Solving Poly Eqns 22: The bridge between BiTri numbers and polygonal subdivisions
Wild Egg Maths
4.6K subscribers
We apply both the binary and ternary operations that we have defined on BiTri roofed polygons in order to capture the array of BiTri numbers interpolated beyond the Catalan numbers and the Fuss numbers. The operations on polygons need to be extended to multisets in a natural way, and then a crucial homomorphsim maps back to polynomials.
Remarkably, this rather soft, abstract approach actually ties us powerful information about the number of ways of subdividing polygons into triangles and quadrilaterals. Combinatorialists would see this as a natural application of generating functions.
NOTE: At around 25:13 I made a silly mistake and had the square operation as a binary, not ternary operation: the relevant equation should read
phi(square(P, Q, R)) = q phi(P) phi(Q) phi(R)
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00:01
hello everyone i'm norman wilbur thanks
00:03
for joining me on this adventure
00:04
where we are trying to bridge
00:08
the gap between solving polynomial
00:11
equations
00:11
and the rich catalan story in
00:15
combinatorics
00:16
and today this bridge really is going to
00:18
come into very clear focus for us so
00:20
this is actually quite an important
00:21
lecture
00:22
we are actually going to be able to
00:24
connect these by tri numbers this array
00:27
that we've been looking at quite a lot
00:28
and polygonal subdivisions and it's
00:31
going to shed light on this fundamental
00:33
question about why these two things
00:35
should be
00:35
related you know why
00:39
should solutions to a cubic equation
00:42
have anything to do with these by tri
00:44
numbers and why
00:46
should these bitry numbers capture these
00:49
polygonal subdivisions of a general
00:51
polygon into
00:52
triangles and quadrilaterals so i remind
00:56
you that the bitry
00:57
array is given by this very succinct
00:59
formula here cmn
01:01
is 2m plus 3n factorial over 1 plus m
01:04
plus 2n factorial m factorial n
01:06
factorial it's a sub
01:08
multinomial coefficient and our indexing
01:11
here is
01:12
starting with uh m equals zero at n
01:14
equals zero
01:15
giving us the catalan column and the
01:18
foos
01:18
row and today we want to
01:22
extend our understanding in the we've
01:24
gained in the last few videos
01:26
on the catalan story and the food story
01:28
we want to extend it to this
01:30
two-dimensional situation so maybe a
01:33
first order of business is to actually
01:34
define
01:36
a concept that we're talking about quite
01:37
a lot so it's nice to have a
01:39
designated name let's agree that i'll
01:41
buy tri
01:42
roofed polygon we're going to introduce
01:45
that phrase
01:46
and what it means it's a polygon which
01:48
is subdivided
01:49
diagonally the way we've been talking
01:51
about so diagonally means
01:53
joining vertices to vertices no interior
01:55
vertices
01:56
into triangles and quadrilaterals
02:01
and also there should be a distinguished
02:04
top edge
02:05
we usually put a little bar over that
02:06
top edge okay that's
02:08
what the roof expresses so in the
02:12
usual language of trees people talk
02:14
about rooted trees or rooted polygons
02:16
maybe
02:18
for us the the special
02:22
nodes not at the bottom it's actually at
02:23
the top so i think roofed which is just
02:25
a small
02:26
perturbation of that is actually a more
02:28
apt
02:29
term okay so let's agree to that by try
02:32
roofed polygons those are the
02:34
things that we're considering and we
02:36
suspect
02:37
we're not sure but we suspect that these
02:39
numbers are counting
02:41
the numbers of such bi-tri-roofed
02:44
polygons
02:44
that's a major conjecture of ours that
02:47
we would like to
02:48
establish we've only really kind of
02:51
looked at it
02:52
in this sort of top left hand corner and
02:55
rather laboriously
02:57
check this out for some special cases
03:00
today we're going to be able to see
03:02
a much more general and powerful
03:04
argument
03:05
so watch carefully it's almost going to
03:07
be like magic
03:09
i'm not actually going to do anything
03:11
really dramatic computationally we're
03:13
just going to
03:14
think about things in a good way
03:16
motivated
03:17
by what we've been doing with catalan
03:19
and foose numbers which is ultimately
03:21
motivated by that exercise
03:24
in this beautiful text concrete
03:25
mathematics by knuth and company
03:29
graham and uh okay
03:33
so we are conjecturing let's follow down
03:35
here we're conjecturing
03:36
the numbers cmn these ones here give us
03:38
a power series solution to a general
03:39
cubic equation first of all that's one
03:41
major conjecture
03:43
and the other one is that they also
03:45
count the number of by
03:46
tri root roofed polygons with
03:49
m triangles and n quadrilaterals and
03:52
that necessarily means with
03:53
one plus m plus two n vertices so this
03:57
is really a bridge
03:58
this by tri story is a bridge from the
04:00
cubic story
04:02
to this combinatorial slash geometrical
04:05
story involving these
04:08
subdivided polygons and the sizes of
04:11
these polygons that are appearing here
04:14
might be just useful to contemplate them
04:17
all right
04:18
so here are the sizes starting in the
04:20
top right corner with the degenerate
04:23
polygon with two sides okay which sort
04:26
of plays a special role
04:28
and then we go down by one and we go
04:30
across by two
04:32
okay so for example this ten thousand
04:34
and ten that's sitting here
04:36
in the sort of third row starting with
04:39
zero so
04:40
down there in that third row and in the
04:43
third
04:45
column that counts the number of 11 gons
04:51
that can be subdivided into three
04:52
triangles and three quadrilaterals
04:55
roofed 11 guns
04:58
we haven't checked that okay but that's
05:00
just the conjecture but that's
05:02
what we're trying to get so this is like
05:04
a huge
05:05
extension really of the catalan story
05:08
all right so um
05:11
how are we going to do this so what
05:13
we're going to do is we're going to take
05:15
our our approach that we've been talking
05:19
about in the last few videos and just
05:20
sort of extend it in the most natural
05:22
way so
05:24
first of all
05:28
we extend the
05:31
binary and ternary
05:35
operations to these
05:38
by tri roofed
05:42
polygons all right so let's grab
05:45
our triangle symbol so
05:49
we are going to take this upside down
05:53
triangle with
05:54
a bar on top and we're going to
05:58
now define p
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and q this binary operation it's just
06:05
what we did before but
06:06
i want to write it in a little bit um
06:09
other way because i'm thinking about
06:12
generalizing this
06:13
to higher numbers so um let's put
06:16
something in there okay this is supposed
06:18
to be p
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and then the triangle the q this is the
06:22
operation that we were talking about
06:23
last time
06:24
okay so this is this is the binary
06:27
operation
06:30
that appends p and
06:33
q to the left and right
06:37
sides of a roofed
06:41
central triangle okay
06:44
so we're just introducing a slight
06:46
variant
06:47
by putting the operator out in front
06:50
okay is that all right that's all now
06:53
and i should say okay that this is now
06:55
we're extending the story now to buy
06:57
tri-roofed polygons okay
07:00
where p and q
07:03
are by tri roofed
07:07
polygons okay so formally we just had
07:10
this triangle operation
07:11
on polygons that were subdivided
07:15
try into triangles okay now we're just
07:18
observing that this operation still
07:21
makes sense if you have
07:22
two polygons which are both subdivided
07:25
into
07:26
triangles and quadrilaterals this one's
07:28
subdivided into triangles and
07:29
quadrilaterals
07:30
this one's subdivided into triangles and
07:31
quadrilaterals then we take our central
07:33
triangle and we
07:34
we rotate these things around and glue
07:36
them to the two sides
07:37
of the central triangle the result is
07:41
another by tri roofed polygon
07:44
okay and similarly
07:48
we're going to take the other operation
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that we had
07:51
okay but now it's going to apply not to
07:54
two
07:55
by tri-roof polygons but to three of
07:57
them okay
07:59
so and similarly this is our ternary
08:02
operation
08:05
now extended
08:08
also to buy try
08:12
roofed polygons
08:17
all right so this world of by tri roof
08:20
polygons
08:21
supports two operations we can
08:25
glue any two of them onto a central
08:27
triangle and we get another bi-tri-roof
08:29
polygon
08:29
we can glue three of them onto a central
08:33
quadrilateral
08:34
p q r and we get another
08:38
by tri-roofed polygon
08:41
okay so this this by tri-roof polygon
08:44
world
08:45
is closed under two operations
08:48
let's have a look to see what this looks
08:50
like uh more concretely
08:56
okay so let's have a look at these by
08:58
tri
09:00
roofed polygons they naturally
09:04
can be organized in an array which
09:06
exactly
09:07
mirrors our by try numbers array
09:10
but which is more elaborate in which the
09:12
entries are not just
09:13
single numbers but actually collections
09:16
or multi-sets
09:17
of these by tri-roofed polygons
09:20
so have a look at this thing here okay
09:23
so
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in this top corner here we just have the
09:26
single
09:28
uh single degenerate two-sided polygon
09:32
as we go down we
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see various triangulations first of a
09:39
triangle
09:40
and then of a quadrilateral then a
09:42
pentagon and so on and we get the usual
09:44
catalan numbers
09:46
and the numbers in red under the bottom
09:49
left corner of each little square
09:51
is a count of the total number in that
09:54
square
09:54
so down here at the bottom with the
09:55
pentagons there are actually five of
09:57
these and i've only shown you two of
09:58
them and then there's three dots but all
10:00
together there's five in that sort of
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box
10:02
and then if we go over to the next uh
10:06
column go over to the next column here
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this one here
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uh that starts with a quadrilateral up
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on top and then we go down by one we get
10:15
a pentagon and the five ways
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of subdividing a pentagon into one
10:19
triangle and one quadrilateral
10:22
this column always has one quadrilateral
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so if we go down one more
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we go to hexagons and there's the 21
10:29
ways
10:29
of subdividing a hexagon into
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one quadrilateral and two triangles and
10:35
then below that the 84 ways
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of subdividing a seven gun i suppose
10:40
into
10:42
one quadrilateral and three
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triangles okay and then the next
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uh column over here we're starting with
10:52
a hexagon
10:53
the three ways of subdividing it into
10:56
two quadrilaterals and then we go down
10:58
one where that means we're adding a
11:00
triangle
11:00
we still have two quadrilaterals but now
11:02
we have one triangle as well and there's
11:03
28 ways of subdividing this
11:05
seven gone then we go down one more now
11:08
we're adding another triangle there's
11:09
two triangles and two quadrilaterals
11:11
altogether 180 ways of
11:13
subdividing a an octagon okay
11:17
and that's the general pattern we
11:20
just keep on adding more and more so i
11:22
want you to think about this entire
11:26
array of these
11:29
these by try roofed
11:32
polygons okay this is an object which is
11:35
really interesting in some sense
11:37
this is a you know a finer object than
11:40
the actual just array of numbers here we
11:42
actually have combinatorial objects
11:45
and happily um well you know
11:49
these are combinatorial objects but we
11:50
do have some very
11:52
simple kind of arithmetic that we can do
11:54
with them like what
11:56
well we can we can form a binary
11:59
operation on any two of them
12:00
that's the our triangle operation and if
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we have any three of them
12:05
we can perform our ternary operation
12:09
okay so let's just just illustrate that
12:11
okay
12:12
so the binary operation
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let's grab two of these things so maybe
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i'll grab this one here
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okay and
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put the top on top and uh maybe uh
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where's another one maybe um
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uh maybe a hexagon
12:32
with with that one there okay there's
12:33
the top okay so those are two things and
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i'm going to put the
12:37
operator on the left okay so what's that
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do
12:40
well i think we all know so there's a
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triangle in the middle
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and then we have to stick the
12:46
quadrilateral on the left-hand side and
12:48
let's
12:48
make sure that the top is sort of being
12:51
glued along
12:52
there so to tilt my head and i guess i
12:54
have to put
12:56
the diagonal in like that and then in
12:58
terms of the hexagon well we have to
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glue that on
13:00
here so there's the hexagon with the top
13:03
again glued there
13:04
and then the other stuff is there like
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that
13:10
okay clear and then we could rewrite
13:13
this if we wanted to it's uh it's gonna
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be a one two three
13:15
four five six seven eight nine gone
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so i guess i could write this uh one two
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three
13:22
four five six seven eight nine something
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like this
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uh where there's a distinguished um
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thing and there's gonna be a triangle
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one two three one
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two three so there's gonna be a
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quadrilateral on that side
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and then this triangle and then our
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hexagon on this side
13:39
and then we have to put the diagonals in
13:41
like like this
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and i suppose like uh
13:46
like this okay
13:50
but with a distinguished roofed top
13:52
there
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so that's an example of taking two
13:56
arbitrary bi-tri
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roofed polygons and performing this
14:01
triangle operator and getting another by
14:03
tri-roof polygon
14:05
okay it's it's a happy situation because
14:09
we have it's essentially like an
14:10
algebraic structure
14:12
but you know it's it's taking place
14:15
in this world of these funny bi-tri
14:18
roof polygons it's important that
14:20
they're roofed right if they weren't
14:22
roofed we wouldn't be able to do this
14:24
there would be a question of well which
14:25
side are we going to glue in but they
14:27
all have distinguished tops so that
14:28
makes it easy to to
14:30
know where the gluing takes place
14:33
okay and then there's also the ternary
14:35
operation
14:37
ternary operation let's just review that
14:40
one
14:40
okay so now we calling it a square or
14:43
quadrilateral
14:44
and um okay so maybe i'll take um
14:48
this one here and i might maybe i'll
14:50
even i'll put uh i'll just use this the
14:53
the degenerate one just to make life a
14:54
little bit easier and uh
14:56
over here i'm going to put a triangle so
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i don't have to work too hard okay
15:00
so i'm going to multiply those three
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things so how do i do that
15:03
well i take our central thing is a
15:06
quadrilateral
15:07
on the right hand side i have to put
15:08
this one here so i'm going to put a
15:09
quadrilateral over here
15:11
glue it along that edge and put the side
15:13
there then
15:14
our degenerate quadrilateral is going to
15:17
end up being glued at the bottom here so
15:19
it's
15:19
basically effectively not doing anything
15:22
you're just sort of gluing an edge to
15:23
that edge so nothing's really changed
15:25
and then on the right hand side we're
15:27
gluing our a triangle
15:29
with its edge like that okay the effect
15:31
is to have a one two three four
15:33
five six seven gone one two three four
15:37
five six seven gone and
15:40
there's a quadrilateral on this side so
15:42
i guess it's like this
15:45
okay and here then there's going to be a
15:48
side like that okay and there's the top
15:53
great okay so those are our binary
15:56
and ternary operations
15:59
okay and now we're going to do the same
16:02
kind of thing
16:03
that was suggested to us by graham
16:06
knuth and padishnik okay in that
16:10
exercise 22 in chapter seven
16:13
we are going to take the multi-set of
16:17
all of these by tri roofed polygons
16:20
and we're going to give that a name so
16:22
let's let a be the entire multiset okay
16:25
a equals
16:28
i might say so if a
16:31
is the entire multiset
16:36
and set of by try
16:41
roofed polygons
16:46
then we're going to write down an
16:48
equation satisfied by a
16:51
a every element in a
16:54
is one of three kinds
16:58
either it's just the
17:01
degenerate two-sided thing which is just
17:04
the single thing in the top
17:06
left corner that's a kind of a special
17:08
case or
17:09
it has it's one of two other
17:12
possibilities so if you have some big
17:14
some big polygon okay and here is its
17:17
top edge
17:18
then that top edge is going to
17:21
be part of a constituent
17:25
polygon which is either going to be a
17:27
triangle
17:28
or a quadrilateral if the starting thing
17:30
is not itself the degenerate
17:32
two-sided thing okay so we're either
17:34
gonna have a triangle
17:35
somewhere or we're gonna have a
17:39
situation where this um
17:44
this top edge is is somehow related to a
17:48
quadrilateral so either a triangle in
17:51
the center
17:52
or quadrilateral in the center
17:56
now what that means in the first case if
17:58
there's a triangle
18:00
then that means that you can write this
18:03
thing
18:03
as essentially the product of
18:06
a p and a q okay and
18:15
what that means is that that every
18:18
element in this multiset
18:19
is of the form triangle every such
18:22
element that has a central triangle
18:24
is a performed triangle um of
18:29
some element in a with some element in a
18:36
okay you know so no matter which
18:40
two elements in a you choose to be p and
18:42
q you can create this
18:44
binary product and you're going to get a
18:46
a new by tribe roof polygon and every
18:48
roof by roofed by tri-roof polygon
18:52
which has a central triangle is going to
18:54
be of that form
18:56
but there's also the ones that are of
18:59
this form
19:00
that have a central quadrilateral and
19:02
they're going to all be of the form
19:05
well something like this where where
19:06
here there's a p and
19:08
maybe here there's a q and here there's
19:09
an r
19:11
okay so they're going to be it's going
19:13
to be the form
19:15
quadrilateral product of a
19:18
with a with a
19:23
this is an identity of
19:26
multisense
19:34
and it's really important to get your
19:35
head around this identity to convince
19:37
yourself that this is going to be the
19:39
case
19:40
what we're doing is we're going a little
19:41
bit beyond uh concrete mathematics here
19:44
in problem 50 which i've also mentioned
19:48
in that text which is very interesting
19:50
it's some kind of generalization of the
19:51
catalan story
19:53
but they're suggesting something that i
19:54
don't really know how to do
19:56
they're also making us um an extension
19:58
to to more general
20:01
subdivided polygons but i don't really
20:02
know how how to to answer their problem
20:05
what i'm doing here is something a
20:06
little bit different i want to
20:09
i want to use both operations okay
20:12
because there are these two
20:13
possibilities for that central
20:15
polygon either it's a triangle or it's a
20:18
a quadrilateral
20:20
or it's not there at all and we have
20:21
this degenerate two
20:23
two uh two gone okay so i'm
20:26
using both the binary and the ternary
20:30
operation combining them both
20:31
to get this identity that a satisfies a
20:35
is the
20:36
entire m set of by try roof polygons
20:42
all right great okay so
20:46
once we've got our head around that what
20:47
are we going to do well what did we do
20:48
last time
20:49
in both the catalan and the foose
20:51
stories what we did is we
20:53
tried to bring this abstract algebra of
20:56
m sets
20:57
down to earth a little bit by connecting
20:59
them with polynomials
21:02
okay we introduced this phi and in the
21:05
triangle case
21:06
phi sort of counted the number of
21:08
triangles it was a little poly
21:10
polynomial with t to the n where n is
21:12
the number of triangles
21:14
and in the fu story we were counting
21:17
quadrilaterals
21:18
and we associated to such a thing a
21:21
little monomial q to the something where
21:24
the something is the number of
21:25
quadrilaterals
21:28
okay now we have to combine those two
21:30
things because we're in this
21:30
two-dimensional situation
21:33
so let's extend our
21:37
let's extend our
21:41
phi mapping mapping
21:46
from now these buy
21:49
tri roofed
21:54
polygons to
21:58
polynomials in
22:02
two variables t and q
22:07
okay so what's our definition going to
22:08
be so phi of
22:11
let's say we have some by tri-roof
22:13
polygon p
22:14
then we're going to define this thing to
22:16
be t to the
22:18
let's say m and q
22:21
to the n where
22:27
p has m triangles
22:31
and n quadrilaterals
22:39
all right so for example
22:42
uh fine let's make up oh sorry maybe
22:48
something like this okay there's a
22:49
triangle uh there's a quadrilateral
22:52
there's another triangle there's a
22:54
quadrilateral okay
22:56
there's the top okay what is 5 this
22:58
thing well we count the number of
22:59
triangles
23:00
there's one there there's one there so
23:02
we have to put t
23:04
squared now we count the number of oh
23:06
sorry that that was
23:07
sort of wrong that that was the triangle
23:09
there there are two triangles now we
23:10
count the number of uh
23:12
quadrilaterals one right there and one
23:14
right there
23:16
so two of them so we put in
23:24
all right so what properties is this
23:26
going to have this is going to have two
23:27
crucial properties
23:29
sorry it's going to have two crucial
23:30
properties
23:32
that if we take
23:36
phi of the triangular product
23:39
of p and q okay
23:43
what are we going to get we're going to
23:45
get
23:49
phi of p
23:52
that's the the uh
23:56
the number of the the monomial of that
23:59
counts triangles and quadrilaterals in
24:01
pay
24:02
times phi of q that's the monomial that
24:05
counts
24:06
triangles and quadrilaterals in q
24:10
okay and they're going to multiply
24:12
together so that the power is add
24:14
so we're effectively just adding the
24:15
number of triangles and the number of
24:17
quadrilaterals but separately but in
24:20
addition
24:20
there's also this central triangle
24:23
that's part of the construction so
24:25
we're upping the number of triangles by
24:27
one we have to put a
24:28
t in front okay
24:32
that's the first property so 5
24:35
triangle of pq is t times 5p
24:40
times 5q and the second property
24:44
is that phi of square
24:49
of p and q
24:54
well um you're still going to have all
24:57
the triangles and quadrilaterals that
24:59
p has and you're still going to have all
25:00
the triangles and quadrilaterals that q
25:02
has
25:03
but now in addition there's now an
25:05
additional central
25:06
quadrilateral which ups the number of
25:09
q's in the expression so we have to put
25:10
a q here
25:15
so when we apply this
25:22
to our well our basic
25:26
fundamental identity to
25:30
our fundamental
25:33
identity let's say star
25:37
we get
25:47
well let's let phi of um
25:51
5a be let's say f
25:55
okay so this is going to be a multi-set
25:58
m set of monomials
26:05
you know the form t to the something q
26:07
to the something
26:13
so what happens when we
26:17
when we apply this phi mapping to this
26:19
okay
26:20
so we'll start with a equals this
26:24
sort of identity kind of fellow times
26:26
triangle
26:27
of uh a a plus
26:31
quadrilateral operation on a a a
26:34
what we're going to get is that this f
26:38
which is just the image of a
26:41
under this mapping so what what is that
26:44
actually
26:44
so that's what you would get when you
26:46
went up to go up to this diagram here
26:47
and you replace
26:48
each one of these polygons with a
26:53
monomial
26:54
okay like this one here would get
26:55
replaced with uh q
26:57
because it's one quadrilateral this one
26:59
get replaced with
27:00
q squared this won't get replaced with q
27:02
squared this one gets replaced with q
27:03
squared etc
27:04
this one gets replaced with um there's
27:08
no three three quadrilaterals so just p
27:11
q
27:11
cubed there'll be you know these 12 q
27:13
cubes etc
27:14
okay down here we've got uh
27:18
i guess one triangle and uh
27:21
two quadrilaterals each one of these
27:23
would we get to get
27:24
each of these 28 would get replaced with
27:26
tq squared there would be 28
27:28
tq squareds in that box
27:31
which we could then just write as 28
27:34
times t q squared so we're going to get
27:36
sort of a
27:37
you know an infinite or an ongoing
27:41
polynomial in ts and q's the image of
27:45
this m set under this
27:46
phi mapping okay and that's f alright so
27:49
now what is
27:50
the claim here that this f which is the
27:52
left-hand side is going to be what on
27:54
the right-hand side
27:55
on the right-hand side well the image of
27:56
this identity thing
27:58
is just one because there's zero
28:01
triangles and zero quadrilaterals so t
28:03
to the zero q to the zero
28:05
is one okay what about this thing here
28:09
a times a so we're multiplying each of
28:11
the elements by a
28:12
um and and each one is always is getting
28:16
multiplied in addition by this by this
28:18
triangle operation so exactly the same
28:20
reasoning that we applied in
28:22
our catalan story tells us that this is
28:25
going to be t
28:25
times f squared
28:30
and exactly the same kind of reasoning
28:32
that we applied
28:33
in the foos lecture we've got a's
28:37
a's a's all multiplying in all possible
28:39
ways
28:40
that's essentially giving us an f cubed
28:44
but all of those terms are going to have
28:46
this additional q in front of them
28:47
because
28:48
the the square or the quadrilateral
28:49
operation always
28:51
gives us additional q as is up here so
28:54
we're going to get
28:54
q times f cubed
29:03
that is a cubic equation
29:06
in f
29:10
a cubic equation
29:13
and that ladies and gentlemen is i think
29:15
why you know
29:16
why there is this intimate connection
29:18
between cubic equations and these
29:20
polygons it's exactly for this
29:22
reason this is a geometrical explanation
29:25
of this connection
29:27
that these polygons and the way they
29:29
multiply are naturally
29:31
inducing this this sort of generic cubic
29:34
equation
29:36
now it's not entirely generic because
29:38
the constant term happens to be one and
29:39
the
29:40
the coefficient of f is actually if you
29:43
bring it to that side is minus one
29:44
but you know aside from that the t and
29:47
the q are kind of
29:48
general aren't they so it's pretty close
29:51
to being a general cubic equation and so
29:54
now we can sort of see why it is that
29:57
our cubic equation story
29:59
ended up with something in this
30:00
direction because there's this natural
30:02
connection between them
30:05
but we can do more with this we can now
30:07
go back
30:09
to our by try array and see if
30:12
it actually does satisfy
30:16
this equation and if it does
30:19
that will basically confirm for us that
30:23
our by tri numbers really are counting
30:26
these by tri roofed polygons
30:34
all right so now i want to see if
30:38
this by try array really does generate
30:41
um a function f or a
30:46
sort of a polynomial f an ongoing
30:48
polynomial f
30:49
that satisfies that cubic equation that
30:51
we just wrote down
30:53
okay so what i'm going to do is to
30:55
create the polynomial is i'm going to
30:57
multiply
30:58
our array here we're just going up to
30:59
this uh case one two three four five
31:02
going up to uh m and n equals five
31:05
and i'm going to multiply by this vector
31:07
one
31:08
t t squared up to t to the fifth on the
31:10
left and by
31:11
this column vector one q q squared down
31:13
to q to the fifth on the right
31:15
and when i do that let me just expand
31:17
what we get
31:19
all right i get this this thing here
31:21
okay
31:22
and um let me actually give it a name
31:25
okay let's call it f
31:27
f equals okay so it's going to have uh
31:31
this is some expression that has i guess
31:33
36 terms
31:36
let me just expand it out just compute
31:39
expand
31:41
there's the 36 terms right each one of
31:43
them corresponding to an entry in this
31:45
table and of course this is not the
31:46
entire buy try right this is just the
31:48
top
31:49
left corner okay but so we're just kind
31:52
of going up to degree
31:54
five and and ignoring higher degrees
31:56
okay so let's
31:57
make this definition so let's get rid of
32:01
that and we'll make this definition
32:03
first of all let's clear our definitions
32:06
definitions clear and that will make
32:08
this definition
32:10
and now
32:13
let's have a look at what's the equation
32:15
that we want to satisfy
32:16
the equation was f equals 1
32:20
plus t
32:24
f squared plus q
32:28
f cubed okay it's an equation involving
32:31
f
32:32
and the variables t and q which are
32:35
involved in f we want to to ask
32:38
is this actually
32:42
true at least up to
32:45
terms of degree five
32:48
in both t and
32:51
q okay that's a really good question
32:54
isn't it
32:56
okay so what we're going to do is we're
32:57
going to take uh let's say take the
32:59
right-hand side here 1 plus t
33:01
f squared plus q f cubed now if we just
33:03
expand this out we're going to get all
33:04
kinds of additional terms because
33:06
we're going to get lots of terms which
33:08
are degree higher
33:10
than 5 in q and t so the way i'm going
33:12
to cut that down i'm going to do it in
33:14
two steps
33:15
um because like you can only have to do
33:18
it in two steps here using this program
33:20
so power series so let's expand this out
33:23
number of terms maybe six let's say
33:27
in terms of q all right expand this out
33:29
in terms of q
33:32
and now i'm going to compute expand this
33:36
out okay get rid of this oh
33:39
this all just means that there's higher
33:41
degree terms and now i'm going to cut
33:43
out all the t terms which are higher so
33:45
power series now in
33:46
t t
33:50
where is it it's down here okay so this
33:51
is what i get
33:55
get this thing here and um plus
33:58
oh t the six let's expand this out
34:02
get rid of this
34:05
and compute expand
34:09
all right so that thing there
34:13
that's what we got right that's what the
34:14
right hand side and now i'm going to
34:18
set this equal to f so i've taken the
34:21
right-hand side i've expanded it out
34:22
i've
34:22
chopped off all terms of degree higher
34:26
than 5 in both
34:27
t and q i'm hoping that this expression
34:29
that i have in front of me is actually
34:31
equal to f
34:32
and i can do that by going to compute
34:34
and check equality
34:36
ladies and gentlemen it is true at one
34:39
stroke
34:40
we establish that these
34:43
these numbers that are appearing here
34:45
really
34:46
are indeed what we think they are
34:49
because
34:51
let you think about it this
34:54
there's real really kind of a uniqueness
34:58
here
34:58
once you have this this thing satisfied
35:01
and you know
35:02
that the first row and the first column
35:04
are are what they should be
35:06
then there's a kind of a uniqueness to
35:08
everything so what this actually
35:10
does convince us is that at least up to
35:12
five okay
35:13
all these terms that we see here in this
35:15
array they do really count
35:17
the number of buy try
35:20
roofed polygons of the appropriate size
35:23
this 10010 is no longer just a
35:26
conjectured number
35:28
of um of 11 gons
35:31
with uh i think three triangles and
35:34
three quadrilaterals it is now
35:35
confirmed we have computed and unchecked
35:39
this
35:40
okay i want you to appreciate how how
35:42
efficient this
35:43
is right compared to to try to count
35:46
these things uh directly
35:48
and basically we've just used some sort
35:49
of soft uh soft thinking we've just
35:52
exploited the algebraic structure of
35:55
this m set
35:56
of by roofed polygons
36:00
and we're quite excited because we can
36:02
imagine ramping this up
36:04
quite considerably