Output the hours at 90 degrees

Question

Today while playing with my kids I noticed that an apparently simple toy in the park hid a challenge.

The wheel has a triangle that points to a number, but also has three circles that point to the numbers every 90 degrees from the first one. So:

Challenge (really simple)

Given an integer between 1 and 12 (the one pointed by the triangle) in any acceptable form, output also in any acceptable form and order the three numbers pointed by the circles (the ones every 90 degrees).

Test cases

In       Out
1        4, 7, 10
2        5, 8, 11
3        6, 9, 12
4        7, 10, 1
5        8, 11, 2
6        9, 12, 3
7        10, 1, 4
8        11, 2, 5
9        12, 3, 6
10       1, 4, 7
11       2, 5, 8
12       3, 6, 9

This is code-golf, so may the shortest code for every language win!

Is this the fourth challenge now based on some activity involving your kids? :P — FlipTack, Dec 29, 2017 at 21:58
@FlipTack I've lost count. :-) But given that I spent most of my free time with my kids, guess where does my inspiration come from... — Charlie, Dec 29, 2017 at 22:17
@Steadybox That would be a meta tag, which is BAD. But yes, we do. (Perhaps a meta question with links would suffice?) — wizzwizz4, Jan 1, 2018 at 10:32

xnor · Accepted Answer · 2017-12-29 21:16:06Z

16

Python 3, 33 bytes

lambda n:{*range(n%3,13,3)}-{n,0}

Try it online!

Python 2, 35 bytes

lambda n:[(n+c)%12+1for c in 2,5,8]

Try it online!

answered Dec 29, 2017 at 21:16

xnor

141k26 gold badges271 silver badges621 bronze badges

Add a comment |

Jonathan Allan · Accepted Answer · 2017-12-29 21:32:53Z

10

Jelly, 8 bytes

12Rṙ’m3Ḋ

A monadic link taking a number and returning a list of numbers.

Try it online! or see all cases.

How?

12Rṙ’m3Ḋ - Link: number, n   e.g. 5
12       - literal twelve         12
  R      - range                  [1,2,3,4,5,6,7,8,9,10,11,12]
    ’    - decrement n            4
   ṙ     - rotate left            [5,6,7,8,9,10,11,12,1,2,3,4]
      3  - literal three          3
     m   - modulo slice           [5,8,11,2]
       Ḋ - dequeue                [8,11,2]

edited Dec 29, 2017 at 21:32

answered Dec 29, 2017 at 21:14

Jonathan Allan

99.7k7 gold badges62 silver badges264 bronze badges

\$\begingroup\$ Alternative solution: 12Rṙm-3Ḋ (output in reverse order) \$\endgroup\$
– DELETE_ME
Dec 30, 2017 at 2:14

Add a comment |

notjagan · Accepted Answer · 2017-12-29 21:09:49Z

8

Python 2, 35 bytes

lambda n:(range(1,13)*2)[n+2:n+9:3]

Try it online!

answered Dec 29, 2017 at 21:09

notjagan

4,2018 silver badges27 bronze badges

Add a comment |

Steadybox · Accepted Answer · 2017-12-29 22:50:56Z

7

R, 29 28 bytes

Thanks to @Giuseppe for saving a byte!

function(n)(n+1:3*3-1)%%12+1

Try it online!

edited Dec 29, 2017 at 22:50

answered Dec 29, 2017 at 21:31

Steadybox

16.4k5 gold badges38 silver badges83 bronze badges

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

6

MATL, 9 bytes

I:I*+12X\

Try it online!

Explanation

Consider input 4 as an exmaple.

I:     % Push [1 2 3]
       % STACK: [1 2 3]
I      % Push 3
       % STACK: [1 2 3], 3
*      % Multiply, element-wise
       % STACK: [3 6 9]
+      % Add implicit input, element-wise
       % STACK: [7 10 13]
12     % Push 12
X\     % 1-based modulus. Implicit display
       % STACK: [7 10 1]

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Dec 29, 2017 at 22:35

Luis Mendo

103k9 gold badges130 silver badges362 bronze badges

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

5

APL+WIN, 13bytes

(⎕⌽⍳12)[3×⍳3]

Explanation:

⎕ Prompt for screen input of indicated time t

⍳12 Create a vector of integers from 1 to 12

⌽ Rotate the vector by t elements front to back

[3×⍳3] Select 3rd, 6th and 9th elements.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Dec 29, 2017 at 22:17

Graham

5,13613 silver badges11 bronze badges

\$\begingroup\$ Quite neat. I like that indexed rotation approach \$\endgroup\$
– Uriel
Dec 30, 2017 at 18:47

Add a comment |

Steadybox · Accepted Answer · 2017-12-29 21:08:59Z

4

C (gcc), 48 bytes

i;f(n){for(i=4;--i;printf("%d ",n%12?:12))n+=3;}

Try it online!

answered Dec 29, 2017 at 21:08

Steadybox

16.4k5 gold badges38 silver badges83 bronze badges

Add a comment |

Arnauld · Accepted Answer · 2017-12-29 23:26:59Z

4

JavaScript (ES6), 29 bytes

Similar to xnor's answer.

n=>[2,5,8].map(k=>(n+k)%12+1)

Demo

let f =

n=>[2,5,8].map(k=>(n+k)%12+1)

for(n = 1; n <= 12; n++) {
  console.log(n + ' -> ' + f(n));
}

answered Dec 29, 2017 at 23:26

Arnauld

181k18 gold badges168 silver badges611 bronze badges

\$\begingroup\$ I had n=>[3,6,9].map(v=>(v+n)%12) then realized that it returns 0, not 12... \$\endgroup\$
– ericw31415
Dec 30, 2017 at 0:47
\$\begingroup\$ @ericw31415 Actually, my 1st approach was n=>[3,6,9].map(v=>(v+n)%12||12) (but that's 31 bytes). \$\endgroup\$
– Arnauld
Dec 30, 2017 at 1:19

Add a comment |

Tom Carpenter · Accepted Answer · 2017-12-30 11:39:30Z

4

Octave, 25 bytes

@(x)[a=1:12 a](3+x:3:9+x)

Try it online!

Fairly simple anonymous function.

We first create an array of [1:12 1:12] - so two copies of the full number set. Then we index in to select the values of x+3, x+6, x+9, where x is the number input.

Octave is 1-indexed, so we can simply select the array elements based on the input (although to be honest 0-indexed would use the same number of bytes here).

This seems to use a method that is unique to the other answers in that by having two copies of the array, we don't have to wrap the indices using modulo.

answered Dec 30, 2017 at 11:39

Tom Carpenter

4,23011 silver badges21 bronze badges

\$\begingroup\$ Very nice, but the "boring" mod approach is shorter! Try it online! \$\endgroup\$
– Giuseppe
Dec 30, 2017 at 12:45
\$\begingroup\$ @Giuseppe lol, I'm sure I tried using mod and couldn't make it shorter. Well done! Feel free to post as an answer. \$\endgroup\$
– Tom Carpenter
Dec 30, 2017 at 13:02

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

Befunge-93, 20 19 18 bytes

852<_@#:.+1%+66+&p

Try it online!

Explanation

852                   Push 8, 5 and 2 onto the stack - the offsets we're going to add.
   <                  Reverse direction, and start the main loop.
852                   Push 2, 5, and 8 onto the stack, but we don't actually want these.
                 p    So we use a "put" operation to drop the top three values.
                &     Read the hour from stdin.
               +      Add it to the topmost offset.
         +1%+66       Mod 12 and add 1 to get it in the range 1 to 12.
        .             Then output the result to stdout.
    _@#:              Exit if the next offset is zero (i.e. nothing more on the stack).
   <                  Otherwise start the main loop again.

This relies on behaviour specific to the reference interpreter: on end-of-file, the & operator returns the last value that was read. That's why we can safely re-read the hour from stdin on each iteration of the loop.

@JoKing Technically it's a bug in the interpreter, another related side-effect being that & can also be used as a kind of one-off random number generator. That is less reliable, though, since it depends on the compiler that was used to build it. It's working on TIO at the moment, but there was a time when Dennis changed to a different version of gcc and we lost that functionality for a while. — James Holderness, Dec 30, 2017 at 13:45

Uriel · Accepted Answer · 2017-12-30 18:46:36Z

3

APL (Dyalog), 12 bytes

1+12|2 5 8∘+

Try it online!

answered Dec 30, 2017 at 18:46

Uriel

13.1k4 gold badges16 silver badges44 bronze badges

Add a comment |

dzaima · Accepted Answer · 2017-12-29 21:28:11Z

2

SOGL V0.12, 9 bytes

3{3+:6«‰P

Try it Here!

answered Dec 29, 2017 at 21:28

dzaima

20.2k2 gold badges39 silver badges75 bronze badges

Add a comment |

Ayb4btu · Accepted Answer · 2017-12-30 10:47:55Z

2

C# (.NET Core), 42 bytes

h=>new[]{(2+h)%12+1,(5+h)%12+1,(8+h)%12+1}

Try it online!

Essentially just a port of many of the other answers to C#.

answered Dec 30, 2017 at 10:47

Ayb4btu

5511 gold badge4 silver badges7 bronze badges

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

2

Japt, 11 bytes

3ÆU±3 uC ªC

Try it

Explanation

Implicit input of integer U. Generate a 3 element array (3Æ) and, for each element, increment U by 3 (U±3), modulo by 12 (uC) and, because 12%12=0, return the result OR 12 (ªC).

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Dec 30, 2017 at 15:51

Shaggy

40.6k4 gold badges32 silver badges90 bronze badges

Add a comment |

mkst · Accepted Answer · 2017-12-30 18:18:00Z

2

K (oK), 11 bytes

Solution:

1+12!2 5 8+

Try it online!

Examples:

1+12!2 5 8+1
4 7 10
1+12!2 5 8+2
5 8 11
1+12!2 5 8+3
6 9 12
1+12!2 5 8+4
7 10 1

Explanation:

This was the first solution that came to mind. Might not be the best or shortest.

1+12!2 5 8+ / the solution
     2 5 8+ / add 2, 5 and 8 to the input
  12!       / apply modulo 12 to the results
1+          / add 1

answered Dec 30, 2017 at 18:18

mkst

5,05511 silver badges19 bronze badges

Add a comment |

QunSyBer · Accepted Answer · 2017-12-31 18:40:33Z

2

GolfScript, 46 bytes

~13,1>:x?:y;0:i;x y 3+12%=x y 6+12%=x y 9+12%=

This is the first time I'm doing code golf, so with my lack of experience I've probably not found the best solution, but I need to start somewhere, right?

Try it online or try all cases

answered Dec 31, 2017 at 18:40

QunSyBer

1212 bronze badges

\$\begingroup\$ Hello, welcome to the site! This looks like a nice first answer :) Unfortunately, I don't know anything about golfscript, but you might be able to get some tips here \$\endgroup\$
– DJMcMayhem
Dec 31, 2017 at 19:20

Add a comment |

MegaTom · Accepted Answer · 2018-01-03 17:27:30Z

2

Brain-Flak, 84 bytes

(()()()){({}<(()(){})(((()()()){}){}<>){(({})){({}[()])<>}{}}<>(([{}]{})<>)>[()])}<>

Try it online!

At least I beat Doorknob's face solution...

Explanation:

LOOP 3 TIMES: (()()()){({}<

  n += 2:
   (()(){})
  push 12:
   (((()()()){}){}<>)
  n mod 12 + 1; pushing to both stacks:
   {(({})){({}[()])<>}{}}<>(([{}]{})<>)

END LOOP: >[()])}<>

answered Jan 3, 2018 at 17:27

MegaTom

4,2061 gold badge19 silver badges24 bronze badges

Add a comment |

Dave · Accepted Answer · 2018-01-05 18:53:31Z

2

Pyth, 12 bytes

%R12+LQ*R3S3

Try it online!

answered Jan 5, 2018 at 18:53

Dave

211 bronze badge

1

\$\begingroup\$ Welcome to the site! :) \$\endgroup\$
– DJMcMayhem
Jan 5, 2018 at 18:54

Add a comment |

FarSeenNomic · Accepted Answer · 2020-11-27 22:31:38Z

2

Wolfram Language (Mathematica) 20 bytes

Mod[#+{3,6,9},12,1]&

The normal modulus operation, threaded over a list, offset by 1.

Try it online!

edited Nov 27, 2020 at 22:31

answered Nov 27, 2020 at 18:42

FarSeenNomic

6243 silver badges5 bronze badges

Add a comment |

SunnyMoon · Accepted Answer · 2020-11-28 18:43:29Z

2

05AB1E, 8 bytes

ƵžS+12%>

At the time of writing this, this submission has the lowest score submitted. Technically I am winning!

Try it online!

How?

ƵžS                   # Push the list [2, 5, 8]
   +                  # Add the input to each number
    12%               # Mod each result by 12
       >              # Increment each
                      # And print the ending list implicitly!

See this tip of Kevin to know why Ƶž is 258!

edited Nov 28, 2020 at 18:43

answered Nov 28, 2020 at 18:38

SunnyMoon

2,64910 silver badges24 bronze badges

Add a comment |

a stone arachnid · Accepted Answer · 2022-02-19 04:13:09Z

2

Vyxal, 8 bytes

⁺ẇf+12%›

Try it online! (All test cases)

Port of SunnyMoon's 05AB1E solution

How it works:

⁺ẇf+12%›
⁺ẇ        The number 258
  f       ...as a list of digits ⟨ 2 | 5 | 8 ⟩
   +      Add the input to each
    12%   Mod-12 each
       ›  Increment each
          (implicit output)

answered Feb 19, 2022 at 4:13

a stone arachnid

3,17012 silver badges21 bronze badges

Add a comment |

naffetS · Accepted Answer · 2022-07-02 23:29:25Z

2

Piet + ascii-piet, 62 bytes (46×2=92 codels)

TABdDTvVvVnNNFJAQUiIiIIAeEeEuUUMTJBCSBvVvVNMLnNnNfFFVBQIMEiCcc

Try Piet online!

answered Jul 2, 2022 at 23:29

naffetS

11.9k2 gold badges16 silver badges47 bronze badges

Add a comment |

Bubbler · Accepted Answer · 2022-07-05 02:07:34Z

Piet + ascii-piet, 45 bytes (5×18=90 codels)

tAaaqreeumccsqqqijlVa rbiqaaueljnvea?_ t?B tt

Try Piet online!

My looping template is getting weirder and weirder... :P

inN 4    [n 4]; force toggle CC at the exit of 4 region
Loop:
  dup CC+    [n flag] switch CC on 3rd iteration so it will exit after the loop
  2 1 roll   [flag n]
  2 +        [flag n+2]
  3 dup dup * + %  [flag (n+2)%12]
  1 + d outN [flag next_n] compute and print the next number
  1 outC     [flag next_n] print the separator (ASCII 1)
  2 1 roll   [next_n flag]
  2 /        [next_n flag/2] divide flag by 2; flag becomes odd after two loops

Οurous · Accepted Answer · 2017-12-29 21:38:24Z

1

Clean, 44 bytes

import StdEnv
@n=[(n+i)rem 12+1\\i<-[2,5,8]]

Try it online!

answered Dec 29, 2017 at 21:38

Οurous

8,0861 gold badge15 silver badges37 bronze badges

Add a comment |

FlipTack · Accepted Answer · 2017-12-29 21:46:55Z

Pushy, 12 bytes

258s{K+12%h_

Try it online!

258            \ Push 258                            
   s           \ Split into digits, yielding [2, 5, 8]
    {K+        \ Add input to each
       12%     \ Modulo each by 12
          h    \ Increment each
           _   \ Print (space separated)

12 bytes

An alternative for the same byte count:

12R{:{;$...#

Try it online!

12R            \ Push range(1, 12), inclusive
   {: ;        \ Input times do:
     {         \   Rotate left
       $       \ While there are items on stack:
        ...    \   Pop the top three
           #   \   Print top item

totallyhuman · Accepted Answer · 2017-12-30 01:16:52Z

1

Haskell, 29 bytes

-2 bytes thanks to Laikoni.

f n=[mod(n+i)12+1|i<-[2,5,8]]

Try it online!

Not very original, no...

edited Dec 30, 2017 at 1:16

answered Dec 30, 2017 at 0:29

totallyhuman

16.2k3 gold badges31 silver badges86 bronze badges

Add a comment |

Endenite · Accepted Answer · 2017-12-30 11:52:32Z

1

Swift, 30 bytes

{n in[2,5,8].map{($0+n)%12+1}}

Try it online!

Port of many other answers to Swift

answered Dec 30, 2017 at 11:52

Endenite

4,05110 silver badges28 bronze badges

Add a comment |

Sean · Accepted Answer · 2017-12-31 20:28:06Z

1

Perl 6, 23 bytes

{map ($_+*)%12+1,2,5,8}

Try it online!

answered Dec 31, 2017 at 20:28

Sean

7,3769 silver badges15 bronze badges

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

1

PHP, 37+1 bytes

while($i++<3)echo(~-$argn+=3)%12+1,_;

Run as pipe with -nR or try it online.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Dec 31, 2017 at 22:21

Titus

14.6k1 gold badge22 silver badges41 bronze badges

Add a comment |

Doorknob · Accepted Answer · 2018-01-01 15:15:33Z

face, 96 94 bytes

(%d
@)\$*,c'$ooiim%*m1*6%+%%%11m!*mn*m~*3!m&!r!&!is!&$pn3!:L+nn1+nn1%nn%+nn1p~>$inw~>~o-!!1?!L

This simply adds two, mods by 12, adds one more, and prints. Then it does that two more times.

Commented version:

(%d
@)

\$*,c'$ooii     ( store format string in $, ip in *, get stdin/out )
m%*m1*6%+%%%11  ( initialize constants, %=12, 1=1 )
m!*mn*m~*       ( malloc space for a counter, input var, and length )
3!m&!r!&!i      ( read into & )
s!&$pn          ( scan into n )
3!:L            ( start of main loop, executed thrice )
  +nn1+nn1      ( add 2 to n )
  %nn%+nn1      ( mod by 12 and add 1 more )
  p~>$in        ( sprintf n into > )
  w~>~o         ( output to stdout )
  -!!1          ( decrement counter )
?!L             ( conditional jump back to loop start )

Try it online! (The trailing newline is required on TIO due to a bug that's been fixed in a newer version of face.)

Creating a variable that holds the value 3 m3*33 allows you to replace the three +nn1s with one +nn3 tio.run/… — user41805, Jan 1, 2018 at 15:56

Output the hours at 90 degrees

46 Answers 46

Python 3, 33 bytes

Python 2, 35 bytes

Jelly, 8 bytes

How?

Python 2, 35 bytes

R, 29 28 bytes

MATL, 9 bytes

Explanation

APL+WIN, 13bytes

C (gcc), 48 bytes

JavaScript (ES6), 29 bytes

Demo

Octave, 25 bytes

Befunge-93, 20 19 18 bytes

APL (Dyalog), 12 bytes

SOGL V0.12, 9 bytes

C# (.NET Core), 42 bytes

Japt, 11 bytes

Explanation

K (oK), 11 bytes

GolfScript, 46 bytes

Brain-Flak, 84 bytes

Pyth, 12 bytes

Wolfram Language (Mathematica) 20 bytes

05AB1E, 8 bytes

How?

Vyxal, 8 bytes

Piet + ascii-piet, 62 bytes (46×2=92 codels)

Piet + ascii-piet, 45 bytes (5×18=90 codels)

Clean, 44 bytes

Pushy, 12 bytes

12 bytes

Haskell, 29 bytes

Swift, 30 bytes

Perl 6, 23 bytes

PHP, 37+1 bytes

face, 96 94 bytes

Your Answer

Sign up or log in

Post as a guest

Not the answer you're looking for? Browse other questions tagged code-golfintegerclock or ask your own question.

Linked

Related

Hot Network Questions

Not the answer you're looking for? Browse other questions tagged
code-golf
integer
clock
or ask your own question.