WEBVTT
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PROFESSOR: I have
our final exam
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schedule from the registrar.
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It comes Tuesday morning
of exam week.
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It's on the web page.
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You won't miss it.
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I want to remind you that the
midterm exam is on Wednesday,
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March 16, that it starts at 9
not at 9:30 so that you're
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less time-limited than you
would be otherwise.
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It's a two-hour exam.
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And it will cover basically up
through chapter eight, which
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is Reed-Solomon codes.
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And you're responsible for
anything that's been
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discussed in class.
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If we haven't discussed
it in class, then
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don't worry about it.
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All right.
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Any questions about any
of those things?
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It's 9 o'clock to 11 o'clock.
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What's the underlined?
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This line up here?
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This says, "This class
goes from 9:30
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to 11.00." Not 11:15.
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That's for me, not for you,
although I've tried to lay off
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some of the responsibility
on you.
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OK, let's continue.
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I hope to finish up chapter
six today, might not
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completely finish it.
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There are really three
topics left to go.
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One is the orthogonality
and inner product
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topic, which I skipped.
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The second is Reed-Muller
codes, which is our main
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objective in this chapter,
a family of useful codes.
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And the last topic is why
making hard decisions
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is not a good idea.
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And so I'll try to say
as much as I can
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about those three things.
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Just to remind you of where
we are, we're in the
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power-limited regime.
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We're trying to design good,
small-signal constellations,
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or moderate-sized signal
constellations now, with
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nominal spectral efficiency,
less than two bits per two
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dimensions.
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The technique we're using is
we're going to start from a
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binary linear block code
in Hamming space.
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And we're going to take the
Euclidean image of that and
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hope it's a good constellation
in Euclidean space.
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And as we were just talking
about before class, the fact
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that a linear code is a subspace
of F2 to the n means
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it's true if and only if,
really, it has the group
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property, which in Euclidean
space leads to a geometrical
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uniformity property.
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We haven't proved that
in its full scope.
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But we have noticed that from
every code word, every code
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word has the same distance
profile to
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all other code words.
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And in particular, the minimum
distance is the minimum weight
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of any non-zero code word.
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That's the main juice we've
squeezed out of
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that at this point.
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So we've now been talking a
little bit about the algebra
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of binary linear block codes.
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We've characterized them
basically by three parameters,
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n, k, d, where n is the length
of the code, k is the
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dimension of the code, d is the
minimum Hamming distance
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of the code.
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In the literature, this is what
you'll mainly find, in
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the n, k, d code.
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We have a subsidiary parameter,
the number of words
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of minimum distance d, which
we're going to need to get the
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error coefficient, or a
union-bound expression.
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This has less prominence
in the literature.
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Given just these numbers, we get
a couple of key parameters
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of our constellation.
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One is its nominal spectral
efficiency, which is 2k over n
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bits per two dimension,
upper bounded by 2.
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Of course, often in the coding
literature, you talk about in
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a more natural quantity,
which is k over n,
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called the code rate.
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Maybe I shouldn't call it cap-R,
because R is used for
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the code rate in bits
per second.
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This means bit per symbol.
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So just let me call this rate.
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But when you're reading the
coding literature, the rate of
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the code means how many
information bits per how many
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transmitted bits.
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And for n, k, d binary linear
block code, it's k over m.
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And the nominal spectral
efficiency is just twice that,
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since we measure it per
two dimensions.
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OK, and even more importantly,
we get the union bound
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estimate in terms of a couple
of simple parameters.
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It's just an error coefficient,
Kb, the number of
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nearest neighbors per bit
times this Q function
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expression, which always has 2Eb
over N_0 in it, multiplied
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by this multiplicative factor,
which we call the coding gain,
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which is just kd over n.
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And this is equal to one
for a 1,1,1 code.
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Now you know coding.
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So anything else?
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Our basic effort is to get
a larger coding gain by
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constructing more complicated
codes.
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This parameter here, which we
need in order to actually plot
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the curve and estimate the
effective coding gain -- the
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effective coding gain is derated
from the nominal
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coding gain by a rule of thumb,
which depends on Kb.
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It's every factor of 2 in Kb
costs you 0.2 dB, roughly in
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the right range, if it's not too
big, all those qualifiers.
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But it's a good engineering
rule of thumb.
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And that's just the number of
nearest neighbors per code
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word divided by k.
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And so our object here is to
see how well we can do.
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And the Reed-Muller codes will
be an infinite family of codes
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that give us a pretty good idea
of what can be achieved
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for a variety of n, k, d that
kind of cover the waterfront.
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That's why I talk about
them first.
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They're all so very simple.
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OK, but first, we forgot to talk
about orthogonality and
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inner products and duality, both
from a geometric point of
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view and from an algebraic
point of view.
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And so I want to go back
and recover that.
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The definition of an inner
product between x and y, where
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x and y are both binary
n-tuples, is, as you'd expect,
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a sort of dot product
expression, a component-wise
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product, the sum over k of xk
yk, where all the arithmetic
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here is in the binary
field, F2.
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And well, this clearly has the
bilinearity properties that
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you expect of an inner product
that's linear in x for a fixed
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y or vice versa.
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But it doesn't turn out to have
the geometric properties
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that you expect.
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We can define orthogonality,
this is another definition, in
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the usual way.
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x and y are said to
be orthogonal --
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do that better.
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x is said to be orthogonal to
y if and only if their inner
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product is 0, which is the same
definition you know from
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real and complex
vector spaces.
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OK.
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But the overall moral I want you
to get from this is while
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this inner product behaves
absolutely as you expect in an
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algebraic sense, it behaves very
different from what you
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expect in a geometric sense.
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So the algebra's fine.
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The geometry is screwy.
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All right.
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That's the catch word
to keep in mind.
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Why is that?
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Where does this inner
product live?
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It's F2 valued, right?
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I'm just doing this sum in
binary space, and the result
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is either 0 or 1.
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So when are two n-tuples
orthogonal?
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Simply if they have an even
number of places in which
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they're both equal to 1.
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Is there a question?
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In particular, suppose I try to
define a norm in the usual
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way, like that.
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Is that going to have
the properties that
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I'd want of a norm?
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No.
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Why?
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Because one of the basic
properties we want of a norm
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is strict positivity, that the
norm of x is equal to 0 if and
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only if x is equal to 0.
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That's clearly not true here.
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What's the requirement for the
inner product of x with itself
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to be equal to 0, in
other words x to be
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orthogonal with itself?
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It just simply has to have
an even number of 1's.
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If it has an even number
of 1's, then this
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so-called norm is 0.
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If there's an odd
number, it's 1.
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So it's perfectly possible for
a vector to be orthogonal to
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itself, a non-zero vector to
be orthogonal to itself.
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And that's basically where all
the trouble comes from in a
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geometric sense.
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AUDIENCE: So which of them would
be [INAUDIBLE] here?
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PROFESSOR: It's mod-2.
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It's in F2.
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All the arithmetic rules
are from F2 which is
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mod-2 rules, correct.
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So, good.
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This means at the most
fundamental level, we don't
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have a Hilbert space here.
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We don't have a projection
theorem.
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Projection theorem is the basic
tool we use in Euclidean
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spaces, more generally,
Hilbert spaces.
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Just say that every vector
can be partitioned.
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In a given space and its
orthogonal space, we can
00:11:19.050 --> 00:11:22.070
express a vector as the sum of
its projection onto the space
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and the projection onto the
orthogonal space, which are
00:11:24.300 --> 00:11:25.980
two orthogonal vectors.
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So it's an orthogonal
decomposition.
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So we have nothing like the
projection theorem here.
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Therefore we have nothing like,
we don't necessarily
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have orthonormal or even
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orthogonal basis for subspaces.
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AUDIENCE: [INAUDIBLE]?
00:11:51.470 --> 00:11:53.510
PROFESSOR: You do have a unique
orthogonal complement.
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I'll get to that in a second.
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But for instance, we might have
that a subspace can be
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orthogonal to itself.
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That's what the problem is.
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For instance, 0, 0, and 1,
1 is a nice, little,
00:12:24.860 --> 00:12:26.330
one-dimensional subspace.
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And what's it's orthogonal
subspace?
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It's itself.
00:12:32.830 --> 00:12:37.120
We may not have an orthogonal
basis for a subspace.
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And for an example of that, I'll
give you our favorite 3,
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2, 2 code, as I'll
now call it.
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It consists of these
four code words.
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A set of generators for this
code consists of any two of
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the non-zero code words.
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You can generate all of the code
words as binary linear
00:12:56.830 --> 00:12:59.490
combinations of any
two of these.
00:12:59.490 --> 00:13:02.900
But no two of these
are orthogonal.
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So there's clearly
no orthogonal
00:13:04.370 --> 00:13:06.090
basis for that code.
00:13:06.090 --> 00:13:09.720
We shouldn't expect to find
orthogonal basis, orthogonal
00:13:09.720 --> 00:13:13.095
decomposition, so all of these
sorts of tools that we relied
00:13:13.095 --> 00:13:16.765
on heavily in 6.450 in
Euclidean spaces.
00:13:19.290 --> 00:13:23.870
OK, so this is just a great
caution to the student.
00:13:26.590 --> 00:13:30.380
Don't expect Hamming space to
have the same geometric
00:13:30.380 --> 00:13:33.490
properties as Euclidean space.
00:13:33.490 --> 00:13:33.970
Yes?
00:13:33.970 --> 00:13:35.220
AUDIENCE: [INAUDIBLE]?
00:13:39.160 --> 00:13:40.360
PROFESSOR: Nothing special.
00:13:40.360 --> 00:13:43.910
It's just finite fields have
a different geometry.
00:13:43.910 --> 00:13:47.570
In F2, there's really only one
geometry you would impose,
00:13:47.570 --> 00:13:48.765
which is the Hamming geometry.
00:13:48.765 --> 00:13:53.010
In other finite fields, there
actually could be more than
00:13:53.010 --> 00:13:55.910
one geometry.
00:13:55.910 --> 00:14:01.510
But let's say the algebraic
properties, however, you can
00:14:01.510 --> 00:14:02.760
still count on.
00:14:13.820 --> 00:14:26.040
For instance, n, k, d
code C has, as we've
00:14:26.040 --> 00:14:30.346
already shown, a basis.
00:14:30.346 --> 00:14:31.795
It has k dimensions.
00:14:31.795 --> 00:14:44.670
It has a basis g1 up to gk of
k, linearly independent,
00:14:44.670 --> 00:14:49.900
though not necessarily
orthogonal basis vectors.
00:14:49.900 --> 00:14:59.910
So we can always write the code
as the set of all u g
00:14:59.910 --> 00:15:06.590
such that u is a k-tuple of
information bits, let's say.
00:15:06.590 --> 00:15:10.740
In other words, this is a
compressed form for the set of
00:15:10.740 --> 00:15:16.820
all binary linear combinations
of these generators, where
00:15:16.820 --> 00:15:27.960
I've written what's called a
generator matrix, g as a k by
00:15:27.960 --> 00:15:33.105
n matrix, whose rows are
these generators.
00:15:37.160 --> 00:15:41.570
And I've multiplied on the left
with a row vector u, if
00:15:41.570 --> 00:15:46.670
I'm doing it in matrix terms.
00:15:46.670 --> 00:15:49.010
You can do this more abstractly
just as a linear
00:15:49.010 --> 00:15:51.620
transformation.
00:15:51.620 --> 00:15:56.100
And side comment, notice that
in coding theory, it's
00:15:56.100 --> 00:15:58.860
conventional to write vectors
as row vectors, whereas in
00:15:58.860 --> 00:16:01.720
every other subject you take,
it's conventional to write
00:16:01.720 --> 00:16:05.140
vectors as column vectors.
00:16:05.140 --> 00:16:05.880
Why is this?
00:16:05.880 --> 00:16:08.170
Is there some deep,
philosophical reason?
00:16:08.170 --> 00:16:08.660
No.
00:16:08.660 --> 00:16:11.900
It's just the way people started
to do it in coding
00:16:11.900 --> 00:16:14.360
theory back at the beginning,
and then everybody has
00:16:14.360 --> 00:16:15.430
followed them.
00:16:15.430 --> 00:16:18.850
I may have even had something
to do with this myself.
00:16:18.850 --> 00:16:21.550
And I'll tell you the reason I
like to write vectors as row
00:16:21.550 --> 00:16:23.222
vectors is I don't have
to write a little
00:16:23.222 --> 00:16:25.350
t up next to them.
00:16:25.350 --> 00:16:27.820
That's the deep philosophical
reason.
00:16:30.480 --> 00:16:32.860
It's like driving on the right
side or the left side.
00:16:32.860 --> 00:16:35.830
Obviously, you could have chosen
how to do it one way or
00:16:35.830 --> 00:16:37.090
another back in the beginning.
00:16:37.090 --> 00:16:41.560
But once you've chosen it,
you better stick with it.
00:16:41.560 --> 00:16:49.550
There's actually some insight
involved here.
00:16:49.550 --> 00:16:52.210
This is sort of associated with
a block diagram, where we
00:16:52.210 --> 00:16:55.700
take k bits, and we run
it through this linear
00:16:55.700 --> 00:16:57.360
transformation.
00:16:57.360 --> 00:17:05.819
And as a result, we get out,
what shall I write, x n bits.
00:17:05.819 --> 00:17:09.599
And in block diagrams, we tend
to take the input bits from
00:17:09.599 --> 00:17:12.940
the left and proceed to
the right, left to
00:17:12.940 --> 00:17:14.050
right kind of thing.
00:17:14.050 --> 00:17:21.450
So this formula reflects this
left to right picture, whereas
00:17:21.450 --> 00:17:26.560
I'd say the deeper reason why
you usually see G u in system
00:17:26.560 --> 00:17:31.710
theory is that G is regarded
as an operator
00:17:31.710 --> 00:17:33.190
that operates on u.
00:17:33.190 --> 00:17:36.750
It's kind of a G of u.
00:17:36.750 --> 00:17:41.240
And so that's why it's more
natural to think of this as
00:17:41.240 --> 00:17:44.500
being a column vector,
because then, we
00:17:44.500 --> 00:17:46.230
have a different picture.
00:17:46.230 --> 00:17:49.270
G is the operator that
somehow transforms u.
00:17:49.270 --> 00:17:51.990
But these are all side
comments, obviously.
00:17:51.990 --> 00:17:56.110
They don't really matter
for anything.
00:17:56.110 --> 00:18:05.645
All right, just remember that
vectors are row vectors.
00:18:09.030 --> 00:18:11.120
OK.
00:18:11.120 --> 00:18:17.720
Let's define the dual code in
the natural way, as the
00:18:17.720 --> 00:18:18.970
orthogonal code.
00:18:24.590 --> 00:18:27.230
Say C dual.
00:18:27.230 --> 00:18:44.840
The definition is that C dual is
the set of all n-tuples y,
00:18:44.840 --> 00:18:49.330
such that x, the inner product
between x and y is 0.
00:18:49.330 --> 00:18:53.130
In other words, y is orthogonal
to x for all the x
00:18:53.130 --> 00:18:57.470
and C. It's the set of all
n-tuples that are orthogonal
00:18:57.470 --> 00:19:02.460
to all the words in C under
our F2 definition of inner
00:19:02.460 --> 00:19:05.390
product orthogonality.
00:19:05.390 --> 00:19:10.170
OK, so that's a natural
definition.
00:19:10.170 --> 00:19:16.270
For example, as I've already
shown you, if C is 0, 0, 1, 1,
00:19:16.270 --> 00:19:20.860
then C dual is 0, 0, 1, 1.
00:19:20.860 --> 00:19:27.860
If C were 0, 0, 0, 1, then C
dual would be 0, 0, what?
00:19:31.400 --> 00:19:32.020
1, 0.
00:19:32.020 --> 00:19:33.990
Thank you.
00:19:33.990 --> 00:19:38.670
Because 1, 1 is not orthogonal
to this, 0, 1 and is not
00:19:38.670 --> 00:19:40.420
orthogonal to this.
00:19:40.420 --> 00:19:44.170
So we simply go through
and pick it out.
00:19:44.170 --> 00:19:49.590
Now as I said, the algebraic
properties of the
00:19:49.590 --> 00:19:52.608
dual code are OK.
00:19:52.608 --> 00:19:56.560
I emphasize they're algebraic.
00:19:56.560 --> 00:20:07.350
If C is an n, k code, in other
words, has dimension k, what
00:20:07.350 --> 00:20:10.070
do you expect the parameters
of C dual to be?
00:20:13.610 --> 00:20:16.155
Its length is what?
00:20:16.155 --> 00:20:17.535
It's n, of course.
00:20:20.070 --> 00:20:22.160
And what's its dimension
going to be?
00:20:22.160 --> 00:20:25.880
If C has dimension k, what do
you guess the dimension of C
00:20:25.880 --> 00:20:27.130
dual is going to be?
00:20:31.350 --> 00:20:34.060
Just guess from Euclidean
spaces, or any --
00:20:34.060 --> 00:20:35.440
it's n minus k.
00:20:35.440 --> 00:20:39.420
The dual space has dimension
n minus k.
00:20:39.420 --> 00:20:40.670
And that holds.
00:20:43.130 --> 00:20:45.930
In the notes, I give two
proofs for this.
00:20:45.930 --> 00:20:49.550
There's the conventional coding
theory textbook proof,
00:20:49.550 --> 00:20:53.920
which involves writing down
a generator matrix for C k
00:20:53.920 --> 00:20:56.950
generators, reducing it to a
canonical form, called the
00:20:56.950 --> 00:21:01.810
systematic form, where there's
some k by k identity matrix
00:21:01.810 --> 00:21:09.490
and a n minus k by k
parity check part.
00:21:09.490 --> 00:21:12.550
Then, by inspection, you can
write down a generator matrix
00:21:12.550 --> 00:21:13.650
for C dual.
00:21:13.650 --> 00:21:17.360
And you find it has dimension
n minus k.
00:21:17.360 --> 00:21:20.660
It's kind of a klutzy proof,
in my opinion.
00:21:20.660 --> 00:21:23.200
A second, more elegant proof,
but one that you don't have
00:21:23.200 --> 00:21:27.740
the background for yet, is to
use simply the fundamental
00:21:27.740 --> 00:21:28.990
theorem of homomorphisms.
00:21:31.510 --> 00:21:33.650
This is in some sense
an image.
00:21:33.650 --> 00:21:37.440
This is the dual of an image,
which is a kernel.
00:21:37.440 --> 00:21:40.800
You work out the dimensions
from that.
00:21:40.800 --> 00:21:47.335
I am still in search of an
elegant, elementary proof.
00:21:47.335 --> 00:21:52.780
And anyone who can come up
with a proof suitable for
00:21:52.780 --> 00:21:54.625
chapter six gets a gold star.
00:21:57.880 --> 00:22:03.160
Believe me, the gold star will
be very valuable to you.
00:22:03.160 --> 00:22:07.110
OK, so exercise for any student
so inclined, give me a
00:22:07.110 --> 00:22:08.000
nice proof of this.
00:22:08.000 --> 00:22:10.930
It's surprisingly hard.
00:22:10.930 --> 00:22:14.480
And I can't say I've spent great
quantities of my life on
00:22:14.480 --> 00:22:17.780
it, but I've spent
some time on it.
00:22:17.780 --> 00:22:23.970
So anyway, the dimensions come
out as I hope you would
00:22:23.970 --> 00:22:27.010
expect, based on your
past experience.
00:22:27.010 --> 00:22:31.510
And I won't give you
a proof in class.
00:22:31.510 --> 00:22:36.330
You get the fundamental duality
relationship, that the
00:22:36.330 --> 00:22:39.600
dual of C dual, what would
you expects that to be?
00:22:42.360 --> 00:22:45.970
C. OK.
00:22:45.970 --> 00:22:53.020
In words, C is the set of all
n-uples that are orthogonal to
00:22:53.020 --> 00:22:56.600
all the n-tuples in C dual.
00:22:59.870 --> 00:22:59.970
OK.
00:22:59.970 --> 00:23:04.610
So this actually means that I
can specify C. If I know C
00:23:04.610 --> 00:23:10.420
dual, I know C. Give me C dual,
the set of all code
00:23:10.420 --> 00:23:14.280
words orthogonal to it
tells me what C is.
00:23:14.280 --> 00:23:23.100
So this implies that I can write
C in the following form.
00:23:23.100 --> 00:23:33.260
C is the set of all, just
emulating this, y n F2 to the
00:23:33.260 --> 00:23:43.780
n such that x, y equals zero
for all x in C dual.
00:23:43.780 --> 00:23:48.470
I probably should have
interchanged x and y for this.
00:23:48.470 --> 00:23:51.666
x such that x, y equals
0 for all --
00:23:55.450 --> 00:23:56.265
this is symmetrical.
00:23:56.265 --> 00:23:58.880
The inner product of x and y is
equal to the inner product
00:23:58.880 --> 00:23:59.590
of y and x.
00:23:59.590 --> 00:24:03.600
So I don't care how
I write it.
00:24:03.600 --> 00:24:08.250
OK, so I can actually
specify a code by a
00:24:08.250 --> 00:24:12.010
set of parity checks.
00:24:12.010 --> 00:24:23.290
Now suppose I have a generator
matrix, call it H, namely a
00:24:23.290 --> 00:24:33.670
set of generators, H1 through
Hn minus k, for C dual.
00:24:33.670 --> 00:24:37.832
I'm going to have a set of
n minus k generators.
00:24:37.832 --> 00:24:42.580
Then I hope it's obvious that
I can test whether x is
00:24:42.580 --> 00:24:50.580
orthogonal to all of C dual by
just checking whether it's
00:24:50.580 --> 00:24:54.720
orthogonal to all of
these generators.
00:24:54.720 --> 00:25:04.060
So I would now have C is the set
of all x n-tuples x such
00:25:04.060 --> 00:25:12.560
that x, H, j equals
zero, all j.
00:25:12.560 --> 00:25:12.880
OK.
00:25:12.880 --> 00:25:16.300
I've just got to test
orthogonality to each of these
00:25:16.300 --> 00:25:18.290
generators.
00:25:18.290 --> 00:25:22.390
In other words, in each
of these is what we
00:25:22.390 --> 00:25:25.400
call a parity check.
00:25:25.400 --> 00:25:29.040
We take the inner product of x
with a certain n-tuple, and we
00:25:29.040 --> 00:25:31.600
ask whether parity checks.
00:25:31.600 --> 00:25:34.350
In other words, we ask whether
the subset of positions in
00:25:34.350 --> 00:25:39.160
which H, j is equal to 1, in
those positions, whether x has
00:25:39.160 --> 00:25:42.610
an even number of 1's.
00:25:42.610 --> 00:25:46.540
Get very concrete about it.
00:25:46.540 --> 00:25:55.752
So writing this out in matrix
form, the test is C is the set
00:25:55.752 --> 00:26:05.990
of x in F2 to the n such that
x h-transpose equals 0.
00:26:08.750 --> 00:26:12.600
That's just a matrix form of
what I've written up there.
00:26:12.600 --> 00:26:15.810
So let me picture
it like this.
00:26:15.810 --> 00:26:23.380
Here the test is, I take
x, which is n bits.
00:26:23.380 --> 00:26:26.320
I put it into what's
called a parity
00:26:26.320 --> 00:26:29.060
checker or syndrome reformer.
00:26:29.060 --> 00:26:32.990
And I ask whether this is 0.
00:26:32.990 --> 00:26:36.470
We call this, in general,
the syndrome.
00:26:36.470 --> 00:26:39.420
And we ask whether it's 0.
00:26:39.420 --> 00:26:42.170
If we get a 0, we say
x is in the code.
00:26:42.170 --> 00:26:44.300
If it's not 0, then x
is not in the code.
00:26:44.300 --> 00:26:45.550
That's the test.
00:26:50.420 --> 00:27:02.070
So when we summarize, we really
have two dual ways of
00:27:02.070 --> 00:27:08.750
characterizing a code, which
you will see in the
00:27:08.750 --> 00:27:10.000
literature.
00:27:12.570 --> 00:27:19.980
We have a generator matrix, we
might give a k by n generator
00:27:19.980 --> 00:27:23.280
matrix for the code.
00:27:23.280 --> 00:27:33.350
And then the code is specified
as C is the set of all U g
00:27:33.350 --> 00:27:43.170
such that U n F2 to the k.
00:27:43.170 --> 00:27:46.380
In other words, this is an
image representation.
00:27:46.380 --> 00:27:49.185
We take C as the image of
a linear transformation.
00:27:52.570 --> 00:27:54.700
We call G a linear
transformation.
00:27:54.700 --> 00:28:04.840
It goes from Fk to F2 to
the k to F2 to the n.
00:28:04.840 --> 00:28:07.150
And then the code is simply
the image of this
00:28:07.150 --> 00:28:09.570
transformation algebraically.
00:28:12.850 --> 00:28:13.410
OK.
00:28:13.410 --> 00:28:21.210
Or we can specify it by means
of a parity check matrix, H,
00:28:21.210 --> 00:28:24.490
which is the generator matrix
of the dual code.
00:28:24.490 --> 00:28:27.660
So this would be the parity
check matrix for the dual code
00:28:27.660 --> 00:28:32.020
G. h is the generator matrix
of the dual code.
00:28:32.020 --> 00:28:35.960
And we ask whether --
00:28:35.960 --> 00:28:44.960
now we specify it as I have up
here, simply the x and F2 to
00:28:44.960 --> 00:28:51.280
the n such that x
H_t equals 0.
00:28:51.280 --> 00:28:57.410
And this is what's called a
kernel representation, because
00:28:57.410 --> 00:29:04.450
it's the kernel of a linear
transformation defined by H2
00:29:04.450 --> 00:29:11.490
from F2 to the end, down
to, this is a m
00:29:11.490 --> 00:29:13.810
by n minus k matrix.
00:29:13.810 --> 00:29:18.750
So the syndrome is actually
an n-minus-k-tuple.
00:29:18.750 --> 00:29:22.000
The elements of the syndrome are
the n minus k individual
00:29:22.000 --> 00:29:23.547
bits, parity check bits.
00:29:23.547 --> 00:29:24.797
OK.
00:29:29.210 --> 00:29:33.640
And sometimes it's more
convenient to characterize the
00:29:33.640 --> 00:29:36.620
code in one way, and sometimes
it's more convenient to
00:29:36.620 --> 00:29:39.650
characterize it in
the other way.
00:29:39.650 --> 00:29:47.560
For instance, we were talking
last time about the n, n minus
00:29:47.560 --> 00:29:53.280
1, 2 single parity check code,
or the even weight code, the
00:29:53.280 --> 00:29:55.640
set of all even weight
n-tuples, which has
00:29:55.640 --> 00:29:57.880
dimension n minus 1.
00:29:57.880 --> 00:30:02.710
And in general, for high-rate
codes especially, it may be
00:30:02.710 --> 00:30:06.890
simpler to give the parity
check representation.
00:30:06.890 --> 00:30:08.525
What is the dual code?
00:30:08.525 --> 00:30:11.970
Let's call this C.
C dual is what?
00:30:19.490 --> 00:30:25.024
C dual is what we call the
n, 1, n repetition code.
00:30:25.024 --> 00:30:27.830
In other words, it has
dimension one.
00:30:27.830 --> 00:30:31.750
It has two code words, the all-0
word and the all-1 word.
00:30:35.290 --> 00:30:40.470
Clearly, the all-1 word is
orthogonal to all of the even
00:30:40.470 --> 00:30:41.450
weight words.
00:30:41.450 --> 00:30:45.090
And vice versa, a word is even
weight if and only if it's
00:30:45.090 --> 00:30:48.620
orthogonal to all 1's.
00:30:48.620 --> 00:30:53.320
So in this case, what is
the generator matrix?
00:30:53.320 --> 00:30:57.560
We had a generator matrix last
time consisting of n minus 1
00:30:57.560 --> 00:31:01.080
weight 2 code words all arranged
in a kind of double
00:31:01.080 --> 00:31:02.300
diagonal pattern.
00:31:02.300 --> 00:31:03.970
That's OK.
00:31:03.970 --> 00:31:07.030
But that's an n minus
1 by n matrix.
00:31:09.850 --> 00:31:13.010
Most people would say it's
easier to say, OK, what's the
00:31:13.010 --> 00:31:15.256
parity check matrix?
00:31:15.256 --> 00:31:22.110
The parity check matrix in this
case, H, is simply a one
00:31:22.110 --> 00:31:26.950
by n matrix consisting
of all one's.
00:31:26.950 --> 00:31:31.470
And what's the characterization
of the code?
00:31:31.470 --> 00:31:34.120
The code consists of all
the words that are
00:31:34.120 --> 00:31:37.290
orthogonal to this.
00:31:37.290 --> 00:31:40.120
And that is why we call it
single parity check code.
00:31:40.120 --> 00:31:41.650
There's one parity check.
00:31:41.650 --> 00:31:44.040
And if you pass the parity
check, you're in the code.
00:31:44.040 --> 00:31:47.050
And if you don't, you're not.
00:31:47.050 --> 00:31:50.850
OK, so we see that these,
first of all, here's an
00:31:50.850 --> 00:31:52.900
example of dual codes.
00:31:52.900 --> 00:31:54.580
Are their dimensions correct?
00:31:54.580 --> 00:31:55.460
They are.
00:31:55.460 --> 00:31:58.420
Is each on characterized
correctly as
00:31:58.420 --> 00:31:59.380
the dual of the other?
00:31:59.380 --> 00:32:00.570
It is.
00:32:00.570 --> 00:32:03.800
So we've passed that.
00:32:03.800 --> 00:32:11.810
This is intended to make point
C. So it's a good example.
00:32:16.470 --> 00:32:24.130
One final thing is suppose I
have a code with generator
00:32:24.130 --> 00:32:30.880
matrix G and another code with
generator matrix H. Are they
00:32:30.880 --> 00:32:34.280
each other's dual
codes are not?
00:32:34.280 --> 00:32:37.570
And the answer is pretty obvious
from all of this.
00:32:37.570 --> 00:32:39.810
Yes, they are.
00:32:39.810 --> 00:32:42.870
Let me a substitute in here,
x is supposed to
00:32:42.870 --> 00:32:44.020
be equal to U g.
00:32:44.020 --> 00:32:58.650
So another requirement is that
UGH_t equals zero for all u.
00:32:58.650 --> 00:33:08.400
And without belaboring the
point, the 5, 2 codes with
00:33:08.400 --> 00:33:14.320
generator matrix G and H, they
are dual codes if and only if
00:33:14.320 --> 00:33:17.150
they have the right dimension,
one is n, k, and the other is
00:33:17.150 --> 00:33:22.800
n, n minus k, and we satisfy
G H_t equals zero.
00:33:22.800 --> 00:33:25.810
Basically, this is the matrix
of inner products of the
00:33:25.810 --> 00:33:31.470
generators of C with the
generators of C dual.
00:33:31.470 --> 00:33:37.470
And if we have k generators
that are all orthogonal to
00:33:37.470 --> 00:33:40.870
these n minus k generators,
then they must be the
00:33:40.870 --> 00:33:43.410
generators of dual codes.
00:33:43.410 --> 00:33:47.060
That's is kind of intuitive
and natural.
00:33:47.060 --> 00:33:51.520
All right, so again, this is a
concise form that you would
00:33:51.520 --> 00:33:53.290
actually, probably most
commonly find in the
00:33:53.290 --> 00:33:55.730
literature.
00:33:55.730 --> 00:33:58.620
It's not hard to get to.
00:33:58.620 --> 00:34:05.100
OK, so in this course, we're
probably going to talk quite a
00:34:05.100 --> 00:34:07.540
bit about orthogonality.
00:34:07.540 --> 00:34:10.920
Duality is very powerful, but
we're not going to be using it
00:34:10.920 --> 00:34:13.460
very much in this course,
I believe.
00:34:13.460 --> 00:34:16.150
I'll mention duality
whenever there's a
00:34:16.150 --> 00:34:18.560
duality property to mention.
00:34:18.560 --> 00:34:20.310
But in general, I'm not going
to spend an awful
00:34:20.310 --> 00:34:22.870
lot of time on it.
00:34:22.870 --> 00:34:25.480
But it's important you know
about it, particularly if you
00:34:25.480 --> 00:34:28.380
were going to go on and do
anything in this subject.
00:34:28.380 --> 00:34:33.540
And at an elementary level, it's
nice to know that we have
00:34:33.540 --> 00:34:36.340
two possible representations,
and one is often going to be
00:34:36.340 --> 00:34:37.360
simpler than the other.
00:34:37.360 --> 00:34:39.032
Use the simple one.
00:34:39.032 --> 00:34:42.420
In general, use the
representation for the
00:34:42.420 --> 00:34:44.540
low-rate code to determine
the high-rate code.
00:34:48.350 --> 00:34:48.900
OK.
00:34:48.900 --> 00:34:50.214
Any questions on this?
00:34:50.214 --> 00:34:51.930
I've now finished up
the preparatory.
00:34:51.930 --> 00:34:52.579
Yeah?
00:34:52.579 --> 00:34:54.710
AUDIENCE: So you're saying
that every basis for
00:34:54.710 --> 00:34:57.160
[INAUDIBLE]?
00:34:57.160 --> 00:34:59.080
PROFESSOR: That's necessary
and sufficient, right.
00:35:02.550 --> 00:35:06.290
OK, Reed-Muller codes.
00:35:06.290 --> 00:35:08.380
Why do I talk about
Reed-Muller codes?
00:35:08.380 --> 00:35:12.480
First of all, they give us an
infinite family of codes, so
00:35:12.480 --> 00:35:17.020
we can see what happens as n
gets large, as k goes from 0
00:35:17.020 --> 00:35:22.313
to n, whose parameters are
sort of representative.
00:35:25.200 --> 00:35:32.900
They aren't necessarily the best
codes that we know of.
00:35:32.900 --> 00:35:37.190
They're very simple, as I will
show, to construct and to
00:35:37.190 --> 00:35:41.660
characterize their parameters,
so we can do all the proofs in
00:35:41.660 --> 00:35:43.950
half an hour here.
00:35:43.950 --> 00:35:46.310
And they're not so bad.
00:35:46.310 --> 00:35:51.060
In terms of the parameters n,
k, d, the Reed-Muller codes,
00:35:51.060 --> 00:35:55.870
up to length 32, are the best
ones we know of, at least for
00:35:55.870 --> 00:35:57.720
their parameters.
00:35:57.720 --> 00:36:03.690
There is no 32, 16 code that's
better than a 32, 16, eight
00:36:03.690 --> 00:36:06.050
Reed-Muller code.
00:36:06.050 --> 00:36:10.990
There's no 32k, eight code that
has k greater than 16,
00:36:10.990 --> 00:36:13.140
all those sorts of things.
00:36:13.140 --> 00:36:17.060
Actually, I'm not 100%
sure of that.
00:36:17.060 --> 00:36:23.620
So for short block lengths,
they're as good as BCH codes,
00:36:23.620 --> 00:36:28.510
or any of the codes that were
discovered subsequently.
00:36:28.510 --> 00:36:37.790
For even longer lengths, up to
64, 128, 256, they are going
00:36:37.790 --> 00:36:42.040
to be slightly sub-optimal in
terms of n, k, d, as we'll see
00:36:42.040 --> 00:36:44.080
in some cases.
00:36:44.080 --> 00:36:48.060
But they still are very good
codes to look at in terms of
00:36:48.060 --> 00:36:50.460
performance versus complexity.
00:36:50.460 --> 00:36:53.190
The decoding algorithm that
I'm going to talk about
00:36:53.190 --> 00:36:55.690
eventually for them is a
trellis-based decoding
00:36:55.690 --> 00:36:58.950
algorithm, and a maximum
likelihood
00:36:58.950 --> 00:37:02.230
decoding algorithm within.
00:37:02.230 --> 00:37:06.710
They can be maximum likelihood
decoded very simply by these
00:37:06.710 --> 00:37:09.700
trellis-based algorithms.
00:37:09.700 --> 00:37:14.880
And that's not true of more
elaborate classes of codes.
00:37:14.880 --> 00:37:17.330
So from a performance versus
complexity point of view,
00:37:17.330 --> 00:37:20.710
they're still good codes to look
at for block lengths up
00:37:20.710 --> 00:37:22.640
to 100, 200.
00:37:22.640 --> 00:37:25.460
As we get up to higher block
lengths, then we'll be talking
00:37:25.460 --> 00:37:28.560
about much more random-like,
non-algebraic codes in the
00:37:28.560 --> 00:37:30.400
final section of the course.
00:37:30.400 --> 00:37:32.390
This is the way you actually
get to pass it.
00:37:32.390 --> 00:37:34.210
You don't worry about n, k, d.
00:37:34.210 --> 00:37:36.640
Right now, we're talking about
moderate complexity, moderate
00:37:36.640 --> 00:37:39.310
performance.
00:37:39.310 --> 00:37:47.690
OK, so they were invented in
1954 independently by Irving
00:37:47.690 --> 00:37:53.050
Reed and, I think it was, David
Muller, D. Muller, in
00:37:53.050 --> 00:37:55.430
two separate papers,
which shows they
00:37:55.430 --> 00:37:57.340
weren't too hard to find.
00:37:57.340 --> 00:37:59.740
As you'll see, they're very
easy to construct.
00:38:03.240 --> 00:38:08.490
They're basically based on a
length-doubling construction.
00:38:12.340 --> 00:38:18.780
So we start off with codes
of length or length 2.
00:38:18.780 --> 00:38:26.040
And then from that, we build up
codes of length 4, 8, 16,
00:38:26.040 --> 00:38:27.240
32, and so forth.
00:38:27.240 --> 00:38:33.120
So in general, their lengths are
equal to a power of 2, and
00:38:33.120 --> 00:38:39.080
m is the parameter that denotes
the power of 2.
00:38:39.080 --> 00:38:42.980
So we only get certain block
lengths, which are equal to
00:38:42.980 --> 00:38:46.130
powers of 2.
00:38:46.130 --> 00:38:51.490
They have a second parameter, r,
whose significance is that
00:38:51.490 --> 00:39:02.010
d is 2 to the m minus r for 0,
less than or equal to r, less
00:39:02.010 --> 00:39:04.520
than or equal to m.
00:39:04.520 --> 00:39:06.920
Or some people, including me,
are going to put the lower
00:39:06.920 --> 00:39:11.420
limit at minus 1, just to be
able to include one more code
00:39:11.420 --> 00:39:13.800
in this family at each length.
00:39:13.800 --> 00:39:16.210
But this is just a
matter of taste.
00:39:16.210 --> 00:39:21.210
You'll see this is a very
special case, the minus 1.
00:39:21.210 --> 00:39:24.830
So let's guess what some of
these codes are going to be.
00:39:24.830 --> 00:39:33.560
So we have two parameters, m,
which can be any integer 0 or
00:39:33.560 --> 00:39:39.250
higher, so the lengths will be
1,2,4, so forth, and r, which
00:39:39.250 --> 00:39:43.940
goes basically from 0 to m,
which means the distances will
00:39:43.940 --> 00:39:53.350
go from 2 to the m down to 1.
00:39:56.000 --> 00:39:58.240
And what are some of the
basic codes that we're
00:39:58.240 --> 00:39:59.490
always going to find?
00:40:02.610 --> 00:40:06.100
We always write Rm
of little rm.
00:40:06.100 --> 00:40:10.570
I'm not sure I completely
approve of how the notation
00:40:10.570 --> 00:40:12.870
goes for these codes, but
that's the way it is.
00:40:12.870 --> 00:40:15.900
So it's this notation.
00:40:15.900 --> 00:40:19.750
That means the Reed-Muller code
with the parameters m and
00:40:19.750 --> 00:40:24.570
r is written Rm of r, m.
00:40:24.570 --> 00:40:25.030
All right.
00:40:25.030 --> 00:40:31.660
So Rm of m, m, this is going
to be a code of length 2 to
00:40:31.660 --> 00:40:34.770
the m and distance 1.
00:40:34.770 --> 00:40:37.810
What do you suppose that's
going to be?
00:40:37.810 --> 00:40:46.110
This is always going to
be the 2 to the m 1.
00:40:46.110 --> 00:40:49.090
Sorry, the distance is 1.
00:40:49.090 --> 00:40:58.230
So it's going to be the 2 to
the m, 1 universe code, in
00:40:58.230 --> 00:41:07.420
other words, simply the set of
all 2 to the m-tuples, which
00:41:07.420 --> 00:41:11.110
has Hamming distance 1.
00:41:11.110 --> 00:41:13.350
OK.
00:41:13.350 --> 00:41:23.170
RM of 0, m, what's
that going to be?
00:41:23.170 --> 00:41:27.810
This is a code now that has
length 2 to the m and minimum
00:41:27.810 --> 00:41:29.660
distance 2 to the m.
00:41:29.660 --> 00:41:31.806
We know what that has to be.
00:41:31.806 --> 00:41:39.980
It has to be the 2 to the m, 1,
2 to the m repetition code,
00:41:39.980 --> 00:41:43.255
a very low-rate code,
dimension one.
00:41:43.255 --> 00:41:44.505
OK.
00:41:47.530 --> 00:41:52.070
And then if we like, we can
go one step further.
00:41:52.070 --> 00:41:54.290
There is a code below
this code.
00:41:54.290 --> 00:41:56.990
This is the highest-rate
code you can get.
00:41:56.990 --> 00:41:59.550
This, however, is not the
lowest-rate code you can get.
00:41:59.550 --> 00:42:07.580
What's the lowest rate code
of length 2 to the m?
00:42:07.580 --> 00:42:11.990
Well, it's one that has
dimension zero and minimum
00:42:11.990 --> 00:42:14.960
distance infinity.
00:42:14.960 --> 00:42:17.220
And I don't think I ever defined
this convention.
00:42:17.220 --> 00:42:20.860
But for the trivial code that
consists of simply the all-0
00:42:20.860 --> 00:42:22.980
word, what's it's minimum
distance?
00:42:22.980 --> 00:42:26.720
Undefined, or infinity,
if you like.
00:42:26.720 --> 00:42:27.970
So this is the trivial code.
00:42:32.540 --> 00:42:36.080
So if we want, we can include
the trivial code in this
00:42:36.080 --> 00:42:39.370
family just by defining
it like this.
00:42:39.370 --> 00:42:42.120
And it works for some things.
00:42:42.120 --> 00:42:43.570
It doesn't work for
all things.
00:42:43.570 --> 00:42:47.030
It doesn't work for
d, for instance.
00:42:47.030 --> 00:42:50.540
This definition holds only
for r between 0 and m.
00:42:50.540 --> 00:42:54.080
It doesn't hold for r equals
minus 1, because there the
00:42:54.080 --> 00:42:57.280
distance is infinite.
00:42:57.280 --> 00:43:05.530
OK, so let's start out and
get even more explicit.
00:43:05.530 --> 00:43:09.220
We want to start with
m equals 0.
00:43:09.220 --> 00:43:17.140
In that case, we have only Rm of
0, 0, and Rm of minus 1, 0.
00:43:17.140 --> 00:43:22.590
And this is going to be the
one by either of these.
00:43:22.590 --> 00:43:26.200
The universe code is equal
to the repetition code.
00:43:26.200 --> 00:43:29.210
It's the 1, 1, 1 code.
00:43:29.210 --> 00:43:33.230
And this is the 1,
0, infinity code.
00:43:33.230 --> 00:43:36.960
And that's really the only two
codes that you can think of
00:43:36.960 --> 00:43:39.600
that have length 1, right?
00:43:39.600 --> 00:43:44.310
This is the one that consists
of 0 and 1, and this is the
00:43:44.310 --> 00:43:47.130
one that consists of 0.
00:43:47.130 --> 00:43:49.350
I don't think there are any
other binary linear block
00:43:49.350 --> 00:43:51.700
codes of length 1.
00:43:51.700 --> 00:43:57.210
So that's a start, not
very interesting.
00:43:57.210 --> 00:44:00.550
Let's go up to length 2.
00:44:00.550 --> 00:44:02.670
So m is going to
be equal to 1.
00:44:02.670 --> 00:44:09.300
Here, Rm of 1, 1 is going
to be length 2.
00:44:09.300 --> 00:44:12.560
And it's going to be the
universe code, so it's only
00:44:12.560 --> 00:44:14.650
going to have distance 1.
00:44:14.650 --> 00:44:19.800
Rm of 0, 1 is going to be length
2, but it's going to be
00:44:19.800 --> 00:44:21.850
the repetition code.
00:44:21.850 --> 00:44:33.010
And Rm of minus 1, 1 is going
to be 2, 0, infinity.
00:44:33.010 --> 00:44:37.140
OK, so there are really the
only three sensible
00:44:37.140 --> 00:44:38.650
codes of length 2.
00:44:38.650 --> 00:44:40.310
This is the only one
of dimension two.
00:44:40.310 --> 00:44:42.490
This is the only one
of dimension zero.
00:44:42.490 --> 00:44:44.880
There are other ones of
dimension one, but they don't
00:44:44.880 --> 00:44:47.800
have minimum distance 2, so
they're not very good for
00:44:47.800 --> 00:44:49.430
coding purposes.
00:44:49.430 --> 00:44:52.950
So this kind of lists all the
good coding codes of length 2.
00:44:55.520 --> 00:44:56.320
All right.
00:44:56.320 --> 00:45:02.920
So now let me introduce the
length-doubling construction.
00:45:05.490 --> 00:45:08.630
Let me make the point, first of
all, that all these codes
00:45:08.630 --> 00:45:11.030
are nested.
00:45:11.030 --> 00:45:12.360
What does that mean?
00:45:12.360 --> 00:45:15.330
That means that his code is a
sub-code of this code, which
00:45:15.330 --> 00:45:18.210
is a sub-code of this code.
00:45:18.210 --> 00:45:20.270
And that's going to be, in
general, a property of
00:45:20.270 --> 00:45:21.720
Reed-Muller codes.
00:45:21.720 --> 00:45:23.140
We're going to get a family.
00:45:23.140 --> 00:45:25.830
And each lower one is going
to be a sub-code of the
00:45:25.830 --> 00:45:30.640
next-higher one, which is going
to be easy to prove
00:45:30.640 --> 00:45:33.170
recursively.
00:45:33.170 --> 00:45:34.620
All right.
00:45:34.620 --> 00:45:38.100
This is the key thing to know
about Reed-Muller codes, how
00:45:38.100 --> 00:45:39.510
do you construct them.
00:45:39.510 --> 00:45:44.740
Once you understand the
construction, then you can
00:45:44.740 --> 00:45:47.320
derive all the properties.
00:45:47.320 --> 00:45:49.190
It's called the u, u plus v
00:45:49.190 --> 00:45:53.770
construction for obvious reasons.
00:45:53.770 --> 00:45:59.870
Apparently, Plotkin was the
first person to show this.
00:45:59.870 --> 00:46:02.560
I think Reed and Muller had two
different constructions,
00:46:02.560 --> 00:46:06.630
and neither one of them was the
u, u plus v construction.
00:46:06.630 --> 00:46:10.025
Reed, in particular, talked
about Boolean functions.
00:46:12.824 --> 00:46:13.610
All right.
00:46:13.610 --> 00:46:17.630
But this is the way I recommend
you to think about
00:46:17.630 --> 00:46:19.600
constructing them.
00:46:19.600 --> 00:46:24.440
And it's simply defined
as follows.
00:46:24.440 --> 00:46:29.660
We assume we've already
constructed the Reed-Muller
00:46:29.660 --> 00:46:31.300
codes of length m minus 1.
00:46:31.300 --> 00:46:34.220
It's a recursive parameter,
m minus 1.
00:46:34.220 --> 00:46:37.100
Now we're going to construct all
the Reed-Muller codes of
00:46:37.100 --> 00:46:40.820
parameter m of length
2 to the m.
00:46:40.820 --> 00:46:42.070
How are we going to do it?
00:46:45.600 --> 00:46:51.780
We want to construct
Rm of r, m.
00:46:51.780 --> 00:46:56.790
And we'll say it's the set of
all code words which consist
00:46:56.790 --> 00:47:02.670
of two halves, a pair of
n-tuples of half the length.
00:47:05.190 --> 00:47:12.200
so the two halves are going to
be u and u plus v, where I
00:47:12.200 --> 00:47:15.410
choose u and u plus
v as follows.
00:47:15.410 --> 00:47:25.170
u is in Rm of r minus 1, m.
00:47:28.740 --> 00:47:30.780
And so what does that mean?
00:47:30.780 --> 00:47:36.930
Sorry, it's got to
be m minus 1.
00:47:36.930 --> 00:47:39.690
So it's got to be
half the length.
00:47:39.690 --> 00:47:46.580
Is that right? r m minus
1, v is in Rm of r
00:47:46.580 --> 00:47:49.270
minus 1, m minus 1.
00:47:49.270 --> 00:47:52.560
Somebody who has the notes,
like my valuable teaching
00:47:52.560 --> 00:47:56.390
assistant, might check whether
I got it right or not.
00:47:56.390 --> 00:47:56.870
Let's see.
00:47:56.870 --> 00:47:59.320
What are going to be the
parameters of these codes?
00:47:59.320 --> 00:48:03.680
They both have n equals
2 to the m minus 1.
00:48:03.680 --> 00:48:08.140
The distance here is 2 to the m
minus 1 minus r minus 1, so
00:48:08.140 --> 00:48:12.800
the distance here is 2 to the
m minus r, which is the
00:48:12.800 --> 00:48:15.170
distance that we want to
achieve for this code.
00:48:17.780 --> 00:48:22.910
And the distance here is 2 to
the m minus r minus 1, which
00:48:22.910 --> 00:48:26.780
is half the distance we want
to achieve for this code.
00:48:26.780 --> 00:48:30.690
So we reach down, if we wanted
to construct now a code of
00:48:30.690 --> 00:48:35.900
length four and distance two,
we would construct it from
00:48:35.900 --> 00:48:39.050
these two codes, the one of half
the length with distance
00:48:39.050 --> 00:48:42.470
2, and the one of half the
length with half the distance,
00:48:42.470 --> 00:48:44.000
distance 1.
00:48:44.000 --> 00:48:49.640
So we would somehow use this to
get a code of length 4 and
00:48:49.640 --> 00:48:51.690
distance 2.
00:48:51.690 --> 00:48:55.110
And we don't know
yet what k is.
00:48:55.110 --> 00:48:55.470
OK.
00:48:55.470 --> 00:49:00.070
So we're going to use these
two codes to construct a
00:49:00.070 --> 00:49:01.090
larger code.
00:49:01.090 --> 00:49:02.450
Is the construction clear?
00:49:07.320 --> 00:49:08.570
All right.
00:49:11.350 --> 00:49:16.650
So let's now derive some
properties from this
00:49:16.650 --> 00:49:21.280
construction, and from the fact
that we've started from a
00:49:21.280 --> 00:49:22.030
set of codes.
00:49:22.030 --> 00:49:24.040
Let's say we start from
length two codes.
00:49:24.040 --> 00:49:25.870
Or we could start
from length one.
00:49:25.870 --> 00:49:29.760
You could satisfy yourself that
this construction applied
00:49:29.760 --> 00:49:32.070
to the length one codes gives
the length two codes.
00:49:32.070 --> 00:49:34.310
I won't go through
that exercise.
00:49:34.310 --> 00:49:40.420
So we have some set of
codes, m minus 1.
00:49:40.420 --> 00:49:42.085
What's the first thing
we notice?
00:49:45.530 --> 00:49:51.510
Obviously, the length is what
we want, because we've put
00:49:51.510 --> 00:49:56.170
together two 2-to-the-m minus
1-tuples, and 2 times 2 to the
00:49:56.170 --> 00:49:57.570
m minus 1 is 2 to the m.
00:49:57.570 --> 00:50:02.055
So we've constructed a set
of 2-to-the-m-tuples.
00:50:02.055 --> 00:50:06.350
The length of the resulting
code is 2 to the m.
00:50:06.350 --> 00:50:12.360
This is Rm r, m, constructed
in this way.
00:50:16.590 --> 00:50:20.430
Second, it's linear.
00:50:20.430 --> 00:50:23.020
It's a linear code.
00:50:23.020 --> 00:50:24.750
All we have to check is
the group property.
00:50:24.750 --> 00:50:27.680
If we add two code words of this
form, we're going to get
00:50:27.680 --> 00:50:31.790
another code word of that form,
from the fact that these
00:50:31.790 --> 00:50:35.420
guys are linear, yes.
00:50:35.420 --> 00:50:35.910
OK.
00:50:35.910 --> 00:50:40.940
So it is a linear code,
that's important.
00:50:40.940 --> 00:50:43.250
Then we might ask, what's
its dimension?
00:50:43.250 --> 00:50:44.540
How many code words are there?
00:50:47.820 --> 00:50:52.110
Well, do I get a unique code
word for every combination of
00:50:52.110 --> 00:51:03.390
u and v. And however you want
to convince yourself, you
00:51:03.390 --> 00:51:05.460
obviously do.
00:51:05.460 --> 00:51:09.960
If I'm given this word here, I
can deduce from it what was u,
00:51:09.960 --> 00:51:11.910
that's simply the first
half of it.
00:51:11.910 --> 00:51:14.060
Subtract u from the second
half of it, and I
00:51:14.060 --> 00:51:15.650
find out what v is.
00:51:15.650 --> 00:51:18.970
So there's a one-to-one map
between all possible pairs, u,
00:51:18.970 --> 00:51:22.860
v, and all possible
new code words.
00:51:22.860 --> 00:51:27.910
And so what that means is, let
me call it the dimension of
00:51:27.910 --> 00:51:32.650
the code with parameters r,
m is simply the sum of the
00:51:32.650 --> 00:51:42.800
dimensions of the codes with
parameters r, m minus 1 and r
00:51:42.800 --> 00:51:46.400
minus 1, m minus 1.
00:51:46.400 --> 00:51:50.670
So for instance, in this
hypothesized code here, if I
00:51:50.670 --> 00:51:53.330
want to know the dimension,
well, I take all possible
00:51:53.330 --> 00:51:56.210
combinations of words
here and words here.
00:51:56.210 --> 00:51:57.060
How many are there?
00:51:57.060 --> 00:51:57.680
There are eight.
00:51:57.680 --> 00:51:59.620
It has dimension three.
00:51:59.620 --> 00:52:04.410
So I'm going to get a 4, 3, 2,
code, which it's not too hard
00:52:04.410 --> 00:52:07.413
to see is the single parity
check code of length 4.
00:52:07.413 --> 00:52:08.663
All right.
00:52:10.470 --> 00:52:11.510
So that's how I do it.
00:52:11.510 --> 00:52:14.805
I simply add up the dimensions
of the two contributing codes.
00:52:19.540 --> 00:52:21.470
Not a very nice formula.
00:52:21.470 --> 00:52:25.160
In the homework, you do a
combinatoric exercise that
00:52:25.160 --> 00:52:28.090
gives you a somewhat more closed
form of the formula.
00:52:28.090 --> 00:52:31.950
But I think this is actually
the most useful one.
00:52:31.950 --> 00:52:35.090
In any case, we eventually get
a table that shows what all
00:52:35.090 --> 00:52:36.340
these things are anyway.
00:52:39.850 --> 00:52:45.630
Now, just as a fine point, I
want to assert that if I start
00:52:45.630 --> 00:52:50.890
out from a set of nested codes,
then I come up with a
00:52:50.890 --> 00:52:55.420
set of nested codes, at the
next highest level.
00:52:55.420 --> 00:52:56.940
That, again, is sort
of obvious.
00:52:56.940 --> 00:53:01.910
If these guys were nested, then
I get the appropriate --
00:53:01.910 --> 00:53:05.360
if I take u, u plus v
from sub-codes, I'm
00:53:05.360 --> 00:53:06.610
going to get a sub-code.
00:53:09.910 --> 00:53:12.300
Look at the notes if you want
more than that little bit of
00:53:12.300 --> 00:53:14.870
hand-waving.
00:53:14.870 --> 00:53:20.490
OK, that's something I need
for the next thing.
00:53:20.490 --> 00:53:22.210
I want to find out what d is.
00:53:28.080 --> 00:53:31.270
What's the minimum
distance here?
00:53:31.270 --> 00:53:34.010
Of course, my objective is
to make it equal to 2
00:53:34.010 --> 00:53:34.960
to the m minus r.
00:53:34.960 --> 00:53:36.690
Did I succeed?
00:53:36.690 --> 00:53:40.620
What are the possibilities
for u, u plus v?
00:53:40.620 --> 00:53:45.210
The possibilities are that
they're both 0, or that this
00:53:45.210 --> 00:53:49.460
one is 0 and this is not equal
to 0, or this is not equal to
00:53:49.460 --> 00:53:53.230
0 and this is 0, or that they're
both not equal to 0.
00:53:53.230 --> 00:53:56.635
And I'm going to consider
those four cases.
00:54:01.430 --> 00:54:04.120
Since every linear code include
the all-0 word, it's
00:54:04.120 --> 00:54:07.490
certainly possible that this
comes out at 0, 0.
00:54:07.490 --> 00:54:10.590
The only possibility for this
to come out 0, 0 is if I
00:54:10.590 --> 00:54:12.260
choose u equals 0.
00:54:12.260 --> 00:54:14.590
Then I have to choose
v equals 0.
00:54:14.590 --> 00:54:17.840
So there's one-code word of
weight 0 in my new code.
00:54:17.840 --> 00:54:19.950
But that's OK.
00:54:19.950 --> 00:54:24.700
If there were two code words
with weight 0, well, then the
00:54:24.700 --> 00:54:25.820
dimension would be wrong.
00:54:25.820 --> 00:54:30.170
This is in effect a proof that
the kernel is just 0, 0.
00:54:30.170 --> 00:54:31.890
And so the dimension is OK.
00:54:31.890 --> 00:54:34.660
It's a one-to-one map.
00:54:34.660 --> 00:54:35.010
All right.
00:54:35.010 --> 00:54:39.140
So I don't really have
to worry about that.
00:54:39.140 --> 00:54:42.280
I'm going to get one all-0
word in my new code.
00:54:42.280 --> 00:54:43.770
I can afford one all-0 word.
00:54:43.770 --> 00:54:46.490
I'm always going to have
to have it anyway.
00:54:46.490 --> 00:54:47.790
It's linear.
00:54:47.790 --> 00:54:49.650
All right, so these are the
more interesting cases.
00:54:49.650 --> 00:54:55.030
Suppose the first half is 0, but
the second half is not 0.
00:54:55.030 --> 00:54:58.150
That implies that u is 0.
00:54:58.150 --> 00:55:02.730
That implies that the second
half is just v,
00:55:02.730 --> 00:55:04.010
which is not 0.
00:55:04.010 --> 00:55:08.340
So v must be a non-zero code
word in this code, which has
00:55:08.340 --> 00:55:12.090
minimum distance 2
to the n minus r.
00:55:12.090 --> 00:55:17.228
So in this case, I prove that
the distance is greater than
00:55:17.228 --> 00:55:20.910
or equal to 2 to
the m minus r.
00:55:20.910 --> 00:55:25.060
Similarly, suppose this one
is not 0, but this is 0.
00:55:25.060 --> 00:55:28.320
OK, if that's 0, it can only be
because I chose u equal to
00:55:28.320 --> 00:55:36.260
some v. and that means the first
half, then, is that v.
00:55:36.260 --> 00:55:39.920
So again, v is in this
code that has
00:55:39.920 --> 00:55:42.090
enough minimum distance.
00:55:42.090 --> 00:55:47.560
So in this case, I proved that
the code word has weight 2 to
00:55:47.560 --> 00:55:48.810
the m minus r.
00:55:51.180 --> 00:55:52.760
And finally, let's take
the case where
00:55:52.760 --> 00:55:55.610
they're both non-zero.
00:55:55.610 --> 00:55:59.570
In that case, u could be an
arbitrary word in this code
00:55:59.570 --> 00:56:04.270
which only has distance
2 to the m r minus 1.
00:56:04.270 --> 00:56:07.210
So the first half is going to
have weight at least 2 to the
00:56:07.210 --> 00:56:09.540
m minus r minus 1, but
that's all I can say
00:56:09.540 --> 00:56:12.000
about the first half.
00:56:12.000 --> 00:56:15.815
But now the second half,
what is this?
00:56:18.320 --> 00:56:22.050
This is a higher-rate
code than this.
00:56:22.050 --> 00:56:25.640
This, by the nesting property,
is a sub-code of this.
00:56:25.640 --> 00:56:31.510
So if I add a word in a sub-code
to a word in the
00:56:31.510 --> 00:56:34.970
code, I'm going to get another
word in this code.
00:56:34.970 --> 00:56:39.950
So u plus v is still in this
Reed-Muller code, still has a
00:56:39.950 --> 00:56:43.710
minimum weight, if it's
non-zero, of 2 to the m
00:56:43.710 --> 00:56:45.015
minus r minus 1.
00:56:47.760 --> 00:56:51.650
So the distance is at least this
in the first half, this
00:56:51.650 --> 00:56:53.770
in the second half.
00:56:53.770 --> 00:56:56.080
And that's, of course,
still good enough.
00:56:59.230 --> 00:56:59.320
OK.
00:56:59.320 --> 00:57:01.860
So by this construction.
00:57:01.860 --> 00:57:08.370
I've assured that I'm going to
get a d greater than or equal
00:57:08.370 --> 00:57:10.860
to 2 to the m minus r.
00:57:10.860 --> 00:57:13.990
And of course, you can easily
find cases of equality, where
00:57:13.990 --> 00:57:16.450
it's only 2 to the m minus r.
00:57:16.450 --> 00:57:20.840
If this has a word of weight 2
to the m minus r, then you can
00:57:20.840 --> 00:57:23.765
clearly set up one like
this that has weight 2
00:57:23.765 --> 00:57:24.690
to the m minus r.
00:57:24.690 --> 00:57:28.430
Just pick one of the
minimum-weight code words as
00:57:28.430 --> 00:57:30.790
v, and u as 0.
00:57:30.790 --> 00:57:33.460
So the minimum distance
is 2 to the m minus r.
00:57:36.740 --> 00:57:39.630
All righty.
00:57:39.630 --> 00:57:44.280
So those are all the
properties we need.
00:57:44.280 --> 00:57:52.200
And then, I like to display
these properties in a tableau
00:57:52.200 --> 00:57:57.255
which you have in the notes,
which goes as follows.
00:58:00.630 --> 00:58:02.670
Let's just start listing
these codes.
00:58:02.670 --> 00:58:06.090
Here are the length 1 ones.
00:58:06.090 --> 00:58:14.370
We only found 2 of them, 1,
1, 1, and 1, 0, infinity.
00:58:14.370 --> 00:58:17.105
So there are two codes
of length 1.
00:58:17.105 --> 00:58:21.790
Now it turns out that if you
combine these according to the
00:58:21.790 --> 00:58:37.570
u, u plus v construction, you
get 2, 1, 2, where the weight
00:58:37.570 --> 00:58:41.100
2, 1 is just the --
00:58:41.100 --> 00:58:45.710
you take the first half is 1,
and the second half is 1.
00:58:45.710 --> 00:58:49.990
So you can build this
in the same way.
00:58:49.990 --> 00:58:53.110
And similarly, we can say just
by definition, we're always
00:58:53.110 --> 00:59:00.600
going to put a universe code at
the top and a trivial code
00:59:00.600 --> 00:59:02.180
at the bottom.
00:59:02.180 --> 00:59:03.750
So now I've listed all
my Reed-Muller
00:59:03.750 --> 00:59:06.520
codes with length 2.
00:59:06.520 --> 00:59:11.850
Now to construct the
ones of length 4.
00:59:11.850 --> 00:59:16.220
Again, I'll put a universe code
at the top, a trivial
00:59:16.220 --> 00:59:18.860
code at the bottom.
00:59:18.860 --> 00:59:27.030
I'll use my construction now to
create a 4, 3, 2 code here.
00:59:27.030 --> 00:59:31.450
I'm just using all
these properties.
00:59:31.450 --> 00:59:36.500
And down here, my construction,
when you combine
00:59:36.500 --> 00:59:39.900
these two things, you always
get a repetition code,
00:59:39.900 --> 00:59:42.550
again, 4, 1, 4.
00:59:42.550 --> 00:59:46.570
And I guess I've hand-waved.
00:59:46.570 --> 00:59:49.980
Exercise for the student,
prove that combining a
00:59:49.980 --> 00:59:53.600
repetition code with a trivial
code under the u, u plus v
00:59:53.600 --> 00:59:57.230
construction always gives a
double length repetition code.
01:00:01.370 --> 01:00:03.340
It's clear.
01:00:03.340 --> 01:00:06.790
v is always the all-0 word.
01:00:06.790 --> 01:00:10.220
u is either all-0 or all-1.
01:00:10.220 --> 01:00:14.400
So we get two words, one of
which is all-0, and one of
01:00:14.400 --> 01:00:18.150
which is all-1, double length.
01:00:18.150 --> 01:00:18.580
All right.
01:00:18.580 --> 01:00:24.590
So now I can just go on
indefinitely, and without a
01:00:24.590 --> 01:00:27.800
great deal of effort.
01:00:27.800 --> 01:00:32.710
Here I find that k is 7, just by
adding up these two things.
01:00:32.710 --> 01:00:37.526
The 8, this gives me an 8, 4.
01:00:37.526 --> 01:00:39.740
4 code.
01:00:39.740 --> 01:00:43.907
This gives me an 8.
01:00:43.907 --> 01:00:48.740
1, 8 code, and similarly
down here.
01:00:48.740 --> 01:00:52.882
And this, I now just
turn the crank.
01:00:52.882 --> 01:00:54.390
16, 16, 1.
01:00:54.390 --> 01:00:55.730
I always put that on top.
01:00:55.730 --> 01:00:59.270
The next one is 16, 15, 2.
01:00:59.270 --> 01:01:04.185
Next one is 16, 11, 4.
01:01:07.360 --> 01:01:12.090
After a while, you don't know
what you're going to get.
01:01:12.090 --> 01:01:14.400
But you get something.
01:01:14.400 --> 01:01:17.630
You've proved that all of
these codes exist, that
01:01:17.630 --> 01:01:19.560
they're all linear.
01:01:19.560 --> 01:01:26.270
They all have the n, k, d that
we've specified, and that
01:01:26.270 --> 01:01:28.820
furthermore, they're nested.
01:01:28.820 --> 01:01:32.590
And a final property, which you
might suspect, looking at
01:01:32.590 --> 01:01:40.860
these tables, is that the dual
of a Reed-Muller code is also
01:01:40.860 --> 01:01:41.780
a Reed-Muller code.
01:01:41.780 --> 01:01:45.140
And they're paired up according
to k and n minus k.
01:01:45.140 --> 01:01:45.900
Let's see.
01:01:45.900 --> 01:01:49.670
15 and 11 is 26.
01:01:49.670 --> 01:01:51.883
11 and five is 16.
01:01:57.770 --> 01:02:00.740
5 and 1 is 6.
01:02:00.740 --> 01:02:05.805
And 32, 1, 32, and so forth.
01:02:08.740 --> 01:02:11.345
Did you see how I proved that
all these codes exist?
01:02:13.980 --> 01:02:17.760
And if I continued, I could get
arbitrarily long codes,
01:02:17.760 --> 01:02:20.870
one of your simple homework
problems is just to do this,
01:02:20.870 --> 01:02:24.640
continue this for 64
and 128, see what
01:02:24.640 --> 01:02:25.910
additional codes you get.
01:02:29.050 --> 01:02:32.700
And so I now have this infinite
family it of that
01:02:32.700 --> 01:02:38.750
kind of covers the space n, k,
d in some sort of sparse way.
01:02:38.750 --> 01:02:42.180
But it indicates how
k and d go with n.
01:02:42.180 --> 01:02:44.810
And we come back to our original
question, how well
01:02:44.810 --> 01:02:47.700
can we do with binary
linear block codes.
01:02:47.700 --> 01:02:50.650
Well, here's some binary linear
block codes, pretty
01:02:50.650 --> 01:02:52.510
close to the best we
can find, actually.
01:02:52.510 --> 01:02:54.160
The ones I've listed
up here are all as
01:02:54.160 --> 01:02:56.320
good as we can find.
01:02:56.320 --> 01:02:59.720
And how well can we do?
01:02:59.720 --> 01:03:03.010
Really, we don't need to know
much to evaluate the
01:03:03.010 --> 01:03:07.830
performance of, say, 32, 6, 8
code, which is now getting to
01:03:07.830 --> 01:03:12.250
be a pretty substantial code,
with 2 to the 16 code words,
01:03:12.250 --> 01:03:17.750
to 65,536 code words.
01:03:17.750 --> 01:03:21.140
So we built a fairly sizable
constellation here in a
01:03:21.140 --> 01:03:24.000
32-dimensional Euclidean
space.
01:03:24.000 --> 01:03:25.850
And how good is it?
01:03:25.850 --> 01:03:29.710
Let's take 32, 16, 8.
01:03:32.350 --> 01:03:40.840
Can I graph its probability
of error per bit, a
01:03:40.840 --> 01:03:42.330
good estimate of it?
01:03:42.330 --> 01:03:44.816
Can I?
01:03:44.816 --> 01:03:46.870
Is there any information
I'm lacking?
01:03:52.090 --> 01:03:57.390
OK, I've been talking too long,
because when I go to the
01:03:57.390 --> 01:03:59.990
class, I'd like a
little response.
01:03:59.990 --> 01:04:02.212
So you were going to
say something?
01:04:02.212 --> 01:04:05.480
AUDIENCE: We don't
have [INAUDIBLE].
01:04:05.480 --> 01:04:12.950
PROFESSOR: OK, let me tell you
that the n, d is approximately
01:04:12.950 --> 01:04:19.226
600-something, so
let's say 630.
01:04:19.226 --> 01:04:21.350
It's probably not
exactly correct.
01:04:21.350 --> 01:04:25.578
In the notes I give a
formula for n, d --
01:04:25.578 --> 01:04:29.830
a formula that I don't
derive, that is known
01:04:29.830 --> 01:04:30.960
for Reed-Muller codes.
01:04:30.960 --> 01:04:34.650
And from that, you can compute
n, d for any of these codes.
01:04:34.650 --> 01:04:37.870
This parameter is m, r.
01:04:37.870 --> 01:04:38.300
All right.
01:04:38.300 --> 01:04:39.570
So I'll give you n, d as well.
01:04:39.570 --> 01:04:45.535
Now can I get the good estimate
for the probability
01:04:45.535 --> 01:04:48.460
of error per bit?
01:04:48.460 --> 01:04:49.950
How do I do that?
01:04:54.380 --> 01:04:55.660
Union-bound estimate.
01:04:55.660 --> 01:04:58.800
All right, so what's it
going to look like?
01:04:58.800 --> 01:05:03.471
What are the two subsidiary
parameters I need?
01:05:03.471 --> 01:05:06.680
One is the coding gain, right?
01:05:06.680 --> 01:05:10.355
What is the nominal coding gain
of this 32, 16, 8 code?
01:05:14.315 --> 01:05:16.965
Come on, this is not a difficult
computation.
01:05:21.550 --> 01:05:24.510
OK, what's k over n?
01:05:24.510 --> 01:05:26.650
1/2?
01:05:26.650 --> 01:05:29.780
I think we can all
do that one.
01:05:29.780 --> 01:05:32.250
What's the nominal
coding gain?
01:05:32.250 --> 01:05:35.820
Take 1/2 times d, 8.
01:05:35.820 --> 01:05:41.440
So now our coding gain is
4, which is what in dB?
01:05:41.440 --> 01:05:44.000
6 dB.
01:05:44.000 --> 01:05:45.530
Wow, gee whiz.
01:05:45.530 --> 01:05:49.280
I've already got a nominal
coding gain of 6 dB.
01:05:49.280 --> 01:05:51.960
Remember, my whole gap was,
depending on how I measured
01:05:51.960 --> 01:05:54.510
it, 9 or 10 or 11 dB.
01:05:54.510 --> 01:05:58.150
This looks like already a
sizable fraction of the gap,
01:05:58.150 --> 01:06:00.660
with just a simple
construction.
01:06:00.660 --> 01:06:04.060
Well, we better pay attention
to this error coefficient as
01:06:04.060 --> 01:06:06.940
well, or the number of
nearest neighbors.
01:06:06.940 --> 01:06:14.260
Kb is, let's just take a rough
estimate here, what's this
01:06:14.260 --> 01:06:15.510
going to be?
01:06:18.240 --> 01:06:19.345
About 40.
01:06:19.345 --> 01:06:22.070
That's good.
01:06:22.070 --> 01:06:25.090
It's just this divided by 16.
01:06:25.090 --> 01:06:27.570
So there are about 40 nearest
neighbors per bit.
01:06:27.570 --> 01:06:29.200
How much is that going
to cost us?
01:06:33.820 --> 01:06:34.018
Not so good.
01:06:34.018 --> 01:06:36.275
Well, it's a little bit more
than 5 factors of 2.
01:06:39.840 --> 01:06:41.660
Maybe 5 and a 1/2 factors.
01:06:41.660 --> 01:06:43.810
It's less than 5 and a 1/2.
01:06:43.810 --> 01:06:47.760
So this very roughly,
something will
01:06:47.760 --> 01:06:49.010
cost me about 1 dB.
01:06:53.850 --> 01:07:00.180
And so I get an effective
coding gain of 5 dB.
01:07:00.180 --> 01:07:04.220
That's my first very
gross estimate.
01:07:04.220 --> 01:07:05.630
All right, well,
it's not 6 dB.
01:07:05.630 --> 01:07:07.850
It's only 5 dB.
01:07:07.850 --> 01:07:10.460
That's still not bad.
01:07:10.460 --> 01:07:14.130
Again, if I really wanted to
know what this was, I would
01:07:14.130 --> 01:07:18.890
write this out as 40 or whatever
it is times Q to the
01:07:18.890 --> 01:07:23.140
square root of four times
2Eb over N_0.
01:07:27.100 --> 01:07:31.170
And can I plot that quantity?
01:07:31.170 --> 01:07:32.770
Yes, if I have MATLAB.
01:07:32.770 --> 01:07:38.900
Or actually, all I need
is my baseline.
01:07:38.900 --> 01:07:50.340
So if this is my baseline, which
went through 9.6 dB at
01:07:50.340 --> 01:07:54.600
10 to the minus 5, then we
remember how to plot that.
01:07:54.600 --> 01:07:57.900
I just take this whole curve
bodily, and I move
01:07:57.900 --> 01:08:02.230
it 6 dB to the left.
01:08:02.230 --> 01:08:07.440
So I get the same curve, this
is not very good, going
01:08:07.440 --> 01:08:08.830
through 3.6 dB.
01:08:12.046 --> 01:08:13.028
Sorry about that.
01:08:13.028 --> 01:08:15.490
How's that?
01:08:15.490 --> 01:08:16.510
Get it way down here.
01:08:16.510 --> 01:08:19.099
But then I also have to raise
it by a factor of 40.
01:08:22.050 --> 01:08:23.910
So it really looks
more like that.
01:08:23.910 --> 01:08:27.370
That's really going to go more
through about 4.6 dB or
01:08:27.370 --> 01:08:29.649
something like that.
01:08:29.649 --> 01:08:33.840
That's what these calculations
are, Ashish took
01:08:33.840 --> 01:08:36.340
you through, I believe.
01:08:36.340 --> 01:08:42.200
And so while 6 dB was my
nominal coding gain, my
01:08:42.200 --> 01:08:45.370
effective coding gain is 5 dB.
01:08:45.370 --> 01:08:47.090
But still, hey, not bad.
01:08:50.090 --> 01:08:53.509
I have very easily been able to
construct a code that gives
01:08:53.509 --> 01:08:56.674
you about 5 dB of coding gain.
01:08:56.674 --> 01:08:58.430
Is there any fly in
this ointment?
01:09:03.400 --> 01:09:05.220
Can I just go out and
build it now?
01:09:08.356 --> 01:09:10.229
I need a decoder.
01:09:10.229 --> 01:09:11.590
Who said that?
01:09:11.590 --> 01:09:13.200
Thank you.
01:09:13.200 --> 01:09:14.100
Good point.
01:09:14.100 --> 01:09:16.580
What's the decoding method
assumed for this code?
01:09:20.270 --> 01:09:20.960
Excuse me?
01:09:20.960 --> 01:09:23.240
AUDIENCE: Just a table
right now.
01:09:23.240 --> 01:09:26.460
PROFESSOR: Just a table,
based on what?
01:09:26.460 --> 01:09:31.790
I get some kind of received
n-tuple, which is actually
01:09:31.790 --> 01:09:35.640
just a random, some kind of
vector in 32-dimensional
01:09:35.640 --> 01:09:40.660
space, 32 numbers,
real numbers.
01:09:40.660 --> 01:09:45.819
And really, I'm assuming minimum
distance decoding.
01:09:45.819 --> 01:09:48.770
So in principle, I want to
compute the distance to each
01:09:48.770 --> 01:09:53.080
of the 2 to the 1/6th,
65,000 code words.
01:09:53.080 --> 01:09:56.930
And actually, nowadays, you
might just do that.
01:09:56.930 --> 01:10:00.460
That's not a formidable task.
01:10:00.460 --> 01:10:02.420
Back when I got into this
business, that would have been
01:10:02.420 --> 01:10:03.995
considered outrageous.
01:10:03.995 --> 01:10:06.610
But nowadays, you could keep
up a pretty good decoding
01:10:06.610 --> 01:10:13.210
rate, even doing 65,000 distance
computations and just
01:10:13.210 --> 01:10:14.260
finding the best one.
01:10:14.260 --> 01:10:15.780
That would do it.
01:10:15.780 --> 01:10:17.030
That would give you
this performance.
01:10:20.030 --> 01:10:22.830
But of course, we are going to
be looking for somewhat more
01:10:22.830 --> 01:10:26.150
efficient decoding schemes
than that.
01:10:26.150 --> 01:10:26.520
All right.
01:10:26.520 --> 01:10:30.850
So at this point, that's the
only fly in the ointment.
01:10:30.850 --> 01:10:36.090
We have a way of getting this
kind of error curve, provided
01:10:36.090 --> 01:10:39.690
that we're willing to do
exhaustive maximum likelihood
01:10:39.690 --> 01:10:41.735
or minimum distance decoding.
01:10:44.600 --> 01:10:47.680
And furthermore, we
can continue.
01:10:47.680 --> 01:10:51.220
It's also instructive to
see what happens as
01:10:51.220 --> 01:10:52.885
we let n go to infinity.
01:10:52.885 --> 01:10:55.200
What are we going to get
with this construction?
01:10:55.200 --> 01:10:58.180
We pretty well know.
01:10:58.180 --> 01:11:02.110
These are all going to be
universe codes, which are not
01:11:02.110 --> 01:11:04.470
very interesting to us.
01:11:04.470 --> 01:11:07.850
They all have a nominal
coding gain of one,
01:11:07.850 --> 01:11:09.380
and are just useless.
01:11:09.380 --> 01:11:13.220
They're basically just
send a bit 32 times,
01:11:13.220 --> 01:11:16.660
send 32 bits, I mean.
01:11:16.660 --> 01:11:20.360
OK, what are these
codes along here?
01:11:20.360 --> 01:11:22.800
Let me start there.
01:11:22.800 --> 01:11:26.890
These are all single
parity check codes.
01:11:26.890 --> 01:11:28.415
What's the nominal
coding gain?
01:11:32.840 --> 01:11:34.460
Well, as we get out here,
what does the
01:11:34.460 --> 01:11:35.710
nominal coding gain approach?
01:11:38.520 --> 01:11:42.210
The code rate approaches 1.
01:11:42.210 --> 01:11:44.790
The code distance stays at 2.
01:11:44.790 --> 01:11:51.800
So the nominal coding gain goes
to 2 or 3 or dB, at a
01:11:51.800 --> 01:11:54.675
rate of 1 or a nominal spectral
efficiency of 2.
01:11:57.460 --> 01:11:58.000
OK.
01:11:58.000 --> 01:12:01.540
Well, these are totally
simple codes, single
01:12:01.540 --> 01:12:03.045
parity check codes.
01:12:03.045 --> 01:12:05.980
And even with that, it looks
like we can get 3
01:12:05.980 --> 01:12:08.860
dB of coding gain.
01:12:08.860 --> 01:12:14.810
But what's the number of
nearest neighbors here?
01:12:14.810 --> 01:12:17.760
Number of nearest neighbors
is just n, n minus 1
01:12:17.760 --> 01:12:21.320
over 2, n choose 2.
01:12:21.320 --> 01:12:28.070
So the number of even dividing
by k, which is n minus 1, we
01:12:28.070 --> 01:12:34.030
still get a Kb that goes
up linearly with n.
01:12:34.030 --> 01:12:35.700
So what's in fact going
to happen to the
01:12:35.700 --> 01:12:36.950
effective coding gain?
01:12:43.490 --> 01:12:47.070
The nominal coding gain will
go up and reach an
01:12:47.070 --> 01:12:48.980
asymptote of 3 dB.
01:12:48.980 --> 01:12:51.850
This is nominal coding gain.
01:12:51.850 --> 01:12:55.590
But somewhere out here, as this
has reached an asymptote,
01:12:55.590 --> 01:12:58.020
the effective coding gain is
always going to be less, and
01:12:58.020 --> 01:13:01.050
it's going to have
to bend over.
01:13:01.050 --> 01:13:03.290
And according to our rule of
thumb, it'll eventually go
01:13:03.290 --> 01:13:10.680
back through 0, because the
cost just keeps going up
01:13:10.680 --> 01:13:12.470
linearly in terms of Kb.
01:13:12.470 --> 01:13:15.260
So the effective coding
gain is not as great.
01:13:15.260 --> 01:13:21.740
It has a peak for some n.
01:13:21.740 --> 01:13:27.310
And so there's some maximum
effective coding gain, which
01:13:27.310 --> 01:13:29.810
again I've left for you
as a homework problem,
01:13:29.810 --> 01:13:32.420
that's less than 3 dB.
01:13:32.420 --> 01:13:35.337
And I'll give you a hint that
it's of the order of 2 dB.
01:13:38.140 --> 01:13:42.080
It's pretty easy to work out in
the five minutes before the
01:13:42.080 --> 01:13:47.410
next class, not a difficult
problem to find out when this
01:13:47.410 --> 01:13:49.000
thing reaches its maximum.
01:13:49.000 --> 01:13:50.960
But still, these are
very simple codes.
01:13:50.960 --> 01:13:54.600
We'll see in a second they
have an extremely simple
01:13:54.600 --> 01:13:56.050
minimum distance decoding
algorithm
01:13:56.050 --> 01:13:58.170
called Wagner decoding.
01:13:58.170 --> 01:14:01.490
It's trivial, not hard to do
minimum distance decoding for
01:14:01.490 --> 01:14:03.680
these codes.
01:14:03.680 --> 01:14:07.400
And so, OK, not hard to get
2 dB of coding gain.
01:14:07.400 --> 01:14:10.050
What happens if we go along
this line here?
01:14:10.050 --> 01:14:13.790
These are all codes of
minimum distance 4.
01:14:13.790 --> 01:14:15.860
Again, start here.
01:14:15.860 --> 01:14:17.890
They're called extended
Hamming codes.
01:14:17.890 --> 01:14:21.270
A Hamming code has minimum
distance three and is suitable
01:14:21.270 --> 01:14:24.320
for hard decisions, single
error correction.
01:14:24.320 --> 01:14:26.450
These codes all have minimum
distance four.
01:14:26.450 --> 01:14:30.230
We've seen that it's always
worthwhile to add an overall
01:14:30.230 --> 01:14:33.440
parity check to get an even
minimum distance, if we're
01:14:33.440 --> 01:14:36.610
looking at Euclidean
space coding gain.
01:14:36.610 --> 01:14:39.220
And so these are actually
slightly better
01:14:39.220 --> 01:14:40.470
than Hamming codes.
01:14:43.170 --> 01:14:45.620
They're called extended Hamming
codes, because they're
01:14:45.620 --> 01:14:48.455
Hamming codes extended by
a single parity check.
01:14:48.455 --> 01:14:52.190
They have one more unit
of minimum distance.
01:14:52.190 --> 01:14:54.240
So again, asymptotically,
these are
01:14:54.240 --> 01:14:55.970
called extended Hamming.
01:14:55.970 --> 01:14:59.490
The nominal coding gain
goes to what now?
01:14:59.490 --> 01:15:01.895
This is, the rate is
again going to 1.
01:15:01.895 --> 01:15:03.670
The distance holds at four.
01:15:03.670 --> 01:15:10.850
So the nominal coding gain
goes to 4, or 6 dB, while
01:15:10.850 --> 01:15:13.750
again, the spectral efficiency
goes to two bits per two
01:15:13.750 --> 01:15:18.460
dimensions, the nominal spectral
efficiency rate to 1.
01:15:18.460 --> 01:15:20.155
But again, you have this
kind of phenomenon.
01:15:20.155 --> 01:15:25.100
And I ask you to work that out
also on the homework, where
01:15:25.100 --> 01:15:29.370
even though the nominal coding
gain plateaus eventually at 6
01:15:29.370 --> 01:15:33.370
dB, there is a maximum effective
coding gain, which
01:15:33.370 --> 01:15:36.130
is something more in the
range of 4 to 5 dB.
01:15:36.130 --> 01:15:37.420
You figure out what it is.
01:15:40.560 --> 01:15:43.445
And there's a limit to how much
effective coding gain you
01:15:43.445 --> 01:15:44.430
can get with these codes.
01:15:44.430 --> 01:15:46.710
Does this all makes sense?
01:15:46.710 --> 01:15:48.185
Are you seeing how
I'm arguing?
01:15:51.780 --> 01:15:56.905
OK, here's another interesting
sequence of codes.
01:16:01.890 --> 01:16:06.880
These are all half-rate codes,
or nominal spectral efficiency
01:16:06.880 --> 01:16:09.280
one bit per two dimensions.
01:16:11.940 --> 01:16:16.170
I briefly mentioned that these
things pair up in duals.
01:16:16.170 --> 01:16:19.750
The 16, 5 code is the dual
code of the 16, 11 codes.
01:16:19.750 --> 01:16:24.210
If you see, there's a symmetry
about rate 1 by 2, such that
01:16:24.210 --> 01:16:26.990
this is k, and this
is n minus k.
01:16:26.990 --> 01:16:30.460
And so you would suspect that
this guy is the dual of this
01:16:30.460 --> 01:16:31.330
guy, which it is.
01:16:31.330 --> 01:16:33.450
This guy is the dual
of this guy.
01:16:33.450 --> 01:16:35.560
This guy is the dual
of this guy.
01:16:35.560 --> 01:16:37.240
And this guy is its own dual.
01:16:37.240 --> 01:16:40.730
It's a self-dual code
of rate 1 by 2 or
01:16:40.730 --> 01:16:43.310
spectral efficiency 1.
01:16:43.310 --> 01:16:45.235
So these are self-dual codes.
01:16:48.700 --> 01:16:52.580
And what does their nominal
coding gain go to?
01:16:57.640 --> 01:17:02.450
Well, the rate is always 1 by 2
times four, that's a nominal
01:17:02.450 --> 01:17:04.090
coding gain of 2.
01:17:04.090 --> 01:17:06.470
This has a nominal
coding gain of 4.
01:17:06.470 --> 01:17:11.280
The next one in line would be
a 128, 64, 16 code, nominal
01:17:11.280 --> 01:17:13.700
coding gain of 8.
01:17:13.700 --> 01:17:15.890
So the nominal coding gain
actually goes to infinity.
01:17:23.450 --> 01:17:25.840
That's pretty good.
01:17:25.840 --> 01:17:28.640
However, what is its
true meaning?
01:17:28.640 --> 01:17:30.930
What we really want to
know is what's the
01:17:30.930 --> 01:17:32.140
effective coding gain.
01:17:32.140 --> 01:17:36.900
And given that the nominal
coding gain goes to infinity,
01:17:36.900 --> 01:17:40.360
is it possible the effective
coding gain can get us all the
01:17:40.360 --> 01:17:41.370
way to the Shannon limit?
01:17:41.370 --> 01:17:44.960
Can we completely close
the gap to capacity?
01:17:44.960 --> 01:17:46.800
As far as I know, this
is an open question.
01:17:46.800 --> 01:17:51.770
I strongly believe that you go
along this sequence through
01:17:51.770 --> 01:17:56.450
maximum likelihood decoding, you
will eventually get to the
01:17:56.450 --> 01:17:59.300
Shannon limit, that is the
Shannon limit for this
01:17:59.300 --> 01:18:02.200
spectral efficiency, which is
not the ultimate Shannon
01:18:02.200 --> 01:18:05.720
limit, but rather is 0, in terms
of Eb over N_0 If you
01:18:05.720 --> 01:18:11.420
remember, for rate 1/2 for rho
equals 1, the Shannon limit on
01:18:11.420 --> 01:18:12.880
Eb over N_0 was 0 dB.
01:18:15.790 --> 01:18:19.480
You may or may not
remember that.
01:18:19.480 --> 01:18:20.420
So there's a question.
01:18:20.420 --> 01:18:26.322
Does this take you all the
way to the Shannon limit?
01:18:29.040 --> 01:18:30.780
And that would be a nice
question for somebody to
01:18:30.780 --> 01:18:31.640
answer someday.
01:18:31.640 --> 01:18:36.250
I think it probably does,
especially in view of the fact
01:18:36.250 --> 01:18:44.850
that if you take this set down
here, again, this will be a
01:18:44.850 --> 01:18:51.080
homework problem, this turns out
to be a set of Euclidean
01:18:51.080 --> 01:18:55.570
images of these codes are
orthogonal signal sets.
01:18:55.570 --> 01:19:00.470
For instance, 32, 6,
16, what is that?
01:19:00.470 --> 01:19:04.610
That's a set of 64 constellation
points in 32
01:19:04.610 --> 01:19:06.940
dimensions.
01:19:06.940 --> 01:19:11.110
And algebraically, it's not hard
to figure out that every
01:19:11.110 --> 01:19:14.070
one of these code words is
orthogonal in a Euclidean
01:19:14.070 --> 01:19:16.870
sense to one another, except for
one that's complementary,
01:19:16.870 --> 01:19:20.510
which is just what you expect in
a bi-orthogonal signal set.
01:19:20.510 --> 01:19:23.850
So these give you bi-orthogonal,
they're called
01:19:23.850 --> 01:19:28.150
bi-orthogonal codes, or
first-order Reed-Muller codes,
01:19:28.150 --> 01:19:35.360
because the parameter r
is 1 for all of these.
01:19:35.360 --> 01:19:38.740
What's happening to the spectral
efficiency here?
01:19:38.740 --> 01:19:41.800
Spectral efficiency goes to 0.
01:19:41.800 --> 01:19:45.330
So these become highly
inefficient from a bandwidth
01:19:45.330 --> 01:19:47.680
point of view, as we already
know about orthogonal,
01:19:47.680 --> 01:19:50.550
bi-orthogonal simplex
signal sets.
01:19:50.550 --> 01:19:52.970
They use up a lot
of bandwidth.
01:19:52.970 --> 01:19:54.760
But what else do we
know about them?
01:19:54.760 --> 01:20:00.880
In this case, we definitely know
that the effective coding
01:20:00.880 --> 01:20:03.575
gain does go to the Shannon
limit, and in this case, to
01:20:03.575 --> 01:20:06.740
the ultimate Shannon limit
for rho equals 0 as
01:20:06.740 --> 01:20:08.780
rho approaches 0.
01:20:08.780 --> 01:20:12.240
So here, there's a proof that
these codes can get you to the
01:20:12.240 --> 01:20:13.500
Shannon limit.
01:20:13.500 --> 01:20:17.530
But as we've already explained
earlier, geometrically, it's
01:20:17.530 --> 01:20:20.930
at the cost of using much more
bandwidth than you probably
01:20:20.930 --> 01:20:22.880
really want to use.
01:20:22.880 --> 01:20:26.160
But here, at least, is one
capacity-approaching set of
01:20:26.160 --> 01:20:29.000
codes, just among these rather
simple Reed-Muller
01:20:29.000 --> 01:20:32.970
codes along this line.
01:20:32.970 --> 01:20:34.940
And of course, we also
see our repetition
01:20:34.940 --> 01:20:37.250
codes, our trivial codes.
01:20:37.250 --> 01:20:39.280
So this is a nice
01:20:39.280 --> 01:20:40.780
representative family of codes.
01:20:40.780 --> 01:20:45.110
And it really does tell you
quite well what to expect.
01:20:45.110 --> 01:20:46.380
Are you looking at your watch?
01:20:46.380 --> 01:20:47.105
Thank you.
01:20:47.105 --> 01:20:50.660
I appreciate the hint.
01:20:50.660 --> 01:20:56.390
So we didn't quite finish
chapter six today.
01:20:56.390 --> 01:20:59.610
Next time, we'll start out with
the penalties of making
01:20:59.610 --> 01:21:03.970
hard decisions, which at first
brush seems like a not
01:21:03.970 --> 01:21:05.710
unreasonable compromise
to make.
01:21:05.710 --> 01:21:09.600
But it actually costs
a serious penalty.
01:21:09.600 --> 01:21:12.360
And that will finish
chapter six.
01:21:12.360 --> 01:21:15.080
I'll do that as briefly
as I can.
01:21:15.080 --> 01:21:18.090
And then we'll get into chapters
seven and eight,
01:21:18.090 --> 01:21:21.890
which is finite fields and
Reed-Solomon codes, which are
01:21:21.890 --> 01:21:25.680
the single great triumph of
algebraic coding theory.
01:21:28.250 --> 01:21:29.780
So OK, that's tomorrow.