Halt with 50% non-Halting

Question

Create a program that halts exactly 50% of the time. Be original. Highest voted answer wins. By exactly I mean that on each run there is a 50% chance of it halting.

Answer 1

Exactly 50% of the time?

OBJ-C

- (void)applicationDidFinishLaunching:(NSNotification*)aNotification {
    BOOL haltedLastRun = [(NSNumber*)[[NSUserDefaults standardUserDefaults] objectForKey:@"halted"] boolValue];
    if (!haltedLastRun) {
        [[NSUserDefaults standardUserDefaults] setObject:[NSNumber numberWithBool:YES] forKey:@"halted"];
        [[NSApplication sharedApplication] terminate:nil];
    }
}

Answer 2

Haskell

Runs for two intervals, each one 1 second long (chosen because 1 second is the SI unit for time). Halts inside 50% of the intervals. So 50% of the running seconds it will not halt, the other 50% it will. Works in GHC only.

import Control.Concurrent (threadDelay)
main = threadDelay 1990000

Answer 3

Shell Script

this script will clobber .md5sum files in the current and child directories.

#!/bin/sh
echo *.md5sum|xargs -n1|head -n1|xargs test -e && exec rm *.md5sum
while ! find . -name '*.md5sum' -print0 |xargs -0r grep 00000000000000
do {
    find . -type f -print|sed -e 's!^\(.*\)$!md5sum "\1" > "\1".md5sum!e'
}
done

Answer 4

GTB

[@r;p;]

I know this isn't code-golf but I decided to golf it anyway.

Answer 5

C++

#include <fstream>
main () {
    int c;
    std::fstream fs;
    fs.open ("myfile.txt", std::fstream::in);
    fs>>c;
    fs.close ();
    fs.open ("myfile.txt", std::fstream::out);
    fs<<c+1;
    fs.close ();
    while (c%2);
    return 0;
}

Each run will halt iff the run before didn't.

Answer 6

Windows Command Script

This script will append code to itself which ultimately alternates 'x' on each run.

call :last
if %x%==1 (
    echo>>%0 set x=0
    exit /b 0
) else (
    echo>>%0 set x=1
)
:nohalt
goto :nohalt
:last
set x=1
[newline here]

Answer 7

Math++

_($rand*2)+2>$
2>$
1

Answer 8

Python 2, 54 bytes

import time;H=[time.time()%2]
for h in H:H+=[h]*int(h)

Try it online!

Halting behaviour is dependent on current time.

Answer 9

Pxem, filename: 7 bytes + content: 0 bytes = 7 bytes, depends on implementation.

• Filename (escaped): \002.r.w.a
• Content: empty.

How it works

• \002.r: pushes one of 0 or 1.
• .w ... .a: while empty || pop!=0; do ... ;done

Try it online!

Answer 10

Thue, 19 bytes

0::=00
10::=
::=
10

Try it online!

TIO's Thue interpreter doesn't seem to be working correctly.

Answer 11

Thunno, \$ 6 \log_{256}(96) \approx \$ 4.94 bytes

2RZw?[

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Explanation

2R      # range(2): push [0, 1]
  Zw    # Random element of ↑
    ?   # If ↑ == 1:
     [  #  Loop forever
        # (Else, exit)

Answer 12

C (GCC)

unsigned long long i,j,k;
r;

main()
{
    srand(time(0));

    i=k=1;
    while (k) {
        k = 0;
        for (j = 1; j <= i; ++j) {
            do {
                r = rand() % 4;
            } while (r == 3);
            
            k += !!r;
        }
        ++i;
    }
}

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It loops while k isn't 0. k is only 0 at the end of the loop if i number of calls to rand() are all divisible by 3. The chance of that equals

$$ \frac {1} {3^i} $$

and the sum of all inverse natural powers of 3 equals 0.5. Of course, i doesn't go from 1 to infinity, but from 1 to ULLONG_MAX. When i reaches ULLONG_MAX, the for loop will loop infinitely as j will always be less than or equal to i (assuming ULL_MAX + 1 is 0).

Since i doesn't quite loop from 1 to infinity, the code will loop slightly less than 50 % of the time, more specifically

$$ 50 \% - \frac {1} {3^{U+1} - 3^{U}} $$

where U is ULLONG_MAX.