made east test series - synchronization - OS

Question

Consider the following proposed solution to Dining Philosopher’s problem to avoid deadlock. The binary semaphore lock is initialized to 1.

Which of the following is correct?

• A. Removing (i) and (ii) will not affect the code. It will work fine.
• B. Removing (iii) and (iv) will not affect the code. It will work fine.
• C. Removing (i), (ii), (iii) and (iv) will not affect the code. It will work fine.
• D. All (i), (ii), (iii) and (iv) are necessary. Removal of any of them will affect the code.

Comments

1. The problem says that this code is proposed to solve deadlock, which I think it doesn’t. Consider this: first philosopher lifts his left and right fork. After that signal(lock) is performed. second philosopher does wait(lock) and enters. now, he can’t lift his right fork because first philosopher is still eating. Also, now the first philosopher can’t enter the signal part for forks because lock is 0. Hence, there is deadlock. But since the question says it’s only a proposed solution, let’s assume that it doesn’t matter if it cannot solve deadlock. Which brings me to the second point.
2. If we remove (i) and (ii), clearly there’s a circular wait and deadlock because every philosopher will lift his left fork and wait for one another. If we remove (iii) and (iv), there will be no hindrance if any philosopher wants to release his forks (unlike before, in first point). So, removing them does change the code.

According to me, answer should be D but the answer key has B. 

Please tell me what’s wrong with my thinking.

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@atulcse Yes, seems like deadlock will happen. Now coming to your second point, I don't think the answer should be D. Say, if the only objective here is to avoid deadlock, removing 3 and 4 should get the job done. 1 and 2 will lead to deadlock only, as you have mentioned. I believe the options could be phrased differently otherwise the question is just wrong.

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iii would create a deadlock. Two philosphers 1 and 2 both run concurrently. one executes upto eat(). next one executes i but blocks on wait(fork[2]). now philospher 1 executes iii. Both wait for each other creating deadlock

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If we remove 1 and 2 then there is chance to deadlock…

Assume every i'th philosopher preempt after wait(fork [i]) ...at the end all process are in waiting state

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This is the explanation.

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Very blurred.

Add with clear handwriting if possible. Will be helpful for anyone checking it later on.