您好,登錄后才能下訂單哦!
這篇文章將為大家詳細講解有關(guān)Python編寫一個哲學家就餐問題,文章內(nèi)容質(zhì)量較高,因此小編分享給大家做個參考,希望大家閱讀完這篇文章后對相關(guān)知識有一定的了解。
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)
關(guān)于Python編寫一個哲學家就餐問題就分享到這里了,希望以上內(nèi)容可以對大家有一定的幫助,可以學到更多知識。如果覺得文章不錯,可以把它分享出去讓更多的人看到。
免責聲明:本站發(fā)布的內(nèi)容(圖片、視頻和文字)以原創(chuàng)、轉(zhuǎn)載和分享為主,文章觀點不代表本網(wǎng)站立場,如果涉及侵權(quán)請聯(lián)系站長郵箱:is@yisu.com進行舉報,并提供相關(guān)證據(jù),一經(jīng)查實,將立刻刪除涉嫌侵權(quán)內(nèi)容。