Python实现哲学家就餐问题实例代码

Python实现哲学家就餐问题实例代码,第1张

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考高分网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考高分网!

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