Creating a falling ball map

Question

Introduction

For the ones who never heard of this game before. You are playing a ball which needs to survive as long as possible. This is done by moving to the left or right, going to the holes. Since the map moves upwards, you need to go downwards to survive longer. If you search for images, you probably know which game I mean.

The Task

Given a positive integer n, output a falling ball map of n layers. Between the layers, there are 4 newlines. The width of the layer is made up of 25 underscore characters, with one hole of length 5. That means that the total width is equal to 30. This is randomly shifted after each layer. An example of a valid layer is:

_______________     __________

The hole can also be at the edges, as if the layer is like a cylinder:

  _________________________   

Note that there are 2 leading spaces and 3 trailing spaces. Making a single hole of width 5.

Test cases

For n=4, this is a valid output:

_______________     __________





______     ___________________





_______________________     __





  _________________________

Note: the holes must be uniformly distributed. Trailing and/or leading newlines are allowed.

This is code-golf, so the submission with the least amount of bytes wins!

Answer 1

Pyth, 19 bytes

VQ.>+*\_25*d5O30*b4

Try it here!

Explanation

VQ.>+*\_25*d5O30*b4  # Q = input

VQ                   # Repeat Q times
     *\_25           # 25 underscores
          *d5        # 5 spaces
    +                # concat the underscores and spaces
  .>         O25     # rotate this string 0-29 times
                *b4  # print the seperating newlines

Answer 2

Ruby, 74 71 59

Thanks to Alex A. for reminding me that being basically one line means I can use the stabby lambda

f=->n{n.times{puts (('_'*25+' '*5)*2)[rand(30),30]+"\n"*5}}

Answer 3

Julia, 72 bytes

n->join([join(circshift(["_"^25*" "^5...],rand(1:30)))"\n"^5 for i=1:n])

This is an anonymous function that accepts an integer and returns a string. To call it, assign it to a variable.

We construct a string of 25 underscores followed by 5 spaces, then splat that into an array of characters. We circularly shift it a random number of times (between 1 and 30), join it back into a string, and tack on 5 newlines. This process is repeated n times, where n is the input. The resulting array is joined and returned.

Try it here

Answer 4

PowerShell 65 Bytes

0..$args[0]|%{-join("_"*25+" "*5)[($r=Random 25)..($r-29)];,""*5}

Run a loop for n iterations. Get a random number and make a underscore string with 5 spaces at the end. Index into the the string as an array to pull starting from a random location. Negative indexing wraps from the end. Join that back up again and append 5 newlines.

Edit: I don't know why I thought I needed the parameter for -Maximum as it's positional. Thanks to TimmyD I shaved some bytes and forgot to get input so that drove me back up to 65.

Answer 5

Pyth, 19 bytes

VQ.>r"5 25_"9O30*4b

Try it here

Thanks to FryAmTheEggman for catching a bug.

Answer 6

SpecBAS - 93 bytes

1 a$="_"*25+" "*5: INPUT n: FOR i=1 TO n: r=1+INT(RND*30): ?a$(r TO)+a$( TO r-1)''''': NEXT i

Makes a string of 25 underscores + 5 spaces, picks a random start point and splices the string at that position and tacks the start of string to end.

The apostrophe forces a newline, so 1 at end of string and 4 empty lines.

Answer 7

MATL, 23 bytes

:"95l25X"5Z"h30YrYS10ct

Try it online!

Explanation

:"           % take input n implicitly. Repeat loop n times
  95l25X"    %   row vector with number 95 ('_') repeated 25 times
  5Z"        %   row vector of 5 blank spaces
  h          %   concatenate vertically
  30Yr       %   random number uniformly distributed on [1,2,...,30]
  YS         %   circularly shift
  10c        %   newline character. Includes another newline when printed
  t          %   duplicate
             % implicitly end loop and display stack contents

Answer 8

Java, 177 140 bytes

-1 byte to Blue moving the i++

-2 bytes to me leaving meaningless whitespace in the original

-34 bytes to Blue for realizing I didn't need the char[] at all

Because, ya know, Java.

n->{String s="";for(;n-->0;){int r=(int)(Math.random()*30);for(int i=0;i<30;){s+=i>r&i<r+6|r>24&i++<=r%25?" ":"_";}s+="\n\n\n\n";}return s;}

Slightly ungolfed:

Function<Integer, String> fallingBall = n -> {
    String s = "";
    for (;n-->0;) {
        int r = (int)(Math.random()*30);
        for (int i=0; i<30; i++) {
            s += i>r&i<r+6|r>24&i<=r%25?" ":"_";
        }
        s += "\n\n\n\n";
    }
    return s;
}

Since Java doesn't have a String 'rotate', I had to take a different approach. I generated a random int 0-29 that represents the start index of the hole (OBO because > is shorter than >=). Each individual character is then decided by the logic

l[index] = (index>random && index<random+6) || (random>24 && index<=random%25) ? ' ' : '_';

This handles the "wrapping" nature of the whole.

Answer 9

Python 3.5 - 109 113 bytes (+4 for printing out 4 new lines between each result):

def g(a,j='_'*25+' '*5):
 for l in range(a):import random;n=random.randrange(len(j)+1);print(j[-n:]+j[:-n]+'\r'*4)

Ungolfed Form:

# j is the starting string with the 25 underscores and then 5 spaces at the end
def g(a,j='_'*25+' '*5):
    import random #Import the random module
    for l in range(a): # Do the following as many times as provided by the user
        n=random.randrange(len(j)+1) Get a random number in the range 0-29
        print(j[-n:]+j[:-n]+'\r'*4) #Shift the string characters right "n" times and print out the results with 4 new lines in between

Basically what is going on is that a string 'j', consisting of 25 underscore followed by 5 spaces, is given as an argument to the function "g". Then, the random module is imported, and, as many times as specified by the user, a random number is chosen in the range 0 => 29, and the string, "j", is shifted right that many times. The result is then printed out after each iteration, with 4 new lines in between each result.

Answer 10

Python, 119 103 bytes

import random
s='_'*20+' '*5
def f(n,a=random.randint(0,25)):[print(s[a:]+s[:a]+3*'\n')for _ in' '*n]