哲學家就餐問題：

哲學家就餐問題是典型的同步問題，該問題描述的是五個哲學家共用一張圓桌，分別坐在五張椅子上，在圓桌上有五個盤子和五個叉子(如下圖)，他們的生活方式是交替的進行思考和進餐，思考時不能用餐，用餐時不能思考。平時，一個哲學家進行思考，饑餓時便試圖用餐，只有在他同時拿到他的盤子左右兩邊的兩個叉子時才能進餐。進餐完畢后，他會放下叉子繼續(xù)思考。請寫出代碼來解決如上的哲學家就餐問題，要求代碼返回“當每個哲學家分別需要進食 n 次”時這五位哲學家具體的行為記錄。

測試用例：

輸入：n = 1 (1<=n<=60，n 表示每個哲學家需要進餐的次數(shù)。)

預期輸出：

[[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:

輸出列表中的每一個子列表描述了某個哲學家的具體行為，它的格式如下：

output[i] = [a, b, c] (3 個整數(shù))

a 哲學家編號。

b 指定叉子：{1 : 左邊, 2 : 右邊}.

c 指定行為：{1 : 拿起, 2 : 放下, 3 : 吃面}。

如 [4,2,1] 表示 4 號哲學家拿起了右邊的叉子。所有自列表組合起來，就完整描述了“當每個哲學家分別需要進食 n 次”時這五位哲學家具體的行為記錄。

代碼實現(xiàn)

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)

總結(jié)

到此這篇關(guān)于Python實現(xiàn)哲學家就餐問題的文章就介紹到這了,更多相關(guān)Python哲學家就餐內(nèi)容請搜索python博客以前的文章或繼續(xù)瀏覽下面的相關(guān)文章希望大家以后多多支持python博客！

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