Distance between the Sun and the nth planet

Question

Challenge

Predict the distance between the Sun and the nth planet when using any formula that gives the same result as the Titius–Bode law: d=(3*2^n+4)/10.

BUT WAIT... there is one restriction:

• Your source code can not include any of the Titius–Bode law's digits
• So, your program can not contain the characters 0, 1, 2, 3, or 4

Input

n a non-negative integer value:

• For Mercury n is -∞ no need to handle this case.
• n=0 for Venus
• n=1 for Earth
• n=2 for Mars, and so on ...

Output

Distance in astronomical unit between the Sun and the nth planet (decimal with at least one decimal place).

Titius–Bode law

Distance in astronomical unit between the Sun and the nth planet is estimated by d=(3*2^n+4)/10.

Example

0          --> 0.7
1          --> 1.0 or 1
11         --> 614.8

Rules

• The input and output can be given in any convenient format.
• No need to handle invalid input values
• Either a full program or a function are acceptable. If a function, you can return the output rather than printing it.
• If possible, please include a link to an online testing environment so other people can try out your code!
• Standard loopholes are forbidden.
• This is code-golf so all usual golfing rules apply, and the shortest code (in bytes) wins.

Answer 1

05AB1E, 7 bytes

oxOÌÌT/

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Explanation

o          # push 2^input
 x         # pop a and push a, 2*a
  O        # sum stack
   ÌÌ      # add 2 twice
     T/    # divide by 10

Answer 2

C (gcc), 22 bytes

-Df(x)=((6<<x)+8.)/''

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Answer 3

Haskell, 29 28 27 25 bytes

s=9-7
d n=(6*s^n+8)/5/s/s

Nothing special, I'm afraid. All allowed digits exactly once!

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Answer 4

JavaScript (Node.js), 20 19 bytes

n=>((6<<n)+8)/5/~-5

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Answer 5

JavaScript (ES6), 20 bytes

Saved 2 bytes thanks to @l4m2

Does not support Mercury.

n=>(5-9-(9-6<<n))/~9

Try it online!

---

JavaScript (ES7), 27 bytes

Supports Mercury.

n=>(k=5,--k+--k*--k**n)/-~9

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Answer 6

Java 8, 20 18 bytes

n->((6<<n)+8f)/''

Port of l4m2's C answer, thanks to Kevin Cruijssen. Try it online here.

Java 8, 30 24 bytes

n->(9-5+(9-6<<n))/(5f+5)

My original approach; admittedly, it's not very creative. Try it online here.

Thanks to Kevin Cruijssen for golfing 6 bytes.

Answer 7

Jelly,  14  8 bytes

-3 thanks to miles (use of æ«, bit-shift) ..and -3 more by golfing from there

6æ«+8H÷⁵

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How?

6æ«+8H÷⁵ - Link: number, N (including -inf)
6        - literal six
 æ«      - bit-shift = 6*(2^N)
   +8    - add eight = 6*(2^N)+8
     H   - halve = 3*(2^N)+4
       ⁵ - literal ten
      ÷  - divide = (3*(2^N)+4)/10

Answer 8

dc, 18 17 bytes

zzrkr^zz+*8z/+A/p

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Assuming n is all that's on the stack, we get the stack depth twice to push 1 and 2 onto the stack. Reverse these and use the 1 to set our precision (k). Reverse so that our stack is now 2 n, exponentiate (^). Use z to get stack depth twice, again pushing 1 and 2 onto the stack. Add them (+) to get 3, multiply (*) by our previous result. Push 8, get the stack depth with z to push 2, divide these (/) to push 4. Add this (+) to the previous result, and divide by 10 (A/). Finally, print.

Knocked off a byte using 8z/ instead of zF+v to make a 4.

Answer 9

Google Sheets, 27 26 25 23 bytes

=((8-5)*(7-5)^A5+9-5)/(5+5)
=(9^.5*(7-5)^A5+9-5)/(5+5)
=(9^.5*(7-5)^A5+9-5)*.7/7
=(7-5)^A5*(.8-.5)+.9-.5

Take input from cell A5.

For a no-digit version (except the A5 is unavoidable),

=(code("")*code("")^a5+code(""))/code("

(there is a trailing newline)

Answer 10

Whispers v2, 325 bytes

> 8
> 8
> 8
> 8
> 9
> 6
> Input
>> 5-6
>> 5-55
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 8
> 7
>> 9+9
>> 9*7
>> 55-56
>> 57⋅58
> 8
> 8
> 8
> 8
> 8
>> 58+55
>> 59+56
>> 66÷65
>> Output 67

Try it online!

Let me outline the problem with using Whispers:

Whispers is a single-expression, line-based programming language. Each line consists of a single mathematical expression. However, if the line starts with >>, the numbers used have different values to what they would normally express.

When a line begins with > (such as > 8) the value returned is the number shown, so that example would return 8 when referenced. Input returns an evaluated line from STDIN.

However, when the line starts with >>, such as >> 5-6, the numbers are interpreted as line references, i.e. the above line would call line 5, then line 6, then subtract the results.

Let's create a program that half-sticks to the rules first. This program doesn't use 0, 1, 2, 3, or 4 to represent those numbers, but does use them as line references. This 'half' solution is 96 bytes:

> 6
> 9
> Input
>> 2-1
>> 1=1
>> 4-5
>> 6*3
>> 7⋅4
>> 6+6
>> 8+9
>> 2+5
>> 10÷11
>> Output 12

Try it online!

Unfortunately, the challenge disallows the characters 0, 1, 2, 3, and 4 from appearing in the program, irregardless of what they represent. So, to work around this, we have to use line references with only the numbers 5, 6, 7, 8 and 9 in their digits. This has the consequence that we cannot use any lines between 10 and 54, which is what the massive chain of > 8 are. In fact, all lines that represent a forbidden line reference has > 8, just for consistency.

So, let's remove those lines, but pretend they're still there, we just can't see them:

> 9
> 6
> Input
>> 5-6
>> 5-55
> 7
>> 9+9
>> 9*7
>> 55-56
>> 57⋅58
>> 58+55
>> 59+56
>> 66÷65
>> Output 67

Which is a very similar structure to the 'half' solution. Here's how this shortened solution works, with an input 11:

> 9		; Line  5: Yield 9
> 6		; Line  6: Yield 6
> Input		; Line  7: Yield a line of input (11)
>> 5-6		; Line  8: Yield 3 	-	9 - 6
>> 5-55		; Line  9: Yield 2	-	9 - 7
⋮
> 7		; Line 55: Yield 7
>> 9+9		; Line 56: Yield 4	-	2 + 2
>> 9*7		; Line 57: Yield 2048 	-	2 ^ 11
>> 55-56	; Line 58: Yield 3	-	7 - 4
>> 57⋅58	; Line 59: Yield 6144	-	2048 ⋅ 3
⋮
>> 58+55	; Line 65: Yield 10	-	3 + 7
>> 59+56	; Line 66: Yield 6148	-	6144 + 4
>> 66÷65	; Line 67: Yield 614.8	-	6148 ÷ 10
>> Output 67	; Line 68: Output 614.8

Answer 11

><>, 27 bytes

l$:?!vll-+$:+!
:*lrl<;n,a+l

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Some heavy abuse of the stack length command (l) command here. Luckily, most of the numbers needed are in ascending order. Takes input through the -v flag

Answer 12

Retina 0.8.2, 54 42 35 bytes

.+
@@@$&$*_@@@@
+`@_
_@@
M`@
.$
.$&

Try it online! Link includes test cases. Edit: Saved 12 bytes thanks to @Leo. Explanation:

.
@@@$&$*_@@@@

Generate a pseudo-binary number using @ as 1 and _ as *2, having 3 as the first digit and 4 as the last digit, thus representing 3*2^n+4.

+`@_
_@@

Convert to unary.

M`@

Convert to decimal.

.$
.$&

Divide by 10.

Answer 13

Mathics, 34 bytes

d[n_]:=((9-6)*(8-6)^n+(9-5))/(5+5)

Try it here!

My first time golfing in Mathics, so there's probably a bit more optimizing to be done.

Answer 14

Symbolic Python, 69 56 bytes

___=_==_
__=-~___
_=__**~-_*-~__+___+___**-___
_/=__-~__

Try it online! Implements (3*2^(n-1)+2)/5

How?

# Input is stored in _ initially

___=_==_         # ___ = True (=1)
__=-~___         # __  = -~1 = 1 + 1 = 2
_=__**~-_*-~__+___+___**-___
# _ = 2**~-_ * -~2 + 1 + 1**-1 = 2**(n-1) * 3 + 1 + 1.0 = 3 * 2**(n-1) + 2.0
_/=__-~__        # _   = _ / (2 -~ 2) = _ / (2 + -~2) = _ / (2 + 2+1) = _ / 5 

# Implicit output of _

Answer 15

Stax, 7 bytes

â╬╫▓♠âε

Run and debug it

PackedStax source encoding makes the character restrictions trivial. Unpacked, ungolfed, and commented this is the same program.

|2  2 to the power
3*  multiply by 3
4+  add 4
A/  divide by 10

Run this one

Answer 16

Python 3, 27 26 bytes

lambda n:((6-9<<n)-~-5)/~9

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---

Python 2, 38 34 31 28 27 bytes

lambda n:((6-9<<n)-9.+5)/~9

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---

Saved:

• -4 bytes, thanks to Giuseppe
• -1 byte, thanks to Jonathan Allan

Answer 17

Jelly, 14 12 bytes

‘⁺
¢*+Ḥ$Ç⁺÷⁵

Try it online!

About time for an explanation

How it works

‘⁺        - Helper link. Can be called monadically, or niladically.
          - When called niladically, `0` is the left argument.
 ⁺        - Perform the previous command twice:
‘         -   Increment the left argument

¢*+Ḥ$Ç⁺÷⁵ - Main link. 1 argument: n
¢*        - Yield 2n
    $     - Perform the next two commands on 2n:
   Ḥ      -   Halve.           Yields 2n-1
  +       -   Add to 2n.       Yields 3⋅2n
     Ç    - Add 2.             Yields 3⋅2n + 2
      ⁺   - Add 2 again.       Yields 3⋅2n + 4
        ⁵ - Yield 10
       ÷  - Divide by 10.      Yields (3⋅2n + 4)/10

Answer 18

C# .NET, 18 bytes

n=>((6<<n)+8f)/''

'' is an unprintable ASCII character with value 20.

Port of @l4m2's C answer.

Try it online.

Answer 19

Ruby, 25 bytes

->n{6-5.6+6%5.7*(9-7)**n}

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Output ±ε due to float imprecision.

Answer 20

Whispers v2, 91 bytes

> Input
>> 9*56
>> 57⋅8
>> 58+7
>> 59÷6
> 5
>> ≺6
>> ≺7
>> ≺8
>> 5÷9
>> Output 65

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Answer 21

Lua, 55 bytes

o=#" "t=o+o function x(m)return((o+t)*t^m+t^t)/(5*t)end

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Answer 22

PowerShell, 38 bytes

(.9-.6)*[math]::pow(8-6,"$args")+.9-.5

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-11 bytes thanks to @mazzy

Answer 23

BRASCA, 29 bytes

ig$}}:a$^A}:a*A}+:l/ml%mn'.on

Try it!

BRASCA only supports whole integers, so I had to spend a couple bytes to at least get one decimal, to pass the test cases.

Explanation

                               - Implicit input
ig                             - Turn 0-9 into the actual numbers and concatenate
  $}}:a$^                      -    2^n
         A}:a*                 - (3*
              A}+              -       +4)
                 :l/           -          /10
                    ml%mn'.on  - Get the first decimal, and print the number.

Answer 24

x86 machine code, 21 20 bytes

6a 03                   push   0x3
58                      pop    eax
d3 e0                   shl    eax,cl
04 04                   add    al,0x4
50                      push   eax
db 04 24                fild   DWORD PTR [esp]
6a 0a                   push   0xa
de 74 24 00             fidiv  WORD PTR [esp+0x0]
58                      pop    eax
58                      pop    eax
c3                      ret

Pretty straightforward, except for two little details:

1. It's enough to add 4 to al - adding it to eax is 1 byte longer
2. [esp+0x0] - explicit displacement equal to 0 is needed to avoid the forbidden byte 0x34 (ASCII 4) in the instruction's encoding; thanks to l4m2 for the idea!