SML Sieve of Eratosthenes - function

I am new to SML. I am trying to create a function that outputs a list of all prime numbers which are smaller than or equal to a given integer n > 2 using the Sieve of Eratosthenes. I am running into a problem however where the output is only showing as [1]. I would like to be able to have an input such as 5 and get [1,3,5] as a result.
This is my code so far, I am very new so I know it is most likely not written correctly.
fun createList(ending) =
let
fun createListX(start, ending) =
if start = ending then []
else start :: createListX(start + 1, ending)
in
createListX(1, ending + 1)
end;
fun removeMult ([], n) = []
| removeMult (x::xs, n) =
if x mod n = 0 then
removeMult(xs, n)
else
x :: removeMult(xs, n);
fun sieve([], primes) = primes
| sieve(n::ns, primes) = sieve(removeMult(ns, n), n :: primes);
fun dosieve(n) = sieve(createList(n-1), []);


Your removeMult function works nicely.
Your sieve function works perfectly too. Too perfectly.
Consider what happens when you call dosieve(10) for instance:
dosieve(10)
sieve(createList(9), [])
sieve([1,2,3,4,5,6,7,8,9], [])
From there:
sieve(removeMult([2, 3, 4, 5, 6, 7, 8, 9], 1), 1 :: [])
sieve([], [1])
[1]
Oops. You removed all multiples of 1, but of course they're all multiples of 1.
Perhaps something like:
fun sieve([], primes) = primes
| sieve(1::ns, primes) = sieve(ns, 1 :: primes)
| sieve(n::ns, primes) = sieve(removeMult(ns, n), n :: primes);

Related

Ocaml function, what is the nature of the problem?

Currently i'm trying
I have a function to calculate the inverse sum of a number
let inverseSum n =
let rec sI n acc =
match n with
| 1 -> acc
| _ -> sI (n - 1) ((1.0 /. float n) +. acc)
in sI n 1.0;;
For example, inverseSum 2 -> 1/2 + 1 = 3/2 = 1.5
I try to test the function with 2 and 5, it's okay:
inverseSum 2;;
inverseSum 5;;
inverseSum 2;;
- : float = 1.5
inverseSum 5;;
- : float = 2.28333333333333321
For the moment, no problem.
After that, I initialize a list which contains all numbers between 1 and 10000 ([1;…;10000])
let initList = List.init 10000 (fun n -> n + 1);;
no problem.
I code a function so that an element of the list becomes the inverse sum of the element
(e.g. [1;2;3] -> [inverseSum 1; inverseSum 2; inverseSum 3])
let rec invSumLst lst =
match lst with
| [] -> []
| h::t -> (inverseSum h) :: invSumLst t;;
and I use it on the list initList:
let invInit = invSumLst initList;;
So far so good, but I start to have doubts from this stage:
I select the elements of invList strictly inferior to 5.0
let listLess5 = List.filter (fun n -> n < 5.0) invInit;;
And I realize the sum of these elements using fold_left:
let foldLess5 = List.fold_left (+.) 0.0 listLess5;;
I redo the last two steps with floats greater than or equal to 5.0
let moreEg5 = List.filter (fun n -> n >= 5.0) invInit;;
let foldMore5 = List.fold_left (+.) 0.0 moreEg5;;
Finally, I sum all the numbers of the list:
let foldInvInit = List.fold_left (+.) 0.0 invInit;;
but at the end when I try to calculate the absolute error between the numbers less than 5, those greater than 5 and all the elements of the list, the result is surprising:
Float.abs ((foldLess5 +. foldMore5) -. foldInvInit);;
Printf.printf "%f\n" (Float.abs ((foldLess5 +. foldMore5) -. foldInvInit));;
Printf.printf "%b\n" ((foldLess5+.foldMore5) = foldInvInit);;
returns:
let foldMore5 = List.fold_left (+.) 0.0 moreEg5;;
val foldMore5 : float = 87553.6762998474733
let foldInvInit = List.fold_left (+.) 0.0 invInit;;
val foldInvInit : float = 87885.8479664799379
Float.abs ((foldLess5 +. foldMore5) -. foldInvInit);;
- : float = 1.45519152283668518e-11
Printf.printf "%f\n" (Float.abs ((foldLess5 +. foldMore5) -. foldInvInit));;
0.000000
- : unit = ()
Printf.printf "%b\n" ((foldLess5+.foldMore5) = foldInvInit);;
false
- : unit = ()
it's probably a rounding problem, but I would like to know where exactly the error comes from?
Because here I am using an interpreter, so I see the error "1.45519152283668518e-11"
But if I used a compiler like ocamlpro, I would just get 0.000000 and false on the terminal and I wouldn't understand anything.
So I would just like to know if the problem comes from one of the functions of the code, or from a rounding made by the Printf.printf function which wrote the result with a "non scientific" notation.

OCaml is showing you the actual results of the operations you performed. The difference between the two sums is caused by the finite precision of floating values.
When adding up a list of large-ish numbers, by the time you reach the end of the list the number is large enough that the lowest-order bits of the new values simply can't be represented. But when adding a list of small-ish numbers, fewer bits are lost.
A system that shows foldLess5 +. foldMore5 as equal to foldInvInit is most likely lying to you for your convenience.

Recursive Haskell function but only does part of it once

I've got a function that needs to work out the minimum change required to break down a certain amount of money.
A user would enter a value of money and the possible denominations, and it would output how many of each denomination would be needed to make up the money eg.
> coinChange 34 [1, 5, 10, 25, 50, 100]
[4,1,0,1,0,0]
Here is what I have so far:
coinChange :: Integer -> [Integer] -> [Integer]
coinChange _ [] = []
coinChange 0 xs = []
coinChange v xs = do
dropWhile (>v) (reverse xs)
createNewList :: Integer -> [Integer]
createNewList 0 = []
createNewList x = 0 : createNewList (x - 1)
I have a separate function called createNewList which I want to call to set up a new list of 0s at the beginning.
However, because I plan on making coinChange a recursive function when working out how much money is left, if I just call makeNewList in coinChange, at the moment it would reset the list because it would be called every recursion.
So my question is, is there a way to make a function call another function only once, before proceeding with the recursion.
Hope I've made it clear, thanks

In general you are allowed to create local helper functions in Haskell. Maybe this example will inspire you:
prepareData x = if x < 0 then -x else x
mainFunction n =
let recursiveLocalFunction x =
if x < 2 then 1 else x * recursiveLocalFunction (x - 1)
in recursiveLocalFunction (prepareData n)

OCaml and creating a list of lists

I am currently trying to make use functions to create:
0 V12 V13 V14
V21 0 V23 V24
V31 V32 0 V34
V41 V42 V43 0
A way that I found to do this was to use theses equations:
(2*V1 - 1)*(2*V2-1) = for spot V(1,2) in the Matrix
(2*V1 - 1)*(2*V3-1) = for spot V(1,3) in the Matrix
etc
Thus far I have:
let singleState state =
if state = 0.0 then 0.0
else
((2.0 *. state) -. 1.0);;
let rec matrixState v =
match v with
| [] -> []
| hd :: [] -> v
| hd :: (nx :: _ as tl) ->
singleState hd *. singleState nx :: matrixState tl;;
My results come out to be:
float list = [-3.; -3.; -3.; -1.]
When they should be a list of lists that look as follows:
0 -1 1 -1
-1 0 -1 1
1 -1 0 -1
-1 1 -1 0
So instead of it making list of lists it is making just one list. I also have trouble figuring out how to make the diagonals 0.
The signatures should look like:
val singleState : float list -> float list list = <fun>
val matrixState : float list list -> float list list = <fun>
and I am getting
val singleState : float -> float = <fun>
val matrixState : float list -> float list = <fun>
Any ideas?

With some fixing up, your function would make one row of the result. Then you could call it once for each row you need. A good way to do the repeated calling might be with List.map.
Assuming this is mostly a learning exercise, it might be good to first make a matrix like this:
V11 V12 V13 V14
V21 V22 V23 V24
V31 V32 V33 V34
V41 V42 V43 V44
I think this will be a lot easier to calculate.
Then you can replace the diagonal with zeroes. Here's some code that would replace the diagonal:
let replnth r n l =
List.mapi (fun i x -> if i = n then r else x) l
let zerorow row (n, res) =
(n - 1, replnth 0.0 n row :: res)
let zerodiag m =
let (_, res) = List.fold_right zerorow m (List.length m - 1, []) in
res

I would prefer to go with an array for your work.
A nice function to use is then Array.init, it works like so,
# Array.init 5 (fun x -> x);;
- : int array = [|0; 1; 2; 3; 4|]
We note that 5 play the role of the size of our Array.
But as you want a matrix we need to build an Array of Array which is achieve with two call of Array.init, the last one nested into the first one,
# Array.init 3 (fun row -> Array.init 3 (fun col -> row+col));;
- : int array array = [|[|0; 1; 2|]; [|1; 2; 3|]; [|2; 3; 4|]|]
Note, I've called my variable row and col to denote the fact that they correspond to the row index and column index of our matrix.
Last, as your formula use a vector of reference V holding value [|V1;V2;V3;V4|], we need to create one and incorporate call to it into our matrix builder, (The value hold on the cell n of an array tab is accessed like so tab.(n-1))
Which finally lead us to the working example,
let vect = [|1;2;3;4|]
let built_matrix =
Array.init 4 (fun row ->
Array.init 4 (fun col ->
if col=row then 0
else vect.(row)+vect.(col)))
Of course you'll have to adapt it to your convenience in order to match this piece of code according to your requirement.
A side note about syntax,
Repeating Array each time can be avoid using some nice feature of OCaml.
We can locally open a module like so,
let built_matrix =
let open Array in
init 4 (fun row ->
init 4 (fun col ->
if col=row then 0
else vect.(row)+vect.(col)))
Even shorter, let open Array in ... can be write as Array.(...), Below a chunk of code interpreted under the excellent utop to illustrate it (and I going to profit of this opportunity to incorporate a conversion of our matrix to a list of list.)
utop #
Array.(
to_list
## map to_list
## init 4 (fun r ->
init 4 (fun c ->
if r = c then 0
else vect.(r)+ vect.(c))))
;;
- : int list list = [[0; 3; 4; 5]; [3; 0; 5; 6]; [4; 5; 0; 7]; [5; 6; 7; 0]]
I hope it helps

OCAML: Difference between let func a x = (a x);; and let func a x = ((fun a -> a) x);;?

I'm a little stuck on this assignment for OCAML. I'm trying to pass in a function and a value as a parameter into another function. For example, I have function called test that takes in (fun x -> x+x) and 3 as parameters. The function test should output 6, since 3 + 3 = 6. I know I can achieve something similar this by completing:
let func x = x + x;;
let a = func;;
let test a x = (a x);;
This way when I input, test a 5, I will get 10.
But when I change the statement to this, I get only the value I placed in for x. How do I get the (fun a -> a) to take in the value x?
let test a x = ((fun a -> a) x);;

fun a -> a is an anonymous identity function, it will always return its parameter. You could say:
let id = fun a -> a;;
id 3;;
=> 3
id (fun x -> x + x);;
=> (fun x -> x + x)
Note that in your
let test a x = ((fun a -> a) x);;
the first a and the other two as are different variables. The first one is never used again later. You can rewrite this line for easier understanding as:
let test a x = ((fun b -> b) x);;
How do I get the (fun a -> a) to take in the value x?
You are, and that's the problem. You're feeding your x to an identity function, and getting x back. What you want to do is feed the x to your a function, and that's what you did in your first attempt:
let func x = x + x;;
let test a x = a x;;
test func 3;;
=> 6

With functional languages in simple cases it is often helpful to write the expression in a form similar to lambda calculus and do the reductions manually (noting which reductions you are using). You can still use OCaml syntax as a simplified version of lambda calculus
So in the case of your example this would become:
let test a x = ((fun a -> a ) x);;
=> let test a x = ((fun b -> b ) x);; (* Variable renamed (alpha equivalence) *)
=> let test a y = ((fun b -> b ) y);; (* Variable renamed (alpha equivalence) *)
let func x = x + x;;
Note that these steps only serve to make sure, that we will later on have no variables with the same name, referring to different values. These steps can be left out, but I personally like working with unique variables much better.
test func 5
=> test (fun x -> x + x) 5 (* Variable func inserted (beta reduction) *)
=> (fun a y -> ((fun b -> b) y) (fun x -> x + x) 5 (* Variable test inserted *)
=> (fun y -> (fun b -> b) y) 5 (* Variable a inserted *)
=> ((fun b -> b) 5 (* Variable y inserted *)
=> 5 (* Variable b inserted *)
The final result is 5. Attempting this at first will seem very unusual and hard, but get's easier very fast. If you do something like this a couple of time you will get much better at understanding common functional patterns and reasoning about your program structure.
Have a look at this article for more examples on this.
Also note, that with a little more effort, this works backward as well. Although this usually is not as helpful as doing it in the same direction as the compiler.

Code Golf: Lights out

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..