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In computer science, dynamic programming is a method of solving problems exhibiting the properties of overlapping subproblems and optimal substructure (described below) that takes much less time than naive methods.
The term was originally used in the 1940s by Richard Bellman to describe the process of solving problems where one needs to find the best decisions one after another. By 1953, he had refined this to the modern meaning. The field was founded as a systems analysis and engineering topic which is recognized by the IEEE. Bellman's contribution is remembered in the name of the Bellman equation, a key result.
The word "programming" in "dynamic programming" has no particular connection to computer programming at all. A program is, instead, the plan for action that is produced. For instance, a finalized schedule of events at an exhibition is sometimes called a program. Programming, in this sense, is finding an acceptable plan of action.
Optimal substructure means that optimal solutions of subproblems can be used to find the optimal solutions of the overall problem. For example, the shortest path to a goal from a vertex in a graph can be found by first computing the shortest path to the goal from all adjacent vertices, and then using this to pick the best overall path, as shown in Figure 1. In general, we can solve a problem with optimal substructure using a three-step process:
To say that a problem has overlapping subproblems is to say that the same subproblems are used to solve many different larger problems. For example, in the Fibonacci sequence, F<sub>3</sub> = F<sub>1</sub> + F<sub>2</sub> and F<sub>4</sub> = F<sub>2</sub> + F<sub>3</sub> — computing each number involves computing F<sub>2</sub>. Because both F<sub>3</sub> and F<sub>4</sub> are needed to compute F<sub>5</sub>, a naïve approach to computing F<sub>5</sub> may end up computing F<sub>2</sub> twice or more. This applies whenever overlapping subproblems are present: a naïve approach may waste time recomputing optimal solutions to subproblems it has already solved.
In order to avoid this, we instead save the solutions to problems we have already solved. Then, if we need to solve the same problem later, we can retrieve and reuse our already-computed solution. This approach is called memoization (not memorization, although this term also fits). If we are sure we won't need a particular solution anymore, we can throw it away to save space. In some cases, we can even compute the solutions to subproblems we know that we'll need in advance.
In summary, dynamic programming makes use of:
Dynamic programming usually takes one of two approaches:
Some programming languages with special extensions [1] can automatically memoize the result of a function call with a particular set of arguments, in order to speed up call-by-name evaluation (this mechanism is referred to as call-by-need). Some languages (e.g., Maple) have automatic memoization builtin. This is only possible for a function which has no side-effects, which is always true in pure functional languages but seldom true in imperative languages.
The steps for using dynamic program goes as follows:
A naive implementation of a function finding the nth member of the Fibonacci sequence, based directly on the mathematical definition:
function fib(n) if n = 0 or n = 1 return 1 else return fib(n − 1) + fib(n − 2)
Notice that if we call, say, <code>fib(5)</code>, we produce a call tree that calls the function on the same value many different times:
In particular, <code>fib(2)</code> was calculated twice from scratch. In larger examples, many more values of <code>fib</code>, or subproblems, are recalculated, leading to an exponential time algorithm.
Now, suppose we have a simple map object, m, which maps each value of <code>fib</code> 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 exponential time:
var m := map(0 → 1, 1 → 1) function fib(n) if map m does not contain key n m[n] := fib(n − 1) + fib(n − 2) return m[n]
This technique of saving values that have already been calculated is called memoization; this is the top-down approach, since we first break the problem into subproblems and then calculate and store values.
In the bottom-up approach we calculate the smaller values of <code>fib</code> first, then build larger values from them. This method also uses linear (O(n)) time since it contains a loop that repeats n − 1 times:
function fib(n) var previousFib := 0, currentFib := 1 repeat n − 1 times var newFib := previousFib + currentFib previousFib := currentFib currentFib := newFib return currentFib
In both these examples, we only calculate <code>fib(2)</code> one time, and then use it to calculate both <code>fib(4)</code> and <code>fib(3)</code>, instead of computing it every time either of them is evaluated.
Consider a checkerboard with n × n squares and a cost-function c(i, j) which returns a cost associated with square i,j (i being the row, j being the column). For instance (on a 5 × 5 checkerboard),
Thus c(1, 3) = 5
Let us say you had a checker that could start at any square on the first rank (i.e., row) 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 diagonally left forward, diagonally right forward, or straight forward. That is, a checker on (1,3) can move to (2,2), (2,3) or (2,4).
This problem exhibits optimal substructure. That is, the solution to the entire problem relies on solutions to subproblems. Let us define a function q(i, j) as
q(i, j) = the minimum cost to reach square (i, j)
If we can find the values of this function for all the squares at rank n, we pick the minimum and follow that path backwards to get the shortest path.
It is easy to see that q(i, j) is equal to the minimum cost to get to any of the three squares below it (since those are the only squares that can reach it) plus c(i, j). For instance:
Now, let us define q(i, j) in little more general terms:
This equation is pretty straightforward. The first line is simply there to make the recursive property simpler (when dealing with the edges, so we need only 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 straightforward recursive code for q(i, j). In the following pseudocode, n is the size of the board, <code>c(i, j)</code> is the cost-function, and <code>min()</code> returns the minimum of a number of values:
function minCost(i, j) if j < 1 or j > n return infinity else if i = 1 return c(i, j) else return min( minCost(i-1, j-1), minCost(i-1, j), minCost(i-1, j+1) ) + c(i, j)
It should be noted that this function just computes the path-cost, not the actual path. We will get to the path soon. This, like the Fibonacci-numbers example, is horribly slow since it spends mountains of time recomputing the same shortest paths over and over. However, we can compute it much faster in a bottom up-fashion if we use a two-dimensional array <code>q[i, j]</code> 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 the actual path is. The path problem we can solve using another array <code>p[i, j]</code>, a predecessor array. This array basically says where paths come from. Consider the following code:
function computeShortestPathArrays() for x from 1 to n q[1, x] := c(1, x) for y from 1 to n q[y, 0] := infinity q[y, n + 1] := infinity for y from 2 to n for x from 1 to n m := min(q[y-1, x-1], q[y-1, x], q[y-1, x+1]) q[y, x] := m + c(y, x) if m = q[y-1, x-1] p[y, x] := -1 else if m = q[y-1, x] p[y, x] := 0 else p[y, x] := 1
Now the rest is a simple matter of finding the minimum and printing it.
function computeShortestPath() computeShortestPathArrays() minIndex := 1 min := q[n, 1] for i from 2 to n if q[n, i] < min minIndex := i min := q[n, i] printPath(n, minIndex)
function printPath(y, x) print(x) print("<-") if y = 2 print(x + p[y, x]) else printPath(y-1, x + p[y, x])