Dynamic Programming

Dynamic programming is a powerful technique for solving a certain class of problems, typically in a more efficient manner than the corresponding recursive strategy. Specifically, when a problem consists of “overlapping subproblems,” a recursive strategy may lead to redundant computation. The corresponding dynamic programming strategy may avoid such waste by addressing and solving the subproblems one at a time in a manner without overlap.

This idea is difficult to understand in the abstract, so let’s consider a couple of examples.

Firstly, let’s write a function to return the nth Fibonacci number: the nth number in the sequence constructed by starting with 0,10, 1 and calculating subsequent numbers as the sum of the previous two numbers, like so:

0,1,1,2,3,5,8,13,21... 0, 1, 1, 2, 3, 5, 8, 13, 21 ...
The Fibonacci sequence may be familiar enough to you that you are able to take a “top down” approach: identifying the recursive relationship simply by considering the definition of the Fibonacci sequence. In other words, a “top down” approach considers the statement “calculating subsequent numbers as the sum of the previous two numbers” and recognizes the relationship f(n)=f(n1)+f(n2)f(n) = f(n-1) + f(n-2).

With 0 and 1 as our base cases, this leads to an implementation in code that looks very much like the mathematical definition of the sequence:

def fib(n):
    if n <= 1:
        return n  # base cases: return 0 or 1 if n is 0 or 1, respectively
    return fib(n - 1) + fib(n - 2)
const fib = (n) => {
  if (n <= 1) return n
  return fib(n - 1) + fib(n - 2)

This is a correct solution, but it poses a problem evident to those who run fib(50) and wait for an answer. The running time of this implementation is O(2n)O(2^n) due to a very large number of redundant recursive executions. When we call fib(n) we recursively call fib(n - 1) and fib(n - 2) which themselves call (i) fib(n - 2) and fib(n - 3); and, (ii) fib(n - 3) and fib(n - 4), respectively.

We can see that there are some redundant calls, and you may recognize that each of those redundant calls trigger trees of further entirely redundant calls. This can be seen more clearly by drawing out the call tree for fib(5), as we do below:

We can see that the entirety of the fib(3) subtree is duplicated below the fib(4) subtree, and the fib(2) subtree occurs three times. As n increases, the size of the redundant call subtrees increases dramatically, while the number of unique calls only grows linearly. Ideally we would only perform the calculations required, giving us 0(n)0(n) running time.

This is a good time to consider that the “top down” approach of recursive problem solving has a counterpart, the unsurprising “bottom up” approach of dynamic programming.

Applying dynamic programming to this problem, we ask ourselves “starting with 0 and 1, how do we build up an answer to f(n)f(n)?” If n were 0 or 1, we could answer immediately. If it were 2, however, we would need to add 0 and 1 to determine the answer: 1. If n were 3, we would need to add the answer to f(2)f(2), that we just determined, to the 1 that we previously determined, giving a total of 2. Following this strategy, we obtain 3, 5, 8, etc., until we have reached the answer for our n.

At any point in time we only need to retain a memory of the previous two calculations, and we never obtain the same sum twice.

An implementation of this strategy might look like:

def fib(n):
    a, b = 0, 1
    for _ in range(n):
        a, b = a + b, a
    return a
const fib = (n) => {
  let a = 0
  let b = 1
  for (let i = 0; i < n; i++) {
    let temp = a
    a = a + b
    b = temp
  return a

With this implementation, we sacrifice some of the elegance and readability of our recursive solution, but gain a much better O(n)O(n) running time and O(1)O(1) space cost.

Let’s now consider a problem where both the recursive and dynamic programming approaches require a little more work to discover.

Given a lattice of height H and width W, how many unique shortest paths exist from the top left corner to bottom right corner?

For instance, consider the lattice of height and width 2:

We can see that the shortest path from top left to bottom right will be of length 4, and that there are 6 unique paths of length 4:

Exploring this problem, we realize that any shortest path must always progress down and to the right—any path that progresses up or left at any point has no chance of being “shortest”. Exploring further, we note that by grouping those paths that start with a right step (shown on the top line above) and those that start with a down step (shown on the bottom line below), we can break this traversal problem into subproblems.

Specifically, the total number of paths along a H×WH \times W lattice is the sum of those along a (H1)×W(H - 1) \times W lattice and a H×(W1)H \times (W - 1) lattice.

Continuing with our 2×22 \times 2 example, the paths starting with a right step lead to the 1×21 \times 2 subproblem with these three solutions:

Looking closely, we see this as a repetition of the right hand portion of our first three solutions to the 2×22 \times 2 problem.

Similarly, the paths starting downward on our 2×22 \times 2 problem lead to the 2×12 \times 1 subproblem with the following three solutions:

This time by looking closely we see that this is a repetition of the bottom portion of our final three solutions to the 2×22 \times 2 problem.

We can say then with some confidence that the total number of paths along a H×WH \times W lattice is the sum of those along a (H1)×W(H - 1) \times W lattice and a H×(W1)H \times (W - 1) lattice. Thus by taking a top down approach to explore the recursive nature of the problem, we’ve identified a recursive relationship: f(h, w) = f(h, w - 1) + f(w, h - 1).

Before we can write a recursive solution to the problem, we must also recognize the base case: when the h or w of our subproblem is 0, we are dealing with a straight line, so the number of paths is simply 1.

Putting our base case and general case together, we obtain a succinct recursive solution:

def num_paths(height, width):
    if height == 0 or width == 0:
        return 1
    return num_paths(height, width - 1) + num_paths(height - 1, width)
const numPaths = (height, width) => {
  if (height === 0 || width === 0) return 1
  return numPaths(height, width - 1) + numPaths(height - 1, width)

Unfortunately, we find ourselves with another O(2n)O(2^n) solution (where n=H+Wn = H + W) due to redundant calls in our overlapping subproblems. For instance, calculating f(3, 2) involves calculating f(2, 2) and f(3, 1), but then in calculating f(2, 3) we redundantly call f(2, 2) once more.

Consider the call tree of num_paths(2, 2)numPaths(2, 2) to convince yourself that the running time is O(2n)O(2^n):

Once again this should be a signal to us that we may be able to find a faster solution by going bottom up, computing and storing the answer to subproblems before we address the core problem.

After some exploration of the problem, you may come to recognize that in order to calculate the number of unique paths to any point on the lattice, we must solve the subproblems of the number of paths to each of the points to the left and above the point in question. In turn, those subproblems can be solved if we first answer the subproblems of the number of paths to each of the points to the left and above those points. The logical conclusion is that if we start with the top leftmost point (which we can say can be reached in only 1 way) we can then iterate through every point of the lattice, row by row, and calculate the paths to that point as the sum of the paths to the points above and to the left, which by the nature of our iteration we have already calculated precisely one time.

We can use a list of lists to store our computed values as we proceed. For a H×WH \times W lattice, we can use a (H+1)×(W+1)(H + 1) \times (W + 1) list of lists, with each entry representing the number of paths that one may take to arrive at that vertex. We may initialize the values to 1, as we know that there is only one way to arrive at a vertex on the top or left edges. Then, iterating through each entry of each row, we can determine the number of paths to that vertex by adding the number of paths to the vertexes directly above and to the left. Finally, we access the value computed in the last entry of the last row of our memo, which represents the number of paths to traverse the entire lattice.

For instance, this is the memo that we will generate in the process of computing f(2, 2) using this strategy:

    [1, 1, 1],
    [1, 2, 3],
    [1, 3, 6]

Again we arrive at our answer 6.

This is what the memo looks like for f(10, 10):

    [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
    [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
    [1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66],
    [1, 4, 10, 20, 35, 56, 84, 120, 165, 220, 286],
    [1, 5, 15, 35, 70, 126, 210, 330, 495, 715, 1001],
    [1, 6, 21, 56, 126, 252, 462, 792, 1287, 2002, 3003],
    [1, 7, 28, 84, 210, 462, 924, 1716, 3003, 5005, 8008],
    [1, 8, 36, 120, 330, 792, 1716, 3432, 6435, 11440, 19448],
    [1, 9, 45, 165, 495, 1287, 3003, 6435, 12870, 24310, 43758],
    [1, 10, 55, 220, 715, 2002, 5005, 11440, 24310, 48620, 92378],
    [1, 11, 66, 286, 1001, 3003, 8008, 19448, 43758, 92378, 184756]

Below is a possible implementation of the dynamic programming strategy we have discussed.

def num_paths_dp(height, width):
    memo = [[1] * (width + 1) for _ in range(0, height + 1)]
    for i, row in enumerate(memo):
        for j, _ in enumerate(row):
            if i == 0 or j == 0:
            memo[i][j] = memo[i - 1][j] + memo[i][j - 1]
    return memo[-1][-1]
const numPathsDp = (height, width) {
  const memo = Array.from(Array(height + 1)).map(
    () => Array.from(Array(width + 1)).map(() => 1)
  for (let i = 1; i < memo.length; i++) {
    const row = memo[i]
    for (let j = 1; j < row.length; j++) {
      memo[i][j] = memo[i - 1][j] + memo[i][j - 1]
  return memo[height][width]

Both the time and space cost for this implementation are O(H×W)O(H \times W), compared to O(2H+W)O(2^{H + W}) time and O(H+W)O(H + W) space previously, making a big time difference as HH and WW increase.

If space is of particular concern, the space cost could be decreased to O(W)O(W) by retaining a memo of the prior row only. This is left as an exercise for the reader.

Practical Algorithms and Data Structures