Selection sorting.
Remember that selection sort works by
finding the smallest element and moving it into the first position,
then finding the second smallest and moving it into the second
position. In general, on the ith iteration, it finds the ith smallest
element and move it to position i. This works as follows:
(selection '(6 4 5 3 9 3))
(cons 3 (selection '(6 4 5 9 3))
(cons 3 (cons 3 (selection '(6 4 5 9))
(cons 3 (cons 3 (cons 4 (selection '(6 5 9))
(cons 3 (cons 3 (cons 4 (cons 5 (selection '(6 9))
(cons 3 (cons 3 (cons 4 (cons 5 (cons 6 (selection '(9))
(cons 3 (cons 3 (cons 4 (cons 5 (cons 6 (cons 9 (selection '())
Which becomes (3 3 4 5 6 9).
The scheme code below does exactly this:
/* put the smallest element at the front of the current list
call selection on the list minus the smallest element */
/* remove the first occurance of atom A from L */
List remove(List lis, int item) {
if (lis.isEmpty()) return lis;
else if (((Integer) lis.get(0)).intValue() == item)
return lis.subList(1,lis.size());
else {
Integer first = (Integer) lis.get(0);
List temp = new ArrayList();
temp.addAll(remove(lis.subList(1,lis.size()),item));
temp.add(0,first);
return temp;
}
}
/* looks for the smallest element in the list
atom A is the current smallest */
int smallest(List lis, int small) {
if (lis.isEmpty()) return small;
else if (((Integer) lis.get(0)).intValue() < small)
return smallest(lis.subList(1,lis.size()),
((Integer) lis.get(0)).intValue());
else return smallest(lis.subList(1,lis.size()), small);
}
List selectionsort(List lis) {
if (lis.isEmpty()) return lis;
else {
Integer first = (Integer) lis.get(0);
int s = smallest(lis.subList(1,lis.size()), first.intValue());
List sorted = new ArrayList();
List removeList = new ArrayList();
removeList.addAll(remove(lis,s));
sorted.addAll(selectionsort(removeList));
sorted.add(0, new Integer(s));
return sorted;
}
}
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 Java
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