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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.

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4
  • 1
    \$\begingroup\$ Can we write a function that prints the result? \$\endgroup\$ Mar 20, 2016 at 11:44
  • \$\begingroup\$ @LegionMammal978 I would prefer a complete program. If this is deemed seriously controversial I will give in of course :) \$\endgroup\$
    – user9206
    Mar 20, 2016 at 11:45
  • \$\begingroup\$ Is there any required order within each of your bullet points? \$\endgroup\$ Mar 20, 2016 at 11:47
  • \$\begingroup\$ @MartinBüttner No, not at all. \$\endgroup\$
    – user9206
    Mar 20, 2016 at 12:56

5 Answers 5

6
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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.
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2
  • \$\begingroup\$ Now it's just getting silly! \$\endgroup\$
    – user9206
    Mar 20, 2016 at 16:37
  • 2
    \$\begingroup\$ "No lists were harmed in the making of this post" \$\endgroup\$ Mar 20, 2016 at 23:30
6
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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
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1
  • 1
    \$\begingroup\$ 18 bytes is outrageous :) \$\endgroup\$
    – user9206
    Mar 20, 2016 at 13:07
4
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Haskell, 61 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.

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2
  • \$\begingroup\$ mapM id is shorter than sequence. \$\endgroup\$
    – xnor
    Mar 20, 2016 at 23:31
  • \$\begingroup\$ @xnor: True. Thanks! \$\endgroup\$
    – nimi
    Mar 21, 2016 at 0:29
2
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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!)

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1
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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.

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