Python实现哲学家就餐问题实例代码

网友投稿 889 2022-05-29

哲学家就餐问题:

哲学家就餐问题是典型的同步问题,该问题描述的是五个哲学家共用一张圆桌,分别坐在五张椅子上,在圆桌上有五个盘子和五个叉子(如下图),他们的生活方式是交替的进行思考和进餐,思考时不能用餐,用餐时不能思考。平时,一个哲学家进行思考,饥饿时便试图用餐,只有在他同时拿到他的盘子左右两边的两个叉子时才能进餐。进餐完毕后,他会放下叉子继续思考。请写出代码来解决如上的哲学家就餐问题,要求代码返回“当每个哲学家分别需要进食 n 次”时这五位哲学家具体的行为记录。

测试用例:

输入:n = 1 (1<=n<=60,n 表示每个哲学家需要进餐的次数。)

预期输出:

[[4,2,1],[4,1,1],[0,1,1],[2,2,1],[2,1,1],[2,0,3],[2,1,2],[2,2,2],[4,0,3],[4,1,2],[0,2,1],[4,2,2],[3,2,1],[3,1,1],[0,0,3],[0,1,2],[0,2,2],[1,2,1],[1,1,1],[3,0,3],[3,1,2],[3,2,2],[1,0,3],[1,1,2],[1,2,2]]

思路:

输出列表中的每一个子列表描述了某个哲学家的具体行为,它的格式如下:

output[i] = [a, b, c] (3 个整数)

a 哲学家编号。

b 指定叉子:{1 : 左边, 2 : 右边}.

Python实现哲学家就餐问题实例代码

c 指定行为:{1 : 拿起, 2 : 放下, 3 : 吃面}。

如 [4,2,1] 表示 4 号哲学家拿起了右边的叉子。所有自列表组合起来,就完整描述了“当每个哲学家分别需要进食 n 次”时这五位哲学家具体的行为记录。

代码实现

123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899import queueimport threadingimport timeimport random  class CountDownLatch:  def __init__(self, count):    self.count = count    self.condition = threading.Condition()  def wait(self):    try:      self.condition.acquire()      while self.count > 0:        self.condition.wait()    finally:      self.condition.release()  def count_down(self):    try:      self.condition.acquire()      self.count -= 1      self.condition.notifyAll()    finally:      self.condition.release()  class DiningPhilosophers(threading.Thread):  def __init__(self, philosopher_number, left_fork, right_fork, operate_queue, count_latch):    super().__init__()    self.philosopher_number = philosopher_number    self.left_fork = left_fork    self.right_fork = right_fork    self.operate_queue = operate_queue    self.count_latch = count_latch    def eat(self):    time.sleep(0.01)    self.operate_queue.put([self.philosopher_number, 0, 3])    def think(self):    time.sleep(random.random())    def pick_left_fork(self):    self.operate_queue.put([self.philosopher_number, 1, 1])    def pick_right_fork(self):    self.operate_queue.put([self.philosopher_number, 2, 1])    def put_left_fork(self):    self.left_fork.release()    self.operate_queue.put([self.philosopher_number, 1, 2])    def put_right_fork(self):    self.right_fork.release()    self.operate_queue.put([self.philosopher_number, 2, 2])    def run(self):    while True:      left = self.left_fork.acquire(blocking=False)      right = self.right_fork.acquire(blocking=False)      if left and right:        self.pick_left_fork()        self.pick_right_fork()        self.eat()        self.put_left_fork()        self.put_right_fork()        break      elif left and not right:        self.left_fork.release()      elif right and not left:        self.right_fork.release()      else:        time.sleep(0.01)    print(str(self.philosopher_number) + ' count_down')    self.count_latch.count_down()  if __name__ == '__main__':  operate_queue = queue.Queue()  fork1 = threading.Lock()  fork2 = threading.Lock()  fork3 = threading.Lock()  fork4 = threading.Lock()  fork5 = threading.Lock()  n = 1  latch = CountDownLatch(5 * n)  for _ in range(n):    philosopher0 = DiningPhilosophers(0, fork5, fork1, operate_queue, latch)    philosopher0.start()    philosopher1 = DiningPhilosophers(1, fork1, fork2, operate_queue, latch)    philosopher1.start()    philosopher2 = DiningPhilosophers(2, fork2, fork3, operate_queue, latch)    philosopher2.start()    philosopher3 = DiningPhilosophers(3, fork3, fork4, operate_queue, latch)    philosopher3.start()    philosopher4 = DiningPhilosophers(4, fork4, fork5, operate_queue, latch)    philosopher4.start()  latch.wait()  queue_list = []  for i in range(5 * 5 * n):    queue_list.append(operate_queue.get())  print(queue_list)

Python

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