1. Algorithms CSCI 235, Spring 2019 Lecture 27 Dynamic Programming II

2. The problem
Problem: Find the best way to multiply n matrices (where best means the one that uses the minimum number of scalar multiplications).
Practical use: Computer Graphics (Long chains of matrices).
Digital Signal processing: applying filters to signals, etc.
Formally: Given <A1, A2, A3, ...An> are n matrices, fully parenthesize the product A1A2A3...An in a way to minimize the number of scalar multiplications.
Exhaustive Search takes prohibitively long:

3. Number of Scalar Multiplications
Multiplying 2 matrices, A1, A2, where
A1 is pxq and
A2 is qxr
takes pqr scalar multiplications.
Example:
2x3
3x1
Number of multiplications = 2x3x1 = 6

4. 1. Characterize structure of optimal solution
Notation: Ai .. j = AiAi+1Ai Aj
An optimal solution of A1 ..n = (A1A2 ...Ak)(Ak+1Ak+2 ...An), 1<=k<n
Cost = cost of computing A1 .. k + cost of computing Ak+1 .. n
+ cost to multiply the 2 results together.
For the solution to be optimal, we must first find the optimal solutions to the subproblems: A1 .. k and Ak+1 .. n
(Why must they be optimal solutions?)

5. Example A1A2A3A4 A1(A2A3A4) (A1A2)(A3A4) (A1A2A3)A4 k=3 k=1 k=2
Note: As we compute the optimal solutions for subproblems, we do the same computation repeatedly (e.g. A2A3). To avoid repeat computations, we will store the values as each one is computed.

6. 2. Recursively define the value of the optimal solution
For subproblem, Ai .. j, 1 <= i <= j <=n
Compute m[i, j] = minimum number of scalar multiplications to compute Ai .. j
Matrix Ai has dimension pi-1 x pi (rows x cols)
Example:
More notation:
pi is number of columns in the ith matrix.
A A2 A3
3x3 3x1 1x4
p0 p1 p1 p2 p2 p3

7. Number of multiplications
If A1 is p0 x p1, A2 is p1 x p2, then A1A2 is p0 x p2
If A3 is p2xp3 then A1A2A3 is p0 x p3
(A1A2 ... Ak) is p0 x pk
M1=(AiAi Ak) is pi-1 x pk
M2=(Ak Aj) is pk x pj
Therefore, the number of scalar multiplications to multiply M1M2 is pi-1pkpj

8. Defining m[i, j] recursively
If i = j, no multiplication needed, so m[i, j] = 0
If i < j, m[i, j] = m[i, k] + m[k+1, j] + pi-1pkpj
Example: m[2, 6], let k = 4:
(A2A3A4)(A5A6)
We will work this out in class.

9. Determining the Best split
We must check each value of k to determine which gives the minimum cost:
Minimum cost of Ai .. j is:
Keep track of minimum cost in table, m[i, j]
Keep track of k value that gives the minimum cost in table, s[i, j] = k

10. 3. Compute the value of the optimal solution
Step 3. Compute the value of the optimal solution from the bottom up.
We start with the smallest subproblems: i = j
m[i, i] = 0
Next compute cost of chains of length 2:
m[i, i+1]
Third, compute cost of chains of length 3 (using values for length 2):
m[i, i+2]
Total number of subproblems to solve: 1 for each i & j, 1 <=i <=j <=n
Combination:

11. Matrix Chain Order Code
Matrix-Chain-Order(p) // p is an array containing pi-1 to pj
n = p.length – // Number of matrices in chain
Let m[1..n, 1..n] and s[1..n – 1, 2..n] be new tables
for i = 1 to n
m[i, i] = 0 // Single matrices take 0 multiplications
for b = 2 to n // b is length of chain
for i = 1 to n – b // all possible starting indices for length b
j = i + b // Ending index of chain of length b
m[i, j] = ∞ // Large value to start to find minimum
for k = i to j - 1 // Try all possible splits of this chain
q = m[i, k] + m[k + 1, j] + pi-1pkpj // smaller chains are
// already computed
if q < m[i, j] // If minimum value, then store it
m[i, j] = q
s[i, j] = k
return m and s

12. Example
Find the best parenthesization for multiplying the following chain: A1A2A3A4, where A1 is 3x3, A2 is 3x2, A3 is 2x2 and A4 is 2x4.
p = <3, 3, 2, 2, 4> j j
m[i, j]
s[i, j] 1 2 3 4 1 2 3 4 i i