## 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.

Pages: 1 2

### 7 Responses to “Imperative Heaps”

1. […] today’s Programming Praxis exercise, our goal is to implement an array-based heap. Let’s get started, […]

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]

3. jpverkamp said

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.

4. […] Pages: 1 2 […]

5. David said

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]

6. Joe A said
# 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])))

7. […] Imperative Heaps (programmingpraxis.com) […]