1. Lecture 7 Topics Dynamic Programming
Reference: Introduction to Algorithm by Cormen
Chapter 15: Dynamic Programming
Data Structure and Algorithm

2. Longest Common Subsequence (LCS)
Biologists need to measure similarity between DNA and thus determine how closely related an organism is to another.
They do this by considering DNA as strings of letters A,C,G,T and then comparing similarities in the strings.
Formally, they look at common subsequences in the strings.
Example X = ABCBDAB, Y=BDCABA
Subsequences may be: ABA, BCA, BCB, BBA
BCBA
BDAB etc.
But the Longest Common Subsequences (LCS) are BCBA and BDAB.
How to find LCS efficiently?
Data Structure and Algorithm

3. Longest Common Subsequence ( Brute Force Approach )
if |X| = m, |Y| = n, then there are 2m subsequences of X; we must compare each with Y (n comparisons)
So the running time of the brute-force algorithm is O(n 2m)
Making it impractical for long sequences.
LCS problem has optimal substructure:
solutions of subproblems are parts of the final solution.
Subproblems: “find LCS of pairs of prefixes of X and Y”
Data Structure and Algorithm

4. LCS: Optimal Substructure
Data Structure and Algorithm

5. LCS: Setup for Dynamic Programming
First we’ll find the length of LCS, along the way we will leave “clues” on finding subsequence.
Define Xi, Yj to be the prefixes of X and Y of length i and j respectively.
Define c[i,j] to be the length of LCS of Xi and Yj
Then the length of LCS of X and Y will be c[m,n]
Data Structure and Algorithm

6. LCS recurrence Notice the issues…This recurrence is exponential.
We don’t know max ahead of time.
The subproblems overlap, to find LCS we need to find LCS of c[i, j-1] and of c[i-1, j]
Data Structure and Algorithm

7. LCS Algorithm
First we’ll find the length of LCS. Later we’ll modify the algorithm to find LCS itself.
Recall we want to let Xi, Yj to be the prefixes of X and Y of length i and j respectively
And that Define c[i,j] to be the length of LCS of Xi and Yj
Then the length of LCS of X and Y will be c[m,n]
Data Structure and Algorithm

8. LCS Recursive Solution
We start with i = j = 0 (empty substrings of x and y)
Since X0 and Y0 are empty strings, their LCS is always empty (i.e. c[0,0] = 0)
LCS of empty string and any other string is empty, so for every i and j: c[0, j] = c[i,0] = 0
Data Structure and Algorithm

9. LCS Recursive Solution
When we calculate c[i,j], we consider two cases:
First case: x[i]=y[j]: one more symbol in strings X and Y matches, so the length of LCS Xi and Yj equals to the length of LCS of smaller strings Xi-1 and Yi-1 , plus 1
Data Structure and Algorithm

10. LCS Recursive Solution
Second case: x[i] != y[j]
As symbols don’t match, our solution is not improved, and the length of LCS(Xi , Yj) is the same as before (i.e. maximum of LCS(Xi, Yj-1) and LCS(Xi-1,Yj)
Data Structure and Algorithm

11. LCS Example We’ll see how LCS algorithm works on the following
X = ABCB
Y = BDCAB
What is the Longest Common Subsequence
of X and Y?
LCS(X, Y) = BCB
X = A B C B
Y = B D C A B
Data Structure and Algorithm

12. LCS Example (0) X = ABCB; m = |X| = 4 Y = BDCAB; n = |Y| = 5j i Yj B D C A B Xi A 1 B 2 3 C 4 B
X = ABCB; m = |X| = 4
Y = BDCAB; n = |Y| = 5
Allocate array c[6,5]
Data Structure and Algorithm

13. LCS Example (1) for i = 1 to m c[i,0] = 0 j 0 1 2 3 4 5 i Yj B D C A BXi A 1 B 2 3 C 4 B
for i = 1 to m c[i,0] = 0
Data Structure and Algorithm

14. LCS Example (2) for j = 0 to n c[0,j] = 0 j 0 1 2 3 4 5 i Yj B D C A BXi A 1 B 2 3 C 4 B
for j = 0 to n c[0,j] = 0
Data Structure and Algorithm

15. LCS Example (3) j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 B 2 3 C 4 BA 1 B 2 3 C 4 B
case i=1 and j=1
A != B
but, c[0,1]>=c[1,0]
so c[1,1] = c[0,1], and b[1,1] =
Data Structure and Algorithm

16. LCS Example (4) j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 B 2 3 C 4 BA 1 B 2 3 C 4 B
case i=1 and j=2
A != D
but, c[0,2]>=c[1,1]
so c[1,2] = c[0,2], and b[1,2] =
Data Structure and Algorithm

17. LCS Example (5) j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 B 2 3 C 4 BA 1 B 2 3 C 4 B
case i=1 and j=3
A != C
but, c[0,3]>=c[1,2]
so c[1,3] = c[0,3], and b[1,3] =
Data Structure and Algorithm

18. LCS Example (6) j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 B 2 3 C 4 BA 1 1 B 2 3 C 4 B
case i=1 and j=4
A = A
so c[1,4] = c[0,3]+1, and b[1,4] =
Data Structure and Algorithm

19. LCS Example (7) j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2 3 C 4 BA 1 1 1 B 2 3 C 4 B
case i=1 and j=5
A != B
this time c[0,5]<c[1,4]
so c[1,5] = c[1, 4], and b[1,5] =
Data Structure and Algorithm

20. LCS Example (8) j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2 1 3 C 4 BA 1 1 1 B 2 1 3 C 4 B
case i=2 and j=1
B = B
so c[2, 1] = c[1, 0]+1, and b[2, 1] =
Data Structure and Algorithm

21. LCS Example (9) j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2 1 1 3 C 4A 1 1 1 B 2 1 1 3 C 4 B
case i=2 and j=2
B != D
and c[1, 2] < c[2, 1]
so c[2, 2] = c[2, 1] and b[2, 2] =
Data Structure and Algorithm

22. LCS Example (10) j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2 1 1 1 3 CA 1 1 1 B 2 1 1 1 3 C 4 B
case i=2 and j=3
B != D
and c[1, 3] < c[2, 2]
so c[2, 3] = c[2, 2] and b[2, 3] =
Data Structure and Algorithm

23. LCS Example (11) j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2 1 1 1 1 3A 1 1 1 B 2 1 1 1 1 3 C 4 B
case i=2 and j=4
B != A
and c[1, 4] = c[2, 3]
so c[2, 4] = c[1, 4] and b[2, 2] =
Data Structure and Algorithm

24. LCS Example (12) j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2 1 1 1 1 2A 1 1 1 B 2 1 1 1 1 2 3 C 4 B
case i=2 and j=5
B = B
so c[2, 5] = c[1, 4]+1 and b[2, 5] =
Data Structure and Algorithm

25. LCS Example (13) j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2 1 1 1 1 2A 1 1 1 B 2 1 1 1 1 2 3 C 1 4 B
case i=3 and j=1
C != B
and c[2, 1] > c[3,0]
so c[3, 1] = c[2, 1] and b[3, 1] =
Data Structure and Algorithm

26. LCS Example (14) j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2 1 1 1 1 2A 1 1 1 B 2 1 1 1 1 2 3 C 1 1 4 B
case i=3 and j= 2
C != D
and c[2, 2] = c[3, 1]
so c[3, 2] = c[2, 2] and b[3, 2] =
Data Structure and Algorithm

27. LCS Example (15) j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2 1 1 1 1 2A 1 1 1 B 2 1 1 1 1 2 3 C 1 1 2 4 B
case i=3 and j= 3
C = C
so c[3, 3] = c[2, 2]+1 and b[3, 3] =
Data Structure and Algorithm

28. LCS Example (16) j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2 1 1 1 1 2A 1 1 1 B 2 1 1 1 1 2 3 C 1 1 2 2 4 B
case i=3 and j= 4
C != A
c[2, 4] < c[3, 3]
so c[3, 4] = c[3, 3] and b[3, 3] =
Data Structure and Algorithm

29. LCS Example (17) j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2 1 1 1 1 2A 1 1 1 B 2 1 1 1 1 2 3 C 1 1 2 2 2 4 B
case i=3 and j= 5
C != B
c[2, 5] = c[3, 4]
so c[3, 5] = c[2, 5] and b[3, 5] =
Data Structure and Algorithm

30. LCS Example (18) j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2 1 1 1 1 2A 1 1 1 B 2 1 1 1 1 2 3 C 1 1 2 2 2 4 B 1
case i=4 and j=1
B = B
so c[4, 1] = c[3, 0]+1 and b[4, 1] =
Data Structure and Algorithm

31. LCS Example (19) j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2 1 1 1 1 2A 1 1 1 B 2 1 1 1 1 2 3 C 1 1 2 2 2 4 B 1 1
case i=4 and j=2
B != D
c[3, 2] = c[4, 1]
so c[4, 2] = c[3, 2] and b[4, 2] =
Data Structure and Algorithm

32. LCS Example (20) j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2 1 1 1 1 2A 1 1 1 B 2 1 1 1 1 2 3 C 1 1 2 2 2 4 B 1 1 2
case i=4 and j= 3
B != C
c[3, 3] > c[4, 2]
so c[4, 3] = c[3, 3] and b[4, 3] =
Data Structure and Algorithm

33. LCS Example (21) j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2 1 1 1 1 2A 1 1 1 B 2 1 1 1 1 2 3 C 1 1 2 2 2 4 B 1 1 2 2
case i=4 and j=4
B != A
c[3, 4] = c[4, 3]
so c[4, 4] = c[3, 4] and b[3, 5] =
Data Structure and Algorithm

34. LCS Example (22) 3 j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2 1 1 1 1A 1 1 1 B 2 1 1 1 1 2 3 C 1 1 2 2 2 4 B 3 1 1 2 2
case i=4 and j=5
B= B
so c[4, 5] = c[3, 4]+1 and b[4, 5] =
Data Structure and Algorithm

35. LCS Algorithm LCS-Length(X, Y) m = length(X), n = length(Y)
for i = 1 to m do
c[i, 0] = 0
for j = 0 to n do
c[0, j] = 0
for j = 1 to n do
if ( xi = = yj ) then
c[i, j] = c[i - 1, j - 1] + 1
else if c[i - 1, j]>=c[i, j - 1] then
c[i, j] = c[i - 1, j]
else c[i, j] = c[i, j - 1]
return c and b
Data Structure and Algorithm

36. LCS Algorithm Running Time
LCS algorithm calculates the values of each entry of the array c[m,n]
So the running time is clearly O(mn) as each entry is done in 3 steps.
Now how to get at the solution?
We use the arrows we created to guide us.
We simply follow arrows back to base case 0
Data Structure and Algorithm

37. 3 Finding LCS j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2 1 1 1 1 2 3A 1 1 1 B 2 1 1 1 1 2 3 C 1 1 2 2 2 3 4 B 1 1 2 2
Data Structure and Algorithm

38. 3 Finding LCS (2) LCS: B C B j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 BA 1 1 1 B 2 1 1 1 1 2 3 C 1 1 2 2 2 3 4 B 1 1 2 2
LCS: B C B
Data Structure and Algorithm

39. Finding LCS (3) Print_LCS (X, i, j) if i = 0 or j = 0 then return
if b[i, j] = “ “ then
Print_LCS (X, i-1, j-1)
Print X[i]
elseif b[i, j] = “ “ then
Print_LCS (X, i-1, j)
else
Print_LCS (X, i, j-1)
Cost: O(m+n)
Data Structure and Algorithm

40. Element of Dynamic Programming
Optimal Substructure
Overlapping Subproblems
Data Structure and Algorithm

41. Optimal Substructure
A problem exhibits optimal substructure if an optimal solution contains optimal solutions to its subproblems.
Build an optimal solution from optimal solutions to subproblems
solutions of subproblems are parts of the final solution.
Example :Longest Common Subsequence -
An LCS contains within it optimal solutions to the prefixes of the two input sequences.
Common with Greedy Solution
Data Structure and Algorithm

42. Overlapping Subproblems
Divide-and-Conquer is suitable when generating brand-new problems at each step of the recursion.
Dynamic-programming algorithms take advantage of overlapping subproblems by solving each subproblem once and then storing the solution in a table where it can be looked up when needed, using constant time per lookup
Data Structure and Algorithm

43. Dynamic VS Greedy
Dynamic programming uses optimal substructure in a bottom-up fashion
First find optimal solutions to subproblems and, having solved the subproblems, we find an optimal solution to the problem
Greedy algorithms use optimal substructure in a top-down fashion
First make a choice – the choice that looks best at the time – and then solving the resulting subproblem
Data Structure and Algorithm