Validate a barcode

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A barcode of EAN-13 symbology consists of 13 digits (0-9). The last digit of this barcode is its check digit. It is calculated by the following means (the barcode 8923642469559 is used as an example):

1. 
   

   Starting from the second digit, sum up all alternating digits and multiply the sum by 3:

   
   
   

   8 9 2 3 6 4 2 4 6 9 5 5 9
    | | | | | |
    9 + 3 + 4 + 4 + 9 + 5 = 34
    |
    34 × 3 = 102
   

   
   


2. 
   

   Then, sum up all of the remaining digits, but do not include the last digit:

   
   
   

   8 9 2 3 6 4 2 4 6 9 5 5 9
   | | | | | |
   8 + 2 + 6 + 2 + 6 + 5 = 29
   

   
   


3. 
   

   Add the numbers obtained in steps 1 and 2 together:

   
   
   

   29 + 102 = 131
   

   
   



4. The number you should add to the result of step 3 to get to the next multiple of 10 (140 in this case) is the check digit.

If the check digit of the barcode matches the one calculated as explained earlier, the barcode is valid.

---

More examples:

6537263729385 is valid.
1902956847427 is valid.
9346735877246 is invalid. The check digit should be 3, not 6.

---

Your goal is to write a program that will:

1. Receive a barcode as its input.


2. Check whether the barcode is valid


3. Return 1 (or equivalent) if the barcode is valid, 0 (or equivalent) otherwise.

---

This is code-golf, so the shortest code in terms of bytes wins.

code-golf arithmetic

share|improve this question

edited 21 mins ago

asked 34 mins ago

Wais Kamal

1375

• 
  
  
  

  
  

  
  
  
  
  
  

  
  why did the last digit (9) become a 4
  – HyperNeutrino
  32 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry, will fix it :)
  – Wais Kamal
  31 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I recommend removing the 13-char check because it's trivial but annoying in certain languages. Up to you though whether you want to do input validation (most people leave it out but you can leave it in)
  – HyperNeutrino
  31 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  OK, I will do that.
  – Wais Kamal
  30 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can we take input as a list of digits? (sorry for all the questions)
  – HyperNeutrino
  30 mins ago
  

  
  
  
  

 | 
show 6 more comments

up vote
2
down vote

favorite

A barcode of EAN-13 symbology consists of 13 digits (0-9). The last digit of this barcode is its check digit. It is calculated by the following means (the barcode 8923642469559 is used as an example):

1. 
   

   Starting from the second digit, sum up all alternating digits and multiply the sum by 3:

   
   
   

   8 9 2 3 6 4 2 4 6 9 5 5 9
    | | | | | |
    9 + 3 + 4 + 4 + 9 + 5 = 34
    |
    34 × 3 = 102
   

   
   


2. 
   

   Then, sum up all of the remaining digits, but do not include the last digit:

   
   
   

   8 9 2 3 6 4 2 4 6 9 5 5 9
   | | | | | |
   8 + 2 + 6 + 2 + 6 + 5 = 29
   

   
   


3. 
   

   Add the numbers obtained in steps 1 and 2 together:

   
   
   

   29 + 102 = 131
   

   
   



4. The number you should add to the result of step 3 to get to the next multiple of 10 (140 in this case) is the check digit.

If the check digit of the barcode matches the one calculated as explained earlier, the barcode is valid.

---

More examples:

6537263729385 is valid.
1902956847427 is valid.
9346735877246 is invalid. The check digit should be 3, not 6.

---

Your goal is to write a program that will:

1. Receive a barcode as its input.


2. Check whether the barcode is valid


3. Return 1 (or equivalent) if the barcode is valid, 0 (or equivalent) otherwise.

---

This is code-golf, so the shortest code in terms of bytes wins.

code-golf arithmetic

share|improve this question

edited 21 mins ago

asked 34 mins ago

Wais Kamal

1375

• 
  
  
  

  
  

  
  
  
  
  
  

  
  why did the last digit (9) become a 4
  – HyperNeutrino
  32 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry, will fix it :)
  – Wais Kamal
  31 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I recommend removing the 13-char check because it's trivial but annoying in certain languages. Up to you though whether you want to do input validation (most people leave it out but you can leave it in)
  – HyperNeutrino
  31 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  OK, I will do that.
  – Wais Kamal
  30 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can we take input as a list of digits? (sorry for all the questions)
  – HyperNeutrino
  30 mins ago
  

  
  
  
  

 | 
show 6 more comments

up vote
2
down vote

favorite

up vote
2
down vote

favorite

A barcode of EAN-13 symbology consists of 13 digits (0-9). The last digit of this barcode is its check digit. It is calculated by the following means (the barcode 8923642469559 is used as an example):

1. 
   

   Starting from the second digit, sum up all alternating digits and multiply the sum by 3:

   
   
   

   8 9 2 3 6 4 2 4 6 9 5 5 9
    | | | | | |
    9 + 3 + 4 + 4 + 9 + 5 = 34
    |
    34 × 3 = 102
   

   
   


2. 
   

   Then, sum up all of the remaining digits, but do not include the last digit:

   
   
   

   8 9 2 3 6 4 2 4 6 9 5 5 9
   | | | | | |
   8 + 2 + 6 + 2 + 6 + 5 = 29
   

   
   


3. 
   

   Add the numbers obtained in steps 1 and 2 together:

   
   
   

   29 + 102 = 131
   

   
   



4. The number you should add to the result of step 3 to get to the next multiple of 10 (140 in this case) is the check digit.

If the check digit of the barcode matches the one calculated as explained earlier, the barcode is valid.

---

More examples:

6537263729385 is valid.
1902956847427 is valid.
9346735877246 is invalid. The check digit should be 3, not 6.

---

Your goal is to write a program that will:

1. Receive a barcode as its input.


2. Check whether the barcode is valid


3. Return 1 (or equivalent) if the barcode is valid, 0 (or equivalent) otherwise.

---

This is code-golf, so the shortest code in terms of bytes wins.

code-golf arithmetic

share|improve this question

edited 21 mins ago

asked 34 mins ago

Wais Kamal

1375

A barcode of EAN-13 symbology consists of 13 digits (0-9). The last digit of this barcode is its check digit. It is calculated by the following means (the barcode 8923642469559 is used as an example):

1. 
   

   Starting from the second digit, sum up all alternating digits and multiply the sum by 3:

   
   
   

   8 9 2 3 6 4 2 4 6 9 5 5 9
    | | | | | |
    9 + 3 + 4 + 4 + 9 + 5 = 34
    |
    34 × 3 = 102
   

   
   


2. 
   

   Then, sum up all of the remaining digits, but do not include the last digit:

   
   
   

   8 9 2 3 6 4 2 4 6 9 5 5 9
   | | | | | |
   8 + 2 + 6 + 2 + 6 + 5 = 29
   

   
   


3. 
   

   Add the numbers obtained in steps 1 and 2 together:

   
   
   

   29 + 102 = 131
   

   
   



4. The number you should add to the result of step 3 to get to the next multiple of 10 (140 in this case) is the check digit.

If the check digit of the barcode matches the one calculated as explained earlier, the barcode is valid.

---

More examples:

6537263729385 is valid.
1902956847427 is valid.
9346735877246 is invalid. The check digit should be 3, not 6.

---

Your goal is to write a program that will:

1. Receive a barcode as its input.


2. Check whether the barcode is valid


3. Return 1 (or equivalent) if the barcode is valid, 0 (or equivalent) otherwise.

---

This is code-golf, so the shortest code in terms of bytes wins.

code-golf arithmetic

code-golf arithmetic

share|improve this question

edited 21 mins ago

asked 34 mins ago

Wais Kamal

1375

share|improve this question

edited 21 mins ago

asked 34 mins ago

Wais Kamal

1375

share|improve this question

share|improve this question

edited 21 mins ago

edited 21 mins ago

edited 21 mins ago

asked 34 mins ago

Wais Kamal

1375

asked 34 mins ago

Wais Kamal

1375

asked 34 mins ago

Wais Kamal

1375

1375

• 
  
  
  

  
  

  
  
  
  
  
  

  
  why did the last digit (9) become a 4
  – HyperNeutrino
  32 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry, will fix it :)
  – Wais Kamal
  31 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I recommend removing the 13-char check because it's trivial but annoying in certain languages. Up to you though whether you want to do input validation (most people leave it out but you can leave it in)
  – HyperNeutrino
  31 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  OK, I will do that.
  – Wais Kamal
  30 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can we take input as a list of digits? (sorry for all the questions)
  – HyperNeutrino
  30 mins ago
  

  
  
  
  

 | 
show 6 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  why did the last digit (9) become a 4
  – HyperNeutrino
  32 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry, will fix it :)
  – Wais Kamal
  31 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I recommend removing the 13-char check because it's trivial but annoying in certain languages. Up to you though whether you want to do input validation (most people leave it out but you can leave it in)
  – HyperNeutrino
  31 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  OK, I will do that.
  – Wais Kamal
  30 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can we take input as a list of digits? (sorry for all the questions)
  – HyperNeutrino
  30 mins ago
  

  
  
  
  

why did the last digit (9) become a 4
– HyperNeutrino
32 mins ago

why did the last digit (9) become a 4
– HyperNeutrino
32 mins ago

Sorry, will fix it :)
– Wais Kamal
31 mins ago

Sorry, will fix it :)
– Wais Kamal
31 mins ago

I recommend removing the 13-char check because it's trivial but annoying in certain languages. Up to you though whether you want to do input validation (most people leave it out but you can leave it in)
– HyperNeutrino
31 mins ago

I recommend removing the 13-char check because it's trivial but annoying in certain languages. Up to you though whether you want to do input validation (most people leave it out but you can leave it in)
– HyperNeutrino
31 mins ago

OK, I will do that.
– Wais Kamal
30 mins ago

OK, I will do that.
– Wais Kamal
30 mins ago

Can we take input as a list of digits? (sorry for all the questions)
– HyperNeutrino
30 mins ago

Can we take input as a list of digits? (sorry for all the questions)
– HyperNeutrino
30 mins ago

 | 
show 6 more comments

6 Answers
6

active

oldest

votes

up vote
2
down vote

Ruby, 45 40 bytes

->nn.gsub(/(.(.?))/,'122').sum%10<1

Try it online!

How it works:

Iterate on the string, triplicating every second character. The sum of the ASCII codes of the string modulo 10 is the same as the checksum we are looking for (the difference is 48*7+48*6*3==1200 so it doesn't matter)

share|improve this answer

edited 11 mins ago

answered 25 mins ago

G B

7,2471327

add a comment | 

up vote
1
down vote

05AB1E, 15 10 bytes

εNÉ·>*}OTÖ

Try it online!

Explanation

 ε } # apply to each digit
 NÉ # is the current index odd?
 ·> # double and increment(yielding 1 or 3)
 * # multiply by the current number
 O # sum all modified digits
 TÖ # is evenly divisible by 10

share|improve this answer

edited 4 mins ago

answered 22 mins ago

Emigna

43.9k431133

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Nice answer! Definitely shorter than what I was working on. I like the x you've used in combination with O to sum the entire stack in one go to combine the *3 and + of the two numbers. As well as the (T% to get the remainder.
  – Kevin Cruijssen
  16 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen: The comparison feels a bit surperflous. Can you think of a counterexample to εNÉ·>*}OTÖ working?
  – Emigna
  11 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hmm, not really. It would be useful to have more test cases than the 4 we have now. Can't really think of anything where S13S13∍*OTÖ would fail right now (so well done golfing 4 bytes). It also seems similar as the Python answer (and Jelly as well I think; I can't really read Jelly all that well).
  – Kevin Cruijssen
  7 mins ago
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

Python 3, 55 bytes

lambda s:not sum(int(i)*d for i,d in zip(s,[1,3]*7))%10

Try it online!

share|improve this answer

answered 25 mins ago

HyperNeutrino

18.7k437147

add a comment | 

up vote
0
down vote

Jelly, 14 bytes

D;0×1,3ẋ7¤S%⁵¬

Try it online!

share|improve this answer

edited 5 mins ago

answered 20 mins ago

HyperNeutrino

18.7k437147

add a comment | 

up vote
0
down vote

Perl 6, 29 bytes

:1[map 3×*+*,0,

Try it online!

share

answered 1 min ago

nwellnhof

4,810920

add a comment | 

up vote
0
down vote

Japt -!, 13 bytes

Ë*(Ev ª3Ãx %A

Try it online!

share

edited 7 secs ago

answered 8 mins ago

Luis felipe De jesus Munoz

3,34111049

add a comment | 

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6 Answers
6

active

oldest

votes

6 Answers
6

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

Ruby, 45 40 bytes

->nn.gsub(/(.(.?))/,'122').sum%10<1

Try it online!

How it works:

Iterate on the string, triplicating every second character. The sum of the ASCII codes of the string modulo 10 is the same as the checksum we are looking for (the difference is 48*7+48*6*3==1200 so it doesn't matter)

share|improve this answer

edited 11 mins ago

answered 25 mins ago

G B

7,2471327

add a comment | 

up vote
2
down vote

Ruby, 45 40 bytes

->nn.gsub(/(.(.?))/,'122').sum%10<1

Try it online!

How it works:

Iterate on the string, triplicating every second character. The sum of the ASCII codes of the string modulo 10 is the same as the checksum we are looking for (the difference is 48*7+48*6*3==1200 so it doesn't matter)

share|improve this answer

edited 11 mins ago

answered 25 mins ago

G B

7,2471327

add a comment | 

up vote
2
down vote

up vote
2
down vote

Ruby, 45 40 bytes

->nn.gsub(/(.(.?))/,'122').sum%10<1

Try it online!

How it works:

Iterate on the string, triplicating every second character. The sum of the ASCII codes of the string modulo 10 is the same as the checksum we are looking for (the difference is 48*7+48*6*3==1200 so it doesn't matter)

share|improve this answer

edited 11 mins ago

answered 25 mins ago

G B

7,2471327

Ruby, 45 40 bytes

->nn.gsub(/(.(.?))/,'122').sum%10<1

Try it online!

How it works:

Iterate on the string, triplicating every second character. The sum of the ASCII codes of the string modulo 10 is the same as the checksum we are looking for (the difference is 48*7+48*6*3==1200 so it doesn't matter)

share|improve this answer

edited 11 mins ago

answered 25 mins ago

G B

7,2471327

share|improve this answer

share|improve this answer

edited 11 mins ago

edited 11 mins ago

edited 11 mins ago

answered 25 mins ago

G B

7,2471327

answered 25 mins ago

G B

7,2471327

answered 25 mins ago

G B

7,2471327

7,2471327

add a comment | 

add a comment | 

up vote
1
down vote

05AB1E, 15 10 bytes

εNÉ·>*}OTÖ

Try it online!

Explanation

 ε } # apply to each digit
 NÉ # is the current index odd?
 ·> # double and increment(yielding 1 or 3)
 * # multiply by the current number
 O # sum all modified digits
 TÖ # is evenly divisible by 10

share|improve this answer

edited 4 mins ago

answered 22 mins ago

Emigna

43.9k431133

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Nice answer! Definitely shorter than what I was working on. I like the x you've used in combination with O to sum the entire stack in one go to combine the *3 and + of the two numbers. As well as the (T% to get the remainder.
  – Kevin Cruijssen
  16 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen: The comparison feels a bit surperflous. Can you think of a counterexample to εNÉ·>*}OTÖ working?
  – Emigna
  11 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hmm, not really. It would be useful to have more test cases than the 4 we have now. Can't really think of anything where S13S13∍*OTÖ would fail right now (so well done golfing 4 bytes). It also seems similar as the Python answer (and Jelly as well I think; I can't really read Jelly all that well).
  – Kevin Cruijssen
  7 mins ago
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

05AB1E, 15 10 bytes

εNÉ·>*}OTÖ

Try it online!

Explanation

 ε } # apply to each digit
 NÉ # is the current index odd?
 ·> # double and increment(yielding 1 or 3)
 * # multiply by the current number
 O # sum all modified digits
 TÖ # is evenly divisible by 10

share|improve this answer

edited 4 mins ago

answered 22 mins ago

Emigna

43.9k431133

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Nice answer! Definitely shorter than what I was working on. I like the x you've used in combination with O to sum the entire stack in one go to combine the *3 and + of the two numbers. As well as the (T% to get the remainder.
  – Kevin Cruijssen
  16 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen: The comparison feels a bit surperflous. Can you think of a counterexample to εNÉ·>*}OTÖ working?
  – Emigna
  11 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hmm, not really. It would be useful to have more test cases than the 4 we have now. Can't really think of anything where S13S13∍*OTÖ would fail right now (so well done golfing 4 bytes). It also seems similar as the Python answer (and Jelly as well I think; I can't really read Jelly all that well).
  – Kevin Cruijssen
  7 mins ago
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

05AB1E, 15 10 bytes

εNÉ·>*}OTÖ

Try it online!

Explanation

 ε } # apply to each digit
 NÉ # is the current index odd?
 ·> # double and increment(yielding 1 or 3)
 * # multiply by the current number
 O # sum all modified digits
 TÖ # is evenly divisible by 10

share|improve this answer

edited 4 mins ago

answered 22 mins ago

Emigna

43.9k431133

05AB1E, 15 10 bytes

εNÉ·>*}OTÖ

Try it online!

Explanation

 ε } # apply to each digit
 NÉ # is the current index odd?
 ·> # double and increment(yielding 1 or 3)
 * # multiply by the current number
 O # sum all modified digits
 TÖ # is evenly divisible by 10

share|improve this answer

edited 4 mins ago

answered 22 mins ago

Emigna

43.9k431133

share|improve this answer

share|improve this answer

edited 4 mins ago

edited 4 mins ago

edited 4 mins ago

answered 22 mins ago

Emigna

43.9k431133

answered 22 mins ago

Emigna

43.9k431133

answered 22 mins ago

Emigna

43.9k431133

43.9k431133

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Nice answer! Definitely shorter than what I was working on. I like the x you've used in combination with O to sum the entire stack in one go to combine the *3 and + of the two numbers. As well as the (T% to get the remainder.
  – Kevin Cruijssen
  16 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen: The comparison feels a bit surperflous. Can you think of a counterexample to εNÉ·>*}OTÖ working?
  – Emigna
  11 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hmm, not really. It would be useful to have more test cases than the 4 we have now. Can't really think of anything where S13S13∍*OTÖ would fail right now (so well done golfing 4 bytes). It also seems similar as the Python answer (and Jelly as well I think; I can't really read Jelly all that well).
  – Kevin Cruijssen
  7 mins ago
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Nice answer! Definitely shorter than what I was working on. I like the x you've used in combination with O to sum the entire stack in one go to combine the *3 and + of the two numbers. As well as the (T% to get the remainder.
  – Kevin Cruijssen
  16 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen: The comparison feels a bit surperflous. Can you think of a counterexample to εNÉ·>*}OTÖ working?
  – Emigna
  11 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hmm, not really. It would be useful to have more test cases than the 4 we have now. Can't really think of anything where S13S13∍*OTÖ would fail right now (so well done golfing 4 bytes). It also seems similar as the Python answer (and Jelly as well I think; I can't really read Jelly all that well).
  – Kevin Cruijssen
  7 mins ago
  
  

  
  
  
  

Nice answer! Definitely shorter than what I was working on. I like the x you've used in combination with O to sum the entire stack in one go to combine the *3 and + of the two numbers. As well as the (T% to get the remainder.
– Kevin Cruijssen
16 mins ago

Nice answer! Definitely shorter than what I was working on. I like the x you've used in combination with O to sum the entire stack in one go to combine the *3 and + of the two numbers. As well as the (T% to get the remainder.
– Kevin Cruijssen
16 mins ago

@KevinCruijssen: The comparison feels a bit surperflous. Can you think of a counterexample to εNÉ·>*}OTÖ working?
– Emigna
11 mins ago

@KevinCruijssen: The comparison feels a bit surperflous. Can you think of a counterexample to εNÉ·>*}OTÖ working?
– Emigna
11 mins ago

Hmm, not really. It would be useful to have more test cases than the 4 we have now. Can't really think of anything where S13S13∍*OTÖ would fail right now (so well done golfing 4 bytes). It also seems similar as the Python answer (and Jelly as well I think; I can't really read Jelly all that well).
– Kevin Cruijssen
7 mins ago

Hmm, not really. It would be useful to have more test cases than the 4 we have now. Can't really think of anything where S13S13∍*OTÖ would fail right now (so well done golfing 4 bytes). It also seems similar as the Python answer (and Jelly as well I think; I can't really read Jelly all that well).
– Kevin Cruijssen
7 mins ago

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Python 3, 55 bytes

lambda s:not sum(int(i)*d for i,d in zip(s,[1,3]*7))%10

Try it online!

share|improve this answer

answered 25 mins ago

HyperNeutrino

18.7k437147

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up vote
0
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Python 3, 55 bytes

lambda s:not sum(int(i)*d for i,d in zip(s,[1,3]*7))%10

Try it online!

share|improve this answer

answered 25 mins ago

HyperNeutrino

18.7k437147

add a comment | 

up vote
0
down vote

up vote
0
down vote

Python 3, 55 bytes

lambda s:not sum(int(i)*d for i,d in zip(s,[1,3]*7))%10

Try it online!

share|improve this answer

answered 25 mins ago

HyperNeutrino

18.7k437147

Python 3, 55 bytes

lambda s:not sum(int(i)*d for i,d in zip(s,[1,3]*7))%10

Try it online!

share|improve this answer

answered 25 mins ago

HyperNeutrino

18.7k437147

share|improve this answer

share|improve this answer

answered 25 mins ago

HyperNeutrino

18.7k437147

answered 25 mins ago

HyperNeutrino

18.7k437147

answered 25 mins ago

HyperNeutrino

18.7k437147

18.7k437147

add a comment | 

add a comment | 

up vote
0
down vote

Jelly, 14 bytes

D;0×1,3ẋ7¤S%⁵¬

Try it online!

share|improve this answer

edited 5 mins ago

answered 20 mins ago

HyperNeutrino

18.7k437147

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up vote
0
down vote

Jelly, 14 bytes

D;0×1,3ẋ7¤S%⁵¬

Try it online!

share|improve this answer

edited 5 mins ago

answered 20 mins ago

HyperNeutrino

18.7k437147

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up vote
0
down vote

up vote
0
down vote

Jelly, 14 bytes

D;0×1,3ẋ7¤S%⁵¬

Try it online!

share|improve this answer

edited 5 mins ago

answered 20 mins ago

HyperNeutrino

18.7k437147

Jelly, 14 bytes

D;0×1,3ẋ7¤S%⁵¬

Try it online!

share|improve this answer

edited 5 mins ago

answered 20 mins ago

HyperNeutrino

18.7k437147

share|improve this answer

share|improve this answer

edited 5 mins ago

edited 5 mins ago

edited 5 mins ago

answered 20 mins ago

HyperNeutrino

18.7k437147

answered 20 mins ago

HyperNeutrino

18.7k437147

answered 20 mins ago

HyperNeutrino

18.7k437147

18.7k437147

add a comment | 

add a comment | 

up vote
0
down vote

Perl 6, 29 bytes

:1[map 3×*+*,0,

Try it online!

share

answered 1 min ago

nwellnhof

4,810920

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up vote
0
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Perl 6, 29 bytes

:1[map 3×*+*,0,

Try it online!

share

answered 1 min ago

nwellnhof

4,810920

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up vote
0
down vote

up vote
0
down vote

Perl 6, 29 bytes

:1[map 3×*+*,0,

Try it online!

share

answered 1 min ago

nwellnhof

4,810920

Perl 6, 29 bytes

:1[map 3×*+*,0,

Try it online!

share

answered 1 min ago

nwellnhof

4,810920

share

share

answered 1 min ago

nwellnhof

4,810920

answered 1 min ago

nwellnhof

4,810920

answered 1 min ago

nwellnhof

4,810920

4,810920

add a comment | 

add a comment | 

up vote
0
down vote

Japt -!, 13 bytes

Ë*(Ev ª3Ãx %A

Try it online!

share

edited 7 secs ago

answered 8 mins ago

Luis felipe De jesus Munoz

3,34111049

add a comment | 

up vote
0
down vote

Japt -!, 13 bytes

Ë*(Ev ª3Ãx %A

Try it online!

share

edited 7 secs ago

answered 8 mins ago

Luis felipe De jesus Munoz

3,34111049

add a comment | 

up vote
0
down vote

up vote
0
down vote

Japt -!, 13 bytes

Ë*(Ev ª3Ãx %A

Try it online!

share

edited 7 secs ago

answered 8 mins ago

Luis felipe De jesus Munoz

3,34111049

Japt -!, 13 bytes

Ë*(Ev ª3Ãx %A

Try it online!

share

edited 7 secs ago

answered 8 mins ago

Luis felipe De jesus Munoz

3,34111049

share

share

edited 7 secs ago

edited 7 secs ago

edited 7 secs ago

answered 8 mins ago

Luis felipe De jesus Munoz

3,34111049

answered 8 mins ago

Luis felipe De jesus Munoz

3,34111049

answered 8 mins ago

Luis felipe De jesus Munoz

3,34111049

3,34111049

add a comment | 

add a comment | 

 

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