Leaping Kangaroos

Question

Backstory

^{Disclaimer: May contain made up information about kangaroos.}

Kangaroos traverse several stages of development. As they grow older and stronger, they can jump higher and longer, and they can jump more times before they get hungry.

In stage 1, the kangaroo is very little and cannot jump at all. Despite this, is constantly requires nourishment. We can represent a stage 1 kangaroo's activity pattern like this.

o

In stage 2, the kangaroo can make small jumps, but not more than 2 before it gets hungry. We can represent a stage 2 kangaroo's activity pattern like this.

 o o
o o o

After stage 2 the kangaroo improves quickly. In each subsequent stage, the kangaroo can jump a bit higher (1 unit in the graphical representation) and twice as many times. For example, a stage 3 kangaroo's activity pattern looks like this.

  o   o   o   o
 o o o o o o o o
o   o   o   o   o

For stage n, the activity pattern consists of 2^n-1 V-shaped jumps of height n.

For example, for stage 4, there are 8 jumps of height 4.

   o     o     o     o     o     o     o     o
  o o   o o   o o   o o   o o   o o   o o   o o
 o   o o   o o   o o   o o   o o   o o   o o   o
o     o     o     o     o     o     o     o     o

Task

Write a full program or a function that takes a positive integer n as input and prints or returns the ASCII art representation of a stage n kangaroo's activity pattern.

Surrounding whitespace and ANSI escape codes are allowed, as long as the pattern looks exactly as depicted above.

If you choose a function that returns the output, it must return a single string or character array that displays the proper output when printed. Returning an array of strings is not allowed.

You can use any printable, non-whitespace character instead of o, as long as it is consistent within the activity pattern and across all patterns in your answer.

This is code-golf; may the shortest answer in bytes win!

Answer 1

05AB1E, 12 10 bytes

Îj¹FÐvû},À

Explanation:

Î              # Push zero and input
 j             # Prepend input - 1 spaces
  ¹F           # Input times do..
    Ð          #   Triplicate the string
     v }       #   Length times do..
      û        #     Palindromize
        ,      #   Pop and print with a newline
         À     #   Rotate the string on to the right

Uses the CP-1252 encoding. Try it online!

Answer 2

MATLAB, 92 90 86 84 bytes

n=input('');p=eye(n)+32;A=repmat([fliplr(p),p,''],1,2^n/2);A(:,n+1:n:end)=[];disp(A)

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eye creates an identity matrix. If we flip it and concatenate the original i.e. [fliplr(p),p] we get (for n=3):

0 0 1 1 0 0
0 1 0 0 1 0
1 0 0 0 0 1

With repmat(...,1,2^n/2) we repeat this 2^(n-1) times and get

0 0 1 1 0 0 0 0 1 1 0 0 0 0 1 1 0 0
0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 ...
1 0 0 0 0 1 1 0 0 0 0 1 1 0 0 0 0 1

From this we just delete the unnecessary columns, with A(:,n+1:n:end)=[];

Answer 3

Charcoal, 14 bytes

ＮλＰ^×λoＦ⁻λ¹‖Ｏ→

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Explanation

Ｎλ inputs an integer into λ. Ｐ^ is a multidirectional print (SE and SW) of ×λo (string multiplication of λ with o). Then Ｆ⁻λ¹ runs a for loop λ - 1 times, in which ‖Ｏ→ reflects the whole thing to the right with overlap.

Answer 4

Python 2, 87 bytes

n=input()
for i in range(n):print''.join(' o'[abs(j%(2*n)-n)==i]for j in range(1,n<<n))

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Uses a formula for the coordinates (i,j) that contain a circle, then joins and prints the grid. There's a lot of golf smell here -- ''.join, two nested ranges, for over exec, so there's likely to be improvements.

Answer 5

Python 2, 83 81 bytes

n=input()
i=0
exec"s=' '*n+'o'+' '*i;i+=1;print(s[i:-1]+s[:i:-1])*2**~-n+s[i];"*n

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Answer 6

Befunge, 98 91 bytes

This uses a , in place of the o, since that enables us to save a couple of bytes.

&::1>\1-:v
+\:v^*2\<_$\1-2*::!+00p*1
:-1_@v0\-g01:%g00:-1<:\p01
 ,:^ >0g10g--*!3g,:#^_$\55+

Try it online!

Explanation

Given the stage number, n, we start by calculating the following three parameters of the pattern:

jump_count = 2 ^ (n - 1)
jump_len   = (n - 1) * 2
width      = (jump_len * jump_count) + 1

The jump_len is normalised to avoid it being zero for a stage 1 kangaroo with:

jump_len += !jumplen    

We can then output the jump pattern by iterating over the x and y coordinates of the output area, and calculating the appropriate charater to output for each location. The y coordinate counts down from n - 1 to 0, and the x coordinate counts down from width - 1 to 0. We determine whether a dot needs to be shown with the following formula:

jump_off = x % jump_len
show_dot = (jump_off == y) or (jump_off == (jump_len-y))

The show_dot boolean is used as a table index to determine actual character to output at each location. To save on space, we use the start of the last line of source as that table, which is why our o character ends up being a ,.

Answer 7

J, 28 25 bytes

' o'{~]_&(](|.,}.)"1)=@i.

Saved 3 bytes thanks to @Conor O'Brien.

This is based on the palindrome trick from @muddyfish's solution.

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Explanation

' o'{~]_&(](|.,}.)"1)=@i.  Input: integer n
                       i.  Form the range [0, 1, ..., n-1]
                     =@    Equality table with itself.
                           Creates an identity matrix of order n
      ]                    Get n
       _&(          )      Repeat n times on x = identity matrix
           (     )"1         For each row
            |.                 Make a reversed copy
               }.              Get a copy with the head removed
              ,                Append them
          ]                  Use that as the new value of x
' o'{~                     Index into the char array

Answer 8

C#, 180, 173 171 bytes

Wont win this, posting for other C# contestants as something they can beat.

n=>{var s=new string[n];for(int a=-1,j=0,i,m=n-1,x=m;j<=m*(Math.Pow(2,n)*n+1);){i=j++%n;s[i]+=x==i?"o":"_";if(i==m&n>1){x+=a;a*=x%m==0?-1:1;}}return string.Join("\n",s);};

complete program:

using System;
public class P
{
    public static void Main()
    {
        Func<int, string> _ = n =>
        {
            var s = new string[n];
            for (int a = -1, j = 0, i, m = n - 1, x = m; j <= m * (Math.Pow(2, n) * n + 1);)
            {
                i = j++ % n;
                s[i] += x‌​ == i ? "o" : "_";
                if (i == m & n > 1)
                {
                    x += a;
                    a *= x % m == 0 ? -1 : 1;
                }
            }
            return string.Join("\n", s);
        };

        Console.Write(_(4));
        Console.ReadKey();
    }
}

---

edit: -7 bytes thanks to @KevinCruijssen

edit: -2 bytes, simplified if

Answer 9

Pyke, 11 bytes

XFd*\o+Q^Vs

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 F          -  for i in range(input)
  d*\o+     -     " "*i+"o"
       Q^   -    ^.lpad(input)
         Vs -   repeat len(^): palindromise()
X           - print(reversed(^))

Answer 10

Haskell, 100 bytes

k 1="o"
k n|n<-n-1,m<-n*2=unlines[[last$' ':['o'|mod c m`elem`[m-r,r]]|c<-[0..m*2^n]]|r<-[n,n-1..0]]

Try it online! Usage: k 3.

Explanation:

Given a row r, a column c and m = 2(n-1) an o is set if c mod m equals r or m-r. The outermost list comprehension sets the range of r from n-1 to 0, the next one sets the range of c from 0 to m*2^(n-1) and the innermost acts as conditional returning 'o' if the above formula is fulfilled and ' ' otherwise. This yields a list of strings which is turned into a single newline separated string by unlines. For n=1 the function produces a division-by-zero error, so this case is handled explicitly in the first line.

Answer 11

J, 58 47 bytes

' o'{&:>~[:(,.}."1)&.>/(2^<:)#<@(|.,.}."1)@=@i.

Saved 11 bytes using the identity matrix idea from @flawr's solution.

Try it online!

A straightforward application of the definition.

Explanation

For n = 3, creates the identity matrix of order n.

1 0 0
0 1 0
0 0 1

Then mirror it to make

0 0 1 0 0
0 1 0 1 0
1 0 0 0 1

Repeat that 2^n-1 times and drop the head of each row on the duplicates

0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0
1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1

Use those values as indices into the char array [' ', 'o'] to output a 2d char array

  o   o   o   o  
 o o o o o o o o 
o   o   o   o   o

Answer 12

Python 2, 115 113 108 98 bytes

lambda n:'\n'.join(map(''.join,zip(*[' '*abs(i)+'o'+~-n*' 'for i in range(-n+1,n-1)*2**~-n])))+'o'

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Using range(-n+1,n-1) to create the absolute number of spaces between the bottom and the o to generate

  o
 o
o
 o

and then appending more copies, rotating 90º everything and appending the last o at bottom right

Answer 13

Pyth, 30 bytes

jC.<V*]+*dtQNh*tQ^2Q*+JUQtP_J^

A program that takes input of an integer and prints the result. Uses a quote mark " instead of o.

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How it works

jC.<V*]+*dtQNh*tQ^2Q*+JUQtP_J^    Program. Input: Q
jC.<V*]+*dtQNh*tQ^2Q*+JUQtP_J^QQ  Implicit input fill
      ]                           Yield a one-element list, A
        *dtQ                      cotaining Q-1 spaces
       +    N                     appended with a quote mark.
             h*tQ^2Q              Yield 1+(Q-1)*2^Q
     *                            Repeat A that many times, giving B
                       UQ         Yield [0, 1, 2, ..., Q-1]
                      J           (Store that in J)
                     +   tP_J     Append the reverse of J, discarding the first and last
                                  elements
                    *        ^QQ  Repeat the above Q^Q times, giving C
    V                             Vectorised map. For each pair [a,b] from B and C:
  .<                               Cyclically rotate a left by b characters
 C                                Transpose
j                                 Join on newlines
                                  Implicitly print

Answer 14

JavaScript (ES6), 83 bytes

f=
n=>` `.repeat(n).replace(/ /g,"$'o$`-$`o$'-".repeat(1<<n-1)+`
`).replace(/-.?/g,``)

<input type=number min=1 oninput=o.textContent=f(this.value)><pre id=o>

Answer 15

Jelly, 11 bytes

ŒḄ¡ḶUz1Ṛa⁶Y

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How?

The printable character used is 0.

Builds upon the method of Dennis's answer to his previous question on the subject of kangaroos.

ŒḄ¡ḶUz1Ṛa⁶Y - Main link: n                      e.g. 3
ŒḄ          - bounce, initial implicit range(n) e.g. [1,2,3,2,1]
  ¡         - repeat n times                    e.g. [1,2,3,2,1,2,3,2,1,2,3,2,1,2,3,2,1]
                  i.e. [1,2,3,2,1] bounced to [1,2,3,2,1,2,3,2,1] bounced to [1,2,3,2,1,2,3,2,1,2,3,2,1,2,3,2,1]
   Ḷ        - lowered range (vectorises)        e.g. [[0],[0,1],[0,1,2],[0,1],[0],[0,1],[0,1,2],[0,1],[0],[0,1],[0,1,2],[0,1],[0],[0,1],[0,1,2],[0,1],[0]]
    U       - upend (vectorises)                e.g. [[0],[1,0],[2,1,0],[1,0],[0],[1,0],[2,1,0],[1,0],[0],[1,0],[2,1,0],[1,0],[0],[1,0],[2,1,0],[1,0],[0]]
     z1     - transpose with filler 1
       Ṛ    - ...and reverse                    e.g. [[1,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1],
                                                      [1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1],
                                                      [0,1,2,1,0,1,2,1,0,1,2,1,0,1,2,1,0]]
        a⁶  - logical and with space character (all non-zeros become spaces)
          Y - join with line feeds              e.g.    0   0   0   0  
                                                       0 0 0 0 0 0 0 0 
                                                      0   0   0   0   0

Answer 16

MATL, 27 bytes

XyPt3LZ)2&Pht4LZ)lGqX"h48*c

Try it out at MATL Online

Answer 17

Perl 6, 104 93 88 bytes

->\n{my @a;@a[$_;$++]="o" for [...] |(n-1,0,n-1)xx 2**n/2;say .join for @a».&{$_//" "}}

Inserts o's into a 2D array, and then prints it.

Answer 18

Python 3, 177 bytes

n=5;f=n-1;w=''
for i in range(n):
 s='';a=0;d='\n'
 if i==f:w='';a=-1;d=''
 for _ in range(2**f):
  s+=' '*(f-i)+'o'+' '*(2*i-1)+w+' '*(n-i-2+a)
 print(s,end=d);w='o'
print('o')

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Answer 19

05AB1E, 16 bytes

L<¹Fû}ð×'o«.BøR»

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Why and how?

                 # Example input of n=2.
L<               # [0,1] (Push [1..a], decrement).
  ¹Fû}           # [0,1,0,1,0] (Palindromize n times).
      ð×'o«      # ['o',' o','o',' o','o'] (Push n spaces, append o's).
           .Bø   # ['o ',' o','o ',' o','o '] (Pad with spaces into 2D array, transpose).
              R» # Reverse, join and print.

Answer 20

Java 8, 254 bytes

Golfed:

n->{if(n==1)return"o";int k,x,y,m=n+n-2;char[][]p=new char[n][m];for(y=0;y<n;++y)for(x=0;x<m;)p[y][x++]=' ';for(k=0;k<m;++k)p[k<n?n-k-1:k-n+1][k]='o';String s="";for(y=0;y<n;++y){for(k=0;k<1<<(n-1);++k)for(x=0;x<m;)s+=p[y][x++];if(y==n-1)s+='o';s+='\n';}

Ungolfed:

import java.util.function.*;

public class LeapingKangaroos {

  public static void main(final String[] args) {
    for (int i = 1; i <= 4; ++i) {
      System.out.println(toString(n -> {
        if (n == 1) {
          return "o";
        }
        int k, x, y, m = (n + n) - 2;
        char[][] p = new char[n][m];
        for (y = 0; y < n; ++y) {
          for (x = 0; x < m;) {
            p[y][x++] = ' ';
          }
        }
        for (k = 0; k < m; ++k) {
          p[k < n ? n - k - 1 : (k - n) + 1][k] = 'o';
        }
        String s = "";
        for (y = 0; y < n; ++y) {
          for (k = 0; k < (1 << (n - 1)); ++k) {
            for (x = 0; x < m;) {
              s += p[y][x++];
            }
          }
          if (y == (n - 1)) {
            s += 'o';
          }
          s += '\n';
        }
        return s;
      } , i));
      System.out.println();
      System.out.println();
    }
  }

  private static String toString(final IntFunction<String> func, final int level) {
    return func.apply(level);
  }

}

Program output:

o

 o o
o o o


  o   o   o   o 
 o o o o o o o o
o   o   o   o   o


   o     o     o     o     o     o     o     o  
  o o   o o   o o   o o   o o   o o   o o   o o 
 o   o o   o o   o o   o o   o o   o o   o o   o
o     o     o     o     o     o     o     o     o

Answer 21

PHP, 157 bytes

for($i=$n=$argv[1],$r=str_repeat;$i>0;)echo$r($r(' ',$i-1).'o'.$r(' ',2*$n-2*$i-1).($i==$n|$i==1?'':'o').$r(' ',$i-2),2**($n-1)).($i--==1&$n!=1?'o':'')."\n";

Ungolfed:

for($i=$n=$argv[1];$i>0;) {

    // Spacing from beginning of pattern to first 'o'   
    $o  = str_repeat(' ',$i-1); 

    // First 'o' for the ascent
    $o .= 'o'; 

    // Spacing between ascent and descent
    $o .= str_repeat(' ',2*$n-2*$i-1); 

    // Second 'o' for the descent, unless we are at the apex or the bottom
    $o .= ($i==$n|$i==1?'':'o'); 

    // Spacing to the end of the pattern
    $o .= str_repeat(' ',$i-2); 

    // Repeat the pattern 2^(n-1) times
    echo str_repeat($o, 2**($n-1)); 

    // Output final 'o' if we are at the bottom in the last pattern
    echo $i--==1&$n!=1?'o':''; 

    // End of line 
    echo "\n"; 

}