Calculate ISBN-13 check digit

Question

Write the shortest code that can calculate the check digit for an ISBN-13 from the first 12 digits.

Requirements:

• Must be a function that accepts a string
• Returns the full ISBN-13 as a string
• Uses the correct algorithm to calculate the digit (shown on the Wikipedia page)

Input
978030640615

Output
9780306406157

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

Haskell, 78 71 66 characters

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

Answer 7

Ruby - 73 65 chars

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

Answer 8

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 9

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 10

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 11

J - 55 45

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

eg

f '978030640615'
9780306406157

old way:

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

Answer 12

Windows PowerShell, 56 67

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

Answer 13

Perl, 53 chars

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

Answer 14

Q, 36 chars

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

Answer 15

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 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

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 18

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 19

Q (44 chars)

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

Answer 20

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 21

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 22

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 23

MATLAB - 82 chars

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

Answer 24

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 25

J, 25

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

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

Answer 26

APL (25)

{⍵,⍕10-10|+/(⍎¨⍵)×12⍴1,3}

Answer 27

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 28

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 29

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.