The output of the following problem should be: P1 -> P3 -> P4 -> P5 -> P0 -> P2 but the output I get when I run my code is: P1 -> P3 -> P4 -> P5 Note: I am sharing my code below. Please tell me what to fix in my code to get the desired output. P Problem: Write a c program that will generate the safe sequence of process execution for the situation given below:(Use Banker’s Algorithm). Note: The code can be implemented in several different ways, but make sure the parameter remains the same as shown below. n = 6; // Number of processes m = 4; // Number of resources int alloc[6][4] = { { 0, 1, 0, 3 }, // P0// Allocation Matrix { 2, 0, 0, 3 }, // P1 { 3, 0, 2, 0 }, // P2 { 2, 1, 1, 5 }, // P3 { 0, 0, 2, 2 }, // P4 {1, 2 , 3, 1 } }; //P5 int max[6][4] = { { 6, 4, 3, 4 }, // P0 // MAX Matrix { 3, 2, 2, 4 }, // P1 { 9, 1, 2, 6 }, // P2 { 2, 2, 2, 8 }, // P3 { 4, 3, 3, 7 }, // P4 { 6, 2, 6, 5 } }; //P5 int avail[4] = { 2, 2, 2, 1 }; //Available resources My code: #include int main() { int n, m; n = 6; // Number of processes m = 4; // Number of resources // Allocation Matrix int allocation[6][4] = { { 0, 1, 0, 3 }, // P0 { 2, 0, 0, 3 }, // P1 { 3, 0, 2, 0 }, // P2 { 2, 1, 1, 5 }, // P3 { 0, 0, 2, 2 }, // P4 {1, 2 , 3, 1 } }; //P5 // MAX Matrix int max[6][4] = { { 6, 4, 3, 4 }, // P0 { 3, 2, 2, 4 }, // P1 { 9, 1, 2, 6 }, // P2 { 2, 2, 2, 8 }, // P3 { 4, 3, 3, 7 }, // P4 { 6, 2 , 6, 5 } }; //P5 //Available resources int avail[4] = { 2, 2, 2, 1 }; //calculating Need Matrix int need[n][m]; for(int i = 0; i < n; i++){ for(int j = 0; j < m; j++) need[i][j] = max[i][j]-allocation[i][j]; } //Banker's Algorithm int k, flag; k = 0; //the variable k is the index of the safe sequence array int completion[20], safe_sequence[20]; /* The completion array stores the processes' completion status. It will store value 0 for an incomplete process and 1 for a complete process */ // Initally all processes are incomplete for(int i = 0; i < n; i++) { completion[i] = 0; } for(int i = 0; i < n; i++) { flag = 0; if(completion[i] == 0) { //Process is incomplete. So execute the Process for(int j = 0; j < m; j++) { if(need[i][j] > avail[j]) //If need is greater than available break out of the loop { flag = 1; break; } } if(flag==0) // Need is less than available { completion[i] = 1; safe_sequence[k] = i; k++; //Calculate new available for(int j = 0; j < m; j++) { avail[j] += allocation[i][j]; } i = i - 1; //loop from the first process } } } printf("Safe Sequence:\n"); for(int i = 0; i < k; i++) { printf("P%d", safe_sequence[i]); if(i < k-1) { printf(" -> "); } } }
Operations
In mathematics and computer science, an operation is an event that is carried out to satisfy a given task. Basic operations of a computer system are input, processing, output, storage, and control.
Basic Operators
An operator is a symbol that indicates an operation to be performed. We are familiar with operators in mathematics; operators used in computer programming are—in many ways—similar to mathematical operators.
Division Operator
We all learnt about division—and the division operator—in school. You probably know of both these symbols as representing division:
Modulus Operator
Modulus can be represented either as (mod or modulo) in computing operation. Modulus comes under arithmetic operations. Any number or variable which produces absolute value is modulus functionality. Magnitude of any function is totally changed by modulo operator as it changes even negative value to positive.
Operators
In the realm of programming, operators refer to the symbols that perform some function. They are tasked with instructing the compiler on the type of action that needs to be performed on the values passed as operands. Operators can be used in mathematical formulas and equations. In programming languages like Python, C, and Java, a variety of operators are defined.
The output of the following problem should be: P1 -> P3 -> P4 -> P5 -> P0 -> P2
but the output I get when I run my code is: P1 -> P3 -> P4 -> P5
Note: I am sharing my code below. Please tell me what to fix in my code to get the desired output. P
Problem:
Write a c program that will generate the safe sequence of process execution for the situation
given below:(Use Banker’s Algorithm).
Note: The code can be implemented in several different ways, but make sure the parameter
remains the same as shown below.
n = 6; // Number of processes
m = 4; // Number of resources
int alloc[6][4] = { { 0, 1, 0, 3 }, // P0// Allocation Matrix
{ 2, 0, 0, 3 }, // P1
{ 3, 0, 2, 0 }, // P2
{ 2, 1, 1, 5 }, // P3
{ 0, 0, 2, 2 }, // P4
{1, 2 , 3, 1 } }; //P5
int max[6][4] = { { 6, 4, 3, 4 }, // P0 // MAX Matrix
{ 3, 2, 2, 4 }, // P1
{ 9, 1, 2, 6 }, // P2
{ 2, 2, 2, 8 }, // P3
{ 4, 3, 3, 7 }, // P4
{ 6, 2, 6, 5 } }; //P5
int avail[4] = { 2, 2, 2, 1 }; //Available resources
My code:
#include <stdio.h>
int main() {
int n, m;
n = 6; // Number of processes
m = 4; // Number of resources
// Allocation Matrix
int allocation[6][4] = { { 0, 1, 0, 3 }, // P0
{ 2, 0, 0, 3 }, // P1
{ 3, 0, 2, 0 }, // P2
{ 2, 1, 1, 5 }, // P3
{ 0, 0, 2, 2 }, // P4
{1, 2 , 3, 1 } }; //P5
// MAX Matrix
int max[6][4] = { { 6, 4, 3, 4 }, // P0
{ 3, 2, 2, 4 }, // P1
{ 9, 1, 2, 6 }, // P2
{ 2, 2, 2, 8 }, // P3
{ 4, 3, 3, 7 }, // P4
{ 6, 2 , 6, 5 } }; //P5
//Available resources
int avail[4] = { 2, 2, 2, 1 };
//calculating Need Matrix
int need[n][m];
for(int i = 0; i < n; i++){
for(int j = 0; j < m; j++)
need[i][j] = max[i][j]-allocation[i][j];
}
//Banker's Algorithm
int k, flag;
k = 0; //the variable k is the index of the safe sequence array
int completion[20], safe_sequence[20]; /* The completion array stores the processes' completion status.
It will store value 0 for an incomplete process and 1 for a
complete process */
// Initally all processes are incomplete
for(int i = 0; i < n; i++) {
completion[i] = 0;
}
for(int i = 0; i < n; i++) {
flag = 0;
if(completion[i] == 0)
{
//Process is incomplete. So execute the Process
for(int j = 0; j < m; j++) {
if(need[i][j] > avail[j]) //If need is greater than available break out of the loop
{
flag = 1;
break;
}
}
if(flag==0) // Need is less than available
{
completion[i] = 1;
safe_sequence[k] = i;
k++;
//Calculate new available
for(int j = 0; j < m; j++)
{
avail[j] += allocation[i][j];
}
i = i - 1; //loop from the first process
}
}
}
printf("Safe Sequence:\n");
for(int i = 0; i < k; i++)
{
printf("P%d", safe_sequence[i]);
if(i < k-1)
{
printf(" -> ");
}
}
}
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