บทนำ

หนาแน่นมดลูกปริมณฑลเป็นอนันต์เมทริกซ์ไบนารีMกำหนดดังนี้ พิจารณา (1-based) ดัชนี(x, y)และแสดงว่าโดยM [x, y]สี่เหลี่ยมย่อยเมทริกซ์ทอดมุม(1, 1)และ(x, y) สมมติว่าค่าทั้งหมดของM [x, y]ยกเว้นM _{x, y} , ค่าที่ดัชนี(x, y)ได้ถูกกำหนดแล้ว แล้วค่าM _{x, y}เป็นแล้วแต่จำนวนใดของ0หรือ1ที่ทำให้ค่าเฉลี่ยของM [x, y]ใกล้ชิดกับ1 / (x + y) ในกรณีที่เสมอกันให้เลือกM_{x, y} = 1

นี่คือ sub-matrix M [20, 20]ด้วยค่าศูนย์แทนที่ด้วยจุดเพื่อความชัดเจน:

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

ตัวอย่างเช่นเรามีM _{1, 1} = 1ที่มุมซ้ายบนเนื่องจาก1 / (1 + 1) = ½และค่าเฉลี่ยของเมทริกซ์ย่อย1 × 1 M [1, 1]เป็น0หรือ1 ; ที่ผูกดังนั้นเราจึงเลือกที่1

พิจารณาแล้วตำแหน่ง(3, 4) เรามี1 / (3 + 4) = 1/7และค่าเฉลี่ยของ sub-เมทริกซ์M [3, 4]คือ1/6ถ้าเราเลือก0และ3/12ถ้าเราเลือกที่ 1 อดีตคือใกล้ชิดกับ1/7ดังนั้นเราจึงเลือกM _{3, 4} = 0

นี่คือ sub-matrix M [800, 800]ในฐานะภาพแสดงโครงสร้างที่ซับซ้อนบางอย่าง

งาน

รับจำนวนเต็มบวกN <1,000เอาท์พุทเมทริกซ์ย่อยN × N M [N, N]ในรูปแบบที่เหมาะสม จำนวนไบต์ต่ำสุดชนะ

R, 158 154 141 ไบต์

แก้ไข: เนื่องจากสิ่งเดียวที่1อยู่ใน2x2submatrix ด้านบนเป็นมุมบนซ้ายM[1,1]เราสามารถเริ่มค้นหา1sเมื่อ{x,y}>1ไม่จำเป็นต้องมีifคำสั่ง

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

การแก้ปัญหาไม่มีประสิทธิภาพสูงเนื่องจากเมทริกซ์ซ้ำสองครั้งสำหรับการทำซ้ำแต่ละครั้ง n=1000ใช้เวลาเพียงไม่ถึงสองชั่วโมงครึ่งในการทำงานและสร้างเมทริกซ์ของ7.6Mb

Ungolfed และอธิบาย

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

ผลผลิตสำหรับ 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

Python 2, 189 ไบต์

ที่นี่ไม่มีเทคนิคบ้า ๆ บอ ๆ มันแค่คำนวณตามที่อธิบายไว้ในบทนำ มันไม่ได้รวดเร็วโดยเฉพาะ แต่ฉันไม่จำเป็นต้องสร้างเมทริกซ์ใหม่เพื่อทำสิ่งนี้

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

คำอธิบาย:

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

สำหรับผู้ที่อยากรู้อยากเห็นต่อไปนี้เป็นเวลา

 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.

เอาต์พุต "Pretty" สำหรับn = 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

แร็กเก็ต 294 ไบต์

(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 ""))
  )

การทดสอบ:

(f 8 8)

เอาท์พุท:

10000000
00000100
00100000
00000000
00001000
01000000
00000000
00000000

Perl, 151 + 1 = 152 ไบต์

วิ่งด้วย-nธง รหัสจะทำงานได้อย่างถูกต้องทุกอื่น ๆซ้ำภายในเช่นเดียวกันของโปรแกรม หากต้องการให้มันทำงานอย่างถูกต้องทุกครั้งให้เพิ่ม 5 ไบต์โดยmy%m;ใส่รหัสลงไป

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""}''

อ่านได้:

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""
}

เอาต์พุตสำหรับอินพุต 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_____________________________________________________________________________________________
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___1________________________________________________________________________________________________
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________________________________1___________________________________________________________________
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________________________1___________________________________________________________________________
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_______________________1____________________________________________________________________________
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______________________1_____________________________________________________________________________
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___________________________________________________________________________________________________1
_____________________1______________________________________________________________________________
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_____________________________________________________________________________________1______________
__________________________________________________________________________________1_________________
____________________1_______________________________________________________________________________
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___________________1________________________________________________________________________________
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_________________________________________________1__________________________________________________
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