I'm making my way through "Programming in Scala" and wrote a quick implementation of the selection sort algorithm. However, since I'm still a bit green in functional programming, I'm having trouble translating to a more Scala-ish style. For the Scala programmers out there, how can I do this using Lists and vals rather than falling back into my imperative ways?
http://gist.github.com/225870
According to the Java 6 documentation, it therefore uses quicksort: The sorting algorithm is a tuned quicksort, adapted from Jon L. Bentley and M. Douglas McIlroy's "Engineering a Sort Function", Software-Practice and Experience, Vol.
In every iteration of selection sort, the minimum element (considering ascending order) from the unsorted subarray is picked and moved to the sorted subarray. Clearly, it is a greedy approach to sort the array.
Example of Selection SortThe smallest number from 5 2 and 1 is 1. So, we replace 10 by 1. The new array is [1,5,2,10] Again, this process is repeated. Finally, we get the sorted array as [1,2,5,10].
Algorithm of the Selection Sort AlgorithmStep 1: Set Min to location 0 in Step 1. Step 2: Look for the smallest element on the list. Step 3: Replace the value at location Min with a different value. Step 5: Continue until the list is sorted.
As starblue already said, you need a function that calculates the minimum of a list and returns the list with that element removed. Here is my tail recursive implementation of something similar (as I believe foldl
is tail recursive in the standard library), and I tried to make it as functional as possible :). It returns a list that contains all the elements of the original list (but kindof reversed - see the explanation below) with the minimum as a head.
def minimum(xs: List[Int]): List[Int] =
(List(xs.head) /: xs.tail) {
(ys, x) =>
if(x < ys.head) (x :: ys)
else (ys.head :: x :: ys.tail)
}
This basically does a fold, starting with a list containing of the first element of xs
If the first element of xs
is smaller than the head of that list, we pre-append it to the list ys
. Otherwise, we add it to the list ys
as the second element. And so on recursively, we've folded our list into a new list containing the minimum element as a head and a list containing all the elements of xs
(not necessarily in the same order) with the minimum removed, as a tail. Note that this function does not remove duplicates.
After creating this helper function, it's now easy to implement selection sort.
def selectionSort(xs: List[Int]): List[Int] =
if(xs.isEmpty) List()
else {
val ys = minimum(xs)
if(ys.tail.isEmpty)
ys
else
ys.head :: selectionSort(ys.tail)
}
Unfortunately this implementation is not tail recursive, so it will blow up the stack for large lists. Anyway, you shouldn't use a O(n^2) sort for large lists, but still... it would be nice if the implementation was tail recursive. I'll try to think of something... I think it will look like the implementation of a fold.
Tail Recursive!
To make it tail recursive, I use quite a common pattern in functional programming - an accumulator. It works a bit backward, as now I need a function called maximum
, which basically does the same as minimum
, but with the maximum element - its implementation is exact as minimum, but using >
instead of <
.
def selectionSort(xs: List[Int]) = {
def selectionSortHelper(xs: List[Int], accumulator: List[Int]): List[Int] =
if(xs.isEmpty) accumulator
else {
val ys = maximum(xs)
selectionSortHelper(ys.tail, ys.head :: accumulator)
}
selectionSortHelper(xs, Nil)
}
EDIT: Changed the answer to have the helper function as a subfunction of the selection sort function.
It basically accumulates the maxima to a list, which it eventually returns as the base case. You can also see that it is tail recursive by replacing accumulator
by throw new NullPointerException
- and then inspect the stack trace.
Here's a step by step sorting using an accumulator. The left hand side shows the list xs
while the right hand side shows the accumulator
. The maximum is indicated at each step by a star.
64* 25 12 22 11 ------- Nil
11 22 12 25* ------- 64
22* 12 11 ------- 25 64
11 12* ------- 22 25 64
11* ------- 12 22 25 64
Nil ------- 11 12 22 25 64
The following shows a step by step folding to calculate the maximum:
maximum(25 12 64 22 11)
25 :: Nil /: 12 64 22 11 -- 25 > 12, so it stays as head
25 :: 12 /: 64 22 11 -- same as above
64 :: 25 12 /: 22 11 -- 25 < 64, so the new head is 64
64 :: 22 25 12 /: 11 -- and stays so
64 :: 11 22 25 12 /: Nil -- until the end
64 11 22 25 12
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