## Archive for the ‘21-point Tic-Tac-Toe’ Category

### turning coffee-stains into theorems

the pink point is “on its own polar”
(in the “polarity of P^2(F_4)” shown here)
& so the Big pink Circle “includes” one
of the little pink circles.

the other colored points of the display
*don’t* have this property… one sees
instead that the Orange Circle “includes”
a blue circle, & the Blue “includes” an
orange. note that the “maps” of 21-space
found in the Orange and Blue Circles
correspond to the *positions taken*
by the orange and blue circles
(respectively): the blue circles
“go across the top and then take the
rightmost point”—the Orange ‘position’,
if you will—and the oranges “go up the
right-hand-side and then take the very middle”
(i.e., the Blue position).

similarly the “turquoise” & “lime”
points (let’s say) are found
respectively on the lime & turquoise
lines.

before you recycle anything
scribble all over it and make copies.
so nothing goes to waste.

### same space, different day

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

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 #

so far the alphabetical version is easier.

### 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).

### 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
of the 4-by-4 “F through U” table
that we’ve created in the lines-beginning-
“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.

### priority claim

here’s another picture of a dualization
of the projective plane on the field
of four elements: $P^2(\Bbb F_4)^*$.

in the first of two tweaks since the last version
published here
, i reversed
the positions of the roots of x^2 + x +1…
i’ve been calling ’em alpha and beta…
reversed their positions, i say,
for the vertical axis (as opposed
to the horizontal:

0a 1a aa ba
0b 1b ab bb
01 11 a1 b1
00 10 a0 b0

as opposed to the earlier

0b 1b ab bb
0a 1a aa ba
01 11 a1 b1
00 10 a0 b0
).
this had the pleasing effect that
*all* the “lines” now had naked-eye
some twisty-looking “lines” once
the alphas and betas got involved;
this was in some part due to the
artificial “symmetry breaking” that
i’d indulged in by *putting alpha
first*. the new version had beta
close to the origin just as often
as alpha… which better describes
their relation in $F_4$.

“tweak” involved swapping (0,0)
and (1,1)… and causing “lines”
actually *looking like lines* to
appear as *broken* lines
(but simultaneously causing the
alpha-points and beta-points
to behave better still).

this looks like a pretty good board
for pee-two-eff-four tic-tac-toe.
(or, ahem, “21-Point Vlorbik”…
he who blowet not his own horn,
that one’s horn shall never be blown).

$P_2(\Bbb F_4)$ Tic Tac Toe

“The Board” is the 21-line
array of letters:

ABCDE
AFGHI
AJKLM
ANOPQ
ARSTU
BFJNR
BGKOS
BHLPT
BIMQU
CFKPU
CGJQT
CHMNS
CILOR
DFLQS
DGMPR
DHJOU
DIKNT
EFMOT
EFLNU
EHKQR
EIJPS
.
two players alternate turns.
in the first turn the first player
chooses any letter from A to U
and “colors” all five copies of
that letter on the board;
the second player then chooses
any *other* letter and “colors”
all five copies (in some other
color… X’s and O’s can be made
to do in a pinch if colored pencils
aren’t available).

players continue alternating turns,
each coloring all five copies of
some previously uncolored letter
at each turn.

play ends if one player… the winner…
has colored *all five letters* of any row.
otherwise play continues until all (21)
letters are used.

if one player… the winner… now has more
“four in a row” lines than the other, so be it;
otherwise the game ends in a draw.

i actually played this for the first time
this morning on the bus with madeline:
a four-to-four draw.

in the thus-far-imaginary computer version,
upon selection of a letter, five dots…
all the same relative position in their
respective “crosses”… light up and
“five in a row” becomes “an entire
cross lights up”. this might be worth
learning to write the code for.

better if somebody else did it, though,
i imagine. i just want a “game designer”
credit and a small piece of every one sold.

• ## (Partial) Contents Page

Vlorbik On Math Ed ('07—'09)
(a good place to start!)