# ASCII-Art n'th Root

## Challenge:

Given two integers $$\a\$$ and $$\b\$$, with lengths $$\A=length(a), B=length(b)\$$, output an ASCII-art of the $$\a^{th}\$$ root of $$\b\$$, including the answer rounded to $$\A\$$ amount of decimal places.
The size of the ASCII-art root also depends on $$\A\$$ and $$\B\$$.

Example:

$$\a=123\$$, $$\b=1234\$$

     ____
123 /1234 = 1.060
\  /
\/


Because $$\B=4\$$, we have four _ above the 1234. Because $$\A=3\$$, we have three / and $$\1.0595772951...\$$ is rounded to 1.060.

: This will not be equal to $$\A\$$, but $$\\left\lceil{\frac{A}{2}}\right\rceil+1\$$ instead.

## Challenge rules:

• I/O format is flexible:
• You're allowed to take the input loose or as a pair of integers; doubles; strings; list of digits; etc.
• You're allowed to print the result directly to STDOUT; return it as a string; return it as a list/array/stream of string-lines; return it as a matrix of characters; etc.
• If input $$\a\$$ has an odd length $$\A\$$, the number is left-aligned in the output. So in the example above you're not allowed to have the second line as  123/1234 = 1.060! (It should be 123<space>/... instead of <space>123/...)
• The = in the output should be surrounded by a single leading/trailing space
• Rounding can be done in any reasonable way (e.g. rounding towards 0, half-up, half-even, banker's rounding, etc.)
• If the rounded result ends with 0s, you're allowed to omit them. (E.g., the 1.060 in the example above may also be 1.06.)
• You can assume $$\a>0\$$ and $$\b\geq0\$$ ($$\a\$$ is positive, $$\b\$$ is non-negative).
• Trailing spaces in the output, and a single trailing newline is fine. Multiple trailing newlines or trailing whitespaces are not.

## General rules:

• This is , so the shortest answer in bytes wins.
Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.
• Standard rules apply for your answer with default I/O rules, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.
• Default Loopholes are forbidden.

## Test cases:

Inputs: a=3, b=100 (A=1, B=3)
___
3 /100 = 4.6
\/

Inputs: a=123, b=1234 (A=3, B=4)
____
123 /1234 = 1.060
\  /
\/

Inputs: a=10, b=10 (A=2, B=2)
__
10/10 = 1.26
\/

Inputs: a=1298123, b=9023847978341 (A=7, B=13)
_____________
1298123 /9023847978341 = 1.0000230
\      /
\    /
\  /
\/

Inputs: a=2500, b=0 (A=4, B=1)
_
2500/0 = 0.0000
\  /
\/

• Related: Square Root of ASCII art. May 9, 2022 at 13:30
• What's the policy on trailing whitespace? May 9, 2022 at 15:42
• @loopywalt Trailing spaces and a single trailing newline is fine. I'll add it. May 9, 2022 at 17:16
• Do you mean to assume that $a$ is a positive integer and $b$ is a nonnegative integer? May 10, 2022 at 8:46
• @GregMartin Yes, isn't that exactly what I stated in rule "You can assume $a>0$ and $b\geq0$ ($a$ is positive, $b$ is non-negative)."? May 10, 2022 at 9:20

# Python, 143 bytes

def f(a,b):
A=len(str(-a))&-2;print(A*" ",len(str(b))*"_"+f"\n{a:<{A}}/{b} = {b**(1/a):.{A}}");p="\\"
while A:A-=2;print(p+A*" "+"/");p=" "+p

Attempt This Online!

• Smart to use len(str(-a)) instead of len(str(a))+1 for -1 byte! :) May 9, 2022 at 17:20

# JavaScript (ES8),  158  156 bytes

Expects (a)(b), where both arguments are strings.

a=>b=>" "[R='repeat'](q=(A=a.length)+A%2)+ ${"_"[R](b.length)}${a.padEnd(q)}/${b} =${(b**=1/a).toFixed(A)}
+(g=p=>q?p+" "[R](q-=2)+/
+g(" "+p):"")\\


Try it online!

# Charcoal, 37 bytes

θ⸿↘⊘⊕Ｌθ↑↗⊘⁺³Ｌθ⟦⭆η_⟧η = ﹪⁺⁺%.ＬθfＸＩη∕¹Ｎ


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

θ


Output the first input.

⸿↘⊘⊕Ｌθ↑↗⊘⁺³Ｌθ⟦⭆η_⟧


Draw the square root sign.

η = ﹪⁺⁺%.ＬθfＸＩη∕¹Ｎ


Output the second input and the formatted result.

# Lua, 205203 188 bytes

a,b=...p,c,s=print,a.rep,a.len d=s(a)r=d%2+d+1 p(c(" ",r)..c("_",s(b)))p(a.format("%s%s/%s = %."..d.."f",a,c(" ",d%2),b,b^(1/a)))for i=1,r//2 do p(c(" ",i-1).."\\"..c(" ",r-i*2-1).."/")end


Attempt This Online! Explanation soon:tm: once I'm out of school. I added an explanation to the footer of the ATO link. • @KevinCruijssen Should be good now. May 9, 2022 at 15:44
• Yep, looks good now. +1 from me. :) May 9, 2022 at 17:18

# Python 3.8 (pre-release), 161 bytes

lambda a,b:f"{' '*(Z:=(A:=len(str(a)))+1&-2)} {'_'*(B:=len(str(b)))}\n{a:<{Z}}/{b} = {b**a**-1:.{A}f}"+"".join(f"\n{' '*i}\\{' '*(Z+2*~i)}/"for i in range(Z//2))


Try it online!

# Julia 1.0, 137 bytes

!x=length("$x") a\b=[" "^-~(X=!a+!a%2)*"_"^!b;"$a"*" "^(r=1:X÷2;!a%2)*"/$b =$(round(b^a^-1,digits=!a))";@. ' '^~-r*"\\"*" "^(X-2r)*'/']


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# Python, 169 bytes

def f(a,b):print(" "*((A:=len(str(a)))-1+(x:=A%2)),"","_"*len(str(b))+f"\n{a}{x*' '}/{b} =",round(b**1/a,A),*("\n"+i*" "+"\\"+(A-i-i-2+x)*" "+"/"for i in range(A//2+x)))

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# C (gcc), 222 218 bytes

-4 bytes thanks to @ceilingcat

#define P;printf(
#define X snprintf(0,0,"%.0f"
A;B;f(double a,double b){A=X,a)P"  %*c",A+A%2,95);for(B=X,b);--B P"_"))P"\n%.0f%*c%.0f = %.*f",a,1+A%2,47,b,A,pow(b,1/a));for(A-=~A&1;B++<A--P"\n%*c%*c",B,92,2+A-B,47));}


Try it online!