In mathematics and computer science, dynamic programming is a method of solving complex problems by breaking them down into simpler steps. It is applicable to problems that exhibit the properties of overlapping subproblems which are only slightly smaller^{[1]} and optimal substructure (described below). When applicable, the method takes much less time than naive methods.
Topdown dynamic programming simply means storing the results of certain calculations, which are then reused later because the same calculation is a subproblem in a larger calculation. Bottomup dynamic programming involves formulating a complex calculation as a recursive series of simpler calculations.
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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, which refers specifically to nesting smaller decision problems inside larger decisions,^{[2]} and the field was thereafter recognized by the IEEE as a systems analysis and engineering topic. Bellman's contribution is remembered in the name of the Bellman equation, a central result of dynamic programming which restates an optimization problem in recursive form.
Originally the word "programming" in "dynamic programming" had no connection to computer programming, and instead came from the term "mathematical programming"^{[3]}  a synonym for optimization. However, nowadays many optimization problems are best solved by writing a computer program that implements a dynamic programming algorithm, rather than carrying out hundreds of tedious calculations by hand. Some of the examples given below are illustrated using computer programs.
Dynamic programming is both a mathematical optimization method, and a computer programming method. In both contexts, it refers to simplifying a complicated problem by breaking it down into simpler subproblems in a recursive manner. While some decision problems cannot be taken apart this way, decisions that span several points in time do often break apart recursively; Bellman called this the "Principle of Optimality". Likewise, in computer science, a problem which can be broken down recursively is said to have optimal substructure.
If subproblems can be nested recursively inside larger problems, so that dynamic programming methods are applicable, then there is a relation between the value of the larger problem and the values of the subproblems.^{[4]} In the optimization literature this relationship is called the Bellman equation.
In terms of mathematical optimization, dynamic programming usually refers to a simplification of a decision by breaking it down into a sequence of decision steps over time. This is done by defining a sequence of value functions V_{1} , V_{2} , ... V_{n} , with an argument y representing the state of the system at times i from 1 to n. The definition of V_{n}(y) is the value obtained in state y at the last time n. The values V_{i} at earlier times i=n1,n2,...,2,1 can be found by working backwards, using a recursive relationship called the Bellman equation. For i=2,...n, V_{i 1} at any state y is calculated from V_{i} by maximizing a simple function (usually the sum) of the gain from decision i1 and the function V_{i} at the new state of the system if this decision is made. Since V_{i} has already been calculated, for the needed states, the above operation yields V_{i 1} for all the needed states. Finally, V_{1} at the initial state of the system is the value of the optimal solution. The optimal values of the decision variables can be recovered, one by one, by tracking back the calculations already performed.
There are two key attributes that a problem must have in order for dynamic programming to be applicable: optimal substructure and overlapping subproblems which are only slightly smaller. When the overlapping problems are, say, half the size of the original problem the strategy is called "divide and conquer" rather than "dynamic programming". This is why merge sort, and quick sort, and finding all matches of a regular expression are not classified as dynamic programming problems.
Optimal substructure means that the solution to a given optimization problem can be obtained by the combination of optimal solutions to its subproblems. Consequently, the first step towards devising a dynamic programming solution is to check whether the problem exhibits such optimal substructure. Such optimal substructures are usually described by means of recursion. For example, given a graph G=(V,E), the shortest path p from a vertex u to a vertex v exhibits optimal substructure: take any intermediate vertex w on this shortest path p. If p is truly the shortest path, then the path p_{1} from u to w and p_{2} from w to v are indeed the shortest paths between the corresponding vertices (by the simple cutandpaste argument described in CLRS). Hence, one can easily formulate the solution for finding shortest paths in a recursive manner, which is what the BellmanFord algorithm does.
Overlapping subproblems means that the space of subproblems must be small, that is, any recursive algorithm solving the problem should solve the same subproblems over and over, rather than generating new subproblems. For example, consider the recursive formulation for generating the Fibonacci series: F_{i} = F_{i1} + F_{i2}, with base case F_{1}=F_{2}=1. Then F_{43} = F_{42} + F_{41}, and F_{42} = F_{41} + F_{40}. Now F_{41} is being solved in the recursive subtrees of both F_{43} as well as F_{42}. Even though the total number of subproblems is actually small (only 43 of them), we end up solving the same problems over and over if we adopt a naive recursive solution such as this. Dynamic programming takes account of this fact and solves each subproblem only once. Note that the subproblems must be only 'slightly' smaller (typically taken to mean a constant additive factor) than the larger problem; when they are a multiplicative factor smaller the problem is no longer classified as dynamic programming (otherwise mergesort and quicksort would be dynamic programming problems).
This can be achieved in either of two ways:^{[citation needed]}
Some programming languages can automatically memoize the result of a function call with a particular set of arguments, in order to speed up callbyname evaluation (this mechanism is referred to as callbyneed). Some languages make it possible portably (e.g. Scheme, Common Lisp or Perl), some need special extensions (e.g. C++, see ^{[5]}). Some languages have automatic memoization built in. In any case, this is only possible for a referentially transparent function.
A mathematical optimization problem that is often used in teaching dynamic programming to economists (because it can be solved by hand^{[6]}) concerns a consumer who lives over the periods t = 0,1,2,...,T and must decide how much to consume and how much to save in each period.
Let c_{t} be consumption in period t, and assume consumption yields utility u(c_{t}) = ln(c_{t}) as long as the consumer lives. Assume the consumer is impatient, so that he discounts future utility by a factor b each period, where 0 < b < 1. Let k_{t} be capital in period t. Assume initial capital is a given amount k_{0} > 0, and suppose that this period's capital and consumption determine next period's capital as , where A is a positive constant and 0 < a < 1. Assume capital cannot be negative. Then the consumer's decision problem can be written as follows:
Written this way, the problem looks complicated, because it involves solving for all the choice variables c_{0},c_{1},c_{2},...,c_{T} and k_{1},k_{2},k_{3},...,k_{T + 1} simultaneously. (Note that k_{0} is not a choice variable—the consumer's initial capital is taken as given.)
The dynamic programming approach to solving this problem involves breaking it apart into a sequence of smaller decisions. To do so, we define a sequence of value functions V_{t}(k), for t = 0,1,2,...,T,T + 1 which represent the value of having any amount of capital k at each time t. Note that V_{T + 1}(k) = 0, that is, there is (by assumption) no utility from having capital after death.
The value of any quantity of capital at any previous time can be calculated by backward induction using the Bellman equation. In this problem, for each t = 0,1,2,...,T, the Bellman equation is
This problem is much simpler than the one we wrote down before, because it involves only two decision variables, c_{t} and k_{t + 1}. Intuitively, instead of choosing his whole lifetime plan at birth, the consumer can take things one step at a time. At time t, his current capital k_{t} is given, and he only needs to choose current consumption c_{t} and saving k_{t + 1}.
To actually solve this problem, we work backwards. For simplicity, the current level of capital is denoted as k. V_{T + 1}(k) is already known, so using the Bellman equation once we can calculate V_{T}(k), and so on until we get to V_{0}(k), which is the value of the initial decision problem for the whole lifetime. In other words, once we know V_{T − j + 1}(k), we can calculate V_{T − j}(k), which is the maximum of ln(c_{T − j}) + bV_{T − j + 1}(Ak^{a} − c_{T − j}), where c_{T − j} is the variable and . It can be shown that the value function at time t = T − j is
where each v_{T − j} is a constant, and the optimal amount to consume at time t = T − j is
which can be simplified to
We see that it is optimal to consume a larger fraction of current wealth as one gets older, finally consuming all current wealth in period T, the last period of life.
Here is 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 return 0 if n = 1 return 1 return fib(n − 1) + fib(n − 2)
Notice that if we call, say, fib(5)
, we produce a call tree that calls the function on the same value many different times:
fib(5)
fib(4) + fib(3)
(fib(3) + fib(2)) + (fib(2) + fib(1))
((fib(2) + fib(1)) + (fib(1) + fib(0))) + ((fib(1) + fib(0)) + fib(1))
(((fib(1) + fib(0)) + fib(1)) + (fib(1) + fib(0))) + ((fib(1) + fib(0)) + fib(1))
In particular, fib(2)
was calculated three times from scratch. In larger examples, many more values of fib
, or subproblems, are recalculated, leading to an exponential time 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 exponential time:
var m := map(0 → 0, 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 topdown approach, since we first break the problem into subproblems and then calculate and store values.
In the bottomup approach we calculate the smaller values of fib
first, then build larger values from them. This method also uses O(n) time since it contains a loop that repeats n − 1 times, however it only takes constant (O(1)) space, in contrast to the topdown approach which requires O(n) space to store the map.
function fib(n) var previousFib := 0, currentFib := 1 if n = 0 return 0 else if n = 1 return 1 repeat n − 1 times var newFib := previousFib + currentFib previousFib := currentFib currentFib := newFib return currentFib
In both these examples, we only calculate fib(2)
one time, and then use it to calculate both fib(4)
and fib(3)
, instead of computing it every time either of them is evaluated.
Consider the problem of assigning values, either zero or one, to the positions of an n x n matrix, n even, so that each row and each column contains exactly n / 2 zeros and n / 2 ones. For example, when n = 4, three possible solutions are:
+     + +     + +     +  0 1 0 1   0 0 1 1   1 1 0 0   1 0 1 0  and  0 0 1 1  and  0 0 1 1   0 1 0 1   1 1 0 0   1 1 0 0   1 0 1 0   1 1 0 0   0 0 1 1  +     + +     + +     +
We ask how many different assignments there are for a given n. There are at least three possible approaches: brute force, backtracking, and dynamic programming. Brute force consists of checking all assignments of zeros and ones and counting those that have balanced rows and columns (n / 2 zeros and n / 2 ones). As there are possible assignments, this strategy is not practical except maybe up to n = 6. Backtracking for this problem consists of choosing some order of the matrix elements and recursively placing ones or zeros, while checking that in every row and column the number of elements that have not been assigned plus the number of ones or zeros are both at least n / 2. While more sophisticated than brute force, this approach will visit every solution once, making it impractical for n larger than six, since the number of solutions is already 116963796250 for n = 8, as we shall see. Dynamic programming makes it possible to count the number of solutions without visiting them all.
We consider boards, where whose k rows contain n / 2 zeros and n / 2 ones. The function f to which memoization is applied maps vectors of n pairs of integers to the number of admissible boards (solutions). There is one pair for each column and its two components indicate respectively the number of ones and zeros that have yet to be placed in that column. We seek the value of (n arguments or one vector of n elements). The process of subproblem creation involves iterating over every one of possible assignments for the top row of the board, and going through every column, subtracting one from the appropriate element of the pair for that column, depending on whether the assignment for the top row contained a zero or a one at that position. If any one of the results is negative, then the assignment is invalid and does not contribute to the set of solutions (recursion stops). Otherwise, we have an assignment for the top row of the board and recursively compute the number of solutions to the remaining board, adding the numbers of solutions for every admissible assignment of the top row and returning the sum, which is being memoized. The base case is the trivial subproblem, which occurs for a board. The number of solutions for this board is either zero or one, depending on whether the vector is a permutation of n / 2 (0,1) and n / 2 (1,0) pairs or not.
For example, in the two boards shown above the sequences of vectors would be
((2, 2) (2, 2) (2, 2) (2, 2)) ((2, 2) (2, 2) (2, 2) (2, 2)) k = 4 0 1 0 1 0 0 1 1 ((1, 2) (2, 1) (1, 2) (2, 1)) ((1, 2) (1, 2) (2, 1) (2, 1)) k = 3 1 0 1 0 0 0 1 1 ((1, 1) (1, 1) (1, 1) (1, 1)) ((0, 2) (0, 2) (2, 0) (2, 0)) k = 2 0 1 0 1 1 1 0 0 ((0, 1) (1, 0) (0, 1) (1, 0)) ((0, 1) (0, 1) (1, 0) (1, 0)) k = 1 1 0 1 0 1 1 0 0 ((0, 0) (0, 0) (0, 0) (0, 0)) ((0, 0) (0, 0), (0, 0) (0, 0))
The number of solutions (sequence A058527 in OEIS) is
Links to the Perl source of the backtracking approach, as well as a MAPLE and a C implementation of the dynamic programming approach may be found among the external links.
Consider a checkerboard with n × n squares and a costfunction 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),
5  6  7  4  7  8 

4  7  6  1  1  4 
3  3  5  7  8  2 
2    6  7  0   
1      5*     
1  2  3  4  5 
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).
5  

4  
3  
2  x  x  x  
1  o  
1  2  3  4  5 
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
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.
Note 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:
5  

4  A  
3  B  C  D  
2  
1  
1  2  3  4  5 
q(A) = min(q(B),q(C),q(D)) + c(A)
Now, let us define q(i, j) in somewhat more general terms:
The first line of this equation is 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 last rank, to provide a base case. The third line, the recursion, is the important part. It is similar to 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, c(i, j)
is the costfunction, and min()
returns the minimum of a number of values:
function minCost(i, j) if j < 1 or j > n return infinity else if i = 5 return c(i, j) else return min( minCost(i+1, j1), minCost(i+1, j), minCost(i+1, j+1) ) + c(i, j)
It should be noted that this function only computes the pathcost, not the actual path. We will get to the path soon. This, like the Fibonaccinumbers 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 bottomup fashion if we store pathcosts in a twodimensional array q[i, j]
rather than using a function. This avoids recomputation; before computing the cost of a path, we check the array q[i, j]
to see if the path cost is already there.
We also need to know what the actual shortest path is. To do this, we use another array p[i, j]
, a predecessor array. This array implicitly stores the path to any square s by storing the previous node on the shortest path to s, i.e. the predecessor. To reconstruct the path, we lookup the predecessor of s, then the predecessor of that square, then the predecessor of that square, and so on, until we reach the starting square. 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[y1, x1], q[y1, x], q[y1, x+1]) q[y, x] := m + c(y, x) if m = q[y1, x1] p[y, x] := 1 else if m = q[y1, 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(y1, x + p[y, x])
In genetics, sequence alignment is an important application where dynamic programming is essential.^{[7]} Typically, the problem consists of transforming one sequence into another using edit operations that replace, insert, or remove an element. Each operation has an associated cost, and the goal is to find the sequence of edits with the lowest total cost.
The problem can be stated naturally as a recursion, a sequence A is optimally edited into a sequence B by either:
The partial alignments can be tabulated in a matrix, where cell (i,j) contains the cost of the optimal alignment of A[1..i] to B[1..j]. The cost in cell (i,j) can be calculated by adding the cost of the relevant operations to the cost of its neighboring cells, and selecting the optimum.
Different variants exist, see SmithWaterman and NeedlemanWunsch.
This article may require cleanup to meet Wikipedia's quality standards. Please improve this article if you can. The talk page may contain suggestions. (January 2010) 
In mathematics and computer science, dynamic programming is a method for solving complex problems by breaking them down into simpler steps. It is applicable to problems exhibiting the properties of overlapping subproblems which are only slightly smaller^{[1]} and optimal substructure (described below). When applicable, the method takes far less time than naive methods.
Topdown dynamic programming simply means storing the results of certain calculations, which are later used again since the completed calculation is a subproblem of a larger calculation. Bottomup dynamic programming involves formulating a complex calculation as a recursive series of simpler calculations.
Contents 
The term dynamic programming 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 refined this to the modern meaning, referring specifically to nesting smaller decision problems inside larger decisions,^{[2]} and the field was thereafter recognized by the IEEE as a systems analysis and engineering topic. Bellman's contribution is remembered in the name of the Bellman equation, a central result of dynamic programming which restates an optimization problem in recursive form.
The word dynamic was chosen by Bellman because it sounded impressive, not because it described how the method worked.^{[3]} The word programming referred to the use of the method to find an optimal program, in the sense of a military schedule for training or logistics. This usage is the same as that in the phrases linear programming and mathematical programming, a synonym for optimization.^{[4]}
Dynamic programming is both a mathematical optimization method and a computer programming method. In both contexts it refers to simplifying a complicated problem by breaking it down into simpler subproblems in a recursive manner. While some decision problems cannot be taken apart this way, decisions that span several points in time do often break apart recursively; Bellman called this the "Principle of Optimality". Likewise, in computer science, a problem which can be broken down recursively is said to have optimal substructure.
If subproblems can be nested recursively inside larger problems, so that dynamic programming methods are applicable, then there is a relation between the value of the larger problem and the values of the subproblems.^{[5]} In the optimization literature this relationship is called the Bellman equation.
In terms of mathematical optimization, dynamic programming usually refers to a simplification of a decision by breaking it down into a sequence of decision steps over time. This is done by defining a sequence of value functions V_{1} , V_{2} , ... V_{n} , with an argument y representing the state of the system at times i from 1 to n. The definition of V_{n}(y) is the value obtained in state y at the last time n. The values V_{i} at earlier times i=n1,n2,...,2,1 can be found by working backwards, using a recursive relationship called the Bellman equation. For i=2,...n, V_{i 1} at any state y is calculated from V_{i} by maximizing a simple function (usually the sum) of the gain from decision i1 and the function V_{i} at the new state of the system if this decision is made. Since V_{i} has already been calculated for the needed states, the above operation yields V_{i 1} for those states. Finally, V_{1} at the initial state of the system is the value of the optimal solution. The optimal values of the decision variables can be recovered, one by one, by tracking back the calculations already performed.
There are two key attributes that a problem must have in order for dynamic programming to be applicable: optimal substructure and overlapping subproblems which are only slightly smaller. When the overlapping problems are, say, half the size of the original problem the strategy is called "divide and conquer" rather than "dynamic programming". This is why mergesort, quicksort, and finding all matches of a regular expression are not classified as dynamic programming problems.
Optimal substructure means that the solution to a given optimization problem can be obtained by the combination of optimal solutions to its subproblems. Consequently, the first step towards devising a dynamic programming solution is to check whether the problem exhibits such optimal substructure. Such optimal substructures are usually described by means of recursion. For example, given a graph G=(V,E), the shortest path p from a vertex u to a vertex v exhibits optimal substructure: take any intermediate vertex w on this shortest path p. If p is truly the shortest path, then the path p_{1} from u to w and p_{2} from w to v are indeed the shortest paths between the corresponding vertices (by the simple cutandpaste argument described in CLRS). Hence, one can easily formulate the solution for finding shortest paths in a recursive manner, which is what the BellmanFord algorithm does.
Overlapping subproblems means that the space of subproblems must be small, that is, any recursive algorithm solving the problem should solve the same subproblems over and over, rather than generating new subproblems. For example, consider the recursive formulation for generating the Fibonacci series: F_{i} = F_{i1} + F_{i2}, with base case F_{1}=F_{2}=1. Then F_{43} = F_{42} + F_{41}, and F_{42} = F_{41} + F_{40}. Now F_{41} is being solved in the recursive subtrees of both F_{43} as well as F_{42}. Even though the total number of subproblems is actually small (only 43 of them), we end up solving the same problems over and over if we adopt a naive recursive solution such as this. Dynamic programming takes account of this fact and solves each subproblem only once. Note that the subproblems must be only 'slightly' smaller (typically taken to mean a constant additive factor^{[citation needed]}) than the larger problem; when they are a multiplicative factor smaller the problem is no longer classified as dynamic programming.
indicates overlapping subproblems.]]
This can be achieved in either of two ways:^{[citation needed]}
Some programming languages can automatically memoize the result of a function call with a particular set of arguments, in order to speed up callbyname evaluation (this mechanism is referred to as callbyneed). Some languages make it possible portably (e.g. Scheme, Common Lisp or Perl), some need special extensions (e.g. C++, see ^{[6]}). Some languages have automatic memoization built in such as tabled Prolog. In any case, this is only possible for a referentially transparent function.
A mathematical optimization problem that is often used in teaching dynamic programming to economists (because it can be solved by hand^{[7]}) concerns a consumer who lives over the periods $t=0,1,2,...,T$ and must decide how much to consume and how much to save in each period.
Let $c\_t$ be consumption in period $t$, and assume consumption yields utility $u(c\_t)=\backslash ln(c\_t)$ as long as the consumer lives. Assume the consumer is impatient, so that he discounts future utility by a factor $b$ each period, where $01\; math>.\; Let$ k\_t$becapitalin\; period$ t$.\; Assume\; initial\; capital\; is\; a\; given\; amount$ k\_00$,\; and\; suppose\; that\; this\; period\text{'}s\; capital\; and\; consumption\; determine\; next\; period\text{'}s\; capital\; as$ k\_\{t+1\}=Ak^a\_t\; \; c\_t$,\; where$ A$is\; a\; positive\; constant\; and$ 01\; math>.\; Assume\; capital\; cannot\; be\; negative.\; Then\; the\; consumer\text{'}s\; decision\; problem\; can\; be\; written\; as\; follows:$1>$
Written this way, the problem looks complicated, because it involves solving for all the choice variables $c\_0,\; c\_1,\; c\_2,\; ...\; ,\; c\_T$ and $k\_1,\; k\_2,\; k\_3,\; ...\; ,\; k\_\{T+1\}$ simultaneously. (Note that $k\_0$ is not a choice variable—the consumer's initial capital is taken as given.)
The dynamic programming approach to solving this problem involves breaking it apart into a sequence of smaller decisions. To do so, we define a sequence of value functions $V\_t(k)$, for $t=0,1,2,...,T,T+1$ which represent the value of having any amount of capital $k$ at each time $t$. Note that $V\_\{T+1\}(k)=0$, that is, there is (by assumption) no utility from having capital after death.
The value of any quantity of capital at any previous time can be calculated by backward induction using the Bellman equation. In this problem, for each $t=0,1,2,...,T$, the Bellman equation is
This problem is much simpler than the one we wrote down before, because it involves only two decision variables, $c\_t$ and $k\_\{t+1\}$. Intuitively, instead of choosing his whole lifetime plan at birth, the consumer can take things one step at a time. At time $t$, his current capital $k\_t$ is given, and he only needs to choose current consumption $c\_t$ and saving $k\_\{t+1\}$.
To actually solve this problem, we work backwards. For simplicity, the current level of capital is denoted as $k$. $V\_\{T+1\}(k)$ is already known, so using the Bellman equation once we can calculate $V\_T(k)$, and so on until we get to $V\_0(k)$, which is the value of the initial decision problem for the whole lifetime. In other words, once we know $V\_\{Tj+1\}(k)$, we can calculate $V\_\{Tj\}(k)$, which is the maximum of $\backslash ln(c\_\{Tj\})\; +\; b\; V\_\{Tj+1\}(Ak^ac\_\{Tj\})$, where $c\_\{Tj\}$ is the choice variable and $Ak^ac\_\{Tj\}\; \backslash ge\; 0$.
Working backwards, it can be shown that the value function at time $t=Tj$ is
where each $v\_\{Tj\}$ is a constant, and the optimal amount to consume at time $t=Tj$ is
which can be simplified to
We see that it is optimal to consume a larger fraction of current wealth as one gets older, finally consuming all remaining wealth in period $T$, the last period of life.
From a dynamic programming point of view, Dijkstra's algorithm for the shortest path problem is a successive approximation scheme that solves the dynamic programming functional equation for the shortest path problem by the Reaching method.^{[8]}^{[9]} ^{[10]}
In fact, Dijkstra's explanation of the logic behind the algorithm,^{[11]} namely
Problem 2. Find the path of minimum total length between two given nodes $P$ and $Q$.
We use the fact that, if $R$ is a node on the minimal path from $P$ to $Q$, knowledge of the latter implies the knowledge of the minimal path from $P$ to $R$.
is a paraphrasing of Bellman's famous Principle of Optimality in the context of the shortest path problem.
Here is 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 return 0 if n = 1 return 1 return fib(n − 1) + fib(n − 2)
Notice that if we call, say, fib(5)
, we produce a call tree that calls the function on the same value many different times:
fib(5)
fib(4) + fib(3)
(fib(3) + fib(2)) + (fib(2) + fib(1))
((fib(2) + fib(1)) + (fib(1) + fib(0))) + ((fib(1) + fib(0)) + fib(1))
(((fib(1) + fib(0)) + fib(1)) + (fib(1) + fib(0))) + ((fib(1) + fib(0)) + fib(1))
In particular, fib(2)
was calculated three times from scratch. In larger examples, many more values of fib
, or subproblems, are recalculated, leading to an exponential time 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 exponential time:
var m := map(0 → 0, 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 topdown approach, since we first break the problem into subproblems and then calculate and store values.
In the bottomup approach we calculate the smaller values of fib
first, then build larger values from them. This method also uses O(n) time since it contains a loop that repeats n − 1 times, however it only takes constant (O(1)) space, in contrast to the topdown approach which requires O(n) space to store the map.
function fib(n) var previousFib := 0, currentFib := 1 if n = 0 return 0 else if n = 1 return 1 repeat n − 1 times var newFib := previousFib + currentFib previousFib := currentFib currentFib := newFib return currentFib
In both these examples, we only calculate fib(2)
one time, and then use it to calculate both fib(4)
and fib(3)
, instead of computing it every time either of them is evaluated. (Note the calculation of the Fibonacci sequence is used to demonstrate dynamic programming. A O(1) formula exists from which an arbitrary term can be calculated, which is more efficient than any dynamic programming technique.)
Consider the problem of assigning values, either zero or one, to the positions of an n × n matrix, n even, so that each row and each column contains exactly n / 2 zeros and n / 2 ones. For example, when n = 4, three possible solutions are
0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \end{bmatrix} \text{ and } \begin{bmatrix} 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 \end{bmatrix} \text{ and } \begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \end{bmatrix}.
We ask how many different assignments there are for a given $n$. There are at least three possible approaches: brute force, backtracking, and dynamic programming. Brute force consists of checking all assignments of zeros and ones and counting those that have balanced rows and columns ($n/2$ zeros and $n/2$ ones). As there are $\backslash tbinom\{n\}\{n/2\}^n$ possible assignments, this strategy is not practical except maybe up to $n=6$. Backtracking for this problem consists of choosing some order of the matrix elements and recursively placing ones or zeros, while checking that in every row and column the number of elements that have not been assigned plus the number of ones or zeros are both at least n / 2. While more sophisticated than brute force, this approach will visit every solution once, making it impractical for n larger than six, since the number of solutions is already 116963796250 for n = 8, as we shall see. Dynamic programming makes it possible to count the number of solutions without visiting them all.
We consider k × n boards, where 1 ≤ k ≤ n, whose $k$ rows contain $n/2$ zeros and $n/2$ ones. The function f to which memoization is applied maps vectors of n pairs of integers to the number of admissible boards (solutions). There is one pair for each column and its two components indicate respectively the number of ones and zeros that have yet to be placed in that column. We seek the value of $f((n/2,\; n/2),\; (n/2,\; n/2),\; \backslash ldots\; (n/2,\; n/2))$ ($n$ arguments or one vector of $n$ elements). The process of subproblem creation involves iterating over every one of $\backslash tbinom\{n\}\{n/2\}$ possible assignments for the top row of the board, and going through every column, subtracting one from the appropriate element of the pair for that column, depending on whether the assignment for the top row contained a zero or a one at that position. If any one of the results is negative, then the assignment is invalid and does not contribute to the set of solutions (recursion stops). Otherwise, we have an assignment for the top row of the k × n board and recursively compute the number of solutions to the remaining (k  1) × n board, adding the numbers of solutions for every admissible assignment of the top row and returning the sum, which is being memoized. The base case is the trivial subproblem, which occurs for a 1 × n board. The number of solutions for this board is either zero or one, depending on whether the vector is a permutation of n / 2 $(0,\; 1)$ and n / 2 $(1,\; 0)$ pairs or not.
For example, in the two boards shown above the sequences of vectors would be
((2, 2) (2, 2) (2, 2) (2, 2)) ((2, 2) (2, 2) (2, 2) (2, 2)) k = 4 0 1 0 1 0 0 1 1 ((1, 2) (2, 1) (1, 2) (2, 1)) ((1, 2) (1, 2) (2, 1) (2, 1)) k = 3 1 0 1 0 0 0 1 1 ((1, 1) (1, 1) (1, 1) (1, 1)) ((0, 2) (0, 2) (2, 0) (2, 0)) k = 2 0 1 0 1 1 1 0 0 ((0, 1) (1, 0) (0, 1) (1, 0)) ((0, 1) (0, 1) (1, 0) (1, 0)) k = 1 1 0 1 0 1 1 0 0 ((0, 0) (0, 0) (0, 0) (0, 0)) ((0, 0) (0, 0), (0, 0) (0, 0))
The number of solutions (sequence A058527 in OEIS) is
Links to the Perl source of the backtracking approach, as well as a MAPLE and a C implementation of the dynamic programming approach may be found among the external links.
Consider a checkerboard with n × n squares and a costfunction 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),
5  6  7  4  7  8 

4  7  6  1  1  4 
3  3  5  7  8  2 
2    6  7  0   
1      *5*     
1  2  3  4  5 
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).
5  

4  
3  
2  x  x  x  
1  o  
1  2  3  4  5 
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
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.
Note 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:
5  

4  A  
3  B  C  D  
2  
1  
1  2  3  4  5 
$q(A)\; =\; min(q(B),q(C),q(D))+c(A)$
Now, let us define q(i, j) in somewhat more general terms:
$q(i,j)=\backslash begin\{cases\}\; \backslash infty\; \&\; j\; <\; 1\; \backslash mbox\{\; or\; \}j\; >\; n\; \backslash \backslash \; c(i,\; j)\; \&\; i\; =\; n\; \backslash \backslash \; \backslash min(q(i1,\; j1),\; q(i1,\; j),\; q(i1,\; j+1))\; +\; c(i,j)\; \&\; \backslash mbox\{otherwise.\}\backslash end\{cases\}$
The first line of this equation is 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 last rank, to provide a base case. The third line, the recursion, is the important part. It is similar to 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, c(i, j)
is the costfunction, and min()
returns the minimum of a number of values:
function minCost(i, j) if j < 1 or j > n return infinity else if i = n return c(i, j) else return min( minCost(i+1, j1), minCost(i+1, j), minCost(i+1, j+1) ) + c(i, j)
It should be noted that this function only computes the pathcost, not the actual path. We will get to the path soon. This, like the Fibonaccinumbers 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 bottomup fashion if we store pathcosts in a twodimensional array q[i, j]
rather than using a function. This avoids recomputation; before computing the cost of a path, we check the array q[i, j]
to see if the path cost is already there.
We also need to know what the actual shortest path is. To do this, we use another array p[i, j]
, a predecessor array. This array implicitly stores the path to any square s by storing the previous node on the shortest path to s, i.e. the predecessor. To reconstruct the path, we lookup the predecessor of s, then the predecessor of that square, then the predecessor of that square, and so on, until we reach the starting square. 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[y1, x1], q[y1, x], q[y1, x+1]) q[y, x] := m + c(y, x) if m = q[y1, x1] p[y, x] := 1 else if m = q[y1, 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(y1, x + p[y, x])
In genetics, sequence alignment is an important application where dynamic programming is essential.^{[3]} Typically, the problem consists of transforming one sequence into another using edit operations that replace, insert, or remove an element. Each operation has an associated cost, and the goal is to find the sequence of edits with the lowest total cost.
The problem can be stated naturally as a recursion, a sequence A is optimally edited into a sequence B by either:
The partial alignments can be tabulated in a matrix, where cell (i,j) contains the cost of the optimal alignment of A[1..i] to B[1..j]. The cost in cell (i,j) can be calculated by adding the cost of the relevant operations to the cost of its neighboring cells, and selecting the optimum.
Different variants exist, see SmithWaterman and NeedlemanWunsch.
The Tower of Hanoi or Towers of Hanoi is a mathematical game or puzzle. It consists of three rods, and a number of disks of different sizes which can slide onto any rod. The puzzle starts with the disks in a neat stack in ascending order of size on one rod, the smallest at the top, thus making a conical shape.
The objective of the puzzle is to move the entire stack to another rod, obeying the following rules:
The dynamic programming solution consists of solving the functional equation
where n denotes the number of disks to be moved, h denotes the home rod, t denotes the target rod, not(h,t) denotes the third rod (neither h nor t), ";" denotes concatenation, and
Note that for n=1 the problem is trivial, namely S(1,h,t) = "move a disk from rod h to rod t".
The number of moves required by this solution is 2^{n} 1. If the objective is to maximize the number of moves (without cycling) then the dynamic programming functional equation is slightly more complicated and 3^{n} 1 moves are required.^{[12]}
The following is a description of the instance of this famous puzzle involving n=2 eggs and a building with H=36 floors:^{[13]}
To derive a dynamic programming functional equation for this puzzle, let the state of the dynamic programming model be a pair s = (n,k), where
For instance, s = (2,6) indicates that 2 test eggs are available and 6 (consecutive) floors are yet to be tested. The initial state of the process is s = (N,H) where N denotes the number of test eggs available at the commencement of the experiment. The process terminates either when there are no more test eggs (n = 0) or when k = 0, whichever occurs first. If termination occurs at state s=(0,k) and k>0, then the test failed.
Now, let
Then it can be shown that^{[14]}
with W(n,1) = 1 for all n>0 and W(1,k) = k for all k. It is easy to solve this equation iteratively by systematically increasing the values of n and k.
An interactive online facility is available for experimentation with this model as well as with other versions of this puzzle (e.g. when the objective is to minimize the expected value of the number of trials.^{[14]}
