Dynamic Programming

• Optimization Problems

• Dynamic Programming Paradigm

• Example: Matrix multiplication

• Principle of Optimality

• Exercise: Trading post problem

Optimization Problems

• In an optimization problem, there are typically many feasible solutions for any input instance I

• For each solution S, we have a cost or value function f(S)

• Typically, we wish to find a feasible solution S such that f(S) is either minimized or maximized

• Thus, when designing an algorithm to solve an optimization problem, we must prove the algorithm produces a best possible solution.

Example Problem

You have six hours to complete as many tasks as possible, all of which are equally important.Task A - 2 hours Task D - 3.5 hoursTask B - 4 hours Task E - 2 hoursTask C - 1/2 hour Task F - 1 hour

How many can you get done?• Is this a minimization or a maximization problem?• Give one example of a feasible but not optimal

solution along with its associated value.• Give an optimal solution and its associated value.

Dynamic Programming

• Dynamic programming is a divide-and-conquer technique at heart

• That is, we solve larger problems by patching together solutions to smaller problems

• Dynamic programming can achieve efficiency by storing solutions to subproblems to avoid redundant computations– We typically avoid redundant computations by

computing solutions in a bottom-up fashion

Efficiency Example: Fibonacci numbers

• F(n) = F(n-1) + F(n-2)– F(0) = 0– F(1) = 1

• Top-down recursive computation is very inefficient– Many F(i) values are computed multiple times

• Bottom-up computation is much more efficient– Compute F(2), then F(3), then F(4), etc. using stored

values for smaller F(i) values to compute next value– Each F(i) value is computed just once

Recursive Computation

F(n) = F(n-1) + F(n-2) ; F(0) = 0, F(1) = 1

Recursive Solution: F(6) = 8

F(1)

F(1) F(0)

F(2)

F(3)

F(1) F(0)

F(2)

F(1)

F(1) F(0)

F(2)

F(3)

F(4)

F(1) F(0)

F(2)

F(1)

F(1) F(0)

F(2)

F(3)

F(4)F(5)

Bottom-up computation

We can calculate F(n) in linear time by storing small values.

F[0] = 0F[1] = 1for i = 2 to n

F[i] = F[i-1] + F[i-2]return F[n]

Moral: We can sometimes trade space for time.

Key implementation steps

• Identify subsolutions that may be useful in computing whole solution– Often need to introduce parameters

• Develop a recurrence relation (recursive solution)– Set up the table of values/costs to be computed

• The dimensionality is typically determined by the number of parameters

• The number of values should be polynomial

• Determine the order of computation of values• Backtrack through the table to obtain complete

solution (not just solution value)

Example: Matrix Multiplication

• Input– List of n matrices to be multiplied together using traditional matrix

multiplication– The dimensions of the matrices are sufficient

• Task– Compute the optimal ordering of multiplications to minimize total

number of scalar multiplications performed

• Observations: – Multiplying an X Y matrix by a Y Z matrix takes X Y Z

multiplications– Matrix multiplication is associative but not commutative

Example Input

• Input: – M1, M2, M3, M4

• M1: 13 x 5• M2: 5 x 89 • M3: 89 x 3 • M4: 3 x 34

• Feasible solutions and their values– ((M1 M2) M3) M4:10,582 scalar multiplications– (M1 M2) (M3 M4): 54,201 scalar multiplications– (M1 (M2 M3)) M4: 2856 scalar multiplications– M1 ((M2 M3) M4): 4055 scalar multiplications– M1 (M2 (M3 M4)): 26,418 scalar multiplications

Identify subsolutions

• Often need to introduce parameters

• Define dimensions to be (d0, d1, …, dn) where matrix Mi has dimensions di-1 x di

• Let M(i,j) be the matrix formed by multiplying matrices Mi through Mj

• Define C(i,j) to be the minimum cost for computing M(i,j)

Develop a recurrence relation

• Definitions– M(i,j): matrices Mi through Mj

– C(i,j): the minimum cost for computing M(i,j)

• Recurrence relation for C(i,j)– C(i,i) = ???– C(i,j) = ???

• Want to express C(i,j) in terms of “smaller” C terms

Set up table of values

• Table– The dimensionality is

typically determined by the number of parameters

– The number of values should be polynomial

C 1 2 3 4

1 0

2 0

3 0

4 0

Order of Computation of Values

• Many orders are typically ok.– Just need to obey some

constraints

• What are valid orders for this table?

C 1 2 3 4

1 0 1 2 3

2 0 4 5

3 0 6

4 0

Representing optimal solutionP 1 2 3 4

1 0 1 1 3

2 0 2 3

3 0 3

4 0

C 1 2 3 4

1 0 5785 1530 2856

2 0 1335 1845

3 0 9078

4 0

P(i,j) records the intermediate multiplication k used to computeM(i,j). That is, P(i,j) = k if last multiplication was M(i,k) M(k+1,j)

Pseudocodeint MatrixOrder()

forall i, j C[i, j] = 0;

for j = 2 to n

for i = j-1 to 1

C(i,j) = mini<=k<=j-1 (C(i,k)+ C(k+1,j) + di-1dkdj)

P[i, j]=k;

return C[1, n];

BacktrackingProcedure ShowOrder(i, j)

if (i=j) write ( “Ai”) ;

else

k = P [ i, j ] ;

write “ ( ” ;

ShowOrder(i, k) ;

write “ ” ;

ShowOrder (k+1, j) ;

write “)” ;

Principle of Optimality

• In book, this is termed “Optimal substructure”• An optimal solution contains within it optimal

solutions to subproblems.• More detailed explanation

– Suppose solution S is optimal for problem P.– Suppose we decompose P into P1 through Pk and that S

can be decomposed into pieces S1 through Sk corresponding to the subproblems.

– Then solution Si is an optimal solution for subproblem Pi

Example 1

• Matrix Multiplication– In our solution for computing matrix M(1,n),

we have a final step of multiplying matrices M(1,k) and M(k+1,n).

– Our subproblems then would be to compute M(1,k) and M(k+1,n)

– Our solution uses optimal solutions for computing M(1,k) and M(k+1,n) as part of the overall solution.

Example 2

• Shortest Path Problem– Suppose a shortest path from s to t visits u– We can decompose the path into s-u and u-t.– The s-u path must be a shortest path from s to

u, and the u-t path must be a shortest path from u to t

• Conclusion: dynamic programming can be used for computing shortest paths

Example 3

• Longest Path Problem– Suppose a longest path from s to t visits u– We can decompose the path into s-u and u-t.– Is it true that the s-u path must be a longest path

from s to u?

• Conclusion?

Example 4: The Traveling Salesman Problem

What recurrence relation will return the optimal solution to the Traveling Salesman Problem?

If T(i) is the optimal tour on the first i points, will this help us in solving larger instances of the problem?

Can we set T(i+1) to be T(i) with the additional point inserted in the position that will result in the shortest path?

No!

T(4) Shortest TourT(5)

Summary of bad examples

• There almost always is a way to have the optimal substructure if you expand your subproblems enough

• For longest path and TSP, the number of subproblems grows to exponential size

• This is not useful as we do not want to compute an exponential number of solutions

When is dynamic programming effective?

• Dynamic programming works best on objects that are linearly ordered and cannot be rearranged– characters in a string

– files in a filing cabinet

– points around the boundary of a polygon

– the left-to-right order of leaves in a search tree.

• Whenever your objects are ordered in a left-to-right way, dynamic programming must be considered.

Efficient Top-Down Implementation

• We can implement any dynamic programming solution top-down by storing computed values in the table– If all values need to be computed anyway,

bottom up is more efficient– If some do not need to be computed, top-down

may be faster

Trading Post Problem

• Input– n trading posts on a river– R(i,j) is the cost for renting at post i and

returning at post j for i < j• Note, cannot paddle upstream so i < j

• Task– Output minimum cost route to get from trading

post 1 to trading post n

Longest Common Subsequence Problem

• Given 2 strings S and T, a common subsequence is a subsequence that appears in both S and T.

• The longest common subsequence problem is to find a longest common subsequence (lcs) of S and T– subsequence: characters need not be contiguous– different than substring

• Can you use dynamic programming to solve the longest common subsequence problem?

Longest Increasing Subsequence Problem

• Input: a sequence of n numbers x1, x2, …, xn.• Task: Find the longest increasing subsequence of

numbers– subsequence: numbers need not be contiguous

• Can you use dynamic programming to solve the longest common subsequence problem?

Book Stacking Problem

• Input– n books with heights hi and thicknesses ti

– length of shelf L

• Task– Assignment of books to shelves minimizing

sum of heights of tallest book on each shelf• books must be stored in order to conform to catalog

system (i.e. books on first shelf must be 1 through i, books on second shelf i+1 through k, etc.)