knapsack | Knapsack splits tests evenly across parallel CI nodes | Continous Integration library
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kandi X-RAY | knapsack Summary
Knapsack splits tests evenly across parallel CI nodes to run fast CI build and save you time.
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knapsack Examples and Code Snippets
Lines of Code : 56License : Permissive (MIT License)
Copydef knapsack_with_example_solution(W: int, wt: list, val: list):
"""
Solves the integer weights knapsack problem returns one of
the several possible optimal subsets.
Parameters
---------
W: int, the total maximum weight for def fractional_knapsack(
value: list[int], weight: list[int], capacity: int
) -> tuple[float, list[float]]:
"""
>>> value = [1, 3, 5, 7, 9]
>>> weight = [0.9, 0.7, 0.5, 0.3, 0.1]
>>> fractional_knapsa def knapsack(capacity: int, weights: list[int], values: list[int], counter: int) -> int:
"""
Returns the maximum value that can be put in a knapsack of a capacity cap,
whereby each weight w has a specific value val.
>>> c Community Discussions
Trending Discussions on knapsack
QUESTION
I've been trying to implement a modified knapsack problem algorithm regarding bioinformatics.
What I have so far is, in my opinion, pretty close to the solution, but the program gets stuck at a certain point.
I have a list of nodes which have mass (of a certain amino-acid), index, and list of nodes that they can get to.
NODE:
...ANSWER
Answered 2021-Jun-03 at 11:40While trying to debug the code, the problem seemed to be in the whole concept of the attribute next in the Node class.
When I printed out all of the Nodes' next lists, I found multiple occurences of the same Node, for example [2,2,2,3,8,...] so when I converted the list to set it didn't get stuck anymore.
Hope this helps someone in the future.
QUESTION
I am trying to solve this problem: There are n customers queuing at post office to wait to send parcels. a[0], a[1], ..., a[n-1] is the list of shipping costs of n customers from the 1st to the nth person. It takes exactly a minute for the postal worker to complete the information needed for a customer to send a parcel. However, all customers are too busy to wait for more than a certain period of time. t[0], t[1], ..., t[n-1] is the list of minutes each of n customers can spend at the post office. Help the postal worker to find a way to serve customers so that the post office can get the largest amount of money, knowing that the staff is allowed to refuse to serve some customers for the profitable reason.)
Example:
- For
a = [10, 20, 5, 12], t = [2, 3, 3, 1], the output should be42. Explanation: The order of the customers is: the 4th person -> the 1st person -> the 2nd person (1-based indexing) - For
a = [5, 1, 3, 2], t = [3, 1, 2, 2], the output should be10. Explanation: Although the 2nd person can wait only 1 minute, this person has to pay the smallest cost. Therefore, the postal worker will not serve this customer. The order of the customers is: the 3rd person -> the 4th person -> the 1st person.
I think it is a variant of the knapsack problem, I can solve it by using brute force but only for small input. Can someone help me to solve this problem? Thanks.
...ANSWER
Answered 2021-May-18 at 22:45If there are no overlapping times, the problem is straightforward just sum up all the shipping costs. The problem becomes non-trivial if there is overlap.
So lets form a tuple of the (time, cost) and sort them first by time and then by cost(descending).
For example for the input:
QUESTION
I am trying to create an array using a struct in c++ which takes two variables i.e value and weight. So I created an array which will have value and weight in one element like this Arr[]={{1,2},{3,4}}...and i want that if i called Arr[0].value and Arr[0].weight then it should return 1 and 2 respectively but I think I'm doing something wrong because im getting many errors
...ANSWER
Answered 2021-May-15 at 03:58Your biggest compile issues are this:
You got to give your Item a default constructor, otherwise, it can't exist in an uninitialized array.
QUESTION
KNAPSACK problem is old, but this has some differences:
At a store that has n items, each item has weight W[i] and value V[i]. A bag can include maximum weight W. Choose some items, which total have max values. And an item can be chosen more than one time.
I only can finish if an item can be chosen one time but more than one time is hard for me. So can you help me!
...ANSWER
Answered 2021-May-08 at 10:00It is called Unbounded Knapsack. You can read about it here
QUESTION
Trying to use GEKKO to solve knapsack optimization problem, but can't reach a satisfactory result. I have input data:
...ANSWER
Answered 2021-Mar-26 at 21:18Try increasing the solver tolerance.
QUESTION
I am working on my version of a genetic algorithm to solve the knapsack problem in C++. I have a map of string to vector like
...ANSWER
Answered 2021-Mar-22 at 06:00For both vector of vectors as well as a map, you can use for each loop! When your map is filled with value
QUESTION
I have a very simple knapsack problem listed below.
The sample of the df looks like this:
...ANSWER
Answered 2021-Mar-18 at 08:51One way to do this would be to re-solve the problem with an additional constraint added to forbid the existing best solution. For example say the solutions to the original problem are stored as player_vars_soln, you would then add a constraint:
prob += pulp.lpSum(player_vars_soln[i]*player_vars[i] for i in range(len(df))) <= 3
QUESTION
I am trying to do a 0-1 knapsack optimization with constraint on the source of items. I took the example from the ortools website (ortool example) and try to add a constraint to only be able to pick one item from each owner in the knapsack.
I have a list of items with associated weights (data['weights']), values (data['values']) and source (data['owns']). I would like to find the best combination of items to put in knapsack knowing that only one item per source can go in the knapsack.
I am not sure how to write the constraint.
If you look at the code below and have 1 knapsack, then the optimal solution should be to take at most 1 item from owner 0, one from owner 1 and one from owner 2 which follows the weight constraint and the uniqueness of item picked (weight below 100).
Here is the code I use (taken from the ortool multiple knapsack example):
...ANSWER
Answered 2021-Mar-13 at 16:00First of all, I would like to thank @sascha who provided the answer!
I attach below the code (I use 2 bins, 3 different owners):
QUESTION
I have a problem and I feel like there should be a well-known algorithm for solving it that's better than just brute force, but I can't think of one, so I'm asking here.
The problem is as follows: given n sorted (from low to high) lists containing m probabilities, choose one index for each list such that the sum of the chosen indexes is less than m. Then, for each list, we flip a coin, where the chance of it landing heads is equal to the probability at the chosen index for that list. Maximize the chance of the coin landing heads at least once.
Are there any algorithms for solving this problem that are better than just brute force?
This problem seems most similar to the knapsack problem, except the value of the items in the knapsack isn't merely a sum of the items in the knapsack. (Written in Python, instead of sum(p for p in chosen_probabilities) it's 1 - math.prod([1 - p for p in chosen_probabilities])) And, there's restrictions on what items you can add given what items are already in the knapsack. For example, if the index = 3 item for a particular list is already in the knapsack, then adding in the item with index = 2 for that same list isn't allowed (since you can only pick one index for each list). So there are certain items that can and can't be added to the knapsack based on what items are already in it.
Linear optimization won't work because the values in the lists don't increase linearly, the final coin probability isn't linear with respect to the chosen probabilities, and our constraint is on the sum of the indexes, rather than the values in the lists themselves. As David has pointed out, linear optimization will work if you use binary variables to pick out the indexes and a logarithm to deal with the non-linearity.
EDIT:
I've found that explaining the motivation behind this problem can be helpful for understanding it. Imagine you have 10 seconds to solve a problem, and three different ways to solve it. You have models of how likely it is that each method will solve the problem, given how many seconds you try that method for, but if you switch methods, you lose all progress on the one you were previously trying. What methods should you try and for how long?
...ANSWER
Answered 2021-Mar-03 at 15:50Maximizing 1 - math.prod([1 - p for p in chosen_probabilities]) is equivalent to minimizing math.prod([1 - p for p in chosen_probabilities]), which is equivalent to minimizing the log of this objective, which is a linear function of 0-1 indicator variables, so you could do an integer programming formulation this way.
I can't promise that this will be much better than brute force. The problem is that math.log(1 - p) is well approximated by -p when p is close to zero. My intuition is that for nontrivial instances it will be qualitatively similar to using integer programming to solve subset sum, which doesn't go particularly well.
If you're willing to settle for a bicriteria approximation scheme (get an answer such that the sum of the chosen indexes is less than m, that is at least as good as the best answer summing to less than (1 − ε) m) then you can round up the probability to multiples of ε and use dynamic programming to get an algorithm that runs in time polynomial in n, m, 1/ε.
QUESTION
I have a program (knapsack, optimized for returning the highest value solution with the least weight) for which I want to use external files for the typedef and struct data.
But I can't get it working, somewhere I am messing up with the pointers. I get an 'multiple definition' error or when I change it, I get an 'xxx not declared' error...
/tmp/ccMy5Yw0.o:(.data+0x0): multiple definition of `item1'
Any help on pointing out my thinking mistake in greatly appreciated. (I compiled online @ https://www.onlinegdb.com/)
It did work when I had everything in one file, but after splitting it in different files I can't get it working...
...ANSWER
Answered 2021-Feb-25 at 11:42As you included data.c in main.c
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On a UNIX-like operating system, using your system’s package manager is easiest. However, the packaged Ruby version may not be the newest one. There is also an installer for Windows. Managers help you to switch between multiple Ruby versions on your system. Installers can be used to install a specific or multiple Ruby versions. Please refer ruby-lang.org for more information.
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