Tips for golfing in Retina

Question

What general tips do you have for golfing in Retina? I'm looking for ideas that can be applied to code golf problems in general that are at least somewhat specific to Retina (e.g. "remove comments" is not an answer). Please post one tip per answer.

For reference, the online compiler is here.

@Sp3000 pointed out there is also Tips for Regex Golf. Answers here should focus specifically on Retina features and not on general regex golfing tips.

Answer 1

Combine loops if possible

In non-trivial computations you'll often find yourself using several loops to process data:

+`stage1
+`stage2
+`stage3

So this runs stage1 until the output converges, then stage2 until the output converges and then stage3 until the output converges.

However, it's always worth examining the stages in detail. Sometimes it's possible to run the loop in an interleaved fashion as stage1, stage2, stage3, stage1, stage2, stage3, ... instead (this depends a lot on what the stages actually do, but sometimes they make completely orthogonal changes or work well as a pipeline). In this case you can save bytes by wrapping them in a single loop:

{`stage1
stage2
}`stage3

If stage1 is the first stage or stage3 is the last stage of the program you can then even omit on of those parentheses as well (which means this can already save bytes for a loop of two stages).

A recent use of this technique can be seen in this answer.

Answer 2

Splitting strings into chunks of equal length n

As in most "normal" languages TMTOWTDI (there's more than one way to do it). I'm assuming here that the input doesn't contain linefeeds, and that "splitting" means splitting it into lines. But there are two quite different goals: if the length of the string isn't a multiple of the chunk length, do you want to keep the incomplete trailing chunk or do you want to discard it?

Keeping an incomplete trailing chunk

In general, there are three ways to go about the splitting in Retina. I'm presenting all three approaches here, because they might make a bigger difference when you try to adapt them to a related problem. You can use a replacement and append a linefeed to each match:

.{n}
$&¶

That's 8 bytes (or a bit less if n = 2 or n = 3 because then you can use .. or ... respectively). This has one issue though: it appends an additional linefeed if the string length is a multiple of the chunk length.

You can also use a split stage, and make use of the fact that captures are retained in the split:

S_`(.{n})

The _ option removes the empty lines that would otherwise result from covering the entire string with matches. This is 9 bytes, but it doesn't add a trailing linefeed. For n = 3 it's 8 bytes and for n = 2 it's 7 bytes. Note that you can save one byte overall if the empty lines don't matter (e.g. because you'll only be processing non-empty lines and getting rid of linefeeds later anyway): then you can remove the _.

The third option is to use a match. With the ! option we can print all the matches. However, to include the trailing chunk, we need to allow for a variable match length:

M!`.{1,n}

This is also 9 bytes, and also won't include a trailing linefeed. This also becomes 8 bytes for n = 3 by doing ..?.?. However note that it reduces to 6 bytes for n = 2 because now we only need ..?. Also note that the M can be dropped if this is the last stage in your program, saving one byte in any case.

Discarding an incomplete trailing chunk

This gets really long if you try to do it with a replacement, because you need to replace the trailing chunk with nothing (if it exists) and also with a split. So we can safely ignore those. Interestingly, for the match approach it's the opposite: it gets shorter:

M!`.{n}

That's 7 bytes, or less for n = 2, n = 3. Again, note that you can omit the M if this is the last stage in the code.

If you do want a trailing linefeed here, you can get that by append |$ to the regex.

Bonus: overlapping chunks

Remember that M has the & option which returns overlapping matches (which is normally not possible with regex). This allows you to get all overlapping chunks (substrings) of a string of a given length:

M!&`.{n}

Answer 3

Arithmetic in Retina 1

In Retina you need to do your arithmetic in unary. However, Retina 1 makes things slightly easier for you in certain cases, principally because of two constructs:

• The repetition operator * accepts an expression as its RHS, whereas the $* operator in Retina 0.8.2 only accepted a character as its RHS
• Retina 1 now includes a $(...) grouping construct, which accepts the length modifier $.(...)

It thus becomes possible to perform some arithmetic operations in both unary or decimal, depending on which is convenient. In particular, the group length construct $.(...) calculates the sum of the lengths of its contents, taking any products using arbitrary precision integers, so it's not limited in value.

Addition

This is straightforward.

• Convert any decimal numbers to unary using *_
• Concatenate the numbers to give a unary result
• Group the sum using $(...) if you intend to multiply it or $.(...) to convert it to decimal

2-argument examples:

• Unary + Unary = Unary: $1$2
• Unary + Unary = Decimal: $.($1$2)
• Unary + Decimal = Unary: $1$2*_
• Unary + Decimal = Decimal: $.($1$2*) (see below)
• Decimal + Decimal = Unary: $1*_$2*_
• Decimal + Decimal = Decimal: $.($1*_$2*)

Multiplication

Note that * normally takes one decimal and one unary parameter and returns a unary result.

• If the numbers are all decimal, add an extra unary parameter of 1
• Convert all but one number to decimal using the length modifier
• Take the product to give a unary result
• Convert the product to decimal by enclosing in $.(...) if desired

2-argument examples:

• Unary * Unary = Unary: $.1*$2
• Unary * Unary = Decimal: $.($.1*$2)
• Decimal * Unary = Unary: $1*$2
• Decimal * Unary = Decimal: $.($1*$2)
• Decimal * Decimal = Unary: $1*$2*_
• Decimal * Decimal = Decimal: $.($1*$2*) (see below)

A Retina replacement string to take the cube of a matched decimal number would be $.(*** (see below).

Further savings

• Only the first non-negative integer contained in the the LHS of a * operator is significant; there is no limit on junk characters (including -s). In particular this makes it easier to use $&.
• If the LHS of a * is indeed $&, then it is not necessary at the start of a $( construct or replacement string or after another *. Example: $.(*__)
• The _ after a * is not necessary at the end of a $( construct or replacement string. Example: $.($1*_$2*)
• It is not necessary to close your $( constructs if they end at the end of the replacement string. Example: $.($1$2
• When adding the constant values 1 or 2 it is shorter to express them in unary.

Answer 4

Using balancing groups to capture the same count twice.

Occasionally you need to capture a balancing group twice, such as to match the substring nice when the letters are equally spaced in the input. The naive approach is to capture the characters between n and i twice, and then use balancing groups to ensure that the same number of character exist between i and c and c and e, like this:

n((.))*i(?<-1>.)*(?(1)^)c(?<-2>.)*(?(2)^)e

However, you can capture into capture group 2 while popping from capture group 1, allowing you to then pop the same count from capture group 2 later, saving a byte:

n(.)*i(?<2-1>.)*(?(1)^)c(?<-2>.)*(?(2)^)e

(Normally you would use (?!) to force the match to fail if the count is too low but obviously in code golf we just use the shortest regex that we know will fail; ^ usually works.)