Double Prime Words

Question

Consider a word/string of length \$n\$, only including the letters A-Z, a-z. A word/string is a double prime word if and only if n is prime and the sum of the letters, s, is also prime, using their numeric position in the alphabet (a=1, B=2, c=3, etc.).

Input can be any combination of upper or lower case alphabetic characters, as there is no numeric difference between a or A.

Output is any appropriate logical format related to your language. i.e. True or False, T or F, 1 or 0, etc. Specifying what format your output will appear is highly appreciated, but not required. (Output need not include n, s, but I include them below as demonstration and example)

Winning condition is shortest code in bytes able to detect if a string is a double prime, fitting both conditions for n and s to be prime. (I've now included cases from all 4 possible situations of n, s.)

Examples

Input -> Output (n, s)

Prime -> True (5, 61)
han -> True (3, 23)
ASK -> True (3, 31)
pOpCoRn -> True (7, 97)
DiningTable -> True (11, 97)
METER -> True (5, 61)

Hello -> False (5, 52)
SMILE -> False (5, 58)
frown -> False (5, 76)

HelpMe -> False (6, 59)
John -> False (4, 47)
TwEnTy -> False (6, 107)

HelloWorld -> False (10, 124)
Donald -> False (6, 50)
telePHONES -> False (10, 119)

A -> False (1, 1) 
C -> False (1, 3) {1 is not prime}
d -> False (1, 4)

Answer 1

Jelly, 12 bytes

ŒuO_64µL,SẒP

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How it works

ŒuO_64µL,SẒP - Main link, takes string s as argument e.g. s = "Prime"
Œu           - Convert to upper case                          "PRIME"
  O          - Convert to ordinals                            [80, 82, 73, 77, 69]
   _64       - Subtract 65 (call this L)                      [16, 18, 9, 13, 5]
      µ      - Start a new link with L as the left argument
       L     - Take the length                                5
         S   - Take the sum                                   61
        ,    - Pair the two values                            [5, 61]
          Ẓ  - Take primality of each                         [1, 1]
           P - Take product                                   1

Answer 2

R, 68 71 bytes

+3 bytes to correct a bug pointed out by Dominic van Essen

`?`=sum;s=?b<-utf8ToInt(scan(,""))%%32;l=?b^0;l-1&5>?c(!s%%1:s,!l%%1:l)

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Notice that to convert both upper and lower case letters to the integers 1...26, we can take the ASCII codepoint modulo 32. sum(!x%%1:x) is a golfy way of counting the number of divisors of x, which will be equal to 2 iff x is prime.

Ungolfed:

`?` = sum                       # shorthand for sum
b = utf8ToInt(scan(, "")) %% 32 # take input and convert to ASCII, then take mod 32
s = sum(b)
l = sum(b^0)                    # l = length(b)
5 > sum(c(!s%%1:s,!l%%1:l))    # sum the number of divisors of s and l, and check whether you get <5.
       & l!=1                   # and that l is not 1

Answer 3

Ruby, 27 59 bytes

->a{[a.size,a.upcase.bytes.map{|i|i-64}.sum].all? &:prime?}

+33 bytes after correcting the solution, thanks to DrQuarius.

Try it online! or Verify all test cases

Answer 4

perl -Mfeature=say -MList::Util=sum -pl, 95 bytes

s/[^a-z]//gi;$m=sum map-64+ord,split//,uc;$_=(1 x y===c)!~/^(11+)\1+$|^1$/&&(1x$m)!~/^(11+)\1$/

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How does it work?

s/[^a-z]//gi;   # Clean the input, remove anything which isn't an ASCII letter.

                          uc;     # Upper case the string
                  split//,        # Split it into individual characters
          -64+ord                 # Calculate its value: 
                                  #           subtract 64 from its ASCII value
       map                        # Do this for each character, return a list
$m=sum                            # Sum the values, and store it in $m

     y===c                        # Returns the length of the input string
(1 x y===c)                       # Length of the input string in unary

/^(11+)\1+$|^1$/                  # Match a string consisting of a composite
                                  # number of 1's, or a single 1
!~                                # Negates the match, so
(1 x y===c)1~/^(11+)\1+$|^1$/     # this is true of the input string (after
                                  # cleaning) has prime length

(1x$m)!~/^(11+)\1+$/              # Similar for the sum of the values --
                                  # note that the value is at least 2, so
                                  # no check for 1.

Combining this, and the program will print 1 on lines which match the conditions, and an empty line for lines which do not match.

Answer 5

05AB1E, 10 bytes

gAIlk>O‚pP

Input as a list of characters.

Try it online or verify all test cases.

Explanation:

g           # Get the length of the (implicit) input-list
 A          # Push the lowercase alphabet
  I         # Push the input-list of characters
   l        # Convert the input to lowercase
    k       # Get the (0-based) index of each character in the alphabet-string
     >      # Increase each by 1 to make them 1-based indices
      O     # Take the sum of that
       ‚    # Pair the length together with this sum
        p   # Check for both whether they're a prime (1 if it's a prime; 0 if not)
         P  # And check if both are truthy by taking the product of the pair
            # (after which the result is output implicitly)

Answer 6

R, 70 bytes

function(s,S=sum,t=S(utf8ToInt(s)%%32))S(!nchar(s)%%1:t)^S(!t%%1:t)==4

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I forced myself not to peek at Robin Ryder's answer before having a shot at this, and (satisfyingly) it turns out that we've used some rather different golfing tricks.

t is the total of all letter indices. This is certain to be greater-than-or-equal-to nchar(s) (it's only equal if the string s is "A" or "a"). So we can use modulo 1:t to test for primality of the string length instead of modulo 1:nchar(s), and there's no need waste characters on a variable declaration to store nchar(s).

Both primality tests sum(!t%%1:t) and sum(!nchar(s)%%1:t) must be equal to 2 if both the sum-of-letter-indices and the string length are prime.
We could check if they're both 2, but this requires ==2 twice (plus a & or equivalent), which seems wasteful. Is it ok to check that the total is 4? The edge-case we need to worry about is if one of them equals 1 and the other 3: this happens for the string "D" (length=1 and character-index=4 with divisors 1,2 and 4). So it's not Ok. Can we multiply them? Also no, because 1 and 4 will again give 4 (think about the string "F").
But - since we know that the string length must be less-than-or-equal to the sum-of-character-indices, we can use exponentiation: the only way to get 4 is 4^1 or 2^2, and since the sum-of-character-indices can't be 1 if the string-length is 4, 2^2 is the only possibility.

So the final, combined check for double-primality is sum(!nchar(s)%%1:t)^sum(!t%%1:t)==4, saving 3 characters compared to testing them separately.

Answer 7

Rockstar, 327 321 319 294 bytes

No built-in for testing primes!
No case conversion!
No way to get the codepoint of a character!

Why do I do these things to myself?! Spent so long just getting the damn thing to work, I'm sure it's far from optimally golfed but it'll do for now.

Outputs 0.25 for true and 0 for false.

F takes N
let D be N
let P be N-1
while P and D-2
let D be-1
let M be N/D
turn M up
let P be N/D-M

give P

G takes I
N's 27
while N
cast N+I in C
if C's S at X
give N

let N be-1

give G taking 64

listen to S
X's 0
T's 0
while S at X
let T be+G taking 96
let X be+1

say F taking T*F taking X

Try it here (Code will need to be pasted in)

Answer 8

Retina 0.8.2, 77 bytes

\W|\d|_

$
¶$`
\G.
1
T`L`l
[t-z]
55$&
[j-z]
55$&
T`_l`ddd
.
$*
A`^(..+)\1+$
¶

Try it online! Link includes test cases. Explanation:

\W|\d|_

Delete anything that isn't a letter.

$
¶$`

Duplicate the letters.

\G.
1

Replace the letters on the first line with 1s, thus taking the length in unary.

T`L`l

Convert the remaining letters to lower case.

[t-z]
55$&
[j-z]
55$&
T`_l`ddd

Convert them to digits that will sum to their numeric position.

.
$*

Convert the digits to unary, thus taking their sum.

A`^(..+)\1+$

Delete any composite values.

¶

Check that both values are still present.

Answer 9

Python 3, 86 78 87 bytes

Saved 8 bytes thanks to ovs!!!
Added 9 bytes to fix a bug kindly pointed out by Robin Ryder.

lambda s:~-len(s)*all(n%i for n in(len(s),sum(ord(c)&31for c in s))for i in range(2,n))

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Returns a truthy or falsey value.

Answer 10

Brachylog, 11 bytes

ḷạ-₉₆ᵐ+ṗ&lṗ

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How it works

ḷạ-₉₆ᵐ+ṗ&lṗ (is the implicit input)
ḷ           to lowercase
 ạ          to list of char codes
  -₉₆ᵐ      minus 96 (so 'a' -> 1)
      +     summed
       ṗ    prime?
        &l  and is the input's length
          ṗ prime?

Answer 11

Wolfram Language (Mathematica), 34 bytes

PrimeQ@*Tr/@(LetterNumber@#&&1^#)&

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-22 bytes from @att

Answer 12

Japt, 16 bytes

Êj ©Uu ¬mc xaI j

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Answer 13

J, ^{27 22} 18 bytes

1*/@p:#,1#.32|3&u:

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-5 bytes thanks to xash

-4 bytes thanks to Dominic van Essen

• 32|3&u: Turn each letter into its index by first converting to its ascii number, the modding by 32.
• 1#. Sum.
• #, Prepend list length.
• 1...p: Are each of those two numbers prime?
• */@ Multiply them together -- are they all prime?

Answer 14

C - 119 108 99 98 bytes (gcc)

@ceilingcat saved another byte!

b,t,e;p(c){for(;--e&&c%e;);c=e==1;}a(char*a){t=0;for(e=b=strlen(a);b;)t+=a[--b]%32;t=p(e)*p(e=t);}

try it online

previously

Many thanks to @DominicvanEssen and @ceilingcat for saving 20 bytes! - and particularly to Dominic for fixing error on n=1 (non-prime)

b,t,e;p(c){for(b=c;--b&&c%b;);c=b==1;}a(char*a){t=0;for(e=b=strlen(a);b;)t+=a[--b]%32;t=p(e)*p(t);}

first attempt below 119 bytes

a(char*a){int t=0,d=strlen(a),e=d;while(d)t+=a[--d]%32;return p(e)*p(t);}
p(int c){int b=c;while(--b&&c%b);return b<2;}

In fact can save 3 bytes by using while(c%--b) in the second routine, but this fails for the case of p(1) e.g. 'a'. or other single characters.

try it online

Answer 15

Scala, 75 74 69 bytes

| =>p(|size)&p(|map(_&95-64)sum)
def p(n:Int)=(2 to n/2)forall(n%_>0)

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Answer 16

Factor, 78 bytes

: d ( s -- ? ) dup [ length ] dip >lower [ 96 - ] map sum [ prime? ] bi@ and ;

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Answer 17

05AB1E, 11 bytes

uÇ64-Op¹gp&

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Bytes removed due to lack of input restrictions

Answer 18

JavaScript (Node.js), 88 bytes

Returns 0 or 1.

s=>(g=k=>n%--k?g(k):k==1)(Buffer(s).map(c=>x+=n<(n+=c>64&(c&=31)<27&&c),x=n=0)|n)&g(n=x)

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Commented

Helper function

g = k =>                   // g is a helper function testing if n is prime
  n % --k ?                //   decrement k; if it does not divide n:
    g(k)                   //     do recursive calls until it does
  :                        //   else:
    k == 1                 //     test whether k = 1

Main function

s =>                       // s = input string
  g(                       // test if the 'sum of the letters' is prime
    Buffer(s).map(c =>     //   for each ASCII code c in s:
      x +=                 //     increment x if ...
        n < (              //       ... n is less than ...
          n +=             //         ... the new value of n:
            c > 64 &       //           if c is greater than 64
            (c &= 31) < 27 //           and c mod 32 is less than 27:
            && c           //             add c mod 32 to n
        ),                 //
      x = n = 0            //     start with x = n = 0
    ) | n                  //   end of map(); yield n
  )                        // end of the first call to g
  & g(n = x)               // 2nd call to g with the 'length' x

Answer 19

Perl 5 -pl, 52 bytes

Uses the prime identification regex from @Abigail's answer

$_.=$".1x s/./1x(31&ord$&)/ge;$_=!/\b((11+)\2+|1)\b/

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Answer 20

Ruby, 50 55 50 bytes

->s{[s.size,s.upcase.sum-64*s.size].all? &:prime?}

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+5 bytes due to a misunderstanding of whether arrays could be considered truthy.

-5 bytes thanks to Razetime, using the nice trick of putting the " &:prime?" at the end instead of doing a ".map(&:prime?)" before the ".all?".

Posted separately because Razetime's solution actually didn't sum the alphabet index but simply the ascii ordinals. It fails for the double prime words "DiningTable" and "METER".

Answer 21

Husk, 12 bytes

&ṗL¹ṗṁȯ-64ca

Try it online! Outputs a truthy number if the word is a double prime word, and 0 otherwise.