Giới thiệu

Các ma trận mật độ chu vi là vô hạn nhị phân ma trận M được xác định như sau. Xem xét một chỉ số ( dựa trên 1) (x, y) và biểu thị bằng M [x, y] ma trận con hình chữ nhật được kéo dài bởi góc (1, 1) và (x, y) . Giả sử rằng tất cả các giá trị của M [x, y] ngoại trừ M _{x, y} , giá trị tại chỉ mục (x, y) , đã được xác định. Khi đó giá trị M _{x, y} là giá trị 0 hoặc 1 đặt giá trị trung bình của M [x, y] gần với 1 / (x + y) . Trong trường hợp hòa, chọn M_{x, y} = 1 .

Đây là ma trận con M [20, 20] với các số 0 được thay thế bằng dấu chấm cho rõ ràng:

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

Ví dụ: chúng ta có M _{1, 1} = 1 ở góc trên bên trái, vì 1 / (1 + 1) = và trung bình của ma trận con 1 × 1 M [1, 1] là 0 hoặc 1 ; đó là cà vạt, vì vậy chúng tôi chọn 1 .

Xem xét sau đó vị trí (3, 4) . Chúng ta có 1 / (3 + 4) = 1/7 và trung bình của ma trận con M [3, 4] là 1/6 nếu chúng ta chọn 0 và 3/12 nếu chúng ta chọn 1 . Cái trước gần với 1/7 hơn , vì vậy chúng tôi chọn M _{3, 4} = 0 .

Dưới đây là ma trận con M [800, 800] dưới dạng hình ảnh, hiển thị một số cấu trúc phức tạp của nó.

Nhiệm vụ

Cho số nguyên dương N <1000 , xuất ra ma trận con N × N M [N, N] , ở bất kỳ định dạng hợp lý nào. Số byte thấp nhất sẽ thắng.

code-golf matrix

— Zgarb
nguồn

3

R, 158 154 141 byte

Chỉnh sửa: Bởi vì chỉ có 1trong lớp con trên 2x2là phía trên bên trái, M[1,1]chúng tôi có thể bắt đầu tìm kiếm 1skhi {x,y}>1không cần ifcâu lệnh.

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

Giải pháp này rất kém hiệu quả vì ma trận được nhân đôi hai lần cho mỗi lần lặp. n=1000chỉ mất chưa đầy hai tiếng rưỡi để chạy và tạo ra ma trận 7.6Mb.

Ung dung và giải thích

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

Đầu ra cho 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
nguồn

1

Python 2, 189 byte

Không có thủ đoạn điên rồ nào ở đây, nó chỉ là tính toán như được mô tả trong phần giới thiệu. Nó không đặc biệt nhanh nhưng tôi không cần tạo bất kỳ ma trận mới nào để làm điều này.

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

Giải trình:

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

Đối với những người tò mò, đây là một số thời gian:

 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.

Đầu ra "Khá" cho 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
nguồn

0

Vợt 294 byte

(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""))

Ung dung:

(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 ""))
  )

Kiểm tra:

(f 8 8)

Đầu ra:

— rnso
nguồn

0

Perl, 151 + 1 = 152 byte

Chạy với -ncờ. Mã sẽ chỉ hoạt động chính xác mỗi lần lặp khác trong cùng một thể hiện của chương trình. Để làm cho nó hoạt động chính xác mọi lúc, hãy thêm 5 byte bằng cách thêm vào my%m;mã.

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""}''

Có thể đọc được

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""
}

Đầu ra cho đầu vào 100:

1___________________________________________________________________________________________________
_____1______________________________________________________________________________________________
__1_________________________________________________________________________________________________
___________________________1________________________________________________________________________
____1_______________________________________________________________________________________________
_1__________________________________________________________________________________________________
_________________________1__________________________________________________________________________
_________________1__________________________________________________________________________________
______________1_____________________________________________________________________________________
____________1_______________________________________________________________________________________
__________1_________________________________________________________________________________________
____________________________________________________________________________________________________
_________1__________________________________________________________________________________________
____________________________________________________________________________________________________
________1___________________________________________________________________________________________
______________________________________________________________________________________1_____________
_________________________________________________________________1__________________________________
_______1____________________________________________________________________________________________
_____________________________________________________________1______________________________________
____________________________________________________1_______________________________________________
______________________________________________1_____________________________________________________
__________________________________________1_________________________________________________________
_______________________________________1____________________________________________________________
____________________________________1_______________________________________________________________
__________________________________1_________________________________________________________________
______1_____________________________________________________________________________________________
____________________________________________________________________________________________________
___1________________________________________________________________________________________________
____________________________________________________________________________________________________
____________________________________________________________________________________________________
____________________________________________________________________________________________________
____________________________________________________________________________________________________
________________________________1___________________________________________________________________
____________________________________________________________________________________________________
________________________1___________________________________________________________________________
____________________________________________________________________________________________________
_______________________1____________________________________________________________________________
____________________________________________________________________________________________________
____________________________________________________________________________________________________
______________________1_____________________________________________________________________________
____________________________________________________________________________________________________
___________________________________________________________________________________________________1
_____________________1______________________________________________________________________________
____________________________________________________________________________________________________
_____________________________________________________________________________________1______________
__________________________________________________________________________________1_________________
____________________1_______________________________________________________________________________
____________________________________________________________________________________________________
________________________________________________________________________________1___________________
______________________________________________________________________________1_____________________
___________________________________________________________________________1________________________
_________________________________________________________________________1__________________________
___________________1________________________________________________________________________________
____________________________________________________________________________________________________
____________________________________________________________________________________________________
_______________________________________________________________________1____________________________
______________________________________________________________________1_____________________________
____________________________________________________________1_______________________________________
___________________________________________________________1________________________________________
__________________________________________________________1_________________________________________
_________________________________________________________1__________________________________________
__________________1_________________________________________________________________________________
____________________________________________________________________________________________________
____________________________________________________________________________________________________
____________________________________________________________________________________________________
________________1___________________________________________________________________________________
____________________________________________________________________________________________________
____________________________________________________________________________________________________
____________________________________________________________________________________________________
____________________________________________________________________________________________________
________________________________________________________1___________________________________________
_______________________________________________________1____________________________________________
____________________________________________________________________________________________________
___________________________________________________1________________________________________________
____________________________________________________________________________________________________
__________________________________________________1_________________________________________________
____________________________________________________________________________________________________
____________________________________________________________________________________________________
_________________________________________________1__________________________________________________
____________________________________________________________________________________________________
________________________________________________1___________________________________________________
____________________________________________________________________________________________________
_____________________________________________1______________________________________________________
____________________________________________________________________________________________________
____________________________________________________________________________________________________
____________________________________________1_______________________________________________________
_______________1____________________________________________________________________________________
____________________________________________________________________________________________________
____________________________________________________________________________________________________
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____________________________________________________________________________________________________
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____________________________________________________________________________________________________
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_________________________________________1__________________________________________________________

— Gabriel Benamy
nguồn

Tính ma trận mật độ chu vi