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

$\text{if} (i\%2==0) \{$

$\quad\text{if} (i<n) \text{ request } R_i;$

$\quad\text{if} (i+2 < n) \text{ request } R_{i+2}; \}$

$\text{else} \{$

$\quad\text{if} (i<n) \text{ request } R_{n-i};$

$\quad\text{if} (i+2 <n) \text{ request } R_{n-i-2}; \}$

In which of the following situations is a deadlock possible?

- $n=40,\: k=26$
- $n=21,\:k=12$
- $n=20,\:k=10$
- $n=41,\:k=19$

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@Arjun Sir @Lakshman Patel RJIT Sir Please correct the question in the PDF. as mentioned by @JAINchiNMay

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311 votes

Best answer

From the resource allocation logic, it's clear that even numbered processes are taking even numbered resources and all even numbered processes share no more than $1$ resource. Now, if we make sure that all odd numbered processes take odd numbered resources without a cycle, then deadlock cannot occur. The "else" case of the resource allocation logic, is trying to do that. But, if $n$ is odd, $R_{n-i}$ and $R_{n-i-2}$ will be even and there is possibility of deadlock, when two processes requests the same $R_i$ and $R_j$. So, only $B$ and $D$ are the possible answers.

Now, in $D$, we can see that $P_0$ requests $R_0$ and $R_2$, $P_2$ requests $R_2$ and $R_4$, so on until, $P_{18}$ requests $R_{18}$ and $R_{20}$. At the same time $P_1$ requests $R_{40}$ and $R_{38}$, $P_3$ requests $R_{38}$ and $R_{36}$, so on until, $P_{17}$ requests $R_{24}$ and $R_{22}$. i.e.; there are no two processes requesting the same two resources and hence there can't be a cycle of dependencies which means, no deadlock is possible.

But for $B$, $P_8$ requests $R_8$ and $R_{10}$ and $P_{11}$ also requests $R_{10}$ and $R_8$. Hence, a deadlock is possible. (Suppose $P_8$ comes first and occupies $R_8$. Then $P_{11}$ comes and occupies $R_{10}$. Now, if $P_8$ requests $R_{10}$ and $P_{11}$ requests $R_8$, there will be deadlock)

Correct Answer: $B$

Now, in $D$, we can see that $P_0$ requests $R_0$ and $R_2$, $P_2$ requests $R_2$ and $R_4$, so on until, $P_{18}$ requests $R_{18}$ and $R_{20}$. At the same time $P_1$ requests $R_{40}$ and $R_{38}$, $P_3$ requests $R_{38}$ and $R_{36}$, so on until, $P_{17}$ requests $R_{24}$ and $R_{22}$. i.e.; there are no two processes requesting the same two resources and hence there can't be a cycle of dependencies which means, no deadlock is possible.

But for $B$, $P_8$ requests $R_8$ and $R_{10}$ and $P_{11}$ also requests $R_{10}$ and $R_8$. Hence, a deadlock is possible. (Suppose $P_8$ comes first and occupies $R_8$. Then $P_{11}$ comes and occupies $R_{10}$. Now, if $P_8$ requests $R_{10}$ and $P_{11}$ requests $R_8$, there will be deadlock)

Correct Answer: $B$

16 votes

Option B is answer No. of resources, n = 21 No. of processes, k = 12 Processes {P0, P1....P11} make the following Resource requests: {R0, R20, R2, R18, R4, R16, R6, R14, R8, R12, R10, R10} For example P0 will request R0 (0%2 is = 0 and 0< n=21). Similarly, P10 will request R10. P11 will request R10 as n - i = 21 - 11 = 10. As different processes are requesting the same resource, deadlock may occur.

13 votes

4 votes

Even numbered process Pi request even numbered resources Ri and Ri+2. Thus they share no more than 1 resource.

Odd numbered process Pi request

Odd numbered resources Rn−i and Rn−i−2 if nn is even

Even numbered resources Rn−i and Rn−i−2 if n is odd.

In this example, for deadlock to occur, two processes must request same set of processes. For this, we need n to be odd so that some odd numbered process will request same set of even numbered resources as some even numbered process.

Thus option A and C are not possible as they have even n.

Now easy way to solve this is by finding the resources of even and odd numbered processes by using the given answer.

Taking option B:

Even numbered processes resource allocation-

P0 R0 , R2

P2 R2, R4

P4 R4, R6

P6 R6, R8

P8 R8, R10

P10 R10 , R12

Odd numbered processes resource allocation-

P1 R20, R18

P3 R18, R16

P5 R16, R14

P7 R14, R12

P9 R12, R10

P11 R10, R8

Odd numbered process Pi request

Odd numbered resources Rn−i and Rn−i−2 if nn is even

Even numbered resources Rn−i and Rn−i−2 if n is odd.

In this example, for deadlock to occur, two processes must request same set of processes. For this, we need n to be odd so that some odd numbered process will request same set of even numbered resources as some even numbered process.

Thus option A and C are not possible as they have even n.

Now easy way to solve this is by finding the resources of even and odd numbered processes by using the given answer.

Taking option B:

Even numbered processes resource allocation-

P0 R0 , R2

P2 R2, R4

P4 R4, R6

P6 R6, R8

P8 R8, R10

P10 R10 , R12

Odd numbered processes resource allocation-

P1 R20, R18

P3 R18, R16

P5 R16, R14

P7 R14, R12

P9 R12, R10

P11 R10, R8

IF YOU SEE BOTH YOU WILL FIND P10 AND P9 REQUESTING SAME RESOURCES WHICH MAY LEAD TO DEADLOCK.****