Count without 3

Question

Background

When I was in elementary school, we used to play a game in math class that goes as follows.

All kids sit in a big circle and take turns counting, starting from 1.

However, the following numbers must be skipped while counting:

• Numbers that are multiples of 3.
• Numbers that have a 3 in its decimal representation.

The first 15 numbers the kids should say are

1 2 4 5 7 8 10 11 14 16 17 19 20 22 25

Whenever somebody gets a number wrong – says a number that isn't in the sequence or skips a number that is – he's removed from the circle. This goes on until there's only one kid left.

Task

You're bad at this game, so you decide to cheat. Write a program or a function that, given a number of the sequence, calculates the next number of the sequence.

You don't have to handle numbers that cannot be represented using your language's native numeric type, provided that your program works correctly up to input 251 and that your algorithm works for arbitrarily large inputs.

Input and output can use any convenient base.

Since you have to conceal your code, it must be as short as possible. In fact, this is code-golf, so the shortest code in bytes wins.

Test cases

  1 ->   2
  2 ->   4
 11 ->  14
 22 ->  25
 29 ->  40
251 -> 254

Answer 1

Brachylog, 10 bytes

<.='e3:I'*

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Explanation

(?)<.                Output > Input
    .=               Assign a value to the Output
    . 'e3            3 cannot be an element of the Output (i.e. one of its digits)
        3:I'*(.)     There is no I such that 3*I = Output

Answer 2

JavaScript (ES6), 30 bytes

f=n=>++n%3*!/3/.test(n)?n:f(n)

Answer 3

J, 24 bytes

3(]0&({$:)~e.&":+.0=|)>:

Straight-forward approach that just iterates forward from input n until it finds the next number that is valid by the rules.

Forms five smileys, $:, :), 0=, =|, and >:.

Usage

   f =: 3(]0&({$:)~e.&":+.0=|)>:
   (,.f"0) 1 2 11 22 29 251
  1   2
  2   4
 11  14
 22  25
 29  40
251 254

Explanation

3(]0&({$:)~e.&":+.0=|)>:  Input: integer n
                      >:  Increment n
3                         The constant 3
 (                   )    Operate dyadically with 3 (LHS) and n+1 (RHS)
                    |       Take (n+1) mod 3
                  0=        Test if equal to 0
             &":            Format both 3 and n+1 as a string
           e.               Test if it contains '3' in str(n+1)
                +.          Logical OR the results from those two tests
  ]                         Right identity, gets n+1
   0&(   )~                 If the result from logical OR is true
       $:                     Call recursively on n+1
      {                       Return that as the result
                            Else act as identity function and return n+1

Answer 4

Python 2, 73 66 43 bytes

Thanks to xnor for telling me I was being silly by using 2 variables, and thanks to Mitch Schwartz too.

x=~input()
while'3'[:x%3]in`x`:x-=1
print-x

Answer 5

Perl, 19 bytes

18 bytes code + 1 for -p.

++$_%3&&!/3/||redo

Usage

perl -pe '++$_%3&&!/3/||redo' <<< 8
10

perl -pe '++$_%3&&!/3/||redo' <<< 11
14

Answer 6

05AB1E, 11 bytes

[>Ð3ås3Ö~_#

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Explanation

               # implicit input
[              # start loop
 >             # increase current number
  Ð            # triplicate
          #    # break loop IF
         _     # logical negation of
   3å          # number has one or more 3's in it
        ~      # OR
     s3Ö       # number % 3 == 0

Answer 7

Java 8, 57 56 55 50 bytes

Thanks to @Numberknot for 1 byte Thanks to @Kevin Cruijssen for 5 bytes

i->{for(;++i%3<1|(i+"").contains("3"););return i;}

This is a Function<Integer, Integer>

Explanation

Naive implementation that simply increments until it reaches an acceptable number.

Test Class

public class CodeGolf {

    public static void main(String[] args) {
        Function<Integer, Integer> countingGame = i->{for(;++i%3<1|(i+"").contains("3"););return i;};
        int val = 1;
        for (int i = 0; i < 10; i++) {
            System.out.print(val + " ");
            val = countingGame.apply(val);
        }
    }

}

Output of Test Class:

1 2 4 5 7 8 10 11 14 16

Answer 8

Python 2, 49 44 42 bytes

f=lambda x:'3'[:~x%3]in`~x`and f(x+1)or-~x

The other Python entry beats this (edit: not any more :-D), but I posted it because I rather like its recursive approach. Thanks to Mitch Schwarz and Erik the Golfer for helping me make this shorter.

Answer 9

Japt, 18 bytes

°U%3*!Us f'3 ?U:ßU

Test it online

I finally have a chance to use ß :-)

How it works

                    // Implicit: U = input integer
°U%3                // Increment U, and take its modulo by 3.
     !Us f'3        // Take all matches of /3/ in the number, then take logical NOT.
                    // This returns true if the number does not contain a 3.
    *               // Multiply. Returns 0 if U%3 === 0  or the number contains a 3.
             ?U     // If this is truthy (non-zero), return U.
               :ßU  // Otherwise, return the result of running the program again on U.
                    // Implicit: output last expression

Answer 10

PowerShell v2+, 46 bytes

for($a=$args[0]+1;$a-match3-or!($a%3)){$a++}$a

Takes input $args[0], adds 1, saves into $a, starts a for loop. The conditional keeps the loop going while either $a-match3 (regex match) -or $a%3 is zero (the ! of which is 1). The loop simply increments $a++. At the end of the loop, we simply place $a on the pipeline, and output via implicit Write-Output happens at program completion.

Examples

PS C:\Tools\Scripts\golfing> 1,2,11,22,29,33,102,251,254|%{"$_ --> "+(.\count-without-three.ps1 $_)}
1 --> 2
2 --> 4
11 --> 14
22 --> 25
29 --> 40
33 --> 40
102 --> 104
251 --> 254
254 --> 256

Answer 11

R, 46 bytes

n=scan()+1;while(!n%%3|grepl(3,n))n=n+1;cat(n)

Answer 12

Haskell, 50 48 bytes

f n=[x|x<-[n..],mod x 3>0,notElem '3'$show x]!!1

Try it on Ideone. Saved 2 bytes thanks to @Charlie Harding.

Alternative: (50 bytes)

g=f.(+1)
f n|mod n 3<1||(elem '3'.show)n=g n|1<3=n

Answer 13

Labyrinth, 117 102 bytes

?       """""""""""_
):_3    (         0/{!@
;  %;:}_';:_3-_10 1
"  1            %;_
""""_""""""""{;;'

Try it online!

Labyrinth is a two-dimensional, stack-based programming language and at junctions, direction is determined by the top of the stack (positive goes right, negative goes left, zero goes straight). There are two main loops in this programs. The first mods the integer input by 3 and increments if 0. The second repeatedly checks if the last digit is 3 (by subtracting 3 and modding by 10) and then dividing by 10 to get a new last digit.

Answer 14

Lua, 58 Bytes

i=...+1while(i%3==0or(i..""):find"3")do i=i+1 end print(i)

Answer 15

Pyke, 13 bytes

Whii3%!3`i`{|

Try it here!

              - i = input
W             - do:
 hi           -  i += 1
   i3%!       -    not (i % 3)
            | -   ^ or V
       3`i`{  -    "3" in str(i)
              - while ^

Answer 16

C#, 56, 51 bytes.

This is surprisingly short for a C# answer!

x=>{while(++x%3<1|(x+"").Contains("3"));return x;};

Answer 17

Pyth, 11 bytes

f&-I`T3%T3h

Try it online: Demonstration or Test Suite

Explanation:

f&-I`T3%T3hQ   implicit Q at the end
f         hQ   find the smallest integer T >= input + 1 which fulfills:
  -I`T3           T is invariant under removing the digit 3
 &                and
       %T3        T mod 3 leaves a positive remainder

Answer 18

Excel, 79 66 bytes

^{It Taylor Rained 13 bytes}

=MIN(LET(r,ROW(A:A),FILTER(r,ISERR(FIND(3,r))*MOD(r,3)*(r>A1)>0)))

Input is in the cell A1. Output is wherever the formula is.

The LET() function allows you to define variables for later reference so ti works in pairs of variable,value. The last argument is unpaired as it is the output.

• r,ROW(A:A) creates an array of 1 to whatever the last row is. In the latest version of Excel, that's 1,048,576.
• g,FILTER(r,~*~*~>0) takes that array and filters just for the values that aren't multiples of 3 and don't have a 3 in them. This done is by multiplying:
  • ISERR(FIND(3,r)) which returns TRUE if if a 3 is not in the number and FALSE if it is
  • MOD(r,3) which returns 0 if it's a multiple of 3 and some positive integer if it isn't
  • (r>A1) creates an array of TRUE / FALSE where the first TRUE value will be the first term that's larger than the input.
  • When you do math on booleans, Excel uses TRUE=1 and FALSE=0 so the formula ~*~*~>0 only returns TRUE when the three pieces are TRUE (no 3 in number), some positive integer (not a multiple of 3), and TRUE (larger than the input).
• MIN(LET(~)) finds the smallest value in the filtered array which only includes the "without 3" numbers larger than the input so the first value in the array is the next value in the sequence.

Answer 19

Python 3, 49 48 bytes

-3 bytes thx to The Thonnu, I missed some spaces and swapping else x+1 and else-~x allowed another space to be taken out, not

+2 bytes bc I didn't know I need to count f= on a recursive function, thx xnor :)

seems clunky but i cant see more golfs, please lmk if you see something!

f=lambda x:f(x+1)if x%3>1or'3'in str(x+1)else-~x

Try it online!

Answer 20

GolfSharp, 43 bytes

m=>r(m,m).w(n=>n%3>0&!n.t().I("3")).a()[1];

Answer 21

Ruby, 47 bytes

i=gets.to_i;i while(i+=1)%3==0||"#{i}"=~/3/;p i

I really feel like this can be golfed further.

Answer 22

MATL, 14 bytes

`Qtt3\wV51-hA~

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Explanation

`       % Do...while
  Q     %   Add 1. Takes input implicitly in the first iteration
  tt    %   Duplicate twice
  3\    %   Modulo 3
  wV    %   Swap, string representation
  51-   %   Subtract 51, which is ASCII for '3'
  h     %   Concatenate
  A~    %   True if any result was 0. That indicates that the number
        %   was a multiple of 3 or had some '3' digit; and thus a 
        %   new iteration is needed

Answer 23

PHP, 60 55 54 46 bytes

Thanks to @user59178 for shaving off a few bytes, @AlexHowansky for a byte, @Titus for another few bytes

for(;strstr($i=++$argv[1],51)|$i%3<1;);echo$i;

Called from command line with -r. Naive method that loops while the number is a multiple of 3, or has 3 in its digits.

Answer 24

Jelly, 11 bytes

D;Æf3ḟµ‘#2ị

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How it works

D;Æf3ḟµ‘#2ị  Main link. Argument: n

      µ      Combine the links to the left into a chain.
       ‘#    Execute the chain for k = n, n + 1, n + 2, ... until n + 1 matches
             were found. Yield the array of all n + 1 matches.
D            Decimal; yield the array of k's decimal digits.
  Æf         Yield the array of k's prime factors.
 ;           Concatenate both.
    3ḟ       Filter false; remove digits and factors from [3].
             This yields [3] (truthy) if neither digits nor factors contain 3,
             [] (falsy) if they do.
         2ị  Extract the second match. (The first match is n.)

Answer 25

PHP, 47 41 bytes

inspired by Xanderhall, but the latest idea finally justifies an own answer.

while(strstr($n+=$n=&$argn%3,51));echo$n;

or

while(strpbrk($n+=$n=&$argn%3,3));echo$n;

This takes advantage from the fact that the input is also from the sequence: For $n%3==1, the new modulo is 2. For $n%3==2, the new modulo is 4-3=1. $n%3==0 never happens.

Run as pipe with -R or try them online.

Answer 26

APL (Dyalog Unicode), 33 28 27 19 bytes^SBCS

1∘+⍣{('3'∊⍕⍺)<×3|⍺}

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-6 thanks to Adám. -8 thanks to ngn.

Old Explanation:

1-⍨g⍣((×3|⊢)>'3'∊⍕)∘(g←+∘1)
                       +∘1  ⍝ curry + with 1, gives the increment function
                            ⍝ increments the left argument so we do not return the number itself
                    (g←   ) ⍝ assign to "g"
                   ∘        ⍝ compose g with the repeat
                 ⍕          ⍝ does parsing the argument to a string...
             '3'∊           ⍝ ...contain '3'?
        3|⊢                 ⍝ residue of a division by 3
      (×   )                ⍝ direction (0 if 0, 1 if greater, ¯1 is lower)
     (      >     )         ⍝ and not (we want the left side to be 1, the right side 0)
   g⍣                       ⍝ repeat "g" (increment) until this function is true ^
1-⍨                         ⍝ afterwards, decrement: inversed -

APL (Dyalog Extended), 23 17 bytes^SBCS

1∘+⍣(3(×⍤|>∊⍥⍕)⊣)

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Thanks to Adám. -6 thanks to ngn.

Old Explanation:

0+⍣(3(×⍤|>∊⍥⍕)⊢)⍢(1+⊢)⊢
0                       ⍝ the left argument (⍺)
 +⍣(3(×⍤|>∊⍥⍕)⊢)        ⍝ the left function (⍺⍺)
                 (1+⊢)  ⍝ the right function (⍵⍵)
                        ⍝     (increments its argument)
                      ⊢ ⍝ the right argument (⍵)
                        ⍝     (just returns the input)
                ⍢       ⍝ under:
                        ⍝     calls (⍵⍵ ⍵) first, which increments the input
                        ⍝     also (⍵⍵ ⍺) which gives 1
                        ⍝     then calls (⍺incremented ⍺⍺ ⍵incremented)
                        ⍝     afterwards, does the opposite of ⍵⍵, and decrements the result
  ⍣                     ⍝ ⍣ fixpoint: repeats the left operation until the right side is truthy
 +                      ⍝ calls + with ⍺incremented and the input (so, 1+input)
   (3(×⍤|>∊⍥⍕)⊢)        ⍝ right operation
    3                   ⍝ on its left, "3"
              ⊢         ⍝ on its right, the current iteration
      ×⍤|               ⍝ divisibility check: × atop |
        |               ⍝     starts with 3|⊢ (residue of ⊢/3)
      ×                 ⍝     then returns the direction (0 if 0, 1 if greater, ¯1 is lower)
          ∊⍥⍕           ⍝ contains 3:
            ⍕           ⍝    stringifies both its arguments (3 and ⊢)
          ∊⍥            ⍝    checks for membership
         >              ⍝ divisibility "and not" contains 3

Answer 27

Turing Machine Code, 390 bytes

0 * * r 0
0 _ _ l 1
1 1 2 l 2
1 2 3 r 0
1 3 4 l 2
1 4 5 l 2
1 5 6 l 2
1 6 7 l 2
1 7 8 l 2
1 8 9 l 2
1 9 0 l 1
1 * 1 l 2
2 * * l 2
2 _ _ r 3
3 1 1 r 4
3 4 4 r 4
3 7 7 r 4
3 2 2 r 5
3 5 5 r 5
3 8 8 r 5
3 _ _ l 1
4 1 1 r 5
4 4 4 r 5
4 7 7 r 5
4 2 2 r 3
4 5 5 r 3
4 8 8 r 3    
5 1 1 r 3
5 4 4 r 3
5 7 7 r 3
5 2 2 r 4
5 5 5 r 4
5 8 8 r 4    
* 0 0 r *
* 6 6 r *
* 9 9 r *
* 3 3 r 0
* _ _ * halt

Try it online.

Answer 28

Perl 6, 27 25 24 bytes

{max $_+1...{!/3/&$_%3}}

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Finds the first number larger than the input that doesn't have a three and has a remainder when moduloed by 3. I was hoping to do something fancy with the condition, like !/3/&*%3 but it doesn't work with the !. :(

Explanation:

{                      }   # Anonymous code block
     $_+1                  # From the input+1
         ...               # Get the series
            {         }    # That ends when
             !/3/            # The number does not contain a 3
                 &           # and
                  $_%3       # The number is not divisible by 3
 max                       # And get the last element of the series

Answer 29

APOL, 51 bytes

v(0 +(⧣ 1));w(|(!(%(⁰ 3)) c(t(⁰) '3')) ∆(0));-(⁰ 1)

Answer 30

GNU sed -E, 148 bytes

Could be easier with a language with a modulo operator. Or division. Or counting. Or some idea about what numbers are. sed has neither, but it can solve the challenge much like a elementary school kid would:

s/^/0123456789-0/
:1
s/((.)(.).*)\2$/\1\3/
:2
s/((.)(.).*)\2-/\1\30/
t2
/-.*3/b1
h
:3
s/((.)(.).*-.*)\3/\1x\2/
s/x0*x0*x//
t3
/x/!{g;b1}
g
s/.*-0*//

Try it online!