Draw the arc of a ball

Question

Draw the parabolic trajectory of a thrown ball.

The input is the ball's initial upward velocity, a positive integer v. Every second, the ball moves 1 space right and v spaces vertically, and then v decreases by 1 to due to gravity. So, the upward velocity eventually steps down from v to 0 and down to -v, finally falling back down to its initial height.

The ball's positions trace a parabola. At horizontal position x, its height is y=x*(2*v+1-x)/2, with (0,0) the ball's initial position at the bottom left.

Output ASCII art of the ball's trajectory with O's on the coordinates it ever occupies. The output should be a single multi-line piece of text, not an animation of the path over time.

The output should have no leading newlines and at most one trailing newline. The bottom line should be flush with the left edge of the screen, i.e. have no extra leading spaces. Trailing spaces are OK. You may assume the output line width fits in the output terminal.

v=1

 OO 
O  O

v=2

  OO  
 O  O 

O    O

v=3

   OO   
  O  O  

 O    O 


O      O

v=4

    OO    
   O  O   

  O    O  


 O      O 



O        O

v=10

          OO          
         O  O         

        O    O        


       O      O       



      O        O      




     O          O     





    O            O    






   O              O   







  O                O  








 O                  O 









O                    O

Related: Bouncing ball simulation

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Answer 1

Charcoal, 18 16 13 bytes

-3 bytes thanks to @Neil!

Ｆ⊕Ｎ«←OＭ⊕ι↓»‖Ｃ

Explanation

Ｆ⊕Ｎ«        »    For ι (implicitly from 0) to (1 + input as number)
       ←O          Print O, with print direction rotated 180 degrees
         Ｍ⊕ι↓     Move 1+ ι units down

                ‖Ｃ Reflect (in the default direction, right), leaving original intact

Try it online! Link is to verbose code.

Answer 2

C, 93 92

(Note, someone got to 87 in the comments)

y,n;f(s){for(y=0;y<=s;){printf("%*c%*c",s-y+1,79,y*2+1,79);for(n=++y;s+1-n&&n--;)puts("");}}

Try it online!

---

Readable:

y,n;f(s){
    for(y=0;y<=s;){
        printf("%*c%*c",s-y+1,79,y*2+1,79);
        for(n=++y;s+1-n&&n--;)puts("");
    }
}

---

Notes:

I can collapse both for loops into just one for loop by iterating the total number of lines outputted, which is given by the formula: n*-~n/2+1

y,n,r;f(s){
    for(r=s,y=n=0;r<s*-~s/2+1;)
        y==n?printf("%*c%*c",s-y+1,79,y*2+1,79),y=0,++n:r++,y++,puts("");
}

But it ends up being even more bytes than just using two seperate for-loops

Answer 3

GNU sed, 41

• Score includes +1 from -r flags to sed.

s/$/OO/
:
s/(\s*) O( *)O$/&\n\1O \2 O/
t

Input is in unary, as a string of spaces - the length of string is the input.

Try it online.

Answer 4

Python 2, 65 bytes

f=lambda n,x=0:(n and f(n-1,x+1)or'')+'\n'*n+' '*x+'O'+'  '*n+'O'

Try it online!

Answer 5

Python 2, 76 bytes

x=input()
for i in range(x):print' '*(x-i),'O'+' '*i*2+'O'+'\n'*(i-x+1and i)

Pretty simple. The i-x+1and i is to prevent a bunch of trailing newlines.

Answer 6

MATL, 19 17 bytes

Q:tqYsQ79Z?PtPv!c

Try it at MATL Online! Or verify all test cases.

Explanation

Q        % Implicitly input v. Add 1
:        % Push [1 2 ... v+1]
tq       % Duplicate and subtract 1: pushes [0 1 ... v]]
Ys       % Cumulative sum: gives [0 1 3 6 ...]
Q        % Add 1: gives [1 2 4 7 ...]
79       % Push 79 (ASCII for 'O')
Z?       % Create sparse matrix from column indices [1 2 3 4 ...],
         % row indices [1 2 4 7 ...], and data 79
P        % Flip vertically
tP       % Duplicate, flip vertically
v        % Concatenate the two matrices vertically
!        % Transpose
c        % Convert to char. Implicitly display. Char 0 is shown as space

Answer 7

05AB1E, 18 14 bytes

Saved 4 bytes thanks to Adnan

ƒ¶N×'ONúRÂJ}.c

Try it online!

Explanation

ƒ                   # for N in [0 ... input]
 ¶N×                # push N newlines
    'O              # push "O"
      Nú            # pad with N spaces in front
        RÂ          # reverese and create a reversed copy
          J         # join everything to a string
           }        # end loop
            .c      # pad lines until centered 

Answer 8

JavaScript (ES6), 98 92 89 84 78 bytes

(-20 bytes thanks to Arnauld!)

f=(v,i=0)=>i>v?"":" "[r="repeat"](v-i)+0+" "[r](2*i)+0+`
`[r](i++<v&&i)+f(v,i)

A recursive solution. This is also my first ever answer in JavaScript, so please be gentle! I am still learning all this neat language has to offer, so golfing tips are very much appreciated. :)

Test Snippet

You may need to scroll to see the entire output.

f=(v,i=0)=>i>v?"":" "[r="repeat"](v-i)+0+" "[r](2*i)+0+`
`[r](i++<v&&i)+f(v,i)

<input id=i min=1 type=number><button onclick=alert(f(document.getElementById("i").value))>Submit</button>

Answer 9

RProgN 2, 37 bytes

x=0xR{y@xy-` *`o` y2**`o...2y{[` };};

Getting in with my kind-of-golfy language before the proper golfy langauges jump in.

Explained

x=              # Set 'x' to the input
0xR{            # For everything between the input and 0
    y@          # Set the iteration value to y, for this function only.
    xy-` *      # Subtract y from x, repeat the string " " that many times.
    `o          # Push an "o" to the stack.
    ` y2**      # Push 2*y " "'s to the stack
    `o          # Push another "o" to the stack
    ...         # Concatenate the parts of this string together, giving us the two balls.
    2y{[` };    # For all numbers between 2 and y, add a newline.
};              #

Try it online!

Answer 10

Retina, 35

• 2 bytes saved thanks to @MartinEnder

Port of my sed answer:

.+
$* OO
+`(\s*) (O *)O$
$&¶$1O $2 O

Try it online.

Answer 11

Retina, 29 19 bytes

 ?
$.`$*¶$&$'O$`$`O

Try it online!

Takes input in unary as a run of spaces. Port of my JavaScript answer. Edit: Saved 10 bytes thanks to @MartinEnder♦.

Answer 12

Bash, 76 bytes

for((n=$1+1;--n;));{
yes ''|head -$n
r=$r›${n}AO
t=›${n}BO$t
}
echo O${r}O$t

Only works in a terminal since it uses ANSI escape sequences. › represents the CSI byte (0x9b).

Test run

$ # The terminal's encoding must be set to ISO-8859-1.
$
$ xxd -g 1 arc.sh
0000000: 66 6f 72 28 28 6e 3d 24 31 2b 31 3b 2d 2d 6e 3b  for((n=$1+1;--n;
0000010: 29 29 3b 7b 0a 79 65 73 20 27 27 7c 68 65 61 64  ));{.yes ''|head
0000020: 20 2d 24 6e 0a 72 3d 24 72 9b 24 7b 6e 7d 41 4f   -$n.r=$r.${n}AO
0000030: 0a 74 3d 9b 24 7b 6e 7d 42 4f 24 74 0a 7d 0a 65  .t=.${n}BO$t.}.e
0000040: 63 68 6f 20 4f 24 7b 72 7d 4f 24 74              cho O${r}O$t
$
$ bash arc.sh 1
 OO
O  O
$ bash arc.sh 2
  OO
 O  O

O    O
$ bash arc.sh 3
   OO
  O  O

 O    O


O      O
$ bash arc.sh 4
    OO
   O  O

  O    O


 O      O



O        O

Answer 13

R, 89 bytes

a=2*v+3
x=matrix(" ",a,v^2+1)
for(k in 0:v)x[c(1-k,k+2)+v,k^2+1]="o"
x[a,]="\n"
cat(x,sep="")

• Create a matrix of spaces (the variable a is the width of this matrix, saving a couple of bytes)
• Fill in "o"s at the required locations, working from the top of the arc downwards and outwards
• Add a newline at the end of each matrix row
• Collapse the matrix down to a single string and print

This is my first attempt at golfing, comments welcome...

Answer 14

Röda, 53 52 bytes

f n{seq 0,n|{|i|["
"*i," "*(n-i),"O"," "*i*2,"O"]}_}

Try it online!

Usage: main { f(5) }

Ungolfed version:

function f(n) {
    seq(0, n) | for i do
        push("\n"*i, " "*(n-i), "O", " "*i*2, "O")
    done
}

Answer 15

Befunge, 75 73 bytes

<vp00:&
1<-1_:v#\+55:g01\-g01g00" O"1\*2g01" O"1p0
#@_\:v>$$:!
1\,:\_^#:-

Try it online!

The first line reads in the velocity, v, and saves a copy in memory. The second line then counts down from v to zero, with the index i, and on each iteration pushes a sequence of character/length pairs onto the stack.

Length  Character
-----------------
1       'O'
i*2     ' '
1       'O'
v-i     ' '
i       LINEFEED

This sequence represents a kind of run-length encoding of the required output in reverse. The last two lines then simply pop these character/length pairs off the stack, outputting length occurrences of each character, until the stack is empty.

Answer 16

Java 8, 129 124 109 bytes

Golfed:

v->{String s="",t="";for(int j,y=0;y<=v;++y){for(j=0;j<v;++j)s+=j<y?"\n":" ";s+="o"+t+"o";t+="  ";}return s;}

Try it online!

Ungolfed:

public class DrawTheArcOfABall {

  public static void main(String[] args) {
    for (int i = 1; i < 6; ++i) {
      System.out.println(f(v -> {
        String s = "", t = "";
        for (int j, y = 0; y <= v; ++y) {
          for (j = 0; j < v; ++j) {
            s += (j < y ? "\n" : " ");
          }
          s += "o" + t + "o";
          t += "  ";
        }
        return s;
      } , i));
      System.out.println();
      System.out.println();
    }
  }

  private static String f(java.util.function.IntFunction<String> f, int v) {
    return f.apply(v);
  }
}

Answer 17

05AB1E, 18 13 bytes

ÝηRO«ð×'O«ζ»

Try it online!

Ý                # [0..n]
 €LRO            # [0.sum(), 0..1.sum(), ..., 0..n-1.sum(), 0..n.sum()]
     «          # Mirror image the array [0, 0..n.sum(), 0]
       ð×'O«     # Push that many spaces with an O appended to it.
            .B   # Pad small elements with spaces to equal largest element length.
              ø» # Transpose and print.

Answer 18

VBA, 124 112 85 88 66 63 59 bytes

For i=0To[A1]:?Space([A1]-i)"O"Space(2*i)"O"String(i,vbCr):Next

_{^{Saved 29 bytes in total thanks to Taylor Scott}}

This must be run the in VBA Immediate window and print the result in the same.

Expanded / Formatted, it becomes:

For i=0 To [A1]
   Debug.Print Space([A1]-i) & "O" & Space(2*i) & "O" & String(i,vbCr)
Next

_{(It turns out that concatenation in a print command is automatic without an operator.)}

Answer 19

Haskell, 69 bytes

r=replicate
f n=[0..n]>>= \a->r a '\n'++r(n-a)' '++'O':r(2*a)' '++"O"

Usage example: f 3 -> " OO\n O O\n\n O O\n\n\nO O". Try it online!.

Answer 20

Jelly, 17 16 bytes

‘Ḷ+\Ṛ⁶ẋ;€”Om0z⁶Y

Try it online!

How?

‘Ḷ+\Ṛ⁶ẋ;€”Om0z⁶Y - Main link: v         e.g. 3
‘                - increment: v+1            4
 Ḷ               - lowered range             [0,1,2,3]
  +\             - reduce with addition      [0,1,3,6]
    Ṛ            - reverse                   [6,3,1,0]
     ⁶           - a space                   ' '
      ẋ          - repeat (vectorises)       ['      ','   ',' ','']
       ;€        - concatenate each with
         ”O      -     an 'O'                ['      O','   O',' O','O']
           m0    - concatenate reflection    ['      O','   O',' O','O','O','O ','O   ','O      ']
             z⁶  - transpose with space fill ['   OO   ','  O  O  ','        ',' O    O ','        ','        ','O      O']
               Y - join with line feeds      ['   OO   \n  O  O  \n        \n O    O \n        \n        \nO      O']
                 - implicit print

Answer 21

JavaScript (ES6), 87 bytes

f=
n=>' '.repeat(n+1).replace(/./g,"$`#$'O$`$`O").replace(/ *#/g,s=>[...s].fill``.join`
`)

<input type=number min=0 oninput=o.textContent=f(+this.value)><pre id=o>

Nonrecursive solution. Indexing requirement was annoying, both in the above and the following 62-byte (I don't know whether it would result in a shorter Retina port) recursive solution:

f=n=>~n?` `.repeat(n)+`OO`+f(n-1).replace(/^ *O/gm,`
$&  `):``

Answer 22

PHP, 76 bytes

for(;$argn>=0;$s.="  ")echo($r=str_repeat)("
",$i++),$r(" ",$argn--),o,$s,o;

Run with echo <v> | php -nR '<code>' or test it online.

loops $argn down from input to 0 and $i up from 0;
prints - in that order - in each iteration

• $i newlines (none in the first iteration)
• left padding: $argn spaces
• left ball: o
• inner padding: 2*$i spaces
• right ball: o

Answer 23

V, 23 19 bytes

2éoÀñYço/^2á O
HPJ>

Try it online!

Explain

2éo            " Insert two 'o's
   Àñ          " <Arg> times repeat
     Y         " Yank the current (top) line.  This is always '\s*oo'
      ço/      " On every line that matches 'o'
         ^     " Go to the first non-whitespace character (the 'o')
          2á   " Append two spaces (between the two 'o's
             O " Add a blank line on top of the current one
H              " Go to the first line
 P             " Paste in front ('\s*oo')
  J            " Join this line with the blank line immediately after it
   >           " Indent once

Answer 24

Perl 5, 48 bytes

47 bytes code + 1 for -n.

$_++;print$/x$-,$"x$_,O,$"x($-++*2),O while$_--

Try it online!

Answer 25

Stacked, 67 63 bytes

args 0#1+:@x:>{!n x\-1-' '*'O'+n 2*' '*+'O'+x 1-n!=n*LF*+out}"!

Initial attempt, 67 bytes

args 0# :@v 1+2*:>[:v:+1+\-2/*' '*'O'+''split]"!fixshape tr rev out

Full program. Generates something like:

('O'
 ' ' 'O'
 ' ' 'O'
 'O')

Which is the padded, transposed, reversed, and outputted.

Answer 26

Batch, 163 bytes

@set l=@for /l %%i in (1,1,%1)do @call
@set s=
%l% set s= %%s%%
@set t=
%l%:c&for /l %%j in (2,1,%%i)do @echo(
:c
@echo %s%O%t%O
@set s=%s:~1%
@set t=  %t%

Answer 27

Ruby, 52 bytes

->x{(0..x).map{|a|$><<$/*a+' '*(x-a)+?O+' '*a*2+?O}}

No trailing newline (allowed by the rules: "at most one trailing newline")

Answer 28

AHK, 93 bytes

m=0
n=1
f=%1%-1
Loop,%1%{
r=%r%{VK20 %f%}O{VK20 %m%}O{`n %n%}
m+=2
n++
f--
}
FileAppend,%r%,*

If I could figure out how to do math inside of repeating keystrokes, that'd be great.
- VK20 equates to a space
- FileAppend outputs to stdout if the filename is *

Answer 29

Haskell, 77 bytes

n!x=[1..n]>>x
-1#i=[]
x#i=(x-1)#(i+1)++x!"\n"++i!" "++'O':(2*x)!" "++"O"
(#0)

Try it online! Usage: (#0) 5

Answer 30

Python 2, 59 bytes

f=lambda n,r='O':(r*n and f(n-1,' '+r))+'\n'*n+r+'  '*n+'O'

Try it online!

Answer 31

Pyth, 26 bytes

VyhQ=+Y+*Zd\O=+ZQ=tQ)j_.tY

Try it Online

Explanation

VyhQ=+Y+*Zd\O=+ZQ=tQ)j_.tY
                            Implicit Q=input() (speed), Z=0 (position), Y=[].
VyhQ                )       For N in [0,...,2Q+2]...
       +*Zd\O               ... prepend Z spaces to 'O'...
    =+Y                     ... add that line to Y...
             =+ZQ=tQ        ... then update position and speed.
                     j_.tY  Combine the lines to form the parabola.

Answer 32

MathGolf, 17 bytes

r{n*íï- *'oï *+_x

Try it online!

Explanation

r                   range(0, n)
 {                  start block or arbitrary length
  n                 newline char, or map array with newlines
   *                pop a, b : push(a*b)
    í               get total number of iterations of for loop
     ï              index of current loop, or length of last loop
      -             pop a, b : push(a-b)
                    space character
        *           pop a, b : push(a*b)
         '          push single character
          o         print TOS without popping
           ï        index of current loop, or length of last loop
                    space character
             *      pop a, b : push(a*b)
              +     pop a, b : push(a+b)
               _    duplicate TOS
                x   reverse int/array/string