Print arrays from the middle out

Question

This is a code-golf question.

Given integers s and n the task is to output all arrays of length n which take values from -s to s. The only twist is that you must output them in the following order.

• The all zeros array of length n.
• All arrays of length n with elements from -1 to 1 excluding any array you have outputted before.
• All arrays of length n with elements from -2 to 2 excluding any array you have outputted before.
• And so on until you get to all arrays of length n with elements from -s to s excluding any array you have outputted before.

You should output one array per line. They can be space or comma separated.

Here is some non-complying python code that outputs the arrays/lists/tuples in the right order.

import itertools
s = 3
n = 2
oldsofar = set()
newsofar = set()
for i in xrange(s):
 for k in itertools.product(range(-i,i+1), repeat = n):
 newsofar.add(k)
 print newsofar - oldsofar
 oldsofar = newsofar.copy()
 print "***"

Extra glory (and an upvote from me) for answers that perform no set subtraction or equivalent.

• LegionMammal978: Can we write a function that prints the result?
• user9206: @LegionMammal978 I would prefer a complete program. If this is deemed seriously controversial I will give in of course :)
• Martin Ender: Is there any required order within each of your bullet points?
• user9206: @MartinBüttner No, not at all.

Answer 1

Jelly, 9 bytes

NRṗμAṀ€Ụị

No list subtraction was used in the making of this post. Try it online!

How it works

NRṗμAṀ€Ụị Main link. Arguments: s, n
N Negate; yield -s.
 R Range; yield [-s, ..., s].
 ṗ Cartesian power; push all vectors of length n of those elements.
 μ Begin a new, monadic link. Argument: L (list of vectors)
 A Compute the absolute values of all vector components.
 Ṁ€ Get the maximum component of each vector.
 Ụ Sort the indices of A according to the maximal absolute value of the
 corresponding vector's components.
 ị Retrieve the vectors of A at those indices.

• user9206: Now it's just getting silly!
• Dennis van Gils: "No lists were harmed in the making of this post"

Answer 2

MATL, 18 bytes

_G2$:iZ^t!|X>4#SY)

First input is s, second is n

This works in current version (15.0.0) of the language.

Try it online!

Explanation

_ % take input s implicitly. Negate to obtain -s
G % push input s again
2$: % inclusive range from -s to s
i % take input n
Z^ % Cartesian power. Gives 2D array, with each result on a row
t! % duplicate and transpose
| % absolute value
X> % maximum of each column 
4#S % sort and push the indices of the sorting
Y) % apply as row indices into the 2D array. Display implicitly

• user9206: 18 bytes is outrageous :)

Answer 3

Haskell, (削除) 61 (削除ここまで) 60 bytes

n#s=[c|b<-[0..s],c<-mapM id$[-b..b]<$[1..n],any((b==).abs)c]

Usage example: 2#2-> [[0,0],[-1,-1],[-1,0],[-1,1],[0,-1],[0,1],[1,-1],[1,0],[1,1],[-2,-2],[-2,-1],[-2,0],[-2,1],[-2,2],[-1,-2],[-1,2],[0,-2],[0,2],[1,-2],[1,2],[2,-2],[2,-1],[2,0],[2,1],[2,2]].

How it works:

 b<-[0..s] -- loop b through 0 .. s
 c<-mapM id$[-b..b]<$[1..n] -- loop c through all lists of length n
 -- made out of the numbers -b .. b
 -- ("[-b..b]<$[1..n]" is "replicate n [-b..b]";
 -- "mapM id" is "sequence")
[c| ,any((b==).abs)c] -- keep c if it contains b or -b

Edit: @xnor pointed out that mapM id is sequence.

• xnor: mapM id is shorter than sequence.
• nimi: @xnor: True. Thanks!

Answer 4

Mathematica, 83 bytes

Print/@Select[Range[-#,b=#]~Tuples~a,Abs@#~MemberQ~b&]&/@Range[0,a=Input[];Input[]]

To use, put in a script and input n then s on separate lines. Prints each array as a curly-bracketed, comma-delimited list (e.g., {-1, 0, 1}). It works by taking every list of length n with numbers between [-cur..cur], and and printing those which include either -cur or cur. It then repeats this for all cur in [0..s]. ^{(This post contains 19 ` characters!)}

Answer 5

JavaScript (SpiderMonkey 30+), 134 bytes

(s,n)=>n?[for(a of f(s,n-1))for(i of Array(s*2+1).keys())[i-n,...a]].sort((a,b)=>g(a)-g(b),g=a=>Math.max(...a,-Math.min(...a))):[[]]

Uses the cartesian-power-and-sort approach, which I thought of separately, but I was recompiling SpiderMonkey at the time so I unable to answer this before @Dennis.