校验位算法 Luhn mod N 与简单和

发布于 2024-10-30 03:38:17 字数 878 浏览 1 评论 0原文

你知道为什么 Luhn mod N 算法为了创建校验位而执行求和吗将每个偶数放置的字符的值加倍，而不是对所有字符进行简单求和？

用伪代码的话来说：

给出：

var s = "some string i want to create check digit";

你知道为什么 Luhn mod N 基本上是这样做的吗：

for(i from s.length-1 to 0)
   if(i is even)
      checkdigit += chr2int(s[i]) * 2;
   else
      checkdigit += chr2int(s[i]);

它们

for(i from s.length-1 to 0)
   checkdigit += chr2int(s[i]);

仍然可以用 mod 操作终止，以使校验位适合一个字符

return int2chr( chr2int('a') + (checkdigit mod 25) );

< em>作为这个问题的旁注，可能对 Luhn 算法的图形表示感兴趣，这使得它更容易理解：

输入图片description here

其实这个是原始的Luhn算法，甚至不需要使用MOD函数。

原文

Do you know why the Luhn mod N algoritm in order to create the check digit performs a sum by doubling the value of each even placed char instead of perfoming a simple sum of all chars?

In pseudo-code words:

given:

var s = "some string i want to create check digit";

do you know why Luhn mod N does basically this:

for(i from s.length-1 to 0)
   if(i is even)
      checkdigit += chr2int(s[i]) * 2;
   else
      checkdigit += chr2int(s[i]);

instead of simply doing a sum

for(i from s.length-1 to 0)
   checkdigit += chr2int(s[i]);

they can still both terminate with a mod operation to make the checkdigit fit into one char

return int2chr( chr2int('a') + (checkdigit mod 25) );

As a side note to this question, to whom it may be interested in a graphic representation of the Luhn algorithm that makes it even more simple to understand:

enter image description here