Play the "bzzt" game

Question

Winner: Aditsu's CJam answer! A whopping 25 bytes! Nice!

You may continue to submit your answer, however you can no longer win. Original post kept for posterity:

The "Bzzt" game is a game where you must count to a number (in this case 500). However, if the number has a 3 in it, or is divisible by 3, you don't say the number. Instead, you say "Bzzt".

Rules:

You can't hard-code the numbers.
The number only has to satisfy at least 1 of the following requirements
- Divisible by 3
- Number contains a 3
Some type of separator is mandatory (12bzzt14 doesn't count)
Score is measured in bytes.
You must count exactly to 500, starting at 1 or 0 (you chose).
The numbers must be output, but it doesn't matter how (e.g., stdout, writing to a text file, etc.).
0 can be divisible by 3 or not divisible. You can choose.
You can output the numbers one at a time (ex, output 1, then 2, then bzzt, then 4, etc) or all at once (e.g., output 1 2 bzzt 4 5).
You must replace the letter 3 with the word "bzzt". This is not case sensitive (bZzt, Bzzt, bzzt are all okay).

This is a code-golf challenge, so the shortest code wins.
This contest ends June 30th 2014 (7 days from posting).

Please clarify. Do I have to output buzz or bzzt if both of the requirements apply? Do I have to output 12bzzt4 or bzzt for 1234? — user344, Commented Jun 23, 2014 at 12:41
I'd say bzzt for 1234. It's a 'common' drinking game here (only we often do it with 7) — Martijn, Commented Jun 23, 2014 at 13:38
"0 can be divisible by 3 or not divisible. You can choose." I really don't think you can choose. 0 mod 3 is 0, that's not really a matter of opinion. — David Conrad, Commented Jun 23, 2014 at 15:54

manatwork · Accepted Answer · 2016-02-09 14:12:27Z

1

jq: 59 bytes

(56 characters code + 3 characters command line option.)

range(501)|select(.%3<1or("\(.)"|test("3"))|not)//"bzzt"

Sample run:

bash-4.3$ jq -n -r 'range(501)|select(.%3<1or("\(.)"|test("3"))|not)//"bzzt"' | head
bzzt
1
2
bzzt
4
5
bzzt
7
8
bzzt

On-line test (Passing -r through URL is not supported – check Raw Output yourself.)

edited Feb 9, 2016 at 14:12

answered Jul 25, 2015 at 16:37

manatwork

20.6k5 gold badges51 silver badges81 bronze badges

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

beeswax, 148 bytes

beeswax was created after the challenge, but why shouldn’t I give the challenge a try anyway? ;)

Not golfed to death yet, but it satisfies both requirements:

 ;pK~gFP{?gp'           <      <
>F B~6-!P~2 .Pfz1~h#>@-"d~9P~:Fd
4 N     >@f>F9P~%~3Kd`bzzt`g?Pp
*0>F3~%'d`bzzt`@Pp
  dN             <            <

Output:

julia> beeswax("bzzt.bswx")         
bzzt                                                     
1                                                        
2                                                        
bzzt                                                     
4                                                        
5                                                        
bzzt                                                     
7                                                        
8                                                        
bzzt                                                     
10                                                       
11                                                       
bzzt                                                     
bzzt                                                     
14                                                       
bzzt                                                     
16                                                       
...
29
bzzt
bzzt
bzzt
bzzt
bzzt
bzzt
bzzt
bzzt
bzzt
bzzt
40
41
...
296
bzzt
298
299
bzzt
bzzt
bzzt
bzzt
bzzt
bzzt
bzzt
bzzt
bzzt
...
bzzt
bzzt
bzzt
bzzt
400
401
bzzt
bzzt
404
...
494
bzzt
496
497
bzzt
499
500
Program finished!

You can clone my beeswax interpreter (written in Julia) from my GitHub repository.

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

1

Groovy, 46 45 bytes

{(0..500).collect{x->x%3<1||x=~/3/?'bzzt':x}}

This is an unnamed closure. Try it Online!

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jan 20, 2017 at 13:04

Gurupad Mamadapur

1,99116 silver badges26 bronze badges

1

\$\begingroup\$ I never programmed in Groovy, but if I read it correctly you can save one by byte changing x%3==0 to x%3<1 (this trick can only be used if the result cannot be negative, which is the case in your 0 to 500 loop). \$\endgroup\$
– Kevin Cruijssen
Commented Jan 20, 2017 at 15:05
\$\begingroup\$ @KevinCruijssen Ah ha, thanks for that. \$\endgroup\$
– Gurupad Mamadapur
Commented Jan 20, 2017 at 15:12

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Bubbler · Accepted Answer · 2018-06-07 00:43:54Z

1

Japt, 21 bytes

501Ç%3«/3/tZ ?Z:"bzzt

Try it online!

Almost direct translation of nderscore's JS submission.

Unpacked & How it works

501oZ{Z%3&&!/3/tZ ?Z:"bzzt

501oZ{                      Map range(501) to a function...
      Z%3&&!/3/tZ             If Z is not a multiple of 3 and Z doesn't have digit 3
                  ?Z          Return Z
                    :"bzzt    Otherwise, return "bzzt"

answered Jun 7, 2018 at 0:43

Bubbler

78.4k5 gold badges149 silver badges469 bronze badges

1

\$\begingroup\$ 20 bytes \$\endgroup\$
– Shaggy
Commented Jun 13, 2018 at 16:49

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Unrelated String · Accepted Answer · 2019-03-05 10:23:41Z

Brachylog, 23 bytes (posted way after the end of the challenge)

500⟦{{f|}∋3∧"bzzt"|}ᵐẉᵐ

Try it online!

    {              }ᵐ      For every
500⟦                       number from 0 to 500,
            "bzzt"         the string "bzzt"
         ∋3∧               if 3 is an element of
     {f }                  the input's factors
       |                   or the input itself,
                  |        the input if neither condition is satisfied.
                     ẉᵐ    Write that all with one line per item.

squid · Accepted Answer · 2019-03-05 12:18:06Z

1

Python 3, 56 bytes (Non-Competing)

for x in range(501):print([x,'bzzt']['3'[:x%3]in'%s'%x])

Try it online!

Shortest I could get it in Python 3. Can't yet comment on the existing Python 3 solution so posting as a separate answer.

answered Mar 5, 2019 at 12:18

squid

1,5021 gold badge7 silver badges15 bronze badges

\$\begingroup\$ Why is this marked non-competing? \$\endgroup\$
– The Fifth Marshal
Commented Sep 21, 2019 at 23:39

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Razetime · Accepted Answer · 2020-09-24 04:25:06Z

1

APL (Dyalog Unicode), 30 bytes

{(0=3|⍵)∨'3'∊⍕⍵:'bzzt'⋄⍵}¨⍳500

Try it online!

edited Sep 24, 2020 at 4:25

answered Sep 24, 2020 at 4:04

Razetime

27.4k3 gold badges29 silver badges72 bronze badges

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Makonede · Accepted Answer · 2020-09-24 17:55:55Z

1

05AB1E, 19 bytes

₄;END3ÖN3å~i.•è‰•},

Try it online!

₄;                   # push 500
  E                  # for N in [0, 1, 2, ..., 500]...
   ND3ÖN3å~          # push N and 1 if either 3 is in N or N is divisible by 3, else 0
           i         # if top of stack is truthy:
            .•è‰•}   # push "bzzt"
                  ,  # output top of stack to STDOUT

answered Sep 24, 2020 at 17:55

Makonede

6,60020 silver badges49 bronze badges

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Fmbalbuena · Accepted Answer · 2023-12-03 01:07:37Z

Vyxal 3 `j`, 14 bytes

500ƛ3∦cḊa["ġ/„

Try it Online!

-9 bytes thanks to @lyxal

-1 byte thanks to @Aaroneous Miller

Explanation (outdated):

500ɾ(nD3c$K3c∨["bzzt",|,⁡‎‎⁪⁡⁪⁠⁪⁡⁪‏⁠‎⁪⁡⁪⁠⁪⁢⁪‏⁠‎⁪⁡⁪⁠⁪⁣⁪‏⁠‎⁪⁡⁪⁠⁪⁤⁪‏‏⁡⁠⁡‌⁢‎‎⁪⁡⁪⁠⁪⁢⁡⁪‏‏⁡⁠⁡‌⁣‎‎⁪⁡⁪⁠⁪⁢⁢⁪‏⁠‎⁪⁡⁪⁠⁪⁢⁣⁪‏‏⁡⁠⁡‌⁤‎‎⁪⁡⁪⁠⁪⁢⁤⁪‏⁠‎⁪⁡⁪⁠⁪⁣⁡⁪‏‏⁡⁠⁡‌⁢⁡‎‎⁪⁡⁪⁠⁪⁣⁢⁪‏⁠‎⁪⁡⁪⁠⁪⁤⁢⁪‏‏⁡⁠⁡‌⁢⁢‎⁪⁪⁠‎⁪⁡⁪⁠⁪⁣⁣⁪‏⁠‎⁪⁡⁪⁠⁪⁣⁤⁪‏⁠‎⁪⁡⁪⁠⁪⁤⁡⁪‏‏⁡⁠⁡‌⁢⁣‎‎⁪⁡⁪⁠⁪⁤⁣⁪‏⁠‎⁪⁡⁪⁠⁪⁤⁤⁪‏⁠‎⁪⁡⁪⁠⁪⁢⁡⁡⁪‏⁠‎⁪⁡⁪⁠⁪⁢⁡⁢⁪‏⁠‎⁪⁡⁪⁠⁪⁢⁡⁣⁪‏⁠‎⁪⁡⁪⁠⁪⁢⁡⁤⁪‏⁠‎⁪⁡⁪⁠⁪⁢⁢⁡⁪‏⁠‎⁪⁡⁪⁠⁪⁢⁢⁢⁪‏‏⁡⁠⁡‌⁢⁤‎‎⁪⁡⁪⁠⁪⁢⁢⁣⁪‏⁠‎⁪⁡⁪⁠⁪⁢⁢⁤⁪‏‏⁡⁠⁡‌
500ɾ                      ## ‎⁡push [1..500]
    (                     ## ‎⁢loop every number
     nD                   ## ‎⁣push the loop variable three times
       3c                 ## ‎⁤Contains 3?
         $   ∨            ## ‎⁢⁡or
          K3c             ## ‎⁢⁢it's divisible by 3?
              ["bzzt",    ## ‎⁢⁣If so, print "bzzt"
                      |,  ## ‎⁢⁤else print the loop variable
💎

Created with the help of Luminespire.

\$\begingroup\$ Try it Online! for 15 bytes \$\endgroup\$
– lyxal
Commented Dec 2, 2023 at 0:27 — lyxal, Commented Dec 2, 2023 at 0:27
\$\begingroup\$ Try it Online! for 14 bytes \$\endgroup\$
– Aaroneous Miller
Commented Dec 2, 2023 at 3:19 — Aaroneous Miller, Commented Dec 2, 2023 at 3:19
\$\begingroup\$ @AaroneousMiller done \$\endgroup\$
– Fmbalbuena
Commented Dec 3, 2023 at 1:07 — Fmbalbuena, Commented Dec 3, 2023 at 1:07

Spedwards · Accepted Answer · 2014-06-23 08:49:24Z

0

Javascript - 109 97

Compressed:

function b(m,c,z){for(i=1;i<=m;i++){if((i+'').indexOf(c)>-1||i%c==0){alert(z);continue}alert(i)}}

Replaced console.log with alert

Uncompressed:

function bzzt(max, contains, z) {
    for(i=1;i<=max;i++) {
        if((i+'').indexOf(contains)>-1||i%contains==0) {
            console.log(z);
            continue;
        }
        console.log(i); // console.log instead of alert for testing purposes
    }
}

max is the number to count to, contains is the number it contains and is divisible by, z is the message, so 'bzzt'

b(500, 3, 'bzzt');

edited Jun 23, 2014 at 8:49

answered Jun 23, 2014 at 7:24

Spedwards

1798 bronze badges

\$\begingroup\$ You know this is about shortest right? The function only makes it longer, and console.log can be replaced with alert. \$\endgroup\$
– Martijn
Commented Jun 23, 2014 at 8:31
\$\begingroup\$ @Martijn Honestly didn't think about alert. And I thought I saw function in the rules. \$\endgroup\$
– Spedwards
Commented Jun 23, 2014 at 8:38
\$\begingroup\$ @Martijn Also, no one would want to run it if it were alerts haha \$\endgroup\$
– Spedwards
Commented Jun 23, 2014 at 8:39
\$\begingroup\$ Haha. You could add comment in the uncompressed //console instead of alert for testingpurposes so people'll get it :) \$\endgroup\$
– Martijn
Commented Jun 23, 2014 at 8:43

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bacchusbeale · Accepted Answer · 2014-06-23 11:26:04Z

0

C 174

int i,j,k;char n[3];
int main(){
while(i++<501){k=0;sprintf(n,"%d",i);
for(j=0;j<3;j++){if(n[j]=='3'){k=1;break;}}
(i%3==0||k)?puts("Bzzt"):printf("%d\n",i);}
return 0;}

answered Jun 23, 2014 at 11:26

bacchusbeale

1,2636 silver badges7 bronze badges

1

\$\begingroup\$ You can save a whole bunch of characters by not returning; not declaring the return type of main; setting k=0 to mean "there's a 3" and writing (i%3&&k):printf(...):puts(...) etc. By the way, you need a newline after "Bzzt". \$\endgroup\$
– David Richerby
Commented Jun 23, 2014 at 22:37
\$\begingroup\$ @David puts put a newline by itself. \$\endgroup\$
– edc65
Commented Jun 25, 2014 at 11:26
\$\begingroup\$ @edc65 Thanks. That being the case, we can save even more: declare n to have length 4 (necessary anyway to avoid a potential buffer overrun from the sprintf), make the other changes I suggested in my first comment and replace the fifth line of the source with puts((i%3&&k)?n:"Bzzt");} \$\endgroup\$
– David Richerby
Commented Jun 25, 2014 at 14:47
\$\begingroup\$ I implemented my suggestions as a new answer. Also, it's much shorter to use strchr to test for the digit 3 in the string. (Another improvement to the for loop: the break is redundant, because it doesn't matter if the number has more than 3 in it and you set k=1 more than once.) \$\endgroup\$
– David Richerby
Commented Jun 25, 2014 at 16:06

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izb · Accepted Answer · 2014-06-26 10:45:12Z

JavaScript, 65

OK, longer than others so far, but I've never tried this before and felt compelled to post...

a='';for(i=0;i-->-500;)a+=(i%3?i:-3);a.replace(/\d*3\d*/g,'bzzt')

Usage:

From a node repl paste the code. Output is by returned value

Output:

'-1-2-bzzt-4-5-bzzt-7-8-bzzt-10-11-bzzt-bzzt-14-bzzt-16-17-bzzt-19-20-bzzt-22-bzzt-bzzt-25-26-bzzt-28-29-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-40-41-bzzt-bzzt-44-bzzt-46-47-bzzt-49-50-bzzt-52-bzzt-bzzt-55-56-bzzt-58-59-bzzt-61-62-bzzt-64-65-bzzt-67-68-bzzt-70-71-bzzt-bzzt-74-bzzt-76-77-bzzt-79-80-bzzt-82-bzzt-bzzt-85-86-bzzt-88-89-bzzt-91-92-bzzt-94-95-bzzt-97-98-bzzt-100-101-bzzt-bzzt-104-bzzt-106-107-bzzt-109-110-bzzt-112-bzzt-bzzt-115-116-bzzt-118-119-bzzt-121-122-bzzt-124-125-bzzt-127-128-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-140-bzzt-142-bzzt-bzzt-145-146-bzzt-148-149-bzzt-151-152-bzzt-154-155-bzzt-157-158-bzzt-160-161-bzzt-bzzt-164-bzzt-166-167-bzzt-169-170-bzzt-172-bzzt-bzzt-175-176-bzzt-178-179-bzzt-181-182-bzzt-184-185-bzzt-187-188-bzzt-190-191-bzzt-bzzt-194-bzzt-196-197-bzzt-199-200-bzzt-202-bzzt-bzzt-205-206-bzzt-208-209-bzzt-211-212-bzzt-214-215-bzzt-217-218-bzzt-220-221-bzzt-bzzt-224-bzzt-226-227-bzzt-229-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-241-242-bzzt-244-245-bzzt-247-248-bzzt-250-251-bzzt-bzzt-254-bzzt-256-257-bzzt-259-260-bzzt-262-bzzt-bzzt-265-266-bzzt-268-269-bzzt-271-272-bzzt-274-275-bzzt-277-278-bzzt-280-281-bzzt-bzzt-284-bzzt-286-287-bzzt-289-290-bzzt-292-bzzt-bzzt-295-296-bzzt-298-299-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-400-401-bzzt-bzzt-404-bzzt-406-407-bzzt-409-410-bzzt-412-bzzt-bzzt-415-416-bzzt-418-419-bzzt-421-422-bzzt-424-425-bzzt-427-428-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-bzzt-440-bzzt-442-bzzt-bzzt-445-446-bzzt-448-449-bzzt-451-452-bzzt-454-455-bzzt-457-458-bzzt-460-461-bzzt-bzzt-464-bzzt-466-467-bzzt-469-470-bzzt-472-bzzt-bzzt-475-476-bzzt-478-479-bzzt-481-482-bzzt-484-485-bzzt-487-488-bzzt-490-491-bzzt-bzzt-494-bzzt-496-497-bzzt-499-500'

ProgramFOX · Accepted Answer · 2014-06-26 12:45:21Z

0

Python3, 62

for i in range(501):print('bzzt'if'3'in str(i) or i%3<1 else i)

edited Jun 26, 2014 at 12:45

ProgramFOX

8,3237 gold badges45 silver badges70 bronze badges

answered Jun 26, 2014 at 12:25

klingt.net

9466 silver badges7 bronze badges

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Adam Speight · Accepted Answer · 2014-06-26 13:30:42Z

0

VB.net (~100c)

1)

Function t()
For c=0 To 500:t+=If(CStr(c).Contains("3")Or(c Mod 3)=0,"Bzzt",c)+" ":Next
End Function
Function t()

2) I'd forgotten about Like operator in vb.

Function t() 
For c = 0To 500:t &=If(CStr(c)Like"*3*"Or(c Mod 3)=0,"Bzzt",c)&" ":Next
End Function

The result is output as the return value of the function.

The numbers must be output, but it doesn't matter how (e.g., stdout, writing to a text file, etc.).

edited Jun 26, 2014 at 13:30

answered Jun 26, 2014 at 12:42

Adam Speight

1,31410 silver badges9 bronze badges

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

0

Scala – 62

for(x<-1 to 501)println(if(x%3<1||(x+""toSet 51))"Bzzt"else x)

toSet is a nice shorthand for contains.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jun 28, 2014 at 21:20

Karol S

1613 bronze badges

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score 0 · Accepted Answer · 2014-08-23 17:22:27Z

0

Haskell - 73

b n|mod n 3<1||'3'`elem`show n="bzzt"|1<2=show n
main=print$map b[0..500]

edited Aug 23, 2014 at 17:22

answered Aug 23, 2014 at 17:11

user344

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Matthew Butterick · Accepted Answer · 2014-08-24 05:57:21Z

0

Racket 80

(for/list([i 501])(if(or(=(modulo i 3)0)(memv #\3(string->list(~a i))))'bzzt i))

answered Aug 24, 2014 at 5:57

Matthew Butterick

4012 silver badges6 bronze badges

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

Pyth, 20

V501?N&%N3-`3`N"bzzt

Explanation:

V501            for N in range(501):
?               _ if _ else _
 N              print(N)
 &              _ and _
  %N3           N mod 3
  -`3`N         Assymetric difference of str(3) and str(N)
 "bzzt          "bzzt" (Final " is implicit)

Essentially, the test is the logical and of N%3, which is only 0, and thus interpreted as false iff N is divisible by 3, with

-`3`N

which is the empty list iff 3 is present in str(N). Setwise minus removes all appearances of characters in the second string from the first string.

If this and returns true, we print N, otherwise we print "bzzt". Having "bzzt" in the else clause saves a character, because there is no need to end-quote the string.

Michael Easter · Accepted Answer · 2014-08-26 00:38:00Z

0

Clojure - 82 chars

Newbie solution: plain and simple. Golfed:

(dotimes[n 501](if(or(.contains(str n)"3")(= 0(mod n 3)))(println "bzzt")(prn n)))

Ungolfed:

(dotimes [n 501] (if (or (.contains (str n) "3") (= 0 (mod n 3)))
                  (println "bzzt")
                  (prn n)))

answered Aug 26, 2014 at 0:38

Michael Easter

6258 silver badges11 bronze badges

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

0

Note: This answer is invalid, because Pyth is too new a language.

Pyth, 20

V501?N&%N3-\3`N"bzzt

Explanation:

V501              for N in range(501):
    ?N                                print(N if
      &%N3                                       N % 3 and
      -\3`N                                      "3" - repr(N)
     "bzzt                                    else "bzzt")

Try it here.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Dec 21, 2014 at 11:40

isaacg

42.1k5 gold badges73 silver badges213 bronze badges

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

0

Swift 2.0, 71

(0...500).map{print($0%3<1||"\($0)".utf8.contains(51) ?"bzzt":"\($0)")}

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jul 25, 2015 at 15:11

Kametrixom

4262 silver badges10 bronze badges

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ApolloZhu · Accepted Answer · 2016-10-30 09:33:24Z

0

Swift 3, 62

(0...500).map{print($0%3<1||"\($0)".contains("3") ?"bzzt":$0)}

or

for n in 0...500{print(n%3<1||"\(n)".contains("3") ?"bzzt":n)}

answered Oct 30, 2016 at 9:33

ApolloZhu

615 bronze badges

\$\begingroup\$ Are all the spaces required? Or is Swift just weird with them? \$\endgroup\$
– Oliver Ni
Commented Nov 2, 2016 at 5:02
\$\begingroup\$ I believe at least I eliminated all unnecessary ones. \$\endgroup\$
– ApolloZhu
Commented Nov 2, 2016 at 6:07

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Stack Exchange Network

Play the "bzzt" game

Winner: Aditsu's CJam answer! A whopping 25 bytes! Nice!

82 Answers 82

jq: 59 bytes

beeswax, 148 bytes

Groovy, 46 45 bytes

Japt, 21 bytes

Unpacked & How it works

Brachylog, 23 bytes (posted way after the end of the challenge)

Python 3, 56 bytes (Non-Competing)

APL (Dyalog Unicode), 30 bytes

05AB1E, 19 bytes

Vyxal 3 `j`, 14 bytes

Explanation (outdated):

Javascript - 109 97

C 174

JavaScript, 65

Usage:

Output:

Python3, 62

Scala – 62

Haskell - 73

Racket 80

Pyth, 20

Clojure - 82 chars

Note: This answer is invalid, because Pyth is too new a language.

Pyth, 20

Swift 2.0, 71

Swift 3, 62

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Winner: Aditsu's CJam answer! A whopping 25 bytes! Nice!

82 Answers 82

jq: 59 bytes

beeswax, 148 bytes

Groovy, 46 45 bytes

Japt, 21 bytes

Unpacked & How it works

Brachylog, 23 bytes (posted way after the end of the challenge)

Python 3, 56 bytes (Non-Competing)

APL (Dyalog Unicode), 30 bytes

05AB1E, 19 bytes

Vyxal 3 j, 14 bytes

Explanation (outdated):

Javascript - 109 97

C 174

JavaScript, 65

Usage:

Output:

Python3, 62

Scala – 62

Haskell - 73

Racket 80

Pyth, 20

Clojure - 82 chars

Note: This answer is invalid, because Pyth is too new a language.

Pyth, 20

Swift 2.0, 71

Swift 3, 62

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Vyxal 3 `j`, 14 bytes

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