Binary Tries (continued)

1. Binary Tries (continued) • split(k). • Similar to split algorithm for unbalanced binary search trees. • Construct S and B on way down the trie. • Follow with a backward cleanup pass over the constructed S and B.

2. x a b a Forward Pass • Suppose you are at node x, which is at level j of the input trie. • If bit j of k is 1, move to root of b and add to level j of S and to level j of B.

3. b x a b Forward Pass • If bit j of k is 0, move to root of a and add to level j of B and to level j of S.

4. a b c d e f g Forward Pass Example S = null B = null k = g.key = 101011

5. S B a b c d e f g Forward Pass Example k = g.key = 101011

6. S B a b c d e f g Forward Pass Example k = g.key = 101011

7. S B a b c d e f g Forward Pass Example k = g.key = 101011

8. S B a b c d e f g Forward Pass Example k = g.key = 101011

9. S B a b c e f g Forward Pass Example d k = g.key = 101011

10. S B a b c d e f Forward Pass Example k = g.key = 101011

11. Backward Cleanup Pass • Retrace path from current nodes in S and B toward roots of respective tries. • Eliminate branch nodes that are roots of subtries that have fewer than 2 dictionary pairs.

12. S B a b c d e f Backward Cleanup Pass Example fis an element node.

13. S B a b c d e Backward Cleanup Pass Example Now backup on B. f

14. S B a b c d e Backward Cleanup Pass Example Now backup on B. f

15. S B a b c d e Backward Cleanup Pass Example Now backup on B. Assume root of d is a branch node. f

16. S B a b c d e Backward Cleanup Pass Example Complexity of split is O(height). f

17. Compressed Binary Tries • No branch node whose degree is 1. • Add a bit# field to each branch node. • bit# tells you which bit of the key to use to decide whether to move to the left or right subtrie.

18. 0 1 0 0 1 0 1 0 0 0 1 0 1 1001 0011 0001 1100 1101 1000 Binary Trie 1 2 3 4 4 bit# field shown in black outside branch node.

19. Compressed Binary Trie 1 0 1 3 2 0 1 0 1 0001 0011 4 4 0 1 0 1 1000 1001 1100 1101 bit# field shown in black outside branch node.

20. 1 0 1 3 2 0 1 0 1 4 4 0 1 0 1 1001 1000 1100 1101 0011 0001 Compressed Binary Trie #branch nodes = n – 1.

21. 1 0 1 3 2 0 1 0 1 4 4 0 1 0 1 1001 1000 1100 1101 0011 0001 Insert Insert 0010.

22. 1 0 1 3 2 0 1 0 1 4 4 0 1 4 0 1 0 1 0010 0011 1100 1001 1000 0001 1101 Insert Insert 0100.

23. 0 1 0011 0010 0001 1000 1001 0100 1101 1100 Insert 1 0 1 2 2 1 0 1 0 3 4 4 0 1 0 1 4 0 1

24. 1 0 1 2 2 1 0 1 0 3 4 4 0 1 0 1 0 1 1100 1001 1000 0010 0011 0001 1101 0100 4 0 1 Delete Delete 0010.

25. 1 0 1 2 2 1 0 1 0 3 4 4 0 1 0 1 0 1 0011 1101 1100 1000 0001 1001 0100 Delete Delete 1001.

26. 1 0 1 2 2 1 0 1 0 3 4 0 1 0 1 0011 1101 1100 1000 0001 0100 Delete

27. Split(k) • Similar to splitting an uncompressed binary trie.

28. Join(S,m,B) • Insert m into B to get B’. • |S| <= 1 or |B’| = 1 handled as special cases as in the case of uncompressed tries. • When |S| > 1 and |B’| > 1, let Smaxbe the largest key in S and let B’min be the smallest key in B’. • Let d be the first bit which is different in Smax andB’min.

29. bit#(S) bit#(B’) 0 1 0 1 a b c d B’ S Cases To Consider • d < min{bit#(S), bit#(B’)} • bit#(S) = bit#(B’) • bit#(S) < bit#(B’) • bit#(S) > bit#(B’)

30. d 0 1 S B’ d < min{bit#(S), bit#(B’)} Bit d ofSmax must be 0.

31. s s 0 1 0 1 a b c d B’ S bit#(S) = bit#(B’) • Not possible, because keys in b are larger than those in c. • However, all keys in S are supposed to be smaller than those in B’.

32. s s b’ 0 1 0 1 0 1 a J(b,B’) a b c d B’ S bit#(S) < bit#(B’) = +

33. b’ s b’ 0 1 0 1 0 1 J(S,c) d a b c d B’ S bit#(S) > bit#(B’) = + Complexity is O(max{height(S), height(B)}). Smax and B’min are found just once.