Multifactorial: Difference between revisions
m (→{{header|Haskell}}: sequence should start from 0 instead of 1) |
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5: [1, 2, 3, 4, 5, 6, 14, 24, 36, 50] |
5: [1, 2, 3, 4, 5, 6, 14, 24, 36, 50] |
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>>> </lang> |
>>> </lang> |
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=={{header|Tcl}}== |
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{{works with|Tcl|8.6}} |
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<lang tcl>package require Tcl 8.6 |
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proc mfact {n m} { |
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set mm [expr {-$m}] |
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for {set r $n} {[incr n $mm] > 1} {set r [expr {$r * $n}]} {} |
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return $r |
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} |
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foreach n {1 2 3 4 5 6 7 8 9 10} { |
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puts $n:[join [lmap m {1 2 3 4 5 6 7 8 9 10} {mfact $m $n}] ,] |
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}</lang> |
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{{out}} |
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<pre> |
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1:1,2,6,24,120,720,5040,40320,362880,3628800 |
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2:1,2,3,8,15,48,105,384,945,3840 |
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3:1,2,3,4,10,18,28,80,162,280 |
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4:1,2,3,4,5,12,21,32,45,120 |
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5:1,2,3,4,5,6,14,24,36,50 |
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6:1,2,3,4,5,6,7,16,27,40 |
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7:1,2,3,4,5,6,7,8,18,30 |
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8:1,2,3,4,5,6,7,8,9,20 |
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9:1,2,3,4,5,6,7,8,9,10 |
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10:1,2,3,4,5,6,7,8,9,10 |
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</pre> |
Revision as of 22:05, 13 November 2012
The factorial of a number, written as is defined as
A generalization of this is the multifactorials where:
- Where the products are for positive integers.
If we define the degree of the multifactorial as the difference in successive terms that are multiplied together for a multifactorial (The number of exclamation marks) then the task is to
- Write a function that given n and the degree, calculates the multifactorial.
- Use the function to generate and display here a table of the first 1..10 members of the first five degrees of multifactorial.
Note: The wikipedia entry on multifactorials gives a different formula. This task uses the Wolfram mathworld definition.
Haskell
<lang haskell>mulfac k = 1:s where s = [1 .. k] ++ zipWith (*) s [k+1..]
-- for single n mulfac1 k n = product [n, n-k .. 1]
main = mapM_ (print . take 10 . tail . mulfac) [1..5]</lang>
Perl 6
<lang perl6>sub mfact($n, :$degree = 1) {
[*] $n, $n - $degree ...^ * <= 0;
}
for 1 .. 5 -> $degree {
say "$degree: ", map &mfact.assuming(:$degree), 1 .. 10;
}</lang>
- Output:
1: 1 2 6 24 120 720 5040 40320 362880 3628800 2: 1 2 3 8 15 48 105 384 945 3840 3: 1 2 3 4 10 18 28 80 162 280 4: 1 2 3 4 5 12 21 32 45 120 5: 1 2 3 4 5 6 14 24 36 50
Python
Python: Iterative
<lang python>>>> from functools import reduce >>> from operator import mul >>> def mfac(n, m): return reduce(mul, range(n, 0, -m))
>>> for m in range(1, 11): print("%2i: %r" % (m, [mfac(n, m) for n in range(1, 11)]))
1: [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800] 2: [1, 2, 3, 8, 15, 48, 105, 384, 945, 3840] 3: [1, 2, 3, 4, 10, 18, 28, 80, 162, 280] 4: [1, 2, 3, 4, 5, 12, 21, 32, 45, 120] 5: [1, 2, 3, 4, 5, 6, 14, 24, 36, 50] 6: [1, 2, 3, 4, 5, 6, 7, 16, 27, 40] 7: [1, 2, 3, 4, 5, 6, 7, 8, 18, 30] 8: [1, 2, 3, 4, 5, 6, 7, 8, 9, 20] 9: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
10: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] >>> </lang>
Python: Recursive
<lang python>>>> def mfac2(n, m): return n if n <= (m + 1) else n * mfac2(n - m, m)
>>> for m in range(1, 6): print("%2i: %r" % (m, [mfac2(n, m) for n in range(1, 11)]))
1: [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800] 2: [1, 2, 3, 8, 15, 48, 105, 384, 945, 3840] 3: [1, 2, 3, 4, 10, 18, 28, 80, 162, 280] 4: [1, 2, 3, 4, 5, 12, 21, 32, 45, 120] 5: [1, 2, 3, 4, 5, 6, 14, 24, 36, 50]
>>> </lang>
Tcl
<lang tcl>package require Tcl 8.6
proc mfact {n m} {
set mm [expr {-$m}] for {set r $n} {[incr n $mm] > 1} {set r [expr {$r * $n}]} {} return $r
}
foreach n {1 2 3 4 5 6 7 8 9 10} {
puts $n:[join [lmap m {1 2 3 4 5 6 7 8 9 10} {mfact $m $n}] ,]
}</lang>
- Output:
1:1,2,6,24,120,720,5040,40320,362880,3628800 2:1,2,3,8,15,48,105,384,945,3840 3:1,2,3,4,10,18,28,80,162,280 4:1,2,3,4,5,12,21,32,45,120 5:1,2,3,4,5,6,14,24,36,50 6:1,2,3,4,5,6,7,16,27,40 7:1,2,3,4,5,6,7,8,18,30 8:1,2,3,4,5,6,7,8,9,20 9:1,2,3,4,5,6,7,8,9,10 10:1,2,3,4,5,6,7,8,9,10