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

05AB1E, 17 bytes

Çžz+b€¦€SDøìO0ÛpP

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Answer 4

Jelly, 15 bytes

O+⁹Bṫ€3µS€;SÆPP

Try it online! or Verify all test cases..

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 5

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 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

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 8

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 9

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 10

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 11

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 12

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 13

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 14

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 15

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 16

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.

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

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 19

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 20

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 21

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.