Binary Slashes Display

Question

(easy mode of Seven Slash Display)

Given a positive integer (or alternatively, a string/list of bits of its binary representation) output it with this diagonal ASCII-art formation:

(output for 53 / 110101)

       \
     /\ \
     \ \
    \ \/
  /\ \
  \ \
 \ \/
\ \
 \

This output has the MSB at the bottom left. Alternatively, you may choose to output with the LSB at the bottom left.

The output may include any extra whitespace as long as that doesn’t impact the presentation of the image.

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

Answer 1

Charcoal, 23 17 bytes

ＦＳ¿Σι↗\«\↑/↗\»‖Ｍ↖

Try it online! Link is to verbose version of code. Takes input as a binary string. Explanation:

ＦＳ

Loop over each digit.

¿Σι

If it's a 1, then...

↗\

... output the bottom half of the 1, finishing with the cursor moved up and right.

«\↑/↗\»

Otherwise, draw the bottom half of the zero, finishing with the cursor moved up and right.

‖Ｍ↖

Reflect to complete the output.

Alternative solution, also 17 bytes:

ＦＳ«Ｆ¬Σι↑\¶/↗\»‖Ｍ↖

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

ＦＳ«

Loop over each digit.

Ｆ¬Σι

If it's a zero, then...

↑\¶/

... draw the \/ part.

↗\

Draw the bottom half of the 1 or complete the bottom half of the zero, finishing with the cursor moved up and right.

»‖Ｍ↖

Reflect to complete the output.

Edit: Saved 6 bytes by using ReflectMirror(:UpLeft); as suggested by @noodleman.

Answer 2

Python 3.8 (pre-release),  263   243   229   226   225   187   168   151  150 bytes

-20 bytes by applying some small modifications.
-14 bytes by emanresu A.
-3 bytes by noodle person.
-75 bytes by ASCII-only.
-1 byte by noodle person.

…Okay, what—

def f(b,c=[]):
 for i in b:c+=[0,1,2][i::2]
 for i in range(l:=len(c),-1,-1):print(" "*i+"  /"[[0,*c][i]]+" \\"[i>0]+" \\"*(i<l)+"  /"[[*c,0,0][i+1]])

Try it online!

Answer 3

J, 78 bytes

' \/'{~[:+&(+|:&.|.)/1 2([*Q=/~([+|.@Q=:i.@#@])*_ 1{~]>.~2-[)"{0,&;(0 1;0){~|.

Try it online!

Accepts input as a list of binary digits. Here's the expanded version I started with, which shows some of the symmetry:

b =: 1 1 0 1 0 1
mask =: 0,;(0 1;0){~|.b

e =: [ + i.@-@#@]
mK =: 2 e mask
mT =: 1 e mask

K =: 2 * (mK *_ 1{~mask) =/ i.#mask
T =: 1 * (mT           ) =/ i.#mask

echo ' \/' {~ (+|:&.|.) K+T

Try it online!

Answer 4

Zsh, 88 bytes

for b;((b))&&L=(\ $^L \\$l)&&l=' \'||{L=('  '$^L '/\'$l '\ \');l=' \/' }
print -rl $L $l

Try it online!

Or add --rcexpandparam to remove the ^ for 86 bytes.

for bit;
    ((bit)) &&
        # prepend all but one of the previous lines with a space,
        # the last line prefixed with a '\'
        lines=(\ $^lines \\$last) &&
        # new last line is ' \'
        last=' \' ||
    {
        
        # prepend all but one of the previous lines with two spaces,
        # the last line prefixed with a '/\',
        # add '\ \' as the center of the zero
        lines=('  '$^lines '/\'$last '\ \')
        # new last line is ' \/'
        last=' \/'
    }
print -rl $lines $last

Answer 5

JavaScript (ES6), 102 bytes

Expects an integer and outputs with the LSB at the bottom left.

f=(n,i,p=`
`,b=n-~n>>i/2)=>b>1?f(n,b/2%2-~i,p+" ")+p+"/ "[b&1|i&1]+s[+!!i]+s+"/  "[b&2|i&1]:p+=s=" \\"

Try it online!

Commented

f = (          // f is a recursive function taking:
  n,           //   n = input integer
  i,           //   i = counter (initially undefined)
  p = `\n`,    //   p = prefix string
  b = n - ~n   //   b = n * 2 + 1 right shifted by floor(i / 2)
      >> i / 2 //
) =>           //
b > 1 ?        // if b is greater than 1:
  f(           //   do a recursive call:
    n,         //     pass n unchanged
    b / 2 % 2  //     add 2 to i if the bit #1 of b is set
    - ~i,      //     or 1 otherwise (*)
               //     (because zeros are twice as large as ones)
    p + " "    //     append a space to the prefix string p
  ) +          //   end of recursive call
  p +          //   append the prefix string p
  "/ "[        //   append a space if:
    b & 1 |    //     the target digit is a one
    i & 1      //     or i is odd (grid misalignment)
  ] +          //   or a "/" otherwise (the top of a zero)
  s[+!!i] +    //   append "\" if i > 0, or a space otherwise
               //   (top segment of previous digit)
  s +          //   append " \" (bottom segment of new digit)
  "/  "[       //   append a space if:
    b & 2 |    //     the target digit is a one
    i & 1      //     or i is odd (grid misalignment)
  ]            //   or a "/" otherwise (the bottom of a zero)
:              // else (end of recursion):
  p +=         //   append p
    s = " \\"  //   followed by " \" (top segment of MSB)

Answer 6

APL+WIN, 66 bytes

Prompts for bit vector. Index origin = 0

n←(⊂3 2⍴'  \  \'),⊂3 3⍴'/\ \ \ \'⋄(-⌽⍳⍴s)⊖(i,i)⍴((i←⍴s)*2)↑n[s←~⎕]

Try it online! Thanks to Dyalog Classic

Answer 7

Perl 5 -F, 104 97 bytes

@;=((($b=$"x(length($;[0].=" \\".'/'x!$_)-3+$_))."/\\","$b\\".' \\'x!$_)[$_..1],@;)for@F;say for@

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Answer 8

Python 3.8 (pre-release), 120 118 bytes

def f(i):
 c=a,=["\n"]
 while i:c[0]+=" \\"+(x:=~i%2)*"/";c=[a+x*"/"+"\\",x*(a+"\\ \\")]+c;a+=-~x*" ";i//=2
 print(*c)

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Output with a leading newline, some trailing whitespaces at the end of each line and with the least significant bit in the bottom left (LSB)

Answer 9

GolfScript, 82 81 76 75 bytes

{49='/\ \  \ \/ \ \ '10/=}%5/zip.3<{' ':a+}%\3>{a\+}%+zip.,.a*:b@*b\+\6+/n*

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LSB at the bottom left.

Answer 10

Pure bash 170

Reading other answers here and re-reading request, there is a version without integer to binary conversion:

l=${1//1};t=${#1};printf -vk '%*s' $[t+${#l}-1];p=;for((c=t;c--;)){
((${1:c:1}))&&echo "$k\ $p"&&p=\\||{
k=${k:1};echo "$k/\ $p"$'\n'"$k\ \\";p=\\/;};k=${k:1};};echo \ \\

Pure bash 204

Full version expecting integer as argument.

for((a=$1;a;a>>=1)){ s=$[a&1]$s;};l=${s//1} t=${#s};printf -vk '%*s' $[t+${#l}-1]
p=;for((c=t;c--;)){ ((${s:c:1}))&&echo "$k\ $p"&&p=\\||{
k=${k:1};echo "$k/\ $p"$'\n'"$k\ \\";p=\\/;};k=${k:1};};echo \ \\

Answer 11

JavaScript (Node.js),  130  99 bytes

f=(n,p=`
`,q=`  `,k=` \\`)=>n?f(n>>1,p+(n&1?` `:`  `),n&1?k:`/\\`)+p+(n&1?q+k:` ${k+k+p+q+k}/`):p+q

Try it online!