Задний план

Вот вызов, связанный с Хэллоуином.

Как вы, наверное, уже видели из моего последнего испытания, мне очень нравится то, что я называю ascii art animations, это не просто рисование рисунка, но рисование рисунка, который прогрессирует. Эта идея пришла ко мне после того, как пару лет назад меня попросили оживить (довольно скучную) презентацию, заставив случайных летучих мышей летать по экрану в Хэллоуин. Излишне говорить, что я должным образом обязан (мне платили за это), но это заставило меня думать, что в жизни есть нечто большее, чем случайные летучие мыши. Вдохновленный этим, я хотел бы предложить этот вызов.

Вызов

Заставь летать летучую мышь вокруг Луны

Вот летучая мышь:

^o^

Вот луна:

     mmm         
   mmmmmmm       
  mmmmmmmmm      
   mmmmmmm       
     mmm     

Вы должны показать каждый этап полета летучих мышей (см. Вывод).

вход

Никто

Выход

      ^o^         
      mmm         
    mmmmmmm       
   mmmmmmmmm      
    mmmmmmm       
      mmm         


      mmm^o^      
    mmmmmmm       
   mmmmmmmmm      
    mmmmmmm       
      mmm         


      mmm         
    mmmmmmm^o^    
   mmmmmmmmm      
    mmmmmmm       
      mmm         


      mmm         
    mmmmmmm       
   mmmmmmmmm^o^   
    mmmmmmm       
      mmm         


      mmm         
    mmmmmmm       
   mmmmmmmmm      
    mmmmmmm^o^    
      mmm         


      mmm         
    mmmmmmm       
   mmmmmmmmm      
    mmmmmmm       
      mmm^o^      


      mmm         
    mmmmmmm       
   mmmmmmmmm      
    mmmmmmm       
      mmm         
      ^o^         


      mmm         
    mmmmmmm       
   mmmmmmmmm      
    mmmmmmm       
   ^o^mmm         


      mmm         
    mmmmmmm       
   mmmmmmmmm      
 ^o^mmmmmmm       
      mmm         


      mmm         
    mmmmmmm       
^o^mmmmmmmmm      
    mmmmmmm       
      mmm         


      mmm         
 ^o^mmmmmmm       
   mmmmmmmmm      
    mmmmmmm       
      mmm         


   ^o^mmm         
    mmmmmmm       
   mmmmmmmmm      
    mmmmmmm       
      mmm         


      ^o^         
      mmm         
    mmmmmmm       
   mmmmmmmmm      
    mmmmmmm       
      mmm   

правила

• Нет чтения из внешних файлов или веб-сайтов
• Вы можете отправить полную программу или функцию
• Лишние пробелы и / или переводы строки меня устраивают
• Стандартные лазейки как обычно запрещены
• Летучая мышь должна закончить назад на вершине луны
• Пожалуйста, не стесняйтесь очищать экран между кадрами, если вы хотите, однако это не является обязательным требованием. Вывод, как показано выше, в порядке
• Поскольку это кодовый гольф, победителем будет ответ с самым низким количеством байтов, хотя любой ответ приветствуется

Образец

Реализована реализация в Python 2, полностью безголовая (620 байт, но только для того, чтобы доказать, что это возможно. Может, позже это удастся).

b='^o^'
m=['                     ',
'         mmm         ',
'       mmmmmmm       ',
'      mmmmmmmmm      ',
'       mmmmmmm       ',
'         mmm         ',
'                     ']
p=[(9,12),(12,15),(14,17),(15,18),(14,17),(12,15),(9,12),(6,9),(4,7),(3,6),(4,7),(6,9),(9,12)]
d=0
for z in range(7):
    x=map(str,m[z])
    c="".join(x[:p[z][0]]) + b + "".join(x[p[z][1]:])+"\n"
    print "\n".join(m[:z]) + "\n" + c+"\n".join(m[z+1:])
for z in range(6)[::-1]:
    x=map(str,m[z])
    c="".join(x[:p[z+6][0]]) + b + "".join(x[p[z+6][1]:])+"\n"
    print "\n".join(m[:z]) + "\n" + c+"\n".join(m[z+1:])

Результат

Хотя @Jonathan явно выигрывает по количеству байтов с Jelly, я собираюсь отметить ответ Brainfuck от @Oyarsa как принятый ответ исключительно потому, что я думаю, что любой, кто действительно может сделать что-то подобное на таком сумасшедшем языке, заслуживает +15 повторений, независимо от того, сколько байт требуется. Это не потому, что у меня есть проблемы с языками игры в гольф. Смотрите мой ответ на вопрос по этому вопросу на мета, если у вас есть какие-либо сомнения. Большое спасибо и уважение всем, кто внес вклад на любом языке.

Brainfuck, 690 байт

Я впервые играю в гольф, так что я уверен, что есть еще много возможностей для улучшения

-[-[-<]>>+<]>-<<+++[>+++++++<-]>>>+>>-[<-->-------]<->++++++++>+>+++++>+>+++++>+>+++++++>+>+++++>+>+++++>+>++++++++>+>+>+>>+++[<++++++>-]<+>+>-->+>+>+[<]>[[<+>>-[>]+[[<]>+[>]<+[<]>>>-]<<[->>+<<]>-]>-[->>>>+<<<<]>++]>>>[[-]<]>>>>[>]>>+>+++++++<<<<[<]+<+<<+[[<]++++[>++++<-]>[<+[>]>[-<+>]<[<]>>-]>[>]++++++++++[>]>>>-[<]<<[<]<]<[<]>->-<<<[-]++++++<+<+++[>++++++<-]+<++++[>++++++++<-]---<-[>+<-----]++<----[>+<----]--<-[>+<---]<-[>++<-----]++++<-[>+<---]++<-[>+<-------]---<----[>+<----]<--[>+<++++++]<+++++[>+++++<-]++++++>[[[>]>[>]>>[-<<+>>]<<[<]<[<]>-]>[>]>[>]+++++++[>+++++++++<-]>-[->+>+>+<<<]++++[>>++++<<-]>>+<<<<[<]>[.>]>>[.>]<[<]>[-]>[-]>[-]<+[--<---[->]<]<[->+>+>+<<<]<<[[->>+<<]<]<]

Попробуйте онлайн

Развернул некоторые для удобства чтения:

-[-[-<]>>+<]>-<<+++[>+++++++<-]>>>+>>-[<-->-------]<->++++++++>+>+++++>+>+++++>+>+++++++>+>+++++>+>+++++>+>++++++++>+>+>+>>+++[<++++++>-]<+>+>-->+>+>+
Sets counters for creation of the m's and spaces
[<]>[[<+>>-[>]+[[<]>+[>]<+[<]>>>-]<<[->>+<<]>-]>-[->>>>+<<<<]>++]
Adds the first counter plus two of the character at the second counter to the end of the
current set of cells removes the first two counters on the list then repeats until
it encounters a 254
>>>[[-]<]>>>>[>]>>+>+++++++<<<<[<]
Removes some excess cells then adds a 7 after the end of the set of cells
+<+<<+[[<]++++[>++++<-]>[<+[>]>[-<+>]<[<]>>-]>[>]++++++++++[>]>>>-[<]<<[<]<]
Adds a newline every 16 characters until 7 newlines are added
<[<]>->-<<<[-]++++++<+<+++[>++++++<-]+<++++[>++++++++<-]---<-[>+<-----]++<----[>+<----]--<-[>+<---]<-[>++<-----]++++<-[>+<---]++<-[>+<-------]---<----[>+<----]<--[>+<++++++]<+++++[>+++++<-]++++++
Removes some excess cells then sets indices for the locations of bats
>[[[>]>[>]>>[-<<+>>]<<[<]<[<]>-]>[>]>[>]+++++++[>+++++++++<-]>-[->+>+>+<<<]++++[>>++++<<-]>>+<<<<[<]>[.>]>>[.>]<[<]>[-]>[-]>[-]<+[--<---[->]<]<[->+>+>+<<<]<<[[->>+<<]<]<]
Loops through the indices replacing the spaces at each index with a bat then printing 
the set of cells then changing the bat back to spaces

Я хотел использовать тот факт, что при добавлении новой строки в начало, каждый этап можно прочитать в обоих направлениях, чтобы получить два разных этапа, но я не мог найти хороший способ сделать это без генерации полных шести с половиной этапов. сразу.

05AB1E , 69 62 60 байт

Сохранено 2 байта благодаря Аднану .

3ð×…^o^)U13FNV0379730vð5y>;ï-y_+×N(Y-12%_Xè'my×NY-12%_y&XèJ,

Попробуйте онлайн!

объяснение

3ð×…^o^)Uсохраняет список [" ","^o^"]в X для последующего использования.

13FNVпетли на протяжении 13 этапов [0 .. 12] и сохраняет текущий индекс итерации в Y .

0379730vпроходит по строкам каждой ступени,
где N - индекс строки, а y - текущее число m .

Мы начинаем с добавления floor(5/(y+1))-(y==0)пробелов в каждую строку с помощью ð5y>;ï-y_+×.

Затем мы определяем, должна ли быть летучая мышь или 3 пробела перед буквами .
Если (-N-Y)%12 == 0это правда, мы добавляем летучую мышь, еще 3 пробела.
Это выражение ( N(Y-12%_Xè) будет размещать летучих мышей поэтапно 0,6-12.

Затем мы помещаем YM с 'my×.

Теперь мы определяем, должна ли быть летучая мышь или 3 пробела после m .
Код NY-12%_y&Xèпоместит летучую мышь, если ((N-Y)%12 == 0) and y!=0это правда, иначе 3 пробела.
Это поместит летучих мышей на стадии 1-5.

Наконец мы объединим всю строку в строку и печати с новой строкой: J,.

JavaScript (ES6), 109 144 140 138 bytes

f=(k=13,b=y=>(y-k)%12?'   ':'^o^')=>k--?[0,3,7,9,7,3,0].map((n,y)=>' '.repeat(5-n/2+!n)+b(y)+'m'.repeat(n)+b(n?-y:.1)).join`
`+`
`+f(k):''

console.log(f());

Animated version

f=(k,b=y=>(y-k)%12?'   ':'^o^')=>[0,3,7,9,7,3,0].map((n,y)=>' '.repeat(5-n/2+!n)+b(-y)+'m'.repeat(n)+b(n?y:.1)).join`
`

var k =0;
setInterval(function() { o.innerHTML = f(k++) }, 150);

<pre id=o></pre>

HTML+JS, 153 149 bytes

n=setInterval(_=>o.innerText=`zzd
zc3e
 b7f
a9g
 l7h
zk3i
zzj`.replace(/\S/g,c=>parseInt(c,36)-n%12-10?`m`.repeat(c)||`   `:`^o^`,n++),1e3)

<pre id=o>

Edit: Saved a bunch of bytes thanks to @RickHitchcock. The boring version that just returns the 13 multiline strings in an array is 132 131 bytes:

_=>[...Array(13)].map((_,n)=>`zza
zl3b
 k7c
j9d
 i7e
zh3f
zzg`.replace(/\S/g,c=>parseInt(c,36)-n%12-10?`m`.repeat(c)||`   `:`^o^`))

Jelly, 76 69 58 bytes

⁾ mṁ11ż“椿¬®µ‘ŒḄ¤Œṙs⁴Y
_4¢ḣ;“^o^”;¢ṫ⁸¤
“½œ0CSbp\I7'ð½‘Ç€Y

TryItOnline

How?

⁾ mṁ11ż“椿¬®µ‘ŒḄ¤Œṙs⁴Y - Link 1, make a moon (niladic)
⁾ m                     - literal " m"
   ṁ                    - reshape like
    11                  - 11 (interpreted as range(11)) -> " m m m m m "
                 ¤      - nilad followed by links as a nilad
       “椿¬®µ‘         - code page indexes [22,3,11,7,8,9]
               ŒḄ       - bounce -> [22,3,11,7,8,9,8,7,11,3,22]
      ż                 - zip -> [[22,' '],[3,'m'],[11,' '],...]
                  Œṙ    - run length decode -> "                      mmm           mmmmmmm        mmmmmmmmm        mmmmmmm           mmm                      "
                    s   - split into chunks of length
                     ⁴  - l6
                      Y - join with line feeds

_4¢ḣ;“^o^”;¢ṫ⁸¤ - Link 2, a moon with a bat placed: endIndex
_4              - subtract 4 (startIndex)
  ¢              - call last link (1) as a nilad
   ḣ            - ḣead to startIndex
    ;           - concatenate
     “^o^”      - literal "^o^"
          ;     - concatenate
              ¤ - nilad followed by links as a nilad
           ¢    - call last link (1) as a nilad
            ṫ⁸  - tail from endIndex

“½œ0CSbp\I7'ð½‘Ç€Y - Main link (niladic)
“½œ0CSbp\I7'𽑠   - code page indexes [10,30,48,67,83,98,112,92,73,55,39,24,10]
                   -    indices to the right of each bat in a constructed moon
               ǀ  - call last link (2) as a monad for each
                 Y - join on line feeds

Python 2, 146 144 138 bytes

-2 bytes thanks to @Willem (use the variable c rather than looping through a map)

m='\n'.join(int(a)*' '+int(b)*'m'+3*' 'for a,b in zip('6643466','0379730'))
for b in" -;L[hrbQ@2' ":c=ord(b)-23;print(m[:c-3]+'^o^'+m[c:])

repl.it

'6643466' is the number of spaces before the moon (with 6 on the empty first and last lines, since the bat will go there).

'0379730' is the number of 'm's in the moon on each line.

The zip unpacks these characters into a and b and makes the moon in the sky, m, with 3 trailing spaces on each line.

The last line then traverses the positions of the bat within the moon, and is effectively:

for c in(9,22,36,53,68,81,91,75,58,41,27,16,9):print(m[:c-3]+'^o^'+m[c:])

but the tuple of positions is encoded as the ordinals of printable characters, with an added 23 (since 9 is not printable). This comes out as " -;L[hrbQ@2' ", and c=ord(b)-23 is used to extract the values.

Autovim, 85 81 bytes

No animation, probably still golfable... Still, not bad for the new kid on the block! (autovim)

ñ9am␛ÿP2xÿP4xyjGpđp2o
%ce 15
®q=8j$p
®z=6␍"_d3hP
ñğyG12PğÿPC^o^␛v^y6@q==o␛6@z==O

To run it:

autovim run ascii-bat.autovim -ni

Explanation

In short, we draw the moon, copy and paste it 12 times, and use two macros: one to append the bat on the first 7 moons, the other to prepend the last 6.

" Draw the moon
ñ                                 " Normal mode
 9am␛                            " Write 9 `m`
      ÿP                          " Duplicate the line
        2x                        " Delete two characters
          ÿP                      " Duplicate the line
            4x                    " Delete 4 characters
              yj                  " Yank two lines
                Gp                " Paste at the end of the file
                  đp              " inverts the current line with the previous
                    2o            " Add two line breaks

%ce 15                            " Center our moon

" Macros
®q=8j$p                           " Macro `q` jumps 8 lines, and pastes
                                  "   at the end
®z=6␍"_d3hP                      " Macro `z` jumps 6 lines and replaces
                                  "   the previous 3 characters by the default

" Draw all 12 moons and add the bats
                                  "   register's content
ñ                                 " Run in normal mode
 ğyG                              " Copy the moon we just drew
    12P                           " Duplicate the moon 12 times
       ğÿP                        " Duplicate the top line (gets 3 centered `m`)
          C^o^␛                  " Change the mmm by a bat
                v^y               " Copy the bat we just drew
                   6@q            " Add the first 6 bats
                      ==          " Center the bat
                        o␛       " Add a line under the bat
                           6@z    " Add the 6 last bats
                              ==  " Align the last bat in the center
                                O " Add a line over the last moon

PHP, 167 bytes

I made a small program in PHP:

<?php foreach([0,3,7,9,7,3,0]as$l)$b.=str_pad(str_repeat('=',$l),15," ",2)."\n";foreach([6,25,43,60,75,89,102,83,65,48,33,19,6]as$x)echo substr_replace($b,'^o^',$x,3);

Here is a more verbose version:

// width of the moon
$moonsizes = [0,3,7,9,7,3,0];
// position where to place the bat
$positions = [6,25,43,60,75,89,102,83,65,48,33,19,6];
// prepare base moon painting
foreach($moonsizes as $size){
    $basepainting .= str_pad(str_repeat('=',$size),15," ",STR_PAD_BOTH)."\n";
}
// loop frames and place bat
foreach($positions as $position) {
    echo substr_replace($basepainting,'^o^',$position,3);
}

This is my first codegolf, if you have any suggestion i'm happy to hear :)

Python 2, 112 bytes

b=['   ','^o^']
for k in range(91):r=k%7;print(b[k/7+r==12]*(r%6>0)+(r*(6-r)*8/5-5)*'m'+b[k/7%12==r]).center(15)

Print the picture. Each line has three parts

• A potential bat on the left
• Some number of m's for the moon
• A potential bat on the right

These parts are concatenated and centered in a box of size 15 for spacing. To avoid the bats shifting the center, a missing bat is three spaces, the same length. For bats on top of or below the moon, the left bar slot is omitted and the right slot is occupied.

There are 91 lines: a 7-line picture for each of 13 pictures. These are counted up via divmod: As k counts from 0 to 91, (k/7, k%7) goes

(0, 0)
(0, 1)
(0, 2)
(0, 3)
(0, 4)
(0, 5)
(0, 6)
(1, 0)
(1, 1)
 ....
(12, 5)
(12, 6)

Taking the first value k/7 to be the picture and the second value r=k%7 as the row number within the picture, this counts up first by picture, then by row number within each picture, both 0-indexed.

The number of m's in the moon changes with row number r=k%7 as [0,3,7,9,7,3,0]. Rather than indexing into this, a formula was shorter. A convenient degree of freedom is that the 0's can be any negative value, as this still gives the empty string when multiplied by m. Fiddling with a parabola and floor-dividing gave a formula r*(6-r)*8/5-5.

Now, we look at selecting whether to draw a bat or empty space on either side. The array b=[' ','^o^'] contains the options.

The bat in row 0 on in picture 0 (on top), in row 1 in picture 1 (on the right), on to in row 6 in picture 6 (on top). So, it's easy to check whether it appears as the row and picture number as being equal, k/7==r. But we also need picture 12 to look like picture 0, so we take the picture number modulo 12 first.

On the left it's similar. The bat appears on the left in rows r=5,4,3,2,1 in pictures 7,8,9,10,11. So, we check whether the row and picture number sum to 12. We also make sure to draw nothing rather than three spaces on rows 0 and 6 -- the right bat slot will draw the bat, and we must not mess up its centering.

C#, 615 582 337 bytes

This is my first (still far too readable) attempt at one of these, so I'll gladly welcome any suggestions to shave off a few hundred bytes! Top of my list right now is a shorter way to create the moon array.

void M(){string[]m={"","      mmm","    mmmmmmm","   mmmmmmmmm","    mmmmmmm","      mmm",""};var b="^o^";for(int x=0;x<13;x++){var a=(string[])m.Clone();int n=x>6?12-x:x;int[] j={0,1,3,6};a[n]=!a[n].Contains("m")?"      "+b:x<=n?a[n]+b:new string(' ',j[Math.Abs(9-x)])+b+a[n].Replace(" ","");foreach(var c in a){Console.WriteLine(c);}}}

Ungolfed (includes loop!)

class Program {
    public static string[] moon = new string[] { "               ", "      mmm      ", "    mmmmmmm    ", "   mmmmmmmmm   ", "    mmmmmmm    ", "      mmm      ", "               " };

    public static string bat = "^o^";

    static void Main(string[] args) {
        while (true) {
            Fly();
        }
    }

    static void Fly() {
        int times = (moon.Length * 2) - 1;
        for (int x = 0; x < times; x++) {
            string[] temp = (string[])moon.Clone(); //create a new array to maintain the original

            int index = x >= moon.Length ? times - x - 1 : x;

            if (!temp[index].Contains("m")) {
                temp[index] = new string(' ', 6) + bat + new string(' ', 6);
            } else if (x <= index) {
                int lastM = temp[index].LastIndexOf('m') + 1;
                temp[index] = temp[index].Insert(lastM, bat);
            } else {
                int firstM = temp[index].IndexOf('m');
                char[] src = temp[index].ToCharArray();
                int i = firstM - bat.Length;
                src[i] = bat[0];
                src[i + 1] = bat[1];
                src[i + 2] = bat[2];

                temp[index] = new string(src);
            }

            for (int y = 0; y < temp.Length; y++) {
                Console.WriteLine(temp[y]);
            }
            Thread.Sleep(100);
            Console.Clear();
        }            
    }
}

Edit:

Took out 21 bytes by removing trailing spaces in the array declaration. Instead of a grid 15 characters wide, each row is only wide enough for the bat to fit. Removed another 12 for the unnecessary string[]args in Main() declaration.

Edit 2:

Rewrote most of the logic, taking out 245 bytes! Includes suggested changes from the comments. Thanks!

From the comments, turned this into a function M() instead of the previous Main() method - so now, this needs to be called externally.

Python 2, 299 300 290 270 bytes

Golfed down to 270 having gained a bit more golfing experience.

Reference implementation golfed down by 321 320 330 bytes. Not pretty or elegant. Just uses brute force string and list slicing. Was fun to get the byte count down though but I think the approach was completely wrong to start with for a serious competitor.

I don't expect this answer to be taken seriously so please no downvotes. I did say in the question that I would try to golf the reference implementation and this is exactly that. Just posted for fun.

c,e,n=' ','m','\n';f,g=c*9+e*3+c*9,c*7+e*7+c*7;h=9,12;i=12,15;j=6,9;k=13,17;l=4,7;m=c*21,f,g,c*6+e*9+c*6,g,f,c*21;p=h,i,k,(15,18),k,i,h,j,l,(3,6),l,j,h;w=0
for z in map(int,'0123456543210'):print n.join(m[:z])+n+m[z][:p[w][0]]+'^o^'+m[z][p[w][1]:]+n+n.join(m[z+1:]);w+=1

Try it online!

Ruby, 164 156 bytes

puts Zlib.inflate Base64.decode64 "eNqlkrUVAAEMQvtMcZP9DfLYvzo3qhNSxQ2GVRC1ad29ar1JXaoHbul16Yig+p6BCKqtFBFU+1IE1famoHrX/jZBvN8e8f5tiE94Ib4SBREwNKH5BNJy2QM="

Very simple program. Can be worked on more. Please leave tips in the comments.

///, 205 bytes

/*/\/\///B/^o^*9/         
*M/mmm*N/MM*O/   *0/OO*n/

*1/O Nm0
*2/NM0
*3/0M9*4/1O2*7/13n*5/34/0B9570MB0
473O NmBO 
O2731ONMBO
75O NmBO 
3n510MB0n
5130B9n51OBM9n5 BNm0 
3n31B273 BNm0 
O27OBM9470B90M0O
410MO

Try it online!

Only if I didn't need tons of trailing spaces...

Mumps, 223 Bytes

This is using InterSystems Cache Mumps -- it allows braces around loops which make it handy for nested loops in a single line.

S X="09121415141209060403040609",R="097679",L="037973",S=-14 F Y=1:1:13 S S=S+2 F C=1:1:7 S F=$E(X,Y*2-1,Y*2),(M,T)="",$P(T," ",21)="" F A=1:1:$E(L,C){S M=M_"m"} S $E(T,$E(R,C))=M S:C=$S(Y<8:Y,1:Y-S) $E(T,F,F+2)="^o^" W T,!

Could probably be golfed more, and I may play with that when I have more time. Add another 9 bytes, and it'll be animated [[ due to the addition of the 'H 1 W # ' --> that halts for a second and clears the screen:

S X="09121415141209060403040609",R="097679",L="037973",S=-14 F Y=1:1:13 S S=S+2 H 1 W # F C=1:1:7 S F=$E(X,Y*2-1,Y*2),(M,T)="",$P(T," ",21)="" F A=1:1:$E(L,C){S M=M_"m"} S $E(T,$E(R,C))=M S:C=$S(Y<8:Y,1:Y-S) $E(T,F,F+2)="^o^" W T,!

Here's an ungolfed / explained version (of the animation version), with the correct 'do' statements and dot loops:

 S X="09121415141209060403040609"
 S R="097679",L="037973",S=-14              ; Initialize Variablex
 F Y=1:1:13 D                               ; Set up our main loop of 13 cels
 . S S=S+2                                  ; set up secondary counter, for start char.
 . H 1                                      ; halt for 1 second
 . W #                                      ; clear screen
 . F C=1:1:7 D                              ; set up our per-line loop
 . . S F=$E(X,Y*2-1,Y*2)                    ; set F to 2-digit "start of bat"
 . . S (M,T)=""                             ; set our "builder" lines to empty strings
 . . S $P(T," ",21)=""                      ; then reset our main 'line string' to 21 spaces.
 . . F A=1:1:$E(L,C){S M=M_"m"}             ; make our 'moon' text.
 . . S $E(T,$E(R,C))=M                      ; and insert our 'moon' into the empty line
 . . S:C=$S(Y<8:Y,1:Y-S) $E(T,F,F+2)="^o^"  ; find the start character for our 'bat' and insert.
 . . W T,!                                  ; and write the line and a CR/LF.

I tried to do the "compress then Base-64 encode" jazz just to see what would happen, but the encoded string ended up a little longer than the program itself! That, and Cache's system calls for the compression and base64 encoding are quite long... for example, here's the system call to decode base64: $System.Encryption.Base64Decode(STRING)

That method would 'bloat' the program to over 300 characters, I think...