Proving that there doesn't exist a binary code with parameters [6,5,4] The 2019 Stack Overflow Developer Survey Results Are InCoding theory (existence of codes with given parameters)Constructing a Linear Binary CodeConstructing a Cyclic code of length p-1 with minimal distance d<=p-1 if p is primeProve there doesn't exist [10,5,4] linear binary and self-dual code.Find the minimal $n$ such that there exists $[n,n-5]$ cyclic binary code with generator polynomial $g(x)=1+x^4+x^5$Generating a binary code with maximized Hamming distanceGive the generator polynomial of a binary cyclic [9, 2] code.Proving that a $(90, 2^78, 5)_2$ code doesn't existIf a binary $(n,M,d)$ code with $d$ even exists, then exists an $(n,M,d)$ code with even weight for all it's codewordsLinear Binary Code doesn't exist, proof by coset weight distribution
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Proving that there doesn't exist a binary code with parameters [6,5,4]
The 2019 Stack Overflow Developer Survey Results Are InCoding theory (existence of codes with given parameters)Constructing a Linear Binary CodeConstructing a Cyclic code of length p-1 with minimal distance d<=p-1 if p is primeProve there doesn't exist [10,5,4] linear binary and self-dual code.Find the minimal $n$ such that there exists $[n,n-5]$ cyclic binary code with generator polynomial $g(x)=1+x^4+x^5$Generating a binary code with maximized Hamming distanceGive the generator polynomial of a binary cyclic [9, 2] code.Proving that a $(90, 2^78, 5)_2$ code doesn't existIf a binary $(n,M,d)$ code with $d$ even exists, then exists an $(n,M,d)$ code with even weight for all it's codewordsLinear Binary Code doesn't exist, proof by coset weight distribution
$begingroup$
I'm new to code theory and I'm having trouble with proving if a code doesn't exist in general.
I found a binary code with the parameters [6,4,4] but I don't really understand why there cant be one with parameters [6,5,4].
It would help a lot having an example of such problems.
Thanks in advance.
coding-theory
$endgroup$
add a comment |
$begingroup$
I'm new to code theory and I'm having trouble with proving if a code doesn't exist in general.
I found a binary code with the parameters [6,4,4] but I don't really understand why there cant be one with parameters [6,5,4].
It would help a lot having an example of such problems.
Thanks in advance.
coding-theory
$endgroup$
$begingroup$
Linear code? Block code? Cyclic code? This question lacks context.
$endgroup$
– Eric Towers
Mar 30 at 17:46
add a comment |
$begingroup$
I'm new to code theory and I'm having trouble with proving if a code doesn't exist in general.
I found a binary code with the parameters [6,4,4] but I don't really understand why there cant be one with parameters [6,5,4].
It would help a lot having an example of such problems.
Thanks in advance.
coding-theory
$endgroup$
I'm new to code theory and I'm having trouble with proving if a code doesn't exist in general.
I found a binary code with the parameters [6,4,4] but I don't really understand why there cant be one with parameters [6,5,4].
It would help a lot having an example of such problems.
Thanks in advance.
coding-theory
coding-theory
asked Mar 30 at 17:41
evgniy tayarovevgniy tayarov
112
112
$begingroup$
Linear code? Block code? Cyclic code? This question lacks context.
$endgroup$
– Eric Towers
Mar 30 at 17:46
add a comment |
$begingroup$
Linear code? Block code? Cyclic code? This question lacks context.
$endgroup$
– Eric Towers
Mar 30 at 17:46
$begingroup$
Linear code? Block code? Cyclic code? This question lacks context.
$endgroup$
– Eric Towers
Mar 30 at 17:46
$begingroup$
Linear code? Block code? Cyclic code? This question lacks context.
$endgroup$
– Eric Towers
Mar 30 at 17:46
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Assuming you mean a linear error-correcting code, a common interpretation of "[6,5,4] code" is a code using $6$-bit codewords to represent $5$-bit messages where any two distinct codewords differ in at least $4$ bits.
There are $2^6 = 64$ possible codewords. We have $2^5 = 32$ messages. In the lattice $0,1^6$, each codeword is surrounded by an open ball of radius $4$ containing no other codewords. One can use the Singleton bound to show that $5 + 4 leq 6 + 1$ if this code is realizable. However, $9 not leq 7$, so this code is not realizable.
The Singleton bound is a very loose bound. Generally, for a realizable [c,m,d] linear error-correcting code, the inequality is $m+d leq c+1$. The Singleton bound also indicates there is no realizable [6,4,4] linear error-correcting code.
There is another way to show there is no [6,5,4] linear error-correcting code. We have $64$ codewords and $32$ messages. By distance 4, there can be no pair of messages assigned to complementary codewords. By the pigeonhole principle, since there are only $32$ such pairs of codewords, any such pair has exactly one message assigned to it. This means there is a message assigned to exactly one of $000000_2$ and $111111_2$. Similarly, there is exactly one message assigned to the codeword pair $000111_2$ and $111000_2$. However, no such assignment is possible since any member of one pair is only distance three from either member of the other pair. beginalign*
d(000000_2, 111000_2) &= 3 text, \
d(000000_2, 000111_2) &= 3 text, \
d(111111_2, 111000_2) &= 3 text, and \
d(111111_2, 000111_2) &= 3 text.
endalign*
Consequently, there is no [6,5,4] linear error-correcting code.
(This counting argument can be adapted to show there is in [6,4,4] linear error-correcting code. Take the codewords in groups of $4$ having matching first four bits. This gives 16 groups each member of which differs from the other members of its group in only two bits, so at most one of those codewords is assigned to a message. There are 16 messages, so each group is assigned exactly one messsage. Then the groups $0000**_2$ and $0001**_2$, where "$*$" represents a bit whose value is either $0$ or $1$ (but this is not critical since either choice gives a codeword in the same group), each have a message, but regardless of which codeword is chosen in each group, those codewords differ in at most three bits.)
$endgroup$
$begingroup$
Thank you for the detailed answer
$endgroup$
– evgniy tayarov
Mar 30 at 18:34
add a comment |
$begingroup$
Assuming you are talking about non-trivial perfect binary codes of the form $[l,x,d]$, we have that the code can correct up to $e=lfloorfracd-12rfloor$ errors. If a non-trivial perfect binary code exists, it must hold that $sum_i=0^ebinomli$ is a power of $2$.
In our case, $d=4$, so $e=lfloorfrac32rfloor=1$ and $l=6$. $binom60+binom61=1+6=7$ which is not a power of $2$, so such code cannot exist, unless it's trivial.
$endgroup$
$begingroup$
Thank you!! This helps alot
$endgroup$
– evgniy tayarov
Mar 30 at 18:35
$begingroup$
This question just seems to be asking about the existence of a code, not about perfect codes. Your argument shows that a code with these parameters is not perfect, but not that it does not exist.
$endgroup$
– Morgan Rodgers
Mar 31 at 14:51
$begingroup$
@MorganRodgers If the OP leaves the people answering the question to guess what they're talking about, we'll fill in answers as we think is necessary. As the original question didn't seem too complicated, I thought they were talking about perfect codes as such a check is simple and short. If they were not, they should've specified that. If they use this answer even though it might be not what they looking for, then they should read up on the subject before asking a question and accepting random answers. Therefore, I don't think my answer is bad, it's just an answer to a bad question.
$endgroup$
– Marc
Mar 31 at 15:53
add a comment |
Your Answer
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2 Answers
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2 Answers
2
active
oldest
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$begingroup$
Assuming you mean a linear error-correcting code, a common interpretation of "[6,5,4] code" is a code using $6$-bit codewords to represent $5$-bit messages where any two distinct codewords differ in at least $4$ bits.
There are $2^6 = 64$ possible codewords. We have $2^5 = 32$ messages. In the lattice $0,1^6$, each codeword is surrounded by an open ball of radius $4$ containing no other codewords. One can use the Singleton bound to show that $5 + 4 leq 6 + 1$ if this code is realizable. However, $9 not leq 7$, so this code is not realizable.
The Singleton bound is a very loose bound. Generally, for a realizable [c,m,d] linear error-correcting code, the inequality is $m+d leq c+1$. The Singleton bound also indicates there is no realizable [6,4,4] linear error-correcting code.
There is another way to show there is no [6,5,4] linear error-correcting code. We have $64$ codewords and $32$ messages. By distance 4, there can be no pair of messages assigned to complementary codewords. By the pigeonhole principle, since there are only $32$ such pairs of codewords, any such pair has exactly one message assigned to it. This means there is a message assigned to exactly one of $000000_2$ and $111111_2$. Similarly, there is exactly one message assigned to the codeword pair $000111_2$ and $111000_2$. However, no such assignment is possible since any member of one pair is only distance three from either member of the other pair. beginalign*
d(000000_2, 111000_2) &= 3 text, \
d(000000_2, 000111_2) &= 3 text, \
d(111111_2, 111000_2) &= 3 text, and \
d(111111_2, 000111_2) &= 3 text.
endalign*
Consequently, there is no [6,5,4] linear error-correcting code.
(This counting argument can be adapted to show there is in [6,4,4] linear error-correcting code. Take the codewords in groups of $4$ having matching first four bits. This gives 16 groups each member of which differs from the other members of its group in only two bits, so at most one of those codewords is assigned to a message. There are 16 messages, so each group is assigned exactly one messsage. Then the groups $0000**_2$ and $0001**_2$, where "$*$" represents a bit whose value is either $0$ or $1$ (but this is not critical since either choice gives a codeword in the same group), each have a message, but regardless of which codeword is chosen in each group, those codewords differ in at most three bits.)
$endgroup$
$begingroup$
Thank you for the detailed answer
$endgroup$
– evgniy tayarov
Mar 30 at 18:34
add a comment |
$begingroup$
Assuming you mean a linear error-correcting code, a common interpretation of "[6,5,4] code" is a code using $6$-bit codewords to represent $5$-bit messages where any two distinct codewords differ in at least $4$ bits.
There are $2^6 = 64$ possible codewords. We have $2^5 = 32$ messages. In the lattice $0,1^6$, each codeword is surrounded by an open ball of radius $4$ containing no other codewords. One can use the Singleton bound to show that $5 + 4 leq 6 + 1$ if this code is realizable. However, $9 not leq 7$, so this code is not realizable.
The Singleton bound is a very loose bound. Generally, for a realizable [c,m,d] linear error-correcting code, the inequality is $m+d leq c+1$. The Singleton bound also indicates there is no realizable [6,4,4] linear error-correcting code.
There is another way to show there is no [6,5,4] linear error-correcting code. We have $64$ codewords and $32$ messages. By distance 4, there can be no pair of messages assigned to complementary codewords. By the pigeonhole principle, since there are only $32$ such pairs of codewords, any such pair has exactly one message assigned to it. This means there is a message assigned to exactly one of $000000_2$ and $111111_2$. Similarly, there is exactly one message assigned to the codeword pair $000111_2$ and $111000_2$. However, no such assignment is possible since any member of one pair is only distance three from either member of the other pair. beginalign*
d(000000_2, 111000_2) &= 3 text, \
d(000000_2, 000111_2) &= 3 text, \
d(111111_2, 111000_2) &= 3 text, and \
d(111111_2, 000111_2) &= 3 text.
endalign*
Consequently, there is no [6,5,4] linear error-correcting code.
(This counting argument can be adapted to show there is in [6,4,4] linear error-correcting code. Take the codewords in groups of $4$ having matching first four bits. This gives 16 groups each member of which differs from the other members of its group in only two bits, so at most one of those codewords is assigned to a message. There are 16 messages, so each group is assigned exactly one messsage. Then the groups $0000**_2$ and $0001**_2$, where "$*$" represents a bit whose value is either $0$ or $1$ (but this is not critical since either choice gives a codeword in the same group), each have a message, but regardless of which codeword is chosen in each group, those codewords differ in at most three bits.)
$endgroup$
$begingroup$
Thank you for the detailed answer
$endgroup$
– evgniy tayarov
Mar 30 at 18:34
add a comment |
$begingroup$
Assuming you mean a linear error-correcting code, a common interpretation of "[6,5,4] code" is a code using $6$-bit codewords to represent $5$-bit messages where any two distinct codewords differ in at least $4$ bits.
There are $2^6 = 64$ possible codewords. We have $2^5 = 32$ messages. In the lattice $0,1^6$, each codeword is surrounded by an open ball of radius $4$ containing no other codewords. One can use the Singleton bound to show that $5 + 4 leq 6 + 1$ if this code is realizable. However, $9 not leq 7$, so this code is not realizable.
The Singleton bound is a very loose bound. Generally, for a realizable [c,m,d] linear error-correcting code, the inequality is $m+d leq c+1$. The Singleton bound also indicates there is no realizable [6,4,4] linear error-correcting code.
There is another way to show there is no [6,5,4] linear error-correcting code. We have $64$ codewords and $32$ messages. By distance 4, there can be no pair of messages assigned to complementary codewords. By the pigeonhole principle, since there are only $32$ such pairs of codewords, any such pair has exactly one message assigned to it. This means there is a message assigned to exactly one of $000000_2$ and $111111_2$. Similarly, there is exactly one message assigned to the codeword pair $000111_2$ and $111000_2$. However, no such assignment is possible since any member of one pair is only distance three from either member of the other pair. beginalign*
d(000000_2, 111000_2) &= 3 text, \
d(000000_2, 000111_2) &= 3 text, \
d(111111_2, 111000_2) &= 3 text, and \
d(111111_2, 000111_2) &= 3 text.
endalign*
Consequently, there is no [6,5,4] linear error-correcting code.
(This counting argument can be adapted to show there is in [6,4,4] linear error-correcting code. Take the codewords in groups of $4$ having matching first four bits. This gives 16 groups each member of which differs from the other members of its group in only two bits, so at most one of those codewords is assigned to a message. There are 16 messages, so each group is assigned exactly one messsage. Then the groups $0000**_2$ and $0001**_2$, where "$*$" represents a bit whose value is either $0$ or $1$ (but this is not critical since either choice gives a codeword in the same group), each have a message, but regardless of which codeword is chosen in each group, those codewords differ in at most three bits.)
$endgroup$
Assuming you mean a linear error-correcting code, a common interpretation of "[6,5,4] code" is a code using $6$-bit codewords to represent $5$-bit messages where any two distinct codewords differ in at least $4$ bits.
There are $2^6 = 64$ possible codewords. We have $2^5 = 32$ messages. In the lattice $0,1^6$, each codeword is surrounded by an open ball of radius $4$ containing no other codewords. One can use the Singleton bound to show that $5 + 4 leq 6 + 1$ if this code is realizable. However, $9 not leq 7$, so this code is not realizable.
The Singleton bound is a very loose bound. Generally, for a realizable [c,m,d] linear error-correcting code, the inequality is $m+d leq c+1$. The Singleton bound also indicates there is no realizable [6,4,4] linear error-correcting code.
There is another way to show there is no [6,5,4] linear error-correcting code. We have $64$ codewords and $32$ messages. By distance 4, there can be no pair of messages assigned to complementary codewords. By the pigeonhole principle, since there are only $32$ such pairs of codewords, any such pair has exactly one message assigned to it. This means there is a message assigned to exactly one of $000000_2$ and $111111_2$. Similarly, there is exactly one message assigned to the codeword pair $000111_2$ and $111000_2$. However, no such assignment is possible since any member of one pair is only distance three from either member of the other pair. beginalign*
d(000000_2, 111000_2) &= 3 text, \
d(000000_2, 000111_2) &= 3 text, \
d(111111_2, 111000_2) &= 3 text, and \
d(111111_2, 000111_2) &= 3 text.
endalign*
Consequently, there is no [6,5,4] linear error-correcting code.
(This counting argument can be adapted to show there is in [6,4,4] linear error-correcting code. Take the codewords in groups of $4$ having matching first four bits. This gives 16 groups each member of which differs from the other members of its group in only two bits, so at most one of those codewords is assigned to a message. There are 16 messages, so each group is assigned exactly one messsage. Then the groups $0000**_2$ and $0001**_2$, where "$*$" represents a bit whose value is either $0$ or $1$ (but this is not critical since either choice gives a codeword in the same group), each have a message, but regardless of which codeword is chosen in each group, those codewords differ in at most three bits.)
edited Mar 30 at 18:44
answered Mar 30 at 18:14
Eric TowersEric Towers
33.8k22370
33.8k22370
$begingroup$
Thank you for the detailed answer
$endgroup$
– evgniy tayarov
Mar 30 at 18:34
add a comment |
$begingroup$
Thank you for the detailed answer
$endgroup$
– evgniy tayarov
Mar 30 at 18:34
$begingroup$
Thank you for the detailed answer
$endgroup$
– evgniy tayarov
Mar 30 at 18:34
$begingroup$
Thank you for the detailed answer
$endgroup$
– evgniy tayarov
Mar 30 at 18:34
add a comment |
$begingroup$
Assuming you are talking about non-trivial perfect binary codes of the form $[l,x,d]$, we have that the code can correct up to $e=lfloorfracd-12rfloor$ errors. If a non-trivial perfect binary code exists, it must hold that $sum_i=0^ebinomli$ is a power of $2$.
In our case, $d=4$, so $e=lfloorfrac32rfloor=1$ and $l=6$. $binom60+binom61=1+6=7$ which is not a power of $2$, so such code cannot exist, unless it's trivial.
$endgroup$
$begingroup$
Thank you!! This helps alot
$endgroup$
– evgniy tayarov
Mar 30 at 18:35
$begingroup$
This question just seems to be asking about the existence of a code, not about perfect codes. Your argument shows that a code with these parameters is not perfect, but not that it does not exist.
$endgroup$
– Morgan Rodgers
Mar 31 at 14:51
$begingroup$
@MorganRodgers If the OP leaves the people answering the question to guess what they're talking about, we'll fill in answers as we think is necessary. As the original question didn't seem too complicated, I thought they were talking about perfect codes as such a check is simple and short. If they were not, they should've specified that. If they use this answer even though it might be not what they looking for, then they should read up on the subject before asking a question and accepting random answers. Therefore, I don't think my answer is bad, it's just an answer to a bad question.
$endgroup$
– Marc
Mar 31 at 15:53
add a comment |
$begingroup$
Assuming you are talking about non-trivial perfect binary codes of the form $[l,x,d]$, we have that the code can correct up to $e=lfloorfracd-12rfloor$ errors. If a non-trivial perfect binary code exists, it must hold that $sum_i=0^ebinomli$ is a power of $2$.
In our case, $d=4$, so $e=lfloorfrac32rfloor=1$ and $l=6$. $binom60+binom61=1+6=7$ which is not a power of $2$, so such code cannot exist, unless it's trivial.
$endgroup$
$begingroup$
Thank you!! This helps alot
$endgroup$
– evgniy tayarov
Mar 30 at 18:35
$begingroup$
This question just seems to be asking about the existence of a code, not about perfect codes. Your argument shows that a code with these parameters is not perfect, but not that it does not exist.
$endgroup$
– Morgan Rodgers
Mar 31 at 14:51
$begingroup$
@MorganRodgers If the OP leaves the people answering the question to guess what they're talking about, we'll fill in answers as we think is necessary. As the original question didn't seem too complicated, I thought they were talking about perfect codes as such a check is simple and short. If they were not, they should've specified that. If they use this answer even though it might be not what they looking for, then they should read up on the subject before asking a question and accepting random answers. Therefore, I don't think my answer is bad, it's just an answer to a bad question.
$endgroup$
– Marc
Mar 31 at 15:53
add a comment |
$begingroup$
Assuming you are talking about non-trivial perfect binary codes of the form $[l,x,d]$, we have that the code can correct up to $e=lfloorfracd-12rfloor$ errors. If a non-trivial perfect binary code exists, it must hold that $sum_i=0^ebinomli$ is a power of $2$.
In our case, $d=4$, so $e=lfloorfrac32rfloor=1$ and $l=6$. $binom60+binom61=1+6=7$ which is not a power of $2$, so such code cannot exist, unless it's trivial.
$endgroup$
Assuming you are talking about non-trivial perfect binary codes of the form $[l,x,d]$, we have that the code can correct up to $e=lfloorfracd-12rfloor$ errors. If a non-trivial perfect binary code exists, it must hold that $sum_i=0^ebinomli$ is a power of $2$.
In our case, $d=4$, so $e=lfloorfrac32rfloor=1$ and $l=6$. $binom60+binom61=1+6=7$ which is not a power of $2$, so such code cannot exist, unless it's trivial.
answered Mar 30 at 18:14
MarcMarc
520211
520211
$begingroup$
Thank you!! This helps alot
$endgroup$
– evgniy tayarov
Mar 30 at 18:35
$begingroup$
This question just seems to be asking about the existence of a code, not about perfect codes. Your argument shows that a code with these parameters is not perfect, but not that it does not exist.
$endgroup$
– Morgan Rodgers
Mar 31 at 14:51
$begingroup$
@MorganRodgers If the OP leaves the people answering the question to guess what they're talking about, we'll fill in answers as we think is necessary. As the original question didn't seem too complicated, I thought they were talking about perfect codes as such a check is simple and short. If they were not, they should've specified that. If they use this answer even though it might be not what they looking for, then they should read up on the subject before asking a question and accepting random answers. Therefore, I don't think my answer is bad, it's just an answer to a bad question.
$endgroup$
– Marc
Mar 31 at 15:53
add a comment |
$begingroup$
Thank you!! This helps alot
$endgroup$
– evgniy tayarov
Mar 30 at 18:35
$begingroup$
This question just seems to be asking about the existence of a code, not about perfect codes. Your argument shows that a code with these parameters is not perfect, but not that it does not exist.
$endgroup$
– Morgan Rodgers
Mar 31 at 14:51
$begingroup$
@MorganRodgers If the OP leaves the people answering the question to guess what they're talking about, we'll fill in answers as we think is necessary. As the original question didn't seem too complicated, I thought they were talking about perfect codes as such a check is simple and short. If they were not, they should've specified that. If they use this answer even though it might be not what they looking for, then they should read up on the subject before asking a question and accepting random answers. Therefore, I don't think my answer is bad, it's just an answer to a bad question.
$endgroup$
– Marc
Mar 31 at 15:53
$begingroup$
Thank you!! This helps alot
$endgroup$
– evgniy tayarov
Mar 30 at 18:35
$begingroup$
Thank you!! This helps alot
$endgroup$
– evgniy tayarov
Mar 30 at 18:35
$begingroup$
This question just seems to be asking about the existence of a code, not about perfect codes. Your argument shows that a code with these parameters is not perfect, but not that it does not exist.
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– Morgan Rodgers
Mar 31 at 14:51
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This question just seems to be asking about the existence of a code, not about perfect codes. Your argument shows that a code with these parameters is not perfect, but not that it does not exist.
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– Morgan Rodgers
Mar 31 at 14:51
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@MorganRodgers If the OP leaves the people answering the question to guess what they're talking about, we'll fill in answers as we think is necessary. As the original question didn't seem too complicated, I thought they were talking about perfect codes as such a check is simple and short. If they were not, they should've specified that. If they use this answer even though it might be not what they looking for, then they should read up on the subject before asking a question and accepting random answers. Therefore, I don't think my answer is bad, it's just an answer to a bad question.
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– Marc
Mar 31 at 15:53
$begingroup$
@MorganRodgers If the OP leaves the people answering the question to guess what they're talking about, we'll fill in answers as we think is necessary. As the original question didn't seem too complicated, I thought they were talking about perfect codes as such a check is simple and short. If they were not, they should've specified that. If they use this answer even though it might be not what they looking for, then they should read up on the subject before asking a question and accepting random answers. Therefore, I don't think my answer is bad, it's just an answer to a bad question.
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– Marc
Mar 31 at 15:53
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$begingroup$
Linear code? Block code? Cyclic code? This question lacks context.
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– Eric Towers
Mar 30 at 17:46