If a system does not employ either a deadlock prevention or deadlock avoidance algorithm then a deadlock situation may occur. In this case- 

• Apply an algorithm to examine state of system to determine whether deadlock has occurred or not.
• Apply an algorithm to recover from the deadlock. For more refer- Deadlock Recovery

Deadlock Avoidance Algorithm/ Bankers Algorithm: 
The algorithm employs several times varying data structures: 
 

• Available – 
  A vector of length m indicates the number of available resources of each type.
• Allocation – 
  An n*m matrix defines the number of resources of each type currently allocated to a process. Column represents resource and resource represent process.
• Request – 
  An n*m matrix indicates the current request of each process. If request[i][j] equals k then process P_i is requesting k more instances of resource type R_j.

This algorithm has already been discussed here 
 

Now, Bankers algorithm includes a Safety Algorithm / Deadlock Detection Algorithm 
The algorithm for finding out whether a system is in a safe state can be described as follows:

Steps of Algorithm: 
 

1. Let Work and Finish be vectors of length m and n respectively. Initialize Work= Available. For i=0, 1, …., n-1, if Request_i = 0, then Finish[i] = true; otherwise, Finish[i]= false.
2. Find an index i such that both 
   a) Finish[i] == false 
   b) Request_i <= Work 
   If no such i exists go to step 4.
3. Work= Work+ Allocation_i 
   Finish[i]= true 
   Go to Step 2.
4. If Finish[i]== false for some i, 0<=i<n, then the system is in a deadlocked state. Moreover, if Finish[i]==false the process P_i is deadlocked.

For example, 
 

1. In this, Work = [0, 0, 0] & 
   Finish = [false, false, false, false, false]
    
2. i=0 is selected as both Finish[0] = false and [0, 0, 0]<=[0, 0, 0].
    
3. Work =[0, 0, 0]+[0, 1, 0] =>[0, 1, 0] & 
   Finish = [true, false, false, false, false].
    
4. i=2 is selected as both Finish[2] = false and [0, 0, 0]<=[0, 1, 0].
5. Work =[0, 1, 0]+[3, 0, 3] =>[3, 1, 3] & 
   Finish = [true, false, true, false, false].
    
6. i=1 is selected as both Finish[1] = false and [2, 0, 2]<=[3, 1, 3].
7. Work =[3, 1, 3]+[2, 0, 0] =>[5, 1, 3] & 
   Finish = [true, true, true, false, false].
    
8. i=3 is selected as both Finish[3] = false and [1, 0, 0]<=[5, 1, 3].
9. Work =[5, 1, 3]+[2, 1, 1] =>[7, 2, 4] & 
   Finish = [true, true, true, true, false].
    
10. i=4 is selected as both Finish[4] = false and [0, 0, 2]<=[7, 2, 4].
     
11. Work =[7, 2, 4]+[0, 0, 2] =>[7, 2, 6] & 
    Finish = [true, true, true, true, true].
     
12. Since Finish is a vector of all true it means there is no deadlock in this example.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.