It’s an interesting problem as it is both an NP-hard (optimization) and NP-complete (decision) problem that has a pseudo-polynomial algorithm, meaning its worst-case time complexity is polynomial in numeric values of the input. The Knapsack problem is actually a combinatorial optimization problem where we need to find an optimal number of objects from a finite set of objects usually based on their properties. To that I say, I will… but I also won’t! I’ll of course try all the algorithms, but since this is a discrete optimization course, I also thought “let me make this even more painful” and try to optimize each algorithm as best I can 🙂
The course is very adamant in trying all the different covered algorithms on all problems for example, if solving Knapsack starts to be too time consuming with DP, use Linear Programming or Branch & Bound with some type of constraint relaxation. As a way for me to brush up on some basics I thought I’d revisit dynamic programming, an often used technique in solving discrete optimization problems, with the Knapsack problem.
I’m currently taking a course on discrete optimization, a field I’ve always thought was cool but never been great at and one which also has a reputation for being particularly hard as you progress deeper. Unfortunately, the times I encountered it in a real situation I never had to actually fully solve it – it was more along the lines of, “hey, this is just like the Knapsack problem, but we can work around it”. Programmatically, we iterate over all the elements available for each knapsack capacity between 1 to W and determine if it can be used to achieve a greater profit.The Knapsack problem is one I’ve encountered a handful of times, both in my studies (courses, homework, whatever…), and in real life. dp = maximum profit we can achieve with a knapsack capacity of i Here, W is the total knapsack capacity, hence our answer would be dp. Thus, our array would be dp, where dp indicates the maximum profit we can achieve with a knapsack capacity of i. This is because we have infinite supply of every element available to us and hence, we don't need to keep a track of which elements have been used. The only difference is we would use a single dimensional array instead of 2-D one used in the classical one. We use dynamic programming approach to solve this problem, similar to what we did in classical knapsack problem. This would be highly inefficient, given the computation time. nCn = 2^n possible combinations of n items. This way, choosing from all combination would mean a time complexity of orderĪs there are total nC0 + nC1 +.
Knapsack dynamic programming update#
MaxProfit = max(maxProfit, profit) // Update max profit. Note the total benefit is (41+41+12+12+2) = 108 with total weight being 57 ( W: // When weight of the combination exceeds Capacity, Break. In this case, the optimal filling will be: Item 3 + 3 + 1 + 1 + 2 Suppose we have three items which is defined by a tuple (weight, benefit). In the Unbounded version of the problem, we are allowed to select one item multiple times, unlike the classical one, where one item is allowed to be selected only once. The objective is to fill the knapsack with items such that we have a maximum profit without crossing the weight limit of the knapsack. Given N items each with an associated weight and value (benefit or profit). Read about the general Knapsack problem here This problem can be solved efficiently using Dynamic Programming. In this case, an item can be used infinite times. This restriction is removed in the new version: Unbounded Knapsack Problem. In the original problem, the number of items are limited and once it is used, it cannot be reused. The objective is the increase the benefit while respecting the bag's capacity. Knapsack problem refers to the problem of optimally filling a bag of a given capacity with objects which have individual size and benefit. Reading time: 30 minutes | Coding time: 10 minutes
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