Chapter 9 Heap Structures CS235102 Data Structures Chapter 9 Heap Structures

Min-Max Heap Deaps Leftist Trees Binomial Heaps Fibonacci Heaps

MIN-MAX Heaps (1/10) Definition A double-ended priority queue is a data structure that supports the following operations: Insert an element with arbitrary key Delete an element with the largest key Delete an element with the smallest key Min heap or Max heap: Only insertion and one of the two deletion operations are supported Min-Max heap: Supports all of the operations just described.

MIN-MAX Heaps (2/10) Definition: A mix-max heap is a complete binary tree such that if it is not empty, each element has a field called key. Alternating levels of this tree are min levels and max levels, respectively. Let x be any node in a min-max heap. If x is on a min (max) level then the element in x has the minimum (maximum) key from among all elements in the subtree with root x. We call this node a min (max) node.

MIN-MAX Heaps (3/10) Insertion into a min-max heap (at a “max” level) If it is smaller/greater than its father (a “min”), then it must be smaller/greater than all “max”/“min” above. So simply check the “min”/“max” ancestors There exists a similar approach at a “min” level

MIN-MAX Heaps (4/10) verify_max Following the nodes the max node i to the root and insert into its proper place MIN-MAX Heaps (4/10) item = 80 i = 3 13 grandparent = 3 #define MAX_SIZE 100 #define FALSE 0 #define TRUE 1 #define SWAP(x,y,t) ((t)=(x), (x)=(y), (y)=(t)) typedef struct { int key; /* other fields */ }element; element heap[MAX_SIZE]; [1] [2] [3] 80 [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] 40

MIN-MAX Heaps (5/10) min_max_insert: Insert item into the min-max heap item.key = 80 5 *n = 13 14 12 complexity: O(log n) parent = 6 7 [1] 7 5 min [2] 70 [3] 80 40 max [4] [5] [6] [7] 30 9 10 7 15 min 45 50 30 max 20 12 10 40 [8] [9] [10] [11] [12] [13] [14]

MIN-MAX Heaps (6/10) Deletion of min element If we wish to delete the element with the smallest key, then this element is in the root. In general situation, we are to reinsert an element item into a min-max-heap, heap, whose root is empty. We consider the two cases: The root has no children Item is to be inserted into the root. The root has at least one child. The smallest key in the min-max-heap is in one of the children or grandchildren of the root. We determine the node k has the smallest key. The following possibilities need to be considered:

MIN-MAX Heaps (7/10) item.key  heap[k].key No element in heap with key smaller than item.key Item may be inserted into the root. item.key  heap[k].key, k is a child of the root Since k is a max node, it has no descendants with key larger than heap[k].key. Hence, node k has no descendants with key larger than item.key. heap[k] may be moved to the root and item inserted into node k.

MIN-MAX Heaps (8/10) item.key  heap[k].key, k is a grandchild of the root In this case, heap[k] may be moved to the root, now heap[k] is seen as presently empty. Let parent be the parent of k. If item.key  heap[parent].key, then interchange them. This ensures that the max node parent contains the largest key in the sub-heap with root parent. At this point, we are faced with the problem of inserting item into the sub-heap with root k. Therefore, we repeat the above process.

delete_min: Delete the minimum element from the min-max heap complexity: O(log n) *n = 11 12 i = 5 1 last = 5 k = 11 5 parent = 2 temp.key = x.key = 12 [1] 9 7 [0] 7 [2] 70 [3] 40 [4] [5] [6] [7] 30 12 9 10 15 45 50 30 20 12 [8] [9] [10] [11] [12]

MIN-MAX Heaps (10/10) Deletion of max element Determine the children of the root which are located on max-level, and find the larger one (node) which is the largest one on the min-max heap We would consider the node as the root of a max-min heap There exist a similar approach (deletion of max element) as we mentioned above max-min heap

Deaps(1/8) Definition The root contains no element The left subtree is a min-heap The right subtree is a max-heap Constraint between the two trees: let i be any node in left subtree, j be the corresponding node in the right subtree. if j not exists, let j corresponds to parent of i i.key <= j.key

Deaps(2/8) i = min_partner(n) = j = max_partner(n) = if j > heapsize j /= 2

Deaps Insert(3/8) } else { public void insert(int x) { i = maxPartner(n); if (x > deap[i]) { deap[n] = deap[i]; maxInsert(i, x); } else minInsert(n, x); } public void insert(int x) { int i; if (++n == 2) { deap[2] = x; return; } if (inMaxHeap(n)) { i = minPartner(n); if (x < deap[i]) { deap[n] = deap[i]; minInsert(i, x); } else maxInsert(n, x);

Deaps(4/8) Insertion Into A Deap

Deaps(5/8)

Deaps(6/8)

Deaps delete min(7/8) public int deleteMin() { int i, j, key = deap[2], x = deap[n--]; // move smaller child to i for (i = 2; 2*i <= n; deap[i] = deap[j], i = j) { j = i * 2; if (j+1 <= n && (deap[j] > deap[j+1]) j++; } // try to put x at leaf i j = maxPartner(i); if (x > deap[j]) { deap[i] = deap[j]; maxInsert(j, x); } else { minInsert(i, x); } return key;

Deaps(8/8)

Leftist Trees(1/7) Support combine (two trees to one)

Leftist Trees(2/7) shortest(x) = 0 if x is an external node, otherwise 1+min(shortest(left(x)),shortest(right(x))}

Leftist Trees(3/7) Definition: shortest(left(x)) >= shortest(right(x))

Leftist Trees(4/7) Algorithm for combine(a, b) assume a.data <= b.data if (a.right is null) then make b be right child of a else combine (a.right, b) if shortest (a.right) > shortest (a.left) then exchange

Leftist Trees(5/7)

Leftist Trees(6/7)

Leftist Trees(7/7)

Binomial Heaps(1/10) Cost Amortization(分期還款) every operation in leftist trees costs O(logn) actual cost of delete in Binomial Heap could be O(n), but insert and combine are O(1) cost amortization charge some cost of a heavy operation to lightweight operations amortized Binomial Heap delete is O(log2n) A tighter bound could be achieved for a sequence of operations actual cost of any sequence of i inserts, c combines, and dm delete in Binomial Heaps is O(i+c+dmlogi)

Binomial Heaps(2/10) Definition of Binomial Heap Node: degree, child ,left_link, right_link, data, parent roots are doubly linked a points to smallest root

Binomial Heaps(3/10)

Binomial Heaps(4/10) Insertion Into A Binomial Heaps make a new node into doubly linked circular list pointed at by a set a to the root with smallest key Combine two B-heaps a and b combine two doubly linked circular lists to one

Binomial Heaps(5/10) Deletion Of Min Element

Binomial Heaps(6/10)

Binomial Heaps(7/10)

Binomial Heaps(8/10)

Binomial Heaps(9/10)

Binomial Heaps(10/10) Trees in B-Heaps is Binomial tree B0 has exactly one node Bk, k > 0, consists of a root with degree k and whose subtrees are B0, B1, …, Bk-1 Bk has exactly 2k nodes actual cost of a delete is O(logn + s) s = number of min-trees in a (original roots - 1) and y (children of the removed node)

Fibonacci Heaps(1/8) Definition delete, delete the element in a specified node decrease key This two operations are followed by cascading cut

Fibonacci Heaps(2/8) Deletion From An F-heap min or not min

Fibonacci Heaps(3/8) Decrease Key if not min, and smaller than parent, then delete

Fibonacci Heap(4/8) To prevent the amortized cost of delete min becomes O(n), each node can have only one child be deleted. If two children of x were deleted, then x must be cut and moved to the ring of roots. Using a flag (true of false) to indicate whether one of x’s child has been cut

Fibonacci Heaps(5/8) Cascading Cut

Fibonacci Heaps(6/8) Lemma Corollary the ith child of any node x in a F-Heap has a degree of at least i – 2, except when i=1 the degree is 0 Corollary Let Sk be the minimum possible number of descendants of a node of degree k, then S0=1, S1=2. From the lemma above, we got (2 comes from 1st child and root)

Fibonacci Heaps(7/8) That’s why the data structure is called Fibonacci Heap

Fibonacci Heaps(8/8) Application Of F-heaps