Imperative Heaps
January 25, 2013
When I made the list of heap algorithms in the previous exercise, I realized that we don’t have an implementation of what is probably the most common form of heaps, the mutable heaps of imperative languages implemented in an array x. The basic idea is to arrange the items in the array so the largest is at index 1, and each item xi is greater than x2i and x2i+1 (that’s a max-heap—a min-heap with the smallest item at the root has the sense of the comparison reversed).
A heap stored sub-array x1..n−1 is unlikely to remain a heap if a new item is placed at xn. The siftup function restores the heap property to x1..n by repeatedly swapping the item at xn with its parent at xn/2 until it reaches its proper place in the heap.
The siftdown operation takes a heap stored in a sub-array x1..n in which the item at x1 is out of place and restores the heap property to x1..n by repeatedly swapping the out-of-order item with its lesser child until it reaches the proper place in the array.
Given the siftup and siftdown operations, sorting an array is easy. First, for each item from the second to the last, sift it up to its proper position, forming a heap in x1..n. Then, swap each item from the last to the second with the first item and sift the new first item down to its proper position in the heap.
Your task is to write functions that implement the siftup, siftdown, and heapsort procedures. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.
Programming Praxis 
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My Haskell solution (see http://bonsaicode.wordpress.com/2013/01/25/programming-praxis-imperative-heaps/ for a version with comments):
import qualified Data.Vector as V (!) :: V.Vector a -> Int -> a v ! i = v V.! (i-1) swap :: Int -> Int -> V.Vector a -> V.Vector a swap i j heap = heap V.// [(i-1, heap ! j), (j-1, heap ! i)] siftup :: Ord a => Int -> V.Vector a -> V.Vector a siftup i heap = let j = div i 2 in if i == 1 || heap ! j <= heap ! i then heap else siftup j $ swap i j heap siftdown :: Ord a => Int -> V.Vector a -> V.Vector a siftdown n = f 1 where f i heap = if 2*i > n || heap ! i <= c then heap else f j $ swap i j heap where (c, j) = minimum [(heap ! x, x) | x <- [2*i, 2*i+1], x <= n] hsort :: Ord a => V.Vector a -> V.Vector a hsort heap = foldr (\i -> siftdown (i - 1) . swap 1 i) (foldl (flip siftup) heap [2..V.length heap]) [2..V.length heap]Went ahead and did this for the previous exercise: Splay heaps redux–imperative model
Personally, I think this one makes more sense (even in a functional language), but to each their own.
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The imperative sort is ideal for languages like Ruby and Python which prefer variable sized arrays over linked structures. I wrote this as a Heap class in Ruby, so the heapsort is just a perfunctory testing function, not an in-place sort.
class ArrayHeap def initialize(seq = nil, &block) @ord = block @heap = [nil] # Cheap way of getting 1-based indexing seq.each {|item| push(item)} unless seq.nil? end def length @heap.length - 1 end def empty? length == 0 end def top return nil if empty? @heap[1] end def pop return nil if empty? # Swap root with element at end @heap[1], @heap[-1] = @heap[-1], @heap[1] min = @heap.pop siftdown return min end def push(item) @heap.push(item) siftup end def inspect "<ArrayHeap " + @heap[1..-1].inspect + ">" end private # Move top of heap down to maintain heap invariant def siftdown i = 1 while 2*i <= length do j = 2*i if j+1 <= length and @ord.call(@heap[j+1], @heap[j]) then j += 1 end break if @ord.call(@heap[i], @heap[j]) @heap[i], @heap[j] = @heap[j], @heap[i] i = j end end # Move bottom of heap up to maintain heap invariant def siftup i = length while i > 1 do break if @ord.call(@heap[i/2], @heap[i]) @heap[i], @heap[i/2] = @heap[i/2], @heap[i] i /= 2 end end end def heapsort(list) h = ArrayHeap.new(list) {|a,b| a<b} r = [] while (item = h.pop) do r << item end return r end irb(main):015:0> heapsort [9,1,4,27,33,21] => [1, 4, 9, 21, 27, 33]# A Python version class Heap: ''' Simple heap impl, with 'min' at the top ''' def __init__(self): self.arr = [None] self.n = 0 def siftup(self, ele): parent = ele / 2 if parent > 0 and self.arr[ele] < self.arr[parent]: self.arr[parent], self.arr[ele] = self.arr[ele], self.arr[parent] self.siftup(parent) def siftdown(self, ele): left = ele * 2 right = ele * 2 + 1 smaller = None arr = self.arr # case where l & r both exist if left <= self.n and right <= self.n: if arr[left] < arr[right]: smaller = left else: smaller = right elif left <= self.n: smaller = left elif right <= self.n: smaller = right if smaller is not None and arr[ele] > arr[smaller]: arr[ele], arr[smaller] = arr[smaller], arr[ele] self.siftdown(smaller) def insert(self, ele): self.n = self.n + 1 self.arr.append(ele) self.siftup(self.n) def get_min(self): tmp = self.arr[1] self.arr[1] = self.arr[self.n] self.n = self.n - 1 self.siftdown(1) return tmp def is_empty(self): return self.n < 1 def heap_sort(lst): heap = Heap() for ele in lst: heap.insert(ele) while not heap.is_empty(): yield heap.get_min() print (list(heap_sort([14, 4, 5, 16, 6, 1, 19, 2, 99, 3, 10, 8, 21])))[…] Imperative Heaps (programmingpraxis.com) […]