Question

A system has n resources $R_0,R_1,....R_1$, and k processes $P_0,P_1,....P_{k-1}$. 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 $R_{n-1-2}$;

In which one of the following situations is a deadlock possible?

Answer 1

According to the given program, in all the cases for even sequences of processes only corresponding sequence number of recource will be allocated except case (b), where n=21, k=12, for $P_{10}\Rightarrow R_{10}$ can be alloted and for $P_{11}\Rightarrow R_{21-11}\Rightarrow R_{10}$ can be alloted.
$P_i\Rightarrow R_i$ alloted and for odd sequence of process. i.e. $P_i\Rightarrow R_{n-1}$ that never conflict in any case except(b), when a same resource is requested by more than one process at the same time may lead to the deadlock.