1. Dynamic Programming

2. Dynamic Programming Dividing a problem into subproblems
Dynamic programming vs divide and conquer
- Dynamic programming : subproblems are overlapped
- Divide and conquer : subproblems are independent
Used for finding optimized solutions

3. Assembly-Line Scheduling
station 1
station 2
station 3
station 4
station 5
station 6
n = 6
Assembly Line 1 7 9 3 4 8 4
enters
exits 8 5 6 4 5 7
Assembly Line 2

4. Assembly-Line Scheduling7 9 3 4 8 4 3 2
enters
exits 4 2 8 5 6 4 5 7
Assembly Line 2

5. Assembly-Line Scheduling7 9 3 4 8 4 3 2 2 3 1 3 4
enters
exits 2 1 2 2 1 4 2 8 5 6 4 5 7
Assembly Line 2

6. Assembly-Line Scheduling7 9 3 4 8 4 40 3 2 2 3 1 3 4
enters
exits 2 1 2 2 1 4 2 41 8 5 6 4 5 7
Assembly Line 2

7. Assembly-Line Scheduling7 9 3 4 8 4 39 3 2 2 3 1 3 4
enters
exits 2 1 2 2 1 4 2 8 5 6 4 5 7
Assembly Line 2

8. Assembly-Line Scheduling
station 1
station 2
station 3
station 4
station 5
station 6
number of cases
= 2n
Assembly Line 1 7 9 3 4 8 4
enters
exits 8 5 6 4 5 7
Assembly Line 2

9. Assembly-Line Scheduling
T1,1
T1,2
T1,3
T1,4
T1,5
T1,6
Assembly Line 1 T
enters
exits
Assembly Line 2
T2,1
T2,2
T2,3
T2,4
T2,5
T2,6

10. Assembly-Line Scheduling
T1,5
T1,6
T = min( T1,6 + 3, T2,6 + 2) 8 4
Divide and Conquer? 3 T 4
exits
T1,6 = min( T1,5 + 4,
T2, ) 1 2
T2,6 = min( T1, ,
T2,5 + 7) 5 7
T2,5
T2,6

11. Assembly-Line Scheduling
T1,5
T1,6
T = min( T1,6 + 3, T2,6 + 2) 8 4
Divide and Conquer? 3 T 4
exits
T1,6 = min( T1,5 + 4,
T2, ) 1 2
T2,6 = min( T1, ,
T2,5 + 7) 5 7
T2,5
T2,6

12. Assembly-Line Scheduling9 12
T1,1
T1,2
T1,3
Assembly Line 1
T1,1 = 9 7 9 3
T2,1 = 12 2 2 3
enters 2 1 4 8 5 6
Assembly Line 2
T2,1
T2,2
T2,3

13. Assembly-Line Scheduling9 18
T1,1
T1,2
T1,3
Assembly Line 1
T1,1 = 9 7 9 3
T2,1 = 12 2 2 3
T1,2 = min(T1,1 + 9,
T2, )
enters 2 1 4 8 5 6
Assembly Line 2
T2,1
T2,2
T2,3 12

14. Assembly-Line Scheduling9 18
T1,1
T1,2
T1,3
Assembly Line 1
T1,1 = 9 7 9 3
T2,1 = 12 2
T1,2 = 18 2 3
enters
T2,2 = min(T1, ,
T2,1 + 5) 2 1 4 8 5 6
Assembly Line 2
T2,1
T2,2
T2,3 12 16

15. Assembly-Line Scheduling9 18 20 22 24 25 32 30 35 37
T1,1
T1,2
T1,3
T1,4
T1,5
T1,6
Assembly Line 1 7 9 3 4 8 4 38 3 2 T 2 3 1 3 4
enters
exits 2 1 2 2 1 4 2 8 5 6 4 5 7
Assembly Line 2
T2,1
T2,2
T2,3
T2,4
T2,5
T2,6 12 16

16. Assembly-Line Scheduling32 30 35 37
T1,5
T1,6
T = T1,6 + 3 ? or T2,6 + 2 ? 8 4 38 3
T1,6 + 3 = T T 4
exits
T2,6 + 2 > T 1 2 5 7
T2,5
T2,6

17. Assembly-Line Scheduling32 30 35 37
T1,5
T1,6
T = T1,6 + 3 ? or T2,6 + 2 ? 8 4 38 3
T1,6 + 3 = T T 4
exits
T2,6 + 2 > T 1 2 5 7
T2,5
T2,6

18. Assembly-Line Scheduling32 30 35 37
T1,5
T1,6
T1,6 = T1,5 + 4 ?
or T2, ? 8 4 38 3 T 4
T1,5 + 4 > T1,6
exits
T2, = T1,6 1 2 5 7
T2,5
T2,6

19. Assembly-Line Scheduling32 30 35 37
T1,5
T1,6
T1,6 = T1,5 + 4 ?
or T2, ? 8 4 38 3 T 4
T1,5 + 4 > T1,6
exits
T2, = T1,6 1 2 5 7
T2,5
T2,6

20. Assembly-Line Scheduling9 18 20 22 24 25 32 30 35 37
T1,1
T1,2
T1,3
T1,4
T1,5
T1,6
Assembly Line 1 7 9 3 4 8 4 38 3 2 T 2 3 1 3 4
enters
exits 2 1 2 2 1 4 2 8 5 6 4 5 7
Assembly Line 2
T2,1
T2,2
T2,3
T2,4
T2,5
T2,6 12 16

21. Assembly-Line Scheduling
T1,5
T1,6
T2,5
T2,6
1. Analyze the problem T
exits
T = min( T1,6 + 3, T2,6 + 2)

22. Assembly-Line Scheduling
T1,5
T1,6
T2,5
T2,6
1. Analyze the problem
2. Find a recursive solution. T
exits
T = min( T1,6 + 3, T2,6 + 2)
T1,6 = min( T1,5 + 4,
T2, )
T2,6 = min( T1, ,
T2,5 + 7)

23. Assembly-Line Scheduling9 18 20 24 32 35
1. Analyze the problem.
T1,1
T1,2
T1,3
T1,4
T1,5
T1,6
2. Find a recursive solution. T
3. Compute the fastest time. 38
T2,1
T2,2
T2,3
T2,4
T2,5
T2,6 12 16 22 25 30 37

24. Assembly-Line Scheduling9 18 20 24 32 35
1. Analyze the problem.
T1,1
T1,2
T1,3
T1,4
T1,5
T1,6
2. Find a recursive solution. T
3. Compute the fastest time. 38
4. Construct the fastest way.
T2,1
T2,2
T2,3
T2,4
T2,5
T2,6 12 16 22 25 30 37

25. Assembly-Line Scheduling9 18 20 24 32 35
1. Analyze the problem.
T1,1
T1,2
T1,3
T1,4
T1,5
T1,6
2. Find a recursive solution. T
3. Compute the fastest time. 38
4. Construct the fastest way.
T2,1
T2,2
T2,3
T2,4
T2,5
T2,6
This algorithm takes time. 12 16 22 25 30 37
Brute force :

26. When can we use Dynamic Programming?
Optimal substructures
Overlapping subproblems

27. Overlapping Subprograms
T1,5
T1,6 T 8 4 3
T1,6
T2,6 T 4
exits
T1,5
T2,5
T1,5
T2,5 1 2 5 7
T1,4
T2,4
T1,4
T2,4
T1,4
T2,4
T1,4
T2,4
T2,5
T2,6

28. Overlapping Subprograms
T1,5
T1,6 T 8 4 3
T1,6
T2,6 T 4
exits
T1,5
T2,5
T1,5
T2,5 1 2 5 7
T1,4
T2,4
T1,4
T2,4
T1,4
T2,4
T1,4
T2,4
T2,5
T2,6

29. Optimal Substructures
Problems should be divided into subproblems.
Optimal solution : from optimal subproblem solutions
One part of the optimal solution to the problem
should be an optimal solution to the subproblem
solutions to the subproblems should be independent

30. Optimal Substructures
1. Shortest Path Problem
2. Longest Simple Path Problem d d s s

31. Optimal Substructures
1. Shortest Path Problem
2. Longest Simple Path Problem d d s s w w

32. Memoization Ordinary - Bottom-up strategy Memoization
- Top-down strategy

33. Memoization Memoization Bottom Up T1,1 T1,2 T1,3 T1,4 T1,5 T1,6 - T2,1

34. Memoization Memoization T T1,6 T2,6 T1,5 T2,5 T1,4 T2,4 … … T1,1 T1,2-
T2,1
T2,2
T2,3
T2,4
T2,5
T2,6
T1,4
T2,4 … …

35. Memoization Memoization T1,3 T1,2 T2,2 T1,1 T2,1 T1,1 T1,2 T1,3 T1,49 -
T2,1
T2,2
T2,3
T2,4
T2,5
T2,6 12
T1,1
T2,1

36. Memoization Memoization T1,3 T1,2 T2,2 T1,1 T2,1 T1,1 T1,2 T1,3 T1,49 18 -
T2,1
T2,2
T2,3
T2,4
T2,5
T2,6 12
T1,1
T2,1

37. Memoization Memoization T1,3 T1,2 T2,2 T1,1 T2,1 T1,1 T2,1 T1,1 T1,29 18 -
T2,1
T2,2
T2,3
T2,4
T2,5
T2,6 12 16
T1,1
T2,1
T1,1
T2,1

38. Memoization Memoization T1,4 T1,3 T2,3 T1,2 T2,2 T1,2 T2,2 T1,1 T2,19 18 20 -
T2,1
T2,2
T2,3
T2,4
T2,5
T2,6 12 16
T1,2
T2,2
T1,2
T2,2
T1,1
T2,1
T1,1
T2,1

39. Memoization Memoization T T1,6 T2,6 T1,5 T2,5 T2,4 T1,4 T1,4 T2,4 … …9 18 20 24 -
T2,1
T2,2
T2,3
T2,4
T2,5
T2,6 12 16 22
T1,4
T2,4 … …
T1,3
T2,3

40. Memoization Memoization T T1,6 T2,6 T1,5 T2,5 T2,4 T1,4 T1,4 T2,4 … …9 18 20 24 32 -
T2,1
T2,2
T2,3
T2,4
T2,5
T2,6 12 16 22 25
T1,4
T2,4 … … … …

41. Memoization Memoization T T1,6 T2,6 T2,5 T1,5 T1,5 T2,5 T2,4 T1,4 T1,49 18 20 24 32 35
T2,1
T2,2
T2,3
T2,4
T2,5
T2,6 12 16 22 25 30 -
T1,4
T2,4 … … … …

42. Memoization Memoization T = 38 T1,6 T2,6 T2,5 T1,5 T1,5 T2,5 T2,4 T1,49 18 20 24 32 35
T2,1
T2,2
T2,3
T2,4
T2,5
T2,6 12 16 22 25 30 37
T1,4
T2,4 … … … …

43. Why Memoization? Ordinary Dynamic Programming (Bottom-Up strategy)
- when all subproblems need to be solved
Memoization
- when some subproblems do not need to be solved

44. Longest Common Subsequence
DNA sequences : composed of four components
– {A, C, G, T}
How similar are they?
S1 = ACCGGTCGTGCGCGGAAGCCGGCCGAA
S2 = GTCGTTCGGAATGCCGTTGCTCTGTAAA

45. Longest Common Subsequence
S1 = ACCGGTCGTGCGCGGAAGCCGGCCGAA
S2 = GTCGTTCGGAATGCCGTTGCTCTGTAAA
Longest Common Sequence :
GTCGTCGGAAGCCGGCCGAA

46. Longest Common Subsequence
X = <x1, x2, …, xm>, Y = <y1, y2, …, yn>
Xi = <x1, x2, …, xi> for i = 1 to m
Yj = <y1, y2, …, yj> for j = 1 to n
Zi,j = <z1, z2, …, zk> for i = 1 to m, j = 1 to n
Longest common subsequence of Xi and Yj

47. Longest Common Subsequence
1. Analyze the problem.
X = <x1, x2, …, xm>, Y = <y1, y2, …, yn>
Xi = <x1, x2, …, xi>
Yj = <y1, y2, …, yj>
Zi,j = <z1, z2, …, zk>
(Longest common subsequence)
Zi-1,j = <z1, z2, …, zk-1> or
<z1, z2, …, zk>
when xi used as zk
otherwise
Zi,j-1 = <z1, z2, …, zk> or <z1, z2, …, zk-1>

48. Longest Common Subsequence
1. Analyze the problem.
X = <x1, x2, …, xm>, Y = <y1, y2, …, yn>
Xi = <x1, x2, …, xi>
Yj = <y1, y2, …, yj>
Zi,j = <z1, z2, …, zk>
(Longest common subsequence)
If xi = yj then zk = xi or zk = yj / / /
Zi-1,j = <z1, z2, …, zk-1> or
<z1, z2, …, zk>
when xi used as zk
otherwise

49. Longest Common Subsequence
1. Analyze the problem.
X = <x1, x2, …, xm>, Y = <y1, y2, …, yn>
Xi = <x1, x2, …, xi>
Yj = <y1, y2, …, yj>
Zi,j = <z1, z2, …, zk>
(Longest common subsequence)
If xi = yj then zk = xi or zk = yj / / /
Zi,j = Zi-1,j or Zi,j-1

50. Longest Common Subsequence
1. Analyze the problem.
X = <x1, x2, …, xm>, Y = <y1, y2, …, yn>
Xi = <x1, x2, …, xi>
Yj = <y1, y2, …, yj>
Zi,j = <z1, z2, …, zk>
(Longest common subsequence)
If xi = yj then /
Zi,j = Zi-1,j or Zi,j-1
Optimal
If xi = yj then
If xi = yj is not used as zk then
Zi,j = Zi-1,j or Zi,j-1
If xi = yj is used as zk then
Zi,j = Zi-1,j-1 + zk

51. Longest Common Subsequence
1. Analyze the problem.
X = <x1, x2, …, xm>, Y = <y1, y2, …, yn>
Xi = <x1, x2, …, xi>
Yj = <y1, y2, …, yj>
Zi,j = <z1, z2, …, zk>
(Longest common subsequence)
If xi = yj then /
Zi,j = Zi-1,j or Zi,j-1
ci,j = max(ci-1,j, ci,j-1)
ci,j = ci-1,j-1 + 1
If xi = yj then
Zi,j = Zi-1,j-1 + zk

52. Longest Common Subsequence
2. Find a recursive solution.
If i = 0 or j = 0 then
ci,j = 0
Else
If xi = yj then /
ci,j = max(ci-1,j, ci,j-1)
If xi = yj then
ci,j = ci-1,j-1 + 1

53. Longest Common Subsequence
3. Compute the fastest time.
C1,1
C1,2
C1,n
C2,1
C2,2
C2,n .
Cm,1
Cm,2
Cm,n
If xi = yj then /
ci,j = max(ci-1,j, ci,j-1)
If xi = yj then
ci,j = ci-1,j-1 + 1

54. Longest Common Subsequence
4. Construct the common
subsequence.
C1,1
C1,2
C1,n
C2,1
C2,2
C2,n .
Cm,1
Cm,2
Cm,n

55. Longest Common Subsequence
4. Construct the common
subsequence.
Ci-2,j-2
Ci-2,j-1
Ci-2,j
Ci-1,j-2
Ci-1,j-1
Ci-1,j
Ci,j-2
Ci,j-1
Ci,j
If xi = yj then
one element of
The longest common subsequence
move to ci-1,j-1
Else
If ci,j = ci-1,j then
move to ci-1,j
If ci,j = ci,j-1 then
If xi = yj then /
ci,j = max(ci-1,j, ci,j-1)
move to ci,j-1
If xi = yj then
ci,j = ci-1,j-1 + 1

56. Dynamic Programming 1. Analyze the problem.
Dividing a problem into subproblems.
For optimization problems.
Optimal substructures / overlapping subproblems
Process of dynamic programmings
1. Analyze the problem.
2. Find a recursive solution.
3. Compute the fastest time/cost.
4. Construct the fastest path.