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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.

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  • 41
    \$\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, 2016 at 20:44
  • 13
    \$\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, 2016 at 21:00
  • 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, 2016 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, 2016 at 3:30
  • 4
    \$\begingroup\$ @MartinEnder damn, missed my free GCC pizza yesterday. \$\endgroup\$
    – Mindwin Remember Monica
    Dec 2, 2016 at 12:04

131 Answers 131

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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, 2017 at 19:48
  • \$\begingroup\$ yes i replace main with f... thank you and a good day \$\endgroup\$
    – user58988
    Apr 12, 2017 at 5:18
0
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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
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Pyt, 1 byte

ɽ

Returns a random 32-bit integer.

Also 1 byte: ɹ, ṛ.

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0
0
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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, 2019 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, 2019 at 2:57
0
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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, ₁, ₂, ₃, ₄, ₅, and ₆, which push abcdefghijklmnopqrstuvwxyz, 10, 100, 256, 26, 95, 1000, 255, and 36, respectively.

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0
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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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2
  • 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, 2021 at 23:51
  • \$\begingroup\$ Also just realized; your two-byter is Þ! in Unicode. :P \$\endgroup\$
    – Makonede
    Mar 2, 2021 at 22:49
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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0
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CSASM v2.2.1.2, 39 bytes

func main:
extern Random.Next
print
ret
end

CSASM has a built-in System.Random object that gets initialized to new System.Random() when CSASM.Core.Sandbox.Main(System.Reflection.MethodInfo, int, string[]) is called.

The default System.Random constructor initializes the seed to Environment.TickCount, which is different every time the executable runs.

extern Random.Next calls Random.Next() on the aforementioned object, which returns a positive int value.

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0
\$\begingroup\$

Thunno, \$ 4 \log_{256}(96) \approx \$ 3 bytes

(actually 3.29 bytes but that doesn't show up on the leaderboard)

9RZw

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Outputs a random integer from 0 to 8:

  • 9R gets range(0, 9) (i.e. [0, 1, ..., 8])
  • Zw gets a random element of that list
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2
  • \$\begingroup\$ This looks like 4 bytes to me. How do you get 3.29 bytes? You realize that coded in binary, it would have to be representable in whole bits at the very least. \$\endgroup\$
    – mbomb007
    Jan 10 at 20:37
  • 1
    \$\begingroup\$ There is a consensus that fractional byte counts are allowed. Since Thunno uses the 96 printable ASCII characters as its codepage, the character count of every Thunno program can be multiplied by 0.82 \$\endgroup\$
    – The Thonnu
    Jan 11 at 7:30
0
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Pip, 3 bytes

RR9

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Prints a random integer from 0 to 9

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0
\$\begingroup\$

Thunno 2, 2 bytes

9ɼ

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Outputs a random integer from 1 to 9.

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