Dynamic programming Implementations and Examples

The main Dynamic_programming page is an excellent overview of the subject for those that already understand the material, but just want to review the details. This page aims to explain the material in as many different ways as possible so that almost any interested student regardless of learning style can grasp the subject.

Fibonacci example

A naive implementation of a function finding the nth member of the Fibonacci sequence, based directly on the mathematical definition, performs much redundant work. Here is that implementation:

The following is wikicode, a proposed pseudocode for BambooWeb articles.

Notice that if we call, say, fib(5), we produce a call tree that calls the function on the same value many different times:

1. fib(5)
2. fib(4) + fib(3)
3. (fib(3) + fib(2)) + (fib(2) + fib(1))
4. ((fib(2) + fib(1)) + (fib(1) + fib(0))) + ((fib(1) + fib(0)) + fib(1))
5. (((fib(1) + fib(0)) + fib(1)) + (fib(1) + fib(0))) + ((fib(1) + fib(0)) + fib(1))

In particular, fib(2) was calculated twice from scratch. In larger examples, many more values of fib, or subproblems, are recalculated, leading to an exponential time (O(2ⁿ)) algorithm.

Now, suppose we have a simple map object, m, which maps each value of fib that has already been calculated to its result, and we modify our function to use it and update it. The resulting function requires only O(n) time instead of O(2ⁿ):

This technique of saving values that have already been calculated is called memoization. We can also reduce the function from using linear (O(n)) space to using constant space by changing our function to calculate smaller values of fib first, then build larger values from them:

A checkerboard example

Lets say you had an checkerboard with n * n squares on it. Along with this checkerboard you get a cost-function c(i, j) which returns a cost associated with square i,j (i being the rank, j being the column). For instance (on a 5 * 5 checkerboard) ,

Thus c(1, 3) = 5

Let's say you had a checker that counld start at any square on the first rank and you wanted to know the shortest path (sum of the costs of the visited squares are at a minimum) to get to the last rank, assuming the checker could move only forward or diagonally left or right forward. That is, a checker on 1,3 can move to 2,2 2,3 or 2,4

This problem exhibits optimal substructure. I.e. the solution to the entire problem relies on solutions to subproblems. Let's define a function q(i, j) as

<math>q(i, j) = \mbox{The minimum cost to reach square (i, j)}<math>

If we can find the values of this function for all the the squares at rank n, we pick the minimum and follow that path backwards to get the shortest path.

It's easy to see that q(i, j) is equal to the minimum cost to get to any of the three squares below it (since that's the only squares that can reach it) plus c(i, j). For instance:

<math>q(A) = \min(q(B),\;q(C),\;q(D))\;+\;c(A)<math>

Now, let's define q(i, j) in little more general terms:

<math>q(i,j)=\left\{\begin{matrix} \infty & j = 0 \mbox{ or }j = n+1 \\ c(i, j) & i = 1 \\ \min(q(i-1, j-1), q(i-1, j), q(i-1, j+1)) + c(i,j) & \mbox{otherwise}\end{matrix}\right.<math>

This equation is pretty straightforward. The first line is simply there to make the recursive property simpler (when dealing with the edges, so we only need one recursion). The second line says what happens in the first rank, so we have something to start with. The third line, the recursion, is the important part. It is basically the same as the A,B,C,D example. From this definition we can make a straight-forward recursive code for q(i, j). In the following pseudocode, n is the size of the board, c(i, j) is the cost-function, and min() returns the minimum of a number of values:

It should be noted that this function just computes the path-cost, not the actual path. We'll get to the path soon. This, like the Fibonacci-numbers example, is horribly slow since it spends mountains of time recomputing the same shorteset paths over and over. However, we can compute it much faster in a bottom up-fashion if we use a two-dimensional array q[i, j] instead of a function. Why do we do that? Simply because when using a function we recompute the same path over and over, and we can choose what values to compute first.

We also need to know what actual path is. The path problem we can solve using another array p[i, j], a predecessor array. This array basically says where path comes from. Consider the following code:

Now the rest is a simple matter of finding the minimum and printing it.

Matrix chain multiplication example

Show how the placement of parentheses affects the number of scalar multiplications required when multiplying a bunch of matrices.

Show how to write a dynamic program to calculate the optimal parentheses placement.

this is such a long example that it might be better to make it its own article.