Hopping numbers

Question

TASK

print integers n, where 12 <= n <= 123456789, and all pairs of consecutive digits in n have the same positive difference between them (e.g. 2468 but not 2469).

NO INPUT.

Output:

12
13
14
15
16
17
18
19
23
24
25
26
27
28
29
34
35
36
37
38
39
45
46
47
48
49
56
57
58
59
67
68
69
78
79
89
123
135
147
159
234
246
258
345
357
369
456
468
567
579
678
789
1234
1357
2345
2468
3456
3579
4567
5678
6789
12345
13579
23456
34567
45678
56789
123456
234567
345678
456789
1234567
2345678
3456789
12345678
23456789
123456789

Rules

1. Standard loopholes apply.
2. no input

shortest code wins.

Credits anarchy golf

Answer 1

Jelly, 12 11 bytes

9œcḊẎIE$ÐfY

Try it online!

How it works

9œcḊẎIE$ÐfY  Main link. No arguments.

9            Set the argument and return value to 9.
   Ḋ         Dequeue; yield [2, ..., 9].
 œc          Take all k-combinations of [1, ..., 9], for each k in [2, ..., 9].
    Ẏ        Concatenate the arrays of k-combinations.
        Ðf   Filter the result by the link to the left.
     IE$         Compute the increments (I) of the combination / digit list and
                 tests if all are equal (E).
          Y  Join the results, separating by linefeeds.

Answer 2

Python 2, 81 bytes

k=71
exec"k+=1;r=range(k/8%9+1,10,k%8+1)\nif r[k/72:]:print`r`[1:k/24+2:3]\n"*576

Try it online!

My solution from anarchy golf. The idea is to iterate over all possible triples of length, start value, and step, which gives sorted outputs. The triple is encoded as a value r from 72 to 647, and the components are extracted as k/72, k/8%9, and k%8. Starting k high enough avoids single-digit numbers from being output.

xsot saved two bytes off this by replacing the range with a hardcoded string of digits '123456789'.

This was written under the constraint of a two second runtime limit. A slower strategy that filters numbers rather than generating these may be shorter.

Answer 3

C, 166 152 Bytes

p,l,d,i=11;main(){for(char s[10];i<=1e9;){sprintf(s,"%d",++i);p=0;for(l=strlen(s);--l>0;){d=s[l]-s[l-1];if(!p)p=d;if(p^d|d<1)break;p=d;}if(!l)puts(s);}}

6 bytes saved thanks to @KevinCruijssen!

8 bytes saved thanks to @JonathanFrech!

Try it online

The fully formatted version of the above code can be seen below.

#include <string.h>
#include <stdio.h>

int main( void )
{
int prev_diff, diff, len;
int num = 11;
char str[10];

while(num<123456789)
    {
    prev_diff = 0;
    sprintf(str,"%d",++num);
    len = strlen(str)-1;
    for( ; len > 0; len-- )
        {
        diff = str[len] - str[len-1];
        if( prev_diff == 0 )
            {
            prev_diff = diff;
            }
        if( prev_diff != diff || diff < 1 )
            {
            break;
            }
        prev_diff = diff;
        }
    if ( len == 0 )
        {
        puts(str);
        }
    }
}

Answer 4

Jelly, 14, 13 bytes

DIµEȧ>0Ȧ
Ç77#

Try it online!

One byte saved thanks to @MrXcoder!

This is extremely inefficient, so it'll time out on TIO, but if it ever finishes, it will produce the correct output. You can try it with smaller numbers here: Try it online!

Explanation:

            # Helper link:
            #
D           # The Digits of 'n'
 I          # The increments (deltas, differences) between digits
  µ         # Give that array as an argument to the rest of this link:
   E        # Are all elements equal?
    ȧ       #   AND
     >0     # Are all elements greater then 0?
       Ȧ    # And is there at least one element?
            # (in the differences, so that numbers < 10 are false)
            #
            # Main link:
            #
 77#        # Return the first 77 'n's for which
Ç           #   The helper link is truthy

Answer 5

Mathematica, 79 bytes

Select[Range@123456789,Min[s=Union@Differences@IntegerDigits@#]>0&&Tr[1^s]==1&]

Try it online! with a lower number because it is very slow

here is another approach that constructs all the numbers in 1sec

Mathematica, 123 bytes

Union[FromDigits/@(F=Flatten)[Table[Partition[#,i,1],{i,2,9}]&/@Select[F[Table[Range[j,9,k],{j,9},{k,9}],1],Tr[1^#]>1&],2]]   

Try it online! all numbers in a sec

Answer 6

05AB1E, 23 bytes

•7=›ζ•FNS¥DËs0›PN11›&&–

Try it online!

---

Replace •7=›ζ• with 7000 to have it finish on TIO, or just hit the "terminate" button before it times out, resulting in the numbers printed up until that point.

Answer 7

Husk, 14 13 bytes

ÖifȯεuẊ≠Ṗ…"19

Prints newline-separated numbers to STDOUT. Try it online!

-1 byte due to inspiration from H.PWiz.

Explanation

ÖifȯεuẊ≠Ṗ…"19
         …"19  The string "19" rangified: "123456789"
        Ṗ      Powerset: ["","1","2","12","3",...,"123456789"]
  fȯ           Filter by function: (input is sublist, say "2469")
      Ẋ≠        Consecutive codepoint differences: [2,2,3]
     u          Remove duplicates: [2,3]
    ε           Is it a one-element list? No, so "2469" is dropped.
Öi             Sort the result by integer value, implicitly print separated by newlines.

Answer 8

Wolfram Language (Mathematica), 76 bytes

Differences@IntegerDigits@#~MatchQ~{a_?Positive..}&&Print@#&~Array~123456789

Try it online!

Answer 9

APL (Dyalog), 37 28 bytes

⍪x/⍨{1=≢∪2-/⍎¨⍕⍵}¨x←11+⍳1E9

Try it online! (with shorter range, due to time-out)

How?

x←11+⍳123456789 - 11, 12... 1e9 into x

¨ - for each

    ⍎¨⍕⍵ - break into digits

    2-/ - get differences list

    ∪ - get unique elements

    1=≢ - length == 1?

x/⍨ - use this as a mask on the range created

⍪ - and columnify

Answer 10

Husk, 14 bytes

ÖifoEẊ≠ftṖ…"19

Try it online!

Answer 11

Java 8, 169 168 145 bytes

v->{byte[]a;for(int i=9,d,l;++i<1e9;System.out.print(l<1?i+"\n":""))for(a=(i+"").getBytes(),d=0,l=a.length;--l>0&&d*(d^(d=a[l]-a[l-1]))<1&d>0;);}

Port of @Jacobinski C answer (after I golfed it a bit).
-23 bytes thanks to @Nevay.

Explanation:

Try it here. (It's a bit too slow near the end, so doesn't print the final number on TIO. It prints the final number locally in about 20 seconds, though.)

v->{                         // Method with empty unused parameter and no return-type
  byte[]a;                   //  Byte-array
  for(int i=9,               //  Index integer, starting at 9
          d,                 //  Difference-integer
          l;                 //  Length integer
      ++i<1e9;               //  Loop (1) from 10 to 1,000,000,000 (exclusive)
      System.out.print(      //    After every iteration: print:
        l<1?                 //     If `l` is 0:
         i+"\n"              //      The current integer + a new-line
        :                    //     Else:
         ""))                //      Nothing
    for(a=(i+"").getBytes(), //   Convert the current item as String to a byte-array
        d=0,                 //   Reset the previous integer to 0
        l=a.length;          //   Set `l` to the length of the byte-array
        --l>0                //   Inner loop (2) from `l-1` down to 0 (exclusive)
        &&d*(d^(d=a[l]-a[l-1]))<1
                             //    As long as the previous difference is either 0
                             //    or the current diff is not equal to the previous diff
        &d>0;                //    and the current diff is larger than 0
    );                       //   End of inner loop (2)
                             //  End of loop (1) (implicit / single-line body)
}                            // End of method

Answer 12

Batch, 210 200 bytes

No optimizations, so very slow - takes about 25 seconds until 12345, so for the complete output you'd have to wait about 3 days.

@set z=@set/a
%z%c=12
:a
@echo %c%
:c
%z%c+=1&if %c%==123456790 exit/b
%z%n=c
%z%d=n%%10-n/10%%10
@if %d% leq 0 goto c
:d
%z%n=n/10
@if %n% leq 9 goto a
%z%e=n%%10-n/10%%10
@if %e%==%d% goto d
@goto c

Answer 13

Python 2, 103 97 95 bytes

-2 bytes thanks to @JonathanFrech

n=11
while n<1e9:
 n+=1;Q=`n`[1:]
 if`n`<Q>1==len({int(a)-int(b)for a,b in zip(`n`,Q)}):print n

Try it online!

Answer 14

05AB1E, 14 bytes

TžhŸʒS¥D0›PsË&

Try it online!

Answer 15

JavaScript (Firefox 30-57), 105 bytes

document.write(`<pre>`+(

_=>[for(x of s=`123456789`)for(y of s)for(z of s)if(x*z<10-y)s.replace(/./g,c=>c<y|(c-y)%z|c-y>x*z?``:c)]

)().join`\n`+`</pre>`);

Loops over lengths from 2 to 10 (x is the index of the last character and therefore 1 less than the length), starting digits from 1 to 9 and step from 1 to 9, then filters on the end digit being less than 10 and if so generates the result by filtering out digits from the digit string.

Answer 16

Python 2, 76 75 bytes

n=9
while n<2e8:
 n+=1
 if`n`in`range(n%10,0,~n%100%11-11)`[-2::-3]:print n

Takes about 3 minutes locally.

Try it online! (modified to print all numbers but the last)

Answer 17

MATL, 17 16 bytes

1e9:"@Vdu0>TX=?@

Try it online! with 1e9 replaced by 1e3 so that it doesn't time out in the online compiler.

Answer 18

Pyth, 21 bytes

Port of Dennis' clever approach.

jjLkf!t{.+Tsm.cS9dtS9

Try it here!

Pyth, 23 bytes

j.f&!tl{K.+jZT*F>R0K77T

This times out on the online interpreter, but works if given enough time and memory.

Try it online with a lower number.

Answer 19

C (gcc), 106 bytes

main(i,j,k,l){for(i=1;i<9;i++)for(j=8;j;j--)for(k=0;k<j/i;puts(""))for(l=0*k++;l<=i;)putchar(57-j+k*l++);}

Try it online!

The ungolfed (prettified) version:

int main() {
  int length, start_with, max_step, step, nth;
  for (length = 2; length <= 9; length++) {
    for (start_with = 1; start_with < 9; start_with++) {
      max_step = (9 - start_with) / (length - 1);
      for (step = 1; step <= max_step; step++) {
        for (nth = 0; nth < length; nth++) {
          putchar('0' + start_with + step * nth);
        }
        puts("");
      }
    }
  }
  return 0;
}

Answer 20

JavaScript (ES6), 121 bytes

Not nearly as short as Neil's answer, but I thought it was still worth posting.

Works by building a powerset of '123456789' where all non-matching entries are truncated and prefixed with 0, sorting the results in numerical order and keeping only the 77 relevant ones.

_=>[...'123456789'].reduce((a,x)=>[...a,...a.map(y=>y[1]-y[0]+(y.slice(-1)-x)?'0':y+x)],['']).sort((a,b)=>a-b).slice(-77)

Demo

let f =

_=>[...'123456789'].reduce((a,x)=>[...a,...a.map(y=>y[1]-y[0]+(y.slice(-1)-x)?'0':y+x)],['']).sort((a,b)=>a-b).slice(-77)

console.log(f().join` `)

Answer 21

JavaScript (ES6), 109 104 bytes

Works by generating all possible numbers: loops through each increment from 8 to 1 (variable i), loops through each starting digit from 8 to 1 (variable j), loops through each digit between j and 10-i (variable k) and generates a string t by appending k to the current t. At each step t is added to the output array.

(o=[],i=9)=>{while(--i){j=9;while(k=--j){t=""+k;while(k<10-i)o.push(t+=k+=i)}}return o.sort((a,b)=>a-b)}

Try it online!

f=

(o=[],i=9)=>{while(--i){j=9;while(k=--j){t=""+k;while(k<10-i)o.push(t+=k+=i)}}return o.sort((a,b)=>a-b)}

;
console.log(f().join("\n"))

Answer 22

Wolfram Language (Mathematica), 71 bytes

Flatten@Array[If[(x=#2+#3 #)<10,FromDigits@Range[#2,x,#3],{}]&,{8,8,8}]

Try it online!

Lightning fast due to constructing rather than selecting the output.

Answer 23

PHP, 85 84 bytes

for(;++$d<9||++$a<9*$d=1;sort($r))for($x=$y=$a+1;10>$y+=$d;)$r[]=$x.=$y;print_r($r);

try it online.

sorting cost 17 bytes. This version prints the results ordered differently:

while(++$d<9||++$a<9*$d=1)for($x=$y=$a+1;10>$y+=$d;)echo"
",$x.=$y;

Answer 24

JavaScript (ES6), 145 bytes

A straight-forward approach with little magic.

Array(123456790).fill().map((x,i)=>(''+i).split('')).map((a,i,d) => {d=a[1]-a[0];d>0&a.every((n,j)=>j<1||n-a[j-1]==d)?console.log(a.join('')):0})

Running the snippet will consume a lot of memory...

Answer 25

Java (OpenJDK 8), 228 174 170 163 bytes

o->{int i=11,x,y,z,l;for(;i++<1e9;){char[]w=(""+i).toCharArray();for(x=w[1]-w[0],y=1,z=0;y<w.length;z+=l>0&l==x?0:1)l=w[y]-w[y++-1];if(z<1)System.out.println(i);}}

Try it online!