Drag Race Countdown

Question

Challenge:

In a hypothetical scenario, the countdown timer for a race has random intervals between the counts, to prevent premature starting, e.g.

3 (0.82 seconds pass), 2 (0.67 seconds pass), 1

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Input:

nothing

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Output:

Write a program (or function) that prints the 3 numbers with a random time interval from 0.50 seconds to 1 second between each count.

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Note:

• The program must output each number (3 , 2 , 1) with the random (any number between 0.50 and 1 to the hundredths; no hard-coding) time interval between each. The precision of the random interval must go out to the hundreds (e.g: 0.52). You are not required to output the interval, only the count.
• As @JoKing clarified, I mean uniformly random (you can use the pseudo-random generator of your language.
• As many people have clarified, I really mean any 2-decimal number between 0.5 and 1. (0.50, 0.51, etc, all the way to 0.98, 0.99, 1)

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This is code-golf, so the program with the lowest byte count wins.

Answer 1

05AB1E, 12 bytes

3LRε=₄D;ŸΩ.W

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3LR          # Push [3,2,1]
   ε         # For each...
    =        # Print it.
     ₄       # Push 1000.
      D      # Duplicate top (1000).
       ;     # Divided by 2 (500).
        Ÿ    # Range from b to a ([1000 .. 500]).
         Ω   # Random pick.
          .W # Wait X ms.

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Try it with debug enabled: Try it online!

Answer 2

SmileBASIC, 64 62 bytes

?3M?2M?1DEF M
M=MILLISEC
WHILE MILLISEC<M+500+RND(501)WEND
END

Unfortunately I can't use WAIT since that only supports intervals of 1/60 of a second (anything less isn't normally useful since input/output only update once per frame)

This requires adjustment depending on the speed of the system it's running on, so it might not be valid (46 bytes):

?3M?2M?1DEF M
FOR I=-66E4-RND(66E4)TO.NEXT
END

Invalid WAIT version (36 bytes):

?3WAIT 30+RND(30)?2WAIT 30+RND(30)?1

Answer 3

R, 46 44 bytes

for an actual countdown:

for(i in 3:1){cat(i)
Sys.sleep(runif(1,.5))}

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printing interval as I initially misunderstood the challenge (46 bytes) Thanks Giuseppe for saving 2 chars.

for(i in 3:1)cat(i,format(runif(1,.5),,2)," ")

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Answer 4

Python 2, 58 bytes

from time import*
for a in'321':print a;sleep(1-time()%.5)

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I created a very simple random number generator that takes the seed time (as many people do).

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Improvements

• From 63 to 58 bytes by Jonathan Allan
• MagicOctopus Urn suggested swapping sleep and count down.

Answer 5

JavaScript (Node.js), 75 65 60 bytes

• thanks to @Shaggy for reducing by 10 bytes
• thanks to @Kevin Cruijssen for reducing by 5 bytes

f=(i=3)=>i&&setTimeout(f,Math.random()*500+500,i-1,alert(i))

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Answer 6

Python 3, 122 bytes

import random as r,time
def w():time.sleep(abs(r.random()-.5)+.5)
print(3,end="");w();print(", 2",end="");w();print(", 1")

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Answer 7

APL+WIN, 37 bytes

3⋄r←⎕dl ↑n←.49+.01×2?51⋄2⋄r←⎕dl 1↓n⋄1

Answer 8

Perl 5, 39 bytes

Gradually tweaked down to 39 bytes, thanks to @jonathan-allan + @xcali.

say-$_+select$a,$a,$a,1-rand.5for-3..-1

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Answer 9

Java 8, 86 bytes

v->{for(int i=0;++i<4;Thread.sleep((int)(Math.random()*500+500)))System.out.print(i);}

Prints without delimiter. If that is not allowed it's +2 bytes by changing print to println (new-line delimiter).

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Prove the intervals are in the correct range of [500, 1000) ms.

Explanation:

v->{                        // Method with empty unused parameter and no return-type
  for(int i=0;++i<4;        //  Loop in range [1,4)
      Thread.sleep((int)(Math.random()*500+500)))
                            //    After every iteration: sleep for [500, 1000) ms randomly
     System.out.print(i);}  //   Print the current number

Answer 10

Chip -wingjj, 33 bytes

0123456e7f s
???????p*9S!ZZZtaABb

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In Chip, we cannot wait for exactly ¹/₁₀₀ of a second, but we can wait for ¹/₂₅₆ of a second, so we use that here.

p, when asked, will pause execution for the stack head (one byte) * ¹/₂₅₆ seconds. On each cycle, we always set the high bit of the stack (¹²⁸/₂₅₆) and set all other stack bits randomly (with the ?'s). This gives an even distribution between 0.50 and 1.00 seconds.

Some of the args, -w and -gjj, specify that the input, instead of using stdin, should be a countdown from 0xFF to 0x00 (then wrapping). We use this to provide the low two bits for counting down. All other output bits remain constant (at the value corresponding to ASCII 0).

Finally, once we are done, we terminate the program with t, preventing a pause after the last number.

Answer 11

[GAWK], 57 bytes

BEGIN{while(a++<2){print 4-a;sleep(rand()/2+0.5)}print 1}

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