{"history":[],"article":{"id":"f33d9232-d1a9-44f3-acd5-b94ed325d2ae","title":"Dynamic programming","slug":"dynamic-programming","content":":::summary\n:::\n\n# \n**Dynamic programming** is both a [mathematical optimization](https://bliptext.com/articles/mathematical-optimization) method and an [algorithmic paradigm](https://bliptext.com/articles/algorithmic-paradigm).\nThe method was developed by [Richard Bellman](https://bliptext.com/articles/richard-bellman) in the 1950s and has found applications in numerous fields, from [aerospace engineering](https://bliptext.com/articles/aerospace-engineering) to [economics](https://bliptext.com/articles/economics).\nIn both contexts it refers to simplifying a complicated problem by breaking it down into simpler sub-problems in a [recursive](https://bliptext.com/articles/recursion) manner.\nWhile some decision problems cannot be taken apart this way, decisions that span several points in time do often break apart recursively.\nLikewise, 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](https://bliptext.com/articles/optimal-substructure).\nIf 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.\nIn the optimization literature this relationship is called the [Bellman equation](https://bliptext.com/articles/bellman-equation).\n# [Figure 1. Finding the shortest path in a graph using optimal substructure; a straight line indicates a single edge; a wavy line indicates a shortest path between the two vertices it connects (among other paths, not shown, sharing the same two vertices); the bold line is the overall shortest path from start to goal. | upload.wikimedia.org/wikipedia/commons/0/03/Shortest///path///optimal///substructure.svg]\n\n# Overview\n\n## Mathematical optimization\nIn terms of mathematical optimization, dynamic programming usually refers to simplifying a decision by breaking it down into a sequence of decision steps over time.\nThis is done by defining a sequence of **value functions** V1, V2, ..., Vn taking y as an argument representing the **[state](https://bliptext.com/articles/state-variable)** of the system at times i from 1 to n.\nThe definition of Vn(y) is the value obtained in state y at the last time n.\nThe values Vi at earlier times i = n −1, n − 2, ..., 2, 1 can be found by working backwards, using a [recursive](https://bliptext.com/articles/recursion) relationship called the [Bellman equation](https://bliptext.com/articles/bellman-equation).\nFor i = 2, ..., n, Vi−1 at any state y is calculated from Vi by maximizing a simple function (usually the sum) of the gain from a decision at time i − 1 and the function Vi at the new state of the system if this decision is made.\nSince Vi has already been calculated for the needed states, the above operation yields Vi−1 for those states.\nFinally, V1 at the initial state of the system is the value of the optimal solution.\nThe optimal values of the decision variables can be recovered, one by one, by tracking back the calculations already performed.\n\n## Control theory\nIn [control theory](https://bliptext.com/articles/control-theory), a typical problem is to find an admissible control which causes the system to follow an admissible trajectory on a continuous time interval that minimizes a [cost function](https://bliptext.com/articles/loss-function)\nThe solution to this problem is an optimal control law or policy, which produces an optimal trajectory and a [cost-to-go function](https://bliptext.com/articles/cost-to-go-function) J^{\\ast}.\nThe latter obeys the fundamental equation of dynamic programming:\na [partial differential equation](https://bliptext.com/articles/partial-differential-equation) known as the [Hamilton–Jacobi–Bellman equation](https://bliptext.com/articles/hamilton-jacobi-bellman-equation), in which and.\nOne finds that minimizing \\mathbf{u} in terms of t, \\mathbf{x}, and the unknown function and then substitutes the result into the Hamilton–Jacobi–Bellman equation to get the partial differential equation to be solved with boundary condition.\nIn practice, this generally requires [numerical techniques](https://bliptext.com/articles/numerical-partial-differential-equations) for some discrete approximation to the exact optimization relationship.\nAlternatively, the continuous process can be approximated by a discrete system, which leads to a following recurrence relation analog to the Hamilton–Jacobi–Bellman equation:\nat the k-th stage of n equally spaced discrete time intervals, and where \\hat{f} and denote discrete approximations to f and \\mathbf{g}.\nThis functional equation is known as the [Bellman equation](https://bliptext.com/articles/bellman-equation), which can be solved for an exact solution of the discrete approximation of the optimization equation.\n\n### Example from economics: Ramsey's problem of optimal saving\nIn economics, the objective is generally to maximize (rather than minimize) some dynamic [social welfare function](https://bliptext.com/articles/social-welfare-function).\nIn Ramsey's problem, this function relates amounts of consumption to levels of [utility](https://bliptext.com/articles/utility).\nLoosely speaking, the planner faces the trade-off between contemporaneous consumption and future consumption (via investment in [capital stock](https://bliptext.com/articles/physical-capital) that is used in production), known as [intertemporal choice](https://bliptext.com/articles/intertemporal-choice).\nFuture consumption is discounted at a constant rate.\nA discrete approximation to the transition equation of capital is given by where c is consumption, k is capital, and f is a [production function](https://bliptext.com/articles/production-function) satisfying the [Inada conditions](https://bliptext.com/articles/inada-conditions).\nAn initial capital stock k_{0} > 0 is assumed.\nLet c_t be consumption in period t, and assume consumption yields [utility](https://bliptext.com/articles/utility) as long as the consumer lives.\nAssume the consumer is impatient, so that he [discounts](https://bliptext.com/articles/discounting) future utility by a factor b each period, where 00, and suppose that this period's [capital](https://bliptext.com/articles/capital-economics) and consumption determine next period's [capital](https://bliptext.com/articles/capital-economics) as, where A is a positive constant and 0 n **return** infinity\n**else if** i = 1\n**return** c(i, j) **else**\n**return** **min**( **min**Cost(i-1, j-1), **min**Cost(i-1, j), **min**Cost(i-1, j+1) ) + c(i, j)\nThis function only computes the path cost, not the actual path.\nWe discuss the actual path below.\nThis, like the Fibonacci-numbers example, is horribly slow because it too exhibits the **overlapping sub-problems** attribute.\nThat is, it recomputes the same path costs over and over.\nHowever, we can compute it much faster in a bottom-up fashion if we store path costs in a two-dimensional array rather than using a function.\nThis avoids recomputation; all the values needed for array are computed ahead of time only once.\nPrecomputed values for are simply looked up whenever needed.\nWe also need to know what the actual shortest path is.\nTo do this, we use another array ; a predecessor array.\nThis array records the path to any square.\nThe predecessor of is modeled as an offset relative to the index (in ) of the precomputed path cost of.\nTo reconstruct the complete path, we lookup the predecessor of, then the predecessor of that square, then the predecessor of that square, and so on recursively, until we reach the starting square.\nConsider the following pseudocode:\n**function** computeShortestPathArrays\n****for**** x ****from**** 1 ****to**** n q[1, x] := c(1, x) ****for**** y ****from**** 1 ****to**** n q[y, 0] := infinity\nq[y, n + 1] := infinity\n****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])\nq[y, x] := m + c(y, x) **if** m = q[y-1, x-1]\np[y, x] := -1\n**else if** m = q[y-1, x] p[y, x] := 0\n**else**\np[y, x] := 1\nNow the rest is a simple matter of finding the minimum and printing it.\n**function** computeShortestPath\ncomputeShortestPathArrays\nminIndex := 1\nmin := q[n, 1]\n**for** i **from** 2 **to** n **if** q[n, i] < min\nminIndex := i min := q[n, i] printPath(n, minIndex)\n**function** **print**Path(y, x) **print**(x)\n**print**(\"<-\")\n**if** y = 2\n**print**(x + p[y, x])\n**else**\nprintPath(y-1, x + p[y, x])\n\n## Sequence alignment\nIn [genetics](https://bliptext.com/articles/genetics), [sequence alignment](https://bliptext.com/articles/sequence-alignment) is an important application where dynamic programming is essential.\nTypically, the problem consists of transforming one sequence into another using edit operations that replace, insert, or remove an element.\nEach operation has an associated cost, and the goal is to find the [sequence of edits with the lowest total cost](https://bliptext.com/articles/edit-distance).\nThe problem can be stated naturally as a recursion, a sequence A is optimally edited into a sequence B by either:\nThe 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].\nThe 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.\nDifferent variants exist, see [Smith–Waterman algorithm](https://bliptext.com/articles/smith-waterman-algorithm) and [Needleman–Wunsch algorithm](https://bliptext.com/articles/needleman-wunsch-algorithm).\n\n## Tower of Hanoi puzzle\nThe **[Tower of Hanoi](https://bliptext.com/articles/tower-of-hanoi)** or **Towers of [Hanoi](https://bliptext.com/articles/hanoi)** is a [mathematical game](https://bliptext.com/articles/mathematical-game) or [puzzle](https://bliptext.com/articles/puzzle).\nIt consists of three rods, and a number of disks of different sizes which can slide onto any rod.\nThe 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.\nThe objective of the puzzle is to move the entire stack to another rod, obeying the following rules:\nThe dynamic programming solution consists of solving the [functional equation](https://bliptext.com/articles/bellman-equation)\nwhere 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\nFor 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).\nThe number of moves required by this solution is 2n − 1.\nIf the objective is to **maximize** the number of moves (without cycling) then the dynamic programming [functional equation](https://bliptext.com/articles/bellman-equation) is slightly more complicated and 3n − 1 moves are required.\n# [A model set of the Towers of Hanoi (with 8 disks) | upload.wikimedia.org/wikipedia/commons/0/07/Tower///of///Hanoi.jpeg]\n\n## Egg dropping puzzle\nThe following is a description of the instance of this famous [puzzle](https://bliptext.com/articles/puzzle) involving N=2 eggs and a building with H=36 floors:\nTo derive a dynamic programming [functional equation](https://bliptext.com/articles/bellman-equation) for this puzzle, let the **state** of the dynamic programming model be a pair s = (n,k), where\nFor instance, s = (2,6) indicates that two test eggs are available and 6 (consecutive) floors are yet to be tested.\nThe initial state of the process is s = (N,H) where N denotes the number of test eggs available at the commencement of the experiment.\nThe process terminates either when there are no more test eggs (n = 0) or when k = 0, whichever occurs first.\nIf termination occurs at state s = (0,k) and k > 0, then the test failed.\nNow, let\nThen it can be shown that\nwith W(n,0) = 0 for all n > 0 and W(1,k) = k for all k.\nIt is easy to solve this equation iteratively by systematically increasing the values of n and k.\n\n### Faster DP solution using a different parametrization\nNotice that the above solution takes O( n k^2 ) time with a DP solution.\nThis can be improved to time by binary searching on the optimal x in the above recurrence, since W(n-1,x-1) is increasing in x while W(n,k-x) is decreasing in x, thus a local minimum of is a global minimum.\nAlso, by storing the optimal x for each cell in the DP table and referring to its value for the previous cell, the optimal x for each cell can be found in constant time, improving it to O( n k ) time.\nHowever, there is an even faster solution that involves a different parametrization of the problem:\nLet k be the total number of floors such that the eggs break when dropped from the kth floor (The example above is equivalent to taking k=37).\nLet m be the minimum floor from which the egg must be dropped to be broken.\nLet f(t,n) be the maximum number of values of m that are distinguishable using t tries and n eggs.\nThen for all t,n \\geq 0.\nLet a be the floor from which the first egg is dropped in the optimal strategy.\nIf the first egg broke, m is from 1 to a and distinguishable using at most t-1 tries and n-1 eggs.\nIf the first egg did not break, m is from a+1 to k and distinguishable using t-1 tries and n eggs.\nTherefore,.\nThen the problem is equivalent to finding the minimum x such that.\nTo do so, we could compute in order of increasing t, which would take O( n x ) time.\nThus, if we separately handle the case of n=1, the algorithm would take time.\nBut the recurrence relation can in fact be solved, giving, which can be computed in O(n) time using the identity for all i \\geq 0.\nSince for all t \\geq 0, we can binary search on t to find x, giving an algorithm.\n\n## Matrix chain multiplication\nMatrix chain multiplication is a well-known example that demonstrates utility of dynamic programming.\nFor example, engineering applications often have to multiply a chain of matrices.\nIt is not surprising to find matrices of large dimensions, for example 100×100.\nTherefore, our task is to multiply matrices A_1, A_2, .... A_n.\nMatrix multiplication is not commutative, but is associative; and we can multiply only two matrices at a time.\nSo, we can multiply this chain of matrices in many different ways, for example:\n((A1 × A2) × A3) × ...\nAn\nA1×(((A2×A3)× ...\n) × An)\n(A1 × A2) × (A3 × ... An)\nand so on.\nThere are numerous ways to multiply this chain of matrices.\nThey will all produce the same final result, however they will take more or less time to compute, based on which particular matrices are multiplied.\nIf 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](https://bliptext.com/articles/matrix-multiplication-algorithm) for purposes of illustration).\nFor example, let us multiply matrices A, B and C. Let us assume that their dimensions are m×n, n×p, and p×s, respectively.\nMatrix A×B×C will be of size m×s and can be calculated in two ways shown below:\nLet us assume that m = 10, n = 100, p = 10 and s = 1000.\nSo, the first way to multiply the chain will require 1,000,000 + 1,000,000 calculations.\nThe second way will require only 10,000+100,000 calculations.\nObviously, the second way is faster, and we should multiply the matrices using that arrangement of parenthesis.\nTherefore, our conclusion is that the order of parenthesis matters, and that our task is to find the optimal order of parenthesis.\nAt 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.\nThe dynamic programming solution is presented below.\nLet'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).\nWe split the chain at some matrix k, such that i <= k < j, and try to find out which combination produces minimum m[i,j].\nThe formula is:\n**if** i = j, m[i,j]= 0\n**if** i < j, m[i,j]= min over all possible values of k (m[i,k]+m[k+1,j] + )\nwhere k ranges from i to j − 1.\nThis formula can be coded as shown below, where input parameter \"chain\" is the chain of matrices, i.e. A_1, A_2, ...\nA_n:\n**function** OptimalMatrixChainParenthesis(chain)\nn = length(chain)\n**for** i = 1, n m[i,i] = 0 // Since it takes no calculations to multiply one matrix\n****for**** len = 2, n ****for**** i = 1, n - len + 1\nj = i + len -1\nm[i,j] = infinity // So that the first calculation updates\n**for** k = i, j-1\nq = m[i, k] + m[k+1, j] +\n**if** q < m[i, j] // The new order of parentheses is better than what we had\nm[i, j] = q // Update\ns[i, j] = k // Record which k to split on, i.e. where to place the parenthesis\nSo far, we have calculated values for all possible\nm[i, j]\n, the minimum number of calculations to multiply a chain from matrix i to matrix j, and we have recorded the corresponding \"split point\"\ns[i, j]\n. For example, if we are multiplying chain\nA1×A2×A3×A4\n, and it turns out that\nm[1, 3] = 100\nand\ns[1, 3] = 2\n, that means that the optimal placement of parenthesis for matrices 1 to 3 is (A_1\\times A_2)\\times A_3 and to multiply those matrices will require 100 scalar calculations.\nThis 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].\nUnraveling the solution will be recursive, starting from the top and continuing until we reach the base case, i.e. multiplication of single matrices.\nTherefore, the next step is to actually split the chain, i.e. to place the parenthesis where they (optimally) belong.\nFor this purpose we could use the following algorithm:\n**function** PrintOptimalParenthesis(s, i, j) **if** i = j print \"A\"i\n**else**\nprint \"(\" PrintOptimalParenthesis(s, i, s[i, j]) PrintOptimalParenthesis(s, s[i, j] + 1, j) print \")\"\nOf course, this algorithm is not useful for actual multiplication.\nThis algorithm is just a user-friendly way to see what the result looks like.\nTo actually multiply the matrices using the proper splits, we need the following algorithm:\n\n# History of the name\nThe term dynamic programming was originally used in the 1940s by [Richard Bellman](https://bliptext.com/articles/richard-bellman) to describe the process of solving problems where one needs to find the best decisions one after another.\nBy 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](https://bliptext.com/articles/ieee) as a [systems analysis](https://bliptext.com/articles/systems-analysis) and [engineering](https://bliptext.com/articles/engineering) topic.\nBellman's contribution is remembered in the name of the [Bellman equation](https://bliptext.com/articles/bellman-equation), a central result of dynamic programming which restates an optimization problem in [recursive](https://bliptext.com/articles/recursion-computer-science) form.\nBellman explains the reasoning behind the term dynamic programming in his autobiography, Eye of the Hurricane: An Autobiography:\n\"I spent the Fall quarter (of 1950) at RAND. My first task was to find a name for multistage decision processes. An interesting question is, 'Where did the name, dynamic programming, come from?' The 1950s were not good years for mathematical research. We had a very interesting gentleman in Washington named Wilson. He was Secretary of Defense, and he actually had a pathological fear and hatred of the word 'research'. I'm not using the term lightly; I'm using it precisely. His face would suffuse, he would turn red, and he would get violent if people used the term research in his presence. You can imagine how he felt, then, about the term mathematical. The RAND Corporation was employed by the Air Force, and the Air Force had Wilson as its boss, essentially. Hence, I felt I had to do something to shield Wilson and the Air Force from the fact that I was really doing mathematics inside the RAND Corporation.\nWhat title, what name, could I choose?\nIn the first place I was interested in planning, in decision making, in thinking.\nBut planning, is not a good word for various reasons.\nI decided therefore to use the word 'programming'.\nI wanted to get across the idea that this was dynamic, this was multistage, this was time-varying.\nI thought, let's kill two birds with one stone.\nLet's take a word that has an absolutely precise meaning, namely dynamic, in the classical physical sense.\nIt also has a very interesting property as an adjective, and that is it's impossible to use the word dynamic in a pejorative sense.\nTry thinking of some combination that will possibly give it a pejorative meaning.\nIt's impossible.\nThus, I thought dynamic programming was a good name.\nIt was something not even a Congressman could object to.\nSo I used it as an umbrella for my activities.\"\nThe word dynamic was chosen by Bellman to capture the time-varying aspect of the problems, and because it sounded impressive.\nThe 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.\nThis usage is the same as that in the phrases [linear programming](https://bliptext.com/articles/linear-programming) and mathematical programming, a synonym for [mathematical optimization](https://bliptext.com/articles/mathematical-optimization).\nThe above explanation of the origin of the term may be inaccurate: According to Russell and Norvig, the above story \"cannot be strictly true, because his first paper using the term (Bellman, 1952) appeared before Wilson became Secretary of Defense in 1953.\"\nAlso, [Harold J. Kushner](https://bliptext.com/articles/harold-j-kushner) stated in a speech that, \"On the other hand, when I asked [Bellman] the same question, he replied that he was trying to upstage [Dantzig's](https://bliptext.com/articles/george-dantzig) linear programming by adding dynamic. Perhaps both motivations were true.\"","current_revision":null,"created_by":"kACnuQmH5k9Nr5VNVEpVeg7XpvUmMctD","created_at":"2025-01-25T18:10:02.572Z","updated_at":"2025-01-25T18:10:02.572Z","search_vector":"'dynam':1 'program':2","search_content":" \n \n\n# \n**Dynamic programming** is both a [mathematical optimization](https://bliptext.com/articles/mathematical-optimization) method and an [algorithmic paradigm](https://bliptext.com/articles/algorithmic-paradigm).\nThe method was developed by [Richard Bellman](https://bliptext.com/articles/richard-bellman) in the 1950s and has found applications in numerous fields, from [aerospace engineering](https://bliptext.com/articles/aerospace-engineering) to [economics](https://bliptext.com/articles/economics).\nIn both contexts it refers to simplifying a complicated problem by breaking it down into simpler sub-problems in a [recursive](https://bliptext.com/articles/recursion) manner.\nWhile some decision problems cannot be taken apart this way, decisions that span several points in time do often break apart recursively.\nLikewise, 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](https://bliptext.com/articles/optimal-substructure).\nIf 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.\nIn the optimization literature this relationship is called the [Bellman equation](https://bliptext.com/articles/bellman-equation).\n# [Figure 1. Finding the shortest path in a graph using optimal substructure; a straight line indicates a single edge; a wavy line indicates a shortest path between the two vertices it connects (among other paths, not shown, sharing the same two vertices); the bold line is the overall shortest path from start to goal. | upload.wikimedia.org/wikipedia/commons/0/03/Shortest///path///optimal///substructure.svg]\n\n# Overview\n\n## Mathematical optimization\nIn terms of mathematical optimization, dynamic programming usually refers to simplifying a decision by breaking it down into a sequence of decision steps over time.\nThis is done by defining a sequence of **value functions** V1, V2, ..., Vn taking y as an argument representing the **[state](https://bliptext.com/articles/state-variable)** of the system at times i from 1 to n.\nThe definition of Vn(y) is the value obtained in state y at the last time n.\nThe values Vi at earlier times i = n −1, n − 2, ..., 2, 1 can be found by working backwards, using a [recursive](https://bliptext.com/articles/recursion) relationship called the [Bellman equation](https://bliptext.com/articles/bellman-equation).\nFor i = 2, ..., n, Vi−1 at any state y is calculated from Vi by maximizing a simple function (usually the sum) of the gain from a decision at time i − 1 and the function Vi at the new state of the system if this decision is made.\nSince Vi has already been calculated for the needed states, the above operation yields Vi−1 for those states.\nFinally, V1 at the initial state of the system is the value of the optimal solution.\nThe optimal values of the decision variables can be recovered, one by one, by tracking back the calculations already performed.\n\n## Control theory\nIn [control theory](https://bliptext.com/articles/control-theory), a typical problem is to find an admissible control which causes the system to follow an admissible trajectory on a continuous time interval that minimizes a [cost function](https://bliptext.com/articles/loss-function)\nThe solution to this problem is an optimal control law or policy, which produces an optimal trajectory and a [cost-to-go function](https://bliptext.com/articles/cost-to-go-function) J^{\\ast}.\nThe latter obeys the fundamental equation of dynamic programming:\na [partial differential equation](https://bliptext.com/articles/partial-differential-equation) known as the [Hamilton–Jacobi–Bellman equation](https://bliptext.com/articles/hamilton-jacobi-bellman-equation), in which and.\nOne finds that minimizing \\mathbf{u} in terms of t, \\mathbf{x}, and the unknown function and then substitutes the result into the Hamilton–Jacobi–Bellman equation to get the partial differential equation to be solved with boundary condition.\nIn practice, this generally requires [numerical techniques](https://bliptext.com/articles/numerical-partial-differential-equations) for some discrete approximation to the exact optimization relationship.\nAlternatively, the continuous process can be approximated by a discrete system, which leads to a following recurrence relation analog to the Hamilton–Jacobi–Bellman equation:\nat the k-th stage of n equally spaced discrete time intervals, and where \\hat{f} and denote discrete approximations to f and \\mathbf{g}.\nThis functional equation is known as the [Bellman equation](https://bliptext.com/articles/bellman-equation), which can be solved for an exact solution of the discrete approximation of the optimization equation.\n\n### Example from economics: Ramsey's problem of optimal saving\nIn economics, the objective is generally to maximize (rather than minimize) some dynamic [social welfare function](https://bliptext.com/articles/social-welfare-function).\nIn Ramsey's problem, this function relates amounts of consumption to levels of [utility](https://bliptext.com/articles/utility).\nLoosely speaking, the planner faces the trade-off between contemporaneous consumption and future consumption (via investment in [capital stock](https://bliptext.com/articles/physical-capital) that is used in production), known as [intertemporal choice](https://bliptext.com/articles/intertemporal-choice).\nFuture consumption is discounted at a constant rate.\nA discrete approximation to the transition equation of capital is given by where c is consumption, k is capital, and f is a [production function](https://bliptext.com/articles/production-function) satisfying the [Inada conditions](https://bliptext.com/articles/inada-conditions).\nAn initial capital stock k_{0} > 0 is assumed.\nLet c_t be consumption in period t, and assume consumption yields [utility](https://bliptext.com/articles/utility) as long as the consumer lives.\nAssume the consumer is impatient, so that he [discounts](https://bliptext.com/articles/discounting) future utility by a factor b each period, where 0 0, and suppose that this period's [capital](https://bliptext.com/articles/capital-economics) and consumption determine next period's [capital](https://bliptext.com/articles/capital-economics) as, where A is a positive constant and 0 n **return** infinity\n**else if** i = 1\n**return** c(i, j) **else**\n**return** **min**( **min**Cost(i-1, j-1), **min**Cost(i-1, j), **min**Cost(i-1, j+1) ) + c(i, j)\nThis function only computes the path cost, not the actual path.\nWe discuss the actual path below.\nThis, like the Fibonacci-numbers example, is horribly slow because it too exhibits the **overlapping sub-problems** attribute.\nThat is, it recomputes the same path costs over and over.\nHowever, we can compute it much faster in a bottom-up fashion if we store path costs in a two-dimensional array rather than using a function.\nThis avoids recomputation; all the values needed for array are computed ahead of time only once.\nPrecomputed values for are simply looked up whenever needed.\nWe also need to know what the actual shortest path is.\nTo do this, we use another array ; a predecessor array.\nThis array records the path to any square.\nThe predecessor of is modeled as an offset relative to the index (in ) of the precomputed path cost of.\nTo reconstruct the complete path, we lookup the predecessor of, then the predecessor of that square, then the predecessor of that square, and so on recursively, until we reach the starting square.\nConsider the following pseudocode:\n**function** computeShortestPathArrays\n****for**** x ****from**** 1 ****to**** n q[1, x] := c(1, x) ****for**** y ****from**** 1 ****to**** n q[y, 0] := infinity\nq[y, n + 1] := infinity\n****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])\nq[y, x] := m + c(y, x) **if** m = q[y-1, x-1]\np[y, x] := -1\n**else if** m = q[y-1, x] p[y, x] := 0\n**else**\np[y, x] := 1\nNow the rest is a simple matter of finding the minimum and printing it.\n**function** computeShortestPath\ncomputeShortestPathArrays\nminIndex := 1\nmin := q[n, 1]\n**for** i **from** 2 **to** n **if** q[n, i] < min\nminIndex := i min := q[n, i] printPath(n, minIndex)\n**function** **print**Path(y, x) **print**(x)\n**print**(\"<-\")\n**if** y = 2\n**print**(x + p[y, x])\n**else**\nprintPath(y-1, x + p[y, x])\n\n## Sequence alignment\nIn [genetics](https://bliptext.com/articles/genetics), [sequence alignment](https://bliptext.com/articles/sequence-alignment) is an important application where dynamic programming is essential.\nTypically, the problem consists of transforming one sequence into another using edit operations that replace, insert, or remove an element.\nEach operation has an associated cost, and the goal is to find the [sequence of edits with the lowest total cost](https://bliptext.com/articles/edit-distance).\nThe problem can be stated naturally as a recursion, a sequence A is optimally edited into a sequence B by either:\nThe 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].\nThe 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.\nDifferent variants exist, see [Smith–Waterman algorithm](https://bliptext.com/articles/smith-waterman-algorithm) and [Needleman–Wunsch algorithm](https://bliptext.com/articles/needleman-wunsch-algorithm).\n\n## Tower of Hanoi puzzle\nThe **[Tower of [Hanoi](https://bliptext.com/articles/hanoi)](https://bliptext.com/articles/tower-of-hanoi)** or **Towers of [Hanoi](https://bliptext.com/articles/hanoi)** is a [mathematical game](https://bliptext.com/articles/mathematical-game) or [puzzle](https://bliptext.com/articles/puzzle).\nIt consists of three rods, and a number of disks of different sizes which can slide onto any rod.\nThe 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.\nThe objective of the puzzle is to move the entire stack to another rod, obeying the following rules:\nThe dynamic programming solution consists of solving the [functional equation](https://bliptext.com/articles/bellman-equation)\nwhere 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\nFor 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).\nThe number of moves required by this solution is 2n − 1.\nIf the objective is to **maximize** the number of moves (without cycling) then the dynamic programming [functional equation](https://bliptext.com/articles/bellman-equation) is slightly more complicated and 3n − 1 moves are required.\n# [A model set of the Towers of Hanoi (with 8 disks) | upload.wikimedia.org/wikipedia/commons/0/07/Tower///of///Hanoi.jpeg]\n\n## Egg dropping puzzle\nThe following is a description of the instance of this famous [puzzle](https://bliptext.com/articles/puzzle) involving N=2 eggs and a building with H=36 floors:\nTo derive a dynamic programming [functional equation](https://bliptext.com/articles/bellman-equation) for this puzzle, let the **state** of the dynamic programming model be a pair s = (n,k), where\nFor instance, s = (2,6) indicates that two test eggs are available and 6 (consecutive) floors are yet to be tested.\nThe initial state of the process is s = (N,H) where N denotes the number of test eggs available at the commencement of the experiment.\nThe process terminates either when there are no more test eggs (n = 0) or when k = 0, whichever occurs first.\nIf termination occurs at state s = (0,k) and k > 0, then the test failed.\nNow, let\nThen it can be shown that\nwith W(n,0) = 0 for all n > 0 and W(1,k) = k for all k.\nIt is easy to solve this equation iteratively by systematically increasing the values of n and k.\n\n### Faster DP solution using a different parametrization\nNotice that the above solution takes O( n k^2 ) time with a DP solution.\nThis can be improved to time by binary searching on the optimal x in the above recurrence, since W(n-1,x-1) is increasing in x while W(n,k-x) is decreasing in x, thus a local minimum of is a global minimum.\nAlso, by storing the optimal x for each cell in the DP table and referring to its value for the previous cell, the optimal x for each cell can be found in constant time, improving it to O( n k ) time.\nHowever, there is an even faster solution that involves a different parametrization of the problem:\nLet k be the total number of floors such that the eggs break when dropped from the kth floor (The example above is equivalent to taking k=37).\nLet m be the minimum floor from which the egg must be dropped to be broken.\nLet f(t,n) be the maximum number of values of m that are distinguishable using t tries and n eggs.\nThen for all t,n \\geq 0.\nLet a be the floor from which the first egg is dropped in the optimal strategy.\nIf the first egg broke, m is from 1 to a and distinguishable using at most t-1 tries and n-1 eggs.\nIf the first egg did not break, m is from a+1 to k and distinguishable using t-1 tries and n eggs.\nTherefore,.\nThen the problem is equivalent to finding the minimum x such that.\nTo do so, we could compute in order of increasing t, which would take O( n x ) time.\nThus, if we separately handle the case of n=1, the algorithm would take time.\nBut the recurrence relation can in fact be solved, giving, which can be computed in O(n) time using the identity for all i \\geq 0.\nSince for all t \\geq 0, we can binary search on t to find x, giving an algorithm.\n\n## Matrix chain multiplication\nMatrix chain multiplication is a well-known example that demonstrates utility of dynamic programming.\nFor example, engineering applications often have to multiply a chain of matrices.\nIt is not surprising to find matrices of large dimensions, for example 100×100.\nTherefore, our task is to multiply matrices A_1, A_2, .... A_n.\nMatrix multiplication is not commutative, but is associative; and we can multiply only two matrices at a time.\nSo, we can multiply this chain of matrices in many different ways, for example:\n((A1 × A2) × A3) × ...\nAn\nA1×(((A2×A3)× ...\n) × An)\n(A1 × A2) × (A3 × ... An)\nand so on.\nThere are numerous ways to multiply this chain of matrices.\nThey will all produce the same final result, however they will take more or less time to compute, based on which particular matrices are multiplied.\nIf 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](https://bliptext.com/articles/matrix-multiplication-algorithm) for purposes of illustration).\nFor example, let us multiply matrices A, B and C. Let us assume that their dimensions are m×n, n×p, and p×s, respectively.\nMatrix A×B×C will be of size m×s and can be calculated in two ways shown below:\nLet us assume that m = 10, n = 100, p = 10 and s = 1000.\nSo, the first way to multiply the chain will require 1,000,000 + 1,000,000 calculations.\nThe second way will require only 10,000+100,000 calculations.\nObviously, the second way is faster, and we should multiply the matrices using that arrangement of parenthesis.\nTherefore, our conclusion is that the order of parenthesis matters, and that our task is to find the optimal order of parenthesis.\nAt 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.\nThe dynamic programming solution is presented below.\nLet'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).\nWe split the chain at some matrix k, such that i <= k < j, and try to find out which combination produces minimum m[i,j].\nThe formula is:\n**if** i = j, m[i,j]= 0\n**if** i < j, m[i,j]= min over all possible values of k (m[i,k]+m[k+1,j] + )\nwhere k ranges from i to j − 1.\nThis formula can be coded as shown below, where input parameter \"chain\" is the chain of matrices, i.e. A_1, A_2, ...\nA_n:\n**function** OptimalMatrixChainParenthesis(chain)\nn = length(chain)\n**for** i = 1, n m[i,i] = 0 // Since it takes no calculations to multiply one matrix\n****for**** len = 2, n ****for**** i = 1, n - len + 1\nj = i + len -1\nm[i,j] = infinity // So that the first calculation updates\n**for** k = i, j-1\nq = m[i, k] + m[k+1, j] +\n**if** q < m[i, j] // The new order of parentheses is better than what we had\nm[i, j] = q // Update\ns[i, j] = k // Record which k to split on, i.e. where to place the parenthesis\nSo far, we have calculated values for all possible\nm[i, j]\n, the minimum number of calculations to multiply a chain from matrix i to matrix j, and we have recorded the corresponding \"split point\"\ns[i, j]\n. For example, if we are multiplying chain\nA1×A2×A3×A4\n, and it turns out that\nm[1, 3] = 100\nand\ns[1, 3] = 2\n, that means that the optimal placement of parenthesis for matrices 1 to 3 is (A_1\\times A_2)\\times A_3 and to multiply those matrices will require 100 scalar calculations.\nThis 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].\nUnraveling the solution will be recursive, starting from the top and continuing until we reach the base case, i.e. multiplication of single matrices.\nTherefore, the next step is to actually split the chain, i.e. to place the parenthesis where they (optimally) belong.\nFor this purpose we could use the following algorithm:\n**function** PrintOptimalParenthesis(s, i, j) **if** i = j print \"A\"i\n**else**\nprint \"(\" PrintOptimalParenthesis(s, i, s[i, j]) PrintOptimalParenthesis(s, s[i, j] + 1, j) print \")\"\nOf course, this algorithm is not useful for actual multiplication.\nThis algorithm is just a user-friendly way to see what the result looks like.\nTo actually multiply the matrices using the proper splits, we need the following algorithm:\n\n# History of the name\nThe term dynamic programming was originally used in the 1940s by [Richard Bellman](https://bliptext.com/articles/richard-bellman) to describe the process of solving problems where one needs to find the best decisions one after another.\nBy 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](https://bliptext.com/articles/ieee) as a [systems analysis](https://bliptext.com/articles/systems-analysis) and [engineering](https://bliptext.com/articles/engineering) topic.\nBellman's contribution is remembered in the name of the [Bellman equation](https://bliptext.com/articles/bellman-equation), a central result of dynamic programming which restates an optimization problem in [recursive](https://bliptext.com/articles/recursion-computer-science) form.\nBellman explains the reasoning behind the term dynamic programming in his autobiography, Eye of the Hurricane: An Autobiography:\n\"I spent the Fall quarter (of 1950) at RAND. My first task was to find a name for multistage decision processes. An interesting question is, 'Where did the name, dynamic programming, come from?' The 1950s were not good years for mathematical research. We had a very interesting gentleman in Washington named Wilson. He was Secretary of Defense, and he actually had a pathological fear and hatred of the word 'research'. I'm not using the term lightly; I'm using it precisely. His face would suffuse, he would turn red, and he would get violent if people used the term research in his presence. You can imagine how he felt, then, about the term mathematical. The RAND Corporation was employed by the Air Force, and the Air Force had Wilson as its boss, essentially. Hence, I felt I had to do something to shield Wilson and the Air Force from the fact that I was really doing mathematics inside the RAND Corporation.\nWhat title, what name, could I choose?\nIn the first place I was interested in planning, in decision making, in thinking.\nBut planning, is not a good word for various reasons.\nI decided therefore to use the word 'programming'.\nI wanted to get across the idea that this was dynamic, this was multistage, this was time-varying.\nI thought, let's kill two birds with one stone.\nLet's take a word that has an absolutely precise meaning, namely dynamic, in the classical physical sense.\nIt also has a very interesting property as an adjective, and that is it's impossible to use the word dynamic in a pejorative sense.\nTry thinking of some combination that will possibly give it a pejorative meaning.\nIt's impossible.\nThus, I thought dynamic programming was a good name.\nIt was something not even a Congressman could object to.\nSo I used it as an umbrella for my activities.\"\nThe word dynamic was chosen by Bellman to capture the time-varying aspect of the problems, and because it sounded impressive.\nThe 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.\nThis usage is the same as that in the phrases [linear programming](https://bliptext.com/articles/linear-programming) and mathematical programming, a synonym for [mathematical optimization](https://bliptext.com/articles/mathematical-optimization).\nThe above explanation of the origin of the term may be inaccurate: According to Russell and Norvig, the above story \"cannot be strictly true, because his first paper using the term (Bellman, 1952) appeared before Wilson became Secretary of Defense in 1953.\"\nAlso, [Harold J. Kushner](https://bliptext.com/articles/harold-j-kushner) stated in a speech that, \"On the other hand, when I asked [Bellman] the same question, he replied that he was trying to upstage [Dantzig's](https://bliptext.com/articles/george-dantzig) linear programming by adding dynamic. Perhaps both motivations were true.\"","revisionCount":0},"hasMore":false,"nextCursor":null}