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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?

1. $n=40,\: k=26$
2. $n=21,\:k=12$
3. $n=20,\:k=10$
4. $n=41,\:k=19$

edited | 9.8k views
+1
Please remove the GATE2018 tag from the question
+40
Deadlock iff Even and odd process request overlaps at some point. It will happen only if n is odd.

So we have only two options left (B) and (D). If you observe for even no. process resource allocation is happening from forward direction but for odd no. resource it is happening from back word direction. But in option (D) we have more then $2*k$ resources so overlap is not possible. Hence overlap will happen only in (B).
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As per the question, all the even processes request only even resources(i.e. Ri & Ri+2) Where as odd processes request: either, odd resources(i.e. Rn-1 & Rn-1-2) when 'n' is even. or, even resources (i.e. Rn-1 & Rn-1-2) when 'n' is odd.

Deadlock can only happen when even and odd processes request overlaps i.e. odd processes have to request even resources which means 'n' must should be odd.

Mathematically,

Let Pn-i be an odd process then,

n-i<=k

or,ceil(n-i)=k here ceil() is ceiling function since the value at max can be equal to k.

i=ceil(n-k)---eqn(i)

i=n-i-2

substituting value of i from eqn(i), we have,

ceil(n-k)=n-i-2

solving further we get,

k=ceil((n+2)/2) where 'n' is odd

since, 'n' must should be odd option A and C can be ignored now coming to option B substituting n=21 we get k=12 In option D substituting n=41 we get k=22.

so option B is correct.For option D to be in deadlock k should be equal to 22.

It is a mathematical way only. You can check it manually too without using the above formula.Like i said for deadlock to happen the the even and odd processes request should overlap .You can run the loops for both the options B and D as for A and  C they can be ignored since, 'n' is even.
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Very Good Question.

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: $B$
by Veteran (431k points)
edited
+7

just a small correction Sir there will be no P19 in case (D) as k=19 so P will be from 0 to 18

+2
Thanks for the correction :)
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+1
very clear explanation arjun sir
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at option, B...how to reach to P8 and P11...just trial and error of any other method?
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check answer given by ishaan sood
+1
Thqqq soo much ..crystal clear explanation...
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great explanation
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Can you plz explain dat more elaborately???
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Thanks sir ji
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How can we tackle such questions in exam.. Seriously!! :(
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. 
by Junior (973 points)

by (429 points)
0

But (d) is also satisfying 1 and 2.

+1 vote

(k-3) processes will take 2 resources and other 3 processes will take 1 resource each.

Total n resources. Deadlock will arise when : (((k-3) * 2) + (3*1)) = n

If k=12, n will be 21.

by (21 points)
0
@dibyajyoti

Thanks
+1
@bikram sir verify this
+1 vote
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

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

by (293 points)
edited by