哲学家进餐问题避免死锁解决方法Python

哲学家进餐问题是一个经典的同步问题，它描述了五个哲学家围坐在一张圆桌前，每个哲学家面前有一碗饭和一只筷子。哲学家的生活方式是交替地进行思考和进餐，当一个哲学家思考时，他不需要任何资源，但是当他想进餐时，他需要两只筷子。问题在于，如果每个哲学家都拿起自己左边的筷子，那么他们将永远无法进餐，因为每个人都在等待右边的筷子。这就是死锁。

解决哲学家进餐问题的方法有很多种，以下是其中的一些：

1. Chandy/Misra解法：这种方法使用了额外的“协调者”进程来协调哲学家们的进餐，从而避免了死锁。具体实现可以参考下面的Python代码：

import threading

class Philosopher(threading.Thread):
    def __init__(self, left_fork, right_fork):
        threading.Thread.__init__(self)
        self.left_fork = left_fork
        self.right_fork = right_fork

    def run(self):
        while True:
            self.left_fork.acquire()
            locked = self.right_fork.acquire(False)
            if locked:
                break
            self.left_fork.release()
        else:
            return
        self.dine()
        self.right_fork.release()
        self.left_fork.release()

    def dine(self):
        print(f"{self.name} is eating")

forks = [threading.Lock() for n in range(5)]
philosophers = [Philosopher(forks[n], forks[(n + 1) % 5]) for n in range(5)]
for p in philosophers:
    p.start()

2. 限制进餐人数：这种方法通过限制同时进餐的哲学家人数来避免死锁。具体实现可以参考下面的Python代码：

import threading

class Philosopher(threading.Thread):
    running = True

    def __init__(self, left_fork, right_fork, count):
        threading.Thread.__init__(self)
        self.left_fork = left_fork
        self.right_fork = right_fork
        self.count = count

    def run(self):
        while self.running:
            self.left_fork.acquire()
            if not self.right_fork.acquire(False):
                self.left_fork.release()
                continue
            self.dine()
            self.right_fork.release()
            self.left_fork.release()

    def dine(self):
        print(f"{self.name} is eating")
        self.count -= 1

forks = [threading.Lock() for n in range(5)]
count = 2
philosophers = [Philosopher(forks[n], forks[(n + 1) % 5], count) for n in range(5)]
Philosopher.running = True
for p in philosophers:
    p.start()
while True:
    try:
        input()
    except KeyboardInterrupt:
        break
Philosopher.running = False
for p in philosophers:
    p.join()

3. 条件变量解法：这种方法使用了条件变量来协调哲学家们的进餐，从而避免了死锁。具体实现可以参考下面的Python代码：

import threading

class Philosopher(threading.Thread):
    def __init__(self, left_fork, right_fork, condition):
        threading.Thread.__init__(self)
        self.left_fork = left_fork
        self.right_fork = right_fork
        self.condition = condition

    def run(self):
        while True:
            with self.condition:
                self.condition.wait()
                self.left_fork.acquire()
                self.right_fork.acquire()
            self.dine()
            self.right_fork.release()
            self.left_fork.release()

    def dine(self):
        print(f"{self.name} is eating")

forks = [threading.Lock() for n in range(5)]
condition = threading.Condition()
philosophers = [Philosopher(forks[n], forks[(n + 1) % 5], condition) for n in range(5)]
with condition:
    condition.notifyAll()
for p in philosophers:
    p.start()
while True:
    try:
        input()
    except KeyboardInterrupt:
        break
for p in philosophers:
    p.join()