### 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

myself, madeline, arlene, and john.

most excitingly, madeline has played

against arlene without my even *asking*

’em to, which has to be considered a good

sign indeed. and now back to the show.

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 ( : start in the middle and work back; “wrap around” from top to bottom and keep going)

2301 (: start in the middle and work forward; same “wraparound” rules)

3210 (: 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 :

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 ,

then (both applied to the “original” JKLM

of this column)… then do column five

using (applied to NOPQ) and

column six with (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).

June 7, 2011 at 11:38 am

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 , and …

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…

isomorphic to the additive group

for the field in question. i’m guessing

this is no coincidence.