32
\$\begingroup\$

Let's play Russian Roulette!

Normally, this would be a race to write the shortest MOD 6 program, but that's not very realistic, as the chance of winning decreases with each click. Here are the rules:

  1. Emulate a real six-shooter:
    • A single bullet is placed into one of the six chambers, and the barrel is spun once, only before playing.
    • The chance of losing on the nth try is 1/6.
    • The chance of losing after n tries is 1/(6-n)
    • You are guaranteed to lose in, at most, 6 tries.
  2. Losing:
    • Display the text *BANG!*
    • Terminate the program.
  3. Winning:
    • "Winning" means the gun does not fire, but the bullet is one chamber closer to the hammer.
    • Display the text *click*
    • Present a "trigger" to the user, along with the ability to terminate the program (e.g. "ctrl+c", see below).
  4. Program specific:
    • Pulling the trigger is some form of user input, including the first try. (This can be a keystroke, a mouse click, whatever; text prompts are not required.)
    • Only one instance of the program is allowed until it is terminated. (Running a new instance of the program is akin to giving the barrel a good spin, i.e. the probability of losing on the next click is reset to 1/6.)

Shortest code wins!

Leaderboard

<style>body { text-align: left !important} #answer-list { padding: 10px; width: 290px; float: left; } #language-list { padding: 10px; width: 290px; float: left; } table thead { font-weight: bold; } table td { padding: 5px; }</style><script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b"> <div id="language-list"> <h2>Shortest Solution by Language</h2> <table class="language-list"> <thead> <tr><td>Language</td><td>User</td><td>Score</td></tr> </thead> <tbody id="languages"> </tbody> </table> </div> <div id="answer-list"> <h2>Leaderboard</h2> <table class="answer-list"> <thead> <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr> </thead> <tbody id="answers"> </tbody> </table> </div> <table style="display: none"> <tbody id="answer-template"> <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr> </tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr> </tbody> </table><script>var QUESTION_ID = 66763; var ANSWER_FILTER = "!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe"; var COMMENT_FILTER = "!)Q2B_A2kjfAiU78X(md6BoYk"; var OVERRIDE_USER = 38512; var answers = [], answers_hash, answer_ids, answer_page = 1, more_answers = true, comment_page; function answersUrl(index) { return "https://api.stackexchange.com/2.2/questions/" + QUESTION_ID + "/answers?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + ANSWER_FILTER; } function commentUrl(index, answers) { return "https://api.stackexchange.com/2.2/answers/" + answers.join(';') + "/comments?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + COMMENT_FILTER; } function getAnswers() { jQuery.ajax({ url: answersUrl(answer_page++), method: "get", dataType: "jsonp", crossDomain: true, success: function (data) { answers.push.apply(answers, data.items); answers_hash = []; answer_ids = []; data.items.forEach(function(a) { a.comments = []; var id = +a.share_link.match(/\d+/); answer_ids.push(id); answers_hash[id] = a; }); if (!data.has_more) more_answers = false; comment_page = 1; getComments(); } }); } function getComments() { jQuery.ajax({ url: commentUrl(comment_page++, answer_ids), method: "get", dataType: "jsonp", crossDomain: true, success: function (data) { data.items.forEach(function(c) { if (c.owner.user_id === OVERRIDE_USER) answers_hash[c.post_id].comments.push(c); }); if (data.has_more) getComments(); else if (more_answers) getAnswers(); else process(); } }); } getAnswers(); var SCORE_REG = /<h\d>\s*([^\n,<]*(?:<(?:[^\n>]*>[^\n<]*<\/[^\n>]*>)[^\n,<]*)*),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/; var OVERRIDE_REG = /^Override\s*header:\s*/i; function getAuthorName(a) { return a.owner.display_name; } function process() { var valid = []; answers.forEach(function(a) { var body = a.body; a.comments.forEach(function(c) { if(OVERRIDE_REG.test(c.body)) body = '<h1>' + c.body.replace(OVERRIDE_REG, '') + '</h1>'; }); var match = body.match(SCORE_REG); if (match) valid.push({ user: getAuthorName(a), size: +match[2], language: match[1], link: a.share_link, }); else console.log(body); }); valid.sort(function (a, b) { var aB = a.size, bB = b.size; return aB - bB }); var languages = {}; var place = 1; var lastSize = null; var lastPlace = 1; valid.forEach(function (a) { if (a.size != lastSize) lastPlace = place; lastSize = a.size; ++place; var answer = jQuery("#answer-template").html(); answer = answer.replace("{{PLACE}}", lastPlace + ".") .replace("{{NAME}}", a.user) .replace("{{LANGUAGE}}", a.language) .replace("{{SIZE}}", a.size) .replace("{{LINK}}", a.link); answer = jQuery(answer); jQuery("#answers").append(answer); var lang = a.language; lang = jQuery('<a>'+lang+'</a>').text(); languages[lang] = languages[lang] || {lang: a.language, lang_raw: lang.toLowerCase(), user: a.user, size: a.size, link: a.link}; }); var langs = []; for (var lang in languages) if (languages.hasOwnProperty(lang)) langs.push(languages[lang]); langs.sort(function (a, b) { if (a.lang_raw > b.lang_raw) return 1; if (a.lang_raw < b.lang_raw) return -1; return 0; }); for (var i = 0; i < langs.length; ++i) { var language = jQuery("#language-template").html(); var lang = langs[i]; language = language.replace("{{LANGUAGE}}", lang.lang) .replace("{{NAME}}", lang.user) .replace("{{SIZE}}", lang.size) .replace("{{LINK}}", lang.link); language = jQuery(language); jQuery("#languages").append(language); } }</script>

\$\endgroup\$
  • 3
    \$\begingroup\$ Your assumptions are wrong - the chance of losing on the n'th try is only ⅙ if you respin the bullets after each shot. Without respinning the chance of losing is ⅙ on the first shot, ⅕ on the second, ¼ on the third... ending in 1 on the 6th. You recognise this with "You are guaranteed to lose in, at most, 6 tries". \$\endgroup\$ – rhialto Dec 16 '15 at 21:52
  • 2
    \$\begingroup\$ @user2956063 you are forgetting that there is a (n-1)/6 chance that you never reach the n:th try, and thus cannot lose. They balance out to a uniform 1/6. \$\endgroup\$ – Jacob Raihle Dec 17 '15 at 10:26
  • 2
    \$\begingroup\$ maybe that's a difference in the way computer scientists and statisticians express probability then - to me "The chance of losing on the nth try is ⅙" say's it's constant - whether n is 1 or 100. \$\endgroup\$ – rhialto Dec 17 '15 at 11:53
  • 3
    \$\begingroup\$ Why was this not titled "Russian Roulette, Reloaded"? \$\endgroup\$ – Hand-E-Food Dec 18 '15 at 4:51
  • 1
    \$\begingroup\$ @user2956063: Your probabilities are conditional. P(lose on shot 2) = ⅙, but P(lose on shot 2 | didn't lose on shot 1) = ⅕. Also, n is (implicitly, I'll grant you) limited to [1,6], so 100 is out. \$\endgroup\$ – Tim Pederick Dec 19 '15 at 5:04

32 Answers 32

3
\$\begingroup\$

Pyth, 23 bytes

VO6"*click*" w;"*BANG!*

Really simple. A random number of iterations 0 - 5 display click and request a line of input, followed by a bang at the end.

\$\endgroup\$
  • 1
    \$\begingroup\$ damn you, pyth! \$\endgroup\$ – Cyoce Dec 18 '15 at 1:55
  • \$\begingroup\$ +3 bytes: the last string should be *BANG!*, not BANG \$\endgroup\$ – ayane Dec 18 '15 at 5:34
14
\$\begingroup\$

Ruby, 51 bytes

[*['*click*']*rand(6),'*BANG!*'].map{|x|gets;$><<x}

Ungolfed:

[
  *(                        # Unwrap the following array into the outer one
    ['*click*'] * rand(6)   # An array of 0-5 clicks, see Array#*
  ),
  '*BANG!*'                 # The End
].map do |x| # Shortest way to iterate I was able to find
  gets       # Await input
  $> << x    # Shove the output string to `stdout`
end          # Return value is an array of several (0-5) `stdout`s. Who cares.

or

(['*click*']*rand(6)<<'*BANG!*').map{|x|gets;$><<x}

Ungolfing left for the readers. Not that difficult

  • Again, kudos to Martin, this time for a trick with $><< as a puts replacement.
  • Doesn't output newlines, but that was not required.
  • The shorter, the simpler. The gist of the required behaviour is to do 0-5 clicks and then fire. For that, outputs are accumulated inside the array.
  • 2 more bytes can be shaved off if outputs like "*click*" are okay (what's required is printed in the end), by replacing $><< with . I wasn't sure if this would still follow the rules.

68 64 bytes

(another approach)

[*0..5].shuffle.find{|x|gets;x<1||puts('*click*')};puts'*BANG!*'

I didn't put much thought into the algorithm (it can possibly be even more compact, but not so clear), but I really like the model inside it:

  • An array emulates a barrel with its elements being chambers' contents. Since only one element is a bullet, rotating it and shuffling it are equivalent.
  • 0 is a bullet. Other numbers are not.
  • find finds a first return value for which the block is neither false nor nil.
  • ||-expression is implicitly returned from the block. It's a short-circuit: it returns its first operand (unless it's nil or false) or a second one (otherwise). So it either returns true (if x<1 or, clearer but longer x == 0) or the return value of puts, while...
  • puts always returns nil. Yep.
  • gets requests input. Merely hitting Enter suffices.
  • Ctrl+C terminates the program
\$\endgroup\$
  • \$\begingroup\$ Crossed out 64 is normal 64? \$\endgroup\$ – Cyoce Dec 18 '15 at 5:06
  • \$\begingroup\$ @Cyoce uhm... yes. Should be, probably. It's been superceded by the above entry, but it is based on a different idea, so I left them, crossed out. \$\endgroup\$ – D-side Dec 18 '15 at 7:51
  • \$\begingroup\$ it was a pun on this \$\endgroup\$ – Cyoce Dec 19 '15 at 7:22
  • \$\begingroup\$ @Cyoce oh, pardon me, first post and all that, not aware of the local lore :) \$\endgroup\$ – D-side Dec 19 '15 at 21:27
9
\$\begingroup\$

JavaScript, 64 bytes

for(i=6;i<7&&prompt();)alert(new Date%i--?"*click*":i="*BANG!*")

Explanation

To pull the trigger enter any text into the prompt. Enter nothing or click cancel to terminate.

for(
  i=6;             // i = number of chambers
  i<7              // i equals "*BANG!*" (not less than 7) if we lost
    &&prompt();    // see if we should do another shot
)
  alert(           // alert the result
    new Date%i--   // use the current time in milliseconds as a random number, this is safe
                   //     to use because the gap between shots is greater than i (max 6ms)
      ?"*click*"   // on win pass "*click*" to alert
      :i="*BANG!*" // on lose alert "*BANG!*" and set i to not less than 7
  )
\$\endgroup\$
  • \$\begingroup\$ "*Bang!*" is not greater than 7. But NaN is not smaller than 7. \$\endgroup\$ – Bergi Dec 16 '15 at 12:38
  • \$\begingroup\$ @Bergi That's true. I reworded the explanation to make it a bit clearer. \$\endgroup\$ – user81655 Dec 16 '15 at 12:43
  • 6
    \$\begingroup\$ @Bergi only in Javascript does that statement come even close to making sense. \$\endgroup\$ – MikeTheLiar Dec 16 '15 at 16:03
7
\$\begingroup\$

Lua, 82 75 bytes

Pretty long, but there's lot of verbose in lua.

for a=math.random(6),1,-1 do io.read()print(a>1 and"*click*"or"*BANG!*")end
\$\endgroup\$
6
\$\begingroup\$

LabVIEW, 46 LabVIEW Primitives

Creates an Array of 0s and one 1, has a loop to wait for clicks and outputs the string. It initially says BANG becuase i forgot to reset the indicator before starting it.

Also note that this is a gif, if if does not play/load for you please reopen the page.

\$\endgroup\$
  • \$\begingroup\$ I don't see where you do the output of "*click*" in case the gun didn't shoot. Also, does it output "bang" or "*BANG!*"? \$\endgroup\$ – Katenkyo Dec 16 '15 at 10:04
  • \$\begingroup\$ this should be a gif but for me it doesnt play, that might be the problem. And yes it only puts bang, that was me not reading properly i change that in a sec \$\endgroup\$ – Eumel Dec 16 '15 at 10:15
  • \$\begingroup\$ stupid me forgot to reinitialize the string to empty before starting so thats why it showes BANG at the start... \$\endgroup\$ – Eumel Dec 16 '15 at 10:25
  • \$\begingroup\$ No problem, I saw the gif now, looks like it works pretty well :) \$\endgroup\$ – Katenkyo Dec 16 '15 at 10:26
5
\$\begingroup\$

Pyth, 31 30 28 bytes

FN.S6 EI!N"*BANG!*"B"*click*

Almost certainly can be improved. Input any number to pull the trigger, blank input to terminate early (with an error).

Explanation:

FN                               for N in...
  .S6                            shuffle(range(6))...
      E                          get a line of input
       I!N                       if N is falsy (i.e. 0)
          "*BANG!*"              implicit output
                   B             break
                    "*click*     else, print click
\$\endgroup\$
  • \$\begingroup\$ Your first one is actually shorter, you don't need the trailing". \$\endgroup\$ – FryAmTheEggman Dec 16 '15 at 14:06
  • \$\begingroup\$ @FryAmTheEggman Oh, right, forgot about that. Thanks! \$\endgroup\$ – Doorknob Dec 16 '15 at 14:31
  • \$\begingroup\$ Also, I've just noticed something silly, FN<any> is still entirely identical to V<any>, that should probably be changed to not confuse new golfers... :P \$\endgroup\$ – FryAmTheEggman Dec 16 '15 at 14:38
  • 1
    \$\begingroup\$ You can simply remove the chars .?. There is no need for an else. \$\endgroup\$ – Jakube Dec 16 '15 at 17:57
  • \$\begingroup\$ @FryAmTheEggman confusing people is awesome. It can be amazing at filtering out the scrubs. \$\endgroup\$ – Cyoce Dec 18 '15 at 8:40
5
\$\begingroup\$

Seriously, 27 25 bytes

"*BANG!*"6J"*click*"nW,X.

No online link because there is no way to do a prompt with piped input. The program can be CTRL-C'd at any time to chicken out terminate.

Explanation:

"*BANG!*"6J"*click*"nW,X.
"*BANG!*"                  push "*BANG!*"
         6J                push a random integer in [0,6) (n)
           "*click*"n      push "*click*" n times
                     W     while loop (implicitly closed at EOF):
                      ,X.    get input and discard, pop and print top of stack
\$\endgroup\$
4
\$\begingroup\$

PHP, 52 bytes

*<?=$n=&rand(0,6-$argi)?click:"BANG!*";$n||@\n?>*

Requires the -F command line option, counted as three. The trigger is pulled by pressing Enter.

Because -F literally runs the script again for every input (I kid you not), die and the like won't actually terminate, so we exit via suppressed runtime error instead, @\n.


Sample Usage

$ php -F primo-roulette.php

*click*
*click*
*BANG!*
$
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4
\$\begingroup\$

Perl 5, 43 bytes

Run with perl -p. Stable bullet variant - i.e. bullet position is decided only once in very beginning.

$i//=0|rand 6;$_=$i--?'*click*':die'*BANG*'
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4
\$\begingroup\$

C, 110 74 72 bytes

Thanks to Dennis for getting rid of the includes and a lot less bytes.

main(r){for(r=time(0)%6;getchar(),r--;)puts("*click*");puts("*BANG!*");}
main(r)
{
    for(r=time(0)%6;getchar(),r--;)
        puts("*click*");
    puts("*BANG!*");
}
\$\endgroup\$
3
\$\begingroup\$

Candy, 36 bytes

About half the program is the text to print out :(

:6H_(=b"*click*"(;):=)"*BANG!*\n"(;)

long form:

getc
digit6 rand range0  # build a range from 0 .. rand#
while
  popA              # these are the *click* instances  
  stack2
  "*click*"
  while
    printChr
  endwhile
  getc
  popA
endwhile
"*BANG!*\n"         # out of luck
while
  printChr
endwhile
\$\endgroup\$
3
\$\begingroup\$

Python 3, 95 bytes

Also my first golf attempt, also in Python 3. I swear Bruce and I aren't the same person.

from random import*
for a in range(randint(0,5)):input();print("*click*")
input();print("*bang*")

Ungolfed:

from random import*
for a in range(randint(0,5)):
    input()
    print("*click*")
input()
print("*bang*")

Generate a random number between 0 and 5 inclusive, print click that many times, then print bang. Press enter/return to pull the trigger.

\$\endgroup\$
  • \$\begingroup\$ Following Bruce's example, you can save a few bytes with from random import* \$\endgroup\$ – wnnmaw Dec 16 '15 at 17:25
  • \$\begingroup\$ Unless there's something else I'm missing, it's one byte of savings. But I'll take it! Thanks! \$\endgroup\$ – Steve Eckert Dec 16 '15 at 17:30
  • \$\begingroup\$ Nice attempt, I used your solution as an inspiration for my python 2 solution ^^ \$\endgroup\$ – basile-henry Dec 16 '15 at 17:52
3
\$\begingroup\$

PlatyPar, 26 25 bytes

6#?;wT"*click*"O;"*BANG!*

Explanation:

6#?;                        ## random integer [0,6)
    w           ;           ## while stack.last
     T                      ## stack.last--
      "*click*"O            ## alert "*click*"
                 "*BANG!*   ## alert "*BANG!*"

Try it online!

\$\endgroup\$
2
\$\begingroup\$

Emacs Lisp, 94 89 bytes

(set'b(%(random)6))(dotimes(a(+ b 1))(read-string"")(message(if(eq a b)"BANG""*click*")))

Ungolfed:

(set 'b (% (random) 6))
(dotimes (a (+ b 1))
  (read-string"")
  (message (if (eq a b) "BANG" "*click*")))
\$\endgroup\$
2
\$\begingroup\$

R, 86 80 77 bytes

As usual, R has awesome features to code golfing but looooooong function names.

sapply(sample(0:5),function(n){!n&&{cat('*BANG!*');q()};readline('*click*')})
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2
\$\begingroup\$

Python 2, 108 104 102 100 98 bytes

My first attempt at golfing:

from random import*
a=[1]+[0]*5
shuffle(a)
for i in a:input();print("*click*","*BANG!*")[i];" "[i]

Maybe I should add that the program doesn't terminate correctly when you lose, it just throws an exception (which results in termination):

Traceback (most recent call last):
  File "russian_roulette.py", line 4, in <module>
    for i in a:input();print("*click*","*BANG!*")[i];" "[i]
IndexError: string index out of range
\$\endgroup\$
  • \$\begingroup\$ Welcome to Programming Puzzles & Code Golf! When you post a code golf answer, please include the language name and the byte count (I edited it in for you here). Thanks! \$\endgroup\$ – ProgramFOX Dec 16 '15 at 16:58
  • \$\begingroup\$ Yes, thanks a lot! I was actually trying to resolve that issue, didn't get to edit it correctly before you did. \$\endgroup\$ – ბიმო Dec 16 '15 at 17:02
  • \$\begingroup\$ I get your byte count as 112, what did you use? \$\endgroup\$ – wnnmaw Dec 16 '15 at 17:22
  • \$\begingroup\$ Also, you can save 2 bytes by doing a=shuffle([1,0,0,0,0,0]) \$\endgroup\$ – wnnmaw Dec 16 '15 at 17:22
  • 1
    \$\begingroup\$ As far as I know shuffle changes the underlying data structure and doesn't return anything. I tried that, thanks anyways. The headline (language and byte count) was written by ProgramFOX. But when I use wc it gives me 109 which one too much, so it's right. How do you count? \$\endgroup\$ – ბიმო Dec 16 '15 at 17:33
2
\$\begingroup\$

Perl 5, 40 bytes

<>,print"*$_*"for(click)x rand 5,'BANG!'

Run without command line options, the trigger is pulled by pressing Enter.

\$\endgroup\$
2
\$\begingroup\$

Python, 81 Bytes

import time
for i in["*click*"]*(int(time.time())%6)+["*BANG!*"]:input();print(i)

inspired by the Ruby(51) and Python solutions

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1
\$\begingroup\$

Common Lisp, 109

(do(g(b(nthcdr(random 6)#1='(t()()()()() . #1#))))(g)(read-char)(princ(if(setf g(pop b))"*BANG!*""*click*")))

Not very competitive, but I like circular lists:

(do (;; auxiliary variable x
     x
     ;; initialize infinite barrel, rotate randomly
     (b (nthcdr (random 6) #1='(t()()()()() . #1#))))

    ;; we end the loop when x is T (a bullet is fired)
    (x)

  ;; press enter to shoot
  (read-char)

  ;; pop from b, which makes b advance down the list. The popped value
  ;; goes into x and is used to select the appropriate string for
  ;; printing.
  (princ
   (if (setf x(pop b))
       "*BANG!*"
       "*click*")))
\$\endgroup\$
1
\$\begingroup\$

Perl 5, 43 bytes

42 bytes + -p command line option. Just press enter to trigger.

$_=0|rand 7-$.<++$i?die"*BANG!*":"*click*"

Thanks to Dom Hastings for his help! Original answer was 67 bytes:

$i++;$a=$i>=int(rand(6));print$_=$a?'*BANG!*':'*click*';last if($a)
\$\endgroup\$
  • \$\begingroup\$ Actually the problem with -p was that it would exit before calling the last print statement, not sure why. I've tried it. Other than that, awesome suggestions, thanks! My knowledge continues to grow... \$\endgroup\$ – Codefun64 Dec 16 '15 at 13:21
  • \$\begingroup\$ @DomHastings Also, unfortunately, for some reason the 0| trick didn't work as expected, but I did shave some bytes off of it, the print statement and the last statement like you suggested. How does it look now? \$\endgroup\$ – Codefun64 Dec 16 '15 at 13:29
  • \$\begingroup\$ @DomHastings Damn, you are good. I recommend putting that as your own answer, since you definitely wrote a smaller program than I (you have 40 bytes compared to my original 67!) \$\endgroup\$ – Codefun64 Dec 16 '15 at 13:41
  • \$\begingroup\$ I appreciate the explanation! Always happy to learn more of my favorite scripting language! I never even knew about the prefex incrementing, that's pretty awesome. Thanks :) \$\endgroup\$ – Codefun64 Dec 16 '15 at 13:50
  • \$\begingroup\$ You're very welcome, glad to have helped! \$\endgroup\$ – Dom Hastings Dec 16 '15 at 13:51
1
\$\begingroup\$

MATL, 41 bytes

6Yr`j?t@=?'*BANG!*'DT.}'*click*'DT]}T.]]x

To pull the trigger, input a non-empty string (such as 'try').

To terminate, input an empty string

Examples

In this case the trigger was pulled once and... bad luck:

>> matl
 > 6Yr`j?t@=?'*BANG!*'DT.}'*click*'DT]}T.]]x
 > 
> try
*BANG!*

In this case the user stopped (note the final empty input) after two lucky pulls:

>> matl
 > 6Yr`j?t@=?'*BANG!*'DT.}'*click*'DT]}T.]]x
 > 
> try
*click*
> try
*click*
> 

Explanation

6Yr                  % random avlue from 1 to 6    
`                    % do...while  
  j                  % input string
  ?                  % if nonempty
    t                % duplicate the orignal random value
    @                % loop iteration index  
    =                % are the equal?
    ?                % if so             
      '*BANG!*'D     % display string
      T.             % unconditional break                                     
    }                % else
      '*click*'D     % display string
      T              % true value, to go on with do...while loop
    ]                % end if               
  }                  % else                                                    
    T.               % unconditional break
  ]                  % end                                                     
]                    % end                                                     
x                    % delete original random value
\$\endgroup\$
1
\$\begingroup\$

Perl 6,  58   53 bytes

for ^6 .pick(*) {get;say <*BANG!* *click*>[?$_];!$_&&last} # 58 bytes

$ perl6 -pe '$///=^6 .pick;$_=$/--??"*click*"!!say("BANG!")&&last' # 52+1= 53 bytes

Press enter to pull the trigger, or ctrl+c to put it down.

\$\endgroup\$
1
\$\begingroup\$

Python 2, 88 84 bytes

This solution is inspired by the python 3 solutions already given. I chose python 2 to remove the print parenthesis even though this changes the behavior of input().

import time
for i in[0]*int(time.time()%6)+[1]:input();print("*click*","*BANG!*")[i]
  • I am using modulo of the time as a random function (good enough for russian roulette)
  • the player input should be "i" then Enter (otherwise input() will throw an error), this trick relies on the fact that the input can be "whatever".
\$\endgroup\$
1
\$\begingroup\$

Ruby, 45+1=46

Not as clever as D-side's but slightly shorter.

With command-line flag p, run

rand(7-$.)<1?(puts'*BANG*';exit):$_='*click*'

The user can pull the trigger with return and leave with control-c. p causes the program to run in a loop, reading lines from STDIN and outputting $_. Each time it runs, it increments $.. So on the first run, it chooses a random positive integer less than 6, then 5, then 4, and so on. On the first 0, we output manually and exit, until then we output implicitly.

(and now I notice that there's already a very similar Perl. Oh well.)

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1
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Perl 5, 69 51 49 bytes

map{<>;print"*click*"}1..rand 6;<>;print"*BANG!*"

There is probably some more golfing potential, I will look into this.

Changes:

  • Saved 8 bytes by removing $l and some semicolons, and 10 bytes by changing <STDIN> to <>
  • Saved 2 bytes thanks to Oleg V. Volkov
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  • 1
    \$\begingroup\$ 49: map{<>;print"*click*"}1..rand 6;<>;print"*BANG!*" \$\endgroup\$ – Oleg V. Volkov Dec 16 '15 at 21:17
  • \$\begingroup\$ @OlegV.Volkov Thanks! I'll edit it in now. \$\endgroup\$ – ASCIIThenANSI Dec 16 '15 at 23:45
0
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VBA, 126 bytes

Golf Version for Minimal Bytes

Sub S()
r=Int(5*Rnd())
e:
a=MsgBox("")
If a=1 Then: If i=r Then MsgBox "*BANG!*" Else: MsgBox "*click*": i=i+1: GoTo e
End Sub

Fun Version that Makes The Buttons more Clear for Increased User Acceptance.

Sub RR()
r = Int(5 * Rnd())
e:
a = MsgBox("Are you Feeling Lucky?", 4)
If a=6 Then: If i=r Then MsgBox "*BANG!*", 16 Else: MsgBox "*click*", 48: i=i+1: GoTo e
End Sub

Some Fun with Custom Forms and You could make a pretty Slick Game in VBA.

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APL, 39/65 bytes

{⍞⊢↑⍵:⍞←'*BANG*'⋄∇1↓⍵⊣⍞←'*click*'}6=6?6

Pretty straightforward answer.

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  • \$\begingroup\$ What do the two byte counts mean? \$\endgroup\$ – Mego Dec 17 '15 at 9:37
0
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C, 180 Bytes

#include<stdio.h>
#include<stdlib.h>
#include<time.h>
int main(){srand(time(NULL));int r,i=6;while(i!=1){getchar();r=rand()%i;if(r){puts("CLICK");}else{puts("BANG");exit(0);}i--;}}

My first attempt at code golf, there's probably a lot of room for improvement :)

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Julia, 71 bytes

b=rand(1:6);while b>0 readline();print(b>1?"*click*":"*BANG!*");b-=1end

Press Enter to fire or Ctrl+C to quit. The latter ends with an InterruptException.

Ungolfed:

# Set an initial bullet location
b = rand(1:6)

while b > 0
    # Read user input
    readline()

    # Check the location
    if b > 1
        print("*click*")
    else
        print("*BANG!*")
    end

    b -= 1
end
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0
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Lua, 73 bytes

q=io.read q()for i=2,math.random(6)do print"*click*"q()end print"*BANG!*"
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