Data Structures/Min and Max Heaps


A heap is an efficient semi-ordered data structure for storing a collection of orderable data. A min-heap supports two operations:

  INSERT(heap, element)
  element REMOVE_MIN(heap)

(we discuss min-heaps, but there's no real difference between min and max heaps, except how the comparison is interpreted.)

This chapter will refer exclusively to binary heaps, although different types of heaps exist. The term binary heap and heap are interchangeable in most cases. A heap can be thought of as a tree with parent and child. The main difference between a heap and a binary tree is the heap property. In order for a data structure to be considered a heap, it must satisfy the following condition (heap property):

If A and B are elements in the heap and B is a child of A, then key(A) ≤ key(B).

(This property applies for a min-heap. A max heap would have the comparison reversed). What this tells us is that the minimum key will always remain at the top and greater values will be below it. Due to this fact, heaps are used to implement priority queues which allows quick access to the item with the most priority. Here's an example of a min-heap:

Binary-heap.png

A heap is implemented using an array that is indexed from 1 to N, where N is the number of elements in the heap.

At any time, the heap must satisfy the heap property

  array[n] <= array[2*n]   // parent element <= left child

and

  array[n] <= array[2*n+1] // parent element <= right child

whenever the indices are in the arrays bounds.

Compute the extreme valueEdit

We will prove that array[1] is the minimum element in the heap. We prove it by seeing a contradiction if some other element is less than the first element. Suppose array[i] is the first instance of the minimum, with array[j] > array[i] for all j < i, and i >= 2. But by the heap invariant array, array[floor(i/2)] <= array[i]: this is a contradiction.

Therefore, it is easy to compute MIN(heap):

  MIN(heap)
     return heap.array[1];

Removing the Extreme ValueEdit

To remove the minimum element, we must adjust the heap to fill heap.array[1]. This process is called percolation. Basically, we move the hole from node i to either node 2i or 2i+1. If we pick the minimum of these two, the heap invariant will be maintained; suppose array[2i] < array[2i+1]. Then array[2i] will be moved to array[i], leaving a hole at 2i, but after the move array[i] < array[2i+1], so the heap invariant is maintained. In some cases, 2i+1 will exceed the array bounds, and we are forced to percolate 2i. In other cases, 2i is also outside the bounds: in that case, we are done.


Therefore, here is the remove algorithm:


#define LEFT(i) (2*i)
#define RIGHT(i) (2*i + 1)  
REMOVE_MIN(heap)
{  
    savemin=arr[1];
    arr[1]=arr[--heapsize];
    i=1;
    while(i<heapsize){
        minidx=i;
        if(LEFT(i)<heapsize && arr[LEFT(i)] < arr[minidx])
            minidx=LEFT(i);
        if(RIGHT(i)<heapsize && arr[RIGHT(i)] < arr[minidx]) 
            minidx=RIGHT(i);
        if(minidx!=i){
            swap(arr[i],arr[minidx]);
            i=minidx;            
        }
        else 
            break;
    }
}

Why does this work?

   If there is only 1 element ,heapsize becomes 0, nothing in the array is valid.
   If there are 2 elements , one min and other max, you replace min with max.
   If there are 3 or more elements say n, you replace 0th element with n-1th element. 
   The heap property is destroyed. Choose the 2 childs of root and check which is the minimum.
   Choose the minimum among them , swap it . Now subtree with swapped child is looses heap property.
   If no violations break.

Inserting a value into the heapEdit

A similar strategy exists for INSERT: just append the element to the array, then fixup the heap-invariants by swapping. For example if we just appended element N, then the only invariant violation possible involves that element, in particular if array[floor(N/2)] > array[N], then those two elements must be swapped and now the only invariant violation possible is between

 array[floor(N/4)]  and  array[floor(N/2)]

we continue iterating until N=1 or until the invariant is satisfied.

INSERT(heap, element)
   append(heap.array, element)
   i = heap.array.length
   while (i > 1)
     {
       if (heap.array[i/2] <= heap.array[i])
         break;
       swap(heap.array[i/2], heap.array[i]);
       i /= 2;
     }

TODOEdit

 Merge-heap: it would take two max/min heap and merge them and return a single heap. O(n) time.
 Make-heap: it would also be nice to describe the O(n) make-heap operation
 Heap sort: the structure can actually be used to efficiently sort arrays
 Treaps: heaps where elements can be deleted by name

Make-heap would make use a function heapify

//Element is a data structure//

Make-heap(Element Arr[],int size) {

   for(j=size/2;j>0;j--)
       {
           Heapify(Arr,size,j);
       }

}

Heapify(Element Arr[],int size,int t) {

   L=2*t;
   R=2*t+1;
   if(L<size )
   {
       mix=minindex(Arr,L,t);
       if(R<=size)
           mix=minindex(Arr,R,mix);
   }
   else
       mix=t;
   if(mix!=t)
   {
       swap(mix,t);
       Heapify(Arr,size,mix);
   }

}

minindex returns index of the smaller element

TreapsEdit

covered in this example in the Algorithms book. Algorithms/Left_rotation

Applications of Priority HeapsEdit

In 2009, a smaller Sort Benchmark was won by OzSort, which has a paper describing lucidly how to use a priority heap as the sorting machine to produce merged parts of large (internally) sorted sections . If a sorted section took M memory and the sorting problem was k x M big, then take sequential sections of each of the k sections of size M/k , at a time, so they fit in M memory ( k * M/k = M ), and feed the first element of each of k sections to make a k sized priority queue, and as the top element is removed and written to an output buffer, take the next element from the corresponding section. This means elements may need to be associated with a label for the section they come from. When a M/k-sized section is exhausted, load in the next M/k sized minisection from the original sorted section stored on disc. Continue until all minisections in each of the k sections on disc have been exhausted.

(As an example of pipelining to fill up disc operational delays, there are twin output buffers, so that once an output buffer is full one gets written the disc while the other is being filled.)

This paper showed that a priority heap is more straightforward than a binary tree, because elements are constantly being deleted, as well as added, as a queuing mechanism for a k way merge, and has practical application for sorting large sets of data that exceed internal memory storage.


Last modified on 13 April 2014, at 19:26