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Bytes	Lang	Time	Link
073	Perl 5 n	240524T221310Z	Xcali
025	Pip P	240524T044235Z	DLosc
078	Arturo	240524T152155Z	chunes
079	Matlab	211105T162846Z	LimaKilo
080	TIBasic	211102T003532Z	Youserna
187	Julia	211031T223824Z	S. Pelle
018	Jelly	211031T185142Z	lynn
097	Excel	211031T175153Z	Axuary
040	J	211031T055544Z	Jonah
024	Vyxal	211031T043606Z	hyper-ne
019	Jelly	211029T023006Z	Unrelate
103	C gcc	211029T164014Z	Noodle9
021	MATL	211029T104127Z	Luis Men
092	C clang	211029T220438Z	AZTECCO
021	Charcoal	211029T083655Z	Neil
084	JavaScript ES6	211029T143308Z	Arnauld
024	05AB1E	211029T104224Z	Kevin Cr
112	Zsh	211029T071914Z	roblogic
072	APL+WIN	211029T104058Z	Graham
068	R	211029T025552Z	Giuseppe
070	Ruby	211029T084759Z	G B
080	Python 3.8 prerelease	211029T080859Z	loopy wa

Perl 5 `-n`, 73 bytes

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

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Pip `-P`, 26 25 bytes

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

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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

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!

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.

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].

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

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.

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

J, 40 bytes

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

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Vyxal, 24 bytes

:±+ʁ:±†vǔd$ʁ:v+∷+2(ṘømÞT

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-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

Jelly, 19 bytes

^þ
ÇḂ+UṚŒḄŒBɓa‘{Ç¬Ḥ

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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

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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.

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!!!

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]

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

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).

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))

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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\$.

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.

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

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

R, 72 68 bytes

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

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Thanks to Robin Ryder for -4 bytes.

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!

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.~~

Perl 5 -n, 73 bytes

Pip -P, 26 25 bytes

Explanation

Arturo, 78 bytes

Matlab, 79 bytes

TI-Basic, 80 bytes

Julia, 336 271 195 187 Bytes

Jelly, 18 bytes

Excel, 97 bytes

Explanation

J, 40 bytes

Vyxal, 24 bytes

Jelly, 19 bytes

Jelly, 25 24 23 22 21 bytes

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

MATL, 22 21 bytes

Explanation

C (clang), 92 bytes

Charcoal, 22 21 bytes

JavaScript (ES6), 87 84 bytes

Note

05AB1E, 25 24 bytes

Zsh, 127 116 112 bytes

APL+WIN, 72 bytes

R, 72 68 bytes

Ruby, 70 bytes

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

Perl 5 `-n`, 73 bytes

Pip `-P`, 26 25 bytes

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