We need to show that this problem has the greedy choice property. To do this, we need to show that any solution X which does not include the greedy choice a does not have get a worse solution after swapping some choice with a. For fractional knapsack, this is very...

c,algorithm,dynamic-programming,knapsack-problem

The problem is that you are assigning infinite weight (cost) to all cases that use 0 tanks -- including, incorrectly, the case where oxy = nit = 0. In that case (i.e. dp[0][0][0]) the weight assigned should be 0, since you can supply 0 oxygen and 0 nitrogen quite comfortably...

javascript,algorithm,dynamic-programming,knapsack-problem

You want to look at each element of the items array. The first element is items[0] and the last element is items[items.length-1]. Therefore, you should change the line for (var i = 1; i <= items.length; i++) to this: for (var i = 0; i < items.length; i++) Now items[i].weight...

Let dp[i] be the maximum profit that can be earned if our budget is i. And cost[j] denote cost of j item and p[j] the profit earned from it. I assume cost[] and profit[] are given. Then following is the code in c++. ( let there be n items )....

Similar question with same code is already asked in StackOverflow But this solution is not memory efficient. Find below your corrected code. public class Algoritmid3 { public static int[] weight; public static int[] value; public static int maxW = 16; public static int n; public static class Node { int...

java,algorithm,out-of-memory,knapsack-problem

You said play-list so I'm assuming you have songs, and a typical song is about 3 minutes so your solution will be about 10 songs. Thus you can divide all your times by 50 and then the typical error for a song will be plus or minus 25 milliseconds and...

You could always use brute force (there may be more efficient approaches, though): %// Data V = [1 2 3 4]; X = 10; %// Generate all combinations of n integers from 0 to X %// (same approach as in http://stackoverflow.com/q/21895335/2586922 ) n = numel(V); c = cell(1,n); [c{:}] =...

You don't have to store all the bit sequences in a matrix, it's unnecessary and will waste way too much memory. You can simply use an integer to denote the current set. The integer will go from 0 to 2^n-1 where n is the number of elements that you can...

algorithm,dynamic-programming,knapsack-problem,partition-problem

This problem is a variation of the Partition Problem, where one set is the elements you add, and the other is the elements you substract. A possible solution is unsurprisingly based on the pseudo-polynomial solution of partition problem: D(V,0) = true D(s,0) = false s != V D(s,i) = false...

The program is incorrect, see this line: if (profit[n] > capacity) return knapsackRecursive(capacity, mass, profit, n-1); Here you compare profit with capacity. You should compare with mass[n]. The rest of the code looks ok for now. Perhaps you better use the maximum of a library and not the "ternary operator",...

python,algorithm,mathematical-optimization,knapsack-problem,greedy

This is an instance of the Knapsack problem. It is NP-hard, so with 45 items, you'll have to use some heuristic algorithm (Hill Climbing, for example) to find an acceptable estimate. To find the optimal solution, you have no other option than to try all possibilities (which is infeasible). Knowledge...

algorithm,dynamic-programming,knapsack-problem

One possibility would be to provide a suitable number of multiplicities of the items. For item i, there can be at most m_i := K / w_i choices of that item, where K denotes the knapsack capacity and w_i denotes the weight of the i-th item. Furthermore, for each weight...

algorithm,dynamic-programming,knapsack-problem

Instead of creating a table of size W*n, where each entry D[x][i] indicates the best value (highest) you can get with at most x weight using the first i items, use the table where now D[x][i] is the minimal weight needed to get to value of x, using the first...

algorithm,dynamic-programming,knapsack-problem

From your DP table we know f[i][w] = the maximum total value of a subset of items 1..i that has total weight less than or equal to w. We can use the table itself to restore the optimal packing: def reconstruct(i, w): # reconstruct subset of items 1..i with weight...

If you were to explore all combinations of addition and subtraction of 5000 numbers, you would have to go through 25000−1 ≈ 1.4⋅101505 alternatives. That's obviously not reasonable. However, since the sum of the numbers is at most 10000, we know that all partial sums (including subtraction) must lie between...

real-time,knapsack-problem,integer-programming

This is the standard knapsack problem which can be solved efficiently using dynamic programming. This has been discussed nicely in the knapsack wiki (and also in mutiple stack overflow posts e.g. here DP algo for knapsack and here recursive knapsack) The c++ implementation measures ~20us on my core i7 @3Ghz...

java,variable-assignment,comparator,knapsack-problem,genetic

You'd be better of having two comparators one comparing based on weight and one based on volume and the use those comparators to sort on demand.

java,recursion,knapsack-problem

Notice that in every recursive call value of W is also getting updated. And we subtract a new weight from leftover weight W only if it is less than W. Otherwise that weight can not be included. This logic is captured here if (wt[n-1] > W) return knapSack(W, wt, val,...

c++,c,algorithm,dynamic-programming,knapsack-problem

To my understanding, you almost have the solution. If the recurrence relation is correct but too inefficient, you just have change the recursion to iteration. Apparently, you already have the array dp which represents the states and their respective values. Basically you should be able to solve fill dp with...

sql,sql-server,sql-server-2008,tsql,knapsack-problem

This is hard coded and has little flexiblity. Took my system 2 minutes to run. But might be helpful, sorry if it isn't. fnGenerate_Numbers is a table function that returns integers within the range of the parameters. Ways to do that. DECLARE @Max INT, @Pens money, @Dvds money, @Pendrives money,...

sql-server,algorithm,tsql,combinatorics,knapsack-problem

Here is a possible solution. I will see if I can finish it tomorrow as it's almost 3 AM here now. The main logic is there. All that's left to be done is to trace back using the prev_w values. Just jump back (starting from the best_price row) till you...

python,algorithm,knapsack-problem

Let's change the meaning of m to the minimum weight for exactly the given value. We then want an initialization like m = [[(float('inf') if j else 0) for j in range(max_v + 1)] for _ in range(len(weights) + 1)] where I'm using positive infinity as a marker of infeasibility...

algorithm,graph,knapsack-problem

First of all, this problem is NP-hard. Here is a reduction from the shortest hamiltonian path in a weighted complete graph to this one. Given a graph, we can assign values of all nodes to 1 and then run binary search over the answer. If there was a polynomial solution...

From what I understand you can basically try to solve it as a normal knapsack problem in multiple iterations, finding the minimal. Now, finding the height of the knapsack is a problem, which means you need multiple iterations. Because you need to solve the knapsack problem to see if a...

c,dynamic-programming,memoization,knapsack-problem

So I know this is a homework question, so I won't give the the correct answer but I can say this: If a function is returning the same answer the second time it is run, it's usually because something hasn't been zeroed. There is some dirt left from the previous...

java,dynamic-programming,knapsack-problem

The recurrence relationship will be the same, it doesn't matter if you are programming it top down or bottom up. Since you have looked upon http://en.wikipedia.org/wiki/Knapsack_problem the recurrence relation will remain the same. Only change is that in general in Knapsack problem you are given a limit (weight limit) and...

c,algorithm,dynamic-programming,pseudocode,knapsack-problem

Knapsack problem can be solved using O(W) space. At each step of the iteration you need only 2 rows - current state of the array m[i] and m[i + 1]. current = 1 int m[2][W] set NONE for all elements of m # that means we are not able to...

dynamic,comparison,knapsack-problem,greedy

Although the question is a bit broadly asked, different criteria for comparison might include: actual running time vs. instance quality, i.e. relative gap to an optimal solution implementation difficulty As practical evaluation of algorithms is an involved subject, I suggest reading A Theoretician's Guide to the Experimental Analysis of Algorithms...

c++,arrays,algorithm,knapsack-problem

This is a variant of the 0-1 knapsack problem. This problem is NP-hard, so there is not much hope to find a solution in polynomial time, but there exists a solution in pseudo-polynomial time which makes this problem rather easy (in the world of complex problems). The algorithm works as...

dynamic-programming,knapsack-problem

Hasty judgment! I am sorry for some people! My problem, I replied: #include <iostream> #include <string> using namespace std; void main(){ // Ali Aghajani 2015/06/08 int i, w, W, n, p[50][2], B[50][50]; string R[50][50]; char end = 'y'; while (end == 'y' || end == 'Y'){ system("cls"); cout << "Enter...