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1. Knapsack Problem Using Opengl Program
2. Knapsack Problem Using Opengl Code
3. Knapsack Problem Using Opengl Command

Knapsack Problem-

Jul 13, 2011 However, as I didn’t find any C# implementation of Knapsack problem I thought of writing a small code that demonstrate how to tackle this problem. Here is the code for that. Static void Main (string args). Item a = new Item (1, 3, 5); Item b = new Item (2, 2, 3); Item c = new Item (3, 1, 4). Knapsack problem using backtracking algorithm with example in ppt, clique problem in java, download free ppt of shell sort, iee paper on y2k38 problem, brute force problem clique program, abstract for network problem notification via sms, microstrip bandpass filter design by insertion loss method.

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-

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-

Knapsack Problem Using Opengl Program

• 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)

(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) }

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) }

Knapsack Problem Using Opengl Code

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-

Knapsack Problem Using Opengl Command

Item-1 and Item-2

To gain better understanding about 0/1 Knapsack Problem,

Next Article-Travelling Salesman Problem

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