**Knapsack Problem-**

You are given the following-

- A knapsack (kind of shoulder bag) with limited weight capacity.
- Few items each having some weight and value.

**The problem states-**

Which items should be placed into the knapsack such that-

- The value or profit obtained by putting the items into the knapsack is maximum.
- And the weight limit of the knapsack does not exceed.

**Knapsack Problem Variants-**

Knapsack problem has the following two variants-

- Fractional Knapsack Problem
- 0/1 Knapsack Problem

In this article, we will discuss about 0/1 Knapsack Problem.

**0/1 Knapsack Problem-**

In 0/1 Knapsack Problem,

- As the name suggests, items are indivisible here.
- We can not take the fraction of any item.
- We have to either take an item completely or leave it completely.
- It is solved using dynamic programming approach.

**Also Read-** **Fractional Knapsack Problem**

**0/1 Knapsack Problem Using Dynamic Programming-**

Consider-

- Knapsack weight capacity = w
- Number of items each having some weight and value = n

0/1 knapsack problem is solved using dynamic programming in the following steps-

**Step-01:**

- Draw a table say ‘T’ with (n+1) number of rows and (w+1) number of columns.
- Fill all the boxes of 0
^{th}row and 0^{th}column with zeroes as shown-

**Step-02:**

Start filling the table row wise top to bottom from left to right.

Use the following formula-

**T (i , j) = max { T ( i-1 , j ) , value _{i} + T( i-1 , j – weight_{i }) }**

Here, T(i , j) = maximum value of the selected items if we can take items 1 to i and have weight restrictions of j.

- This step leads to completely filling the table.
- Then, value of the last box represents the maximum possible value that can be put into the knapsack.

**Step-03:**

To identify the items that must be put into the knapsack to obtain that maximum profit,

- Consider the last column of the table.
- Start scanning the entries from bottom to top.
- On encountering an entry whose value is not same as the value stored in the entry immediately above it, mark the row label of that entry.
- After all the entries are scanned, the marked labels represent the items that must be put into the knapsack.

**Time Complexity-**

- Each entry of the table requires constant time θ(1) for its computation.
- It takes θ(nw) time to fill (n+1)(w+1) table entries.
- It takes θ(n) time for tracing the solution since tracing process traces the n rows.
- Thus, overall θ(nw) time is taken to solve 0/1 knapsack problem using dynamic programming.

**PRACTICE PROBLEM BASED ON 0/1 KNAPSACK PROBLEM-**

**Problem-**

For the given set of items and knapsack capacity = 5 kg, find the optimal solution for the 0/1 knapsack problem making use of dynamic programming approach.

Item |
Weight |
Value |

1 | 2 | 3 |

2 | 3 | 4 |

3 | 4 | 5 |

4 | 5 | 6 |

**OR**

Find the optimal solution for the 0/1 knapsack problem making use of dynamic programming approach. Consider-

n = 4

w = 5 kg

(w1, w2, w3, w4) = (2, 3, 4, 5)

(b1, b2, b3, b4) = (3, 4, 5, 6)

**OR**

A thief enters a house for robbing it. He can carry a maximal weight of 5 kg into his bag. There are 4 items in the house with the following weights and values. What items should thief take if he either takes the item completely or leaves it completely?

Item |
Weight (kg) |
Value ($) |

Mirror | 2 | 3 |

Silver nugget | 3 | 4 |

Painting | 4 | 5 |

Vase | 5 | 6 |

**Solution-**

**Given-**

- Knapsack capacity (w) = 5 kg
- Number of items (n) = 4

**Step-01:**

- Draw a table say ‘T’ with (n+1) = 4 + 1 = 5 number of rows and (w+1) = 5 + 1 = 6 number of columns.
- Fill all the boxes of 0
^{th}row and 0^{th}column with 0.

**Step-02:**

Start filling the table row wise top to bottom from left to right using the formula-

**T (i , j) = max { T ( i-1 , j ) , value _{i} + T( i-1 , j – weight_{i }) }**

**Finding T(1,1)-**

We have,

- i = 1
- j = 1
- (value)
_{i}= (value)_{1}= 3 - (weight)
_{i}= (weight)_{1}= 2

Substituting the values, we get-

T(1,1) = max { T(1-1 , 1) , 3 + T(1-1 , 1-2) }

T(1,1) = max { T(0,1) , 3 + T(0,-1) }

T(1,1) = T(0,1) { Ignore T(0,-1) }

T(1,1) = 0

**Finding T(1,2)-**

We have,

- i = 1
- j = 2
- (value)
_{i}= (value)_{1}= 3 - (weight)
_{i}= (weight)_{1}= 2

Substituting the values, we get-

T(1,2) = max { T(1-1 , 2) , 3 + T(1-1 , 2-2) }

T(1,2) = max { T(0,2) , 3 + T(0,0) }

T(1,2) = max {0 , 3+0}

T(1,2) = 3

**Finding T(1,3)-**

We have,

- i = 1
- j = 3
- (value)
_{i}= (value)_{1}= 3 - (weight)
_{i}= (weight)_{1}= 2

Substituting the values, we get-

T(1,3) = max { T(1-1 , 3) , 3 + T(1-1 , 3-2) }

T(1,3) = max { T(0,3) , 3 + T(0,1) }

T(1,3) = max {0 , 3+0}

T(1,3) = 3

**Finding T(1,4)-**

We have,

- i = 1
- j = 4
- (value)
_{i}= (value)_{1}= 3 - (weight)
_{i}= (weight)_{1}= 2

Substituting the values, we get-

T(1,4) = max { T(1-1 , 4) , 3 + T(1-1 , 4-2) }

T(1,4) = max { T(0,4) , 3 + T(0,2) }

T(1,4) = max {0 , 3+0}

T(1,4) = 3

**Finding T(1,5)-**

We have,

- i = 1
- j = 5
- (value)
_{i}= (value)_{1}= 3 - (weight)
_{i}= (weight)_{1}= 2

Substituting the values, we get-

T(1,5) = max { T(1-1 , 5) , 3 + T(1-1 , 5-2) }

T(1,5) = max { T(0,5) , 3 + T(0,3) }

T(1,5) = max {0 , 3+0}

T(1,5) = 3

**Finding T(2,1)-**

We have,

- i = 2
- j = 1
- (value)
_{i}= (value)_{2}= 4 - (weight)
_{i}= (weight)_{2}= 3

Substituting the values, we get-

T(2,1) = max { T(2-1 , 1) , 4 + T(2-1 , 1-3) }

T(2,1) = max { T(1,1) , 4 + T(1,-2) }

T(2,1) = T(1,1) { Ignore T(1,-2) }

T(2,1) = 0

**Finding T(2,2)-**

We have,

- i = 2
- j = 2
- (value)
_{i}= (value)_{2}= 4 - (weight)
_{i}= (weight)_{2}= 3

Substituting the values, we get-

T(2,2) = max { T(2-1 , 2) , 4 + T(2-1 , 2-3) }

T(2,2) = max { T(1,2) , 4 + T(1,-1) }

T(2,2) = T(1,2) { Ignore T(1,-1) }

T(2,2) = 3

**Finding T(2,3)-**

We have,

- i = 2
- j = 3
- (value)
_{i}= (value)_{2}= 4 - (weight)
_{i}= (weight)_{2}= 3

Substituting the values, we get-

T(2,3) = max { T(2-1 , 3) , 4 + T(2-1 , 3-3) }

T(2,3) = max { T(1,3) , 4 + T(1,0) }

T(2,3) = max { 3 , 4+0 }

T(2,3) = 4

**Finding T(2,4)-**

We have,

- i = 2
- j = 4
- (value)
_{i}= (value)_{2}= 4 - (weight)
_{i}= (weight)_{2}= 3

Substituting the values, we get-

T(2,4) = max { T(2-1 , 4) , 4 + T(2-1 , 4-3) }

T(2,4) = max { T(1,4) , 4 + T(1,1) }

T(2,4) = max { 3 , 4+0 }

T(2,4) = 4

**Finding T(2,5)-**

We have,

- i = 2
- j = 5
- (value)
_{i}= (value)_{2}= 4 - (weight)
_{i}= (weight)_{2}= 3

Substituting the values, we get-

T(2,5) = max { T(2-1 , 5) , 4 + T(2-1 , 5-3) }

T(2,5) = max { T(1,5) , 4 + T(1,2) }

T(2,5) = max { 3 , 4+3 }

T(2,5) = 7

Similarly, compute all the entries.

After all the entries are computed and filled in the table, we get the following table-

- The last entry represents the maximum possible value that can be put into the knapsack.
- So, maximum possible value that can be put into the knapsack = 7.

**Identifying Items To Be Put Into Knapsack-**

Following Step-04,

- We mark the rows labelled “1” and “2”.
- Thus, items that must be put into the knapsack to obtain the maximum value 7 are-

** Item-1 and Item-2**

To gain better understanding about 0/1 Knapsack Problem,

**Next Article-** **Travelling Salesman Problem**

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