dynamic programing?

Answer 1

dynamic programming in many ways is already there but it depends really on what abstraction level we are talking. Are we talking about program that can dynamically adapt to its problem domain and generate logic on the fly to handle changes in it or we are just talking on the techniques or tools level where we can allow for future improvements or changes in the program and when its time the code changes the behavior of program to incorporate those changes( we are in the stages of reaching this).

Answer 2

There is really nothing dynamic about it. You just save off results from one computation you don't have to make the same resource consuming computation later. It is used a lot instead of recursion where the same calculations may be made repeatedly.

Answer 3

Dynamic programming is a method for reducing the runtime of algorithms exhibiting the properties of overlapping subproblems and optimal substructure.
http://en.wikipedia.org/wiki/Dynamic_Programming
Check this out to know more

Good luck.

Answer 4

In computer science, dynamic programming is a method for reducing the runtime of algorithms exhibiting the properties of overlapping subproblems and optimal substructure, described below.

Mathematician Richard Bellman invented dynamic programming in 1953. The field was founded as a systems analysis and engineering topic which is recognized by the IEEE.

Contents [hide]
1 Overview
2 Examples
2.1 Fibonacci sequence
2.2 Checkerboard
3 Algorithms that use dynamic programming
4 See also
5 External links
6 References



[edit]
Overview

Figure 1. Finding the shortest path in an acyclic graph using optimal substructure; a wavy line indicates a shortest path between the two vertices it connectsOptimal substructure means that optimal solutions of subproblems can be used to find the optimal solutions of the overall problem. For example, the shortest path to a goal from a vertex in an acyclic graph can be found by first computing the shortest path to the goal from all adjacent vertices, and then using this to pick the best overall path, as shown in Figure 1. In general, we can solve a problem with optimal substructure using a three-step process:

Break the problem into smaller subproblems.
Solve these problems optimally using this three-step process recursively.
Use these optimal solutions to construct an optimal solution for the original problem.
The subproblems are, themselves, solved by dividing them into sub-subproblems, and so on, until we reach some simple case that is easy to solve.


Figure 2. The subproblem graph for the Fibonacci sequence. That it is not a tree but a DAG indicates overlapping subproblems.To say that a problem has overlapping subproblems is to say that the same subproblems are used to solve many different larger problems. For example, in the Fibonacci sequence, F3 = F1 + F2 and F4 = F2 + F3 — computing each number involves computing F2. Because both F3 and F4 are needed to compute F5, a naïve approach to computing F5 may end up computing F2 twice or more. This applies whenever overlapping subproblems are present: a naïve approach may waste time recomputing optimal solutions to subproblems it has already solved.

In order to avoid this, we instead save the solutions to problems we have already solved. Then, if we need to solve the same problem later, we can retrieve and reuse our already-computed solution. This approach is called memoization (not memorization, although this term also fits). If we are sure we won't need a particular solution anymore, we can throw it away to save space. In some cases, we can even compute the solutions to subproblems we know that we'll need in advance.

In summary, dynamic programming makes use of:

Overlapping subproblems
Optimal substructure
Memoization
Dynamic programming usually takes one of two approaches:

Top-down approach: The problem is broken into subproblems, and these subproblems are solved and the solutions remembered, in case they need to be solved again. This is recursion and memoization combined together.
Bottom-up approach: All subproblems that might be needed are solved in advance and then used to build up solutions to larger problems. This approach is slightly better in stack space and number of function calls, but it is sometimes not intuitive to figure out all the subproblems needed for solving given problem.
Originally, the term dynamic programming only applied to solving certain kinds of operational problems outside the area of computer science, just as linear programming did. In this context, it has no particular connection to programming at all; the name is a coincidence. The term was also used in the 1940s by Richard Bellman, an American mathematician, to describe the process of solving problems where one needs to find the best decisions one after another.

Some functional programming languages, most notably Haskell, 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). This is only possible for a function which has no side-effects, which is always true in Haskell but seldom true in imperative languages.

[edit]
Examples
[edit]
Fibonacci sequence
A naive implementation of a function finding the nth member of the Fibonacci sequence, based directly on the mathematical definition:

function fib(n)
if n = 0 or n = 1
return n
else
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 twice 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 → 1, 1 → 1)
function fib(n)
if n not in keys(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 this case, we can also reduce the function from using linear (O(n)) space to using constant space by changing our function to use a bottom-up approach which calculates smaller values of fib first, then build larger values from them:

function fib(n)
var previousFib := 1, currentFib := 1
repeat n − 1 times
var newFib := previousFib + currentFib
previousFib := currentFib
currentFib := newFib
return currentFib
This bottom-up version is close to the simple imperative loop that might be used to compute the Fibonacci function in an introductory computer science course.

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.

[edit]
Checkerboard
Consider a checkerboard with n × n squares and a cost-function c(i, j) which returns a cost associated with square i,j (i being the row, j being the column). For instance (on a 5 × 5 checkerboard),

+---+---+---+---+---+
5 | 6 | 7 | 4 | 7 | 8 |
+---|---|---|---|---+
4 | 7 | 6 | 1 | 1 | 4 |
+---|---|---|---|---+
3 | 3 | 5 | 7 | 8 | 2 |
+---|---|---|---|---+
2 | 2 | 6 | 7 | 0 | 2 |
+---|---|---|---|---+
1 | 7 | 3 | 5 | 6 | 1 |
+---+---+---+---+---+
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 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 forward or diagonally left or right 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

q(i, j) = the minimum cost to reach square (i, j)
If we can find the values of this function for all the squares at rank n, we pick the minimum and follow that path backwards to get the shortest path.

It is easy to see that q(i, j) is equal to the minimum cost to get to any of the three squares below it (since those are the only squares that can reach it) plus c(i, j). For instance:

+---+---+---+---+---+
5 | | | | | |
+---|---|---|---|---+
4 | | | A | | |
+---|---|---|---|---+
3 | | B | C | D | |
+---|---|---|---|---+
2 | | | | | |
+---|---|---|---|---+
1 | | | | | |
+---+---+---+---+---+
1 2 3 4 5

Now, let us define q(i, j) in little more general terms:


This equation is pretty straightforward. The first line is simply there to make the recursive property simpler (when dealing with the edges, so we need only one recursion). The second line says what happens in the first rank, so we have something to start with. The third line, the recursion, is the important part. It is basically the same as the A,B,C,D example. From this definition we can make a straightforward recursive code for q(i, j). In the following pseudocode, n is the size of the board, c(i, j) is the cost-function, 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 = 1
return c(i, j)
else
return min( minCost(i-1, j-1), minCost(i-1, j), minCost(i-1, j+1) ) + c(i, j)
It should be noted that this function just computes the path-cost, not the actual path. We will get to the path soon. This, like the Fibonacci-numbers example, is horribly slow since it spends mountains of time recomputing the same shortest paths over and over. However, we can compute it much faster in a bottom up-fashion if we use a two-dimensional array q[i, j] instead of a function. Why do we do that? Simply because when using a function we recompute the same path over and over, and we can choose what values to compute first.

We also need to know what the actual path is. The path problem we can solve using another array p[i, j], a predecessor array. This array basically says where paths come from. Consider the following code:

function computeShortestPathArrays()
for x from 1 to n
q[1, x] := c(1, x)
for y from 1 to n
q[y, 0] := infinity
q[y, n + 1] := infinity
for y from 2 to n
for x from 1 to n
m := min(q[y-1, x-1], q[y-1, x], q[y-1, x+1])
q[y, x] := m + c(y, x)
if m = q[y-1, x-1]
p[y, x] := -1
else if m = q[y-1, x]
p[y, x] := 0
else
p[y, x] := 1
Now the rest is a simple matter of finding the minimum and printing it.

function computeShortestPath()
computeShortestPathArrays()
minIndex := 1
min := q[n, 1]
for i from 2 to n
if q[n, i] < min
minIndex := i
min := q[n, i]
printPath(n, minIndex)
function printPath(y, x)
print(x)
print("<-")
if y = 2
print(x + p[y, x])
else
printPath(y-1, x + p[y, x])



[edit]
Algorithms that use dynamic programming
Many string algorithms including longest-common subsequence problem
The Cocke-Younger-Kasami (CYK) algorithm which determines whether and how a given string can be generated by a given context-free grammar
The use of transposition tables and refutation tables in computer chess
The Viterbi algorithm
The Earley algorithm
The Needleman-Wunsch and other sequence alignment algorithms used in bioinformatics
Levenshtein distance (edit distance)
Bellman-Ford algorithm
Dijkstra's algorithm Single source and destination shortest path algorithm
Floyd's All-Pairs shortest path algorithm
Optimizing the order for chain matrix multiplication
subset sum algorithm
knapsack problem
The dynamic time warping algorithm for computing the global distance between two time series
De Boor algorithm for evaluating B-spline curves.
Duckworth-Lewis method for resolving the problem when games of cricket are interrupted

Answer 5

what about it

Answer 6

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