Lua, 531 509 488 487 464 424 405 404 byte
Ai muốn một đệ trình lớn? \ o /
Chỉnh sửa: Đã cải thiện nó, nhưng không biết chơi golf nữa, vì vậy ... những lời giải thích đang được thêm vào :)
Đã lưu ~ 60 byte với sự trợ giúp của @ KennyLau
chơi golf nhỏ cắt thêm một byte bằng cách đổi tên athành Yđể ngăn chuyển đổi thập lục phân nội tuyến
function f(m)c={}t=table.concat::z::c[#c+1]={}p=c[#c]Y={}for i=1,#m do k=m[i]p[#p+1]=t(k)Y[i]={}for j=1,#k
do v=m[i%#m+1]l=j%#k+1w=m[(i-2)%#m+1]h=(j-2)%#k+1Y[i][j]=v[l]+k[l]+w[l]+v[h]+v[j]+w[h]+w[j]+k[h]end
end s=''for i=1,#m do k=m[i]for j=1,k do
x=Y[i][j]k[j]=k[j]>0 and((x<2or x>3)and 0or 1)or (x==3 and 1or 0)end
s=s..t(k)end for i=1,#c do
if(s==t(c[i]))then return#c>i and-1or i-1
end end goto z end
Bị đánh cắp
function f(m) -- takes a 2D array of 0 and 1s as input
c={} -- intialise c -> contains a copy of each generation
t=table.concat -- shorthand for the concatenating function
::z:: -- label z, used to do an infinite loop
c[#c+1]={} -- initialise the first copy
p=c[#c] -- initialise a pointer to this copy
a={} -- initialise the 2D array of adjacency
for i=1,#m -- iterate over the lines of m
do
k=m[i] -- shorthand for the current line
p[#p+1]=t(k]) -- saves the current line of m as a string
a[i]={} -- initialise the array of adjacency for the current line
for j=1,#k -- iterate over each row of m
do
-- the following statements are used to wraps at borders
v=m[i%#m+1] -- wrap bottom to top
l=j%#k+1 -- wrap right to left
w=m[(i-2)%#m+1] -- wrap top to bottom
h=(j-2)%#k+1 -- wrap left to right
a[i][j]= v[l] -- living cells are 1 and deads are 0
+k[l] -- just add the values of adjacent cells
+w[l] -- to know the number of alive adjacent cells
+v[h]
+v[j]
+w[h]
+w[j]
+k[h]
end
end
s='' -- s will be the representation of the current generation
for i=1,#m -- iterate over each line
do
k=m[i] -- shorthand for the current line
for j=1,#k -- iterate over each row
do
x=a[i][j] -- shorthand for the number of adjacent to the current cell
-- the next line change the state of the current cell
k[j]=k[j]>0 -- if it is alive
and((x<2 -- and it has less than 2 adjacent
or x>3) -- or more than 3 adjacent
and 0 -- kill it
or 1) -- else let it alive
or -- if it is dead
(x==3 -- and it has 3 adjacent
and 1 -- give life to it
or 0) -- else let it dead
end
s=s..t(k) -- save the representation of the current line
end
for i=1,#c -- iterate over all the generation done until now
do
if(s==t(c[i])) -- if the representation of the current generation
then -- is equal to one we saved
return#c>i -- check if it is the latest generation
and-1 -- if it isn't, it means we are in a loop -> return -1
or i-1 -- if it is, we did 2 generations without changing
-- -> return the number of generation
end
end
goto z -- if we reach that point, loop back to the label z
end
Các trường hợp thử nghiệm
Đây là một số trường hợp thử nghiệm
function f(m)c={}t=table.concat::z::c[#c+1]={}p=c[#c]a={}for i=1,#m do k=m[i]p[#p+1]=t(k)a[i]={}for j=1,#k
do v=m[i%#m+1]l=j%#k+1w=m[(i-2)%#m+1]h=(j-2)%#k+1
a[i][j]=v[l]+k[l]+w[l]+v[h]+v[j]+w[h]+w[j]+k[h]end
end s=''for i=1,#m do k=m[i]for j=1,k do
x=a[i][j]k[j]=k[j]>0 and((x<2or x>3)and 0or 1)or (x==3 and 1or 0)end
s=s..t(k)end for i=1,#c do
if(s==t(c[i]))then return#c>i and-1or i-1
end end goto z end
print(f({{0,0,0},{0,1,1},{0,1,0}}))
print(f({{0,1,0,0},{0,1,0,0},{0,1,0,0},{0,0,0,0}}))
-- 53 generation, 15x15, takes 50-100 ms on a bad laptop
print(f({{0,0,0,0,1,1,0,1,0,0,0,0,1,0,0},
{0,1,1,0,1,1,1,1,1,1,1,0,0,0,0},
{0,1,1,1,0,1,0,1,0,0,0,0,1,0,0},
{0,0,1,0,1,1,1,0,0,1,1,1,0,1,1},
{1,1,0,0,1,1,1,0,1,1,0,0,0,1,0},
{0,0,0,0,1,1,0,1,0,0,0,0,1,0,0},
{0,1,1,0,1,1,1,1,1,1,1,0,0,0,0},
{0,1,1,1,0,1,0,1,0,0,0,0,1,0,0},
{0,0,1,0,1,1,1,0,0,1,1,1,0,1,1},
{1,1,0,0,1,1,1,0,1,1,0,0,0,1,0},
{0,0,1,0,1,1,1,0,0,1,1,1,0,1,1},
{1,1,0,0,1,1,1,0,1,1,0,0,0,1,0},
{0,0,0,0,1,1,0,1,0,0,0,0,1,0,0},
{0,0,0,0,1,1,0,1,0,0,0,0,1,0,0},
{0,1,1,0,1,1,1,1,1,1,1,0,0,0,0}}))
-- Glider on a 15x14 board
-- 840 distinct generation
-- loop afterward -> return -1
-- takes ~4-5 seconds on the same bad laptop
print(f({{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{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},
{0,0,0,0,0,1,0,0,0,0,0,0,0,0,0},
{0,0,0,1,1,1,0,0,0,0,0,0,0,0,0},
{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},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}}))