Tree Heap Priority Queue
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1 Tree Heap Priority Queue Rivers May 23, 2018 Rivers Rivers Second Open Topic May 23, / 33
2 Contents Section 1 Contents Rivers ( @smail.nju.end.cn) Rivers Second Open Topic May 23, / 33
3 Contents Contents Pseudocode C source code Rivers ( @smail.nju.end.cn) Rivers Second Open Topic May 23, / 33
4 Contents Build-Complete- Rivers Rivers Second Open Topic Binary-Tree-From- May 23, / 33 Contents 1 Why is array preferred? Pseudocode C source code 2 Those elementary things 3 How to implement Tree Heap Max-Heapify Build-MaxHeap- FromTree Heapsort Heap Priority Queue HeapMax Extract-HeapMax Heap-Increase-Key MaxHeap-Insert 4 Easter egg
5 Why is array preferred? Section 2 Why is array preferred? Rivers ( @smail.nju.end.cn) Rivers Second Open Topic May 23, / 33
6 Why is array preferred? Parent(i) 1 return i/2 Left(i) 1 return 2i Right(i) 1 return 2i A[1] = A[heap-size] 2 heap-size = heap-size 1 All of them is Θ(1) How can we implement these operations with tree form? Can we maintain these amazing time costs? 1 A[i] = key Rivers ( @smail.nju.end.cn) Rivers Second Open Topic May 23, / 33
7 Those elementary things Section 3 Those elementary things Rivers ( @smail.nju.end.cn) Rivers Second Open Topic May 23, / 33
8 Those elementary things Structure of Tree Node key parent left right Parent(T ) 1 return T.parent Left(T ) 1 return T.left Right(T ) 1 return T.right Rivers ( @smail.nju.end.cn) Rivers Second Open Topic May 23, / 33
9 Those elementary things Parent(i) 1 return i/2 Left(i) 1 return 2i Right(i) 1 return 2i A[1] = A[heap-size] 2 heap-size = heap-size 1 1 A[i] = key Rivers ( @smail.nju.end.cn) Rivers Second Open Topic May 23, / 33
10 Those elementary things Figure: Heap Node Number and Direction of Path Rivers Rivers Second Open Topic May 23, / 33
11 Those elementary things { HeapTree H Total Last FindLast(H) 1 H.Last = H.HeapTree 2 index = 0 3 while (1 << index) H.total 4 Direction[index] = ((H.Total) >> index)&1 5 index = index index = index 2 7 while index 0 8 if Direction[index] == 1 9 Last = Last.right 10 else 11 Last = Last.left 12 index = index 1 Rivers ( @smail.nju.end.cn) Rivers Second Open Topic May 23, / 33
12 Those elementary things Find-Node(H, N) 1 result = H.HeapTree 2 index = 0 3 while (1 << index) N 4 Direction[index] = (N >> index)&1 5 index = index index = index 2 7 while index 0 8 if Direction[index] == 1 9 result = result.right 10 else 11 result = result.left 12 index = index 1 One way to get random access to the element in a complete binary tree with time cost Θ(log N) Rivers ( @smail.nju.end.cn) Rivers Second Open Topic May 23, / 33
13 Tree 2 Heap Section 4 Tree 2 Heap Rivers ( @smail.nju.end.cn) Rivers Second Open Topic May 23, / 33
14 Tree 2 Heap MaxHeapify Subsection 1 MaxHeapify Rivers ( @smail.nju.end.cn) Rivers Second Open Topic May 23, / 33
15 Tree 2 Heap MaxHeapify MaxHeapify MaxHeapify(H) 1 Left = H.left 2 Right = H.right 3 if Left NIL and Left.key > H.key 4 largest = Left Right(H)=H.right 5 else Left(H)=H.left 6 largest = H 7 if Right NIL and Right.key >largest.key 8 largest = Right 9 if largest H 10 exchange largest.key and H.key 11 MaxHeapify(largest) Rivers ( @smail.nju.end.cn) Rivers Second Open Topic May 23, / 33
16 Tree 2 Heap Build-MaxHeap-FromTree Subsection 2 Build-MaxHeap-FromTree Rivers ( @smail.nju.end.cn) Rivers Second Open Topic May 23, / 33
17 Tree 2 Heap Build-MaxHeap-FromTree Build-MaxHeap-FromTree Build-MaxHeap-FromTree(T ) 1 if T == NIL 2 return 3 Build-MaxHeap-FromTree(T.left) 4 Build-MaxHeap-FromTree(T.right) 5 MaxHeapify(T ) Recurrence T (n) = T (αn) + T ((1 α)n) + Θ(log n) with 1/3 α 2/3 Solution T (n) = O(n)? Rivers ( @smail.nju.end.cn) Rivers Second Open Topic May 23, / 33
18 Tree 2 Heap Build-MaxHeap-FromTree Proof. 1 Claim that T (n) dn T (n) = T (αn) + T ((1 α)n) + c log n dαn + d(1 α)n + c log n = dn + c log n dn 2 Claim that T (n) d 1 n d 2 log n T (n) = T (αn) + T ((1 α)n) + c log n d 1 n d 2 log αn d 2 log(1 α)n + c log n = d 1 n d 2 log n (d 2 log α + d 2 log(1 α) + d 2 log n c log n) We only needs d 2 to be big enough to overwhelm c log n Rivers ( @smail.nju.end.cn) Rivers Second Open Topic May 23, / 33
19 Tree 2 Heap Heapsort Subsection 3 Heapsort Rivers ( @smail.nju.end.cn) Rivers Second Open Topic May 23, / 33
20 Tree 2 Heap Heapsort Heapsort Heapsort(H) 1 Build-MaxHeap-FromTree(H) 2 for i = H.total downto 1 3 exchange H.HeapTree.key and Last.key 4 A[i] = Last.key 5 if H.Last.parent.right == NIL 6 H.Last.parent.left = NIL 7 else 8 H.Last.parent.right = NIL 9 H.total = H.total 1 10 FindLast(H) 11 MaxHeapify(H.HeapTree) Rivers ( @smail.nju.end.cn) Rivers Second Open Topic May 23, / 33
21 Heap 2 Priority Queue Section 5 Heap 2 Priority Queue Rivers ( @smail.nju.end.cn) Rivers Second Open Topic May 23, / 33
22 Heap 2 Priority Queue HeapMax and Extract-HeapMax Subsection 1 HeapMax and Extract-HeapMax Rivers ( @smail.nju.end.cn) Rivers Second Open Topic May 23, / 33
23 Heap 2 Priority Queue HeapMax and Extract-HeapMax HeapMax(H) return H.HeapTree.key Extract-HeapMax(H) 1 if H.HeapTree == NIL 2 error heap underflow 3 max = H.HeapTree.key 4 H.HeapTree.key = H.Last.key 5 if H.Last.parent.right == NIL 6 H.Last.parent.left = NIL 7 else 8 H.Last.parent.right = NIL 9 H.Total = H.Total 1 10 FindLast(H) 11 MaxHeapify(H.HeapTree) Rivers ( @smail.nju.end.cn) Rivers Second Open Topic May 23, / 33
24 Heap 2 Priority Queue Heap-Increase-Key and MaxHeap-Insert Subsection 2 Heap-Increase-Key and MaxHeap-Insert Rivers ( @smail.nju.end.cn) Rivers Second Open Topic May 23, / 33
25 Heap 2 Priority Queue Heap-Increase-Key and MaxHeap-Insert Heap-Increase-Key Heap-Increase-Key(H, i, key) 1 target = Find-Node(H, i) 2 if key < target.key 3 error new key is smaller than current key 4 while target H.HeapTree and target.parent.key < target.key 5 exchange target.parent.key and target.key 6 target = target.parent Rivers ( @smail.nju.end.cn) Rivers Second Open Topic May 23, / 33
26 Heap 2 Priority Queue Heap-Increase-Key and MaxHeap-Insert MaxHeap-Insert MaxHeap-Insert(H, key) 1 H.Total = H.Total parent = Find-Node(H, H.Total/2 ) 3 if parent.left == NIL 4 parent.left = Create( ) 5 H.Last = parent.left 6 else 7 parent.right = Create( ) 8 H.Last = parent.right 9 Heap-Increase-Key(H, H.Total, key) Rivers ( @smail.nju.end.cn) Rivers Second Open Topic May 23, / 33
27 Easter Egg Section 6 Easter Egg Rivers ( @smail.nju.end.cn) Rivers Second Open Topic May 23, / 33
28 Easter Egg Build-Complete-Binary-Tree-From-Array Subsection 1 Build-Complete-Binary-Tree-From-Array Rivers ( @smail.nju.end.cn) Rivers Second Open Topic May 23, / 33
29 Easter Egg Build-Complete-Binary-Tree-From-Array Build-Complete-Binary- Tree-From-Array(T, A, n) 1 if n == 0 2 T = NIL 3 T = CreateTNode(A[0]) 4 Enqueue(Q, T ) 5 index = 1 6 while index < n 7 Parent = Dequeue(Q) 8 Left = CreateTNode(A[index]) 9 Parent.left = Left 10 index = index Enqueue(Q, Left) 11 if index < n 12 Right = CreateTNode (A[index]) 13 Parent.right = Right 14 index = index Enqueue(Q, Right) Loop invariant? Rivers ( @smail.nju.end.cn) Rivers Second Open Topic May 23, / 33
30 Easter Egg Build-MaxHeap-From-Array Subsection 2 Build-MaxHeap-From-Array Rivers ( @smail.nju.end.cn) Rivers Second Open Topic May 23, / 33
31 Easter Egg Build-MaxHeap-From-Array Build-MaxHeap-From-Array(H, A, n) 1 Build-Complete-Binary-Tree-From-Array (H.HeapTree, A, n) 2 Build-MaxHeap-FromTree(H.HeapTree) Θ(n) + Θ(n) Rivers ( @smail.nju.end.cn) Rivers Second Open Topic May 23, / 33
32 Easter Egg Build-MaxHeap-From-Array A more intuitive Build-MaxHeap-From-Array Build-MaxHeap-From-Array(H, A, n) 1 for i = 1 to n 2 Heap-Insert(H, A[i]) Analysis of time cost n log i = log n! = Θ(n log n) i=1 Rivers ( @smail.nju.end.cn) Rivers Second Open Topic May 23, / 33
33 Easter Egg Build-MaxHeap-From-Array Rivers Rivers Second Open Topic May 23, / 33
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