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Binomial Heaps

Binomial Heaps are a collection of min trees with a circular linked list representation. They support efficient operations like insert, delete min, and merge. The complexity of remove min operation is O(log n) due to pairwise tree combination.

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Binomial Heaps

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  1. Binomial Heaps

  2. 1 4 6 5 5 2 3 7 9 9 9 4 5 8 7 6 6 8 Min Binomial Heap • Collection of min trees.

  3. Node Structure • Degree • Number of children. • Child • Pointer to one of the node’s children. • Null iff node has no child. • Sibling • Used for circular linked list of siblings. • Data

  4. A 1 4 6 5 5 2 3 7 9 9 9 4 5 8 7 6 6 8 Binomial Heap Representation • Circular linked list of min trees. Degree fields not shown.

  5. A 1 4 6 5 5 2 3 7 10 9 9 9 4 5 8 7 6 6 8 Insert 10 • Update min-element pointer if necessary. • Add a new single-node min tree to the collection.

  6. 4 1 7 5 B 3 7 9 A 8 Meld • Combine the 2 top-level circular lists. • Set min-element pointer.

  7. Meld 4 1 7 5 3 C 7 9 8

  8. Remove Min • Empty binomial heap=> fail.

  9. A 1 4 6 5 5 2 3 7 10 9 9 9 4 5 8 7 6 6 8 Nonempty Binomial Heap • Remove a min tree. • Reinsert subtrees of removed min tree. • Update binomial heap pointer.

  10. a b c d e d A Remove Min Tree • Same as remove a node from a circular list. • No next node => empty after remove. • Otherwise, copy next-node data and remove next node.

  11. 4 1 7 5 3 7 9 B A subtrees 8 Reinsert Subtrees • Combine the 2 top-level circular lists. • Same as in meld operation.

  12. Update Binomial Heap Pointer • Must examine roots of all min trees to determine the min value.

  13. Complexity Of Remove Min • Remove a min tree. • O(1). • Reinsert subtrees. • O(1). • Update binomial heap pointer. • O(s), where s is the number of min trees in final top-level circular list. • s = O(n). • Overall complexity of remove min is O(n).

  14. Enhanced Remove Min • During reinsert of subtrees, pairwise combine min trees whose roots have equal degree.

  15. 1 4 6 5 5 10 2 3 7 9 9 9 4 5 8 7 6 6 8 Pairwise Combine Examine the s = 7 trees in some order. Determined by the 2 top-level circular lists.

  16. 1 4 6 5 5 10 2 3 7 9 9 9 4 5 8 4 3 tree table 2 7 6 6 1 0 8 Pairwise Combine Use a table to keep track of trees by degree.

  17. 1 4 6 5 5 10 2 3 7 9 9 9 4 5 8 4 3 tree table 2 7 6 6 1 0 8 Pairwise Combine

  18. 1 4 6 5 5 10 2 3 7 9 9 9 4 5 8 4 3 tree table 2 7 6 6 1 0 8 Pairwise Combine Combine 2 min trees of degree 0. Make the one with larger root a subtree of other.

  19. 10 1 4 6 5 5 2 3 7 9 9 9 4 5 8 4 3 tree table 2 7 6 6 1 0 8 Pairwise Combine Update tree table.

  20. 10 1 4 6 5 5 2 3 7 9 9 9 4 5 8 4 3 tree table 2 7 6 6 1 0 8 Pairwise Combine Combine 2 min trees of degree 1. Make the one with larger root a subtree of other.

  21. 10 1 5 4 6 5 2 3 9 7 9 9 4 5 8 4 3 tree table 2 7 6 6 1 0 8 Pairwise Combine Update tree table.

  22. 10 1 5 4 6 5 2 3 9 7 9 9 4 5 8 4 3 tree table 2 7 6 6 1 0 8 Pairwise Combine

  23. 10 1 5 4 6 5 2 3 9 7 9 9 4 5 8 4 3 tree table 2 7 6 6 1 0 8 Pairwise Combine Combine 2 min trees of degree 2. Make the one with larger root a subtree of other.

  24. 1 4 6 5 2 3 7 10 5 9 9 4 5 8 9 4 3 tree table 2 7 6 6 1 0 8 Pairwise Combine Combine 2 min trees of degree 3. Make the one with larger root a subtree of other.

  25. 4 2 1 6 5 10 9 5 3 4 5 7 9 9 7 6 6 8 4 3 tree table 2 8 1 0 Pairwise Combine Update tree table.

  26. 4 2 1 6 5 10 9 5 3 4 5 7 9 9 7 6 6 8 4 3 tree table 2 8 1 0 Pairwise Combine

  27. 4 2 1 6 5 10 9 5 3 4 5 7 9 9 7 6 6 8 4 3 tree table 2 8 1 0 Pairwise Combine Create circular list of remaining trees.

  28. Complexity Of Remove Min • Create and initialize tree table. • O(MaxDegree). • Done once only. • Examine s min trees and pairwise combine. • O(s). • Collect remaining trees from tree table, reset table entries to null, and set binomial heap pointer. • O(MaxDegree). • Overall complexity of remove min. • O(MaxDegree + s).

  29. B0 … Bk-1 B2 B1 B0 Binomial Trees • Bkis degree k binomial tree. • Bk, k > 0, is:

  30. B3 B2 B0 B1 Examples

  31. … Bk-1 B2 B1 B0 B0 Number Of Nodes In Bk N0 = 1 • Nk = number of nodes in Bk. • Bk, k > 0, is: • Nk = N0 + N1 + N2 + …+ 1 • = 2k.

  32. B3 B2 B1 Equivalent Definition • Bk, k > 0, is two Bk-1s. • One of these is a subtree of the other.

  33. Nk And MaxDegree • N0 = 1 • Nk = 2Nk-1 = 2k. • If we start with zero elements and perform operations as described, then all trees in all binomial heaps are binomial trees. • So, MaxDegree = O(log n).

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