JavaScript (ES6),10097 93 bytes

Saved 4 bytes thanks to @edc65

s=>s.replace(/./g,(c,i)=>" '.:"[(i?q>' '&q!=".":c>"'")+(q=c,s[++i]?s[i]>"'":c>' '&c!=".")*2])

How it works

This decides on the character we need to insert by performing some calculations on the chars before and after the current one. We sum:

• If it's the first char, and it has a dot on the bottom, 2;
• Otherwise, if the one before it has a dot on top, 2.
• If it's the last char, and it has a dot on top, 1;
• Otherwise, if the one after it has a dot on the bottom, 1.

This sums nicely to 0 for a space, 1 for', 2 for., and 3 for:.

Test snippet

f=s=>s.replace(/./g,(c,i)=>" '.:"[(i?q>' '&q!=".":c>"'")+(q=c,s[++i]?s[i]>"'":c>' '&c!=".")*2])

Try it out:<input id=A type="text" value=".'  '.::  :.'. . ::.'  '. . .::'  :.'."><button>Run</button><pre id=B></pre>

• \$\begingroup\$Well done. Save 4:s=>s.replace(/./g,(c,i)=>" '.:"[(i?q>' '&q!=".":c>"'")+(q=c,s[++i]?s[i]>"'":c>' '&c!=".")*2]) (flip 2 parts so I can increase i, less regexp and more simple test, save prev c in q)\$\endgroup\$

  edc65

  – edc65

  2016-03-10 22:55:08 +00:00

  CommentedMar 10, 2016 at 22:55
• \$\begingroup\$@edc65 Thanks for the tip!\$\endgroup\$

  ETHproductions

  – ETHproductions

  2016-03-11 01:15:07 +00:00

  CommentedMar 11, 2016 at 1:15

Jelly,3230 29 bytes

,Ṛe€"“':“.:”µṪ€;"ṚU1¦ZḄị“'.:

Note the trailing space.Try it online! orverify all test cases.

Background

We begin by considering the input string (e.g.,:..:'.) and its reverse.

:..:'..':..:

For each character in the top row, we check if it belongs to':, and for each character of the bottom row if it belongs to.:. This gives the 2D array of Booleans

100110101111

which is the matrix from the question, with reversed bottom row.

We remove the last Boolean of each row, reverse the order of the rows, prepend the Booleans in their original order, and finally reverse the top row.

100110    10011    10111    010111    111010101111    10111    10011    110011    110011

This yields the rotated matrix from the question.

Finally, we consider each column of Booleans a binary number and index into'.: to obtain the appropriate characters.

332031    ::. :'

How it works

,Ṛe€"“':“.:”µṪ€;"ṚU1¦ZḄ‘ị“'.:   Main link. Argument: S (string) Ṛ                              Reverse S.,                               Form a pair of S and S reversed.     “':“.:”                    Yield ["':" ".:"].  e€"                           For each character in S / S reversed, check if it                                is an element of "':" / ".:".                                Yield the corresponding 2D array of Booleans.            µ                   Begin a new, monadic chain.                                Argument: A (2D array of Booleans)             Ṫ€                 Pop the last Boolean of each list.                 Ṛ              Yield the reversed array of popped list.               ;"               Prepend the popped items to the popped lists.                  U1¦           Reverse the first list.                     Z          Zip to turn top and bottom rows into pairs.                      Ḅ         Convert each pair from base 2 to integer.                        “'.:    Yield "'.: ".                       ị        Retrieve the characters at the corr. indices.

Retina, 66

• 2 bytes saved thanks to @daavko
• 4 bytes saved thanks to @randomra

:1e\.1f'0e 0fT`h`Rh`^.|.$(.)(\d)$2$1e1:e0'f0 f1.

Explanation

Starting with input:

: ''. :

The first 4 stages construct the matrix, using1/e for true and0/f for false for the top/bottom rows, respectively. Top and bottom rows are interlaced together. This would yield a string like:

e1f0e0e0f1f0e1

However, these 4 stages also effectively move the lower row 1 to the left, simply by reversing the order of the letters and digits:

1e0f0e0e1f0f1e

TheTransliteration stage reverses hex digits for the first and last characters only, i.e. replaces0-9a-f withf-a9-0. This has the effect of moving the bottom-left character up to the top row and the top-right character down to the bottom row:

ee0f0e0e1f0f11

The next stage then swaps every letter-digit pair, thereby moving the upper row 1 to the right. Previously this was(\D)(\d), but it turns out that(.)(\d) is sufficient because the substitutions always happen left-to-right and so the final two digits won't be erroneously matched by this, because penultimate character will have been already substituted. The matrix has now been fully rotated as required:

e0e0f0e1e0f1f1

The final 4 stages then translate back to the original format:

'' :'..

Try it online.

All testcases, one per line,m added toT line to allow separate treatment of each input line.

Perl,7069646361 60 bytes

Includes +2 for-lp

Run with the input string on STDIN, e.g.

perl -lp rotatedots.pl <<< ":..:'."

rotatedots.pl:

y/'.:/02/r=~/./;y/.':/01/;$_=$'.2*chop|$&/2 .$_;y;0-3; '.:

Explanation

y/'.:/02/r                                        Construct bottom row but                                                  with 2's instead of 1's                                                  Return constructed value                                                  (for now assume space                                                  becomes 0 too)          =~/./                                   Match first digit on bottom                                                  row into $&. $' contains                                                  the rest of the bottom row                y/.':/01/                         Convert $_ to top row                                                  (again assume space                                                  becomes 0 too)                             $'.2*chop            Remove last digit from                                                  the top row, multiply by 2                                                  and append to bottom row                                       $&/2 .$_   Divide removed digit by                                                  2 and prepend it to the                                                  top row                          $_=         |           "or" the top and bottom                                                  row together. The ASCII                                                  values of 0,1,2,3 have                                                  00,01,10,11 as their last                                                  two bits.y;0-3; '.:                  Convert the smashed together top and bottom rows                            to the corresponding representation characters.                            Drop the final ; since it is provided by -p                            (after a newline which doesn't matter here)

Space is not converted in the above code. For the calculations/2 and*2 it will behave like and become0. In the other positions it will be part of the "or", but the 1 bits of space are a subset of the one bits of0 and will have the same effect as0 if or-ed with any of the digits. Only if the character it is or-ed with is a space will it remain a space instead of becoming a0. But that's ok since0 would have been converted back to space anyways.

Python 3,145141 130 bytes

def f(s):a=[i in"':"for i in s]+[i in".:"for i in s][::-1];return''.join(" '.:"[i+2*j]for i,j in zip([a[-1]]+a,a[-2:len(s)-2:-1]))

Explanation

The golfed solution use the following property of zip :zip('ABCD', 'xy') --> Ax Bysozip(a[:l],a[l:]) can be replace byzip(a,a[l:]) and that allow to remove the definition ofl

def f(s): l=len(s)-1 #                ┌───── unfold input string :  123  -> 123456 #                │                             654 #  ──────────────┴────────────────────────────── a=[i in"':"for i in s]+[i in".:"for i in s][::-1] # ─────────┬─────────   ────────────┬─────────── #          │                        └──── generate the second row and reverse it #          └─────────── generate the first row  return''.join(" '.:"[i+2*j]for i,j in zip([a[-1]]+a[:l],a[l:-1][::-1])) #             ──────┬──────           ─┬    ────────────┬─────────── #                   │                  │                └──── rotate and create first/second new row :  123456  -> 612345  -> 612 #                   │                  │                                                                                      543 #                   │                  └ group pair of the first and second row : 612 -> (6,5),(1,4),(2,3) #                   │                                                             543 #                   └─────────── replace pair by symbol

Results

>>> f(".'  '.::  :.'. . ::.'  '. . .::'  :.'.")"' ' .:.''..'.'. ..:' ' .'. ...'''..'.'">>> f(".....''''''")":...  '''':"

• \$\begingroup\$You can save a few bytes by putting the last three lines on one line, separated by semicolons.\$\endgroup\$

  mbomb007

  – mbomb007

  2016-03-11 21:00:49 +00:00

  CommentedMar 11, 2016 at 21:00

Pyth, 66 bytes

KlQJ.nCm@[,1Z,Z1,ZZ,1 1)%Cd5Qjkm@" .':"id2Ccs[:JKhK<JtK>JhK:JtKK)K

Try it here!

Explanation

This can be broken down in 3 parts:

• Convert the input into a flat array of ones and zeros.
• Do the rotation.
• Convert it back into ASCII.

Convert input

This is fairly trivial. Each character is mapped in the following way:

  -> (0,0). -> (0,1)' -> (1,0): -> (1,0)

First one is a whitespace.
We get a list of 2-tuples which we transpose to get the 2 rows of the matrix which then gets flattened.

Code

KlQJ.nCm@[,1Z,Z1,ZZ,1 1)%Cd5Q   # Q = inputKlQ                             # save the width of the matrix in K (gets used later)       m                    Q   # map each character d                        %Cd5    # ASCII-code of d modulo 5        @[,1Z,Z1,ZZ,1 1)        # use this as index into a lookup list   J.nC                         # transpose, flatten and assign to J

Rotate

We have the matrix as flat array inJ and the width of the matrix inK. The rotation can be described as:

J[K] + J[:K-1] + J[K+1:] + J[K-1]

Code

s[:JKhKJhK:JtKK)    # J = flat array, K = width of matrixs[                  )    # Concat all results in this list  :JKhK                  # J[K]       JhK          # J[K+1:]               :JtKK     # J[K-1]

Convert it back

jkm@" .':"id2Cc[)K    # [) = resulting list of the step above              c[)K    # chop into 2 rows             C        # transpose to get the 2-tuples back  m                   # map each 2-tuple d          id2         # interpret d as binary and convert to decimal   @" .':"            # use that as index into a lookup string to get the correct charjk                    # join into one string

Pyth,38 36

L,hb_ebsXCyc2.>syCXzJ" .':"K.DR2T1KJ

2 bytes thanks to Jakube!

Try it here or run theTest Suite.

Explanation:

L,hb_eb         ##  Redefine the function y to take two lists                ##  and return them but with the second one reversed                ##  Uses W to apply a function only if it's first argument is truthyXzJ" .':"K.DR2T ##  Does a translation from the string " .':" to                ##  .DR2T which is [0,1,2,3...,9] mapped to divmod by 2                ##  (which is [0,0],[0,1],[1,0],[1,1], then some extra, unused values)                ##  we also store the string and the list for later use in J and K.>syC ... 1     ##  zip the lists to get the bits on top and below as two separate lists                ##  apply the function y from before, flatten and rotate right by 1Cyc2            ##  split the list into 2 equal parts again, then apply y and zip againsX ... KJ       ##  apply the list to string transformation from above but in reverse                ##  then flatten into a string

• \$\begingroup\$Seems like I did it way too complicated ^^ Would you mind adding an explanation?\$\endgroup\$

  Denker

  – Denker

  2016-03-10 19:53:59 +00:00

  CommentedMar 10, 2016 at 19:53
• 1
  \$\begingroup\$@DenkerAffe Was in the middle of adding one :) Added!\$\endgroup\$

  FryAmTheEggman

  – FryAmTheEggman

  2016-03-10 19:56:25 +00:00

  CommentedMar 10, 2016 at 19:56
• \$\begingroup\$Did the same approach as you. Two things I've noticed: this lambdaL,hb_eb is one byte shorter, and.DR2T creates also the Cartesian product and a few more pairs, but doesn't and in a digit and helps saving a space.\$\endgroup\$

  Jakube

  – Jakube

  2016-03-11 16:57:47 +00:00

  CommentedMar 11, 2016 at 16:57

Python 3,166154153150146138137135132 127 bytes

Edit: I've borrowed the use ofzip fromErwan's Python answer at the end of the function.and their idea to use[::-1] reversals, though I put in my own twist. As it turns out, reversals were not a good idea for my function. I have changed my use offormat for further golfing. Moveda andb directly intozip for further golfing (ungolfing remains unchanged because the separation ofa andb there is useful for in avoiding clutter in my explanation)

Edit: Borrowed(some number)>>(n)&(2**something-1) fromthis answer by xnor on the Music Interval Solver challenge. The clutter that iszip(*[divmod(et cetera, 2) for i in input()]) can probably be golfed better, though I do like the expediency it grants from using two tuplest andv.

t,v=zip(*[divmod(708>>2*(ord(i)%5)&3,2)for i in input()])print("".join(" '.:"[i+j*2]for i,j in zip((v[0],*t),(*v[1:],t[-1]))))

Ungolfed:

def rotate_dots(s):    # dots to 2 by len(s) matrix of 0s and 1s (but transposed)    t = []    v = []    for i in s:        m = divmod(708 >> 2*(ord(i)%5) & 3, 2)            # ord(i)%5 of each char in . :' is in range(1,5)            # so 708>>2 * ord & 3 puts all length-2 01-strings as a number in range(0,4)            # e.g. ord(":") % 5 == 58 % 5 == 3            # 708 >> 2*3 & 3 == 0b1011000100 >> 6 & 3 == 0b1011 == 11            # divmod(11 & 3, 2) == divmod(3, 2) == (1, 1)            # so, ":" -> (1, 1)        t.append(m[0])        v.append(m[1])    # transposing the matrix and doing the rotations    a = (v[0], *t)          # a tuple of the first char of the second row                             # and every char of the first row except the last char    b = (v[1:], t[-1])      # and a tuple of every char of the second row except the first                            # and the last char of the first row    # matrix to dots    z = ""    for i, j in zip(a, b):        z += " '.:"[i + j*2]    # since the dots are binary                                # we take " '.:"[their binary value]    return z

Ruby,166 163 bytes

->s{a=s.tr(f=" .':",t='0-3').chars.map{|x|sprintf('%02b',x).chars}.transpose;a[1]+=[a[0].pop];a[0]=[a[1].shift]+a[0];a.transpose.map{|x|x.join.to_i 2}.join.tr t,f}

Yuck...transpose is too long.

Tricks used here:

• sprintf('%02b',x) to convert"0","1","2","3" into"00","01","10", and"11" respectively. Surprisingly, the second argument doesnot have to be converted into an integer first.

• The rotation is done viaa[1].push a[0].pop;a[0].unshift a[1].shift;, which I thought was at least a little clever (if not excessively verbose in Ruby). The symmetry is aesthetically nice, anyway :P

• \$\begingroup\$May I suggest to golf it a little?->s{a=s.tr(f=" .':",'001').chars;b=s.tr(f,'0101').chars;b<<a.pop;([b.shift]+a).zip(b).map{|x|x.join.to_i 2}.join.tr'0-3',f}\$\endgroup\$

  manatwork

  – manatwork

  2016-03-11 07:54:28 +00:00

  CommentedMar 11, 2016 at 7:54
• \$\begingroup\$Finally the ☕ made its effect. Was looking for this all morning:.map{|x|x.join.to_i 2}.join.tr'0-3',f →.map{|x|f[x.join.to_i 2]}*''\$\endgroup\$

  manatwork

  – manatwork

  2016-03-11 09:44:33 +00:00

  CommentedMar 11, 2016 at 9:44

Javascript ES6 125 bytes

q=>(n=[...q].map(a=>(S=` .':`).indexOf(a))).map((a,i)=>(i?n[i-1]&2:n[0]&1&&2)|((I=n[i+1])>-1?I&1:n[i]&2&&1)).map(a=>S[a]).join``

I map each character to a two digit binary equivalent

 : becomes 3   11 ' becomes 2   10 . becomes 1   01   becomes 0   00

and I'm thinking of them as being one on top of the other

3212021 becomes11010101010001

I save that to n

For each character (0-3) of n, I check it's neighbors, adding the highest order bit of the left neighbor to the lowest order bit of the right neighbor.if i==0 (first character) I use it's own lower order bit instead of the left neighbor's higher order bit.

if n[i+1]>-1 it means we got 0,1,2,3 so when that's false we hit the last element.

When that happens I use the character's own highest order bit instead of the right neighbor's lower bit

map that back to .': land and join that array back together

MATL,40 39 bytes

' ''.:'tjw4#mqBGnXKq:QKEh1Kq:K+hv!)XBQ)

Try it online!The linked version hasv repaced by&v, because of changes in the language after this answer was posted.

' ''.:'               % pattern string. Will indexed into, twice: first for reading                       % the input and then for generating the ouputt                     % duplicate this stringj                     % input stringw                     % swap4#m                   % index of ocurrences of input chars in the pattern stringqB                    % subtract 1 and convert to binay. Gives 2-row logical arrayGnXKq:QKEh1Kq:K+hv!   % (painfully) build two-column index for rotation)                     % index into logical array to perform the rotationXBQ                   % transform each row into 1, 2, 3 or 4)                     % index into patter string. Implicitly display

Uiua, 19 bytes

⍜(♭⍜⊣⇌⍉⋯⨂" '.:")°↻₁

try it!

Perhaps the only task on this site where Uiua can out-golf Jelly byten seven bytes. This task is extremely well suited to Uiua because the entire operation can be expressed with a single⍜ under. For those out of the loop, under takes two functions and applies the first one to some value, then applies the second one, and finally applies an inverse transformation of the first. Here, the first function does the following:

♭⍜⊣⇌⍉⋯⨂" '.:"      ⨂" '.:"  # zero-indexed position of each in this string     ⋯         # each converted to binary (little-endian)    ⍉          # transpose (giving our 2xN array) ⍜⊣⇌           # reverse the last row♭              # flatten the array

From here, all that has to be done is°↻₁ to cycle this flattened form one to the right, and the above transformation is undone, resulting in the desired string.

This is possible because Uiua's implementation of under allows very advanced use-cases like this because each operation stores context which it can use to undo itself in the end. For example, when the array is flattened, Uiua keeps track of the original shape of the array and molds it back while inverting.

Perl,144142137 131 bytes

y/.':/1-3/;s/./sprintf'%02b ',$&/ge;@a=/\b\d/g;@b=(/\d\b/g,pop@a);@a=(shift@b,@a);say map{substr" .':",oct"0b$a[$_]$b[$_]",1}0..@a

Byte added for the-n flag.

Pretty much the same algorithm asmy Ruby answer, just shorter, because... Perl.

y/.':/1-3/;                         # transliterate [ .':] to [0123]s/./sprintf'%02b ',$&/ge;           # convert each digit to 2-digit binary@a=/\b\d/g;                         # grab the 1st digit of each pair@b=(/\d\b/g,                        # 2nd digit of each pairpop@a);                             # push the last element of a to b@a=(shift@b,@a);                    # unshift the first element of b to asay                                 # output...map{                                # map over indices of a/bsubstr" .':",oct"0b$a[$_]$b[$_]",1  # convert back from binary, find right char}0..@a                              # @a is length of a

Obnoxiously,@a=(shift@b,@a) is shorter thanunshift@a,shift@b.

Alas, these are the same length:

y/ .':/0-3/;s/./sprintf'%02b ',$&/ge;s/./sprintf'%02b ',index" .':",$&/ge;

Thanks toTon Hospel for 5 bytes andmsh210 for a byte!

• \$\begingroup\$Can you use..@a instead of..$#a? (Perhapsoct dies or returns 0 or something. I haven't tried it.)\$\endgroup\$

  msh210

  – msh210

  2016-03-10 23:08:33 +00:00

  CommentedMar 10, 2016 at 23:08
• \$\begingroup\$No need to convert space to 0. It will evaluate as 0 for the sprintf anyways. Also get rid of the parenthesis in the regex. If there is no capture the whole match is returned for a//g\$\endgroup\$

  Ton Hospel

  – Ton Hospel

  2016-03-11 13:30:36 +00:00

  CommentedMar 11, 2016 at 13:30
• \$\begingroup\$@TonHospel Thanks, incorporated those into the answer (although obviously yours still completely blows mine out of the water).\$\endgroup\$

  Doorknob

  – Doorknob

  2016-03-11 23:06:37 +00:00

  CommentedMar 11, 2016 at 23:06
• \$\begingroup\$Thatsprintf is soooo long.map$_%2,/./g andmap$_/2|0,//g almost has to be shorter (untested)\$\endgroup\$

  Ton Hospel

  – Ton Hospel

  2016-03-12 00:22:14 +00:00

  CommentedMar 12, 2016 at 0:22

Common Lisp,253 247 bytes

(defun R(i &aux(n(length i))(u #1=(make-list n))(b #1#))(dotimes(k n)(setf(elt u k)(find #2=(elt i k)"':")(elt b k)(find #2#".:")))(map'string(lambda(u b)(cond((and u b)#\:)(u #\')(b #\.)(t #\ )))`(,(car b),@(butlast u))`(,@(cdr b),(nth(1- n)u))))

Try it online!

Tests included in TIO.

(defun R (i &aux (n (length i))  ; take the length of input                 (u #1=(make-list n))   ; create the top (up) list                 (b #1#))  ; create the bottom list  (dotimes (k.. n)    (setf (elt u k) (find #2=(elt i k) "':") ; set the k'th element of the top list to a non-falsey value if k'th element of input is #\' or #\:          (elt b k) (find #2# ".:"); set the k'th element of the bottom list to a non-falsey value if k'th element of input is #\. or #\:(map 'string  ; collect the results into a string     (lambda (u b) ; for each element of the rotated up and bottom lists       (cond        ((and u b) #\:) ; if both top and bottom are non-falsey add #\:        (u #\')         ; if only top is non-falsey add #\'        (b #\.)         ; if only bottom is non-falsey add #\.        (t #\ )))       ; if both are falsey add '#\ ' (#\Space)     `(,(car b),@(butlast u))   ; first element of the bottom and the all but last element of the top     `(,(cdr b),(nth(1- n)u)))) ; all but first element of the bottom and the last element of the top

Saved 6 bytes using#=notation for(make-list n)and(elt i k)

Python 3,105 104 bytes

s=" .':"a=[*map(s.find,input())]b=a[0]*2,*a,a[-1]//2print(*map(lambda x,y:s[x&2|y&1],b,b[2:]),sep='')

Try it online!

Dyalog APL, 53 bytes

{' ''.:'⌷⍨⊂1+2⊥⊖⌽@2⊢2(≢⍵)⍴¯1⌽,⌽@2⊖2 2⊤⌊7÷⍨32-⍨⎕UCS⍵}­⁡​‎⁠⁠⁠‎⁡⁠⁣⁤⁣‏⁠‎⁡⁠⁣⁤⁤‏⁠‎⁡⁠⁤⁡⁡‏⁠‎⁡⁠⁤⁡⁢‏⁠‎⁡⁠⁤⁡⁣‏⁠‏​⁡⁠⁡‌⁢​‎‎⁡⁠⁣⁣⁣‏⁠‎⁡⁠⁣⁣⁤‏⁠‎⁡⁠⁣⁤⁡‏⁠‎⁡⁠⁣⁤⁢‏⁠‏​⁡⁠⁡‌⁣​‎‎⁡⁠⁣⁢⁣‏⁠‎⁡⁠⁣⁢⁤‏⁠‎⁡⁠⁣⁣⁡‏⁠‎⁡⁠⁣⁣⁢‏⁠‏​⁡⁠⁡‌⁤​‎‎⁡⁠⁣⁡⁣‏⁠‎⁡⁠⁣⁡⁤‏⁠‎⁡⁠⁣⁢⁡‏⁠‎⁡⁠⁣⁢⁢‏⁠‏​⁡⁠⁡‌⁢⁡​‎‎⁡⁠⁣⁡⁢‏⁠‏​⁡⁠⁡‌⁢⁢​‎‎⁡⁠⁢⁤⁣‏⁠‎⁡⁠⁢⁤⁤‏⁠‎⁡⁠⁣⁡⁡‏‏​⁡⁠⁡‌⁢⁣​‎‎⁡⁠⁢⁣⁣‏⁠‎⁡⁠⁢⁣⁤‏⁠‎⁡⁠⁢⁤⁡‏⁠‎⁡⁠⁢⁤⁢‏‏​⁡⁠⁡‌⁢⁤​‎‎⁡⁠⁢⁢⁡‏⁠‎⁡⁠⁢⁢⁢‏⁠‎⁡⁠⁢⁢⁣‏⁠‎⁡⁠⁢⁢⁤‏⁠‎⁡⁠⁢⁣⁡‏⁠‎⁡⁠⁢⁣⁢‏‏​⁡⁠⁡‌⁣⁡​‎‎⁡⁠⁢⁡⁡‏⁠‎⁡⁠⁢⁡⁢‏⁠‎⁡⁠⁢⁡⁣‏⁠‎⁡⁠⁢⁡⁤‏⁠‏​⁡⁠⁡‌⁣⁢​‎‎⁡⁠⁤⁤‏‏​⁡⁠⁡‌⁣⁣​‎‎⁡⁠⁤⁢‏⁠‎⁡⁠⁤⁣‏‏​⁡⁠⁡‌⁣⁤​‎‎⁡⁠⁣⁤‏⁠‎⁡⁠⁤⁡‏‏​⁡⁠⁡‌⁤⁡​‎‎⁡⁠⁢‏⁠‎⁡⁠⁣‏⁠‎⁡⁠⁤‏⁠‎⁡⁠⁢⁡‏⁠‎⁡⁠⁢⁢‏⁠‎⁡⁠⁢⁣‏⁠‎⁡⁠⁢⁤‏⁠‎⁡⁠⁣⁡‏⁠‎⁡⁠⁣⁢‏⁠‎⁡⁠⁣⁣‏‏​⁡⁠⁡‌­                                              ⎕UCS⍵   # ‎⁡Get codepoint                                          32-⍨        # ‎⁢subtact 32 (space (32) being the smallest of the 4 chars)                                      ⌊7÷⍨            # ‎⁣divide by 7 and floor                                  2 2⊤                # ‎⁤Encode each of those as binary (2 digits)                                 ⊖                    # ‎⁢⁡Flip it over                              ⌽@2                     # ‎⁢⁢Replace 2nd row with itself reversed                          ¯1⌽,                        # ‎⁢⁣Ravel and shift to the right                                                      # ‎⁢⁣This gives the conveyor belt shift                    2(≢⍵)⍴                            # ‎⁢⁤Reshape that to 2xN (the original input length)                ⌽@2⊢                                  # ‎⁣⁡Re-reverse 2nd row again               ⊖                                      # ‎⁣⁢And flip back over             2⊥                                       # ‎⁣⁣Decode the binary number           1+                                         # ‎⁣⁤Add 1 for indexing's sake ' ''.:'⌷⍨⊂                                           # ‎⁤⁡And finally select from magic string💎

Created with the help ofLuminespire.

Jelly, 26 bytes

“.': ”ðṖiⱮd2j@/ṙ1ŒHUż¥/Ḅịḷ

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Granted thatis 2016 jelly

Newer answers could be shorter

-- lyxal

Just barely! Jelly changed a lot between 2016 and 2019, but the main byte saver here overDennis' solution is just not having two separate string constants. There might be a shorter way to perform the core logic as well, but this kind of thing isn't really Jelly's forte--either rearranging multi-dimensional arrays or computing things under some transformation.

        iⱮ                    Find the 1-index of every character in“.': ” Ṗ                      the string ".': " without the trailing space.          d2                  Divmod the indices by 2.              /               Reduce the list of [div, mod] pairs by            j@                sticking the accumulator between the div and mod.“.': ” ṖiⱮd2j@/               The result is the 's backwards followed by the .s.               ṙ1             Rotate that 1 to the left,                 ŒH           split it in half,                    ż /       pair the elements of the halves                   U ¥        with the first half (un-)reversed,                       Ḅ      convert each pair from binary,“.': ”ð                 ịḷ    and index into ".': ".

• 1
  \$\begingroup\$Hey, at least it isn't a 10 byte difference any more :p\$\endgroup\$

  lyxal

  – lyxal♦

  2025-09-29 13:56:40 +00:00

  CommentedSep 29 at 13:56

JavaScript, 311 bytes

Can probably be improved alot:

a=(s=prompt()).length-1;o=s[0]==":"||s[0]=="."?s[1]==":"||s[1]=="."?":":"'":s[1]==":"||s[1]=="."?".":" ";for(i=1;i<a;i++)o+=s[i-1]=="'"||s[i-1]==":"?s[i+1]=="."||s[i+1]==":"?":":"'":s[i+1]=="."||s[i+1]==":"?".":" ";alert(o+=s[a]==":"||s[a]=="'"?s[a-1]==":"||s[a-1]=="'"?":":"'":s[a-1]==":"||s[a-1]=="'"?".":" ")

• \$\begingroup\$Maybe set something tos[i-1]? That could save some bytes.\$\endgroup\$

  Riker

  – Riker

  2016-03-10 20:08:46 +00:00

  CommentedMar 10, 2016 at 20:08
• \$\begingroup\$Same withs[i+1].\$\endgroup\$

  Riker

  – Riker

  2016-03-10 20:13:08 +00:00

  CommentedMar 10, 2016 at 20:13
• 1
  \$\begingroup\$Try using ES6 arrow functions and a lookup, also using< instead of== might save you quite a few bytes. You may also want to checkoutTips for Golfing in JS andTips for golfing in ES6\$\endgroup\$

  Downgoat

  – Downgoat

  2016-03-10 20:39:06 +00:00

  CommentedMar 10, 2016 at 20:39
• 1
  \$\begingroup\$@Downgoat How can you use< instead of==\$\endgroup\$

  Jens Renders

  – Jens Renders

  2016-03-10 20:42:28 +00:00

  CommentedMar 10, 2016 at 20:42

JavaScript (ES6),237210204188182 178 bytes

Credit to@Downgoat for saving 16 bytes in the 188-byte revision

Update: I had a brainwave and reduced the first operation ons to a singlemap call instead of two separate ones

s=>(r=" .':",a=[],s=[...s].map(c=>(t=('00'+r.indexOf(c).toString(2)).slice(-2),a.push(t[0]),t[1])),a.splice(0,0,s.shift()),s.push(a.pop()),a.map((v,i)=>r[+('0b'+v+s[i])]).join``)

Pretty Print & Explanation

s => (  r = " .':", // Map of characters to their (numerical) binary representations (e.g. r[0b10] = "'")  a = [],     // extra array needed  // Spread `s` into an array  s = [...s].map(c => (    // Map each character to a `0`-padded string representation of a binary number, storing in `t`    t = ('00' + r.indexOf(c).toString(2)).slice(-2)),    // Put the first character of `t` into `a`    a.push(t[0]),    // Keep the second character for `s`    t[1]  )),  // Put the first character of `s` in the first index of `a`  a.splice(0,0,s.shift()),  // Append the last character of `a` to `s`  s.push(a.pop(),  // Rejoin the characters, alternating from `a` to `s`, representing the rotated matrix, and map them back to their string representation  // Use implicit conversion of a binary number string using +'0b<num>'  a.map((v,i) => r[+('0b' + v + s[i])]).join``)

• 1
  \$\begingroup\$does:s=>(r=" .':",a=[],s=[...s].map(c=>('00'+r.indexOf(c).toString(2)).slice(-2)).map(n=>(a.push(n[0]),n[1]),a.splice(0,0,s.shift()),s.push(a.pop()),a.map((v,i)=>r[parseInt(v+s[i],2)]).join``) work?\$\endgroup\$

  Downgoat

  – Downgoat

  2016-03-10 23:28:29 +00:00

  CommentedMar 10, 2016 at 23:28
• \$\begingroup\$Sorry for not replying to this earlier, didn't see the notification - my developer tools are giving me an "illegal character" exception\$\endgroup\$

  RevanProdigalKnight

  – RevanProdigalKnight

  2016-03-11 13:25:53 +00:00

  CommentedMar 11, 2016 at 13:25
• \$\begingroup\$Got it to work as you put it - apparently when I copied it there were some extra invisible characters in it that didn't show up in the browser developer tools.\$\endgroup\$

  RevanProdigalKnight

  – RevanProdigalKnight

  2016-03-11 13:40:58 +00:00

  CommentedMar 11, 2016 at 13:40

Vyxal, 44 bytes

f`:'. `f⁺↑b2ẇĿ2ẇÞT÷ḣ„ṫ$„p‟JWÞT⁺↑b2ẇ`:'. `fĿ∑

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Yuck.

Python 3, 124 bytes

def f(a,s=" .':",j="".join):q=j(bin(4+s.find(i))for i in a);return j(s[int(j(x),2)]for x in zip(q[4]+q[3::5],q[9::5]+q[-2]))

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How it works :

• q="".join(bin(4+s.find(i))for i in a) store an unpacked version of our starting stringa by using binary conversion

Ifa is equal to

ABCDEFGH

thenq is equal to---ae---bf---cg---dh with a to h equal to"1" if there is a dot in the corresponding emplacement and"0" otherwise (and--- equal to0b1)

• zip(q[4]+q[3::5],q[9::5]+q[-2]) reshapesq to have :(eabcd),(fghd) (the lastd in the first part is ignored since the second part is shorter)

• s[int("".join(x),2)] reverse the process converting back the binary number into their dot version. Result :

EABCFGHD

05AB1E, 41 bytes

Â)εS„':„.:‚Nès夈¨}R¯øí˜2äćR¸ìζJC" '.:"sè

Try it online! Outputs as a list of characters. Link includes a footer to format the output.

Â)εS„':„.:‚Nès夈¨}R¯øí˜2äćR¸ìζJC"..."sè  # trimmed program                                       è  # push character...                                          # (implicit) s...                                       è  # in...                                 "..."s   # literal...                                       è  # at...                                      s   # (implicit) indices in...                              ζ           # list of elements with same indices in each element of...                            ¸             # list of...                          ć               # first element of... )                                        # list of...                                         # implicit input... )                                        # and...                                          # implicit input...                                         # reversed...  ε                                       # with each element replaced by...              å¤                          # is...             s                            # (implicit) each element of...   S                                      # list of characters of...                                          # (implicit) current element in map...              å                           # in...            ès                            # element in...          ‚                               # list of...    „':                                   # literal...          ‚                               # and...       „.:                                # literal...            è                             # at index...           N                              # current index in map...                 ¨                        # excluding the last element...                ˆ                         # and append...               ¤                          # last element of...              å¤                          # is...             s                            # (implicit) each element of...   S                                      # list of characters of...                                          # (implicit) current element in map...              å                           # in...            ès                            # element in...          ‚                               # list of...    „':                                   # literal...          ‚                               # and...       „.:                                # literal...            è                             # at index...           N                              # current index in map...                ˆ                         # to global array...                   R                      # reversed...                      í                   # with...                    ¯                     # global array...                     ø                    # with each element paired with corresponding element in...                    ¯                     # global array...                      í                   # prepended...                       ˜                  # flattened...                         ä                # split into...                        2                 # literal...                         ä                # equal pieces...                           R              # reversed...                             ì            # with... )                                        # list of...                                         # implicit input... )                                        # and...                                          # implicit input...                                         # reversed...  ε                                       # with each element replaced by...              å¤                          # is...             s                            # (implicit) each element of...   S                                      # list of characters of...                                          # (implicit) current element in map...              å                           # in...            ès                            # element in...          ‚                               # list of...    „':                                   # literal...          ‚                               # and...       „.:                                # literal...            è                             # at index...           N                              # current index in map...                 ¨                        # excluding the last element...                ˆ                         # and append...               ¤                          # last element of...              å¤                          # is...             s                            # (implicit) each element of...   S                                      # list of characters of...                                          # (implicit) current element in map...              å                           # in...            ès                            # element in...          ‚                               # list of...    „':                                   # literal...          ‚                               # and...       „.:                                # literal...            è                             # at index...           N                              # current index in map...                ˆ                         # to global array...                   R                      # reversed...                      í                   # with...                    ¯                     # global array...                     ø                    # with each element paired with corresponding element in...                    ¯                     # global array...                      í                   # prepended...                       ˜                  # flattened...                         ä                # split into...                        2                 # literal...                         ä                # equal pieces...                          ć               # excluding the first element...                             ì            # prepended...                                          # (implicit) with each element...                               J          # joined...                                          # (implicit) with each element...                                C         # converted from binary to decimal

Excel, 139 bytes

=LET(e,LEN(A1),q,SEQUENCE(e),x,CODE(MID(A1,q,1))-32,l,x>7,u,x>l*14,CONCAT(MID(" '.:",IF(q=1,l,INDEX(u,q-1))+IF(q=e,u,INDEX(l,q+1))*2+1,1)))

Link to Spreadsheet

How it works

=LET(*Assignments*e,LEN(A1),              'e = length of A1q,SEQUENCE(e),          'q = (1 .. e)x,CODE(MID(A1,q,1))-32, 'x = list of ASCII codes of each character - 32l,x>7,                  'l = array of lower numbers = 1 if x>7 (.:)u,x>l*14,               'u = array of upper numbers = 1 if x>l*14 (':)*Result*IF(q=1,l,INDEX(u,q-1)) +   'String indices = if q = 1 then l[1] else u[q-1] IF(q=e,u,INDEX(l,q+1))*2+1 '               + (if q=e then u[e] else l[q+1]) * 2 + 1CONCAT(MID(" '.:",~,1)))   'Concatenate all the results

J, 55 bytes

' .'':'&([{~([:g@($$_1|.,)g=.|.@{:1}_2{.])&.(|:@#:)@i.)

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Python 3,294287 283 bytes

Waaayyyyyy too long, but I will try to golf of some bytes:

z=input()x=len(z)M=[0,1,2,3]for Q in M:z=z.replace(":'. "[Q],"11100100"[Q*2:Q*2+2])a=[]b=[]for X in range(x):a+=[z[X*2]];b+=[z[X*2+1]]b=b[1:]+[a.pop()]c=[b[0]]+az=""for X in range(len(c)): y=c[X]+b[X] for Q in M:y=y.replace("11100100"[Q*2:Q*2+2],":'. "[Q]) z+=yprint(z)

Lua, 139 bytes

print(((...):gsub(".",{[" "]="NN@",["."]="YN@",["'"]="NY@",[":"]="YY@"}):gsub("(.)@(.?)","%2%1"):gsub("..",{NN=" ",NY=".",YN="'",YY=":"})))

Usage:

$ lua conveyor.lua ".'  '.::  :.'. . ::.'  '. . .::'  :.'."' ' .:.''..'.'. ..:' ' .'. ...'''..'.'

• code-golf
• string