6.4. Base Case and Recursive Formulation

For our last example of dynamic programming to this week, now we look at the problem of the efficiently multiplying the sequence of matrices. So, as you probably know to multiply two matrices A and B, we need compatible dimensions, so if we have A and B, then we need that the number of columns in A must match the number of rows in B.

In order to multiply two matrices A and B, the dimensions must meet specific criteria: the number of columns in the first matrix (A) must equal the number of rows in the second matrix (B). For instance, if matrix A has dimensions m x n and matrix B has n x p, the resulting product will have dimensions m x p. This dimensional compatibility is crucial for the multiplication to be valid.

Think of this like a recipe: if you need 1 cup of sugar (representing columns in matrix A) for every batch of cookies you want to bake (representing the rows in matrix B), you can only bake a certain number of batches (resulting matrix). If your sugar can only fill 2 cups (dimensions), you need to limit how many batches you can bake at once.

Matrix multiplication is not commutative, we do not have this in general. So, it is important that the order is the same, but within that in which sequence I do this simplification does not matter. So, it is associative, I can bracket it by as A times B followed by C or A times B followed by C, either way I could do two matrix multiplications and what associativity means is that the answer will not change.

Matrix multiplication is associative, meaning that when multiplying multiple matrices, the way you group them does not change the final product. For example, (A * B) * C will yield the same result as A * (B * C). However, it is important to note that matrix multiplication is not commutative, so A * B does not necessarily equal B * A.

Consider stacking boxes to move them. You can group the boxes in different ways before moving them. Whether you group (Box A with Box B) first and then add Box C later, or form a group of (Box B and Box C) first, you will still end up with the same total number of boxes at the end, but if you swap the order of boxes, the arrangement might not fit in the truck.

So, you can see that there can be a dramatic difference in the complexity depending on whether you, the bracketed is A B followed by C or A followed by B C.

The sequence in which matrices are multiplied can affect the computational complexity significantly. For example, if multiplying matrices A, B, and C, first computing B * C may yield a larger intermediate matrix and subsequently require more operations when multiplying with A, compared to first computing A * B, which produces a smaller matrix for the final multiplication.

This concept is similar to solving a puzzle. If you pick the right pieces that fit together easily first, you can complete it faster. However, if you start with pieces that don’t fit well together and create a larger mess, it takes longer to finish the puzzle.

So, our goal is to find an optimum way to compute this product, what is the sequence of basic operations multiplying two matrices together there we need to perform to get the minimum overall cost.

In dynamic programming, we seek to determine the least expensive way to multiply a chain of matrices. This involves exploring all possible orderings of multiplication to find which combination yields the fewest computational steps. The challenge is to ensure at each step that we adhere to the rules of matrix dimensions while evaluating various groupings of the matrices.

Imagine planning a road trip where you need to visit multiple cities. The goal is to find the shortest possible route that visits all the cities. By considering all possible routes, you can calculate which combination gets you to your destination quickest, similar to finding the optimal matrix multiplication sequence.

In other words, the cost for multiplying M 1 to M n in terms of total number of operations should be the minimum value of k between 1 and n.

When calculating the cost of matrix multiplication in a chain from M1 to Mn, we must consider all possible positions (k) to split the multiplication into two parts. The overall cost will be the sum of the costs for the two parts plus the cost of multiplying the resulting matrices. The objective is to find that k which gives the minimum possible cost.

Consider planning meals for a week. You could group meals in various ways, planning for individual days. By testing each grouping, you can ensure you’ll spend the least amount of time cooking each week, similar to testing each k value to minimize computation costs.

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