Python实现哲学家就餐问题实例代码

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}.

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

import queue
import threading
import time
import 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)

总结

到此这篇关于Python实现哲学家就餐问题的文章就介绍到这了,更多相关Python哲学家就餐内容请搜索python博客以前的文章或继续浏览下面的相关文章希望大家以后多多支持python博客！