Convert version string to pack_format

Question

When making a Minecraft data/resource pack, you need to include a pack.mcmeta file, which specifies information about it. pack.mcmeta contains a description, but it also contains a pack_format number, which tells Minecraft which versions this pack is for.

Your task is to take a version string, and output what pack_format number matches the string.

Your input must be a string, and you can assume it's either 1.x or 1.x.y.

As of when this challenge was posted, the conversion goes like so:

1.6.1 - 1.8.9 -> 1
1.9 - 1.10.2 -> 2
1.11 - 1.12.2 -> 3
1.13 - 1.14.4 -> 4
1.15 - 1.16.1 -> 5
1.16.2 - 1.16.5 -> 6
1.17 -> 7

These are all the possible inputs, and what they need to be mapped to:

1.6.1 -> 1
1.6.2 -> 1
1.6.4 -> 1
1.7.2 -> 1
1.7.4 -> 1
1.7.5 -> 1
1.7.6 -> 1
1.7.7 -> 1
1.7.8 -> 1
1.7.9 -> 1
1.7.10 -> 1
1.8 -> 1
1.8.1 -> 1
1.8.2 -> 1
1.8.3 -> 1
1.8.4 -> 1
1.8.5 -> 1
1.8.6 -> 1
1.8.7 -> 1
1.8.8 -> 1
1.8.9 -> 1
1.9 -> 2
1.9.1 -> 2
1.9.2 -> 2
1.9.3 -> 2
1.9.4 -> 2
1.10 -> 2
1.10.1 -> 2
1.10.2 -> 2
1.11 -> 3
1.11.1 -> 3
1.11.2 -> 3
1.12 -> 3
1.12.1 -> 3
1.12.2 -> 3
1.13.1 -> 4
1.13.2 -> 4
1.14 -> 4
1.14.1 -> 4
1.14.2 -> 4
1.14.3 -> 4
1.14.4 -> 4
1.15 -> 5
1.15.1 -> 5
1.15.2 -> 5
1.16 -> 5
1.16.1 -> 5
1.16.2 -> 6
1.16.3 -> 6
1.16.4 -> 6
1.16.5 -> 6
1.17 -> 7

This is code golf, so the shortest answer wins. Good luck!

Answer 1

JavaScript (ES6),  43  42 bytes

s=>([,x,y]=s.split`.`,x*x+x*3+~~y)/51^x>16

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How?

Given an input string 1.x.y, we compute:

$$\left\lfloor\frac{x^2+3x+y}{51}\right\rfloor$$

Because we're using ~~y, we implicitly assume \$y=0\$ if the input string is just 1.x.

This gives the correct result for all versions except the last one, which gives \$6\$ instead of \$7\$. This is fixed by XOR'ing with \$1\$ if \$x>16\$.

Answer 2

JavaScript (ES7), 31 bytes

s=>s.slice(2)**2/45+.2|s[3]%1.2

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TIO setup is based on Arnauld's answer.

Ignore the 1. part and treat the input as a floating point number. For example, 1.16.1 becomes 16.1. This formula (find out by some brute force search) works quite good (48/52 correctness).

$$\left\lfloor\frac{s^2}{45}+0.2\right\rfloor$$

And with some manually fix... It finally works.

So if you are interesting in how it works: I don't know why it work either.

Answer 3

Excel, 46 44 bytes

-2 bytes change first item from 6.1 to 0

=MATCH(MID(A1,3,9)*1,{0,9,11,13,15,16.2,17})

Since they are all 1.x or 1.x.y, this basically matches everything from the 3rd character on.

Answer 4

Jelly, 16 bytes

ṫ3V×⁵>“Ym²Ɱƙʋ‘S‘

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ṫ3V×⁵>“Zn³Ɲɱȥ‘S‘   Main Link; take the string as required
ṫ3                 Remove the first two characters
  V                Eval it
   ×⁵              Multiply it by 10
     >“Zn³Ɲɱȥ‘     Compare (>) to each of [89, 109, 130, 149, 161, 169]
              S    Sum
               ‘   Increment

Answer 5

Japt, 18 bytes

Port of hyper-neutrino's solution so be sure to upvote them too.

Ò"Ym¡©"¬mc x<¢*A

Try it (Includes all test cases)

Ò"..."¬mc x<¢*A     :Implicit input of string U
Ò                   :Negate the bitwise NOT of
 "..."              :String containing the characters at codepoints 89, 109, 130, 149, 161 & 169
      ¬             :Split
       m            :Map
        c           :  Codepoint
          x         :Reduce by addition
           <        :After checking each is less than
            ¢       :  Slice the first 2 characters off U
             *A     :  Multiply by 10

Answer 6

Raku, 69 bytes

[+] Version.new($v) <<>=<<(v1.6,v1.9,v1.11,v1.13,v1.15,v1.16.2,v1.17)

Like Perl, Raku has native support for version string comparison. See Version for more info.

Explanation

• Version.new($v) converts from Str to Version type (it's a shame v$v doesn't work)
• <<>=<< hyper-operator converts major versions to Bools
• [+] counts True values

Answer 7

Python 3, 94 60 57 bytes

lambda s:sum(x<=float(s[2:])for x in(0,9,11,13,15,16.2,17))

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This function takes the string s and returns the corresponding pack_format.

-34 bytes thanks to Command Master (using lambda and sum instead of a basic loop + a stopping condition)

-3 bytes thanks to tsh (flipping the conditional, changing the list structure, replacing float with eval)

Answer 8

C (gcc), ^{67 \$\cdots\$ 60} 58 bytes

Saved a byte thanks to ceilingcat!!!
Saved 4 6 bytes thanks to tsh!!!
Saved 2 bytes thanks to rtpax!!!

y;f(p){p=(~-sscanf(p,"1.%d.%d",&p,&y)*y+p*p+3*p)/51+p/17;}

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Uses a variation of Arnauld's formula from his JavaScript answer.

Answer 9

05AB1E, 18 bytes

¦¦T*•BÎΣÚÁÿ•Ƶζв›O>

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¦¦T*•...•Ƶζв›O>  # trimmed program
             O   # sum of all elements of...
            ›    # is...
                 # implicit input...
¦                # excluding the first character...
 ¦               # excluding the first character...
   *             # times...
  T              # 10...
            ›    # greater than...
                 # (implicit) each element of...
           в     # list of base-10 values of base...
         Ƶζ      # 170...
           в     # digits of...
    •...•        # 12728408213639...
              >  # plus 1
                 # implicit output

Answer 10

Java, 93 bytes

s->{var i=Float.valueOf(s.substring(2));return i<9?1:i<11?2:i<13?3:i<15?4:i<16.2?5:i<17?6:7;}

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Answer 11

Vyxal, 23 bytes

L2$"iE₀*`-AViu}`C44+>∑›

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Answer 12

Python 3, 72 bytes

f=lambda p,n=0:float(p[2:])<[9,11,13,15,16.2,17,18][n]and n+1or f(p,n+1)

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Answer 13

Charcoal, 32 29 bytes

ＩＬΦＩ⪪”#｜⊞u⦄,<q'Ｚχ=” ¬›ιＩ✂θ²χ¹

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

     ”...”              Compressed string `0 9 11 13 15 16.2 17`
    ⪪                   Split on spaces
   Ｉ                    Cast to float
  Φ                     Filtered where
             ι          Current value
           ¬›           Is not greater than
               ✂θ²χ¹    Slice of input string
              Ｉ         Cast to float
 Ｌ                      Take the length
Ｉ                       Cast to string
                        Implicitly print

Prefixing 0 to the compressed string or post-incrementing both cost 1 byte. I've stuck with prefixing 0 as it used to be shorter with the previous method. Edit: Switched to a less-than-or-equals comparison which costs a byte but saves 4 bytes from the compressed string.

Answer 14

BRASCA 0.5.2, 30 bytes

^{The G operator was added on the same day as the challenge was posted, but not specifically for this challenge.}

The program assumes input as 1.x.y at all times. Ex. 1.16 must be input as 1.16.0.

The math is from Arnauld's JS answer, go upvote it!

C'.iGX$:16S>[7n@]x::*$3*++51S/

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Explanation

The second number will be called X, the third number will be called Y.

C                               - Set implicit output to number mode
   i                            - Turn "0-9" from char into the numbers 0-9
 '. G                           - Split stack by "."
     X                          - Remove the first number, it's not needed.
      $                         - Put X on top of the stack
       :16S>[7n@]x              - If it's more than 16, the version is 1.17.0, so print 7 and terminate.
                                - Calculate the final number:
                  ::*           -   (X**2
                     $3*+       -        + X*3
                         +      -              + Y)
                          51S/  -                  /51
<implicit>                      - Output the number

Answer 15

Python 3, 66 bytes

lambda i:sum([float(i[2:])>x+8.2 for x in [.7,2,4,6.2,7.9,8.3]])+1

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Explanation:

• Our input is i. Remove the 1. at the start with i[2:].
• Convert the rest to a number with float.
• For each x in the list, check if this is greater than x+8.2.
  • Why 8.2? Well, all of the finishing versions (1.8.9, 1.10.2, etc.) are greater than 8, thus saving some bytes.
  • In addition, since two of them end with .2, I can subtract that as well, saving 2 more bytes.
• Since True and False are treated as 1 and 0 by the arithmetic operators, we can use sum to count the number of Trues in the list.
• This is now off by one, so we add 1.