Archive for June, 2011


the three directions associated
with “arrows’… 001, 010, and 100
in the binary code… are here
colored yellow, red, and blue
(it didn’t reproduce as well as
one would’ve liked).

these so-called “primary” colors
(for pigments; for colored lights
the rules are different) blend
along the appropriate cross-sections
of the diagram (the easy-to-see
ones with the arrows attached)…
*or* the sides of the triangle:
yellow and red form orange,
yellow and blue form green,
red and blue form purple.

blending *opposite* colors…
the primary/”secondary” pairs
yellow/purple, red/green, blue/orange…
gives a muddy brown (“mahogany”
sez crayola) and again the appropriate
cross-sections of the cube (and lines
of the “fano plane” triangle) associate
the appropriate “blends” with the
pairs-of-pigments combining to form them.

finally, there’s the weird cross-section
(the 3-5-6 arc in the fano diagram)
consisting of all three secondaries.
on the cube diagram, this appears
as a tetrahedron-through-the-origin
rather than a plane-through-the-origin.

colored pencils (with erasers) rock.
we now return to zeerox-friendly B&W.

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probably my best-looking version
of P^2(\Bbb F_4)^* so far
(but is graph paper “cheating”?).
[* this paragraph has no footnotes;
that’s a “star” standing for “dual”
rather than an asterisk: “the *dual* of
the two-dimensional projective space on
the field of four elements”]

meanwhile, i’ve played 21-tac-toe
right on the board a couple times
(versus self and madeline).

once one had memorized the board
(all 21 “lines”… the five circles in any
of the 21 “cross” diagrams here
are arranged in the pattern of a line
in the “big space” consisting of
the “crosses” [points of the space]),
the game would begin to very-closely-
-resemble “ordinary” tic-tac-toe:
just sketch a single cross and start
alternating X’s and O’s in the usual way.

all five points of any line wins;
otherwise count sets of four-on-a-line.

the last point isn’t played
(since there are an odd #
and the first player *already*
has an advantage).

so far the alphabetical version is easier.

post deleted

blag.

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 ( \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).

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.