Module2-OptimalBST Module2-OptimalBS.ppt

17/9/2024
Advanced Algorithms
Course Code: 21CS7E12
Module 2
TEXT BOOK
Introduction to Algorithms- Thomas H. Cormen, Charles E.
Leiserson, Ronald L. Rivest, Clifford Stein PHI, 3rd
Edition, 2009
1
Savita S

Optimal Binary Search Tree
Translate text from English to Latvian.
Lookup Operation
By building a binary search tree with n
English words as keys and their Latvian
equivalents as satellite data.
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Savita S

1.Preface
 OBST is one special kind of advanced tree.
 It focus on how to reduce the cost of the search of
the BST.
 It may not have the lowest height !
 It needs 3 tables to record probabilities, cost, and
root.
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2.Premise
 It has n keys (representation k1,k2,…,kn) in sorted order
(so that k1<k2<…<kn), and we wish to build a binary
search tree from these keys. For each ki ,we have a
probability pi that a search will be for ki.
 In contrast of, some searches may be for values not in ki,
and so we also have n+1 “dummy keys” d0,d1,…,dn
representating not in ki.
 In particular, d0 represents all values less than k1, and dn
represents all values greater than kn, and for i=1,2,…,n-1,
the dummy key di represents all values between ki and
ki+1.
＊ The dummy keys are leaves (external nodes), and the
data keys mean internal nodes.
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3.Formula & Prove
 The case of search are two situations, one
is success, and the other, without saying, is
failure.
 We can get the first statement :
(i=1~n) ∑ pi + (i=0~n) ∑ qi = 1
Success Failure
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 Because we have probabilities of searches for each key
and each dummy key, we can determine the expected
cost of a search in a given binary search tree T. Let us
assume that the actual cost of a search is the number of
nodes examined, i.e., the depth of the node found by the
search in T,plus1. Then the expected cost of a search in
T is : (The second statement)
 E[ search cost in T]
= (i=1~n) ∑ pi ． (depthT(ki)+1)
+ (i=0~n) ∑ qi ． (depthT(di)+1)
=1 + (i=1~n) ∑ pi ． depthT(ki)
+ (i=0~n) ∑ qi ． depthT(di)
Where depthT denotes a node’s depth in the tree T.
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k2
k1 k4
k3 k5
k2
k1 k5
k4
k3
d0 d1
d2 d3 d4 d5
d0 d1
d2 d3
d4
d5
Figure (a)
Figure (b)
i 0 1 2 3 4 5
pi 0.15 0.10 0.05 0.10 0.20
qi 0.05 0.10 0.05 0.05 0.05 0.10
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 By Figure (a), we can calculate the expected search cost node by node:
Node# Depth probability cost
k1 1 0.15 0.30
k2 0 0.10 0.10
k3 2 0.05 0.15
k4 1 0.10 0.20
K5 2 0.20 0.60
d0 2 0.05 0.15
d1 3 0.10 0.30
d2 3 0.05 0.20
d3 3 0.05 0.20
d4 3 0.05 0.20
d5 3 0.10 0.40
Cost=
Probability *
(Depth+1)
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 And the total cost = (0.30 + 0.10 + 0.15 +
0.20 + 0.60 + 0.15 + 0.30 + 0.20 + 0.20 +
0.20 + 0.40 ) = 2.80
 So Figure (a) costs 2.80 ,on another, the
Figure (b) costs 2.75, and that tree is really
optimal.
 We can see the height of (b) is more than
(a) , and the key k5 has the greatest search
probability of any key, yet the root of the
OBST shown is k2.(The lowest expected
cost of any BST with k5 at the root is 2.85)
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Step1:The structure of an OBST
 To characterize the optimal substructure of
OBST, we start with an observation about
subtrees. Consider any subtree of a BST. It
must contain keys in a contiguous range ki,
…,kj, for some 1≦i ≦j ≦n. In addition, a
subtree that contains keys ki,…,kj must also
have as its leaves the dummy keys di-1 ,
…,dj.
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 We need to use the optimal substructure to show
that we can construct an optimal solution to the
problem from optimal solutions to subproblems.
Given keys ki ,…, kj, one of these keys, say kr (I r
≦
j), will be the root of an optimal subtree containing
≦
these keys. The left subtree of the root kr will contain
the keys (ki ,…, kr-1) and the dummy keys( di-1 ,…, dr-
1), and the right subtree will contain the keys (kr+1 ,…,
kj) and the dummy keys( dr ,…, dj). As long as we
examine all candidate roots kr, where I r j, and we
≦ ≦
determine all optimal binary search trees containing
ki ,…, kr-1 and those containing kr+1 ,…, kj , we are
guaranteed that we will find an OBST.
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 There is one detail worth nothing about “empty”
subtrees. Suppose that in a subtree with keys
ki,...,kj, we select ki as the root. By the above
argument, ki ‘s left subtree contains the keys ki,…,
ki-1. It is natural to interpret this sequence as
containing no keys. It is easy to know that
subtrees also contain dummy keys. The
sequence has no actual keys but does contain the
single dummy key di-1. Symmetrically, if we select
kj as the root, then kj‘s right subtree contains the
keys, kj+1 …,kj; this right subtree contains no
actual keys, but it does contain the dummy key dj.
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Step2: A recursive solution
 We are ready to define the value of an optimal
solution recursively. We pick our subproblem
domain as finding an OBST containing the keys
ki,…,kj, where i 1, j n, and j i-1. (It is when
≧ ≦ ≧
j=i-1 that ther are no actual keys; we have just
the dummy key di-1.)
 Let us define e[i,j] as the expected cost of
searching an OBST containing the keys ki,…, kj.
Ultimately, we wish to compute e[1,n].
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 The easy case occurs when j=i-1. Then
we have just the dummy key di-1. The
expected search cost is e[i,i-1]= qi-1.
 When j 1, we need to select a root k
≧ rfrom
among ki,…,kj and then make an OBST
with keys ki,…,kr-1 its left subtree and an
OBST with keys kr+1,…,kj its right subtree.
By the time, what happens to the expected
search cost of a subtree when it becomes
a subtree of a node? The answer is that
the depth of each node in the subtree
increases by 1.
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 By the second statement, the excepted
search cost of this subtree increases by the
sum of all the probabilities in the subtree. For
a subtree with keys ki,…,kj let us denote this
sum of probabilities as
w (i , j) = (l=i~j) ∑ pl + (l=i-1~j) ∑ ql
Thus, if kr is the root of an optimal subtree
containing keys ki,…,kj, we have
E[i,j]= pr + (e[i,r-1]+w(i,r-1))+(e[r+1,j]+w(r+1,j))
Nothing that w (i , j) = w(i,r-1)+ pr +w(r+1,j)
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 We rewrite e[i,j] as
e[i,j]= e[i,r-1] + e[r+1,j]+w(i,j)
The recursive equation as above assumes that
we know which node kr to use as the root. We
choose the root that gives the lowest expected
search cost, giving us our final recursive
formulation:
E[i,j]=
case1: if i j,i r j
≦ ≦ ≦
E[i,j]=min{e[i,r-1]+e[r+1,j]+w(i,j)}
case2: if j=i-1; E[i,j]= qi-1
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 The e[i,j] values give the expected search
costs in OBST. To help us keep track of
the structure of OBST, we define root[i,j],
for 1 i j n, to be the index r for which k
≦ ≦ ≦ r
is the root of an OBST containing keys ki,
…,kj.
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Step3: Computing the expected
search cost of an OBST
 We store the e[i.j] values in a table e[1..n+1, 0..n].
The first index needs to run to n+1rather than n
because in order to have a subtree containing only
the dummy key dn, we will need to compute and
store e[n+1,n]. The second index needs to start from
0 because in order to have a subtree containing only
the dummy key d0, we will need to compute and
store e[1,0]. We will use only the entries e[i,j] for
which j i-1. we also use a table root[i,j], for recording
≧
the root of the subtree containing keys ki,…, kj. This
table uses only the entries for which 1 i j n.
≦ ≦ ≦
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 We will need one other table for efficiency.
Rather than compute the value of w(i,j)
from scratch every time we are computing
e[i,j] ----- we tore these values in a table
w[1..n+1,0..n]. For the base case, we
compute w[i,i-1] = qi-1 for 1 i n.
≦ ≦
 For j I, we compute :
≧
w[i,j]=w[i,j-1]+pi+qi
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OPTIMAL—BST(p,q,n)
 For i 1 to n+1
do e[i,i-1] qi-1
do w[i,i-1] qi-1
For l 1 to n
do for i 1 to n-l +1
do j i+l-1
e[i,j] ∞
w[i,j] w[i,j-1]+pj+qj
For r i to j
do t e[i,r-1]+e[r+1,j]+w[i,j]
if t<e[i,j]
then e[i,j] t
root [i,j] r
Return e and root
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2.75
1.75 2.00
1.25
1.20 1.30
0.90
0.70 0.60 0.90
0.45 0.40
0.25
0.30
0.50
0.05 0.10 0.05 0.05 0.05 0.10
e
0
1
2
3
4
5 1
2
2
2
1
1
0
3 3
4
4
5
5
5
3
4
6
6
0.05 0.05 0.05 0.05
0.10 0.10
0.15
1.00
0.70 0.80
0.50
0.30
0.35
0.45
0.35
0.20
0.25
0.30
0.60
0.50
0.55
w
root
1
1
1
1
2
2
2
2
2
2
2
2
3
3
3
4
4
4
4
4
5
5
5
5
5
The tables e[i,j], w[i,j], and root [i,j]computed by Optimal-BST
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Exercise
i 0 1 2 3 4 5 6 7
pi 0.04 0.06 0.08 0.02 0.10 0.12 0.14
qi 0.06 0.06 0.06 0.06 0.05 0.05 0.05 0.05
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i 10 20 30 40
pi 3 3 1 1
qi 2 3 1 1 1
Example
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Module2-OptimalBST Module2-OptimalBS.ppt

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