Multiplicity Test

Question

Use any programming language to display numbers between 1 and 99 (including both) in such a way, so that:

• the numbers are separated by single space,
• if a number is divisible by 3, it should be in parentheses,
• if a number is divisible by 4, it should be in square brackets,
• if a number is divisible by both 3 and 4, it should be in both parentheses and square brackets (with square brackets closer to the number).

Your program should display exactly:

1 2 (3) [4] 5 (6) 7 [8] (9) 10 11 ([12]) 13 14 (15) [16] 17 (18) 19 [20] (21) 22 23 ([24]) 25 26 (27) [28] 29 (30) 31 [32] (33) 34 35 ([36]) 37 38 (39) [40] 41 (42) 43 [44] (45) 46 47 ([48]) 49 50 (51) [52] 53 (54) 55 [56] (57) 58 59 ([60]) 61 62 (63) [64] 65 (66) 67 [68] (69) 70 71 ([72]) 73 74 (75) [76] 77 (78) 79 [80] (81) 82 83 ([84]) 85 86 (87) [88] 89 (90) 91 [92] (93) 94 95 ([96]) 97 98 (99)

Answer 1

05AB1E, 23 bytes

-1 byte thanks to Kevin Cruijssen

тGND4Öi…[ÿ]}N3Öi…(ÿ)]ðý

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Answer 2

Python 2, 68 65 60 bytes

i=0
exec"i+=1;print'('[i%3:]+`[i][i%4:]or i`+')'[i%3:],;"*99

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Answer 3

R, 61 bytes

"+"=c
r=""+""
cat(paste0(r+"(",r+""+"[",1:99,r+""+"]",r+")"))

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Answer 4

Jelly, 21 20 bytes

³Ṗµ3,4ṚƬḍד([“])”j)K

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How it works

³Ṗµ3,4ṚƬḍד([“])”j)K  Main link. No arguments.

³                     Set the return value to 100.
 Ṗ                    Pop; yield [1, ..., 99].
  µ               )   Map the chain in between over [1, ..., 9]; for each integer k:
   3,4                    Set the return value to [3, 4].
      ṚƬ                  Reverse until a loop is reached. Yields [[3, 4], [4, 3]].
        ḍ                 Test k for divisibility by [[3, 4], [4, 3]], yielding a
                          matrix of Booleans.
         ד([“])”         Repeat the characters of [['(', '['], [']', ')']] as many
                          times as the Booleans indicate.
                 j        Join the resulting pair of strings, separated by k.
                   K  Join the resulting array of strings, separated by spaces.

Answer 5

D, 110 bytes

import std.stdio;void f(){for(int i;i<99;)write(++i%3?"":"(",i%4?"":"[",i,i%4?"":"]",i%3?"":")",i%99?" ":"");}

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Ported from @HatsuPointerKun's C++ answer.

Answer 6

Charcoal, 30 bytes

⪫ＥＥＥ⁹⁹Ｉ⊕ι⎇﹪⊕κ⁴ι⪫[]ι⎇﹪⊕κ³ι⪫()ι 

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

    ⁹⁹                          Literal 99
   Ｅ                            Map over implicit range
        ι                       Current value
       ⊕                        Incrementd
      Ｉ                         Cast to string
  Ｅ                             Map over list of strings
            κ                   Current index
           ⊕                    Incremented
             ⁴                  Literal 4
          ﹪                     Modulo
              ι                 Current value
                []              Literal string `[]`
                  ι             Current value
               ⪫                Join i.e wrap value in `[]`
         ⎇                      Ternary
 Ｅ                              Map over list of strings
                      κ         Current index
                     ⊕          Incremented
                       ³        Literal 3
                    ﹪           Modulo
                        ι       Current value
                          ()    Literal string `()`
                            ι   Current value
                         ⪫      Join i.e wrap value in `()`
                   ⎇            Ternary
                                Literal space
⪫                               Join list
                                Implicitly print

Answer 7

J, 54 53 bytes

1 byte less thanks to @Jonah

(*stdout(3|.0=4 3 1 3 4&|,1=":)#3|.']) ([',":)@>i.100

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Answer 8

C, C++, 136 133 131 129 128 124 bytes

-5 bytes thanks to Zacharý and inspired by write() function in D language ( see Zacharý answer )

-2 bytes thanks to mriklojn

-12 bytes for the C version thanks to mriklojn

-4 bytes thanks to ceilingcat

#include<cstdio>
void f(){for(int i=0;i++<99;)printf("(%s%d%s%s%s"+!!(i%3),i%4?"":"[",i,i%4?"":"]",i%3?"":")",i%99?" ":"");}

C Specific Optimization : 115 bytes

#include<stdio.h>
i;f(){for(;i++<99;)printf("(%s%d%s%s%s"+!!(i%3),i%4?"":"[",i,i%4?"":"]",i%3?"":")",i%99?" ":"");}

Answer 9

Powershell, 60 bytes

"$(1..99|%{($_,"($_)","[$_]","([$_])")[!($_%3)+2*!($_%4)]})"

Explanation:

• the array with 4 elements: $_, "($_)", "[$_]", "([$_])"
• and the index: [!($_%3)+2*!($_%4)]
• repeat for each number
• convert the result to a string

Less golfed Test script:

$f = {

$r = 1..99|%{
    ($_, "($_)", "[$_]", "([$_])") [!($_%3)+2*!($_%4)]
}
"$($r)"

}

$expected = '1 2 (3) [4] 5 (6) 7 [8] (9) 10 11 ([12]) 13 14 (15) [16] 17 (18) 19 [20] (21) 22 23 ([24]) 25 26 (27) [28] 29 (30) 31 [32] (33) 34 35 ([36]) 37 38 (39) [40] 41 (42) 43 [44] (45) 46 47 ([48]) 49 50 (51) [52] 53 (54) 55 [56] (57) 58 59 ([60]) 61 62 (63) [64] 65 (66) 67 [68] (69) 70 71 ([72]) 73 74 (75) [76] 77 (78) 79 [80] (81) 82 83 ([84]) 85 86 (87) [88] 89 (90) 91 [92] (93) 94 95 ([96]) 97 98 (99)'
$result = &$f
$result-eq$expected
$result

Output:

True
1 2 (3) [4] 5 (6) 7 [8] (9) 10 11 ([12]) 13 14 (15) [16] 17 (18) 19 [20] (21) 22 23 ([24]) 25 26 (27) [28] 29 (30) 31 [32] (33) 34 35 ([36]) 37 38 (39) [40] 41 (42) 43 [44] (45) 46 47 ([48]) 49 50 (51) [52] 53 (54) 55 [56] (57) 58 59 ([60]) 61 62 (63) [64] 65 (66) 67 [68] (69) 70 71 ([72]) 73 74 (75) [76] 77 (78) 79 [80] (81) 82 83 ([84]) 85 86 (87) [88] 89 (90) 91 [92] (93) 94 95 ([96]) 97 98 (99)

Answer 10

MathGolf, 41 40 34 29 bytes

♀({îû)(î+╫îa_'(\')ßyΓî34α÷ä§ 

NOTE: It has a trailing space

Only my second MathGolf answer..
-5 bytes thanks to @JoKing.

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Explanation:

♀(             # Push 99 (100 decreased by 1)
  {            # Start a loop, which implicitly loops down to (and excluding) 0
   û)(         #  Push string ")("
      î+       #  Append the 1-indexed index
        ╫      #  Rotate the string once towards the right
   îa          #  Push the 1-indexed index of the loop, wrap in a list
   _           #  Duplicate it
    '(        '#  Push string "("
      \        #  Swap the top two items of the stack
       ')     '#  Push string ")"
         ßy    #  Wrap all three into a list, and join them
   Γ           #  Wrap all four into a list
               #  (We now have a list [N, "(N)", "[N]", "([N])"], where N is the index)
   î           #  Push the 1-indexed index of the loop
    34         #  Push 3 and 4 to the stack
      α        #  Wrap all three into a list
       ÷       #  Check for each if the index is divisible by it
               #  (resulting in either [0,0], [0,1], [1,0], or [1,1]
        ä      #  Convert from binary to integer
               #  (resulting in either 0, 1, 2, or 3
         §     #  Push the string at that index from the array
               #  Push a space
               # (After the loop, output the entire stack joined together implicitly)

Answer 11

Haskell, 77 bytes

unwords[f"()"3$f"[]"4$show n|n<-[1..99],let f(x:r)m s|mod n m<1=x:s++r|1<3=s]

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I wonder if show[n] can be used to shorten things up, so far to no avail.

Answer 12

Lua, 161 123 bytes

b=""for i=1,99 do;c,d=i%3==0,i%4==0;b=b..(c and"("or"")..(d and"["or"")..i..(d and"]"or"")..(c and")"or"").." "end;print(b)

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Ungolfed:

b = ""
for i = 1, 99 do
    c = 1 % 3 == 0
    d = 1 % 4 == 0
    a = ""
    if c then
        a = a .. "("
    end
    if d then
        a = a .. "["
    end
    a = a .. i
    if d then
        a = a .. "]"
    end
    if c then
        a = a .. ")"
    end
    b = b .. a .. " "
end
print(b)

Answer 13

Python 2, 105 97 88 86 85 84 bytes

x=1
while x<100:print((('%s','(%s)')[x%3<1],'[%s]')[x%4<1],'([%s])')[x%12<1]%x,;x+=1

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Answer 14

C (gcc), 84 bytes

main(i){while(99/i++)printf("%s%s%d%s%s ","("+i%3,"["+i%4,i-1,"]"+i%4,")"+i%3);}

There's a null byte at the beginning of each "bracket string".

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Answer 15

PowerShell, 67 62 bytes

"$(1..99|%{'('*!($x=$_%3)+'['*!($y=$_%4)+$_+']'*!$y+')'*!$x})"

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Basically a FizzBuzz using string multiplication times Boolean variables (implicitly cast to 1 or 0). Those strings are left on the pipeline and gathered within a script block inside quotes. Since the default $OutputFieldSeparator for an array is spaces, this implicitly gives us space-delimited array elements.

Answer 16

C#, 124 117 123 bytes

-5 bytes thanks to Kevin Cruijssen

x=>{for(int i=0;i++<99;)System.Console.Write((i%3<1?"(":"")+(i%4<1?"[":"")+i+(i%4<1?"]":"")+(i%3<1?")":"")+(i>98?"":" "));}

Test with :

Action<int> t = x=>{for(int i=0;i++<99;)System.Console.Write((i%3<1?"(":"")+(i%4<1?"[":"")+i+(i%4<1?"]":"")+(i%3<1?")":"")+(i>98?"":" "));}
t.Invoke(0);
Console.ReadKey();

Answer 17

Red, 99 97 bytes

repeat n 99[a: b: c: d:""if n% 3 = 0[a:"("c:")"]if n% 4 = 0[b:"["d:"]"]prin rejoin[a b n d c" "]]

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Answer 18

Ruby, 72 66 bytes

p [*1..99].map{|x|a=x
a="[#{x}]"if x%4<1
a="(#{a})"if x%3<1
a}*' '

Thanks to @jonathan-frech and @conor-obrien for additional trimming.

Answer 19

PowerShell, 98 82 74 67 63 62 bytes

A whopping -31 bytes thanks to @Veskah -5 bytes thanks to @ASCII-only

(1..99|%{(($a=($_,"[$_]")[!($_%4)]),"($a)")[!($_%3)]})-join' '

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I'm still not quite sure what I've done here.

Answer 20

perl -E, 60 bytes

$,=$";say map$_%12?$_%3?$_%4?$_:"[$_]":"($_)":"([$_])",1..99

Some bytes can be saved if we can use newlines between the numbers: in that case, we can remove the $,=$";, change the map into a for loop, while moving the say into the loop.

Answer 21

Perl 6, 51 48 bytes

put {$_%3??$^a!!"($a)"}(++$_%4??$_!!"[$_]")xx 99

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Answer 22

Batch, 145 bytes

@set s=
@for /l %%i in (1,1,99)do @set/an=%%i,b=n%%4,p=n%%3&call:c
@echo%s%
:c
@if %b%==0 set n=[%n%]
@if %p%==0 set n=(%n%)
@set s=%s% %n%

The code falls through into the subroutine but the string has already been printed by this point so the code executes harmlessly.

Answer 23

PHP 103

for(;$i<1e2;$i++)$a.=$i%12==0?"([$i]) ":($i%3==0?"($i) ":($i%4==0?"[$i] ":"$i "));echo substr($a,5,-1);

https://www.ideone.com/SBAuWp

Answer 24

Clean, 100 bytes

import StdEnv,Text

join" "[if(n/3*3<n)m("("+m+")")\\n<-[1..99],m<-[if(n/4*4<n)(""<+n)("["<+n<+"]")]]

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Answer 25

sfk, 225 bytes

for n from 1 to 99 +calc -var #(n)/3+1/3 -dig=0 +calc -var #text*3-#(n) +setvar t +calc -var #(n)/4 -dig=0 +calc -var #text*4-#(n) +xed -var _0_#(t)\[#(n)\]_ _*_#(t)#(n)_ +xed _0*_([part2])_ _?*_[part2]_ +xed "_\n_ _" +endfor

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Answer 26

Bash, 61 bytes

-14 bytes, thanks to Dennis

seq 99|awk '{ORS=" ";x=$1%4?$1:"["$1"]";print$1%3?x:"("x")"}'

explanation

Pretty straightforward:

• seq produces 1..99
• we pipe that into awk with the output record separator (ORS) set to space so the output is a single line.
• the main awk body just adds "[]" when the number is divisible by 4, and then adds, on top of that, "()" when divisible by 3.

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Answer 27

JavaScript (Node.js), 57 bytes

f=(n=99)=>n&&f(n-1)+(s=n%4?n:`[${n}]`,n%3?s:`(${s})`)+' '

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Changed to community cuz the optimize rely too much on it

Answer 28

PHP, 65 bytes

while($i++<99)echo$i%4?$i%3?$i:"($i)":($i%3?"[$i]":"([$i])")," ";

or

while($i++<99)echo"("[$i%3],"["[$i%4],$i,"]"[$i%4],")"[$i%3]," ";

(requires PHP 5.5 or later)

Run with -nr or try them online.

Answer 29

Python 2, 78 bytes

i=0
exec"i+=1;u=i%3/-2*(i%4/-3-1);print'([%0d])'[u:7-u:1+(i%3<1<=i%4)]%i,;"*99

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I envisioned this cool approach slicing '([%0d])' but I can't get the expressions any shorter.

Answer 30

Java 8, 92 91 bytes

-1 byte thanks to @Dana

i->{for(;i++<99;)out.printf((i>1?" ":"")+(i%12<1?"([%d])":i%3<1?"(%d)":i%4<1?"[%d]":i),i);}

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Alternative solution, 82 bytes (with trailing space in the output - not sure if that's allowed):

i->{for(;i++<99;)out.printf((i%12<1?"([%d])":i%3<1?"(%d)":i%4<1?"[%d]":i)+" ",i);}

Explanation:

for(;i++<99;) - a for loop that goes from the value of i (reused as input, taken to be 0 in this case) to 99

out.printf(<part1>+<part2>,i); - formats the string before immediately printing it to stdout with the value of i

where <part1> is (i>1?" ":"") - prints the space before printing the number unless that number is 1, in which case it omits the space

and <part2> is (i%12<1?"([%d])":i%3<1?"(%d)":i%4<1?"[%d]":i) - if i is divisible by both 3 and 4, i has both square and round brackets around it; else if i is divisible by 3, i has round brackets; else if i is divisible by 4, i has square brackets; else, i has no brackets.