# Dynamic Programming Optimization Problems Dynamic Programming Paradigm

## Presentation on theme: "Dynamic Programming Optimization Problems Dynamic Programming Paradigm"— Presentation transcript:

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 hours Task B - 4 hours Task E - 2 hours Task 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(0) F(2) F(3) F(4) F(5) Implementing it as a recursive procedure is easy but slow! We keep calculating the same value over and over!

Bottom-up computation
We can calculate F(n) in linear time by storing small values. F[0] = 0 F[1] = 1 for i = 2 to n F[i] = F[i-1] + F[i-2] return F[n] Moral: We can sometimes trade space for time. Dynamic programming is a technique for efficiently computing recurrences by storing partial results. Once you understand dynamic programming, it is usually easier to reinvent certain algorithms than try to look them up! Dynamic programming is best understood by looking at a bunch of different examples.

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: Feasible solutions and their values
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 Recurrence relation for C(i,j) C(i,j) = mink=i to j-1 ( C(i,k)+ C(k+1,j) + di-1dkdj) The last multiplication is between matrices M(i,k) and M(k+1,j) C(i,i) = 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 5 6 Diagonals order ok

Representing optimal solution
C 1 2 3 4 5785 1530 2856 1335 1845 9078 P 1 2 3 4 P(i,j) records the intermediate multiplication k used to compute M(i,j). That is, P(i,j) = k if last multiplication was M(i,k) M(k+1,j)

Pseudocode int 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];

Backtracking Procedure 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 Conclusion?
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(5) Shortest Tour T(4)

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

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?