In[]:=

UndirectedGraph[SPMultiwaySystem[{b[{{1,2,1},{1,None,1},{1,1,1}}]},10,"StatesGraph"]]

Out[]=

In[]:=

Length/@FindCycle[%8,{4,20},10]

Out[]=

{12,14,14,16,16,16,16,18,18,18}

Center objects seem to characterize the large loops:

Empty square always stays in its quadrant

In[]:=

UndirectedGraph[SPMultiwaySystem[{b[{{1,2},{1,None}}]},10,"StatesGraph"]]

Out[]=

In[]:=

UndirectedGraph[SPMultiwaySystem[{b[{{1,2},{3,None}}]},10,"StatesGraph"]]

Out[]=

In[]:=

UndirectedGraph[SPMultiwaySystem[{b[{{1,1,1},{1,1,None}}]},10,"StatesGraph"]]

Out[]=

In[]:=

UndirectedGraph[SPMultiwaySystem[{b[{{1,1,1},{1,2,None}}]},10,"StatesGraph"]]

Out[]=

Loop on left has empty square on left; right has it on right.

In[]:=

UndirectedGraph[SPMultiwaySystem[{b[{{1,1,1},{2,2,None}}]},20,"StatesGraph"]]

## Only 1s and None’s

Only 1s and None’s

In this case, the None’s act as coordinate markers...

## Limiting Behavior [from a particular initial condition]

Limiting Behavior [from a particular initial condition]

When is it ergodic, i.e. it visits all states?

## All Possible Initial Conditons

All Possible Initial Conditons

The invariant is the parity of the permutation of all 16 squares plus the parity of the taxicab distance (number of rows plus number of columns) of the empty square from the lower right corner. This is an invariant because each move changes both the parity of the permutation and the parity of the taxicab distance. In particular, if the empty square is in the lower right corner then the puzzle is solvable if and only if the permutation of the remaining pieces is even.