Merge Sort
Mergesort divides the list in half everytime,
calling itself on each half. As the recursion unwinds, it merges the
sorted list into a new
sorted list. For example, assume we have the following:
(mergesort '(6 4 5 7 8 9 3 4))
/ \
(mergesort '(6 4 5 7)) (mergesort '(8 9 3 4))
/ \ / \
(mergesort '(6 4)) (mergesort '(5 7)) (mergesort '(8 9)) (mergesort '(3 4))
At the lowest tree leves, the two element get sorted and the recursion unwinds.
(mergesort '(6 4 5 7 8 9 3 4))
/ \
(mergesort '(6 4 5 7)) (mergesort '(8 9 3 4))
/ \ / \
(4 6) (5 7) (8 9)) (3 4))
The next level of recursion:
(mergesort '(6 4 5 7 8 9 3 4))
/ \
(4 5 6 7)) (3 4 8 9)
/ \ / \
(4 6) (5 7) (8 9)) (3 4))
And the final level with merges into the overall list
(3 4 4 5 6 7 8 9)
/ \
(4 5 6 7) (3 4 8 9)
Of course, this is not the actual order that things happen but it give
the overall algorithm flavor.
In scheme, we can write functions to split lists, and merge sorted
lists. Once we have these, the mergesort itself is fairly short.
NOT YET IMPLEMENTED IN CPP
def mergesort(lis)
if lis.length == 0 then
[]
elsif lis.length == 1 then
lis
elsif lis.length == 2 then
mergelists ([lis[0]], lis[1..lis.length-1])
else
mergelists(mergesort(split(lis)[0]),mergesort(split(lis)[1..lis.length-1][0]))
end
end
def mergelists(lis, m) # assume L and M are sorted lists
if lis.length==0 then
m
elsif m.length==0 then
lis
elsif lis[0] < m[0] then
[lis[0]] + mergelists(lis[1..lis.length-1], m)
else
[m[0]] + mergelists(lis, m[1..m.length-1])
end
end
def sub(lis, start, stop, ctr) # extract elements start to stop into a list
if lis.length== 0 then
lis
elsif ctr < start then
sub(lis[1..lis.length-1], start, stop, ctr+1)
elsif ctr > stop then
[]
else
[lis[0]] + sub(lis[1..lis.length-1],start, stop, ctr+1)
end
end
def split(lis) # split the list in half:, ; returns ((first half)(second half))
len=lis.length
if len == 0 then
[lis]+[lis]
elsif len == 1 then
[lis] + [[]]
else
[sub(lis, 1, len//2, 1)] + [sub(lis, (len//2)+1, len, 1)]
end
end