use the source luke

the gospel i preach: simple things first.

basic arithmetic, for example, turns out
to possess a bottomless depth… and will
reveal new secrets to any investigator
that can bring it new eyes.

the easiest way to get new eyes is of course
to find a student.

[
or rather to have students
provided for you as i did at indiana u,
ohio dominican, columbus state cc, ohio state u,
and capital u at various times. finding students
outside ivory tower walls is a new challenge for me.
i’m recruiting. and at reasonable rates too but so far
i haven’t really got anything you could dignify with
the name of “a plan”. or for that matter any students.
]

but the easiest way to find the depth
is to lend one’s own eyes to old texts.
and start simple.

better to understand arithmetic,
for example, one could and should
investigate (as soon as possible)
modular arithmetic.
the most familiar example
(“clock arithmetic”) uses the
hour-numbers marked on a clockface.
replacing “12” with “0” gives
the commonest mathclass
representation of “the integers
mod 12”:
{0′,1′,2′,3′,4′,5′,6′,7′,8′,9′,10′,11′}.
i’ve “primed” the numerals
representing the elements
of Z_12 (the set of integers-mod-12)
here to emphasize a point…
namely that they are *not*
integers. 3’+5′ = 8′ all right
like anybody would expect but
we also have… what would just
look *horribly wrong* without
the “primes” 8’+7′ = 3′
(because starting at eight
o’clock and waiting seven hours
takes us to three o’clock…
we “throw out multiples of
twelve” to do arithmetic mod 12.)
it turns out you can multiply too.
very important this. turns out if
you use a *prime* number in place
of twelve you get something called
a finite field. manymany
very simple examples to be found
of phenomena whose depth remains
unsounded. the ever-growing subject
of “crypto” more or less *begins*
with the study of these fields
from what i’ve been given to understand.

but that’s not the subject of today’s ramble.
which is counting permutations. and starting simple.

there’s one permututation of one thing.
A
that’s it.

there are two permutations of two things.
AB, BA
pretty simple still.
and uninstructive.

things get much more interesting
when three things are considered.
ABC
ACB
BAC
BCA
CAB
CBA
.

this is the key example in understanding
what goes on. but the key *exercise*
is “write out the permuations of
the elements of {W, X, Y, Z}”.
(one will have been given a solution to
“write out the permutations of
the elements of {A, B, C, D}”
already in the body of the text.)
ABCD BACD
ABDC BADC…
ACBD .
ACDB .
ADBC .
ADCB

how does it work? well.

the “reason” that the
six permutations of
the elements of the set
{A, B, C} are six
(ABC, ACB, BAC, BCA, CAB, CBA)
is that 6 = 3*2*1.
(where “star” denotes “times”:
5*3 =15 etcetera).
the point here being that we can
imagine writing our letters
one-at-a-time into three “spots”
“_ _ _”
with *three* choices for the first letter
and *two* (we are, thus far implicitly,
but now by explicit decree, not allowing
ourselves to repeat letters [and moreover
each letter *must* be used]) choices
for the next. finally, there’s one letter
left… our only “choice”. the “3” (A, B, or C),
the “2” (don’t reuse the one you picked!)
and the “1” (whatever’s “left over”)
in 6 =3*2*1 are now all accounted for.

the same reasoning tells us than in the case of four letters
one will have 4*3*2*1 = 24 possible permutations.

one-fourth *of* ’em will start with “A”;
one-fourth with “B”; so on. six each.
so one might write down

A______B______C______D______
A______B______C______D______
A______B______C______D______
A______B______C______D______
A______B______C______D______
A______B______C______D______

first. (the underlines denote
*blank space*… the very model
of “harder than it looks” in my book).

then “go down the A’s column”.
there are six entries…
and B, C, and D
(the “unused” letters so far)…
will each have the same chance
of appearing next: twice each
as it turns out. so fill in two B’s,
two C’s, and two D’s.

AB_____B______C______D______
AB_____B______C______D______
AC_____B______C______D______
AC_____B______C______D______
AD_____B______C______D______
AD_____B______C______D______

likewise for the B’s…
“go down” the column
distributing “missing letters”
(in this case A, C, and D).
and the C’s… finally the D’s.

we’re now at
AB_____BA_____CA_____DA_____
AB_____BA_____CA_____DA_____
AC_____BC_____CB_____DB_____
AC_____BC_____CB______DB_____
AD_____BD_____CD_____DC_____
AD_____BD_____CD_____DC_____

finally go down each column one more time.
there are two of each “string” so far.
in the first, AB, add the remaining letters
first in their “natural” order (ABCD)
then in the opposite order (ABDC).
continue with this pattern:
two missing letters in each case,
first in the natural then in the opposite
order. go through the whole list.
voila. all twentyfour permutations
of four things. in alphabetic order
no less.

i’ve taught this procedure to dozens.
one-at-a-time, too. somehow seeing me
do it in class doesn’t do a thing *for* ’em
sometimes. the student has to move the pencil
(and wants an expert standing by
to bail out at the first sign of trouble).

sure it’s tedious. but it’s just incredibly necessary.
you can’t do *anything* with certain sections of
very common textbooks until students can rattle off
lists like this.

one way or another. certainly there’s no need
to learn *my* algorithm if you’ve got one
of your own that works. i’d love to have you show me.
better if you show the whole class.

there is nothing for “knowing what you’re talking about”
quite like “being able to write it down”.