# Problem

Let's define a generalized Cantor set by iteratively deleting some rational length segments from the middle of all intervals that haven't yet been deleted, starting from a single continuous interval.

Given the relative lengths of segments to delete or not, and the number of iterations to do, the problem is to write a program or function that outputs the relative lengths of the segments that have or have not been deleted after n iterations.

Example: Iteratively delete the 4th and 6th eighth

# Input:

n – number of iterations, indexed starting from 0 or 1

l – list of segment lengths as positive integers with gcd(l)=1 and odd length, representing the relative lengths of the parts that either stay as they are or get deleted, starting from a segment that doesn't get deleted. Since the list length is odd, the first and last segments never get deleted. For example for the regular Cantor set this would be [1,1,1] for one third that stays, one third that gets deleted and again one third that doesn't.

# Output:

Integer list o, gcd(o)=1, of relative segment lengths in the nth iteration when the segments that weren't deleted in the previous iteration are replaced by a scaled down copy of the list l. The first iteration is just [1]. You can use any unambiguous output method, even unary.

# Examples

n=0, l=[3,1,1,1,2] →                 [1]
n=1, l=[3,1,1,1,2] →     [3,    1,    1,    1,    2]
n=2, l=[3,1,1,1,2] → [9,3,3,3,6,8,3,1,1,1,2,8,6,2,2,2,4]

n=3, l=[5,2,3]     → [125,50,75,100,75,30,45,200,75,30,45,60,45,18,27]
n=3, l=[1,1,1]     → [1,1,1,3,1,1,1,9,1,1,1,3,1,1,1]


You can assume the input is valid. This is , so the shortest program measured in bytes wins.

• Would it be acceptable to input and output the indices of non-deleted segments instead of the lengths? For instance, [0, 1, 2, 4, 6, 7] instead of [3, 1, 1, 1, 2]?
– user48543
May 22 '18 at 15:06
• @Mnemonic it's not too far from unary, so I'd say it's fine.
– Angs
May 22 '18 at 15:25
• Could you add one (or multiple) test case(s) for even-sized input-lists? May 23 '18 at 7:09
• @KevinCruijssen the input lists are guaranteed to be odd-sized
– Angs
May 23 '18 at 7:18

# Jelly,  15 13  12 bytes

-2 thanks to Dennis (using a Link rather than a chain allows right to be used implicitly by ¡; No need to wrap the 1 in a list due to the fact that Jelly prints lists of one item the same as the item)
-1 thanks to Erik the Outgolfer (use Ɗ to save the newline from using Ç)

1×€³§JḤ$¦ẎƊ¡  A full program printing a list in Jelly format (so [1] is printed as 1) Try it online! ### How? 1×€³§JḤ$¦ẎƊ¡ - Main link: segmentLengths; iterations
1            - literal 1 (start with a single segment of length 1)
¡ - repeat...
- ...times: implicitly use chain's right argument, iterations
Ɗ  - ...do: last 3 links as a monad (with 1 then the previous output):
³         - (1) program's 3rd argument = segmentLengths
×€          -  1  multiply €ach (e.g. [1,2,3] ×€ [1,2,1] = [[1,4,3],[2,4,2],[3,6,3]])
¦    -  2  sparse application...
$- (2) ...to: indices: last two links as a monad: J - (2) range of length = [1,2,3,...,numberOfLists] Ḥ - (2) double [2,4,6,...] (note: out-of bounds are ignored by ¦) § - (2) ...of: sum each (i.e. total the now split empty spaces) Ẏ - 3 tighten (e.g. [[1,2,3],4,[5,6,7]] -> [1,2,3,4,5,6,7]) - implicit print  # Python 2, 12010710410310099 89 bytes f=lambda n,l:n and[x*y for i,x in enumerate(l)for y in[f(n-1,l),[sum(l)**~-n]][i%2]]or[1]  Try it online! Saved • -10 bytes, thanks to Neil • 89 bytes – Neil May 22 '18 at 21:47 • @Neil, Thanks :) May 23 '18 at 7:00 # R, 94 bytes f=function(n,a)"if"(n,unlist(Map(function(g,i)g(i),c(c,sum),split(m<-a%o%f(n-1,a),row(m)))),1)  Try it online! # Haskell, 76 58 bytes l%0=[1] l%n=do(x,m)<-l%(n-1)zipcycle[l,[sum l]];map(*x)m  Try it online! The function (%) takes the list of line lengths l as first argument and the number of iterations n as second input. Thanks to Angs and Ørjan Johansen for -18 bytes! • You should be able to save at least 7 bytes by switching to a recursion on n and dropping # altogether – Angs May 23 '18 at 23:59 • Independently of @Angs 's suggestion, the original % can be shortened to l%a=do(x,m)<-zip a$a>>[l,[sum l]];(*x)<$>m . May 24 '18 at 18:09 ## JavaScript (Firefox 42-57), 80 bytes f=(n,l,i=0)=>n--?[for(x of l)for(y of(i^=1)?f(n,l):[eval(l.join+)**n])x*y]:[1]  Needs those specific versions because it uses both array comprehensions and exponentiation. # JavaScript (Node.js), 71 bytes l=>f=(n,c=1,e)=>n--?''+l.map(_=>(e=!e)?f(n,c*_):eval(l.join+)**n*c):c  Try it online! # Java 10, 261 bytes L->n->{if(n<1){L.clear();L.add(1);}else if(n>1){var C=new java.util.ArrayList<Integer>(L);for(int l=C.size(),i,x,t,s;n-->1;)for(i=x=0;i<L.size();){t=L.remove(i);if(i%2<1)for(;i%-~l<l;)L.add(i,C.get((i++-x)%l)*t);else{x++;s=0;for(int c:C)s+=c;L.add(i++,t*s);}}}}  Modifies the input-List instead of returning a new one to save bytes. Try it online. L->n->{ // Method with List and integer parameters and no return-type if(n<1){ // If n is 0: L.clear(); // Remove everything from the List L.add(1);} // And only add a single 1 // Else-if n is 1: Leave the List as is else if(n>1){ // Else-if n is 2 or larger: var C=new java.util.ArrayList<Integer>(L); // Create a copy of the input-List for(int l=C.size(), // Set l to the size of the input-List i,x,t,s; // Index and temp integers n-->1;) // Loop n-1 times: for(i=x=0; // Reset x to 0 i<L.size();){ // Inner loop i over the input-List t=L.remove(i); // Remove the current item, saving its value in t if(i%2<1) // If the current iteration is even: for(;i%-~l<l;) // Loop over the copy-List L.add(i,C.get((i++-x)%l)*t); // And add the values multiplied by t // at index i to the List L else{ // Else (the current iteration is odd): x++; // Increase x by 1 s=0;for(int c:C)s+=c; // Calculate the sum of the copy-List L.add(i++,t*s);}}}} // Add this sum multiplied by t // at index i to the List L  # Jelly, 13 bytes Ø1××S¥ƭ€³Ẏ$¡Ṗ


Try it online!

Full program. Outputs 1 instead of [1]. Annoyingly, ḋ doesn't work like ×S¥ in this context, and ƭ doesn't work well with nilads. >_<

# APL (Dyalog Classic), 20 bytes

{(∊⊢×⍵(+/⍵)⍴⍨≢)⍣⍺,1}


Try it online!

# K (ngn/k), 27 bytes

{x{,/y*(#y)#x}[(y;+/y)]/,1}


Try it online!

{ } is a function with arguments x and y

(y;+/y) a pair of y and its sum

{ }[(y;+/y)] projection (aka currying or partial application) of a dyadic function with one argument. x will be (y;+/y) and y will be the argument when applied.

,1 singleton list containing 1

x{ }[ ]/ apply the projection x times

(#y)#x reshape to the length of the current result, i.e. alternate between the outer y and its sum

y* multiply each element of the above with the corresponding element of the current result

,/ concatenate

# Ruby, 73 bytes

->a,l{r=[1];a.times{s=p;r=r.flat_map{|x|(s=!s)?l.map{|y|x*y}:x*l.sum}};r}


Try it online!

# Pyth, 20 bytes

us.e?%k2*bsQ*LbQGE]1


Input is segment array l, then iterations n. Try it online here, or verify all the test cases at once here.

us.e?%k2*bsQ*LbQGE]1   Implicit, Q=1st arg (segment array), E=2nd arg (iterations)
u                E     Execute E times, with current value G...
]1   ... and initial value [1]:
.e            G        Map G, with element b and index k:
*bsQ               Multiply b and the sum of Q {A}
*LbQ           Multiply each value of Q by b {B}
?%k2                   If k is odd, yield {A}, otherwise yield {B}
s                       Flatten the resulting nested array