1. 1 Pertemuan 20 Binomial Heap Matakuliah: T0026/Struktur Data Tahun: 2005 Versi: 1/1

2. 2 Learning Outcomes Pada akhir pertemuan ini, diharapkan mahasiswa akan mampu : Mahasiswa dapat merumuskan program modular untuk mengimplementasikan Binomial dan fibonacci heap

3. 3 Outline Materi Why binomial heaps Binomial trees Binomial max heaps Insert Merging two binomial heaps Delete max

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7. 7 Why Binomial Heaps? The growth rates for MaxHeap operations are: –find-max() is O(1) –insert(k ), delete-max() is O(lgn) –merge(heap1, heap2) –2 methods: “append” the heaps and then use fast-build- heap. O(n) where n is the total number of elements “append” the heaps and then percolate the nodes in the smaller heap –If number elements in small heap proportional to n, the worst case time complexity is O(n lg n)

8. 8 Why Binomial Heaps? Binomial Heaps have improved runtime for some of the primary operations of heaps: –find-max() is O(1) –delete-max() is O(lgn) –insert(k ) amortized O(1) and worst case is O(lg n) –merge(heap1, heap2) with a total of n elements is O(lgn) – Example of a Binomial heap shown below B1B1 B3B3

9. 9 Binomial Trees The i th binomial tree, B i, i  0 has a root with i children, B 0, …, B i-1 B 0 B 1 B 2 B 3... B i n 1 2 4 8... 2 i depth 0 1 2 3... i = lgn B1B1 B3B3 B2B2 B0B0

10. 10 The Number of Nodes The number of nodes at depth 0,1,2, …, i are B4B4 0 1 2 3 4 depth n

11. 11 Binomial (Maximum) Heap Is a collection of binomial trees of distinct sizes each of which has the heap property. The roots of the binomial trees are connected (as a doubly linked list) in the order of increasing size. There is a pointer to the largest key. 35 23 21 914 1 20 5 17 8 B1B1 B3B3 29 B0B0

12. 12 Structure of a binomial heap There is a relation between the binary representation of a number, and the structure of a binomial heap Any number n can be represented uniquely as a binary number with  lg n  +1 bits The representation is:

13. 13 Structure of a binomial heap Any binomial heap with n items can be represented with at most  lg n  +1 binomial trees Let b i = 1 if B i is in the binomial heap and b i = 0 otherwise Let |B i | denote the number of nodes in the binomial tree

14. 14 Merging same size binomial heaps into one binomial tree Link trees - Make the root of the tree with the smaller max value, the i+1 child of the binomial tree with the larger max value. –O(1) 15 5 7 2 12 2 9 3 +  15 5 7 2 12 2 9 3

15. 15 Insert New Note: The correspondence between “count” and inserting a new key into a binomial heap 1) Convert item to be inserted into a B* 0 2) Set i to 0 3) If H includes a B i A) Remove B i from H. B) Link B i with B* i to form B* i+1 C) Set i to i +1 D) Repeat step 3 4) link B* i with H and update max pointer. Worst case time is O(lgn)

16. 16 B1B1 B2B2 B0B0 B4B4 4 trees 23 nodes Example for insertion i = 0. Remove B 0 and link it with B 0 * getting B 1 * B* 0 merge with B* 1 i =0 i = 1. Remove B 1, link B 1 * with B 1 getting B 2 * B* 2 i =1 B4B4 B2B2 B1B1

17. 17 Example Continued I = 2. Remove B 2, link B 2 * with B 2 getting B 3 * B* 3 i =2 B4B4 i = 3. H does not include B 3. link B* 3 with B 4. Binomial heap has 2 binomial trees and 24 nodes.. B* 3 B4B4 i =3 2 trees 24 nodes

18. 18 Doing a sequence of insertions Insert 10 10 Insert 20 20 10

19. 19 Doing a sequence of insertions Insert 3 20 10 3

20. 20 Doing a sequence of insertions First 3 is removed Then 3 and 8 are joined Then the two B1s are joined Insert 8 20 8 10 3 20 10 3

21. 21 Doing a sequence of insertions Insert 30 20 8 10 3 30

22. 22 Doing a sequence of insertions Insert 15 20 8 10 3 30 15

23. 23 Amortized analysis for insert new Assume n=2 k inserts For n/2 inserts the binomial heap does not have a B 0 and 1 link is done For n/4 inserts the binomial heap has B 0 but no B 1 and 2 links are done For n/8 inserts the binomial heap has B 0 and B 1 but not B 2 and 3 links are done... For n/2 k inserts the binomial heap has B 0, B 1,…B k-2 and either not B k-1 or B k-1 and k links are done The total time is 1*(n/2)+2*(n/4)+…k*1+k*1=O(n) So amortized cost for insertion is O(1)

24. 24 Merge 2 Binomial Heaps, H 1 and H 2 into Result Note: The correspondence between adding two binary numbers and merging two binomial heaps Merge H 1 and H 2 into a new binomial heap, result. stage 0: 1. If neither contain B 0 do nothing 2. If only one binomial heap includes a B 0 put it into the result. 3. If both include B 0, link them into B 1 and save for next stage. (like carry bit in binary addition)

25. 25 Merge 2 Binomial Heaps, H 1 and H 2 into Result stage i (i = 1,…,  lg n  ): There may be 0 to 3 B i ‘s, one from H 1, one from H 2 and one from the previous stage. 1. If no B i do nothing 2. If there is only one B i put it into the result. 3. Otherwise link two B i ‘s into B i+1 and save for next stage. 4. If there is still a B i put it in the result. If there is a saved B, add to result There are exactly  lg n  +1 stages. So worst case growth rate is O(lgn)

26. 26 deleting-max 1) Remove the B i containing the max from H, joining remaining parts into a new binomial heap H 1. 2) Remove root of B i. link the binomial subtrees of the root into a new binomial heap H 2. 3) Merge H 1 and H 2. Time: 1) O(1) 2) Since B i has i children there are i links. B i contains 2 i   n nodes, so i  lg n and the worst case time is O(lg n) 3) O(lg n) TOTAL O(lg n)