Other points could be discussed. First, the parameter selection in the inverse problem is a trial-and-error experiment. In most cases, the procedure becomes user-dependent and time-consuming. In the simulated brain phantom, synthesizedclinical patient and preclinical physical phantom experiments, an efficient GA-based strategy was developed to address this challenge. It should be noted that the parameters in the preclinical pig leg experiment were empirically determined. To address this problem, other automatic methods can be used, e.g., a task-driven method [40] and a deep learning (DL) method [41].翻译

还有其他可以讨论的问题。首先，逆问题中的参数选择是一个试错实验。在大多数情况下，该过程变得依赖于用户且耗时。在模拟的脑幻像、合成的临床患者和临床前物理幻像实验中，开发了一种有效的基于遗传算法的策略来解决这个挑战。值得注意的是，临床前猪腿实验中的参数是经验性确定的。为了解决这个问题，可以使用其他的自动方法，例如，任务驱动方法[40]和深度学习（DL）方法[41]。

相关问题

翻译：With the given constraints, the problem is two-fold. How should a BS determine the optimal access price? How should each UE choose the BS to associate with in response to the block-varying price? First, UE interest must be discussed. Offloading to a certain BS requires an access payment of the offlfloaded UE based on the announced access price vector p. Second, on the BS side, a larger load will increase the access revenue, but spectrum acquisition for the BS will also increase. Therefore, in the user association procedure, all BSs determine their optimal access prices to maximize their benefifits in terms of cell load, congestion and UE payments. Then, using the announced price vector, each UE device maximizes the profifit of its utility function minus the access payment. In this section, the mathematical representation of the proposed user association and resource allocation procedure will be discussed.

在给定的约束条件下，问题是双重的。一个基站应该如何确定最优的接入价格？每个用户设备应该如何选择要关联的基站，以应对块变化的价格？

首先，必须讨论用户设备的兴趣。将数据卸载到某个基站需要根据宣布的接入价格向卸载用户设备支付一定的接入费用。其次，在基站方面，较大的负载将增加接入收入，但基站的频谱获取也会增加。因此，在用户关联过程中，所有基站确定其最优的接入价格，以在单元负载、拥塞和用户设备支付方面最大化其利益。然后，使用宣布的价格向量，每个用户设备最大化其效用函数减去接入费用的收益。本节将讨论所提出的用户关联和资源分配过程的数学表示。

Consider a problem to find the shortest path from a given starting node s to any other nodes in the (undirected) graph. The length of a path in the graph is the number of edges included. Complete the function shortest_path(adj_list, s), which takes the adjacency list of the graph and a starting node as input, and returns a list containing the length of the shortest path from the starting node to all other nodes. Note: If there is no path from the starting node to a node, set the length of shortest path as float('inf'). You can use any Python built-in package to implement the stack/queue/priority queue if needed. Hint: Consider the three graph traversal strategies (DFS, BFS and PFS) discussed in the class. Which one is applicable in this problem? For example: Test Result adj_list = [[], [2, 3], [1, 4], [1], [2]] d = shortest_path(adj_list, 0) print(d) d = shortest_path(adj_list, 2) print(d) [0, inf, inf, inf, inf] [inf, 1, 0, 2, 1]

To find the shortest path from a given starting node s to any other nodes in the (undirected) graph, we can use Breadth-First Search (BFS) algorithm. The basic idea of BFS is to explore all the vertices at distance 1 from the current vertex before moving on to vertices at distance 2.

Here is the implementation of the shortest_path function using BFS algorithm:

from collections import deque

def shortest_path(adj_list, s):
    n = len(adj_list)
    visited = [False] * n
    distance = [float('inf')] * n
    distance[s] = 0
    queue = deque([s])
    
    while queue:
        u = queue.popleft()
        visited[u] = True
        
        for v in adj_list[u]:
            if not visited[v]:
                visited[v] = True
                distance[v] = distance[u] + 1
                queue.append(v)
    
    for i in range(n):
        if not visited[i]:
            distance[i] = float('inf')
    
    return distance

In the above code, we first initialize two lists: visited and distance. The visited list is used to keep track of the visited nodes and the distance list is used to store the shortest distance from the starting node to all other nodes. We initialize all the distances as infinity except the distance of the starting node which is set to 0.

We then use a deque (double-ended queue) to implement the BFS algorithm. We start by adding the starting node to the queue. Then, while the queue is not empty, we remove a vertex u from the front of the queue and mark it as visited. We then iterate over all the neighbors v of u and if v is not visited, we mark it as visited, update its distance from the starting node and add it to the end of the queue.

Finally, we check if there are any nodes that were not visited during the BFS traversal and set their distance as infinity. We then return the distance list.

Let's use the above code to solve the given example:

adj_list = [[], [2, 3], [1, 4], [1], [2]]
d = shortest_path(adj_list, 0)
print(d) # Output: [0, inf, inf, inf, inf]

d = shortest_path(adj_list, 2)
print(d) # Output: [inf, 1, 0, 2, 1]

In the first test case, the starting node is 0 and there are no edges connected to it. Hence, the distance to all other nodes is infinity.

In the second test case, the starting node is 2 and the shortest path to node 2 is 0 (itself). The shortest path to node 1 is 1 (through node 2), the shortest path to node 3 is 2 (through nodes 2 and 1), and the shortest path to node 4 is 1 (through node 2).