Gfilui

Question

This challenge is based on a sandbox post by user48538. Since he is no longer active on this site, I took over this challenge.

apgsearch, the distributed soup search program for Conway's Game of Life and the search results database, Catagolue use apgcodes to classify and denote patterns. apgcodes themselves use the extended Wechsler format, an extension of a pattern notation developed by Allan Wechsler in 1992.

The following examples and the images are taken from LifeWiki.

A string of n characters in the set matching the regex [0-9a-v] denotes a strip of five rows, n columns wide. Each character denotes five cells in a vertical column corresponding to the bitstrings [00000, 10000, 01000... 00010, 10010, 01010, 11010... 11111].

For instance, 27deee6 corresponds to a heavyweight spaceship:

27deee6

$$ beginbmatrix
colorGray0 & 1 & 1 & colorGray0 & colorGray0 & colorGray0 & colorGray0 \
1 & 1 & colorGray0 & 1 & 1 & 1 & 1 \
colorGray0 & 1 & 1 & 1 & 1 & 1 & 1 \
colorGray0 & colorGray0 & 1 & 1 & 1 & 1 & colorGray0 \
colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0
endbmatrix $$

The character z separates contiguous five-row strips.

For example, 0ca178b96z69d1d96 corresponds to a 31-bit still life:

enter image description here

$$ beginbmatrix
colorGray0 & colorGray0 & colorGray0 & 1 & 1 & colorGray0 & 1 & 1 & colorGray0 \
colorGray0 & colorGray0 & 1 & colorGray0 & 1 & colorGray0 & 1 & colorGray0 & 1 \
colorGray0 & 1 & colorGray0 & colorGray0 & 1 & colorGray0 & colorGray0 & colorGray0 & 1 \
colorGray0 & 1 & 1 & colorGray0 & colorGray0 & 1 & 1 & 1 & colorGray0 \
colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 \
colorGray0 & 1 & 1 & 1 & 1 & 1 & colorGray0 & colorGray0 & colorGray0 \
1 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & 1 & colorGray0 & colorGray0 \
1 & colorGray0 & 1 & colorGray0 & 1 & colorGray0 & 1 & colorGray0 & colorGray0 \
colorGray0 & 1 & 1 & colorGray0 & 1 & 1 & colorGray0 & colorGray0 & colorGray0 \
colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0
endbmatrix $$

The characters w and x are used to abbreviate 00 and 000, respectively.

So w33z8kqrqk8zzzx33 corresponds to a trans-queen bee shuttle:

enter image description here

(10 blank rows omitted)

enter image description here

$$ beginbmatrix
colorGray0 & colorGray0 & 1 & 1 & colorGray0 & colorGray0 & colorGray0 \
colorGray0 & colorGray0 & 1 & 1 & colorGray0 & colorGray0 & colorGray0 \
colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 \
colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 \
colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 \
colorGray0 & colorGray0 & colorGray0 & 1 & colorGray0 & colorGray0 & colorGray0 \
colorGray0 & colorGray0 & 1 & 1 & 1 & colorGray0 & colorGray0 \
colorGray0 & 1 & colorGray0 & colorGray0 & colorGray0 & 1 & colorGray0 \
1 & colorGray0 & 1 & 1 & 1 & colorGray0 & 1 \
colorGray0 & 1 & 1 & 1 & 1 & 1 & colorGray0 \
colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 \
colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 \
colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 \
colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 \
colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 \
colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 \
colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 \
colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 \
colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 \
colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 \
colorGray0 & colorGray0 & colorGray0 & 1 & 1 & colorGray0 & colorGray0 \
colorGray0 & colorGray0 & colorGray0 & 1 & 1 & colorGray0 & colorGray0 \
colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 \
colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 \
colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0
endbmatrix $$

Finally, the symbols matching the regex y[0-9a-z] correspond to runs of between 4 and 39 consecutive 0s.

A good example is 31a08zy0123cko, corresponding to a ship on quadpole:

enter image description here

$$ beginbmatrix
1 & 1 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 \
1 & colorGray0 & 1 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 \
colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 \
colorGray0 & colorGray0 & 1 & colorGray0 & 1 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 \
colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 \
colorGray0 & colorGray0 & colorGray0 & colorGray0 & 1 & colorGray0 & 1 & colorGray0 & colorGray0 & colorGray0 \
colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & 1 & 1 & colorGray0 & colorGray0 & colorGray0 \
colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & 1 & 1 & colorGray0 \
colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & 1 & colorGray0 & 1 \
colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & colorGray0 & 1 & 1
endbmatrix $$

The Challenge

Write a program or a function to parse a string of the extended Wechsler format, defined above, and print (or return) the pattern corresponding to this string.

You may assume that the string is nonempty, and does not start or end with z.

You may use any reasonable output format, e.g., a string, a matrix, a 2d array. You may use any two values to represent 0 and 1, given that you declare them in the answer.

You may omit the trailing zeroes lines in the output, or add extra trailing zeroes lines. You may also add/omit trailing zeroes on each line, as long as all lines have the same length in the output.

You may return the transpose of the array, if that is more convenient.

This is code-golf, so the shortest code wins.

Test cases

153 => [[1, 1, 1], [0, 0, 1], [0, 1, 0], [0, 0, 0], [0, 0, 0]]
27deee6 => [[0, 1, 1, 0, 0, 0, 0], [1, 1, 0, 1, 1, 1, 1], [0, 1, 1, 1, 1, 1, 1], [0, 0, 1, 1, 1, 1, 0], [0, 0, 0, 0, 0, 0, 0]]
0ca178b96z69d1d96 => [[0, 0, 0, 1, 1, 0, 1, 1, 0], [0, 0, 1, 0, 1, 0, 1, 0, 1], [0, 1, 0, 0, 1, 0, 0, 0, 1], [0, 1, 1, 0, 0, 1, 1, 1, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 1, 1, 1, 1, 0, 0, 0], [1, 0, 0, 0, 0, 0, 1, 0, 0], [1, 0, 1, 0, 1, 0, 1, 0, 0], [0, 1, 1, 0, 1, 1, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0]]
w33z8kqrqk8zzzx33 => [[0, 0, 1, 1, 0, 0, 0], [0, 0, 1, 1, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0], [0, 0, 1, 1, 1, 0, 0], [0, 1, 0, 0, 0, 1, 0], [1, 0, 1, 1, 1, 0, 1], [0, 1, 1, 1, 1, 1, 0], [0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 1, 1, 0, 0], [0, 0, 0, 1, 1, 0, 0], [0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0]]
31a08zy0123cko => [[1, 1, 0, 0, 0, 0, 0, 0, 0, 0], [1, 0, 1, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 1, 0, 1, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 1, 0, 1, 0, 0, 0], [0, 0, 0, 0, 0, 1, 1, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 1, 1, 0], [0, 0, 0, 0, 0, 0, 0, 1, 0, 1], [0, 0, 0, 0, 0, 0, 0, 0, 1, 1]]
o5995ozes88sezw33 => [[0, 1, 1, 1, 1, 0], [0, 0, 0, 0, 0, 0], [0, 1, 0, 0, 1, 0], [1, 0, 1, 1, 0, 1], [1, 0, 0, 0, 0, 1], [0, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 1], [1, 1, 0, 0, 1, 1], [1, 1, 1, 1, 1, 1], [0, 1, 0, 0, 1, 0], [0, 0, 1, 1, 0, 0], [0, 0, 1, 1, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0]]
y3343x6bacy6cab6x343zkk8yy8kkzgo8gywg8ogz0123yw321zzgo4syws4ogzgh1yy1hgz221yy122zy3c2cx6d53y635d6xc2c => [[0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1], [0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0], [1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0], [1, 1, 0, 1, 0, 0, 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@Mr.Xcoder The LifeWiki page does not say anything about that. But you can assume here that the string does not start or end with z. — yesterday
Not really a Game of Life pattern, but just for fun: vv99f60vv99f60uv11320uv19rqz33y133y11322310132231 — yesterday
"You may return the transpose of the array" - if that's OK, what about other symmetries like upside-down etc...? — 20 hours ago
Two of the first three answers fail for some of these: 1yx1, 1yy1, 1yw1 and 1yz1. Maybe they should be tests. — 19 hours ago

score 6 · Answer 1 · 2018-08-05 15:34:37Z

JavaScript (ES8), 197 bytes

Takes input as a string. Returns an array of strings with '#' and spaces. The output may include extra (but consistent) trailing spaces on each line.

s=>[x='x',...s.replace(/w|x|y./g,s=>''.padEnd(s<x?2:s>x?P(s[1],36)+4:3),P=parseInt)].map(c=>[!++w,1,2,3,4].map(i=>c<x?o[y+i-5]+=' #'[P(c,36)>>i&1]:o=[...++y&&o,'']),w=y=o=)&&o.map(r=>r.padEnd(w))

Try it online! (prettified output)

How?

Global variables

The character "x" is used several times, so it's worth storing it into the variable x.

The function parseInt is used twice, so it's worth storing it into the variable P.

y is the row index, initialized to 0.

w keeps track of an upper bound of the width which is used to pad the final output.

o[Â ] is the output array, initially empty.

Pre-processing of repeated zeros

We first replace all patterns "w", "x" and "yX" in the input string with the appropriate number of spaces. These spaces will later be interpreted as "0".

s.replace(
 /w|x|y./g,
 s => ''.padEnd(
 // w --> 2
 s < x ? 2 :
 // yX --> value of X + 4
 s > x ? P(s[1], 36) + 4 :
 // x --> 3
 3
 )
)

Decoding

We split the resulting string, prepend an initial "x" and iterate 5 times (with i = 0 to 4) on each character c:

If c is lower than "x", we append the corresponding pattern to the next 5 rows.
```
o[y + i - 5] += ' #'[P(c, 36) >> i & 1]
```

If c is greater than or equal to "x", we allocate 5 new empty strings in o[Â ] and add 5 to y. This is triggered by the initial "x" that was added at the beginning of the string, or by any "z" in the original content.
```
o = [...++y && o, '']
```

Padding

Finally, we pad each string in o[Â ] with spaces so that they all have w characters.

o.map(r => r.padEnd(w))

score 5 · Answer 2 · 2018-08-06 03:35:05Z

05AB1E, 148 132 bytes

I'm a highschool student and this was my first time for both golfing and using 05AB1E, so comments are appreciated!

"yy"Ã‚Â¾38ÃƒÂ—:'yÃƒÂ°:Ã…Â¾hAÃ‚Â«DUsÃƒÂ°Ã‚Â«#ÃŽÂµN>Ã¢Â‰Â iDÃÂ½Xsk4+sÃ‚Â¦s0sÃƒÂ—ÃƒÂ¬}}'wÃ‚Â¾2ÃƒÂ—:'xÃ‚Â¾3ÃƒÂ—:J'zÃƒÂ°:ÃƒÂ°Ã‚Â«#DÃƒÂ©ÃŽÂ¸gUÃŽÂµDgXs-0sÃƒÂ—Ã‚Â«}vNUyvNÃ¢Â‚Â„+Ã¢Â‚Â„*X+VDykD32Ã¢Â€Â¹ibDg5s-0sÃƒÂ—ÃƒÂ¬RX0QiÃ‚Â¶ÃƒÂ¬}YÃƒÂ°Ã‚Â«ÃƒÂ¬Ã‹Â†]Ã‚Â¯ÃŽÂµ#ÃŽÂ¸J

~~Try it online!~~

Try it online!

Outputs the transposed matrix as multiple output lines of 1s and 0s. If you want pretty-printed output, add J'1'Ã¢Â–Âˆ:'0'.: at the end.

Will add extra zeros.

Explanation

(I'm calling "rows" and "columns" opposite of what you might expect because it generates the transposed matrix)

The basic algorithm is:

Replace "yy" with 38 0s

Split on "y" and and expand the 0-runs.

Replace "w" and "x"

Figure out the longest column (that is, the longest string between z's) and pad all the other columns so they are that length. (This is necessary because of how the algorithm below works)

Split on z

At this point, the input string is an array of columns where each column is a string of [0-9a-v], where each column is the same length.

The algorithm to get it into the output format is to
1. Convert the characters to numbers by using indexOf in a lookup string
2. Convert the characters to binary and then pad to length 5
3. If it is the first column, add a linebreak before the binary number
4. Add a prefix to the beginning of the binary string that stores the row and then the column of the character.
5. Push the binary string with prefix to 05AB1E's "global array" register/variable

Sort the global array. The prefix string, which determines the sort order, will make sure everything ends up in the right order and the linebreaks are in the right places.

Remove the prefix string from each element of the global array

Join the array with "" and print it.

There are some other minor details you can see below in the expanded code. Everything after the tabs at the end of a line is a comment and can be ignored. (This comment scheme is not part of 05AB1E, by the way. I did it this way because it looked nice.) Lines that have comments starting with "@" are for debugging purposes and can be omitted without changing the final output.

"yy"Ã‚Â¾38ÃƒÂ—: Replace "yy" with 38 0's. (y is the 34th character in the index, so need to add 34+4 = 38). Ã‚Â¾ is initialized to zero, which saves one byte from typing '038
'yÃƒÂ°: Replace "y" with " " 
"Replaced input Stage1 = "?D, @DEBUG

Ã…Â¾hAÃ‚Â« Pushes [0-9a-z]
DU Set X to [0-9a-z] and leave [0-9a-z] on the stack.
s Swap top two elements. Now the stack is [0-9a-z],input

ÃƒÂ°Ã‚Â« Append a space so the below splitter works even if there are no other spaces.
# Split input on " ", which "y" has been replaced with
ÃŽÂµ Map. This map replaces the y runs
 "Element = "?D, @DEBUG
 "N = "?N, @DEBUG
 N>Ã¢Â‰Â i If N != 0 
 "Replacing!", @DEBUG
 DÃÂ½ Get the first letter
 X Push X
 s Swap
 k X (which is [0-9a-z]) .indexOf(firstLetter)
 "indexOf run token = "?D, @DEBUG
 4+ Add 4 to the index, now becoming the number of zeros to add
 s Swap. Now top is the string between runs
 Ã‚Â¦ Remove the first character
 s Swap again. Now the top is the number of zeros to add.
 0sÃƒÂ—ÃƒÂ¬ Add that many zeros to the start of the element.
 "Element Replaced = "?D, @DEBUG
 }
}
'wÃ‚Â¾2ÃƒÂ—: Need to do this replacement after the y replacement so stuff like yw gets parsed properly. Again, Ã‚Â¾ is zero.
'xÃ‚Â¾3ÃƒÂ—:
J Join with ""
"Replaced input Stage2 = "?D, @DEBUG
'zÃƒÂ°: Replace "z" with " "
ÃƒÂ°Ã‚Â« Append a space so the below splitter works even if there are no other spaces.
# Split on space. (Can't do split on z because it splits on z+)
DÃƒÂ©ÃŽÂ¸gU Get max length and store it in X
"Max length = "?X, @DEBUG
ÃŽÂµ Map
 DgXs- max column length - this column length
 "Zeros to add = "?D, @DEBUG
 0sÃƒÂ—Ã‚Â« Add 0s to the end of the column so it is the same length as the longest one.
}
"Replaced input Stage3 = "?D, @DEBUG
"0 Stack top = "?D, @SDEBUG
"0 Stack top-1 = "?sD,s @SDEBUG
v pop and loop over top element, which is array of column code
 NU Store the column number in X
 yv Loop over each character in the column
 NÃ¢Â‚Â„+Ã¢Â‚Â„*X+V Set Y to (rowNumber+1000)*1000+X, where X is the column number. This result is used for sorting. 
 "1 Stack top = "?D, @SDEBUG
 "1 Stack top = "?D, @SDEBUG
 D Duplicate top element (which is now [0-9a-z])
 y Push y (special loop variable)
 "Character = "?D, @DEBUG
 k Push indexOf y in [0-9a-z]
 "Index = "?D, @DEBUG
 "2.1 Stack top = "?D, @SDEBUG
 "2.1 Stack top-1 = "?sD,s @SDEBUG
 D Duplicate index.
 32Ã¢Â€Â¹i index < 32. This uses up one copy of index
 b Push the index in binary. This uses the second copy of index. 
 Dg Get the length of the binary number
 5s- 5 - length
 0sÃƒÂ—ÃƒÂ¬ Add 5-length 0s to the start of the binary number
 "Binary = "?D, @SDEBUG
 R Reverse the binary string. This gives the proper transposed output. 
 X0Qi If X (which is set to the column number) == 0
 Ã‚Â¶ÃƒÂ¬ Stick a linebreak at the beginning of the binary string
 } 
 Y Push Y (the sort prefix).
 ÃƒÂ° Push a space character
 Ã‚Â«ÃƒÂ¬ This gives "<row prefix> <binary representation of row>" 
 "2.5 Stack top = "?D, @SDEBUG
 "2.5 Stack top-1 = "?sD,s @SDEBUG
 "Pushing '"?D?"'", @DEBUG
 Ã‹Â† Push "<row prefix>|<binary representation of row>" to global array
 "2.6 Stack top = "?D, @SDEBUG
 "2.6 Stack top-1 = "?sD,s @SDEBUG
 ] Close all blocks
Ã‚Â¯ Push global array to stack and sort it. 
"Sorted Data = "?D, @DEBUG
ÃŽÂµ Map. This block removes the row numbers 
 # Split on spaces
 ÃŽÂ¸ Get the last element

J Join with ""
'1'Ã¢Â–Âˆ:'0'.: @DEBUG pretty print

Hello and welcome to PPCG; very nice first post! Regarding your byte count: you are right. TIO even handily displays that the 148 characters are represented by 148 bytes using the 05AB1E code page. — 7 hours ago
One small note: Most golfing languages implement implicit printing; I think you can ommit your last ,. — 7 hours ago
I looked it over and made some improvements. The behavior will only be different (theoretically) for things with more than 1000 rows or columns, at which point the smaller program will do something unexpected. — 6 hours ago
Not the easiest challenge to start on! Nice job, I imagine some of the 05AB1E stalwart will give some handy pointers when they come by. — 3 hours ago
Welcome to PPCG and the world of 05AB1E. I don't have that much time right now, and I've only started using 05AB1E a few months ago, so I'm not that much of a pro yet either, but here some small things to golf: "yy" can be Ã¢Â€Âžyy (05AB1E has both a 2-char and 3-char string builtin). >Ã¢Â‰Â can be Ã„Â€ (truthified, basically 0 remains 0 and everything else becomes 1). }} can be ] (close all loops and if-elses). I'm sure some of the pro 05AB1E golfers are able to golf more substantial things than me, possibly even halving the byte-count. — 2 hours ago

score 3 · Answer 3 · 2018-08-05 22:14:47Z

APL (Dyalog Unicode), 87 80 bytes

thangs to H.PWiz for saving 7 bytes

aÃ¢Â†ÂÃ¢ÂŽÂ•D,Ã¢ÂŽÂ•AÃ¢Â‹Â„,Ã¢Â†Â“Ã¢Â†Â‘Ã¢ÂŠÂ–(5/2)Ã¢ÂŠÂ¤aÃ¢ÂÂ³Ã¢ÂÂµÃ‚Â¨'Z'(Ã¢Â‰Â Ã¢ÂŠÂ†Ã¢ÂŠÂ¢)('Y.|W|X'Ã¢ÂŽÂ•R'0'/Ã¢ÂÂ¨0::2+'WX'Ã¢ÂÂ³Ã¢ÂÂµÃ¢Â‹Â„4+aÃ¢ÂÂ³1Ã¢ÂŠÂƒÃ¢ÂÂµÃ¢ÂÂµ.Match)Ã¢ÂÂž

Try it online!

NB: Takes the input as an upper case string.

With pretty printed output

Explanation

aÃ¢Â†ÂÃ¢ÂŽÂ•D,Ã¢ÂŽÂ•A a is the string '0123...89ABCD...XYZ'
('Y.|W|X'Ã¢ÂŽÂ•R'0'/Ã¢ÂÂ¨0::2+'WX'Ã¢ÂÂ³Ã¢ÂÂµÃ¢Â‹Â„4+aÃ¢ÂÂ³1Ã¢ÂŠÂƒÃ¢ÂÂµÃ¢ÂÂµ.Match) replaces X W and Yx with the corresponding number of '0's
'Z'(Ã¢Â‰Â Ã¢ÂŠÂ†Ã¢ÂŠÂ¢) splits the string at each 'Z'
Ã¢ÂŠÂ–(5/2)Ã¢ÂŠÂ¤aÃ¢ÂÂ³Ã¢ÂÂµÃ‚Â¨ decodes each substring (explained below)
,Ã¢Â†Â“Ã¢Â†Â‘ reshapes and pads the result, returning a nested boolean array

Ã¢ÂŠÂ–(5/2)Ã¢ÂŠÂ¤aÃ¢ÂÂ³Ã¢ÂÂµÃ‚Â¨
 Ã‚Â¨ Ã¢ÂÂ apply to each substring
 aÃ¢ÂÂ³Ã¢ÂÂµ Ã¢ÂÂ return indecies of each charactor in a (i.e. '0'=>0, '9'=>9, 'B'=>11
 (5/2)Ã¢ÂŠÂ¤ Ã¢ÂÂ turn each of these into a 5-bit binary column
 Ã¢ÂŠÂ– Ã¢ÂÂ reverse each column

Here is a 7 byte saving: aÃ¢Â†ÂÃ¢ÂŽÂ•D,Ã¢ÂŽÂ•AÃ¢Â‹Â„,Ã¢Â†Â“Ã¢Â†Â‘Ã¢ÂŠÂ–(5/2)Ã¢ÂŠÂ¤aÃ¢ÂÂ³Ã¢ÂÂµÂ¨'Z'(Ã¢Â‰Â Ã¢ÂŠÂ†Ã¢ÂŠÂ¢)('Y.|W|X'Ã¢ÂŽÂ•R'0'/Ã¢ÂÂ¨0::2+'WX'Ã¢ÂÂ³Ã¢ÂÂµÃ¢Â‹Â„4+aÃ¢ÂÂ³1Ã¢ÂŠÂƒÃ¢ÂÂµÃ¢ÂÂµ.Match)Ã¢ÂÂž. The error catch also doesn't seem very golfy — 12 hours ago
Thanks @H.PWiz. Yeah I don't like the error catch either but otherwise I was using two regex replaces. — 12 hours ago

score 2 · Answer 4 · 2018-08-05 14:36:07Z

Perl 6, 156 142 bytes

14 bytes saved thanks to Jo King. (Also fixed a little bug with parsing the y and added a prettifier.)

Fixed the buggy parsing of y[wxy].

s:g/y(.)/0000 x:36(~$0)/;s:g/x/w0/;s:g/w/00/;s/$/z/;map 
.flatmap::32($_).fmt("%05b").flip.comb,[Z] .split("z").map: (*~0 x.chars).comb

Try it online!

The line break is there just to make the text fit on the screen. It is not a part of the program.

How does it work

It is an anonymous function that takes a mutable string. (This makes using the function a little bit peculiar, because you can give it only variables, not literals.) After some work, it returns a list of lists containing 0's and 1's, with the same meaning as in the original post.

The input string comes in in the variable $_. We start by using a series of substitution statements on it in order to get rid of all of those shorthands for various numbers of zeroes. First, we need to sort out the y, because in the case of yx or yw, the w (or x) does not constitute a shorthand by itself. We search for y(.) (y and one character, which it remembers) and replace it by 000"0"x:36(~$0): the three zeroes are copied verbatim, then we convert the next character from base 36 to base 10 (:36(~$0)) and add that many more zeroes. Then, we replace the w's using s:g/w/00/, and the x's using s:g/x/000/. Finally, with s/$/z/, we add a z onto the end, adding a whole lot of empty lines to the bottom. (We'll see the reason later.)

The rest is just a big map statement. We're mapping over .split("z").map: (*~0 x.chars).comb}), which is the input string (without zero shorthands), splitted into lines at z, with each line being first padded with 0 x.chars (tons of zeroes, namely as many as is the total length of the input string) on the right and then broken down into a list of individual characters (.comb). Finally, we transpose it with [Z] (reduce with zip). Zipping ends as soon as the shortest list is depleted, which results in all lines having the same length. (The number of useless trailing zeroes on the right is equal to the length of the shortest line. Also, this transposition trick fails for a "matrix" with only one row. That's why we forcibly added another row at the end before.)

Now, we just map over the rows (columns of the original matrix) and replace each character encountered with the corresponding 5 bits. That is done using :32($_) (base 32 to base 10) .fmt("%05b") (format as a bitstring of width 5, padded with zeroes) .flip (reverse the string, because the LSB is in the top row, not the bottom) .comb (break the string into a list of characters). We have used .flatmap, which flattens the resulting list (otherwise we would get a list of lists in each column). The transpose of the result is then implicitly returned.

(I feel a little bit bad because of abusing the possibility of trailing zeroes so hard. But it reduced the bytecount quite considerably :Ã¢Â€Â”).)

@JonathanAllan, you're right, thank you. I'll fix it in a moment. (I first thought for some reason that there could be only y[0-9a-v] and when I found that it's not the case, I just added a quick (and wrong) fix.) — 19 hours ago

score 2 · Answer 5 · 2018-08-05 20:57:57Z

Jelly, 66 bytes

^Yowza!

ÃƒÂ˜BiÃ¢Â±Â®+
Ã…Â’uÃ…Â“Ã¡Â¹Â£Ã¢ÂÂ¾YYj38Ã…Â»Ã¡Â¹Â£Ã¢Â€ÂYÃ‚ÂµÃ¡Â¸Â¢ÃƒÂ§3;)FÃ¡Â¹Â£Ã¢Â€ÂWj2Ã¡Â¹Â£Ã¢Â€ÂXj3Ã¢Â€Â0Ã¡ÂºÂ‹$Ã‚Â¹OÃ†Â‘?Ã¢Â‚Â¬FÃ¡Â¹Â£Ã¢Â€ÂZÃ‚ÂµÃˆÂ¯Ã¢Â€Â0ÃƒÂ§31BÃ¡Â¸ÂŠÃ¢Â‚Â¬UZ)Ã¡ÂºÂŽz0

A monadic link which yields a transposed version as a list of lists (add Z to the end to transpose back).

Try it online! Or see the test-suite (with pretty printed output).

Chas Brown 3,510216 · Answer 6 · 2018-08-05 19:09:07Z

Python 2, 249 bytes

def f(s,Z='0'):
 while 'y'in s:i=s.find('y');s=s[:i]+4*Z+Z*int(s[i+1],36)+s[i+2:]
 t=s.replace('w',2*Z).replace('x',3*Z).split('z')
 return map(''.join,zip(*[[(Z*5+bin(int(c,32))[2:])[-5:][::-1]for c in r]+[Z*5]*(max(map(len,t))-len(r))for r in t]))

Try it online!

There is a superfluous space in while 'y'in. I think the subscript [-5:][::-1] can be golfed to [:-6:-1]. The return statement can be put on the previous line. — 7 hours ago

score 6 · Answer 7 · 2018-08-05 15:34:37Z

JavaScript (ES8), 197 bytes

Takes input as a string. Returns an array of strings with '#' and spaces. The output may include extra (but consistent) trailing spaces on each line.

s=>[x='x',...s.replace(/w|x|y./g,s=>''.padEnd(s<x?2:s>x?P(s[1],36)+4:3),P=parseInt)].map(c=>[!++w,1,2,3,4].map(i=>c<x?o[y+i-5]+=' #'[P(c,36)>>i&1]:o=[...++y&&o,'']),w=y=o=)&&o.map(r=>r.padEnd(w))

Try it online! (prettified output)

How?

Global variables

The character "x" is used several times, so it's worth storing it into the variable x.

The function parseInt is used twice, so it's worth storing it into the variable P.

y is the row index, initialized to 0.

w keeps track of an upper bound of the width which is used to pad the final output.

o[Â ] is the output array, initially empty.

Pre-processing of repeated zeros

We first replace all patterns "w", "x" and "yX" in the input string with the appropriate number of spaces. These spaces will later be interpreted as "0".

s.replace(
 /w|x|y./g,
 s => ''.padEnd(
 // w --> 2
 s < x ? 2 :
 // yX --> value of X + 4
 s > x ? P(s[1], 36) + 4 :
 // x --> 3
 3
 )
)

Decoding

We split the resulting string, prepend an initial "x" and iterate 5 times (with i = 0 to 4) on each character c:

If c is lower than "x", we append the corresponding pattern to the next 5 rows.
```
o[y + i - 5] += ' #'[P(c, 36) >> i & 1]
```

If c is greater than or equal to "x", we allocate 5 new empty strings in o[Â ] and add 5 to y. This is triggered by the initial "x" that was added at the beginning of the string, or by any "z" in the original content.
```
o = [...++y && o, '']
```

Padding

Finally, we pad each string in o[Â ] with spaces so that they all have w characters.

o.map(r => r.padEnd(w))

score 5 · Answer 8 · 2018-08-06 03:35:05Z

05AB1E, 148 132 bytes

I'm a highschool student and this was my first time for both golfing and using 05AB1E, so comments are appreciated!

"yy"Ã‚Â¾38ÃƒÂ—:'yÃƒÂ°:Ã…Â¾hAÃ‚Â«DUsÃƒÂ°Ã‚Â«#ÃŽÂµN>Ã¢Â‰Â iDÃÂ½Xsk4+sÃ‚Â¦s0sÃƒÂ—ÃƒÂ¬}}'wÃ‚Â¾2ÃƒÂ—:'xÃ‚Â¾3ÃƒÂ—:J'zÃƒÂ°:ÃƒÂ°Ã‚Â«#DÃƒÂ©ÃŽÂ¸gUÃŽÂµDgXs-0sÃƒÂ—Ã‚Â«}vNUyvNÃ¢Â‚Â„+Ã¢Â‚Â„*X+VDykD32Ã¢Â€Â¹ibDg5s-0sÃƒÂ—ÃƒÂ¬RX0QiÃ‚Â¶ÃƒÂ¬}YÃƒÂ°Ã‚Â«ÃƒÂ¬Ã‹Â†]Ã‚Â¯ÃŽÂµ#ÃŽÂ¸J

~~Try it online!~~

Try it online!

Outputs the transposed matrix as multiple output lines of 1s and 0s. If you want pretty-printed output, add J'1'Ã¢Â–Âˆ:'0'.: at the end.

Will add extra zeros.

Explanation

(I'm calling "rows" and "columns" opposite of what you might expect because it generates the transposed matrix)

The basic algorithm is:

Replace "yy" with 38 0s

Split on "y" and and expand the 0-runs.

Replace "w" and "x"

Figure out the longest column (that is, the longest string between z's) and pad all the other columns so they are that length. (This is necessary because of how the algorithm below works)

Split on z

At this point, the input string is an array of columns where each column is a string of [0-9a-v], where each column is the same length.

The algorithm to get it into the output format is to
1. Convert the characters to numbers by using indexOf in a lookup string
2. Convert the characters to binary and then pad to length 5
3. If it is the first column, add a linebreak before the binary number
4. Add a prefix to the beginning of the binary string that stores the row and then the column of the character.
5. Push the binary string with prefix to 05AB1E's "global array" register/variable

Sort the global array. The prefix string, which determines the sort order, will make sure everything ends up in the right order and the linebreaks are in the right places.

Remove the prefix string from each element of the global array

Join the array with "" and print it.

There are some other minor details you can see below in the expanded code. Everything after the tabs at the end of a line is a comment and can be ignored. (This comment scheme is not part of 05AB1E, by the way. I did it this way because it looked nice.) Lines that have comments starting with "@" are for debugging purposes and can be omitted without changing the final output.

"yy"Ã‚Â¾38ÃƒÂ—: Replace "yy" with 38 0's. (y is the 34th character in the index, so need to add 34+4 = 38). Ã‚Â¾ is initialized to zero, which saves one byte from typing '038
'yÃƒÂ°: Replace "y" with " " 
"Replaced input Stage1 = "?D, @DEBUG

Ã…Â¾hAÃ‚Â« Pushes [0-9a-z]
DU Set X to [0-9a-z] and leave [0-9a-z] on the stack.
s Swap top two elements. Now the stack is [0-9a-z],input

ÃƒÂ°Ã‚Â« Append a space so the below splitter works even if there are no other spaces.
# Split input on " ", which "y" has been replaced with
ÃŽÂµ Map. This map replaces the y runs
 "Element = "?D, @DEBUG
 "N = "?N, @DEBUG
 N>Ã¢Â‰Â i If N != 0 
 "Replacing!", @DEBUG
 DÃÂ½ Get the first letter
 X Push X
 s Swap
 k X (which is [0-9a-z]) .indexOf(firstLetter)
 "indexOf run token = "?D, @DEBUG
 4+ Add 4 to the index, now becoming the number of zeros to add
 s Swap. Now top is the string between runs
 Ã‚Â¦ Remove the first character
 s Swap again. Now the top is the number of zeros to add.
 0sÃƒÂ—ÃƒÂ¬ Add that many zeros to the start of the element.
 "Element Replaced = "?D, @DEBUG
 }
}
'wÃ‚Â¾2ÃƒÂ—: Need to do this replacement after the y replacement so stuff like yw gets parsed properly. Again, Ã‚Â¾ is zero.
'xÃ‚Â¾3ÃƒÂ—:
J Join with ""
"Replaced input Stage2 = "?D, @DEBUG
'zÃƒÂ°: Replace "z" with " "
ÃƒÂ°Ã‚Â« Append a space so the below splitter works even if there are no other spaces.
# Split on space. (Can't do split on z because it splits on z+)
DÃƒÂ©ÃŽÂ¸gU Get max length and store it in X
"Max length = "?X, @DEBUG
ÃŽÂµ Map
 DgXs- max column length - this column length
 "Zeros to add = "?D, @DEBUG
 0sÃƒÂ—Ã‚Â« Add 0s to the end of the column so it is the same length as the longest one.
}
"Replaced input Stage3 = "?D, @DEBUG
"0 Stack top = "?D, @SDEBUG
"0 Stack top-1 = "?sD,s @SDEBUG
v pop and loop over top element, which is array of column code
 NU Store the column number in X
 yv Loop over each character in the column
 NÃ¢Â‚Â„+Ã¢Â‚Â„*X+V Set Y to (rowNumber+1000)*1000+X, where X is the column number. This result is used for sorting. 
 "1 Stack top = "?D, @SDEBUG
 "1 Stack top = "?D, @SDEBUG
 D Duplicate top element (which is now [0-9a-z])
 y Push y (special loop variable)
 "Character = "?D, @DEBUG
 k Push indexOf y in [0-9a-z]
 "Index = "?D, @DEBUG
 "2.1 Stack top = "?D, @SDEBUG
 "2.1 Stack top-1 = "?sD,s @SDEBUG
 D Duplicate index.
 32Ã¢Â€Â¹i index < 32. This uses up one copy of index
 b Push the index in binary. This uses the second copy of index. 
 Dg Get the length of the binary number
 5s- 5 - length
 0sÃƒÂ—ÃƒÂ¬ Add 5-length 0s to the start of the binary number
 "Binary = "?D, @SDEBUG
 R Reverse the binary string. This gives the proper transposed output. 
 X0Qi If X (which is set to the column number) == 0
 Ã‚Â¶ÃƒÂ¬ Stick a linebreak at the beginning of the binary string
 } 
 Y Push Y (the sort prefix).
 ÃƒÂ° Push a space character
 Ã‚Â«ÃƒÂ¬ This gives "<row prefix> <binary representation of row>" 
 "2.5 Stack top = "?D, @SDEBUG
 "2.5 Stack top-1 = "?sD,s @SDEBUG
 "Pushing '"?D?"'", @DEBUG
 Ã‹Â† Push "<row prefix>|<binary representation of row>" to global array
 "2.6 Stack top = "?D, @SDEBUG
 "2.6 Stack top-1 = "?sD,s @SDEBUG
 ] Close all blocks
Ã‚Â¯ Push global array to stack and sort it. 
"Sorted Data = "?D, @DEBUG
ÃŽÂµ Map. This block removes the row numbers 
 # Split on spaces
 ÃŽÂ¸ Get the last element

J Join with ""
'1'Ã¢Â–Âˆ:'0'.: @DEBUG pretty print

Hello and welcome to PPCG; very nice first post! Regarding your byte count: you are right. TIO even handily displays that the 148 characters are represented by 148 bytes using the 05AB1E code page. — 7 hours ago
One small note: Most golfing languages implement implicit printing; I think you can ommit your last ,. — 7 hours ago
I looked it over and made some improvements. The behavior will only be different (theoretically) for things with more than 1000 rows or columns, at which point the smaller program will do something unexpected. — 6 hours ago
Not the easiest challenge to start on! Nice job, I imagine some of the 05AB1E stalwart will give some handy pointers when they come by. — 3 hours ago
Welcome to PPCG and the world of 05AB1E. I don't have that much time right now, and I've only started using 05AB1E a few months ago, so I'm not that much of a pro yet either, but here some small things to golf: "yy" can be Ã¢Â€Âžyy (05AB1E has both a 2-char and 3-char string builtin). >Ã¢Â‰Â can be Ã„Â€ (truthified, basically 0 remains 0 and everything else becomes 1). }} can be ] (close all loops and if-elses). I'm sure some of the pro 05AB1E golfers are able to golf more substantial things than me, possibly even halving the byte-count. — 2 hours ago

score 3 · Answer 9 · 2018-08-05 22:14:47Z

APL (Dyalog Unicode), 87 80 bytes

thangs to H.PWiz for saving 7 bytes

aÃ¢Â†ÂÃ¢ÂŽÂ•D,Ã¢ÂŽÂ•AÃ¢Â‹Â„,Ã¢Â†Â“Ã¢Â†Â‘Ã¢ÂŠÂ–(5/2)Ã¢ÂŠÂ¤aÃ¢ÂÂ³Ã¢ÂÂµÃ‚Â¨'Z'(Ã¢Â‰Â Ã¢ÂŠÂ†Ã¢ÂŠÂ¢)('Y.|W|X'Ã¢ÂŽÂ•R'0'/Ã¢ÂÂ¨0::2+'WX'Ã¢ÂÂ³Ã¢ÂÂµÃ¢Â‹Â„4+aÃ¢ÂÂ³1Ã¢ÂŠÂƒÃ¢ÂÂµÃ¢ÂÂµ.Match)Ã¢ÂÂž

Try it online!

NB: Takes the input as an upper case string.

With pretty printed output

Explanation

aÃ¢Â†ÂÃ¢ÂŽÂ•D,Ã¢ÂŽÂ•A a is the string '0123...89ABCD...XYZ'
('Y.|W|X'Ã¢ÂŽÂ•R'0'/Ã¢ÂÂ¨0::2+'WX'Ã¢ÂÂ³Ã¢ÂÂµÃ¢Â‹Â„4+aÃ¢ÂÂ³1Ã¢ÂŠÂƒÃ¢ÂÂµÃ¢ÂÂµ.Match) replaces X W and Yx with the corresponding number of '0's
'Z'(Ã¢Â‰Â Ã¢ÂŠÂ†Ã¢ÂŠÂ¢) splits the string at each 'Z'
Ã¢ÂŠÂ–(5/2)Ã¢ÂŠÂ¤aÃ¢ÂÂ³Ã¢ÂÂµÃ‚Â¨ decodes each substring (explained below)
,Ã¢Â†Â“Ã¢Â†Â‘ reshapes and pads the result, returning a nested boolean array

Ã¢ÂŠÂ–(5/2)Ã¢ÂŠÂ¤aÃ¢ÂÂ³Ã¢ÂÂµÃ‚Â¨
 Ã‚Â¨ Ã¢ÂÂ apply to each substring
 aÃ¢ÂÂ³Ã¢ÂÂµ Ã¢ÂÂ return indecies of each charactor in a (i.e. '0'=>0, '9'=>9, 'B'=>11
 (5/2)Ã¢ÂŠÂ¤ Ã¢ÂÂ turn each of these into a 5-bit binary column
 Ã¢ÂŠÂ– Ã¢ÂÂ reverse each column

Here is a 7 byte saving: aÃ¢Â†ÂÃ¢ÂŽÂ•D,Ã¢ÂŽÂ•AÃ¢Â‹Â„,Ã¢Â†Â“Ã¢Â†Â‘Ã¢ÂŠÂ–(5/2)Ã¢ÂŠÂ¤aÃ¢ÂÂ³Ã¢ÂÂµÂ¨'Z'(Ã¢Â‰Â Ã¢ÂŠÂ†Ã¢ÂŠÂ¢)('Y.|W|X'Ã¢ÂŽÂ•R'0'/Ã¢ÂÂ¨0::2+'WX'Ã¢ÂÂ³Ã¢ÂÂµÃ¢Â‹Â„4+aÃ¢ÂÂ³1Ã¢ÂŠÂƒÃ¢ÂÂµÃ¢ÂÂµ.Match)Ã¢ÂÂž. The error catch also doesn't seem very golfy — 12 hours ago
Thanks @H.PWiz. Yeah I don't like the error catch either but otherwise I was using two regex replaces. — 12 hours ago

score 2 · Answer 10 · 2018-08-05 14:36:07Z

Perl 6, 156 142 bytes

14 bytes saved thanks to Jo King. (Also fixed a little bug with parsing the y and added a prettifier.)

Fixed the buggy parsing of y[wxy].

s:g/y(.)/0000 x:36(~$0)/;s:g/x/w0/;s:g/w/00/;s/$/z/;map 
.flatmap::32($_).fmt("%05b").flip.comb,[Z] .split("z").map: (*~0 x.chars).comb

Try it online!

The line break is there just to make the text fit on the screen. It is not a part of the program.

How does it work

It is an anonymous function that takes a mutable string. (This makes using the function a little bit peculiar, because you can give it only variables, not literals.) After some work, it returns a list of lists containing 0's and 1's, with the same meaning as in the original post.

The input string comes in in the variable $_. We start by using a series of substitution statements on it in order to get rid of all of those shorthands for various numbers of zeroes. First, we need to sort out the y, because in the case of yx or yw, the w (or x) does not constitute a shorthand by itself. We search for y(.) (y and one character, which it remembers) and replace it by 000"0"x:36(~$0): the three zeroes are copied verbatim, then we convert the next character from base 36 to base 10 (:36(~$0)) and add that many more zeroes. Then, we replace the w's using s:g/w/00/, and the x's using s:g/x/000/. Finally, with s/$/z/, we add a z onto the end, adding a whole lot of empty lines to the bottom. (We'll see the reason later.)

The rest is just a big map statement. We're mapping over .split("z").map: (*~0 x.chars).comb}), which is the input string (without zero shorthands), splitted into lines at z, with each line being first padded with 0 x.chars (tons of zeroes, namely as many as is the total length of the input string) on the right and then broken down into a list of individual characters (.comb). Finally, we transpose it with [Z] (reduce with zip). Zipping ends as soon as the shortest list is depleted, which results in all lines having the same length. (The number of useless trailing zeroes on the right is equal to the length of the shortest line. Also, this transposition trick fails for a "matrix" with only one row. That's why we forcibly added another row at the end before.)

Now, we just map over the rows (columns of the original matrix) and replace each character encountered with the corresponding 5 bits. That is done using :32($_) (base 32 to base 10) .fmt("%05b") (format as a bitstring of width 5, padded with zeroes) .flip (reverse the string, because the LSB is in the top row, not the bottom) .comb (break the string into a list of characters). We have used .flatmap, which flattens the resulting list (otherwise we would get a list of lists in each column). The transpose of the result is then implicitly returned.

(I feel a little bit bad because of abusing the possibility of trailing zeroes so hard. But it reduced the bytecount quite considerably :Ã¢Â€Â”).)

@JonathanAllan, you're right, thank you. I'll fix it in a moment. (I first thought for some reason that there could be only y[0-9a-v] and when I found that it's not the case, I just added a quick (and wrong) fix.) — 19 hours ago

score 2 · Answer 11 · 2018-08-05 20:57:57Z

Jelly, 66 bytes

^Yowza!

ÃƒÂ˜BiÃ¢Â±Â®+
Ã…Â’uÃ…Â“Ã¡Â¹Â£Ã¢ÂÂ¾YYj38Ã…Â»Ã¡Â¹Â£Ã¢Â€ÂYÃ‚ÂµÃ¡Â¸Â¢ÃƒÂ§3;)FÃ¡Â¹Â£Ã¢Â€ÂWj2Ã¡Â¹Â£Ã¢Â€ÂXj3Ã¢Â€Â0Ã¡ÂºÂ‹$Ã‚Â¹OÃ†Â‘?Ã¢Â‚Â¬FÃ¡Â¹Â£Ã¢Â€ÂZÃ‚ÂµÃˆÂ¯Ã¢Â€Â0ÃƒÂ§31BÃ¡Â¸ÂŠÃ¢Â‚Â¬UZ)Ã¡ÂºÂŽz0

A monadic link which yields a transposed version as a list of lists (add Z to the end to transpose back).

Try it online! Or see the test-suite (with pretty printed output).

Chas Brown 3,510216 · Answer 12 · 2018-08-05 19:09:07Z

Python 2, 249 bytes

def f(s,Z='0'):
 while 'y'in s:i=s.find('y');s=s[:i]+4*Z+Z*int(s[i+1],36)+s[i+2:]
 t=s.replace('w',2*Z).replace('x',3*Z).split('z')
 return map(''.join,zip(*[[(Z*5+bin(int(c,32))[2:])[-5:][::-1]for c in r]+[Z*5]*(max(map(len,t))-len(r))for r in t]))

Try it online!

There is a superfluous space in while 'y'in. I think the subscript [-5:][::-1] can be golfed to [:-6:-1]. The return statement can be put on the previous line. — 7 hours ago

Decrypt the extended Wechsler format

The Challenge

Test cases

The Challenge

Test cases

The Challenge

Test cases

The Challenge

Test cases

6 Answers 6

JavaScript (ES8), 197 bytes

How?

Global variables

Pre-processing of repeated zeros

Decoding

Padding

05AB1E, 148 132 bytes

Explanation

APL (Dyalog Unicode), 87 80 bytes

Explanation

Perl 6, 156 142 bytes

How does it work

Jelly, 66 bytes

Python 2, 249 bytes

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6 Answers 6

6 Answers 6

JavaScript (ES8), 197 bytes

How?

Global variables

Pre-processing of repeated zeros

Decoding

Padding

JavaScript (ES8), 197 bytes

How?

Global variables

Pre-processing of repeated zeros

Decoding

Padding

JavaScript (ES8), 197 bytes

How?

Global variables

Pre-processing of repeated zeros

Decoding

Padding

JavaScript (ES8), 197 bytes

How?

Global variables

Pre-processing of repeated zeros

Decoding

Padding

05AB1E, 148 132 bytes

Explanation

05AB1E, 148 132 bytes

Explanation

05AB1E, 148 132 bytes

Explanation

05AB1E, 148 132 bytes

Explanation

APL (Dyalog Unicode), 87 80 bytes

Explanation

APL (Dyalog Unicode), 87 80 bytes

Explanation

APL (Dyalog Unicode), 87 80 bytes

Explanation

APL (Dyalog Unicode), 87 80 bytes

Explanation

Perl 6, 156 142 bytes

How does it work

Perl 6, 156 142 bytes

How does it work

Perl 6, 156 142 bytes

How does it work

Perl 6, 156 142 bytes

How does it work

Jelly, 66 bytes

Jelly, 66 bytes

6 Answers
6

6 Answers
6

6 Answers
6