Code Golf: Lights out

Answer 1

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').

Answer 2

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} }
}

Answer 3

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

Answer 4

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""

Answer 5

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

Answer 6

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..