Braille graphics

Question

Cut a boolean matrix in 4x2 blocks and render them as Braille characters U+2800...U+28FF.

[[0,1,0,0,1,0],
 [1,0,0,0,0,0],
 [1,0,0,0,1,0],
 [1,1,1,1,0,0]]

⣎⣀⠅

Pad with 0-s if dimensions are not multiples of 4 and 2.

[[0,1,0],
 [1,0,0],
 [1,1,1]]

⠮⠄

Usual golfing rules apply, flexible on input format. Output should either have the structure of a matrix or look like a matrix, e.g. list of strings; single string with newlines.

Hint: chr(0x2800 + 128*b7 + 64*b6 + 32*b5 + 16*b4 + 8*b3 + 4*b2 + 2*b1 + b0) is the dot pattern

b0 b3
b1 b4
b2 b5
b6 b7

Larger test:

[[0,0,1,1,1,1,0,0,0,0,0,0,0,1,1,0,0,0,0,1,1,1,0],
 [0,1,1,1,1,1,1,0,0,0,0,0,1,1,1,0,0,0,1,1,1,1,1],
 [0,1,1,0,0,1,1,1,0,0,0,1,1,1,1,0,0,1,1,0,0,0,1],
 [1,1,0,0,0,0,1,1,0,0,0,0,0,0,1,0,0,1,1,0,0,1,1],
 [1,1,0,0,0,1,1,0,0,0,0,0,0,0,1,0,0,1,1,1,0,1,0],
 [1,1,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,1,0,0,0],
 [1,1,0,1,1,1,1,1,1,0,0,0,0,0,1,0,0,0,0,1,0,0,0],
 [1,1,0,1,1,1,1,1,0,0,1,1,0,0,1,0,0,1,1,1,1,1,1],
 [1,1,0,1,1,1,1,0,0,1,1,1,1,0,1,0,1,1,1,1,1,1,0],
 [1,1,0,1,0,1,1,0,1,1,0,1,1,0,1,0,0,0,0,1,1,0,0],
 [1,1,0,0,0,1,1,0,1,0,0,0,1,0,1,1,0,0,0,1,1,0,0],
 [1,1,0,0,0,1,1,0,1,0,0,0,1,0,1,1,1,0,0,1,1,0,0],
 [0,1,1,0,1,1,1,0,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0],
 [0,1,1,1,1,1,0,0,1,1,1,1,1,0,0,1,1,0,1,1,1,0,0],
 [0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,0,1,1,1,1,0]]

⣰⠟⠻⣦⠀⠠⠾⡇⢠⡞⢛⡆
⣿⢠⣬⣥⠄⣀⠀⡇⢈⣻⣈⡀
⣿⠘⢹⡇⡞⠙⡇⣧⡉⢹⡏⠀
⠘⠷⠟⠁⠳⠾⠃⠘⠇⠾⠧⠀

Answer 1

Jelly,  31  30 bytes

sz0Z
ç€2ZF€ç€8Zœ?@€€-36Ḅ+⁽$ṁỌY

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How?

sz0Z - Link 1, split & right-pad with zeros: list, items; number, chunkSize
s    - split items into chunks of length chunkSize
 z0  - transpose with filler zero
   Z - transpose

ç€2ZF€ç€8Zœ?@€€-36Ḅ+⁽$ṁỌY - Main link: list of lists of numbers (1s & 0s), M
ç€2                       - call the last link (1) as a dyad for €ach (left=M, right=2)
                          -  ((left,right) bits read left-right then top-bottom)
   Z                      - transpose the resulting list of lists of lists
                          -  ((left, right) bits read top-bottom then left-right)
    F€                    - flatten €ach
      ç€8                 - call the last link (1) as a dyad for €ach (left=^, right=8)
         Z                - transpose the resulting list of lists of lists
                          -  ("blocks" each and all read left-right top-to bottom)
               -36        - literal -36
             €€           - for €ach (block-wise row) for €ach (block)
          œ?@             -   lexicographical permutation with reversed arguments
                          -    (get the permutation at index -36 (modular) in a list of
                          -     all permutations of the indexes sorted lexicographically.
                          -     That is the 8!-36 = 40284th - equivalently the values at
                          -     indexes [8,7,6,4,2,5,3,1])
                  Ḅ       - convert from binary list to integer (vectorises)
                    ⁽$ṁ   - base 250 literal = 10240
                   +      - add
                       Ọ  - cast to character (vectorises)
                        Y - join with newlines
                          - implicit print

Answer 2

JavaScript ES7 210 207 201 200 198 194 185 183 bytes

a=>eval('for(y=0,c="";A=a[y];y+=4,c+=`\n`)for(x=0;A[x]+1;x+=2)c+=String.fromCharCode(10240+eval("for(N=k=0;k<6;k++)N+=(g=(X,Y=3)=>(a[Y+y]||0)[X+x]|0)(k>2,k%3)*2**k")|g(0)+g(1)*2<<6)')

4 bytes saved thanks to ngn

3 bytes saved thanks to Luke

How it works

I'm going to split the code up into parts and speak about them separately:

for(y=x=0, c=""; a[y]; x+=2)
    !((a[y] || [])[x]+1) && (y+=4,x=0,c+=`\n`)

This is where every variable gets declared. x and y is the position of the "cursor" (the upper left edge of the current braille character). The x coordinate increases by 2 every iteration, and it stops, when there's no row with the index y (a[x] returns undefined if it doesn't exist, which gets converted to false).

There are multiple tricks in the second row. (a[y] || [])[x] ensures that looking up the value at the (x, y) position doesn't throw an error. The && is the usual and operator, and it only checks the right side of the expression, if the left was true. This can be translated to

if (!((a[y] || [])[x] + 1)) 
    y+=4,x=0,c+=`\n`

The next part:

c+=String.fromCharCode(10240+eval("for(N=k=0;k<6;k++)N+=(g=(x,y)=>(a[y]||[])[x]||0)(~~(k/3)+x,k%3+y)*2**k,N")+g(x,y+3)*64+g(x+1,y+3)*128)

String.fromCharCode simply converts the passed number to a unicode character with the same character code. The expression in the parentheses calculates the index of the Braille character:

for(N=k=0;k<6;k++)N+=(g=(x,y)=>(a[y]||[])[x]||0)(~~(k/3)+x,k%3+y)*2**k

Goes through the position in the

1 4
2 5
3 6

order, multiplies the values at those positions with 2ⁱ, where i is the index and adds them together. The

g=(x,y)=>(a[y]||[])[x]||0

part declares a lambda function called g, which given an x and y coordinate returns either the value at the (x, y) position or 0 if the position lies outside the bounds of the array.

+g(x,y+3)*64+g(x+1,y+3)*128

This part adds up the last two positions with the correct weights using the function defined a bit earlier.

Last but not least, the

a=>eval('...')

part has 2 functions. It defines an anonymous lambda and ensures, that the for loop doesn't cause any problems (a single line lambda such as this one can't contain a single for loop only, an evaluation circumvents this).

Answer 3

Mathematica, 126 110 97 90

FromCharacterCode[10240+ListCorrelate[2^{{0,3},{1,4},{2,5},{6,7}},#,1,0][[;;;;4,;;;;2]]]&

---

This solution takes advantage of ListCorrelate to convolve a (reversed) kernel over a matrix, which is essentially a sliding matrix multiplication (or dot product). See a visual explanation here. Padding is done by using 0 as the fourth argument. In the following example, we expect the result to match the hint above:

ListCorrelate[
  2^{{0, 3}, {1, 4}, {2, 5}, {6, 7}},
  {{b0, b3}, {b1, b4}, {b2, b5}, {b6, b7}}
]

(* returns {{b0 + 2 b1 + 4 b2 + 8 b3 + 16 b4 + 32 b5 + 64 b6 + 128 b7}} *)

Note that ListConvolve is not any shorter, since the third argument would be -1.

Since this applies the kernel at every position of the matrix, we just need to extract the elements in every fourth row and second column. We use shorthands for Span and Part: [[;;;;4,;;;;2]].

Helpfully, FromCharacterCode can take a matrix of character codes and return a list of strings.

---

This solution returns a list of strings, which is one of the allowed output formats. Simply prepend Column@ for the output to “look like a matrix.”

---

You can play around with this in a free online Mathematica notebook. Go here, click Create a New Notebook, wait a moment, paste in this code, then press shift+enter.

m1={{0,1,0,0,1,0},{1,0,0,0,0,0},{1,0,0,0,1,0},{1,1,1,1,0,0}};
m2={{0,1,0},{1,0,0},{1,1,1}};
m3={{0,0,1,1,1,1,0,0,0,0,0,0,0,1,1,0,0,0,0,1,1,1,0},{0,1,1,1,1,1,1,0,0,0,0,0,1,1,1,0,0,0,1,1,1,1,1},{0,1,1,0,0,1,1,1,0,0,0,1,1,1,1,0,0,1,1,0,0,0,1},{1,1,0,0,0,0,1,1,0,0,0,0,0,0,1,0,0,1,1,0,0,1,1},{1,1,0,0,0,1,1,0,0,0,0,0,0,0,1,0,0,1,1,1,0,1,0},{1,1,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,1,0,0,0},{1,1,0,1,1,1,1,1,1,0,0,0,0,0,1,0,0,0,0,1,0,0,0},{1,1,0,1,1,1,1,1,0,0,1,1,0,0,1,0,0,1,1,1,1,1,1},{1,1,0,1,1,1,1,0,0,1,1,1,1,0,1,0,1,1,1,1,1,1,0},{1,1,0,1,0,1,1,0,1,1,0,1,1,0,1,0,0,0,0,1,1,0,0},{1,1,0,0,0,1,1,0,1,0,0,0,1,0,1,1,0,0,0,1,1,0,0},{1,1,0,0,0,1,1,0,1,0,0,0,1,0,1,1,1,0,0,1,1,0,0},{0,1,1,0,1,1,1,0,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0},{0,1,1,1,1,1,0,0,1,1,1,1,1,0,0,1,1,0,1,1,1,0,0},{0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,0,1,1,1,1,0}};

MatrixToBraille := Column@
  FromCharacterCode[10240+ListCorrelate[2^{{0,3},{1,4},{2,5},{6,7}},#,1,0][[;;;;4,;;;;2]]]&

MatrixToBraille/@{m1,m2,m3}

Then you should see this:

Answer 4

Dyalog APL, 133 122 114 112 101 100 98 95 94 93 90 88 86 bytes

Assumes ⎕IO←0

{C⍴{⎕UCS 10240+2⊥(∊S⌷⍨⍵+⍳¨A)[⍎¨⍕76531420]}¨(,b)/,⍳⍴b←{0 0≡A|⍵}¨⍳⍴S←⍵↑⍨A×C←⌈(⍴⍵)÷A←4 2}

-8 9 12 bytes thanks to @Adám in chat

-2 bytes thanks to @ngn

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How (input is ⍵)?

• A←4 2, store the vector 4 2 in variable A
• (⍴⍵)÷, the dimensions of ⍵ divided by A
• ⌈, ceiling
• C←, stored in C
• A×, multiplied by A
• ⍵↑⍨, fit ⍵ to those dimensions
• S←, stored in S
• ⍳⍴, indexes of S
• {0 0≡A|⍵}¨, 1 where the top left of a cell is, 0 everywhere else
• (,b)/,⍳⍴b←, truthy indexes
• {⎕UCS 10240+2⊥(∊S⌷⍨⍵+⍳¨A)[⍎¨⍕76531420]}¨, turn each element into braille
• C⍴, reshape the dimensions to C

Answer 5

Python 2 + drawille, 141 125 120 116 bytes

Saved 16 bytes thanks to ngn and L3viathan

Saved 5 bytes thanks to L3viathan

Saved 4 bytes thanks to ngn

from drawille import*
def a(d,c=Canvas(),e=enumerate):[c.set(j,i)for i,x in e(d)for j,y in e(x)if y];print c.frame()

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tio does not have drawille installed so it does not work

Answer 6

JavaScript, 136 bytes

a=>(b=a.map(x=>[]),a.map((l,i)=>l.map((c,j)=>b[i>>2][j>>1]|=c<<'01263457'[i%4+j*4%8])),b.map(l=>l.map(c=>String.fromCharCode(10240+c))))

Thanks to ngn, using bit shifts save 4 bytes.

Answer 7

APL (Dyalog), 57 54 bytes*

-3 thanks to OP. Prompts for Boolean matrix. Prints character matrix.

1↓⎕UCS{2⊥40,⌽(,⍉3↑⍵),⊢⌿⍵}⌺(2 2⍴4 2)⊢0⍪⍣3⍪∘0⍣3⊢⎕,0

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⎕,0 append a zero on right (ignored if even number of columns)

⊢ yield that (to separate 3 and ⎕)

⍪∘0⍣3 append zeros on bottom thrice (because ⌺ drops partial windows)

0⍪⍣3 stack zeros on top thrice (because ⌺ starts in the top left corner)

⊢ yield that (separates the parenthesis and the 0)

{…}⌺(2 2⍴4 2) on each 4-row 2-column window, with 4-row vertical and 2 row horizontal step:

 ⊢⌿⍵ last row (lit. vertical right reduction); [b6,b7]

 (…), prepend:

  3↑ take three rows; [[b0,b3],[b1,b4],[b2,b5]]

  ⍉ transpose; [[b0,b1,b2],[b3,b4,b5]]

  , ravel; [b0,b1,b2,b3,b4,b5]

 now we have [b0,b1,b2,b3,b4,b5,b6,b7]

 ⌽ reverse; [b7,b6,b5,b4,b3,b2,b1,b0]

 40, prepend 40 (for 40×2⁹ = 10240); [40,b7,b6,b5,b4,b3,b2,b1,b0]

 2⊥ evaluate as base-2 (binary)

⎕UCS convert to character

1↓ drop the first row (all-zero because of ⌺'s padding)

---

* In Classic, counting ⌺ as ⎕U233A.

Answer 8

Python 3, 168 165 161 bytes

def f(m):
 while m:
  r,m,s=[*zip(*m)],m[4:],''
  while r:s+=chr(10240+sum(q<<int(w)for(q,w)in zip((r[0]+(0,)*3)[:4]+(r+[()])[1],'01263457')));r=r[2:]
  print(s)

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Answer 9

Haskell, 145 bytes

(a!b)c=take b$c++repeat a
r([]:_)=[]
r l=['⠀'..]!!(sum.zipWith(*)[1,8,2,16,4,32,64,128]$l>>=0!2):r(drop 2<$>l)
b[]=[]
b s=r(([]!4)s):b(drop 4s)

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Answer 10

Python 3, 169 bytes

a=[]
y=0
for l in eval(input()):
 y-=1;a+=y%4//3*[-~len(l)//2*[10240]];x=0
 for v in l:a[-1][x//2]|=v<<(6429374>>y%4*6+x%2*3&7);x+=1
for l in a:print(*map(chr,l),sep='')

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Answer 11

Perl 5, 164 bytes

163 bytes of code + 1 flag -p

@a=eval}{for(;$r<@a;$r+=4){for($c=0;$c<@{$a[0]};$c+=2){$n="0b";map$n.=0|$a[$r+3][$c+$_],1,0;for$y(1,0){map$n.=0|$a[$r+$_][$c+$y],2,1,0}$\.=chr 0x2800+oct$n}$\.=$/}

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Takes each row comma separated on one line.

Answer 12

APL (Dyalog), 53 bytes

4 2∘{⎕ucs 2⊥40⍪(⍎¨⍕76531420)∘.⊃,¨⍺∘↑¨↓∘⍵¨⍺∘ר⍳⌈⍺÷⍨⍴⍵}

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