🚀 go-pugleaf

>I'm trying to solve the "picture puzzle" and it *seems* impossible...

There are configurations that are impossible. The key is to click down at
the bottom of the screen and start again with a new configuration until you
get a random starting point which is solvable. A friend of mine came up with
the following analysis:

An analysis of the 7th guest picture box puzzle:

Each segment in the puzzle is described in the hints as a "nine sided cube"
(whatever that is :-). A description which is better from the mathematical
standpoint is as a 3x3 matrix with each cell containing a number between
0 and 8 (inclusive). The desired state of this matrix is:

   0 1 2
   3 4 5
   6 7 8

The operations you are allowed to perform are incrementing or decrementing
any row or column modulo 9. (Actually decrementing is not required since the
same result can be achieved by incrementing by 8).

Another way to state this is as an initial (random) configuration consisting
of the cell values a thru i and an increment count for each row or column.

   C1 C2 C3
R1  a  b  c
R2  d  e  f
R3  g  h  i

We need to find the (integer) increment count values (C1 thru C3 and R1 thru
R3) that will satisfy the following equations:

(a + C1 + R1) mod 9 = 0
(b + C2 + R1) mod 9 = 1
(c + C3 + R1) mod 9 = 2
(d + C1 + R2) mod 9 = 3
(e + C2 + R2) mod 9 = 4
(f + C3 + R2) mod 9 = 5
(g + C1 + R3) mod 9 = 6
(h + C2 + R3) mod 9 = 7
(i + C3 + R3) mod 9 = 8

This is an overspecified set of equations (9 equations and only 6 unknowns).
If you pick six equations involving all six unknowns, such as:

(a + C1 + R1) mod 9 = 0
(b + C2 + R1) mod 9 = 1
(c + C3 + R1) mod 9 = 2
(d + C1 + R2) mod 9 = 3
(g + C1 + R3) mod 9 = 6
(h + C2 + R3) mod 9 = 7

and generate a working solution, then it is clearly possible to find a value
for one of the missing components (e, f, or i) which will *not* satisfy
the solution.

This means there *are* impossible states this puzzle can get into (although
I can't necessarily prove any given state is impossible).

This also means you can simply enumerate all possible values of C1..C3 and
R1..R3 and see if one works (if you have a computer to go through them all).
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