What is my volume?

Question

Given an input integer, output the volume corresponding to the value. The value will always be in the range \$[0,100]\$.

Examples

If the input \$=0\$:

 _/|
|  |\/
|_ |/\
  \|

If the input \$>0\$ and \$<33\$:

 _/|
|  |\
|_ |/
  \|

If the input \$\ge33\$ and \$<66\$:

 _/| \
|  |\ \
|_ |/ /
  \| /

If the input \$\ge66\$, \$\le100\$:

 _/| \ \
|  |\ \ \
|_ |/ / /
  \| / /

Rules

• Input/output can be given by any convenient method.
• You can print it to STDOUT or return it as a function result.
• Either a full program or a function are acceptable.
• Any amount of extraneous whitespace is permitted, provided the characters line up appropriately.
• Standard loopholes are forbidden.
• This is code-golf so all usual golfing rules apply, and the shortest code (in bytes) wins.

Answer 1

JavaScript (ES6), 98 bytes

n=>` _/|2
|  |\\0
|_ |/1
  \\|3`.replace(/\d/g,k=>n?(k&1?' /':' \\').repeat(n/33^n>98):['/\\'[k]])

Try it online!

How?

We use the following template:

 _/|2
|  |\0
|_ |/1
  \|3

If \$n=0\$ (volume muted):

• \$0\$ is replaced with / and \$1\$ is replaced with \
• the other digits are removed

If \$n\neq 0\$:

• odd digits are replaced with the string " /" repeated \$k\$ times
• even digits are replaced with the string " \" repeated \$k\$ times

where \$k\$ is defined as:

$$k=\cases{\left\lfloor n/33\right\rfloor&\text{$n<99$}\\ 2&\text{$n\ge 99$} }$$

Answer 2

05AB1E, 65 64 bytes

„ /SõD‚„\ ‚D4δ∍D)T•4a֙镓| 0\
1/“ÅвJrI33÷IĀ+©è‡∊'_1•uтÄ•22в®è‚ǝ

When I started I thought I would be able to make this pretty short with the approach I had in mind, but it had some annoying edge cases to fix.. Can definitely be golfed, though. Will take another look later on.

Try it online or verify some more test cases.

Explanation:

„ /              # Push string " /"
   S             # Convert it to a list of characters: [" ","/"]
õD‚              # Push an empty string "", duplicate it, and pair them together: ["",""]
„\               # Push string "\ "
   Â             # Bifurcate it (shorter for duplicate and reverse copy)
    ‚            # Pair them together: ["\ ","\ "]
D                # Duplicate it
 4δ∍             # Extend both inner strings to size 4: ["\ \ ","\ \ "]
D                # Duplicate it
)                # Wrap all lists on the stack into a list:
                 #  [[" ","/"],["",""],["\ ","\ "],["\ \ ","\ \ "],["\ \ ","\ \ "]]
T                # Push 10
•4aÖ™é•          # Push compressed integer 17523821317
“| 0\
1/“              # Push string "| 0\\n1/"
   Åв            # Convert the integer to custom base-"| 0\\n1/" 
     J           # And join these characters together to a single string:
                 #  "  /| 1\n
                 #   |  |\0"
r                # Reverse the items on the stack (list, 10, string to string, 10, list)
 I33÷            # Push the input, and integer-divide it by 33
     IĀ          # Push the input again, and truthy it (0 if 0; else 1)
       +         # And add them together (0→0; 1..32→1; 33..65→2; 66..98→3; 98..100→4)
        ©        # Store it in variable `®` (without popping)
         è       # And index it into the list we created at the start
          ‡      # Then transliterate the ["1","0"] of 10 to the strings in this pair
∊                # Now vertically mirror the entire string
                 #  (i.e. "  /|  \n
                 #         |  |\/" becomes:
                 #   "  /|  \n
                 #    |  |\/\n
                 #    |  |/\\n
                 #      \|  "
    •/Tδ•        # Push compressed integer 7095187
         ₂в      # Converted to base-26 as list: [15,13,17,21,21]
           ®     # Push variable `®` again
            è    # Index it into this list
   1         ‚   # Pair it with a leading 1
 '_           ǝ '# And insert a "_" at those 0-based positions inside the string
                 # (after which the result is output implicitly)

See this 05AB1E tips of mine (section How to compress large integers? and How to compress integer lists?) to understand why •4aÖ™é• is 17523821317; •/Tδ• is 7095187, and •/Tδ•₂в is [15,13,17,21,21].

Answer 3

Python 2, 111 101 90 bytes

n=input()
for s in'\ _/|','\|  |\/','/|_ |/\\','/  \|':print s[1:6+0**n],s[:n>32],s[:n/66]

Try it online!

_{-11 bytes, thanks to xnor}

Answer 4

Lua, 308 259 246 242 219 180 bytes

i=io.read("*n")
print(i<1 and[[ _/|
|  |\/
|_ |/\
  \|]]or i<33 and[[ _/|
|  |\
|_ |/
  \|]]or i<66 and[[ _/| \
|  |\ \
|_ |/ /
  \| /]]or[[ _/| \ \
|  |\ \ \
|_ |/ / /
  \| / /]])

Try it online!

New to code golf and coding in general.
Thanks to all comments for their help!

Answer 5

PowerShell, 113 bytes

param($n)' _/|'+($y=' \'*($x=($n-ge33)+($n-ge66)))
'|  |\'+($y,'/')[!$n]
'|_ |/'+(($z=' /'*$x),'\')[!$n]
"  \|$z"

Try it online!

Answer 6

Perl 5 (-p), 86 bytes

inspired from @Arnauld's solution

$_=' _/|3
|  |\\1
|_ |/0
  \|2'=~s,\d,$_?" $&"x($_/33-$_/99):$&<2&&$&^1,ger;y;0213;//\

Try it online!

Answer 7

Charcoal, 40 bytes

ＮθＦ AＦ›θ℅ι \Ｃ¹¦¹⸿\Ｆ¬θ/‖Ｍ↓⸿←Ｐ↑⁴←↖¹←_↑² _/

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

Ｎθ

Input the volume.

Ｆ AＦ›θ℅ι \

Output a space and a \ for each character ordinal in space and A that the volume is greater than. In other words, output once for inputs of 33 or more and twice for inputs of 66 or more.

Ｃ¹¦¹

Copy the \s one square diagonally.

⸿\

Output an extra \ at the start of the second line.

Ｆ¬θ/

If the volume was zero then add a /.

‖Ｍ↓

Reflect to complete the volume.

⸿←Ｐ↑⁴←↖¹←_↑² _/

Draw the speaker symbol.

Answer 8

PowerShell, 94 92 bytes

param($n)33,66-le$n|%{$y+=' \';$z+=' /'}
" _/|$y
|  |\$y"+'/'*!$n
"|_ |/$z"+'\'*!$n
"  \|$z"

Try it online!

Answer 9

Canvas, 32 27 26 bytes

┤＃Mi‾‟⁸？⁸‾｛÷ｕ２ｍ２＼×＋］ ¶/＋｝═

Try it here!

Answer 10

Python 3, 109 bytes

lambda n:'\n'.join(i[:5+0**n]+(n//33*(' '+j))[:4]for i,j in zip([' _/|','|  |\\/','|_ |/\\','  \|'],'\\\//'))

Try it online!

Answer 11

Keg, -hd 167 bytes

¿:0=[` _/|\n|  |\\/\n|_ |/\\\n  \\|`|:1\!"∂[` _/|\n|  |\\\n|_ |/\n  \\|`|:\!B"∂[` _/| \\\n|  |\\ \\\n|_ |/ /\n  \\| /`|` _/| \\ \\\n|  |\\ \\ \\\n|_ |/ / /\n  \\| / /`

Try it online!

Definitely golfable by using variables and string formatting. That's what Imma go do now. Anyhow, it acts as a sort of switch statement to determine which range the volume is in and prints the corresponding string.

Nevermind. Variables only make the program longer.

Answer 12

Lua, 196 bytes

i=tonumber(io.read())
t=[[ _/|%s
|  |\%s
|_ |/%s
  \|%s]]
t=(i==0 and t:format("","/","\\","")) or t
while i>=33 do t=t:format(" \\%s"," \\%s"," /%s"," /%s");i=i-34 end
t=t:gsub("%%s","")
print(t)

Try it online!

A bit shorter (below 200) than the other Lua answer.