Doll room puzzle "I'm flippnig out" can be impossible at times.

[RavenMac]

I was doing a play through since I hadn't played it through steam yet and got to the picture flipping puzzle. You press a button either above or below each column, or left or right each row to change the cells or each column or row corresponding to the button pressed and try to get to the completed picture.

Without much thought, I got fairly close and was off by 2 cells of the picture on my first try. I decided to try again and got only one cell off. Try again and one cell again. I tried a couple things, but nothing that I could do would increment one cell by just one. So I started to look online.

I found a couple walkthroughs and all of them said it was randomly generated, but one also pointed out that it could start out as impossible if only one cell isn't matching and to reset the puzzle. So I spend some more time at it, but couldn't finish it; I tried for about 20 - 30 minutes.

I started looking online again and through some site found out that there is some sort of bug with the game running on ScummVM that makes the random start impossible. I checked in the steamapps folder to see if this was the case and it was.

I tried a couple things to try to get it not impossible. I saved and reloaded, restart and reload, leave a room and come back but none of these worked. I eventually closed the game and reloaded the game in the doll room and finally got a solvable puzzle. After over an hour I got this finished.

I remember playing the game on dosbox before and never ran into any problems (except the microscope puzzle, but that's the same with everyone). I don't know where to submit a bug report, so I'm leaving my little rant here. So if anyone else has the problem, they may be able to get it done.

[Leost]

Super necro post, but just to confirm, yes, sometimes the starting position is impossible. I wasn't sure so I did some of the math on it as follows:

If we consider each position of the game board as being in a particular row or column of a matrix we can solve this with relatively simple matrix math. To make things simple I solved a particular board as best I could and my solution was complete except the upper right square had the image seen in the top middle. For this I considered the images in the correct order as being equal to the numbers 1-9. I then considered that each square could be given a value of 0 if it had the correct image or a value of the difference between zero and the image's value in the original 1-9 scheme if wrong. For instance in my case the entire board could be thought of as zero with the upper right square being negative 1 (since it should have the 3 image in it, but it actually had the 2 image, so it was 1 low from the target 0 end value).

Anyway, with that in place I considered the buttons as the coefficients of the matrix. The buttons along the top all increment the pictures in their column forward or +1 in value. The bottom row all decrements or -1 the cells in their column. The buttons alone the left hand edge all increment their row and the right hand edge all decrement them.

One last thing is that we have 12 desired solutions (how many time to hit each button) but only 9 known values which is the value of the picture, but we can fix that. We can just ignore the bottom row of buttons. If you think about what they do we can get the same result from the other 9 buttons. Since each button on the bottom just decrements one column by 1 in each position we can instead decrement all three rows using the right hand buttons one each and then increment 2 of the columns by one using 2 of the top buttons. The result is the same as if you hit just the one bottom row button you originally wanted. But for us this means we can get a nice square matrix.

Ok, in English that maybe doesn't make a ton of sense. I had a matrix equalion:

bx=y

y = [0,0,-1,0,0,0,0,0,0]

b= 
{{ 1 0 0 1 -1 0 0 0 0 -1 0 0}
 { 0 1 0 1 -1 0 0 0 0 0 -1 0}
 { 0 0 1 1 -1 0 0 0 0 0 0 -1}
 { 1 0 0 0 0 1 -1 0 0 -1 0 0}
 { 0 1 0 0 0 1 -1 0 0 0 -1 0}
 { 0 0 1 0 0 1 -1 0 0 0 0 -1}
 { 1 0 0 0 0 0 0 1 -1 -1 0 0}
 { 0 1 0 0 0 0 0 1 -1 0 -1 0}
 { 0 0 1 0 0 0 0 1 -1 0 0 -1}}

I can't imagine how the formatting of that is going to get mangled but... the info is there.

Anyway, that is easily plugged into an online solver and it spat out that there was no solution. You cannot flip a single cell. I assume you also can't flip exactly two . So every time you start this puzzle you have basically a 1/3 chance of getting a solvable game. That's not too bad, but it's a shame the game didn't simulate flipping rows and columns a bunch rather than what they seem to have done which is just to have randomly assigned each cell a value.