大门|大门-CS-2016（第1组）|问题60

考虑下面提出的临界截面问题的解决方案。有n个过程：P0…Pn−1.在代码中，函数pmax返回一个不小于其任何参数的整数。对于所有i，t[i]被初始化为零。

关于上述解决方案，以下哪项是正确的？ （A） 在任何时候，最多有一个过程处于关键部分 （B） 满足有界等待条件 （C） 进度条件满足 （D） 它不会导致僵局 答复: （A） 说明：

Mutual exclusion  is satisfied:
All other processes j started before i must have value (i.e. t[j]) 
less than the value of process i (i.e. t[i])  as function pMax() 
return a integer not smaller  than any of its arguments. So if anyone 
out of the processes j have positive value will be executing in its 
critical section as long as the condition t[j] > 0 && t[j] <= t[i] within 
while will persist. And when  this j process comes out of its critical 
section, it sets t[j] = 0;  and next process will be selected in for loop.
So when i process reaches to its critical section none of the  processes j 
which started earlier before process i  is in its critical section. This 
ensure that only one process is executing its critical section at a time. 

Deadlock and progress are  not satisfied:  
while (t[j] != 0 && t[j] <=t[i]); because of this condition deadlock is 
possible when value of j process becomes equals to the value of process i 
(i.e t[j] == t[i]).  because of the deadlock progress is also not possible 
(i.e. Progress == no deadlock) as no one process is able to make progress  
by stopping other process. 

Bounded waiting is also not satisfied: 
In this case both deadlock and bounded waiting to be arising from the same 
reason as if t[j] == t[i] is possible then starvation is possible means 
infinite waiting.

德伦德拉·辛格对此做出了解释。 这个问题的小测验