Digital Search Trees & Binary Tries Analog of radix sort to searching. Keys are binary bit strings. Fixed length 0110, 0010, 1010, 1011. Variable length 01, 00, 101, 1011. Application IP routing, packet classification, firewalls. IPv4 32 bit IP address; |key| <= 32. IPv6 128 bit IP address; |key| <= 128. Digital Search Tree Assume fixed number of bits. Not empty => Root contains one dictionary pair (any pair). All remaining pairs whose key begins with a 0 are in the left subtree. All remaining pairs whose key begins with a 1 are in the right subtree. Left and right subtrees are digital search trees on remaining bits.
Example Start with an empty digital search tree and insert a pair whose key is 0110. 0110 Now, insert a pair whose key is 0010. 0110 0010 Example Now, insert a pair whose key is 1001. 0110 0010 0110 0010 1001 Example Now, insert a pair whose key is 1011. 0110
0110 0010 1001 0010 1001 1011 Example Now, insert a pair whose key is 0000. 0110 0110 0010 0010 1001 1011
0000 1001 1011 Search/Insert/Delete 0110 0010 0000 1001 1011 Complexity of each operation is O(#bits in a key). #key comparisons = O(height). Expensive when keys are very long. Binary Trie Information Retrieval. At most one key comparison per operation.
Fixed length keys. Branch nodes. Left and right child pointers. No data field(s). Element nodes. No child pointers. Data field to hold dictionary pair. Example 0 1 0 0 0 0001 1 0011 1100
0 0 1000 1 1 1001 At most one key comparison for a search. Variable Key Length Left and right child fields. Left and right pair fields. Left pair is pair whose key terminates at root of left subtree or the single pair that might otherwise be in the left subtree. Right pair is pair whose key terminates at root of right subtree or the single pair that might otherwise be in the right subtree. Field is null otherwise.
Example 0 null 1 0 00 01100 10 0 0 0000 001 1000
11111 101 1 00100 001100 At most one key comparison for a search. Fixed Length Insert 0 0 0 0001 1 0011 Insert 0111.
1 0 1 0111 1100 0 0 1000 1 1 1001 Zero compares. Fixed Length Insert 0 0
0 0001 1 0011 Insert 1101. 1 0 1 0111 1100 0 0 1000 1
1 1001 Fixed Length Insert 1100 0 0 0 0001 1 0011 Insert 1101. 1 0 1 0111 1
0 0 1000 0 1 1001 0 Fixed Length Insert 0 0 0 0001 1 0011 Insert 1101.
1 0 1 0111 1 0 0 1000 0 1 1001 0 1100 One compare. 1
1101 Fixed Length Delete 0 0 0 0001 1 0011 Delete 0111. 1 0 1 0111 1 0
0 1000 0 1 1001 0 1100 1 1101 Fixed Length Delete 0 1 0 0 0 0001
1 0011 Delete 0111. 1 0 0 1000 0 1 1001 0 1100 One compare. 1
1101 Fixed Length Delete 0 1 0 0 0 0001 1 0011 Delete 1100. 1 0 0 1000
0 1 1001 0 1100 1 1101 Fixed Length Delete 0 1 0 0 0 0001 1
0011 Delete 1100. 0 0 1000 1 0 1 1001 1 1101 Fixed Length Delete 1101 0
1 0 0 0 0001 1 0011 Delete 1100. 0 0 1000 1 0 1 1001
Fixed Length Delete 1101 0 1 0 0 0 0001 1 0011 Delete 1100. 0 0 1000 1
1001 1 Fixed Length Delete 0 1 0 0 0 0001 1 0011 Delete 1100. 1101 0
0 1000 1 1 1001 One compare. Fixed Length Join(S,m,B) Convert 3-way join to 2-way join Insert m into B to get B. S empty => B is answer; done. S is element node => insert S element into B; done; B is element node => insert B element into S; done; If you get to this step, the roots of S and B are branch nodes.
Fixed Length Join(S,m,B) S has empty right subtree. S a B b c J(S,B) J(a,b) c J(X,Y) join X and Y, all keys in X < all in Y. Fixed Length Join(S,m,B) S has nonempty right subtree. Left subtree of B must be empty, because all keys in B > all keys in S.
S a B b c Complexity = O(height). J(S,B) a J(b,c)
