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

Input:

nothing

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.

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)

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

Hi LordColus, and welcome to PPCG! This seems like a good first challenge. For future challenges, we recommend going through the sandbox first to iron out all the details. — user48543, May 1 '18 at 3:05
As I said in a previous comment that has been deleted, specifying "uniformly random" is fine if you are not too stringent with precision. As it stands now, the pause times must be uniform with precision of two decimals (or is it at least two decimals?). Does that mean the distribution should be uniform on the set 0.5, 0.51, 0.52, ..., 1, or can it be any floating point (possibly with more than two decimals) beween 0.5 and 1? — Luis Mendo, May 2 '18 at 10:07
@mbomb007 Same... why did this get closed again? It's basically count from 3 to 1 with two .50-1.00 second waits in between. It's really not complicated. — Magic Octopus Urn, May 2 '18 at 17:01

Magic Octopus Urn · Accepted Answer · 2018-05-01 16:47:49Z

5

05AB1E, 12 bytes

3LRε=₄D;ŸΩ.W

Try it online!

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.

Try it with debug enabled: Try it online!

answered May 1 '18 at 16:47

Magic Octopus Urn

19.9k4 gold badges51 silver badges130 bronze badges

add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

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

JayCe · Accepted Answer · 2018-05-02 02:21:15Z

2

R, 46 44 bytes

for an actual countdown:

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

Try it online!

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)," ")

Try it online!

edited May 2 '18 at 2:21

answered May 1 '18 at 17:16

JayCe

2,8157 silver badges15 bronze badges

\$\begingroup\$ I believe runif() by default has the left and right endpoints as 0 and 1 respectively, so runif(1,.5) should work the same for -2 bytes in both. \$\endgroup\$ – Giuseppe May 1 '18 at 19:45
\$\begingroup\$ Good catch thanks @Giuseppe. \$\endgroup\$ – JayCe May 1 '18 at 20:01

add a comment |

Neil · Accepted Answer · 2018-05-03 03:56:05Z

2

Python 2, 58 bytes

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

Try it online!

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

Improvements

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

edited May 3 '18 at 3:56

answered May 1 '18 at 18:23

Neil

2,4774 silver badges26 bronze badges

\$\begingroup\$ 1-time()%.5 should do the trick. (You need [3,2,1] by the way) \$\endgroup\$ – Jonathan Allan May 1 '18 at 20:38
\$\begingroup\$ Also for a in'321' saves another two \$\endgroup\$ – Jonathan Allan May 1 '18 at 20:50
\$\begingroup\$ @JonathanAllan Very good point, updated. \$\endgroup\$ – Neil May 1 '18 at 22:21
\$\begingroup\$ This sleeps once before the countdown too. I think you need the print statement before the sleep. \$\endgroup\$ – Magic Octopus Urn May 2 '18 at 17:05
\$\begingroup\$ @MagicOctopusUrn Agreed, updated. \$\endgroup\$ – Neil May 3 '18 at 3:56

add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

1

APL+WIN, 37 bytes

3⋄r←⎕dl ↑n←.49+.01×2?51⋄2⋄r←⎕dl 1↓n⋄1

edited Jun 17 '20 at 9:04

Community♦

1

answered May 1 '18 at 19:17

Graham

3,8869 silver badges9 bronze badges

add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

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).

Try it online.
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

Community · Accepted Answer · 2020-06-17 09:04:33Z

1

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")

Try it online!

edited Jun 17 '20 at 9:04

Community♦

1

answered May 1 '18 at 17:15

hakr14

1,5555 silver badges10 bronze badges

6

\$\begingroup\$ You need the imports \$\endgroup\$ – Jonathan Allan May 1 '18 at 20:34
\$\begingroup\$ On scoring imported functions \$\endgroup\$ – mbomb007 May 2 '18 at 15:30
\$\begingroup\$ Whoops! It's been fixed. \$\endgroup\$ – hakr14 May 2 '18 at 16:06

add a comment |

DanielIndie · Accepted Answer · 2018-05-02 16:50:44Z

1

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))

Try it online!

edited May 2 '18 at 16:50

answered May 1 '18 at 17:07

DanielIndie

1,2203 silver badges5 bronze badges

\$\begingroup\$ 55 bytes \$\endgroup\$ – Shaggy May 1 '18 at 21:42
1

\$\begingroup\$ @Shaggy will it really be random if you're using Date ? (for the second time the random occured) \$\endgroup\$ – DanielIndie May 2 '18 at 3:27
1

\$\begingroup\$ Why *1000%500+500? You can just use *500+500. \$\endgroup\$ – Kevin Cruijssen May 2 '18 at 7:08
\$\begingroup\$ With the updates to the spec, it probably won't be random enough but it might be worth asking for clarification. \$\endgroup\$ – Shaggy May 2 '18 at 9:27

add a comment |

steve · Accepted Answer · 2018-05-02 18:23:24Z

1

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

Try it online!

edited May 2 '18 at 18:23

answered May 1 '18 at 19:36

steve

2,2962 gold badges12 silver badges24 bronze badges

1

\$\begingroup\$ would 1-rand(.5) work? \$\endgroup\$ – Jonathan Allan May 1 '18 at 20:56
\$\begingroup\$ Good idea. Can then use 1-rand.5 too. \$\endgroup\$ – steve May 1 '18 at 21:00
1

\$\begingroup\$ Cut four more bytes by removing the parens and changing the counter to be negative. Try it online! \$\endgroup\$ – Xcali May 2 '18 at 17:07

add a comment |

Phlarx · Accepted Answer · 2018-05-03 18:03:15Z

Chip `-wingjj`, 33 bytes

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

Try it online!

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.

hakr14 · Accepted Answer · 2021-01-16 05:59:48Z

1

Pyth, 12 bytes

V_S3N.d-1O.5

Try it online!

Python 3 translation:

V_S3         | for N in range(1, 4)[::-1]:
    N        |     print(N)
     .d-1O.5 |     time.sleep(1 - random.uniform(0, 0.5))

edited Jan 16 at 5:59

answered Jan 16 at 5:51

hakr14

1,5555 silver badges10 bronze badges

add a comment |

xigoi · Accepted Answer · 2021-01-16 11:47:41Z

1

Nim, 70 bytes

import os,random
randomize()
for i in[3,2,1]:echo i;sleep 500.rand+500

Try it online!

answered Jan 16 at 11:47

xigoi

1,1311 silver badge10 bronze badges

add a comment |

Razetime · Accepted Answer · 2021-01-16 16:41:42Z

Jelly, ²¹ 18 bytes

3RṚðṄṛ50r³÷³XœS@ð/

Try it online!

-3 bytes from xigoi.

Explanation

from caird coinheringaahing.

3RṚðṄṛ50r³÷³XœS@ð/ - Main link. Takes no arguments on the left
3R                 - Generate the range [1, 2, 3]
  Ṛ                - Reverse it
   ð            ð  - Define a dyadic chain f(a, b):
    Ṅ              -   Print a
     ṛ             -   Replace a with b
      50r³         -   Yield a range [50, 51, ..., 100]
          ÷³       -   Divide each by 100; Yields [0.5, 0.51, ..., 1]
            X      -   Take a random element t
             œS@   -   Sleep for t seconds then return b
                 / - Reduce the range by the dyadic chain.
                     Remember, f(a, b) = b (with some side effects), so this runs:
                     f(f(3, 2), 1) = f(2, 1) = 1, printing a each time (3, 2, 1)

18 bytes: 3RṚðṄṛ50r³÷³XœS@ð/ (also fixes the problem that your program can't generate 0.5) — xigoi, Jan 16 at 11:39

steve · Accepted Answer · 2018-05-02 21:30:22Z

0

[GAWK], 57 bytes

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

Try it online!

answered May 2 '18 at 21:30

steve

2,2962 gold badges12 silver badges24 bronze badges

add a comment |

Stack Exchange Network

current community

your communities

more stack exchange communities

Drag Race Countdown

Challenge:

Input:

Output:

Note:

14 Answers 14

05AB1E, 12 bytes

SmileBASIC, 64 62 bytes

R, 46 44 bytes

Python 2, 58 bytes

APL+WIN, 37 bytes

Java 8, 86 bytes

Python 3, 122 bytes

JavaScript (Node.js), 75 65 60 bytes

Perl 5, 39 bytes

Chip `-wingjj`, 33 bytes

Pyth, 12 bytes

Nim, 70 bytes

Jelly, ²¹ 18 bytes

Explanation

[GAWK], 57 bytes

Your Answer

Not the answer you're looking for? Browse other questions tagged code-golf random or ask your own question.

Hot Network Questions

Drag Race Countdown

Challenge:

Input:

Output:

Note:

14 Answers 14

05AB1E, 12 bytes

SmileBASIC, 64 62 bytes

R, 46 44 bytes

Python 2, 58 bytes

APL+WIN, 37 bytes

Java 8, 86 bytes

Python 3, 122 bytes

JavaScript (Node.js), 75 65 60 bytes

Perl 5, 39 bytes

Chip -wingjj, 33 bytes

Pyth, 12 bytes

Nim, 70 bytes

Jelly, 21 18 bytes

Explanation

[GAWK], 57 bytes

Your Answer

Sign up or log in

Post as a guest

Not the answer you're looking for? Browse other questions tagged code-golf random or ask your own question.

Related

Hot Network Questions

Chip `-wingjj`, 33 bytes

Jelly, ²¹ 18 bytes