![]() ![]() So, our main task is to maximize the value i.e., $\sum_$ And the bag has a limitation of maximum weight ($W$). $x_i$ is the number of $i$ kind of items we have picked. There are different kinds of items ($i$) and each item $i$ has a weight ($w_i$) and value ($v_i$) associated with it. This problem is commonly known as the knapsack or the rucksack problem. So, you need to choose items to put in your bag such that the overall value of items in your bag maximizes. You are also provided with a bag to take some of the items along with you but your bag has a limitation of the maximum weight you can put in it. Each item has a different value and weight. you woke up on some mysterious island and there are different precious items on it. * Takes necessary inputs and initializes for solving Private int V //table to store results of sub-problems Private int w,v //weights and values of items Private int n,W //number of items and maximum capacity So item2 and item3 were selected to have total value of 9. ![]() Retracing our matrix, we get the following elements (marked with red). While creating the array, we had an option to select the element or reject it on the basis of the value we were obtaining. So the maximum value that we can get is 9, but what about the elements ? We can get the selected elements by tracing back the array. We get this matrix after filling the last row If we select this item, we'll get a value 5, and if we reject this item, then also we have 5, from above row. Now for item3 having weight 4, we have to copy the value till knapsack limit 3, on knapsack weight 4, we have an option of selecting/rejecting this item. ![]() Filling the second row in similar fashion. for knapsack of size 3, we have an option of selecting item1 or item2, since value of item2 is greater, we select item2 whose value is 4.įor knapsack size 4, we can have both item1 as well as item2, hence total value will be 5. Now for item 2 having weight 3, since weight 1, 2 knapsack cannot accomodate this item, for 1 and 2 size, we'll have 1 element only. Now for knapsack of size 1, we can have 1 item of size 1, and since we have only one item of each type, for higher weights knapsack also, we can accomodate only 1 item of type 1. To solve this problem through dynamic approach, we'll take a 2d array, whose first dimensions will represent each item, and second dimension will represent weight of knapsack.įor weight(0), or knapsack of size 0, we can have 0 items of weight 0, 0 items of weight 3, 0 items of weight 4 and 0 items of weight 5. But this method has a exponential time complexity. Lets say our knapsack has size W and these are the elements available to us.Ī recursive approach can be, select/deselect first element, then for each selection, select/deselect second element. In this problem, we have a size of knapsack (W), n elements and each element e i, has a value v i and weight w i. Fractional knapsack problem is rather easy and can be solved using greedy approach, whereas 0/1 knapsack problem is quite tricky and requires a dynamic approach. In fractional knapsack, we are allowed to select a fraction of the element, and in 0/1 knapsack, we can either select that element, or completely reject it. In both the problems, aim is the same, but there's a little difference. 01 knapsack full#Now our knapsack is full and we have the maximum value products. We are left with 1Kg space, here we can put 1Kg silver. Since we have to maximize the value, we have to take 4Kg of gold as its value is maximum. Say you have a knapsack which can handle 5KG weight, and you have 4Kg gold, 3kg silver and 7Kg bronze. You need to select those items which fits in the knapsack and their value is maximum. ![]() There are couple of items with different value and space. In a knapsack problem, you have a knapsack which has a limited space available. Before diving into the main 0/1 knapsack problem, lets first know about the original knapsack problem. ![]()
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