Is it possible to validate IMEI Number? - c

Question

For a mobile shop application, I need to validate an IMEI number. I know how to validate based on input length, but is their any other mechanism for validating the input number? Is there any built-in function that can achieve this?
Logic from any language is accepted, and appreciated.

Answer 1

A search suggests that there isn't a built-in function that will validate an IMEI number, but there is a validation method using the Luhn algorithm.
General process:
Input IMEI: 490154203237518
Take off the last digit, and remember it: 49015420323751 & 8. This last digit 8 is the validation digit.
Double each second digit in the IMEI: 4 18 0 2 5 8 2 0 3 4 3 14 5 2 (excluding the validation digit)
Separate this number into single digits: 4 1 8 0 2 5 8 2 0 3 4 3 1 4 5 2 (notice that 18 and 14 have been split).
Add up all the numbers: 4+1+8+0+2+5+8+2+0+3+4+3+1+4+5+2 = 52
Take your resulting number, remember it, and round it up to the nearest multiple of ten: 60.
Subtract your original number from the rounded-up number: 60 - 52 = 8.
Compare the result to your original validation digit. If the two numbers match, your IMEI is valid.
The IMEI given in step 1 above is valid, because the number found in step #7 is 8, which matches the validation digit.

Answer 2

According to the previous answer from Karl Nicoll i'm created this method in Java.
public static int validateImei(String imei) {
//si la longitud del imei es distinta de 15 es invalido
if (imei.length() != 15)
return CheckImei.SHORT_IMEI;
//si el imei contiene letras es invalido
if (!PhoneNumber.allNumbers(imei))
return CheckImei.MALFORMED_IMEI;
//obtener el ultimo digito como numero
int last = imei.charAt(14) - 48;
//duplicar cada segundo digito
//sumar cada uno de los digitos resultantes del nuevo imei
int curr;
int sum = 0;
for (int i = 0; i < 14; i++) {
curr = imei.charAt(i) - 48;
if (i % 2 != 0){
// sum += duplicateAndSum(curr);
// initial code from Osvel Alvarez Jacomino contains 'duplicateAndSum' method.
// replacing it with the implementation down here:
curr = 2 * curr;
if(curr > 9) {
curr = (curr / 10) + (curr - 10);
}
sum += curr;
}
else {
sum += curr;
}
}
//redondear al multiplo de 10 superior mas cercano
int round = sum % 10 == 0 ? sum : ((sum / 10 + 1) * 10);
return (round - sum == last) ? CheckImei.VALID_IMEI_NO_NETWORK : CheckImei.INVALID_IMEI;
}

Answer 3

IMEI can start with 0 digit. This is why the function input is string.
Thanks for the method #KarlNicol
Golang
func IsValid(imei string) bool {
digits := strings.Split(imei, "")
numOfDigits := len(digits)
if numOfDigits != 15 {
return false
}
checkingDigit, err := strconv.ParseInt(digits[numOfDigits-1], 10, 8)
if err != nil {
return false
}
checkSum := int64(0)
for i := 0; i < numOfDigits-1; i++ { // we dont need the last one
convertedDigit := ""
if (i+1)%2 == 0 {
d, err := strconv.ParseInt(digits[i], 10, 8)
if err != nil {
return false
}
convertedDigit = strconv.FormatInt(2*d, 10)
} else {
convertedDigit = digits[i]
}
convertedDigits := strings.Split(convertedDigit, "")
for _, c := range convertedDigits {
d, err := strconv.ParseInt(c, 10, 8)
if err != nil {
return false
}
checkSum = checkSum + d
}
}
if (checkSum+checkingDigit)%10 != 0 {
return false
}
return true
}

Answer 4

I think this logic is not right because this working only for the specific IMEI no - 490154203237518 not for other IMEI no ...I implement the code also...
var number = 490154203237518;
var array1 = new Array();
var array2 = new Array();
var specialno = 0 ;
var sum = 0 ;
var finalsum = 0;
var cast = number.toString(10).split('');
var finalnumber = '';
if(cast.length == 15){
for(var i=0,n = cast.length; i<n; i++){
if(i !== 14){
if(i == 0 || i%2 == 0 ){
array1[i] = cast[i];
}else{
array1[i] = cast[i]*2;
}
}else{
specialno = cast[14];
}
}
for(var j=0,m = array1.length; j<m; j++){
finalnumber = finalnumber.concat(array1[j]);
}
while(finalnumber){
finalsum += finalnumber % 10;
finalnumber = Math.floor(finalnumber / 10);
}
contno = (finalsum/10);
finalcontno = Math.round(contno)+1;
check_specialno = (finalcontno*10) - finalsum;
if(check_specialno == specialno){
alert('Imei')
}else{
alert('Not IMEI');
}
}else{
alert('NOT imei - length not matching');
}
//alert(sum);

Answer 5

According to the previous answer from Karl Nicoll i'm created this function in Python.
from typing import List
def is_valid_imei(imei: str) -> bool:
def digits_of(s: str) -> List[int]:
return [int(d) for d in s]
if len(imei) != 15 or not imei.isdecimal():
return False
digits = digits_of(imei)
last = digits.pop()
for i in range(1, len(digits), 2):
digits[i] *= 2
digits = digits_of(''.join(map(str, digits)))
return (sum(digits) + last) % 10 == 0

Answer 6

This one is working for me
try:
immie_procesed = list(
map(int, str(request.data.decode("utf-8").split(',')[0])))
last_number = immie_procesed[len(immie_procesed) - 1]
val = sum([sum(map(int, str(int(v)*2))) if i % 2 else int(v)
for i, v in enumerate(immie_procesed[: -1])])
round_value = math.ceil(val / 10) * 10
validation_value = round_value - val
if (validation_value == last_number):
IMEI = request.data.decode("utf-8").split(',')[0]
else:
return "INVALID IMEI"
except:
return 'Something goes wrong in validation process'

Answer 7

I don't believe there are any built-in ways to authenticate an IMEI number. You would need to verify against a third party database (googling suggests there are a number of such services, but presumably they also get their information from more centralised sources).