Exponentiation Sequence

Question

The oldest Polish salt mine, located in Bochnia*, was started in year 1248, which we can consider a magical number. We can see that it's equal to 4 digits from the sequence of exponentiations: $2^0\textrm{, }2^1\textrm{, }2^2\textrm{ and }2^3$ .

As the date is actually 4 digits from the sequence, we could make it longer. We could repeat the process until we reach infinity. The sequence would look like this, if we limit it to number 2048

124816326412825651210242048

To make it look a bit better, we can separate the numbers:

1|2|4|8|16|32|64|128|256|512|1024|2048

Let's try a custom, longer sequence than the date. Let's say, we want it to have 5 digits - there are more than one possibility:

24816
81632
64128

Or 3 digit ones:

124
248
816

We could also add the 3 digit numbers to this, but let's say, that a sequence must have at least two numbers.

^{* There is no information about this on the English Wikipedia. If you enter the Polish version - then there is. If you visit the mine, the workers will also tell you, that it started in 1248.}

The challenge

Create a exponentiation sequence like in examples above with 2 as the base.

Given a number from range 2-27, output all possible parts of the sequence (The 2048 one or larger if you want) with amount of digits equal to the input. You cannot cut a number, so output like 481 is invalid, because 16 is cut in half.

Rules:

Standard loopholes are forbidden.
You can assume the input is a number inside the range.
Your program can accept inputs larger than the range (28+), but that won't increase/decrease score.
Spaces in output are ignored. You can output like 124 or like 4 8 16.
Different possibilities should be separated by any character from the list: ,./| or a line feed.
You can output as an array.
Every possibility should include at least 2 different numbers.
You must output a part of the sequence, you cannot mix numbers that aren't next to each other, like: 14.
Hardcoded output isn't allowed, however, you can hardcode a string/number/array containing the full sequence.
Input 27 should return the full 2048 sequence.
As already mentioned before, do not cut numbers. Ex. 16 must stay 16 - you can't use 481 - you must use 4816.
EDIT: I might have said something wrong there; 2048 is the last number which your program should support, you can add support for larger int's.

Test cases

Input: 2

12, 24, 48

Input: 3

124, 248, 816

Input: 4

1248, 4816, 1632, 3264

Input: 5

24816, 81632, 64128

Input: 27

124816326412825651210242048

And later numbers...

If I made a mistake in any of the test cases, tell me or edit the question.

This is code-golf, so the shortest code in bytes wins!

So this is only with 2 as the base, correct? Could you clarify that in the question? I'm not sure if it's implied by "Sequence of Exponentiations," but even if it is, I'm sure there are people out there like me who don't know that. — cole, Sep 12, 2017 at 16:56
@cole Actually, yes, it's only with 2. Thanks for mentioning! — RedClover, Sep 12, 2017 at 16:56
No worries; like I said, I was pushing it. Some challenge authors can be incredibly flexible on the output format so, for the sake of a byte or 2, it's worth asking ;) (Note: That should not be interpreted as a suggestion!) — Shaggy, Sep 13, 2017 at 22:27
In the intro, you should capitalize Polish. "polish" is a different English word. — Peter Cordes, Sep 17, 2017 at 3:45

Emigna · Accepted Answer · 2017-09-13 13:46:53Z

8

05AB1E, 12 11 10 bytes

Supports the sequence up to 2^95 = 39614081257132168796771975168

₃ÝoŒʒg≠}Jù

Try it online!

Explanation

₃Ý            # push range [0 ... 95]
  o           # raise 2 to the power of each
   Œ          # get a list of all sublists
    ʒ         # filter, keep elements that satisfy:
     g        # length
      ≠       # false (not equal to 1)
       }      # end filter
        J     # join each
         ù    # keep numbers of length matching the input

Saved 1 byte thanks to Erik the Outgolfer
Saved 1 byte thanks to Riley

edited Sep 13, 2017 at 13:46

answered Sep 13, 2017 at 11:22

Emigna

52.9k5 gold badges42 silver badges162 bronze badges

\$\begingroup\$ X› can be ≠ \$\endgroup\$
– Erik the Outgolfer
Sep 13, 2017 at 12:31
\$\begingroup\$ @EriktheOutgolfer: Oh yeah. I always forget that one exist now. Thanks :) \$\endgroup\$
– Emigna
Sep 13, 2017 at 12:41
\$\begingroup\$ Can Y₃Ým be ₃Ýo? \$\endgroup\$
– Riley
Sep 13, 2017 at 13:35
\$\begingroup\$ @Riley: Yep. I had that before, but for some reason I must have changed it. Thanks :) \$\endgroup\$
– Emigna
Sep 13, 2017 at 13:46
\$\begingroup\$ Trying the code now (much late after challenge end)... and your solution seems to return tons of empty arrays for me... Am I doing something wrong? \$\endgroup\$
– RedClover
Nov 13, 2017 at 19:33

| Show 1 more comment

score 6 · Accepted Answer · 2017-09-12 20:33:11Z

6

Pyth, 22 21 20 17 bytes

fqQlTjLkt#.:^L2yT

Try it Online

Explanation

fqQlTjLkt#.:^L2yT
            ^L2yT  Get the powers of 2 up to 2^20
        t#.:       Get all consecutive sequences of at least 2
     jLk           Concatenate each
fqQlT              Get the ones whose length is the input

edited Sep 12, 2017 at 20:33

answered Sep 12, 2017 at 17:20

user48543

\$\begingroup\$ 21 bytes: fqQlTmjkdftlT.:m^2d12 \$\endgroup\$
– Mr. Xcoder
Sep 12, 2017 at 17:22
1

\$\begingroup\$ 20 bytes - fqQlTjLkftlT.:m^2d12 \$\endgroup\$
– Mr. Xcoder
Sep 12, 2017 at 17:26
\$\begingroup\$ 19 bytes - fqQlTjLkftT.:m^2d12 \$\endgroup\$
– Jakube
Sep 12, 2017 at 19:42
\$\begingroup\$ 18 bytes - fqQlTjLkftT.:^L2yT \$\endgroup\$
– Jakube
Sep 12, 2017 at 19:45
1

\$\begingroup\$ 17 bytes - fqQlTjLkt#.:^L2yT \$\endgroup\$
– Jakube
Sep 12, 2017 at 19:49

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Jonathan Allan · Accepted Answer · 2017-09-12 19:07:55Z

Jelly, 19 18 16 bytes

^{There may be a shorter solution now that we may use any cut-off (not just 2048), although this change to the specification has allowed a one byte save from this implementation by moving to a cut-off of 32768.

--yep...}

-2 bytes thanks to Erik the Outgolfer (use of V to allow implicit right argument of the filter and tightening)
--yes it is very similar to his inefficient one now; go upvote his!

⁴Ḷ2*Ẇṫ17VDL$⁼¥Ðf

A monadic link taking a number and returning a list of numbers.

Try it online!

How?

⁴Ḷ2*Ẇṫ17VDL$⁼¥Ðf - Link: number, n        e.g. 3
⁴                - literal sixteen             16
 Ḷ               - lowered range               [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
  2              - literal two                 2
   *             - exponentiate                [1,2,4,8,16,32,...,32768]
    Ẇ            - all sublists                [[1],[2],...,[1,2],[2,4],...,[1,2,4],...]
      17         - literal seventeen           17
     ṫ           - tail from index             [[1,2],[2,4],...,[1,2,4],...]]
        V        - evaluate as Jelly code      [12,24,...,124,...]
              Ðf - filter keep:
             ¥   -   last two links as a dyad
           $     -     last two links as a monad:
         D       -       decimal list (of entry) (i.e. 816 -> [8,1,6] or 24 -> [2,4])
          L      -       length                  (i.e. 816 -> 3, or 24 -> 2)
            ⁼    -   equals (n)                  (i.e. 816 -> 1, or 24 -> 0)
                 - ...resulting in             [816, 124, 248]

Funnily enough I was just trying to use V and it'll work for 16 rather than 1000: ⁴Ḷ2*Ẇṫ17VDL$⁼¥Ðf. — Jonathan Allan, Sep 12, 2017 at 18:52
@EriktheOutgolfer they may now be similar but I am of the mind we should both keep them I'm sure you came up with yours independently and I'm sure I'd have found the eval trick too (since I was looking at exactly that, just needed to get the chaining right). — Jonathan Allan, Sep 12, 2017 at 19:09
@EriktheOutgolfer I have made the assumption you are male, and prefer to be referred to as such, but actually do not know either as fact; do let me know if you prefer a different pronoun! — Jonathan Allan, Sep 12, 2017 at 19:13

nwellnhof · Accepted Answer · 2018-03-20 18:45:21Z

4

Perl 6, 62 59 bytes

{grep *.comb==$_,map {[~] 2 X**[...] $_},combinations 12,2}

Try it online!

edited Mar 20, 2018 at 18:45

answered Sep 12, 2017 at 18:41

nwellnhof

10.5k1 gold badge18 silver badges38 bronze badges

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Community · Accepted Answer · 2020-06-17 09:04:33Z

Japt, 22 20 19 16 bytes

Supports input up to 639 but gaps start appearing in the sequence after 234 (See the full list of supported input ranges here). Outputs an array of strings.

IÆIo!²ãX m¬lUÃ¤c

Test it

I (64) could be replaced with L (100) but we'd be getting into scientific notation and precision inaccuracies. Filtering those out would, obviously, increase the byte count and only raise the maximum input to 736.

                     :Implicit input of integer U
I                    :64
 Æ                   :Map each X in [0,64)
  Io                 :  Range [0,64)
    !²               :  Raise 2 to the power of each
      ãX             :  Subsections of length X
         m           :  Map
          ¬          :    Join
           lU        :  Filter elements of length U
             Ã       :End map
              ¤      :Slice off the first 2 elements
               c     :Flatten

Rod · Accepted Answer · 2017-09-12 17:36:01Z

3

Python 2, 105 bytes

lambda l,r=range:[x for x in[''.join(`2**n`for n in r(i,j+2))for i in r(13)for j in r(i,11)]if len(x)==l]

Try it online!

answered Sep 12, 2017 at 17:36

Rod

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Community · Accepted Answer · 2020-06-17 09:04:33Z

Husk, 18 17 bytes

Output is separated by newlines

fo=⁰LmṁsftQ↑12¡D1

Try it online!

How?

           ↑12¡D1    The sequence [1,2,4...2048]
              ¡      Repeatedly apply function, collecting results in a list
               D     double
                1    initially applying to 1
           ↑12       Take the first 12 elements
          Q          Get all sublists
        ft           With a length greater than 1
     mṁs             Convert each list into a string, e.g [4,8,16] -> "4816"
fo=⁰L                Keep only those whose length is equal to the input

Erik the Outgolfer · Accepted Answer · 2017-09-12 18:46:39Z

3

Jelly, 16 bytes

ȷḶ2*ẆṫȷḊVDL$⁼¥Ðf

Try it online!

Note: very inefficient. Returns a list of numbers.

answered Sep 12, 2017 at 18:46

Erik the Outgolfer

40.5k5 gold badges41 silver badges121 bronze badges

\$\begingroup\$ Tio doesn't seem to be able to parse this code... It always exceeds 60 seconds... \$\endgroup\$
– RedClover
Sep 12, 2017 at 18:50
1

\$\begingroup\$ @Soaku Try replacing the ȷs (i.e. 1000) with 20s (lower upper limit). \$\endgroup\$
– Erik the Outgolfer
Sep 12, 2017 at 18:51
1

\$\begingroup\$ @Soaku it works in theory - it just times out because it is very inefficient. \$\endgroup\$
– Jonathan Allan
Sep 12, 2017 at 18:53
1

\$\begingroup\$ @Soaku I meant on my comment. I have already replaced them there, and there's output [12, 24, 48]. \$\endgroup\$
– Erik the Outgolfer
Sep 12, 2017 at 18:55
1

\$\begingroup\$ @Soaku Why not go for the most if you can without extra cost? ;) \$\endgroup\$
– Erik the Outgolfer
Sep 13, 2017 at 10:53

| Show 5 more comments

Arnauld · Accepted Answer · 2017-09-13 11:03:50Z

JavaScript (ES7), 102 100 bytes

Prints all matching sub-sequences with alert().

l=>[...1e11+''].map((_,k,a)=>a.map((_,x)=>(s=(g=n=>x<=k|n<k?'':g(n-1)+2**n)(x)).length-l||alert(s)))

Demo

NB: This snippet is buffering the results and printing them to the console for user-friendliness.

let f =

l=>[...1e11+''].map((_,k,a)=>a.map((_,x)=>(s=(g=n=>x<=k|n<k?'':g(n-1)+2**n)(x)).length-l||alert(s)))

alert = s => res += ' ' + s;

for(l = 2; l <= 27; l++) {
  res = '';
  f(l);
  console.log('L = ' + l + ' -->' + res);
}

jferard · Accepted Answer · 2017-09-13 18:38:02Z

3

Haskell, 72 67 bytes

f n=[s|i<-[0..99],j<-[i+1..99],s<-[show.(2^)=<<[i..j]],length s==n]

Try it online!

Saved 5 bytes thanks to Laikoni

I used a limit of 99 because 2^99 has a length > 27.

edited Sep 13, 2017 at 18:38

answered Sep 12, 2017 at 18:07

jferard

1,7946 silver badges7 bronze badges

\$\begingroup\$ It's returning extra cases for an input of 27 though. \$\endgroup\$
– Jonathan Allan
Sep 12, 2017 at 18:10
\$\begingroup\$ You can replace 99 with 11, so it becomes more valid. Although, I did not state that numbers > 2048 are invalid. I stated only, that 2048 is the minimum range. \$\endgroup\$
– RedClover
Sep 12, 2017 at 18:13
\$\begingroup\$ @JonathanAllan I think it's still correct: "The 2048 one or larger if you want" I took the 633825300114114700748351602688 sequence, because it guarantees that there won't be any other solution (in the range 2-27). Actually, I think a 45 limit would be sufficient, because length$(show$2^44)++(show$2^45)==28. \$\endgroup\$
– jferard
Sep 12, 2017 at 18:27
\$\begingroup\$ @jferard that was actually edited in after your answer (it actually stated "limited to 2048" before). I can shorten mine in this case too. \$\endgroup\$
– Jonathan Allan
Sep 12, 2017 at 18:30
1

\$\begingroup\$ @JonathanAllan Yes, I actually realised that some of rules in the question are wrong and misleading just after posting this anwer. \$\endgroup\$
– RedClover
Sep 12, 2017 at 18:41

| Show 3 more comments

ZaMoC · Accepted Answer · 2017-09-12 21:42:43Z

Mathematica, 122 bytes

(s=#;FromDigits@F[f/@#]&/@Select[Subsequences[Array[2^#&,99,0]],l@#>1&&(l=Length)@(F=Flatten)[(f=IntegerDigits)/@#]==s&])&

Input

[27]

Output

{879609302220817592186044416, 134217728268435456536870912, 524288104857620971524194304, 163843276865536131072262144, 204840968192163843276865536, 256512102420484096819216384, 641282565121024204840968192, 163264128256512102420484096, 124816326412825651210242048}

Input [1000]
Output  1441151880758558722882303761517117445764607523034234881152921504606846976230584300921369395246116860184273879049223372036854775808184467440737095516163689348814741910323273786976294838206464147573952589676412928295147905179352825856590295810358705651712118059162071741130342423611832414348226068484722366482869645213696944473296573929042739218889465931478580854784377789318629571617095687555786372591432341913615111572745182864683827230223145490365729367654460446290980731458735308812089258196146291747061762417851639229258349412352483570327845851669882470496714065569170333976494081934281311383406679529881638685626227668133590597632773712524553362671811952641547425049106725343623905283094850098213450687247810566189700196426901374495621121237940039285380274899124224247588007857076054979824844849517601571415210995964968969903520314283042199192993792198070406285660843983859875843961408125713216879677197516879228162514264337593543950336158456325028528675187087900672316912650057057350374175801344

Steadybox · Accepted Answer · 2017-09-12 22:05:47Z

2

C, 170 bytes

i,j;f(n){char t[99],s[12][5]={"1"};for(i=j=1;i<12;)sprintf(s+i++,"%d",j*=2);for(i=0;i<12;++i,strlen(t)-n||j>1&&puts(t))for(j=*t=0;strlen(t)<n&&j+i<12;)strcat(t,s+i+j++);}

Try it online!

Unrolled:

i,j;
f(n)
{
    char t[99], s[12][5] = {"1"};
    for (i=j=1; i<12;)
        sprintf(s+i++, "%d", j*=2);
    for (i=0; i<12; ++i, strlen(t)-n || j>1 && puts(t))
        for (j=*t=0; strlen(t)<n && j+i<12;)
            strcat(t, s+i+j++);
}

edited Sep 12, 2017 at 22:05

answered Sep 12, 2017 at 21:54

Steadybox

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ngn · Accepted Answer · 2021-11-30 11:01:16Z

K (ngn/k), 33 bytes

{(x=#:')#,/{,/'$x':|2/=12}'2+!11}

Try it online!

Generates all possible sublists, then selects those of the proper length. Returns a list of strings.

{...}'2+!11 generate 2..12 (the valid sublist lengths for input from 2..27), and run the code in {...} on each of those numbers
- |2/=12 generate the sequence 1 2 4 8 16 32 64 128 256 512 1024 2048
- ,/'$x': take x-length rolling windows, converting each to strings, then concatenate them (e.g., 1 2 4 8 => (1 2;2 4;4 8) => ((,"1";,"2");(,"2";,"4");(,"4";,"8")) => ("12";"24";"48"))
(x=#:')#,/ flatten the potential strings, keeping those of the desired length

flodel · Accepted Answer · 2017-09-13 01:02:33Z

1

R, 99 bytes

function(n)for(i in 1:11)for(j in i:11+1)if(sum(nchar(x<-2^(0:11))[i:j])==n)cat(x[i:j],"\n",sep="")

Try it online!

answered Sep 13, 2017 at 1:02

flodel

2,39715 silver badges16 bronze badges

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Xcali · Accepted Answer · 2017-09-15 00:37:11Z

1

Perl 5, 76 bytes

75 bytes of code + 1 for -a

for$i(0..10){$/='',(map$/.=2**$_,$i..$_)&&$F[0]-length$/||say$/for$i+1..11}

Try it online!

edited Sep 15, 2017 at 0:37

answered Sep 15, 2017 at 0:13

Xcali

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G B · Accepted Answer · 2018-10-12 11:09:03Z

0

Ruby, 94 bytes

->n{a=*b=1;a<<b*=2until b>3**n;(2..n).flat_map{|x|a.each_cons(x).map &:join}.grep /^.{#{n}}$/}

Try it online!

edited Oct 12, 2018 at 11:09

answered Oct 12, 2018 at 11:01

G B

21.3k1 gold badge19 silver badges51 bronze badges

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Stack Exchange Network

Exponentiation Sequence

The challenge

Test cases

16 Answers 16

05AB1E, 12 11 10 bytes

Pyth, 22 21 20 17 bytes

Explanation

Jelly, 19 18 16 bytes

How?

Perl 6, 62 59 bytes

Japt, 22 20 19 16 bytes

Python 2, 105 bytes

Husk, 18 17 bytes

How?

Jelly, 16 bytes

JavaScript (ES7), 102 100 bytes

Demo

Haskell, 72 67 bytes

Mathematica, 122 bytes

C, 170 bytes

K (ngn/k), 33 bytes

R, 99 bytes

Perl 5, 76 bytes

Ruby, 94 bytes

Your Answer

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code-golf
sequence
number-theory
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Exponentiation Sequence

The challenge

Test cases

16 Answers 16

05AB1E, 12 11 10 bytes

Pyth, 22 21 20 17 bytes

Explanation

Jelly, 19 18 16 bytes

How?

Perl 6, 62 59 bytes

Japt, 22 20 19 16 bytes

Python 2, 105 bytes

Husk, 18 17 bytes

How?

Jelly, 16 bytes

JavaScript (ES7), 102 100 bytes

Demo

Haskell, 72 67 bytes

Mathematica, 122 bytes

C, 170 bytes

K (ngn/k), 33 bytes

R, 99 bytes

Perl 5, 76 bytes

Ruby, 94 bytes

Your Answer

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Hot Network Questions

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