Pretty Printing Parentheses [closed]

Question

The worst part about long lines or code or math equations are the parentheses. Nobody wants to read through 2x(3y+4*abs(x^[2*e^x])-5{3(x+y)-5})!

So, the goal of this challenge is to make it at least slightly easier to read these parentheses-filled beasts! For the above math equation, the output should be:

2x (
  3y+4*abs (
    x^ [
      2*e^x
    ]
  )
  -5 {
    3 (
      x+y
    )
    -5
  }
)

A parenthesis is defined as either of these characters: ()[]{}. Indents should each be two spaces, and there should be one space of seperation between the expression and the parenthesis: 2x (, x^ [ as some examples.

I would like to debut the Atomic Code Golf metric in this question. Take out all separators in your code (()[]{}. ; to name some common ones), and then take the length of your program if every reserved word and operator were 1 character long. That is your program's score. if x**(3.2)==b[3]:print x is 11 points: if|x|**|3|.|2|==|b|3|print|x, or ix*3.2=b3px. Any quick questions about the metric can be posted here; larger questions should be discussed in meta using the link above.

Lowest score wins. Good luck, have fun!

Answer 1

C, 139 137 chars, 67 tokens

i,n;                                                      // 2 tokens
main(c){                                                  // 2 tokens
        while(~(c=getchar()))                             // 5 tokens
                n=strchr("([{}])",c)?                     // 11 tokens
                        n=c%4!=1,                         // 7 tokens
                        i+=n*4-2,                         // 7 tokens
                        printf("\n%*s%c"+n,n?1:i,"",c)    // 15 tokens
                :                                         // 1 token
                        !printf("\n%*s%3$c"+!n*4,i,"",c); // 17 tokens
}

Logic:

• Read until EOF (detect it because ~EOF==0).
• If parenthesis, check open/close (for c in ")]}", c%4==1), update indentation (i), print with some whitespace.
• If not, print the character, possibly prefixed by newline and indentation (n is true after parenthesis).

Answer 2

Python, 102(?) Tokens

(Please correct if I didn't count correctly)

def f(t):
#1  2 3
    l,s,p,c,e,n=0,"",1," ",")]}","\n"
#   4 5 6 7 8 9ab cd e fgh ijklm no p
    for i in t:
#   q   r s  t
        if i in"([{": l+=2;s+=c+i+n+l*c
#       u  v w xyzAB  CD E FG HIJKLMNOP
        elif i in e: l-=2;s+=n+l*c+i+p*(n+l*c);p=1
#       Q    R S  T  UV W XY Z123456789 abcde  fgh
        else:
#       i
            if len(s) and s[-1]in e: s+=n+l*c
#           j  k   l  m   n  o p  q  rs tuvwx
            s+=i;p=0
#           yz A BCD
    return s
#   E      F

Answer 3

Perl, 86 char

57 tokens? (21+25+5+3+1+2?) Newlines are significant.

s![[{()}\]]!1-(3&ord$&)?" $&
"."  "x++$x:$/.($y="  "x--$x)."$&
$y"!ge;s/
 *
/
/g;print

Answer 4

JavaScript (V8), 76 74

(a,d)=>(d=0)||a.replaceAll(/[(\[{]/g,b=>` ${b}\n`+'  '.repeat(++d)).replaceAll(/[)\]}]/g,b=>`\n${(f='  '.repeat(--d))+b}\n${f}`)

Try it online!

(a,d)=>               pas argument a and d (a is the string, d is just as decleration)
 (d=0)||               set d to 0
 a.replaceAll(         replace everything in a that matches....
   /[(\[{]/g,            [,(,{   with....
   b=>                   pass matches into function
     ` ${b}\n`             space, bracket(the match) and newline
     +'  '.repeat(++d)     increment d and repeat the 2 spaces d times
   )
 .replaceAll(/[)\]}]/g,  match all ),],} and replace them with....
    b=>                  pass matches into function
      `\n${(f='  '.repeat(--d))+b}\n${f}`   newline, decrement d and repeat the 2 spaces d times, closing bracket(the match), newline and once again the repeated spaces
)

results in

#a#,#d#=>#d#=#0#||#a#replaceAll#/#[#(#\#[#{#]#/#g#b#=>#`# #$#{#b#}#\#n#`#+#'# # #'#repeat#++#d#replaceAll#/#[#)#\#]#}#]#/#g#b#=>#`#\#n#$#{#f#=#'# # #'#r#e#p#e#a#t#--#d#+#b#}#\#n#$#{#f#}#`#