Pali-n-drome this List

The challenge here is to extend an implementation of palindrome given the following as inputs:

• n > 1 and a list l.

Your program must palindrome the list both vertically and horizontally, that is to say it must first palindrome the list itself, then each element in the list after; or the other way around. Before palindromization, all elements are ensured to be equal length. The palindrome action is then to be performed n times in sequence until the desired output is met. The easiest way to show the expected outputs is just to run through a few examples:

One iteration performed on [123,456,789]:

First you palindromize the list to [123,456,789,456,123].

• While this is not a palindrome if joined together, it is a palindrome in terms of the list.
• [a,b,c] became [a,b,c,b,a], so the LIST was palindromized.

Then, you palindromize each list element [12321,45654,78987,45654,12321].

This is how each iteration is performed, it's essentially an omnidirectional palindrome.

Given n=1 and l=[123,456,789]:

12321
45654
78987
45654
12321


Given n=2 and l=[123,456,789]

123212321
456545654
789878987
456545654
123212321
456545654
789878987
456545654
123212321


Given n=1 and l=[3,2,1]:

3
2
1
2
3


Given n=2 and l=["hat","mad"," a "]:

hatahatah
a a a a
hatahatah
a a a a
hatahatah


Given n=2 and l=[" 3 ","2000"," 100"]:

 3   3 3   3
2000002000002
100 00100 001
2000002000002
3   3 3   3
2000002000002
100 00100 001
2000002000002
3   3 3   3


Given n=4 and l=["3 ","20","1 "]:

3 3 3 3 3 3 3 3 3
20202020202020202
1 1 1 1 1 1 1 1 1
20202020202020202
3 3 3 3 3 3 3 3 3
20202020202020202
1 1 1 1 1 1 1 1 1
20202020202020202
3 3 3 3 3 3 3 3 3
20202020202020202
1 1 1 1 1 1 1 1 1
20202020202020202
3 3 3 3 3 3 3 3 3
20202020202020202
1 1 1 1 1 1 1 1 1
20202020202020202
3 3 3 3 3 3 3 3 3
20202020202020202
1 1 1 1 1 1 1 1 1
20202020202020202
3 3 3 3 3 3 3 3 3
20202020202020202
1 1 1 1 1 1 1 1 1
20202020202020202
3 3 3 3 3 3 3 3 3
20202020202020202
1 1 1 1 1 1 1 1 1
20202020202020202
3 3 3 3 3 3 3 3 3
20202020202020202
1 1 1 1 1 1 1 1 1
20202020202020202
3 3 3 3 3 3 3 3 3


Given n=3 and l=["_|__","__|_","___|"]:

_|___|_|___|_|___|_|___|_
__|_|___|_|___|_|___|_|__
___|_____|_____|_____|___
__|_|___|_|___|_|___|_|__
_|___|_|___|_|___|_|___|_
__|_|___|_|___|_|___|_|__
___|_____|_____|_____|___
__|_|___|_|___|_|___|_|__
_|___|_|___|_|___|_|___|_
__|_|___|_|___|_|___|_|__
___|_____|_____|_____|___
__|_|___|_|___|_|___|_|__
_|___|_|___|_|___|_|___|_
__|_|___|_|___|_|___|_|__
___|_____|_____|_____|___
__|_|___|_|___|_|___|_|__
_|___|_|___|_|___|_|___|_


Given n=2 and l=["---|---","__|","___|","____|"]:

---|-----|-----|-----|---
__|   |__   __|   |__
___|   |___ ___|   |___
____| |____ ____| |____
___|   |___ ___|   |___
__|   |__   __|   |__
---|-----|-----|-----|---
__|   |__   __|   |__
___|   |___ ___|   |___
____| |____ ____| |____
___|   |___ ___|   |___
__|   |__   __|   |__
---|-----|-----|-----|---


Rules

• n will always be greater than 1.
• l will always have more than 1 element.
• All elements of l are the same length.
• This is shortest solution will be marked as winner.
• This would be a better challenge if we didn't have to pad elements. – mbomb007 Mar 10 '17 at 17:01
• @JonathanAllan it's an omnidirectional palindrome, or 2D palindrome you could say. I've updated the description; also, the padding prevents a few odd fringe cases where a smaller string is already a palindrome. – Magic Octopus Urn Mar 10 '17 at 17:07
• @JonathanAllan it is in terms of the list, if you are looking at the LIST as the item to be palindromized. Just like [@1,@2,@1] is also a palindrome when looking at it as a list, not by the elements... – Magic Octopus Urn Mar 10 '17 at 17:08
• @JonathanAllan yeah, essentially, you can look at it like that if you want. – Magic Octopus Urn Mar 10 '17 at 17:11
• Last example requires padding. – Jonathan Allan Mar 10 '17 at 17:20

05AB1E, 4 bytes

Note that if only a single iteration was required (n=1), then the program would be the palindrome û€û.

Fû€û


Try it online

F       Do n times
û      Palindromize the list
€û    Palindromize each element in the list


If padding the input was still a required part of the program (11 bytes):

€R.B€RIFû€û


I couldn't find a shorter way to right-justify. Left-justification and centering were all easy, but this was longer for some reason. Using E or ² instead of I also works.

Python 2, 71 63 bytes

lambda x,n,f=lambda x:x+x[-2::-1]:eval('f(map(f,'*n+x+'))'*n)


Try it online!

Assign a palindrome function to f, generate and evaluate the following pattern (for n=4)
f(map(f,f(map(f,f(map(f,f(map(f,<input>))))))))

• I think you mean assign. I don't think assing is a verb, lol. – mbomb007 Mar 10 '17 at 18:08
• @mbomb007 welp, time to get more coffee~ – Rod Mar 10 '17 at 18:18

Jelly, 6 bytes

ŒḄŒB$¡  Dyadic link, or full program taking the list and n. Try it online! Using both versions of Lynn's fantastic built-in "bounce". ŒḄŒB$¡ - Main link: l, n
¡ - repeat n times
$- last two links as a monad (firstly with l then the result...) ŒḄ - bounce ("palindromise") the list ŒB - bounce the elements  Python 2, 64 bytes h=lambda a:a+a[-2::-1] f=lambda a,n:n and f(h(map(h,a)),n-1)or a  Try it online! - footer prints each of the elements of the resulting list, one per line, a "pretty print". h is the palindomisation function, it appends to the input, all the elements of a list from the last but one, index -2, to the start in steps of size -1. f calls h with the result of calling h on each element in turn, reduces n by one and calls itself until n reaches 0, at which point a is the finished product. • ...and I am still forgetting the f= for recursive functions, one day I'll remember. – Jonathan Allan Mar 10 '17 at 18:18 APL, 15 bytes (Z¨Z←⊢,1↓⌽)⍣⎕⊢⎕  Explanation: • (...)⍣⎕⊢⎕: read the list and N as input, and run N times: • ⊢,1↓⌽: the list, followed by the tail of the reversed list • Z←: store this function in Z • Z¨: and apply it to each element of the list as well Test:  (Z¨Z←⊢,1↓⌽)⍣⎕⊢⎕ ⎕: 'hat' 'mad' ' a ' ⎕: 2 ┌─────────┬─────────┬─────────┬─────────┬─────────┬─────────┬─────────┬─────────┬─────────┐ │hatahatah│madamadam│ a a a a │madamadam│hatahatah│madamadam│ a a a a │madamadam│hatahatah│ └─────────┴─────────┴─────────┴─────────┴─────────┴─────────┴─────────┴─────────┴─────────┘  Groovy, 66 bytes {x,n->f={z->z+z[z.size()-2..0]};n.times{x=f(x).collect{f(it)}};x}  Haskell, 51 bytes x%n=iterate((++)<*>reverse.init)x!!n x?n=(%n)<$>x%n


Usage example: ["123","456","789"] ? 1 -> ["12321","45654","78987","45654","12321"]. Try it online!.

(++)<*>reverse.init makes a palindrome out of a list, iterate(...)x repeats this again and again and collects the intermediate results in a list, !!n picks the nth element of this list. (%n)<\$>x%n makes a n-palindrom of each element of the n-palindrome of x.

JavaScript (ES6), 87 bytes

f=(n,l,r=l=>[...a].reverse().slice(1))=>n--?f(l.concat(r(l)).map(s=>s+r(s).join),n):l


Pip, 25 bytes

24 bytes of code, +1 for -l flag.

Lq{gM:_.@>RV_gAL:@>RVg}g


Takes the list as command-line arguments and the number n from stdin. Try it online!

Explanation

                          g is list of cmdline args (implicit)
Lq{                   }   Read a line of input and loop that many times:
_.@>RV_             Lambda function: take all but the first character (@>) of the
reverse (RV) of the argument (_), and concatenate that (.) to
the argument (_)
gM:                    Map this function to g and assign the result back to g
@>RVg    Take all but the first element of the reverse of g
gAL:         Append that list to g and assign the result back to g
g  After the loop, print g (each item on its own line due to -l)