Introdução

A matriz de densidade do perímetro é uma matriz binária infinita M definida da seguinte forma. Considere um índice ( com base em 1) (x, y) e denote por M [x, y] a sub-matriz retangular estendida pelo canto (1, 1) e (x, y) . Suponha que todos os valores de M [x, y], exceto M _{x, y} , o valor no índice (x, y) , já tenham sido determinados. Então o valor M _{x, y} é o valor 0 ou 1 que coloca o valor médio de M [x, y] mais próximo de 1 / (x + y) . Em caso de empate, escolha M_{x, y} = 1 .

Esta é a sub-matriz M [20, 20] com zeros substituídos por pontos para maior clareza:

1 . . . . . . . . . . . . . . . . . . .
. . . . . 1 . . . . . . . . . . . . . .
. . 1 . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . .
. . . . 1 . . . . . . . . . . . . . . .
. 1 . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . 1 . .
. . . . . . . . . . . . . . 1 . . . . .
. . . . . . . . . . . . 1 . . . . . . .
. . . . . . . . . . 1 . . . . . . . . .
. . . . . . . . . . . . . . . . . . . .
. . . . . . . . . 1 . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . .
. . . . . . . . 1 . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . .
. . . . . . . 1 . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . .

Por exemplo, temos M _{1, 1} = 1 no canto superior esquerdo, pois 1 / (1 + 1) = ½ , e a média da sub-matriz 1 × 1 M [1, 1] é 0 ou 1 ; isso é empate, então escolhemos 1 .

Considere então a posição (3, 4) . Temos 1 / (3 + 4) = 1/7 , e a média da sub-matriz M [3, 4] é 1/6 se escolhermos 0 e 3/12 se escolhermos 1 . O primeiro está mais próximo de 1/7 , então escolhemos M _{3, 4} = 0 .

Aqui está a sub-matriz M [800, 800] como uma imagem, mostrando parte de sua intrincada estrutura.

A tarefa

Dado um número inteiro positivo N <1000 , produza a sub-matriz N × N M [N, N] , em qualquer formato razoável. A menor contagem de bytes vence.

code-golf matrix

— Zgarb
fonte

3

R, 158 154 141 bytes

Edit: Como o único 1na 2x2submatriz superior é a parte superior esquerda M[1,1], podemos iniciar a busca por 1squando, {x,y}>1portanto, não há necessidade da ifdeclaração.

M=matrix(0,n<-scan(),n);M[1]=1;for(i in 2:n)for(j in 2:n){y=x=M[1:i,1:j];x[i,j]=0;y[i,j]=1;d=1/(i+j);M[i,j]=abs(d-mean(x))>=abs(d-mean(y))};M

A solução é altamente ineficiente, pois a matriz é duplicada duas vezes para cada iteração. n=1000demorou pouco menos de duas horas e meia para rodar e produz uma matriz de 7.6Mb.

Ungolfed e explicou

M=matrix(0,n<-scan(),n);                        # Read input from stdin and initialize matrix with 0s
M[1]=1;                                         # Set top left element to 1
for(i in 2:n){                                  # For each row    
    for(j in 2:n){                              # For each column
        y=x=M[1:i,1:j];                         # Generate two copies of M with i rows and j columns
        x[i,j]=0;                               # Set bottom right element to 0
        y[i,j]=1;                               # Set bottom right element to 1
        d=1/(i+j);                              # Calculate inverse of sum of indices
        M[i,j]=abs(d-mean(x))>=abs(d-mean(y))   # Returns FALSE if mean(x) is closer to d and TRUE if mean(y) is
    }
};
M                                               # Print to stdout

Saída para n=20

      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] [,15] [,16] [,17] [,18] [,19] [,20]
[1,]     1    0    0    0    0    0    0    0    0     0     0     0     0     0     0     0     0     0     0     0
[2,]     0    0    0    0    0    1    0    0    0     0     0     0     0     0     0     0     0     0     0     0
[3,]     0    0    1    0    0    0    0    0    0     0     0     0     0     0     0     0     0     0     0     0
[4,]     0    0    0    0    0    0    0    0    0     0     0     0     0     0     0     0     0     0     0     0
[5,]     0    0    0    0    1    0    0    0    0     0     0     0     0     0     0     0     0     0     0     0
[6,]     0    1    0    0    0    0    0    0    0     0     0     0     0     0     0     0     0     0     0     0
[7,]     0    0    0    0    0    0    0    0    0     0     0     0     0     0     0     0     0     0     0     0
[8,]     0    0    0    0    0    0    0    0    0     0     0     0     0     0     0     0     0     1     0     0
[9,]     0    0    0    0    0    0    0    0    0     0     0     0     0     0     1     0     0     0     0     0
[10,]    0    0    0    0    0    0    0    0    0     0     0     0     1     0     0     0     0     0     0     0
[11,]    0    0    0    0    0    0    0    0    0     0     1     0     0     0     0     0     0     0     0     0
[12,]    0    0    0    0    0    0    0    0    0     0     0     0     0     0     0     0     0     0     0     0
[13,]    0    0    0    0    0    0    0    0    0     1     0     0     0     0     0     0     0     0     0     0
[14,]    0    0    0    0    0    0    0    0    0     0     0     0     0     0     0     0     0     0     0     0
[15,]    0    0    0    0    0    0    0    0    1     0     0     0     0     0     0     0     0     0     0     0
[16,]    0    0    0    0    0    0    0    0    0     0     0     0     0     0     0     0     0     0     0     0
[17,]    0    0    0    0    0    0    0    0    0     0     0     0     0     0     0     0     0     0     0     0
[18,]    0    0    0    0    0    0    0    1    0     0     0     0     0     0     0     0     0     0     0     0
[19,]    0    0    0    0    0    0    0    0    0     0     0     0     0     0     0     0     0     0     0     0
[20,]    0    0    0    0    0    0    0    0    0     0     0     0     0     0     0     0     0     0     0     0

— Billywob
fonte

1

Python 2, 189 bytes

Não há truques malucos aqui, é apenas um cálculo, conforme descrito na introdução. Não é particularmente rápido, mas não preciso criar nenhuma nova matriz para fazer isso.

n=input()
k=[n*[0]for x in range(n)]
for i in range(1,-~n):
 for j in range(1,-~n):p=1.*i*j;f=sum(sum(k[l][:j])for l in range(i));d=1./(i+j);k[i-1][j-1]=0**(abs(f/p-d)<abs(-~f/p-d))
print k

Explicação:

n=input()                                     # obtain size of matrix  
k=[n*[0]for x in range(n)]                    # create the n x n 0-filled matrix
for i in range(1,-~n):                        # for every row:
  for j in range(1,-~n):                      # and every column:
    p=1.*i*j                                  # the number of elements 'converted' to float
    f=sum(sum(k[l][:j])for l in range(i))     # calculate the current sum of the submatrix
    d=1./(i+j)                                # calculate the goal average
    k[i-1][j-1]=0**(abs(f/p-d)<abs(-~f/p-d))  # decide whether cell should be 0 or 1
print k                                       # print the final matrix

Para os curiosos, aqui estão alguns horários:

 20 x  20 took 3 ms.
 50 x  50 took 47 ms.
100 x 100 took 506 ms.
250 x 250 took 15033 ms.
999 x 999 took 3382162 ms.

Saída "bonita" para n = 20:

1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

— Kade
fonte

0

Raquete 294 bytes

(define(g x y)(if(= 1 x y)1(let*((s(for*/sum((i(range 1(add1 x)))(j(range 1(add1 y)))#:unless(and(= i x)(= j y)))
(g i j)))(a(/ s(* x y)))(b(/(add1 s)(* x y)))(c(/ 1(+ x y))))(if(<(abs(- a c))(abs(- b c)))0 1))))
(for((i(range 1(add1 a))))(for((j(range 1(add1 b))))(print(g i j)))(displayln""))

Ungolfed:

(define(f a b)  
  (define (g x y)
    (if (= 1 x y) 1
        (let* ((s (for*/sum ((i (range 1 (add1 x)))
                             (j (range 1 (add1 y)))
                             #:unless (and (= i x) (= j y)))
                    (g i j)))
               (a (/ s (* x y)))
               (b (/ (add1 s) (* x y)))
               (c (/ 1 (+ x y))))
          (if (< (abs(- a c))
                 (abs(- b c)))
              0 1))))
  (for ((i (range 1 (add1 a))))
    (for ((j (range 1 (add1 b))))
      (print (g i j)))
    (displayln ""))
  )

Teste:

(f 8 8)

Resultado:

— rnso
fonte

0

Perl, 151 + 1 = 152 bytes

Corra com a -nbandeira. O código só funcionará corretamente em qualquer outra iteração na mesma instância do programa. Para que ele funcione corretamente todas as vezes, adicione 5 bytes acrescentando my%m;o código.

for$b(1..$_){for$c(1..$_){$f=0;for$d(1..$b){$f+=$m{"$d,$_"}/($b*$c)for 1..$c}$g=1/($b+$c);print($m{"$b,$c"}=abs$f-$g>=abs$f+1/($b*$c)-$g?1:_).$"}say""}''

Legível:

for$b(1..$_){
    for$c(1..$_){
        $f=0;
        for$d(1..$b){
            $f+=$m{"$d,$_"}/($b*$c)for 1..$c
        }
        $g=1/($b+$c);
        print($m{"$b,$c"}=abs$f-$g>=abs$f+1/($b*$c)-$g?1:_).$"
    }
    say""
}

Saída para entrada de 100:

1___________________________________________________________________________________________________
_____1______________________________________________________________________________________________
__1_________________________________________________________________________________________________
___________________________1________________________________________________________________________
____1_______________________________________________________________________________________________
_1__________________________________________________________________________________________________
_________________________1__________________________________________________________________________
_________________1__________________________________________________________________________________
______________1_____________________________________________________________________________________
____________1_______________________________________________________________________________________
__________1_________________________________________________________________________________________
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________1___________________________________________________________________________________________
______________________________________________________________________________________1_____________
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___1________________________________________________________________________________________________
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________________________1___________________________________________________________________________
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_______________________1____________________________________________________________________________
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_________________________________________1__________________________________________________________

— Gabriel Benamy
fonte

Calcular a matriz de densidade do perímetro