Horizontally mirror a brainflak program

Question

Your task is to write a program or function which takes a pure brainflak program as input (assume it contains only balanced ()[]{}<>), and output a visually mirrored copy of it.

If you reversed a brainflak program as a string, say ({}()) (add one to an input), you would get ))(}{(, which is not valid.

Instead, you should mirror it, by reversing the direction of the parentheses/brackets/braces/angle-brackets, like what you would see if the program was placed in front of a mirror: ((){}).

Test cases (??? means anything is okay):

()               -> ()
({}[{}])         -> ([{}]{})
()[]             -> []()
<(){}[]>         -> <[]{}()>
([{<()[]{}<>>}]) ->([{<<>{}[]()>}])
(a)              -> ???
{>               -> ???
))(}{(           -> ???
()↵[]            -> ???

This is a code golf challenge, shortest answer per language wins.

I'd recommend removing the invalid test cases as they add nothing to the challenge, and instead add some edge cases, or something maybe a little longer — caird coinheringaahin g, Nov 9, 2019 at 15:48
Maybe something like ([{<()[]{}<>>}]) or similarly convoluted? — caird coinheringaahin g, Nov 9, 2019 at 15:51
A nice brain-flak term to describe the inputs: a pure brain-flak program contains only (){}[]<> and is perfectly balanced. — DJMcMayhem, Nov 9, 2019 at 17:26

Neil · Accepted Answer · 2019-11-09 17:18:09Z

10

Charcoal, 3 bytes

Ｓ‖Ｔ

Try it online! Link is to verbose version of code. Explanation: Ｓ reads the input and implicitly echos it, while ‖ is the mirroring operator, which normally just reverses the input, but the Ｔ modifies it to mirror the characters at the same time.

answered Nov 9, 2019 at 17:18

Neil

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3

\$\begingroup\$ +1 right primitive for the job \$\endgroup\$
– Jonah
Nov 9, 2019 at 17:21

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xnor · Accepted Answer · 2019-11-09 20:58:49Z

Python 3, 47 bytes

lambda x:x[::-1].translate('>]<[)('*20+'}_{'*2)

Try it online!

Python has a convenient string method translate that accepts a translation string s, and maps each character c to s[ord(c)]. So, we just need to make a translation string with the right characters at the positions of ASCII values [40, 41, 60, 62, 91, 93, 123, 125] of ()<>[]{}.

The ideal would be to make the table like '????????'*16, putting the correct character for each ASCII value modulo 8. Unfortunately, the values [40, 41, 60, 62, 91, 93, 123, 125] are not distinct modulo 8, and the first modulus making them distinct is 18, which would mean something like '??????????????????'*6.

However, conveniently, the lowest 6 ASCII values [40, 41, 60, 62, 91, 93] are distinct modulo 6. That lets us handle them with '>]<[)('*20, translating ASCII values 0 through 119. For the remaining two, 123 and 125, we use '}_{'*2 to hit the next 6 ASCII values 120 to 125. Python 3 lets us stop there, unlike Python 2, which requires the translation string to be length 256 exactly.

Community · Accepted Answer · 2020-06-17 09:04:33Z

7

JavaScript (Node.js), 51 47 44 bytes

Saved 3 bytes thanks to @xnor

s=>Buffer(s).map(c=>c^68%c/3%8).reverse()+''

Try it online!

How?

For each pair of characters, we can switch from the 1st to the 2nd character (and vice versa) by XOR'ing the ASCII code with \$1\$, \$2\$ or \$6\$.

 char 1 | code | XOR | code | char 2
--------+------+-----+------+--------
   '('  |   40 |  1  |   41 |  ')'
   '<'  |   60 |  2  |   62 |  '>'
   '['  |   91 |  6  |   93 |  ']'
   '{'  |  123 |  6  |  125 |  '}'

Given an ASCII code \$c\$, we can turn it into the correct XOR value by using the following function:

$$f(c)=\left\lfloor((68 \bmod c)/3)\bmod 8\right\rfloor$$

   c  | 68 mod c |   / 3  | mod 8 | floor
------+----------+--------+-------+-------
   40 |    28    |  9.333 | 1.333 |   1
   41 |    27    |    9   |   1   |   1
------+----------+--------+-------+-------
   60 |     8    |  2.667 | 2.667 |   2
   62 |     6    |    2   |   2   |   2
------+----------+--------+-------+-------
   91 |    68    | 22.667 | 6.667 |   6
   93 |    68    | 22.667 | 6.667 |   6
  123 |    68    | 22.667 | 6.667 |   6
  125 |    68    | 22.667 | 6.667 |   6

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 9, 2019 at 16:19

Arnauld

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\$\begingroup\$ Why Buffer(s) instead of s.split``? \$\endgroup\$
– Rydwolf Programs
Nov 9, 2019 at 16:26
\$\begingroup\$ @RedwolfPrograms [...s] would be shorter than using split. But this answer is working on ASCII codes, not on characters. \$\endgroup\$
– Arnauld
Nov 9, 2019 at 16:31
\$\begingroup\$ Oh, I get it now. (I've also never though of using the ... operator on strings, that's handy to know) \$\endgroup\$
– Rydwolf Programs
Nov 9, 2019 at 16:32
2

\$\begingroup\$ I brute forced some expressions for the XOR value table and got 68%c/3%8, which seems to work for 3 bytes shorter: TIO \$\endgroup\$
– xnor
Nov 9, 2019 at 21:21

Add a comment |

Arnauld · Accepted Answer · 2019-11-10 13:52:06Z

Jelly, 13 12 bytes

Saved 1 byte thanks to @JonathanAllan

O&80%15‘^OỌṚ

Try it online!

How?

The idea here is to XOR the ASCII codes to turn each character into its mirrored counterpart, like I did in my JS answer. But we use a formula that is bit golfier in Jelly:

$$f(c)=((c\text{ & }80)\bmod 15)+1$$

where \$\text{&}\$ is a bitwise AND.

 char | code | and 80 | mod 15 |  + 1 | XOR code | new char
------+------+--------+--------+------+----------+----------
  '(' |   40 |    0   |    0   |   1  |    41    |   ')'
  ')' |   41 |    0   |    0   |   1  |    40    |   '('
------+------+--------+--------+------+----------+----------
  '<' |   60 |   16   |    1   |   2  |    62    |   '>'
  '>' |   62 |   16   |    1   |   2  |    60    |   '<'
------+------+--------+--------+------+----------+----------
  '[' |   91 |   80   |    5   |   6  |    93    |   ']'
  ']' |   93 |   80   |    5   |   6  |    91    |   '['
  '{' |  123 |   80   |    5   |   6  |   125    |   '}'
  '}' |  125 |   80   |    5   |   6  |   123    |   '{'

Commented

O&80%15‘^OỌṚ - a monadic link taking a string, e.g. "(<>[])"
O            - convert the input to ASCII codes --> [40, 60, 62, 91, 93, 41]
 &80         - bitwise AND with 80              --> [ 0, 16, 16, 80, 80,  0]
    %15      - modulo 15                        --> [ 0,  1,  1,  5,  5,  0]
       ‘     - increment                        --> [ 1,  2,  2,  6,  6,  1]
         O   - input to ASCII codes again       --> [40, 60, 62, 91, 93, 41]
        ^    - bitwise XOR                      --> [41, 62, 60, 93, 91, 40]
          Ọ  - convert back to characters       --> ")><][("
           Ṛ - reverse                          --> "([]<>)"

You can save one if you avoid the register use by taking advantage of chaining rules and repeating O like so: O&80%15‘^OỌU (side note: personally I go for Ṛ when U's depth action is not required, so would do O&80%15‘^OỌṚ) — Jonathan Allan, Nov 9, 2019 at 23:45

Wisław · Accepted Answer · 2019-11-09 15:50:48Z

5

05AB1E, 4 bytes

º2äθ

Try it online! or verify all test cases

Explanation

º    | Mirror (i.e. "({}[{}])" -> "({}[{}])([{}]{})"
 2ä  | Split into two pieces
   θ | Take the last piece

answered Nov 9, 2019 at 15:50

Wisław

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dzaima · Accepted Answer · 2019-11-11 11:26:21Z

5

Canvas, 1 byte

↔

Try it here!

Surprised that this is the first 1-byte answer.

answered Nov 11, 2019 at 11:26

dzaima

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Galen Ivanov · Accepted Answer · 2019-11-09 19:07:15Z

4

J, 24 bytes

|.rplc(;"0|.)@'([{<>}])'

Try it online!

Test cases by Jonah.

answered Nov 9, 2019 at 19:07

Galen Ivanov

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Neil · Accepted Answer · 2019-11-09 17:24:40Z

2

Retina 0.8.2, 20 bytes

O^$`.

T`([{<>}])`Ro

Try it online! Link includes test cases. Explanation:

O`.

Sort individual characters...

... by substitution value (constant empty string, therefore keeps original order)...

... with reversed sort order (i.e. reversing the string).

T`([{<>}])

Substitute each character of ([{<>}])...

`Ro

... with the matching character from the reverse of that string, thus mirroring those characters.

answered Nov 9, 2019 at 17:24

Neil

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1

\$\begingroup\$ Note that in Retina 1 you can save 4 bytes by using V` to reverse the string instead of O`.. \$\endgroup\$
– Neil
Nov 9, 2019 at 17:31

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Jonah · Accepted Answer · 2019-11-09 18:12:03Z

2

J, ³² 25 bytes

'([{<)]}>'|.@([{~8|4+i.)]

Try it online!

Quick explanation:

Find the index of each input char within the string '([{<)]}>'.
Add 4 to it to get its mirror
Take that mod 8 to handle wrapping around to the beginning
Use the result to index back into '([{<)]}>'
And reverse that result

edited Nov 9, 2019 at 18:12

answered Nov 9, 2019 at 17:23

Jonah

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\$\begingroup\$ 24 bytes using rplc \$\endgroup\$
– Galen Ivanov
Nov 9, 2019 at 18:50
\$\begingroup\$ @GalenIvanov, v nice! i think it's different enough in both concept and implementation that you should feel free to post separately. \$\endgroup\$
– Jonah
Nov 9, 2019 at 19:01
\$\begingroup\$ Ok, thanks! I'll post it separately. \$\endgroup\$
– Galen Ivanov
Nov 9, 2019 at 19:07

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

2

Japt, 12 bytes

Blatant port of Arnauld's JS solution so go upvote him.

ÔcÈ^68%X/3%8

Try it

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 10, 2019 at 0:33

Shaggy

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hyper-neutrino · Accepted Answer · 2019-11-09 15:48:43Z

1

Python 3, 61 bytes

lambda x,s="([<{}>])":"".join(s[7-s.find(c)]for c in x[::-1])

Try it online!

answered Nov 9, 2019 at 15:48

hyper-neutrino♦

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Galen Ivanov · Accepted Answer · 2019-11-09 18:03:34Z

1

Icon, 67 bytes

procedure f(s)
r:="reverse"
return map(r(s),d:="([{<>}])",r(d))
end

Try it online!

answered Nov 9, 2019 at 18:03

Galen Ivanov

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\$\begingroup\$ Cool idea for finding the mirroring. Is Icon a language worth looking into? \$\endgroup\$
– Jonah
Nov 9, 2019 at 18:50
1

\$\begingroup\$ @Jonah Thank you! Yes, I think so, but keep in mind it's an old language. Icon is one of the first languages to have generators. A cool concept is the "goal directed evaluation" - if some expression fails, Icon automatically backtracks and generates another value(s) if there's a generator involved. \$\endgroup\$
– Galen Ivanov
Nov 9, 2019 at 18:59

Add a comment |

Jonathan Allan · Accepted Answer · 2019-11-09 23:53:51Z

Jelly, 13 bytes

This approach has many 13-byte solutions, although I can golf Arnauld's 13 byter to 12 bytes.

QṢiⱮịQṢṭ2/FƲṚ

Try it online!

How?

QṢiⱮịQṢṭ2/FƲṚ - Link: list of characters, P
Q             - de-duplicate (P)
 Ṣ            - sort (call this X)
   Ɱ          - map across (p in) P with:
  i           -   first index of (p) in (X)
           Ʋ  - last four links as a monad - i.e. f(P):
     Q        -   de-duplicate (P)
      Ṣ       -   sort
        2/    -   2-wise reduce with:
       ṭ      -     tack
          F   -   flatten (i.e. pair-wise reversal of sorted unique values)
    ị         - (left) index into (right)
            Ṛ - reverse

Jo King · Accepted Answer · 2019-11-13 00:23:58Z

1

Brain-Flak, 302 bytes

{(({}(<()>))<({}[(((()()()()()){}){}){}])>{[()](<()>)}{}<({}[()])>{()(<()>)}{}<({}[(((()()())){}{}){}()])>{[()()](<()>)}{}<({}[()()])>{()()(<()>)}{}<({}[(((()()()){}()){}){}()])>{[()()](<()>)}{}<({}[()()])>{()()(<()>)}{}<({}[(((()()()()()){})){}{}])>{[()()](<()>)}{}<({}[()()])>{()()(<()>)}{}<>)<>{}}<>

Try it online!

This is (currently) very poorly golfed, but I thought there ought to be at least one Brain-Flak answer to this question.

answered Nov 13, 2019 at 0:23

Jo King♦

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Galen Ivanov · Accepted Answer · 2019-11-09 19:06:03Z

0

Red, 56 bytes

func[s][foreach c reverse s[prin select"[][()({}{<><"c]]

Try it online!

answered Nov 9, 2019 at 19:06

Galen Ivanov

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Nick Kennedy · Accepted Answer · 2019-11-09 21:40:17Z

0

Jelly, 15 bytes

“[({})]”©iⱮUịUɼ

Try it online!

A monadic link that reverses the input and input brackets of all their types.

answered Nov 9, 2019 at 21:40

Nick Kennedy

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Xcali · Accepted Answer · 2019-11-09 23:40:09Z

0

Perl 5 `-p`, 32 bytes

$_=reverse y/(){}[]<>/)(}{][></r

Try it online!

answered Nov 9, 2019 at 23:40

Xcali

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girobuz · Accepted Answer · 2019-11-12 22:00:16Z

0

C (gcc), 85 84 64 60 bytes

60 bytes thanks to Neil

p;f(char*s){for(p=strlen(s);p--;)putchar(s[p]^68%s[p]/3%8);}

Try it online!

edited Nov 12, 2019 at 22:00

answered Nov 10, 2019 at 21:21

girobuz

3992 silver badges8 bronze badges

\$\begingroup\$ 60 bytes \$\endgroup\$
– Neil
Nov 12, 2019 at 11:49

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Stack Exchange Network

Horizontally mirror a brainflak program

18 Answers 18

Charcoal, 3 bytes

Python 3, 47 bytes

JavaScript (Node.js), 51 47 44 bytes

How?

Jelly, 13 12 bytes

How?

Commented

05AB1E, 4 bytes

Canvas, 1 byte

J, 24 bytes

Retina 0.8.2, 20 bytes

J, ³² 25 bytes

Japt, 12 bytes

Python 3, 61 bytes

Icon, 67 bytes

Jelly, 13 bytes

How?

Brain-Flak, 302 bytes

Red, 56 bytes

Jelly, 15 bytes

Perl 5 `-p`, 32 bytes

C (gcc), 85 84 64 60 bytes

Your Answer

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code-golf
brain-flak
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Horizontally mirror a brainflak program

18 Answers 18

Charcoal, 3 bytes

Python 3, 47 bytes

JavaScript (Node.js), 51 47 44 bytes

How?

Jelly, 13 12 bytes

How?

Commented

05AB1E, 4 bytes

Canvas, 1 byte

J, 24 bytes

Retina 0.8.2, 20 bytes

J, 32 25 bytes

Japt, 12 bytes

Python 3, 61 bytes

Icon, 67 bytes

Jelly, 13 bytes

How?

Brain-Flak, 302 bytes

Red, 56 bytes

Jelly, 15 bytes

Perl 5 -p, 32 bytes

C (gcc), 85 84 64 60 bytes

Your Answer

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Not the answer you're looking for? Browse other questions tagged code-golfbrain-flak or ask your own question.

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Hot Network Questions

J, ³² 25 bytes

Perl 5 `-p`, 32 bytes

Not the answer you're looking for? Browse other questions tagged
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