问题
Given the following pseudocode for the bubble-sort
procedure bubbleSort( A : list of sortable items )
repeat
swapped = false
for i = 1 to length(A) - 1 inclusive do:
/* if this pair is out of order */
if A[i-1] > A[i] then
/* swap them and remember something changed */
swap( A[i-1], A[i] )
swapped = true
end if
end for
until not swapped
end procedure
Here is the code for Bubble Sort as Scala
def bubbleSort[T](arr: Array[T])(implicit o: Ordering[T]) {
import o._
val consecutiveIndices = (arr.indices, arr.indices drop 1).zipped
var hasChanged = true
do {
hasChanged = false
consecutiveIndices foreach { (i1, i2) =>
if (arr(i1) > arr(i2)) {
hasChanged = true
val tmp = arr(i1)
arr(i1) = arr(i2)
arr(i2) = tmp
}
}
} while(hasChanged)
}
This is the Haskell implementation:
bsort :: Ord a => [a] -> [a]
bsort s = case _bsort s of
t | t == s -> t
| otherwise -> bsort t
where _bsort (x:x2:xs) | x > x2 = x2:(_bsort (x:xs))
| otherwise = x:(_bsort (x2:xs))
_bsort s = s
Is it possible to formulate this as a monoid or semigroup?
回答1:
I'm using my phone with a poor network connection, but here goes.
tl;dr bubblesort is insertion sort is the monoidal "crush" for the monoid of ordered lists with merging.
Ordered lists form a monoid.
newtype OL x = OL [x]
instance Ord x => Monoid (OL x) where
mempty = OL []
mappend (OL xs) (OL ys) = OL (merge xs ys) where
merge [] ys = ys
merge xs [] = xs
merge xs@(x : xs') ys@(y : ys')
| x <= y = x : merge xs' ys
| otherwise = y : merge xs ys'
Insertion sort is given by
isort :: Ord x => [x] -> OL x
isort = foldMap (OL . pure)
because insertion is exactly merging a singleton list with another list. (Mergesort is given by building a balanced tree, then doing the same foldMap.)
What has this to do with bubblesort? Insertion sort and bubblesort have exactly the same comparison strategy. You can see this if you draw it as a sorting network made from compare-and-swap boxes. Here, data flows downward and lower inputs to boxes [n] go left:
| | | |
[1] | |
| [2] |
[3] [4]
| [5] |
[6] | |
| | | |
If you perform the comparisons in the sequence given by the above numbering, cutting the diagram in / slices, you get insertion sort: the first insertion needs no comparison; the second needs comparison 1; the third 2,3; the last 4,5,6.
But if, instead, you cut in \ slices...
| | | |
[1] | |
| [2] |
[4] [3]
| [5] |
[6] | |
| | | |
...you are doing bubblesort: first pass 1,2,3; second pass 4,5; last pass 6.
来源:https://stackoverflow.com/questions/21877572/can-you-formulate-the-bubble-sort-as-a-monoid-or-semigroup