### 银行家算法示例题目及解答
#### 示例场景描述
考虑一个系统中有五个进程 \( P_0 \),\( P_1 \),\( P_2 \),\( P_3 \),\( P_4 \) 和三种类型的资源 R1, R2, R3。当前系统的状态如下:
| Process | Max | Allocation | Need |
|---------|--------|------------|----------|
| \( P_0 \)| 7 5 3 | 0 1 0 | 7 4 3 |
| \( P_1 \)| 3 2 2 | 2 0 0 | 1 2 2 |
| \( P_2 \)| 9 0 2 | 3 0 2 | 6 0 0 |
| \( P_3 \)| 2 2 2 | 2 1 1 | 0 1 1 |
| \( P_4 \)| 4 3 3 | 0 0 2 | 4 3 1 |
当前可用的资源数量为 Available = (3, 3, 2).
#### 安全序列检测过程
为了判断该状态下是否存在安全序列,按照银行家算法进行安全性检查。
1. 初始化工作列表 Work 设置为 Available 的初始值 (3, 3, 2).
2. 创建 Finish 数组用于记录每个进程的状态,默认全部设置为 false.
3. 寻找满足条件 `Need[i] <= Work` 并且尚未完成 (`Finish[i]=false`) 的进程 i.
对于上述表格中的数据:
- **第一次迭代**
- 检查所有进程的需求是否小于等于 Work 值。
- 发现 \( P_1 \)'s Need(1, 2, 2) 小于等于 Work(3, 3, 2)[^2].
- 更新 Work += Allocation of \( P_1 \)=Work+(2, 0, 0)=(5, 3, 2).
- 标记 \( P_1 \) 已经完成(Finish\[1\]=true).
- **第二次迭代**
- 继续寻找符合条件的新进程...
- 找到 \( P_3 \)'s Need(0, 1, 1)<=Work(5, 3, 2).
- 更新 Work+=Allocation of \( P_3 \)=Work+(2, 1, 1)=(7, 4, 3).
- 标记 \( P_3 \) 已经完成(Finish\[3\]=true).
重复此操作直到找到完整的安全序列或者无法再找到任何可运行的进程为止。最终得到的安全序列为 `<P1,P3,...>` (具体取决于后续几步的结果)。如果能够成功构建这样一个序列,则说明此时处于安全状态;反之则表示存在潜在死锁风险[^4].
```python
def banker_algorithm(max_resources, allocated_resources, available_resources):
need_matrix = [[max_res - alloc for max_res, alloc in zip(max_row, alloc_row)]
for max_row, alloc_row in zip(max_resources, allocated_resources)]
n_processes = len(max_resources)
finish = [False] * n_processes
work = list(available_resources)
safe_sequence = []
while False in finish:
found = False
for i in range(n_processes):
if not finish[i] and all([need <= w for need, w in zip(need_matrix[i], work)]):
work = [w + a for w, a in zip(work, allocated_resources[i])]
finish[i] = True
safe_sequence.append(f"P{i}")
found = True
if not found:
break
return "Safe Sequence:" + ", ".join(safe_sequence) if all(finish) else "No Safe Sequence Found"
# Example usage with given data points
print(banker_algorithm(
[(7, 5, 3), (3, 2, 2), (9, 0, 2), (2, 2, 2), (4, 3, 3)],
[(0, 1, 0), (2, 0, 0), (3, 0, 2), (2, 1, 1), (0, 0, 2)],
(3, 3, 2)))
```