1

// 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]

// Add your functions below:
let i = batch.length - 1
const validateCred = () => {
    for(i; i >= 0; i--) {
        for (var j = batch[i].length - 1; j >= 0; j--) {
           console.log(batch[i][j]);
        }
    }
}
validateCred();

I need help on a small project on codecademy to create a credit-card validator. The idea is to check the digits from right to left in the arrays. After that I need to multiply every other index on the credit card numbers by 2 if its greater than 9 after doubling subtract 9. All of the elements in the card should equal 100. If they do then the card number is valid. I was trying to do this with a nested for loop but I am struggling on checking the number for every other index.

shreyasm-dev
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Basil Sheppard
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  • Simplify the task into multiple individual parts. 1) Write a function to check _one_ "card array" at a time and return a boolean, eg. `function isValidLunCode(number) { .. return true/false; }`. 2) Once this is implemented, call the function multiple times, as needed to check all cards. – user2864740 May 23 '20 at 20:13
  • (Oops, that's not a "LUN" code, anyway.. same thing: break down the task.) – user2864740 May 23 '20 at 20:20
  • Ok ill break it into multiple functions after work Ill post later tonight on the changes. Thank you for the help – Basil Sheppard May 23 '20 at 20:22
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    https://codecademy-content.s3.amazonaws.com/PRO/independent-practice-projects/credit-card-checker/diagrams/cc+validator+diagram+1.svg – Basil Sheppard May 23 '20 at 20:25
  • They want you to check every other index of each card then perform the said calculation and check. – Basil Sheppard May 23 '20 at 20:26
  • You should be reading those values in sets of 2, or using modulo to work with even and odd indexes. Although I don't see the point of iterating from the right in the example. The end result doesn't really change. – user120242 May 23 '20 at 21:24
  • simple implementation using reduce: `const validateCred = numList => numList.reverse().reduce((acc,n,i)=> acc+(i%2==0?n:(n*2<=9?n*2:n*2-9)) ,0) % 10 === 0; console.log(batch.map(numList=>validateCred(numList)))` – user120242 May 23 '20 at 22:22

1 Answers1

0

The code should be self-explanatory. Since you are a beginner, you can find the standard array methods here: 

// All valid credit card numbers
const valids =[ 
 [4, 5, 3, 9, 6, 7, 7, 9, 0, 8, 0, 1, 6, 8, 0, 8],
 [5, 5, 3, 5, 7, 6, 6, 7, 6, 8, 7, 5, 1, 4, 3, 9],
 [3, 7, 1, 6, 1, 2, 0, 1, 9, 9, 8, 5, 2, 3, 6],
 [6, 0, 1, 1, 1, 4, 4, 3, 4, 0, 6, 8, 2, 9, 0, 5],
 [4, 5, 3, 9, 4, 0, 4, 9, 6, 7, 8, 6, 9, 6, 6, 6]
]
// All invalid credit card numbers
const invalids = [ 
  [4, 5, 3, 2, 7, 7, 8, 7, 7, 1, 0, 9, 1, 7, 9, 5],
  [5, 7, 9, 5, 5, 9, 3, 3, 9, 2, 1, 3, 4, 6, 4, 3],
  [3, 7, 5, 7, 9, 6, 0, 8, 4, 4, 5, 9, 9, 1, 4],
  [6, 0, 1, 1, 1, 2, 7, 9, 6, 1, 7, 7, 7, 9, 3, 5],
  [5, 3, 8, 2, 0, 1, 9, 7, 7, 2, 8, 8, 3, 8, 5, 4]
]

// Can be either valid or invalid
const mysterious = [
  [3, 4, 4, 8, 0, 1, 9, 6, 8, 3, 0, 5, 4, 1, 4],
  [5, 4, 6, 6, 1, 0, 0, 8, 6, 1, 6, 2, 0, 2, 3, 9],
  [6, 0, 1, 1, 3, 7, 7, 0, 2, 0, 9, 6, 2, 6, 5, 6, 2, 0, 3],
  [4, 9, 2, 9, 8, 7, 7, 1, 6, 9, 2, 1, 7, 0, 9, 3],
  [4, 9, 1, 3, 5, 4, 0, 4, 6, 3, 0, 7, 2, 5, 2, 3]
]

function validate_credit_card( code ) {
  result = code.reverse();
  result = result.map( double_odd_positions );
  result = result.map( substract_nine_from_numbers_above_nine );
  result = result.reduce( add, 0);
  result = result % 10 === 0;
  return result;
}

function double_odd_positions( number, index ) {
  if( index % 2 === 0 ) return number;
  else return number * 2;
}

function substract_nine_from_numbers_above_nine (number) {
  if( number > 9 ) return number - 9;
  else return number;
}
function add(number, accumulator) {
  return accumulator + number;
}

console.log('valids:');
valids.forEach( code => {
  console.log( validate_credit_card(code) )
})

console.log('invalids:');
invalids.forEach( code => {
  console.log( validate_credit_card(code) )
})
console.log('mysterious:');
mysterious.forEach( code => {
  console.log( validate_credit_card(code) )
})
Rob Monhemius
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  • I’d use a single loop/function for this problem - there are only two if statements needed (double this index?, if doubled subtract 9?) and the loop is a direct translation from the diagram/algorithm. Can’t see the forest for the trees and all that with over-decomposition and a “functional” approach that does not inherently simplify the problem. I’d not expect a novice coder to be able to easily follow this code as it depends on tracking multiple intermediate data states (including a double reversal). 0.02 – user2864740 May 24 '20 at 06:01
  • @user2864740 imo this is easier to follow/change/extend/debug than a single clump of logic. You could write the same thing in a single reducer. If you did that you would have the exact same intermediate data states in your head, I just made it as explicit as I could. – Rob Monhemius May 24 '20 at 08:01
  • That is very clean code. Thank you for your help I hope as time goes by I will learn to code in such a orderly and neat fashion. – Basil Sheppard May 24 '20 at 21:33
  • I’m might jump on my computer later and post an alternative implementation simply to counter such statements about “intermediate states”.. by following the algorithm ‘verbatim’, at most one index/value (and running total) needs to be considered at a time using a single pass. – user2864740 May 24 '20 at 21:53