I found what he's talking about.
For (E,N,D,X,1,C) (our start record) in (E,1) there is always a record (e,1,d,x,a,b) where
D-d = N-1
and
a = N
I found what he's talking about.
For (E,N,D,X,1,C) (our start record) in (E,1) there is always a record (e,1,d,x,a,b) where
D-d = N-1
and
a = N
Also in this cell the X's are the same as the (a,b)=(1,c) cell. Then we can go straight to that cell through
(E,N,D,X,A,B)
to
(knowing (d-D = N-1 =N+D-1 = d)
(also since d-x=a we have: N+D-1-X = a
(E, 1, N+D-1, X, N+D-1-X, ?)
You could then just calculate B however you want.