# Deadlock Detection Algorithm in Operating System

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 has occurred or not.
• Apply an algorithm to recover from the deadlock. For more refer- Deadlock Recovery

The algorithm employs several time 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 Pi is requesting k more instances of resource type Rj.

We treat rows in the matrices Allocation and Request as vectors, we refer them as Allocationi and Requesti.

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 Requesti = 0, then Finish[i] = true; otherwise, Finish[i]= false.
2. Find an index i such that both
a) Finish[i] == false
b) Requesti <= Work
If no such i exists go to step 4.
3. Work= Work+ Allocationi
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 Pi 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 = 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 = 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 = false and [2, 0, 2]<=[3, 1, 3].
7. Work =[3, 1, 3]+[2, 0, 2] =>[5, 1, 5] &
Finish = [true, true, true, false, false].
8. i=3 is selected as both Finish = false and [1, 0, 0]<=[5, 1, 5].
9. Work =[5, 1, 5]+[2, 1, 1] =>[7, 2, 6] &
Finish = [true, true, true, true, false].
10. i=4 is selected as both Finish = false and [0, 0, 2]<=[7, 2, 6].
11. Work =[7, 2, 6]+[0, 0, 2] =>[7, 2, 8] &
Finish = [true, true, true, true, true].
12. Since Finish is a vector of all true it means there is no deadlock in this example.

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