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Introduction to Algorithms
6.046J/18.401J
LECTURE 15
Dynamic Programming
• Longest common
subsequence
• Optimal substructure
• Overlapping subproblems

Prof. Charles E. Leiserson
November 7, 2005 Copyright © 2001-5 by Erik D. Demaine and Charles E. Leiserson L15.1

Dynamic programming
Design technique, like divide-and-conquer.
Example: Longest Common Subsequence (LCS)
• Given two sequences x[1 . . m] and y[1 . . n], find
a longest subsequence common to them both.


Dynamic programming
“a” not “the”


Dynamic programming
x: A B C B D A B

y: B D C A B A


Dynamic programming
x: A B C B D A B
BCBA =
LCS(x, y)
y: B D C A B A
functional notation,
but not a function

Brute-force LCS algorithm
Check every subsequence of x[1 . . m] to see
if it is also a subsequence of y[1 . . n].


Brute-force LCS algorithm
Check every subsequence of x[1 . . m] to see
if it is also a subsequence of y[1 . . n].
Analysis
• Checking = O(n) time per subsequence.
• 2m subsequences of x (each bit-vector of
length m determines a distinct subsequence
of x).
Worst-case running time = O(n2m)
= exponential time.

Towards a better algorithm
Simplification:
1. Look at the length of a longest-common
subsequence.
2. Extend the algorithm to find the LCS itself.


Simplification:
subsequence.
Notation: Denote the length of a sequence s
by | s |.


Recursive formulation
Theorem.
c[i–1, j–1] + 1 if x[i] = y[j],
c[i, j] = max{c[i–1, j], c[i, j–1]} otherwise.


Theorem.
c[i–1, j–1] + 1 if x[i] = y[j],
Proof. Case x[i] = y[ j]:
1 2 i m
x: L
1 2 = j n
y: L


Theorem.
c[i–1, j–1] + 1 if x[i] = y[j],
Proof. Case x[i] = y[ j]:
1 2 i m
x: L
1 2 = j n
y: L
Let z[1 . . k] = LCS(x[1 . . i], y[1 . . j]), where c[i, j]
= k. Then, z[k] = x[i], or else z could be extended.
Thus, z[1 . . k–1] is CS of x[1 . . i–1] and y[1 . . j–1].

Proof (continued)
Claim: z[1 . . k–1] = LCS(x[1 . . i–1], y[1 . . j–1]).
Suppose w is a longer CS of x[1 . . i–1] and
y[1 . . j–1], that is, | w | > k–1. Then, cut and
paste: w || z[k] (w concatenated with z[k]) is a
common subsequence of x[1 . . i] and y[1 . . j]
with | w || z[k] | > k. Contradiction, proving the
claim.


Proof (continued)
Claim: z[1 . . k–1] = LCS(x[1 . . i–1], y[1 . . j–1]).
Suppose w is a longer CS of x[1 . . i–1] and
y[1 . . j–1], that is, | w | > k–1. Then, cut and
paste: w || z[k] (w concatenated with z[k]) is a
common subsequence of x[1 . . i] and y[1 . . j]
with | w || z[k] | > k. Contradiction, proving the
claim.
Thus, c[i–1, j–1] = k–1, which implies that c[i, j]
= c[i–1, j–1] + 1.
Other cases are similar.

Dynamic-programming
hallmark #1

Optimal substructure
An optimal solution to a problem
(instance) contains optimal
solutions to subproblems.


Dynamic-programming
hallmark #1

Optimal substructure
An optimal solution to a problem
(instance) contains optimal
solutions to subproblems.

If z = LCS(x, y), then any prefix of z is
an LCS of a prefix of x and a prefix of y.


Recursive algorithm for LCS
LCS(x, y, i, j)
if x[i] = y[ j]
then c[i, j] ← LCS(x, y, i–1, j–1) + 1
else c[i, j] ← max{ LCS(x, y, i–1, j),
LCS(x, y, i, j–1)}


Recursive algorithm for LCS
LCS(x, y, i, j)
if x[i] = y[ j]
then c[i, j] ← LCS(x, y, i–1, j–1) + 1
else c[i, j] ← max{ LCS(x, y, i–1, j),
LCS(x, y, i, j–1)}
Worst-case: x[i] ≠ y[ j], in which case the
algorithm evaluates two subproblems, each
with only one parameter decremented.


Recursion tree
m = 3, n = 4: 3,4
3,4

2,4
2,4 3,3
3,3

1,4
1,4 2,3
2,3 2,3
2,3 3,2
3,2

1,3
1,3 2,2
2,2 1,3
1,3 2,2
2,2


Recursion tree
m = 3, n = 4: 3,4
3,4

2,4
2,4 3,3
3,3

1,4
1,4 2,3
2,3 2,3
2,3 3,2
3,2 m+n

1,3
1,3 2,2
2,2 1,3
1,3 2,2
2,2

Height = m + n ⇒ work potentially exponential.


Recursion tree
m = 3, n = 4: 3,4
3,4

2,4
2,4 same 3,3
3,3
subproblem
1,4
1,4 2,3
2,3 2,3
2,3 3,2
3,2 m+n

1,3
1,3 2,2
2,2 1,3
1,3 2,2
2,2

Height = m + n ⇒ work potentially exponential.,
but we’re solving subproblems already solved!

Dynamic-programming
hallmark #2

Overlapping subproblems
A recursive solution contains a
“small” number of distinct
subproblems repeated many times.


Dynamic-programming
hallmark #2

Overlapping subproblems
A recursive solution contains a
“small” number of distinct
subproblems repeated many times.

The number of distinct LCS subproblems for
two strings of lengths m and n is only mn.


Memoization algorithm
Memoization: After computing a solution to a
subproblem, store it in a table. Subsequent calls
check the table to avoid redoing work.


LCS(x, y, i, j)
if c[i, j] = NIL
then if x[i] = y[j]
then c[i, j] ← LCS(x, y, i–1, j–1) + 1 same
else c[i, j] ← max{ LCS(x, y, i–1, j), as
before
LCS(x, y, i, j–1)}


LCS(x, y, i, j)
if c[i, j] = NIL
then if x[i] = y[j]
then c[i, j] ← LCS(x, y, i–1, j–1) + 1 same
else c[i, j] ← max{ LCS(x, y, i–1, j), as
before
LCS(x, y, i, j–1)}
Time = Θ(mn) = constant work per table entry.
Space = Θ(mn).

Dynamic-programming
algorithm
IDEA: A B C B D A B
Compute the 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0
table bottom-up.
B 0 0 1 1 1 1
0 0 1 1 1 1 1 1
1 1
D 0 0 1 1 1 2
0 0 1 1 1 2 2 2
2 2
C 0 0 1
0 0 1 2 2 2 2 2
2 2 2 2 2
A 0 1 1
0 1 1 2 2 2 3 3
2 2 2 3 3
B 0 1 2
0 1 2 2 3 3 3 4
2 3 3 3 4
A 0 1 2
0 1 2 2 3 3 4 4
2 3 3 4 4

Dynamic-programming
algorithm
IDEA: A B C B D A B
0 0 0 0 0 0 0 0
0 0
table bottom-up.
B 0 0 1 1 1 1
0 0 1 1 1 1 1 1
1 1
Time = Θ(mn). D 0 0 1 1 1 2 2 2
0 0 1 1 1 2 2 2
C 0 0 1
0 0 1 2 2 2 2 2
2 2 2 2 2
A 0 1 1
0 1 1 2 2 2 3 3
2 2 2 3 3
B 0 1 2
0 1 2 2 3 3 3 4
2 3 3 3 4
A 0 1 2
0 1 2 2 3 3 4 4
2 3 3 4 4

Dynamic-programming
algorithm
IDEA: A B C B D A B
0 0 0 0 0 0 0 0
0 0
table bottom-up.
B 0 0 1 1 1 1
0 0 1 1 1 1 1 1
1 1
Time = Θ(mn). D 0 0 1 1 1 2 2 2
0 0 1 1 1 2 2 2
Reconstruct
C 0 0 1
0 0 1 2 2 2 2 2
2 2 2 2 2
LCS by tracing
backwards. A 0 1 1
0 1 1 2 2 2 3 3
2 2 2 3 3
B 0 1 2
0 1 2 2 3 3 3 4
2 3 3 3 4
A 0 1 2
0 1 2 2 3 3 4 4
2 3 3 4 4

Dynamic-programming
algorithm
IDEA: A B C B D A B
0 0 0 0 0 0 0 0
0 0
table bottom-up.
B 0 0 1 1 1 1
0 0 1 1 1 1 1 1
1 1
Time = Θ(mn). D 0 0 1 1 1 2 2 2
0 0 1 1 1 2 2 2
Reconstruct
C 0 0 1
0 0 1 2 2 2 2 2
2 2 2 2 2
LCS by tracing
backwards. A 0 1 1
0 1 1 2 2 2 3 3
2 2 2 3 3
Space = Θ(mn). B 0 1 2
0 1 2 2 3 3 3 4
2 3 3 3 4
Exercise: A 0 1 2
0 1 2 2 3 3 4 4
2 3 3 4 4
O(min{m, n}).

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