Divisible strings

Question

Task

Given a string s, output a truthy value if the ASCII code of each letter is divisible by the length of s, and a falsey otherwise.

Input/Output

Input is a nonempty string containing only ASCII [32-126]. Output is a standard truthy/falsey value. Note that you can switch the values, for example returning 0/False if divisible and vice versa

Test cases

Input         Output

Hello         False       (72 101 108 108 111), 5
lol           True        (108 111 108), 3
Codegolf      False       (67 111 100 101 103 111 108 102), 8
A             True        (65), 1
nope          False       (110 111 112 101),4
8  8          True        (56 32 32 56), 4

Answer 1

Aceto, 21 bytes

&L
|%o 1
€l|!
rM@dp

Explanation

We read a string and €xplode it, push the stack length and Memorize that.

Then, after setting the exception catch point (@), we always duplicate the top stack element, negate (!) it, and m|rror horizontally if we get a truthy value (string has ended; we popped a 0). Otherwise, we get the ordinal of the character, Load the memorized value and do modulo (%). If this is truthy, we m|rror again.

Finally, we raise an exception (&) to land back in front of the d, for our next character.

If we mirrored, then we eventually land on p, printing the top-most element of the stack. In one of the two cases of mirroring, we will have pushed a 1 before.

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I don't see much potential to golf this down further; there's only one space character used, and 3 newlines. Perhaps one or two bytes could be saved by making it a 16x16 in two lines.

Answer 2

Clojure, 41 chars

(every? #(= 0 (mod (int %) (count x))) x)

Removing spaces after comment 37 chars

(every? #(= 0(mod(int %)(count x)))x) 

Answer 3

MAWP, 34 33 24 23 bytes

`|_=M0=A0/[M%{0:.}?`]1:

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Thanks to @Razetime for saving 9 bytes!

Explanation:

`        Remove starting 1 on stack
|        Push input on stack as ASCII codes
_=M      Set variable M to length of stack (length of input)
0=A      Set variable A to 0
0/       Push 0 and cycle stack
[        Start of loop
M%       Modulo by M
{0:.}    If not 0 then print 0 and terminate
?`       If 0 then pop value
]        End of loop
1:       Print 1

Answer 4

Brachylog, 8 bytes

ạfᵐ∋ᵛ~l?

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ạfᵐ∋ᵛ~l?
ạ        characters to integer
 fᵐ      find all factors
   ∋ᵛ    every list of factors contain …
     ~l? the length of the input

Alternative version,

⟨ạzl⟩%ᵛ0
⟨fhg⟩    forks! fA & gB ∧ [A, B]h
 ạzl     zip the code blocks with the length;
          [[108, 3], [111, 3], [108, 3]]
     %ᵛ0 every list must be 0 after modulo

Answer 5

NARS2000 (12 characters)

(∧/0=⍴|⎕ucs)

Fork ⍴|⎕ucs finds residue of each ASCII int value when divided by length of string. ∧/0= checks if all residue are zero.

Answer 6

Python 3.8 (pre-release), 63 49 bytes

lambda h:print({0}=={*[ord(i)%len(h)for i in h]})

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* in the {} constructs a set which I think is more byte-efficient than using set(), which I didn't know.

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Python 3.8 (pre-release), 42? bytes

lambda h:{0}=={*[ord(i)%len(h)for i in h]}

I can't get TIO to output the True or False. This works with IDLE or Jupyter. Try it online.

Answer 7

PowerShell Core, 30 bytes

!(($args|?{(++$l)})|?{+$_%$l})

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Answer 8

Java, 41 bytes

s->s.chars().allMatch(i->i%s.length()==0)

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Answer 9

Husk, 6 bytes

Λo¦L¹c

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Explanation

Λo¦L¹c   translates = Λo¦L⁰c⁰
Λ     ⁰  all of input:
 o   c   charcode
  ¦      divides
   L⁰    length of input

Answer 10

APOL, 23 bytes

!(x(ƒ(i %(↶(∋) l(⋒)))))

Explanation:

!(           Not
  x(         Any list item is true
    ƒ(       List-builder for
      i      Input
      %(     Modulo
        ↶(   ASCII codepoint
          ∋  Loop item
       )
       l(    Length of
         ⋒   Loop iterator (what the loop is looping through)
       )
     )
    )
  )
)

Answer 11

Thunno 2 B, 3 bytes

lḊp

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Explanation

lḊp  # Implicit input
     # Convert to charcodes
l    # Length of input
 Ḋ   # Are they divisible
  p  # Product of the list
     # Implicit output

Answer 12

Vyxal 2.4.1 K, 3 bytes

LḊΠ

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I too can play the K flag game, especially given that I invented it.

Answer 13

jq, 38 characters

length as$a|explode|map(.%$a)|all(.<1)

Sample run:

bash-5.2$ jq -R 'length as$a|explode|map(.%$a)|all(.<1)' <<< 'Hello'
false

bash-5.2$ jq -R 'length as$a|explode|map(.%$a)|all(.<1)' <<< 'lol'
true

Try it online! / Try all test cases online!

Answer 14

Excel, 66 bytes

=SUM(MOD(CODE(MID(A1,ROW(OFFSET(A1,0,0,LEN(A1))),1))/LEN(A1),1))=0

Input is in the cell A1.

ROW(OFFSET(A1,0,0,LEN(A1))) gives us an array of values from 1 to the length of the input.
MID(A1,ROW(~),1) pulls out each character of the input, one at a time.
CODE(MID(~) converts those characters to their decimal ASCII equivalent.
MOD(CODE(~)/LEN(A1),1) returns just the decimal portion of those codes divided by the input length.
SUM(MOD(~))=0 adds up all those decimal portions and returns TRUE if there weren't any (SUM=0 means everything divided nicely) and FALSE if there were.

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I thought this would have to be an array formula but it seems to work without that. Color me surprised.

Answer 15

T-SQL, 65 bytes

Returns 1 for true, 0 for false

SELECT-1/~max(ascii(substring(@,number,1))%len(@))FROM spt_values

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Answer 16

Go, 76 bytes

func f(a string)int{for _,v:=range a{if int(v)%len(a)>0{return 0}}
return 1}

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Answer 17

x86-16 machine code, 10 bytes

00000000: 880e 0601 acd4 0ae1 fbc3                 ..........

Listing:

88 0E 0106      MOV  BYTE PTR[CHR_LOOP+2], CL   ; move len to second byte of AAM opcode 
            CHR_LOOP: 
AC              LODSB                           ; AL = next char 
D4 0A           AAM  CL                         ; ZF = (AL % CL == 0) 
E1 FB           LOOPZ CHR_LOOP                  ; loop if CX > 0 AND ZF = 1 
C3              RET                             ; return to caller

As a callable function. Input: string in SI, length in CX. Output: ZF = 1 if Truthy, ZF = 0 if Falsey.

Output of DOS test program:

Answer 18

C (clang), 149 148 bytes

#include<stdio.h>
int main(){char s[100];scanf("%s",s);int j;int k=strlen(s);for(int i=0;i<k;i++){if(s[i]%k>0){printf("0");return 0;}};printf("1");}

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Answer 19

Factor + math.unicode, 33 bytes

[ dup length '[ _ mod 0 = ] ∀ ]

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