Lecture: Heaps and Priority Queues

Motivations:

1. Need a data structure that returns the maximum-keyed element of a set, that is, a data structure called a priority queue. The key’s are the priorities.

   (We could implement a priority queue using an AVL tree, but we want to be more space efficient.)

2. Want to do in-place sorting in O(n log(n)).

3. Want to be able to change the key (priority) associated with an element.

Heaps

        __16__
       /      \
     14        10
    /  \      /  \
   /    \    /    \
  8      7  9      3
 / \    /
2   4  1

Idea: store elements in an array left-to-right, one level at a time, top-to-bottom.

          0  1  2  3  4  5  6  7  8  9 
        [16,14,10, 8, 7, 9, 3, 2, 4, 1]

Def. A heap is an array viewed as a nearly complete binary tree. Given an array A, the root of the tree is stored at A[0] and for a node stored at index i, the left child is stored at index 2i+1 and the right child is stored at index 2(i+1). (Not to be confused with the use of “heap” to mean a computer’s main memory.)

left(i) = 2*i + 1

right(i) = 2*(i + 1)

parent(i) =  floor( (i-1) / 2 )

Def. A max heap is a heap in which for every node other than the root, A[i] ≤ A[parent(i)]

This is called the heap property or max-heap property. One can instead create a min-heap by flipping this around.

Overview of the Heap operations

• build_max_heap creates a heap from an unordered array in O(n).
• maximum returns maximum element in O(1).
• extract_max removes the max in O(log(n)).
• sortInPlace sorts an array in-place in O(n log(n)).
• increase_key updates the key of an element in the heap in O(log(n)).
• insert adds a new element to a heap in O(log(n)).

Heap class:

public class Heap<E> {
        ArrayList<E> data;
        BiPredicate<E,E> lesseq;
        ...
}

max_heapify helper function

Many of the heap operations need to turn an array that is almost a max heap, except for one element, into a max heap. The max_heapify operation moves the element at position i into the correct position.

The tree rooted at i is not a max heap, but the subtrees left(i) and right(i) are max heaps.

Example: 4 is less than one of its children so we swap it with the larger child and repeat.

         ___16___
        /        \
      *4*         10
     /  \        /  \
    /    \      9    3
   14     7
  / \    /
 2   8  1

becomes

         ___16___
        /        \
      14          10
     /  \        /  \
    /    \      9    3
  *4*     7
  / \    /
 2   8  1

becomes

         ___16___
        /        \
      14          10
     /  \        /  \
    /    \      9    3
   8      7
  / \    /
 2  *4* 1

Java declaration for max_heapify:

void max_heapify(int i, int heap_length);

What is the time complexity of max_heapify? O(log(n))

build_max_heap method

void build_max_heap() {
    int last_parent = data.size() / 2 - 1;
    for (int i = last_parent; i != -1; --i) {
        max_heapify(i, data.size());
    }
}

Why does this procedure work? What is the loop invariant? Answer: the invariant is that the trees rooted at positions from i+1 to the end are max heaps.

What is the time complexity? Answer: O(n log(n)) is the easy answer, but not tight.

The tight upper bound is O(n) which can be obtained by observing that the time for max_heapify depends on the height of the node, and build_max_heap calls `max_heapify many times on nodes with a low height.

Consider how many nodes there can be in an n-element heap at each height. The worst cast is a complete tree. For example,

   n = 7

      _o_        height 2, 1 node
     /   \
    o     o      height 1, 2 nodes
   / \   / \
  o   o o   o    height 0, 4 nodes

Height 0: n / 2     =  n / 2^(0+1)

Height 1: n / 4     =  n / 2^(1+1)

Height 2: n / 8     =  n / 2^(2+1)

In general, there are at most n / 2^(h+1) nodes at a given height h.

So we can sum these up, from h=0 to log(n), with O(h) cost for each:

sum from h=0 to log(n) of (n / 2^(h+1)) * O(h) = O(n * sum from h=0 to log(n) of h / 2^h)

recall formula A.8:

sum from k=0 to ∞ of (k * x^k) = x / (1 - x)², for

x

< 1

let x = 1/2:

sum from k=0 to ∞ of (k / 2^k) = (1/2) / (1 - (1/2))² = (1/2) / (1 \times 1 - 2\times 1/2 + 1/4) = (1/2) / (1/4) = 2

so we get

sum from h=0 to log(n) of (h / 2^h)  < 2

Thus

  O( n * sum from h=0 to log(n) of (h / 2^h))
= O( n * 2 )
= O(n).

maximum method

public E maximum() {
    return data.get(0);
}

extract_max method

Idea: first record the max element, which is at data[0], then we need to delete that element. But deleting the first element of an array is expensive. So we move the last element of the heap to data[0], shrink the array, and re-establish the max heap property with max_heapify.

E extract_max() {
    E max = data.get(0);
    data.set(0, data.get(data.size()-1));
    data.remove(data.size()-1);
    max_heapify(0, data.size());
    return max;
}

sortInPlace method

Idea: swap the max (the root) with the last element, call max_heapify, then shrink the heap by 1. Similar to extract_max but does a swap instead of move.

static <E> void sortInPlace(ArrayList<E> A, 
                            BiPredicate<E,E> lessThanOrEqual)
{
    Heap<E> H = new Heap<E>(lessThanOrEqual);
    H.data = A;
    H.build_max_heap();
    for (int i  = H.data.size() - 1; i != 0; --i) {
        swap(H.data, 0, i);
        H.max_heapify(0, i);
    }
}

Time complexity: The for loop executes n times, and max_heapify is O(log(n)), so we have O(n log(n)).

increase_key method (used by insert)

The key of the object at position i has increased. How should we move it to get a valid heap? Example: Change the key of 9 to 20.

     ___16___
    /        \
  14          10
 /  \        /  \
8    7     *9*    3

Student exercise: come up with the algorithm for increase_key

Answer: the idea is to propagate the element up. Continuing the above example,

     ___16___
    /        \
  14          10
 /  \        /  \
8    7    *20*   3

becomes

     ___16___
    /        \
  14         *20*
 /  \        /  \
8    7      10    3

becomes

     __*20*__
    /        \
  14          16
 /  \        /  \
8    7      10    3

insert method

void insert(E k) {
    if (data.size() + 1 ≥ data.size()) {
        ArrayList<E> d = new ArrayList<>((data.size() + 1) * 2);
        for (E e : data)
            d.add(e);
        data = d;
    }
    data.add(k);
    increase_key(data.size() - 1);
}

Priority Queues

• implement using a Heap
• the priorities are the keys
• push: implement with insert
• pop: implement with extract_max

In Java:

class PriorityQueue<E> {
    Heap<E> heap;
    PriorityQueue(BiPredicate<E,E> lessThanOrEqual) {
        heap = new Heap<>(lessThanOrEqual);
    }
    void push(E key) {
        heap.insert(key);
    }
    E pop() {
        return heap.extract_max();
    }
}