Modified GATE 2010 question

Question

A system has $n$ resources $R_0, \dots,R_{n-1}$, and $k$ processes $P_0, \dots, P_{k-1}$ ($n > k$). The implementation of the resource request logic of each process $P_i$ is as follows:

if (i%2==0) {
    if (i<n) request $R_i$;
    if (i+2 < n) request $R_{i+2}$
}

else {
    if (i<n) request $R_{n-1}$;
    if (i+2 <n) request $R_{n-i-2}$;
}

Is it deadlock free? If not, what is the minimum value of $n + k$ for which deadlock is possible?

Comments

I think

It is not deadlock free. And minimum value of n + k should be 6. As the number of process = 3 from 0 to 2 and resources = 3 from 0 to 2.

And the order of execution will be P2 -> P1 -> P0.

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@Shubhanshu Yes it wont be deadlock free. But in your example, I can't understand how there is a deadlock.

P2 is executed first, requests for R2, and receives it. Now P2 is not waiting for any other resource, so it will be able to complete. Right?