Dynamic Programming
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 1
Dynamic Programming
Coming up
Assembly-line scheduling
Matrix-chain multiplication
Elements of dynamic programming
Longest common subsequence
Optimal binary search trees
(Chap 15)
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 2
First Dynamic Programming Example:
Assembly-line
Scheduling
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 3
Assembly-line Scheduling
Station S1,1 Station S1,2 Station S1,3 Station S1,4 … Station S1,n-1 Station S1,n
Assembly
line 1
a1,1
e1
t1,1
chassis
enters
a1,3
t1,2
a1,n-1
a1,4
t1,3
a1,n
t1,n-1
x1
t2,n-1
x2
Completed
auto exits
…
t2,1
e2
Assembly
line 2
a1,2
a2,1
t2,2
a2,2
t2,3
a2,3
a2,4
a2,n-1
a2,n
Station S2,1 Station S2,2 Station S2,3 Station S2,4 … Station S2,n-1 Station S2,n
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 4
Assembly-line Scheduling
Problem: To minimize the total processing time for one auto.
Station S1,1 Station S1,2 Station S1,3 Station S1,4 … Station S1,n-1 Station S1,n
a1,1
x1
a1,2
a1,3
a1,4
…
a1,n-1
a1,n
x1
t1,1
chassis
enters
t2,n-1
t2,3
a2,1
a2,2
a2,3
Completed
auto exits
a2,4
…
a2,n-
a2,n
1
Station S2,1 Station S2,2 Station S2,3 Station S2,4 … Station S2,n-1 Station S2,n
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 5
Assembly-line Scheduling
To find the fastest
ways, we may
inspect the speeds of
all possible ways.
But … there are
2n possible ways !!!
It takes (2n) time.
Solve it by
dynamic-programming.
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 6
Assembly-line Scheduling (Step 1)
Step 1. Characterize the structure of an optimal solution
(the fastest way through the factory)
The fastest way through the factory =
The faster one of either
a1,n
The fastest way
to reach S1,j-1
x1
or
The fastest way
to reach S2,j-1
x2
a2,j-1
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 7
Assembly-line Scheduling (Step 1)
Step 1. Characterize the structure of an optimal solution
(the fastest way through the factory)
The fastest way to reach station S1,j =
The faster one of either
a1,j-1
The fastest way
to reach S1,j-1
a1,j
a1,j
or
The fastest way
to reach S2,j-1
t2,n-1
a2,j-1
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 8
Assembly-line Scheduling (Step 2)
Step 2. Recursively define an optimal solution (fastest way through the
factory), in terms of the optimal solutions to subproblems
Let f* denote the fastest time through the factory.
Let fi[j] denote the fastest time to reach and get through Si,j.
Then,
f*= min (f1[n] + x1, f2[n] + x2)
f1(j) =
f2(j) =
e1+a1,1
if j=1
min (f1[j-1] + a1,j, f2[j-1] + t2,j-1+ a1,j) if j>1
e2+a2,1
if j=1
min (f2[j-1] + a2,j, f1[j-1] + t1,j-1+ a2,j) if j>1
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
Most of the subproblems are
visited more than
once.
Any clever
method to handle
them?
4. Dynamic Programming - 9
Assembly-line
Scheduling (Step 3)
FASTEST-WAY (a,t,e,x,n)
1
2
f1[1]  e1 + a1,1
f2[1]  e2 + a2,1
Step 3. Compute the value of
3
4
for j  2 to n
do if f1[j-1] + a1,j  f2[j-1] + t2,j-1+
a1,j
then f1[j]  f1[j-1] + a1,j
l1[j]  1
else f1[j]  f2[j-1] + t2,j-1 + a1,j
l1[j]  2
an optimal solution in
a bottom-up fashion
5
Let li[j] be the line number, 1 or 2, 6
whose station j-1 is used in 7
the solution.
8
eg. l1[5] = 2 means that the fastest
9
do if f2[j-1] + a2,j  f1[j-1] + t1,j-1+
way to reach station 5 of
a2,j
line 1 should pass through
10
then f2[j]  f2[j-1] + a2,j
station 4 of line 2.
11
l2[j]  2
Let l* be the line whose station n 12
else f2[j]  f1[j-1] + t1,j-1 + a2,j
is used in the solution.
13
l2[j]  1
eg. l*=2 means that the fastest way 14 if
f1[n] + x1  f2[n] + x2
involves station n of line 2. 15 then f*  f [n]+x
1
1
16
l*  1
CS3381 Des & Anal of Alg (2001-2002 SemA)
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City Univ of HK / Dept of CS / Helena Wong
17 else f*  f2[n]+x4.2Dynamic Programming - 10
Assembly-line Scheduling (Step 3)
What we are doing, is:
• Starting at the bottom level,
continuously filling in the tables (f1[], f2[]
and l1[], l2[]), and
• Looking up the tables to compute new
results for next higher level
Exercise: what is the complexity of this
algorithm?
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 11
Assembly-line Scheduling (Step 4)
Step 4. Construct an optimal solution from computed information.
PRINT-STATIONS()
1 i  l*
2 print “line ” i “, station ” n
3 for j  n downto 2
4
do i  li[j]
5
print “line ” i “, station ” j-1
line 1, station 6
line 2, station 5
line 2, station 4
line 1, station 3
line 2, station 2
line 1, station 1
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 12
Assembly-line Scheduling
What we have done is indeed
Dynamic Programming
Now it is time to test your
memory:
Dynamic Programming
Step ? Characterize the structure of an optimal solution
Eg. Study the structure of the fastest way through the factory.
Step ? Recursively define the value of an optimal solution
Step ? Compute the value of an optimal solution in a bottom-up fashion
Step ? Construct an optimal solution from computed information.
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 13
Second Dynamic Programming Example:
Matrix-Chain
Multiplication
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 14
Matrix-Chain Multiplication
5
5
5
5
5
5
5
5
5
5
5
5
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 15
Matrix-Chain Multiplication
Can we freely multiply
matrices of any
different shapes?
No. They must be compatible:
To multiply A and B, number of
columns of A must be equal to
number of rows of B.
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 16
Matrix-Chain Multiplication
We can also
multiply them
in different
ways:
However, we must consider
the different efficiencies of
the different ways.
(parenthesizations).
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 17
Matrix-Chain Multiplication
5x3x4=60 multiplications
5 5
5
5
5 5
5
5
5 5
5
5
4x5x2=40 multiplications
5
4x3x2= multiplications
Total =
multiplications
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
5x3x2= multiplications
Total =
http://www.cs.cityu.edu.hk/~helena
multiplications
4. Dynamic Programming - 18
Matrix-Chain Multiplication
A
B
5 5
5
5
5 5
5
5
5 5
5
5
5
…
100
…
Another Example:
…
C
…
5
…
B
50
…
…
…
100
…
A
…
…
10
(AB):
(BC):
((AB)C):
(A(BC)):
Number of scalar multiplications
= No. of rows of A *
No. of columns of A *
or no. of
No. of columns of B
rows of B
Size of resultant matrix:
No. of rows
= No. of rows of A
No. of columns = No. of columns of B
…
No. of multiplication = 10 * 100 * 5 = 5000
No. of multiplication = 100 * 5 * 50 = 25000
No. of multiplication = 5000 + 10 * 5 * 50 =
No. of multiplication = 25000 + 10 * 100 * 50 =
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 19
Matrix-Chain Multiplication
So before
multiplying the 3
matrices, we better
decide the optimal
parenthesization.
But how to decide?
Shall we check all the
parenthesizations and
find out the optimal one?
If we are multiplying 100
matrices, it seems there
are a lot of candidates.
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
Mmm, Let’s see
how many
parenthesizations
we’d need to check
…
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 20
Matrix-Chain Multiplication
n-1 ways
Suppose we need to compute A1A2A3…An
Our question is “How many parenthesizations we need to check”.
We may compute in many ways:
1st way: A1(A2…An)
but, there are different methods to compute for (A2…An).
2nd way: (A1A2)(A3…An) but, there are different methods to compute for (A3…An).
..
n-1th way: (A1…An-1)An
)An
but, there are different methods to compute for (A1…An-1).
Total no. of methods =
Let’s handle in kth way, then
sum of the methods for all n-1 ways.
no. of methods for (A1..An)
= no. of methods for (A1..Ak)(Ak+1An) Let P(n) = no. of alternative parenthesizations
of n matrices, then
= (no. of methods for k matrices) x
(no. of methods for n-k matrices)
1
if n=1
P(n) =
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
 k=1 to n-1 P(k) P(n-k)
http://www.cs.cityu.edu.hk/~helena
if n>1
4. Dynamic Programming - 21
Matrix-Chain Multiplication
P(n) =
1
if n=1
 k=1 to n-1 P(k) P(n-k)
if n>1
Hence,
P(1) = 1
P(2) = P(1)*P(1) = 1
P(3) = P(1)*P(2)+P(2)*P(1) = 2
P(4) = P(1)*P(3)+P(2)*P(2)+P(3)*P(1) = 5
P(5) = P(1)*P(4)+P(2)*P(3)+P(3)*P(2)+P(4)*P(1) = 14
P(6) = P(1)*P(5)+P(2)*P(4)+P(3)*P(3)+P(4)*P(2)+P(5)*P(1) =
…
This function increases very fast [(2n)].
So it is not feasible to check all possible
parenthesizations.
We’ll solve it using dynamic programming.
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 22
Matrix-Chain Multiplication (Step 1)
Step 1. The structure of an optimal parenthesization
For ..AiA2A3…Aj..
Let ( Ai…Ak )( Ak+1…Aj ) be the optimal parenthesization.
Then Ai…Ak
must be the optimal parenthesization for Ai..Ak.
And Ak+1…Aj must be the optimal parenthesization for Ak+1..Aj.
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 23
Matrix-Chain Multiplication (Step 2)
Step 2. A recursive solution
Define m[i,j] as minimum number of scalar multiplications for Ai…Aj.
And let ps-1 be the number of columns of As-1 (= number of rows of As).
Then,
m[i,j] =
0
if i=j
min i  k < j {m[i,k]+m[k+1,j]+pi-1pkpj } if i<j
Then, each m[i,j] problem can be solved by making use
of the solutions of smaller problems.
Indeed there are not many problems: one problem for
each choice of i and j satisfying 1  i  j  n.
But their solutions are to be visited many times.
=> Any clever method?
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 24
Matrix-Chain Multiplication (Step 3)
Step 3. Compute an optimal
solution from bottom-up
0
m[i,j] =
m[1,2]
= m[1,1]+m[2,2]+30x35x15
=
m[2,3]
= m[2,2]+m[3,3]+35x15x5
=
if i=j
min i  k < j {m[i,k]+m[k+1,j]+pi-1pkpj } if i<j
For chain length = 2:
Example:
col x row
A1 35x30
A2 15x35
A3 5x15
A4 10x5
A5 20x10
A6 25x20
For chain length = 1:
m
1
2
j 3
4
5
6
i
3
1
2
4
5 6
0
15750 0
7875 2625 0
9375 4375 750 0
11875 7125 2500 1000 0
15125 10500 5375 3500 5000 0
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
m[1,3] = min
m[1,1] + m[2,3] + 30x35x5
=
m[1,2] + m[3,3] + 30x15x5
=
m[1,4] = min
http://www.cs.cityu.edu.hk/~helena
m[2,2] + m[3,4] + 35x15x10
=
m [2,3] + m[4,4] + 35x5x10
=
4. Dynamic Programming - 25
Matrix-Chain Multiplication (Step 3)
Step 3. Compute an optimal solution from bottom-up
O()
()
Let s[i,j] be the value of k where optimal splitting occurs.
A1
A2
A3
A4
A5
A6
col x row
35x30
15x35
5x15
10x5
20x10
25x20
m
1
2
j 3
4
5
6
i
s 1 2 3 4 5 6
1
2 1
j 3 1 2
4 3 3 3
5 3 3 3 4
6 3 3 3 5 5
i
3
1
2
4
5 6
0
15750 0
7875 2625 0
9375 4375 750 0
11875 7125 2500 1000 0
15125 10500 5375 3500 5000 0
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
MATRIX-CHAIN-ORDER
1 for
2
3 for
4
5
6
7
8
9
10
11
i  1 to n
m[i,i]  0
len  2 to n //len is chain length
for i  1 to n - len + 1
j  i + len - 1
m[i,j]  
for k  i to j-1
q  m[i,k]+m[k+1,j]+pi-1pkpj
if q < m[i,j]
then m[i,j]  q
s[i,j]  k
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 26
Matrix-Chain Multiplication (Step 4)
Step 4. Constructing an optimal solution
A recursive printing procedure:
Recall that
s[i,j] = the value of k where optimal splitting occurs.
PRINT-OPTIMAL-PARENS(i,j)
1 if i=j then print “A”I
2 else print “(“
3
PRINT-OPTIMAL-PARENS(i,s[i,j])
4
PRINT-OPTIMAL-PARENS(s[i,j]+1,j)
5
print “)“
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
i
s 1 2 3 4 5 6
1
2 1
j 3 1 2
4 3 3 3
5 3 3 3 4
6 3 3 3 5 5
Sample
((A1(A2A3))((A4A5)A6))
4. Dynamic Programming - 27
Elements of Dynamic Programming:
2 Key Ingredients
&
Memoization
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 28
Elements of Dynamic Programming
Comparison:
Divide & Conquer:
Break up into smaller problems
Dynamic Programming: Solve all smaller problems but only reuse
optimal subproblem solutions
But when should
we apply dynamic
programming?
Optimizatio
n Problems
Bottom-up
Table-based
approach
Solution
2 key ingredients:
Optimal Substructure
Overlapping
Subproblems
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 29
Elements of Dynamic Programming
2 key ingredients of Dynamic Programming:
Optimal Substructure
For any problem, if an optimal solution consists of optimal
solutions to its subproblems, then, it exhibits optimal
substructure.
Overlapping Subproblems
If a recursive algorithm revisits the same subproblem over
and over again, we say that the subproblems are
overlapping.
If both ingredients happen, it is a good clue that dynamic
programming might apply.
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 30
Elements of Dynamic Programming
Memoization
- a variation of dynamic programming.
• Uses a top-down (recursive) strategy: After computing solutions to
subproblems, store in the table. Subsequent calls do table lookup.
• Memorization? Memoization?
MEMOIZED-MATRIX-CHAIN()
1 for i  1 to n
2
for j  i to n
3
m[i,j]  
4 return LOOKUP-CHAIN(i,j)
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
LOOKUP-CHAIN(i,j)
1 if m[i,j] < 
2
return m[i,j]
3 if i=j
4
then m[i,j]  0
5
else for k  i to j-1
6
q  LOOKUP-CHAIN(i,k)
+ LOOKUP-CHAIN(k+1,j)
+ pi-1pkpj
7
if q < m[i,j]
8
then m[i,j]  q
9 return m[i,j]
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 31
Elements of Dynamic Programming
When Memoization outperforms bottom-up
dynamic programming?
For the some cases when some subproblems need
not be solved at all.
Eg. In the assembly line, some stations’ delivery
paths are blocked.
Then Memoization is faster.
Otherwise, a bottom-up dynamic programming is
usually more efficient:
No overhead for recursion
Less overhead for maintaining the table.
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 32
Third Dynamic Programming Example:
Longest Common
Subsequence
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 33
Longest Common Subsequence
Suppose we have 2 strings:
S1=ACCGGTCGAGTGCGCGGAAGCCGGCCGAA
S2=GTCGTTCGGAATGCCGTTGCTCTGTAAT
The Longest Common Subsequence is
GTCGTCGGAAGCCGGCCGAA
The Longest Common Subsequence problem :
Given 2 sequences X=<x1,x2,…xm> and Y=<y1,y2,…yn>, find
a maximum-length common subsequence (LCS) of X and Y.
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 34
Longest Common Subsequence
Optimal substructure of LCS:
Let Z=<z1,z2,..zk> be any LCS of X=<x1,x2,…xm> and Y=<y1,y2,…yn>.
Let’s define the notations: Xr=<x1,x2,..xr>, Yr=<y1,y2,..yr>, Zr=<z1,z2,..zr>.
Then, if xm=yn, then Zk-1 is an LCS of the pair Xm-1 and Yn-1.
if xmyn, then Z is an LCS of the pair Xm-1 and Yn,
or
Z is an LCS of the pair Xm and Yn-1,
or
Z is an LCS of both pairs.
We can work out a recursive formula for the problem:
let c[i,j] = length of an LCS of Xi and Yj, then
C[i,j] =
0
if i=0 or j=0
c[i-1,j-1]+1
if i,j>0 and xi=yj
Max(c[i,j-1],c[i-1,j]) if i,j>0 and xi yj
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
Note that in this
case we skip the
computation of
the subproblems
c[i,j-1] and c[i1,j]
4. Dynamic Programming - 35
Longest Common Subsequence
To finish the solution by
dynamic programming?
You may like to try out
yourselves,
Or, read Chap 15.4.
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 36
Forth Dynamic Programming Example:
Optimal binary
search trees
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 37
Optimal binary search trees
A binary search tree for dictionary lookup:
Let K=<k1,k2,..kn> be n words (distinct
keys) in sorted order
Let d0,d1,..dn be n “dummy keys”
representing values not in K
k2
k1
d0
k4
d1
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
Define the cost of a search as the
number of nodes examined in a search.
(depth of the key + 1)
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
k3
d2
Let pi = probability of searching k1
qi = probability representing di
http://www.cs.cityu.edu.hk/~helena
d5
d5
k4
k3
d2
d4
Expected cost
of an arbitrary
k5 search = 2.75
k1
d1
k5
d3
k2
d0
Expected cost
of an arbitrary
search = 2.8
d4
d3
Optimal binary
search tree
4. Dynamic Programming - 38
Optimal binary search trees
The Optimal Binary Search Tree problem :
For a given set of probabilities of the keys and dummy
nodes, construct an optimal binary search tree.
Optimal substructure:
For an optimal binary search tree with root kr as the root, the
left sub-tree and the right sub-tree are also optimal binary
search trees.
kr
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 39
Optimal binary search trees
We can work out a recursive
formula for the problem:
Let e[i,j] = expected cost of searching
an optimal binary search tree
containing ki,..,kj.
When j = i-1, it corresponds to searching
di-1, then e[i,i-1]=qi-1.
e[i,j] =
k2
k1
d0
k5
d1
d5
k4
k3
d2
d4
d3
qi-1
if j=i-1
Minir j { e[i,r-1] + e[r+1,j] + s=i to jps +  s=i-1 to jqs }
if i j
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 40
Optimal binary search trees
To finish the solution by
dynamic programming?
You may like to try out
yourselves,
Or, read Chap 15.5.
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 41
Dynamic Programming
Summary
Assembly-line scheduling
Matrix-chain multiplication
Elements of dynamic programming
Optimal Substructure
Overlapping Subproblems
Memoization
Longest common subsequence
Optimal binary search trees
CS3381 Des & Anal of Alg (2001-2002 SemA)
City Univ of HK / Dept of CS / Helena Wong
http://www.cs.cityu.edu.hk/~helena
4. Dynamic Programming - 42