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哲学家就餐问题是典型的同步问题#xff0c;该问题描述的是五个哲学家共用一张圆桌#xff0c;分别坐在五张椅子上#xff0c;在圆桌上有五个盘子和五个叉子#xff08;如下图#xff09;#xff0c;他们的生活方式是交替的进行思考和进餐
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 1n60n 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(blockingFalse)
right self.right_fork.acquire(blockingFalse)
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
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本文标题: Python实现哲学家就餐问题实例代码
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