ID generator with no matching contiguous pairs

Question

Given 2 inputs (m=minimum starting id, n=number of ids), generate a sequential list of ids, starting at >=m, made up of the digits 0-9 where no 2 contiguous numbers are the same, e.g. 1232 is ok, 1233 is not (contains 2 '3's next to each other).

Example

for m=985, n=6, below shows what ids would be generated/skipped

...
985 - ok, 1st
986 - ok, 2nd
987 - ok, 3rd
# skip 0988 (contains '88')
989 - ok, 4th
# Skip 099x (contains '99')
# Skip 100x (contains '00')
1010 - ok, 5th
# Skip 1011 (contains '11')
1012 - ok, 6th
...

Output

Each ID should be printed on a new line, as for the above sequence:

985
986
987
989
1010
1012

Rules

Standard code-golf rules, lowest byte count wins

Answer 1

Brachylog, 11 10 bytes

{≤ṫẹ~ḅẉ}ᶠ⁾

Input is a list of two numbers. Try it online!

Explanation

The built-in ḅ takes a list or string, like "1000220", and splits it into blocks of equal adjacent elements, like ["1","000","22","0"]. In this program, I apply the ~ operator to it, so it works in reverse: it takes a list of strings, checks that each string consists of repetitions of a single character and neighboring strings have different characters, and concatenates the list. The predicate ≤ enumerates numbers starting from the first input in increasing order, and I check a condition on them, printing those that satisfy it and stopping when I've found enough.

{≤ṫẹ~ḅẉ}ᶠ⁾  Input is a pair, say [M=988,N=3].
{      }ᶠ⁾  Apply this predicate to M and compute the first N results.
 ≤          Take a number that is greater than or equal to M (first up is 988).
  ṫ         Convert it to string: "988"
   ẹ        Split it into substrings of length 1: ["9","8","8"]
     ~ḅ     Apply ḅ in reverse: fails, try next number.
       ẉ    If ḅ succeeds, print the resulting string and a newline.
            This counts as a result of the predicate.

Answer 2

05AB1E, 9 bytes

µÐÔQi=¼}>

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Exlpanation

µ           # loop until counter equals n
 Ð          # triplicate current value (initially m)
  Ô         # connected uniqueified on the copy at the top of the stack
   Q        # compare to unmodified for equality
    i  }    # if true
     =      # print current value while leaving it on the stack
      ¼     # increment the counter
        >   # increment current value

Answer 3

Java 8, 83 bytes

(m,n)->{for(;n>0;m++)if(!(m+"").matches(".*(.)\\1.*")){System.out.println(m);n--;}}

Explanation:

Try it online.

(m,n)->{                      // Method with two integer parameters and no return-type
  for(;n>0;                   //  Loop as long as `n` is larger than 0
      m++)                    //    After every iteration: increase `m` by 1
    if(!(m+"").matches(".*(.)\\1.*")){
                              //   If there are no repeated adjacent digits:
      System.out.println(m);  //    Print `m`
      n--;}}                  //    And decrease `n` by 1

Answer 4

Jelly, 6 bytes

DIẠµ#Y

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How it works?

DIẠµ#Y – Full program. Argument: two integers, X and Y.

   µ#  – Return the first Y integers higher than or equal to X that satisfy:
 I     – The increments ...
D      – ... Of their base-10 digits ...
  Ạ    – ... Are all non-0. 
     Y – Join the result by newlines.

Answer 5

PowerShell, 59 bytes

param($m,$n)for(;$n){if("$m"-notmatch"(.)\1"){$m;$n--}$m++}

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Basically similar to the other answers. Loops so long as we have numbers left to print (for(;$n)), tests whether we have a regex match against a double-digit and if not it puts that on the pipeline and decreases $n. Then we increment $m and loop again. The elements are picked up from the pipeline and the implicit Write-Output gives us a newline-separated output for free.

Answer 6

R, 111 92 71 bytes

function(m,n)while(F<n){if(!grepl("(.)\\1",m)){cat(m,'
');F=F+1}
m=m+1}

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Uses grepl to search for repeated digits.

Answer 7

Jelly, 9 bytes

DŒgZLỊµ#Y

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If not for the output restriction, Y can be removed.

Answer 8

C, 73 bytes

k;f(s,n){for(;n--;printf("%d\n",s++))for(k=s;k;)k=k%10-k/10%10?k/10:++s;}

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Answer 9

Perl 6, 56 bytes

{.put for grep({none .comb Zeq.comb.skip},$^a..*)[^$^b]}

Try it

Expanded:

{  # bare block lambda with placeholder params $a $b

  .put for

  grep(

    {
      none          # create a none junction (True if all are False)

        .comb       # the input split into digits
        Zeq         # Zip using &infix:«eq» operator
        .comb.skip  # the input split into digits starting at second one
    },

    $^a .. *        # a Range starting at the first argument

  )[ ^ $^b ]        # take the first $b values
}

Answer 10

Retina, 34 bytes

.0A`
"$+"{\/(.)\1/{`.+
*
)C`
.+
*

Try it online! Takes n and m as input on separate lines. Explanation:

.0A`

Turn off automatic output and delete n from the work buffer.

"$+"{

Repeat n times.

\

Print the value of m at the end of the following looped group.

/(.)\1/{`

Loop while there are contiguous digits.

.+
*

Convert to unary.

)C`

Count the number of empty strings, which is one more than the number of characters, thus adding 1 and converting back to decimal. This ends the inner loop.

.+
*

Having printed m, add 1 again in the same way. (The last line doesn't need a C` because it's the default stage type for the last line.) The outer loop is implicitly ended.

Note that converting to unary and back to decimal is a little slow; for 39 bytes, a version that doesn't do any conversion:

.0A`
"$+"{\/(.)\1/+`.+
$.(*__
.+
$.(*__

Try it online! Explanation: $.( calculates the length of the rest of the substitution, conveniently without actually expanding it; since the length of *_ is implicitly the matched value and the length of _ is of course 1 this just increments the value.

Answer 11

Perl 5.10.0 + -n, 40 39 bytes

for(;$i;$_++){!/(.)\1/&&$i--&&say}$i=$_

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-1 byte thanks to Xcali

Input on two lines, first n, then m. Make sure that there is NO newline after m:

(echo '6'; echo -n '985') | perl -nE 'for(;$i;$_++){!/(.)\1/&&$i--&&say}$i=$_'

Answer 12

Perl 5, -ln 33 bytes

Enter 2 lines on STDIN, first starting id then count

#!/usr/bin/perl -ln
$n-=!/(.)\1/&&say,$_++while$n.=<>

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Answer 13

Pyth, 12 bytes

j.f.A.+jZTEQ

Try it here

Answer 14

Python 2, 80 73 bytes

^{-7 bytes thanks to Mr. Xcoder}

m,n=input()
while n:
 if~-sum(i*2in`m`for i in`99**9`):print m;n-=1
 m+=1

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Answer 15

Haskell, 70 bytes

-19 bytes thanks to nimi.

s#n=unlines$map show$take n$filter(and.(zipWith(/=)=<<tail).show)[s..]

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17 bytes dedicated to separating by newlines.

Answer 16

Stax, 9 8 bytes^CP437

ç@F6╕↔┤ú

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Happen to coincide with @Mr.Xcoder's algorithm in his Jelly answer.

-1 byte per comment by @recursive.

Explanation

Uses the unpacked format to explain.

{$:-|A}gn
{     }gn    Generator, generate given number of values, using block as a filter
                 And implicit output, one element on a line.
             In Stax, only 0 and [] is falsy.
              
 $           Convert the number to string
  :-         Difference between contiguous digit
    |A       All of them are truthy (may use `:*` as well)

Answer 17

Haskell, 94 93 91 bytes

-1 byte thanks to Laikoni
-2 bytes thanks to Zgarb

import Data.List
a!0=[]
a!b|all(null.tail)$group$show a=show a++'\n':(a+1)!(b-1)|c<-a+1=c!b

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First Haskell golf.

Answer 18

JavaScript (ES6), 50 bytes

Takes input in currying syntax (m)(n).

m=>g=n=>n?/(.)\1/.test(m++)?g(n):~-m+`
`+g(n-1):''

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Answer 19

C# (.NET Core), 91 bytes

m=>n=>{for(;n-->0;System.Console.WriteLine(m++))for(int k=m;k>0;k=k%10!=k/10%10?k/10:++m);}

Port of @Steadybox' C answer.

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Answer 20

Perl 5 (-nl), 42 bytes

$l=<>;do{!/(.)\1/&&$l--&&say}while++$_&&$l

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Answer 21

AWK, 90 bytes

{for(j=$1;k<$2;j++){for(a=b=d;++b<=(n=split(j,A,""))&&a!=c=A[b];a=c);if(n<b&&++k)print j}}

Try it online!

This is a lot uglier than I thought it would be. I discovered that AWK only has back-substitution in its gensub function which would make it not very efficient to use for this application.

So this is a very straightforward approach of incrementing an ID counter, splitting the counter at each character, looping through the characters to see if any character is the same as the previous character. If no repeated characters are found, print the ID counter and increment the number of IDs printed.

I will try to refrain from making a statement along the lines of "AWK should have the ability to do more robust regular expressions", but it will be difficult.

Answer 22

Bash + GNU utilities, 35

seq $1 inf|egrep -v '(.)\1'|sed $2q

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Answer 23

Ruby, 47 bytes

->s,n{/(.)\1/!~s&&p(s)&&n-=1;s.next!;n>0&&redo}

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If you don't like the quote marks, it's 50 bytes

->s,n{puts s if/(.)\1/!~s&&n-=1;s.next!;n>0&&redo}

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Answer 24

Pip, 22 bytes

--aLb{W`(.)\1`N++a0Pa}

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Explanation

--aLb{W`(.)\1`N++a0Pa}
                        a,b are command-line arguments
--a                     Decrement a
   Lb{               }  Loop b times:
       `(.)\1`           Regex matching a character followed by itself
      W       N++a       While that regex matches in incremented a:
                  0       No-op
                   Pa    When the while loop exits, a is a valid ID; print it