Primenary Strings

Question

A Primenary (binary-prime) string is one which, when written as a binary grid, every row and column has a prime total.

That's quite a vague explanation, so let's break it down with a worked example...

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For this example we'll use the string bunny:

First, find the ASCII code point of each character and its binary representation:

Char | ASCII | Binary

b      98      1100010
u      117     1110101
n      110     1101110
n      110     1101110
y      121     1111001

Take these binary values, from top to bottom, and arrange them into grid (adding leading zeros if necessary):

1 1 0 0 0 1 0
1 1 1 0 1 0 1
1 1 0 1 1 1 0
1 1 0 1 1 1 0
1 1 1 1 0 0 1

Then, count the number of 1s in each row and column:

1 1 0 0 0 1 0   > 3
1 1 1 0 1 0 1   > 5
1 1 0 1 1 1 0   > 5
1 1 0 1 1 1 0   > 5
1 1 1 1 0 0 1   > 5

v v v v v v v

5 5 2 3 3 3 2

If, and only if, every single total is prime (such as here) then the string is a valid binary-prime.

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The Challenge

Your task is to create a function or program which, when given a string, returns/outputs truthy if the string is primenary, and falsy otherwise.

Rules/Details

• You may assume that the string's characters will always be in the ASCII range 33-126 (inclusive).
• The string will not be empty.
• A primenary string does not have to have a prime length - for example, W1n* is valid, despite having 4 characters.
• This is code-golf, so the shortest answer (in bytes) wins - but all submissions are welcome.
• Standard loopholes are banned.

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Test Cases

'husband'     -> True
'HOTJava'     -> True
'COmPaTIBILE' -> True
'AuT0HACk'    -> True

'PPCW'        -> False
'code-golf'   -> False
'C++'         -> False
'/kD'         -> False

'HI'          -> False
'A'           -> False

There is also a working, but incredibly verbose Python example on repl.it that you can test your solution against.

Answer 1

MATL, 10 bytes

BtXsw!shZp

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This is the ideal language for the job. It's pretty much a literal transliteration of the challenge specification.

Bt % Converts input to binary matrix, duplicate
Xs  % Sum columns (alternative X version to prevent defaulting to sum along first non-singleton dimension, thanks @Jonathan Allan)
w! % Get the duplicate to the top of the stack, transpose
s  % Sum again
h  % Concatenate horizontally
Zp % Check primality element-wise. Display implicitly.

Since any zero makes a MATL array falsy as per meta, nothing else is needed - basically, an implicit A is called on ? (if).

Answer 2

Jelly, 13 12 11 bytes

OBUZ;$S€ÆPẠ

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How?

OBUZ;$S€ÆPẠ - Main link: word                  e.g. ha!
O           - cast to ordinals                 e.g. [104,97,33]
 B          - convert to binary                e.g. [[1,1,0,1,0,0,0],[1,1,0,0,0,0,1],[1,0,0,0,0,1]]
  U         - reverse each entry (say "b")     e.g. [[0,0,0,1,0,1,1],[1,0,0,0,0,1,1],[1,0,0,0,0,1]]
     $      - last two links as a monad
   Z        - transpose                        e.g. [[0,1,1],[0,0,0],[0,0,0],[1,0,0],[0,0,0],[1,1,1],[1,1]]
    ;       - concatenate with "b"             e.g. [[0,1,1],[0,0,0],[0,0,0],[1,0,0],[0,0,0],[1,1,1],[1,1],[0,0,0,1,0,1,1],[1,0,0,0,0,1,1],[1,0,0,0,0,1]]
      S€    - sum €ach                         e.g. [2,0,0,1,0,3,2,3,3,2]
        ÆP  - is prime (1 if prime, 0 if not)  e.g. [1,0,0,0,0,1,1,1,1,1]
          Ạ - all truthy?                      e.g. 0

Answer 3

Mathematica, 75 bytes

And@@Join@@PrimeQ@{+##&@@#,+##&@@@#}&@IntegerDigits[ToCharacterCode@#,2,7]&

Unnamed function taking a string as input and returning True or False.

ToCharacterCode@# converts the input into the list of its ASCII values; IntegerDigits[...,2,7] turns each value into the list of its bits, padded to length 7 if necessary. So now we have a 2D array and we want all its row sums and column sums; lo and behold, the character-spasm {+##&@@#,+##&@@@#}&@... does exactly that (it applies the +##&, "sum all arguments", function to the list of vectors in the first coordinate using @@, and to each vector as its own list of integers in the second coordinate using @@@). Then we just check whether the results are PrimeQ, flatten the list with Join@@, and take the And of all those values.

Answer 4

05AB1E, 17 bytes

Çžz+b€¦€SDøìO0ÛpP

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Answer 5

Jelly, 15 bytes

O+⁹Bṫ€3µS€;SÆPP

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Explanation

O+⁹Bṫ€3µS€;SÆPP  Main link. Input: string z
O                Ordinal, get ASCII value of each char
  ⁹              Nilad representing 256
 +               Add 256 to each ordinal value
   B             Binary digits of each
    ṫ€3          Tail, take each list of digits from the 3rd value to the end
                 These are the last seven digits of each
       µ         Start a new monadic chain
        S€       Sum each list of digits by rows
           S     Sum by column
          ;      Concatenate
            ÆP   Test if each is prime, 1 if true else 0
              P  Product

Answer 6

Ruby -rprime, 100 bytes

->s{a=s.bytes.map{|b|b.digits 2}
a.all?{|r|r.sum.prime?}&([0]*7).zip(*a).all?{|c|c.count(1).prime?}}

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Explanation

->s{
    a=s.bytes                       # Get byte values from string
             .map{|b|b.digits 2}    # For each, map it to its binary digits
                                    #   (least significant digits first)
a.all?{|r|r.sum.prime?}             # Each character has a prime number of 1's?
    &                               # Bit-and (because it saves bytes here)
    ([0]*7).zip(*a)                 # Zip bit array with an all-zero array
                                    #   (If we don't, then uneven array lengths
                                    #   cause some columns to not be returned.)
    .all?{|c|c.count(1).prime?}     # All columns have a prime number of 1's?
                                    #   (We use count instead of sum here because
                                    #   zip pads columns with trailing nils, which
                                    #   can't be added to numbers via sum.)
}

Answer 7

Python 3, 209 189 180 171 160 bytes

Thanx squid for -9 bytes :)

def p(s):n=s.count('1');return(n>1)*all(n%i for i in range(2,n))
def f(s):t=[f'{ord(c):07b}'for c in s];return all(map(p,t+[[u[j]for u in t]for j in range(7)]))

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Answer 8

Pyth, 37 bytes

L.AmP_sdb=sMM.[L\0lh.MlZQ=.BMQ&yZy.TZ

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---

                                 Code | Explanation
--------------------------------------+----------------------------------------------------------------
L.AmP_sdb=sMM.[L\0lh.MlZQ=.BMQ&yZy.TZ | Full code
L                                     | Define function y(b):
   m    b                             |   For each d in b:
    P_sd                              |     Is the sum of the elements of the list prime?
 .A                                   |   Return whether all elements of the resulting list are truthy
                         =   Q        | Assign the following to Q:
                          .BMQ        |   The list of binary strings for each character in the input
         =             Z              | Assign the following to Z:
               L             Q        |   For every element in Q:
             .[ \0                    |     Pad with 0 on the left
                  lh.MlZQ             |     To the length of the longest element in Q
            M                         |   For each element in the resulting list:
          sM                          |     Convert each character to an integer
                              &yZ     | Print y(Z) AND
                                 y.TZ |   y( <Transpose of Z> )

Answer 9

Brachylog, 14 bytes

ạḃᵐB↔ᵐz₁,B+ᵐṗᵐ

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Outputs through success or failure. (In the case of success, a list of all column and row sums is available through the output variable.

   B              The variable B is
ạ                 the codepoints of the input
 ḃᵐ               converted to lists of binary digits,
    ↔ᵐ            which with each list reversed
      z₁          then zipped without cycling
        ,B        and concatenated with B
          +ᵐ      has elements which all sum to
            ṗᵐ    prime numbers.

Answer 10

O5AB1E, 12 bytes

Çžy+bø€SOp¦W

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This is my first code golf so go easy :)

Ç              % Converts the implicit input into ascii values
 žy+           % Adds 128 to each value, inspired by Emigna as a way to pad zeros
    b          % Convert all values into bits
     ø         % Transpose
      €SO      % Sum each string of binary digits created
         p¦    % Check if each element is prime and cut the first element out (adding 128 makes it equal to the number of characters)
           W   % Take the minimum value to effectively "and" all the elements

Answer 11

K (oK), 40 33 bytes

Solution:

&/{2=+/d=_d:x%!x}'+/'m,+m:(7#2)\'

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Explanation:

Half is creating the matrix, the other half is the primality check.

&/{2=+/d=_d:x%!x}'+/'m,+m:(7#2)\' / the solution
                                ' / apply to each (')
                               \  / decode
                          (   )   / do this together
                           7#2    / 7#2 => 2 2 2 2 2 2 2
                        m:        / save as m
                       +          / transpose
                     m,           / append to m
                  +/'             / sum (+/) each (')
                 '                / apply to each
  {             }                 / lambda taking implicit x
              !x                  / range 0..x
            x%                    / x divided by ...
          d:                      / save as d
         _                        / floor
       d=                         / equal to d?
     +/                           / sum (+/)
   2=                             / equal to 2?
&/                                / minimum

Answer 12

Python 3, 228 227 225 bytes

Not a great answer, I wasn't able to golf it as much as I would've liked, but I spent so long on it I feel I should post it. Suggestions on cutting bytes would be greatly appreciated.

r=range
n=[format(ord(c),"08b")for c in input()]
n=map(lambda s:s.count("1"),n+["".join([f[1]for f in filter(lambda e:e[0]%8<1,enumerate("X"*-~i+"".join(n)))][1:])for i in r(8)])
print(all(all(s%d for d in r(2,s))for s in n))

Edit 1: replaced e[0]%8==0 with e[0]%8<1, losing a byte. Thanks Flp.Tkc!

Edit 2: replacing (i+1) with -~i, losing two additional bytes. Thanks Erik for exposing how bad my bit-level knowledge is :) While testing this revision, I discovered that kappa is valid... make of that what you will.

Answer 13

Perl, 151 121 111 + 3 = 114 bytes

Run with -lF. The program will only function correctly for the first input. Terminate the program and rerun for your next input.

Thanks to @Dada for letting me know the // after F were redundant. An additional byte can be removed (for 112) by piping the input in via echo -n, but I feel that that's technically adding more code, so YMMV.

for$c(@a=map{sprintf"%07b",ord}@F){$b[$_].=substr$c,$_,1 for 0..6}s/0//g,$d|=/^1?$|^(11+?)\1+$/ for@a,@b;say!$d

Readable:

                                     #Implicitly split input into characters in @F array
for$c(@a=map{sprintf"%07b",ord}@F)  #Convert @F to 7-bit binary as @a, then loop through it                        
    $b[$_].=substr$c,$_,1 for 0..6   #Transpose @a's bits into @b
}
s/0//g,$d|=/^1?$|^(11+?)\1+$/ for@a,@b; #Remove any zeros, then run through composite regex
say!$d                          #If all composite regex checks fail, then it's fully prime.

Answer 14

Groovy, 151 137 bytes

{p={x->x<3||(2..(x**0.5)).every{x%it}};y={it.every{p(it.count("1"))}};x=it.collect{0.toString((int)it,2) as List};y(x)&&y(x.transpose())}

No primality check in groovy...

p={x->x<3||(2..(x**0.5)).every{x%it}}; - Closure for primality testing.

y={it.every{p(it.count("1"))}}; - Closure to ensure that all counts of "1" for a passed binary 2D array are prime.

x=it.collect{0.toString((int)it,2) as List}; - Coversion from string to binary array.

y(x)&&y(x.transpose()) - For all prime-validated sums in the main matrix and the transposed matrix, ensure that they return true.

Answer 15

JavaScript (Node.js), 149 146 ... 134 130 129 bytes

x=>[...x].map(y=>a=[...a.map(n=>y.charCodeAt()&2**i++?++z&&-~n:n,z=i=0),z],a=[...Array(7)])&&!a.some(n=>(P=r=>n%--r?P(r):~-r)(n))

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Explanation

x=>                        // Main function:
 [...x].map(               //  For each y in x:
  y=>
   a=[...a.map(            //   For each i in range(0, len(a)):
    n=>                   
     y.charCodeAt()&2**i++ //    If y AND 2**i is not zero:
     ?++z&&-~n:n,          //     z = z + 1; a[i] = a[i] + 1 (1 if a[i] is undefined)
    z=i=0                  //   Initially z = 0
   ),z],                   //   Then push z at the end of a
  a=[...Array(7)]          //  Initially a = [undefined] * 7
 )&&!a.some(               //  Then for each n in a:
  n=>(
   P=r=>                   //   Composite function:
    n%--r?                 //    If n % (r - 1) == 0 or r == 1:
     P(r)                  //     Return P(r - 1)
    :~-r                   //    Else: Return r - 2
  )(n)                     //   Starting from r = n
 )                         //  Return whether the composite function returns 0 for all n.

How does it even work!?

• y.charCodeAt()&2**i
  • We need this code to return the corresponding bit of y.charCodeAt() if 0 <= i < 7, and 0 otherwise.
  • When i < 7, the code apparently works as usual.
  • When 7 <= i <= 32, since the corresponding bit of y.charCodeAt() is 0 anyway, the result is 0 as expected.
  • When 32 < i < 1024, since int32(2**i) == 0, the result is 0 as expected.
  • When 1024 <= i, we have 2**i == Infinity, and since int32(Infinity) == 0, the result is 0 as expected.
• (P=r=>n%--r?P(r):~-r)(n)
  • For simplicity we let R = --r = r - 1.
  • This helper function terminates when n % R == 0 or n % R is NaN.
    • n % R == 0: R is a factor of n.
      • If R == 1, then n is prime because all 1 < R < n can't divide n. Return 0 (falsy).
      • If R == -1, then n == 0. Return -2 (truthy).
      • Otherwise, return R - 1 where R - 1 > 0 (truthy).
    • n % R is NaN: Invalid modular calculation.
      • If R == 0: n == 1. Return -1 (truthy).
      • If n is NaN: R is NaN. Return -1 (truthy).
  • As a result, only when R == 1 can this function return a falsy value, indicating n is prime.

Answer 16

Perl 5 -MList::Util=all,sum -pF, 96 92 bytes

$_=all{//;all{$'%$_}2..$_-1}(map{$_=sprintf'%07b',ord;y/1//}@F),map{sum map{s/.//;$&}@F}0..6

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Answer 17

Clojure, 180 bytes

#(let[S(for[i %](for[j[1 2 4 8 16 32 64]](min(bit-and(int i)j)1)))A apply](not-any?(fn[i](or(= i 1)(seq(for[d(range 2 i):when(=(mod i d)0)]d))))(into(for[s S](A + s))(A map + S))))

There might be a shorter way of generating bit lists and also the primality test.

Answer 18

Python 3, 164 bytes

import numpy;a=numpy.array([list(f'{ord(_):07b}')for _ in input()]).astype(int);print(all([(v>1)*all(v%i for i in range(2,v))for v in set(a.sum(0))|set(a.sum(1))]))

Answer 19

Ruby 2.7 -rprime, 95 bytes

->s{a=s.bytes.map{[*@1.digits(2),0][..6]}
(a.map(&:sum)+a.transpose.map(&:sum)).all?(&:prime?)}

No TiO link because TiO still runs Ruby 2.5.5.

Explanation

Pretty simple. The first line gets the binary digits of each character as an array padded out to seven digits, which really ought to be easier:

a = s.bytes.map { [*@1.digits(2), 0][..6] }

Check out that numbered block parameter (@1) and beginless range (..6) hotness.

The second line sums the rows and columns and tests if they're all prime:

(a.map(&:sum) + a.transpose.map(&:sum)).all?(&:prime?)

Answer 20

PHP, 173 bytes

for($r=1;$b=substr_count($t[$i]=sprintf('%07b',ord($argv[1][$i++])),1);)$r&=$b==2|$b%2&$b>2;for(;$k++<7;){for($b=$j=0;$t[++$j];$b+=$t[$j][$k-1]);$r&=$b==2|$b%2&$b>2;}echo$r;

Test it online

Answer 21

JavaScript, 234 bytes

f=z=>(z=[...z].map(v=>v.charCodeAt(0))).map(v=>v.toString(2).replace(/0/g,"").length).every((p=v=>{for(i=2;i<v;i++){if(v%i===0){return 0}};return v>1}))&&[...""+1e6].map((v,i)=>z.reduce((a,e)=>!!(e&Math.pow(2,i))+a,0)).every(v=>p(v))

We get the horizontal values by converting the number to binary, removing the zeros using a string replacement, and then counting the 1s. The vertical sums are obtained by looping 1 to 7 and using a bitwise AND with 2 raised to the nth power.