Knapsack | 4 implementations of the 0-1 knapsack problem

 by   patrickherrmann Java Version: Current License: MIT

kandi X-RAY | Knapsack Summary

kandi X-RAY | Knapsack Summary

Knapsack is a Java library. Knapsack has no bugs, it has no vulnerabilities, it has a Permissive License and it has low support. However Knapsack build file is not available. You can download it from GitHub.

4 implementations of the 0-1 knapsack problem and a comparison of their effectiveness.

            kandi-support Support

              Knapsack has a low active ecosystem.
              It has 30 star(s) with 20 fork(s). There are 4 watchers for this library.
              It had no major release in the last 6 months.
              There are 1 open issues and 0 have been closed. On average issues are closed in 1036 days. There are 1 open pull requests and 0 closed requests.
              It has a neutral sentiment in the developer community.
              The latest version of Knapsack is current.

            kandi-Quality Quality

              Knapsack has 0 bugs and 33 code smells.

            kandi-Security Security

              Knapsack has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.
              Knapsack code analysis shows 0 unresolved vulnerabilities.
              There are 0 security hotspots that need review.

            kandi-License License

              Knapsack is licensed under the MIT License. This license is Permissive.
              Permissive licenses have the least restrictions, and you can use them in most projects.

            kandi-Reuse Reuse

              Knapsack releases are not available. You will need to build from source code and install.
              Knapsack has no build file. You will be need to create the build yourself to build the component from source.
              Knapsack saves you 114 person hours of effort in developing the same functionality from scratch.
              It has 288 lines of code, 24 functions and 8 files.
              It has medium code complexity. Code complexity directly impacts maintainability of the code.

            Top functions reviewed by kandi - BETA

            kandi has reviewed Knapsack and discovered the below as its top functions. This is intended to give you an instant insight into Knapsack implemented functionality, and help decide if they suit your requirements.
            • Solves the capacity of the sketch with the given capacity
            • Transforms a list of subsets into a list of subsets
            • Get the total value for a list item
            • Get the weight of a list of items
            • Runs the Knapsack solution
            • Creates a comparator for items by their ratio
            • Get the ratio
            • Searches the capacity of the equation
            • Tracks the entire table
            • Get cell
            • Returns a string representation of the items
            • Build a comparator for items by label
            • Solves the greedy solution
            • Main entry point
            Get all kandi verified functions for this library.

            Knapsack Key Features

            No Key Features are available at this moment for Knapsack.

            Knapsack Examples and Code Snippets

            No Code Snippets are available at this moment for Knapsack.

            Community Discussions


            To which Knapsack-problem variation does this problem correspond?
            Asked 2022-Mar-19 at 17:06
            Let us imagine that I have to fill my knapsack with items under constraints:
            • Each item has an associated weight wi and profit pi
            • With a maximum total weight Wmax
            Knowing that:
            • There are categories of items and I have to choose exactly one item from each category
            • Of course, the aim is to choose items to maximise the sum of the profits
            Example : Wmax=400 Books Books weights Books profits Food Food weights Food profits The Bible 500 25 Cheese 80 120 The little prince 150 5 Banana 250 200

            Here, the best solution is (The little prince, Banana)

            I have a similar problem and I'd like to find out the best way to code it but I can't figure out what version/ variation of the probleme this is, is it a known variation ?



            Answered 2022-Mar-19 at 17:06

            I’m not sure if there’s an existing variation that matches yours, but it’s easy to draw inference from the classical variant and solve this.

            Classic variant has 2D dynamic programming (DP[N][W], where N and W are number of items and max weight).

            In this variant, since we can only pick one of each category, you can use 3D DP like dp[i][w][j], which denotes the maximum value you can get from the first i items with weight w and j is a 0/1 int denoting whether an item from category number j has been selected or not.

            I’ll leave the implementation, since the recursive relation is relatively simple and quite similar to the classic variant.



            Maximize the number of items bought
            Asked 2022-Mar-08 at 20:21

            We are given an array Ai denoting the price of item i.

            We have k amount of currency and we can buy the first item n times, second item n-1 times, third item n-2 times and so on. Find the maximum number of items which can be bought.



            Answered 2022-Mar-08 at 20:21

            If you want to maximize the number of items you can buy you don't have a Knapsack problem (in a Knapsack problem each item has some value or weight and you want to maximize the overall weight of the items you pack). A greedy solution should work for your problem. Just sort the items according to their price (you have to remember to keep track of how many times you can buy each item, e.g. using pairs) and then buy the cheapest available until you run out of money. This should give you an O(nlog(n)) solution.



            Using Iterative deepening DFS for knapsack similar problem
            Asked 2022-Mar-03 at 08:38

            I'm solving a knapsack similar problem: which is to print out the first combination of objects that has value above a number but weight below a limit.

            I have tried:



            Answered 2022-Feb-28 at 07:11

            I tried this and it works for some samples:



            MongoDB - bulk updating values in arrays of nested objects
            Asked 2022-Feb-21 at 19:46

            I have a collection in MongoDB that represents the chores kids needs to do each day and whether they are done. Here is an example of a document.



            Answered 2022-Feb-21 at 15:44

            You could use the update method with aggregation pipeline to update the chores array. You would need to use the following pipeline to achieve the desired results:



            what is the maximum possible number of monsters you can defeat?
            Asked 2021-Dec-18 at 09:28

            Problem Statement:

            While playing an RPG game, you were assigned to complete one of the hardest quests in this game. There are n monsters you'll need to defeat in this quest. Each monster i is described with two integer numbers - poweri and bonusi. To defeat this monster, you'll need at least poweri experience points. If you try fighting this monster without having enough experience points, you lose immediately.

            You will also gain bonusi experience points if you defeat this monster. You can defeat monsters in any order. The quest turned out to be very hard - you try to defeat the monsters but keep losing repeatedly. Your friend told you that this quest is impossible to complete. Knowing that, you're interested, what is the maximum possible number of monsters you can defeat?


            The first line contains an integer, n, denoting the number of monsters.

            The next line contains an integer, e, denoting your initial experience.

            Each line i of the n subsequent lines (where 0 ≤ i < n) contains an integer, poweri, which represents power of the corresponding monster.

            Each line i of the n subsequent lines (where 0 ≤ i < n) contains an integer, bonusi, which represents bonus for defeating the corresponding monster.

            Sample cases:

            Input 2 123 78 130 10 0

            Output 2

            Output description

            Initial experience level is 123 points. Defeat the first monster having power of 78 and bonus of 10. Experience level is now 123+10=133. Defeat the second monster.

            What I have tried:



            Answered 2021-Aug-07 at 17:35

            You can just sort the monsters from lowest to highest power required and defeat them in that order.



            Problems with a branch and bound knapsack algorithm
            Asked 2021-Dec-14 at 13:51

            I am working on a knapsack optimization exercise written in Python. The goal is to fill a knapsack with limited weight capacity with items with a certain value and weight resulting maximizing the total value of the knapsack. I ran into a problem with setting the variables when a better solution is found.

            I've created a simplified version of the code without any constraints, demonstrating my problem. Since there are no constraints, the first solution is always the best solution since that is the solution where all items are taken.

            There is an if statement in the code that sets the max_value and best_taken variables in case a better solution is found.

            However: When the print statement at the end of the code prints the max_value and best_taken variables, the max_value shows the correct value (the sum of the first indices in the list of lists assigned to the items variable). The best_taken value always ends up as [0,0] where a value of [1,1] (=take both items) is expected.

            What am I doing wrong?



            Answered 2021-Dec-14 at 13:51

            You need to copy local_taken. Assigning an array to an attribute just adds a reference to the same (mutable) array

            self.best_taken = self.local_taken[:]



            How can I correct objective function error in OR Tools in Python?
            Asked 2021-Dec-02 at 15:06

            I'm working on a variation of the 0-1 knapsack problem using OR tools. I've managed to solve one type of variation, however, on this attempt, I'm having difficulty specifically with the objective function. Here is the code that I'm using. Basically, I want to sum each of the variables associated with an item in the objective function, and then if the item is selected the x[i,j] binary variable will be set to 1:



            Answered 2021-Dec-02 at 14:48

            most likely you data is a numpy.float(). Cast it to float in SetCoefficient.



            Counting the number of ways to make up a string
            Asked 2021-Oct-22 at 00:51

            I have just started learning dynamic programming and was able to do some of the basic problems, such as fibbonaci, the knapsack and a few more problems. Coming across the problem below, I got stuck and do not know how to proceed forward. What confuses me is what would be the base case in this case, and the overlapping problems. Not knowing this prevents me from developing a relation. They are not as apparent in this example as they were in the previous ones I have solved thus far.

            Suppose we are given some string origString, a string toMatch and some number maxNum greater than or equal to 0. How can we count in how many ways it is possible to take maxNum number of nonempty and nonoverlapping substrings of the string origString to make up the string toMatch?


            If origString = "ppkpke", and toMatch = "ppke"

            maxNum = 1: countWays("ppkpke", "ppke", 1) will give 0 because toMatch is not a substring of origString.

            maxNum = 2: countWays("ppkpke", "ppke", 2) will give 4 because 4 different combinations of 2 substring made up of "ppkpke" can make "ppke". Those strings are "ppk" & "e", "pp" & "ke" , "p" & "pke" (excluding "p") and "p" & "pke" (excluding "k")



            Answered 2021-Oct-12 at 07:28

            As an initial word of caution, I’d say that although my solution happens to match the expected output for the tiny test set, it is very likely wrong. It’s up to you to double-check it on other examples you may have etc.

            The algorithm walks the longer string and tries to spread the shorter string over it. The incremental state of the algorithm consists of tuples of 3 elements:

            1. long string coordinate i (origString[i] == toMatch[j])
            2. short string coordinate j (origString[i] == toMatch[j])
            3. number of ways we made it into that^^^ state

            Then we just walk along the strings over and over again, using stored, previously discovered state, and sum up the total number(s) of ways each state was achieved — in the typical dynamic programming fashion.

            For a state to count as a solution, j must be at the end of the short string and the number of iterations of the dynamic algorithm must be equivalent to the number of substrings we wanted at that point (because each iteration added one substring).

            It is not entirely clear to me from the assignment whether maxNum actually means something like “exactNum”, i.e. exactly that many substrings, or whether we should sum across all lower or equal numbers of substrings. So the function returns a dictionary like { #substrings : #decompositions }, so that the output can be adjusted as needed.



            How does the O(n sqrt(n)) algorithm work that lists all possible sums given an array of numbers?
            Asked 2021-Oct-20 at 15:39

            I came across one algorithm problem in the book Competitive programming handbook Integer partitions, Knapsack.

            The problem is: Given an array of integers whose sum is n. Output all possible sums that can be formed by using a subset of integers.

            The dynamic programming solution is understandable. Define dp(x,k) as a bool function that is true when we can use the first k numbers to sum up to x, and false otherwise. Then dp(x,k) = dp(x,k-1) | dp(x-A[k],k-1). The complexity is O(n^2).

            However, the book described another solution that leverages one fact that the number of distinct numbers in the array is O(sqrt (n)), and then mentioned a solution by grouping similar numbers together. How does it work? The book seems hard to follow.



            Answered 2021-Oct-20 at 15:39

            I agree that the author should have been more explicit. My reconstruction is that, given an element a and an (n+1)-element bit array indicating which sums can be made by subsets of the previously considered elements, we can calculate in linear time the minimum number of copies of a that we need to make a particular sum, then threshold by how many copies of a we actually have. Python implementation below.



            Confusing Output for 0/1 Knapsack Problem
            Asked 2021-Oct-09 at 08:47

            I'm learning Dynamic Programming, and have the book Grokking Algorithms

            The example given there uses these images/values:

            When I run the code below, which is supposed to implement the 0/1 Knapsack Problem in Python, I get the following output:



            Answered 2021-Oct-09 at 08:47
            | 0 |  0   |  0   |  0   |  0   |
            | 0 | 1500 | 1500 | 1500 | 1500 |  guitar
            | 0 | 1500 | 1500 | 2000 | 3500 |  laptop
            | 0 | 1500 | 1500 | 2000 | 3500 |  stereo


            Community Discussions, Code Snippets contain sources that include Stack Exchange Network


            No vulnerabilities reported

            Install Knapsack

            You can download it from GitHub.
            You can use Knapsack like any standard Java library. Please include the the jar files in your classpath. You can also use any IDE and you can run and debug the Knapsack component as you would do with any other Java program. Best practice is to use a build tool that supports dependency management such as Maven or Gradle. For Maven installation, please refer For Gradle installation, please refer .


            For any new features, suggestions and bugs create an issue on GitHub. If you have any questions check and ask questions on community page Stack Overflow .
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