Luhn算法逻辑

Question

我目前正在读Codecademy的全部堆栈工程师课程，到目前为止，我一直对此非常满意，发现新事物，自己解决问题，但这是我前进过程中的一个严重障碍，因为我似乎无法用这种逻辑来识别问题所在。我不想质疑卢恩的算法，但我真的需要澄清一下.

因此，我的问题是，算法将我的所有数组返回为有效，我的代码如下所示(codecademy提供的数组)：

// All valid credit card numbers
const valid1 = [4, 5, 3, 9, 6, 7, 7, 9, 0, 8, 0, 1, 6, 8, 0, 8];
const valid2 = [5, 5, 3, 5, 7, 6, 6, 7, 6, 8, 7, 5, 1, 4, 3, 9];
const valid3 = [3, 7, 1, 6, 1, 2, 0, 1, 9, 9, 8, 5, 2, 3, 6];
const valid4 = [6, 0, 1, 1, 1, 4, 4, 3, 4, 0, 6, 8, 2, 9, 0, 5];
const valid5 = [4, 5, 3, 9, 4, 0, 4, 9, 6, 7, 8, 6, 9, 6, 6, 6];

// All invalid credit card numbers
const invalid1 = [4, 5, 3, 2, 7, 7, 8, 7, 7, 1, 0, 9, 1, 7, 9, 5];
const invalid2 = [5, 7, 9, 5, 5, 9, 3, 3, 9, 2, 1, 3, 4, 6, 4, 3];
const invalid3 = [3, 7, 5, 7, 9, 6, 0, 8, 4, 4, 5, 9, 9, 1, 4];
const invalid4 = [6, 0, 1, 1, 1, 2, 7, 9, 6, 1, 7, 7, 7, 9, 3, 5];
const invalid5 = [5, 3, 8, 2, 0, 1, 9, 7, 7, 2, 8, 8, 3, 8, 5, 4];

// Can be either valid or invalid
const mystery1 = [3, 4, 4, 8, 0, 1, 9, 6, 8, 3, 0, 5, 4, 1, 4];
const mystery2 = [5, 4, 6, 6, 1, 0, 0, 8, 6, 1, 6, 2, 0, 2, 3, 9];
const mystery3 = [6, 0, 1, 1, 3, 7, 7, 0, 2, 0, 9, 6, 2, 6, 5, 6, 2, 0, 3];
const mystery4 = [4, 9, 2, 9, 8, 7, 7, 1, 6, 9, 2, 1, 7, 0, 9, 3];
const mystery5 = [4, 9, 1, 3, 5, 4, 0, 4, 6, 3, 0, 7, 2, 5, 2, 3];

// An array of all the arrays above
const batch = [valid1, valid2, valid3, valid4, valid5, invalid1, invalid2, invalid3, invalid4, invalid5, mystery1, mystery2, mystery3, mystery4, mystery5];

我的函数实现了算法：

const validateCred = arr => {

    let checkSum = 0;
    let ifEvenDouble = 0;
    arr.push(checkSum);

    //Iterate through array, double what is needed

    for(let i = arr.length - 2; i >= 0; i--){
      console.log(ifEvenDouble);

      //If ifEvenDouble is even, we are at the 'other' cell

        if((ifEvenDouble % 2) === 0){
          let doubled = arr[i] * 2;

          //If doubled digit is greater than 9, store sum of individual digits
          //Convert the doubled number to a string then extract each member and convert back to number for calculation, add to checkSum and skip to next iteration, otherwise, add arr[i]

          let newDigit = 0;
          if(doubled > 9){
            newDigit = Number(doubled.toString()[0]) + Number(doubled.toString()[1]);
            //Add doubled & split digit to total and continue the loop
            checkSum += newDigit;
            ifEvenDouble++;
            continue;
          }
          //Add doubled digit less than 9 to total and continue the loop
          checkSum += doubled;
          ifEvenDouble++;
          continue;
        }

        //Add current array member to total
        checkSum += arr[i];
        ifEvenDouble++;

    }//End for loop

    console.log(checkSum);
    const checkDigit = (checkSum * 9) % 10;
    const totalSum = checkDigit + checkSum;

    if(totalSum % 10 === 0){
      console.log('Valid');
      return true;
    } else {
      console.log('Invalid');
      return false;
    }
};

validateCred(invalid1); // -> Output: Valid

据我所知，我的totalSum是总是将是10的倍数，如果我从10减去我的单位数字，把它加到我的checkSum中，总是会给我10的倍数。我错了吗？

编辑:我已经尝试过调试它，但是我做的越多，就越远离核心算法。

编辑(2)：所以，感谢下面的人，我认为我的问题是生成我自己的支票数字，而不是使用已经提供的数字？我感到困惑的是，从维基百科的页面上可以看出：

“用于计算支票数字的示例:假设一个帐号"7992739871”的示例，其中将添加一个支票数字，使其为7992739871x‘

然后他们用x以外的数字进行所有的计算，我认为这是现在的主要混淆。

Answer 1

你的算法太复杂了。维基百科简洁地描述了它，只需实现以下三个步骤

1. 从最右边的数字(不包括选中的数字)向左移动，是每秒钟数字的两倍。检查数字既不加倍，也不包括在此计算中；第一个数字加倍是位于支票数字左侧的数字。如果加倍运算的结果大于9(例如，8×2= 16)，则添加结果的数字(例如，16: 1+6= 7，18: 1+8= 9)或从结果中相应地减去9(例如，16: 16−9= 7，18: 18−9= 9)。
2. 取所有数字之和(包括检查数字)。
3. 如果总模10等于0(如果总数以零结尾)，那么根据Luhn公式，这个数字是有效的；否则它是无效的。

我还认为你误解了支票数字是什么。您似乎将它作为0附加到数组中，并试图在最后计算它。数字里已经有了--这是最后一个数字。

​

const validateCred = arr => {

   let doubleIt = true;
   let sum = 0;
   // From the rightmost digit excluding check digit...
   for(let i = arr.length - 2; i >= 0; i--){
        
        if(doubleIt){
          let doubled = arr[i] * 2;
         
          if(doubled > 9){
            doubled -= 9
          }
          sum += doubled
        }
        else {
          sum += arr[i]
        }
        doubleIt = !doubleIt;

    }

    // Add the check digit to the sum
    sum += arr[arr.length-1];

    // If sum is divisible by 10 it is valid
    if(sum % 10 === 0){
      console.log('Valid');
      return true;
    } else {
      console.log('Invalid');
      return false;
    }
};

const invalid1 = [4, 5, 3, 2, 7, 7, 8, 7, 7, 1, 0, 9, 1, 7, 9, 5];
const valid1 = [4, 5, 3, 9, 6, 7, 7, 9, 0, 8, 0, 1, 6, 8, 0, 8];
validateCred(invalid1);
validateCred(valid1);

​

你出错的地方主要围绕着支票数字的使用。您似乎在计算它，而它只是数组中的最后一个元素。下面的代码片段更接近您的原始代码，只是不需要计算检查数字。

​

const validateCred = arr => {

    let ifEvenDouble = 0;
   let checkSum=0
    //Iterate through array, double what is needed

    for(let i = arr.length - 2; i >= 0; i--){

      //If ifEvenDouble is even, we are at the 'other' cell

        if((ifEvenDouble % 2) === 0){
          let doubled = arr[i] * 2;

          //If doubled digit is greater than 9, store sum of individual digits
          //Convert the doubled number to a string then extract each member and convert back to number for calculation, add to checkSum and skip to next iteration, otherwise, add arr[i]

          let newDigit = 0;
          if(doubled > 9){
            newDigit = Number(doubled.toString()[0]) + Number(doubled.toString()[1]);
            //Add doubled & split digit to total and continue the loop
            checkSum += newDigit;
            ifEvenDouble++;
            continue;
          }
          //Add doubled digit less than 9 to total and continue the loop
          checkSum += doubled;
          ifEvenDouble++;
          continue;
        }

        //Add current array member to total
        checkSum += arr[i];
        ifEvenDouble++;

    }//End for loop

    const checkDigit = arr[arr.length-1]
    const totalSum = checkDigit + checkSum;

    if(totalSum % 10 === 0){
      console.log('Valid');
      return true;
    } else {
      console.log('Invalid');
      return false;
    }
};


const invalid1 = [4, 5, 3, 2, 7, 7, 8, 7, 7, 1, 0, 9, 1, 7, 9, 5];
const valid1 = [4, 5, 3, 9, 6, 7, 7, 9, 0, 8, 0, 1, 6, 8, 0, 8];
validateCred(invalid1);
validateCred(valid1);

​