Static Dictionaries

1
Static Dictionaries

• Collection of items.
• Each item is a pair.
• (key, element)
• Pairs have different keys.
• Operations are
• initialize/create
• get (search)

2
Hashing

• Perfect hashing (no collisions).
• Minimal perfect hashing (space n).
• CHD (compress, hash, and displace) algorithm.
• O(n) time to construct the perfect or minimal
  perfect hash function.
• O(1) search time.
• Bothelo, Belazzougui Dietzfelbinger. Compress,
  hash, and displace.  17th European Symposium on
  Algorithms, 2009.

3
Search Tree

• Hashing not efficient for extended operations
  such as range search and nearest match.
• Will examine a binary search tree structure for
  static dictionaries.
• Each item/key/element has an estimated access
  frequency (or probability).

4
Example
Key Probability
a 0.8
b 0.1
c 0.1
a lt b lt c
Cost 0.8 1 0.1 2 0.1 3 1.3
Cost 0.8 2 0.1 1 0.1 2 1.9
5
Search Types

• Successful.
• Search for a key that is in the dictionary.
• Terminates at an internal node.
• Unsuccessful.
• Search for a key that is not in the dictionary.
• Terminates at an external/failure node.

6
Internal And External Nodes

• A binary tree with n internal nodes has n 1
  external nodes.
• Let s1, s2, , sn be the internal nodes, in
  inorder.
• key(s1) lt key(s2) lt lt key(sn).
• Let key(s0) infinity and key(sn1) infinity.
• Let f0, f1, , fn be the external nodes, in
  inorder.
• fi is reached iff key(si) lt search key lt
  key(si1).

7
Cost Of Binary Search Tree

• Let pi probability for key(si).
• Let qi probability for key(si) lt search key lt
  key(si1).
• Sum of ps and qs 1.
• Cost of tree S0 lt i lt n qi (level(fi) 1)
• S1lt i lt n pi
  level(si)
• Cost weighted path length.

8
Brute Force Algorithm

• Generate all binary search trees with n internal
  nodes.
• Compute the weighted path length of each.
• Determine tree with minimum weighted path length.
• Number of trees to examine is O(4n/n1.5).
• Brute force approach is impractical for large n.

9
Dynamic Programming

• Keys are a1 lt a2 lt lt an.
• Let Ti j least cost tree for ai1, ai2, , aj.
• T0n least cost tree for a1, a2, , an.

T2,7

• Ti j includes pi1, pi2, , pj and qi, qi1, ,
  qj.

10
Terminology

• Ti j least cost tree for ai1, ai2, , aj.
• ci j cost of Ti j
• Si lt u lt j qu (level(fu) 1)
• Si lt u lt j pu level(su).
• ri j root of Ti j.
• wi j weight of Ti j
• sum of ps and qs in Ti j
• pi1 pi2 pj qi qi1 qj

T2,7
11
i j

• Ti j includes pi1, pi2, , pj and qi, qi1, ,
  qj.
• Ti i includes qi only.

• ci i cost of Ti i 0.
• ri i root of Ti i 0.
• wi i weight of Ti i
• sum of ps and qs in Ti i
• qi

12
i lt j

• Ti j least cost tree for ai1, ai2, , aj.
• Ti j includes pi1, pi2, , pj and qi, qi1, ,
  qj.
• Let ak, i lt k lt j, be in the root of Ti j.

• L includes pi1, pi2, , pk-1 and qi, qi1, ,
  qk-1.
• R includes pk1, pk2, , pj and qk, qk1, , qj.

13
cost(L)

• L includes pi1, pi2, , pk-1 and qi, qi1, ,
  qk-1.
• cost(L) weighted path length of L when viewed
  as a stand alone binary search tree.

14
Contribution To cij

• ci j Si lt u lt j qu (level(fu) 1)
• Si lt u lt j pu level(su).
• When L is viewed as a subtree of Ti j , the level
  of each node is 1 more than when L is viewed as a
  stand alone tree.
• So, contribution of L to cij is cost(L) wi k-1.

15
cij

• Contribution of L to cij is cost(L) wi k-1.
• Contribution of R to cij is cost(R) wkj.
• cij cost(L) wi k-1 cost(R) wkj pk
• cost(L) cost(R) wij

16
cij

• cij cost(L) cost(R) wij
• cost(L) cik-1
• cost(R) ckj
• cij cik-1 ckj wij
• Dont know k.
• cij mini lt k lt jcik-1 ckj wij

17
cij

• cij mini lt k lt jcik-1 ckj wij
• rij k that minimizes right side.

18
Computation Of c0n And r0n

• Start with ci i 0, ri i 0, wi i qi, 0 lt i
  lt n (zero-key trees).
• Use cij mini lt k lt jcik-1 ckj wij to
  compute cii1, ri i1, 0 lt i lt n 1 (one-key
  trees).
• Now use the equation to compute cii2, ri i2, 0
  lt i lt n 2 (two-key trees).
• Now use the equation to compute cii3, ri i3, 0
  lt i lt n 3 (three-key trees).
• Continue until c0n and r0n(n-key tree) have been
  computed.

19
Computation Of c0n And r0n
20
Complexity

• cij mini lt k lt jcik-1 ckj wij
• O(n) time to compute one cij.
• O(n2) cijs to compute.
• Total time is O(n3).
• May be reduced to O(n2) by using
• cij min ri,j-1 lt k lt ri1,j
  cik-1 ckj wij

21
Construct T0n

• Root is r0n.
• Suppose that r0n 10.

• Construct T09 and T10,n recursively.
• Time is O(n).