### the rest of the story

page two!
and a word from our sponsor:
MEdZ… math ed zine… is starved
for money and attention so please
give all you can. comments are welcome.
particularly if anybody actually *plays*
this thing. so far, i’ve played against
against arlene without my even *asking*
’em to, which has to be considered a good

i left off here a few hours ago:

ABCDE
AFGHI
AJKLM
ANOPQ
ARSTU
BFJNR
BGKOS
BHLPT
BIMQU
CF
CG
CH
CI
DF
DG
DH
DI
EF
EG
EH
EI

and i’ll go ahead and claim that unguided
common sense, together with a certain devotion
to alphabetical order, can make this much seem
pretty much “necessary”. the first letter, A,
has to be in five lines, so put ’em first.
then fill in these five lines in alpha order
(“because we can”: this is *the easiest way
to remember*… it’s what alpha order is *for*!).
as for the B’s.
well, there’s four left to account for,
and they have to be in different lines.
4 B’s vertically. what next?
well, there’ll be 4 C’s, 4 D’s, and 4 E’s
coming up sooner or later… and *none*
of ’em sharing a line (because our first
line, ABCDE, is the *only* time *any* pair
from {A, B, C, D, E} appear together
[“two points determine a unique line”
(just as each pair of lines “intersect”
in a unique point… “duality”)]).

so what about F? well, that works, so shove it in.
but G can’t be in another line *with* F, so it’ll
have to go… hmm… underneath. and say! this’ll
continue! H has to go under *that*…

and i just end up transposing the 4-by-4 array
of “letters higher than E” just like i said last time.
finally (so far), the whole “FGHI-transpose” bit
also appears to follow naturally from the
“because we can” principle: each of these
letters must be used once-each-with each of
A, B, C, D, and E. why *not* in alpha order?

okay. end of recap. begin serious work.
consider
BFJNR
BGKOS
BHLPT
BIMQU
.
consider the *column* FGHI
(i’m dropping the “transpose” notation
and will simply count on the reader to
perform transposes… rows into columns…
as necessary).

we’ve simply *repeated* FGHI
(as a column) three more times.
we’re going to do something similar
with the *other* columns of letters
appearing in the part of the board
i’ve quoted just now:
JKLM, NOPQ, and RSTU.

BFJNR
BGKOS
BHLPT
BIMQU

first the JKLM’s. we need for
*each* of C, D, and E to be
matched (in a line with) each
of J, K, L, and M. however,
we must *not* reuse the “natural”
(alphabetical) order since, first
of all, that would cause a
repeated “FJ” (in the partial-lines
BFJ.. and CFJ..).

so some *permutation* of JKLM
will have to be used to match up
against the C’s… and a *different*
permutation against the D’s…

now. something i guess from experience.
i haven’t even *tried* to prove it, but
i’m guessing that the permutations that
work best… here *and* in the next three
columns (which fills in the board altogether)
take the forms
0123 (identity; we’ve used this)
1032 ( $\mu\rho$: start in the middle and work back; “wrap around” from top to bottom and keep going)
2301 ($\mu$: start in the middle and work forward; same “wraparound” rules)
3210 ($\rho$: rho for reverse (mu was for middle)).
these are pretty easily memorized
once you put the “visual learner” spin
on it by actually waving your pencil
over the appropriate letters while saying
the names. anyhow, it works for me…
and that’s the *order* i like for the
first lot (the first “non-trivial” lot;
repeating the FGHI columns verbatim counts
for me as part of this process, but with
the *identity* permutation applied repeatedly)
so we get
BFJNR
BGKOS
BHLPT
BIMQU
CFK
CGJ
CHM
CIL,
permuting JKLM according to $\mu\rho$:
the 1032 pattern here becomes KJML.

if you made it to here with clarity, we’re good.
fill in the rest of the third column with $\mu$,
then $\rho$ (both applied to the “original” JKLM
of this column)… then do column five
using $\mu, \rho, \mu\rho$ (applied to NOPQ) and
column six with $\rho, \mu\rho, \mu$ (to RSTU).

you should get

ABCDE
AFGHI
AJKLM
ANOPQ
ARSTU
BFJNR
BGKOS
BHLPT
BIMQU
CFKPU
CGJQT
CHMNS
CILOR
DFLQS
DGMPR
DHJOU
DIKNT
EFMOT
EFLNU
EHKQR
EIJPS
.
i think.
“twenty-one line Vlorbik”
(aka 21-point tic-tac-toe:
the alphabetical version).

#### 1 Comment

1. the same strategy works to produce
alphabetical versions of other
projective planes: i wrote out
P^2(F_3) and P^2(F_5) with
surprising ease yesterday.

there are p^2 + p^1 + p^0
points in each case, with p+1
points on a line. set down a
column of p A’s followed by
(p-1) each of B, C,… up to the (p+1)th.
then write out a p-by-p square
of the remaining letters, in alpha order,
next to the A’s. then *transpose*
the p-by-p square and put the result
next to the B’s.

(with “A” as the point at infinity
and “B” as the point of the *line*
at infinity associated with ordinary
(“affine”) lines of slope zero, these
lines correspond respectively to the
vertical and horizontal lines of my
*diagrammatic* versions of the
projective spaces. and we get the
*rest* of the lines (finite non-zero
slopes) by permuting the columns
of the transposed matrix repeatedly.

in the case of *prime* order, p,
one has *cyclic* permutations
(of order p).

in the case of F_4 the permutations
(producing appropriate columns for our
list of alphabetical “lines” in the “board”)…
what i’ve called $\mu, \rho$, and $\mu\rho$
can written in (the much more convenient-
-for-calculation) “cycle” notation as
(03)(24), (02)(14), and (01)(23).
together with (), the identity element,
these again form a group…
in fact, a “klein four” group…