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6.3.1 Heapsort

6.3.1 Heapsort. Idea: two phases: 1. Construction of the heap 2. Output of the heap For ordering number in an ascending sequence: use a Heap with reverse order: the maximum number should be at the root (not the minimum). Heapsort is an in-situ-Procedure.

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6.3.1 Heapsort

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  1. 6.3.1 Heapsort Idea:two phases: 1. Construction of the heap 2. Output of the heap For ordering number in an ascending sequence: use a Heap with reverse order: the maximum number should be at the root (not the minimum). Heapsort is an in-situ-Procedure

  2. Remembering Heaps: change the definition Heap with reverse order: • For each node x and each successor y of x the following holds: m(x)  m(y), • left-complete, which means the levels are filled starting from the root and each level from left to right, • Implementation in an array, where the nodes are set in this order (from left to right) .

  3. Second Phase: Heap 2. Output of the heap: take n-times the maximum (in the root, deletemax) and exchange it with the element at the end of the heap.  Heap is reduced by one element and the subsequence of ordered elements at the end of the array grows one element longer. cost: O(n log n). Heap Ordered elements Ordered elements

  4. First Phase: 1. Construction of the Heap: simple method: n-times insert Cost: O(n log n). making it better: consider the array a[1 … n ] as an already left-complete binary tree and let sink the elements in the following sequence ! a[n div 2] … a[2] a[1] (The elements a[n] … a[n div 2 +1] are already at the leafs.) HH The leafs of the heap

  5. Formally: heap segment an array segment a[ i..k ] ( 1 ik <=n ) is said to be a heapsegment when following holds:         for all j from {i,...,k}     m(a[ j ]) m(a[ 2j ])     if 2j k and   m(a[ j ]) m(a[ 2j+1])  if 2j+1 k If a[i+1..n] is already a heap segment we can convert a[i…n] into a heap segment by letting a[i] sink.

  6. Cost calculation Be k = [log n+1]+ - 1. (the height of the complete portion of the heap) cost: For an element at level j from the root: k – j. alltogether: {j=0,…,k} (k-j)•2j = 2k•{i=0,…,k} i/2i =2 • 2k = O(n).

  7. advantage: The new construction strategy is more efficient ! Usage: when only the m biggest elements are required: 1. construction in O(n) steps. 2. output of the m biggest elements in O(m•log n) steps. total cost: O( n + m•log n).

  8. Addendum: Sorting with search trees Algorithm: • Construction of a search tree (e.g. AVL-tree) with the elements to be sorted by n insert opeartions. • Output of the elements in InOrder-sequence.  Ordered sequence. cost: 1. O(n log n) with AVL-trees, 2. O(n). in total: O(n log n). optimal!

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