1. Dynamic Programming
Dr. Yingwu Zhu
Chapter 15

2. Dynamic Programming (DP)
It is used, when the solution can be recursively described in terms of solutions to subproblems (optimal substructure)
Algorithm finds solutions to subproblems and stores them in memory for later use
More efficient than “brute-force methods”, which solve the same subproblems over and over again

3. DP
An algorithm design technique for optimization problems: often minimizing or maximizing.
Like divide-and-conquer, DP solves problems by combining solutions to subproblems.
Unlike divide-and-conquer, subproblems are not independent.
Subproblems may share subsubproblems,
However, solution to one subproblem may not affect the solutions to other subproblems of the same problem. (More on this later.)
DP reduces computation by
Solving subproblems in a bottom-up fashion.
Storing solution to a subproblem the first time it is solved.
Looking up the solution when subproblem is encountered again.
Key: determine structure of optimal solutions

4. Steps in DP Characterize structure of an optimal solution.
Define value of optimal solution recursively.
Compute optimal solution values either top-down with caching or bottom-up in a table.
Construct an optimal solution from computed values.
We’ll study these with the help of examples.

5. Computing a binomial coefficient by DP
Binomial coefficients are coefficients of the binomial formula:
(a + b)n = C(n,0)anb C(n,k)an-kbk C(n,n)a0bn
Recurrence: C(n,k) = C(n-1,k) + C(n-1,k-1) for n > k > 0
C(n,0) = 1, C(n,n) = 1 for n  0

6. Computing a binomial coefficient by DP
Binomial coefficients are coefficients of the binomial formula:
(a + b)n = C(n,0)anb C(n,k)an-kbk C(n,n)a0bn
Recurrence: C(n,k) = C(n-1,k) + C(n-1,k-1) for n > k > 0
C(n,0) = 1, C(n,n) = 1 for n  0
Value of C(n,k) can be computed by filling a table:
k k .
n C(n-1,k-1) C(n-1,k)
n C(n,k)

7. Computing C(n,k): pseudocode and analysis
Time efficiency: Θ(nk)
Space efficiency: Θ(nk)

8. More Space Efficient
We can achieve Θ(k)

9. Properties of a problem that can be solved with DP
Simple Subproblems
We should be able to break the original problem to smaller subproblems that have the same structure
Optimal Substructure of the problems
The solution to the problem must be a composition of subproblem solutions
Subproblem Overlap
Optimal solutions to unrelated problems can contain subproblems in common

10. 0-1 Knapsack problem
Given a knapsack with maximum capacity W, and a set S consisting of n items
Each item i has some weight wi and benefit value bi (all wi , bi and W are integer values)
Problem: How to pack the knapsack to achieve maximum total value of packed items?

11. 0-1 Knapsack problem: a picture
Weight
Benefit value wi bi
Items 3 2
This is a knapsack
Max weight: W = 20 4 3 5 4
W = 20 8 5 9 10

12. 0-1 Knapsack problem Problem, in other words, is to find
The problem is called a “0-1” problem, because each item must be entirely accepted or rejected.
Just another version of this problem is the “Fractional Knapsack Problem”, where we can take fractions of items.
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13. 0-1 Knapsack problem: brute-force approach
Let’s first solve this problem with a straightforward algorithm
Since there are n items, there are 2n possible combinations of items.
We go through all combinations and find the one with the most total value and with total weight less or equal to W
Running time will be O(2n)
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14. 0-1 Knapsack problem: brute-force approach
Can we do better?
Yes, with an algorithm based on dynamic programming
We need to carefully identify the subproblems
Let’s try this:
If items are labeled 1..n, then a subproblem
would be to find an optimal solution for
Sk = {items labeled 1, 2, .. k}
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15. Defining a Subproblem
If items are labeled 1..n, then a subproblem would be to find an optimal solution for Sk = {items labeled 1, 2, .. k}
This is a valid subproblem definition.
The question is: can we describe the final solution (Sn ) in terms of subproblems (Sk)?
Unfortunately, we can’t do that. Explanation follows….
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16. ? Defining a Subproblem wi bi
Weight
Benefit wi bi
Item # ? 1 2 3
Max weight: W = 20
For S4:
Total weight: 14;
total benefit: 20 S4 2 3 4 S5 3 4 5 4 5 8 5 9 10
w1 =2
b1 =3
w2 =4
b2 =5
w3 =5
b3 =8
w5 =9
b5 =10
Solution for S4 is not part of the solution for S5!!!
For S5:
Total weight: 20
total benefit: 26
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17. Defining a Subproblem (continued)
As we have seen, the solution for S4 is not part of the solution for S5
So our definition of a subproblem is flawed and we need another one!
Let’s add another parameter: w, which will represent the exact weight for each subset of items
The subproblem then will be to compute B[k,w]
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18. Recursive Formula for subproblems
It means, that the best subset of Sk that has total weight w is one of the two:
1) the best subset of Sk-1 that has total weight w, or
2) the best subset of Sk-1 that has total weight w-wk plus the item k
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19. Recursive Formula
The best subset of Sk that has the total weight w, either contains item k or not.
First case: wk>w. Item k can’t be part of the solution, since if it was, the total weight would be > w, which is unacceptable
Second case: wk <=w. Then the item k can be in the solution, and we choose the case with greater value
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20. 0-1 Knapsack Algorithm for w = 0 to W B[0,w] = 0 for i = 0 to n
B[i,0] = 0
if wi <= w // item i can be part of the solution
if bi + B[i-1,w-wi] > B[i-1,w]
B[i,w] = bi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
else B[i,w] = B[i-1,w] // wi > w
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21. Remember that the brute-force algorithm
Running time
for w = 0 to W
B[0,w] = 0
for i = 0 to n
B[i,0] = 0
< the rest of the code >
O(W)
Repeat n times
O(W)
What is the running time of this algorithm?
O(n*W)
Remember that the brute-force algorithm
takes O(2n)
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22. Example Let’s run our algorithm on the following data:
n = 4 (# of elements)
W = 5 (max weight)
Elements (weight, benefit):
(2,3), (3,4), (4,5), (5,6)
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23. Example (2) i 1 2 3 4 W 1 2 3 4 5
for w = 0 to W
B[0,w] = 0
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24. Example (3) i 1 2 3 4 W 1 2 3 4 5
for i = 0 to n
B[i,0] = 0
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25. Example (4) Items: 1: (2,3) 2: (3,4) 3: (4,5) 4: (5,6) i=1 bi=3 wi=21 2 3 4 W
i=1
bi=3
wi=2
w=1
w-wi =-1 1 2 3 4 5
if wi <= w // item i can be part of the solution
if bi + B[i-1,w-wi] > B[i-1,w]
B[i,w] = bi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
else B[i,w] = B[i-1,w] // wi > w
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26. Example (5) Items: 1: (2,3) 2: (3,4) 3: (4,5) 4: (5,6) i=1 bi=3 wi=21 2 3 4 W
i=1
bi=3
wi=2
w=2
w-wi =0 1 2 3 3 4 5
if wi <= w // item i can be part of the solution
if bi + B[i-1,w-wi] > B[i-1,w]
B[i,w] = bi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
else B[i,w] = B[i-1,w] // wi > w
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27. Example (6) Items: 1: (2,3) 2: (3,4) 3: (4,5) 4: (5,6) i=1 bi=3 wi=21 2 3 4 W
i=1
bi=3
wi=2
w=3
w-wi=1 1 2 3 3 3 4 5
if wi <= w // item i can be part of the solution
if bi + B[i-1,w-wi] > B[i-1,w]
B[i,w] = bi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
else B[i,w] = B[i-1,w] // wi > w
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28. Example (7) Items: 1: (2,3) 2: (3,4) 3: (4,5) 4: (5,6) i=1 bi=3 wi=21 2 3 4 W
i=1
bi=3
wi=2
w=4
w-wi=2 1 2 3 3 3 4 3 5
if wi <= w // item i can be part of the solution
if bi + B[i-1,w-wi] > B[i-1,w]
B[i,w] = bi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
else B[i,w] = B[i-1,w] // wi > w
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29. Example (8) Items: 1: (2,3) 2: (3,4) 3: (4,5) 4: (5,6) i=1 bi=3 wi=21 2 3 4 W
i=1
bi=3
wi=2
w=5
w-wi=2 1 2 3 3 3 4 3 5 3
if wi <= w // item i can be part of the solution
if bi + B[i-1,w-wi] > B[i-1,w]
B[i,w] = bi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
else B[i,w] = B[i-1,w] // wi > w
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30. Example (9) Items: 1: (2,3) 2: (3,4) 3: (4,5) 4: (5,6) i=2 bi=4 wi=31 2 3 4 W
i=2
bi=4
wi=3
w=1
w-wi=-2 1 2 3 3 3 4 3 5 3
if wi <= w // item i can be part of the solution
if bi + B[i-1,w-wi] > B[i-1,w]
B[i,w] = bi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
else B[i,w] = B[i-1,w] // wi > w
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31. Example (10) Items: 1: (2,3) 2: (3,4) 3: (4,5) 4: (5,6) i=2 bi=4 wi=31 2 3 4 W
i=2
bi=4
wi=3
w=2
w-wi=-1 1 2 3 3 3 3 4 3 5 3
if wi <= w // item i can be part of the solution
if bi + B[i-1,w-wi] > B[i-1,w]
B[i,w] = bi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
else B[i,w] = B[i-1,w] // wi > w
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32. Example (11) Items: 1: (2,3) 2: (3,4) 3: (4,5) 4: (5,6) i=2 bi=4 wi=31 2 3 4 W
i=2
bi=4
wi=3
w=3
w-wi=0 1 2 3 3 3 3 4 4 3 5 3
if wi <= w // item i can be part of the solution
if bi + B[i-1,w-wi] > B[i-1,w]
B[i,w] = bi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
else B[i,w] = B[i-1,w] // wi > w
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33. Example (12) Items: 1: (2,3) 2: (3,4) 3: (4,5) 4: (5,6) i=2 bi=4 wi=31 2 3 4 W
i=2
bi=4
wi=3
w=4
w-wi=1 1 2 3 3 3 3 4 4 3 4 5 3
if wi <= w // item i can be part of the solution
if bi + B[i-1,w-wi] > B[i-1,w]
B[i,w] = bi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
else B[i,w] = B[i-1,w] // wi > w
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34. Example (13) Items: 1: (2,3) 2: (3,4) 3: (4,5) 4: (5,6) i=2 bi=4 wi=31 2 3 4 W
i=2
bi=4
wi=3
w=5
w-wi=2 1 2 3 3 3 3 4 4 3 4 5 3 7
if wi <= w // item i can be part of the solution
if bi + B[i-1,w-wi] > B[i-1,w]
B[i,w] = bi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
else B[i,w] = B[i-1,w] // wi > w
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35. Example (14) Items: 1: (2,3) 2: (3,4) 3: (4,5) 4: (5,6) i=3 bi=5 wi=41 2 3 4 W
i=3
bi=5
wi=4
w=1..3 1 2 3 3 3 3 3 4 4 4 3 4 5 3 7
if wi <= w // item i can be part of the solution
if bi + B[i-1,w-wi] > B[i-1,w]
B[i,w] = bi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
else B[i,w] = B[i-1,w] // wi > w
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36. Example (15) Items: 1: (2,3) 2: (3,4) 3: (4,5) 4: (5,6) i=3 bi=5 wi=41 2 3 4 W
i=3
bi=5
wi=4
w=4
w- wi=0 1 2 3 3 3 3 3 4 4 4 3 4 5 5 3 7
if wi <= w // item i can be part of the solution
if bi + B[i-1,w-wi] > B[i-1,w]
B[i,w] = bi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
else B[i,w] = B[i-1,w] // wi > w
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37. Example (15) Items: 1: (2,3) 2: (3,4) 3: (4,5) 4: (5,6) i=3 bi=5 wi=41 2 3 4 W
i=3
bi=5
wi=4
w=5
w- wi=1 1 2 3 3 3 3 3 4 4 4 3 4 5 5 3 7 7
if wi <= w // item i can be part of the solution
if bi + B[i-1,w-wi] > B[i-1,w]
B[i,w] = bi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
else B[i,w] = B[i-1,w] // wi > w
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38. Example (16) Items: 1: (2,3) 2: (3,4) 3: (4,5) 4: (5,6) i=3 bi=5 wi=41 2 3 4 W
i=3
bi=5
wi=4
w=1..4 1 2 3 3 3 3 3 3 4 4 4 4 3 4 5 5 5 3 7 7
if wi <= w // item i can be part of the solution
if bi + B[i-1,w-wi] > B[i-1,w]
B[i,w] = bi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
else B[i,w] = B[i-1,w] // wi > w
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39. Example (17) Items: 1: (2,3) 2: (3,4) 3: (4,5) 4: (5,6) i=3 bi=5 wi=41 2 3 4 W
i=3
bi=5
wi=4
w=5 1 2 3 3 3 3 3 3 4 4 4 4 3 4 5 5 5 3 7 7 7
if wi <= w // item i can be part of the solution
if bi + B[i-1,w-wi] > B[i-1,w]
B[i,w] = bi + B[i-1,w- wi]
else
B[i,w] = B[i-1,w]
else B[i,w] = B[i-1,w] // wi > w
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40. Knapsack Problem and Memory Functions
Bottom-up is classic solutions to DP problems
Idea: keep solutions to subproblems in the table, fill the table entries
Drawback: some unnecessary entries are filled, inefficient!
Memory functions are used to avoid the drawbacks
How? Lookup table + top-down solution
Memoization

41. Knapsack Problem and Memory Functions
ALGORITHM MFKnapsack(i, j) //W[1..n], V[1..n] //T[0..n, 0..w] is the lookup table, initialized with -1 for each entry except for //row 0 and column 0 initialized with 0s If T[i, j] < 0 if j < W[i] value  MFKnapsack(i-1, j) else value  max (MFKnapsack(i-1,j), MFKnapsack(i-1,j-W[i]) + V[i]) T[i,j]  value return T[i,j]

42. Comments
This algorithm only finds the max possible value that can be carried in the knapsack
To know the items that make this maximum value, an addition to this algorithm is necessary
We can define another subproblem to solve the problem
B(w): the maximum value for a knapsack of capacity w given the n items
What’s the optimal substructure?
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43. Longest Common Subsequence (LCS)
Application: comparison of two DNA strings
Ex: X= {A B C B D A B }, Y= {B D C A B A}
Longest Common Subsequence:
X = A B C B D A B
Y = B D C A B A
Brute force algorithm would compare each subsequence of X with the symbols in Y
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44. LCS
Problem: Given 2 sequences, X = x1,...,xm and Y = y1,...,yn, find a common subsequence whose length is maximum.
springtime ncaa tournament basketball
printing north carolina snoeyink
Subsequence need not be consecutive, but must be in order.

45. LCS Algorithm
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)
Notice that the 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”
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46. LCS Algorithm
First we’ll find the length of LCS. Later we’ll modify the algorithm to find LCS itself.
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]
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47. 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
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48. 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
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49. 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)
Why not just take the length of LCS(Xi-1, Yj-1) ?
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50. LCS Length Algorithm LCS-Length(X, Y)
1. m = length(X) // get the # of symbols in X
2. n = length(Y) // get the # of symbols in Y
3. for i = 1 to m c[i,0] = 0 // special case: Y0
4. for j = 1 to n c[0,j] = 0 // special case: X0
5. for i = 1 to m // for all Xi
6. for j = 1 to n // for all Yj
7. if ( Xi == Yj )
8. c[i,j] = c[i-1,j-1] + 1
9. else c[i,j] = max( c[i-1,j], c[i,j-1] )
10. return c
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51. LCS Example
We’ll see how LCS algorithm works on the following example:
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
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52. LCS Example (0) ABCB BDCAB X = ABCB; m = |X| = 4j 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[5,4]
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53. LCS Example (1) ABCB BDCAB for i = 1 to m c[i,0] = 0j i Yj B D C A B Xi A 1 B 2 3 C 4 B
for i = 1 to m c[i,0] = 0
for j = 1 to n c[0,j] = 0
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54. LCS Example (2) ABCB BDCAB j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 B 2 3 CA 1 B 2 3 C 4 B
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
else c[i,j] = max( c[i-1,j], c[i,j-1] )
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55. LCS Example (3) ABCB BDCAB j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 B 2 3 CA 1 B 2 3 C 4 B
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
else c[i,j] = max( c[i-1,j], c[i,j-1] )
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56. LCS Example (4) ABCB BDCAB j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 B 2 3A 1 1 B 2 3 C 4 B
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
else c[i,j] = max( c[i-1,j], c[i,j-1] )
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57. LCS Example (5) ABCB BDCAB j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2A 1 1 1 B 2 3 C 4 B
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
else c[i,j] = max( c[i-1,j], c[i,j-1] )
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58. LCS Example (6) ABCB BDCAB j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2A 1 1 1 B 2 1 3 C 4 B
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
else c[i,j] = max( c[i-1,j], c[i,j-1] )
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59. LCS Example (7) ABCB BDCAB j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2A 1 1 1 B 2 1 1 1 1 3 C 4 B
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
else c[i,j] = max( c[i-1,j], c[i,j-1] )
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60. LCS Example (8) ABCB BDCAB j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2A 1 1 1 B 2 1 1 1 1 2 3 C 4 B
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
else c[i,j] = max( c[i-1,j], c[i,j-1] )
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61. LCS Example (10) ABCB BDCAB 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 4 B
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
else c[i,j] = max( c[i-1,j], c[i,j-1] )
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62. LCS Example (11) ABCB BDCAB 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 4 B
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
else c[i,j] = max( c[i-1,j], c[i,j-1] )
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63. LCS Example (12) ABCB BDCAB 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 4 B
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
else c[i,j] = max( c[i-1,j], c[i,j-1] )
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64. LCS Example (13) ABCB BDCAB 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 4 B 1
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
else c[i,j] = max( c[i-1,j], c[i,j-1] )
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65. LCS Example (14) ABCB BDCAB 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 4 B 1 1 2 2
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
else c[i,j] = max( c[i-1,j], c[i,j-1] )
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66. LCS Example (15) 3 ABCB BDCAB j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1A 1 1 1 B 2 1 1 1 1 2 3 C 1 1 2 2 2 3 4 B 1 1 2 2
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
else c[i,j] = max( c[i-1,j], c[i,j-1] )
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67. LCS Algorithm Running Time
LCS algorithm calculates the values of each entry of the array c[m,n]
So what is the running time?
O(m*n)
since each c[i,j] is calculated in constant time, and there are m*n elements in the array
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68. How to find actual LCS
So far, we have just found the length of LCS, but not LCS itself.
We want to modify this algorithm to make it output Longest Common Subsequence of X and Y
Each c[i,j] depends on c[i-1,j] and c[i,j-1]
or c[i-1, j-1]
For each c[i,j] we can say how it was acquired:
For example, here
c[i,j] = c[i-1,j-1] +1 = 2+1=3 2 2 2 3
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69. How to find actual LCS - continued
Remember that
So we can start from c[m,n] and go backwards
Whenever c[i,j] = c[i-1, j-1]+1, remember x[i] (because x[i] is a part of LCS)
When i=0 or j=0 (i.e. we reached the beginning), output remembered letters in reverse order
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70. Finding LCS 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 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
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71. Finding LCS (2) 3 B C B LCS (reversed order): B C Bj i Yj B D C A B Xi A 1 1 1 B 2 1 1 1 1 2 3 C 1 1 2 2 2 3 4 B 1 1 2 2 B C B
LCS (reversed order):
B C B
(this string turned out to be a palindrome)
LCS (straight order):
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