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

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.

probably my best-looking version

of 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.

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).

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

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.

here’s another picture of a dualization

of the projective plane on the field

of four elements: .

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

symmetry. the earlier drafts had

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 .

anyhow, the second, more radical

“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).

Tic Tac Toe

Official Rules (all rights reserved).

“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.