0/1 Knapsack using Least Cost Branch and Bound

Given N items with weights W[0..n-1], values V[0..n-1] and a knapsack with capacity C, select the items such that: 

1. The sum of weights taken into the knapsack is less than or equal to C.
2. The sum of values of the items in the knapsack is maximum among all the possible combinations.

Examples:

Input: N = 4, C = 15, V[]= {10, 10, 12, 18}, W[]= {2, 4, 6, 9} 
Output:
Items taken into the knapsack are 
1 1 0 1 
Maximum profit is 38 
Explanation:
1 in the output indicates that the item is included in the knapsack while 0 indicates that the item is excluded. 
Since the maximum possible cost allowed is 15, the ways to select items are: 
(1 1 0 1) -> Cost = 2 + 4 + 9 = 15, Profit = 10 + 10 + 18 = 38. 
(0 0 1 1) -> Cost = 6 + 9 = 15, Profit = 12 + 18 = 30 
(1 1 1 0) -> Cost = 2 + 4 + 6 = 12, Profit = 32 
Hence, maximum profit possible within a cost of 15 is 38.
Input: N = 4, C = 21, V[]= {18, 20, 14, 18}, W[]= {6, 3, 5, 9} 
Output:
Items taken into the knapsack are 
1 1 0 1 
Maximum profit is 56 
Explanation:
Cost = 6 + 3 + 9 = 18 
Profit = 18 + 20 + 18 = 56 

Approach:
In this post, the implementation of Branch and Bound method using Least cost(LC) for 0/1 Knapsack Problem is discussed.
Branch and Bound can be solved using FIFO, LIFO and LC strategies. The least cost(LC) is considered the most intelligent as it selects the next node based on a Heuristic Cost Function. It picks the one with the least cost. 
As 0/1 Knapsack is about maximizing the total value, we cannot directly use the LC Branch and Bound technique to solve this. Instead, we convert this into a minimization problem by taking negative of the given values. 
Follow the steps below to solve the problem: 

1. Sort the items based on their value/weight(V/W) ratio.
2. Insert a dummy node into the priority queue.
3. Repeat the following steps until the priority queue is empty:
   • Extract the peek element from the priority queue and assign it to the current node.
   • If the upper bound of the current node is less than minLB, the minimum lower bound of all the nodes explored, then there is no point of exploration. So, continue with the next element. The reason for not considering the nodes whose upper bound is greater than minLB is that, the upper bound stores the best value that might be achieved. If the best value itself is not optimal than minLB, then exploring that path is of no use. 
     
   • Update the path array.
   • If the current node’s level is N, then check whether the lower bound of the current node is less than finalLB, minimum lower bound of all the paths that reached the final level. If it is true, update the finalPath and finalLB. Otherwise, continue with the next element.
   • Calculate the lower and upper bounds of the right child of the current node.
   • If the current item can be inserted into the knapsack, then calculate the lower and upper bound of the left child of the current node.
   • Update the minLB and insert the children if their upper bound is less than minLB.

Illustration:
N = 4, C = 15, V[]= {10 10 12 18}, W[]= {2 4 6 9} 

👁 0 1 Knapsack using Least Cost Branch and Bound

Left branch and right branch at i^th level stores the maximum obtained including and excluding the i^th element.
Below image shows the state of the priority queue after every step: 

👁 priority queue for 0 1 knapsack problem using Least Cost Branch and Bound

Below is the implementation of the above approach:

Items taken into the knapsack are : 
1 1 0 1 
Maximum profit is : 38