嗨...我可以获得有关3重量法的更多信息 [英] Hi... Can I get more info on the the 3 weight method

问题描述

此信息的参考或来源确实会有所帮助。



3重量法

3重量方法也是被称为银行业使用的电子资金转帐路由号码方案。它是一个基于mod 10的系统，但它不是作为一个整数操作，而是在点积（即数字n的各个数字）上运行。



所以，给定一个8位数的路由号码，校验位c = n·（7,3,9,7,3,9,7,3），mod 10.换句话说，让ai为第i位n（最重要到最不重要）。然后



c = 7a1 + 3a2 + 9a3 + 7a4 + 3a5 + 9a6 + 7a7 + 3a8 mod 10


$ b $然后将bc连接到8位数路由，形成一个9位数的路由号码



该方案基于乘法模10产生10位小数的排列这一事实如果乘法因子是集合{1,3,7,9}的数字，则为数字;但如果因子为5或偶数，则只有十进制数字的子集。此系统无法检测相差5的相邻换位数。



3-Weight方法不使用简单的模块缩减，因此不需要c = 10-c保持平等。



读者应该发现以下检查方程式成立（其中a9是-c）：



0 = 7a1 + 3a2 + 9a3 + 7a4 + 3a5 + 9a6 + 7a7 + 3a8 + a9 mod 10



3重量计划如下所示。正如读者所看到的，它可以很容易地推广到任意长度的数字。但是，如果读者需要一个系统用于非常长的数据集，或者二进制数据集考虑使用CRC或Alder校验和，或基于散列的校验和。



我尝试了什么：



是



我试过但我还没有除了本网站之外，我们已成功获取此信息的来源。之前是否研究过3重量法？请指出我正确的方向，以便我可以阅读更多相关信息。

A reference or a source of this information would really help.

3-Weight Method
The 3-Weight Methods is also known as the Electronic Funds Transfer Routing Number Scheme used by the banking industry. It is a mod 10 based system, but rather than operating on n as a whole number, it operates on a dot product (i.e., the individual digits of the number n).

So, given an 8 digit routing number, the check digit c = n · (7, 3, 9, 7, 3, 9, 7, 3), mod 10. Stated another way, let ai be the ith digit of n (most significant to least significant). Then

c = 7a1 + 3a2 + 9a3 + 7a4 + 3a5 + 9a6 + 7a7 + 3a8 mod 10

c is then concatenated to the 8 digit route, forming a 9 digit routing number

This scheme is based on the fact that multiplication modulo 10 yields a permutation of the 10 decimal digits if the multiplication factor is a digit of the set { 1, 3, 7, 9 }; but only a subset of the decimal digits if the factor is 5 or an even digit. This system cannot detect adjacent transpositions of digits that differ by 5.

The 3-Weight method does not use simple modular reductions, so c = 10 - c is not needed to hold the equality.

The reader should find the following check equation holds true (where a9 is -c):

0 = 7a1 + 3a2 + 9a3 + 7a4 + 3a5 + 9a6 + 7a7 + 3a8 + a9 mod 10

The 3-Weight Scheme is presented below. As the reader can see, it generalizes to a arbitrary length of digits easily. However, if the reader desires a system for very long datasets, or binary data sets consider using a CRC or Alder checksums, or a hash-based checksum.

What I have tried:

yes

I have tried but I have not been successful in obtaining the source of this information apart from this website. Has the 3 - weight method been studied before? Kindly point me in the right direction so that I can read more about it.

推荐答案

公式相对容易计算校验位。



路由号码是一个8位数字，数字9是校验位。在我的例子中，我将使用变量d1-d9来表示各个数字



计算校验位：

The formula is relatively easy to calculate the check digit.

The routing number is an 8 digit number, and digit number 9 is the check digit. In my example I will be using the variables d1-d9 to represent the individual digits

To calculate the check digit:

int d9 = 3(d1+d4+d7) + 7(d2+d5+d8) + 1(d3+d6+d9) MOD 10

验证号码：

To validate the number:

bool IsValid = (3(d1+d4+d7) + 7(d2+d5+d8) + 1(d3+d6+d9) MOD 10  = 0)

更多信息： ABA路由转接号码 - 维基百科 [ ^ ]

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