Find your dog's age

Question

Task

Take the (integer) number of human years that the dog has lived, \$n\$, as input and return its age in dog years, \$d\$, to two decimal places.
The number of human years, \$n\$, will be between \$1\$ and \$122\$ inclusive: \$n \in [1,122], n \in \Bbb{N}\$.

According to BBC's Science Focus Magazine, the correct function for determining your dog's age is: $$ n = 16 \times \ln(d) + 31 $$ Where

\$n\$ is the age in human years

\$d\$ is the age in dog years

Rules

The final output in dog years must be a decimal, rounded to 2 decimal places.

Answer 1

Io, 114 bytes

While you all are enjoying rounding built-ins in your language, Io doesn't have any of those convenient rounding built-ins...

method(x,I :=((((x-31)/16)exp*100)round asString);I=if(I size<3,"0","").. I;I asMutable atInsertSeq(I size-2,"."))

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Answer 2

Perl 5, 36 31 bytes

printf"%.2f",1.0644944**($_-31)

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Command line one-liner examples:

echo 52 | perl -ne 'printf"%.2f\n",2.71828**(($_-31)/16)'
echo 52 | perl -ne 'printf"%.2f\n",1.06449445**($_-31)'
echo 52 | perl -ne 'printf"%.2f",1.06449445**($_-31)'
echo 52 | perl -ne 'printf"%.2f",1.0644944**($_-31)'

for a in {1..122};do echo "Man: $a Dog: "`echo $a|perl -ne'printf"%.2f\n",1.0644944**($_-31)'`;done

Answer 3

Jelly, 10 bytes

_31÷⁴Æeær2

A monadic Link accepting a float which yields a float. (As a full program a decimal input causes the result to be printed.)

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How?

_31÷⁴Æeær2 - Link: float, n
 31        - 31
_          - subtract         -> n-31
    ⁴      - 16
   ÷       - divide           -> (n-31)/16
     Æe    - exp(x)           -> exp((n-31)/16)
         2 - 2
       ær  - round to 10^(-y) -> round(exp((n-31)/16), 2)

Answer 4

Integral, 14 11 Bytes

My first Integral answer!

x▼◄w‼5*↕u*‼

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An alternative: ▲w‼!◄5*↕u*‼ (11 bytes)

Explanation

x              Push 32
 ▼             x-1
  ◄            Subtraction
   w           Push 16
    ‼          Divide
     5         e ^ a
      *        100
       ↕       Multiplicaiton
        u      Round to the closest integer
         *     100
          ‼    Division

Answer 5

Orst, 9 bytes

31FᎽGć2Š

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As bytes, this is

F3 F1 1B FF 30 1C 64 F2 42

Or, encoded as UTF-8,

óñÿ0dòB

which can be run by omitting the -u flag (filled by default on TIO). Note the not shown 0x1b and 0x1c bytes

How it works

31FᎽGć2Š - Full program. n is pushed to the stack
31       - Push 31
  F      - Subtract; n-31
   Ꮍ     - Push 16
    G    - Divide; (n-31)÷16
     ć   - Exp; exp((n-31)÷16)
      2Š - Round to 2 decimal places

Answer 6

APL+WIN, 13 bytes

Prompts for input of n:

⍎2⍕*(⎕-31)÷16

Try it online! Courtesy of Dyalog Classic

Answer 7

Vyxal s, 23 bytes

31-16/W∆ehS\.€ḣh2wi\.$W

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From this answer two stupid features of Vyxal:

• Why \$ln(x)^{-1}\$ needs to take a list instead of digit?
• Seriously it cannot round a digit to n decimal places, so that part is handrolled.

Answer 8

JavaScript (V8), 33 32 bytes

a=>Math.exp(-~a/16-2).toFixed(2)

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Thanks to @my pronoun is monicareinstate for -1 byte

Answer 9

cQuents, 24 bytes

:b$)/100
:R100*x_+$/16-2

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For some reason, rounding works on the whole line. I would do /100/R100*, but floats went all 9y on me.

Answer 10

R, 28 bytes

round(exp((scan()-31)/16),2)

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There seems to be some controversy about whether trailing zeros must be printed. If so, the following 35 byte solution works:

sprintf("%.2f",exp((scan()-31)/16))

Answer 11

bc, 40 bytes

scale=2
define f(n){return e((n-31)/16)}

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Answer 12

C (gcc), 35 44 38 36 bytes

Saved a byte thanks to my pronoun is monicareinstate!!!
Saved 2 bytes thanks to Arnauld!!!
Added 10 bytes to fix a bug kindly pointed out by Kjetil S.

f(n){printf("%.2f",exp(-~n/16.-2));}

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Answer 13

Japt, ¹² 11 bytes

-1 byte thanks to @Shaggy

Me°U/G-2 x2

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Explanation

Me°U/G-2 x2
Me            //  exp(
  °U          //      (U + 1)
    /G-2      //              / 16 - 2
              //  )
          x2  //  round to 2 decimal digits

Answer 14

Python 3, 52 36 35 bytes

Saved a whopping 16 17 bytes thanks to Jonathan Allan!!!

lambda n:round(1.0644944**(n-31),2)

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Answer 15

MathGolf, 19 bytes

T-☻/e▬♪*i♂‼/%5≥+♀∩*

If only MathGolf had a round builtin.. :/
6 bytes are used for the actual formula, 13 for rounding to 2 decimals, haha.

I/O both as a float.

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Explanation:

T-            # Subtract 31 from the (implicit) input-float
              #  i.e. 50.0 → 19.0
  ☻/          # Divide it by 16
              #  → 1.1875
    e▬        # Push 2.718281828459045 to the power this value
              #  → 3.2788737679386735
♪*            # Multiply it by 1000
              #  → 3278.8737679386735
  i           # Convert it to an integer, truncating the decimal values
              #  → 3278
   ♂          # Push 10
    ‼         # Apply the following two commands separated to the stack:
     /        #  Integer-division
      %       #  Modulo
              #   → 327 and 8
       5≥     # Check if the modulo-10 result (the last digit) is >= 5
              # (1 if truthy; 0 if falsey)
              #   → 327 and 1
         +    # Add that to the division-by-10 result
              #   → 328
          ♀∩  # Push 1/100
            * # Multiply that to the integer
              #   → 3.28
              # (note: `♀/` cannot be used, because it would act as integer division)
              # (after which the entire stack is output implicitly as result)

Answer 16

Rust, 38 bytes

|n|print!("{:.2}",((n-31.)/16.).exp())

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Answer 17

Charcoal, 20 bytes

﹪%.2fＸ¹·⁰⁶⁴⁴⁹⁴⁴⁵⁻Ｎ³¹

Try it online! Link is to verbose version of code. Numeric constant stolen from @JonathanAllan. Explanation:

                Ｎ      Input as a number
               ⁻ ³¹    Subtract literal 31
     Ｘ¹·⁰⁶⁴⁴⁹⁴⁴⁵        Raise literal 1.06449445 to that power
﹪%.2f                   Format using literal format string `%.2f`
                        Implicitly print

Answer 18

Zsh, 50 bytes

zmodload zsh/mathfunc
printf %.2f $[16*log($1)+31]

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Gives "dog years" per the original refs and examples. For example, the "ScienceFocus" article gives 2 examples, and the "Cell Systems" paper gives 1 example, as tabulated:
_{dog age | equivalent human age
actual years | approx. "dog years"
1 | ~30
4 | ~52
12 | ~70}

Do programmers ever read requirements?

Answer 19

05AB1E, 12 bytes

>16/Ížrsm2.ò

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>16/Ížrsm2.ò  # full program
          .ò  # round...
     žr       # 2.718281828459045...
       sm     # to the power of...
              # implicit input...
>             # plus 1...
   /          # divided by...
 16           # literal...
    Í         # minus 2...
          .ò  # to...
         2    # literal...
          .ò  # decimal places
              # implicit output

Answer 20

HP‑41C series, 18 B

You must place \$n\$ on top of the stack first (a. k. a. the X register). Then you can XEQ (execute) the following program:

01♦LBL "Y"         5 Bytes    global label requires 4 + (length of string) Bytes
   NULL            1 Byte     invisible Null byte before numbers
02 31              2 Bytes    X ≔ 31;  Y ≔ 𝘯
03 −               1 Byte     X ≔ 𝘯 − 31
   NULL            1 Byte     invisible Null byte before numbers
04 16              2 Bytes    place the value 16 on top of the stack; X ≔ 16
05 ∕               1 Byte     X ≔ (𝘯 − 31) ∕ 16
06 E↑X             1 Byte     𝘦 raised to the power of X
07 FIX 2           2 Bytes    show (up to) two place after the radix point
08 RND             1 Byte     round the value
09 RTN             1 Byte     `RTN` does not affect local label search

The fixed decimal point display is always rounded. However, the display settings do not affect the internally stored numbers. To ensure all digits after the second decimal place are zero, too, RND (round) is applied.