Optimal Binary
Search Tree
Optimal(理想)
+
Binary (二元 )
+
Search (搜尋)
+
Tree (樹)
=
OBST
Preface of OBST
It 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.

Purpose
In order to promote the efficiency of the
search , we can let the cost of the search
minimum or mean the average of compare
minimum
It has n keys in sorted order
(representation k1,k2,…,kn ) ( k1<k2<…<kn)
 some searches may be for values not in ki,
so we also have n+1 “dummy keys”
d0,d1,…,dn representating not in ki.
d0 = all values < k1.
dn = all values > kn.

APPLE
Bird
Cat
Dog
Egg
Fish
dummy keys (d0,d1,…,dn)
Keys
Grape
(k1,k2,..kn)
 For
each key ki, a pi which means
the probability searching for ki.
 For each key di, a qi which means the
probability searching for di.
elements
kn = key
 pn = probability of searching kn
 dn = dummy key
 qn = probability of searching dn

Useful statement
Success probability = (i=1~n) ∑ pi
Failure probability = (i=0~n) ∑ qi
(i=1~n) ∑ pi + (i=0~n) ∑ qi = 1
Success cost = (i=1~n) ∑ pi * (depth(ki)+1)
Failure cost= (i=0~n) ∑ qi * (depth(di)+1)
E[search cost in ]
= Success+ Failure
=(i=1~n) ∑ pi * (depth(ki)+1)
+(i=0~n) ∑ qi * (depth(di)+1)
= (i=1~n) ∑ pi+(i=0~n) ∑ qi
+(i=1~n) ∑ pi * depth(ki)
+(i=0~n) ∑ qi * depth(di)
= 1+(i=1~n) ∑ pi * depth(ki)
+(i=0~n) ∑ qi * depth(di)
Example#1
K2
K2
K1
d0
d0
d1
0
1
d1
K5
K3
d2
i
K1
K4
d3
d4
2
3
d5
4
K5
K3
d4
5
pi
0.15 0.1 0.05 0.1
0.2
qi
0.05 0.1 0.05 0.05 0.05 0.1
d2
d5
K4
d3
Cost=Probability * (Depth+1)
K2
K4
K1
d0 d1 K3 K5
d2 d3 d4 d5
i
pi
qi
0
0.05
1
0.15
0.1
2
0.1
0.05
3
0.05
0.05
4
0.1
0.05
5
0.2
0.1
K1=2*0.15=0.3
K2=1*0.1 =0.1
K3=3*0.05=0.15
K4=2*0.1 =0.2
K5=3*0.2 =0.6
d0=3*0.05 =0.15
d1=3*0.1 =0.3
d2=4*0.05 =0.2
d3=4*0.05 =0.2
d4=4*0.05 =0.2
d5=4*0.1 =0.4
all cost=2.8
K2
K1
Cost=Probability * (Depth+1)
K5
d0 d1 K4
d5
K3 d4
d2 d3
i
pi
qi
0
1
2
3
4
5
0.05
0.15
0.1
0.1
0.05
0.05
0.05
0.1
0.05
0.2
0.1
K1=2*0.15
K2=1*0.1
K3=4*0.05
K4=3*0.1
K5=2*0.2
d0=3*0.05
d1=3*0.1
d2=5*0.05
d3=5*0.05
d4=4*0.05
d5=3*0.1
=0.3
=0.1
=0.2
=0.3
=0.4
=0.15
=0.3
=0.25
=0.25
=0.2
=0.3
all cost=2.75
Picture#1=2.8
Picture#2=2.75
SO
Picture#1 cost more than Picture#2
Picture#2 is better.
The depth of Picture#1 is 3
The depth of Picture#2 is 4