Sroan 有一个私人的保险箱,密码是 7 个 不同的数字。 Guess #1: 9062437 G...

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问题

Sroan 有一个私人的保险箱,密码是 7 个 不同的数字。 Guess #1: 9062437 Guess #2: 8593624 Guess #3: 4286915 Guess #4: 3450982 Sroan 说: 你们 4 个人每人都猜对了位置不相邻的两个数字。 (只有 “位置及其对应的数字” 都对才算对) 问:密码是什么?

思考

Translating the problem

Okay, I need to clarify the problem: Sroan has a private safe with a 7-digit password, and there are four guesses. Each guess has exactly two digits in the correct positions, but those digits cannot be in adjacent positions.

The challenge now is figuring out which two digits from each guess are correct and figuring out the password from all those clues.

Hmm, it looks like I'll need to analyze the positions of the digits across all guesses carefully to figure this out. Let’s see...

Solving digit puzzle

This is a tricky one! We’ve got four guesses and need to figure out a 7-digit password where two digits per guess are correct and in the right position—but they’re not adjacent to each other.

The digits in the password are distinct, and we know that across the four guesses, each guess has exactly two correct digits in the right spots. I’ll need to go through each guess, compare positions, and identify which two digits match, keeping in mind that they are not next to each other. Let’s break this down carefully!

Analyzing the password guess

Alright, we've got a systematic approach here where we need to figure out the positions of correct digits across the guesses, making sure no two correct digits are adjacent.

To find this, I'll carefully analyze each guess and see which positions could be correct. For instance, looking at frequency: for position 1, the possible digits are 9, 8, 4, and 3. I'll need to check for conflicts and consistency across all guesses while adhering to the constraints. Let’s narrow things down step-by-step to match the criteria!

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Determining password positions

I’m trying to figure out a consistent assignment for the password’s digits by looking at the frequencies of digits in each position across the guesses. It seems that position 5 is forced to be 9, as it appears twice in guesses 3 and 4.

Now, we need to distribute the remaining digits through the positions. Each position must have one correct digit, and each guess contributes two correct digits. I'll go through each guess, carefully eliminating impossible combinations based on adjacency rules, and assign the digits that satisfy all constraints. This will require balancing all the positions while conforming to all the given rules.

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Adjusting position assignments

I’m trying different combinations to tackle the assignment problem. For Case B, if f4 = 1 and f6 = 2, I start by trying a multiset {1,2,3,4} for the remaining positions, but I run into duplicates. For example, pos4 and pos6 can't both be 2. I then try switching things up—maybe Case C, where f4 = 2, f6 = 1, with similar frequency and constraints. The goal is to avoid digit duplicates while assigning positions that follow allowable sets and non-adjacency rules.

Testing different assignments

I’m adjusting assignments to ensure non-repeating digits while sticking to constraints. For instance, I tested positions like pos1 = 3, pos2 = 1, pos3 = 4, pos7 = 1, ensuring all digits are distinct. Luckily, no duplicates emerged across p1-p7! I then carefully checked non-adjacency for each guess, particularly ensuring that if guesses share a digit, their positions are far apart. Some matches worked, while others didn't. The process feels like fitting pieces in a puzzle, and it’s all about finding combinations that don't break any rules!

Reasoned for 3m 0s

回答

我们下面证明唯一符合条件的密码是 4053927

下面给出一种解题思路(证明过程较复杂,这里只介绍主要思路与关键结论):

  1. 令密码的 7 个数字分别为 p_1 p_2 p_3 p_4 p_5 p_6 p_7。四个猜测分别为
  • 猜1:9  0  6  2  4  3  7
  • 猜2:8  5  9  3  6  2  4
  • 猜3:4  2  8  6  9  1  5
  • 猜4:3  4  5  0  9  8  2

每个猜测中,只有“数字与其位置”同时正确才算命中,而且每个猜测命中了两个位置,且这两个位置绝不相邻。

  1. 注意到各位置各猜测给出的数字均不同,除位置5外只有猜3和猜4在该位给出同一数字 9。若 p_5 ≠ 9,则所有 7 个位置最多各为 1 个命中,共 7 个;但四人共应命中 8 个(4×2),故 p_5 必须取 9,此时猜3和猜4在位置5均正确(计 2 次命中)。

  2. 于是除 p_5 外其余 6 位各恰有 1 次命中。进一步分析每一位上各猜测给出的数字,可以推出:

    • p_1只能取 8、4、或 3(分别对应猜2、猜3、猜4,但不能取 9,因为 9已用在 p_5);
    • 同理 p_2、p_3、p_4、p_6、p_7 分别只能从各猜的数字中选出,且不同猜测的命中分布必须满足:猜1、猜2各命中 2 个位置,而猜3、猜4各除 p_5 外只命中 1 个且不能与 p_5 邻接(即不能在 p_4 或 p_6)。
  3. 经仔细排列、排除(例如必须防止因选取相同数字而重复,同时还要保证同一猜测命中的两个位置不连续),最后得到唯一满足条件的分配是:

    • p_1 = 4(取自猜3的数字 4),
    • p_2 = 0(取自猜1的数字 0),
    • p_3 = 5(取自猜4的数字 5),
    • p_4 = 3(取自猜2的数字 3),
    • p_5 = 9(强制),
    • p_6 = 2(取自猜2的数字 2),
    • p_7 = 7(取自猜1的数字 7)。

验证各猜测:

  • 猜1:仅在 p_2(0)和 p_7(7)命中(位置2与7不相邻);
  • 猜2:仅在 p_4(3)和 p_6(2)命中(位置4与6不相邻);
  • 猜3:在 p_1(4)和 p_5(9)命中(位置1与5不相邻);
  • 猜4:在 p_3(5)和 p_5(9)命中(位置3与5不相邻)。

因此密码确定为 4053927。


答案:4053927

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