I want to make a function that finds numbers divisible by x and y in range w to z. For example this one works:
T = []
for i in range(1500,2701):
if (i % 7 == 0) and (i % 5 == 0):
T.append(str(i))
print('\n'.join(T))
Question is can I make all from above in function where I can change range and divisible by:
x=input("range from")
y=input("range to")
w=input("divisible by")
z=input("divisible by")
def divisible(x,y,w,z):
for i in range(x,y):
if (i % w == 0) and (i % z == 0):
T.append(str(i))
return T
print('\n'.join(T))
def divisible(x,y,w,z):
T=[]
for i in range(x,y):
if (i % w == 0) and (i % z == 0):
T.append(str(i))
return ('\n'.join(T))
print(divisible(1500,2701,5,7))
These two attempts doesn't work, any idea why? I'm beginner and any help is more than welcome.
Indentation matters in Python. You're returning T right after you find the first divisible number. You have to return when the loop ends. Change to:
def divisible(x,y,w,z):
for i in range(x,y):
if (i % w == 0) and (i % z == 0):
T.append(str(i))
return T
Now is working
x=int(input("range from"))
y=int(input("range to"))
w=int(input("divisible by"))
z=int(input("divisible by"))
def divisible(x,y,w,z):
T=[]
for i in range(x,y):
if (i % w == 0) and (i % z == 0):
T.append(str(i))
return T
print(divisible(x,y,w,z))
I've heard that any recursive algorithm can always be expressed by using a stack. Recently, I've been working on programs in an environment with a prohibitively small available call stack size.
I need to do some deep recursion, so I was wondering how you could rework any recursive algorithm to use an explicit stack.
For example, let's suppose I have a recursive function like this
function f(n, i) {
if n <= i return n
if n % i = 0 return f(n / i, i)
return f(n, i + 1)
}
how could I write it with a stack instead? Is there a simple process I can follow to convert any recursive function into a stack-based one?
If you understand how a function call affects the process stack, you can understand how to do it yourself.
When you call a function, some data are written on the stack including the arguments. The function reads these arguments, does whatever with them and places the result on the stack. You can do the exact same thing. Your example in particular doesn't need a stack so if I convert that to one that uses stack it may look a bit silly, so I'm going to give you the fibonacci example:
fib(n)
if n < 2 return n
return fib(n-1) + fib(n-2)
function fib(n, i)
stack.empty()
stack.push(<is_arg, n>)
while (!stack.size() > 2 || stack.top().is_arg)
<isarg, argn> = stack.pop()
if (isarg)
if (argn < 2)
stack.push(<is_result, argn>)
else
stack.push(<is_arg, argn-1>)
stack.push(<is_arg, argn-2>)
else
<isarg_prev, argn_prev> = stack.pop()
if (isarg_prev)
stack.push(<is_result, argn>)
stack.push(<is_arg, argn_prev>)
else
stack.push(<is_result, argn+argn_prev>)
return stack.top().argn
Explanation: every time you take an item from the stack, you need to check whether it needs to be expanded or not. If so, push appropriate arguments on the stack, if not, let it merge with previous results. In the case of fibonacci, once fib(n-2) is computed (and is available at top of stack), n-1 is retrieved (one after top of stack), result of fib(n-2) is pushed under it, and then fib(n-1) is expanded and computed. If the top two elements of the stack were both results, of course, you just add them and push to stack.
If you'd like to see how your own function would look like, here it is:
function f(n, i)
stack.empty()
stack.push(n)
stack.push(i)
while (!stack.is_empty())
argi = stack.pop()
argn = stack.pop()
if argn <= argi
result = argn
else if n % i = 0
stack.push(n / i)
stack.push(i)
else
stack.push(n)
stack.push(i + 1)
return result
You can convert your code to use a stack like follows:
stack.push(n)
stack.push(i)
while(stack.notEmpty)
i = stack.pop()
n = stack.pop()
if (n <= i) {
return n
} else if (n % i = 0) {
stack.push(n / i)
stack.push(i)
} else {
stack.push(n)
stack.push(i+1)
}
}
Note: I didn't test this, so it may contain errors, but it gives you the idea.
Your particular example is tail-recursive, so with a properly optimising compiler, it should not consume any stack depth at all, as it is equivalent to a simple loop. To be clear: this example does not require a stack at all.
Both your example and the fibonacci function can be rewritten iteratively without using stack.
Here's an example where the stack is required, Ackermann function:
def ack(m, n):
assert m >= 0 and n >= 0
if m == 0: return n + 1
if n == 0: return ack(m - 1, 1)
return ack(m - 1, ack(m, n - 1))
Eliminating recursion:
def ack_iter(m, n):
stack = []
push = stack.append
pop = stack.pop
RETURN_VALUE, CALL_FUNCTION, NESTED = -1, -2, -3
push(m) # push function arguments
push(n)
push(CALL_FUNCTION) # push address
while stack: # not empty
address = pop()
if address is CALL_FUNCTION:
n = pop() # pop function arguments
m = pop()
if m == 0: # return n + 1
push(n+1) # push returned value
push(RETURN_VALUE)
elif n == 0: # return ack(m - 1, 1)
push(m-1)
push(1)
push(CALL_FUNCTION)
else: # begin: return ack(m - 1, ack(m, n - 1))
push(m-1) # save local value
push(NESTED) # save address to return
push(m)
push(n-1)
push(CALL_FUNCTION)
elif address is NESTED: # end: return ack(m - 1, ack(m, n - 1))
# old (m - 1) is already on the stack
push(value) # use returned value from the most recent call
push(CALL_FUNCTION)
elif address is RETURN_VALUE:
value = pop() # pop returned value
else:
assert 0, (address, stack)
return value
Note it is not necessary here to put CALL_FUNCTION, RETURN_VALUE labels and value on the stack.
Example
print(ack(2, 4)) # -> 11
print(ack_iter(2, 4))
assert all(ack(m, n) == ack_iter(m, n) for m in range(4) for n in range(6))
print(ack_iter(3, 4)) # -> 125
Locked. This question and its answers are locked because the question is off-topic but has historical significance. It is not currently accepting new answers or interactions.
The challenge
The shortest code by character count to solve the input lights out board.
The lights out board is a 2d square grid of varying size composed of two characters - . for a light that is off and * for a light that is on.
To solve the board, all "lights" have to be turned off. Toggling a light (i.e. turning off when it is on, turning on when it is off) is made 5 lights at a time - the light selected and the lights surround it in a + (plus) shape.
"Selecting" the middle light will solve the board:
.*.
***
.*.
Since Lights Out! solution order does not matter, the output will be a new board with markings on what bulbs to select. The above board's solution is
...
.X.
...
Turning off a light in a corner where there are no side bulbs to turn off will not overflow:
...
..*
.**
Selecting the lower-right bulb will only turn off 3 bulbs in this case.
Test cases
Input:
**.**
*.*.*
.***.
*.*.*
**.**
Output:
X...X
.....
..X..
.....
X...X
Input:
.*.*.
**.**
.*.*.
*.*.*
*.*.*
Output:
.....
.X.X.
.....
.....
X.X.X
Input:
*...*
**.**
..*..
*.*..
*.**.
Output:
X.X.X
..X..
.....
.....
X.X..
Code count includes input/output (i.e full program).
Perl, 172 characters
Perl, 333 251 203 197 190 172 characters. In this version, we randomly push buttons until all of the lights are out.
map{$N++;$E+=/\*/*1<<$t++for/./g}<>;
$C^=$b=1<<($%=rand$t),
$E^=$b|$b>>$N|($%<$t-$N)*$b<<$N|($%%$N&&$b/2)|(++$%%$N&&$b*2)while$E;
die map{('.',X)[1&$C>>$_-1],$_%$N?"":$/}1..$t
Haskell, 263 characters (277 and 285 before edit) (according to wc)
import List
o x=zipWith4(\a b c i->foldr1(/=)[a,b,c,i])x(f:x)$tail x++[f]
f=0>0
d t=mapM(\_->[f,1>0])t>>=c t
c(l:m:n)x=map(x:)$c(zipWith(/=)m x:n)$o x l
c[k]x=[a|a<-[[x]],not$or$o x k]
main=interact$unlines.u((['.','X']!!).fromEnum).head.d.u(<'.').lines
u=map.map
This includes IO code : you can simply compile it and it works.
This method use the fact that once the first line of the solution is determined, it is easy to determine what the other lines should look like. So we try every solution for the first line, and verify that the all lights are off on the last line, and this algorithm is O(n²*2^n)
Edit : here is an un-shrunk version :
import Data.List
-- xor on a list. /= works like binary xor, so we just need a fold
xor = foldr (/=) False
-- what will be changed on a line when we push the buttons :
changeLine orig chg = zipWith4 (\a b c d -> xor [a,b,c,d]) chg (False:chg) (tail chg ++ [False]) orig
-- change a line according to the buttons pushed one line higher :
changeLine2 orig chg = zipWith (/=) orig chg
-- compute a solution given a first line.
-- if no solution is given, return []
solution (l1:l2:ls) chg = map (chg:) $ solution (changeLine2 l2 chg:ls) (changeLine l1 chg)
solution [l] chg = if or (changeLine l chg) then [] else [[chg]]
firstLines n = mapM (const [False,True]) [1..n]
-- original uses something equivalent to "firstLines (length gris)", which only
-- works on square grids.
solutions grid = firstLines (length $ head grid) >>= solution grid
main = interact $ unlines . disp . head . solutions . parse . lines
where parse = map (map (\c ->
case c of
'.' -> False
'*' -> True))
disp = map (map (\b -> if b then 'X' else '.'))
Ruby, 225 221
b=$<.read.split
d=b.size
n=b.join.tr'.*','01'
f=2**d**2
h=0
d.times{h=h<<d|2**d-1&~1}
f.times{|a|e=(n.to_i(2)^a^a<<d^a>>d^(a&h)>>1^a<<1&h)&f-1
e==0&&(i=("%0*b"%[d*d,a]).tr('01','.X')
d.times{puts i[0,d]
i=i[d..-1]}
exit)}
F#, 672 646 643 634 629 628 chars (incl newlines)
EDIT: priceless: this post triggered Stackoverflow's human verification system. I bet it's because of the code.
EDIT2: more filthy tricks knocked off 36 chars. Reversing an if in the second line shaved off 5 more.
Writing this code made my eyes bleed and my brain melt.
The good: it's short(ish).
The bad: it'll crash on any input square larger than 4x4 (it's an O(be stupid and try everything) algorithm, O(n*2^(n^2)) to be more precise). Much of the ugliness comes from padding the input square with zeroes on all sides to avoid edge and corner cases.
The ugly: just look at it. It's code only a parent could love. Liberal uses of >>> and <<< made F# look like brainfuck.
The program accepts rows of input until you enter a blank line.
This code doesn't work in F# interactive. It has to be compiled inside a project.
open System
let rec i()=[let l=Console.ReadLine()in if l<>""then yield!l::i()]
let a=i()
let m=a.[0].Length
let M=m+2
let q=Seq.sum[for k in 1..m->(1L<<<m)-1L<<<k*M+1]
let B=Seq.sum(Seq.mapi(fun i s->Convert.ToInt64(String.collect(function|'.'->"0"|_->"1")s,2)<<<M*i+M+1)a)
let rec f B x=function 0L->B&&&q|n->f(if n%2L=1L then B^^^(x*7L/2L+(x<<<M)+(x>>>M))else B)(x*2L)(n/2L)
let z=fst<|Seq.find(snd>>(=)0L)[for k in 0L..1L<<<m*m->let n=Seq.sum[for j in 0..m->k+1L&&&(((1L<<<m)-1L)<<<j*m)<<<M+1+2*j]in n,f B 1L n]
for i=0 to m-1 do
for j=0 to m-1 do printf"%s"(if z&&&(1L<<<m-j+M*i+M)=0L then "." else "X")
printfn""
F#, 23 lines
Uses brute force and a liberal amount of bitmasking to find a solution:
open System.Collections
let solve(r:string) =
let s = r.Replace("\n", "")
let size = s.Length|>float|>sqrt|>int
let buttons =
[| for i in 0 .. (size*size)-1 do
let x = new BitArray(size*size)
{ 0 .. (size*size)-1 } |> Seq.iter (fun j ->
let toXY n = n / size, n % size
let (ir, ic), (jr, jc) = toXY i, toXY j
x.[j] <- ir=jr&&abs(ic-jc)<2||ic=jc&&abs(ir-jr)<2)
yield x |]
let testPerm permutation =
let b = new BitArray(s.Length)
s |> Seq.iteri (fun i x -> if x = '*' then b.[i] <- true)
permutation |> Seq.iteri (fun i x -> if x = '1' then b.Xor(buttons.[i]);() )
b |> Seq.cast |> Seq.forall (fun x -> not x)
{for a in 0 .. (1 <<< (size * size)) - 1 -> System.Convert.ToString(a, 2).PadLeft(size * size, '0') }
|> Seq.pick (fun p -> if testPerm p then Some p else None)
|> Seq.iteri (fun i s -> printf "%s%s" (if s = '1' then "X" else ".") (if (i + 1) % size = 0 then "\n" else "") )
Usage:
> solve ".*.
***
.*.";;
...
.X.
...
val it : unit = ()
> solve "**.**
*.*.*
.***.
*.*.*
**.**";;
..X..
X.X.X
..X..
X.X.X
..X..
val it : unit = ()
> solve "*...*
**.**
..*..
*.*..
*.**.";;
.....
X...X
.....
X.X.X
....X
C89, 436 characters
Original source (75 lines, 1074 characters):
#include <stdio.h>
#include <string.h>
int board[9][9];
int zeroes[9];
char presses[99];
int size;
int i;
#define TOGGLE { \
board[i][j] ^= 4; \
if(i > 0) \
board[i-1][j] ^= 4; \
if(j > 0) \
board[i][j-1] ^= 4; \
board[i+1][j] ^= 4; \
board[i][j+1] ^= 4; \
presses[i*size + i + j] ^= 118; /* '.' xor 'X' */ \
}
void search(int j)
{
int i = 0;
if(j == size)
{
for(i = 1; i < size; i++)
{
for(j = 0; j < size; j++)
{
if(board[i-1][j])
TOGGLE
}
}
if(memcmp(board[size - 1], zeroes, size * sizeof(int)) == 0)
puts(presses);
for(i = 1; i < size; i++)
{
for(j = 0; j < size; j++)
{
if(presses[i*size + i + j] & 16)
TOGGLE
}
}
}
else
{
search(j+1);
TOGGLE
search(j+1);
TOGGLE
}
}
int main(int c, char **v)
{
while((c = getchar()) != EOF)
{
if(c == '\n')
{
size++;
i = 0;
}
else
board[size][i++] = ~c & 4; // '.' ==> 0, '*' ==> 4
}
memset(presses, '.', 99);
for(c = 1; c <= size; c++)
presses[c * size + c - 1] = '\n';
presses[size * size + size] = '\0';
search(0);
}
Compressed source, with line breaks added for your sanity:
#define T{b[i][j]^=4;if(i)b[i-1][j]^=4;if(j)b[i][j-1]^=4;b[i+1][j]^=4;b[i][j+1]^=4;p[i*s+i+j]^=118;}
b[9][9],z[9],s,i;char p[99];
S(j){int i=0;if(j-s){S(j+1);T S(j+1);T}else{
for(i=1;i<s;i++)for(j=0;j<s;j++)if(b[i-1][j])T
if(!memcmp(b[s-1],z,s*4))puts(p);
for(i=1;i<s;i++)for(j=0;j<s;j++)if(p[i*s+i+j]&16)T}}
main(c){while((c=getchar())+1)if(c-10)b[s][i++]=~c&4;else s++,i=0;
memset(p,46,99);for(c=1;c<=s;c++)p[c*s+c-1]=10;p[s*s+s]=0;S(0);}
Note that this solution assumes 4-byte integers; if integers are not 4 bytes on your system, replace the 4 in the call to memcmp with your integer size. The maximum sized grid this supports is 8x8 (not 9x9, since the bit flipping ignores two of the edge cases); to support up to 98x98, add another 9 to the array sizes in the declarations of b, z and p and the call to memset.
Also note that this finds and prints ALL solutions, not just the first solution. Runtime is O(2^N * N^2), where N is the size of the grid. The input format must be perfectly valid, as no error checking is performed -- it must consist of only ., *, and '\n', and it must have exactly N lines (i.e. the last character must be a '\n').
Ruby:
class Array
def solve
carry
(0...(2**w)).each {|i|
flip i
return self if solved?
flip i
}
end
def flip(i)
(0...w).each {|n|
press n, 0 if i & (1 << n) != 0
}
carry
end
def solved?
(0...h).each {|y|
(0...w).each {|x|
return false if self[y][x]
}
}
true
end
def carry
(0...h-1).each {|y|
(0...w).each {|x|
press x, y+1 if self[y][x]
}
}
end
def h() size end
def w() self[0].size end
def press x, y
#presses = (0...h).map { [false] * w } if #presses == nil
#presses[y][x] = !#presses[y][x]
inv x, y
if y>0 then inv x, y-1 end
if y<h-1 then inv x, y+1 end
if x>0 then inv x-1, y end
if x<w-1 then inv x+1, y end
end
def inv x, y
self[y][x] = !self[y][x]
end
def presses
(0...h).each {|y|
puts (0...w).map {|x|
if #presses[y][x] then 'X' else '.' end
}.inject {|a,b| a+b}
}
end
end
STDIN.read.split(/\n/).map{|x|x.split(//).map {|v|v == '*'}}.solve.presses
Lua, 499 characters
Fast, uses Strategy to find a quicker solution.
m={46,[42]=88,[46]=1,[88]=42}o={88,[42]=46,[46]=42,[88]=1}z={1,[42]=1}r=io.read
l=r()s=#l q={l:byte(1,s)}
for i=2,s do q[#q+1]=10 l=r()for j=1,#l do q[#q+1]=l:byte(j)end end
function t(p,v)q[p]=v[q[p]]or q[p]end
function u(p)t(p,m)t(p-1,o)t(p+1,o)t(p-s-1,o)t(p+s+1,o)end
while 1 do e=1 for i=1,(s+1)*s do
if i>(s+1)*(s-1)then if z[q[i]]then e=_ end
elseif z[q[i]]then u(i+s+1)end end
if e then break end
for i=1,s do if 42==q[i]or 46==q[i]then u(i)break end u(i)end end
print(string.char(unpack(q)))
Example input:
.....
.....
.....
.....
*...*
Example output:
XX...
..X..
X.XX.
X.X.X
...XX
Some of these have multiple answers. This seems to work but it's not exactly fast.
Groovy: 790 chracters
bd = System.in.readLines().collect{it.collect { it=='*'}}
sl = bd.collect{it.collect{false}}
println "\n\n\n"
solve(bd, sl, 0, 0, 0)
def solve(board, solution, int i, int j, prefix) {
/* println " ".multiply(prefix) + "$i $j"*/
if(done(board)) {
println sl.collect{it.collect{it?'X':'.'}.join("")}.join("\n")
return
}
if(j>=board[i].size) {
j=0; i++
}
if(i==board.size) {
return
}
solve(board, solution, i, j+1, prefix+1)
flip(solution, i, j)
flip(board, i, j)
flip(board, i+1, j)
flip(board, i-1, j)
flip(board, i, j+1)
flip(board, i, j-1)
solve(board, solution, i, j+1, prefix+1)
}
def flip(board, i, j) {
if(i>=0 && i<board.size && j>=0 && j<board[i].size)
board[i][j] = !board[i][j]
}
def done(board) {
return board.every { it.every{!it} }
}
For Haskell, here's a 406 376 342 character solution, though I'm sure there's a way to shrink this. Call the s function for the first solution found:
s b=head$t(b,[])
l=length
t(b,m)=if l u>0 then map snd u else concat$map t c where{i=[0..l b-1];c=[(a b p,m++[p])|p<-[(x,y)|x<-i,y<-i]];u=filter((all(==False)).fst)c}
a b(x,y)=foldl o b[(x,y),(x-1,y),(x+1,y),(x,y-1),(x,y+1)]
o b(x,y)=if x<0||y<0||x>=r||y>=r then b else take i b++[not(b!!i)]++drop(i+1)b where{r=floor$sqrt$fromIntegral$l b;i=y*r+x}
In its more-readable, typed form:
solution :: [Bool] -> [(Int,Int)]
solution board = head $ solutions (board, [])
solutions :: ([Bool],[(Int,Int)]) -> [[(Int,Int)]]
solutions (board,moves) =
if length solutions' > 0
then map snd solutions'
else concat $ map solutions candidates
where
boardIndices = [0..length board - 1]
candidates = [
(applyMove board pair, moves ++ [pair])
| pair <- [(x,y) | x <- boardIndices, y <- boardIndices]]
solutions' = filter ((all (==False)) . fst) candidates
applyMove :: [Bool] -> (Int,Int) -> [Bool]
applyMove board (x,y) =
foldl toggle board [(x,y), (x-1,y), (x+1,y), (x,y-1), (x,y+1)]
toggle :: [Bool] -> (Int,Int) -> [Bool]
toggle board (x,y) =
if x < 0 || y < 0 || x >= boardSize || y >= boardSize then board
else
take index board ++ [ not (board !! index) ]
++ drop (index + 1) board
where
boardSize = floor $ sqrt $ fromIntegral $ length board
index = y * boardSize + x
Note that this is a horrible breadth-first, brute-force algorithm.
F#, 365 370, 374, 444 including all whitespace
open System
let s(r:string)=
let d=r.IndexOf"\n"
let e,m,p=d+1,r.ToCharArray(),Random()
let o b k=m.[k]<-char(b^^^int m.[k])
while String(m).IndexOfAny([|'*';'\\'|])>=0 do
let x,y=p.Next d,p.Next d
o 118(x+y*e)
for i in x-1..x+1 do for n in y-1..y+1 do if i>=0&&i<d&&n>=0&&n<d then o 4(i+n*e)
printf"%s"(String m)
Here's the original readable version before the xor optimization. 1108
open System
let solve (input : string) =
let height = input.IndexOf("\n")
let width = height + 1
let board = input.ToCharArray()
let rnd = Random()
let mark = function
| '*' -> 'O'
| '.' -> 'X'
| 'O' -> '*'
| _ -> '.'
let flip x y =
let flip = function
| '*' -> '.'
| '.' -> '*'
| 'X' -> 'O'
| _ -> 'X'
if x >= 0 && x < height && y >= 0 && y < height then
board.[x + y * width] <- flip board.[x + y * width]
let solved() =
String(board).IndexOfAny([|'*';'O'|]) < 0
while not (solved()) do
let x = rnd.Next(height) // ignore newline
let y = rnd.Next(height)
board.[x + y * width] <- mark board.[x + y * width]
for i in -1..1 do
for n in -1..1 do
flip (x + i) (y + n)
printf "%s" (String(board))
Python — 982
Count is 982 not counting tabs and newlines. This includes necessary spaces. Started learning python this week, so I had some fun :). Pretty straight forward, nothing fancy here, besides the crappy var names to make it shorter.
import re
def m():
s=''
while 1:
y=raw_input()
if y=='':break
s=s+y+'\n'
t=a(s)
t.s()
t.p()
class a:
def __init__(x,y):
x.t=len(y);
r=re.compile('(.*)\n')
x.z=r.findall(y)
x.w=len(x.z[0])
x.v=len(x.z)
def s(x):
n=0
for i in range(0,x.t):
if(x.x(i,0)):
break
def x(x,d,c):
b=x.z[:]
for i in range(1,x.v+1):
for j in range(1,x.w+1):
if x.c():
break;
x.z=b[:]
x.u(i,j)
if d!=c:
x.x(d,c+1)
if x.c():
break;
if x.c():
return 1
x.z=b[:]
return 0;
def y(x,r,c):
e=x.z[r-1][c-1]
if e=='*':
return '.'
elif e=='x':
return 'X'
elif e=='X':
return 'x'
else:
return '*'
def j(x,r,c):
v=x.y(r+1,c)
x.r(r+1,c,v)
def k(x,r,c):
v=x.y(r-1,c)
x.r(r-1,c,v)
def h(x,r,c):
v=x.y(r,c-1)
x.r(r,c-1,v)
def l(x,r,c):
v=x.y(r,c+1)
x.r(r,c+1,v)
def u(x,r,c):
e=x.z[r-1][c-1]
if e=='*' or e=='x':
v='X'
else:
v='x'
x.r(r,c,v)
if r!=1:
x.k(r,c)
if r!=x.v:
x.j(r,c)
if c!=1:
x.h(r,c)
if c!=x.w:
x.l(r,c)
def r(x,r,c,l):
m=x.z[r-1]
m=m[:c-1]+l+m[c:]
x.z[r-1]=m
def c(x):
for i in x.z:
for j in i:
if j=='*' or j=='x':
return 0
return 1
def p(x):
for i in x.z:
print i
print '\n'
if __name__=='__main__':
m()
Usage:
*...*
**.**
..*..
*.*..
*.**.
X.X.X
..X..
.....
.....
X.X..