Stirling numbers of the second kind
Stirling numbers of the second kind, or Stirling partition numbers, are the number of ways to partition a set of n objects into k non-empty subsets. They are closely related to Bell numbers, and may be derived from them.
You are encouraged to solve this task according to the task description, using any language you may know.
Stirling numbers of the second kind obey the recurrence relation:
S2(n, 0) and S2(0, k) = 0 # for n, k > 0 S2(n, n) = 1 S2(n + 1, k) = k * S2(n, k) + S2(n, k - 1)
- Task
- Write a routine (function, procedure, whatever) to find Stirling numbers of the second kind. There are several methods to generate Stirling numbers of the second kind. You are free to choose the most appropriate for your language. If your language has a built-in, or easily, publicly available library implementation, it is acceptable to use that.
- Using the routine, generate and show here, on this page, a table (or triangle) showing the Stirling numbers of the second kind, S2(n, k), up to S2(12, 12). it is optional to show the row / column for n == 0 and k == 0. It is optional to show places where S2(n, k) == 0 (when k > n).
- If your language supports large integers, find and show here, on this page, the maximum value of S2(n, k) where n == 100.
- See also
- Related Tasks
ALGOL 68
Uses the LONG LONG INT mode of Algol 68g which allows large precision integers. As the default precision of LONG LONG INT is too small, the precision is specified via a pragmatic comment. <lang algol68>BEGIN
# show some Stirling numbers of the second kind #
# specify the precision of LONG LONG INT, somewhat under 160 digits are # # needed for Stirling numbers of the second kind with n, k = 100 # PR precision 160 PR MODE SINT = LONG LONG INT;
# returns a triangular matrix of Stirling numbers up to max n, max n # PROC make s2 = ( INT max n )REF[,]SINT: BEGIN REF[,]SINT s2 := HEAP[ 0 : max n, 0 : max n ]SINT; FOR n FROM 0 TO max n DO FOR k FROM 0 TO max n DO s2[ n, k ] := 0 OD OD; FOR n FROM 0 TO max n DO s2[ n, n ] := 1 OD; FOR n FROM 0 TO max n - 1 DO FOR k FROM 1 TO n DO s2[ n + 1, k ] := k * s2[ n, k ] + s2[ n, k - 1 ]; OD OD; s2 END # make s2 # ; # task requirements: # # print Stirling numbers up to n, k = 12 # BEGIN INT max stirling = 12; REF[,]SINT s2 = make s2( max stirling ); print( ( "Stirling numbers of the second kind:", newline ) ); print( ( " k" ) ); FOR k FROM 0 TO max stirling DO print( ( whole( k, -10 ) ) ) OD; print( ( newline, " n", newline ) ); FOR n FROM 0 TO max stirling DO print( ( whole( n, -2 ) ) ); FOR k FROM 0 TO n DO print( ( whole( s2[ n, k ], -10 ) ) ) OD; print( ( newline ) ) OD END; # find the maximum Stirling number with n = 100 # BEGIN INT max stirling = 100; REF[,]SINT s2 = make s2( max stirling ); SINT max 100 := 0; FOR k FROM 0 TO max stirling DO IF s2[ max stirling, k ] > max 100 THEN max 100 := s2[ max stirling, k ] FI OD; print( ( "Maximum Stirling number of the second kind with n = 100:", newline ) ); print( ( whole( max 100, 0 ), newline ) ) END
END</lang>
- Output:
Stirling numbers of the second kind: k 0 1 2 3 4 5 6 7 8 9 10 11 12 n 0 1 1 0 1 2 0 1 1 3 0 1 3 1 4 0 1 7 6 1 5 0 1 15 25 10 1 6 0 1 31 90 65 15 1 7 0 1 63 301 350 140 21 1 8 0 1 127 966 1701 1050 266 28 1 9 0 1 255 3025 7770 6951 2646 462 36 1 10 0 1 511 9330 34105 42525 22827 5880 750 45 1 11 0 1 1023 28501 145750 246730 179487 63987 11880 1155 55 1 12 0 1 2047 86526 611501 1379400 1323652 627396 159027 22275 1705 66 1 Maximum Stirling number of the second kind with n = 100: 7769730053598745155212806612787584787397878128370115840974992570102386086289805848025074822404843545178960761551674
Factor
<lang factor>USING: combinators.short-circuit formatting io kernel math math.extras prettyprint sequences ; RENAME: stirling math.extras => (stirling) IN: rosetta-code.stirling-second
! Tweak Factor's in-built stirling function for k=0
- stirling ( n k -- m )
2dup { [ = not ] [ nip zero? ] } 2&& [ 2drop 0 ] [ (stirling) ] if ;
"Stirling numbers of the second kind: n k stirling:" print "n\\k" write 13 dup [ "%8d" printf ] each-integer nl
<iota> [
dup dup "%-2d " printf [0,b] [ stirling "%8d" printf ] with each nl
] each nl
"Maximum value from the 100 _ stirling row:" print 100 <iota> [ 100 swap stirling ] map supremum .</lang>
- Output:
Stirling numbers of the second kind: n k stirling: n\k 0 1 2 3 4 5 6 7 8 9 10 11 12 0 1 1 0 1 2 0 1 1 3 0 1 3 1 4 0 1 7 6 1 5 0 1 15 25 10 1 6 0 1 31 90 65 15 1 7 0 1 63 301 350 140 21 1 8 0 1 127 966 1701 1050 266 28 1 9 0 1 255 3025 7770 6951 2646 462 36 1 10 0 1 511 9330 34105 42525 22827 5880 750 45 1 11 0 1 1023 28501 145750 246730 179487 63987 11880 1155 55 1 12 0 1 2047 86526 611501 1379400 1323652 627396 159027 22275 1705 66 1 Maximum value from the 100 _ stirling row: 7769730053598745155212806612787584787397878128370115840974992570102386086289805848025074822404843545178960761551674
Fōrmulæ
In this page you can see the solution of this task.
Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text (more info). Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for transportation effects more than visualization and edition.
The option to show Fōrmulæ programs and their results is showing images. Unfortunately images cannot be uploaded in Rosetta Code.
Go
<lang go>package main
import (
"fmt" "math/big"
)
func main() {
limit := 100 last := 12 s2 := make([][]*big.Int, limit+1) for n := 0; n <= limit; n++ { s2[n] = make([]*big.Int, limit+1) for k := 0; k <= limit; k++ { s2[n][k] = new(big.Int) } s2[n][n].SetInt64(int64(1)) } var t big.Int for n := 1; n <= limit; n++ { for k := 1; k <= n; k++ { t.SetInt64(int64(k)) t.Mul(&t, s2[n-1][k]) s2[n][k].Add(&t, s2[n-1][k-1]) } } fmt.Println("Stirling numbers of the second kind: S2(n, k):") fmt.Printf("n/k") for i := 0; i <= last; i++ { fmt.Printf("%9d ", i) } fmt.Printf("\n--") for i := 0; i <= last; i++ { fmt.Printf("----------") } fmt.Println() for n := 0; n <= last; n++ { fmt.Printf("%2d ", n) for k := 0; k <= n; k++ { fmt.Printf("%9d ", s2[n][k]) } fmt.Println() } fmt.Println("\nMaximum value from the S2(100, *) row:") max := new(big.Int).Set(s2[limit][0]) for k := 1; k <= limit; k++ { if s2[limit][k].Cmp(max) > 0 { max.Set(s2[limit][k]) } } fmt.Println(max) fmt.Printf("which has %d digits.\n", len(max.String()))
}</lang>
- Output:
Stirling numbers of the second kind: S2(n, k): n/k 0 1 2 3 4 5 6 7 8 9 10 11 12 ------------------------------------------------------------------------------------------------------------------------------------ 0 1 1 0 1 2 0 1 1 3 0 1 3 1 4 0 1 7 6 1 5 0 1 15 25 10 1 6 0 1 31 90 65 15 1 7 0 1 63 301 350 140 21 1 8 0 1 127 966 1701 1050 266 28 1 9 0 1 255 3025 7770 6951 2646 462 36 1 10 0 1 511 9330 34105 42525 22827 5880 750 45 1 11 0 1 1023 28501 145750 246730 179487 63987 11880 1155 55 1 12 0 1 2047 86526 611501 1379400 1323652 627396 159027 22275 1705 66 1 Maximum value from the S2(100, *) row: 7769730053598745155212806612787584787397878128370115840974992570102386086289805848025074822404843545178960761551674 which has 115 digits.
Java
<lang java> import java.math.BigInteger; import java.util.HashMap; import java.util.Map;
public class SterlingNumbersSecondKind {
public static void main(String[] args) { System.out.println("Stirling numbers of the second kind:"); int max = 12; System.out.printf("n/k"); for ( int n = 0 ; n <= max ; n++ ) { System.out.printf("%10d", n); } System.out.printf("%n"); for ( int n = 0 ; n <= max ; n++ ) { System.out.printf("%-3d", n); for ( int k = 0 ; k <= n ; k++ ) { System.out.printf("%10s", sterling2(n, k)); } System.out.printf("%n"); } System.out.println("The maximum value of S2(100, k) = "); BigInteger previous = BigInteger.ZERO; for ( int k = 1 ; k <= 100 ; k++ ) { BigInteger current = sterling2(100, k); if ( current.compareTo(previous) > 0 ) { previous = current; } else { System.out.printf("%s%n(%d digits, k = %d)%n", previous, previous.toString().length(), k-1); break; } } } private static Map<String,BigInteger> COMPUTED = new HashMap<>(); private static final BigInteger sterling2(int n, int k) { String key = n + "," + k; if ( COMPUTED.containsKey(key) ) { return COMPUTED.get(key); } if ( n == 0 && k == 0 ) { return BigInteger.valueOf(1); } if ( (n > 0 && k == 0) || (n == 0 && k > 0) ) { return BigInteger.ZERO; } if ( n == k ) { return BigInteger.valueOf(1); } if ( k > n ) { return BigInteger.ZERO; } BigInteger result = BigInteger.valueOf(k).multiply(sterling2(n-1, k)).add(sterling2(n-1, k-1)); COMPUTED.put(key, result); return result; }
} </lang>
- Output:
Stirling numbers of the second kind: n/k 0 1 2 3 4 5 6 7 8 9 10 11 12 0 1 1 0 1 2 0 1 1 3 0 1 3 1 4 0 1 7 6 1 5 0 1 15 25 10 1 6 0 1 31 90 65 15 1 7 0 1 63 301 350 140 21 1 8 0 1 127 966 1701 1050 266 28 1 9 0 1 255 3025 7770 6951 2646 462 36 1 10 0 1 511 9330 34105 42525 22827 5880 750 45 1 11 0 1 1023 28501 145750 246730 179487 63987 11880 1155 55 1 12 0 1 2047 86526 611501 1379400 1323652 627396 159027 22275 1705 66 1 The maximum value of S2(100, k) = 7769730053598745155212806612787584787397878128370115840974992570102386086289805848025074822404843545178960761551674 (115 digits, k = 28)
Julia
<lang julia>using Combinatorics
const s2cache = Dict()
function stirlings2(n, k)
if haskey(s2cache, Pair(n, k)) return s2cache[Pair(n, k)] elseif n < 0 throw(DomainError(n, "n must be nonnegative")) elseif n == k == 0 return one(n) elseif n == 0 || k == 0 return zero(n) elseif k == n - 1 return binomial(n, 2) elseif k == 2 return 2^(n-1) - 1 end
ret = k * stirlings2(n - 1, k) + stirlings2(n - 1, k - 1) s2cache[Pair(n, k)] = ret return ret
end
function printstirling2table(kmax)
println(" ", mapreduce(i -> lpad(i, 10), *, 0:kmax))
sstring(n, k) = begin i = stirlings2(n, k); lpad(k > n && i == 0 ? "" : i, 10) end
for n in 0:kmax println(rpad(n, 2) * mapreduce(k -> sstring(n, k), *, 0:kmax)) end
end
printstirling2table(12) println("\nThe maximum for stirling2(100, _) is: ", maximum(k-> stirlings2(BigInt(100), BigInt(k)), 1:100))
</lang>
- Output:
0 1 2 3 4 5 6 7 8 9 10 11 12 0 1 1 0 1 2 0 1 1 3 0 1 3 1 4 0 1 7 6 1 5 0 1 15 25 10 1 6 0 1 31 90 65 15 1 7 0 1 63 301 350 140 21 1 8 0 1 127 966 1701 1050 266 28 1 9 0 1 255 3025 7770 6951 2646 462 36 1 10 0 1 511 9330 34105 42525 22827 5880 750 45 1 11 0 1 1023 28501 145750 246730 179487 63987 11880 1155 55 1 12 0 1 2047 86526 611501 1379400 1323652 627396 159027 22275 1705 66 1 The maximum for stirling2(100, _) is: 7769730053598745155212806612787584787397878128370115840974992570102386086289805848025074822404843545178960761551674
Perl
<lang perl>use strict; use warnings; use bigint; use feature 'say'; use feature 'state'; no warnings 'recursion'; use List::Util qw(max);
sub Stirling2 {
my($n, $k) = @_; my $n1 = $n - 1; return 1 if $n1 == $k; return 0 unless $n1 && $k; state %seen; return ($seen{"{$n1}|{$k}" } //= Stirling2($n1,$k ) * $k) + ($seen{"{$n1}|{$k-1}"} //= Stirling2($n1,$k-1))
}
my $upto = 12; my $width = 1 + length max map { Stirling2($upto+1,$_) } 0..$upto+1;
say 'Unsigned Stirling2 numbers of the second kind: S2(n, k):'; print 'n\k' . sprintf "%${width}s"x(1+$upto)."\n", 0..$upto;
for my $row (1..$upto+1) {
printf '%-3d', $row-1; printf "%${width}d", Stirling2($row, $_) for 0..$row-1; print "\n";
}
say "\nMaximum value from the S2(100, *) row:"; say max map { Stirling2(101,$_) } 0..100;</lang>
- Output:
Unsigned Stirling2 numbers of the second kind: S2(n, k): n\k 0 1 2 3 4 5 6 7 8 9 10 11 12 0 1 1 0 1 2 0 1 1 3 0 1 3 1 4 0 1 7 6 1 5 0 1 15 25 10 1 6 0 1 31 90 65 15 1 7 0 1 63 301 350 140 21 1 8 0 1 127 966 1701 1050 266 28 1 9 0 1 255 3025 7770 6951 2646 462 36 1 10 0 1 511 9330 34105 42525 22827 5880 750 45 1 11 0 1 1023 28501 145750 246730 179487 63987 11880 1155 55 1 12 0 1 2047 86526 611501 1379400 1323652 627396 159027 22275 1705 66 1 Maximum value from the S2(100, *) row: 7769730053598745155212806612787584787397878128370115840974992570102386086289805848025074822404843545178960761551674
Phix
<lang Phix>include mpfr.e
constant lim = 100,
lim1 = lim+1, last = 12
sequence s2 = repeat(0,lim1) for n=1 to lim1 do
s2[n] = mpz_inits(lim1) mpz_set_si(s2[n][n],1)
end for mpz {t, m100} = mpz_inits(2) for n=1 to lim do
for k=1 to n do mpz_set_si(t,k) mpz_mul(t,t,s2[n][k+1]) mpz_add(s2[n+1][k+1],t,s2[n][k]) end for
end for printf(1,"Stirling numbers of the second kind: S2(n, k):\n n k:") for i=0 to last do
printf(1,"%5d ", i)
end for printf(1,"\n--- %s\n",repeat('-',last*10+5)) for n=0 to last do
printf(1,"%2d ", n) for k=1 to n+1 do mpfr_printf(1,"%9Zd ", s2[n+1][k]) end for printf(1,"\n")
end for for k=1 to lim1 do
mpz s100k = s2[lim1][k] if mpz_cmp(s100k,m100) > 0 then mpz_set(m100,s100k) end if
end for printf(1,"\nThe maximum S2(100,k): %s\n",shorten(mpz_get_str(m100)))</lang>
- Output:
Stirling numbers of the second kind: S2(n, k): n k: 0 1 2 3 4 5 6 7 8 9 10 11 12 --- ------------------------------------------------------------------------------------------------------------------------------ 0 1 1 0 1 2 0 1 1 3 0 1 3 1 4 0 1 7 6 1 5 0 1 15 25 10 1 6 0 1 31 90 65 15 1 7 0 1 63 301 350 140 21 1 8 0 1 127 966 1701 1050 266 28 1 9 0 1 255 3025 7770 6951 2646 462 36 1 10 0 1 511 9330 34105 42525 22827 5880 750 45 1 11 0 1 1023 28501 145750 246730 179487 63987 11880 1155 55 1 12 0 1 2047 86526 611501 1379400 1323652 627396 159027 22275 1705 66 1 Maximum value from the S2(100, *) row: 7769730053598745155...3545178960761551674 (115 digits)
Python
<lang Python> computed = {}
def sterling2(n, k): key = str(n) + "," + str(k)
if key in computed.keys(): return computed[key] if n == k == 0: return 1 if (n > 0 and k == 0) or (n == 0 and k > 0): return 0 if n == k: return 1 if k > n: return 0 result = k * sterling2(n - 1, k) + sterling2(n - 1, k - 1) computed[key] = result return result
print("Stirling numbers of the second kind:") MAX = 12 print("n/k".ljust(10), end="") for n in range(MAX + 1): print(str(n).rjust(10), end="") print() for n in range(MAX + 1): print(str(n).ljust(10), end="") for k in range(n + 1): print(str(sterling2(n, k)).rjust(10), end="") print() print("The maximum value of S2(100, k) = ") previous = 0 for k in range(1, 100 + 1): current = sterling2(100, k) if current > previous: previous = current else: print("{0}\n({1} digits, k = {2})\n".format(previous, len(str(previous)), k - 1)) break </lang>
- Output:
Stirling numbers of the second kind: n/k 0 1 2 3 4 5 6 7 8 9 10 11 12 0 1 1 0 1 2 0 1 1 3 0 1 3 1 4 0 1 7 6 1 5 0 1 15 25 10 1 6 0 1 31 90 65 15 1 7 0 1 63 301 350 140 21 1 8 0 1 127 966 1701 1050 266 28 1 9 0 1 255 3025 7770 6951 2646 462 36 1 10 0 1 511 9330 34105 42525 22827 5880 750 45 1 11 0 1 1023 28501 145750 246730 179487 63987 11880 1155 55 1 12 0 1 2047 86526 611501 1379400 1323652 627396 159027 22275 1705 66 1 The maximum value of S2(100, k) = 7769730053598745155212806612787584787397878128370115840974992570102386086289805848025074822404843545178960761551674 (115 digits, k = 28)
Raku
(formerly Perl 6)
<lang perl6>sub Stirling2 (Int \n, Int \k) {
((1,), { (0, |@^last) »+« (|(@^last »*« @^last.keys), 0) } … *)[n;k]
}
my $upto = 12;
my $mx = (1..^$upto).map( { Stirling2($upto, $_) } ).max.chars;
put 'Stirling numbers of the second kind: S2(n, k):'; put 'n\k', (0..$upto)».fmt: "%{$mx}d";
for 0..$upto -> $row {
$row.fmt('%-3d').print; put (0..$row).map( { Stirling2($row, $_) } )».fmt: "%{$mx}d";
}
say "\nMaximum value from the S2(100, *) row:"; say (^100).map( { Stirling2 100, $_ } ).max;</lang>
- Output:
Stirling numbers of the second kind: S2(n, k): n\k 0 1 2 3 4 5 6 7 8 9 10 11 12 0 1 1 0 1 2 0 1 1 3 0 1 3 1 4 0 1 7 6 1 5 0 1 15 25 10 1 6 0 1 31 90 65 15 1 7 0 1 63 301 350 140 21 1 8 0 1 127 966 1701 1050 266 28 1 9 0 1 255 3025 7770 6951 2646 462 36 1 10 0 1 511 9330 34105 42525 22827 5880 750 45 1 11 0 1 1023 28501 145750 246730 179487 63987 11880 1155 55 1 12 0 1 2047 86526 611501 1379400 1323652 627396 159027 22275 1705 66 1 Maximum value from the S2(100, *) row: 7769730053598745155212806612787584787397878128370115840974992570102386086289805848025074822404843545178960761551674
REXX
Some extra code was added to minimize the displaying of the column widths. <lang>/*REXX program to compute and display Stirling numbers of the second kind. */ parse arg lim . /*obtain optional argument from the CL.*/ if lim== | lim=="," then lim= 12 /*Not specified? Then use the default.*/ olim= lim /*save the original value of LIM. */ lim= abs(lim) /*only use the absolute value of LIM. */ numeric digits max(9, 2*lim) /*(over) specify maximum number in grid*/ @.=
do j=0 for lim+1 @.j.j = 1; if j==0 then iterate /*define the right descending diagonal.*/ @.0.j = 0; @.j.0 = 0 /* " " zero values. */ end /*j*/
max#.= 0 /* [↓] calculate values for the grid. */
do n=0 for lim+1; np= n + 1 do k=1 for lim; km= k - 1 @.np.k = k * @.n.k + @.n.km /*calculate a number in the grid. */ max#.k= max(max#.k, @.n.k) /*find the maximum value for a column. */ max#.b= max(max#.b, @.n.k) /*find the maximum value for all rows. */ end /*k*/ end /*n*/ /* [↓] only show the maximum value ? */ do k=0 for lim+1 /*find max column width for each column*/ max#.a= max#.a + length(max#.k) end /*k*/
w= length(max#.b) /*calculate max width of all numbers. */ if olim<0 then do; say 'The maximum value (which has ' w " decimal digits):"
say max#.b /*display maximum number in the grid. */ exit /*stick a fork in it, we're all done. */ end
wi= max(3, length(lim+1) ) /*the maximum width of the grid's index*/ say 'row' center('columns', max(9, max#.a + lim), '═') /*display header of the grid.*/
do r=0 for lim+1; $= /* [↓] display the grid to the term. */ do c=0 for lim+1 until c>=r /*build a row of grid, 1 col at a time.*/ $= $ right(@.r.c, length(max#.c) ) /*append a column to a row of the grid.*/ end /*c*/ say right(r,wi) strip(substr($,2), 'T') /*display a single row of the grid. */ end /*r*/ /*stick a fork in it, we're all done. */</lang>
- output when using the default input:
row ══════════════════════════════columns══════════════════════════════ 0 1 1 0 1 2 0 1 1 3 0 1 3 1 4 0 1 7 6 1 5 0 1 15 25 10 1 6 0 1 31 90 65 15 1 7 0 1 63 301 350 140 21 1 8 0 1 127 966 1701 1050 266 28 1 9 0 1 255 3025 7770 6951 2646 462 36 1 10 0 1 511 9330 34105 42525 22827 5880 750 45 1 11 0 1 1023 28501 145750 246730 179487 63987 11880 1155 55 1 12 0 1 2047 86526 611501 1379400 1323652 627396 159027 22275 1705 66 1
- output when using the input of: -100
The maximum value (which has 115 decimal digits): 7769730053598745155212806612787584787397878128370115840974992570102386086289805848025074822404843545178960761551674
Sidef
<lang ruby>func S2(n, k) { # Stirling numbers of the second kind
stirling2(n, k)
}
const r = (0..12)
var triangle = r.map {|n| 0..n -> map {|k| S2(n, k) } } var widths = r.map {|n| r.map {|k| (triangle[k][n] \\ 0).len }.max }
say ('n\k ', r.map {|n| "%*s" % (widths[n], n) }.join(' '))
r.each {|n|
var str = ('%-3s ' % n) str += triangle[n].map_kv {|k,v| "%*s" % (widths[k], v) }.join(' ') say str
}
with (100) {|n|
say "\nMaximum value from the S2(#{n}, *) row:" say { S2(n, _) }.map(^n).max
}</lang>
- Output:
n\k 0 1 2 3 4 5 6 7 8 9 10 11 12 0 1 1 0 1 2 0 1 1 3 0 1 3 1 4 0 1 7 6 1 5 0 1 15 25 10 1 6 0 1 31 90 65 15 1 7 0 1 63 301 350 140 21 1 8 0 1 127 966 1701 1050 266 28 1 9 0 1 255 3025 7770 6951 2646 462 36 1 10 0 1 511 9330 34105 42525 22827 5880 750 45 1 11 0 1 1023 28501 145750 246730 179487 63987 11880 1155 55 1 12 0 1 2047 86526 611501 1379400 1323652 627396 159027 22275 1705 66 1 Maximum value from the S2(100, *) row: 7769730053598745155212806612787584787397878128370115840974992570102386086289805848025074822404843545178960761551674
Alternatively, the S2(n,k) function can be defined as: <lang ruby>func S2((0), (0)) { 1 } func S2(_, (0)) { 0 } func S2((0), _) { 0 } func S2(n, k) is cached { S2(n-1, k)*k + S2(n-1, k-1) }</lang>
zkl
<lang zkl>fcn stirling2(n,k){
var seen=Dictionary(); // cache for recursion if(n==k) return(1); // (0.0)==1 if(n<1 or k<1) return(0); z1,z2 := "%d,%d".fmt(n-1,k), "%d,%d".fmt(n-1,k-1); if(Void==(s1 := seen.find(z1))){ s1 = seen[z1] = stirling2(n-1,k) } if(Void==(s2 := seen.find(z2))){ s2 = seen[z2] = stirling2(n-1,k-1) } k*s1 + s2; // k is first to cast to BigInt (if using BigInts)
}</lang> <lang zkl>// calculate entire table (cached), find max, find num digits in max N,mx := 12, [1..N].apply(fcn(n){ [1..n].apply(stirling2.fp(n)) }).flatten() : (0).max(_); fmt:="%%%dd".fmt("%d".fmt(mx.numDigits + 1)).fmt; // "%9d".fmt println("Stirling numbers of the second kind: S2(n,k):"); println("n\\k",[0..N].pump(String,fmt)); foreach row in ([0..N]){
println("%3d".fmt(row), [0..row].pump(String, stirling2.fp(row), fmt));
}</lang>
- Output:
Stirling numbers of the second kind: S2(n,k): n\k 0 1 2 3 4 5 6 7 8 9 10 11 12 0 1 1 0 1 2 0 1 1 3 0 1 3 1 4 0 1 7 6 1 5 0 1 15 25 10 1 6 0 1 31 90 65 15 1 7 0 1 63 301 350 140 21 1 8 0 1 127 966 1701 1050 266 28 1 9 0 1 255 3025 7770 6951 2646 462 36 1 10 0 1 511 9330 34105 42525 22827 5880 750 45 1 11 0 1 1023 28501 145750 246730 179487 63987 11880 1155 55 1 12 0 1 2047 86526 611501 1379400 1323652 627396 159027 22275 1705 66 1
GNU Multiple Precision Arithmetic Library
<lang zkl>var [const] BI=Import("zklBigNum"); // libGMP N=100; S2100:=[BI(2)..N].apply(stirling2.fp(BI(N))).reduce(fcn(m,n){ m.max(n) }); println("Maximum value from the S2(%d,*) row (%d digits):".fmt(N,S2100.numDigits)); println(S2100);</lang>
- Output:
Maximum value from the S2(100,*) row (115 digits): 7769730053598745155212806612787584787397878128370115840974992570102386086289805848025074822404843545178960761551674