Java implements guessing the number AB

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Preface

前几天刚好看了《登录圆鱼洲》，对里面的一个游戏环节"猜数字AB"很感兴趣，所以想用Java来实现这个小游戏。

The rule is: select four non-repeating numbers from 0 to 9, and then let the participants guess. If a certain number is guessed correctly and the corresponding position is also correct, it is expressed as 1A. If a certain number is guessed correctly but the corresponding position is correct, it will be expressed as 1A. If the position is wrong, it is expressed as 1B. For example, the number written down is 1234, and you guess 1245, it is expressed as 2A1B; participants judge whether the guessed number is correct by how many A and how many B. When the values ​​and positions of the four numbers are correct and The challenge is successful if you guess no more than 10 times!

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Code

代码如下：(思路简单，代码有点粗糙，如果有更简单的，请大佬贴在讨论区一起学习)

import java.util.Scanner;
import java.util.List;
import java.util.Random;
import java.util.HashSet;
import java.util.ArrayList;
import java.util.Arrays;
public class GuessNumberAB {
    
    

    public static void main(String[] args) {
    
    

        Scanner scanner = new Scanner(System.in);
        Scanner scanner2 = new Scanner(System.in);
        Random random = new Random();
        String[] num = {
    
    "0", "1", "2", "3", "4", "5", "6", "7", "8", "9"};
        HashSet<String> set = new HashSet<>();
        while(set.size() != 4){
    
    
            set.add(num[random.nextInt(10)]);
        }
        ArrayList<String> target = new ArrayList<>(set);
        while(true){
    
    
            System.out.print("是否进行游戏(y:是 , n否)");
            String isGame = scanner.next();
            if ("n".equals(isGame)){
    
    
                System.out.println("游戏结束!");
                break;
            }else if ("y".equals(isGame)){
    
    
                String guessNum = null;
                String[] arr=null;
                    int times = 0;
                    while (times++ < 10){
    
    
                        System.out.print("输入你的答案不重复的四位数：");
                        guessNum = scanner2.next();
                        arr = guessNum.trim().split("");
                        if ((arr.length == 4)&& (isRepeat(arr))) {
    
    
                            String tar = isReal(arr, target);
                            System.out.printf("第%d轮，结果为%s\n",times,tar);
                            if("4A0B".equals(tar)){
    
    
                                System.out.println("您猜对了！恭喜");
                                return;
                            }
                        }else System.out.println("必须输入4个不重复的数,再次输入");
                    }
                    if (times>10){
    
    
                        System.out.println("游戏失败! target="+set);
                        return;
                    }
            }else{
    
    
                System.out.println("输入错误，请重新输入");
            }
        }

    }

    public static boolean isRepeat(String[] arr){
    
      //判断每次输入答案是否有重复
        HashSet<String> set = new HashSet<>(Arrays.asList(arr));
        return set.size() == arr.length;
    }

    private static String isReal(String[] arr,List<String> target) {
    
    
        String result = "%dA%dB";
        int A = 0;  //判断A值
        for (int i=0;i<arr.length;i++){
    
    
            if (target.get(i).equals(arr[i]))
                A++;
        }
        int B = 0; 
        for (String a:arr){
    
    
            for (String b: target){
    
    
                if (a.equals(b)){
    
    
                    B++;
                    break;
                }
            }
        } //这里B表示所有数值相同的个数，减去位置和数值都相同的A,就是位置不同数值相同的个数
        return String.format(result, A,(B-A));
    }
}

Personal problem-solving ideas for the game

When I first played this game, I had no idea at all. I could only guess one by one, purely relying on luck. After playing it a few times, I gradually developed a method that was slightly better than blind guessing. Let me record it:
I divided the numbers into Three copies of (1,2,3,4), (5,6,7,8) and (0,9) guess (1,2,3,4) and (5,6,7,8 respectively in the first two rounds) ) In this way, the approximate distribution of the target numbers can be obtained; then the uncertain numbers are determined by the determined target number/non-target number. When four numbers are obtained, the order is determined by the previously determined positions. Let’s use this problem-solving idea by executing code.

step1: 1234结果为0A1B,说明1234只有1个是目标数

step2: 5678结果为0A2B，说明5678里面有两个目标数，所以(0,9) 里面必须有一个目标数，那么第三轮可以用(0,9) 与只用一个目标数的(1,2,3,4)中的两个数进行猜测；随机选(0,9,1,2)

step3: (0,9,1,2) 结果是2B，说明(1,2)有一个目标数，而(1,2,3,4)只有一个目标数，故而(3,4)就是非目标数，那么第四轮就可以用(3,4)非目标数与(5,6,7,8)其中两个进行猜测，随机选择(3,4,5,6)

step4: (3,4,5,6)结果是没有一个目标数，所以(5,6,7,8)中的两个目标数为(7,8), (5,6)是非目标数；现在的非目标数有(3,4,5,6), 目标数有(7,8), (1,2)与(0,9)各一个。 选用(3,1,4,5)判断出(1,2)中的目标数

step5: 如图：1不是目标数，所以(1,2)中的目标数为2,现在确定的目标数有(2,7,8),非目标数有(1,3,4,5,6)；再用(1,3,9,4)确定(0,9)中的目标数

step6: 如图：说明9是目标数，0非目标数，四个目标数分别为(2,7,8,9)；接下来就是判断顺序(用非目标数猜测时应该顺序尽量与原来调换)

判断顺序：

首先四个数分成四个位置abcd，(2,7,8,9)中，由第一轮中2不能在b，由第三轮2不能在d; 由第二轮7不能在c;由第三轮9不能在b, 第六轮9不能在c; 由第二轮8不能在d; 所以如图：

因为bd不能为2，ac有一个为2，所以假设a=2，那么d就只能是9，因为bc都不能为9，又因为c不能为7，在剩下的bc中只能b=7,所以最后c=8; 那么可以猜测2789, 将2789再与前几轮作对比看看是否有错。
结果：
那么如果先选择了c=2，并不能得到唯一的结果，所以不可取。

我们也可以选择假设a=9(因为只能ad其中一个为9)，那么d只能=7，对于2，在剩下的bc中b不能为2，所以b=8,c=2; 此时猜测结果为 9827；9827由正确结果可以在第七轮显示为0A4B，所以推断a不能为9，故d=9,在此情况下有(2,7,8,9)和(7,8,2,9)和 (8,7,2,9)三种情况，在最坏运气时可以刚好第10轮得到结果，挑战成功！