Dynamic programming

Dynamic programming is both a mathematical optimization method and a computer programming method. The method was developed by Richard Bellman in the 1950s and has found applications in numerous fields, from aerospace engineering to economics.
In both contexts it refers to simplifying a complicated problem by breaking it down into simpler sub-problems 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. Likewise, in computer science, if a problem can be solved optimally by breaking it into sub-problems and then recursively finding the optimal solutions to the sub-problems, then it is said to have optimal substructure.
If sub-problems 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 sub-problems. In the optimization literature this relationship is called the Bellman equation.
Overview.
Mathematical optimization.
In terms of mathematical optimization, dynamic programming usually refers to simplifying 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" taking "y" as an argument 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" = "n" −1, "n" − 2, ..., 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 a decision at time "i" − 1 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.
Control theory.
In control theory, a typical problem is to find an admissible control formula_1 which causes the system formula_2 to follow an admissible trajectory formula_3 on a continuous time interval formula_4 that minimizes a cost function
The solution to this problem is an optimal control law or policy formula_6, which produces an optimal trajectory formula_3 and a cost-to-go function formula_8. The latter obeys the fundamental equation of dynamic programming:
a partial differential equation known as the Hamilton–Jacobi–Bellman equation, in which formula_10 and formula_11. One finds the minimizing formula_12 in terms of formula_13, formula_14, and the unknown function formula_15 and then substitutes the result into the Hamilton–Jacobi–Bellman equation to get the partial differential equation to be solved with boundary condition formula_16. In practice, this generally requires numerical techniques for some discrete approximation to the exact optimization relationship.
Alternatively, the continuous process can be approximated by a discrete system, which leads to a following recurrence relation analog to the Hamilton–Jacobi–Bellman equation:
at the formula_18-th stage of formula_19 equally spaced discrete time intervals, and where formula_20 and formula_21 denote discrete approximations to formula_22 and formula_23. This functional equation is known as the Bellman equation, which can be solved for an exact solution of the discrete approximation of the optimization equation.
Example from economics: Ramsey's problem of optimal saving.
In economics, the objective is generally to maximize (rather than minimize) some dynamic social welfare function. In Ramsey's problem, this function relates amounts of consumption to levels of utility. Loosely speaking, the planner faces the trade-off between contemporaneous consumption and future consumption (via investment in capital stock that is used in production), known as intertemporal choice. Future consumption is discounted at a constant rate formula_24. A discrete approximation to the transition equation of capital is given by
where formula_26 is consumption, formula_18 is capital, and formula_22 is a production function satisfying the Inada conditions. An initial capital stock formula_29 is assumed.
Let formula_30 be consumption in period , and assume consumption yields utility formula_31 as long as the consumer lives. Assume the consumer is impatient, so that he discounts future utility by a factor each period, where formula_32 be capital in period . Assume initial capital is a given amount formula_33, and suppose that this period's capital and consumption determine next period's capital as formula_34, where is a positive constant and formula_35 subject to formula_36 for all formula_37
Written this way, the problem looks complicated, because it involves solving for all the choice variables formula_38. (The capital formula_39 is not a choice variable—the consumer's initial capital is taken as given.)
The dynamic programming approach to solve this problem involves breaking it apart into a sequence of smaller decisions. To do so, we define a sequence of "value functions" formula_40, for formula_41 which represent the value of having any amount of capital at each time . There is (by assumption) no utility from having capital after death, formula_42.
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 formula_37, the Bellman equation is
This problem is much simpler than the one we wrote down before, because it involves only two decision variables, formula_30 and formula_47. Intuitively, instead of choosing his whole lifetime plan at birth, the consumer can take things one step at a time. At time , his current capital formula_48 is given, and he only needs to choose current consumption formula_30 and saving formula_47.
To actually solve this problem, we work backwards. For simplicity, the current level of capital is denoted as . formula_51 is already known, so using the Bellman equation once we can calculate formula_52, and so on until we get to formula_53, which is the "value" of the initial decision problem for the whole lifetime. In other words, once we know formula_54, we can calculate formula_55, which is the maximum of formula_56, where formula_57 is the choice variable and formula_58.
Working backwards, it can be shown that the value function at time formula_59 is
where each formula_61 is a constant, and the optimal amount to consume at time formula_59 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 , the last period of life.
Computer programming.
There are two key attributes that a problem must have in order for dynamic programming to be applicable: optimal substructure and overlapping sub-problems. If a problem can be solved by combining optimal solutions to "non-overlapping" sub-problems, the strategy is called "divide and conquer" instead. This is why merge sort and quick sort 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 sub-problems. 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 it can be split into sub-paths "p1" from "u" to "w" and "p2" from "w" to "v" such that these, in turn, are indeed the shortest paths between the corresponding vertices (by the simple cut-and-paste argument described in "Introduction to Algorithms"). Hence, one can easily formulate the solution for finding shortest paths in a recursive manner, which is what the Bellman–Ford algorithm or the Floyd–Warshall algorithm does.
"Overlapping" sub-problems means that the space of sub-problems must be small, that is, any recursive algorithm solving the problem should solve the same sub-problems over and over, rather than generating new sub-problems. For example, consider the recursive formulation for generating the Fibonacci series: "F""i" = "F""i"−1 + "F""i"−2, 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 sub-trees of both "F"43 as well as "F"42. Even though the total number of sub-problems 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 sub-problem only once.
This can be achieved in either of two ways:
Some programming languages 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 make it possible portably (e.g. Scheme, Common Lisp, Perl or D). Some languages have automatic memoization built in, such as tabled Prolog and J, which supports memoization with the "M." adverb. In any case, this is only possible for a referentially transparent function. Memoization is also encountered as an easily accessible design pattern within term-rewrite based languages such as Wolfram Language.
Bioinformatics.
Dynamic programming is widely used in bioinformatics for the tasks such as sequence alignment, protein folding, RNA structure prediction and protein-DNA binding. The first dynamic programming algorithms for protein-DNA binding were developed in the 1970s independently by Charles DeLisi in USA and Georgii Gurskii and Alexander Zasedatelev in USSR. Recently these algorithms have become very popular in bioinformatics and computational biology, particularly in the studies of nucleosome positioning and transcription factor binding.
Examples: Computer algorithms.
Dijkstra's algorithm for the shortest path problem.
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.
In fact, Dijkstra's explanation of the logic behind the algorithm, namely
is a paraphrasing of Bellman's famous Principle of Optimality in the context of the shortest path problem.
Fibonacci sequence.
Using dynamic programming in the calculation of the "n"th member of the Fibonacci sequence improves its performance greatly. Here is a naïve implementation, based directly on the mathematical definition:
 function fib(n)
 if n <= 1 return n
 return fib(n − 1) + fib(n − 2)
Notice that if we call, say, codice_1, we produce a call tree that calls the function on the same value many different times:
In particular, codice_7 was calculated three times from scratch. In larger examples, many more values of codice_8, or "subproblems", are recalculated, leading to an exponential time algorithm.
Now, suppose we have a simple map object, "m", which maps each value of codice_8 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 (but requires O("n") space):
 var m := map(0 → 0, 1 → 1)
 function fib(n)
 if "key n is not in "map m 
 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 codice_8 first, then build larger values from them. This method also uses O("n") time since it contains a loop that repeats n − 1 times, but it only takes constant (O(1)) space, in contrast to the top-down approach which requires O("n") space to store the map.
 function fib(n)
 if n = 0
 return 0
 else
 var previousFib := 0, currentFib := 1
 repeat n − 1 times "// loop is skipped if n = 1"
 var newFib := previousFib + currentFib
 previousFib := currentFib
 currentFib := newFib
 return currentFib
In both examples, we only calculate codice_7 one time, and then use it to calculate both codice_12 and codice_13, instead of computing it every time either of them is evaluated.
The above method actually takes formula_65 time for large n because addition of two integers with formula_66 bits each takes formula_66 time. (The "n"th fibonacci number has formula_66 bits.) Also, there is a closed form for the Fibonacci sequence, known as Binet's formula, from which the formula_19-th term can be computed in approximately formula_70 time, which is more efficient than the above dynamic programming technique. However, the simple recurrence directly gives the matrix form that leads to an approximately formula_71 algorithm by fast matrix exponentiation.
A type of balanced 0–1 matrix.
Consider the problem of assigning values, either zero or one, to the positions of an matrix, with even, so that each row and each column contains exactly zeros and ones. We ask how many different assignments there are for a given formula_19. For example, when , four possible solutions are
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 ( zeros and ones). As there are formula_74 possible assignments, this strategy is not practical except maybe up to formula_75.
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 . While more sophisticated than brute force, this approach will visit every solution once, making it impractical for larger than six, since the number of solutions is already 116,963,796,250 for  = 8, as we shall see.
Dynamic programming makes it possible to count the number of solutions without visiting them all. Imagine backtracking values for the first row – what information would we require about the remaining rows, in order to be able to accurately count the solutions obtained for each first row value? We consider boards, where , whose formula_18 rows contain formula_77 zeros and formula_77 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 zeros and ones that have yet to be placed in that column. We seek the value of formula_79 (formula_19 arguments or one vector of formula_19 elements). The process of subproblem creation involves iterating over every one of formula_82 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 formula_83 and formula_84 pairs or not.
For example, in the first two boards shown above the sequences of vectors would be
The number of solutions is
Links to the MAPLE implementation of the dynamic programming approach may be found among the external links.
Checkerboard.
Consider a checkerboard with "n" × "n" squares and a cost function codice_14 which returns a cost associated with square codice_15 (codice_16 being the row, codice_17 being the column). For instance (on a 5 × 5 checkerboard),
Thus codice_18
Let us say there was a checker that could start at any square on the first rank (i.e., row) and you wanted to know the shortest path (the sum of the minimum costs at each visited rank) 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 codice_19 can move to codice_20, codice_21 or codice_22.
This problem exhibits optimal substructure. That is, the solution to the entire problem relies on solutions to subproblems. Let us define a function codice_23 as
Starting at rank codice_24 and descending to rank codice_25, we compute the value of this function for all the squares at each successive rank. Picking the square that holds the minimum value at each rank gives us the shortest path between rank codice_24 and rank codice_25.
The function codice_23 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 codice_14. For instance:
Now, let us define codice_23 in somewhat more general terms:
The first line of this equation deals with a board modeled as squares indexed on codice_25 at the lowest bound and codice_24 at the highest bound. The second line specifies what happens at the first rank; providing a base case. The third line, the recursion, is the important part. It represents the codice_33 terms in the example. From this definition we can derive straightforward recursive code for codice_34. In the following pseudocode, codice_24 is the size of the board, codice_14 is the cost function, and codice_37 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)
This function only computes the path cost, not the actual path. We discuss the actual path below. This, like the Fibonacci-numbers example, is horribly slow because it too exhibits the overlapping sub-problems attribute. That is, it recomputes the same path costs over and over. However, we can compute it much faster in a bottom-up fashion if we store path costs in a two-dimensional array codice_38 rather than using a function. This avoids recomputation; all the values needed for array codice_38 are computed ahead of time only once. Precomputed values for codice_15 are simply looked up whenever needed.
We also need to know what the actual shortest path is. To do this, we use another array codice_41; a "predecessor array". This array records the path to any square codice_42. The predecessor of codice_42 is modeled as an offset relative to the index (in codice_38) of the precomputed path cost of codice_42. To reconstruct the complete path, we lookup the predecessor of codice_42, then the predecessor of that square, then the predecessor of that square, and so on recursively, 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[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])
Sequence alignment.
In genetics, sequence alignment is an important application where dynamic programming is essential. 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 Smith–Waterman algorithm and Needleman–Wunsch algorithm.
Tower of Hanoi puzzle.
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
For n=1 the problem is trivial, namely S(1,h,t) = "move a disk from rod h to rod t" (there is only one disk left).
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.
Egg dropping puzzle.
The following is a description of the instance of this famous puzzle involving N=2 eggs and a building with H=36 floors: 
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 two 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
with "W"("n",0) = 0 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".
Faster DP solution using a different parametrization.
Notice that the above solution takes formula_88 time with a DP solution. This can be improved to formula_89 time by binary searching on the optimal formula_90 in the above recurrence, since formula_91 is increasing in formula_90 while formula_93 is decreasing in formula_90, thus a local minimum of formula_95 is a global minimum. Also, by storing the optimal formula_90 for each cell in the DP table and referring to its value for the previous cell, the optimal formula_90 for each cell can be found in constant time, improving it to formula_98 time. However, there is an even faster solution that involves a different parametrization of the problem:
Let formula_18 be the total number of floors such that the eggs break when dropped from the formula_18th floor (The example above is equivalent to taking formula_101).
Let formula_102 be the minimum floor from which the egg must be dropped to be broken.
Let formula_103 be the maximum number of values of formula_102 that are distinguishable using formula_13 tries and formula_19 eggs.
Then formula_107 for all formula_108.
Let formula_109 be the floor from which the first egg is dropped in the optimal strategy.
If the first egg broke, formula_102 is from formula_111 to formula_109 and distinguishable using at most formula_113 tries and formula_114 eggs.
If the first egg did not break, formula_102 is from formula_116 to formula_18 and distinguishable using formula_113 tries and formula_19 eggs.
Therefore, formula_120.
Then the problem is equivalent to finding the minimum formula_90 such that formula_122.
To do so, we could compute formula_123 in order of increasing formula_13, which would take formula_125 time.
Thus, if we separately handle the case of formula_126, the algorithm would take formula_127 time.
But the recurrence relation can in fact be solved, giving formula_128, which can be computed in formula_129 time using the identity formula_130 for all formula_131.
Since formula_132 for all formula_133, we can binary search on formula_13 to find formula_90, giving an formula_136 algorithm.
Matrix chain multiplication.
Matrix chain multiplication is a well-known example that demonstrates utility of dynamic programming. For example, engineering applications often have to multiply a chain of matrices. It is not surprising to find matrices of large dimensions, for example 100×100. Therefore, our task is to multiply matrices . Matrix multiplication is not commutative, but is associative; and we can multiply only two matrices at a time. So, we can multiply this chain of matrices in many different ways, for example:
and so on. There are numerous ways to multiply this chain of matrices. They will all produce the same final result, however they will take more or less time to compute, based on which particular matrices are multiplied. If matrix A has dimensions m×n and matrix B has dimensions n×q, then matrix C=A×B will have dimensions m×q, and will require m*n*q scalar multiplications (using a simplistic matrix multiplication algorithm for purposes of illustration).
For example, let us multiply matrices A, B and C. Let us assume that their dimensions are m×n, n×p, and p×s, respectively. Matrix A×B×C will be of size m×s and can be calculated in two ways shown below:
Let us assume that m = 10, n = 100, p = 10 and s = 1000. So, the first way to multiply the chain will require 1,000,000 + 1,000,000 calculations. The second way will require only 10,000+100,000 calculations. Obviously, the second way is faster, and we should multiply the matrices using that arrangement of parenthesis.
Therefore, our conclusion is that the order of parenthesis matters, and that our task is to find the optimal order of parenthesis.
At this point, we have several choices, one of which is to design a dynamic programming algorithm that will split the problem into overlapping problems and calculate the optimal arrangement of parenthesis. The dynamic programming solution is presented below.
Let's call m[i,j] the minimum number of scalar multiplications needed to multiply a chain of matrices from matrix i to matrix j (i.e. Ai × ... × Aj, i.e. i<=j). We split the chain at some matrix k, such that i <= k < j, and try to find out which combination produces minimum m[i,j].
The formula is:
 if i = j, m[i,j]= 0
 if i < j, m[i,j]= min over all possible values of k 
where "k" ranges from "i" to "j" − 1.
This formula can be coded as shown below, where input parameter "chain" is the chain of matrices, i.e. :
 function OptimalMatrixChainParenthesis(chain)
 n = length(chain)
 for i = 1, n
 m[i,i] = 0 "// Since it takes no calculations to multiply one matrix"
 for len = 2, n
 for i = 1, n - len + 1
 j = i + len -1
 m[i,j] = infinity "// So that the first calculation updates"
 for k = i, j-1
 if q < m[i, j] "// The new order of parentheses is better than what we had"
 m[i, j] = q "// Update"
 s[i, j] = k "// Record which k to split on, i.e. where to place the parenthesis"
So far, we have calculated values for all possible , the minimum number of calculations to multiply a chain from matrix "i" to matrix "j", and we have recorded the corresponding "split point". For example, if we are multiplying chain , and it turns out that and , that means that the optimal placement of parenthesis for matrices 1 to 3 is and to multiply those matrices will require 100 scalar calculation.
This algorithm will produce "tables" "m"[, ] and "s"[, ] that will have entries for all possible values of i and j. The final solution for the entire chain is m[1, n], with corresponding split at s[1, n]. Unraveling the solution will be recursive, starting from the top and continuing until we reach the base case, i.e. multiplication of single matrices.
Therefore, the next step is to actually split the chain, i.e. to place the parenthesis where they (optimally) belong. For this purpose we could use the following algorithm:
 function PrintOptimalParenthesis(s, i, j)
 if i = j
 print "A"i
 else
 print "(" 
 PrintOptimalParenthesis(s, i, s[i, j]) 
 PrintOptimalParenthesis(s, s[i, j] + 1, j) 
 print ")"
Of course, this algorithm is not useful for actual multiplication. This algorithm is just a user-friendly way to see what the result looks like.
To actually multiply the matrices using the proper splits, we need the following algorithm:
 function MatrixChainMultiply(chain from 1 to n) // returns the final matrix, i.e. A1×A2×... ×An
 OptimalMatrixChainParenthesis(chain from 1 to n) // this will produce s[ . ] and m[ . ] "tables"
 OptimalMatrixMultiplication(s, chain from 1 to n) // actually multiply
 function OptimalMatrixMultiplication(s, i, j) // returns the result of multiplying a chain of matrices from Ai to Aj in optimal way
 if i < j
 // keep on splitting the chain and multiplying the matrices in left and right sides
 LeftSide = OptimalMatrixMultiplication(s, i, s[i, j])
 RightSide = OptimalMatrixMultiplication(s, s[i, j] + 1, j)
 return MatrixMultiply(LeftSide, RightSide) 
 else if i = j
 return Ai // matrix at position i
 else 
 print "error, i <= j must hold"
 function MatrixMultiply(A, B) // function that multiplies two matrices
 if columns(A) = rows(B) 
 for i = 1, rows(A)
 for j = 1, columns(B)
 C[i, j] = 0
 for k = 1, columns(A)
 C[i, j] = C[i, j] + A[i, k]*B[k, j] 
 return C 
 else 
 print "error, incompatible dimensions."
History.
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, 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.
Bellman explains the reasoning behind the term "dynamic programming" in his autobiography, "Eye of the Hurricane: An Autobiography":
The word "dynamic" was chosen by Bellman to capture the time-varying aspect of the problems, and because it sounded impressive. 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 mathematical optimization.
The above explanation of the origin of the term is lacking. As Russell and Norvig in their book have written, referring to the above story: "This cannot be strictly true, because his first paper using the term (Bellman, 1952) appeared before Wilson became Secretary of Defense in 1953." Also, there is a comment in a speech by Harold J. Kushner, where he remembers Bellman. Quoting Kushner as he speaks of Bellman: "On the other hand, when I asked him the same question, he replied that he was trying to upstage Dantzig's linear programming by adding dynamic. Perhaps both motivations were true."