ASCII-Art n'th Root

Question

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ドル.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 code-golf, 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.
• If possible, please add a link with a test for your code (e.g. TIO).
• Also, adding an explanation for your answer is highly recommended.

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 =わ 1.0000230
\ /
 \ /
 \ /
 \/
Inputs: a=2500, b=0 (A=4, B=1)
 _
2500/0 = 0.0000
\ /
 \/

• Kevin Cruijssen: Related: Square Root of ASCII art.
• loopy walt: What's the policy on trailing whitespace?
• Kevin Cruijssen: @loopywalt Trailing spaces and a single trailing newline is fine. I'll add it.
• Greg Martin: Do you mean to assume that \$a\$ is a positive integer and \$b\$ is a nonnegative integer?
• Kevin Cruijssen: @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)."?

Answer 1

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

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• Kevin Cruijssen: Smart to use len(str(-a)) instead of len(str(a))+1 for -1 byte! :)

Answer 2

JavaScript (ES8), (削除) 158 (削除ここまで) 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):"")`\\`

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Answer 3

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.

Answer 4

Lua, (削除) 205 (削除ここまで) (削除) 203 (削除ここまで) 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! (削除) Explanation soon:tm: once I'm out of school. (削除ここまで) I added an explanation to the footer of the ATO link. enter image description here

• twentysix: @KevinCruijssen Should be good now.
• Kevin Cruijssen: Yep, looks good now. +1 from me. :)

Answer 5

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))

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Answer 6

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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Answer 7

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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Answer 8

C (gcc), (削除) 222 (削除ここまで) 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));}

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