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Deadlock means something specific: there are two (or more) processes that are currently blocked waiting for each other. In an unsafe state you can also be in a situation where there might be a deadlock sometime in the future, but it hasn't happened yet because one or both of the processes haven't actually started waiting. Consider the following example: ...


10

I agree that no deadlock is possible here. If there are three or fewer processes, there clearly cannot be a deadlock because there are enough resources for every process to just hold two resources the whole time. So any deadlock must have at least four participating processes. To participate in a deadlock, a process must hold at least one resource. Further,...


7

Just to expound on what Wandering Logic was saying. Say I have two threads that both need access to X and Y, and have no synchronization and no mechanism to fix deadlock. This is unsafe, as one could lock X and the other Y and then neither could proceed. But it isn't guaranteed. Thread 1 Thread 2 Lock X Lock Y OS ...


7

You should first state the deadlock freedom property and the starvation freedom property more precisely. I use the definition in the Book: The Art of Multiprocessor Programming; Section 2.2. Freedom from Deadlock If some thread attempts to acquire the lock, then some thread (not necessarily the thread referred to in the if statement; emphasis added) will ...


6

You are correct. Consider the simple synchronization algorithm which denies entry to all processes. In this case we have both deadlock and bounded waiting, since any process $p$ is not bypassed by some process $p'$ before entering the critical section, so you could say bounded waiting is satisfied with the constant function $f=0$.


5

I think you are asking about expressivity of concurrent programming languages. This is a deep and not well-understood field. For example you say that "the $\pi$-calculus [...] has the power to implement almost any synchronization primitive I've ever heard of". It is well known that the $\pi$-calculus cannot implement broadcasting (see e.g. here). The $\pi$-...


4

Righties: will never try to acquire the left fork before they have the right one. Lefties: will never try to acquire the right fork before they have the left one. Deadlock: Assume that you reached deadlock, this means that none of the philosophers can eat (so each philosopher holds at most one fork, and cannot pick up the other one). Sine you have both ...


4

All 4 conditions must be satisfied at the same time. Hold and Wait Non-preemption of resources Mutual Exclusion Circular wait In case of single instance of resources: Cycle in resource allocation graph represents deadlock. In case of multiple instance of resources: Cycle in RAG doesn't mean deadlock. You must check in the same way as you did. Let me ...


4

Resources that can only be used by a single process at a time are usually protected by a lock. If the process that currently holds the lock attempts to acquire it a second time, what happens depends on the type of lock. If the lock is reentrant, then it can be acquired multiple times by the same process. The process will need to call the release function as ...


4

Almost every modern processor has special memory instructions built in specifically to deal with this problem. For example, many processors have a swap instruction. This atomically swaps a value between a local variable in a thread and a global variable that is shared between threads. (Under the covers the "local variable" will be stored in a machine ...


4

the dining philosopher problem seems to be somewhat of a pedagogical "toy" example of concurrency and (dead)locking concepts for educational purposes. however it is studied seriously in some literature and there are some performance analyses of its probability of deadlock scenarios. here are some examples of quantitative models and performance analysis and ...


4

Agent A consumes agentSem, and produces tobacco and paper. That might make Smoker I smoke but he is late: Smoker II already took the paper and Smoker III took tobacco. Now, all the smokers are stuck, and the agents as well.


3

Your attempt "If a process is unable to change its state indefinitely because the resources requested by it are being used by another waiting process, then the system is said to be in a deadlock." is close, but not quite there. A deadlock is a situation where one "process" (not a "process" as used in software development, but something more general) is ...


3

I don't know what algorithm the philosophers follow, but let's assume that deadlock is reached if all philosophers try to start eating at the very same time. Since your time is discrete, under any reasonable probability distribution over the "thinking times", it will necessarily (almost surely) eventually happen that all philosophers try to start eating at ...


3

you can say there is no deadlock if there is a safe sequence for the completion, after $P1$ request is granted there must be a safe sequence where by which every process is completed. After P1 request is granted | P1 | P2 | P3 | Max Requirement | 8 | 7 | 5 | Current Allocation | 4 | 1 | 3 | ...


3

It seems that deadlock prevention and deadlock avoidance are two names for the same concept. Indeed, the Wikipedia section on deadlock avoidance has been marked as redundant. While the distinction might be taken from the literature, some people at least are arguing that this distinction is superfluous. See the paper The classification of deadlock prevention ...


3

The question in the given situation is equivalent to the question: is it possible to create or to design a system such that m processes sharing n resources of the same type may enter a deadlock state subject to the following: each process needs less than or equal to n resources, and the total number of needed resources is less than or equal to m+n (“For a ...


3

You simply need to place a total order on the accounts and always lock the accounts in that order. You define a < operator on Accounts and then change the beginning like this: mutex lock1 = getlock(from) mutex lock2 = getlock(to) if (from < to) mutex lockSwap = lock1 lock1 = lock2 lock2 = lockSwap It doesn't matter what order you choose, ...


3

Preventing deadlocks can be achieve by removing one of the Coffman condition, one being the circular wait condition. For this we need to build a partial ordering of the resources. From Wikipedia a partial ordered (poset) set is : A (non-strict) partial order is a binary relation "≤" over a set P which is reflexive, antisymmetric, and transitive, i.e., ...


3

I did some thinking and then searching and all RAG conditions of this example leads to conclusion that there is no deadlock (altho there is a cycle here). Beside after searching I found this where you can find this: Which clearly stands Graph With A Cycle But No Deadlock Are you sure that you have your graph correct here?


3

The idea is that the process is not terminated: only the current operation is cancelled. The process reacts to the cancellation by trying again. In this model, it is assumed that each process runs something like the following pseudocode: begin_transaction: try: acquire(lock_1); acquire(lock_2); … release(lock_2); ...


3

In the following, I treat the first two statements acquire(lock1) acquire(lock2) as "trying", the middle two statements withdraw(from, amount) deposit(to, amount) as the "critical section", and the last two statements release(lock2) release(lock1) as "exiting". This algorithm does not satisfy the bounded waiting property. Consider the following execution: $...


3

Simply to put Deadlock avoidance: you employ some methods to avoid the deadlock ,but prevention is more restrictive than avoidance. Deadlock detection: Here only the detection only takes place whether the current state is in deadlock or not.


3

No, starvation-free doesn't imply bounded waiting. For instance, consider a procedure that never even attempts to acquire any lock; but the amount of time it takes is variable and can be arbitrarily long. Then there is no bound on the amount of time it might take to complete its operation. Here is another example of how it can fail. Starvation-free means ...


3

The dining philosopher's problem is a worst case scenario used to illustrate issues in algorithms. It's a test case. You probably won't find a static one-to-one mapping between it and a real world case. However, similar situations can occur dynamically. If at any point the graph of actors waiting for resources forms a cycle, you are stuck in a deadlock. ...


2

To understand this problem you have to understand the rules for the safety checks. I am not going to give a detailed explanation of these but you can go here to read more about it. I will try to give an explanation of why C request 1 is safe but A request 1 is not. There is a value associated with a process and a resource called need. The need is an n-tuple ...


2

It's fixed. Your example violates the protocol. If the process needs mutually exclusive access to both the tape drive and the disk drive, then it needs to take locks in increasing order, i.e. first the tape drive then the disk drive. If it wants to use the disk drive before the tape drive, that's fine, it will just be holding the tape drive lock in the ...


2

The question is not clear about the details, but assuming said resources are always locked-then-accessed sequentially, then a deadlock is not possible to occur. Since the locks are always obtained in the same order, precedence would prevent any cross-lock scenarios. Once a transaction T gained the lock for the first resource then a) it's guaranteed that no ...


2

A little remark first: The bounded waiting (BW) property is defined with respect to algorithms, which is in turn defined as a set of concrete executions. Thus, we cannot conclude whether an algorithm satisfies BW if a specific execution does. For your question: For the specific execution in which only $P_1$ takes steps, BW is satisfied. However, it is not ...


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