well first i have to apologize since this isn''t really gamedev/programming related, but i figured since there are so many smart people here asking won''t hurt I have a little problem with a riddle, which i don''t think is possible to solve at all, but i don''t know for sure. The riddle goes something like this: there are 4 switches, aligned in a square. goal is to bring them all in the same position, but you never see them and don''t know their initial positions (on or off). You only have have 3 buttons: A will change the state of one randomly chosen switch, B will change the state of two randomly chosen switches (with both being in the same row or column), and C will change the state of two randomly chosen switches that are in a diagonal line. Now the riddle asks for the shortest code (series of A, B, C buttons to be pressed) that is guaranteed to bring all switches in the same position, no matter what their initial position was. Now i''m pretty sure this isn''t solvable, since you never know anything about the current state of the switches and its almost completely random which switches do change when you hit a button so you will never reach a position that is guaranteed to produce the desired goal in the next step, so already the first step when trying to solve this thing (or proof it''s solvability) with induction fails. I also wrote some piece of code that just tries one ''code'' after the other..it''s running since two hours now, currently at a codelength of 19, and still no goal. Well, i''m not asking for a solution to this problem, i just want to know if it is possible to solve at all( and if so, why?), or if it''s unsolvable and i can stop wasting time with it Thanks Runicsoft -- latest attraction: obfuscated Brainfuck Interpreter in SML ( This post was made entirely from re-cycled electrons )

I think

CBCACBC

will do it: if you operate the buttons in the above sequence then I think one of the arrangements of lights will be all the same (not necessarily the final position, there''s no sequence that guarantees that). And I think this is shortest.

thanks for all the answers, but i still think it''s not possible to solve
as i said, i already wrote a program, which i had test all sequences of upto length 20 without success so far. (would be a bit hard to do it by hand.. )
the CBCACBC unfortunately doesn''t help, since it doesn''t put all switches in the same position, and even if it would put them in the same arrangement all the time (which it doesn''t always do according to my tests) it would be no better then any random initial arrangment
but thanks anyway

Runicsoft -- latest attraction: obfuscated Brainfuck Interpreter in SML
( This post was made entirely from re-cycled electrons )

> the CBCACBC unfortunately doesn''t help, since it doesn''t put
> all switches in the same position,

Hm, I think it does. Didn''t bother with a program just logic like so:

Initially either 4, 3 or 2 are the same
4: Already all the same, i.e. this sequnce does it after 0 buttons

2: then either:
* diagonal pairs the same, then all the same after C (done after 1 button)
* horizontally or vertically the same:
** after C have vertically or horizontally the same (swapped from above)
** after CB have either 4 the same (done after 2 buttons) or have diagonals the same, so CBC makes all 4 the same (done after 3 buttons)

3: after pressing CBC still odd number lit, i.e. still 3 the same. After CBCA even number the same so either 4 the same (done after 4 buttons), or 2 the same. If 2 then repeating CBC results in all the same after 5, 6 or 7 buttons.

If you object to it being solved after 0 then insert an extra button at the start, e.g.

ACBCACBC

Then it''s got to be the same after 1, 2, 3, 4, 5, 6, 7 or 8 button presses.

As noted earlier it''s only possible to solve this way. There''s no way you can produce a fixed sequence that always makes them all the same at the end: I think the ending arrangement is as random as the initial arrangement for a fixed sequence, so any sequence is as good or bad as any other at producing a correct pattern at the end.