• Knapsack Problem Using Opengl Programming
• Knapsack Problem Using Opengl Program

Jul 30, 2020 The knapsack problem can be solved by using different methods of computational algorithms. But here we will solve this problem by using a genetic algorithm. So, we could put valuable items in the knapsack. Example: Let us consider, we have four items and their weights and values are given:in the below table: ITEM. WEIGHT (Kg) VALUE. Aug 13, 2020 Therefore, a 0-1 knapsack problem can be solved in using dynamic programming. It should be noted that the time complexity depends on the weight limit of. Although it seems like it’s a polynomial-time algorithm in the number of items, as W increases from say 100 to 1,000 ( to ), processing goes from bits to bits.

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A knapsack is a bag. And the knapsack problem deals with the putting items to the bag based on the value of the items. It aim is to maximise the value inside the bag. In 0-1 Knapsack you can either put the item or discard it, there is no concept of putting some part of item in the knapsack.

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. 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.

Sample Problem

Weight distribution

The maximum value is 65 so we will put the item 2 and 3 in the knapsack.

PROGRAM FOR 0-1 KNAPSACK PROBLEM

Output

Knapsack problem/Continuous You are encouraged to solve this task according to the task description, using any language you may know. A thief burgles a butcher's shop, where he can select from some items. The knapsack problem is a way to solve a problem in such a way so that the capacity constraint of the knapsack doesn't break and we receive maximum profit. In the next article, we will see it’s the first approach in detail to solve this problem. 0/1 knapsack problem knapsack problem in alogo. Analysis and design of algorithms.

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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-

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)

Best free online train simulator. (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) }

Knapsack Problem Using Opengl Programming

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}

Simulation games for macos. 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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Knapsack Problem Using Opengl Program