DYNAMIC PROGRAMMING:
Matrix Chain Multiplication & Optimal
Triangulation
CSC 252
Algorithms
Haniya Aslam
Presentation Overview
 Understanding dynamic programming
 Dynamic programming vs. Recursion and
Demand & Conquer
 Matrix chain multiplication
 Optimal polygon triangulation
 Acknowledgements
Dynamic Programming
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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:
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Start at the bottom
Solve small sub-problems
Store solutions
Reuse previous results for solving larger sub-problems
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}
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.
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.]
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.
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 + 30*15*5 = 18,000
 How to optimize:
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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).
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 = 4
(A1A2A3A4)(A5A6)
Total cost of A1..6 = cost of A1..4 plus total
cost of multiplying these two matrices
together.
Matrix Chain Multiplication
Overlapping Sub-Problems
 Overlapping sub-problems helps in reducing the running
time considerably.
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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
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Compute cost for chain of length n
A1..n
Each level relies on smaller sub-strings
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.
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.
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.
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.
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.
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 lesserweight triangulation of either sub-polygon would
contradict the minimality of the eight of T.
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 Δvi1vkvj 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
Acknowledgments
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http://www-cse/uta/edu/~holder/courses/cse5311/lectures/18/node18.html
http://www.middlebury.edu/~dickerso/ccsc/ugcg.html
http://www.eecs.harvard.edu/~nr/cs152/readings/dynamic.html
http://www.catalase.com/dprog.htm
http://mail.informs.org/classes/dynamic/node1.html
http://cse.hanyang.ac.kr/~jmchoi/c…6-2/algorithm/classnote/node6.html
http://people.bu.edu/rlynch/cs566/sld002.htm