:
35656
</pre>
=={{header|Factor}}==
{{works with|Factor|0.98}}
<syntaxhighlight lang="text">USING: formatting fry io kernel math math.factorials
math.functions math.parser math.ranges sequences ;
IN: rosetta-code.left-factorials
Line 1,074 ⟶ 1,234:
: main ( -- ) part1 part2 part3 ;
MAIN: main</
{{out}}
<pre>
Line 1,112 ⟶ 1,272:
!10000 35656
</pre>
=={{header|Forth}}==
{{works with|Gforth 0.7.3}}
This solution inspired by the Fortran one.
<
CREATE S #DIGITS ALLOT S #DIGITS ERASE VARIABLE S#
CREATE F #DIGITS ALLOT F #DIGITS ERASE VARIABLE F#
Line 1,171 ⟶ 1,323:
dup REPORT
dup F F# @ rot B* F# !
1+ REPEAT drop ;</
{{out}}
<pre>$ gforth left-factorials.fs -e 'GO bye'
Line 1,213 ⟶ 1,365:
Because this calculation won't get far, no attempt is made to save intermediate results (such as the factorial numbers) nor develop the results progressively even though they are to be produced in sequence. Each result is computed from the start, as per the specified formulae.
For output, to have the exclamation mark precede the number without a gap, format sequence <code>"!",I0</code> will do, the <code>I0</code> format code being standardised in F90. However, this produces varying-length digit sequences, which will mean that the following output changes position likewise. Rather than use say <code>I20</code> for the result and have a wide gap, code <code>I0</code> will do, and to start each such number in the same place, the code <code>T6</code> will start it in column six, far enough along not to clash with the first number on the line, given that it will not be large. <
CONTAINS !The usual suspects.
INTEGER*8 FUNCTION FACT(N) !Factorial, the ordinary.
Line 1,250 ⟶ 1,402:
WRITE (6,1) I,LFACT(I)
END DO
END</
Output:
Line 1,271 ⟶ 1,423:
</pre>
Obviously, one could proceed using the services of some collection of "bignum" routines, and then the code would merely depict their uses for this problem. Since the task is to produce consecutive values, all that need be done is to maintain a S value holding the accumulated sum, and a F value for the successive factorials to be added into S. The only difficulty is to arrange the proper phasing of the starting values so that the calculation will work. Since only one multiply and one addition is needed per step, explicit code might as well be used, as follows: <
INTEGER ENUFF,BASE !Some parameters.
PARAMETER (BASE = 10, ENUFF = 40000) !This should do.
Line 1,337 ⟶ 1,489:
END DO !If there is one, as when N > BASE.
END DO !On to the next result.
END !Ends with a new factorial that won't be used.</
Output: achieved in a few seconds. A larger BASE would give a faster calculation, but would complicate the digit count.
<pre>
Line 1,376 ⟶ 1,528:
{{trans|C}}
{{libheader|GMP}}
<
#include "gmp.bi"
Line 1,418 ⟶ 1,570:
Print
Print "Press any key to quit"
Sleep</
{{out}}
Line 1,454 ⟶ 1,606:
!9000 has 31678 digits
!10000 has 35656 digits
</pre>
=={{header|Fōrmulæ}}==
{{FormulaeEntry|page=https://formulae.org/?script=examples/Left_factorials}}
'''Solution'''
The following function calculates the left factorial directly by its definition:
[[File:Fōrmulæ - Left factorials 08.png]]
However, the following is much faster:
[[File:Fōrmulæ - Left factorials 01.png]]
'''Test case 1. Showing left factorials from zero to ten'''
[[File:Fōrmulæ - Left factorials 02.png]]
[[File:Fōrmulæ - Left factorials 03.png]]
'''Test case 2. Showing left factorials from 20 to 110, by tens'''
[[File:Fōrmulæ - Left factorials 04.png]]
[[File:Fōrmulæ - Left factorials 05.png|858px]]
'''Test case 3. Showing length of left factorials, from 1,000 to 10,000 by thousands'''
[[File:Fōrmulæ - Left factorials 06.png]]
[[File:Fōrmulæ - Left factorials 07.png]]
=={{header|Frink}}==
Frink contains efficient algorithms for calculating and caching factorials and this program will work for arbitrarily-large numbers.
<syntaxhighlight lang="frink">leftFactorial[n] :=
{
sum = 0
for k = 0 to n-1
sum = sum + k!
return sum
}
println["Zero through ten"]
for n = 0 to 10
println["$n\t" + leftFactorial[n]]
println["\n20 through 110"]
for n = 20 to 110 step 10
println["$n\t" + leftFactorial[n]]
println["\nlength of 1000 through 10000"]
for n = 1000 to 10000 step 1000
println["$n has " + length[toString[leftFactorial[n]]] + " digits"]</syntaxhighlight>
{{out}}
<pre>
Zero through ten
0 0
1 1
2 2
3 4
4 10
5 34
6 154
7 874
8 5914
9 46234
10 409114
20 through 110
20 128425485935180314
30 9157958657951075573395300940314
40 20935051082417771847631371547939998232420940314
50 620960027832821612639424806694551108812720525606160920420940314
60 141074930726669571000530822087000522211656242116439949000980378746128920420940314
70 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314
80 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314
90 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314
100 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314
110 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314
length of 1000 through 10000
1000 has 2565 digits
2000 has 5733 digits
3000 has 9128 digits
4000 has 12670 digits
5000 has 16322 digits
6000 has 20062 digits
7000 has 23875 digits
8000 has 27749 digits
9000 has 31678 digits
10000 has 35656 digits
</pre>
=={{header|Go}}==
<
import (
Line 1,501 ⟶ 1,748:
}
fmt.Println()
}</
{{out}}
<pre>
Line 1,520 ⟶ 1,767:
=={{header|Haskell}}==
<
leftFact = scanl (+) 0 fact
fact :: [Integer]
fact = scanl (*) 1 [1 ..]
main :: IO ()
Line 1,537 ⟶ 1,784:
, ""
, "length of 1,000 ~ 10,000 by thousands:"
, show $
, ""
]</
{{Out}}
<pre>0 ~ 10:
Line 1,563 ⟶ 1,810:
{{trans|D}}
The following works in both languages:
<syntaxhighlight lang="text">procedure main()
every writes(lfact(0 | !10)," ")
write()
Line 1,578 ⟶ 1,825:
every (i := !n, r +:= .f, f *:= .i)
return r
end</
{{out}}
Line 1,604 ⟶ 1,851:
This could be made more efficient (in terms of machine time), is there a practical application for this? The more efficient machine approach would require a more specialized interface or memory dedicated to caching.
<
Task examples:
<
0 0
1 1
Line 1,641 ⟶ 1,888:
8000 27749
9000 31678
10000 35656</
=={{header|Java}}==
<
public class LeftFac{
Line 1,676 ⟶ 1,923:
}
}
}</
{{out}}
<pre>!0 = 0
Line 1,715 ⟶ 1,962:
'''Using builtin arithmetic''':
<
reduce range(1; .+1) as $i
# state: [i!, !i]
([1,0]; .[1] += .[0] | .[0] *= $i)
| .[1];</
'''Using BigInt.jq''':
Line 1,728 ⟶ 1,975:
To compute the lengths of the decimal representation without having to recompute !n,
we also define left_factorial_lengths(gap) to emit [n, ( !n|length) ] when n % gap == 0.
<
# integer input
Line 1,748 ⟶ 1,995:
| (.[1] | tostring | length) as $lf
| if $i % gap == 0 then .[2] += [[$i, $lf]] else . end)
| .[2];</
'''The specific tasks''':
<
(10000 | long_left_factorial_lengths(1000) | .[] | "\(.[0]): length is \(.[1])")</
{{out}}
(scrollable)
<div style="overflow:scroll; height:200px;">
<
0: 0
1: 1
Line 1,789 ⟶ 2,036:
8000: length is 27749
9000: length is 31678
10000: length is 35656</
</div>
Line 1,795 ⟶ 2,042:
{{works with|Julia|0.6}}
<
@show leftfactorial.(0:10)
@show ndigits.(leftfactorial.(big.(1000:1000:10_000)))</
{{out}}
Line 1,805 ⟶ 2,052:
=={{header|Kotlin}}==
<
import java.math.BigInteger
Line 1,827 ⟶ 2,074:
for (i in 1000..10000 step 1000)
println("!${i.toString().padEnd(5)} has ${leftFactorial(i).toString().length} digits")
}</
{{out}}
Line 1,865 ⟶ 2,112:
!10000 has 35656 digits
</pre>
=={{header|Lambdatalk}}==
The code can be tested in this wiki page: http://lambdaway.free.fr/lambdawalks/?view=left_factorial
<syntaxhighlight lang="scheme">
'''1) defining !n'''
{def !n // the main function's name
{def !n.mem {A.new 0 1 2}} // initializing a global array
// memorizing the computed values of !n
{def !n.set // computing and storing the values
{lambda {:n}
{A.set! :n // assign at n
{long_mult :n {A.get {- :n 1} {!n.mem}} } // product of n and computed value at n-1
{!n.mem}}}} // in the global array
{def !n.get // getting a value
{lambda {:n}
{A.get :n {if {W.equal? {A.get :n {!n.mem}} undefined} // if it doesn't exist
then {!n.set :n} // then compute it
else {!n.mem}}}}} // else get it from the global array
{lambda {:n} // the main function's body
{if {< :n 2} // if n=0 and n=1
then :n // then return 0 or 1
else {S.reduce long_add // apply add_long to
1 {S.map !n.get {S.serie 1 {- :n 1}}}}}}} // the sequence of computed values
'''2) the task'''
A) computing !n from 2 to 10 takes about 4ms
{S.map {lambda {:n} {br}!n(:n) = {!n :n}}
{S.serie 0 10}}
->
!n(0) = 0
!n(1) = 1
!n(2) = 2
!n(3) = 4
!n(4) = 10
!n(5) = 34
!n(6) = 154
!n(7) = 874
!n(8) = 5914
!n(9) = 46234
!n(10) = 409114
B) computing !n from 20 to 110 with step 10 takes about 25ms
{S.map {lambda {:n} {br}!n(:n) = {!n :n}}
{S.serie 20 110 10}}
->
!n(20) = 128425485935180314
!n(30) = 9157958657951075573395300940314
!n(40) = 20935051082417771847631371547939998232420940314
!n(50) = 620960027832821612639424806694551108812720525606160920420940314
!n(60) = 141074930726669571000530822087000522211656242116439949000980378746128920420940314
!n(70) = 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314
!n(80) = 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314
!n(90) = 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314
!n(100) = 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314
!n(110) = 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314
C) computing !n from 1000 to 1000 with step 1000 takes about 87 seconds
1) compute n! from 1 to 10000 and store in MEM // about 86 seconds
{def foo {!n 10000}} -> foo
2) take 10 values from MEM // about 1ms
{S.map {lambda {:n}
{br}Digits of !n(:n) = {W.length {A.get {- :n 1} {MEM}}}}
{S.serie 1000 10000 1000}}
->
Digits of !n(1000) = 2565
Digits of !n(2000) = 5733
Digits of !n(3000) = 9128
Digits of !n(4000) = 12670
Digits of !n(5000) = 16322
Digits of !n(6000) = 20062
Digits of !n(7000) = 23875
Digits of !n(8000) = 27749
Digits of !n(9000) = 31678
Digits of !n(10000) = 35656
</syntaxhighlight>
=={{header|Lua}}==
Takes about five seconds...
<
require("bc")
Line 1,895 ⟶ 2,228:
for i = 1000, 10000, 1000 do
print("!" .. i .. " contains " .. #leftFac(i) .. " digits")
end</
{{out}}
<pre>!0 = 0
Line 1,930 ⟶ 2,263:
=={{header|Maple}}==
<
seq(left_factorial(i), i = 1 .. 10);
seq(left_factorial(i), i = 20 .. 110, 10);
seq(length(left_factorial(i)), i = 1000 .. 10000, 1000);</
{{out}}
<pre>0, 1, 3, 9, 33, 153, 873, 5913, 46233, 409113
Line 1,941 ⟶ 2,274:
2565, 5733, 9128, 12670, 16322, 20062, 23875, 27749, 31678, 35656</pre>
=={{header|Mathematica}}/{{header|Wolfram Language}}==
<
Print["left factorials 0 through 10:"]
Print[left /@ Range[0, 10] // TableForm]
Line 1,948 ⟶ 2,281:
Print[left /@ Range[20, 110, 10] // TableForm]
Print["Digits in left factorials 1,000 through 10,000, by thousands:"]
Print[Length[IntegerDigits[left[#]]] & /@ Range[1000, 10000, 1000] // TableForm]</
{{out}}
<pre>left factorials 0 through 10:
Line 1,985 ⟶ 2,318:
27749
31678
35656</pre>
=={{header|Maxima}}==
<syntaxhighlight lang="maxima">
l_factorial(n):=sum(k!,k,0,n-1)$
/* Test cases */
makelist(l_factorial(i),i,0,10);
makelist(l_factorial(i),i,20,110,10);
</syntaxhighlight>
{{out}}
<pre>
[0,1,2,4,10,34,154,874,5914,46234,409114]
[128425485935180314,9157958657951075573395300940314,20935051082417771847631371547939998232420940314,620960027832821612639424806694551108812720525606160920420940314,141074930726669571000530822087000522211656242116439949000980378746128920420940314,173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314,906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314,16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314,942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314,145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314]
</pre>
=={{header|Nim}}==
{{trans|Python}}
<
proc lfact: iterator: BigInt =
Line 2,015 ⟶ 2,363:
echo "Digits in 1,000 through 10,000 (inclusive) by thousands:"
for i in lfact().slice(1_000, 10_000, 1_000):
echo " ", ($i).len</
{{out}}
<pre>first 11:
Line 2,054 ⟶ 2,402:
=={{header|Oforth}}==
<
{{out}}
Line 2,082 ⟶ 2,430:
=={{header|PARI/GP}}==
<
apply(lf, [0..10])
apply(lf, 10*[2..11])
forstep(n=1000,1e4,1000,print1(#digits(lf(n))", "))</
{{out}}
<pre>%1 = [0, 1, 2, 4, 10, 34, 154, 874, 5914, 46234, 409114]
Line 2,096 ⟶ 2,444:
If performance is a concern, this will run over 100x faster by replacing the line "use bigint" with "use Math::GMP qw/:constant/" (after installing that module).
<
use 5.010;
use strict;
Line 2,120 ⟶ 2,468:
printf "!%d has %d digits.\n", $_, length leftfact($_) for map $_*1000, 1..10;
</syntaxhighlight>
Since I copied the printf format strings from the
the output from the code above is identical to the output of the
=={{header|Phix}}==
{{trans|Lua}}
{{libheader|Phix/mpfr}}
<!--<syntaxhighlight lang="phix">(phixonline)-->
<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>
<span style="color: #008080;">include</span> <span style="color: #004080;">mpfr</span><span style="color: #0000FF;">.</span><span style="color: #000000;">e</span>
<span style="color: #004080;">sequence</span> <span style="color: #000000;">lf_list</span>
<span style="color: #008080;">procedure</span> <span style="color: #000000;">init</span><span style="color: #0000FF;">(</span><span style="color: #004080;">integer</span> <span style="color: #000000;">n</span><span style="color: #0000FF;">)</span>
<span style="color: #004080;">mpz</span> <span style="color: #000000;">f</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">mpz_init</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">)</span>
<span style="color: #000000;">lf_list</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">repeat</span><span style="color: #0000FF;">(</span><span style="color: #000000;">f</span><span style="color: #0000FF;">,</span><span style="color: #000000;">n</span><span style="color: #0000FF;">+</span><span style="color: #000000;">1</span><span style="color: #0000FF;">)</span>
<span style="color: #008080;">for</span> <span style="color: #000000;">i</span><span style="color: #0000FF;">=</span><span style="color: #000000;">1</span> <span style="color: #008080;">to</span> <span style="color: #000000;">n</span> <span style="color: #008080;">do</span>
<span style="color: #000000;">f</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">mpz_init_set</span><span style="color: #0000FF;">(</span><span style="color: #000000;">f</span><span style="color: #0000FF;">)</span>
<span style="color: #7060A8;">mpz_mul_si</span><span style="color: #0000FF;">(</span><span style="color: #000000;">f</span><span style="color: #0000FF;">,</span><span style="color: #000000;">f</span><span style="color: #0000FF;">,</span><span style="color: #000000;">i</span><span style="color: #0000FF;">)</span>
<span style="color: #000000;">lf_list</span><span style="color: #0000FF;">[</span><span style="color: #000000;">i</span><span style="color: #0000FF;">+</span><span style="color: #000000;">1</span><span style="color: #0000FF;">]</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">f</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">procedure</span>
<span style="color: #008080;">function</span> <span style="color: #000000;">lf</span><span style="color: #0000FF;">(</span><span style="color: #004080;">integer</span> <span style="color: #000000;">n</span><span style="color: #0000FF;">,</span> <span style="color: #004080;">bool</span> <span style="color: #000000;">len</span><span style="color: #0000FF;">=</span><span style="color: #004600;">false</span><span style="color: #0000FF;">)</span>
<span style="color: #000080;font-style:italic;">-- Returns left factorial of n, or it's length, as a string</span>
<span style="color: #004080;">mpz</span> <span style="color: #000000;">sumf</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">mpz_init</span><span style="color: #0000FF;">(</span><span style="color: #000000;">0</span><span style="color: #0000FF;">)</span>
<span style="color: #008080;">for</span> <span style="color: #000000;">k</span><span style="color: #0000FF;">=</span><span style="color: #000000;">1</span> <span style="color: #008080;">to</span> <span style="color: #000000;">n</span> <span style="color: #008080;">do</span> <span style="color: #7060A8;">mpz_add</span><span style="color: #0000FF;">(</span><span style="color: #000000;">sumf</span><span style="color: #0000FF;">,</span><span style="color: #000000;">sumf</span><span style="color: #0000FF;">,</span><span style="color: #000000;">lf_list</span><span style="color: #0000FF;">[</span><span style="color: #000000;">k</span><span style="color: #0000FF;">])</span> <span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
<span style="color: #008080;">return</span> <span style="color: #008080;">iff</span><span style="color: #0000FF;">(</span><span style="color: #000000;">len</span><span style="color: #0000FF;">?</span><span style="color: #7060A8;">sprintf</span><span style="color: #0000FF;">(</span><span style="color: #008000;">"%d"</span><span style="color: #0000FF;">,</span><span style="color: #7060A8;">mpz_sizeinbase</span><span style="color: #0000FF;">(</span><span style="color: #000000;">sumf</span><span style="color: #0000FF;">,</span><span style="color: #000000;">10</span><span style="color: #0000FF;">))</span>
<span style="color: #0000FF;">:</span><span style="color: #7060A8;">shorten</span><span style="color: #0000FF;">(</span><span style="color: #7060A8;">mpz_get_str</span><span style="color: #0000FF;">(</span><span style="color: #000000;">sumf</span><span style="color: #0000FF;">)))</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">function</span>
<span style="color: #000080;font-style:italic;">-- Main procedure</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>
<span style="color: #000000;">init</span><span style="color: #0000FF;">(</span><span style="color: #000000;">10000</span><span style="color: #0000FF;">)</span>
<span style="color: #008080;">for</span> <span style="color: #000000;">i</span><span style="color: #0000FF;">=</span><span style="color: #000000;">0</span> <span style="color: #008080;">to</span> <span style="color: #000000;">10</span> <span style="color: #008080;">do</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;">"!%d = %s\n"</span><span style="color: #0000FF;">,{</span><span style="color: #000000;">i</span><span style="color: #0000FF;">,</span><span style="color: #000000;">lf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">i</span><span style="color: #0000FF;">)})</span> <span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
<span style="color: #008080;">for</span> <span style="color: #000000;">i</span><span style="color: #0000FF;">=</span><span style="color: #000000;">20</span> <span style="color: #008080;">to</span> <span style="color: #000000;">110</span> <span style="color: #008080;">by</span> <span style="color: #000000;">10</span> <span style="color: #008080;">do</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;">"!%d = %s\n"</span><span style="color: #0000FF;">,{</span><span style="color: #000000;">i</span><span style="color: #0000FF;">,</span><span style="color: #000000;">lf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">i</span><span style="color: #0000FF;">)})</span> <span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
<span style="color: #008080;">for</span> <span style="color: #000000;">i</span><span style="color: #0000FF;">=</span><span style="color: #000000;">1000</span> <span style="color: #008080;">to</span> <span style="color: #000000;">10000</span> <span style="color: #008080;">by</span> <span style="color: #000000;">1000</span> <span style="color: #008080;">do</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;">"!%d contains %s digits\n"</span><span style="color: #0000FF;">,{</span><span style="color: #000000;">i</span><span style="color: #0000FF;">,</span><span style="color: #000000;">lf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">i</span><span style="color: #0000FF;">,</span><span style="color: #004600;">true</span><span style="color: #0000FF;">)})</span> <span style="color: #008080;">end</span> <span style="color: #008080;">for</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;">"complete (%3.2fs)\n"</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>
<!--</syntaxhighlight>-->
{{out}}
<pre>
Line 2,226 ⟶ 2,524:
!30 = 9157958657951075573395300940314
!40 = 20935051082417771847631371547939998232420940314
!50 = 62096002783282161263...25606160920420940314 (63 digits)
!60 = 14107493072666957100...78746128920420940314 (81 digits)
!70 = 17363951180298752669...09216528920420940314 (99 digits)
!80 = 90608958798769534653...22336528920420940314 (117 digits)
!90 = 16695570072624210767...42336528920420940314 (137 digits)
!100 = 94278623976582657916...42336528920420940314 (156 digits)
!110 = 14572298106158529700...42336528920420940314 (177 digits)
!1000 contains 2565 digits
!2000 contains 5733 digits
!3000 contains 9128 digits
!4000 contains 12670 digits
!5000 contains
!6000 contains 20062 digits
!7000 contains 23875 digits
Line 2,243 ⟶ 2,541:
!9000 contains 31678 digits
!10000 contains 35656 digits
complete (0.
</pre>
=={{header|PicoLisp}}==
<
(cache '(NIL) N
(if (> 2 N) 1
Line 2,264 ⟶ 2,562:
(prinl "length of 1000 - 10000")
(pril (mapcar 'length (mapcar '!n (range 1000 10000 1000))))
</syntaxhighlight>
{{out}}
<syntaxhighlight lang="text">0-10
1
1
Line 2,300 ⟶ 2,598:
31678
35656
</syntaxhighlight>
=={{header|PL/I}}==
Line 2,307 ⟶ 2,605:
Results are shown for the first 11 integers, as required;
then for the integers from 20 through 30 only, because factorials for n = 40 and larger are not possible.
<
declare n fixed binary;
declare (s, f) fixed (31);
Line 2,329 ⟶ 2,627:
end;
end left_factorials;</
{{out}}
<pre>
Line 2,358 ⟶ 2,656:
=={{header|PowerShell}}==
{{works with|PowerShell|4.0}}
<syntaxhighlight lang="powershell">
function left-factorial ([BigInt]$n) {
[BigInt]$k, [BigInt]$fact = ([BigInt]::Zero), ([BigInt]::One)
Line 2,385 ⟶ 2,683:
else {"!$i has $digits digit"}
}
</syntaxhighlight>
<b>Output:</b>
<pre>
Line 2,423 ⟶ 2,721:
!9000 has 31678 digits
!10000 has 35656 digits
</pre>
=={{header|Prolog}}==
{{works with|SWI Prolog}}
<syntaxhighlight lang="prolog">leftfact(N):-
leftfact(N, 0, 0, 1).
leftfact(N, N, _, _):-
!.
leftfact(N, M, L, F):-
((M =< 10 ; (M =< 110, 0 is M mod 10)) ->
writef("!%w = %w\n", [M, L])
;
(0 is M mod 1000 ->
number_string(L, S),
string_length(S, Len),
writef("length of !%w is %w\n", [M, Len])
;
true)),
L1 is L + F,
M1 is M + 1,
F1 is F * M1,
leftfact(N, M1, L1, F1).
main:-
leftfact(10001).</syntaxhighlight>
{{out}}
<pre>
!0 = 0
!1 = 1
!2 = 2
!3 = 4
!4 = 10
!5 = 34
!6 = 154
!7 = 874
!8 = 5914
!9 = 46234
!10 = 409114
!20 = 128425485935180314
!30 = 9157958657951075573395300940314
!40 = 20935051082417771847631371547939998232420940314
!50 = 620960027832821612639424806694551108812720525606160920420940314
!60 = 141074930726669571000530822087000522211656242116439949000980378746128920420940314
!70 = 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314
!80 = 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314
!90 = 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314
!100 = 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314
!110 = 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314
length of !1000 is 2565
length of !2000 is 5733
length of !3000 is 9128
length of !4000 is 12670
length of !5000 is 16322
length of !6000 is 20062
length of !7000 is 23875
length of !8000 is 27749
length of !9000 is 31678
length of !10000 is 35656
</pre>
=={{header|Python}}==
<
def lfact():
Line 2,440 ⟶ 2,798:
print(lf)
print('Digits in 1,000 through 10,000 (inclusive) by thousands:\n %r'
% [len(str(lf)) for lf in islice(lfact(), 1000, 10001, 1000)] )</
{{out}}
Line 2,466 ⟶ 2,824:
{{Trans|Haskell}}
<syntaxhighlight lang
from itertools import (accumulate, chain, count, islice)
Line 2,474 ⟶ 2,832:
# leftFact :: [Integer]
def leftFact():
'''Left factorial series defined in terms
of the factorial series.
return accumulate(
chain([0], fact()), add
)
Line 2,482 ⟶ 2,842:
# fact :: [Integer]
def fact():
'''
'''
return
chain([1], count(1)), mul
)
#
# main :: IO ()
def main():
Line 2,512 ⟶ 2,873:
#
# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
Line 2,518 ⟶ 2,879:
'''Function composition.'''
return lambda f: lambda x: g(f(x))
# scanl :: (b -> a -> b) -> b -> [a] -> [b]
def scanl(f):
'''scanl is like reduce, but
intermediate values, building from the left.
'''
def go(a):
def g(xs):
return accumulate(chain([a], xs), f)
return g
return go
Line 2,560 ⟶ 2,915:
if __name__ == '__main__':
main()</
{{Out}}
<pre>Terms 0 thru 10 inclusive:
Line 2,579 ⟶ 2,934:
Digit counts for terms 1k through 10k (inclusive) by k:
[2565, 5733, 9128, 12670, 16322, 20062, 23875, 27749, 31678, 35656]</pre>
=={{header|Quackery}}==
<syntaxhighlight lang="quackery"> [ 1 swap times [ i 1+ * ] ] is ! ( n --> n )
[ 0 swap times [ i ! + ] ] is !n ( n --> n )
say "First 11 left factorials:" cr
11 times [ i^ !n echo sp ] cr
cr
say "20 through 110 (inclusive) by tens:" cr
10 times [ i^ 2 + 10 * !n echo cr ]
cr
say "Digits in 1,000 through 10,000 by thousands:" cr
10 times [ i^ 1+ 1000 * !n number$ size echo cr ]
cr</syntaxhighlight>
{{out}}
<pre>First 11 left factorials:
0 1 2 4 10 34 154 874 5914 46234 409114
20 through 110 (inclusive) by tens:
128425485935180314
9157958657951075573395300940314
20935051082417771847631371547939998232420940314
620960027832821612639424806694551108812720525606160920420940314
141074930726669571000530822087000522211656242116439949000980378746128920420940314
173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314
906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314
16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314
942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314
145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314
Digits in 1,000 through 10,000 by thousands:
2565
5733
9128
12670
16322
20062
23875
27749
31678
35656</pre>
=={{header|R}}==
===Imperative solution===
<syntaxhighlight lang="rsplus">library(gmp)
left_factorial <- function(n) {
Line 2,613 ⟶ 3,013:
cat("!",n," has ",digit_count(left_factorial(n))," digits\n", sep = "")
}
</syntaxhighlight>
{{out}}
<pre>
Line 2,646 ⟶ 3,046:
!8000 has 27749 digits
!9000 has 31678 digits
!10000 has 35656 digits</pre>
===Vectorization solution===
Due to vectorization, these sorts of problems are R's bread and butter. The only challenge comes from making sure that R plays nice with objects from the gmp library.
<syntaxhighlight lang="rsplus">library(gmp)
leftFact <- function(numbs)
{
#As we will never actually use the numeric values of our outputs, we will
#immediately coerce them to characters. For technical reasons to do with
#nchar misbehaving, this also makes task 3 much easier.
#As task 1 will demonstrate, the n=0 special case is covered.
sapply(numbs, function(n) as.character(sum(factorialZ(seq_len(n)-1))))
}
printer <- function(inputs) print(data.frame(Value = leftFact(inputs), row.names = paste0("!", inputs)))
#Task 1
printer(0:10)
#Task 2
printer(seq(20, 110, by = 10))
#Task 3
inputs<-seq(1000, 10000, by = 1000)
print(data.frame(Digits = sapply(leftFact(inputs), nchar), row.names = paste0("!", inputs)))</syntaxhighlight>
{{out}}
<pre>> printer(0:10)
Value
!0 0
!1 1
!2 2
!3 4
!4 10
!5 34
!6 154
!7 874
!8 5914
!9 46234
!10 409114
> printer(seq(20, 110, by = 10))
Value
!20 128425485935180314
!30 9157958657951075573395300940314
!40 20935051082417771847631371547939998232420940314
!50 620960027832821612639424806694551108812720525606160920420940314
!60 141074930726669571000530822087000522211656242116439949000980378746128920420940314
!70 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314
!80 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314
!90 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314
!100 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314
!110 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314
> print(data.frame(Digits = sapply(leftFact(inputs), nchar), row.names = paste0("!", inputs)))
Digits
!1000 2565
!2000 5733
!3000 9128
!4000 12670
!5000 16322
!6000 20062
!7000 23875
!8000 27749
!9000 31678
!10000 35656</pre>
=={{header|Racket}}==
<
(define ! (let ((rv# (make-hash))) (λ (n) (hash-ref! rv# n (λ () (if (= n 0) 1 (* n (! (- n 1)))))))))
Line 2,667 ⟶ 3,127:
"Display the length (in decimal digits) of the left factorials for:"
"1,000, 2,000 through 10,000 (inclusive), by thousands."
(pretty-format (for/list ((i (in-range 1000 10001 1000))) (add1 (order-of-magnitude (!n i))))))</
{{out}}
Line 2,687 ⟶ 3,147:
1,000, 2,000 through 10,000 (inclusive), by thousands.
'(2565 5733 9128 12670 16322 20062 23875 27749 31678 35656)</pre>
=={{header|Raku}}==
(formerly Perl 6)
Implement left factorial as a prefix !. Note that this redefines the core prefix ! (not) function.
<syntaxhighlight lang="raku" line>sub prefix:<!> ($k) { (constant l = 0, |[\+] 1, (|[\*] 1..*))[$k] }
$ = !10000; # Pre-initialize
.say for ( 0 … 10, 20 … 110 ).hyper(:4batch).map: { sprintf "!%d = %s", $_, !$_ };
.say for (1000, 2000 … 10000).hyper(:4batch).map: { sprintf "!%d has %d digits.", $_, chars !$_ };</syntaxhighlight>
{{out}}
<pre>!0 = 0
!1 = 1
!2 = 2
!3 = 4
!4 = 10
!5 = 34
!6 = 154
!7 = 874
!8 = 5914
!9 = 46234
!10 = 409114
!20 = 128425485935180314
!30 = 9157958657951075573395300940314
!40 = 20935051082417771847631371547939998232420940314
!50 = 620960027832821612639424806694551108812720525606160920420940314
!60 = 141074930726669571000530822087000522211656242116439949000980378746128920420940314
!70 = 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314
!80 = 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314
!90 = 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314
!100 = 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314
!110 = 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314
!1000 has 2565 digits.
!2000 has 5733 digits.
!3000 has 9128 digits.
!4000 has 12670 digits.
!5000 has 16322 digits.
!6000 has 20062 digits.
!7000 has 23875 digits.
!8000 has 27749 digits.
!9000 has 31678 digits.
!10000 has 35656 digits.</pre>
If you would rather not override prefix ! operator and you can live with just defining lazy lists and indexing into them, this should suffice; (and is in fact very slightly faster than the first example since it avoids routine dispatch overhead):
<syntaxhighlight lang="raku" line>constant leftfact = 0, |[\+] 1, (|[\*] 1..*);
$ = leftfact[10000]; # Pre-initialize
.say for ( 0 … 10, 20 … 110 ).hyper(:4batch).map: { sprintf "!%d = %s", $_, leftfact[$_] };
.say for (1000, 2000 … 10000).hyper(:4batch).map: { sprintf "!%d has %d digits.", $_, chars leftfact[$_] };</syntaxhighlight>
Same output.
=={{header|REXX}}==
Programmer's note: this REXX version automatically adjusts the number of decimal digits (precision) after the factorial is calculated.
This is possible because there will be a number of trailing zeros which allow a floating point number to be converted to an integer.
<syntaxhighlight lang="rexx">/*REXX program computes/display the left factorial (or its dec. width) of N (or a range)*/
bot= abs(bot)
w= length(top)
do j=bot to top by inc
if
end /*j*/ /* [↑] show either L! or # of digits*/
exit 0 /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
L!: procedure; parse arg x .; if x<3 then return x;
return $
{{out|output|text= when using the input of: <tt> 0 10 </tt>}}
<pre>
left ! of 0 ───► 0
Line 2,725 ⟶ 3,239:
left ! of 10 ───► 409114
</pre>
<pre>
left ! of 20 ───► 128425485935180314
Line 2,738 ⟶ 3,252:
left ! of 110 ───► 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314
</pre>
<pre>
left ! of 1000 ───► 2565 digits
Line 2,753 ⟶ 3,267:
=={{header|Ring}}==
<
a = leftFact(0,10,1)
see "" + a + nl
Line 2,769 ⟶ 3,283:
else see "" + i + " " + leftFact + nl ok
next
</syntaxhighlight>
=={{header|RPL}}==
For small values of n, the built-in number type can support the job:
≪ '''IF''' DUP '''THEN''' 0 1 ROT 1 - '''FOR''' k k FACT + '''NEXT END''' ≫ ‘<span style="color:blue">LFACT</span>’ STO
≪ { } 0 10 '''FOR''' n n <span style="color:blue">LFACT</span> + '''NEXT''' ≫ EVAL
'''Output:'''
1: { 0 1 2 4 10 34 154 874 5914 46234 409114 }
For 20 through 110 (inclusive) by tens, we need a kind of BigInt library if using a RPL version from the previous century. <code>ADDbig</code> and <code>MULbig</code> are defined at [[Long multiplication#RPL|Long multiplication]].
Calculation is optimized by using the recursive formula: <code>!(n+1) = !n * n + 1</code>
≪ "1" SWAP
'''WHILE''' DUP 1 > '''REPEAT'''
1 - DUP →STR ROT <span style="color:blue">MULbig</span> "1" <span style="color:blue"><span style="color:blue">ADDbig</span></span> SWAP
'''END''' DROP
≫ ‘<span style="color:blue">LFACTbig</span>’ STO
≪ 20 110 '''FOR''' n n <span style="color:blue">LFACTbig</span> 10 '''STEP''' ≫ EVAL
{{out}}
<pre>
10: "128425485935180314"
9: "9157958657951075573395300940314"
8: "20935051082417771847631371547939998232420940314"
7: "620960027832821612639424806694551108812720525606160920420940314"
6: "141074930726669571000530822087000522211656242116439949000980378746128920420940314"
5: "173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314"
4: "906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314"
3: "16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314"
2: "942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314"
1: "145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314"
</pre>
=={{header|Ruby}}==
<
f, lf = 1, 0
1.step do |n|
Line 2,779 ⟶ 3,324:
f *= n
end
end</
'''Test:'''
<
thousands = 1000.step(10_000, 1000)
Line 2,792 ⟶ 3,337:
puts "!#{n} has #{lf.to_s.size} digits"
end
end</
{{out}}
<pre>!0 = 0
Line 2,828 ⟶ 3,373:
=={{header|Run BASIC}}==
<
a = lftFct(20,110,10)
a = lftFct(1000,10000,1000)
Line 2,847 ⟶ 3,392:
end if
next i
end function</
<pre>------ From 0 --To-> 10 Step 1 -------
0 1
Line 2,886 ⟶ 3,431:
=={{header|Rust}}==
<syntaxhighlight lang="rust">
#[cfg(target_pointer_width = "64")]
type USingle = u32;
Line 3,010 ⟶ 3,555:
}
}
</syntaxhighlight>
{{out}}
<pre>
Line 3,044 ⟶ 3,589:
!9000 has 31678 digits.
!10000 has 35656 digits.
</pre>
===Faster alternative===
This solution uses the arbitrary precision integers from the rug crate,
which uses GMP under the covers.
<syntaxhighlight lang="rust">// [dependencies]
// rug = "1.9"
fn left_factorials() -> impl std::iter::Iterator<Item = rug::Integer> {
use rug::Integer;
let mut factorial = Integer::from(1);
let mut next = Integer::from(0);
let mut n = 1;
std::iter::from_fn(move || {
let result = next.clone();
next += &factorial;
factorial *= n;
n += 1;
Some(result)
})
}
fn main() {
let mut lf = left_factorials().take(10001).enumerate();
println!("Left factorials 0 through 10:");
for (i, n) in lf.by_ref().take(11) {
println!("!{} = {}", i, n);
}
println!("Left factorials 20 through 110, by tens:");
for (i, n) in lf.by_ref().take(100).skip(9).step_by(10) {
println!("!{} = {}", i, n);
}
println!("Lengths of left factorials 1000 through 10000, by thousands:");
for (i, n) in lf.skip(1000 - 111).step_by(1000) {
println!("length of !{} = {}", i, n.to_string().len());
}
}</syntaxhighlight>
{{out}}
<pre>
Left factorials 0 through 10:
!0 = 0
!1 = 1
!2 = 2
!3 = 4
!4 = 10
!5 = 34
!6 = 154
!7 = 874
!8 = 5914
!9 = 46234
!10 = 409114
Left factorials 20 through 110, by tens:
!20 = 128425485935180314
!30 = 9157958657951075573395300940314
!40 = 20935051082417771847631371547939998232420940314
!50 = 620960027832821612639424806694551108812720525606160920420940314
!60 = 141074930726669571000530822087000522211656242116439949000980378746128920420940314
!70 = 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314
!80 = 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314
!90 = 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314
!100 = 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314
!110 = 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314
Lengths of left factorials 1000 through 10000, by thousands:
length of !1000 = 2565
length of !2000 = 5733
length of !3000 = 9128
length of !4000 = 12670
length of !5000 = 16322
length of !6000 = 20062
length of !7000 = 23875
length of !8000 = 27749
length of !9000 = 31678
length of !10000 = 35656
</pre>
=={{header|Scala}}==
<
// this part isn't really necessary, it just shows off Scala's ability
Line 3,067 ⟶ 3,686:
}
}
</syntaxhighlight>
{{out}}
<pre>
Line 3,110 ⟶ 3,729:
and (iota 5 100 2) produces a list (100 102 104 106 108)
<
(import (scheme base) ;; library imports in R7RS style
(scheme write)
Line 3,138 ⟶ 3,757:
(lambda (i) (show i (string-length (number->string (left-factorial i)))))
(iota 10 1000 1000))
</syntaxhighlight>
=={{header|Seed7}}==
<
include "bigint.s7i";
Line 3,176 ⟶ 3,795:
end for;
writeln;
end func;</
{{out}}
Line 3,208 ⟶ 3,827:
=={{header|Sidef}}==
Built-in:
<syntaxhighlight lang="ruby">say 20.of { .left_factorial }</syntaxhighlight>
Straightforward:
<
^
}</
<
^
}</
A
<
static cached = 0
static factorial = 1
static leftfact = 0
if (n < cached) {
cached = 0
Line 3,229 ⟶ 3,851:
leftfact = 0
}
while (n > cached) {
leftfact += factorial
factorial *= ++cached
}
leftfact
}</
Completing the task:
<
printf("!%d = %s\n",
}
for
printf("!%d has %d digits.\n",
}</
{{out}}
Line 3,279 ⟶ 3,901:
!9000 has 31678 digits.
!10000 has 35656 digits.
</pre>
=={{header|Standard ML}}==
<syntaxhighlight lang="sml">
(* reuse earlier factorial calculations in dfac, apply to listed arguments in cumlfac *)
(* example: left factorial n, is #3 (dfac (0,n-1,1,1) ) *)
(* output list contains (number, factorial, left factorial) *)
(* tested in PolyML *)
val store = ref 0;
val rec dfac = fn
(from,to,acc,cm) => if from = to then (from,acc,cm) else (store:=(from+1)*acc;dfac (from+1,to,!store,!store+cm ) );
val rec cumlfac = fn
(x::y::rm) => x :: cumlfac ( dfac (#1 x, #1 y, #2 x, #3 x) :: rm ) |
rm =>rm ;
val arguments = List.tabulate (10,fn 0=>(0,1,1)|i=>(i,0,0)) @
List.tabulate (10,fn i=> (10*i+19,0,0) ) @
List.tabulate ( 10,fn i=> (1000*i+999,0,0));
val result = (~1,0,0)::(cumlfac arguments);
(* done *)
(* display: *)
List.app (fn triple :int*int*int =>
print(Int.toString (1+ #1 triple ) ^ " : " ^ Int.fmt StringCvt.DEC (#3 triple ) ^" \n" )
) (List.take(result,21) ) ;
List.app (fn triple :int*int*int =>
print( Int.toString (1+ #1 triple ) ^ " : " ^ Int.toString (size(Int.toString (#3 triple ))) ^" \n" ) ) (List.drop(result,21) );
</syntaxhighlight>
{{out}}
<pre>
time poly --script thisscript
0 : 0
1 : 1
2 : 2
3 : 4
4 : 10
5 : 34
6 : 154
7 : 874
8 : 5914
9 : 46234
10 : 409114
20 : 128425485935180314
30 : 9157958657951075573395300940314
40 : 20935051082417771847631371547939998232420940314
50 : 620960027832821612639424806694551108812720525606160920420940314
60 : 141074930726669571000530822087000522211656242116439949000980378746128920420940314
70 : 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314
80 : 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314
90 : 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314
100 : 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314
110 : 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314
1000 : 2565
2000 : 5733
3000 : 9128
4000 : 12670
5000 : 16322
6000 : 20062
7000 : 23875
8000 : 27749
9000 : 31678
10000 : 35656
(CPU 2.1Ghz:) 0.36 real 0.29 user 0.08 sys
</pre>
Line 3,285 ⟶ 3,976:
{{libheader|AttaSwift BigInt}}
<
func factorial<T: BinaryInteger>(_ n: T) -> T {
Line 3,321 ⟶ 4,012:
for i in stride(from: BigInt(2000), through: 10_000, by: 1000) {
print("!\(i) = \((!i).description.count) digit number")
}</
{{out}}
Line 3,360 ⟶ 4,051:
!10000 = 35656 digit number</pre>
=={{header|Tcl}}==
<
set s 0
for {set i [set f 1]} {$i <= $n} {incr i} {
Line 3,443 ⟶ 4,066:
for {set i 1000} {$i <= 10000} {incr i 1000} {
puts "!$i has [string length [leftfact $i]] digits"
}</
{{out}}
<pre>
Line 3,477 ⟶ 4,100:
!9000 has 31678 digits
!10000 has 35656 digits
</pre>
=={{header|Wren}}==
{{libheader|Wren-fmt}}
{{libheader|Wren-big}}
<syntaxhighlight lang="wren">import "./fmt" for Fmt
import "./big" for BigInt
var lfacts = List.filled(12, BigInt.zero)
var lfact = BigInt.one
var sum = BigInt.zero
for (i in 1..10) {
sum = sum + lfact
lfacts[i] = sum
lfact = lfact * i
}
System.print("Left factorials from 0 to 10:")
for (i in 0..10) System.write(" %(lfacts[i])")
for (i in 11..110) {
sum = sum + lfact
if (i%10 == 0) lfacts[i/10] = sum
lfact = lfact * i
}
System.print("\n\nLeft factorials from 20 to 110 by tens:")
for (i in 2..11) Fmt.print(" !$-3d -> $s", i * 10, lfacts[i])
for (i in 111..10000) {
sum = sum + lfact
if (i%1000 == 0) lfacts[i/1000] = sum
lfact = lfact * i
}
System.print("\nLengths of left factorals from 1000 to 10000 by thousands:")
for (i in 1..10) Fmt.print(" !$-5d -> $5s", i * 1000, lfacts[i].toString.count)</syntaxhighlight>
{{out}}
<pre>
Left factorials from 0 to 10:
0 1 2 4 10 34 154 874 5914 46234 409114
Left factorials from 20 to 110 by tens:
!20 -> 128425485935180314
!30 -> 9157958657951075573395300940314
!40 -> 20935051082417771847631371547939998232420940314
!50 -> 620960027832821612639424806694551108812720525606160920420940314
!60 -> 141074930726669571000530822087000522211656242116439949000980378746128920420940314
!70 -> 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314
!80 -> 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314
!90 -> 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314
!100 -> 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314
!110 -> 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314
Lengths of left factorals from 1000 to 10000 by thousands:
!1000 -> 2565
!2000 -> 5733
!3000 -> 9128
!4000 -> 12670
!5000 -> 16322
!6000 -> 20062
!7000 -> 23875
!8000 -> 27749
!9000 -> 31678
!10000 -> 35656
</pre>
=={{header|zkl}}==
{{trans|D}}
<
fcn leftFact(n){
[1..n].reduce(fcn(p,n,rf){ p+=rf.value; rf.set(rf.value*n); p },
BN(0),Ref(BN(1)));
}</
<
lfs:=[20..111,10].apply(leftFact);
println(("\n20 through 110 (inclusive) by tens:\n" +
Line 3,493 ⟶ 4,180:
println("Digits in 1,000 through 10,000 by thousands:\n",
[0d1_000..0d10_000, 1000].pump(List,fcn(n){leftFact(n).toString().len()}));</
{{out}}
<pre>
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