Checkered grid with X mark

Question

Challenge

Given two integer values \$a \ge 2\$ and \$0 \le b < a\$, generate a \$(2a-1) \times (2a-1)\$ matrix consisting of the integers 0, 1, and 2 as follows:

Create a checkerboard of 0s and 1s of the given size, corners being 0.
If \$b > 0\$, overwrite the checkerboard with 2s in the X shape at the center, each leg being of length \$b\$ (not counting the center).

^{This pattern is directly modeled from Art Puzzle event grids (where \$a\$ is fixed at 3, \$b=0,1,2\$ represent easy/normal/hard grids, and the 0, 1, 2s in the output represent easy/normal/hard minipuzzles respectively). Blame it if you don't like the edge case of \$b=0\$ :P}

Standard code-golf rules apply. The shortest code in bytes wins.

Test cases

a = 2, b = 0
[[0, 1, 0],
 [1, 0, 1],
 [0, 1, 0]]

a = 2, b = 1
[[2, 1, 2],
 [1, 2, 1],
 [2, 1, 2]]

a = 3, b = 0
[[0, 1, 0, 1, 0],
 [1, 0, 1, 0, 1],
 [0, 1, 0, 1, 0],
 [1, 0, 1, 0, 1],
 [0, 1, 0, 1, 0]]

a = 3, b = 1
[[0, 1, 0, 1, 0],
 [1, 2, 1, 2, 1],
 [0, 1, 2, 1, 0],
 [1, 2, 1, 2, 1],
 [0, 1, 0, 1, 0]]

a = 3, b = 2
[[2, 1, 0, 1, 2],
 [1, 2, 1, 2, 1],
 [0, 1, 2, 1, 0],
 [1, 2, 1, 2, 1],
 [2, 1, 0, 1, 2]]

a = 4, b = 0
[[0, 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, 0, 1],
 [0, 1, 0, 1, 0, 1, 0],
 [1, 0, 1, 0, 1, 0, 1],
 [0, 1, 0, 1, 0, 1, 0]]

a = 4, b = 1
[[0, 1, 0, 1, 0, 1, 0],
 [1, 0, 1, 0, 1, 0, 1],
 [0, 1, 2, 1, 2, 1, 0],
 [1, 0, 1, 2, 1, 0, 1],
 [0, 1, 2, 1, 2, 1, 0],
 [1, 0, 1, 0, 1, 0, 1],
 [0, 1, 0, 1, 0, 1, 0]]

a = 4, b = 2
[[0, 1, 0, 1, 0, 1, 0],
 [1, 2, 1, 0, 1, 2, 1],
 [0, 1, 2, 1, 2, 1, 0],
 [1, 0, 1, 2, 1, 0, 1],
 [0, 1, 2, 1, 2, 1, 0],
 [1, 2, 1, 0, 1, 2, 1],
 [0, 1, 0, 1, 0, 1, 0]]

a = 4, b = 3
[[2, 1, 0, 1, 0, 1, 2],
 [1, 2, 1, 0, 1, 2, 1],
 [0, 1, 2, 1, 2, 1, 0],
 [1, 0, 1, 2, 1, 0, 1],
 [0, 1, 2, 1, 2, 1, 0],
 [1, 2, 1, 0, 1, 2, 1],
 [2, 1, 0, 1, 0, 1, 2]]

\$\begingroup\$ Can we output the matrix flattened? \$\endgroup\$
– emanresu A
Commented Oct 29, 2021 at 8:51 — emanresu A, Commented Oct 29, 2021 at 8:51
\$\begingroup\$ @emanresuA No, the grid structure must be identifiable. \$\endgroup\$
– Bubbler
Commented Oct 29, 2021 at 9:31 — Bubbler, Commented Oct 29, 2021 at 9:31

Giuseppe · Accepted Answer · 2021-10-29 11:03:00Z

6

R, 72 68 bytes

function(a,b)`[<-`(diag(2*a-1)*0+0:1,cbind(c(x<--b:b,-x),x)+a,2*!!b)

Try it online!

Thanks to Robin Ryder for -4 bytes.

edited Oct 29, 2021 at 11:03

answered Oct 29, 2021 at 2:55

Giuseppe

28.8k3 gold badges31 silver badges105 bronze badges

\$\begingroup\$ 71 bytes by avoiding the if. \$\endgroup\$
– Robin Ryder
Commented Oct 29, 2021 at 8:25
3

\$\begingroup\$ I think this works for 68 bytes. \$\endgroup\$
– Robin Ryder
Commented Oct 29, 2021 at 8:30
\$\begingroup\$ @RobinRyder - Nice trick with [<-! Is that already in the R tips? I need to go back over some older answers to see if I can save some more bytes using that trick... \$\endgroup\$
– Dominic van Essen
Commented Oct 29, 2021 at 14:46
1

\$\begingroup\$ @DominicvanEssen I think it's buried in vlo's list of tips so it's tough to find, but it's also not super apparent how to use it from that description \$\endgroup\$
– Giuseppe
Commented Oct 29, 2021 at 14:52

Add a comment |

Neil · Accepted Answer · 2021-10-29 18:30:52Z

Charcoal, 22 21 bytes

ＵＯ±Ｎ01¶10‖Ｏ⌈¿ＩηＰX×2⊕η

Try it online! Link is to verbose version of code. Explanation:

ＵＯ±Ｎ01¶10

Draw a chequerboard of size a with 0s on its major diagonals above and to the left of the current position.

‖Ｏ⌈

Reflect it to produce a chequerboard of size 2a-1, leaving the current position in the middle of the chequerboard. (Note that while drawing the chequerboard below and right would save a byte, there is no single byte to reflect up and left, and Charcoal's up reflect is buggy and moves the current position incorrectly anyway.)

¿Ｉη

Test whether b is zero.

ＰX×2⊕η

If not then draw arms of 2s of length b+1 (including the centre).

loopy walt · Accepted Answer · 2021-10-29 08:58:24Z

4

Python 3.8 (pre-release), 82, 80 bytes (@ovs)

lambda a,b:(R:=range(1-a,a))and[[1&x+y|(x*x==y*y<=b*b>0)*2for x in R]for y in R]

Try it online!

Old version

Nothing too fancy, still learning the ropes. ~~Should run on older python3s, too.~~

edited Oct 29, 2021 at 8:58

answered Oct 29, 2021 at 8:08

loopy walt

16.8k2 gold badges11 silver badges69 bronze badges

\$\begingroup\$ A normal function is a byte shorter because you can reuse the range: tio.run/##TY/LCsIwEEX3/… \$\endgroup\$
– ovs
Commented Oct 29, 2021 at 8:37
\$\begingroup\$ @ovs Wow, I genuinely didn't know ranges can be used multiple times, thought they're single shot like generators. This being python3.8 we can walruss it instead of making a proper function, saves another byte. \$\endgroup\$
– loopy walt
Commented Oct 29, 2021 at 8:48
\$\begingroup\$ Could be 74 bytes, if this is valid ? : a=5 b=2 r=range(1-a,a) [[2*(ll==cc<=b*b>0)|(1&l+c)for c in r]for l in r] \$\endgroup\$
– Malo
Commented Nov 3, 2021 at 13:18
\$\begingroup\$ @Malo thanks for the suggestion. I'm rather new to this site myself, but as I understand it it is required to provide a proper interface for the inputs. Your snippet appears to hard code them? \$\endgroup\$
– loopy walt
Commented Nov 3, 2021 at 21:21

Add a comment |

Kevin Cruijssen · Accepted Answer · 2021-10-29 14:48:18Z

05AB1E, 25 24 bytes

Dи2.ý01ŽNœSΛĀ+Y87S.Λ.º.∊

Can probably be shortened with a different approach, but I wanted to try and use the Canvas builtin for this challenge.

-1 byte thanks to @ovs.

Try it online.

Explanation:

D            # Duplicate the first (implicit) input-integer `a`
 и           # Create a list with `a` amount of `a`s
  2.ý        # Intersperse it with 2: [a,2,a,2,a,2,...,a]
     01      # Push "01"
       ŽNœ   # Push compressed integer 6020
          S  # Convert it to a list of digits: [6,0,2,0]
           Λ # Use the Canvas builtin with these three options:
             #  Length: [a,2,a,2,a,2,...,a]
             #  Characters to draw: "01"
             #  Directions: [6,0,2,0]
Ā            # Check if the (implicit) second input-integer `b` is NOT 0
 +           # Add this check to the (implicit) second input-integer `b`
             # (b=0 will remain 0; b≥1 is now b+1)
  Y          # Push "2"
   87S       # Push list [8,7]
      .Λ     # Use the modifiable Canvas builtin with these three options:
             #  Length: b+1 (or 0 if b=0)
             #  Characters to draw: "2"
             #  Directions: [8,7]
.º           # Intersected mirror the result towards the left
  .∊         # Intersected mirror the result downwards
             # (after which the result is output implicitly)

See this 05AB1E tip of mine (section How to compress large integers?) to understand why ŽNœ is 6020.

The Canvas Builtin uses three arguments to draw a shape:

Length of the lines we want to draw
Character/string to draw
The direction to draw in, where each digit represents a certain direction:

Here a step-by-step explanation of how it draws for example inputs \$a=5,b=2\$ with these steps:

Preparation: Dи2.ý makes the list [5,2,5,2,5,2,5,2,5].

Dи2.ý01ŽNœSΛ for this input will have the following Canvas arguments:

Length: [5,2,5,2,5,2,5,2,5]
Characters: "01"
Directions: [6,0,6,2], which translate to \$[←,↑,→,↑]\$

Step 1: Draw 5 characters ("01010") in direction 6←:

Step 2: Draw 2-1 characters ("1") in direction 0↑:

1
01010

Step 3: Draw 5-1 characters ("0101") in direction 2→:

10101
01010

Step 4: Draw 2-1 characters ("0") in direction 0↑:

    0
10101
01010

Step 5: Draw 5-1 characters ("1010") in direction 6←:

01010
10101
01010

Step 6: Draw 2-1 characters ("1") in direction 0↑:

Step 7: Draw 5-1 characters ("0101") in direction 2→:

Step 8: Draw 2-1 characters ("0") in direction 0↑:

Step 9: Draw 5-1 characters ("1010") in direction 6←:

We then use the Canvas builtin again with three new arguments: Ā+Y87S.Λ

Length: \$b+1\$ (or \$0\$ if \$b=0\$, because with length 1, direction 8 will draw a "2" at the position it was at before the reset unfortunately)
Characters to draw: "2"
Direction: [8,7], where the 8 is a special direction that resets the starting position (to the bottom-left we started at) and 7 translates to \$↖\$

Draw \$3\$ (\$b=2\$) characters ("222") in direction 7↖:

After which we mirror this multi-line string both horizontally and vertically, after which it is output implicitly as our result.

See this 05AB1E tip of mine for an in-depth explanation of the Canvas builtin.

@ovs Ah, of course. The _- was a leftover when I thought I needed -1 instead of 0 when I doubled the _ before subtracting, but apparently only 1 is giving the wrong result, <=0 and >1 is fine. Thanks. — Kevin Cruijssen, Commented Oct 29, 2021 at 14:45

Luis Mendo · Accepted Answer · 2021-10-29 23:40:23Z

MATL, 22 21 bytes

_Zvt&+oy&=bitg+>&+>E+

Inputs are a, then b.

Try it online!

Explanation

This uses inputs 3, 1 as an example. The stack contents are shown in each step. The stack is shown upside down, with the top (i.e. most recent) element down.

_Zvt % Implicit input: a. Symmetric range [a a-1 ... 2 1 2 ... a-1 a]. Duplicate
     % STACK: [3 2 1 2 3],
              [3 2 1 2 3]
&+   % Duplicate. Add to itself transposed, elementwise with broadcast
     % STACK  [3 2 1 2 3],
              [6 5 4 5 6
               5 4 3 4 5
               4 3 2 3 4
               5 4 3 4 5
               6 5 4 5 6]
o    % Modulo 2, elementwise. Gives checkered matrix
     % STACK: [3 2 1 2 3],
              [0 1 0 1 0
               1 0 1 0 1
               0 1 0 1 0
               1 0 1 0 1
               0 1 0 1 0
y    % Duplicate second-top element
     % STACK: [3 2 1 2 3],
              [0 1 0 1 0
               1 0 1 0 1
               0 1 0 1 0
               1 0 1 0 1
               0 1 0 1 0],
              [3 2 1 2 3]
&=   % Equal? to itself transposed, elementwise with broadcast. Gives "X" matrix
     % STACK: [3 2 1 2 3],
              [0 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 0 0 0 1]
               0 1 0 1 0
               0 0 1 0 0
               0 1 0 1 0
               1 0 0 0 1]
b    % Bubble up third-top element
     % STACK: [0 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 0 0 0 1]
               0 1 0 1 0
               0 0 1 0 0
               0 1 0 1 0
               1 0 0 0 1],
              [3 2 1 2 3]   
i    % Input: b
     % STACK: [0 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 0 0 0 1]
               0 1 0 1 0
               0 0 1 0 0
               0 1 0 1 0
               1 0 0 0 1],
              [3 2 1 2 3],
               1
tg+  % Duplicate, convert to logical, add. This converts positive b into b+1
     % STACK: [0 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 0 0 0 1]
               0 1 0 1 0
               0 0 1 0 0
               0 1 0 1 0
               1 0 0 0 1],
              [3 2 1 2 3],
               2
>    % Greater than?, elementwise
     % STACK: [0 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 0 0 0 1]
               0 1 0 1 0
               0 0 1 0 0
               0 1 0 1 0
               1 0 0 0 1],
              [1 0 0 0 1]
&+   % Add to itself transposed, elementwise with broadcast
     % STACK: [0 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 0 0 0 1]
               0 1 0 1 0
               0 0 1 0 0
               0 1 0 1 0
               1 0 0 0 1],
              [2 1 1 1 2
               1 0 0 0 1
               1 0 0 0 1
               1 0 0 0 1
               2 1 1 1 2]
>    % Greater than?, elementwise. Equivalent to "and not". This trims the "X" matrix
     % STACK: [0 1 0 1 0
               1 0 1 0 1
               0 1 0 1 0
               1 0 1 0 1
               0 1 0 1 0],
              [0 0 0 0 0]
               0 1 0 1 0
               0 0 1 0 0
               0 1 0 1 0
               0 0 0 0 0],
E+   % Multiply by 2, element-wise. Add, element-wise. Implicit display
     % STACK: [0 1 0 1 0
               1 2 1 2 1
               0 1 2 1 0
               1 2 1 2 1
               0 1 0 1 0]

@UnrelatedString Sadly I forgot to put my dinner in the oven so I've only just got around to restoring the tie. — Neil, Commented Oct 29, 2021 at 18:32
@Neil Nice to see three very different languages with the same score! — Luis Mendo, Commented Oct 29, 2021 at 19:53

G B · Accepted Answer · 2021-10-29 09:28:03Z

3

Ruby, 70 bytes

->a,b{(r=1-a...a).map{|x|r.map{|y|x*x==y*y&&x*x<=b*b&&b>0?2:(x+y)%2}}}

Try it online!

edited Oct 29, 2021 at 9:28

answered Oct 29, 2021 at 8:47

G B

22.8k1 gold badge20 silver badges53 bronze badges

Add a comment |

Unrelated String · Accepted Answer · 2021-10-31 05:43:00Z

Jelly, 19 bytes

^þ
ÇḂ+UṚŒḄŒBɓa‘{Ç¬Ḥ

Try it online!

With some inspiration from hyper-neutrino's Vyxal solution.

^þ     Monadic helper link: 
 þ     Table the (range from 1 to the) argument with itself by
^      XOR.
  Ḃ    The parities of the cells form a checkerboard,
  ¬    and the diagonal consists of zeroes while the rest is truthy.

ÇḂ+UṚŒḄŒBɓa‘{Ç¬Ḥ    Dyadic main link:
             Ç¬     Generate an identity matrix of size
         ɓ          b
           ‘        + 1
          a {       or 0 if b is 0,
               Ḥ    then double it.
  +                 Add that to (the top left corner of)
ÇḂ                  an a by a checkerboard,
   UṚ               rotate the result 180 degrees,
     ŒḄŒB           and palindromize it in both dimensions.

Jelly, 25 24 23 22 21 bytes

^þ
ÇḂ+ŒḄŒBðÇUoḶ>¥*¬ḤU

Try it online!

That feels better. \$a\$ on the left, \$b\$ on the right.

^þ     Monadic helper link: 
 þ     Table the (range from 1 to the) argument with itself by
^      XOR.
  Ḃ    The parities of the cells form a checkerboard,
  ¬    and the diagonal consists of zeroes while the rest is truthy.

ÇḂ+ŒḄŒBðÇUoḶ>¥*¬ḤU    Dyadic main link:
        Ç             Generate an a by a matrix of zeroes on the diagonal and nonzeroes elsewhere.
         U            Mirror it,
          o           and replace zeroes with corresponding elements from:
           Ḷ ¥        for each 0 .. a-1,
            >         is it greater than b?
              *       Raise each to the bth power,
               ¬      logically negate,
                Ḥ     double,
                 U    and mirror back.
  +    ð              Add that to
ÇḂ                    an a by a checkerboard,
   ŒḄŒB               then palindromize in both dimensions.

Jonah · Accepted Answer · 2021-10-31 16:17:30Z

3

J, 40 bytes

((2|+/~)@]+2**@[*=/~@]*(>:]*=/~))|@i:@<:

Try it online!

edited Oct 31, 2021 at 16:17

answered Oct 31, 2021 at 5:55

Jonah

33.8k4 gold badges40 silver badges94 bronze badges

1

\$\begingroup\$ Your a is off by one and the shape of 2's is incorrect. \$\endgroup\$
– Bubbler
Commented Oct 31, 2021 at 7:53

Add a comment |

lynn · Accepted Answer · 2021-10-31 21:30:20Z

Jelly, 18 bytes

_þµṂ>Ɠa~¤×¬Ḥ+ḂŒBŒḄ

Try it online!

-1 from caird, who noted that I could get rid of a backtick by reading b from stdin.

                For inputs (a, b):
_þµ              Make a subtraction table over [1..a], and operate on that.
    Ṃ             Get the minimum row: this is always [1-a, 2-a ... 0].
     >Ɠa~¤        Which elements are greater than (b and ~b)?
                  For example for (5, 2): Ṃ    → [-4, -3, -2, -1,  0]
                                          Ṃ>-3 → [ 0,  0,  1,  1,  1]
          ×¬      Multiply this vector by the a×a identity matrix "¬table"
            Ḥ     and double the result (generating the 2s)
             +Ḃ   then add to an a×a checkerboard of 1s "table % 2":

                    0 0 0 0 0   0 1 0 1 0   0 1 0 1 0
                    0 0 0 0 0   1 0 1 0 1   1 0 1 0 1
                    0 0 2 0 0 + 0 1 0 1 0 → 0 1 2 1 0
                    0 0 0 2 0   1 0 1 0 1   1 0 1 2 1
                    0 0 0 0 2   0 1 0 1 0   0 1 0 1 2

ŒBŒḄ              Finally, palindromize both ways to make the full output.

\$\begingroup\$ 18 bytes as a full program taking b on STDIN \$\endgroup\$
– caird coinheringaahin g ♦
Commented Oct 31, 2021 at 19:12 — caird coinheringaahin g, Commented Oct 31, 2021 at 19:12

roblogic · Accepted Answer · 2021-10-29 12:53:54Z

2

Zsh, 127 116 112 bytes

R=({1..$[c=2*$1-1]})
for i ($R)(for j ($R)S+=$[(i==j)|(j+i==c+1)?$2==0?0:${$((i-$1))/\-/}>$2?0:2:(i+j)%2]
<<<$S)

Attempt this Online! _{116 bytes} _{127 bytes}

_{I used bits of the X without Y problem.}
_{@pxeger saved 4 bytes!}

Description

Using `a = $1` and `b = $2` as defined in the question...

R=({1..X})               define array R as a sequence from 1 to expression X
c=$2*$1-1                size of matrix
for i.. for j..          iterate rows, columns 
S+=$[...]                add a digit to S, per the logic in [...]
(i==j)|(j+i==c+1)?...    are we on a diagonal? if so, do ...
:(i+j)%2                 otherwise, fill the matrix with 1s and 0s
$2==0?0                  if b is 0, then default to 0 on the diagonal
:${$((i-$1))/\-/}>$2?0:2 if abs(i-a) > b then put 0 on the diagonal, otherwise 2

edited Oct 29, 2021 at 12:53

answered Oct 29, 2021 at 7:19

roblogic

4,01218 silver badges23 bronze badges

1

\$\begingroup\$ -4 bytes by (a): appending characters to a variable instead of printfing, and then printing the whole variable per line, and implicitly resetting it for each iteration of i using a (subshell); and (b): extracting {1..$c} into an array variable \$\endgroup\$
– pxeger
Commented Oct 29, 2021 at 10:08

Add a comment |

Arnauld · Accepted Answer · 2021-10-29 17:34:55Z

2

JavaScript (ES6), 87 84 bytes

Expects (a)(b).

a=>b=>[...Array(a+--a)].map((_,y,A)=>A.map(_=>x+y&1|(x*x/b<=b&x*x--==y*y)*2,y-=x=a))

Try it online!

Note

We do x*x/b<=b rather than x*x<=b*b so that x*x/b evaluates to NaN when both \$x\$ and \$b\$ are \$0\$. This makes the condition fail as expected for this edge case without the need for an explicit test on \$b\$.

edited Oct 29, 2021 at 17:34

answered Oct 29, 2021 at 14:33

Arnauld

198k20 gold badges179 silver badges650 bronze badges

Add a comment |

Noodle9 · Accepted Answer · 2021-10-30 19:47:30Z

2

C (gcc), ^{118 \$\cdots\$ 106} 103 bytes

i;f(a,b){for(a*=2,i=--a*a;i--;)printf("%d%c",!b|b<abs(a/2-i/a)||i/a-i%a&&i/a+1+i%a-a?i%2:2,i%a?32:10);}

Try it online!

Saved 3 bytes thanks to ceilingcat!!!

edited Oct 30, 2021 at 19:47

answered Oct 29, 2021 at 16:40

Noodle9

20.3k3 gold badges21 silver badges46 bronze badges

Add a comment |

hyperneutrino · Accepted Answer · 2021-10-31 06:24:24Z

Vyxal, 24 bytes

:±+ʁ:±†vǔd$ʁ:v+∷+2(ṘømÞT

Try It Online!

-3 bytes thanks to lyxal
-1 byte thanks to Unrelated String

:±+ʁ:±†vǔd$ʁ:v+∷+2(ṘømÞT     Full Program; take in order b, a
:                            duplicate b
 ±                           sign
  +                          add (0 if b is 0, otherwise b+1)
   ʁ                         range (0..x-1)
    :                        duplicate
     ±                       sign ([0, 1, 1, ..., 1] of length 0 / b+1)
      †                      logical NOT - [1, 0, 0, ..., 0]
       vǔ                    vectorized rotate; identity matrix of size 0 / b+1
         d                   double; [2, 0, 0], [0, 2, 0], [0, 0, 2]
          $                  swap a to TOS
           ʁ                 range; 0..a-1
            :                duplicate
             v+              add each; addition table; [2, 3, 4], [3, 4, 5], [4, 5, 6]
               ∷             parity; checkerboard of size a with 0 in top-left
                +            add the two matrices together
                  2(....)    repeat twice
                    Ṙ        reverse (flip vertical)
                     øm      palindromize (reflect vertical; do not copy middle)
                       ÞT    transpose

\$\begingroup\$ already posted this -1 in chat but you didn't notice \$\endgroup\$
– Unrelated String
Commented Oct 31, 2021 at 6:16 — Unrelated String, Commented Oct 31, 2021 at 6:16
\$\begingroup\$ @UnrelatedString oh thanks \$\endgroup\$
– hyperneutrino ♦
Commented Oct 31, 2021 at 6:24 — hyperneutrino, Commented Oct 31, 2021 at 6:24

Axuary · Accepted Answer · 2021-10-31 17:51:53Z

2

Excel, 97 bytes

=LET(x,SEQUENCE(A1*2-1)-A1,y,TRANSPOSE(x),IF(MOD(x+y,2),1,IF((x-y)*(x+y)+(B1=0)+(ABS(x)>B1),,2)))

Link to Spreadsheet

Explanation

=LET(x,SEQUENCE(A1*2-1)-A1, # [-a..a] vertical
y,TRANSPOSE(x),             # [-a..a] horizontal
IF(MOD(x+y,2),1,            # if x+y mod 2 <> 0 then 1
  IF((x-y)*(x+y)+(B1=0)+(ABS(x)>B1),,2)))  
                   # else if (x<>y and x<>-y) or b = 0 or abs(x) > b then 0 else 2

answered Oct 31, 2021 at 17:51

Axuary

3,4766 silver badges11 bronze badges

Add a comment |

S. Pellegrino · Accepted Answer · 2021-11-01 06:50:51Z

2

Julia, 336 271 195 187 Bytes

v4, and thanks for the help!

function x(a,b)
s=(2*a-1)
h = zeros(Int8,(s,s))
h[2:2:s^2].=1
if b!=0;h[a,a]=2end
for i in range(1,stop=min(b,a-1)) 
h[a+i,a+i]=2
h[a-i,a-i]=2
h[a-i,a+i]=2
h[a+i,a-i]=2
end
return(h)
end

old:

function cross(a,b)
    h = zeros(Int8,((2*a-1),(2*a-1)))
    h[2:2:(2*a-1)^2] .= 1
    c = (Int8((2*a-1)/2+0.5), Int8((2*a-1)/2+0.5))
    h[c[1],c[2]] = 2
    for i in range(1,stop=min(b,a-1))
        h[c[1]+i,c[2]+i] = 2
        h[c[1]-i,c[2]-i] = 2
        h[c[1]-i,c[2]+i] = 2
        h[c[1]+i,c[2]-i] = 2
    end
    return(h)
end

edited Nov 1, 2021 at 6:50

answered Oct 31, 2021 at 22:38

S. Pellegrino

213 bronze badges

5

\$\begingroup\$ Welcome to Code Golf, and nice first answer! Is it possible to golf this by changing the name of the function (cross \$\to\$ x, perhaps), and removing a lot of the whitespace (e.g. spaces around =, indentation, etc.)? Be sure to check out our Tips for golfing in Julia page for ways you can golf your program \$\endgroup\$
– caird coinheringaahin g ♦
Commented Oct 31, 2021 at 22:58
1

\$\begingroup\$ Also, guessing from the code, h[c[1],c[2]] = 2 should be run only when b is nonzero. \$\endgroup\$
– Bubbler
Commented Oct 31, 2021 at 23:39
\$\begingroup\$ You could use a variable for (2*a-1). And Int8((2*a-1)/2+0.5) is simply a, so you can erase c and just use a for all those c[1] and c[2] :) \$\endgroup\$
– Bubbler
Commented Nov 1, 2021 at 6:32
\$\begingroup\$ 135 bytes \$\endgroup\$
– MarcMush
Commented Nov 8, 2021 at 9:58

Add a comment |

DLosc · Accepted Answer · 2024-05-24 16:55:35Z

Pip `-P`, 26 25 bytes

QP R*R(%$+GMCa)+2*EYb+SGb

Attempt This Online!

Explanation

QP R*R(%$+GMCa)+2*EYb+SGb
             a             ; First command-line arg
           MC              ; Make a grid of that size and map to each
                           ; coordinate pair:
        $+G                ;  Sum the pair
       %                   ;  Mod 2
      (       )+           ; To that array, add the following array,
                           ; overlapping them from the top left corner:
                    b      ;  Second command-line arg
                     +SGb  ;  Add its sign (0 -> 0, positive x -> x+1)
                  EY       ;  Identity matrix of that size
                2*         ;  Multiply each element by 2
                           ; This gives us the bottom-right quadrant of our
                           ; desired result
     R                     ; Reverse the rows
   R*                      ; Reverse the elements within each row
                           ; This gives us the top-left quadrant, which
                           ; we can then...
QP                         ; Quad-palindromize

Yousername · Accepted Answer · 2021-11-02 00:35:32Z

1

TI-Basic, 80 bytes

Prompt A,B
identity(2A-1→[A]
For(I,1,2A-1
For(J,1,2A-1
remainder(I+J,2)+2max(abs({I,J}-A)≤B and B)max(I={J,2A-J→[A](I,J
End
End

+1 byte if not ran on a TI-84+/SE with the 2.53 MP OS or above by replacing remainder(I+J,2) with 2fPart(.5(I+J)).

Output is stored in [A].

answered Nov 2, 2021 at 0:35

Yousername

4,08010 silver badges24 bronze badges

Add a comment |

LimaKilo · Accepted Answer · 2021-11-05 16:28:46Z

1

Matlab, 79 bytes

m=zeros(2*a-1);
m(2:2:end)=1;
for i=a-b:a+b
m(i,i)=2*(b>0);end
max(m,flipud(m))

Only make one diagonal of twos and generate the antidiagonal by maxing with the flipped matrix. The 2*(b>0) was my shortest way to take care of the b=0 edge case.

answered Nov 5, 2021 at 16:28

LimaKilo

3111 silver badge5 bronze badges

Add a comment |

chunes · Accepted Answer · 2024-05-24 15:21:55Z

1

Arturo, 78 bytes

$[a,b][r:..1-a,a-1map r'x->map r'y->(/\/\=x^2y*y(x*x)=<b^2b>0)?->2->abs%x+y,2]

Try it!

answered May 24 at 15:21

chunes

26.8k3 gold badges24 silver badges53 bronze badges

Add a comment |

Graham · Accepted Answer · 2021-10-29 11:03:13Z

APL+WIN, 72 bytes

Prompts for a then b:

(s⊖(s←.5×b-a)⌽z↑(b≠1)×2×m∨⌽m←n∘.=(n←⍳b←(2×b←⎕)+1))+(z←a,(a←(2×⎕)-1))⍴0 1

Explanation:

(z←a,(a←(2×⎕)-1))⍴0 1 create the 0 1 checkerboard matrix defined by a

2×m∨⌽m←n∘.=(n←⍳b←(2×b←⎕)+1)) create the 2s diagonals defined by b

(b≠1)× filter the case where b=0

z↑ pad the diagonals matrix to the same size as the checkerboard matrix

s⊖(s←.5×b-a)⌽ centre the diagonals in the padded diagonals matrix. 

The two matrices are then summed together.

Try it online! Thanks to Dyalog Classic

AZTECCO · Accepted Answer · 2021-10-29 22:25:26Z

C (clang), 92 bytes

z,i;f(a,b){for(i*=z=i=2*a-1;i;)printf("\n%d"+!!(i--%z),i%~z&&i%~-z||(i/z-~-a)/~b|!b?i%2:2);}

Try it online!

Function printing to stdout.
Leading line can be removed at the cost of 1 Byte.

z : side length 
i : characters counter(backwards)
f(a,b){
for(i*=z=i=2*a-1;i;)    - initialize variables and itetate backwards until 0
printf("\n%d"+!!(i--%z) - prints:
         > a newline if i%z == 0
           > the number obtained by expression below at each iteration (decrements i)

i%~z&&i%~-z        - false on diagonals
||(i/z-~-a)/~b|!b  -  false on horizontal window
     * for example f(3,1)
    
  01110    11111   11111
  10101    00000   10101
  11011 || 00000 = 11011
  10101    00000   10101
  01110    11111   11111

?i%2:2    - draw board on 1's
                 2's on zeros

Xcali · Accepted Answer · 2024-05-24 22:13:10Z

0

Perl 5 `-n`, 73 bytes

$b=<>;//,say map{($'+$_)%2||2*!!$b*($b>=abs)*(abs==abs$')}@;for@;=-$_..$_

Try it online!

answered May 24 at 22:13

Xcali

16.2k2 gold badges15 silver badges39 bronze badges

Add a comment |

Stack Exchange Network

Checkered grid with X mark

Challenge

Test cases

22 Answers 22

R, 72 68 bytes

Charcoal, 22 21 bytes

Python 3.8 (pre-release), 82, 80 bytes (@ovs)

05AB1E, 25 24 bytes

MATL, 22 21 bytes

Explanation

Ruby, 70 bytes

Jelly, 19 bytes

Jelly, 25 24 23 22 21 bytes

J, 40 bytes

Jelly, 18 bytes

Zsh, 127 116 112 bytes

JavaScript (ES6), 87 84 bytes

Note

C (gcc), ^{118 \$\cdots\$ 106} 103 bytes

Vyxal, 24 bytes

Excel, 97 bytes

Explanation

Julia, 336 271 195 187 Bytes

Pip `-P`, 26 25 bytes

Explanation

TI-Basic, 80 bytes

Matlab, 79 bytes

Arturo, 78 bytes

APL+WIN, 72 bytes

C (clang), 92 bytes

Perl 5 `-n`, 73 bytes

Your Answer

Not the answer you're looking for? Browse other questions tagged
code-golf
matrix
or ask your own question.

Linked

Hot Network Questions

Checkered grid with X mark

Challenge

Test cases

22 Answers 22

R, 72 68 bytes

Charcoal, 22 21 bytes

Python 3.8 (pre-release), 82, 80 bytes (@ovs)

05AB1E, 25 24 bytes

MATL, 22 21 bytes

Explanation

Ruby, 70 bytes

Jelly, 19 bytes

Jelly, 25 24 23 22 21 bytes

J, 40 bytes

Jelly, 18 bytes

Zsh, 127 116 112 bytes

JavaScript (ES6), 87 84 bytes

Note

C (gcc), 118 \$\cdots\$ 106 103 bytes

Vyxal, 24 bytes

Excel, 97 bytes

Explanation

Julia, 336 271 195 187 Bytes

Pip -P, 26 25 bytes

Explanation

TI-Basic, 80 bytes

Matlab, 79 bytes

Arturo, 78 bytes

APL+WIN, 72 bytes

C (clang), 92 bytes

Perl 5 -n, 73 bytes

Your Answer

Sign up or log in

Post as a guest

Not the answer you're looking for? Browse other questions tagged code-golfmatrix or ask your own question.

Linked

Related

Hot Network Questions

C (gcc), ^{118 \$\cdots\$ 106} 103 bytes

Pip `-P`, 26 25 bytes

Perl 5 `-n`, 73 bytes

Not the answer you're looking for? Browse other questions tagged
code-golf
matrix
or ask your own question.