DynamicProgramming.ppt

Analysis of Algorithms
CS 477/677
Dynamic Programming
Instructor: George Bebis
(Chapter 15)

2
Dynamic Programming
• An algorithm design technique (like divide and
conquer)
• Divide and conquer
– Partition the problem into independent subproblems
– Solve the subproblems recursively
– Combine the solutions to solve the original problem

3
Dynamic Programming
• Applicable when subproblems are not independent
– Subproblems share subsubproblems
E.g.: Combinations:
– A divide and conquer approach would repeatedly solve the
common subproblems
– Dynamic programming solves every subproblem just once and
stores the answer in a table
n
k
n-1
k
n-1
k-1
= +
n
1
n
n
=1 =1

4
Example: Combinations
+
=
=
=
=
=
=
+ +
+ + + +
+
+ + + + +
+
+
+
+
+ + +
+ + + + + + +
+ +
+
+
+
+
+
+
+
+
+
3
3
Comb (3, 1)
2
Comb (2, 1)
1
Comb (2, 2)
Comb (3, 2)
Comb (4,2)
2
Comb (2, 1)
1
Comb (2, 2)
Comb (3, 2)
1
1
Comb (3, 3)
Comb (4, 3)
Comb (5, 3)
2
Comb (2, 1)
1
Comb (2, 2)
Comb (3, 2)
1
1
Comb (3, 3)
Comb (4, 3)
1
1
1
Comb (4, 4)
Comb (5, 4)
Comb (6,4)
n
k
n-1
k
n-1
k-1
= +

5
Dynamic Programming
• Used for optimization problems
– A set of choices must be made to get an optimal
solution
– Find a solution with the optimal value (minimum or
maximum)
– There may be many solutions that lead to an optimal
value
– Our goal: find an optimal solution

6
Dynamic Programming Algorithm
1. Characterize the structure of an optimal
solution
2. Recursively define the value of an optimal
solution
3. Compute the value of an optimal solution in a
bottom-up fashion
4. Construct an optimal solution from computed
information (not always necessary)

7
Assembly Line Scheduling
• Automobile factory with two assembly lines
– Each line has n stations: S1,1, . . . , S1,n and S2,1, . . . , S2,n
– Corresponding stations S1, j and S2, j perform the same function
but can take different amounts of time a1, j and a2, j
– Entry times are: e1 and e2; exit times are: x1 and x2

8
• After going through a station, can either:
– stay on same line at no cost, or
– transfer to other line: cost after Si,j is ti,j , j = 1, . . . , n - 1

9
• Problem:
what stations should be chosen from line 1 and which
from line 2 in order to minimize the total time through the
factory for one car?

10
One Solution
• Brute force
– Enumerate all possibilities of selecting stations
– Compute how long it takes in each case and choose
the best one
• Solution:
– There are 2n possible ways to choose stations
– Infeasible when n is large!!
1 0 0 1 1
1 if choosing line 1
at step j (= n)
1 2 3 4 n
0 if choosing line 2
at step j (= 3)

11
1. Structure of the Optimal Solution
• How do we compute the minimum time of going through
a station?

12
• Let’s consider all possible ways to get from the
starting point through station S1,j
– We have two choices of how to get to S1, j:
• Through S1, j - 1, then directly to S1, j
• Through S2, j - 1, then transfer over to S1, j
a1,j
a1,j-1
a2,j-1
t2,j-1
S1,j
S1,j-1
S2,j-1
Line 1
Line 2

13
• Suppose that the fastest way through S1, j is
through S1, j – 1
– We must have taken a fastest way from entry through S1, j – 1
– If there were a faster way through S1, j - 1, we would use it instead
• Similarly for S2, j – 1
a1,j
a1,j-1
a2,j-1
t2,j-1
S1,j
S1,j-1
S2,j-1
Optimal Substructure
Line 1
Line 2

14
• Generalization: an optimal solution to the
problem “find the fastest way through S1, j” contains
within it an optimal solution to subproblems: “find
the fastest way through S1, j - 1 or S2, j – 1”.
• This is referred to as the optimal substructure
property
• We use this property to construct an optimal
solution to a problem from optimal solutions to
subproblems

15
2. A Recursive Solution
• Define the value of an optimal solution in terms of the optimal
solution to subproblems

16
2. A Recursive Solution (cont.)
• Definitions:
– f* : the fastest time to get through the entire factory
– fi[j] : the fastest time to get from the starting point through station Si,j
f* = min (f1[n] + x1, f2[n] + x2)

17
• Base case: j = 1, i=1,2 (getting through station 1)
f1[1] = e1 + a1,1
f2[1] = e2 + a2,1

18
• General Case: j = 2, 3, …,n, and i = 1, 2
• Fastest way through S1, j is either:
– the way through S1, j - 1 then directly through S1, j, or
f1[j - 1] + a1,j
– the way through S2, j - 1, transfer from line 2 to line 1, then through S1, j
f2[j -1] + t2,j-1 + a1,j
f1[j] = min(f1[j - 1] + a1,j ,f2[j -1] + t2,j-1 + a1,j)
a1,j
a1,j-1
a2,j-1
t2,j-1
S1,j
S1,j-1
S2,j-1
Line 1
Line 2

19
e1 + a1,1 if j = 1
f1[j] =
min(f1[j - 1] + a1,j ,f2[j -1] + t2,j-1 + a1,j) if j ≥ 2
e2 + a2,1 if j = 1
f2[j] =
min(f2[j - 1] + a2,j ,f1[j -1] + t1,j-1 + a2,j) if j ≥ 2

20
3. Computing the Optimal Solution
f* = min (f1[n] + x1, f2[n] + x2)
• Solving top-down would result in exponential
running time
f1[j]
f2[j]
1 2 3 4 5
f1(5)
f2(5)
f1(4)
f2(4)
f1(3)
f2(3)
2 times
4 times
f1(2)
f2(2)
f1(1)
f2(1)

21
3. Computing the Optimal Solution
• For j ≥ 2, each value fi[j] depends only on the
values of f1[j – 1] and f2[j - 1]
• Idea: compute the values of fi[j] as follows:
• Bottom-up approach
– First find optimal solutions to subproblems
– Find an optimal solution to the problem from the
subproblems
f1[j]
f2[j]
1 2 3 4 5
in increasing order of j

22
Example
e1 + a1,1, if j = 1
f1[j] = min(f1[j - 1] + a1,j ,f2[j -1] + t2,j-1 + a1,j) if j ≥ 2
f* = 35[1]
f1[j]
f2[j]
1 2 3 4 5
9
12 16[1]
18[1] 20[2]
22[2]
24[1]
25[1]
32[1]
30[2]

23
FASTEST-WAY(a, t, e, x, n)
1. f1[1] ← e1 + a1,1
2. f2[1] ← e2 + a2,1
3. for j ← 2 to n
4. do if f1[j - 1] + a1,j ≤ f2[j - 1] + t2, j-1 + a1, j
5. then f1[j] ← f1[j - 1] + a1, j
6. l1[j] ← 1
7. else f1[j] ← f2[j - 1] + t2, j-1 + a1, j
8. l1[j] ← 2
9. if f2[j - 1] + a2, j ≤ f1[j - 1] + t1, j-1 + a2, j
10. then f2[j] ← f2[j - 1] + a2, j
11. l2[j] ← 2
12. else f2[j] ← f1[j - 1] + t1, j-1 + a2, j
13. l2[j] ← 1
Compute initial values of f1 and f2
Compute the values of
f1[j] and l1[j]
f2[j] and l2[j]
O(N)

24
FASTEST-WAY(a, t, e, x, n) (cont.)
14. if f1[n] + x1 ≤ f2[n] + x2
15. then f* = f1[n] + x1
16. l* = 1
17. else f* = f2[n] + x2
18. l* = 2
the fastest time through the
entire factory

25
4. Construct an Optimal Solution
Alg.: PRINT-STATIONS(l, n)
i ← l*
print “line ” i “, station ” n
for j ← n downto 2
do i ←li[j]
print “line ” i “, station ” j - 1
f1[j]/l1[j]
f2[j]/l2[j]
1 2 3 4 5
9
12 16[1]
18[1] 20[2]
22[2]
24[1]
25[1]
32[1]
30[2]
l* = 1

26
Matrix-Chain Multiplication
Problem: given a sequence A1, A2, …, An,
compute the product:
A1  A2  An
• Matrix compatibility:
C = A  B C=A1  A2  Ai  Ai+1  An
colA = rowB coli = rowi+1
rowC = rowA rowC = rowA1
colC = colB colC = colAn

27
MATRIX-MULTIPLY(A, B)
if columns[A]  rows[B]
then error “incompatible dimensions”
else for i  1 to rows[A]
do for j  1 to columns[B]
do C[i, j] = 0
for k  1 to columns[A]
do C[i, j]  C[i, j] + A[i, k] B[k, j]
rows[A]
rows[A]
cols[B]
cols[B]
i
j
j
i
A B C
* =
k
k
rows[A]  cols[A]  cols[B]
multiplications

28
Matrix-Chain Multiplication
• In what order should we multiply the matrices?
A1  A2  An
• Parenthesize the product to get the order in which
matrices are multiplied
• E.g.: A1  A2  A3 = ((A1  A2)  A3)
= (A1  (A2  A3))
• Which one of these orderings should we choose?
– The order in which we multiply the matrices has a
significant impact on the cost of evaluating the product

29
Example
A1  A2  A3
• A1: 10 x 100
• A2: 100 x 5
• A3: 5 x 50
1. ((A1  A2)  A3): A1  A2 = 10 x 100 x 5 = 5,000 (10 x 5)
((A1  A2)  A3) = 10 x 5 x 50 = 2,500
Total: 7,500 scalar multiplications
2. (A1  (A2  A3)): A2  A3 = 100 x 5 x 50 = 25,000 (100 x 50)
(A1  (A2  A3)) = 10 x 100 x 50 = 50,000
Total: 75,000 scalar multiplications
one order of magnitude difference!!

30
Matrix-Chain Multiplication:
Problem Statement
• Given a chain of matrices A1, A2, …, An, where
Ai has dimensions pi-1x pi, fully parenthesize the
product A1  A2  An in a way that minimizes the
number of scalar multiplications.
A1  A2  Ai  Ai+1  An
p0 x p1 p1 x p2 pi-1 x pi pi x pi+1 pn-1 x pn

31
What is the number of possible
parenthesizations?
• Exhaustively checking all possible
parenthesizations is not efficient!
• It can be shown that the number of
parenthesizations grows as Ω(4n/n3/2)
(see page 333 in your textbook)

32
1. The Structure of an Optimal
Parenthesization
• Notation:
Ai…j = Ai Ai+1  Aj, i  j
• Suppose that an optimal parenthesization of Ai…j
splits the product between Ak and Ak+1, where
i  k < j
Ai…j = Ai Ai+1  Aj
= Ai Ai+1  Ak Ak+1  Aj
= Ai…k Ak+1…j

33
Ai…j = Ai…k Ak+1…j
• The parenthesization of the “prefix” Ai…k must be an
optimal parentesization
• If there were a less costly way to parenthesize Ai…k, we
could substitute that one in the parenthesization of Ai…j
and produce a parenthesization with a lower cost than
the optimum  contradiction!
• An optimal solution to an instance of the matrix-chain
multiplication contains within it optimal solutions to
subproblems

34
• Subproblem:
determine the minimum cost of parenthesizing
Ai…j = Ai Ai+1  Aj for 1  i  j  n
• Let m[i, j] = the minimum number of
multiplications needed to compute Ai…j
– full problem (A1..n): m[1, n]
– i = j: Ai…i = Ai  m[i, i] = 0, for i = 1, 2, …, n

35
• Consider the subproblem of parenthesizing
Ai…j = Ai Ai+1  Aj for 1  i  j  n
= Ai…k Ak+1…j for i  k < j
• Assume that the optimal parenthesization splits
the product Ai Ai+1  Aj at k (i  k < j)
m[i, j] =
min # of multiplications
to compute Ai…k
# of multiplications
to compute Ai…kAk…j
min # of multiplications
to compute Ak+1…j
m[i, k] m[k+1,j]
pi-1pkpj
m[i, k] + m[k+1, j] + pi-1pkpj

36
m[i, j] = m[i, k] + m[k+1, j] + pi-1pkpj
• We do not know the value of k
– There are j – i possible values for k: k = i, i+1, …, j-1
• Minimizing the cost of parenthesizing the product
Ai Ai+1  Aj becomes:
0 if i = j
m[i, j] = min {m[i, k] + m[k+1, j] + pi-1pkpj} if i < j
ik<j

37
3. Computing the Optimal Costs
0 if i = j
ik<j
• Computing the optimal solution recursively takes
exponential time!
• How many subproblems?
– Parenthesize Ai…j
for 1  i  j  n
– One problem for each
choice of i and j
 (n2)
1
1
2 3 n
2
3
n
j
i

38
3. Computing the Optimal Costs (cont.)
0 if i = j
ik<j
• How do we fill in the tables m[1..n, 1..n]?
– Determine which entries of the table are used in computing m[i, j]
Ai…j = Ai…k Ak+1…j
– Subproblems’ size is one less than the original size
– Idea: fill in m such that it corresponds to solving problems of
increasing length

39
3. Computing the Optimal Costs (cont.)
0 if i = j
ik<j
• Length = 1: i = j, i = 1, 2, …, n
• Length = 2: j = i + 1, i = 1, 2, …, n-1
1
1
2 3 n
2
3
n
Compute rows from bottom to top
and from left to right
m[1, n] gives the optimal
solution to the problem
i
j

40
Example: min {m[i, k] + m[k+1, j] + pi-1pkpj}
m[2, 2] + m[3, 5] + p1p2p5
m[2, 3] + m[4, 5] + p1p3p5
m[2, 4] + m[5, 5] + p1p4p5
1
1
2 3 6
2
3
6
i
j
4 5
4
5
m[2, 5] = min
• Values m[i, j] depend only
on values that have been
previously computed
k = 2
k = 3
k = 4

41
Example min {m[i, k] + m[k+1, j] + pi-1pkpj}
Compute A1  A2  A3
• A1: 10 x 100 (p0 x p1)
• A2: 100 x 5 (p1 x p2)
• A3: 5 x 50 (p2 x p3)
m[i, i] = 0 for i = 1, 2, 3
m[1, 2] = m[1, 1] + m[2, 2] + p0p1p2 (A1A2)
= 0 + 0 + 10 *100* 5 = 5,000
m[2, 3] = m[2, 2] + m[3, 3] + p1p2p3 (A2A3)
= 0 + 0 + 100 * 5 * 50 = 25,000
m[1, 3] = min m[1, 1] + m[2, 3] + p0p1p3 = 75,000 (A1(A2A3))
m[1, 2] + m[3, 3] + p0p2p3 = 7,500 ((A1A2)A3)
0
0
0
1
1
2
2
3
3
5000
1
25000
2
7500
2

42
Matrix-Chain-Order(p)
O(N3)

43
4. Construct the Optimal Solution
• In a similar matrix s we
keep the optimal
values of k
• s[i, j] = a value of k
such that an optimal
parenthesization of
Ai..j splits the product
between Ak and Ak+1
k
1
1
2 3 n
2
3
n
j

44
• s[1, n] is associated with
the entire product A1..n
– The final matrix
multiplication will be split
at k = s[1, n]
A1..n = A1..s[1, n]  As[1, n]+1..n
– For each subproduct
recursively find the
corresponding value of k
that results in an optimal
parenthesization
1
1
2 3 n
2
3
n
j

45
• s[i, j] = value of k such that the optimal
parenthesization of Ai Ai+1  Aj splits the
product between Ak and Ak+1
3 3 3 5 5 -
3 3 3 4 -
3 3 3 -
1 2 -
1 -
-
1
1
2 3 6
2
3
6
i
j
4 5
4
5
• s[1, n] = 3  A1..6 = A1..3 A4..6
• s[1, 3] = 1  A1..3 = A1..1 A2..3
• s[4, 6] = 5  A4..6 = A4..5 A6..6

46
4. Construct the Optimal Solution (cont.)
3 3 3 5 5 -
3 3 3 4 -
3 3 3 -
1 2 -
1 -
-
1
1
2 3 6
2
3
6
i
j
4 5
4
5
PRINT-OPT-PARENS(s, i, j)
if i = j
then print “A”i
else print “(”
PRINT-OPT-PARENS(s, i, s[i, j])
PRINT-OPT-PARENS(s, s[i, j] + 1, j)
print “)”

47
Example: A1  A6
3 3 3 5 5 -
3 3 3 4 -
3 3 3 -
1 2 -
1 -
-
1
1
2 3 6
2
3
6
i
j
4 5
4
5
PRINT-OPT-PARENS(s, i, j)
if i = j
then print “A”i
else print “(”
PRINT-OPT-PARENS(s, i, s[i, j])
PRINT-OPT-PARENS(s, s[i, j] + 1, j)
print “)”
P-O-P(s, 1, 6) s[1, 6] = 3
i = 1, j = 6 “(“ P-O-P (s, 1, 3) s[1, 3] = 1
i = 1, j = 3 “(“ P-O-P(s, 1, 1)  “A1”
P-O-P(s, 2, 3) s[2, 3] = 2
i = 2, j = 3 “(“ P-O-P (s, 2, 2)  “A2”
P-O-P (s, 3, 3)  “A3”
“)”
“)”
( ( ( A4 A5 ) A6 ) )
A1 ( A2 A3 ) )
…
(
s[1..6, 1..6]

48
Memoization
• Top-down approach with the efficiency of typical dynamic
programming approach
• Maintaining an entry in a table for the solution to each
subproblem
– memoize the inefficient recursive algorithm
• When a subproblem is first encountered its solution is
computed and stored in that table
• Subsequent “calls” to the subproblem simply look up that
value

49
Memoized Matrix-Chain
Alg.: MEMOIZED-MATRIX-CHAIN(p)
1. n  length[p] – 1
2. for i  1 to n
3. do for j  i to n
4. do m[i, j]  
5. return LOOKUP-CHAIN(p, 1, n)
Initialize the m table with
large values that indicate
whether the values of m[i, j]
have been computed
Top-down approach

50
Memoized Matrix-Chain
Alg.: LOOKUP-CHAIN(p, i, j)
1. if m[i, j] < 
2. then return m[i, j]
3. if i = j
4. then m[i, j]  0
5. else for k  i to j – 1
6. do q  LOOKUP-CHAIN(p, i, k) +
LOOKUP-CHAIN(p, k+1, j) + pi-1pkpj
7. if q < m[i, j]
8. then m[i, j]  q
9. return m[i, j]
Running time is O(n3)

51
Dynamic Progamming vs. Memoization
• Advantages of dynamic programming vs.
memoized algorithms
– No overhead for recursion, less overhead for
maintaining the table
– The regular pattern of table accesses may be used to
reduce time or space requirements
• Advantages of memoized algorithms vs.
dynamic programming
– Some subproblems do not need to be solved

53
Elements of Dynamic Programming
• Optimal Substructure
– An optimal solution to a problem contains within it an
optimal solution to subproblems
– Optimal solution to the entire problem is build in a
bottom-up manner from optimal solutions to
subproblems
• Overlapping Subproblems
– If a recursive algorithm revisits the same subproblems
over and over  the problem has overlapping
subproblems

54
Parameters of Optimal Substructure
• How many subproblems are used in an optimal
solution for the original problem
– Assembly line:
– Matrix multiplication:
• How many choices we have in determining
which subproblems to use in an optimal solution
– Assembly line:
– Matrix multiplication:
One subproblem (the line that gives best time)
Two choices (line 1 or line 2)
Two subproblems (subproducts Ai..k, Ak+1..j)
j - i choices for k (splitting the product)

55
Parameters of Optimal Substructure
• Intuitively, the running time of a dynamic
programming algorithm depends on two factors:
– Number of subproblems overall
– How many choices we look at for each subproblem
• Assembly line
– (n) subproblems (n stations)
– 2 choices for each subproblem
• Matrix multiplication:
– (n2) subproblems (1  i  j  n)
– At most n-1 choices
(n) overall
(n3) overall

56
Longest Common Subsequence
• Given two sequences
X = x1, x2, …, xm
Y = y1, y2, …, yn
find a maximum length common subsequence
(LCS) of X and Y
• E.g.:
X = A, B, C, B, D, A, B
• Subsequences of X:
– A subset of elements in the sequence taken in order
A, B, D, B, C, D, B, etc.

57
Example
X = A, B, C, B, D, A, B X = A, B, C, B, D, A, B
Y = B, D, C, A, B, A Y = B, D, C, A, B, A
• B, C, B, A and B, D, A, B are longest common
subsequences of X and Y (length = 4)
• B, C, A, however is not a LCS of X and Y

58
Brute-Force Solution
• For every subsequence of X, check whether it’s
a subsequence of Y
• There are 2m subsequences of X to check
• Each subsequence takes (n) time to check
– scan Y for first letter, from there scan for second, and
so on
• Running time: (n2m)

59
Making the choice
X = A, B, D, E
Y = Z, B, E
• Choice: include one element into the common
sequence (E) and solve the resulting
subproblem
X = A, B, D, G
Y = Z, B, D
• Choice: exclude an element from a string and
solve the resulting subproblem

60
Notations
• Given a sequence X = x1, x2, …, xm we define
the i-th prefix of X, for i = 0, 1, 2, …, m
Xi = x1, x2, …, xi
• c[i, j] = the length of a LCS of the sequences
Xi = x1, x2, …, xi and Yj = y1, y2, …, yj

61
A Recursive Solution
Case 1: xi = yj
e.g.: Xi = A, B, D, E
Yj = Z, B, E
– Append xi = yj to the LCS of Xi-1 and Yj-1
– Must find a LCS of Xi-1 and Yj-1  optimal solution to a
problem includes optimal solutions to subproblems
c[i, j] = c[i - 1, j - 1] + 1

62
A Recursive Solution
Case 2: xi  yj
e.g.: Xi = A, B, D, G
Yj = Z, B, D
– Must solve two problems
• find a LCS of Xi-1 and Yj: Xi-1 = A, B, D and Yj = Z, B, D
• find a LCS of Xi and Yj-1: Xi = A, B, D, G and Yj = Z, B
• Optimal solution to a problem includes optimal
solutions to subproblems
c[i, j] = max { c[i - 1, j], c[i, j-1] }

63
Overlapping Subproblems
• To find a LCS of X and Y
– we may need to find the LCS between X and Yn-1 and
that of Xm-1 and Y
– Both the above subproblems has the subproblem of
finding the LCS of Xm-1 and Yn-1
• Subproblems share subsubproblems

64
3. Computing the Length of the LCS
0 if i = 0 or j = 0
c[i, j] = c[i-1, j-1] + 1 if xi = yj
max(c[i, j-1], c[i-1, j]) if xi  yj
0 0 0 0 0 0
0
0
0
0
0
yj:
xm
y1 y2 yn
x1
x2
xi
j
i
0 1 2 n
m
1
2
0
first
second

65
Additional Information
0 if i,j = 0
c[i, j] = c[i-1, j-1] + 1 if xi = yj
0 0 0 0 0 0
0
0
0
0
0
yj:
D
A C F
A
B
xi
j
i
0 1 2 n
m
1
2
0
A matrix b[i, j]:
• For a subproblem [i, j] it
tells us what choice was
made to obtain the
optimal value
• If xi = yj
b[i, j] = “ ”
• Else, if
c[i - 1, j] ≥ c[i, j-1]
b[i, j] = “  ”
else
b[i, j] = “  ”
3
3 C
D
b & c:
c[i,j-1]
c[i-1,j]

66
LCS-LENGTH(X, Y, m, n)
1. for i ← 1 to m
2. do c[i, 0] ← 0
3. for j ← 0 to n
4. do c[0, j] ← 0
5. for i ← 1 to m
6. do for j ← 1 to n
7. do if xi = yj
8. then c[i, j] ← c[i - 1, j - 1] + 1
9. b[i, j ] ← “ ”
10. else if c[i - 1, j] ≥ c[i, j - 1]
11. then c[i, j] ← c[i - 1, j]
12. b[i, j] ← “↑”
13. else c[i, j] ← c[i, j - 1]
14. b[i, j] ← “←”
15.return c and b
The length of the LCS if one of the sequences
is empty is zero
Case 1: xi = yj
Case 2: xi  yj
Running time: (mn)

67
Example
X = A, B, C, B, D, A
Y = B, D, C, A, B, A
0 if i = 0 or j = 0
c[i, j] = c[i-1, j-1] + 1 if xi = yj
0 1 2 6
3 4 5
yj B D A
C A B
5
1
2
0
3
4
6
7
D
A
B
xi
C
B
A
B
0 0 0
0 0 0
0
0
0
0
0
0
0
0

0

0

0 1 1 1
1 1 1

1 2 2

1

1 2 2

2

2
1

1

2

2 3 3

1 2

2

2

3

3

1

2

3

2 3 4
1

2

2

3 4

4
If xi = yj
b[i, j] = “ ”
Else if
c[i - 1, j] ≥ c[i, j-1]
b[i, j] = “  ”
else
b[i, j] = “  ”

68
4. Constructing a LCS
• Start at b[m, n] and follow the arrows
• When we encounter a “ “ in b[i, j]  xi = yj is an element
of the LCS
0 1 2 6
3 4 5
yj B D A
C A B
5
1
2
0
3
4
6
7
D
A
B
xi
C
B
A
B
0 0 0
0 0 0
0
0
0
0
0
0
0
0

0

0

0 1 1 1
1 1 1

1 2 2

1

1 2 2

2

2
1

1

2

2 3 3

1 2

2

2

3

3

1

2

3

2 3 4
1

2

2

3 4

4

69
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)
Initial call: PRINT-LCS(b, X, length[X], length[Y])
Running time: (m + n)

70
Improving the Code
• What can we say about how each entry c[i, j] is
computed?
– It depends only on c[i -1, j - 1], c[i - 1, j], and
c[i, j - 1]
– Eliminate table b and compute in O(1) which of the
three values was used to compute c[i, j]
– We save (mn) space from table b
– However, we do not asymptotically decrease the
auxiliary space requirements: still need table c

71
Improving the Code
• If we only need the length of the LCS
– LCS-LENGTH works only on two rows of c at a time
• The row being computed and the previous row
– We can reduce the asymptotic space requirements by
storing only these two rows

DynamicProgramming.ppt

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