IBAN

October 25, 2019

This is tedious, but not hard in a language that supports big-integer arithmetic. First we convert the string to a large integer:;

(define (iban str)
  (let-values (((start end) (split 4 (string->list str))))
    (let loop ((cs (append end start)) (zs (list)))
      (cond ((null? cs)
              (string->number
                (list->string
                  (flatten
                    (reverse zs)))))
            ((char-alphabetic? (car cs))
              (loop (cdr cs)
                    (cons (string->list
                            (number->string
                              (- (char->integer (car cs)) 55)))
                          zs)))
            (else (loop (cdr cs) (cons (car cs) zs)))))))

Then it is easy to validate an IBAN or generate a hash code:

(define (valid? str) (= 1 (modulo (iban str) 97)))
(define (gen str) (- 98 (modulo (iban str) 97)))

 

Here are the example problems:

> (valid? "GB82WEST12345698765432")

#t
> (gen “GB00WEST12345698765432”)

82

You can run the program at https://ideone.com/k7S3Rg, but that calculates the answers incorrectly because the version of Chicken Scheme at ideone doesn’t compute big integers properly.

Pages: 1 2

Posted by programmingpraxis
Filed in Exercises
8 Comments »

8 Responses to “IBAN”

  1. Alex B said
    October 25, 2019 at 11:43 AM

    Python 3.8 solution:

    #! /usr/bin/env python3

# IBAN
# https://programmingpraxis.com/2019/10/25/iban/

IBAN_CHARS = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789'

class IBAN(object):
    """International Bank Account Number"""

    def __init__(self, iban):
        self.iban = self.normalise(iban)
        self.country_code = self.iban[:2]
        self.hash_code = self.iban[2:4]
        self.account_details = self.iban[4:]

    def __repr__(self):
        return f"IBAN.IBAN('{self.iban}')"

    def __str__(self):
        chunks = [self.iban[i:i+4] for i in range(0, len(self.iban), 4)]
        return ' '.join(chunks)

    def __eq__(self, other):
        return self.iban == other.iban

    def is_valid(self):
        istr = self.iban[4:] + self.iban[:4]
        istr = self.alpha_to_num(istr)
        chk = int(istr) % 97
        return chk == 1

    @classmethod
    def from_details(cls, country_code, details):
        iban = cls.normalise(details + country_code) + '00'
        istr = cls.alpha_to_num(iban)
        chk = 98 - (int(istr) % 97)
        iban = iban[-4:-2] + str(chk) + iban[:-4]
        return cls(iban)

    @staticmethod
    def normalise(iban):
        return ''.join(i for c in iban if (i := c.upper()) in IBAN_CHARS)

    @staticmethod
    def alpha_to_num(iban):
        return ''.join(c if c.isdigit() else str(ord(c)-55) for c in iban)

    Used as follows:

    >>> import IBAN
>>> pp_iban = IBAN.IBAN('GB82 WEST 1234 5698 7654 32')
>>> pp_iban
IBAN.IBAN('GB82WEST12345698765432')
>>> str(pp_iban)
'GB82 WEST 1234 5698 7654 32'
>>> pp_iban.is_valid()
True
>>> pp_iban.country_code
'GB'
>>> pp_iban.hash_code
'82'
>>> pp_iban.account_details
'WEST12345698765432'
>>> bad_iban = IBAN.IBAN('GB82 WEST 1234 5698 7654 36')
>>> bad_iban.is_valid()
False
>>> new_iban = IBAN.IBAN.from_details('GB', 'West 123456-9876-5432')
>>> new_iban.is_valid()
True
>>> new_iban == pp_iban
True
>>>
  2. chaw said
    October 25, 2019 at 2:53 PM

    Here is a simple solution in R7R6 Scheme and a few helpers from
    popular SRFIs.

    It emphasizes clarity over efficiency and makes rather liberal use of
    potentially expensive string operations and conversions. On the other
    hand, given the fixed size and format of IBAN, it is a Theta(1)
    solution.

    It also allows human-friendly spaces (and potentially other
    punctuation) in the IBANs.

    (import (scheme base)
        (scheme write)
        (scheme char)
        (only (srfi 1) split-at append-map iota filter)
        (only (srfi 8) receive))

(define iban-sample "GB82 WEST 1234 5698 7654 32")
(define iban-sample-1 "GB82:WEST:1234-5698-7654-32")

;; 0..9 -> "0".."9"; A .. Z -> "10" .. "35"; a .. z -> "10" .. "35"; else "".
(define iban-char->number-string
  (let ((letter-values (map cons
                            (map integer->char
                                 (iota 26 (char->integer #\A)))
                            (map number->string
                                 (iota 26 10)))))
    (lambda (c)
      (cond ((char-numeric? c) (string c))
            ((assv (char-upcase c) letter-values) => cdr)
            (else "")))))

(define (iban->number str)
  (receive (pre suf) (split-at (map iban-char->number-string
                                    (string->list str))
                               4)
    (string->number (apply string-append (append suf pre)))))

(define (iban-hash-valid? str)
  (= 1 (remainder (iban->number str)
                  97)))

(define (iban-hash str)
  (receive (pre suf) (split-at (filter (lambda (c)
                                         (or (char-numeric? c)
                                             (char-alphabetic? c)))
                                       (string->list str))
                               2)
    (- 98
       (remainder (iban->number (apply string
                                       (append pre
                                               '(#\0 #\0)
                                               (cddr suf))))
                  97))))

(define (iban-demo str)
  (display (list str
               (iban-hash str)
               (iban-hash-valid? str)))
  (newline))

(for-each iban-demo (list iban-sample iban-sample-1))

    Output:

    (GB82 WEST 1234 5698 7654 32 82 #t)
(GB82:WEST:1234-5698-7654-32 82 #t)

  3. matthew said
    October 25, 2019 at 8:05 PM

    I’m not going to try to compete with Alex’s nicely engineered solution, so here’s a Python two-liner (it would go into one, but it feels like cheating to go over 80 chars in a line):

    >>> chars = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"
>>> def iban(s): return int("".join(str(chars.index(c)) for c in s[4:]+s[:4])) % 97
... 
>>> iban('GB82WEST12345698765432')
1
>>> 98-iban('GB00WEST12345698765432')
82
  4. matthew said
    October 25, 2019 at 9:15 PM

    And in fact, we can use the Python string -> int conversion with base 36 and get an under 80 chars one-liner (I got this from the Rosetta Code IBAN page):

    def iban(s): return int("".join(str(int(c,36)) for c in s[4:]+s[:4])) % 97
  5. Globules said
    October 26, 2019 at 5:40 AM

    A Haskell version.

    import Control.Monad ((>=>))
import Data.List (lookup)
import Data.Maybe (maybe)
import Numeric (readDec)
import System.Environment (getArgs)
import Text.Printf (printf)

-- Return a "broken out" IBAN, consisting of a two character country code, two
-- check digits and the remaining BBAN (Basic Bank Account Number).  Return
-- nothing if the IBAN doesn't have enough characters.
ibanSplit :: String -> Maybe (String, String, String)
ibanSplit (cc0:cc1:cd0:cd1:bban) = Just ([cc0, cc1], [cd0, cd1], bban)
ibanSplit _                      = Nothing

-- A mapping from IBAN characters to their corresponding decimal strings.
digits :: [(Char, String)]
digits = zip (['0'..'9'] ++ ['A'..'Z']) $ map show [0..]

-- Convert a string consisting of the characters '0' through '9' and 'A' through
-- 'Z' to an integer.  Return nothing if the string contains any characters not
-- in those ranges.
toInt :: String -> Maybe Integer
toInt cs = case readDec . concat <$> mapM (`lookup` digits) cs of
             Just [(n, "")] -> Just n
             _              -> Nothing

-- Validate a broken out IBAN.  Return true if all the characters are valid and
-- the check digits are correct, otherwise return false.
ibanValidate' :: (String, String, String) -> Bool
ibanValidate' (cc, cd, bban) = maybe False remOk (toInt $ bban ++ cc ++ cd)
  where remOk n = n `rem` 97 == 1

-- Validate an IBAN.  Return true if all the characters are valid and the check
-- digits are correct, otherwise return false.
ibanValidate :: String -> Bool
ibanValidate = maybe False ibanValidate' . ibanSplit

-- Given a broken out IBAN return the IBAN containing generated check digits.
ibanChecksum' :: (String, String, String) -> Maybe String
ibanChecksum' (cc, _, bban) = newIban <$> toInt (bban ++ cc ++ "00")
  where newIban n = cc ++ checkDigits n ++ bban
        checkDigits n = printf "%02d" $ 98 - (n `rem` 97)

-- Given an IBAN return a new IBAN containing generated check digits.
ibanChecksum :: String -> Maybe String
ibanChecksum = ibanSplit >=> ibanChecksum'

ibanTest :: String -> IO ()
ibanTest iban = do
  printf "%s: valid? %s\n" iban (show $ ibanValidate iban)
  printf "%s: new check digits: %s\n" iban (show $ ibanChecksum iban)

main :: IO ()
main = getArgs >>= mapM_ ibanTest

    $ ./iban GB82WEST12345698765432 GBXXWEST12345698765432 G@82WEST12345698765432
GB82WEST12345698765432: valid? True
GB82WEST12345698765432: new check digits: Just "GB82WEST12345698765432"
GBXXWEST12345698765432: valid? False
GBXXWEST12345698765432: new check digits: Just "GB82WEST12345698765432"
G@82WEST12345698765432: valid? False
G@82WEST12345698765432: new check digits: Nothing
  6. Steve said
    October 26, 2019 at 5:07 PM

    When I tried to run your solution in Gambit Scheme, I got the following:

    (define-syntax let-values
    (syntax-rules ()
    ((_ () f1 f2 …) (let () f1 f2 …))
    ((_ ((fmls1 expr1) (fmls2 expr2) …) f1 f2 …)
    (let-values-help fmls1 () () expr1 ((fmls2 expr2) …) (f1 f2 …)))))
    *** ERROR IN (console)@56.17 — Ill-formed expression

    3> (define-syntax let-values-help
    (syntax-rules ()
    ((_ (x1 . fmls) (x …) (t …) e m b)
    (let-values-help fmls (x … x1) (t … tmp) e m b))
    ((_ () (x …) (t …) e m b)
    (call-with-values
    (lambda () e)
    (lambda (t …)
    (let-values m (let ((x t) …) . b)))))
    ((_ xr (x …) (t …) e m b)
    (call-with-values
    (lambda () e)
    (lambda (t … . tmpr)
    (let-values m (let ((x t) … (xr tmpr)) . b)))))))
    *** ERROR IN (console)@61.17 — Ill-formed expression

  7. Daniel said
    February 18, 2020 at 1:54 AM

    Here’s a solution in C.

    #include <stdio.h>
#include <stdlib.h>
#include <string.h>

int calc_hash(char* iban) {
  int result = 0;
  int n = strlen(iban);
  for (int i = 0; i < n; ++i) {
    char c = iban[(i + 4) % n];
    if (c >= 'A' && c <= 'Z') {
      result = (result * 100) + (10 + c - 'A');
    } else if (c >= '0' && c <= '9') {
      result = (result * 10) + (c - '0');
    } else {
      // Unknown character
    }
    result %= 97;
  }
  return result;
}

int main(int argc, char* argv[]) {
  if (argc != 2) {
    return EXIT_FAILURE;
  }
  int hash = calc_hash(argv[1]);
  printf("%d\n", hash);
  return EXIT_SUCCESS;
}

    Example usage:

    $ ./iban "GB82 WEST 1234 5698 7654 32"
1

$ expr 98 - $(./iban "GB00 WEST 1234 5698 7654 32")
82
  8. Daniel said
    February 18, 2020 at 1:58 AM

    calc_hash is not a fitting name for the function above. I think that reflects my original intent, although the function calculates a remainder, not a hash.

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