### twentyone point tic-tac-toe: drawing the board (alphabetical version)

what we need is, first,
{A, B, C, … , U}…
or any other such set
(of twentyone objects).

next we seek to find a certain collection
of twentyone 5-sets taken from our
“alphabet” (i’ll start calling letters
points sooner or later, so let
me go ahead and start now).

and, since i’ll be writing ’em
out as “lines” of letters here (
ABCDE
rather than
{A, B, C, D, E},
for instance),
i’ll call these 5-sets lines
(there *is* a more “geometric” reason for
the terminology but it need not concern us).

the properties defining an acceptable collection
are as follows.
1. there are 21 lines of five points each
2. for every distinct pair {x, y} of points
(i.e., pairs-of-letters chosen from A through U
[there are of course ${21\choose 2} =210$
such “combinations” of letters]) there exists
exactly one line containing both x and y.
3. for every distinct pair {l, m} of lines
(e.g., {ABCDE, DFLQS}… there are 210 such pairs again)
there exists exactly one “point on both lines”
(letter belonging to both “strings”; D in the example).

and that’s it. and i like to suppose that,
given these constraints, anybody but outright
logic-phobics would figure out a way to get it.
just start trying and see what happens.

but i’m here to spoil all the fun by discussing
*my* favorite way… so, you know… the usual
disclaimer. feel free to quit reading and fiddle
it out. it’s a great exercise.

drawing the board

top row:
ABCDE
now. every point has to belong
to exactly five lines… so “A”
has to belong to five lines.
write these out downward.
then fill in the rest of the letters
in the usual alphabetical order to get
ABCDE
AFGHI
AJKLM
ANOPQ
ARSTU
: so far, so good. now.
for the other letters (B,C, D, & E)
of the top row, we’ve got one line each
already and will need four more. so,
writing *downward* again, fill in
4 B’s, 4 C’s, 4 D’s, and 4 E’s.
then (two “steps” displayed as one
here… again…) add the *transpose*
of the 4-by-4 “F through U” table
that we’ve created in the lines-beginning-
-with-A already (“transposing” here means
“swapping rows for columns”)

ABCDE
AFGHI
AJKLM
ANOPQ
ARSTU
BFJNR
BGKOS
BHLPT
BIMQU
C
C
C
C
D
D
D
D
E
E
E
E

then you break out your cut-and-paste feature
and redo the whole thing with “FGHI-transpose”
written in along the edge three (more) times:

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

that’s the easy part. and enough for now,
i guess. i got this bit essentially by
accident without even thinking much
about alphabetical order. i just filled
in one of my many pictures of P^2(F_4)
more or less at random and started copying.
the initial result has been memory-holed
as far as this blog is concerned.
suffice it to say that when i went to
“understand” it… well enough to *memorize*
it… i found the current one to be an improvement.