java - 猜测游戏逻辑

我正在尝试创建一个游戏，其中AI会通过一系列问题来猜测用户正在思考的数字。

例：

Is your number greater than 50? (y/n)
n
Is your number greater than 25? (y/n)
y
Is your number greater than 38? (y/n)
y
Is your number greater than 44? (y/n)
n
Is your number greater than 41? (y/n)
y
Is your number greater than 43? (y/n)
y
The number you're thinking of is 44.

我了解您必须根据用户输入的内容设置新的upperLimit和lowerLimit，然后返回新限制的平均值。我的代码运行不正常。

如果我想到数字59，我的代码会做什么。

Is your number greater than 50? (y/n)
y
Is your number greater than 75? (y/n)
n
Is your number greater than 62? (y/n)
n
Is your number greater than 56? (y/n)
y
Is your number greater than 59? (y/n)
n
Is your number greater than 57? (y/n)
y
Is your number greater than 58? (y/n)
y
Is your number greater than 58? (y/n)
y
Is your number greater than 58? (y/n)
y
Is your number greater than 58? (y/n)
y
Is your number greater than 58? (y/n)

任何建议/技巧/帮助将不胜感激！

我的代码：

public void play(int lowerLimit, int upperLimit) {
    instructions(lowerLimit, upperLimit);

    while (lowerLimit != upperLimit) {
        if(isGreaterThan(average(lowerLimit, upperLimit))) {
            lowerLimit = average(lowerLimit, upperLimit);
        } else {
            upperLimit = average(lowerLimit, upperLimit);
        }
    }
    System.out.println("The number you're thinking of is " + lowerLimit);
}

public boolean isGreaterThan(int value) {
    System.out.println("Is your number greater than " + value + "? (y/n)");
    String answer = reader.nextLine();
    return answer.equals("y");
}

public int average(int firstNumber, int secondNumber) {
    return (firstNumber + secondNumber) / 2;
}

最佳答案

试试这个lowerLimit = average（lowerLimit，upperLimit）+ 1;

解释很简单。您一直在问的问题是数字是否大于X。如果是，则X + 1应该成为下限，因为显然数字不能为X。另一端不能应用相同的数字。间隔，因为如果您的数字是X，则问题的答案是否更大是否，因此upperLimit应该保持X。