33
\$\begingroup\$

An EAN-8 barcode includes 7 digits of information and an 8th checksum digit.

The checksum is calculated by multiplying the digits by 3 and 1 alternately, adding the results, and subtracting from the next multiple of 10.

For example, given the digits 2103498:

Digit:        2   1   0   3   4   9   8
Multiplier:   3   1   3   1   3   1   3
Result:       6   1   0   3  12   9  24

The sum of these resulting digits is 55, so the checksum digit is 60 - 55 = 5


The Challenge

Your task is to, given an 8 digit barcode, verify if it is valid - returning a truthy value if the checksum is valid, and falsy otherwise.

  • You may take input in any of the following forms:
    • A string, 8 characters in length, representing the barcode digits
    • A list of 8 integers, the barcode's digits
    • A non-negative integer (you can either assume leading zeroes where none are given, i.e. 1 = 00000001, or request input with the zeroes given)
  • Builtins that compute the EAN-8 checksum (i.e, take the first 7 digits and calculate the last) are banned.
  • This is , so the shortest program (in bytes) wins!

Test Cases

20378240 -> True
33765129 -> True
77234575 -> True
00000000 -> True

21034984 -> False
69165430 -> False
11965421 -> False
12345678 -> False
\$\endgroup\$
  • \$\begingroup\$ Related to Luhn algorithm for verifying credit card numbers, possibly a dupe. \$\endgroup\$ – xnor Nov 15 '17 at 17:58
  • 1
    \$\begingroup\$ This question is not actually about a bar code (which is the black-white striped thing), but about the number encoded by a barcode. The number can exist without a bar code, and the bar code can encode other things than EANs. Maybe just "Is my EAN-8 valid" is a better title? \$\endgroup\$ – Paŭlo Ebermann Nov 15 '17 at 19:49
  • 2
    \$\begingroup\$ @PaŭloEbermann doesn't quite have the same ring to it... \$\endgroup\$ – FlipTack Nov 15 '17 at 20:17
  • 7
    \$\begingroup\$ When reading about barcodes, I expect some image reading (or at least a bit-string), not verifying a checksum. \$\endgroup\$ – Paŭlo Ebermann Nov 15 '17 at 20:19
  • \$\begingroup\$ Strongly related, since an ISBN-13 is an EAN. \$\endgroup\$ – Olivier Grégoire Nov 16 '17 at 14:55

57 Answers 57

5
\$\begingroup\$

Jelly, 7 bytes

s2Sḅ3⁵ḍ

Try it online!

How it works

s2Sḅ3⁵ḍ  Main link. Argument: [a,b,c,d,e,f,g,h] (digit array)

s2       Split into chunks of length 2, yielding [[a,b], [c,d], [e,f], [g,h]].
  S      Take the sum of the pairs, yielding [a+c+e+g, b+d+f+h].
   ḅ3    Convert from ternary to integer, yielding 3(a+c+e+g) + (b+d+f+h).
     ⁵ḍ  Test if the result is divisible by 10.
\$\endgroup\$
13
\$\begingroup\$

JavaScript (ES6), 41 40 38 bytes

Saved 2 bytes thanks to @ETHProductions and 1 byte thanks to @Craig Ayre.

s=>s.map(e=>t+=e*(i^=2),t=i=1)|t%10==1

Takes input as a list of digits.

Determines the sum of all digits, including the checksum.

If the sum is a multiple of 10, then it's a valid barcode.

Test Cases

let f=

s=>s.map(e=>t+=e*(i^=2),t=i=1)|t%10==1

console.log(f([2,0,3,7,8,2,4,0]));
console.log(f([3,3,7,6,5,1,2,9]));
console.log(f([7,7,2,3,4,5,7,5]));
console.log(f([0,0,0,0,0,0,0,0]));

console.log(f([2,1,0,3,4,9,8,4]));
console.log(f([6,9,1,6,5,4,3,0]));
console.log(f([1,1,9,6,5,4,2,1]));
console.log(f([1,2,3,4,5,6,7,8]));

\$\endgroup\$
  • \$\begingroup\$ I was going to say you could save 3 bytes by switching from pre-recursion to post-recursion with g=([n,...s],i=3,t=0)=>n?g(s,4-i,t+n*i):t%10<1, but you may have found a better way... \$\endgroup\$ – ETHproductions Nov 15 '17 at 16:45
  • \$\begingroup\$ Thanks, @ETHproductions, I've changed to map, which I think works better since input can be a list of digits instead of a string. \$\endgroup\$ – Rick Hitchcock Nov 15 '17 at 16:48
  • \$\begingroup\$ Perhaps save another byte with s=>s.map(e=>t+=e*(i=4-i),t=i=1)&&t%10==1? \$\endgroup\$ – ETHproductions Nov 15 '17 at 16:48
  • \$\begingroup\$ Yes, brilliant, thanks : ) \$\endgroup\$ – Rick Hitchcock Nov 15 '17 at 16:50
  • \$\begingroup\$ Great solution! Could you replace && with | to output 1/0 since truthy/falsy is allowed? \$\endgroup\$ – Craig Ayre Nov 15 '17 at 17:22
10
\$\begingroup\$

Python 2, 64 48 35 29 bytes

mypetlion saved 19 bytes

lambda x:sum(x[::2]*2+x)%10<1

Try it online!

\$\endgroup\$
  • \$\begingroup\$ lambda x:sum(x[::2]*3+x[1::2])%10<1 For 35 bytes. \$\endgroup\$ – mypetlion Nov 15 '17 at 17:37
  • 2
    \$\begingroup\$ lambda x:sum(x[::2]*2+x)%10<1 For 29 bytes. \$\endgroup\$ – mypetlion Nov 15 '17 at 18:51
8
\$\begingroup\$

Jelly, 8 bytes

m2Ḥ+µS⁵ḍ

Try the test suite.

Jelly, 9 bytes

JḂḤ‘×µS⁵ḍ

Try it online or Try the test suite.

How this works

m2Ḥ+µS⁵ḍ ~ Full program.

m2       ~ Modular 2. Return every second element of the input.
  Ḥ      ~ Double each.
   +µ    ~ Append the input and start a new monadic chain.
     S   ~ Sum.
      ⁵ḍ ~ Is divisible by 10?
JḂḤ‘×µS⁵ḍ  ~ Full program (monadic).

J          ~ 1-indexed length range.
 Ḃ         ~ Bit; Modulo each number in the range above by 2.
  Ḥ        ~ Double each.
   ‘       ~ Increment each.
    ×      ~ Pairwise multiplication with the input.
     µ     ~ Starts a new monadic chain.
      S    ~ Sum.
       ⁵ḍ  ~ Is the sum divisible by 10?

The result for the first 7 digits of the barcode and the checksum digit must add to a multiple of 10 for it to be valid. Thus, the checksum is valid iff the algorithm applied to the whole list is divisible by 10.

\$\endgroup\$
  • \$\begingroup\$ Still 9 bytes but with consistent values: JḂḤ‘×µS⁵ḍ \$\endgroup\$ – HyperNeutrino Nov 15 '17 at 16:59
  • \$\begingroup\$ @HyperNeutrino Thanks, I knew there was an atom for this! \$\endgroup\$ – Mr. Xcoder Nov 15 '17 at 17:00
  • \$\begingroup\$ Also 9 bytes: JḂaḤ+µS⁵ḍ :P \$\endgroup\$ – HyperNeutrino Nov 15 '17 at 17:04
  • \$\begingroup\$ @HyperNeutrino Well there are a lot of alternatives :P \$\endgroup\$ – Mr. Xcoder Nov 15 '17 at 17:05
  • 1
    \$\begingroup\$ 8 bytes or 8 characters? m2Ḥ+µS⁵ḍ is 15 bytes in UTF-8, unless I've calculated it wrong. \$\endgroup\$ – ta.speot.is Nov 18 '17 at 8:31
7
\$\begingroup\$

MATL, 10 bytes

Thanks to @Zgarb for pointing out a mistake, now corrected.

IlhY"s10\~

Try it online! Or verify all test cases.

Explanation

Ilh     % Push [1 3]
Y"      % Implicit input. Run-length decoding. For each entry in the
        % first input, this produces as many copies as indicated by
        % the corresponding entry of the second input. Entries of
        % the second input are reused cyclically
s       % Sum of array
10\     % Modulo 10
~       % Logical negate. Implicit display
\$\endgroup\$
7
\$\begingroup\$

Befunge-98 (PyFunge), 16 14 bytes

Saved 2 bytes by skipping the second part using j instead of ;s, as well as swapping a ~ and + in the first part to get rid of a + in the second.

~3*+~+6jq!%a+2

Input is in 8 digits (with leading 0s if applicable) and nothing else.

Outputs via exit code (open the debug dropdown on TIO), where 1 is true and 0 is false.

Try it online!

Explanation

This program uses a variety of tricks.

First of all, it takes the digits in one by one through their ASCII values. Normally, this would require subtracting 48 from each value as we read it from the input. However, if we don't modify it, we are left with 16 (3+1+3+1+3+1+3+1) extra copies of 48 in our sum, meaning our total is going to be 768 greater than what it "should" be. Since we are only concerned with the sum mod 10, we can just add 2 to the sum later. Thus, we can take in raw ASCII values, saving 6 bytes or so.

Secondly, this code only checks if every other character is an EOF, because the input is guaranteed to be only 8 characters long.

Thirdly, the # at the end of the line doesn't skip the first character, but will skip the ; if coming from the other direction. This is better than putting a #; at the front instead.

Because the second part of our program is only run once, we don't have to set it up so that it would skip the first half when running backwards. This lets us use the jump command to jump over the second half, as we exit before executing it going backwards.

Step by step

Note: "Odd" and "Even" characters are based on a 0-indexed system. The first character is an even one, with index 0.

~3*+~+      Main loop - sum the digits (with multiplication)
~           If we've reached EOF, reverse; otherwise take char input. This will always
                be evenly indexed values, as we take in 2 characters every loop.
 3*+        Multiply the even character by 3 and add it to the sum.
    ~       Then, take an odd digit - we don't have to worry about EOF because
                the input is always 8 characters.
     +      And add it to the sum.
      6j    Jump over the second part - We only want to run it going backwards.

        q!%a+2    The aftermath (get it? after-MATH?)
            +2    Add 2 to the sum to make up for the offset due to reading ASCII
          %a      Mods the result by 10 - only 0 if the bar code is valid
         !        Logical not the result, turning 0s into 1s and anything else into 0s
        q         Prints the top via exit code and exits
\$\endgroup\$
6
\$\begingroup\$

C,  78  77 bytes

i,s,c,d=10;f(b){for(i=s=0,c=b%d;b/=d;)s+=b%d*(3-i++%2*2);return(d-s%d)%d==c;}

Try it online!

C (gcc), 72 bytes

i,s,c,d=10;f(b){for(i=s=0,c=b%d;b/=d;)s+=b%d*(i++%2?:3);b=(d-s%d)%d==c;}

Try it online!

\$\endgroup\$
6
\$\begingroup\$

Wolfram Language (Mathematica), 26 21 bytes

10∣(2-9^Range@8).#&

Try it online!

Takes input as a list of 8 digits.

How it works

2-9^Range@8 is congruent modulo 10 to 2-(-1)^Range@8, which is {3,1,3,1,3,1,3,1}. We take the dot product of this list with the input, and check if the result is divisible by 10.

Wolfram Language (Mathematica), 33 bytes and non-competing

Check[#~BarcodeImage~"EAN8";1,0]&

Try it online!

Takes input as a string. Returns 1 for valid barcodes and 0 for invalid ones.

How it works

The best thing I could find in the way of a built-in (since Mathematica is all about those).

The inside bit, #~BarcodeImage~"EAN8";1, generates an image of the EAN8 barcode, then ignores it entirely and evaluates to 1. However, if the barcode is invalid, then BarcodeImage generates a warning, which Check catches, returning 0 in that case.

\$\endgroup\$
  • 3
    \$\begingroup\$ Are you doing the calculation by hand because it's shorter, or because Wolfram doesn't yet have a ValidateEAN8BarCode() function somewhere in its standard library? \$\endgroup\$ – Mark Nov 16 '17 at 0:44
  • 1
    \$\begingroup\$ @Mark Mathematica can't validate the barcode directly, but I just found BarcodeImage, which generates the image of the barcode, and validates the barcode in the process. So Check[#~BarcodeImage~"EAN8";0,1]<1& would work (but it's longer). \$\endgroup\$ – Misha Lavrov Nov 16 '17 at 0:50
5
\$\begingroup\$

Java 8, 58 56 55 bytes

a->{int r=0,m=1;for(int i:a)r+=(m^=2)*i;return r%10<1;}

-2 bytes indirectly thanks to @RickHitchcock, by using (m=4-m)*i instead of m++%2*2*i+i after seeing it in his JavaScript answer.
-1 byte indirectly thanks to @ETHProductions (and @RickHitchcock), by using (m^=2)*i instead of (m=4-m)*i.

Explanation:

Try it here.

a->{              // Method with integer-array parameter and boolean return-type
  int r=0,        //  Result-sum
      m=1;        //  Multiplier
  for(int i:a)    //  Loop over the input-array
    r+=           //   Add to the result-sum:
       (m^=2)     //    Either 3 or 1,
       *i;        //    multiplied by the digit
                  //  End of loop (implicit / single-line body)
  return r%10<1;  //  Return if the trailing digit is a 0
}                 // End of method
\$\endgroup\$
  • 1
    \$\begingroup\$ You can save another byte with a trick @ETHProductions showed me: change m=4-m to m^=2. \$\endgroup\$ – Rick Hitchcock Nov 16 '17 at 18:39
  • \$\begingroup\$ @RickHitchcock Ah, of course.. I use ^=1 pretty often in answers when I want to alter between 0 and 1. ^=2 works in this case to alter between 1 and 3. Nice trick, and thanks for the comment to mention it. :) \$\endgroup\$ – Kevin Cruijssen Nov 16 '17 at 18:49
4
\$\begingroup\$

05AB1E, 14 bytes

θ¹¨3X‚7∍*O(T%Q

Try it online!

Needs leading 0s, takes list of digits.

\$\endgroup\$
  • \$\begingroup\$ Seems to fail on 3100004 (should be truthy). \$\endgroup\$ – Zgarb Nov 15 '17 at 16:15
  • \$\begingroup\$ @Zgarb You're missing a 0 there. \$\endgroup\$ – Erik the Outgolfer Nov 15 '17 at 16:15
  • \$\begingroup\$ Oh, it takes a string? Okay then, my bad. \$\endgroup\$ – Zgarb Nov 15 '17 at 16:17
  • \$\begingroup\$ @Zgarb Well, you can omit the quotes, but yes, you do need the leading 0. This answer actually uses number functions on strings, one of the features of 05AB1E. \$\endgroup\$ – Erik the Outgolfer Nov 15 '17 at 16:20
  • \$\begingroup\$ @Mr.Xcoder The question isn't very clear on that, I'll add another code which does handle for that below. \$\endgroup\$ – Erik the Outgolfer Nov 15 '17 at 16:23
4
\$\begingroup\$

Pyth, 8 bytes

!es+*2%2

Verify all the test cases!

Pyth, 13 bytes

If we can assume the input always has exactly 8 digits:

!es.e*bhy%hk2

Verify all the test cases!


How does this work?

!es+*2%2 ~ Full program.

      %2 ~ Input[::2]. Every second element of the input.
    *2   ~ Double (repeat list twice).
   +     ~ Append the input.
  s      ~ Sum.
 e       ~ Last digit.
!        ~ Logical NOT.
!es.e*sbhy%hk2 ~ Full program.

               ~ Convert the input to a String.
   .e          ~ Enumerated map, storing the current value in b and the index in k.
          %hk2 ~ Inverted parity of the index. (k + 1) % 2.
        hy     ~ Double, increment. This maps odd integers to 1 and even ones to 3.
      b        ~ The current digit.
     *         ~ Multiply.
  s            ~ Sum.
 e             ~ Last digit.
!              ~ Logical negation.

If the sum of the first 7 digit after being applied the algorithm is subtracted from 10 and then compared to the last digit, this is equivalent to checking whether the sum of all the digits, after the algorithm is applied is a multiple of 10.

\$\endgroup\$
  • \$\begingroup\$ Seems to fail on 3100004 (should be truthy). \$\endgroup\$ – Zgarb Nov 15 '17 at 16:13
  • \$\begingroup\$ @Zgarb Wait should we do 3*3+1*1+0*3+... or 0*3+3*1+1*0..? I thought we are supposed to do the former \$\endgroup\$ – Mr. Xcoder Nov 15 '17 at 16:14
  • \$\begingroup\$ In the new spec, leading digits are added to ensure there are exactly 8 (if I understand correctly). \$\endgroup\$ – Zgarb Nov 15 '17 at 16:16
  • \$\begingroup\$ @Zgarb Ok, fixed. \$\endgroup\$ – Mr. Xcoder Nov 15 '17 at 16:18
4
\$\begingroup\$

Haskell, 40 38 bytes

a=3:1:a
f x=mod(sum$zipWith(*)a x)10<1

Try it online!

Takes input as a list of 8 integers. A practical example of using infinite lists.

Edit: Saved 2 bytes thanks to GolfWolf

\$\endgroup\$
4
\$\begingroup\$

Retina, 23 22 bytes

-1 byte thanks to Martin Ender!

(.).
$1$1$&
.
$*
M`
1$

Try it online!

Explanation

Example input: 20378240

(.).
$1$1$&

Replace each couple of digits with the first digit repeated twice followed by the couple itself. We get 2220333788824440

.
$*

Convert each digit to unary. With parentheses added for clarity, we get (11)(11)(11)()(111)(111)...

M`

Count the number of matches of the empty string, which is one more than the number of ones in the string. (With the last two steps we have basically taken the sum of each digit +1) Result: 60

1$

Match a 1 at the end of the string. We have multiplied digits by 3 and 1 alternately and summed them, for a valid barcode this should be divisible by 10 (last digit 0); but we also added 1 in the last step, so we want the last digit to be 1. Final result: 1.

\$\endgroup\$
  • 2
    \$\begingroup\$ I think you can drop the . on the match stage and match 1$ at the end. \$\endgroup\$ – Martin Ender Nov 16 '17 at 10:46
  • \$\begingroup\$ @MartinEnder very nice, I'll do that, thanks! \$\endgroup\$ – Leo Nov 16 '17 at 21:32
3
\$\begingroup\$

PowerShell, 85 bytes

param($a)(10-(("$a"[0..6]|%{+"$_"*(3,1)[$i++%2]})-join'+'|iex)%10)%10-eq+"$("$a"[7])"

Try it online! or Verify all test cases

Implements the algorithm as defined. Takes input $a, pulls out each digit with "$a"[0..6] and loops through them with |%{...}. Each iteration, we take the digit, cast it as a string "$_" then cast it as an int + before multiplying it by either 3 or 1 (chosen by incrementing $i modulo 2).

Those results are all gathered together and summed -join'+'|iex. We take that result mod 10, subtract that from 10, and again take the result mod 10 (this second mod is necessary to account for the 00000000 test case). We then check whether that's -equal to the last digit. That Boolean result is left on the pipeline and output is implicit.

\$\endgroup\$
  • \$\begingroup\$ Seems to fail on 3100004 (should be truthy). \$\endgroup\$ – Zgarb Nov 15 '17 at 16:14
  • \$\begingroup\$ @Zgarb Works for me? Try it online! \$\endgroup\$ – AdmBorkBork Nov 15 '17 at 16:31
  • \$\begingroup\$ Ah ok, I tested it without the quotes. \$\endgroup\$ – Zgarb Nov 15 '17 at 16:33
  • \$\begingroup\$ @Zgarb Ah, yeah. Without the quotes, PowerShell will implicitly cast as an integer, stripping the leading zero(s). \$\endgroup\$ – AdmBorkBork Nov 15 '17 at 16:36
3
\$\begingroup\$

Jelly, 16 bytes

ż3,1ṁ$P€SN%⁵
Ṫ=Ç

Try it online!

takes input as a list of digits

\$\endgroup\$
  • \$\begingroup\$ Nitpick: your TIO timeouts. Also, 16 bytes. \$\endgroup\$ – Erik the Outgolfer Nov 15 '17 at 15:37
  • \$\begingroup\$ @EriktheOutgolfer Wait what how. It works when I put the D in the footer. And yay thanks! :D \$\endgroup\$ – HyperNeutrino Nov 15 '17 at 15:39
  • \$\begingroup\$ @EriktheOutgolfer Am I doing something wrong? Your 16-byter appears to be invalid? \$\endgroup\$ – HyperNeutrino Nov 15 '17 at 15:41
  • \$\begingroup\$ Maybe, it works kinda differently, but yours seems a bit invalid too...specifically I think the last line should be DµṪ=Ç. \$\endgroup\$ – Erik the Outgolfer Nov 15 '17 at 15:42
  • 1
    \$\begingroup\$ Seems to fail on 3100004 (should be truthy). \$\endgroup\$ – Zgarb Nov 15 '17 at 16:15
3
\$\begingroup\$

APL (Dyalog), 14 bytes

Equivalent with streetster's solution.

Full program body. Prompts for list of numbers from STDIN.

0=10|+/⎕×8⍴3 1

Try it online!

Is…

0= zero equal to

10| the mod-10 of

+/ the sum of

⎕× the input times

8⍴3 1 eight elements cyclically taken from [3,1]

?

\$\endgroup\$
  • 1
    \$\begingroup\$ You mean APL can't do it in one character from something like Ancient Sumerian or Linear B? \$\endgroup\$ – Mark Nov 17 '17 at 0:20
  • \$\begingroup\$ train: 0=10|-/+2×+/ \$\endgroup\$ – ngn Nov 18 '17 at 18:44
3
\$\begingroup\$

05AB1E, 9 bytes

3X‚7∍*OTÖ

Try it online!

3X‚7∍*OTÖ    # Argument a
3X‚          # Push [3, 1]
   7∍        # Extend to length 7
     *       # Multiply elements with elements at same index in a
      O      # Total sum
       TÖ    # Divisible by 10
\$\endgroup\$
  • \$\begingroup\$ Nice! First thing I thought was "extend to length" when I saw this, haven't used that one yet. \$\endgroup\$ – Magic Octopus Urn Nov 17 '17 at 22:17
  • \$\begingroup\$ 31×S*OTÖ for 8 bytes. × just pushes 31 n number of times. When you multiply, it automatically drops the extra 31's. \$\endgroup\$ – Magic Octopus Urn Nov 17 '17 at 22:21
  • \$\begingroup\$ @MagicOctopusUrn That seems to fail on the 6th testcase 69165430 -> 1 \$\endgroup\$ – kalsowerus Nov 17 '17 at 23:25
3
\$\begingroup\$

J, 17 bytes

-10 bytes thanks to cole

0=10|1#.(8$3 1)*]

Try it online!

This uses multiplication of equal sized lists to avoid the zip/multiply combo of the original solution, as well as the "base 1 trick" 1#. to add the products together. The high level approach is similar to the original explanation.

original, 27 bytes

0=10|{:+[:+/[:*/(7$3 1),:}:

Try it online!

explained

0 =                                        is 0 equal to... 
    10 |                                   the remainder when 10 divides...
         {: +                              the last input item plus...
              [: +/                        the sum of...
                    [: */                  the pairwise product of...
                          7$(3 1) ,:       3 1 3 1 3 1 3 zipped with...
                                     }:    all but the last item of the input
\$\endgroup\$
  • \$\begingroup\$ 0=10|1#.(8$3 1)*] should work for 17 bytes (does the same algorithm, too). I'm pretty sure that in the beta you can have a hook ended on the right side with a noun, so 0=10|1#.]*8$3 1 may work for 15 (I'd check on tio but it seems to be down?) \$\endgroup\$ – cole Nov 16 '17 at 8:06
  • \$\begingroup\$ @cole, I love this improvement. I've learned about and forgotten the 1#. trick like 2 or 3 times... thanks for reminding me. Oh btw the 15 byte version did not work in TIO. \$\endgroup\$ – Jonah Nov 16 '17 at 14:06
3
\$\begingroup\$

C (gcc), 84 82 72 61 54 bytes

c;i;f(x){for(i=c=0;x;x/=10)c+=(1+2*i++%4)*x;c=c%10<1;}

-21 bytes from Neil

-7 bytes from Nahuel Fouilleul

Try it online!

Developed independently of Steadybox's answer

'f' is a function that takes the barcode as an int, and returns 1 for True and 0 for False.

  • f stores the last digit of x in s (s=x%10),

  • Then calculates the sum in c (for(i=c=0;x;x/=10)c+=(1+2*i++%4)*x;)

    • c is the sum, i is a counter

    • for each digit including the first, add 1+2*i%4 times the digit (x%10) to the checksum and increment i (the i++ in 3-2*i++%4)

      • 1+2*i%4 is 1 when i is even and 0 when i is odd
  • Then returns whether the sum is a multiple of ten, and since we added the last digit (multiplied by 1), the sum will be a multiple of ten iff the barcode is valid. (uses GCC-dependent undefined behavior to omit return).

\$\endgroup\$
  • \$\begingroup\$ I think (x%10) can just be x as you're taking c%10 later anyway. Also I think you can use i<8 and then just test whether c%10 is zero at the end. \$\endgroup\$ – Neil Nov 15 '17 at 16:30
  • \$\begingroup\$ @Neil Thanks! That got -10 bytes. \$\endgroup\$ – pizzapants184 Nov 15 '17 at 16:49
  • \$\begingroup\$ In fact I think s is unnecessary: c;i;f(x){for(i=c=0;i<8;x/=10)c+=(1+2*i++%4)*x;return c%10<1;} \$\endgroup\$ – Neil Nov 15 '17 at 16:59
  • \$\begingroup\$ the tio link is 61 bytes but in the answer it's 72, also don't know why x=c%10<1 or c=c%10<1 instead of return c%10<1 still works \$\endgroup\$ – Nahuel Fouilleul Nov 16 '17 at 13:03
  • \$\begingroup\$ also i<8 can be replaced by x \$\endgroup\$ – Nahuel Fouilleul Nov 16 '17 at 13:07
3
\$\begingroup\$

C, 63 bytes

i;s=0;c(int*v){for(i=0;i<8;i++){s+=v[i]*3+v[++i];}return s%10;}

Assumes that 0 is true and any other value is false.

+3 bytes for better return value

i;s=0;c(int*v){for(i=0;i<8;i++){s+=v[i]*3+v[++i];}return s%10==0;}

Add ==0 to the return statement.

Ungolfed

int check(int* values)
{
    int result = 0;
    for (int index = 0; index < 8; index++)
    {
        result += v[i] * 3 + v[++i]; // adds this digit times 3 plus the next digit times 1 to the result
    }
    return result % 10 == 0; // returns true if the result is a multiple of 10
}

This uses the alternative definition of EAN checksums where the check digit is chosen such that the checksum of the entire barcode including the check digit is a multiple of 10. Mathematically this works out the same but it's a lot simpler to write.

Initialising variables inside loop as suggested by Steadybox, 63 bytes

i;s;c(int*v){for(i=s=0;i<8;i++){s+=v[i]*3+v[++i];}return s%10;}

Removing curly brackets as suggested by Steadybox, 61 bytes

i;s;c(int*v){for(i=s=0;i<8;i++)s+=v[i]*3+v[++i];return s%10;}

Using <1 rather than ==0 for better return value as suggested by Kevin Cruijssen

i;s=0;c(int*v){for(i=0;i<8;i++){s+=v[i]*3+v[++i];}return s%10<1;}

Add <1 to the return statement, this adds only 2 bytes rather than adding ==0 which adds 3 bytes.

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  • \$\begingroup\$ You can save two bytes by removing the {} after the for. Also, function submissions have to be reusable, so you need to initialize s inside the function (just change i;s=0; to i,s; and i=0; to i=s=0;). \$\endgroup\$ – Steadybox Nov 16 '17 at 20:44
  • \$\begingroup\$ @Steadybox How can I remove the curly brackets? \$\endgroup\$ – Micheal Johnson Nov 16 '17 at 21:01
  • \$\begingroup\$ There's only one statement inside them. When there are no curly brackets after for, the loop body will be the next statement. for(i=0;i<8;i++){s+=v[i]*3+v[++i];} is the same as for(i=0;i<8;i++)s+=v[i]*3+v[++i];. \$\endgroup\$ – Steadybox Nov 16 '17 at 22:32
  • \$\begingroup\$ @Steadybox Oh of course. That's one of the quirks of C syntax that I usually forget about, because when writing normal code I always include the curly brackets even if they're unnecessary, because it makes the code more readable. \$\endgroup\$ – Micheal Johnson Nov 16 '17 at 22:51
  • \$\begingroup\$ In your true/false answer, instead of +3 by adding ==0 it can be +2 by using <1 instead. :) \$\endgroup\$ – Kevin Cruijssen Nov 17 '17 at 9:05
2
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JavaScript (Node.js), 47 bytes

e=>eval(e.map((a,i)=>(3-i%2*2)*a).join`+`)%10<1

Although there is a much shorter answer already, this is my first attempt of golfing in JavaScript so I'd like to hear golfing recommendations :-)

Testing

let f=

e=>eval(e.map((a,i)=>(3-i%2*2)*a).join`+`)%10<1

console.log(f([2,0,3,7,8,2,4,0]));
console.log(f([3,3,7,6,5,1,2,9]));
console.log(f([7,7,2,3,4,5,7,5]));
console.log(f([0,0,0,0,0,0,0,0]));

console.log(f([2,1,0,3,4,9,8,4]));
console.log(f([6,9,1,6,5,4,3,0]));
console.log(f([1,1,9,6,5,4,2,1]));
console.log(f([1,2,3,4,5,6,7,8]));

Alternatively, you can Try it online!

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2
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Perl 5, 37 32 + 1 (-p) bytes

s/./$-+=$&*(--$|*2+1)/ge;$_=/0$/

-5 bytes thanks to Dom Hastings. 37 +1 bytes was

$s+=$_*(++$i%2*2+1)for/./g;$_=!!$s%10

try it online

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  • 1
    \$\begingroup\$ Had a little play with this and thought I'd share a useful trick: --$| toggles between 1 and 0 so you can use that instead of ++$i%2 for an alternating boolean! Also, all that matters is that the total ($s) matches /0$/, managed to get 33 bytes combining those changes with s///: Try it online! (-l is just for visibility) \$\endgroup\$ – Dom Hastings Nov 16 '17 at 12:01
  • \$\begingroup\$ yes i though to s/./(something with $&)/ge and to /0$/ match but not the two combined. \$\endgroup\$ – Nahuel Fouilleul Nov 16 '17 at 12:36
2
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Brainfuck, 228 Bytes

>>>>++++[<++>-]<[[>],>>++++++[<++++++++>-]<--[<->-]+[<]>-]>[->]<<<<[[<+>->+<]<[>+>+<<-]>>[<+>-]<<<<<]>>>>[>>[<<[>>+<<-]]>>]<<<++++[<---->-]+++++[<++<+++>>-]<<[<[>>[<<->>-]]>[>>]++[<+++++>-]<<-]<[[+]-<]<++++++[>++[>++++<-]<-]>>+.

Can probably be improved a fair bit. Input is taken 1 digit at a time, outputs 1 for true, 0 for false.

How it works:

>>>>++++[<++>-]<

Put 8 at position 3.

[[>],>>++++++[<++++++++>-]<--[<->-]+[<]>-]

Takes input 8 times, changing it from the ascii value to the actual value +2 each time. Inputs are spaced out by ones, which will be removed, to allow for easier multiplication later.

>[->]

Subtract one from each item. Our tape now looks something like

0 0 0 0 4 0 4 0 8 0 7 0 6 0 2 0 3 0 10 0 0
                                         ^

With each value 1 more than it should be. This is because zeros will mess up our multiplication process.

Now we're ready to start multiplying.

<<<<

Go to the second to last item.

[[<+>->+<]<[>+>+<<-]>>[<+>-]<<<<<]

While zero, multiply the item it's at by three, then move two items to the left. Now we've multiplied everything we needed to by three, and we're at the very first position on the tape.

>>>>[>>[<<[>>+<<-]]>>]

Sum the entire list.

<<<++++[<---->-]

The value we have is 16 more than the actual value. Fix this by subtracting 16.

+++++[<++<+++>>-]

We need to test whether the sum is a multiple of 10. The maximum sum is with all 9s, which is 144. Since no sum will be greater than 10*15, put 15 and 10 on the tape, in that order and right to the right of the sum.

<<[<[>>[<<->>-]]>[>>]++[<+++++>-]<<-]

Move to where 15 is. While it's non-zero, test if the sum is non-zero. If it is, subtract 10 from it. Now we're either on the (empty) sum position, or on the (also empty) ten position. Move one right. If we were on the sum position, we're now on the non-zero 15 position. If so, move right twice. Now we're in the same position in both cases. Add ten to the ten position, and subtract one from the 15 position.

The rest is for output:

<[[+]-<]<++++++[>++[>++++<-]<-]>>+.

Move to the sum position. If it is non-zero (negative), the barcode is invalid; set the position to -1. Now add 49 to get the correct ascii value: 1 if it's valid, 0 if it's invalid.

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2
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Java 8, 53 bytes

Golfed:

b->(3*(b[0]+b[2]+b[4]+b[6])+b[1]+b[3]+b[5]+b[7])%10<1

Direct calculation in the lambda appears to the shortest solution. It fits in a single expression, minimizing the lambda overhead and removing extraneous variable declarations and semicolons.

public class IsMyBarcodeValid {

  public static void main(String[] args) {
    int[][] barcodes = new int[][] { //
        { 2, 0, 3, 7, 8, 2, 4, 0 }, //
        { 3, 3, 7, 6, 5, 1, 2, 9 }, //
        { 7, 7, 2, 3, 4, 5, 7, 5 }, //
        { 0, 0, 0, 0, 0, 0, 0, 0 }, //
        { 2, 1, 0, 3, 4, 9, 8, 4 }, //
        { 6, 9, 1, 6, 5, 4, 3, 0 }, //
        { 1, 1, 9, 6, 5, 4, 2, 1 }, //
        { 1, 2, 3, 4, 5, 6, 7, 8 } };
    for (int[] barcode : barcodes) {
      boolean result = f(b -> (3 * (b[0] + b[2] + b[4] + b[6]) + b[1] + b[3] + b[5] + b[7]) % 10 < 1, barcode);
      System.out.println(java.util.Arrays.toString(barcode) + " = " + result);
    }
  }

  private static boolean f(java.util.function.Function<int[], Boolean> f, int[] n) {
    return f.apply(n);
  }
}

Output:

[2, 0, 3, 7, 8, 2, 4, 0] = true
[3, 3, 7, 6, 5, 1, 2, 9] = true
[7, 7, 2, 3, 4, 5, 7, 5] = true
[0, 0, 0, 0, 0, 0, 0, 0] = true
[2, 1, 0, 3, 4, 9, 8, 4] = false
[6, 9, 1, 6, 5, 4, 3, 0] = false
[1, 1, 9, 6, 5, 4, 2, 1] = false
[1, 2, 3, 4, 5, 6, 7, 8] = false
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2
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QBasic, 54 52 bytes

Ugh, the boring answer turned out to be the shortest:

INPUT a,b,c,d,e,f,g,h
?(3*a+b+3*c+d+3*e+f+3*g+h)MOD 10=0

This inputs the digits comma-separated. My original 54-byte solution, which inputs one digit at a time, uses a "nicer" approach:

m=3
FOR i=1TO 8
INPUT d
s=s+d*m
m=4-m
NEXT
?s MOD 10=0
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2
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C# (.NET Core), 65 62 bytes

b=>{int s=0,i=0,t=1;while(i<8)s+=b[i++]*(t^=2);return s%10<1;}

Try it online!

Acknowledgements

-3 bytes thanks to @KevinCruijssen and the neat trick using the exclusive-or operator.

DeGolfed

b=>{
    int s=0,i=0,t=1;

    while(i<8)
        s+=b[i++]*(t^=2); // exclusive-or operator alternates t between 3 and 1.

    return s%10<1;
}

C# (.NET Core), 53 bytes

b=>(3*(b[0]+b[2]+b[4]+b[6])+b[1]+b[3]+b[5]+b[7])%10<1

Try it online!

A direct port of @Snowman's answer.

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  • \$\begingroup\$ For your first answer: b=>{int s=0,i=0,t=1;while(i<8)s+=b[i++]*(t^=2);return s%10<1;} (62 bytes), or alternatively with a foreach, also 62 bytes: b=>{int s=0,t=1;foreach(int i in b)s+=i*(t^=2);return s%10<1;} (which is a port of my Java 8 answer). \$\endgroup\$ – Kevin Cruijssen Nov 17 '17 at 8:58
1
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MATLAB/Octave, 32 bytes

@(x)~mod(sum([2*x(1:2:7),x]),10)

Try it online!

I'm going to post this despite the other Octave answer as I developed this code and approach without looking at the other answers.

Here we have an anonymous function which takes the input as an array of 8 values, and return true if a valid barcode, false otherwise..

The result is calculated as follows.

              2*x(1:2:7)
             [          ,x]
         sum(              )
     mod(                   ,10)
@(x)~
  1. Odd digits (one indexed) are multiplied by 2.
  2. The result is prepended to the input array, giving an array whose sum will contain the odd digits three times, and the even digits once.
  3. We do the sum which will also include the supplied checksum within our sum.
  4. Next the modulo 10 is performed. If the checksum supplied was valid, the sum of all multiplied digits including the checksum value would end up being a multiple of 10. Therefore only a valid barcode would return 0.
  5. The result is inverted to get a logical output of true if valid.
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1
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Excel, 37 bytes

Interpreting "A list of 8 integers" as allowing 8 separate cells in Excel:

=MOD(SUM(A1:H1)+2*(A1+C1+E1+G1),10)=0
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  • \$\begingroup\$ =MOD(SUM((A1:H1)+2*(A1+C1+E1+G1)),10)=0 this formula exist in Excel? \$\endgroup\$ – RosLuP Nov 23 '17 at 21:40
  • \$\begingroup\$ @RosLuP, not predefined, no. But Modulo, Sum, + etc do ;-) \$\endgroup\$ – Wernisch Nov 24 '17 at 10:15
  • \$\begingroup\$ I want only to say that seems that in APL goes well doing first y= (A1:H1)+2*(A1+C1+E1+G1), and after the sum and the mod; in APL not goes well first sum(A1:H1) etc something as (1,2,3)+4=(5,6,7) and than sum(5,6,7)=18; note that sum(1,2,3)=6 and 6+4=10 different from 18. But possible I make error in something \$\endgroup\$ – RosLuP Nov 24 '17 at 16:37
  • \$\begingroup\$ @RosLuP, Apologies, missed the changed ()s in your comment. \$\endgroup\$ – Wernisch Nov 24 '17 at 16:43
  • \$\begingroup\$ Problem is how Excel interprets =(A1:H1): This is not handled as an array. Is invalid if placed in any column not in A-H range. If placed in a column in A-H, returns the value for that column only. (Formula in % results in %: C2 --> C1 H999 --> H1 K1 --> #VALUE!) \$\endgroup\$ – Wernisch Nov 24 '17 at 16:59
1
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Ruby, 41 Bytes

Takes an array of integers. -6 bytes thanks to Jordan.

->n{n.zip([3,1]*4){|x,y|$.+=x*y};$.%10<1}
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  • \$\begingroup\$ Nice! FWIW you don’t need map here at all: zip takes a block. You can save a couple more bytes by using $. instead of initializing s: ->n{n.zip([3,1]*4){|x,y|$.+=x*y};$.%10<1} \$\endgroup\$ – Jordan Nov 16 '17 at 6:44
1
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TI-Basic (83 series), 18 bytes

not(fPart(.1sum(2Ans-Ans9^cumSum(binomcdf(7,0

Takes input as a list in Ans. Returns 1 for valid barcodes and 0 for invalid ones.

A port of my Mathematica answer. Includes screenshot, in lieu of an online testing environment:

barcode screenshot

Notable feature: binomcdf(7,0 is used to generate the list {1,1,1,1,1,1,1,1} (the list of probabilities that from 7 trials with success probability 0, there will be at most N successes, for N=0,1,...,7). Then, cumSum( turns this into {1,2,3,4,5,6,7,8}.

This is one byte shorter than using the seq( command, though historically the point was that it's also significantly faster.

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