是否有一个算法来解决一个5位数的引脚，它不会重复出现在字符串中？ [英] Is there an algorithm for solving a 5 digit pin that doesn't repeat making it in a string ?

问题描述

您好......



我对字符串有疑问...



我来了从一个逃生室，他们有一个5键的密码，不能重复所以是5！ = 120种可能性所以我问自己是否可以找到一个字符串，在不到120x5 = 600次按下按钮时有所有可能性。



例如：3针组合= 3！ = 6次尝试

123 132 213 231 312 321 ...但是我们可以做到123121321所以9次按下不是6x3 = 18

它有一个算法，或者我们可以为更多不重复的字符做一个程序吗？



谢谢你，抱歉我的英文不好！



Vlad Pineta

[已删除] @ gmail.com



我尝试过：



我试图在互联网上找到一个算法，但我没有找到任何东西

Hello...

I have a question about strings...

I came from an escape room and they had a 5 button password that can't repeat so were 5! = 120 possibilities so i've asked myself if i can find a string that have all the possibilities in less than 120x5=600 button presses.

Example: 3 pin combination = 3! = 6 tries
123 132 213 231 312 321 ...but we can do it 123121321 so 9 presses not 6x3=18
It's there an algorith for this or can we do a program that does this for more characters that don't repeat ?

Thank you and sorry for my bad english !

Vlad Pineta
[DELETED]@gmail.com

What I have tried:

I tried to find an algorithm on internet but i didn't found anything

推荐答案

停止尝试在互联网上找到算法并思考问题。

首先生成完整序列字符串：

Stop trying to "find an algorithm on internet" and think about the problem.
Start by generating the "full sequence" string:

123124125132134...

每次添加新序列时，请查看它是否包含在现有字符串中。如果是的话，不要添加它。



BTW：永远不要在任何论坛发布你的电子邮件地址，除非你真的喜欢垃圾邮件！如果有人回复你，你会收到一封电子邮件通知你。

And each time you go to add a new sequence see if it's "contained" in the existing string. If it is, don't add it.

BTW: Never post your email address in any forum, unless you really like spam! If anyone replies to you, you will receive an email to let you know.

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