Dynamic Programming Code
Straightforward top-down rod cutting
•
CUT-ROD(p, n)
– If n == 0
• Return 0
– q = -1
– For I = 1 to n
• q = max (q, p[i] + CUT-ROD(p, n-i))
– Return q
Memoization in rod cutting
•
MEMOIZED-CUT-ROD(p, n)
– let r[0..n] be a new array
– for I = 0 to n
• r[i] = -1
– return MEMOIZED-CUT-ROD-AUX(p, n, r)
•
MEMOIZED-CUT-ROD-AUX (p, n, r)
– if r[n] >= 0
• return r[n]
– if n == 0
• q=0
– else
• q = -1
• For I = 1 to n
– q = max (q, p[i] + MEMOIZED-CUT-ROD-AUX(p, n-i, r))
– r[n] = q
– return q
Bottom up memoization
• BOTTOM-UP-CUT-ROD(p, n)
– let r[0..n] be a new array
– r[0] = 0
– for j = 1 to n
• q = -1
• for i = 1 to j
– q = max (q, p[i] + r[j – i])
• r[j] = q
– return r[n]
– How do we know where to make the cuts?
Bottom up memoization
• EXTENDED-BOTTOM-UP-CUT-ROD(p, n)
– let r[0..n] and s[0..n] be a new array
– r[0] = 0
– for j = 1 to n
• How do you print the solution?
• q = -1
• for i = 1 to j
– if q < p[i] + r[j – 1]
» q = p[i] + r[j – 1]
» s[j] = i
• r[j] = q
– return r[n]
Bottom up memoization
• EXTENDED-BOTTOM-UP-CUT-ROD(p, n)
– let r[0..n] and s[0..n] be a new array
– r[0] = 0
– for j = 1 to n
• PRINT-CUT-ROD-SOLUTION(p, n)
• q = -1
• (r, s) = EXT-BUCR(p, n)
• for i = 1 to j
• while n > 0
– if q < p[i] + r[j – 1]
• print s[n]
» q = p[i] + r[j – 1]
• n = n – s[n]
» s[j] = i
• r[j] = q
– return r[n]
LCS
Theorem
Let Z = z1, . . . , zk be any LCS of X and Y .
1. If xm = yn, then zk = xm = yn and Zk-1 is an LCS of Xm-1 and Yn-1.
2. If xm  yn, then either zk  xm and Z is an LCS of Xm-1 and Y .
3.
or zk  yn and Z is an LCS of X and Yn-1.
LCS-LENGTH (X, Y)
1. m ← length[X]
2. n ← length[Y]
3. for i ← 1 to m
4.
do c[i, 0] ← 0
5. for j ← 0 to n
6.
do c[0, j ] ← 0
7. for i ← 1 to m
8.
do for j ← 1 to n
9.
do if xi = yj
10.
then c[i, j ] ← c[i1, j1] + 1
11.
12.
else if c[i1, j ] ≥ c[i, j1]
13.
then c[i, j ] ← c[i 1, j ]
14.
15.
else c[i, j ] ← c[i, j1]
16.
17. return c
LCS-LENGTH (X, Y)
1. m ← length[X]
2. n ← length[Y]
3. for i ← 1 to m
4.
do c[i, 0] ← 0
5. for j ← 0 to n
6.
do c[0, j ] ← 0
7. for i ← 1 to m
8.
do for j ← 1 to n
9.
do if xi = yj
10.
then c[i, j ] ← c[i1, j1] + 1
11.
b[i, j ] ← “ ”
12.
else if c[i1, j ] ≥ c[i, j1]
13.
then c[i, j ] ← c[i 1, j ]
14.
b[i, j ] ← “↑”
15.
else c[i, j ] ← c[i, j1]
16.
b[i, j ] ← “←”
17. return c and b
PRINT-LCS (b, X, i, j)
1. if i = 0 or j = 0
2.
then return
3. if b[i, j ] = “ ”
4.
then PRINT-LCS(b, X, i1, j1)
5.
print xi
6.
elseif b[i, j ] = “↑”
7.
then PRINT-LCS(b, X, i1, j)
8. else PRINT-LCS(b, X, i, j1)
Matrix-chain Multiplication
coitweb.uncc.edu/~ras/courses/Matrix-Mult.ppt
•
Suppose we have a sequence or chain A1, A2,
…, An of n matrices to be multiplied
– That is, we want to compute the product
A1A2…An
•
There are many possible ways
(parenthesizations) to compute the product
11-11
Matrix-chain Multiplication
•
…contd
Example: consider the chain A1, A2, A3, A4 of
4 matrices
– Let us compute the product A1A2A3A4
•
There are 5 possible ways:
1.
2.
3.
4.
5.
(A1(A2(A3A4)))
(A1((A2A3)A4))
((A1A2)(A3A4))
((A1(A2A3))A4)
(((A1A2)A3)A4)
11-12
Matrix-chain Multiplication
•
…contd
To compute the number of scalar
multiplications necessary, we must know:
– Algorithm to multiply two matrices
– Matrix dimensions
•
Can you write the algorithm to multiply two
matrices?
11-13
Algorithm to Multiply 2 Matrices
Input: Matrices Ap×q and Bq×r (with dimensions p×q and q×r)
Result: Matrix Cp×r resulting from the product A·B
MATRIX-MULTIPLY(Ap×q , Bq×r)
1. for i ← 1 to p
2.
for j ← 1 to r
3.
C[i, j] ← 0
4.
for k ← 1 to q
5.
C[i, j] ← C[i, j] + A[i, k] · B[k, j]
6. return C
Scalar multiplication in line 5 dominates time to compute
CNumber of scalar multiplications = pqr
11-14
Matrix-chain Multiplication
•
•
…contd
Example: Consider three matrices A10100,
B1005, and C550
There are 2 ways to parenthesize
– ((AB)C) = D105 · C550
•
•
AB  10·100·5=5,000 scalar multiplications
DC  10·5·50 =2,500 scalar multiplications
Total:
7,500
– (A(BC)) = A10100 · E10050
•
•
BC  100·5·50=25,000 scalar multiplications
AE  10·100·50 =50,000 scalar multiplications
Total:
75,000
11-15
Matrix-chain Multiplication
•
…contd
Matrix-chain multiplication problem
– Given a chain A1, A2, …, An of n matrices, where
for i=1, 2, …, n, matrix Ai has dimension pi-1pi
– Parenthesize the product A1A2…An such that the
total number of scalar multiplications is
minimized
•
Brute force method of exhaustive search
takes time exponential in n
11-16
Dynamic Programming Approach
•
The structure of an optimal solution
– Let us use the notation Ai..j for the matrix that
results from the product Ai Ai+1 … Aj
– An optimal parenthesization of the product
A1A2…An splits the product between Ak and Ak+1
for some integer k where1 ≤ k < n
– First compute matrices A1..k and Ak+1..n ; then
multiply them to get the final matrix A1..n
11-17
Dynamic Programming Approach
…contd
– Key observation: parenthesizations of the
subchains A1A2…Ak and Ak+1Ak+2…An must also be
optimal if the parenthesization of the chain
A1A2…An is optimal (why?)
– That is, the optimal solution to the problem
contains within it the optimal solution to
subproblems
11-18
Dynamic Programming Approach …contd
•
Recursive definition of the value of an
optimal solution
– Let m[i, j] be the minimum number of scalar
multiplications necessary to compute Ai..j
– Minimum cost to compute A1..n is m[1, n]
– Suppose the optimal parenthesization of Ai..j
splits the product between Ak and Ak+1 for some
integer k where i ≤ k < j
11-19
Dynamic Programming Approach …contd
– Ai..j = (Ai Ai+1…Ak)·(Ak+1Ak+2…Aj)= Ai..k · Ak+1..j
– Cost of computing Ai..j = cost of computing Ai..k +
cost of computing Ak+1..j + cost of multiplying Ai..k
and Ak+1..j
– Cost of multiplying Ai..k and Ak+1..j is pi-1pk pj
– m[i, j ] = m[i, k] + m[k+1, j ] + pi-1pk pj
k<j
– m[i, i ] = 0 for i=1,2,…,n
for i ≤
11-20
Dynamic Programming Approach …contd
– But… optimal parenthesization occurs at one
value of k among all possible i ≤ k < j
– Check all these and select the best one
0
m[i, j ] =
min {m[i, k] + m[k+1, j ] + pi-1pk pj }
if i=j
if i<j
i ≤ k< j
11-21
Dynamic Programming Approach …contd
•
•
•
To keep track of how to construct an optimal
solution, we use a table s
s[i, j ] = value of k at which Ai Ai+1 … Aj is split
for optimal parenthesization
Algorithm: next slide
– First computes costs for chains of length l=1
– Then for chains of length l=2,3, … and so on
– Computes the optimal cost bottom-up
11-22
Algorithm to Compute Optimal Cost
Input: Array p[0…n] containing matrix dimensions and n
Result: Minimum-cost table m and split table s
MATRIX-CHAIN-ORDER(p[ ], n)
for i ← 1 to n
Takes O(n3) time
m[i, i] ← 0
2) space
Requires
O(n
for l ← 2 to n
for i ← 1 to n-l+1
j ← i+l-1
m[i, j] ← 
for k ← i to j-1
q ← m[i, k] + m[k+1, j] + p[i-1] p[k] p[j]
if q < m[i, j]
m[i, j] ← q
s[i, j] ← k
return m and s
11-23
Constructing Optimal Solution
•
•
Our algorithm computes the minimum-cost
table m and the split table s
The optimal solution can be constructed
from the split table s
– Each entry s[i, j ]=k shows where to split the
product Ai Ai+1 … Aj for the minimum cost
11-24
Example
•
•
Show how to multiply this
matrix chain optimally
Solution on the board
–
–
Minimum cost 15,125
Optimal parenthesization
((A1(A2A3))((A4 A5)A6))
Matrix Dimension
A1
30×35
A2
35×15
A3
15×5
A4
5×10
A5
10×20
A6
20×25
11-25
LPS (problem set 4)
• A palindrome is a nonempty string over some
alphabet that reads the same forward and
backward. Examples of palilndromes are all
strings of length 1, "civic", "racecar", and
"aibohphobia" (fear of palindromes).
• Give an efficient algorithm to find the longest
palindrome that is a subsequence of a given input
string. For example, given the input "character",
your algorithm should return "carac". What is the
running time of your algorithm?