please answer question #33 to answer the question. please use (Algorithm 5.7) for the Backtracking Algorithm for the 0-1 Knapsack Problemplease use java language for coding java file- should start with your Last Name, Like SmithClubApp, PetrowskiTravelLogApp, for your classshort paragraph should include-please label the question with the answer as well -description of the java file-what it does?-what is the input data? expected output, samples of both 33. Use the Backtracking algorithm for the 0-1 Knapsack problem (Algorithm5.7) to maximize the profit for the following problem instance. Show theactions step by step.PiiPiWiWi1$202102$303$3575W = 94$12345$313 Algorithm 5.7The Backtracking Algorithm for the 0-1 Knapsack ProblemProblem: Let n items be given, where each item has a weight and a profit.The weights and profits are positive integers. Furthermore, let a positiveinteger W be given. Determine a set of items with maximum total profit,under the constraint that the sum of their weights cannot exceed W.Inputs: Positive integers n and W; arrays w and p, each indexed from 1to n, and each containing positive integers sorted in nonincreasing orderaccording to the values of p[i]/w[i].Outputs: an array bestset indexed from 1 to n, where the values of bestset[i]is "yes" if the ith item is included in the optimal set and is "no" otherwise;an integer maxprofit that is the maximum profit.BACKTRACKINGvoid knapsack (index i,1int profit, int weight)if (weight W && profit > maxprofit){\}maxprofitnumbest = i;profit;bestset include;if (promising (i)){include [i+1] = "yes";This set is best// so far.// Set numbest to// number of itemsconsidered. Set//bestset to this// solution.// Include w[i+1].knapsack (i+1, profit + p[i+1], weight + w[i + 1]);include [i+1]="no";knapsack (i+1, profit, weight);// Do not include// w[i + 1].}}bool promising (index i){index j, k;int totweight;float bound;if (weight > W)return false;else{j = i + 1;bound profit;totweight weight;// Node is promising onlyif we should expand toits children. There mustbe some capacity left for// the children.while (j <=n&& totweight + w[j] < = W){\totweight totweight + w[j];bound bound + p[j];Grab as many items aspossible.j++;}k = j;if (k <=n)bound = bound (Wtotweight) p[k]/w[k];Use k for consistency// with formula in text.*return bound > maxprofit;// Grab fraction of kth// item.}}

The 0-1 Knapsack problem aims to maximize the total profit by selecting items with given weights and…

Answered: 33. Use the Backtracking algorithm for…

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please answer question #33 to answer the question. please use (Algorithm 5.7) for the Backtracking Algorithm for the 0-1 Knapsack Problem please use java language for coding java file - should start with your Last Name, Like SmithClubApp, PetrowskiTravelLogApp, for your class short paragraph should include- please label the question with the answer as well -description of the java file -what it does? -what is the input data? expected output, samples of both

33. Use the Backtracking algorithm for the 0-1 Knapsack problem (Algorithm
5.7) to maximize the profit for the following problem instance. Show the
actions step by step.
Pi
i
Pi
Wi
Wi
1
$20
2
10
2
$30
3
$35
7
5
W = 9
4
$12
3
4
5
$3
1
3

Algorithm 5.7
The Backtracking Algorithm for the 0-1 Knapsack Problem
Problem: Let n items be given, where each item has a weight and a profit.
The weights and profits are positive integers. Furthermore, let a positive
integer W be given. Determine a set of items with maximum total profit,
under the constraint that the sum of their weights cannot exceed W.
Inputs: Positive integers n and W; arrays w and p, each indexed from 1
to n, and each containing positive integers sorted in nonincreasing order
according to the values of p[i]/w[i].
Outputs: an array bestset indexed from 1 to n, where the values of bestset[i]
is "yes" if the ith item is included in the optimal set and is "no" otherwise;
an integer maxprofit that is the maximum profit.
BACKTRACKING
void knapsack (index i,
1
int profit, int weight)
if (weight W && profit > maxprofit){\
}
maxprofit
numbest = i;
profit;
bestset include;
if (promising (i)){
include [i+1] = "yes";
This set is best
// so far.
// Set numbest to
// number of items
considered. Set
//bestset to this
// solution.
// Include w[i+1].
knapsack (i+1, profit + p[i+1], weight + w[i + 1]);
include [i+1]
=
"no";
knapsack (i+1, profit, weight);
// Do not include
// w[i + 1].
}
}
bool promising (index i)
{
index j, k;
int totweight;
float bound;
if (weight > W)
return false;
else{
j = i + 1;
bound profit;
totweight weight;
// Node is promising only
if we should expand to
its children. There must
be some capacity left for
// the children.
while (j <=n&& totweight + w[j] < = W){\
totweight totweight + w[j];
bound bound + p[j];
Grab as many items as
possible.
j++;
}
k = j;
if (k <=n)
bound = bound (Wtotweight) p[k]/w[k];
Use k for consistency
// with formula in text.
*
return bound > maxprofit;
// Grab fraction of kth
// item.
}
}

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