Esse desafio é sobre a construção de um tabuleiro de xadrez no qual o tamanho do quadrado, em vez de ser constante no tabuleiro, segue uma certa sequência não decrescente, conforme descrito abaixo.

O quadro é definido iterativamente. Uma placa de tamanho é ampliada para tamanho estendendo-a para baixo e para a direita por uma "camada" de quadrados do tamanho , onde é o maior divisor de não excede . Os quadrados na diagonal são sempre da mesma cor.$n \times n$$(n+k)\times(n+k)$$k$$k$$n$$\sqrt{n}$

Especificamente, considere o quadro com as cores representadas como #e +.

1. Inicialize o tabuleiro de xadrez para

   #
   

2. Até agora, o quadro tem tamanho . O único divisor de é e não excede . Então, pegamos e estendemos o quadro adicionando uma camada de quadrados de tamanho , com na diagonal:$1\times 1$$1$$1$$\sqrt{1}$$k=1$$1$#

   #+
   +#
   

3. A placa criada até agora tem tamanho . Os divisores de são 1 , 2 e o divisor máximo não excede √$2 \times 2$$2$$1,2$$\sqrt{2}$ é$1$. Então, novamente,$k=1$, e o quadro é estendido para

   #+#
   +#+
   #+#
   

4. O tamanho é $3 \times 3$ . $k=1$ . Estender-se para

   #+#+
   +#+#
   #+#+
   +#+#
   

5. O tamanho é $4 \times 4$ . Agora $k=2$ , porque $2$ é o divisor máximo de $4$ não excede . Estenda com uma camada de espessura , formada por quadrados de tamanho , com cor na diagonal:$\sqrt 4$$2$$2\times 2$#

   #+#+##
   +#+###
   #+#+++
   +#+#++
   ##++##
   ##++##
   

6. O tamanho é $6 \times 6$ . Agora $k=2$ . Estenda para o tamanho $8 \times 8$ . Agora $k=2$ . Estenda para o tamanho $10 \times 10$ . Agora $k=2$ . Estenda para o tamanho $12 \times 12$ . Agora $k=3$ . Estenda para o tamanho $15$ :

   #+#+##++##++###
   +#+###++##++###
   #+#+++##++#####
   +#+#++##++##+++
   ##++##++##+++++
   ##++##++##+++++
   ++##++##++#####
   ++##++##++#####
   ##++##++##++###
   ##++##++##+++++
   ++##++##++##+++
   ++##++##++##+++
   ###+++###+++###
   ###+++###+++###
   ###+++###+++###
   

Observe como os quadrados adicionados mais recentemente, de tamanho $3 \times 3$ , têm lados que coincidem parcialmente com os dos quadrados adicionados anteriormente, de tamanho $2 \times 2$ .

A sequência formada pelos valores de $k$ não é decrescente:

1 1 1 2 2 2 2 3 3 3 3 4 4 4 6 6 6 6 6 6 ...

e não parece estar no OEIS. No entanto, sua versão cumulativa, que é a sequência de tamanhos do quadro, é A139542 (obrigado a @Arnauld por perceber).

O desafio

Entrada : um número inteiro positivo $S$ representando o número de camadas no quadro. Se preferir, você também pode obter $S-1$ vez de $S$ como entrada ( $0$ indexado); ver abaixo.

Saída : uma representação em arte ASCII de uma placa com $S$ camadas.

• A saída pode ser através de STDOUT ou um argumento retornado por uma função. Nesse caso, pode ser uma string com novas linhas, uma matriz de caracteres 2D ou uma matriz de cadeias.

• Você pode escolher sempre dois caracteres para representar o quadro.

• Você pode escolher consistentemente a direção do crescimento. Ou seja, em vez das representações acima (que crescem para baixo e para a direita), você pode produzir qualquer uma de suas versões refletidas ou giradas.

• O espaço à direita ou à esquerda é permitido (se a saída for através de STDOUT), desde que o espaço não seja um dos dois caracteres usados ​​para o quadro.

• Opcionalmente, você pode usar a entrada " $0$ indexada "; ou seja, tome como entrada $S-1$ , que especifica uma placa com $S$ camadas.

O menor código em bytes vence.

Casos de teste

1:

#

3:

#+#
+#+
#+#

5:

#+#+##
+#+###
#+#+++
+#+#++
##++##
##++##

6:

#+#+##++
+#+###++
#+#+++##
+#+#++##
##++##++
##++##++
++##++##
++##++##

10:

#+#+##++##++###+++
+#+###++##++###+++
#+#+++##++#####+++
+#+#++##++##+++###
##++##++##+++++###
##++##++##+++++###
++##++##++#####+++
++##++##++#####+++
##++##++##++###+++
##++##++##+++++###
++##++##++##+++###
++##++##++##+++###
###+++###+++###+++
###+++###+++###+++
###+++###+++###+++
+++###+++###+++###
+++###+++###+++###
+++###+++###+++###

15:

#+#+##++##++###+++###+++####++++####
+#+###++##++###+++###+++####++++####
#+#+++##++#####+++###+++####++++####
+#+#++##++##+++###+++#######++++####
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++##++##++#####+++###+++++++####++++
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25:

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Geléia , 40 31 bytes

1SÆD>Ðḟ½ƊṀṭƲ³¡Äż$Ḷ:Ḃ^þ`ʋ/€ḷ""/Y

Experimente online!

$S-1$

Sem o final Y, isso retorna uma lista de listas de números inteiros, mas está fora de especificação para esse desafio.

Explicação

Este programa funciona em três etapas.

1. $k$$k$
2. $k$
3. Percorra a lista de tabuleiros de xadrez, substituindo sempre a seção superior esquerda do próximo quadro pelo quadro existente.

Estágio 1

1                 | Start with 1
           Ʋ³¡    | Loop through the following the number of times indicated by the first argument to the program; this generates a list of values of k
 S                | - Sum
        Ɗ         | - Following three links as a monad 
  ÆD              |   - List of divisors
    >Ðḟ½          |   - Exclude those greater than the square root
         Ṁ        |   - Maximum
          ṭ       | - Concatenate this to the end of the current list of values of k 
              Äż$ | Zip the cumulative sum of the values of k with the values

Etapa 2

      ʋ/€ | For each pair of k and cumulative sum, call the following as a dyad with the cumulative sum of k as the left argument and k as the right (e.g. 15, 3)
Ḷ         | - Lowered range [0, 1 ... , 13, 14]
 :        | - Integer division by k [0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4]
  Ḃ       | - Mod 2 [0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0]
   ^þ`    | - Outer product using xor function and same argument to both side

Etapa 3

   /  | Reduce using the following:
ḷ""   | - Replace the top left portion of the next matrix with the current one
    Y | Finally join by newlines

Tela , 34 32 bytes

０#０⁸［#+¶+#ｘｘ＊ｙｘ＋ｍ⤢αｍ；ｎｌｗ√｛ｙ；％‽²Ｘ

Experimente aqui!

Python 2 , 217 215 212 bytes

def f(x):
 b=['1'];n=1
 for i in range(x):P=max(j*(n%j<(j<=n**.5))for j in range(1,1+n));n+=P;b=[l+P*`j/P%2^i%2`for j,l in enumerate(b)];s=len(b[0]);b+=[((v*P+`1^int(v)`*P)*s)[:s]for v in b[0][len(b):]]
 return b

Experimente online!

Indexado com 0, usa 0e 1como caracteres

Python 2 , 184 178 176 169 bytes

def h(j,a=['1'],R=range):
 for i in R(j):L=len(a);k=max(x for x in R(1,L+1)if(x*x<=L)>L%x);a=[a[m]+k*`(i+m/k)%2`for m in R(L)]+[((`i%2`*k+`~i%2`*k)*L)[:L+k]]*k
 return a

Experimente online!

Utiliza 1, 0para #, -; usa 0-indexing.

JavaScript (ES7), 164 bytes

$0$#$1$+

n=>(b=[1],g=(a,w,d=w**.5|0)=>b[n]?a:w%d?g(a,w,d-1):g(a.concat(Array(d).fill(b.push(d)&&i++)),w+d))([0],i=1).map((_,y,a)=>a.map((_,x)=>(x/b[v=a[x>y?x:y]]^y/b[v])&1))

Experimente online!

Carvão , 37 bytes

ＦＮ«≔⊕⌈Φ₂⊕Ｌυ¬﹪Ｌυ⊕κηＦη«ＰL⭆⊞Ｏυω§#+÷⁻κμη↙

Experimente online! Link é a versão detalhada do código. 1 indexado. A saída cresce para baixo e para a esquerda (para baixo e para a direita custa um byte extra, mas pode crescer para a mesma contagem de bytes). Explicação:

ＦＮ«

$S$

≔⊕⌈Φ₂⊕Ｌυ¬﹪Ｌυ⊕κη

$k\leq\sqrt{n+1}$$n=0$$k=1$

Ｆη«

$k$

ＰL⭆⊞Ｏυω§#+÷⁻κμη

#+#⊞Ｏυω$n$

↙

Mova para baixo e para a esquerda, pronto para a próxima linha.

05AB1E , 43 42 bytes

$G©ÐX‚ˆÑʒ®>t‹}àDU+}¯εÝ`θ÷ɨDδ^}RζεðKζðδK€θ

Inspirado por @NickKennedy 'resposta Jelly s , e a parte posterior ζεðKζðδK€θé uma porta de @Emigna ' resposta 05AB1E s aqui .

Retorna uma matriz em 0vez de #e em 1vez de +.

$[2,n]$J,--no-lazy

Explicação:

$                # Push 1 and the input
 G               # Loop the input - 1 amount of times:
  ©              #  Store the top of the stack in variable `r` (without popping)
   Ð             #  And triplicate the top as well
    X‚           #  Pair it with variable `X` (which is 1 by default)
      ˆ          #  And pop and store this pair in the global array
    Ñ            #  Get the divisors of the integer we triplicated
     ʒ    }à     #  Get the highest divisor which is truthy for:
         ‹       #   Where the divisor integer is smaller than
      ®>t        #   the square root of `r+1`
            DU   #  Store a copy of this largest filtered divisor as new variable `X`
              +  #  And add it to the triplicated integer
 }¯              # After the loop: push the global array
   ε             # Map each pair to:
    Ý θ          #  Convert the first value in the pair to a list in the range [0,n]
     `           #  and push both this list and the second value to the stack
       ÷         # Integer-divide each value in the list by the second value
        É        # Check for each value if it's even (1 if even; 0 if odd)
         ¨       # Remove the last item
          Dδ     # Loop double vectorized over this list:
            ^    #  And XOR the values with each other
   }R            # After the map: reverse the list of digit-matrices
     ζ           # Zip/transpose; swapping rows and columns, with a space as filler
      ε          # map each matrix to:
       ðK        #  Remove all spaces from the current matrix
         ζ       #  Zip/transpose with a space as filler again
          ðδK    #  Deep remove all spaces
             €θ  #  Then only leave the last values of each row
                 # (after which the resulting matrix of 0s and 1s is output implicitly)

Haskell, 149 146 bytes

(iterate g["#"]!!)
g b|let e=(<$[1..d]);l=length b;d=last[i|i<-[1..l],i*i<=l,mod l i<1];m="+#"++m=(e$take(l+d)$e=<<'#':m)++zipWith(++)(e=<<e<$>m)b

Este é 0 indexado, retorna uma lista de strings e cresce para cima e para a esquerda.

Experimente online!

(iterate g["#"]!!)                    -- start with ["#"], repeatedly add a layer
                                      -- (via function 'g'), collect all results in
                                      -- a list and index it with the input number

g b | let                             -- add a single layer to chessboard 'b'

 l=length b                           -- let 'l' be the size of 'b'
 d=last[i|i<-[1..l],i*i<=l,mod l i<1] -- let 'd' be the size of the new layer
 e=(<$[1..d])                         -- let 'e' be a functions that makes 'd'
                                      --   copies of it's argument
 m="#+"++m                            -- let 'm' be an infinite string of "+#+#+..."

 =                                    -- return
              zipWith(++)             --   concatenate pairwise
                         (e=<<e<$>m)  --   a list of squares made by expanding each
                                      --   char in 'm' to size 'd'-by-'d'
                                    b --   and 'b' (zipWith truncates the infinite
                                      --   list of squares to the length of 'b')
                                      --
           ++                         --   and prepend
                                      --
(e$take(l+d)$e=<<'#':m)               --   the top layer, i.e. a list of 'd' strings
                                      --   each with the pattern 'd' times '#'
                                      --   followed by 'd' times '+', etc., each
                                      --   shortened to the correct size of 'l'+'g'

Perl 6 , 156 144 155 155 154 bytes

+11 para corrigir um erro relatado por nimi.

{$!=-1;join "
",(1,{my \k=max grep $_%%*,1.. .sqrt;++$!;flat .kv.map(->\i,\l {l~($!+i/k)%2+|0 x k}),substr(($!%2 x k~1-$!%2 x k)x$_,0,$_+k)xx k}...*)[$_]}

Basicamente baseado na solução Python de Chas Brown . Leva S com indexação zero. Saídas 0e 1.

Experimente online!