Case 3: 2 1 3

Matrix 3:

Bag size

	0kg	1kg	2kg	3kg	4kg	5kg
0 (no item)	0	0	0	0	0	0
2 (item2)	0	A5	B5	C5	D5	E5
1 (item2/1)	0	F5	G5	H5	I5	J5
3 (item2/1/3)	0	K5	L5	M5	N5	O5

A5: show the math calculation

B5:

O5:

Table 3 (steal or not):

Bag size

	0kg	1kg	2kg	3kg	4kg	5kg
0 (no item)	0	0	0	0	0	0
2 (item2)	0	A6	B6	C6	D6	E6
1 (item2/1)	0	F6	G6	H6	I6	J6
3 (item2/1/3)	0	K6	L6	M6	N6	O6

Analysis for Case 3 (2, 1, 3):

0/1 Knapsack problem: dynamic programming There are 3 items in your house. The thief will come to your house. His maximum bag size is 5kg. The thief wants to maximize the profit, but he can not take more than 5kg.For each item, he can steal (1) or not steal (0). Item1: value =$5, weight =3kg Item2: value =$3, weight =2kg Item3: value =$4, weight =1kg No coding is needed. 1. Please fill in the matrix of bag size (column) and item (row) for the case of 2,1,3 And show the math calculation process for each cell in the matrix (bag size, item) 2. Also fill in the table of steal or not steal 3. show the traverse process in the matrix and the analysis for the optimal solution. The optimal solution is that the thief steals item 1 and item 3 with profit of 9 and weight of 4kg. The solution for each case should look like this