如何检测 python 中的模型死锁

Question

如何编写模型检测程序？

流程、资源、请求和发布列表：

A req R
A req S
A rel R
A rel S
B req S
B req R
B rel S
B rel R

无法重新安排流程中的顺序（在上面的示例中，A 必须始终在请求 S 之前先请求 R），但是可以重新安排跨流程的顺序（B 可以在 A 请求 R 之前请求 S 和 R）

每个排序都将输出带有标题、排序以及订单是否导致死锁的指示。

输出我想要的结果：-

ORDER 1: A req R, A req S, A rel R, A rel S, B req S, B req R, B rel S, B rel R
ORDER 2: A req R, A req S, A rel R, B req S, A rel S, B req R, B rel S, B rel R
...
ORDER X: A req R, B req S, A req S, B req R DEADLOCK!

测试用例输入：-

案例一：-

A req R A req S A rel R A rel S B req S B req T B rel S B rel T C req T 
C req R C rel T C rel R

案例2：-

A req R A req S A rel R A rel S B req T B rel T C req S C rel S D req U 
D req S D req T D rel U D rel S D rel T E req T E req V E rel T E rel V 
F req W F req S F rel W F rel S G req V G req U G rel V G rel U

from itertools import permutations


class Resource:
    def __init__(self, name):
        self.name = name
        self.holder = None
        self.waiting_list = []

    def request(self, process):
        if self.holder is None:
            self.holder = process
            return True
        else:
            self.waiting_list.append(process)
            return False

    def release(self):
        if len(self.waiting_list) == 0:
            self.holder = None
        else:
            next_process = self.waiting_list.pop(0)
            self.holder = next_process
            return next_process


class Process:
    def __init__(self, name):
        self.name = name
        self.resources = []

    def add_resource(self, resource):
        self.resources.append(resource)

    def request_resources(self):
        for resource in self.resources:
            if not resource.request(self):
                return False
        return True

    def release_resources(self):
        for resource in self.resources:
            next_process = resource.release()
            if next_process is not None:
                next_process.request_resources()


def simulate_ordering(processes):
    for ordering in permutations(processes):
        for process in ordering:
            if not process.request_resources():
                break
        else:
            for process in reversed(ordering):
                process.release_resources()
            return True
    return False


def detect_deadlock(processes):
    if simulate_ordering(processes):
        print("No deadlock detected.")
    else:
        print("Deadlock detected!")


if __name__ == "__main__":
    # File 1 input
    r = Resource("R")
    s = Resource("S")
    t = Resource("T")
    processes = [
        Process("A"),
        Process("B"),
        Process("C")
    ]
    for process in processes:
        process.add_resource(r)
        process.add_resource(s)
        process.add_resource(t)
    processes[0].add_resource(t)
    processes[0].add_resource(r)
    processes[1].add_resource(s)
    processes[1].add_resource(r)
    processes[2].add_resource(t)
    processes[2].add_resource(r)

    detect_deadlock(processes)

    # File 2 input
    r = Resource("R")
    s = Resource("S")
    t = Resource("T")
    u = Resource("U")
    v = Resource("V")
    w = Resource("W")
    processes = [
        Process("A"),
        Process("B"),
        Process("C"),
        Process("D"),
        Process("E"),
        Process("F"),
        Process("G")
    ]
    for process in processes:
        process.add_resource(r)
        process.add_resource(s)
        process.add_resource(t)
        process.add_resource(u)
        process.add_resource(v)
        process.add_resource(w)
    processes[0].add_resource(r)
    processes[0].add_resource(s)
    processes[1].add_resource(t)
    processes[2].add_resource(s)
    processes[3].add_resource(u)
    processes[3].add_resource(s)
    processes[3].add_resource(t)
    processes[4].add_resource(t)
    processes[4].add_resource(v)
    processes[5].add_resource(w)
    processes[5].add_resource(s)
    processes[6].add_resource(v)
    processes[6].add_resource(u)

    detect_deadlock(processes)

上面的程序打印输出如下

Deadlock detected!

现在，我怎样才能像上面提到的那样打印结果。请指导我打印输出。谢谢。