GATE CSE 2010 | Question: 46

Question

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?

• A. $n=40,\: k=26$
• B. $n=21,\:k=12$
• C. $n=20,\:k=10$
• D. $n=41,\:k=19$

Comments

Please remove the GATE2018 tag from the question

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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)

For deadlock to happen,

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.

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In the code, if the first 'if' statement is true then do we have to execute second 'if 'statement also?

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why deadlock happens only when even and odd processes are overlapping but here we can see in all the cases every odd process are overlapping and evens are also.why not deadlock in that case???

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Sree14  According to my understanding ,  Every even or odd processes overlap by atmost only one resource

R2 : 2 , 4                             R1 : n-1 , n-3

R4 : 4 , 6                             R3 : n-3 , n-5

R10 : 10 , 12                       R5 : n-5 , n-7

So, even though P2 takes R4 but releases it after its use then P4 will use R4.

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Very good analysis

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there is a mistake in go book in this question ,
In the second "if" of the "else" part instead of Rn−i−2 it shows Rn-

@go_editor

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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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we can solve by taking small value just like below solution 

https://gateoverflow.in/2348/gate-cse-2010-question-46?show=417063#a417063

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In the gate previous PYQ's pdf, the condition for odd processes is missing. Please update that in the pdf.

Answer 1

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: $B$

Comments

Is there any method with out brute force method to solve these type of problems ?

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Such a nice explanation.

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Mind blowing explanation !!

Answer 2

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. 

Comments

how every process is requesting only 1 resource?

according to the code it is requesting 2 resources...

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@Abhineet Singh

its like Give one resource and then prempt and go to the other process and give it another resource

Answer 3

The answer

Comments

@Ishaan Sood

But (d) is also satisfying 1 and 2.

 

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option D is not satisfying, for D k=19 and n=41, so (41-2)/2=19.5, so it is not satisfying the k>=(n-2)/2 condition

Answer 4

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.​​​​​​​

Comments

Even process p0 and p1 are requesting for resource R2 in every case isn't that a deadlock?