1. Chapter 9 Heap Structures
CS Data Structures
Chapter 9 Heap Structures

2. Min-Max Heap
Deaps
Leftist Trees
Binomial Heaps
Fibonacci Heaps

3. 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.

4. 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.

5. 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

6. 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

7. 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]

8. 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:

9. 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.

10. 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.

11. 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]

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

13. 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

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

15. 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);

16. Deaps(4/8)
Insertion Into A Deap

17. Deaps(5/8)

18. Deaps(6/8)

19. 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;

20. Deaps(8/8)

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

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

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

24. 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

25. Leftist Trees(5/7)

26. Leftist Trees(6/7)

27. Leftist Trees(7/7)

28. 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)

29. 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

30. Binomial Heaps(3/10)

31. 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

32. Binomial Heaps(5/10)
Deletion Of Min Element

33. Binomial Heaps(6/10)

34. Binomial Heaps(7/10)

35. Binomial Heaps(8/10)

36. Binomial Heaps(9/10)

37. 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)

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

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

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

41. 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

42. Fibonacci Heaps(5/8)
Cascading Cut

43. 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)

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

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