8 adjacent squares

Question

List the coordinates of 8 squares adjacent to (0, 0) in a 2d grid. Namely,

[(1,-1),(1,0),(1,1),(0,1),(-1,1),(-1,0),(-1,-1),(0,-1)]

Order does not matter. The goal is to store the points into a list. Shortest answer wins.

Answer 1

J - 12 char

The most natural representation of a list of pairs is a list of boxes containing each pair.

   }.4|.,{;~i:1
+---+----+---+---+-----+----+----+----+
|0 1|1 _1|1 0|1 1|_1 _1|_1 0|_1 1|0 _1|
+---+----+---+---+-----+----+----+----+

The magic happens in {, called Catalogue. It is essentially a generalized Cartesian product. The }.4|. part is for removing the 0 0 pair.

Answer 2

Bash + grep 35

eval echo\ {-1..1},{-1..1}\;|grep 1

I know this doesn't beat the other bash answer but someone had to think of grep

Answer 3

Ruby 29 28 26

(a=-1,1,0).product(a)[0,8]

Bonus solution:

Ruby has many nice builtin array operators, but some of them have ridiculously long names.

[*[-1,1,0].repeated_permutation(2)].take 8

Answer 4

Python 43 32

l=zip([1,-1,0]*3,[-1,1]*3+[0]*2)

Inspired by @primo

Answer 5

GolfScript (17 chars)

9,4-{[.3/(\3%(]}%

Since the question calls for the points to be "listed in natural format", this leaves an array on the stack. To pretty-print it, append p, as in the online demo.

Answer 6

Haskell 32

c=[0,1,-1]
_:l=[(a,b)|a<-c,b<-c]

The list is l (L).

Answer 7

Brainfuck, 1093 characters.

++++++++[>+>++>+++>++++>+++++>++++++>+++++++>++++++++>+++++++++>++++++++++>+++++++++++>++++++++++++>+++++++++++++>++++++++++++++>+++++++++++++++>++++++++++++++++<<<<<<<<<<<<<<<<-]>>>>>>>>>>>+++.---<<<<<<<<<<<>>>>>.<<<<<>>>>>>+.-<<<<<<>>>>>>----.++++<<<<<<>>>>>>---.+++<<<<<<>>>>>>+.-<<<<<<>>>>>+.-<<<<<>>>>>>----.++++<<<<<<>>>>>.<<<<<>>>>>>+.-<<<<<<>>>>>>----.++++<<<<<<>>>>>>.<<<<<<>>>>>+.-<<<<<>>>>>>----.++++<<<<<<>>>>>.<<<<<>>>>>>+.-<<<<<<>>>>>>----.++++<<<<<<>>>>>>+.-<<<<<<>>>>>+.-<<<<<>>>>>>----.++++<<<<<<>>>>>.<<<<<>>>>>>.<<<<<<>>>>>>----.++++<<<<<<>>>>>>+.-<<<<<<>>>>>+.-<<<<<>>>>>>----.++++<<<<<<>>>>>.<<<<<>>>>>>---.+++<<<<<<>>>>>>+.-<<<<<<>>>>>>----.++++<<<<<<>>>>>>+.-<<<<<<>>>>>+.-<<<<<>>>>>>----.++++<<<<<<>>>>>.<<<<<>>>>>>---.+++<<<<<<>>>>>>+.-<<<<<<>>>>>>----.++++<<<<<<>>>>>>.<<<<<<>>>>>+.-<<<<<>>>>>>----.++++<<<<<<>>>>>.<<<<<>>>>>>---.+++<<<<<<>>>>>>+.-<<<<<<>>>>>>----.++++<<<<<<>>>>>>---.+++<<<<<<>>>>>>+.-<<<<<<>>>>>+.-<<<<<>>>>>>----.++++<<<<<<>>>>>.<<<<<>>>>>>.<<<<<<>>>>>>----.++++<<<<<<>>>>>>---.+++<<<<<<>>>>>>+.-<<<<<<>>>>>+.-<<<<<>>>>>>>>>>>>---.+++<<<<<<<<<<<<.

EDIT 209 characters

++++++++++[>+++++++++>++++>+++++>++++<<<<-]>+.>.>-.>++++.<.<+.>>.<<-.>.>.<-.<+.>>.<<-.>+.>.+.-<.<+.>>.<<-.>-.>.<+.<+.>>.<<-.>-.>.+.<+.<+.>>-.<<-.>>+.<.>-.<.<+.>>.<<-.>>+.<.>-.<-.<+.>>.<<-.>>+.<+.>-.+.<.<+.<++.

Original by Benjamin - i gave it another shot as well and got to 209 characters as well.

Answer 8

Perl, 43 bytes

for$x(-1..1){$x|$_&&push@r,[$x,$_]for-1..1}

The result is stored in array @r. Explicitly it can be defined with 58 bytes:

@r=([-1,-1],[-1,0],[-1,1],[0,-1],[0,1],[1,-1],[1,0],[1,1])

Answer 9

J, 16 characters

   }.4|.,<@,"0/~i:1

Usage:

   }.4|.,<@,"0/~i:1
┌───┬────┬───┬───┬─────┬────┬────┬────┐
│0 1│1 _1│1 0│1 1│_1 _1│_1 0│_1 1│0 _1│
└───┴────┴───┴───┴─────┴────┴────┴────┘

Pretty ugly - especially dropping the (0, 0). There's probably a better way.

Answer 10

Mathematica 23

Most@Tuples[{-1,1,0},2]
(*
{{-1, -1}, {-1, 1}, {-1, 0}, {1, -1}, {1, 1}, {1, 0}, {0, -1}, {0, 1}}
*)

Answer 11

Brainfuck, 217 209 characters.

Only a bit worse than Java! 5x less than the other BF solution.

++++++++++[>+++++++++>++++>+++++>++++<<<<-]>+.>.>-.>++++.+.<.<+.>>-.<<-.>.>.<-.<+.>>.<<-.>+.>.<.<+.>>.<<-.>-.>.<+.<+.>>.<<-.>>+.<.>-.<.<+.>>.<<-.>>+.<.>-.<-.<+.>>.<<-.>>+.<+.>-.+.<.<+.>>-.<<-.>-.>.+.<+.<+.<++.

Answer 12

Mathematica, 44 characters

Outer[List,d=Range@3-2,d]~Flatten~1~Drop~{5}

Well. Generates all 9 coordinates, flattens the list and drops the centre.

Answer 13

Python 50:

l=[(a,b)for a in[1,0,-1]for b in[1,0,-1]];l.pop(4)

Answer 14

Bash, 25 characters

Storing the list of coordinates as a bash array a:

a=({-1..1},{-1..1});a[4]=

Output:

$ a=({-1..1},{-1..1});a[4]=
$ echo ${a[@]}
-1,-1 -1,0 -1,1 0,-1 0,1 1,-1 1,0 1,1
$

Answer 15

F# - 41 36 bytes

[for i in 0..7->i*2/5-1,(i*5/4)%3-1]

Answer 16

Python - 57 54 56

r=range(-1,2);list({(i,j) for i in r for j in r}-{(0,0)})

Improved to

r=[-1,0,1];list({(i,j) for i in r for j in r}-{(0,0)})

with storage of the variable

r=[-1,0,1];l=list({(i,j) for i in r for j in r}-{(0,0)})

Answer 17

C++, 60

Let's do this an interesting way. Hint: "2" is ASCII 50.

char x[]="11123332",y[8],i;for(i=8;i--;)y[(i+2)%8]=x[i]-=50;

Answer 18

Perl 6: 27 characters

Most simply:

$/=(-1..1 X -1..1)[^4,5..*]

Get the permutations of -1, 0, and 1 with itself, and hard-code skipping (0, 0). But that's no fun! I prefer things to be more complex: (50 characters)

$/=map {[.re,.im]},grep *.abs>0,(-1..1 X+ -i,0i,i)

Answer 19

Python - 37 bytes

a=[(i*2/5-1,-i%3-1)for i in range(8)]

An alternative at 40:

a=zip([0,1,-1]*3,[0]*3+[1]*3+[-1]*3)[1:]

And a 42 that's also interesting:

a=0,1,-1;a=zip(sum(zip(a,a,a),()),a*3)[1:]

Answer 20

Java 207 204

Way too much, I know, but its the best I could do right now...
I might think of something better...

class A{public static void main(String...a){int[][] l=new int[8][2];for(int x=-1,y,i=0;x<2;x++){for(y=-1;y<2;y++){if(x!=0||y!=0){l[i][0]=x;l[i][1]=y;System.out.print("("+l[i][0]+","+l[i][1]+")");i++;}}}}}

Sensibly:

class A {
    public static void main(String...a) {
        int[][] l = new int[8][2];
        for (int x = -1, y, i = 0; x < 2; x++) {
            for (y = -1; y < 2; y++) {
                if (x != 0 || y != 0) {
                    l[i][0] = x;
                    l[i][1] = y;
                    System.out.print("("+l[i][0]+","+l[i][1]+")");
                    i++;
                }
            }
        }
    }
}

Answer 21

APL (13 characters)

1↓,∘.,⍨0 1 ¯1

Explanation:

1↓           ⍝ Drop the first element of
  ,          ⍝ an array that comes from the matrix formed by
   ∘.,⍨      ⍝ concatenating each element with every other element
       0 1 ¯1⍝ in this array [0,1,-1]

Answer 22

PYG (27):

P(*ItPr((-1,0,1),repeat=2))

Nothing all that special, really.

In regular python:

from itertools import product
for i in product((-1,0,1),repeat=2): print i

Answer 23

C, 73

main(i){for(i=0;i<9;i++)if(i-4)printf("%s(%d,%d)",i?",":"",i%3-1,i/3-1);}

Answer 24

Mathematica: 41

This works most of the time ;)

Union[RandomInteger[2,{99,2}]-1]~Drop~{5}

Equivalently:

Drop[Union@RandomInteger[2,{99,2}]-1,{5}]

Answer 25

Matlab: 53

Similar to my other answer:

unique(round(2*rand(99,2))-1,'rows');ans([1:4,6:9],:)

I couldn't index on the output of unique directly, and I couldn't find a way around that...

Answer 26

Haskell 97

Too much for Haskell and definitely not as good as the other Haskell answer, but here's a different, albeit sucky, approach:

take 8 (zip (replicate 3 (-1) ++ replicate 3 1 ++ replicate 3 0) (concat (replicate 3 [-1,1,0])))

:D

Answer 27

Scala, 56

Self-explanatory.

def x=List(-1,0,1);for{i<-x;j<-x;if i!=0|j!=0}yield(i,j)

Ungolfed:

def x=List(-1,0,1)
for{
  i <- x
  j <- x
  if i!=0 | j!=0
} yield(i,j)

Answer 28

Perl, 42 characters:

@r=map[split","],grep/1/,glob"{-1,0,1},"x2

Uses the shell-like glob functionality to generate the list of strings ("-1,-1","-1,0","-1,1","0,-1","0,0","0,1","1,-1","1,0","1,1"), discards strings that don't contain the character "1" (i.e. drop "0,0"), then splits the strings on comma to get the desired data structure.

41 characters:

@r=map[split x],grep/1/,glob"{-1,0,1}x"x2

abuses barewords: the strings are ("-1x-1","-1x0", etc.) and the split is done on the letter x instead.

Answer 29

JavaScript (ES5) 52 bytes

[[n=-1,n],[n,0],[n,1],[0,n],[0,1],[1,n],[1,0],[1,1]]

---

Javascript (ES5) (algorithmic) 57 bytes

for(a=-2,x=[];++a<2;)for(b=-2;++b<2;)a|b&&x.push([a,b]);x

Answer 30

Fortran: 101 96

Fortran supports complex variables, so we just store the 8 values as a set of complex numbers and get ourselves a list of tuples.

complex x(8);x=reshape([(1,-1),(1,0),(1,1),(0,1),(-1,1),(-1,0),(-1,-1),(0,-1)],[8]);print*,x;end

Printing this to look nice might be a pain in the butt because printing the real and aimag components adds a lot of bytes. But we can deal with floats.

---

Old answer:

complex x(8);k=1;do i=-1,1;do j=-1,1;if((j==0).and.(i==0))cycle;x(k)=cmplx(i,j);k=k+1;enddo;enddo;end

This abuses implicit typing (as usual).

Answer 31

SQL 109 (that far)

create table a(x int);insert into a values(-1),(0),(1);select a.x,b.x y from a,a b where not(a.x=0 and b.x=0)

if you want to retest it, you can prepend drop table if exists a;

Answer 32

C# 140 chars

Shorter version:

List<string> a=new List<string>();for(int i=0,m=-1;i<3;i++,m++){for(int j=0,n=-1;j<3;j++,n++){if(m==0 && n==0)continue;array.Add(m+","+n);}}

Easy to read version:

List<string> array = new List<string>();
for (int i = 0, m = -1; i < 3; i++, m++)
{
    for(int j = 0, n = -1; j < 3; j++, n++)
    {
         if (m == 0 && n == 0) continue;
         array.Add(m + "," + n);
    }
}