Output the text "Code Bowling"
Python 3, 82 87 88
Thanks to @JonathanAllan for +1 score
print("g!#$%&'*n+,./012i34579;<l=>?@ADEwFGHIJKLoMNOPQRSBTUVWXYZ ^_`abchejkmqsuvdyz{|}~ \x6f C"[::-8])
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Nothing special, just slicing and skipping characters. The string is reversed so you can't remove characters and have the original string being printed.
Glypho, 94
The source file is encoded in CP437.
␀␀!"☺☺#$☻♥♥☻♦♣♦♣%♠♠&•◘•◘'○○(◙♂◙♂♀♪♪♀♫☼♫☼►◄◄►↕↕)*‼¶¶‼§§+,▬↨↨▬↑↓↑↓→←→←∟∟-.↔▲▲↔/▼▼▼⌂Ç⌂Çüééüââ01ää23àååàçêçê4ëë5èïèïîî67ìÄÄì8ÅÅ9ÉÉ:;æÆÆæô<=ôöòòöûùûù>ÿÿÿÖÖ?@ÜÜAB¢££¢¥₧¥₧CƒƒDáíáíóóEFúññúѪѪGººHI¿¿J⌐¬⌐¬K½½½¼¼LM¡««¡N»»»░░OP▒▒QR▓││▓┤╡┤╡S╢╢T╖╕╖╕U╣╣V║╗║╗╝╜╜╝W╛╛╛┐X┐┐Y└└└┴┬┴┬├─├─Z┼┼┼╞╞[\╟╟]^╚╔╔╚╩╦╩╦_╠╠`═╬═╬╧╨╨╧╤╥╥╤a╙╙╙╘╘bc╒╒de╓╓fg╫╪╪╫┘┌┌┘█▄█▄▌hi▌▐j▐▐▀αα▀ßΓßΓπΣπΣkσσσµlµµτmnτΦΘΘΦΩδΩδo∞∞∞φpφφεεqr∩≡≡∩±±st≥≤≤≥u⌠⌠⌠⌡⌡vw÷xy÷≈°°≈∙∙z{·|}·√ⁿⁿ√~²²²
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Explanation
Glypho is quite useful for challenges like this because it doesn't care about the actual characters being used at all. Instead it looks at each chunk of 4 characters and the command being used is determined by the pattern these four characters make:
0000 n
0001 i
0010 >
0011 \
0012 1
0100 <
0101 d
0102 [
0110 +
0111 o
0112 *
0120 -
0121 ]
0122 !
0123 e
That means we can just solve the problem and then fill in unique characters in each quartet with the printable characters and all the repeated ones with some of the "exotic" characters, which are ignored by the scoring. Glypho is sufficiently verbose that a normal handwritten solution contains enough unique characters to fit all the 94 printable ones inside. In fact, I ended up golfing it down until it had exactly 94, just so that I could use unique exotic characters for the repeated ones (hopefully, to make it harder to reduce the program).
The shorthand form of the above program is this:
11+d*d*d+d+1+1+dd1+o
d+11+d*d1+*1+-+do
11+d*d1+d**do
1+o
11+d*d*d+o
<o
ddo
11+d*d++o
111++d-<+ddo
<-+do
<1+1+o
1-+1-+o
Where each line prints one of the characters.
I've used this Retina script to convert it to Glypho using 0123. Afterwards, I've just filled in the characters in place of the digits.
In theory it might be possible to reduce this further, if someone managed to golf down the shorthand program and then managed to recombine the characters such that the right patterns show up, but I'm not sure how to prove or disprove that this is possible. If anyone manages to form a valid solution from a subset of my program, please let me know so I can delete the answer until it's fixed. Until then, I'll have to assume that this is valid.
><>, 92 94
YAY, I DID IT!
␑ is the exotic character \x11 with decimal value 17. The program exits with an error after printing the result (it's a stack underflow). I had to manage which mirrors and directional commands I used carefully, since I can only use each one once.
>"vCoUV␑3`h]Z_X
/a[f+
#$%&'r(!
.0456~8?
:;<=@9l)
ABDEFcGe
HIJKL*
MNOPQ7
RSTWYd
bgijk1
mnqst-
uwxyz2
{|} ,
\p^
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The core program:
>"vCoUV␑3`h]Z_X
/a[f+
r !
~ ?
9 )
c e
*
7
d
1
-
2
,
\p^Explanation:
Push the string vCoUV␑3`h]Z_X > (execution wraps). Move downward. Reverse the stack and remove the v.
Push 9 and c (12). Multiply to get 108 (l). Push 7. Push d (13), subtract 1, divide by 2 to get 6. Put l at (x,y) of (13,6), which is below the e. I could've done this way shorter, but this is lengthened so I have room to put more ASCII characters.
Move up into a loop. Push length of stack with l. If greater than 14, output character, otherwise create a new stack with top 10 elements, add 15 then reflect execution right then up and output. This is used to prevent the extra spaces and > from being printed at the end, as well as turning the random-looking string into what needs to be printed. Continue looping.
If any of the lines is shorter (by removing a character), the program will no longer work, either because the vertical parts of the program no longer line up, or because the string to print isn't correct anymore. Attempting to remove a character from in front of each line will change the stack depth and where the l is placed, also causing problems.
Fish