使用 Repast Simphony 的意外结果
Unexpected results using Repast Simphony
我需要使用 Repast SimphonyJava 开发 Iterated Prisoner Dilemma 版本=64=]作为模拟器。
想法是每个 Player 都是一个 代理 ，我们有一个 Player 的 n x n 网格，不能感动。每个 Player 必须与 4 个邻居（北部、南部、西部和东部）进行比赛，根据每轮 4 场不同比赛的结果找到最佳策略。
由于 Repast Simphony 中没有用于在代理之间交换消息的内置系统，我不得不实施某种解决方法来处理代理的同步（A vs B和 B vs A 应该算作同一回合，这就是为什么他们需要同步）。
这是通过将每一轮视为：
Player i 为4个敌人中的每一个选择下一步行动Player i 向 4 个敌人中的每一个发送正确的移动Player i 等待4个敌人回复
根据我对 Repast Simphony 的理解，计划的方法是顺序的（没有代理级并行性），这意味着我被迫以不同于发送一个（以较低的优先级安排，以确保在开始等待之前完成所有发送）。
这里的问题是，尽管收到了所有 4 条预期消息（至少这是打印的内容），但一旦等待方法启动，它就会报告收到的元素少于 4 个。
这里的代码取自 Player class:
// myPoint is the location inside the grid (unique, agents can't move and only one per cell is allowed)
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + ((myPoint == null) ? 0 : myPoint.hashCode());
return result;
}
// Returns enemy's choice in the previous round
private byte getLastPlay(Player enemy) {
return (neighbors.get(enemy)[1]) ? COOPERATE : DEFECT;
}
// Elements are saved as (player, choice)
private void receivePlay(Player enemy, byte play) {
System.out.println(this + " receives (" + play + ") from " + enemy);
while (!playSharedQueue.add(new Object[] { enemy, play })){
// This doesn't get printed, meaning that the insertion is successful!
System.out.println(this + " failed inserting");
}
}
@ScheduledMethod(start = 1, interval = 1, priority = 10)
public void play() {
System.out.println(this + " started playing");
// Clear previous plays
playSharedQueue.clear();
for (Player enemy : neighbors.keySet()) {
// properties[0] = true if we already played together
// properties[1] = true if enemy choose to cooperate on the previous round
Boolean[] properties = neighbors.get(enemy);
// Choose which side we take this time
byte myPlay;
if (properties[0]) {
// First time that we play, use memory-less strategy
myPlay = (Math.random() <= strategy[0]) ? COOPERATE : DEFECT;
// Report that we played
properties[0] = false;
neighbors.put(enemy, properties);
} else {
// We already had a round, use strategy with memory
byte enemyLastPlay = enemy.getLastPlay(this);
// Choose which side to take based on enemy's previous decision
myPlay = (Math.random() <= strategy[(enemyLastPlay) == COOPERATE ? 1 : 2]) ? COOPERATE : DEFECT;
}
// Send my choice to the enemy
System.out.println(this + " sent (" + myPlay + ") to " + enemy);
enemy.receivePlay(this, myPlay);
}
}
// Waits for the results and processes them
@ScheduledMethod(start = 1, interval = 1, priority = 5)
public void waitResults() {
// Clear previous score
lastPayoff = 0;
System.out.println(this + " waits for results [" + playSharedQueue.size() + "]");
if (playSharedQueue.size() != 4) {
// Well, this happens on the first agent :(
System.exit(1);
}
// ... process ...
}
这是控制台输出，因此您可以看到所有内容似乎都已发送和接收，没有任何问题（使用 3 x 3 网格）：
Player[2, 0] started playing
Player[2, 0] sent (0) to Player[2, 1]
Player[2, 1] receives (0) from Player[2, 0]
Player[2, 0] sent (0) to Player[2, 2]
Player[2, 2] receives (0) from Player[2, 0]
Player[2, 0] sent (0) to Player[0, 0]
Player[0, 0] receives (0) from Player[2, 0]
Player[2, 0] sent (0) to Player[1, 0]
Player[1, 0] receives (0) from Player[2, 0]
Player[1, 2] started playing
Player[1, 2] sent (1) to Player[2, 2]
Player[2, 2] receives (1) from Player[1, 2]
Player[1, 2] sent (1) to Player[0, 2]
Player[0, 2] receives (1) from Player[1, 2]
Player[1, 2] sent (1) to Player[1, 0]
Player[1, 0] receives (1) from Player[1, 2]
Player[1, 2] sent (1) to Player[1, 1]
Player[1, 1] receives (1) from Player[1, 2]
Player[0, 2] started playing
Player[0, 2] sent (1) to Player[2, 2]
Player[2, 2] receives (1) from Player[0, 2]
Player[0, 2] sent (1) to Player[0, 0]
Player[0, 0] receives (1) from Player[0, 2]
Player[0, 2] sent (1) to Player[0, 1]
Player[0, 1] receives (1) from Player[0, 2]
Player[0, 2] sent (1) to Player[1, 2]
Player[1, 2] receives (1) from Player[0, 2]
Player[0, 1] started playing
Player[0, 1] sent (1) to Player[2, 1]
Player[2, 1] receives (1) from Player[0, 1]
Player[0, 1] sent (1) to Player[0, 0]
Player[0, 0] receives (1) from Player[0, 1]
Player[0, 1] sent (1) to Player[0, 2]
Player[0, 2] receives (1) from Player[0, 1]
Player[0, 1] sent (1) to Player[1, 1]
Player[1, 1] receives (1) from Player[0, 1]
Player[1, 0] started playing
Player[1, 0] sent (0) to Player[2, 0]
Player[2, 0] receives (0) from Player[1, 0]
Player[1, 0] sent (0) to Player[0, 0]
Player[0, 0] receives (0) from Player[1, 0]
Player[1, 0] sent (0) to Player[1, 1]
Player[1, 1] receives (0) from Player[1, 0]
Player[1, 0] sent (0) to Player[1, 2]
Player[1, 2] receives (0) from Player[1, 0]
Player[1, 1] started playing
Player[1, 1] sent (0) to Player[2, 1]
Player[2, 1] receives (0) from Player[1, 1]
Player[1, 1] sent (0) to Player[0, 1]
Player[0, 1] receives (0) from Player[1, 1]
Player[1, 1] sent (0) to Player[1, 0]
Player[1, 0] receives (0) from Player[1, 1]
Player[1, 1] sent (0) to Player[1, 2]
Player[1, 2] receives (0) from Player[1, 1]
Player[2, 2] started playing
Player[2, 2] sent (0) to Player[2, 0]
Player[2, 0] receives (0) from Player[2, 2]
Player[2, 2] sent (0) to Player[2, 1]
Player[2, 1] receives (0) from Player[2, 2]
Player[2, 2] sent (0) to Player[0, 2]
Player[0, 2] receives (0) from Player[2, 2]
Player[2, 2] sent (0) to Player[1, 2]
Player[1, 2] receives (0) from Player[2, 2]
Player[0, 0] started playing
Player[0, 0] sent (1) to Player[2, 0]
Player[2, 0] receives (1) from Player[0, 0]
Player[0, 0] sent (1) to Player[0, 1]
Player[0, 1] receives (1) from Player[0, 0]
Player[0, 0] sent (1) to Player[0, 2]
Player[0, 2] receives (1) from Player[0, 0]
Player[0, 0] sent (1) to Player[1, 0]
Player[1, 0] receives (1) from Player[0, 0]
Player[2, 1] started playing
Player[2, 1] sent (1) to Player[2, 0]
Player[2, 0] receives (1) from Player[2, 1]
Player[2, 1] sent (1) to Player[2, 2]
Player[2, 2] receives (1) from Player[2, 1]
Player[2, 1] sent (1) to Player[0, 1]
Player[0, 1] receives (1) from Player[2, 1]
Player[2, 1] sent (1) to Player[1, 1]
Player[1, 1] receives (1) from Player[2, 1]
Player[2, 2] waits for results [1]
正如你在最后一行看到的，playSharedQueue.size() 是 1，我真的不明白为什么。
如果方法调用是 waitResults()methos is invoked after the 9play()` 执行的顺序，并且假设每个正确发送 4 条消息，我找不到该大小仍然为 1 的原因。
当然，一切都是顺序的意味着没有 synchronization 问题，即使我在使用 LinkedBlockingQueue 而不是 HashSet 时遇到了同样的问题。
你们对此有什么提示吗？
一段时间后我再次打开代码，发现我犯了一个简单但严重的错误：
@ScheduledMethod(start = 1, interval = 1, priority = 10)
public void play() {
System.out.println(this + " started playing");
// Clear previous plays
playSharedQueue.clear();
执行 playSharedQueue.clear(); 以清除之前的结果，但由于调用是连续的，第二个玩家将在第一个玩家向他发送他的游戏后调用它，因此游戏被丢弃。
移动 waitResults 末尾的那一行解决了它。
我需要使用 Repast SimphonyJava 开发 Iterated Prisoner Dilemma 版本=64=]作为模拟器。
想法是每个 Player 都是一个 代理 ，我们有一个 Player 的 n x n 网格，不能感动。每个 Player 必须与 4 个邻居（北部、南部、西部和东部）进行比赛，根据每轮 4 场不同比赛的结果找到最佳策略。
由于 Repast Simphony 中没有用于在代理之间交换消息的内置系统，我不得不实施某种解决方法来处理代理的同步（A vs B和 B vs A 应该算作同一回合，这就是为什么他们需要同步）。
这是通过将每一轮视为：
Playeri 为4个敌人中的每一个选择下一步行动Playeri 向 4 个敌人中的每一个发送正确的移动Playeri 等待4个敌人回复
根据我对 Repast Simphony 的理解，计划的方法是顺序的（没有代理级并行性），这意味着我被迫以不同于发送一个（以较低的优先级安排，以确保在开始等待之前完成所有发送）。
这里的问题是，尽管收到了所有 4 条预期消息（至少这是打印的内容），但一旦等待方法启动，它就会报告收到的元素少于 4 个。
这里的代码取自 Player class:
// myPoint is the location inside the grid (unique, agents can't move and only one per cell is allowed)
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + ((myPoint == null) ? 0 : myPoint.hashCode());
return result;
}
// Returns enemy's choice in the previous round
private byte getLastPlay(Player enemy) {
return (neighbors.get(enemy)[1]) ? COOPERATE : DEFECT;
}
// Elements are saved as (player, choice)
private void receivePlay(Player enemy, byte play) {
System.out.println(this + " receives (" + play + ") from " + enemy);
while (!playSharedQueue.add(new Object[] { enemy, play })){
// This doesn't get printed, meaning that the insertion is successful!
System.out.println(this + " failed inserting");
}
}
@ScheduledMethod(start = 1, interval = 1, priority = 10)
public void play() {
System.out.println(this + " started playing");
// Clear previous plays
playSharedQueue.clear();
for (Player enemy : neighbors.keySet()) {
// properties[0] = true if we already played together
// properties[1] = true if enemy choose to cooperate on the previous round
Boolean[] properties = neighbors.get(enemy);
// Choose which side we take this time
byte myPlay;
if (properties[0]) {
// First time that we play, use memory-less strategy
myPlay = (Math.random() <= strategy[0]) ? COOPERATE : DEFECT;
// Report that we played
properties[0] = false;
neighbors.put(enemy, properties);
} else {
// We already had a round, use strategy with memory
byte enemyLastPlay = enemy.getLastPlay(this);
// Choose which side to take based on enemy's previous decision
myPlay = (Math.random() <= strategy[(enemyLastPlay) == COOPERATE ? 1 : 2]) ? COOPERATE : DEFECT;
}
// Send my choice to the enemy
System.out.println(this + " sent (" + myPlay + ") to " + enemy);
enemy.receivePlay(this, myPlay);
}
}
// Waits for the results and processes them
@ScheduledMethod(start = 1, interval = 1, priority = 5)
public void waitResults() {
// Clear previous score
lastPayoff = 0;
System.out.println(this + " waits for results [" + playSharedQueue.size() + "]");
if (playSharedQueue.size() != 4) {
// Well, this happens on the first agent :(
System.exit(1);
}
// ... process ...
}
这是控制台输出，因此您可以看到所有内容似乎都已发送和接收，没有任何问题（使用 3 x 3 网格）：
Player[2, 0] started playing
Player[2, 0] sent (0) to Player[2, 1]
Player[2, 1] receives (0) from Player[2, 0]
Player[2, 0] sent (0) to Player[2, 2]
Player[2, 2] receives (0) from Player[2, 0]
Player[2, 0] sent (0) to Player[0, 0]
Player[0, 0] receives (0) from Player[2, 0]
Player[2, 0] sent (0) to Player[1, 0]
Player[1, 0] receives (0) from Player[2, 0]
Player[1, 2] started playing
Player[1, 2] sent (1) to Player[2, 2]
Player[2, 2] receives (1) from Player[1, 2]
Player[1, 2] sent (1) to Player[0, 2]
Player[0, 2] receives (1) from Player[1, 2]
Player[1, 2] sent (1) to Player[1, 0]
Player[1, 0] receives (1) from Player[1, 2]
Player[1, 2] sent (1) to Player[1, 1]
Player[1, 1] receives (1) from Player[1, 2]
Player[0, 2] started playing
Player[0, 2] sent (1) to Player[2, 2]
Player[2, 2] receives (1) from Player[0, 2]
Player[0, 2] sent (1) to Player[0, 0]
Player[0, 0] receives (1) from Player[0, 2]
Player[0, 2] sent (1) to Player[0, 1]
Player[0, 1] receives (1) from Player[0, 2]
Player[0, 2] sent (1) to Player[1, 2]
Player[1, 2] receives (1) from Player[0, 2]
Player[0, 1] started playing
Player[0, 1] sent (1) to Player[2, 1]
Player[2, 1] receives (1) from Player[0, 1]
Player[0, 1] sent (1) to Player[0, 0]
Player[0, 0] receives (1) from Player[0, 1]
Player[0, 1] sent (1) to Player[0, 2]
Player[0, 2] receives (1) from Player[0, 1]
Player[0, 1] sent (1) to Player[1, 1]
Player[1, 1] receives (1) from Player[0, 1]
Player[1, 0] started playing
Player[1, 0] sent (0) to Player[2, 0]
Player[2, 0] receives (0) from Player[1, 0]
Player[1, 0] sent (0) to Player[0, 0]
Player[0, 0] receives (0) from Player[1, 0]
Player[1, 0] sent (0) to Player[1, 1]
Player[1, 1] receives (0) from Player[1, 0]
Player[1, 0] sent (0) to Player[1, 2]
Player[1, 2] receives (0) from Player[1, 0]
Player[1, 1] started playing
Player[1, 1] sent (0) to Player[2, 1]
Player[2, 1] receives (0) from Player[1, 1]
Player[1, 1] sent (0) to Player[0, 1]
Player[0, 1] receives (0) from Player[1, 1]
Player[1, 1] sent (0) to Player[1, 0]
Player[1, 0] receives (0) from Player[1, 1]
Player[1, 1] sent (0) to Player[1, 2]
Player[1, 2] receives (0) from Player[1, 1]
Player[2, 2] started playing
Player[2, 2] sent (0) to Player[2, 0]
Player[2, 0] receives (0) from Player[2, 2]
Player[2, 2] sent (0) to Player[2, 1]
Player[2, 1] receives (0) from Player[2, 2]
Player[2, 2] sent (0) to Player[0, 2]
Player[0, 2] receives (0) from Player[2, 2]
Player[2, 2] sent (0) to Player[1, 2]
Player[1, 2] receives (0) from Player[2, 2]
Player[0, 0] started playing
Player[0, 0] sent (1) to Player[2, 0]
Player[2, 0] receives (1) from Player[0, 0]
Player[0, 0] sent (1) to Player[0, 1]
Player[0, 1] receives (1) from Player[0, 0]
Player[0, 0] sent (1) to Player[0, 2]
Player[0, 2] receives (1) from Player[0, 0]
Player[0, 0] sent (1) to Player[1, 0]
Player[1, 0] receives (1) from Player[0, 0]
Player[2, 1] started playing
Player[2, 1] sent (1) to Player[2, 0]
Player[2, 0] receives (1) from Player[2, 1]
Player[2, 1] sent (1) to Player[2, 2]
Player[2, 2] receives (1) from Player[2, 1]
Player[2, 1] sent (1) to Player[0, 1]
Player[0, 1] receives (1) from Player[2, 1]
Player[2, 1] sent (1) to Player[1, 1]
Player[1, 1] receives (1) from Player[2, 1]
Player[2, 2] waits for results [1]
正如你在最后一行看到的，playSharedQueue.size() 是 1，我真的不明白为什么。
如果方法调用是 waitResults()methos is invoked after the 9play()` 执行的顺序，并且假设每个正确发送 4 条消息，我找不到该大小仍然为 1 的原因。
当然，一切都是顺序的意味着没有 synchronization 问题，即使我在使用 LinkedBlockingQueue 而不是 HashSet 时遇到了同样的问题。
你们对此有什么提示吗？
一段时间后我再次打开代码，发现我犯了一个简单但严重的错误：
@ScheduledMethod(start = 1, interval = 1, priority = 10)
public void play() {
System.out.println(this + " started playing");
// Clear previous plays
playSharedQueue.clear();
执行 playSharedQueue.clear(); 以清除之前的结果，但由于调用是连续的，第二个玩家将在第一个玩家向他发送他的游戏后调用它，因此游戏被丢弃。
移动 waitResults 末尾的那一行解决了它。