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Program for Banker’s Algorithm | Set 1 (Safety Algorithm)

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The banker’s algorithm is a resource allocation and deadlock avoidance algorithm that tests for safety by simulating the allocation for the predetermined maximum possible amounts of all resources, then makes an “s-state” check to test for possible activities, before deciding whether allocation should be allowed to continue. The following Data structures are used to implement the Banker’s Algorithm: Let ‘n’ be the number of processes in the system and ‘m’ be the number of resource types. Available : 

  • It is a 1-d array of size ‘m’ indicating the number of available resources of each type.
  • Available[ j ] = k means there are ‘k’ instances of resource type Rj

Max :

  • It is a 2-d array of size ‘n*m’ that defines the maximum demand of each process in a system.
  • Max[ i, j ] = k means process Pi may request at most ‘k’ instances of resource type Rj.

Allocation :

  • It is a 2-d array of size ‘n*m’ that defines the number of resources of each type currently allocated to each process.
  • Allocation[ i, j ] = k means process Pi is currently allocated ‘k’ instances of resource type Rj

Need :

  • It is a 2-d array of size ‘n*m’ that indicates the remaining resource need of each process.
  • Need [ i, j ] = k means process Pi currently allocated ‘k’ instances of resource type Rj
  • Need [ i, j ] = Max [ i, j ] – Allocation [ i, j ]

Allocationi specifies the resources currently allocated to process Pi and Needi specifies the additional resources that process Pi may still request to complete its task. Banker’s algorithm consist of Safety algorithm and Resource request algorithm Safety Algorithm

The algorithm for finding out whether or not a system is in a safe state can be described as follows:

  1. Let Work and Finish be vectors of length ‘m’ and ‘n’ respectively. Initialize: Work= Available Finish [i]=false; for i=1,2,……,n
  2. Find an i such that both a) Finish [i]=false b) Need_i<=work 
    if no such i exists goto step (4)
  3. Work=Work + Allocation_i Finish[i]= true goto step(2)
  4. If Finish[i]=true for all i, then the system is in safe state.

Safe sequence is the sequence in which the processes can be safely executed. A state is safe if the system can allocate resources to each process (up to its maximum) in some order and still avoid a deadlock. More formally, ” a system is in a safe state only if there exists a safe sequence” .If a system has a safe sequence, it implies that there is no deadlock present. The sequence guarantees that all processes will complete their execution without getting into a circular wait situation. It is possible to have multiple safe sequences in a system. This occurs when there are multiple ways to order the execution of processes such that no deadlock will occur. Each safe sequence represents a different order in which processes can be executed while avoiding deadlocks. The existance of multiple safe sequences allows the system to have flexibility in scheduling processes and allocating resources. Different scheduling algorithms or resource allocation can depend on factors such as system efficiency, resource utilization, fairness or other criteria that the operating system aims to optimize.

In this post, implementation of Safety algorithm of Banker’s Algorithm is done. 

C++




// C++ program to illustrate Banker's Algorithm
#include<iostream>
using namespace std;
 
// Number of processes
const int P = 5;
 
// Number of resources
const int R = 3;
 
// Function to find the need of each process
void calculateNeed(int need[P][R], int maxm[P][R],
                   int allot[P][R])
{
    // Calculating Need of each P
    for (int i = 0 ; i < P ; i++)
        for (int j = 0 ; j < R ; j++)
 
            // Need of instance = maxm instance -
            //                    allocated instance
            need[i][j] = maxm[i][j] - allot[i][j];
}
 
// Function to find the system is in safe state or not
bool isSafe(int processes[], int avail[], int maxm[][R],
            int allot[][R])
{
    int need[P][R];
 
    // Function to calculate need matrix
    calculateNeed(need, maxm, allot);
 
    // Mark all processes as infinish
    bool finish[P] = {0};
 
    // To store safe sequence
    int safeSeq[P];
 
    // Make a copy of available resources
    int work[R];
    for (int i = 0; i < R ; i++)
        work[i] = avail[i];
 
    // While all processes are not finished
    // or system is not in safe state.
    int count = 0;
    while (count < P)
    {
        // Find a process which is not finish and
        // whose needs can be satisfied with current
        // work[] resources.
        bool found = false;
        for (int p = 0; p < P; p++)
        {
            // First check if a process is finished,
            // if no, go for next condition
            if (finish[p] == 0)
            {
                // Check if for all resources of
                // current P need is less
                // than work
                int j;
                for (j = 0; j < R; j++)
                    if (need[p][j] > work[j])
                        break;
 
                // If all needs of p were satisfied.
                if (j == R)
                {
                    // Add the allocated resources of
                    // current P to the available/work
                    // resources i.e.free the resources
                    for (int k = 0 ; k < R ; k++)
                        work[k] += allot[p][k];
 
                    // Add this process to safe sequence.
                    safeSeq[count++] = p;
 
                    // Mark this p as finished
                    finish[p] = 1;
 
                    found = true;
                }
            }
        }
 
        // If we could not find a next process in safe
        // sequence.
        if (found == false)
        {
            cout << "System is not in safe state";
            return false;
        }
    }
 
    // If system is in safe state then
    // safe sequence will be as below
    cout << "System is in safe state.\nSafe"
         " sequence is: ";
    for (int i = 0; i < P ; i++)
        cout << safeSeq[i] << " ";
 
    return true;
}
 
// Driver code
int main()
{
    int processes[] = {0, 1, 2, 3, 4};
 
    // Available instances of resources
    int avail[] = {3, 3, 2};
 
    // Maximum R that can be allocated
    // to processes
    int maxm[][R] = {{7, 5, 3},
                     {3, 2, 2},
                     {9, 0, 2},
                     {2, 2, 2},
                     {4, 3, 3}};
 
    // Resources allocated to processes
    int allot[][R] = {{0, 1, 0},
                      {2, 0, 0},
                      {3, 0, 2},
                      {2, 1, 1},
                      {0, 0, 2}};
 
    // Check system is in safe state or not
    isSafe(processes, avail, maxm, allot);
 
    return 0;
}


Java




// Java program to illustrate Banker's Algorithm
import java.util.*;
 
class GFG
{
 
// Number of processes
static int P = 5;
 
// Number of resources
static int R = 3;
 
// Function to find the need of each process
static void calculateNeed(int need[][], int maxm[][],
                int allot[][])
{
    // Calculating Need of each P
    for (int i = 0 ; i < P ; i++)
        for (int j = 0 ; j < R ; j++)
 
            // Need of instance = maxm instance -
            //                 allocated instance
            need[i][j] = maxm[i][j] - allot[i][j];
}
 
// Function to find the system is in safe state or not
static boolean isSafe(int processes[], int avail[], int maxm[][],
            int allot[][])
{
    int [][]need = new int[P][R];
 
    // Function to calculate need matrix
    calculateNeed(need, maxm, allot);
 
    // Mark all processes as infinish
    boolean []finish = new boolean[P];
 
    // To store safe sequence
    int []safeSeq = new int[P];
 
    // Make a copy of available resources
    int []work = new int[R];
    for (int i = 0; i < R ; i++)
        work[i] = avail[i];
 
    // While all processes are not finished
    // or system is not in safe state.
    int count = 0;
    while (count < P)
    {
        // Find a process which is not finish and
        // whose needs can be satisfied with current
        // work[] resources.
        boolean found = false;
        for (int p = 0; p < P; p++)
        {
            // First check if a process is finished,
            // if no, go for next condition
            if (finish[p] == false)
            {
                // Check if for all resources of
                // current P need is less
                // than work
                int j;
                for (j = 0; j < R; j++)
                    if (need[p][j] > work[j])
                        break;
 
                // If all needs of p were satisfied.
                if (j == R)
                {
                    // Add the allocated resources of
                    // current P to the available/work
                    // resources i.e.free the resources
                    for (int k = 0 ; k < R ; k++)
                        work[k] += allot[p][k];
 
                    // Add this process to safe sequence.
                    safeSeq[count++] = p;
 
                    // Mark this p as finished
                    finish[p] = true;
 
                    found = true;
                }
            }
        }
 
        // If we could not find a next process in safe
        // sequence.
        if (found == false)
        {
            System.out.print("System is not in safe state");
            return false;
        }
    }
 
    // If system is in safe state then
    // safe sequence will be as below
    System.out.print("System is in safe state.\nSafe"
        +" sequence is: ");
    for (int i = 0; i < P ; i++)
        System.out.print(safeSeq[i] + " ");
 
    return true;
}
 
// Driver code
public static void main(String[] args)
{
    int processes[] = {0, 1, 2, 3, 4};
 
    // Available instances of resources
    int avail[] = {3, 3, 2};
 
    // Maximum R that can be allocated
    // to processes
    int maxm[][] = {{7, 5, 3},
                    {3, 2, 2},
                    {9, 0, 2},
                    {2, 2, 2},
                    {4, 3, 3}};
 
    // Resources allocated to processes
    int allot[][] = {{0, 1, 0},
                    {2, 0, 0},
                    {3, 0, 2},
                    {2, 1, 1},
                    {0, 0, 2}};
 
    // Check system is in safe state or not
    isSafe(processes, avail, maxm, allot);
}
}
 
// This code has been contributed by 29AjayKumar


Python3




# Python3 program to illustrate
# Banker's Algorithm
 
# Number of processes
P = 5
 
# Number of resources
R = 3
 
# Function to find the need of each process
def calculateNeed(need, maxm, allot):
 
    # Calculating Need of each P
    for i in range(P):
        for j in range(R):
             
            # Need of instance = maxm instance -
            # allocated instance
            need[i][j] = maxm[i][j] - allot[i][j]
 
# Function to find the system is in
# safe state or not
def isSafe(processes, avail, maxm, allot):
    need = []
    for i in range(P):
        l = []
        for j in range(R):
            l.append(0)
        need.append(l)
         
    # Function to calculate need matrix
    calculateNeed(need, maxm, allot)
 
    # Mark all processes as infinish
    finish = [0] * P
     
    # To store safe sequence
    safeSeq = [0] * P
 
    # Make a copy of available resources
    work = [0] * R
    for i in range(R):
        work[i] = avail[i]
 
    # While all processes are not finished
    # or system is not in safe state.
    count = 0
    while (count < P):
         
        # Find a process which is not finish
        # and whose needs can be satisfied
        # with current work[] resources.
        found = False
        for p in range(P):
         
            # First check if a process is finished,
            # if no, go for next condition
            if (finish[p] == 0):
             
                # Check if for all resources
                # of current P need is less
                # than work
                for j in range(R):
                    if (need[p][j] > work[j]):
                        break
                     
                # If all needs of p were satisfied.
                if (j == R - 1):
                 
                    # Add the allocated resources of
                    # current P to the available/work
                    # resources i.e.free the resources
                    for k in range(R):
                        work[k] += allot[p][k]
 
                    # Add this process to safe sequence.
                    safeSeq[count] = p
                    count += 1
 
                    # Mark this p as finished
                    finish[p] = 1
 
                    found = True
                 
        # If we could not find a next process
        # in safe sequence.
        if (found == False):
            print("System is not in safe state")
            return False
         
    # If system is in safe state then
    # safe sequence will be as below
    print("System is in safe state.",
              "\nSafe sequence is: ", end = " ")
    print(*safeSeq)
 
    return True
 
# Driver code
if __name__ =="__main__":
     
    processes = [0, 1, 2, 3, 4]
 
    # Available instances of resources
    avail = [3, 3, 2]
 
    # Maximum R that can be allocated
    # to processes
    maxm = [[7, 5, 3], [3, 2, 2],
            [9, 0, 2], [2, 2, 2],
            [4, 3, 3]]
 
    # Resources allocated to processes
    allot = [[0, 1, 0], [2, 0, 0],
             [3, 0, 2], [2, 1, 1],
             [0, 0, 2]]
 
    # Check system is in safe state or not
    isSafe(processes, avail, maxm, allot)
 
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)


C#




// C# program to illustrate Banker's Algorithm
using System;
 
class GFG
{
     
// Number of processes
static int P = 5;
 
// Number of resources
static int R = 3;
 
// Function to find the need of each process
static void calculateNeed(int [,]need, int [,]maxm,
                int [,]allot)
{
    // Calculating Need of each P
    for (int i = 0 ; i < P ; i++)
        for (int j = 0 ; j < R ; j++)
 
            // Need of instance = maxm instance -
            //             allocated instance
            need[i,j] = maxm[i,j] - allot[i,j];
}
 
// Function to find the system is in safe state or not
static bool isSafe(int []processes, int []avail, int [,]maxm,
            int [,]allot)
{
    int [,]need = new int[P,R];
 
    // Function to calculate need matrix
    calculateNeed(need, maxm, allot);
 
    // Mark all processes as infinish
    bool []finish = new bool[P];
 
    // To store safe sequence
    int []safeSeq = new int[P];
 
    // Make a copy of available resources
    int []work = new int[R];
    for (int i = 0; i < R ; i++)
        work[i] = avail[i];
 
    // While all processes are not finished
    // or system is not in safe state.
    int count = 0;
    while (count < P)
    {
        // Find a process which is not finish and
        // whose needs can be satisfied with current
        // work[] resources.
        bool found = false;
        for (int p = 0; p < P; p++)
        {
            // First check if a process is finished,
            // if no, go for next condition
            if (finish[p] == false)
            {
                // Check if for all resources of
                // current P need is less
                // than work
                int j;
                for (j = 0; j < R; j++)
                    if (need[p,j] > work[j])
                        break;
 
                // If all needs of p were satisfied.
                if (j == R)
                {
                    // Add the allocated resources of
                    // current P to the available/work
                    // resources i.e.free the resources
                    for (int k = 0 ; k < R ; k++)
                        work[k] += allot[p,k];
 
                    // Add this process to safe sequence.
                    safeSeq[count++] = p;
 
                    // Mark this p as finished
                    finish[p] = true;
 
                    found = true;
                }
            }
        }
 
        // If we could not find a next process in safe
        // sequence.
        if (found == false)
        {
            Console.Write("System is not in safe state");
            return false;
        }
    }
 
    // If system is in safe state then
    // safe sequence will be as below
        Console.Write("System is in safe state.\nSafe"
        +" sequence is: ");
    for (int i = 0; i < P ; i++)
            Console.Write(safeSeq[i] + " ");
 
    return true;
}
 
// Driver code
static public void Main ()
{
    int []processes = {0, 1, 2, 3, 4};
 
    // Available instances of resources
    int []avail = {3, 3, 2};
 
    // Maximum R that can be allocated
    // to processes
    int [,]maxm = {{7, 5, 3},
                    {3, 2, 2},
                    {9, 0, 2},
                    {2, 2, 2},
                    {4, 3, 3}};
 
    // Resources allocated to processes
    int [,]allot = {{0, 1, 0},
                    {2, 0, 0},
                    {3, 0, 2},
                    {2, 1, 1},
                    {0, 0, 2}};
 
    // Check system is in safe state or not
    isSafe(processes, avail, maxm, allot);
 
    }
}
 
// This code has been contributed by ajit.


Javascript




// Number of processes
const P = 5;
 
// Number of resources
const R = 3;
 
// Function to find the need of each process
function calculateNeed(need, maxm, allot) {
  // Calculating Need of each P
  for (let i = 0; i < P; i++) {
    for (let j = 0; j < R; j++) {
      // Need of instance = maxm instance -
      //                    allocated instance
      need[i][j]=(maxm[i][j] - allot[i][j]);
    }
  }
}
 
// Function to find the system is in safe state or not
function isSafe(processes, avail, maxm, allot) {
  let need = new Array(P);
  for(let i=0;i<P;i++) need[i]= new Array(R);
  let finish = [];
 
  // Function to calculate need matrix
  calculateNeed(need, maxm, allot);
 
  // Mark all processes as infinish
  for (let i = 0; i < processes.length; i++) {
    finish.push(false);
  }
 
  // To store safe sequence
  let safeSeq = [];
 
  // Make a copy of available resources
  let work = [];
  for (let i = 0; i < R; i++) {
    work.push(avail[i]);
  }
 
  // While all processes are not finished
  // or system is not in safe state.
  let count = 0;
  while (count < P) {
    // Find a process which is not finish and
    // whose needs can be satisfied with current
    // work[] resources.
    let found = false;
    for (let p = 0; p < P; p++) {
      // First check if a process is finished,
      // if no, go for next condition
      if (finish[p] == false) {
        // Check if for all resources of
        // current P need is less
        // than work
        let j;
        for (j = 0; j < R; j++) {
          if (need[p][j] > work[j]) {
            break;
          }
        }
 
        // If all needs of p were satisfied.
        if (j == R) {
          // Add the allocated resources of
          // current P to the available/work
          // resources i.e.free the resources
          for (let k = 0; k < R; k++) {
            work[k] += allot[p][k];
          }
 
          // Add this process to safe sequence.
          safeSeq[count] = p;
          count++;
 
          // Mark this p as finished
          finish[p] = true;
 
          found = true;
        }
      }
    }
 
    // If we could not find a next process in safe
    // sequence.
    if (found == false) {
      console.log("System is not in safe state");
      return false;
    }
  }
 
  // If system is in safe state then
  // safe sequence will be as below
  console.log("System is in safe state.\nSafe sequence is: ");
  for (let i = 0; i < P; i++) {
    console.log(safeSeq[i] + " ");
  }
 
  return true;
}
 
let processes = [0, 1, 2, 3, 4];
 
// Available instances of resources
let avail = [3, 3, 2];
 
// Maximum R that can be allocated to processes
let maxm = [
  [7, 5, 3],
  [3, 2, 2],
  [9, 0, 2],
  [2, 2, 2],
  [4, 3, 3],
];
 
// Resources allocated to processes
let allot = [
  [0, 1, 0],
  [2, 0, 0],
  [3, 0, 2],
  [2, 1, 1],
  [0, 0, 2],
];
 
// Check system is in safe state or not
isSafe(processes, avail, maxm, allot);
 
// 7 4 3
// 1 2 2
// 6 0 0
// 0 1 1
// 4 3 1
 
// This code is contributed by ishankhandelwals.


Output:

System is in safe state.
Safe sequence is: 1 3 4 0 2

Illustration : Considering a system with five processes P0 through P4 and three resources types A, B, C. Resource type A has 10 instances, B has 5 instances and type C has 7 instances. Suppose at time t0 following snapshot of the system has been taken: We must determine whether the new system state is safe. To do so, we need to execute Safety algorithm on the above given allocation chart. banker's algorithm Following is the resource allocation graph: BankersExecuting safety algorithm shows that sequence < P1, P3, P4, P0, P2 > satisfies safety requirement.

Time complexity = O(n*n*m)

  1. where n = number of processes and m = number of resources.


Last Updated : 27 Sep, 2023
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