用python实现银行家算法

好的，我可以为您解答这个问题。银行家算法是一种避免死锁的算法，它用于分配资源给进程。下面是用Python实现银行家算法的一种方法：

首先，我们需要定义一些变量，包括进程数、资源数、可用资源、每个进程已分配的资源和所需的资源。

# 进程数
n = 5
# 资源数
m = 3

# 可用资源
available = [3, 3, 2]

# 每个进程已分配的资源
allocation = [[0, 1, 0],
              [2, 0, 0],
              [3, 0, 2],
              [2, 1, 1],
              [0, 0, 2]]

# 每个进程所需的资源
max_need = [[7, 5, 3],
            [3, 2, 2],
            [9, 0, 2],
            [2, 2, 2],
            [4, 3, 3]]

然后我们需要计算出每个进程还需要的资源量，即所需资源减去已分配资源。

# 计算每个进程还需要的资源量
need = []
for i in range(n):
    need.append([max_need[i][j] - allocation[i][j] for j in range(m)])

接下来，我们定义一个函数来检查是否存在安全序列。如果存在安全序列，则返回True，否则返回False。

def is_safe(need, available, allocation):
    # 计算进程还需要的资源量
    work = available.copy()
    finish = [False] * n
    safe_seq = []

    while False in finish:
        found = False
        for i in range(n):
            if not finish[i] and all(need[i][j] <= work[j] for j in range(m)):
                found = True
                finish[i] = True
                safe_seq.append(i)
                work = [work[j] + allocation[i][j] for j in range(m)]
        if not found:
            return False

    return True

最后，我们调用is_safe函数来检查是否存在安全序列。

if is_safe(need, available, allocation):
    print("存在安全序列")
else:
    print("不存在安全序列")

完整代码如下：

# 进程数
n = 5
# 资源数
m = 3

# 可用资源
available = [3, 3, 2]

# 每个进程已分配的资源
allocation = [[0, 1, 0],
              [2, 0, 0],
              [3, 0, 2],
              [2, 1, 1],
              [0, 0, 2]]

# 每个进程所需的资源
max_need = [[7, 5, 3],
            [3, 2, 2],
            [9, 0, 2],
            [2, 2, 2],
            [4, 3, 3]]

# 计算每个进程还需要的资源量
need = []
for i in range(n):
    need.append([max_need[i][j] - allocation[i][j] for j in range(m)])

def is_safe(need, available, allocation):
    # 计算进程还需要的资源量
    work = available.copy()
    finish = [False] * n
    safe_seq = []

    while False in finish:
        found = False
        for i in range(n):
            if not finish[i] and all(need[i][j] <= work[j] for j in range(m)):
                found = True
                finish[i] = True
                safe_seq.append(i)
                work = [work[j] + allocation[i][j] for j in range(m)]
        if not found:
            return False

    return True

if is_safe(need, available, allocation):
    print("存在安全序列")
else:
    print("不存在安全序列")

希望对您有所帮助！