Create a checkerboard matrix

Question

Take a positive integer n as input, and output a n-by-n checkerboard matrix consisting of 1 and 0.

The top left digit should always be 1.

Test cases:

n = 1
1

n = 2
1 0
0 1

n = 3
1 0 1
0 1 0
1 0 1

n = 4
1 0 1 0
0 1 0 1
1 0 1 0
0 1 0 1

Input and output formats are optional. Outputting the matrix as a list of lists is accepted.

Answer 1

Jelly, 4 bytes

52 seconds!

+€ḶḂ

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Answer 2

MATL, 5 bytes

:otYT

Try it at MATL online!

Explanation

Consider input 4 as an example.

:    % Implicit input, n. Push range [1 2 ... n]
     %   STACK: [1 2 3 4]
o    % Parity, element-wise
     %   STACK: [1 0 1 0]
t    % Duplicate
     %   STACK: [1 0 1 0], [1 0 1 0]
YT   % Toeplitz matrix with two inputs. Implicit display
     %   STACK: [1 0 1 0;
     %           0 1 0 1;
     %           1 0 1 0;
     5           0 1 0 1]

Answer 3

Japt, 6 bytes

ÆÇ+X v

Test it online! (Uses -Q flag for easier visualisation)

Explanation

 Æ   Ç   +X v
UoX{UoZ{Z+X v}}  // Ungolfed
                 // Implicit: U = input number
UoX{          }  // Create the range [0...U), and map each item X to
    UoZ{     }   //   create the range [0...U), and map each item Z to
        Z+X      //     Z + X
            v    //     is divisible by 2.
                 // Implicit: output result of last expression

An interesting thing to note is that v is not a "divisible by 2" built-in. Instead, it's a "divisible by X" built-in. However, unlike most golfing languages, Japt's functions do not have fixed arity (they can accept any number of right-arguments). When given 0 right-arguments, v assumes you wanted 2, and so acts exactly like it was given 2 instead of nothing.

Answer 4

V, 16, 15 bytes

Ài10À­ñÙxñÎÀlD

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Hexdump:

00000000: c069 3130 1bc0 adf1 d978 f1ce c06c 44    .i10.....x...lD

Answer 5

Haskell, 50 41 39 38 bytes

Thanks to nimi and xnor for helping to shave off a total of 9 10 bytes

f n=r[r"10",r"01"]where r=take n.cycle

Alternately, for one byte more:

(!)=(.cycle).take
f n=n![n!"10",n!"01"]

or:

r=flip take.cycle
f n=r[r"10"n,r"01"n]n

Probably suboptimal, but a clean, straightforward approach.

Answer 6

APL (Dyalog), 8 bytes

~2|⍳∘.+⍳

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Explanation

Let's call the argument n.

⍳∘.+⍳

This creates a matrix

1+1 1+2 1+2 .. 1+n
2+1 2+2 2+3 .. 2+n
...
n+1 n+2 n+3 .. n+n

Then 2| takes modulo 2 of the matrix (it vectorises) after which ~ takes the NOT of the result.

Answer 7

Mathematica, 25 bytes

1-Plus~Array~{#,#}~Mod~2&

Answer 8

JavaScript ES6, 55 54 51 46 bytes

Saved 1 byte thanks to @Neil

Saved 2 bytes thanks to @Arnauld

n=>[...Array(n)].map((_,i,a)=>a.map(_=>++i&1))

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This outputs as an array of arrays. JavaScript ranges are pretty unweildy but I use [...Array(n)] which generates an array of size n

Answer 9

Retina, 33 30 bytes

.+
$*
1
$_¶
11
10
T`10`01`¶.+¶

Try it online! Explanation: The first stage converts the input to unary using 1s (conveniently!) while the second stage turns the value into a square. The third stage inverts alternate bits on each row while the last stage inverts bits on alternate rows. Edit: Saved 3 bytes thanks to @MartinEnder.

Answer 10

MATL, 7 bytes

:t!+2\~

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Explanation:

         % Implicit input (n)
:        % Range from 1-n, [1,2,3]
 t       % Duplicate, [1,2,3], [1,2,3]
  !      % Transpose, [1,2,3], [1;2;3]
   +     % Add        [2,3,4; 3,4,5; 4,5,6]
    2    % Push 2     [2,3,4; 3,4,5; 4,5,6], 2
     \   % Modulus    [0,1,0; 1,0,1; 0,1,0]
      ~  % Negate     [1,0,1; 0,1,0; 1,0,1]

_{Note: I started solving this in MATL after I posted the challenge.}

Answer 11

Java (OpenJDK 8), 80 77 bytes

-3 bytes thanks to Kevin Cruijssen

j->{String s="1";for(int i=1;i<j*j;s+=i++/j+i%j&1)s+=1>i%j?"\n":"";return s;}

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Oh look, a semi-reasonable length java answer, with lots of fun operators.

lambda which takes an int and returns a String. Works by using the row number and column number using / and % to determine which value it should be, mod 2;

Ungolfed:

j->{
    String s="1";
    for(int i=1; i<j*j; s+= i++/j + i%j&1 )
        s+= 1>i%j ? "\n" : "";
    return s;
}

Answer 12

Brachylog, 15 bytes

^₂⟦₁%₂ᵐ;?ḍ₎pᵐ.\

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Explanation

Example Input: 4

^₂               Square:                            16
  ⟦₁             1-indexed Range:                   [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16]
    %₂ᵐ          Map Mod 2:                         [1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0]
       ;?ḍ₎      Input-Chotomize:                   [[1,0,1,0],[0,1,0,1],[1,0,1,0],[0,1,0,1]]
           pᵐ.   Map permute such that..
             .\  ..the output is its own transpose: [[1,0,1,0],[0,1,0,1],[1,0,1,0],[0,1,0,1]]

Answer 13

Clojure, 36 bytes

#(take %(partition % 1(cycle[1 0])))

Yay, right tool for the job.

Answer 14

05AB1E, 9 7 bytes

-2 bytes thanks to Emigna

LDÈD_‚è

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Explanation

LDÈD_‚sè» Argument n
LD        Push list [1 .. n], duplicate
  ÈD      Map is_uneven, duplicate
    _     Negate boolean (0 -> 1, 1 -> 0)
     ‚    List of top two elements of stack
      è   For each i in [1 .. n], get element at i in above created list
          In 05AB1E the element at index 2 in [0, 1] is 0 again

Answer 15

Charcoal, 8 bytes

ＵＯＮ10¶01

Try it online! Explanation: This roughly translates to the following verbose code (unfortunately the deverbosifier is currently appending an unnecessary separator):

Oblong(InputNumber(), "10\n01");

Answer 16

C, 69 67 63 bytes

Thanks to @Kevin Cruijssen for saving two bytes and @ceilingcat for saving four bytes!

i,j;f(n){for(i=n;i--;puts(""))for(j=n;j;)printf("%d",j--+i&1);}

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Answer 17

Pyth, 9 bytes

VQm%+hdN2

Try this!

another 9 byte solution:

mm%+hdk2Q

Try it!

Answer 18

J, 9 bytes

<0&=$&1 0

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Answer 19

Octave, 24 bytes

@(n)~mod((a=(1:n))+a',2)

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---

Or the same length:

@(n)mod(toeplitz(1:n),2)

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Answer 20

Mathematica, 23 bytes

ToeplitzMatrix@#~Mod~2&

Answer 21

R, 38 37 bytes

n=scan();(matrix(1:n,n,n,T)+1:n-1)%%2

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-1 byte thanks to Giuseppe

Takes advantage of R's recycling rules, firstly when creating the matrix, and secondly when adding 0:(n-1) to that matrix.

Answer 22

Swi-Prolog, 142 bytes.

t(0,1).
t(1,0).
r([],_).
r([H|T],H):-t(H,I),r(T,I).
f([],_,_).
f([H|T],N,B):-length(H,N),r(H,B),t(B,D),f(T,N,D).
c(N,C):-length(C,N),f(C,N,1).

Try online - http://swish.swi-prolog.org/p/BuabBPrw.pl

It outputs a nested list, so the rules say:

• t() is a toggle, it makes the 0 -> 1 and 1 -> 0.
• r() succeeds for an individual row, which is a recursive check down a row that it is alternate ones and zeros only.
• f() recursively checks all the rows, that they are the right length, that they are valid rows with r() and that each row starts with a differing 0/1.
• c(N,C) says that C is a valid checkerboard of size N if the number of rows (nested lists) is N, and the helper f succeeds.

Test Cases:

Answer 23

PHP, 56 bytes

Output as string

for(;$i<$argn**2;)echo++$i%2^$n&1,$i%$argn?"":"
".!++$n;

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PHP, 66 bytes

Output as 2 D array

for(;$i<$argn**2;$i%$argn?:++$n)$r[+$n][]=++$i%2^$n&1;print_r($r);

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Answer 24

QBIC, 19 bytes

[:|?[b|?(a+c+1)%2';

Explanation

[:|         FOR a = 1 to b (b is read from cmd line)
?           PRINT - linsert a linebreak in the output
[b|         FOR c = 1 to b
?(a+c+1)%2  PRINT a=c=1 modulo 2 (giving us the 1's and 0's
';            PRINT is followed b a literal semi-colon, suppressing newlines and 
              tabs. Printing numbers in QBasic adds one space automatically.

Answer 25

Brachylog, 19 bytes

⟦₁Rg;Rz{z{++₁%₂}ᵐ}ᵐ

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Answer 26

CJam, 17 bytes

{_[__AAb*<_:!]*<}

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Returns a list (TIO link has formatted output).

Answer 27

Bash + rs, 42

eval echo \$[~{1..$1}+{1..$1}\&1]|rs $1 $1

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Answer 28

Cheddar, 38 bytes

n->(|>n).map(i->(|>n).map(j->i+j+1&1))

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Answer 29

///, 87 bytes + input

/V/\\\///D/VV//*/k#D#k/k#D&k/k&DVk/k\D/SD/#/r
DSkk/10DSk/1D&/V#rV#0r;VV0;VVV1;V\D/r/S/&[unary input in asterisks]

Try it online! (input for 4)

Unary input in 1s, 95 bytes + input

/V/\\\///D/VV//&1/k#&D&|D/#k/k#D&k/k&DVk/k\D/SD/#/r
DSkk/10DSk/1D&/V#rV#0r;VV0;VVV1;V\D/r/S/&&[unary input in ones]|

Try it online! (input for 8)

How does this work?

• V and D are to golf \/ and // respectively.

• /*/k#/ and /&1/k#&//&|// separate the input into the equivalent of 'k#'*len(input())

• /#k//k#//&k/k&//\/k/k\// move all the ks to the /r/S/ block

• Ss are just used to pad instances where ks come after /s so that they don't get moved elsewhere, and the Ss are then removed

• #s are then turned into r\ns

• The string of ks is turned into an alternating 1010... string

• The r\ns are turned into 1010...\ns

• Every pair of 1010...\n1010\n is turned into 1010...\01010...;\n

• Either 0; or 1; are trimmed off (because the 01010... string is too long by 1)

Answer 30

Python, 73 63 62 66 bytes

Saved 4 bytes thanks to officialaimm

r=range(input());print[''.join([`(x+y+1)%2`for x in r])for y in r]

Answer 31

Mathematica, 28 bytes

Cos[+##/2Pi]^2&~Array~{#,#}&

Pure function taking a positive integer as input and returning a 2D array. Uses the periodic function cos²(πx/2) to generate the 1s and 0s.

For a little more fun, how about the 32-byte solution

Sign@Zeta[1-+##]^2&~Array~{#,#}&

which uses the locations of the trivial zeros of the Riemann zeta function.

Answer 32

q/kdb+, 37 16 bytes

Solution:

{x#x#'(1 0;0 1)}

Example:

q){x#x#'(1 0;0 1)}1
1
q){x#x#'(1 0;0 1)}2
1 0
0 1
q){x#x#'(1 0;0 1)}3
1 0 1
0 1 0
1 0 1
q){x#x#'(1 0;0 1)}4
1 0 1 0
0 1 0 1
1 0 1 0
0 1 0 1

Explanation:

2nd version is much simpler, and thus shorter and faster (4x). Create a 2-item list containing 01... and 10... to the length of the input, then take 'x' number of items from this new list.

{              } / lambda function
      (1 0;0 1)  / 2-item list of (0;1) and (1;0)
   x#'           / take 'x' items from each list, if x=3 then (1 0 1;0 1 0)
 x#              / take 'x' items from *this* list

Notes:

I've re-written this twice during this edit, went from 37->25->24->16 bytes. Now it's a little more competitive.

Edits:

• -21 bytes with complete re-write...

Answer 33

braingasm, 21 bytes

;$[$[>+o:<]p[>+<]10.]

Explanation:

;                       Read a number from stdin.
 $[                 ]   That many times...
   $[     ]               That many times...
     >+o:<                  Increase the next cell and print its value modulo 2.
           p[   ]       If the input was even...
             >+<          Increase the next cell once more.
                 10.    Print a newline.

Answer 34

J, 41 bytes

I know there's a J answer at 9, but I'm pleased to get anything working, even if it's not a tacit programming best possible..

ch=:monad define
2|(2$y)$(1+i.y),(i.y)
)

It takes a number y and generates two lists of numbers 1,2,3,..y and 0,1,2,3,..y-1 to make the offset first and second row, appends them into one long list, reshapes that into a y,y matrix (wrapping around when it runs out), and then does modulo 2 on all the elements to make them a 1 or 0.

   ch 1
1

   ch 2
1 0
0 1

   ch 3
1 0 1
0 1 0
1 0 1

   ch 4
1 0 1 0
0 1 0 1
1 0 1 0
0 1 0 1

   ch 5
1 0 1 0 1
0 1 0 1 0
1 0 1 0 1
0 1 0 1 0
1 0 1 0 1

Answer 35

Haskell, 43 bytes

h=1:0:h
l=h:(0:h):l
s x=take x$map(take x)l

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This creates the infinite checkerboard matrix and saves it to l. Then our function s chops off a square.

Answer 36

Mathematica, 28 bytes

Table[Mod[i+j+1,2],{i,#},{j,#}]&

Answer 37

Julia, 38 bytes

n->map(x->x%2,(1:n).+transpose(0:n-1))

The usual approach

Answer 38

Python 2, 72 70 61 54 bytes

-7 bytes thanks to Leaky Nun. -1 byte thanks to Wheat Wizard.

lambda n:[[i-~j&1for i in range(n)]for j in range(n)]

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Answer 39

Actually, 15 bytes

r;╗⌠╜+u⌠2@%⌡M⌡M

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Answer 40

Python, 43 bytes

lambda n:[('10'*n)[i:i+n]for i in range(n)]

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Anonymous function that outputs a list like ['1010', '0101', '1010', '0101'].

Answer 41

tcl, 82

time {set a "";time {set a $a[expr [incr i]%2]} $n;puts $a;if \!($n%2) incr\ i} $n

demo

Still a looser, but I think i can be golfed down a little bit further.

Answer 42

C#, 145 Bytes

int u=1;int n=int.Parse(b.Text);for(int i=0;i<n;i++){if(i%2==0){u=1;}else{u=0;}for(int j=0;j<n;j++){t.Text+=u.ToString();u=1-u;}t.Text+="\r\n";}[enter link description here][1]

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Answer 43

Bash & coreutils, 82 bytes

x=$1;yes 10|tr -d \\n|dd bs=1 count=$[x*x*2]|fold -w$[x*2-1]|grep -om$x "^.\{$x\}"

Takes one command-line integer as input. Apparently the spaces between are optional.

Answer 44

Common Lisp, SBCL, 81 bytes

version 1:

(#1=dotimes(i(set'a(read)))(#1#(j a)(format t"~:[1~;0~] "(oddp(+ j i))))(terpri))

version 2:

(lambda(n)(#1=dotimes(i n)(#1#(j n)(format t"~:[1~;0~] "(oddp(+ j i))))(terpri)))

Two loops, and inside I print either 1 and space or zero and space based on parity of (+ j i). (terpri) for newline.

It's long, I know:/

Answer 45

C# (.NET Core), 88 bytes

n=>{var r=new int[n,n];for(int i=0,j;i<n;i++)for(j=0;j<n;j++)r[i,j]=(i+j+1)%2;return r;}

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I can't believe this is the shortest way to initialize a 2D array of integers, there must be another way...

Answer 46

Perl 5, 49 bytes

$_=10x(($n=<>)/2).1x($n%2);say&&y/01/10/while$n--

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Answer 47

C (gcc), 60 59 bytes

i;f(n){for(i=n*n;i--;)printf(i%n?"%d":"%d\n",i%n+i/n+1&1);}

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