Dynamic programming

Dynamic Programming
Dr. AMIT KUMAR @JUET

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

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

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
= +

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

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)

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

• 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

• 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?

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)

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

• 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,ja1,j-1
a2,j-1
t2,j-1
S1,jS1,j-1
S2,j-1
Line 1
Line 2

• 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,ja1,j-1
a2,j-1
t2,j-1
S1,jS1,j-1
S2,j-1
Optimal Substructure
Line 1
Line 2

• 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

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

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)

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

• 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,ja1,j-1
a2,j-1
t2,j-1
S1,jS1,j-1
S2,j-1
Line 1
Line 2

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

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 times4 times
f1(2)
f2(2)
f1(1)
f2(1)

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

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]

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)

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

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

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

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

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

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!!

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

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)

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

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

• 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

• 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

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

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

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

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

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

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

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

• 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

• 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

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 “)”

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]

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

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

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)

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

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

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)

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

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.

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

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)

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

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

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

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] }

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

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

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]

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)

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 63 4 5
yj B D AC A B
5
1
2
0
3
4
6
7
D
A
B
xi
C
B
A
B
0 0 00 0 00
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] = “  ”

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 63 4 5
yj B D AC A B
5
1
2
0
3
4
6
7
D
A
B
xi
C
B
A
B
0 0 00 0 00
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

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)

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

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

Dynamic programming

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