1. Binomial Tree B k-1 B0B0 BkBk B0B0 B1B1 B2B2 B3B3 B4B4 Adapted from: Kevin Wayne B k : a binomial tree B k-1 with the addition of a left child with another binomial tree B k-1

2. Binomial Tree B0B0 B1B1 B2B2 B3B3 B4B4 B1B1 B k-1 BkBk B2B2 B0B0 Number of nodes with respect to k? N(B o ) = 1 N(B k ) = 2 N(B k-1 ) = 2 k

3. Binomial Tree B0B0 B1B1 B2B2 B3B3 B4B4 Height? H(B o ) = 1 H(B k ) = 1 + H(B k-1 ) = k B1B1 B k-1 BkBk B2B2 B0B0

4. Binomial Tree B0B0 B1B1 B2B2 B3B3 B4B4 Degree of root node? k, each time we add another binomial tree B1B1 B k-1 BkBk B2B2 B0B0

5. Binomial Tree B0B0 B1B1 B2B2 B3B3 B4B4 What are the children of the root? k binomial trees: B k-1, B k-2, …, B 0 B1B1 B k-1 BkBk B2B2 B0B0

6. Binomial Tree B4B4 depth 2 depth 3 depth 4 depth 0 depth 1 Why is it called a binomial tree?

7. Binomial Tree B k has nodes at depth i. B4B4 depth 2 depth 3 depth 4 depth 0 depth 1

8. Binomial Heap Binomial heap Vuillemin, 1978. Sequence of binomial trees that satisfy binomial heap property: – each tree is min-heap ordered – top level: full or empty binomial tree of order k – which are empty or full is based on the number of elements B4B4 B0B0 B1B1 55 45 32 30 24 2322 50 483117 44 82910 6 37 318

9. Binomial Heap Like our “set” data structure from last time, except binomial tree heaps instead of arrays B4B4 B0B0 B1B1 55 45 32 30 24 2322 50 483117 44 82910 6 37 318 A 0 : [18] A 1 : [3, 7] A 2 : empty A 3 : empty A 3 : [6, 8, 29, 10, 44, 30, 23, 22, 48, 31, 17, 45, 32, 24, 55] N = 19 # trees = 3 height = 4 binary = 10011

10. Binomial Heap: Properties B4B4 B0B0 B1B1 55 45 32 30 24 2322 50 483117 44 82910 6 37 318 N = 19 # trees = 3 height = 4 binary = 10011 How many heaps? O(log n) – binary number representation

11. Binomial Heap: Properties B4B4 B0B0 B1B1 55 45 32 30 24 2322 50 483117 44 82910 6 37 318 N = 19 # trees = 3 height = 4 binary = 10011 Where is the max/min? Must be one of the roots of the heaps

12. Binomial Heap: Properties B4B4 B0B0 B1B1 55 45 32 30 24 2322 50 483117 44 82910 6 37 318 N = 19 # trees = 3 height = 4 binary = 10011 Runtime of max/min? O(log n)

13. Binomial Heap: Properties B4B4 B0B0 B1B1 55 45 32 30 24 2322 50 483117 44 82910 6 37 318 N = 19 # trees = 3 height = 4 binary = 10011 Height? floor(log 2 n) - largest tree = B log n - height of that tree is log n

14. Binomial Heap: Union How can we merge two binomial tree heaps of the same size (2 k )? – connect roots of H' and H'' – choose smaller key to be root of H H'' 55 45 32 30 24 2322 50 483117 44 82910 6 H' Runtime? O(1)

15. Binomial Heap: Union 55 45 32 30 24 2322 50 483117 44 82910 6 37 318 33 28 15 25 712 How can we combine/merge binomial heaps (i.e. a combination of binomial tree heaps)?

16. Binomial Heap: Union 0011 1001+ 0111 11 1 1 0 1 19 + 7 = 26 55 45 32 30 24 2322 50 483117 44 82910 6 37 318 41 33 28 15 25 712 + Go through each tree size starting at 0 and merge as we go

17. Binomial Heap: Union 55 45 32 30 24 2322 50 483117 44 82910 6 37 318 41 33 28 15 25 712 +

18. Binomial Heap: Union 55 45 32 30 24 2322 50 483117 44 82910 6 37 3 41 33 28 15 25 7 + 12 18 12

19. 55 45 32 30 24 2322 50 483117 44 82910 6 37 3 41 33 28 15 25 7 + 12 18 25 37 7 3 18 12 18 12

20. 55 45 32 30 24 2322 50 483117 44 82910 6 37 3 41 33 28 15 25 7 12 + 18 25 37 7 3 41 283325 37 157 3 18 12 18 12

21. 55 45 32 30 24 2322 50 483117 44 82910 6 37 3 41 33 28 15 25 7 + 18 12 41 283325 37 157 3 12 18 25 37 7 3 41 283325 37 157 3 18 12

22. 55 45 32 30 24 2322 50 483117 44 82910 6 37 3 41 33 28 15 25 7 + 18 12 41 283325 37 157 3 12 18 25 37 7 3 41 283325 37 157 3 55 45 32 30 24 2322 50 483117 44 82910 6 18 12

23. Binomial Heap: Union Analogous to binary addition Running time? n Proportional to number of trees in root lists 2 O(log 2 N) n O(log N) 0011 1001+ 0111 11 1 1 0 1 19 + 7 = 26

24. 3 37 618 55 45 32 30 24 2322 50 483117 44 82910 H Binomial Heap: Delete Min/Max We can find the min/max in O(log n). How can we extract it? Hint: B k consists of binomial trees: B k-1, B k-2, …, B 0

25. 3 37 618 55 45 32 30 24 2322 50 483117 44 82910 H Binomial Heap: Delete Min Delete node with minimum key in binomial heap H. n Find root x with min key in root list of H, and delete n H'  broken binomial trees n H  Union(H', H)

26. 3 37 618 55 45 32 30 24 2322 50 483117 44 82910 H Binomial Heap: Delete Min Delete node with minimum key in binomial heap H. n Find root x with min key in root list of H, and delete n H'  broken binomial trees n H  Union(H', H) Running time? O(log N)

27. 3 37 618 55 x 32 30 24 2322 50 483117 44 82910 H Binomial Heap: Decrease Key Just call Decrease-Key/Increase-Key of Heap n Suppose x is in binomial tree B k n Bubble node x up the tree if x is too small Running time: O(log N) n Proportional to depth of node x depth = 3

28. Binomial Heap: Delete Delete node x in binomial heap H n Decrease key of x to -  n Delete min Running time: O(log N)

29. Binomial Heap: Insert Insert a new node x into binomial heap H n H'  MakeHeap(x) n H  Union(H', H) Running time. O(log N) 3 37 618 55 45 32 30 24 2322 50 483117 44 82910 H x H'

30. Build-Heap Call insert n times Runtime? Can we get a tighter bound? O(n log n)

31. Build-Heap Call insert n times Consider inserting n numbers n how many times will B 0 be empty? n how many times will we need to merge with B 0 ? n how many times will we need to merge with B 1 ? n how many times will we need to merge with B 2 ? n … n how many times will we need to merge with B log n ? timescost

32. Build-Heap Call insert n times Consider inserting n numbers n how many times will B 0 be empty? n/2O(1) n how many times will we need to merge with B 0 ? n/2O(1) n how many times will we need to merge with B 1 ? n/4O(1) n how many times will we need to merge with B 2 ? n/8O(1) n … n how many times will we need to merge with B log n ? 1O(1) timescost Runtime?Θ(n)

33. Heaps

34. Fibonacci Heaps Similar to binomial heap A Fibonacci heap consists of a sequence of heaps More flexible Heaps do not have to be binomial trees More complicated

35. Heaps Should you always use a Fibonacci heap?

36. Heaps Extract-Max and Delete are O(n) worst case Constants can be large on some of the operations Complicated to implement

37. Heaps Can we do better?