Lowest digit addition generator

Question

A digit addition generator of an integer n is any integer x that satisfy the equation x + s(x) = n, with s(x) being the sum of the digits of x. (We will work under base 10 for convenience.)

For example, a digit addition generator for 29 would be 19, because 19 + (1 + 9) = 29. Some numbers have more than one generator. An example might be 216, which has generators of 198 and 207.

Your objective is to generate the sequence a_n where a_i is the lowest digit addition generator of every non-negative integer i, and anything other than a non-negative integer if there is none for i. The non-negative terms in your result should match the sequence A096234. You may find this paper related to the challenge.

Fewest bytes win; standard rules apply.

Answer 1

Vyxal gM, 28 bits^v2, 3.5 bytes

'∑+?=

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Outputs an empty list for no result. I think (spoiler I did) I ninja'd someone in the process of writing this lol.

Explained

'∑+?=
'      # filter the range [0, n] by:
 ∑+    #   sum of digits + x
   ?=  #   equals input? 
# g flag prints minimum. 

Answer 2

05AB1E, 7 bytes

Ý.ΔDªOQ

Outputs -1 if there is no result.

Try it online or verify a few more test cases.

Explanation:

Ý        # Push a list in the range [0, (implicit) input-integer]
 .Δ      # Find the first value of this list that's truthy for, or -1 if none are:
   D     #  Duplicate the current value
    ª    #  Convert the first value to a list of digits,
         #  and append the duplicated number to this list
     O   #  Sum the list together
      Q  #  Check whether it's equal to the (implicit) input-integer
         # (after which the result is output implicitly)

Answer 3

PARI/GP, 44 bytes

f(n,i)=if(i>n,x,i+sumdigits(i)-n,f(n,i+1),i)

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Returns x when there is no solution.

Answer 4

Japt, 8 bytes

Outputs -1 for no result.

ôÈ+ìxÃbU

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(Eventually) outputs undefined for no result.

@¶X+ìx}a

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ôÈ+ìxÃbU     :Implicit input of integer U
ô            :Range [0,U]
 È           :Pass each through the following function
  +          :  Add
   ì         :  Convert to digit array
    x        :  Reduce by addition
     Ã       :End function
      bU     :First 0-based index of U

@¶X+ìx}a     :Implicit input of integer U
@            :Function taking an integer X as argument
 ¶           :  Test U for equality with
  X+         :    Add to X
    ì        :      Convert X to digit array
     x       :      Reduce by addition
      }      :End function
       a     :Get the first integer >=0 that returns true

Answer 5

APL (Dyalog Extended), 21 bytes

Anoymous tacit prefix function. Requires 0-indexing (⎕IO←0).

⍸⊢<\⍤=0,⍨⍳+1⊥¨⍎¨∘⍕¨∘⍳

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⍸ where do we find that

⊢ the argument

<\ is the first (lit. cum. right-ass. less red.)

⍤ that:

 = equals

0,⍨ zero appended to

⍳ the range

+ plus

1⊥ the sum (lit. base-1 eval.)

¨ of each of

⍎ the evaluation

¨ of each

⍤ of:

 ⍕ the stringification

 ¨ of each

 ⍤ of:

  ⍳ the range

Answer 6

C (clang), 58 bytes

loops forever if there's no solution

d,s;f(*i,n){for(*i=s=0;n-s;)for(s=d=++*i;d;d/=10)s+=d%10;}

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C (clang), 70 bytes

For comparison, this returns negative values when there's no solution.

g,o,l;f(*i,n){for(*i=g=-n;g++;*i=n+l?*i:-g)for(l=o=g;o;o/=10)l+=o%10;}

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Answer 7

C++ (gcc), ^{71 \$\cdots\$ 66} 64 bytes

[](int&n){for(int m=n,d=n=0,t;m-d;)for(d=t=++n;t;t/=10)d+=t%10;}

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Saved 6 bytes thanks to c--!!!

Answer 8

JavaScript, 49 bytes

Loops forever, in theory, if there's no result but, in practice, throws a recursion error.

n=>(g=x=>n-eval([x,...x+``].join`+`)?g(++x):x)`0`

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Answer 9

Raku, 34 bytes

{first 0..$^n: {$_+.comb.sum==$n}}

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Returns the first element of the range 0 through the input number $^n/$n such that the element $_ plus the sum of its digits .comb.sum is equal to the input number. If there is no such number, Nil is returned.

Answer 10

R, 45 46 bytes

Edit: +1 byte to correctly handle input of zero, but then -1 byte thanks pajonk for removal of useless accidental space

\(n){while(n-F-sum(F%/%10^(0:F)%%10))F=F+1;+F}

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Tests increasing integers until it finds a solution, so loops forever if no solution exists.

Answer 11

Pyth, 8 bytes

x|msajdT

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Returns -1 for no result.

Explanation

x|msajdTdQQQ    # implicitly add dQQQ
                # implicitly assign Q = eval(input())
  m      Q      # map lambda d over range(Q):
     jdT        #   list the digits of d
    a   d       #   append d to this list
   s            #   sum the list
 |        Q     # short circuiting or, does nothing unless the input is 0, then [] -> 0
x          Q    # find the first index of Q in the mapped list (or xors with 0 for 0), returns -1 if not found

Answer 12

Arturo,  40  35 bytes

$->x[0while[x<>+∑digits<=<=]->1+]

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Causes an infinite loop when there is no result.

$->x[                  ; a function taking an integer x
    0                  ; push 0 to stack
    while[x<>          ; while x doesn't equal...
        <=<=           ; duplicate top of stack twice
        +∑digits       ; digit sum then add
    ]                  ; end while condition
    ->1+               ; increment top of stack (while body)
]                      ; end function

Answer 13

Elixir, 74 72 bytes

import Enum
c=& &1-?0
r=&find(1..&1,fn x->&1==x+sum map'#{x}',c end)||-1

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Answer 14

Wolfram Language (Mathematica), 89 53 49 44 bytes

#&@@Pick[r=Range@#,r+Tr/@IntegerDigits@r,#]&

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Thanks to @ovs and @att!

Answer 15

Julia, 42 37 bytes

~n=argmax(x->x+sum(digits(x))==n,0:n)

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-5 bytes thanks to MarcMush: use argmax instead of findfirst

Answer 16

Java, 85 bytes

n->{for(int i=-1;i++<n;)if((i+"").chars().map(d->d-48).sum()+i==n)return i;return-1;}

Outputs -1 if there is no result.

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Explanation:

n->{                   // Method with integer as both parameter and return-type
  for(int i=-1;i++<n;) //  Loop `i` in the range (-1,n]:
    if((i+"")          //   Convert `i` to a String
       .chars()        //   Then to an IntStream of codepoints
       .map(d->d-48)   //   Then to an IntStream of digits
       .sum()          //   And sum those digits together
             +i        //   Add integer `i`
               ==n)    //   And if it's equal to input `n`:
      return i;        //    Return `i` as result
  return-1;}           //  If no result is found, return `-1` instead

Using a recursive function which throws a StackOverflowError when there is no result is also 85 bytes:

n->f(n,0)int f(int n,int i){return(i+"").chars().map(d->d-48).sum()+i==n?i:f(n,i+1);}

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Answer 17

Charcoal, 13 bytes

ＮθＩ⌊Φ⊕θ⁼θ⁺ιΣι

Try it online! Link is to verbose version of code. Takes i as an input and outputs None if no generator exists. Explanation: Brute force approach.

Ｎθ              First input as an integer
      θ         First input
     ⊕          Incremented
    Φ           Filter on implicit range
          ι     Current value
         ⁺      Plus
            ι   Current value
           Σ    Digit sum
       ⁼        Equals
        θ       First input
   ⌊            Take the minimum
  Ｉ             Cast to string
                Implicitly print

23 bytes for a less inefficient version:

ＮθＩ⌊Φ…·⁻θ×⁹Ｌθθ⁼θ⁺ιΣ⌈⟦⁰ι

Try it online! Link is to verbose version of code. Explanation: Starts checking for generators starting from i minus 9 times the number of digits of i.

28 bytes for a more efficient version:

ＮθＩ⌊ΦＥＬθ⁻⁻θ﹪×⁵θ⁹×⁹ι⁼θ⁺ιΣ⌈⟦⁰ι

Try it online! Link is to verbose version of code. Explanation: As above, but only checks for generators which when doubled are equivalent to i modulo 9 (because i equals the generator plus its digit sum but both the generator and the digit sum are equivalent modulo 9) although it actually computes the maximum possible generator by subtracting 5i reduced modulo 9 from i.

Answer 18

C# (.NET Core), 73 72 bytes

n=>{for(int i=-1;i++<n;)if((i+"").Sum(d=>d-'0')+i==n)return i;return-1;}

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This is a C# port of this answer. All the kudos goes to Kevin Cruijssen.

I really like that .NET's sum is taking a lambda to avoid the common map followed by sum!

Answer 19

Python 3, 63 bytes

g=lambda n:min(x for x in range(n)if x+sum(map(int,str(x)))==n)

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Answer 20

Python 3, 56 bytes

f=lambda n,o=0:(o+sum(map(int,str(o)))==n)*o or f(n,o+1)

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Answer 21

Fig, \$11\log_{256}(96)\approx\$ 9.054 bytes

[KFax'=#x+S

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[            # The first item from
   a         # the range from 1 to
    x        # the input,
  F  '       # filtered by:
          S  #   the digit sum of n
         +   #   plus n
      =      #   is equal to
       #x    #   the program's input,
 K           # sorted (smallest to largest).

Answer 22

Swift, 84  80 bytes

{n in(0...n).map{($0,"\($0)".reduce($0){$0+$1.hexDigitValue!})}.first{$1==n}!.0}

Takes an Int in and returns an Int.

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Ungolfed:

{ (n: Int) -> Int in
  (0...n) // a sequence of all integers from 0 to n, inclusive
    // the parens are needed to avoid parsing as 0...(n.map)
    .map { i in // transform each one
      ( // return a tuple of:
        i, // 0. the integer itself
        "\(i)".reduce(i) { $0 + $1.hexDigitValue! } // 1. the digit sum plus the number
        // technically, wholeNumberValue would be more correct than hexDigitValue
      )
    }
    .first { $1 == n }! // find the first one where the sum is the input
    .0 // and return the number that gave that sum
}

Answer 23

Perl 5 -MList::Util=sum -pa, 29 bytes

$_=0;$_++while"@F"-sum$_,/./g

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Answer 24

Desmos, 66 bytes

I=[0...n]
f(n)=I[[i+∑_{k=0}^imod(floor(i/10^k),10)fori=I]=n].min

Returns undefined if no solution exists.

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Try It On Desmos! - Prettified

Answer 25

Ruby, 46 bytes

1.step{|c|p (1..c).find{|x|x+x.digits.sum==c}}

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Answer 26

Thunno 2 M, 5 bytes

æDS+=

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Explanation

æDS+=  # Implicit input
æ      # Filter [1..input] by:
 D     #  Duplicate
  S    #  Sum digits
   +   #  Add
    =  #  Equals input?
       # Take the minimum
       # Implicit output

Answer 27

Haskell, 52 bytes

a n=find((==n).ap(+)(sum.map digitToInt.show))[1..n]

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Answer 28

Scala, 71 bytes

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n=>(0 to n).find(i=>(i.toString.map(_.asDigit).sum+i)==n).getOrElse(-1)