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In this lecture we are going to talk about
what do we mean by weight distribution of
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a linear block code and then we are going
to talk about how is the error correcting
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capability and error detecting capability
of a linear block code dependent on the minimum
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distance of a code So we will continue basically
our discussion on distance properties that
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we have started last time So this is one example
of a linear block code where number of information
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bits is 3 and number of coded bits is 6 This
is a list of 2 k codewords which is 8 codewords
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message bits and these are their corresponding
codewords So these are the from 0 0 0 to 1
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1 1 these are our 2 k message bits and corresponding
to each of our message bits these are the
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corresponding codewords Ok Now let us look
at what is the weight distribution of these
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codewords So these codewords this is all zero
codeword so the weight Hamming weight for
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this is basically 0 What about this This codeword
has 3 1's So Hamming weight is 3 this codeword
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has three 1's
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So its Hamming weight is 3 This codeword has
four 1's So the Hamming weight is 4 This codeword
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has three 1's so Hamming weight is 3 This
one similarly has Hamming weight 4 this one
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Hamming weight 4 and this one has Hamming
weight 3 Now what is the minimum distance
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of the code As you recall we define the minimum
distance of the code as minimum weight of
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a non-zero codeword So what is the minimum
weight of the non-zero codeword in this case
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Its 3 so minimum distance of this code is
3 So let a i denotes the number of codewords
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in C with Hamming weight i So if you look
here if I so I will use a 0 to denote number
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of codewords which have Hamming weight 0 and
that number is 1 Do we have any codeword with
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Hamming weight 1 No So a 1 is going to be
0 What about a 2 How many codewords we have
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with Hamming weight 2 Again that's 0 What
about a 3 That's basically 1 2 3 4We have
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4 codewords with Hamming weight 3 a 4 1 2
3 Ok we don't have any codeword with Hamming
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weight 5 or Hamming weight 6 And you can do
a quick check the number of codewords should
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add up to number of codewords that we have
which is 8 1 plus 4 plus 3 Ok So we are denoting
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by a i the number of codewords in this linear
block code with Hamming weight i Now this
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set which describes how many codewords we
have of particular weight this is basically
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known as weight distribution of a linear block
code see So for this block code the weight
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distribution is given by this This completely
specifies the weight distribution of this
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particular 6 3 linear block code And since
we have said a linear block code will have
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an all zero codeword so a 0 will be 1 and
sum of all these codewords they should all
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add up to total number of codewords which
is 2 to the power k
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I just worked out this example for the 6 3
code we have shown in the previous slide and
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I showed you that in this particular example
a 0 is 1 a 3 is 4 a 4 is 3 Rest all others
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are 0 And I also showed you that the minimum
distance of this code is 3 because minimum
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weight of a non-zero codeword in this example
is 3 Now the probability of undetected error
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for a linear block code over a binary symmetric
channel is basically related to the weight
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distribution of the code So for a 6 3 linear
block code and so when does an when does a
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undetected error happens An undetected error
happens if let's say you send one particular
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codeword and at the receiver you receive some
other codeword So without loss of generality
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let's assume that we sent a all zero codeword
And at the receiver you received any other
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non-zero codeword So if I send an all zero
codeword at the transmitter and at the receiver
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you receive any other non-zero codeword then
that will be the case of undetected error
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So you can see basically that's why I have
written it as so what is the probability when
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you are sending an all zero codeword what
is the probability of getting another codeword
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of weight a i or weight i What is the probability
that when I am sending an all zero codeword
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and you receive a codeword which has weight
i Now that probability is given by since we
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are considering a binary symmetric channel
now recall what happens in binary symmetric
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channel two inputs 0 and 1 two outputs 0 and
1 and what is the crossover probability That
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is basically given by p So with probability
p 0 can get flipped to 1 1 can get flipped
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to 0 And the probability of correct detection
is 1 minus p So you are sending a codeword
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which is an n-bit tuple Now what's a probability
that you are sending an all zero codeword
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of all zero bits you receive another codeword
of weight i Now that probability is given
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by p raised to power i This will happen when
i bits get flipped and n minus i bits do not
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get flipped So that probability is given by
p raised to power i into 1 minus p raised
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to power n minus i and how many such codewords
exist That number is given by a i So the probability
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of getting a weight i codeword at the receiver
when you send an all zero codeword that probability
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is basically given by this Ok Now an undetected
error will happen if the receiver receives
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any non-zero codeword
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So I have to sum up this probability for all
i going from 1 to n So this is my overall
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undetected error probability if I send a linear
block code over a binary symmetric channel
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So for the example that I have considered
I know the weight distribution so if I plug
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that in here what I get is so there were 4
codewords with weight 3 so this is 4 p raised
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to power 3 And what was n n is 6 So 6 minus
i which is 3 in this case it's 3 So first
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term that I will get is this The next term
corresponding to these codewords is given
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so there are 3 codewords of weight 4 Probability
of 4 bits getting flipped is p raised to power
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4 and probability of the other 2 bits not
getting flipped is 1 minus p whole square
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And since p is typically small I mean I can
approximate it for small p I can approximate
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this undetected error probability as 4 times
p q because this will be close to 1 and since
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p is 4 small number p raised to power 4 will
be a small number So this will be roughly
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equal to 4 into p raised to power 3 This is
for the case when p is small So you can see
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in general so in this particular example the
undetected probability basically varies as
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p raised to power 3 which is basically same
as n minus k In general we can show that that
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undetected probability is dependent on how
many parity bits that we have so the more
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the number of parity bits lesser will be the
undetected error probability So we can make
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undetected probability go small by increasing
the number of parity bits Now see how a codeword
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with minimum distance d min we know that any
error pattern or weight less than equal to
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d min minus 1 is not going to change that
codeword into any other valid codeword So
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in other words if there is an error pattern
of weight d min minus 1 or less then it cannot
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change a valid codeword into another valid
codeword What does that mean It means that
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we can actually detect any error pattern of
weight up to d min minus 1So all error patterns
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of weight d min minus 1 or fewer errors are
basically detectable and this is also known
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as random error correcting capability of a
linear block code
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Now take an example of a repetition code that
we did in the first class So let's say we
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have a rate one half repetition code So then
for 0 we are sending 0 0 and for 1 we are
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sending 1 1 Now let's assume because of error
in the channel some of the bits got flipped
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So let's say this what we received when we
let's say what we received was 1 0 If you
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receive 1 0 can you detect So what is the
minimum distance first answer this question
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What is the minimum distance of this code
this rate one half repetition code We can
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see the minimum distance is 2 Minimum distance
of this code is 2 So according to this we
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should be able all error patterns of weight
1 So let's take an example Let's say we received
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1 0 can you detect the error Yes we can because
since itâ€™s a rate one half repetition code
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what we expect to receive either 0 0 or 1
1 if we transmit these codewords over a binary
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symmetric channel But what we have received
is 1 0 which is neither 0 0 nor 1 1 So we
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are able to detect single error So to repeat
basically if you have a linear block code
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whose minimum distance is d min You will be
able to detect all errors random errors of
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error pattern up to d min minus 1
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Next we are going to show how is the error
detecting capability error correcting capability
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of a linear block code related to the minimum
distance of a code So if we have a linear
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block code C whose minimum distance is d min
where d min satisfies this relation d min
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is greater than equal to 2 t plus 1 where
t is an integer and its less than an integer
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and it is less than equal to 2 t plus 2 If
d min satisfies this relation and if we have
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a linear block code with minimum distance
d min then it is capable of correcting all
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error patterns up to weight t So let us prove
this result Let us assume the codeword that
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is transmitted is given by v and what we received
is say tuple r Let us assume there is another
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codeword w which is not same as v Now we know
from triangular inequality that Hamming distance
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between v and w will be less than equal to
Hamming distance between v and r plus Hamming
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distance between r and w Now let us assume
that the error pattern has weight t hat And
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what's r r is nothing but v plus this error
pattern correct So the Hamming distance between
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v and r is going to be the weight of this
error pattern and which we are denoting by
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t dash Now since v and w are valid codewords
so the Hamming distance between v and w will
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be at least equal to the minimum distance
of the code So the Hamming distance between
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v and w is greater than equal to minimum distance
of the code and in the beginning we defined
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that our minimum distance is at least 2 t
plus 1 So from these two we can write that
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Hamming distance between v and w is greater
than equal to 2 t plus 1 Now from the triangular
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inequality we know that Hamming distance between
r and w this we can see from here this relationship
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basically triangular inequality what we have
is the Hamming distance between v and w to
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be less than equal to Hamming distance between
r and w plus Hamming distance between r and
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v right Now this we can write as we can bring
this here and we can bring this here what
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we can write this as let us say we can write
this this relation in this particular form
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Ok
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Now what is this quantity Hamming distance
between v and w The Hamming distance between
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v and w is at least equal to 2 t plus 1 And
what is Hamming distance between the transmitted
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codeword and the received codeword This is
we denote it by t dash So then Hamming distance
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between r and w is given by 2 t plus 1 minus
t dash Now as long as your error pattern is
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less than equal to t the weight of error pattern
is less than equal to t in that case the Hamming
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distance between r and w will be you can plug
that value of t here and what we will get
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is Hamming distance between r and w is greater
than equal to t plus 1 which is greater than
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equal to t where as the Hamming distance between
transmitted codeword and the received codeword
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is t hat which is less than equal to t What
does it mean It means that the received codeword
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is closer to v than any other codeword w So
what will be your maximum likelihood decoder
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for binary symmetric channel will decide in
favor of It will decide in favor of v So you
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will correctly decode this received sequence
to be v and this was our transmitted codeword
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So you will not make an error So what we have
shown here is as long as your error pattern
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has weight up to t those error patterns are
correctable provided the minimum distance
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of your code is d min and it satisfies this
relationship So the minimum distance of the
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code is at least 2 t plus 1 and it is less
than equal to 2 t plus 2 then it can correct
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all error patterns of weight t or less So
as we can see here the received codeword is
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closer to v than any other codeword w so it
will decide in favor of v and this r will
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be decoded as v
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Next we are going to show that if there exists
an error pattern of weight greater than equal
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to t plus 1 then our decoder whose minimum
distance is at least 2 t plus 1 but less than
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2 t plus 2 this decoder will make an error
In other words it would not be able to correct
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this error pattern of weight t plus 1 So for
all error patterns of weight l if l is at
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least t plus 1 then our maximum likelihood
decoder may not be able to correctly decode
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or correct that error So let's prove this
If v and w are 2 codewords and let's assume
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that the Hamming distance between v and w
is equal to the minimum distance of the code
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which is denoted by t min And let e 1 and
e 2 are two error patterns which satisfies
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these 3 properties and what are these 3 properties
The sum of e 1 and e 2 is the same as v plus
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w The second property is e 1 and e 2 they
do not have any overlapping 1s So weight of
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e 1 plus e 2 can be written as weight of e
1 plus weight of e 2 And we will show that
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if there is an error pattern of weight l where
l is at least t plus 1 then our maximum likelihood
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decoder will make an error in decoding So
the way we have chosen our error pattern weight
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of e 1 plus weight of e 2 is given by weight
of e 1 plus e 2 this is from 2 and from 1
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we know e 1 plus e 2 is nothing but v plus
w so this is same as weight of v plus w and
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this is nothing but this is Hamming distance
between v and w and we have said the Hamming
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distance between v and w is the minimum distance
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So this is equal to the minimum distance Now
let us assume that we transmitted this codeword
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v and what we received is r So this v got
corrupted by this error pattern e 1 which
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has Hamming weight of at least t plus 1 Now
we will repeat the same exercise we will try
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to find out the Hamming distance of this received
codeword from the correct transmitted codeword
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v and Hamming distance between the received
codeword and any other codeword w So if we
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calculate the Hamming distance between w and
the received codeword we know that Hamming
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distance between w and r is nothing but Hamming
weight of w and r And what is r r is my received
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codeword v plus e 1 So I can write this as
w plus v plus e 1 Now what is w plus v From
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1 I have w plus v is same as e 1 plus e 2
So then this is e 1 plus e 2 plus e 1 So e
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1 plus e 1 will be 0 So this will be e 2 weight
of e 2 And what is weight of e 2 From this
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relation we can see weight of e 1 plus weight
of e 2 is d min So weight of e 2 is d min
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minus weight of e 1 So this we can write as
weight of e 2 as d min minus weight of e 1
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So d min is less than equal to 2 t plus 1
and weight of e 1 is at least t plus 1 So
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weight of e 2 will be less than 2 t plus 2
minus t plus 1 which is t plus 1
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So the Hamming distance between w and r is
less than t plus 1 And what is the Hamming
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distance between v and r This is weight of
e 1 Ok and what is weight of e 1 Weight of
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e 1 is given by l which is at least t plus
1 So what we have shown here is weight of
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w Hamming distance between w and r is less
than t plus 1 where as Hamming distance between
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v and r is greater than equal to t plus 1
So what we have shown is Hamming distance
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between w and r is less than equal to Hamming
distance between received codeword r and the
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true codeword which was actually transmitted
which is v So in this case the maximum likelihood
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decoder will decode in favor of w and not
v and will make a mistake So through this
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construction we have shown that if your error
pattern is of weight t plus 1 then you are
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not guaranteed to correct that error So from
this and the previous result we can conclude
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that if we have a block code with minimum
distance d min which satisfies relationship
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that d min lies between 2 t plus 1 and 2 t
plus 2 then this linear block code with minimum
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distance d min should be able to correct all
error patterns up to weight t where t is given
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by this So this t is known as random error
correcting capability of the linear block
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code
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Next we are going to prove a result which
is as follows So if we have an n k linear
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block code whose minimum distance is given
by t min then we can show where d min lies
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between 2 t plus 1 and 2 t plus 2 then we
can show that all end tuples of weight t or
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less can be used as coset leader in our standard
array So we are going to prove this result
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using method of contradiction Now let's say
so how method of contradiction work We will
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say let's say they are all error patterns
or weight up to t let's say they are not coset
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leaders Let's say we will assume a scenario
where there are 2 such end tuples with weight
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up to t which are not coset leaders In other
words they lie in the same coset or same row
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And then later on we will show that that is
not possible So that's how this method of
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contradiction will work so minimum distance
of the code is d min so minimum weight of
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the code is also d min Let x and y are 2 n-tuples
of weight t or less Now weight of x plus y
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will be less than equal to weight of x plus
weight of y Why because there might be some
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overlapping 1s at some locations of this n-tuple
x and y and we are given that the weight of
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x and weight of y is at most t so then weight
of x plus weight of y will be less than equal
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to 2 t and this is less than minimum distance
because minimum distance of code is at least
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2 t plus 1
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Now let us assume that these x and y which
are error patterns of weight t or less let
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us assume that they are not coset leaders
If they are not coset leaders let us assume
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they are in the same coset they are in the
same row So if we assume x and y are in the
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same row or same coset then x plus y must
be a codeword Why this is so If you recall
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your standard array we had something like
this First row first column was all zero vector
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and then we had other codewords And then we
had error pattern let's say e 2 This was e
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2 plus v 2 Like like this was e 2 plus v 2
k If you look at any 2 elements in the same
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coset or same row and if you add them up what
do you get Let's add this and this what do
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we get e 2 plus e 2 plus v 2 we will get v
2 If we add this and this we will get v 2
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plus v 2 k which is another codeword v s So
if we take any two elements in the same coset
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and we add them up we are going to get a non
zero codeword So if x and y are in the same
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coset then x plus y must be a codeword This
is impossible Why Because if x plus y is a
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codeword then what is the minimum distance
of x plus y x plus y minimum distance of that
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must be d min But what is the what is the
weight of x plus y we just showed in this
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bullet that weight of x plus y is less than
d min That means weight of x plus y is less
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than d min If weight of x plus y is less than
d min then x plus y cannot be a non zero codeword
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because the weight of a non-zero codeword
should be at least d min So our assumption
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that x and y are in the same coset is wrong
In other words then x and y must be in different
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cosets different rows and we can always make
these x and y as coset leaders So this proves
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our result that all n-tuples of weight n of
weight t or less can be used as coset leaders
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in the standard array and we know that uh
if we use them as coset leaders we those are
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our correctable error patterns
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Next I am going to show you a result which
is as follows So if you have a n k linear
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block code whose minimum distance is d min
and if all n-tuples of weight t or less are
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already used as coset leader then there is
at least 1 n-tuple of weight t plus 1 which
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cannot be used as coset leader So this essentially
is going to show us again the same result
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that any weight pattern of error pattern of
weight t plus 1 is not guaranteed to be corrected
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So how do we prove it So let's assume v is
the minimum weight codeword of C and we have
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2 n-tuples x and y which satisfies these following
conditions First x plus y is equal to v and
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x and y do not have any component common So
they do not have 1s common in same position
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So from the definition x and y must be in
the same coset Why Because we know if two
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elements are in the same coset and if we add
them sum is a valid codeword So x plus y is
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equal to v which is a valid codeword then
x and y must be in the same coset
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So that's what I said from definition x and
y must be in the same coset because x plus
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y is v which is a valid codeword And we know
that if we add any two elements in a coset
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their sum is a valid codeword And similarly
weight of x plus weight of y is equal to weight
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of v And we have chosen v to be the minimum
distance codeword so this is given by d min
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Now if we choose our y to have a weight of
t plus 1 then we can see from here d min is
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greater than equal to 2 t plus 1 but less
than equal to 2 t plus 2 So from this and
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using the fact that d min lies between 2 t
plus 1 and 2 t plus 2 using these 2 results
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what we get is weight of x can be t or t plus
1 So therefore if we choose x to be our coset
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leader then we cannot choose y as our coset
leader You can see because x and y are in
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the same coset and weight of x is t or t plus
1 whereas weight of y is t plus 1 So I will
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choose x as my coset leader And if I choose
x as my coset leader then I cannot choose
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y as my coset leader which proves my result
which says that if all end tuples of weight
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t or less are used as coset leaders then there
exist at least one error pattern of weight
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t plus 1 which cannot be used as coset leader
and if this error pattern of weight t plus
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1 cannot be put as coset leader then this
is not a correctable error pattern So with
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this I will conclude my lecture on random
error correcting and random error detecting
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properties of block codes Thank you