1. CSC 252 Algorithms Haniya Aslam
DYNAMIC PROGRAMMING: Matrix Chain Multiplication & Optimal Triangulation
CSC 252
Algorithms
Haniya Aslam

2. Presentation Overview
Understanding dynamic programming
Dynamic programming vs. Recursion and Demand & Conquer
Matrix chain multiplication
Optimal polygon triangulation
Acknowledgements

3. Dynamic Programming A generalization of iteration and recursion.
“Dynamic Programming is recursion’s somewhat neglected cousin. …(It) is the basis of comparison and alignment routines.
Bottom-up design:
Start at the bottom
Solve small sub-problems
Store solutions
Reuse previous results for solving larger sub-problems

4. Dynamic Programming cont.
Fibonacci:
function Fibonacci(n: integer) : integer;
var
i : index; sum, interm1, interm2: integer;
begin
interm1:= 0; {F0}
interm2:= 1; {F1}
for i:=3 to n do
sum :=interm1 + interm2;
interm1:=interm2;
interm2:= sum;
end {for}
Fibonacci := sum;
end {Fibonacci}

5. Dynamic Programming vs. Recursion and Divide & Conquer
In a recursive program, a problem of size n is solved by first solving a sub-problem of size n-1.
In a divide & conquer program, you solve a problem of size n by first solving a sub-problem of size k and another of size k-1, where 1 < k < n.
In dynamic programming, you solve a problem of size n by first solving all sub-problems of all sizes k, where k < n.

6. Matrix Chain Multiplication
Given : a chain of matrices {A1,A2,…,An}.
Once all pairs of matrices are parenthesized, they can be multiplied by using the standard algorithm as a sub-routine.
A product of matrices is fully parenthesized if it is either a single matrix or the product of two fully parenthesized matrix products, surrounded by parentheses. [Note: since matrix multiplication is associative, all parenthesizations yield the same product.]

7. Matrix Chain Multiplication cont.
For example, if the chain of matrices is {A, B, C, D}, the product A, B, C, D can be fully parenthesized in 5 distinct ways:
(A ( B ( C D ))),
(A (( B C ) D )),
((A B ) ( C D )),
((A ( B C )) D),
((( A B ) C ) D ).
The way the chain is parenthesized can have a dramatic impact on the cost of evaluating the product.

8. Matrix Chain Multiplication Optimal Parenthesization
Example: A[30][35], B[35][15], C[15][5]
minimum of A*B*C
A*(B*C) = 30*35*5 + 35*15*5 = 7,585
(A*B)*C = 30*35* *15*5 = 18,000
How to optimize:
Brute force – look at every possible way to parenthesize : Ω(4n/n3/2)
Dynamic programming – time complexity of Ω(n3) and space complexity of Θ(n2).

9. Matrix Chain Multiplication Structure of Optimal Parenthesization
For n matrices, let Ai..j be the result of AiAi+1….Aj
An optimal parenthesization of AiAi+1…An splits the product between Ak and Ak+1 where 1  k < n.
Example, k = (A1A2A3A4)(A5A6)
Total cost of A1..6 = cost of A1..4 plus total
cost of multiplying these two matrices
together.

10. Matrix Chain Multiplication Overlapping Sub-Problems
Overlapping sub-problems helps in reducing the running time considerably.
Create a table M of minimum Costs
Create a table S that records index k for each optimal sub-problem
Fill table M in a manner that corresponds to solving the parenthesization problem on matrix chains of increasing length.
Compute cost for chains of length 1 (this is 0)
Compute costs for chains of length 2
A1..2, A2..3, A3..4, …An-1…n
Compute cost for chain of length n
A1..n
Each level relies on smaller sub-strings

11. Optimal Polygon Triangulation
A triangulation of a polygon is a set of T chords of the polygon that divide the polygon into disjoint triangles. In a triangulation, no chords intersect (except at end-points) and the set T of chords is maximal: every chord not in T intersects some chord in T. The sides of triangles produced by the triangulation are either chords in the triangulation or sides of the polygon. Every triangulation of an n-vertex convex polygon has n-3 chords and divides the polygon into n-2 triangles.
a b.

12. Optimal Polygon Triangulation cont.
In the optimal polygon triangulation problem , we are given a polygon P = {v0, v1, v2, …., vn-1} and a weight function w defined on triangles formed by sides and chords of P.
The problem is to find a triangulation that minimizes the sum of the weights of the triangles in the triangulation.
This problem, like matrix chain multiplication, uses parenthesization.
A full parenthesization corresponds to a
full binary tree also called a parse tree.

13. Parenthesization in Triangulation
Above is the parse tree for the triangulation of a polygon.The internal nodes of the parse tree are the chords of the triangulation plus the side v0v6, which is the root.

14. Triangulation and Matrix Chain Multiplication
Since a fully parenthesized product of n matrices corresponds to a parse tree with n leaves, it therefore also corresponds to a triangulation of an (n+1)-vertex polygon.
Each matrix Ai in a product of A1A2…An corresponds to side vi-1vi of an (n+1)-vertex polygon.
The matrix chain is actually a special case of the optimal triangulation problem.

15. Triangulation and Matrix Chain Multiplication cont.
Given a matrix chain product A1A2….AN, we define an (n+1)-vertex convex polygon p = {v0,v1,….,vn}. If matrix Ai has dimensions pi-1 x pi, for I = 1,2,…,n, the weight function for the triangulation is defined as
w(Δvivjvk) = pipjpk
An optimal triangulation of P with respect to this weight function gives the parse tree for an optimal parenthesization of A1A2….An.

16. Substructure of an optimal triangulation
Given: an optimal triangulation T of an (n+1)-vertex polygon P that includes the triangle Δv0vkvn, where 1  k  n-1.
The weight of T is the sum of the weights of Δv0vkvn and triangles in the triangulation of the two sub-polygons {v0,v1,….vk} and {vk, vk+1,….vn}.
The triangulation of the sub-polygons determined by T , therefore, must be optimal, since a lesser-weight triangulation of either sub-polygon would contradict the minimality of the eight of T.

17. T[i,j] = min {t[i,k] + t[k+1,j] + w(Δvi-1vkvj)}
A Recursive Solution
Given, for 1  I  j  n:
T[i,j] is the weight of an optimal triangulation of the polygon {vi-1, vi, …..vj} [Since m[i,j] is the minimum cost of computing the matrix chain sub-product AiAi+1….Aj]
In the case of a 2-vertex polygon: t[i,i] = 0 for i = 1,2,…,n.
To minimize over all vertices vk, where k = i,i+1,…j-1, the weight of Δvi-1vkvj plus the weights of the optimal triangulation of the polygons {vi-1, vi,….,vk} and {vk,vk+1,…,vj}.
The recursive solution then is:
If i < j,
T[i,j] = min {t[i,k] + t[k+1,j] + w(Δvi-1vkvj)}
i k j-1

18. Acknowledgments