Let’s address each of your questions one by one based on the context you provided and the existing literature on mobile crowdsensing (MCS) and reinforcement learning:
Addressing Selfishness in Cell Selection
1. Is addressing selfishness fair or unfair?
Addressing the selfish behavior in MCS by limiting the amount of data one can collect aims to ensure fairness and effective data distribution across different cells. It prevents certain cells (or agents) from monopolizing the rewards, which could otherwise result in skewed data collection and potential overfitting of models to specific regions or scenarios. Although highly capable individuals or cells might earn less than they potentially could without restrictions, this approach helps maintain a broad and representative dataset, which is crucial for generalized MCS applications. It can be seen as fair in the context of the overall system’s health and data integrity, even if it might feel unfair from the perspective of individual high performers.
Modification to Algorithm for Addressing Selfishness
2. Any necessary modifications to the pseudocode?
Your pseudocode needs to explicitly include the condition checking for past selection frequency of each cell. If a cell exceeds a certain threshold in terms of how often it has been selected, the algorithm should either:
- Skip selecting that cell for a defined period or,
- Diminish the rewards associated with selecting that cell, thus making it less attractive to the algorithm.
You can introduce a decay factor for the reward associated with a cell each time it is chosen, or an “ineligibility” period during which the cell cannot be selected again.
Suitable Multi-agent Reinforcement Learning Algorithm
3. What multi-agent reinforcement learning method to use?
Considering the need for both competitive and cooperative dynamics between cells, algorithms that balance exploration (discovering new or less frequently used cells) and exploitation (leveraging known good cells) are essential. Nash Q-learning, as presented in your initial algorithm, is suitable as it accounts for the equilibrium strategy where each agent’s action is optimal given the other agents’ strategies. Enhancements could involve:
- Deep Multi-Agent Reinforcement Learning methods that use neural networks to approximate the Q-values, allowing for better generalization across a large state space.
- Counterfactual Multi-Agent (COMA) Learning which can be used to differentiate the effect of an agent’s action from the actions of other agents, fostering more cooperative strategies among cells.
Python Code Implementation
4. Simple Python code for the proposed scenario
Here’s a basic structure of what the Python code might look like, focusing on implementing a check for past selection frequency to adjust the selection strategy:
import numpy as np
class CellAgent:
def __init__(self, num_cells, threshold):
self.num_cells = num_cells
self.threshold = threshold
self.selection_count = np.zeros(num_cells)
self.q_values = np.random.rand(num_cells)
def update_selection(self, selected_cell):
self.selection_count[selected_cell] += 1
if self.selection_count[selected_cell] > self.threshold:
self.q_values[selected_cell] *= 0.9 # Reduce Q-value by 10%
def select_cell(self):
available_cells = np.where(self.selection_count <= self.threshold)[0]
if available_cells.size > 0:
return np.argmax(self.q_values[available_cells])
else:
return np.argmax(self.q_values) # Fallback to the best available option if all exceed threshold
def simulate_selection(self):
for _ in range(100): # Simulate 100 selections
cell = self.select_cell()
self.update_selection(cell)
print(f"Cell {cell} selected, count: {self.selection_count[cell]}")
# Example usage
agent = CellAgent(num_cells=10, threshold=5)
agent.simulate_selection()
This Python code defines a simple reinforcement learning agent managing a fixed number of cells, using a threshold to limit selections and decrease the attractiveness of a cell after it reaches a certain number of selections. This is a basic approach and would need to be expanded significantly for real-world application, including incorporating actual MCS data and a more sophisticated learning algorithm.
树形DP
由于是一个有向无环图且入度为 $0$ 的点只有 $1$ 个，可以用两次 $DFS $解决：
-
第一次 $DFS$ 从入度为 $0$ 的点出发，找出每个点往下走的最长的链的长度，要么就是 $f[u] $，要么是子节点 $f[j] + 1$，取最大值。
-
找到每个节点的最长的链的长度后，第二次 $DFS$ 从入度为 $0$ 的点出发，依次找到子节点中满足 $f[j] = f[u] - 1$ 的结点的最小值（题目要求字典序最小），继续递归。
-
在第二次 $DFS$ 时，边递归边存入结果，最后输出答案。
附C++，Java，Go代码
C++代码：
#include <iostream>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
const int N = 10010;
int h[N], e[N], ne[N];
int d[N], f[N];
vector<int> res;
int n, idx;
void add(int a, int b)
{
e[idx] = b;
ne[idx] = h[a];
h[a] = idx ++;
}
void dfs(int u, int fa)
{
f[u] = 1;
for (int i = h[u]; i != -1; i = ne[i]) {
int j = e[i];
if (j == fa) continue;
dfs(j, u);
f[u] = max(f[u], f[j] + 1);
}
}
void dfs_g(int u, int fa)
{
int minv = 1e9;
for (int i = h[u]; i != -1; i = ne[i]) {
int j = e[i];
if (j == fa) continue;
if (f[j] == f[u] - 1)
minv = min(minv, j);
}
if (minv != 1e9) { // 不是叶子节点时
res.push_back(minv);
dfs_g(minv, u);
}
}
int main()
{
cin >> n;
memset(h, -1, sizeof h);
for (int i = 0; i < n; i ++)
{
int k, x;
cin >> k;
while (k -- )
{
cin >> x;
add(i, x);
d[x] ++;
}
}
int g = -1;
for (int i = 0; i < n; i ++)
if (d[i] == 0)
g = i;
dfs(g, -1);
cout << f[g] << endl; // 最长的链的长度
res.push_back(g);
dfs_g(g, -1);
for (int x : res) cout << x << " ";
return 0;
}
Java代码：
import java.util.*;
public class Main {
static final int N = 10010;
static int[] h = new int[N];
static int[] e = new int[N], ne = new int[N];
static int[] d = new int[N]; // 入度
static int[] f = new int[N]; // 存每个点的最长的链的长度
static List<Integer> res = new ArrayList<>();
static int n, idx;
static void add(int a, int b) {
e[idx] = b;
ne[idx] = h[a];
h[a] = idx ++;
}
static void dfs(int u, int fa) {
f[u] = 1;
for (int i = h[u]; i != -1; i = ne[i]) {
int j = e[i];
if (j == fa) continue;
dfs(j, u);
f[u] = Math.max(f[u], f[j] + 1);
}
}
static void dfs_g(int u, int fa) {
int minv = 1_000_000_000;
for (int i = h[u]; i != -1; i = ne[i]) {
int j = e[i];
if (j == fa) continue;
if (f[j] == f[u] - 1)
minv = Math.min(minv, j);
}
if (minv != 1_000_000_000) { // 不是叶子节点时
res.add(minv);
dfs_g(minv, u);
}
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
n = sc.nextInt();
Arrays.fill(h, -1);
for (int i = 0; i < n; i ++) {
int k = sc.nextInt();
while (k -- > 0) {
int x = sc.nextInt();
add(i, x);
d[x] ++;
}
}
int g = -1;
for (int i = 0; i < n; i ++)
if (d[i] == 0)
g = i;
dfs(g, -1);
System.out.println(f[g]); // 最长的链的长度
res.add(g);
dfs_g(g, -1);
for (int x : res)
System.out.printf("%d ", x);
}
}
Go代码：
package main
import (
. "fmt"
)
const N = 10010
var (
h = make([] int, N)
e = make([] int, N)
ne = make([] int, N)
d = make([] int, N)
f = make([] int, N)
res []int
n, idx int
)
func add(a, b int) {
e[idx] = b
ne[idx] = h[a]
h[a] = idx
idx ++
}
func dfs(u, fa int) {
f[u] = 1
for i := h[u]; i != -1; i = ne[i] {
j := e[i]
if j == fa {
continue
}
dfs(j, u)
f[u] = max(f[u], f[j] + 1)
}
}
func dfs_g(u, fa int) {
minv := 1_000_000_000
for i := h[u]; i != -1; i = ne[i] {
j := e[i]
if j == fa {
continue
}
if f[j] == f[u] - 1 {
minv = min(minv, j)
}
}
if minv != 1_000_000_000 {
res = append(res, minv)
dfs_g(minv, u)
}
}
func main() {
Scan(&n)
for i := 0; i < n; i ++ {
h[i] = -1
}
for i := 0; i < n; i ++ {
var k, x int
Scan(&k)
for j := 0; j < k; j ++ {
Scan(&x)
add(i, x)
d[x]++
}
}
g := -1
for i := 0; i < n; i ++ {
if d[i] == 0 {
g = i
}
}
dfs(g, -1)
Println(f[g])
res = append(res, g)
dfs_g(g, -1)
for _, x := range res {
Printf("%d ", x)
}
}
func max(a, b int) int { if a > b { return a }; return b }
func min(a, b int) int { if a < b { return a }; return b }