Knapsack Problem – Dynamic Programming

Dynamic Programming

Dynamic Programming is one very useful technique for solving problems like the Knapsack problem.

The basic idea of Dynamic programming is to break down a problem in recursive subproblems. Each subproblem must be easy to solve, and its solution stored in some data structure to be used again.

In a schematic drawing, it’s something like this:

 

DynamicScheme.jpg

Dynamic programming is not a general procedure to be applied in any problem. The problem has to allow it. Richard Bellman, the creator of Dynamic Programming, called it  “Principle of optimallity”. It is like saying “this method works in the problems it works, and doesn’t works in the problems it doesn’t works”. But mathematicians love this kind of concept, because they can ask: which problems has the principle of optimality? Is the Knapsack one of them?

Since this method is not general, the programmer has to design a new method for every different problem and also figure out if the problem has the principle of optimality.

But, despite all these observations, it is a useful way of thinking.


 

The Algorithm

The detailed algorithm can be found in Wikipedia  (https://en.wikipedia.org/wiki/Dynamic_programming). I’ll sketch here the main ideias.

Suppose the problem is the same of previous post.
Max Weight = 14.
List of itens:
ValueWeight

1) It is easy to compute the solution with 1 item for a range of weights.

Array of values
Dyn01.JPG
Item 1, with weight 8 and value 10,000. If we vary the max weight of the knapsack from 0 to MaxW, the value array will be Zero before we have max weight = 8, and 10,000 after it.

 Array of picked itens:

Dyn02.JPG
“i1” refers to item 1

 


2) We increase the number of itens, and compare current solution versus solution with new item.
Step2_details
With items 1 and 2: if the max weight is 6, we pick item 2. When max weight is 8, we compare solution with weight 8 from item 1 (10,000) against solution with item 1 and weight 2 plus item 2 (0) and weight 6 (3,000). It is better to keep the 10,000 value solution.
Step2_details2.JPG
When maxWeight is 14, we compare current solution with item 1 (10,000) versus solution with item 1 and weight 8 (10,000) plus item 2: 13,000
Complete array with 2 items.
Step2.JPG
Evaluation arrays with itens 1 and 2
Proceeding this way, we get the full evaluation matrix.
Step5.JPG
Evaluation array with 5 items.
The optimum is to pick itens 1, 3 and 4, and the value of those items is 14,000.
Note that, in order to get the optimum, I had to calculate every solution for 1 to number of itens and 1 to number weigths. Thus, the computational effort to calculate it is proportional to N*maxW.
The problem happens when N and W are large. It needs a lot of computation processing.
Knapsack is a pseudo polynomial time algorithm. Polynomial because it is proportional to N. But pseudo since it is not quite true it is proportional to N – it is also proportional to W, and W can be very very large. One day I’ll post a more detailed way of thinking of this.

Divide and conquer
Other question I had when studying dynamic programming for the first time was: what is the difference of this method compared to divide and conquer methods?
I think this discussion deserves a post apart, but just to give a simple explanation, divide and conquer has a “horizontal” structure, with subproblems independent of each other, while dynamic programming has a “nested” structure, with subproblems dependent of other subproblems.

Computation implementation
An Excel- VBA implementation of dynamic programming applied to the knapsack problem can be found at https://github.com/asgunzi/KnapsackDynamicProgrammingVBA.
It uses the Random case generator of previous post and generates a solution for this case.
ScreenDynProg.JPG

Other Applications
Dynamic programming can be applied in several other cases. Cutting stock problems are also a classic application.
One less known application is in the optimization of forest multiproducts. We have a forest of eucalyptus. Each tree has to be cut in several pieces. Each piece has a different diameter (since the tree is larger in the base). A product is a combination of diameter and lenght. Given a demand, how can I minimize the waste and maximize the use of the forest?

 

In the next posts, I’ll show alternative methods for solving the Knapsack Problem.

 

Arnaldo Gunzi.

 

 

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Author: Arnaldo Gunzi

"Navegadores antigos tinham uma frase gloriosa: Navegar é preciso, viver não é preciso"

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