Knapsack Problem-

You are given the following-

Fractional Knapsack Problem Solution in C and Java. The same approach we are using in our program. We have taken an array of structures named Item. Each Item has value & weight. We are calculating density= value/weight for each item and sorting the i tems array in the order of decreasing density. We add values from the top of the array to. If using a simple sort algorithm (selection, bubble) then the complexity of the whole problem is O(n2). If using quick sort or merge sort then the complexity of the whole problem is O(nlogn). Java code for Greedy Three. Firstly, you define class KnapsackPackage. This class has properties are: weight, value and corresponding cost of each package. Write a program in C/C using OpenGL to show a ball is shrinking. Try this code, run before writing. #include // for MS Windows #include // GLUT, include glu.h and gl.h #include // Global variable GLfloat angle = 0.0f; // Current rotational angle of the shapes /. Initialize OpenGL Graphics./ void.

• 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-

1. Fractional Knapsack Problem
2. 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.

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)

Knapsack Problem Using Opengl Programming

(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?

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) }

Knapsack Problem Using Opengl Function

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,

Knapsack Problem Using Opengl Programming

• 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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A list of items is given, each item has its own value and weight. Items can be placed in a knapsack whose maximum weight limit is W. The problem is to find the weight that is less than or equal to W, and value is maximized.

There are two types of Knapsack problem.

• 0 – 1 Knapsack
• Fractional Knapsack

For the 0 – 1 Knapsack, items cannot be divided into smaller pieces, and for fractional knapsack, items can be broken into smaller pieces.

Here we will discuss the fractional knapsack problem.

The time complexity of this algorithm is O(n Log n).

Input and Output

Algorithm

Input − maximum weight of the knapsack, list of items and the number of items

Output: The maximum value obtained.

EXample

Output

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