Is this a valid PZN?

Question

Is this a valid PZN?

A Pharmazentralnummer, or PZN, is an eight-digit code which uniquely identifies medications, medical products, and similar items in the German healthcare system. Examples include ibuprofen tablets, insulin injectors, toothpaste, warm-and-cold compresses, all sorts of things. When a doctor issues a prescription, it's almost always using a PZN, which then is fulfilled by the Apotheke.

The challenge

Write a program or function that takes an eight-digit code as input, and returns a truthy if it is a valid PZN, or falsy if it is not.

A PZN consists of seven code-digits, and a check-digit. Each code-digit is multiplied by a weight based on its position, from 1 to 7, then each value is summed up, and reduced modulo 11. The result is checked against the check-digit. If they are equal, then the PZN is valid.

When reducing the value modulo 11, the result could be 10. In that case, the PZN isn't valid.

Example: PZN 13724857.

The code-digits are 1, 3, 7, 2, 4, 8, 5, and the check-digit is 7.

Step 1: Multiply each code-digit by its weight.

The weight is the position of the code-digit in the PZN, from 1 to 7.

Digit | Weight | Value
1     | 1      | 1*1 = 1
3     | 2      | 3*2 = 6
7     | 3      | 7*3 = 21
2     | 4      | 2*4 = 8
4     | 5      | 4*5 = 20
8     | 6      | 8*6 = 48
5     | 7      | 5*7 = 35

Step 2: Add all the weighted values together, and reduce modulo 11

The total of the weighted values is 1 + 6 + 21 + 8 + 20 + 48 + 35 = 139

Then reduce the total modulo 11: 139 mod 11 = 7

The computed result is the same as our check-digit 7, so the PZN is valid.

Test cases

PZN      | Is valid?
00000000 | True
13724857 | True
08108754 | True
04861880 | True
16086334 | True
01349035 | True
00000001 | False
00000002 | False
06765774 | False
08108759 | False
00100010 | False

Rules

• You can assume that the input will always be a code precisely eight digits long - you don't need to account for grossly invalid inputs (e.g. with letters or spaces).
• However, your solution must accept a code that is precisely eight digits long, including leading zeros.

Finally, this is code-golf - the shortest solution by number of bytes wins!

Answer 1

K (ngn/k), 17 bytes

It takes a digit array.

{*|x=11!+/1+&7#x}
             7#x     First 7 digits
          1+&        a b c d ... ->  1 for a times, 2 for b times, ...
        +/           Inner product between first 7 digits and 1 2 3 4 5 6 7
     11!             Mod 11
 *|x=                Same as the last digit?               

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Answer 2

Nekomata + -e, 8 bytes

7R¢ɔ∙11¦

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7R¢ɔ∙11¦
7R          [1, 2, 3, 4, 5, 6, 7]
  ¢ɔ        Append 10
    ∙       Dot product with the input
     11¦    Check if the result is divisible by 11

Answer 3

Python, 42 bytes

lambda d:int(d[6::-1].hex())//.99%11==d[7]

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Expects input as a byte string.

How?

This abuses the hex method of bytes objects to implement a poor man's base 100 conversion. The old divide by base-1 trick is then used to get partial sums over the digits (we had to reverse their order for this to come out right). To get the sum over digits we would normally have done mod 99 but as we have to do mod 11 in the end anyway we can save it here.

Answer 4

Python 3.8,  50  47 bytes

-1 thanks to l4m2^* leading to two more.

^{Could save at least two bytes by outputting a falsey value for valid and a truthy value for invalid; potentially quite a few using a recursive function.}

lambda d,i=0:sum(v*(i:=i+5)for v in d)%11==d[7]

An unnamed function that accepts the list of digits and returns True or False.

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How?

Call the weighted sum of the first seven digits \$S\$ and the final digit \$h\$.

A PZN is valid iff \$S \equiv h \mod 11\$

We want an equivalent condition using all eight digits in the sum, i.e. \$S+8h\$, so we can do

for v in d

rather than

for v in d[:7]

So we find \$m\$ such that \$m(S+8h) \equiv h \mod 11\$.

Substituting in \$S \equiv h \mod 11\$ we have \$m(h+8h) \equiv h \mod 11\$, so \$9mh \equiv h \mod 11\$ thus \$9m \equiv 1 \mod 11\$, which is satisfied by \$m=5\$ since \$45 \equiv 1 \mod 11\$.

This gives us

lambda d,i=0:sum(v*(i:=i+1)for v in d)*5%11==d[7]

For 49 bytes, as suggested by l4m2.

Now we can move the multiplication inside our weighted sum to give the code at the top.

---

My original 50 byte answer

lambda d,i=0:sum(v*(i:=i+1)+3*d[7]for v in d)%11<1

An unnamed function that accepts the list of digits and returns True or False.

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Note that \$8\times 3 + 8 = 32 \equiv -1 \mod 11 \$

---

^* arnauld also made a fairly trivial 49er that avoids much real trickery (just make the right multiplier for each digit!):

lambda d,i=1:sum(v*~-((i:=i+1)%9)for v in d)%11<1

Answer 5

Uiua ^SBCS, 16 bytes

=◿11/+×⇌¯⇡¯7:°⊂⇌

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Answer 6

JavaScript (ES6), 40 bytes

Expects an array of 8 digits and returns a Boolean value.

a=>!a.reduce(p=>p%11+a[i++]*~(~i%9),i=0)

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Commented

a =>           // a[] = input array
!a.reduce(p => // for each value in a[] with accumulator p:
  p % 11 +     //   reduce p modulo 11
  a[i++] *     //   add the next value from a[] multiplied by
  ~(~i % 9),   //   1 .. 7 for the first 7 elements
               //   or -1 for the last one
  i = 0        //   starting with p = 0 and i = 0
)              // end of reduce() -> turned into a Boolean

Answer 7

05AB1E, 8 bytes

¨ƶO11%Qθ

Input as a list of digits.

Try it online or verify all test cases.

Explanation:

¨         # Remove the last digit of the (implicit) input-list
 ƶ        # Multiply each value by their 1-based index
  O       # Sum
   11%    # Modulo-11
      Q   # Check for each digit in the (implicit) input-list whether it equals this
       θ  # Pop and only leave the last check
          # (which is output implicitly as result)

Answer 8

APL+WIN, 23 bytes

Prompts for number as a string of digits.

(7↓n)=11|+/(⍳7)×7↑n←⍎¨⎕

Try it online! Thanks to Dyalog Classic

Answer 9

MATL, 12 11 bytes

0&)7:*s11\=

Try it online! Or verify all test cases.

Explanation

       % Implicit input: numeric vector
0&)    % 2-output indexing (modular, 1-based) with 0: pushes the last element,
       % then a vector with the remaining elements
7:     % Push [1 2 ... 7]
*      % Multiply, element-wise
s      % Sum
11\    % Modulo 11
=      % Equal?
       % Implicit display

Answer 10

Python 3, 60 bytes

lambda a:sum([x*-~i for i,x in enumerate(a[:-1])])%11==a[-1]

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Answer 11

C (gcc), 51 bytes

s,i;f(int*l){for(i=s=0;i<7;)s+=*l++*++i;s=s%11-*l;}

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function returning true if invalid, false if valid PZN code.

Answer 12

PowerShell Core, 43 bytes

$args[0..6]|%{$s+=$_*++$i}
$s%11-eq$args[7]

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Answer 13

Japt -h, 11 bytes

Takes input as a digit array.

ǶUxÈ*ÒYÃuB

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Answer 14

JavaScript (Node.js), 40 bytes

a=>1>a.reduce((p,v,i)=>p-~i*v,a[7]*2)%11

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From Arnauld's. 8th digit * 8 + 8th digit * 2 = 8th digit * (-1), modulo 11

Answer 15

Wolfram Language (Mathematica), 31 bytes

Mod[Most@#.Range@7,11]==Last@#&

Try all test cases online! Unnamed function taking a list of eight digits as input and returning True or False.

Answer 16

Google Sheets, 58 bytes

=SORT(MOD(A1,10)=MOD(SUM(MID(A1,ROW(1:7),1)*ROW(1:7)),11))

Answer 17

TI-BASIC, 29 bytes

Input A
ʟA(8)=remainder(Σ(IʟA(I),I,1,7),11

This program takes an array of digits as inputs, and outputs 1 (for true) or 0 (for false). It uses a summation to multiply the digits by their weights and add them together. It then checks if the result modulo eleven is equal to the last digit.

Example input and output:

Answer 18

Jelly, 8 bytes

ṖḋJ%11⁼Ṫ

A monadic Link that accepts the list of eight digits and yields 1 (truthy) if a valid PZN or 0 (falsey) otherwise.

Try it online! Or see the test-suite.

How?

Direct implementation of the specification.

ṖḋJ%11⁼Ṫ - Link: list of eight digits, A = [a,b,c,d,e,f,g,h]
Ṗ        - pop {A} -> P = [a,b,c,d,e,f,g]
  J      - indices {A} -> I = [1,2,3,4,5,6,7,8]
 ḋ       - {P} dot product {I} -> S = 1×a+2×b+3×c+4×d+5×e+6×f+7×g+8×0
   %11   - {S} mod 11
       Ṫ - tail {A} -> h
      ⁼  - {S mod 11} equals {h}?

Answer 19

R, 28 24 bytes

\(d)sum(d*c(1:7,-1))%%11

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Takes input as a vector of digits. Outputs swapped truthy (non-zero) / falsy (zero).

Answer 20

Common Lisp, 92 bytes

Input is a list of numbers

(defun p(n)(=(mod(reduce'+(mapcar(lambda(x y)(* x y))n'(1 2 3 4 5 6 7 0)))11)(car(last n))))

Answer 21

Arturo, 42 bytes

$->a[0x:∑map a\[0..6]=>[&*<=1+]a\7=x%11]

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Explanation

$->a[]            ; a function taking a list a
0                 ; push zero, our multiplier, to stack
x:                ; let x be...
∑                 ; sum of...
map a\[0..6]=>[]  ; first seven numbers of input mapped to...
&                 ; current digit...
*                 ; multiplied by...
<=                ; extra copy of...
1+                ; top of stack plus one
a\7=              ; is the last number of input equal to...
x%11              ; x modulo eleven?

Answer 22

Charcoal, 16 bytes

⁼﹪Σ⭆…θ⁷×ι⊕κ¹¹﹪Ｎχ

Try it online! Takes input as a string and outputs a Charcoal boolean, i.e. - for a valid PZN, nothing if not. Explanation:

     θ              Input string
    …               Truncated to length
      ⁷             Literal integer `7`
   ⭆                Map over characters and join
        ι           Current character
       ×            Repeated by
          κ         Current index
         ⊕          Incremented
  Σ                 Take the digital sum
 ﹪                  Modulo
           ¹¹       Literal integer `11`
⁼                   Is equal to
              Ｎ     Input as a number
             ﹪      Modulo
               χ    Predefined variable `10`
                    Implicitly print

13 bytes by taking input as an array of integers:

⁼⊟θ﹪ΣＥθ×ι⊕κ¹¹

Try it online! Link is to verbose version of code. Explanation:

  θ             Input array
 ⊟              Remove last element
⁼               Equals
      θ         Remainder of input array
     Ｅ          Map over elements
        ι       Current element
       ×        Multiplied by
          κ     Current index
         ⊕      Incremented
    Σ           Take the sum
   ﹪            Modulo
           ¹¹   Literal integer `11`
                Implicitly print

Answer 23

J, 15 bytes

{:=11|1#.7$[*#\

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{:=11|1#.7$[*#\
             #\   NB. 1,2,3,...,8
            *     NB. multiplication
           [      NB. input digit array
         7$       NB. first 7 digits
      1#.         NB. sum
   11|            NB. mod 11
  =               NB. equals
{:                NB. last digit of input

Answer 24

AWK, 36 bytes

{for(;++i<NF;)s+=$i*++x}$0=s%11==$NF

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{for(;++i<NF;) # each digit
s+=            # sum
$i*++x}        # digit times location
$0=            # set output
s%11==$NF      # sum mod 11 = last digit

Prints 1 if valid or nothing.

Answer 25

Perl 5 -MList::Util=sum -pl, 34 bytes

$_=(pop@F)==(sum map$_*++$i,@F)%11

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Answer 26

Haskell, 41 bytes

f x=mod(x!!7*2+sum(zipWith(*)[1..]x))11<1

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