human
Double Knapsack | Dynamic Programming | w1_r represents remaining capacity of 1 st knapsack w2_r represents remaining capacity of 2 nd knapsack i represents index of the array arr we are working on.
Base case.
Variables to store the result of three parts of recurrence relation.
Store the state in the 3D array.
Driver code.
Input array.
3D array to store states of DP.
Number of elements in the array.
Capacity of knapsacks.
Function to be called"
Solution in Python:

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What is a code to double Knapsack and the other steps required?
assistant
#Here is code to double Knapsack and the other steps required.
 def maxWeight ( arr , n , w1_r , w2_r , i ) :
	if i == n :
		return 0
	if dp [ i ] [ w1_r ] [ w2_r ] != - 1 :
		return dp [ i ] [ w1_r ] [ w2_r ]
	fill_w1 , fill_w2 , fill_none = 0 , 0 , 0
	if w1_r >= arr [ i ] :
		fill_w1 = arr [ i ] + maxWeight ( arr , n , w1_r - arr [ i ] , w2_r , i + 1 )
	if w2_r >= arr [ i ] :
		fill_w2 = arr [ i ] + maxWeight ( arr , n , w1_r , w2_r - arr [ i ] , i + 1 )
	fill_none = maxWeight ( arr , n , w1_r , w2_r , i + 1 )
	dp [ i ] [ w1_r ] [ w2_r ] = max ( fill_none , max ( fill_w1 , fill_w2 ) )
	return dp [ i ] [ w1_r ] [ w2_r ]
if __name__ == " _ _ main _ _ " :
	arr = [ 8 , 2 , 3 ]
	maxN , maxW = 31 , 31
	dp = [ [ [ - 1 ] * maxW ] * maxW ] * maxN
	n = len ( arr )
	w1 , w2 = 10 , 3
	print ( maxWeight ( arr , n , w1 , w2 , 0 ) )