Calculate ISBN-13 check digit

Question

Write a function which, given the first 12 digits of an ISBN-13 code, will calculate the entire ISBN via calculating and appending an appropriate check digit.

Your function's input is a string containing the first 12 digits of the ISBN. Its output is a string containing all 13 digits.

Formal specification

Write a function which, when given a string s consisting entirely of exactly 12 decimal digits (and no other characters), returns a string t with the following properties:

• t consists of exactly 13 decimal digits (and no other characters);
• s is a prefix of t;
• the sum of all digits in odd positions in t (i.e. the first, third, fifth, etc.), plus three times the sum of all digits in even positions in t (i.e. the second, fourth, sixth, etc), is a multiple of 10.

Example / test case

Input
978030640615

Output
9780306406157

Victory condition

As a code-golf challenge, the shortest answer wins.

Answer 1

Golfscript - 25 chars

{...+(;2%+{+}*3-~10%`+}:f

Whole program version is only 19 chars

...+(;2%+{+}*3-~10%

Check back here for analysis later. Meanwhile check out my old uninspired answer

Golfscript - 32 chars

Similar to the luhn number calculation

{.{2+}%.(;2%{.+}%+{+}*~)10%`+}:f

Analysis for 978030640615

{...}:f this is how you define the function in golfscript
.       store an extra copy of the input string
        '978030640615' '978030640615'
{2+}%   add 2 to each ascii digit, so '0'=>50, I can get away with this instead
        of {15&}% because we are doing mod 10 math on it later
        '978030640615' [59 57 58 50 53 50 56 54 50 56 51 55]
.       duplicate that list
        '978030640615' [59 57 58 50 53 50 56 54 50 56 51 55] [59 57 58 50 53 50 56 54 50 56 51 55]
(;      trim the first element off
        '978030640615' [59 57 58 50 53 50 56 54 50 56 51 55] [57 58 50 53 50 56 54 50 56 51 55]
2%      select every second element
        '978030640615' [59 57 58 50 53 50 56 54 50 56 51 55] [57 50 50 54 56 55]
{.+}%   double each element by adding to itself
        '978030640615' [59 57 58 50 53 50 56 54 50 56 51 55] [114 100 100 108 112 110]
+       join the two lists together
        '978030640615' [59 57 58 50 53 50 56 54 50 56 51 55 114 100 100 108 112 110]
{+}*    add up the items in the list
        '978030640615' 1293
~       bitwise not
        '978030640615' -1294
)       add one
        '978030640615' -1293            
10%     mod 10
        '978030640615' 7
`       convert to str
        '978030640615' '7'
+       join the strings
        '9780306406157'

Answer 2

Python - 44 chars

f=lambda s:s+`-sum(map(int,s+s[1::2]*2))%10`

Python - 53 chars

def f(s):d=map(int,s);return s+`-sum(d+d[1::2]*2)%10`

Answer 3

Haskell - 54 characters

i s=s++show(sum[-read[c]*m|c<-s|m<-cycle[1,3]]`mod`10)

This requires support for parallel list comprehensions, which is supported by GHC (with the -XParallelListComp flag) and Hugs (with the -98 flag).

Answer 4

APL (27 characters)

F←{⍵,⍕10|10-(12⍴1 3)+.×⍎¨⍵}

I'm using Dyalog APL as my interpreter. Here's a quick explanation, mostly from right to left (within the function definition, F←{ ... }):

• ⍎¨⍵: Execute/evaluate (⍎) each (¨) character given in the right argument (⍵).
• (12⍴1 3): Reshape (⍴) the vector 1 3 into a 12-element vector (repeating to fill the gaps).
• +.×: Take the dot product (+.×) of its left argument ((12⍴1 3)) and its right argument (⍎¨⍵).
• 10-: Subtract from 10.
• 10|: Find the remainder after division by 10.
• ⍕: Format the number (i.e., give a character representation).
• ⍵,: Append (,) our calculated digit to the right argument.

Answer 5

PHP - 86 85 82 chars

function c($i){for($a=$s=0;$a<12;)$s+=$i[$a]*($a++%2?3:1);return$i.(10-$s%10)%10;}

Re-format and explanation:

function c($i){                     // function c, $i is the input

    for($a=$s=0;$a<12;)             // for loop x12 - both $a and $s equal 0
                                    // notice there is no incrementation and
                                    // no curly braces as there is just one
                                    // command to loop through

        $s+=$i[$a]*($a++%2?3:1);    // $s (sum) is being incremented by
                                    // $ath character of $i (auto-casted to
                                    // int) multiplied by 3 or 1, depending
                                    // wheter $a is even or not (%2 results
                                    // either 1 or 0, but 0 == FALSE)
                                    // $a is incremented here, using the
                                    // post-incrementation - which means that
                                    // it is incremented, but AFTER the value
                                    // is returned

    return$i.(10-$s%10)%10;         // returns $i with the check digit
                                    // attached - first it is %'d by 10,
                                    // then the result is subtracted from
                                    // 10 and finally %'d by 10 again (which
                                    // effectively just replaces 10 with 0)
                                    // % has higher priority than -, so there
                                    // are no parentheses around $s%10
}

Answer 6

Windows PowerShell, 57

filter i{$_+(990-($_-replace'(.)(.)','+$1+3*$2'|iex))%10}

Answer 7

Haskell, 78 71 66 characters

i s=s++(show$mod(2-sum(zipWith(*)(cycle[1,3])(map fromEnum s)))10)

Answer 8

Ruby - 73 65 chars

f=->s{s+((2-(s+s.gsub(/.(.)/,'\1')*2).bytes.inject(:+))%10).to_s}

Answer 9

Python - 91, 89

0123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890
|         |         |         |         |         |         |         |         |         |
 def c(i):return i+`(10-(sum(int(x)*3for x in i[1::2])+sum(int(x)for x in i[::2]))%10)%10`

Answer 10

C# (94 characters)

string I(string i){int s=0,j=0;for(;j<12;)s+=(i[j]-48)*(j++%2<1?1:3);return i+((10-s%10)%10);}

With linebreaks/whitespace for readability:

string I(string i) 
{ 
    int s = 0, j = 0;
    for (; j < 12; )
        s += (i[j] - 48) * (j++ % 2 < 1 ? 1 : 3); 
    return i + ((10 - s % 10) % 10); 
}

Tested on several ISBNs from books on my shelf, so I know it's working!

Answer 11

J - 55 45 38

f=:3 :'y,":10|10-10|+/(12$1 3)*"."0 y'

eg

f '978030640615'
9780306406157

old way:

f=:,":@(10(10&|@-)(10&|@+/@((12$1 3)*(i.12)&(".@{))))

Answer 12

Perl, 53 chars

sub i{$_=shift;s/(.)(.)/$s+=$1+$2*3/ge;$_.(10-$s)%10}

Answer 13

C# – 89 77 characters

string I(string s){return s+(9992-s.Sum(x=>x-0)-2*s.Where((x,i)=>i%2>0).Sum(x=>x-0))%10;}

Formatted for readability:

string I(string s)
{
    return s +
            (9992
            - s.Sum(x => x - 0)
            - 2 * s.Where((x, i) => i%2 > 0).Sum(x => x - 0)
            ) % 10;
}

We do not multiply by one or three, we just add everything, plus we add all even-placed characters one more time, multiplied by two.

9992 is large enough so that the sum of all ASCII characters is less than that (so that we may mod by 10 and be sure the result is positive, no need to mod by 10 twice), and is not divisible by zero because we add up all those extra 2*12*48 (twelve ASCII digits, weighed by 1 and 3) == 1152, which allows us to spare one extra character (instead of twice subtracting 48, we subtract 0 just to convert from char to int, but instead of 990, we need to write 9992).

But then again, even though much less beautiful ;-), this old-school solution gets us to 80 characters (but this is almost C-compatible):

string T(string i){int s=2,j=0;for(;j<12;)s+=i[j]*(9-j++%2*2);return i+s%10;}

Answer 14

Ruby - 80 characters

def f s;s+(10-s.bytes.zip([1,3]*6).map{|j,k|(j-48)*k}.inject(:+)%10).to_s[0];end

Answer 15

Q, 36 chars

{x,-3!10-mod[;10]sum(12#1 3)*"I"$'x}

Answer 16

dc, 44 chars

[d0r[I~3*rI~rsn++lndZ0<x]dsxx+I%Ir-I%rI*+]sI

Invoke as lIx, e.g:

dc -e'[d0r[I~3*rI~rsn++lndZ0<x]dsxx+I%Ir-I%rI*+]sI' -e '978030640615lIxp'

Answer 17

D - 97 characters

auto f(string s){int n;foreach(i,c;s)n+=((i&1)*2+1)*(c-48);return s~cast(char)((10-n%10)%10+48);}

Formatted more legibly:

auto f(string s)
{
    int n;

    foreach(i, c; s)
        n += ((i & 1) * 2 + 1) * (c - 48);

    return s ~ cast(char)((10 - n % 10) % 10 + 48);
}

The verbosity of D's cast operator definitely makes it harder to write obsessively short code though.

Answer 18

Java - 161 Characters :(

int b[]=new int[a.length];
int d=0,n=0,j=1;
for(char c:a.toCharArray())b[d++]=Integer.valueOf(c+"");
for(int i:b)n+=(j++%2==0)?(i*3):(i*1);
return a+(10-(n%10));

Answer 19

Perl 6, 29 bytes

{$_~-:1[.comb «*»(1,3)]%10}

Try it online!

Answer 20

Python 2, 78 76 bytes

lambda n:n+`10-(sum(int(a)+3*int(b)for a,b in zip(n[::2],n[1::2]))%10or 10)`

Try it online!

Takes a string as an argument.

Explanation:

Using python slice notation, converts a string into a list of character pairs. ("978030640615" -> [("9","7"), ("8", "0"), ("3", "0"), ("6", "4"), ("0", "6"), ("1", "5")] )

For that list of pairs, converts each item into an integer and returns a+3b.

Sums all the results.

Gets the sum modulo 10, OR 10 if the remainder is 0. (This prevents the final digit from being 10 instead of 0.)

Removes the remainder from 10 to get the check digit.

Converts the calculated check digit to a string via deprecated backtick expression.

Returns the original number plus the calculated check digit.

Edit:

Saved 2 byes by removing spaces (thanks Jo King!).

Answer 21

APL (Dyalog Unicode), 18 bytes^SBCS

Anonymous tacit prefix function taking string as argument. Using Bubbler's approach.

⊢,∘⍕10|⍎¨+.×9 7⍴⍨≢

Try it online!

≢ length of argument (12)

9 7⍴⍨ cyclically reshape [9,7] to that length

+.× dot product of the following with that:

⍎¨ `evaluate each character

10| mod-10 of that

,∘⍕ prepend the following the the stringification of that:

⊢ the unmodified argument

Answer 22

dc, 25 bytes

dn[A~9z^8+*rd0<M+]dsMxA%p

Try it online!

I know there's already a dc answer here, but 25<44 so I guess I feel 19 bytes of okay about it. This uses the fact that 8+9^z is equivalent to either -3 or -1 mod 10 depending on whether z is even or odd. So I use A~ to break the number into digits on the stack, but as I build the stack I multiply each digit by 8+9^z where z is the current stack size. Then I add them all as the function stack unrolls, and print the last digit.

Answer 23

Q (44 chars)

f:{x,string 10-mod[;10]0+/sum@'2 cut"I"$/:x}

Answer 24

Scala 84

def b(i:String)=i+(10-((i.sliding(2,2).map(_.toInt).map(k=>k/10+k%10*3).sum)%10)%10)

Testing:

val isbn="978030640615"
b(isbn)

Result:

"9780306406157"

Answer 25

C, 80 79 characters

The function modifies the string in place, but returns the original string pointer to satisfy the problem requirements.

s;char*f(char*p){for(s=2;*p;s+=7**p++)s+=9**p++;*p++=48+s%10;*p=0;return p-13;}

Some explanation: Instead of subtracting 48 (the ASCII value of the digit 0) from each input character, the accumulator s is initialized so that it is modulo 10 equal to 48+3*48+48+3*48...+48+3*48 = 24*48 = 1152. The step 10-sum can be avoided by accumulating s by subtraction instead of addition. However, the module operator % in C would not give a usable result if s was negative, so instead of using s-= the multipliers 3 and 1 are replaced by -3=7 modulo 10 and -1=9 modulo 10, respectively.

Test harness:

#include <stdio.h>
#define N 12
int main()
{
     char b[N+2];
     fgets(b, N+1, stdin);
     puts(f(b));
     return 0;
}

Answer 26

Groovy 75, 66 chars

i={int i;it+(10-it.inject(0){t,c->t+(i++&1?:3)*(c as int)}%10)%10}

use:

String z = "978030640615"
println i(z)

-> 9780306406157

Answer 27

APL (25)

{⍵,⍕10-10|+/(⍎¨⍵)×12⍴1,3}

Answer 28

Tcl, 79

proc f c {lmap a\ b [split $c {}] {incr d [expr $a+3*$b]};list $c[expr -$d%10]}

Answer 29

Tcl, 87 bytes

proc C s {lmap c [split $s ""] {incr t [expr [incr i]%2?$c:3*$c]}
list $s[expr -$t%10]}

Try it online!

Answer 30

Tcl, 64 bytes

proc C s {list $s[expr ([regsub -all (.)(.) $s {-\1-3*\2}])%10]}

Try it online!

Answer 31

Husk, 16 bytes

S:(→sΣz¤*or;¢"97

Try it online!

Explanation

Instead of subtracting the sum from 10 I use negated coefficients, the rest is pretty self-explanatory:

S:(              )  -- apply the following function and append result to itself:
   →s               -- mod 10 / last digit
     Σ              -- sum
      z      ¢"97   -- zip input with ['9','7','9','7'...] under:
       ¤ (r;)       --   convert both chars to number, then
        *           --   multiply

Answer 32

Stax, 11 10 bytes

╣Γ♠☼ù≤m╒X₧

Run and debug it

This is a complete program instead of a function, as stax does not implement functions.

Unpacked, ungolfed, and commented, it looks like this.

q   print input without popping and without newline
E   explode into array of decimal digits
97E [9,7] (this will become the repetition counts)
:B  repeat elements; repetition counts wrap around to cover all input digits
|+  get sum of array with repeated digts
A%  modulus 10
P   print

Run this one

Answer 33

J, 21 bytes

,10":@|1#."."+*9 7$~#

Try it online!

How it works

,10":@|1#."."+*9 7$~#
                    #  Take length of input
               9 7$~   Repeat 9 7 that many times
          "."+*        Multiply (elementwise) to the digits
       1#.             Sum
 10   |                Modulo 10
   ":@                 Format to string
,                      Append to the end of the input string

The negative of a + 3b + c + 3d + ... modulo 10 is equal to 9a + 7b + 9c + 7d + ... modulo 10.

Answer 34

Powershell, 55 53 52 bytes

Inspired by Joey's answer. I'm impressed with her regex.

"$args"+($args-replace'(.)(.)','+40-$1-3*$2'|iex)%10

Powershell, 55 bytes, whithout regex

param($s)$s[1,3,5,7,9,11*2+0..11]|%{$r+=58-$_};$s+$r%10

Explanation:

• the index range = (the even positions 1,3,5,7,9,11 repeated 2 times + all positions 0..11)
• $r+=58-$_ makes a sum of supplements up to 10+[char]'0'
• $s+$r%10 outputs the result

Test script:

$f = {

param($s)$s[1,3,5,7,9,11*2+0..11]|%{$r+=58-$_};$s+$r%10

}

@(
    ,("978030640615" , "9780306406157")

) | % {
    $s,$expected = $_
    $result = &$f $s
    "$($result-eq$expected)"
    $s
    $result
}

Output:

True
978030640615
9780306406157

Answer 35

MBASIC, 106 bytes

1 INPUT T$:FOR I=1 TO 12 STEP 2:O=O+VAL(MID$(T$,I,1)):E=E+3*VAL(MID$(T$,I+1,1)):NEXT:PRINT 10-(O+E) MOD 10

It ain't pretty, but it works.

? 978030640615
7

Answer 36

APL(NARS), chars 38, bytes 76

{⍵,10∣10-10∣+/w,2×w[v/⍨∼2∣v←⍳≢w←⍎¨⍕⍵]}

I had some other test for see better and get definition from Wikipedia, and examples in the page: https://isbn-information.com/check-digit-for-the-13-digit-isbn.html test:

  h←{⍵,10∣10-10∣+/w,2×w[v/⍨∼2∣v←⍳≢w←⍎¨⍕⍵]}
  h 978186197271
978186197271 2 
  h 978168197271
978168197271 8 
  h 978186197371      
978186197371 9 
  h 978186197375      
978186197375 7 

the formula i use would be in one not APL language if w is the argument =>

f(w)==w,(10-(sum(digit(w),2*digitsEVENINDICES(w))mod 10))mod 10

comment:

{⍵,10∣10-10∣+/w,2×w[v/⍨∼2∣v←⍳≢w←⍎¨⍕⍵]}
                   v/⍨∼2∣v←⍳≢w←⍎¨⍕⍵] call w the array of digits of ⍵, return even index
               2×w[                  consider all the digit even of w and multipy for 2
             w,                      put above that array, the array w
           +/                        sum all
      10-10∣                          10 - (above sum)mod 10
    10∣                               above mod 10
 {⍵,                                 add as digit to the number of argument, and return all that

Answer 37

Kotlin, 120 108 83 77 69 bytes

{s->s+(10-(s.map{it.toInt()-48}.chunked(2).sumBy{it[0]+3*it[1]})%10)}

Try it online!

Thanks to @ovs for removing 8 bytes!!

Answer 38

><>, 24 bytes

0{:9)68**-:nl2%2*7+*+a%!

Try it online!

Explanation

0                      !  # initiates sum (final digit) as 0
 {                        # get the next digit
  :9)68**-                # convert from char to int for any iteration except the last
          :n              # print the digit
            l2%2*7+*      # multiply the digit by either 9 or 7 alternately
                    +     # add to sum
                     a%   # mod sum by 10

Answer 39

JavaScript, 57 bytes

f=s=>s+(10-s.split``.reduce((a,b,i)=>a+b*(3-i%2*2),0)%10)

You can invoke it like this:

f("978030640615")

Effectively, it goes through each (digit, index) pair (b, i), and works out i % 2 (is i odd?), multiplies it by 2, and subtracts it from 3. Take a look at this:

For odd numbers:

• (i % 2) is 1
• ((i % 2) * 2) is 2
• 3 - ((i % 2) * 2) is 1

For even numbers:

• (i % 2) is 0
• ((i % 2) * 2) is 0
• 3 - ((i % 2) * 2) is 3

And so we have now generated the correct multipliers. From there it is simple, just add the current accumulator value, initialised at 0, to the current character multiplied by this generated multiplier. JavaScript automatically casts string to numbers when you do this multiplication.

Calculating the last digit is done by taking the last digit (by computing the value modulo 10), and subtracting this from 10.

Answer 40

MATLAB - 82 chars

function c(i)
[i num2str(mod(10-mod(sum(str2num(i(:)).*repmat([1;3],6,1)),10),10))]

Answer 41

R, 147 characters

f=function(v){s=as.numeric(strsplit(v,"")[[1]]);t=0;for(i in 1:12)if(i%%2==0)t=t+s[i]*3 else t=t+s[i];paste(v,(10-(t%%10))%%10,collapse="",sep="")}

Usage:

f("978030640615")
[1] "9780306406157"

Answer 42

Python

>>>r=input()
>>>a=0
>>>for x in r[::2]:
    a+=int(x)
>>>for x in r[1::2]:
    a+=int(x)*3
>>>a=(10-(a%10))%10
print(r+str(a))

Answer 43

J, 25

,[:":10|0(-+/)"."0*1 3$~#

   f =: ,[:":10|0(-+/)"."0*1 3$~#
   f '978030640615'
9780306406157

Answer 44

Javascript: 73 71 chars

function _(b){for(i=s=0;12>i;)s+=((i*2&2)+1)*b[i++];return b+(10-s%10)}

Usage:

_('978030640615') //"9780306406157"

Answer 45

Clojure 264 -> 241 characters:

(defn s [a b] (let [c (count a)] (if (= 0 c) b (s (rest a) (+ b (* (if (= 0 (m c 2)) 1 3) (first a)))))))
(defn m [a b] (let [f (mod a b)] (if (= f 10) 0 f)))
(defn c [a] (str a (m (- 10 (m (s (map #(read-string %) (map str a)) 0) 10)) 10)))

Usage

(c "978030640615")

Non minified:

(defn sum [collection total]
  (let [len (count collection)]
    (if (= 0 len)
         total
         (sum (rest collection)
           (+ total
            (* (if (= 0 (mod len 2)) 1 3)
             (first collection)))))))

(defn mods [a b]
  (let [f (mod a b)]
    (if (= f 10) 
      0
      f)))


(defn create [isbn] 
  (str isbn
       (mods (- 10 (mods (sum (map #(read-string %) (map str isbn)) 0) 10)) 10))) 

Usage:

(create "978030640615")

New to Clojure so I have to believe that can be knocked down.