
Some interesting patterns ...

1. Look at "CORNERS.BRD" (choose option #5).  To solve this puzzle, just
   push all the buttons that are lit up at the beginning.

PROBLEM: Are there any other boards that satisfy this property?


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2. Look at "ALL.BRD" (choose option #5).  To solve this puzzle, push all the
   buttons (A through Y).  There is a quicker way, though.
   ( the quicker sequence has 9 moves )

   Another equivalent pair of solutions exists for "CORNERS.BRD".  To solve,
   as stated in problem #1, simply push all the buttons that are lit up
   at the beginning.  An alternative is to push the buttons that form
   a "*" except M (total 16 moves).  Significantly, the latter solution 
   has the same number of moves as the former.

PROBLEM: What other sequences are equivalent (aside from trivial sequences 
         where a single button is pressed an even number of times)?


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3. On first inspection, it would seem that the most complicated 1 colour
   board would be the one with all the buttons pushed (ALL.BRD).  However,
   in view of problem #2, we know that this board can be solved another
   way (in 9 moves).

   We can call this 9 move sequence an "inversion" of the board and the
   resulting board the "complement" of the original because its solution 
   consists of the buttons not included in the solution of the original.

   It follows from this that any board with 24 buttons pushed can be solved
   by inverting it (9 moves) and then executing a 10th move corresponding
   to the one button that was not pressed in the generation of the board.

   In this way, we can find (by induction) an upper bound on the minimum
   number of moves required to solve the most complicated board:

     - 25 moves to make ... max 9 to solve
     - 24 moves to make ... max 10 to solve
     ...
     - 17 moves to make ... max 17 to solve

   Obviously, if one of the 9 inversion buttons was not included in the 
   original board generation sequence, the number to solve can be reduced 
   by two by not pressing the button during inversion (and not pressing
   it afterwards either).

   In that case, we must conclude that the maximum complexity of a board
   is not more than 17 moves, and the any board of complexity 17 must
   contain all 9 inversion buttons in its generation.

PROBLEM: Can this maximum complexity be reduced?  By how much? (must prove)


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4. The above discussion was applied to boards that can be solved.  That is
   equivalent to boards that can be generated by legal moves.  It is
   believed that there are impossible-to-generate boards.  An example of
   this is the "A" board in which the only lit button is the "A".

   If this board is indeed impossible, then any board that can be generated
   starting with the "A" is also impossible.  Therefore, "BF" would also
   be impossible (pushing "A" would create "BF" from "A").
   
   Less obviously, "B" (and any symmetric board such as "D", "F" etc.) would 
   also be impossible.  This is because should "B" have a solution, then 
   so would "F" (by symmetry) and the "superposition" of these two solutions
   would solve "BF".  Pushing A would then create the "A" board which
   is believed to be impossible.

   To continue the list of impossible boards, consider "C".  Pushing A and
   B will lead to the position "FG".  But "G" is readily solved:

       B   D
     F G   I J
           N
     P Q R
       V

   leaving "F" which is impossible by hypothesis.

   To complete the list of impossible boards, consider "CH".  This is
   solved:

     A B   

     K L M 
       Q   S   
         W X Y 

   Executing this sequence will convert "C" to "H" and vice versa, so
   that if "C" is impossible, then so is "H".

   "M" can be solved:

     A B
         H
     K   M N
     P       T
       V W   Y

   To summarize:

   If "A" is impossible, then so are the "singletons":

     A B C D E
     F   H   J
     K L   N O
     P   R   T
     U V W X Y

   and the only soluble singletons are:

       G   I
         M
       Q   S    

   
   REMARK: The impossibility of "B" does not necessarily imply that 
           "A" is impossible (by the above considerations) although 
           "C" and "H" must be impossible like "B".  It is therefore
           conceivable that "A" has a solution but "B" etc. do not.

   It is useful to be able to convert one singleton into another.
   For example, the following sequence:

      B C D
        H

   converts "A" to "ME" and M can be removed as described earlier.
   Therefore, any corner singleton can be converted to any other.
   It also follows that any board consisting of 2 or 4 corner
   singletons can be solved.

   Can a similar conversion be done between "B" and say, "K".  The
   answer is yes.  Pushing A,F does so, with a G left over (and 
   can be removed).  We will call G the "remainder".

   And even more trivially, pushing C converts "B" to "D" with a 
   remainder "CH" (we've seen that solved already).

   Now, if there exists a conversion from "B" to "F", then we will
   have found a solution to "A" (which is the same as "BF").

   If not (and I suspect not), then the following singletons are 
   equivalent:

       B   D                                C    
                                        F   H   J
     K L   N O        and so are
                     (by rotation)      P   R   T
       V   X                                W

   Linking any light on the left to any light on the right will lead
   (immediately) to a solution of "A".


PROBLEMS: a. Solve "A" or prove that it cannot be solved.

          b. Solve "B" or prove that it cannot be solved 
             (for example by proving that "A" cannot be solved).


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