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You need to produce output that is non-deterministic.

In this case, this will be defined to mean that the output will not always be the same result.

Rules:

  • A pseudo-random number generator that always has the same seed does not count.

  • You can rely on the program being run at a different (unknown) time each execution.

  • Your code's process id (if it's not fixed by the interpreter) can be assumed to be non-deterministic.

  • You may rely on web-based randomness.

  • Your code may not take non-empty input. Related meta post.

  • The program is not required to halt, but the output must be displayed.

Leaderboard

function answersUrl(a){return"https://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+a+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function commentUrl(a,b){return"https://api.stackexchange.com/2.2/answers/"+b.join(";")+"/comments?page="+a+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+COMMENT_FILTER}function getAnswers(){jQuery.ajax({url:answersUrl(answer_page++),method:"get",dataType:"jsonp",crossDomain:!0,success:function(a){answers.push.apply(answers,a.items),answers_hash=[],answer_ids=[],a.items.forEach(function(a){a.comments=[];var b=+a.share_link.match(/\d+/);answer_ids.push(b),answers_hash[b]=a}),a.has_more||(more_answers=!1),comment_page=1,getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:"get",dataType:"jsonp",crossDomain:!0,success:function(a){a.items.forEach(function(a){a.owner.user_id===OVERRIDE_USER&&answers_hash[a.post_id].comments.push(a)}),a.has_more?getComments():more_answers?getAnswers():process()}})}function getAuthorName(a){return a.owner.display_name}function process(){var a=[];answers.forEach(function(b){var c=b.body;b.comments.forEach(function(a){OVERRIDE_REG.test(a.body)&&(c="<h1>"+a.body.replace(OVERRIDE_REG,"")+"</h1>")});var d=c.match(SCORE_REG);d?a.push({user:getAuthorName(b),size:+d[2],language:d[1],link:b.share_link}):console.log(c)}),a.sort(function(a,b){var c=a.size,d=b.size;return c-d});var b={},c=1,d=null,e=1;a.forEach(function(a){a.size!=d&&(e=c),d=a.size,++c;var f=jQuery("#answer-template").html();f=f.replace("{{PLACE}}",e+".").replace("{{NAME}}",a.user).replace("{{LANGUAGE}}",a.language).replace("{{SIZE}}",a.size).replace("{{LINK}}",a.link),f=jQuery(f),jQuery("#answers").append(f);var g=a.language;g=jQuery("<a>"+g+"</a>").text(),b[g]=b[g]||{lang:a.language,lang_raw:g,user:a.user,size:a.size,link:a.link}});var f=[];for(var g in b)b.hasOwnProperty(g)&&f.push(b[g]);f.sort(function(a,b){return a.lang_raw.toLowerCase()>b.lang_raw.toLowerCase()?1:a.lang_raw.toLowerCase()<b.lang_raw.toLowerCase()?-1:0});for(var h=0;h<f.length;++h){var i=jQuery("#language-template").html(),g=f[h];i=i.replace("{{LANGUAGE}}",g.lang).replace("{{NAME}}",g.user).replace("{{SIZE}}",g.size).replace("{{LINK}}",g.link),i=jQuery(i),jQuery("#languages").append(i)}}var QUESTION_ID=101638,ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe",COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk",OVERRIDE_USER=34718,answers=[],answers_hash,answer_ids,answer_page=1,more_answers=!0,comment_page;getAnswers();var SCORE_REG=/<h\d>\s*([^\n,<]*(?:<(?:[^\n>]*>[^\n<]*<\/[^\n>]*>)[^\n,<]*)*),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/,OVERRIDE_REG=/^Override\s*header:\s*/i;
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  • 36
    \$\begingroup\$ @mbomb007 In C there are many things that are simply "undefined" behaviour. Any given interpreter is allowed to do whatever it wants in any situation. For all we know, gcc might order you a pizza if you try to overflow a signed integer on a rainy Tuesday, but will make a trout jump out of your screen on all other days. So you wouldn't really ever know if it's actually deterministic or not in any given implementation. \$\endgroup\$ – Martin Ender Nov 30 '16 at 20:44
  • 12
    \$\begingroup\$ @MartinEnder I'm not sure if that matters. We define languages here by their implementation, not by the specification (as languages without an implementation is not allowed) \$\endgroup\$ – Nathan Merrill Nov 30 '16 at 21:00
  • 2
    \$\begingroup\$ @MartinEnder Yeah, I agree with Nathan. \$\endgroup\$ – mbomb007 Nov 30 '16 at 21:01
  • 8
    \$\begingroup\$ Note that undefined behaviour in C often leads to crashes, and crashes on UNIX and Linux lead to core files which contain the process ID inside them. That would seem to comply with the question as currently worded. \$\endgroup\$ – user62131 Nov 30 '16 at 21:23
  • 5
    \$\begingroup\$ Unless I misunderstood, the question did not ask for code that takes advantage of undefined behavior. It asks for code that takes advantage of defined behavior to guarantee non-determinism. \$\endgroup\$ – WGroleau Dec 1 '16 at 3:30

109 Answers 109

1 2 3
4
0
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JavaScript, 8 bytes

Date.now

This evaluates to a function whose return value is nondeterministic, and conveniently doesn't actually need to be bound to anything specific to work.

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2
  • \$\begingroup\$ Basically, this is the same as the Date answer. \$\endgroup\$ – Ismael Miguel Nov 30 '16 at 23:20
  • \$\begingroup\$ @IsmaelMiguel I did have the idea of returning a builtin first; I just returned the wrong builtin. \$\endgroup\$ – Neil Dec 1 '16 at 0:31
0
\$\begingroup\$

Batch, 6 4 bytes

time

Needs to be run with <nul which is apparently permissible.

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2
0
\$\begingroup\$

Swift 3, 13 bytes

print(Date())
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0
\$\begingroup\$

QBIC, 4 bytes

?_r|

This generates and prints a random number between 0 and 10. QBIC uses the classical RANDOMIZE TIMER at the start of execution to set the RNG.

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0
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Actually, 1 byte

G

Try it online!

Good old rand(). In Python 3, 2500 random bits from the system's cryptographically-secure randomness source (getrandom() for Linux, /dev/urandom for *NIX, or CryptGenRandom for Windows) are used for the seed, falling back on the current UNIX time (as precisely as possible for the given platform - at minimum, 1-second precision) if such a source of randomness is not available.

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0
\$\begingroup\$

Microscript, 2 bytes

r9

Produces a random integer on [0,8]

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0
\$\begingroup\$

Mouse2002, 7 bytes

&RAND !
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0
\$\begingroup\$

Microscript II, 1 byte

R

Produces a random 64-bit float on [0,1).

Another one-byte solution would be C, which produces a new continuation object, whose timestamp (which the reference implementation includes in its string representation) will be nondeterministic.

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0
\$\begingroup\$

HSPAL, 18 bytes

26FFFF
400000
120000

Prints a random 16-bit integer.

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0
\$\begingroup\$

Math++, 5 bytes

$rand

Produces a random 64-bit float

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0
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ASP VBScript, 43 bytes

<%=createobject("scriptlet.typelib").guid%>

Outputs a unique ID like {52076580-3151-4EE7-AAFD-D975CD141EE4} based on current date/time.

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0
\$\begingroup\$

C, 22 bytes

f(){putchar(time(0));}
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2
  • \$\begingroup\$ OP indicates in the comments that a full program is not needed and a function will suffice. Might I suggest replacing main with f to save 3 bytes? \$\endgroup\$ – Albert Renshaw Apr 11 '17 at 19:48
  • \$\begingroup\$ yes i replace main with f... thank you and a good day \$\endgroup\$ – user58988 Apr 12 '17 at 5:18
0
\$\begingroup\$

Chip-8, 4 bytes

0xCFFF 'RND vF,FF
0xFF18 'LD ST,vF

This plays a sound for between 0 and 4.25 seconds. The seed always starts at 0, but it updates during the display interrupt which has a tiny chance of happening before the randomizer call, I hope.

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0
\$\begingroup\$

Pyt, 1 byte

ɽ

Returns a random 32-bit integer.

Also 1 byte: ɹ, .

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2
0
\$\begingroup\$

Gol><>, 3 bytes

x1h

The 'x' is a randomizer for the direction of the pointer, the 1 pushes a 1, and the h outputs and halts.

Try it online!

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2
  • \$\begingroup\$ You can also do Sxh to print a random float between 0 and 1 \$\endgroup\$ – Jo King Mar 1 '19 at 2:31
  • \$\begingroup\$ @JoKing I didn't think of that, thanks for pointing that out. that works as well. Though I don't think there is a smaller version of this \$\endgroup\$ – KrystosTheOverlord Mar 1 '19 at 2:57
0
\$\begingroup\$

Pxem (Filename: 5 bytes; Contents: 0 byte)

  • Filename: x.r.n
  • File content is empty.

Explaination

  • x pushes an ascii code of the character (120) as a signed integer.
  • .r pops 120 to push back one of 0-119.
  • .n pops the integer to output as a decimal.
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0
\$\begingroup\$

05AB1E, 2 bytes

Any program matching /^[ATт₁-₆]Ω|ž[a-g]$/ will give nondeterministic output. There are, of course, infinitely many programs that can do this, but these are all the two-byters. Unfortunately, I could not find a one-byter... :(

Try them online!

Explanation

The ones starting with ž are constants. The ones used here are time constants - ža, žb, žc, žd, že, žf, and žg are hours, minutes, seconds, microseconds, day, month, and year, respectively. The ones ending in Ω pick a random element (for lists), character (for strings), or digit (for integers) from top of stack. The one-byte commands that I could find that push something with more than one distinct element/character/digit were A, T, т, , , , , , and , which push abcdefghijklmnopqrstuvwxyz, 10, 100, 256, 26, 95, 1000, 255, and 36, respectively.

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0
\$\begingroup\$

ThumbGolf, 4 bytes

Machine code:

de78 de20

Assembly:

        // Include ThumbGolf wrapper macros
        .include "thumbgolf.inc"
        .globl main
        .thumb_func
main:
        // Read random integer into r0
        rand    r0 // udf #0170
        // Print to stdout in base 10
        puti    r0 // udf #0040
        // Not required to exit, so just crash

Internally, ThumbGolf uses /dev/urandom, so this is guaranteed to be nondeterministic.

Alternative version, 2 bytes

Machine code:

de21

Assembly:

        // Include ThumbGolf wrapper macros
        .include "thumbgolf.inc"
        .globl main
        .thumb_func
main:
        // print address of argv
        // printf("%i", (int)argv)
        puti    r1 // udf #0041
        // exit by crash

Prints the memory address of argv which is almost guaranteed to be dynamically allocated, and therefore, nondeterministic.

However, it is context sensitive unlike the 4 byte solution.

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1
  • 1
    \$\begingroup\$ Non-competing status shouldn't be used nowadays. Answers in languages newer than the question are allowed, obviously considering that you don't hardcode the answer to the challenge into the language's interpreter and submit a 0-byter (or similar). That is already a standard loophole. \$\endgroup\$ – Makonede Feb 6 at 23:51
0
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Phooey, TIO version, 4 bytes

<2$i

Try it online!

Abuses a bug to print the frame pointer.

This is because the tape does not wrap, so the tape just underflows to the stack.

<2 moves the pointer 16 bytes back, $i prints as an integer.

Very dependent on the version of the binary.

Phooey, proper version, 5 4 bytes

~t$i

Try it online!

~t stores the Unix time to the tape, and I already explained $i.

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