We all know what a quine is. An inverted quine is a non-empty program that prints the inverse of its source code without reading its source code, and consists solely of printable-ASCII characters (space through ~
).
Here, “inverse of the source code” means the following: the output of your program must contain each printable ASCII character (m − c) times, where c is the number of times said character occurs in your code, and m is the maximum number of times any character is repeated in your code.
(In other words: your code + your output = permutation of m times all of printable-ASCII.)
For example, if your program is 12345
, then m = 1, and you should output any permutation of this string:
!"#$%&'()*+,-./06789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~
If your program is AAB
, then m = 2, and you should output any permutation of:
!!""##$$%%&&''(())**++,,--..//00112233445566778899::;;<<==>>??@@BCCDDEEFFGHHIIJJKKLLMMNNOOPPQQRRSSTTUUVVWWXXYYZZ[[\\]]^^__``aabbccddeeffgghhiijjkkllmmnnooppqqrrssttuuvvwwxxyyzz{{||}}~~
Note how there are two missing A
s, and one missing B
.
Although a program that contains all the printable ASCII characters and outputs nothing is a valid inverse quine (satisfying m = 1), such an answer would not be very competitive, given its length.
You are to write a program that is an inverted quine, as described in the above paragraph. As this is code-golf, the shortest program in bytes will win. Good luck!
duplicated for every repeated character in the source code
it makes all the difference in the challenge \$\endgroup\$11234512345
? \$\endgroup\$duplicated for every repeated character in the source code
means or if the submission is valid for that criterion, because the OP hasn't addressed quite a few of the questions here. \$\endgroup\$