Solutions to the Missionaries and Cannibals problem
==================================================

Note that at the start we really only have two choices:
either send to cannibals in the boat or send one missionary
and one cannibal (sending one person is not useful, the only
thing that can happen next is that that person comes right 
back again.)  A solution can be found with either start.

Notice too that there is usually very little choice
once you get started.  Remember, returning to a situation
you have already had is wasteful.  Some comments about choosing 
are indicated below. 

M denotes a missionary
C denotes a cannibal
<> denotes the boat

near side of river  far side of river   comments about choices
==================  =================   ======================

MMM CCC <>                              other option listed below
MMM C               <> CC               someone has to bring the boat back
MMM CC  <>             C                send M across leaves trouble next trip
MMM                 <> CCC              someone has to bring the boat back
MMM C   <>             CC               must send 2M's if any
M   C               <> MM  CC           must send equal # of M's and C's
-----------------------------           now we can just reverse everything
MM  CC  <>             M   C            
    CC              <> MMM C
    CCC <>             MMM
    C               <> MMM CC
    CC  <>             MMM C
                    <> MMM CC

Here is a second solution that starts differently:

MMM CCC <>                              here's the other start
MM  CC              <> C   M            can't send C back, so M goes
MMM CC  <>             C                sending M's leads to trouble
MMM                 <> CCC              hmm. looks just like above now
MMM C   <>             CC
M   C               <> CC  MM           
-----------------------------           now we can either reverse this
MM  CC  <>             C   M            
    CC              <> C   MMM
    CCC <>                 MMM
    C               <> CC  MMM
M   C   <>             CC  MM
                    <> CCC MMM

note that we could mix and match tops and bottoms, so we have actually
found 4 solutions!