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Permutations by swapping: Difference between revisions
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([1,0,3,2],1)
([1,0,2,3],-1)
</pre>
=={{header|Icon}} and {{header|Unicon}}==
Works in both languages.
{{trans|Python}}
<lang unicon>procedure main(A)
every write("Permutations of length ",n := !A) do
every p := permute(n) do write("\t",showList(p[1])," -> ",right(p[2],2))
end
procedure permute(n)
items := [[]]
every (j := 1 to n, new_items := []) do {
every item := items[i := 1 to *items] do {
if *item = 0 then put(new_items, [j])
else if i%2 = 0 then
every k := 1 to *item+1 do {
new_item := item[1:k] ||| [j] ||| item[k:0]
put(new_items, new_item)
}
else
every k := *item+1 to 1 by -1 do {
new_item := item[1:k] ||| [j] ||| item[k:0]
put(new_items, new_item)
}
}
items := new_items
}
suspend (i := 0, [!items, if (i+:=1)%2 = 0 then 1 else -1])
end
procedure showList(A)
every (s := "[") ||:= image(!A)||", "
return s[1:-2]||"]"
end</lang>
Sample run:
<pre>
->pbs 3 4
Permutations of length 3
[1, 2, 3] -> -1
[1, 3, 2] -> 1
[3, 1, 2] -> -1
[3, 2, 1] -> 1
[2, 3, 1] -> -1
[2, 1, 3] -> 1
Permutations of length 4
[1, 2, 3, 4] -> -1
[1, 2, 4, 3] -> 1
[1, 4, 2, 3] -> -1
[4, 1, 2, 3] -> 1
[4, 1, 3, 2] -> -1
[1, 4, 3, 2] -> 1
[1, 3, 4, 2] -> -1
[1, 3, 2, 4] -> 1
[3, 1, 2, 4] -> -1
[3, 1, 4, 2] -> 1
[3, 4, 1, 2] -> -1
[4, 3, 1, 2] -> 1
[4, 3, 2, 1] -> -1
[3, 4, 2, 1] -> 1
[3, 2, 4, 1] -> -1
[3, 2, 1, 4] -> 1
[2, 3, 1, 4] -> -1
[2, 3, 4, 1] -> 1
[2, 4, 3, 1] -> -1
[4, 2, 3, 1] -> 1
[4, 2, 1, 3] -> -1
[2, 4, 1, 3] -> 1
[2, 1, 4, 3] -> -1
[2, 1, 3, 4] -> 1
->
</pre>
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