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.

• xnor: Is a list of strings OK?
• Stewie Griffin: Yes, that's OK.
• beaker: Related.
• BradC: Your examples show spaces between numbers on the same row, is that required, so as to look more like a square?
• Stewie Griffin: @BradC it's not required. The first approach here is valid.

Answer 1

Jelly, 4 bytes

52 seconds!

+€ḶḂ

Try it online!

• Erik the Outgolfer: "52 seconds!" like I'm not used to it...
• Magic Octopus Urn: Do ya'll have, like, a beeper, you wear 24/7 for new PPCG challenges?
• Erik the Outgolfer: @carusocomputing Those who have a faster internet connection are usually the lucky ones that will win.

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

Haskell, (削除) 50 (削除ここまで) (削除) 41 (削除ここまで) (削除) 39 (削除ここまで) 38 bytes

Thanks to nimi and xnor for helping to shave off a total of (削除) 9 (削除ここまで) 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.

• nimi: concat.repeat is cycle: n!l=take n$cycle l. If you go pointfree it saves one more byte: (!)=(.cycle).take.
• Julian Wolf: Lovely! I knew there was a builtin for that, but couldn't remember the name for the life of me
• xnor: I was going to suggest f n|r<-take n.cycle=r[r"10",r"01"] or similar. but Haskell seems to infer the wrong type for r? It works with explicit typing f n|r<-take n.cycle::[a]->[a]=r[r"10",r"01"].
• xnor: @JulianWolf Haskell seems to have trouble inferring polymorphic types
• xnor: @zbw I thought this was the case but using NoMonomorphismRestriction didn't help. Nor did Rank2Types or RankNTypes. Do you know what's going on there?

Answer 4

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 5

V, (削除) 16 (削除ここまで), 15 bytes

Ài10À­ñÙxñÎÀlD

Try it online!

Hexdump:

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

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

JavaScript ES6, (削除) 55 (削除ここまで) (削除) 54 (削除ここまで) (削除) 51 (削除ここまで) 46 bytes

Saved 1 byte thanks to @Neil

Saved 2 bytes thanks to @Arnauld

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

Try it online!

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

• Neil: It's still a byte shorter to use the index parameters: n=>[...Array(n)].map((_,i,a)=>a.map((_,j)=>(i+j+1)%2))
• Downgoat: @Neil huh, I never thought to use the third parameter in map, thanks!
• Downgoat: @Arnauld thanks! that inspired me to save 5 more bytes!

Answer 8

J, 9 bytes

<0&=$&1 0

Try it online!

Answer 9

05AB1E, (削除) 9 (削除ここまで) 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

• Emigna: You can cut the » as list-of-lists output is okay and you can also remove s.
• Erik the Outgolfer: Explanation is a bit irrelevant.

Answer 10

Bash + rs, 39

• 3 bytes saved thanks to @roblogic

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

Try it online!

• roblogic: you can do simply rs 1ドル, saving 3 bytes

Answer 11

Mathematica, 25 bytes

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

Answer 12

Clojure, 36 bytes

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

Yay, right tool for the job.

Answer 13

Retina, (削除) 33 (削除ここまで) 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.

• Martin Ender: $`1$' is just $_.
• Neil: 01 is just d. (Post is too old for me to bother editing it.)

Answer 14

Java (OpenJDK 8), (削除) 80 (削除ここまで) 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;}

Try it online!

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;
}

• Kevin Cruijssen: You can remove the space to save a byte. The challenge states the output format is flexible. Oh, and you can save two more bytes by changing (i++/j+i%j)%2 to i++/j+i%j&1 so you won't need those parenthesis. Which make the total 1 byte shorter than my nested for-loop solution (n->{String r="";for(int i=0,j;i++<n;r+="\n")for(j=0;j<n;r+=j+++i&1);return r;}), so +1 from me. :)
• PunPun1000: @KevinCruijssen Yeah I was still waiting on a response on the space. I didn't think about & having higher precedence than % and &1 == %2

Answer 15

Arturo, (削除) 39 (削除ここまで) (削除) 36 (削除ここまで) 32 bytes

$[n][map n'x->map n'y->%x+y+1 2]

Try it

-3 thanks to alephalpha's simpler method

Prettified and rearranged for clarity:

$[n][ ; a function taking an argument n
 map n 'x -> ; map over 1..n, assign current elt to x
 map n 'y -> ; map over 1..n again, assign current elt to y
 (x+y+1)%2 ; what it looks like
] ; end function

Answer 16

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.}

• Luis Mendo: Equivalent, and shorter: :&+o~
• Stewie Griffin: Still learning :-) I'll update tomorrow. I liked your other approach too :-)
• Sanchises: This is what I came up with, too. And hey, you only use the pure MATL instruction set, not those pesky Y-modified instructions @LuisMendo uses.
• Luis Mendo: @Sanchises Pesky, huh? :-P

Answer 17

Brachylog, 19 bytes

⟦1Rg;Rz{z{++1%2}m}m

Try it online!

Answer 18

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 19

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 20

Pyth, 9 bytes

VQm%+hdN2

Try this!

another 9 byte solution:

mm%+hdk2Q

Try it!

Answer 21

///, 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円...;\n

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

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: enter image description here

• Joao-3: Can't you strip off all newlines?

Answer 23

Nekomata, 5 bytes

o+→2%

Attempt This Online!

o+→2%
o+ Create an (input x input) matrix where the element at (i,j) is i+j
 → Increment
 2% Mod 2

Answer 24

AWK, 57 bytes

{for(;i++<1ドル^2;1ドル%2||i%1ドル||x=!x)printf(x=!x)(i%1ドル?FS:RS)}

Attempt This Online!

Answer 25

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 26

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 27

CJam, 17 bytes

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

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Returns a list (TIO link has formatted output).

• Esolanging Fruit: out-golfed
• Erik the Outgolfer: @Challenger5 Sorry you can't outgolf with deleted answer.

Answer 28

Mathematica, 23 bytes

ToeplitzMatrix@#~Mod~2&

Answer 29

Octave, 24 bytes

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

Try it online!

---

Or the same length:

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

Try it online!

Answer 30

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.