Square root of ASCII art

Question

You are working as an intern for a mathematician who really hates TeX, LaTeX, etc. So much so that he has decided to abandon all typesetting whatsoever and make you do all the formatting in ASCII. You got tired of this after a while and decided to start automating parts of it, starting with square roots.

Here's how a square root is made:

###  this is the input
###
###

_____  put a row of underscores stretching one character off either side on top
 ### 
 ### 
 ### 

   _____  put a diagonal of slashes in front going the height of the input
  / ### 
 /  ### 
/   ### 

      _____ put a diagonal of backslashes in front of that, going up half of the input's height rounded up
     / ###
 \  /  ###
  \/   ###

And that's it!

Rules

You are to make a program or function that takes in a string, list of strings (i.e. lines), or array of characters, and outputs the input transformed according to the description above (not necessarily by the exact same order or process)

You may assume that the input is rectangular if preferred. Trailing whitespace is neither required nor disallowed.

This is code-golf, so the shortest answer in bytes wins.

Examples:

Input:
40
Output:
  ____
\/ 40

Input:
  ____
\/ 40
Output:
   ________
  /   ____
\/  \/ 40 

Input:
/| 
 | 
 | 
_|_
Output:
      _____
     / /|
    /   |
\  /    |
 \/    _|_

Input:
#  # 
#  # 
#####
   # 
   # 
Output:
        _______
       / #  #  
      /  #  #  
\    /   ##### 
 \  /       #  
  \/        #  

Answer 1

Python 2, 196 bytes

x=Q=input()
l=len(x)
k=(l+1)/2
q=l+k
x=[[' ']*(q+1)+list(y)for y in x]
for i in range(k):x[i+l/2][i]='\\'
for j in range(l):x[j][q-j-1]='/'
print'\n'.join([' '*q+'_'*(2+len(Q[0]))]+map(''.join,x))

Try it online!

-2 bytes thanks to Step Hen

-13 bytes thanks to Jonathan Allan

Answer 2

SOGL V0.12, 27 bytes

1w⁄2+╔*00žl»╚;lH╚@Κ№↕h┼№↕;┼

Try it Here! (→ added for ease-of-use; the program expects the input on stack)

Answer 3

Charcoal, 32 bytes

ＷＳ⊞υιＰ⪫υ¶↖Ｐ×_⁺²⌈ＥυＬι↙↙Ｌυ↑↖÷⁺¹Ｌυ²

Try it online! Link is to verbose version of code. 29-byte version that assumes rectangular input:

ＷＳ⊞υιＰ⪫υ¶↖Ｐ×_⁺²Ｌθ↙↙Ｌυ↑↖÷⁺¹Ｌυ²

Answer 4

Python 3, 138 147 Bytes

def f(l):h=len(l);w=len(l[0]);c=int(h/2);print('\n'.join([(h*2-c)*' '+w*'_']+[(i*' '+'\\'+(h-i-1)*2*' '+'/'+i*' ')[c:]+s for(i,s)in enumerate(l)]))

Variable 'l' is a list of strings, each string a line. Readable Version:

def f(l):
  height = len(l)
  width = len(l[0])
  half_height_floor = int(height / 2)

  print((height * 2 - half_height_floor) * ' ' + width * '_')

  for (index, line) in enumerate(l):
    #build a V based on the location of the line
    #surrounding V whitespace
    outer_space = index * ' '

    #inner V whitespace
    inner_space = (height - index - 1) * 2 * ' ' #inner v space

    #complete V
    v = outer_space + '\\' + inner_space + '/' + outer_space

    #left half_height_floor chars removed
    v_chopped = v[half_height_floor:]

    print(v_chopped + line)

Forms a square root symbol with a complete V then shaves off the left accordingly.

Try it Online!

Answer 5

Python 2,  131  130 bytes

x=input()
n=len(x)
s=" "
for v in[s*2*n+"_"*(2+len(x[0]))]+[s*i+"\\"+s*2*(n+~i)+"/"+s*-~i+r for i,r in enumerate(x)]:print v[n/2:]

A full program taking a list of lines as input with the rectangular-only allowance (actually that the first line is one of the longest).

Try it online!

Answer 6

Java 8, 244 bytes

A very long solution, but probably close to the shortest for Java. This lambda takes lines of input as a String[] and returns a String. All lines must have the same length.

Based on the example outputs, I assumed that there's no need to append a space to each line in the input, so the program doesn't.

Thanks to Jonathan Allan for reminding me about the ~ operator.

l->{int h=l.length,w=l[0].length(),a=h/2,i=w+3,x=h+a+h%2;String s="",t=s;for(;--i>0;)t+="_";for(;i++<x;)s+=" ";t=s+t;for(i=0;i<h;)t+="\n"+s.substring(0,i<a?x+~i:i-a)+(i<a?"":"\\"+s.substring(0,(h+~i)*2))+"/"+s.substring(0,i+1)+l[i++];return t;}

Try It Online

Ungolfed

l -> {
    int
        h = l.length,
        w = l[0].length(),
        a = h / 2,
        i = w + 3,
        x = h + a + h % 2
    ;
    String
        s = "",
        t = s
    ;
    for (; --i > 0; )
        t += "_";
    for (; i++ < x; )
        s += " ";
    t = s + t;
    for (i = 0; i < h; )
        t +=
            "\n"
            + s.substring(0, i < a ? x + ~i : i - a)
            + (i < a ? "" : "\\" + s.substring(0, (h + ~i) * 2))
            + "/"
            + s.substring(0, i + 1)
            + l[i++]
        ;
    return t;
}

Acknowledgments

• -2 bytes thanks to Kevin Cruijssen

Answer 7

Japt, 46 bytes

l *2
£Vç hY'\ h~Y'/ +S+XÃuVç +'_p2+Ug l¹msV/4

Leading newline is part of the program. Input and output is an array of strings representing lines.

Try it online! using the -R flag to join the resulting array with newlines.

Answer 8

Perl 5, 177 185 160 bytes

$s=$f=int((@t=<>)*1.5+.5);print" "x$f;print"_"x length$t[0];print"_
";$b=-int(@t/2);for(0..$#t){--$s;print$_==$s?"/":$_==$b?"\\":" " for 0..$f;print$t[$_];++$b}

Try it online!

Changelog:

• needed more bytes to fix a bug (it assumed a square input first)
• fixed another bug and used some tips from comments (thanks Dada!)

Answer 9

JavaScript (ES6), 140 bytes

Takes input as an array of strings / returns an array of strings.

a=>[a[0].replace(/./g,'_'),...a].map((r,y)=>[...a,...a].map((_,x)=>x-y+1|y-.5<l/2?l*2-x-y?' ':'/':'\\',c=y?' ':'_').join``+c+r+c,l=a.length)

Test cases

let f =

a=>[a[0].replace(/./g,'_'),...a].map((r,y)=>[...a,...a].map((_,x)=>x-y+1|y-.5<l/2?l*2-x-y?' ':'/':'\\',c=y?' ':'_').join``+c+r+c,l=a.length)

console.log(
  f([
    '40'
  ]).join`\n`
)

console.log(
  f([
    '  ____',
    '\\/ 40 '
  ]).join`\n`
)

console.log(
  f([
    '/| ',
    ' | ',
    ' | ',
    '_|_'
  ]).join`\n`
)

console.log(
  f([
    '#  # ',
    '#  # ',
    '#####',
    '   # ',
    '   # '
  ]).join`\n`
)

Answer 10

C++, 291 bytes

The function assumes that all the strings in the vector passed as parameter have the same length

#include<vector>
#include<string>
#define S std::string
#define T size()
void f(std::vector<S>&c){c.insert(c.begin(),S(c[0].T+1,'_'));int i,j=0;for(i=1;i<c.T;++i){c[i]='/'+S(i,' ')+c[i];if(i>=c.T/2)c[i]='\\'+S((c.T-i-1)*2,' ')+c[i];}for(auto&a:c)j=j>a.T?j:a.T;for(auto&a:c)a=S(j-a.T,' ')+a;}

Answer 11

Dyalog APL, 95 bytes

{((' '/⍨y+x),'_'/⍨2+⊃⌽⍴⍵)⍪⍉(x-y)↓⍉(⊖(x(-x)↑⌽y y⍴'\',' '\⍨y←⌈x÷2),x x⍴'/',' '\⍨x←⊃⍴⍵),' ',⍵,' '}

Try it online!

Answer 12

C, 485 bytes

This program takes up to 999 characters from standard input and reads them into an array. It prints them 1 at a time to standard output with the changes your challenge indicated. It assumes the input is rectangular.

#include<stdio.h>
#define p(a)putc(a,stdout);
#define P j+j/2+1
a[999],i,j,k,l,m,n,q;char c;pad(a){m=P;if(a&&!k){m-=1;}for(n=0;n!=m;n++){q=32;if((!a||k)&&n==c){c--;q=47;}else if((P-c+1)>j/2+1&&(P)/2-n==c-2){q=92;}p(q);}}int main(){k=i=j=0;x:if(read(0,&c,1)){if('\n'==(a[i++]=c)){if(!j){l=i;}j++;}goto x;}i--;if('\n'==a[i-1]){i--;}else{j++;}c=P-2;for(;k!=i;k++){if(!k||a[k]==10){if(a[k]==10){p(10);}pad(1);if(!k){l++;while(l-->0){p(95);}p(10);pad(0);}if(a[k]==10){continue;}}p(a[k]);}}

Answer 13

Perl 5, 159 bytes

@a=map{$m=(y///c)>$m?y///c:$m;$_}<>;$_=$"x($l=@a/2-.5).'\\/'.$"x@a;for$i(1..@a){$a[-$i]=$_.$a[-$i];s| \\|\\ |;s|/ | /|;$i>$l&&y/\\/ /}chop;say$_.'_'x++$m,$/,@a

Try it online!