CS331 Advanced Data Structures – Spring 2016
Optimal Binary Search Tree Problem
The Problem : Given n key values k1 , k2 , . . . , kn and probabilities p1 , p2 , . . . , pn and q0 , q1 , . . . , qn
where
pi = probability of key ki being searched for
qi = probability of value v between ki and ki+1 being searched for (i.e, a failed
search)
determine an optimal binary tree that will minimize the expected number of nodes visited.
The Recurrence Relation : Let
e(i, j) =
w(i, j) =
expected value for optimal BST containing keys ki through kj ,
j
X
pl +
l=i
j
X
qj .
l=i−1
Then
(
e(i, j) =
qi−1 ,
j = i − 1,
min (e(i, r − 1) + e(r + 1, j) + w(i, j), j ≥ i
i≤r<j
(
w(i, j) =
qi−1 ,
j = i − 1,
w(i, j − 1) + pj + qj , j ≥ i
The Code :
int optimalBST(double [] p, double [] q)
{
int n = p.length()-1;
int [][] e = new int[n+2][n+1];
int [][] w = new int[n+2][n+1];
for(int i=1; i<=n+1; i++)
e[i][i-1] = w[i][i-1] = q[i-1];
for(int b=0; b<n; b++)
// b = distance from main diagonal
for(int i=1; i<=n-b; i++) {
int j = i+b;
w[i][j] = w[i][j-1] + p[j] + q[j];
e[i][j] = e[i][i-1] + e[i+1][j] + w[i][j];
for(int r=i+1; r<=j; r++) {
tmp = e[i][r-1] + e[r+1][j] + w[i][j];
if (tmp < e[i][j])
e[i][j] = tmp;
}
}
return e[1][n];
}
The Time Complexity : T (n) = O(n3 ).