Mixing
Binary codes that can correct one error and what is encoded/decoded has rate “arbitrarily close” to $1$
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Task:
Show that there exist binary codes that can correct one error and that have rate arbitrarily close to $1$.
This is asking for an existence proof, so either by contruction or using some well-known result, but I do not know where to even start with this. The statement seems so simple but one word is throwing me off: "arbitrarily close", which means that we can make the rate as close to $1$ as we like. We would have to come up with some code that has better and better rate if we make it larger, maybe the rate would be some fraction of the form $frac{a-1}{a}$ for some natural number $a$. I think this corresponds to puncturing so-called "convolutional code"
Does anybody have a hint or tip?
coding-theory
asked Jan 10 at 14:27
Wesley StrikWesley Strik
1,946423
$endgroup$
add a comment |
$begingroup$
Task:
Show that there exist binary codes that can correct one error and that have rate arbitrarily close to $1$.
This is asking for an existence proof, so either by contruction or using some well-known result, but I do not know where to even start with this. The statement seems so simple but one word is throwing me off: "arbitrarily close", which means that we can make the rate as close to $1$ as we like. We would have to come up with some code that has better and better rate if we make it larger, maybe the rate would be some fraction of the form $frac{a-1}{a}$ for some natural number $a$. I think this corresponds to puncturing so-called "convolutional code"
Does anybody have a hint or tip?
coding-theory
asked Jan 10 at 14:27
Wesley StrikWesley Strik
1,946423
$endgroup$
$begingroup$
Can you explain for a code what being close to $1$ means?
$endgroup$
– Levent
Jan 10 at 14:29
$begingroup$
Every bit of information is useful and there is no redundant information.
$endgroup$
– Wesley Strik
Jan 10 at 14:30
$begingroup$
@Levent not for the code itself, for the bitrate of what is encoded by code can get arbitrarily close to not having redundancy. I guess it is meant encoded/unencoded can get arbitrarily close to 1.
$endgroup$
– mathreadler
Jan 10 at 14:41
$begingroup$
That's indeed what is meant, the question is in a way poorly phrased
$endgroup$
– Wesley Strik
Jan 10 at 14:46
add a comment |
$begingroup$
Task:
Show that there exist binary codes that can correct one error and that have rate arbitrarily close to $1$.
This is asking for an existence proof, so either by contruction or using some well-known result, but I do not know where to even start with this. The statement seems so simple but one word is throwing me off: "arbitrarily close", which means that we can make the rate as close to $1$ as we like. We would have to come up with some code that has better and better rate if we make it larger, maybe the rate would be some fraction of the form $frac{a-1}{a}$ for some natural number $a$. I think this corresponds to puncturing so-called "convolutional code"
Does anybody have a hint or tip?
coding-theory
asked Jan 10 at 14:27
Wesley StrikWesley Strik
1,946423
$endgroup$
Task:
Show that there exist binary codes that can correct one error and that have rate arbitrarily close to $1$.
This is asking for an existence proof, so either by contruction or using some well-known result, but I do not know where to even start with this. The statement seems so simple but one word is throwing me off: "arbitrarily close", which means that we can make the rate as close to $1$ as we like. We would have to come up with some code that has better and better rate if we make it larger, maybe the rate would be some fraction of the form $frac{a-1}{a}$ for some natural number $a$. I think this corresponds to puncturing so-called "convolutional code"
Does anybody have a hint or tip?
coding-theory
coding-theory
asked Jan 10 at 14:27
Wesley StrikWesley Strik
1,946423
asked Jan 10 at 14:27
Wesley StrikWesley Strik
1,946423
edited Jan 10 at 14:53
Wesley Strik
asked Jan 10 at 14:27
Wesley StrikWesley Strik
1,946423
asked Jan 10 at 14:27
Wesley StrikWesley Strik
1,946423
asked Jan 10 at 14:27
Wesley StrikWesley Strik
1,946423
1,946423
$begingroup$
Can you explain for a code what being close to $1$ means?
$endgroup$
– Levent
Jan 10 at 14:29
$begingroup$
Every bit of information is useful and there is no redundant information.
$endgroup$
– Wesley Strik
Jan 10 at 14:30
$begingroup$
@Levent not for the code itself, for the bitrate of what is encoded by code can get arbitrarily close to not having redundancy. I guess it is meant encoded/unencoded can get arbitrarily close to 1.
$endgroup$
– mathreadler
Jan 10 at 14:41
$begingroup$
That's indeed what is meant, the question is in a way poorly phrased
$endgroup$
– Wesley Strik
Jan 10 at 14:46
add a comment |
$begingroup$
Can you explain for a code what being close to $1$ means?
$endgroup$
– Levent
Jan 10 at 14:29
$begingroup$
Every bit of information is useful and there is no redundant information.
$endgroup$
– Wesley Strik
Jan 10 at 14:30
$begingroup$
@Levent not for the code itself, for the bitrate of what is encoded by code can get arbitrarily close to not having redundancy. I guess it is meant encoded/unencoded can get arbitrarily close to 1.
$endgroup$
– mathreadler
Jan 10 at 14:41
$begingroup$
That's indeed what is meant, the question is in a way poorly phrased
$endgroup$
– Wesley Strik
Jan 10 at 14:46
$begingroup$
Can you explain for a code what being close to $1$ means?
$endgroup$
– Levent
Jan 10 at 14:29
$begingroup$
Can you explain for a code what being close to $1$ means?
$endgroup$
– Levent
Jan 10 at 14:29
$begingroup$
Every bit of information is useful and there is no redundant information.
$endgroup$
– Wesley Strik
Jan 10 at 14:30
$begingroup$
Every bit of information is useful and there is no redundant information.
$endgroup$
– Wesley Strik
Jan 10 at 14:30
$begingroup$
@Levent not for the code itself, for the bitrate of what is encoded by code can get arbitrarily close to not having redundancy. I guess it is meant encoded/unencoded can get arbitrarily close to 1.
$endgroup$
– mathreadler
Jan 10 at 14:41
$begingroup$
@Levent not for the code itself, for the bitrate of what is encoded by code can get arbitrarily close to not having redundancy. I guess it is meant encoded/unencoded can get arbitrarily close to 1.
$endgroup$
– mathreadler
Jan 10 at 14:41
$begingroup$
That's indeed what is meant, the question is in a way poorly phrased
$endgroup$
– Wesley Strik
Jan 10 at 14:46
$begingroup$
That's indeed what is meant, the question is in a way poorly phrased
$endgroup$
– Wesley Strik
Jan 10 at 14:46
add a comment |
1 Answer
1
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oldest
votes
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My suggestion would be to look at Hamming codes of increasing length to see the rate converging to $1$.
answered Jan 10 at 14:55
Andreas CarantiAndreas Caranti
56.5k34395
$endgroup$
$begingroup$
Yep, that's a good example, I didn't know there were other Hamming codes besides (8,4) and (7,4)
$endgroup$
– Wesley Strik
Jan 10 at 17:14
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
My suggestion would be to look at Hamming codes of increasing length to see the rate converging to $1$.
answered Jan 10 at 14:55
Andreas CarantiAndreas Caranti
56.5k34395
$endgroup$
$begingroup$
Yep, that's a good example, I didn't know there were other Hamming codes besides (8,4) and (7,4)
$endgroup$
– Wesley Strik
Jan 10 at 17:14
add a comment |
$begingroup$
My suggestion would be to look at Hamming codes of increasing length to see the rate converging to $1$.
answered Jan 10 at 14:55
Andreas CarantiAndreas Caranti
56.5k34395
$endgroup$
$begingroup$
Yep, that's a good example, I didn't know there were other Hamming codes besides (8,4) and (7,4)
$endgroup$
– Wesley Strik
Jan 10 at 17:14
add a comment |
$begingroup$
My suggestion would be to look at Hamming codes of increasing length to see the rate converging to $1$.
answered Jan 10 at 14:55
Andreas CarantiAndreas Caranti
56.5k34395
$endgroup$
My suggestion would be to look at Hamming codes of increasing length to see the rate converging to $1$.
answered Jan 10 at 14:55
Andreas CarantiAndreas Caranti
56.5k34395
answered Jan 10 at 14:55
Andreas CarantiAndreas Caranti
56.5k34395
answered Jan 10 at 14:55
Andreas CarantiAndreas Caranti
56.5k34395
answered Jan 10 at 14:55
Andreas CarantiAndreas Caranti
56.5k34395
56.5k34395
$begingroup$
Yep, that's a good example, I didn't know there were other Hamming codes besides (8,4) and (7,4)
$endgroup$
– Wesley Strik
Jan 10 at 17:14
add a comment |
$begingroup$
Yep, that's a good example, I didn't know there were other Hamming codes besides (8,4) and (7,4)
$endgroup$
– Wesley Strik
Jan 10 at 17:14
$begingroup$
Yep, that's a good example, I didn't know there were other Hamming codes besides (8,4) and (7,4)
$endgroup$
– Wesley Strik
Jan 10 at 17:14
$begingroup$
Yep, that's a good example, I didn't know there were other Hamming codes besides (8,4) and (7,4)
$endgroup$
– Wesley Strik
Jan 10 at 17:14
add a comment |
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$begingroup$
Can you explain for a code what being close to $1$ means?
$endgroup$
– Levent
Jan 10 at 14:29
$begingroup$
Every bit of information is useful and there is no redundant information.
$endgroup$
– Wesley Strik
Jan 10 at 14:30
$begingroup$
@Levent not for the code itself, for the bitrate of what is encoded by code can get arbitrarily close to not having redundancy. I guess it is meant encoded/unencoded can get arbitrarily close to 1.
$endgroup$
– mathreadler
Jan 10 at 14:41
$begingroup$
That's indeed what is meant, the question is in a way poorly phrased
$endgroup$
– Wesley Strik
Jan 10 at 14:46