Opposite of the digital root!

Question

^{Also known as the [analog root]}

(Opposite of the digital root!) ;)

The digital root of a number is the continuous summation of its digits until it is a single digit, for example, the digital root of 89456 is calculated like this:

8 + 9 + 4 + 5 + 6 = 32

3 + 2 = 5

The digital root of 89456 is 5.

Given a digit as input via STDIN, print/return all of the possible two digit numbers that have that digital root. If you need it to, it can include itself, e.g. 05

These are all of the possible inputs and outputs:

(You get to choose whether or not to include the leading zero for the digit itself)

I/O

0 => 0 or 00 or nothing

1 => 01 and/or 1, 10, 19, 28, 37, 46, 55, 64, 73, 82, 91 - Make sure that 1 does not return 100

2 => 02 and/or 2, 11, 20, 29, 38, 47, 56, 65, 74, 83, 92

3 => 03 and/or 3, 12, 21, 30, 39, 48, 57, 66, 75, 84, 93

4 => 04 and/or 4, 13, 22, 31, 40, 49, 58, 67, 76, 85, 94

5 => 05 and/or 5, 14, 23, 32, 41, 50 ,59, 68, 77, 86, 95

6 => 06 and/or 6, 15, 24, 33, 42, 51, 60, 69, 78, 87, 96

7 => 07 and/or 7, 16, 25, 34, 43, 52, 61, 70, 79, 88, 97

8 => 08 and/or 8, 17, 26, 35, 44, 53, 62, 71, 80, 89, 98

9 => 09 and/or 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99

No standard loopholes, and it's code-golf, so the shortest answer in bytes wins.

Congrats to Heeby Jeeby Man on his amazing 46 byte brain-flak answer!

Answer 1

JavaScript (ES6), 27 31 30 bytes

Returns 0 for 0 or an array of solutions otherwise.

n=>n&&[...1e9+''].map(_=>n+=9)

Demo

let f =

n=>n&&[...1e9+''].map(_=>n+=9)

for(n = 0; n < 10; n++) {
  console.log(n, JSON.stringify(f(n)));
}

Answer 2

05AB1E, 13 12 9 bytes

-3 bytes thanks to Adnan

тL<ʒSOSOQ

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Explanation

тL<ʒSOSOQ   Main link. Argument n
тL<         List from 1 to 100, then decrement to get 0 to 99
   ʒ        Filter
    SOSO    Sum of all chars, twice
        Q   Compare to input

Answer 3

Jelly, 8 bytes

11Ḷ×9+ȧ@

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Different algorithm than my other answer.

Answer 4

Brain-Flak, 46 bytes

{((()()()()()){}){(({}[()])<(({}{}[]))>)}}{}{}

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Explanation

This answer uses an idea from Megatom's answer namely using the stack height as the difference between the loop counter and the increment. Like previous answers this answer has a large outer loop to catch all the zeros. Inside the loop we push 10 to act as a counter, we then start another nested loop. In this loop we decrement the counter by 1

({}[()])

Then we pop the top two items, which are the counter and the last item we calculated. We add these to the stack height in order to counter balance the decrementation, we then push this twice, once for output and once so that it can be consumed to calculate the next result. Pushing things twice means we accidentally push an additional value which needs to be removed at the end of execution.

The reason this just barely beats Megatom is Megatom's answer is forced to get its stack heights while the last result is still on the stack. This means they are forced to use a rather expensive [()] to decrease the total by one. By moving the duplicate to the end of the loop I am able to avoid having to use [()] at the cost of an additional {} at the very end of the program. If Megatom were to use this strategy his answer would look like:

{<>((()()()()()){}){((({}[()])<>{}[]))<>}}<>{}

also 46 bytes.

Brain-Flak, 52 bytes

{((()()()()()){}){({}[()]<(({})((()()())){}{})>)}}{}

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Explanation

The main outerloop makes a special case for input of zero. If zero is input we jump over the entire loop, pop zero and then output nothing. Otherwise we enter the loop. Here we push loop 10 times each time adding 9 to the top of the stack, keeping old values. Since 9 preserves digital sums this will get us the next value. Once the loop has expired we use the zero it generated to exit the loop which is then popped by the {} at the end.

Brain-Flak, 56 bytes

{([(((()()())){}{})]){({}()<(({})<(({}{}))>)>)}}{}({}{})

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Explanation

This version works very similarly to the last one, except we loop 9 times instead of 10 leaving out the original value. In order to do this we have to rearrange the way we handle memory a bit. All of the bytes we might have saved using this method get put into cleanup.

Answer 5

Haskell, 21 bytes

f takes an integer and returns a list of integers.

f d=[d,d+9..99^0^0^d]

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• Starts with the digit d and generates the range with every 9th number up to a bound of 99, except for the tricky case of 0.
• To stop early for 0, uses that the power 0^d==1 for 0 and ==0 for all other digits. Thus 99^0^0^d gives 1 for 0 but 99 for anything else.

Answer 6

Pyke, 6 bytes

ITV
9+

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ITV\n9+ - if input: (don't print anything for 0 case)
 TV\n9+ -  repeat 10 times:
   \n   -    print ^
     9+ -   ^ += 9

Answer 7

Python 2, 29 bytes

lambda n:n and range(n,100,9)

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Answer 8

Brachylog, 12 bytes

0g|g{t+₉}ᵃ¹⁰

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Explanation

0g               Input = 0, Output = [0]
  |              Or
   g{   }ᵃ¹⁰     Accumulate 10 times, starting with [Input]
     t+₉         Take the last element, add 9

Answer 9

Bash, 31 27 bytes

seq $1 9 $(($1?99:0))|xargs

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previous

eval echo {$1..$(($1?99:0))..9}

Answer 10

Dyalog APL, 15 bytes

{(×⍵)/+\⍵,10⍴9}

How?

⍵,10⍴9 - concatenate input with 10 9s (⍵ 9 9 9 9 9 9 9 9 9 9).

+\ - cumulative sum.

(×⍵)/ - expand signum times - where signum gives 1 for 1-9 and 0 for 0.

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Dyalog APL, 24 bytes

{⍵/⍨⎕=(⍵≠0)×1+9|⍵-1}⍳100

Requires ⎕IO←0.

How?

                      ⍳100  ⍝ 0 .. 99
              1+9|⍵-1      ⍝ digit sum (⍵-1 mod 9 + 1)
        (⍵≠0)×             ⍝ edge case for 0
     ⎕=                    ⍝ equals to the input
 ⍵/⍨                       ⍝ compress with the range

Answer 11

Brain-Flak, 48 bytes

{<>((()()()()()){}){(({}[()])<>[][()]({}))<>}}<>

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I may add an explanation later.

Answer 12

Mathematica, 25 bytes

If[#==0,0,Range[#,99,9]]&

works for 0

Answer 13

Jelly, 12 bytes

⁵²Ḷµ,³%9EµÐf

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How It Works

⁵²Ḷµ,³%9EµÐf
⁵             - literal 10
 ²            - square
  R           - lowered range: 0 to 99 inclusive.
   µ     µÐf  - filter based on:
    ,³          - element and input
      %9        - mod 9
        E       - are equal

Answer 14

Actually, 18 bytes

╗2╤DR⌠╜-9@%Y⌡░╜;)I

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Explanation:

╗2╤DR⌠╜-9@%Y⌡░╜;)I
╗                   save input to register 0
 2╤DR               range(1, 100)
     ⌠╜-9@%Y⌡░      elements in range where function returns truthy:
      ╜-              subtract from input
        9@%           mod 9
           Y          is equal to 0
              ╜;)   push a copy of the input on the top and the bottom of the stack
                 I  if input is truthy, return the filtered range, else return the input (special-cases 0)

Answer 15

PHP, 41 Bytes

prints underscore separated values

for(;100>$a=&$argn;$a+=$a?9:ERA)echo$a._;

ERA is the shortest constant in PHP with the value 131116. You can replace it with the boring alternative 100 or end the program with die

Online Version

Answer 16

Brain-Flak, 54 52 bytes

{<>((((()()())){}{})()){({}<(({})<>({}))><>[()])}}<>

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My first foray with Brain-Flak, and I think I've done pretty well. Anyone with more experience have advice?

How It Works:

{ Don't do anything if input is 0
  <>((((()()())){}{})()) Switch to other stack and add 9 and 10
                         10 is the counter, 9 is to add to the current num
  { While counter
     (
       {} Pop the counter
       <(({})<>({}))> Get a copy of the 9, switch to the other stack and add it to a copy of the top of it. Use <...> to make it return 0
       <>[()] Switch to the other stack and decrement the counter
     ) 
  }
}<> Switch to the stack with the values on it

Answer 17

Jelly, 12 bytes

ȷ2ḶDS$ÐL⁼¥Ðf

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Answer 18

PHP, 35

print_r(range($argn,!!$argn*99,9));

Creates the range [$argn, 100) with a step of 9 as array and prints it. If the input is 0 it creates the range [0,0] => array(0).

Answer 19

Python, 48 51 bytes

3 bytes saved thanks to @WheatWizard

lambda n:[x for x in range(100)if~-n==~-x%9or x==n]

Answer 20

R, 23 bytes

pryr::f(x+0:(10*!!x)*9)

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The TIO link uses function(x) instead of pryr::f, since TIO doesn't have the pryr package installed.

Answer 21

Pyke, 6 bytes (old version)

Working commit

TXU#sq

Explanation:

TX     -   10**2
  U    -  range(^)
   #   - filter(^)
    s  -   digital_root(^)
     q -  ^==input

Answer 22

Ruby, 25 bytes

->a{[*a.step(99,9)]*a|[]}

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Answer 23

C (gcc), 55 bytes

f() need not actually be called with any argument; the n is just there instead of outside the function to save a byte.

f(n){for(scanf("%d",&n);n&&n<100;n+=9)printf("%d ",n);}

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Answer 24

Charcoal, 14 11 bytes

Ｉ∧Ｎ⁺Ｉθ×⁹…¹¹

Try it online! Link is to verbose version of code. Edit: Saved 2 bytes by not printing anything for zero input and 1 byte by using vectorising operations 3 bytes thanks to @ASCII-only. Explanation:

         ¹¹ Literal 11
        …   Range
       ⁹    Literal 9
      ×     Vector multiply
     θ      (First) input
    Ｉ       Cast to number
   ⁺        Vector add
  Ｎ         Input digit as a number
 ∧          Logical AND
Ｉ           Cast to string
            Implicitly print on separate lines

Answer 25

Pyt, 11 bytes

?56+ř⁻9*+:;

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?               implicit input; if input is truthy:
 56+            push 11
    ř⁻          řangify; decrement
      9*        multiply by 9
        +       add input
         :;     otherwise, do nothing; implicit print

Answer 26

Vyxal H, 6 bytes

ʁ'∑∑?=

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Port of kalsowerus's 05AB1E answer.

A port of Kip the Malamute's Pyt answer would be 8 bytes:

:[11ʁ9*+

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Explanation

ʁ'∑∑?=  # H flag pushes 100
ʁ'      # Filter range(100) by:
  ∑∑    #  Digit sum twice
    ?=  #  Equals the input

:[11ʁ9*+  # Implicit input
:         # Duplicate
 [        # If input is not 0:
  11ʁ     #  Push range(11)
     9*   #  Multiply by 9
       +  #  Add to input
          # (Otherwise, do nothing)

Answer 27

Julia 0.6, 18 bytes

I use a ternary to catch the 0 case, and a range n:9:99 to create the numbers. In julia a range is an AbstractVector and can be used in place of an actual Vector of numbers in most cases, but it will just print as 1:9:91 which doesn't satisfy the challenge, so I wrap it in [_;] to collect the contents into a Vector.

n->n>0?[n:9:99;]:0

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Answer 28

Perl 5, 25 + 1 (-n) = 26 bytes

say;!$_||($_+=9)>99||redo

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Answer 29

Clojure, 33 bytes

(fn[n](if(> n 0)(range n 100 9)))

Answer 30

Clojure, 38 bytes

(defn f[n](if(pos? n)(range n 100 9)))

or as anonymous function which is 29 bytes

(#(if(pos? %)(range % 100 9))n)

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thanks @steadybox