Penafian: Kisah yang diceritakan dalam pertanyaan ini sepenuhnya fiktif, dan diciptakan semata-mata untuk tujuan memberikan intro.

Saya seorang petani jahat, dan untuk menaikkan harga gandum di daerah saya, saya telah memutuskan untuk membakar ladang semua petani di sekitar saya. Saya benar-benar ingin melihat ladang terbakar (sehingga saya bisa menggunakan tawa jahat saya dan menggosok tangan saya bersama-sama dengan gembira), tetapi saya juga tidak ingin ketahuan menonton, jadi saya perlu Anda untuk mensimulasikan lapangan menjadi dibakar untukku.

Tugas Anda:

Tulis sebuah program atau fungsi yang mengambil sebagai input bidang, dan mengembalikan tahapan-tahapannya membakar sampai seluruh bidang menjadi abu. Bagian spesifik dari bidang yang terbakar diwakili oleh bilangan bulat yang mewakili intensitas nyala api. Api dimulai pada "1" dan bergerak ke "2" dan kemudian "3", dan seterusnya. Begitu api mencapai "4", api akan menangkap area yang berdekatan secara langsung (tidak diagonal) yang mudah terbakar. Setelah mencapai "8", ia terbakar pada iterasi berikutnya, dan berubah menjadi abu, diwakili oleh "A". Ketika suatu daerah belum disentuh oleh api, itu diwakili oleh "0". Misalnya, jika bidangnya terlihat seperti ini:

100
000

Program Anda harus menampilkan ini:

100
000

200
000

300
000

410
100

520
200

630
300

741
410

852
520

A63
630

A74
741

A85
852

AA6
A63

AA7
A74

AA8
A85

AAA
AA6

AAA
AA7

AAA
AA8

AAA
AAA

Jika mau, Anda dapat mengganti simbol di atas dengan seperangkat simbol yang Anda pilih, asalkan konsisten dan berbeda satu sama lain.

Memasukkan:

Posisi awal bidang, dalam bentuk standar apa pun, seperti string yang dibatasi baru seperti di atas.

Keluaran:

Bidang di setiap iterasi saat dibakar, baik sebagai array, atau sebagai string yang dibatasi oleh beberapa karakter.

Kasus uji:

0301
000A
555
 |
 v
0301
000A
555

1412
010A
666

2523
020A
777

3634
030A
888

4745
141A
AAA

5856
252A
AAA

6A67
363A
AAA

7A78
474A
AAA

8A8A
585A
AAA

AAAA
6A6A
AAA

AAAA
7A7A
AAA

AAAA
8A8A
AAA

AAAA
AAAA
AAA

Mencetak:

Ini adalah kode-golf , skor terendah dalam byte menang!

APL (Dyalog) , 52 byte *

Mengasumsikan ⎕IO←0yang default pada banyak sistem. Mengambil lapangan menggunakan 0 untuk slot kosong, 1 untuk bidang tidak terbakar, 2 untuk api baru, 5 untuk menyebarkan api, dan 10 untuk abu. Input harus minimal 3 × 3, yang tidak masalah karena baris dan kolom tambahan mungkin diisi dengan nol (spasi dalam format OP).

{}{1=c←4⊃r←,⍵:1+4∊r/⍨9⍴⍳2⋄10⌊c+×c}⌺3 3⍣{⍺≡⎕←⍵}

Cobalah online!

Format saya membuatnya sulit untuk memeriksa kebenarannya, jadi di sini adalah versi dengan pemrosesan pra dan pasca yang ditambahkan untuk menerjemahkan dari dan ke format OP.

⍣{... } ulangi sampai:

 ⍺ generasi penerus bangsa

 ≡ identik dengan

 ⎕←⍵ generasi saat ini, dikeluarkan

{... }⌺3 3 ganti setiap sel dengan hasil fungsi ini diterapkan pada lingkungan Moore:

 ,⍵ ravel (meratakan) argumen; memberikan daftar sembilan elemen

 r← tetapkan ke r

 4⊃ pilih elemen keempat; pusat, yaitu nilai sel asli

 c← ditugaskan untuk c

 1= Apakah sama dengan itu?

 : jika demikian, maka:

  ⍳2 pertama ke ɩ ntegers; 0 1

  9⍴ r eshape sampai panjang sembilan; 0 1 0 1 0 1 0 1 0

  r/⍨ gunakan itu untuk menyaring r (ini hanya mendapat tetangga orthogonal)

  4∊ Apakah empat anggota itu? (yaitu apakah akan ada lima generasi berikutnya?)

  1+ tambahkan satu; 1 jika tidak terbakar atau 2 jika terbakar

 ⋄ lain (yaitu nilai saat ini adalah 0 atau ≥ 2)

  ×c signum dari c

  c+ c plus itu (yaitu bertambah satu jika terbakar)

  10⌊ minimum of ten and that (as ash doesn't burn on)

* In Dyalog Classic, using ⎕U233A instead of ⌺.

Python 3, 232 bytes

def f(j):
 k=[[int(min(9,j[x][y]+(j[x][y]>0)or 3in(lambda k,x,y:[k[i][j]for i,j in[[x-1,y],[x+1,y],[x,y-1],[x,y+1]]if(-1<i<len(k))and(-1<j<len(k[i]))])(j,x,y)))for y in range(len(j[x]))]for x in range(len(j))]
 if k!=j:print(k);f(k)

Try it online!

-3 bytes thanks to officialaimm by merging the other lambda into f (looks messy but saves bytes and that's all we care about)
-8 bytes thanks to Mr. Xoder
-26 bytes thanks to ovs
-6 bytes thanks to ppperry

JavaScript (ES6), 217 210 207 204 193 192 190 bytes

Saved 2 bytes thanks to @Shaggy's suggestion of using 9 as A.

f=F=>[F,...(t=[],y=0,g=x=>(r=F[y])?(x||(t[y]=[]),r[x]+1)?(t[y][x]=r[x]<8?r[x]+(r[x]>0|[r[x-1],r[x+1],F[y-1]&&F[y-1][x],F[y+1]&&F[y+1][x]].includes(3)):9,g(x+1)):g(0,y++):t)(0)+""!=F?f(t):[]]

// test code
test=s=>{
  var test = document.querySelector("#in").value.split`\n`.map(e => Array.from(e).map(e => parseInt(e)));
  var out = f(test);
  document.querySelector("#out").innerHTML = out.map(e => e.map(e => e.join``).join`\n`).join`\n\n`;
};window.addEventListener("load",test);

<textarea id="in" oninput="test()">0301&#10;0009&#10;555</textarea><pre id="out"></pre>

Uses 9 instead of A. Input as a 2D array of integers. Output as an array of such arrays.

Simulating the World (in Emoji), 1407 bytes?

Don't you love using an explorable explanation as a programming language? The downside to this is there's usually not a very well defined program, so in this case, I'm using the JSON it exports. (if you have any better ideas, let me know)

{"meta":{"description":"","draw":1,"fps":1,"play":true},"states":[{"id":0,"icon":"0","name":"","actions":[{"sign":">=","num":1,"stateID":"4","actions":[{"stateID":"1","type":"go_to_state"}],"type":"if_neighbor"}],"description":""},{"id":1,"icon":"1","name":"","description":"","actions":[{"stateID":"2","type":"go_to_state"}]},{"id":2,"icon":"2","name":"","description":"","actions":[{"stateID":"3","type":"go_to_state"}]},{"id":3,"icon":"3","name":"","description":"","actions":[{"stateID":"4","type":"go_to_state"}]},{"id":4,"icon":"4","name":"","description":"","actions":[{"stateID":"5","type":"go_to_state"}]},{"id":5,"icon":"5","name":"","description":"","actions":[{"stateID":"6","type":"go_to_state"}]},{"id":6,"icon":"6","name":"","description":"","actions":[{"stateID":"7","type":"go_to_state"}]},{"id":7,"icon":"7","name":"","description":"","actions":[{"stateID":"8","type":"go_to_state"}]},{"id":8,"icon":"8","name":"","description":"","actions":[{"stateID":"9","type":"go_to_state"}]},{"id":9,"icon":"A","name":"","description":"","actions":[]}],"world":{"update":"simultaneous","neighborhood":"neumann","proportions":[{"stateID":0,"parts":100},{"stateID":1,"parts":0},{"stateID":2,"parts":0},{"stateID":3,"parts":0},{"stateID":4,"parts":0},{"stateID":5,"parts":0},{"stateID":6,"parts":0},{"stateID":7,"parts":0},{"stateID":8,"parts":0},{"stateID":9,"parts":0}],"size":{"width":9,"height":9}}}

Try it here or here:

<iframe width="100%" height="450" src="http://ncase.me/simulating/model/?remote=-Kr2X939XcFwKAunEaMK" frameborder="0"></iframe>

Retina, 103 96 88 bytes

^
¶
;{:`

T`0d`d
(?<=¶(.)*)0(?=4|.*¶(?<-1>.)*(?(1)_)4|(?<=40|¶(?(1)_)(?<-1>.)*4.*¶.*))
1

Try it online! Uses 9 for ash; this can be changed at a cost of 4 bytes using T`1-8`2-8A. Edit: Saved 6 bytes thanks to @MartinEnder. Explanation:

^
¶

Add a separator so that the outputs don't run into each other. (Also helps when matching below.)

;{:`

Don't print the final state (which is the same as the previous state which has already been printed). Repeat until the pass does not change the state. Print the current state before each pass.

T`0d`d

Advance the intensity of all the fire.

(?<=¶(.)*)0(?=4|.*¶(?<-1>.)*(?(1)_)4|(?<=40|¶(?(1)_)(?<-1>.)*4.*¶.*))
1

Light unlit fields as appropriate. Sub-explanation:

(?<=¶(.)*)

Measure the column number of this unlit field.

0

Match the unlit field.

(?=4

Look for a suitable field to the right.

  |.*¶(?<-1>.)*(?(1)_)4

Look for a suitable field in the same column (using a balancing group) in the line below. Note that if the input could be guaranteed rectangular then this could be simplified to |.*¶(?>(?<-1>.)*)4 for a saving of 3 bytes.

  |(?<=40

Look for a suitable field to the left. (Since we're looking from the right-hand side of the field, we see the unlit field as well.)

      |¶(?(1)_)(?<-1>.)*4.*¶.*))

Look for a suitable field in the same column in the line above. Since this is a lookbehind and therefore a right-to-left match, the balancing group condition has to appear before the columns that have been matched by the balancing group.

Perl 5, 365 bytes

@a=map{[/./g]}<>;do{say@$_ for@a;say;my@n=();for$r(0..$#a){$l=$#{$a[$r]};for(0..$l){$t=$a[$r][$_];$n[$r][$_]=($q=$n[$r][$_])>$t?$q:$t==9?9:$t?++$t:0;if($t==4){$n[$r-1][$_]||=1if$r&&$_<$#{$a[$r-1]};$n[$r+1][$_]||=1if$r<$#a&&$_<$#{$a[$r+1]};$n[$r][$_-1]||=1if$_;$n[$r][$_+1]||=1if$_<$l}}}$d=0;for$r(0..$#a){$d||=$a[$r][$_]!=$n[$r++][$_]for 0..$#{$a[$r]}}@a=@n}while$d

Try it online!

Uses '9' instead of 'A' to indicated a burned out location.

Explained

@a=map{[/./g]}<>;   # split input into a 2-D array

do{
say@$_ for@a;say;   # output the current state
my@n=();            # holds the next iteration as it's created
for$r(0..$#a){      # loop through rows
  $l=$#{$a[$r]};    # holder for the length of this row
  for(0..$l){
    $t=$a[$r][$_];  # temporary holder for current value
    $n[$r][$_]=($q=$n[$r][$_])>$t?$q:$t==9?9:$t?++$t:0; #update next iteration
    if($t==4){      # ignite the surrounding area if appropriate
      $n[$r-1][$_]||=1if$r&&$_<$#{$a[$r-1]};
      $n[$r+1][$_]||=1if$r<$#a&&$_<$#{$a[$r+1]};
      $n[$r][$_-1]||=1if$_;
      $n[$r][$_+1]||=1if$_<$l
    }
  }
}
$d=0;              # determine if this generation is different than the previous
for$r(0..$#a){$d||=$a[$r][$_]!=$n[$r++][$_]for 0..$#{$a[$r]}}
@a=@n              # replace master with new generation
}while$d

Haskell, 162 bytes

import Data.List
i x|x<'9'=succ x|1<2=x
f('3':'@':r)="30"++f r
f(x:r)=x:f r
f e=e
a%b=a.b.a.b
g=map(i<$>).(transpose%map(reverse%f))
h x|y<-g x,y/=x=x:h y|1<3=[x]

Try it online! Usage: h takes a field as a list of lines and returns a list of fields. An unburned field is indicated by @ and ash by 9, the different fires are the digits 1 to 8.

• f manages the spreading of the fire from left to right by replacing all unburned @ fields which are right to a burning 3 field with 0.
• i increments each digit as long as it is less than 9.
• g applies f to each line, then reverses the line, applies f again and reverses back. Then the list of of lines is transposed and again on each line and its reverse f is applied.
• h applies g to the input until it no longer changes and collects the results.

C (gcc), 308 305 299 297 295 291 bytes

#define F B[i][v]
m(A,B,k,y,m,U,i,v)char**B;{do{for(i=k=y=0;i<A;puts(""),++i)for(v=0;v<(m=strlen(B[i]));F=(U=F)>48&&F<56?F+1:F>55?65:(i+1<A&&B[i+1][v]==51||v+1<m&&B[i][v+1]==51||v-1>-1&&B[i][v-1]==52||i-1>-1&&B[i-1][v]==52)&&U<49?49:F,putchar(U),k+=U<49||U>64,++y,++v);puts("");}while(k-y);}

This program defines a function which takes two inputs, a pointer to an array of strings preceded by its length, as allowed by this I/O default. Outputs to STDOUT with a trailing newline.

Try it online!

Octave, 72 69 bytes

a=input(''),do++a(a&a<9);a+=imdilate(a==4,~(z=-1:1)|~z')&~a,until a>8

The input is taken as a 2D array of numbers, and empty spots marked with Inf. 'A' has been replaced with 9. Intermediate results (as array of numbers) implicitly printed .

Try it online!

Explanation:

In a loop the function imdilate (morphological image dilation) from the image package is used to simulate fire propagation.

Python 2, 325 Bytes

def f(x):
 while{i for m in x for i in m}>{9,''}:
    yield x;i=0
    for l in x:
     j=0
     for c in l:
        if 0<c<9:x[i][j]+=1
        j+=1
     i+=1
    i=0
    for l in x:
     j=0
     for c in l:
        if c==0 and{1}&{x[k[1]][k[2]]==4for k in[y for y in[[i,i-1,j],[i<len(x)-1,i+1,j],[j,i,j-1],[j<len(l)-1,i,j+1]]if y[0]]}:x[i][j]+=1
        j+=1
     i+=1
 yield x

f takes the input as a 2D array of integers, and empty spots marked with ''. 'A' has been replaced with 9. The function outputs a generator of all the fields over time in the same format.

Try it online!

Octave, 212 bytes

function r(f)b=48;f-=b;c=-16;l=f~=-c;d=17;p=@(x)disp(char(x+b));g=@()disp(' ');p(f);g();while~all(any(f(:)==[0 d c],2))m=f>0&l;f(m)+=1;k=conv2(f==4,[0 1 0;1 0 1;0 1 0],'same')&~m&l;f(k)+=1;f(f>8&l)=d;p(f);g();end

To run, specify a character array such as:

f = ['0301','000A','555 '];

... then do:

r(f)

Explanation of the code to follow...

Try it online!

^{Note: I tried to run this code with tio.run, but I wasn't getting any output. I had to use another service.}

PHP, 226 212 210 209 185 177 bytes

for($f=file(m);$f!=$g;print"
")for($g=$f,$y=0;$r=$f[$y++];)for($x=-1;~$c=$r[++$x];$f[$y-1][$x]=$c=="0"?strstr($g[$y-2][$x].$r[$x-1].$f[$y][$x].$r[$x+1],51)/3:min(++$c,9))echo$c;

takes input with a trailing newline from a file named m; 9 for ashes.

Run with -nr or try it online.

first approach: PHP 7.0, 209 bytes

for($f=$g=file(m);trim(join($f),"A
");print"
".join($f=$g))for($y=-1;(a&$c=$r[++$x])||$r=$f[$y-=$x=-1];)for(+$c&&$g[$y][$x]=++$c<9?$c:A,$a=4;$a--;$q=$p)$c-4|($w=&$g[$y+$p=[1,0,-1][$a]])[$q+=$x]!="0"||$w[$q]=1;

takes input with a trailing newline from a file named m.

Run with -nr or try it online.

PHP version notes (for old approach)

• for PHP older than 7.0, replace everything after $c-4| with $g[$y+$p=[1,0,-1][$a]][$q+=$x]!="0"||$g[$y+$p][$q]=1;
• for PHP older than 5.5, replace [1,0,-1][$a] with $a%2*~-($a&2)
• for PHP newer than 7.0, replace a&$c with ""<$c, +$c with 0<$c and $c-4 with $c!=4

Octave, 419 312 bytes

M=input('') %take a matrix M as input
[a, b]=size(M); %size M for later
while mean(mean(M))!=9 %while the whole matrix M isn't ashes
M=M+round(M.^0.001); %if there is a nonzero int add 1 to it
for i=1:a %loop over all values of M
for j=1:b
if(M(i,j)==4) %if a value of 4 is found, ignite the zeros around it
if(M(max(i-1,1),j)==0) M(i-1,j)++;end
if(M(min(i+1,a),j)==0) M(i+1,j)++;end
if(M(i,max(j-1,1))==0) M(i,j-1)++;end      
if(M(i,min(j+1,b))==0) M(i,j+1)++;end
elseif(M(i,j)==10) M(i,j)--;
end
end
end
M %display the matrix
end

Try it online!

This is my version it works, so now I still need to golf it. I think it can be a lot shorter if I find a way to find the indices of the 4 in a matrix, but I don't know how.
PS: A is a 9 in my code.

Stencil (∊-mode), 22 bytes

S≡8:'A'⋄0::S⋄(3∊N)⌈S+s

Try it online!

Just like in the test input, use integers separated by spaces for 0-8, ' ' for blank and 'A' for A. Remember to add trailing blanks too.