Permutations/Rank of a permutation: Difference between revisions
Permutations/Rank of a permutation (view source)
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12 items, permute 422 is 0 1 2 3 4 5 9 8 10 7 6 11
12 items, permute 1212 is 0 1 2 3 4 6 10 5 9 7 8 11
</pre>
=={{header|Ring}}==
<lang ring>
# Project : Permutations/Rank of a permutation
# Date : 2017/10/21
# Author : Gal Zsolt (~ CalmoSoft ~)
# Email : <calmosoft@gmail.com>
list = [0, 1, 2, 3]
for perm = 0 to 23
str = ""
for i = 1 to len(list)
str = str + list[i] + ", "
next
see nl
str = left(str, len(str)-2)
see "(" + str + ") -> " + perm
nextPermutation(list)
next
func nextPermutation a
elementcount = len(a)
if elementcount < 1 then return ok
pos = elementcount-1
while a[pos] >= a[pos+1]
pos -= 1
if pos <= 0 permutationReverse(a, 1, elementcount)
return ok
end
last = elementcount
while a[last] <= a[pos]
last -= 1
end
temp = a[pos]
a[pos] = a[last]
a[last] = temp
permutationReverse(a, pos+1, elementcount)
func permutationReverse a, first, last
while first < last
temp = a[first]
a[first] = a[last]
a[last] = temp
first += 1
last -= 1
end
</lang>
Output:
<pre>
(0, 1, 2, 3) -> 0
(0, 1, 3, 2) -> 1
(0, 2, 1, 3) -> 2
(0, 2, 3, 1) -> 3
(0, 3, 1, 2) -> 4
(0, 3, 2, 1) -> 5
(1, 0, 2, 3) -> 6
(1, 0, 3, 2) -> 7
(1, 2, 0, 3) -> 8
(1, 2, 3, 0) -> 9
(1, 3, 0, 2) -> 10
(1, 3, 2, 0) -> 11
(2, 0, 1, 3) -> 12
(2, 0, 3, 1) -> 13
(2, 1, 0, 3) -> 14
(2, 1, 3, 0) -> 15
(2, 3, 0, 1) -> 16
(2, 3, 1, 0) -> 17
(3, 0, 1, 2) -> 18
(3, 0, 2, 1) -> 19
(3, 1, 0, 2) -> 20
(3, 1, 2, 0) -> 21
(3, 2, 0, 1) -> 22
(3, 2, 1, 0) -> 23
</pre>
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