try ?s to pr ed pe ng ui nf re nz ys to p?
looked at solution.
i was stumped by FZ-NG; BG-ND pair. i was wrong
to assume message ended as ND, it ends DX. So
BG-DX is a pair. i will try to continue pretending
i didn't see the solution.
LEMMA:
The BG-ND pair was an impossibility given FZ-NG.
PROOF:
If FZ-NG and BG-ND were both squares, then F, N and
B would be in the same row and Z and G would be in
the same row. so Z and G would be in different col;s.
by FZ-NG however, Z is in G's col.
Better proof not both square: G =c N ... interrupted
If only one were a square, say FZ-NG, since G and N
are in diff. col and row, BG cannot be in the same
row or col and and G rotate into N.
ERROR: however BG-ND could be a square is FZ-NG were
a v. line (but not an h line):
F
N B
Z
G D
If both were in aligned row and col, that would make
6 different letters in a row and col.
This proof is valid for all AB-CD and ED-CF, where the
letters represent arbitrary, distinct characters.
Since CD-AB is also a subst, then EB-AF are also blocked,
for all E, F distinct from A,B,C,D. Hence one codon
blocks 2x21x20 others.
This could be called an obstruction method.
LEMMA
AB-CD obstructs EC-DF
PROOF:
Let A =r C signify that A and C are in the same row, and
A =c D signify that A and D are in the same col. The signs
=r and =c are reflex, sym, transitive,
and X =i Y for i both r and c
implies X and Y are the same symbols, X=Y (unsubscribed
equality sign).
If AB-CD and EC-DF are both squares, then A =r C and D =c A
from the first pair, and D =c C from the second. Combining
A =c D and D =c C across common D, A =c C. so A=C, contradicing
distinctness.
If A =r B then all A, B, C and D share a row, and hence all
E, C, D and F share a row. If they are distinct, this would make
6 letters in a row, a contradiction. A =c B is similar.
LEMMA
AB-CD obstructs EA-BF.
PROOF
If AB-CD is a square, so is CD-AB. Apply the previous, EA-BF is
obstructed. If it is a line, so is CD-AB, and so on.
COR
AB-CD obstructs xC-Dy, xD-Cy, xB-Ay, and xA-By, for all x,y
distinct from A, B, C and D.
OBVIOUS
AB-CD obstructs AB-xy unless x=C and y=D, likewise CD-xy is
obstructed unless x=A and y=B.