Python实现哲学家就餐问题实例代码
(编辑：jimmy 日期: 2024/11/2 浏览：3 次 )
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
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