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Do you know the optical effect of a tridimensional hand painted whit horizontal lines? Examples

This challenge consists of making something like that effect with ascii, and transforming one 2d input into a "3d" output.

The algorithm

To perform this transformation, you first replace all 1 with a ¯ and all 0 with a _. In order to make things more realistic, you should replace a ¯ that does not have another ¯ before it with /, and a ¯ that does not have another ¯ after it with \.

Some examples:

Input: 001111100 Output: __/¯¯¯\__ 
Input: 0110 1111 Output: _/\_ /¯¯\

^ In this case, there are multiple lines, so apply this to all lines.

Input: 000000000000000000000000000000000 000000111100000000000001111100000 000000111110000000000111111000000 000000111111000000011111100000000 000000111111000001111110000000000 000000011111100011111100000000000 000000111111111111111100000000000 000000111101111111011110000000000 000000111100111110011110000000000 000000111111111111111110000000000 000000111111111111111110000000000 000000001111111111110000000000000 000000000001111100000000000000000 000000000000000000000000000000000 Output: _________________________________ ______/¯¯\_____________/¯¯¯\_____ ______/¯¯¯\__________/¯¯¯¯\______ ______/¯¯¯¯\_______/¯¯¯¯\________ ______/¯¯¯¯\_____/¯¯¯¯\__________ _______/¯¯¯¯\___/¯¯¯¯\___________ ______/¯¯¯¯¯¯¯¯¯¯¯¯¯¯\___________ ______/¯¯\_/¯¯¯¯¯\_/¯¯\__________ ______/¯¯\__/¯¯¯\__/¯¯\__________ ______/¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯\__________ ______/¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯\__________ ________/¯¯¯¯¯¯¯¯¯¯\_____________ ___________/¯¯¯\_________________ _________________________________

Exceptions / rules:

  • The input will never have a single positive cell in a row (e.g. 00100)

  • You can consider other characters for the input. However, it should only be two characters and not the same characters that the output uses. For instance:

Valid input: 0001111000 # two characters, different from the output Valid input: aaiiaiiaaa # two characters, different from the output Valid input: ,€€€€,,,,, # two characters, different from the output Invalid input: 0001223000 # four different characters are used. Invalid input: ___1111___ # invalid, because the output uses underscores. Invalid input: ///\\\\/// # both slash and backslash are used by the output.

  • The output must use the four characters described above and only those four. Alternatively, you may use - instead of the macron (¯)

  • The macron (upper character) has a codepoint of 175, but you may count it as one byte.

This is , so the shortest code in bytes wins.

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    \$\begingroup\$ I like the test cases, but could you also describe the algorithm? (if I'm understanding correctly, change 1s to macrons, 0s to underscores, _¯ to _/, and ¯_ to \_) \$\endgroup\$ Commented May 17, 2021 at 14:59
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    \$\begingroup\$ This isn't "ascii-art" as a macron (¯) isn't ASCII. Additionally, you should provide an explanation as when to convert to / and `. It seems to be at the first and last 1`s of each row, but that isn't clear \$\endgroup\$ Commented May 17, 2021 at 15:03
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    \$\begingroup\$ I think this is a cool challenge, but unfortunately it's underspecified right now. A description of the actual algorithm to generate the output is needed. Also, ^; it's fine to include that but it can't be ascii-art in that case and I'd recommend allowing answers to include it for 1 byte, or just use - or something \$\endgroup\$ Commented May 17, 2021 at 15:07
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    \$\begingroup\$ I've fixed the spec for you since I think it's quite obvious what you mean, so I've reopened this. Please let me know if anything is incorrect with this. For the future, I would definitely recommend using the Sandbox; it's a good way to get great ideas like this into clearly formatted great challenges. +1 to this though! \$\endgroup\$ Commented May 17, 2021 at 15:34
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    \$\begingroup\$ The test case 1111 -> /¯¯\ does not match the spec from the first paragraph. It implies there is another rule like "for the first and last characters, convert ¯ to / and \ . Is that correct? \$\endgroup\$ Commented May 17, 2021 at 15:41

21 Answers 21

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sed 4.2.2, Score 25

26 bytes, but scoring ¯ as 1, as per note in challenge.

s|1(1*)1|/\1\\|g y/01/_¯/

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0
12
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Jelly, 16 bytes

Ż<ƝoḤU)⁺ị“\/Ø-_”

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Uses hyphens instead of macrons. Takes a good deal of inspiration from hyper-neutrino's answer, so don't forget to send her some votes. Feels like there's got to be some way to shave a byte off the string at the end, with the right replacement for oḤ...

 ) For each line, ⁺ twice: Ż prepend a 0, Ɲ then for each pair of neighboring elements < is the second greater than the first? o Replace zeroes with corresponding elements of Ḥ the line doubled, U and reverse. ị Modular 1-index into “\/Ø-_ "\/Ø-_". (0 -> _, 1 -> \, 2 -> /, 4 -> -)
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11
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Retina 0.8.2, 26 19 bytes

1(1*)1 /$.1$*¯\ 0 _

Try it online! Link includes test cases. Edit: Saved 7 bytes with inspiration from @DigitalTrauma. Explanation:

1(1*)1 /$.1$*¯\

Replace a run of 1s with a run of ¯s between / and \.

0 _

Replace 0s with _s.

(Note that Retina uses the ISO-8859-1 code page, so ¯ is 1 byte anyway.)

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    \$\begingroup\$ @DigitalTrauma Thanks, I've actually used a more Retina-y approach for the same byte count. \$\endgroup\$ Commented May 17, 2021 at 17:46
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Python 3.8 (pre-release), 80 bytes (but actually 79 because ¯ is counted as a 1 byte char in this challenge)

import re while r:=re.sub:print(r(*"0_",r(*"1¯",r("1(1*)1",r"/\1\\",input()))))

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Because we love re.sub :)

Old solution : 94 bytes (but actually 93, you know the song ...)

I don't know why but I thought a pure vanilla python solution would beat a solution using regex ... I was wrong.

while r:=str.replace:print(r(r(r(r(f"0{input()}0","10","\\0"),"01","0/")[1:-1],*"1¯"),*"0_"))

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Nothing too crazy here, just a few str.replace, and some str.replace and even more str.replace

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8
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Perl 5 (-p -Mutf8), 25 bytes, score 24

s|1(1*)1|/\1\\|g;y;01;_¯

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Same as Digital Trauma's sed answer, except the trick using semicolon delimiter the last character can be removed.

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8
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///, 80 bytes

/0/_//1/a//aaa/a-a//-aa/--a//aa-/a--//-a-/---//a-/\\\/-//-a/-\\\\//aa/\\\/\\\\/

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Uses - instead of the macron, as it is allowed.

I'm marking it as 80 bytes because of the required newline at the end of the input.

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6
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Jelly, 18 bytes

ŒgḤ1¦€N0¦€)ị“-/\_”

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Inputs a matrix of 0 and 1s, outputs a list of lines. Uses - instead of a macron

How it works

ŒgḤ1¦€N0¦€)ị“-/\_” - Main link. Takes a matrix M on the left ) - Over each row R in M: Œg - Group adjacent equal elements in R € - Over each group G in R: Ḥ - Double: 1¦ - The first value € - Over each group G in R: N - Negate: 0¦ - The last value For zeros, these are left unchanged. For ones, the first becomes 2 and the last -1 “-/\_” - Yield the string "-/\_" ị - Index, 1 based and modularly. This means that: 1 or -3 -> - 2 or -2 -> / 3 or -1 -> \ 4 or 0 -> _
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    \$\begingroup\$ o.O shouldiblamecaching.com \$\endgroup\$ Commented May 17, 2021 at 15:31
  • \$\begingroup\$ (+1) how did you answer? \$\endgroup\$ Commented May 17, 2021 at 15:31
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    \$\begingroup\$ @ophact I didn't reload the page \$\endgroup\$ Commented May 17, 2021 at 15:32
  • \$\begingroup\$ @cairdcoinheringaahing but, once, I was writing an answer to an SO question which was closed while I was typing. Inspecting and trying to make the button clickable did not seem to work and so I could not post. \$\endgroup\$ Commented May 17, 2021 at 15:33
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Jelly, 18 bytes

Ż;0ṡ3Ḅ)ị“_./__\-_”

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Uses - instead of the macron

Ż;0ṡ3Ḅ)ị“_./__\-_” Main Link; takes a matrix of bits and outputs a list of lines ) For each row Ż Prepend 0 ;0 Append 0 ṡ3 Get overlapping slices of length 3 Ḅ Convert from binary // basically, the center of the 3-slice is the character itself // and we need the left and right context to determine if it needs // to be changed. So ___, __-, -__, and -_- (0, 1, 4, 5) become _, // _-- (3) becomes /, --_ (6) becomes \, and --- (7) stays as - ị“_./_.\-_” index into "_./_.\-_", so 1=_ 3=/ 4=_ 5=_ 6=\ 7=- 0=_ // note that Jelly is 1-indexed and indexes wrap around
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6
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JavaScript (ES6), 54 bytes*

* by counting the macron as 1 byte, as allowed in this challenge

s=>s.replace(/./g,(c,i)=>'_/¯\\'[c*=!+s[i+1]-~s[i-1]])

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5
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Excel (Insider Beta), 139 96 bytes

=LET(s,LAMBDA(t,x,y,SUBSTITUTE(t,x,y)),s(s(s(s(s(A2,101,"\_/"),10,"\_"),0&1,"_/"),0,"_"),1,"-"))

Had to spend quite a few bytes shortening the SUBSTITUTE function. Originally, I thought you had to adapt to any two characters which is what I did below.

=LET(s,LAMBDA(t,x,y,SUBSTITUTE(t,x,y)),a,LEFT(A2),b,LEFT(s(s(A2,a,"")," ","")),s(s(s(s(s(A2,b&a&b,"\_/"),b&a,"\_"),a&b,"_/"),a,"_"),b,"-"))
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4
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05AB1E, 24 bytes

εγεS¬·0ǝR¬(0ǝR}˜"_¯/\"sè

Try it online! Takes input as a list of lines of ones and zeros and outputs as a list of lists of characters in each line.

εγεS¬·0ǝR¬(0ǝR}˜"..."sè # trimmed program # implicit input... ε # with each element replaced by... è # list of characters in... "..."s # literal... è # with indices in... ˜ s # flattened... γ # list of groups of consecutive equal elements in... # (implicit) current element in map... ε # with each element replaced by... R # reversed... R # reversed... S # list of characters in... # (implicit) current element in map... ǝ # with element at index... 0 # literal... ǝ # replaced with... ¬ # first element of... S # list of characters in... # (implicit) current element in map... · # doubled... ǝ # with element at index... 0 # literal... ǝ # replaced with... ¬ # first element of... R # reversed... S # list of characters in... # (implicit) current element in map... ǝ # with element at index... 0 # literal... ǝ # replaced with... ¬ # first element of... S # list of characters in... # (implicit) current element in map... · # doubled... ( # negated } # exit map # (implicit) exit map # implicit output

R} can also be a }í with no change in functionality: Try it online!

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4
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Pip -rl, 21 "bytes"

gR+X1'/.TM_.'\TRt"¯_"

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Explanation

g With -r flag, g is a list of all lines from stdin R In each line, replace +X1 regex match of one or more 1's with this callback function: TM_ Trim the first and last characters from the match '/. Prepend / .'\ Append \ TR Transliterate t 10 "¯_" into ¯_ Autoprint, one list element per line (-l flag)
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4
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AWK, 56 bytes, 54 by the rules

gsub(0,"_")gsub(1,"/\\")gsub(/\\\//,"¯")gsub("/¯","/")

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Substitutes: all 0 to _; all 1 to /\; all \/ to ¯; and, finally, all /¯ to /.

AWK, 52 bytes, 51 by the rules

gsub("01","0/")gsub(10,"\\0")gsub(0,"_")gsub(1,"¯")

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Thanks to DLsoc for a 52 bytes/51 characters version.

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  • \$\begingroup\$ @DLosc Nice! The simpler, the better! Thanks for sharing that. I'll edit my answer, giving credit to your code. I tried using another algorithm, but with no success. \$\endgroup\$ Commented May 18, 2021 at 15:54
  • \$\begingroup\$ Unfortunately the 52/51 byte one fails on the 2nd example: Try it online!. \$\endgroup\$ Commented Oct 30, 2024 at 5:55
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Vyxal aṠD, 21 bytes

-1 from Aaron Miller

ƛĠƛ⌊ḣ$NpṫdJ`_¯\/`$İ;f

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Heavily inspired by @caird coinheringaahing's Jelly answer.
Outputs a list of lines containing a list of characters in each line. Uses the actual macron!

ƛĠƛ⌊ḣ$NpṫdJ`_¯\/`$İ;f ƛ For each line of (implicit) input... Ġƛ ; For each group of consecutive characters... ⌊ Convert each character to an integer (so double and negate work) N Negate... ḣ$ p the first element of the group. d Double... ṫ J the last element of the group. İ Index each element... `_¯\/`$ in "_¯\/" (negative indices start from the back) f Flatten the list of groups to a lists of chars
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    \$\begingroup\$ 21 bytes \$\endgroup\$ Commented May 18, 2021 at 17:27
  • \$\begingroup\$ You may as well use the D flag and add an overline... \$\endgroup\$ Commented May 31, 2021 at 5:51
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J, 48 44 43 41 bytes

'_^/\'{~(>./@,+:@g&.|.,:3*g=.1=2-/\,&0)"1

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3
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Mathematica, 74 bytes

StringReplace[#,{"101"->"\_/","01"->"_/","10"->"¯\\","0"->"_","1"->"¯"}]&
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  • \$\begingroup\$ StringReplace@{"101"->"\\_/","01"->"_/","10"->"¯\\","0"->"_","1"->"¯"} \$\endgroup\$ Commented Dec 31, 2021 at 11:34
3
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05AB1E, 17 bytes

γε¬i¦¦…/ÿ\]JT„¯_‡

I/O as a multiline string.

Try it online.

Explanation:

γ # Split the (implicit) input-string into equal adjacent parts # i.e. "0110\n1111" → ["0","11","0","\n","1111"] ε # Map each part to: ¬ # Get its first character (without popping) i # If this is a 1: ¦¦ # Remove two characters from this string …/ÿ\ # Surround it with leading "/" and trailing "\" ] # Close both the if-statement and map # → ["0","/\","0","\n","/11\"] J # Join everything back together # → "0/\0\n/11\" T ‡ # Then transliterate the characters "1" and "0" „¯_ # to the characters "¯" and "_" # → "_/\_\n/¯¯\" # (after which the result is output implicitly)
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2
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Charcoal, 20 bytes

ＷＳ⟦⪫Ｅ⪪ι0∧κ⪫/\ׯ⁻Ｌκ²_

Try it online! Link is to verbose version of code. Explanation:

ＷＳ⟦

Loop over each input string and print each output on a new line until an empty line is reached. (String array input format would have saved a byte.)

⪫Ｅ⪪ι0∧κ⪫/\ׯ⁻Ｌκ²_

Split each string on 0s, then replace each (non-empty) run of 1s with a run of ¯s wrapped in / and \, finally joining with _s.

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2
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Stax, 19 bytes

√E6∙²δ♪₧♂─Ç,áR0Z◄@╖

Run and debug it

A regex based solution similar to the Pip and sed answers. Takes a full multiline string.

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2
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JavaScript (Node.js), 71 65 61 bytes

n=>n.replace(/./g,(e,i)=>+e?+n[i+1]?+n[i-1]?'-':'/':'\\':'_')

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How it works

Replaces every letter. For each letter in question: if the numerical representation is falsy (0) then replace with _, otherwise use a simple ternary to find the correct character to use. Uses - instead of macron, you know, just because. Saved 6 bytes by reorganizing and removing assignment to unused a variable. Then saved 4 thanks to a username by entirely getting rid of variables.

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  • \$\begingroup\$ Why the b? -4 bytes \$\endgroup\$ Commented May 18, 2021 at 5:33
  • \$\begingroup\$ @Ausername thanks, did not realize that b was unused. \$\endgroup\$ Commented May 18, 2021 at 5:36
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APL(Dyalog Unicode), 23 bytes SBCS

'_\¯/'⌷⍨∘⊂⊢×{5|2⊥⍵}⌺3⍤1

Try it on APLgolf!

This function operates with boolean arrays and needs 0 as the index origin (⎕IO←0).

Explanation

 ⍤1 Rank 1 (for each row) ⌺3 Stencil (e.g. 0 1 1 0 → (0 0 1) (0 1 1) (1 1 0) (1 0 0)) { } direct function ⍵ right argument (triplet) 2⊥ Decode (from binary) | Modulo 5 {5|2⊥⍵} transforms 1 1 0 → 1, 1 1 1 → 2, 0 1 1 → 3 × Multiply ⊢ right argument (initial array) ⌷⍨∘⊂ Select elements from the left argument by indices on the right '_\¯/'
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