# Output every sublist ... eventually

You will be given as input an infinite stream of positive integers.

Your task is to write a program which outputs an infinite sequence of lists with two requirements:

• All lists in the output are finite sub-lists of the input stream.
• Every sub-list must be output eventually by your program.

Sub-lists are not necessarily contiguous, so [1,1] is a sub-list of [0,1,0,1,0,..., and should be included in your output if your input was [0,1,0,1,0,....

The empty list [] is a sub-list of every possible input, so it should always be output at some point.

Aside from this, you can do what you want. You can output them in whatever order suits you, you can output any sub-list any (positive) number of times. It's up to you to figure out what's best.

This is so answers will be scored in bytes with fewer bytes being the goal.

## IO

Outputting an infinite sequence is done as per the defaults of the sequence tag wiki.

Input of an infinite sequence is basically the dual of those rules. You may:

• Take the input as a black-box function which takes $$\n\$$ and outputs the $$\n\$$th term
• Take the input as a black-box function which takes $$\n\$$ and outputs all terms up to the $$\n\$$th term
• Take the input as a lazy list, generator or stream.
• Take the input by repeatedly querying STDIN each time receiving the next term of the sequence.
• must order be preserved? or is [3,2,1] a sublist of [1,2,3..., and if so, are [1,2,3] and [3,2,1] both required outputs? etc Jan 16 at 19:59
• @thejonymyster Order must be preserved. Jan 16 at 20:09

-4 bytes thanks to Wheat Wizard!

Takes an infinite list as input and returns each finite sublist an infinite number of times.

f(a:l)=[]:do x<-f l;[a:x,x]


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• 27 bytes Jan 16 at 15:02
• @WheatWizard thanks a lot, at some point I'll remember to try this myself ;)
– ovs
Jan 16 at 15:40

# 05AB1E, 1 byte

æ


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# 05AB1E no builtin, 5 bytes

[DNÏ,


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Takes the input as an 05AB1E infinite lazy list

[     infinite loop
D    create a copy of the infinite list
N    iteration index (starting from 0)
Ï    elements of the list for which the corresponding element of the index is 1
,    print

• "element of the index" - is it using the bits of the index? Or what? Jan 16 at 16:14
• @pxeger It is using the digits of the decimal representation and takes the elements where the number has a 1
– ovs
Jan 16 at 16:25

# Python, 43 bytes

lambda s,n:[s(b)for b in range(n)if n>>b&1]

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Outputs the nth sublist.

# Python, 55 bytes

n=0
while[print([s(b)for b in range(n)if n>>b&1])]:n+=1

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Takes input using a black-box function predefined under the name s, which takes a number and outputs the corresponding 0-indexed value of the sequence. Shown in the ATO link using the powers of 2 [1, 2, 4, 8, .... Outputs infinitely to STDOUT.

Uses the binary representation of each non-negative integer n to produce a sub-list; 1s in the binary representation correspond to indices where the list's value is included in the output.

A bug fix for -12 bytes (!):

By using n itself rather than the bit length of n to determine how many bits to select, we also get, pretty much for free, the ability to sometimes output sublists which don't include the first item.

Because n is always greater than or equal to its bit length (besides for 0), we will effectively select some 0 bits which form part of its "padding", but which allow us to skip some of the first few elements.

When n is 0, this is the only number with no 1s in its binary. As a result, by initialising n to 0, we output an empty list at the start, satisfying that requirement at no byte cost.

# Python, 63 62 bytes

def f(s,*e):
yield e
for x in f(s,next(s)):yield from{x,e+x}

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This is a version based on @ovs' clever Haskell answer, but with some changes to make it work in Python where we don't have infinite lists. It might be improvable.

-1 thanks to Command Master

Thanks to loopy walt, this version now outputs every sub-list exactly once, for the same byte count.

• 62 bytes for the second version, using an iterator as an input instead of a function Jan 16 at 17:58
• @CommandMaster damn, I don't know why I dismissed that option. Thanks! Jan 16 at 18:00
• Same byte count but perhaps cleaner (output every sublist exactly once). ato.pxeger.com/… Jan 17 at 15:05

# JavaScript (Node.js), 30 bytes

a=>n=>a(n).filter(_=>(n/=2)&1)


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Input array as a function which output first n elements. Output n-th element.

# R, 73 68 bytes

Or R>=4.1, 61 bytes by replacing the word function with a \.

Edit: -5 bytes thanks to @Giuseppe.

function(g)while(T<-T+1)for(i in 1:2^T)show(g(T)[!i%/%2^(T:1-1)%%2])


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Explanation:

1. Take a black-box function g as an argument.
2. Increment T in each iteration - how many numbers to read from g.
3. In the loop, calculate all the sub-sequences of the current vector (and some more):
• take all numbers from 1 to 2^T,
• convert to binary,
• use as logical indices, which numbers from g(T) to print.
• 68 bytes by accepting a black-box that returns the first n values (could use F instead of T) Jan 17 at 19:32
• Alternately, 56 bytes using combn; here we do need to use T because combn(3,1) returns matrix(c(1,2,3)), though this output format might be a mild stretch of the rules. Jan 17 at 19:33
• @Giuseppe, thanks - I never took input using generating function in R, so I went with the familiar scan. As per using combn - I also have thought of that, but outputting as matrices is too far-fetched for me. Jan 17 at 19:42

# Scala, 89 bytes

_.scanLeft(Seq[Int]())(_:+_)flatMap(l=>l.indices.toSet.subsets.map(_.toSeq.sorted map l))


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Welp, this turned out much longer than I expected. Time to reconsider this approach. Input is a LazyList[Int]. It outputs all the sublists up to the current integer for every integer.

_                            //The input (infinite list)
.scanLeft                   //Scan left over the sequence,
(Seq[Int]())(_:+_)        //building an infinite list of subsequences
flatMap(l=>                 //Map each of those subsequences l to its sublists
l.indices.toSet.subsets   //Get all subsets of the indices
.map(
_.toSeq.sorted          //Sort the indices to ensure sublists are in order
map l)                  //Map the index to the corresponding element in l
)


# [Ruby], 64 56 48 bytes

->g{1.step{|n|n.times{p *g[n].combination(_1)}}}

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• saved 8 Bucks thanks to @ovs

takes a black box function returning first n elements.

• You can use n.times instead of (0..n).map and p *... instead of ....map{p _1} to save a couple bytes: Attempt it online!
– ovs
Jan 17 at 10:27
• Thanks @ovs for the "couple" of bytes, and also for the working link Jan 17 at 14:27

# Charcoal, 21 bytes

ＷＳ«Ｅ⌕Ａ⮌⍘Ｌυ²1§υκＤ⎚Ｄ⊞υι


Try it online! Link is to verbose version of code. Explanation:

ＷＳ«


Repeatedly input the next integer as a string. (This will actually stop on a blank line, which is what I use to allow the linked example to halt, but for positive integers it will keep prompting for input.)

Ｅ⌕Ａ⮌⍘Ｌυ²1§υκ


Convert the length of the predefined empty list to binary and find the positions of the 1s counting the least significant bit as 0. Output the elements of the predefined empty list with matching positions.

Ｄ⎚Ｄ


Output the canvas double-spaced. (Charcoal's default output format lists one element per line, so it's not easy to distinguish consecutive lists otherwise, although if you were running it interactively then the Enter input: prompt would do the trick.)

⊞υι


Append the next term of the input stream to the predefined empty list.

# Pari/GP, 35 bytes

g->i->[g(j)|j<-[0..i],bittest(i,j)]


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Take the input as a black-box function which takes $$\n\$$ (0-indexed) and outputs the $$\n\$$th term.

# Julia 1.0, 45 bytes

!(x,a=[[]])=for i=x a=[@show a;vcat.(a,i)]end


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Input is an iterator. Output has a bit of unnecessary stuff but hopefully it's acceptable

# Vyxal, 1 byte (non-competing)

ṗ


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Non-competing as this builtin and many others were recently given infinite list support because of this challenge.

# Pip-p, 18 bytes

lPBlW1P*lAL:_AEqMl


Reads values from stdin, outputs sublists to stdout. Each sublist will be printed infinitely many times. Attempt This Online!

### Explanation

We track every sublist of the inputs-thus-far in l. At each iteration, we read a new input, append it to each of the existing sublists, append that list to the list of existing sublists, and output the whole thing.

lPBlW1P*lAL:_AEqMl
l                   Variable preset to empty list
PBl                Push the empty list to l; l is now [[]]
W1              While 1 (loop forever):
Ml   Map this function to each sublist in l:
q      Read a line of stdin
_AE       and append it to the sublist
lAL:         Append those results to l in-place
P*             Print each sublist in l


# Haskell + hgl, 2 bytes

sS


I just finished adding builtins having to do with subsequences to hgl. sS gets all the subsequences of the input and works just fine on infinite lists.

It outputs subsequences in order of the index of their last element last element. That is first it outputs the empty list, then it outputs all sequences that end in the 0th element, then all sequences that end in the 1st element, then all sequences that end in the 2nd etc.

# Haskell + hgl, 21 bytes

jn<<mM(ap[i,p]<p)<+cg


Since it's no fun to just post a built-in here's one that solves the task without sS or related functions. First we make jn<<mM(ap[i,p]<p). This takes a finite list and outputs all subsequences.

To do this it uses mM, (mapM or traverse if you are a Haskell user) which is a little complicated, but it's equivalent to sQ<m (sequence.map if your a Haskell user) so I'll explain it like that.

First m(ap[i,p]<p) will replace every element with a list of a singleton containing itself and an empty list:

>>> m(ap[i,p]<p)[1,2,3]
[[[],[1]],[[],[2]],[[],[3]]]


Then sQ basically does the generalized cartesian product. It takes n lists as input and gives all the ways to take 1 item from each list. Since we have two possibilities in each list, the integer singleton or any empty list, this gives us basically binary numbers of n digits with 0 being the empty list and 1 being the value.

>>> mM(ap[i,p]<p)[1,2,3]
[[[],[],[]],[[],[],[3]],[[],[2],[]],[[],[2],[3]],[[1],[],[]],[[1],[],[3]],[[1],[2],[]],[[1],[2],[3]]]


Now we want to get rid of the singletons so we use m jn

>>> jn<<mM(ap[i,p]<p)\$[1,2,3]
[[],[3],[2],[2,3],[1],[1,3],[1,2],[1,2,3]]


Now we have all the subsequences of the input! However this stalls out on infinite lists.

Our fix here is to use cg. cg gives all the contiguous subsequences of the input, possibly infinite. So we first run cg and then run this finite subsequence on every result.

Unlike the first solution this

## Reflections

Nice to see that we used no 3 byte functions anywhere. Still

• ap[i,p]<p is a little long for my liking (<p)<ap could probably be it's own function. I think it would have use beyond this challenge.