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A system has n resources R0,…,Rn-1, and k processes P0,…..Pk-1. The implementation of the resource request logic of each process Pi. is as follows: if (i%2 == 0) { if (i

A system has n resources R0,…,Rn-1, and k processes P0,…..Pk-1. The implementation of the resource request logic of each process Pi. is as follows: if (i%2 == 0) { if (i<n) request Ri ; if (i+2<n)request Ri+2; } else { if (i<n) request Rn-i; if (i+2<n)request Rn-i-2; } In which one of the following situations is a deadlock possible?   Correct Answer n = 21, k = 12

Concept:

 A deadlock is a situation where a set of processes are blocked because each process is holding a resource and waiting for another resource acquired by some other process. This makes a cycle.

Explanation:

When n is even:

Even processes(P0, P2, P4...) Requests the even resources(R0, R2,.....) and odd processes (P1, P3,...) requests odd resources(Rn-i, Rn-i-2,.....)

Then, the system never going to deadlock

When n is odd:

Even processes(P0, P2, P4...) Requests the even resources (R0, R2,.....) and odd processes (P1, P3,...) requests also even resources(Rn-i, Rn-i-2,.....)

Then, the system may fall into deadlock

Option 1:  deadlock not possible

Here, n=40 is even

so, the system will never fall into deadlock

Option 2: deadlock possible

Here, n= 21, is odd, k=12

processes Requests
P0 R0, R2
P1 R20, R18
P2 R2, R4
P3 R18, R16
P4 R4, R6
P5 R16, R14
P6 R6, R8
P7 R14, R12
P8 R8, R10
P9 R12, R10
P10 R10, R12
P11 R10, R8

Deadlock is possible here because a cycle is detected here

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Option 3:  deadlock not possible

Here, n=20 is even

so, the system will never fall into deadlock

Option 4: deadlock not possible

n = 41(odd), k = 19

P18 requests R18 and R20

P17 Requests R24 and R22

so, the system will never fall into deadlock

because k!=21

if k=21 then deadlock possible

P19 requests R22 and R20

P20 Requests R20 and R22

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