Finding maximal antichain in poset of binary stringsMin-cut Max-flow $Rightarrow$ Dilworth's theoremThe...
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Finding maximal antichain in poset of binary strings
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Define the partial order on a set $X$ of all binary strings of length n to be $xRy$ if and only if $x=y$ or $x$ has an odd number of ones and $x$ and $y$ are adjacent vertices on the hypercube $Q_n$, ie differ in exactly one place. I'm able to show this is a partial order easily enough, but I'm struggling to find a maximal antichain $A$ on this poset, and I'm also struggling to find a minimum chain covering with $|A|$ chains.
I've written out the partial order for $n=3$ but still am not getting anywhere, even with finding an antichain.
combinatorics combinatorial-geometry
$endgroup$
add a comment |
$begingroup$
Define the partial order on a set $X$ of all binary strings of length n to be $xRy$ if and only if $x=y$ or $x$ has an odd number of ones and $x$ and $y$ are adjacent vertices on the hypercube $Q_n$, ie differ in exactly one place. I'm able to show this is a partial order easily enough, but I'm struggling to find a maximal antichain $A$ on this poset, and I'm also struggling to find a minimum chain covering with $|A|$ chains.
I've written out the partial order for $n=3$ but still am not getting anywhere, even with finding an antichain.
combinatorics combinatorial-geometry
$endgroup$
1
$begingroup$
If $x<y$, you must have $x$ has an odd number of ones and $y$ has an even number of ones. Two strings are guaranteed to be unrelated if they have the same parity...does this help you find a large antichain?
$endgroup$
– Mike Earnest
Mar 19 at 3:48
$begingroup$
From what I gather after this, the largest collection of unrelated strings will be the strings with all even parity, which is exactly $frac{n}{2}$. Do you think this heading in the right direction? Thanks for the tip.
$endgroup$
– EJJJJ
Mar 19 at 3:50
1
$begingroup$
Yes, though I think you mean $2^n/2$. Half the strings in $Q_n$ have even parity. Now you just need to partition $Q_n$ into that many chains. I think with this new info you should go back to drawing pictures; if you find the chain partition by hand for $n=1,2,3,4$, you will likely see a pattern.
$endgroup$
– Mike Earnest
Mar 19 at 3:56
$begingroup$
You're totally right, I did mean $frac{2^n}{2}$, thanks for catching that. I really appreciate it!
$endgroup$
– EJJJJ
Mar 19 at 3:56
add a comment |
$begingroup$
Define the partial order on a set $X$ of all binary strings of length n to be $xRy$ if and only if $x=y$ or $x$ has an odd number of ones and $x$ and $y$ are adjacent vertices on the hypercube $Q_n$, ie differ in exactly one place. I'm able to show this is a partial order easily enough, but I'm struggling to find a maximal antichain $A$ on this poset, and I'm also struggling to find a minimum chain covering with $|A|$ chains.
I've written out the partial order for $n=3$ but still am not getting anywhere, even with finding an antichain.
combinatorics combinatorial-geometry
$endgroup$
Define the partial order on a set $X$ of all binary strings of length n to be $xRy$ if and only if $x=y$ or $x$ has an odd number of ones and $x$ and $y$ are adjacent vertices on the hypercube $Q_n$, ie differ in exactly one place. I'm able to show this is a partial order easily enough, but I'm struggling to find a maximal antichain $A$ on this poset, and I'm also struggling to find a minimum chain covering with $|A|$ chains.
I've written out the partial order for $n=3$ but still am not getting anywhere, even with finding an antichain.
combinatorics combinatorial-geometry
combinatorics combinatorial-geometry
asked Mar 19 at 3:04
EJJJJEJJJJ
456
456
1
$begingroup$
If $x<y$, you must have $x$ has an odd number of ones and $y$ has an even number of ones. Two strings are guaranteed to be unrelated if they have the same parity...does this help you find a large antichain?
$endgroup$
– Mike Earnest
Mar 19 at 3:48
$begingroup$
From what I gather after this, the largest collection of unrelated strings will be the strings with all even parity, which is exactly $frac{n}{2}$. Do you think this heading in the right direction? Thanks for the tip.
$endgroup$
– EJJJJ
Mar 19 at 3:50
1
$begingroup$
Yes, though I think you mean $2^n/2$. Half the strings in $Q_n$ have even parity. Now you just need to partition $Q_n$ into that many chains. I think with this new info you should go back to drawing pictures; if you find the chain partition by hand for $n=1,2,3,4$, you will likely see a pattern.
$endgroup$
– Mike Earnest
Mar 19 at 3:56
$begingroup$
You're totally right, I did mean $frac{2^n}{2}$, thanks for catching that. I really appreciate it!
$endgroup$
– EJJJJ
Mar 19 at 3:56
add a comment |
1
$begingroup$
If $x<y$, you must have $x$ has an odd number of ones and $y$ has an even number of ones. Two strings are guaranteed to be unrelated if they have the same parity...does this help you find a large antichain?
$endgroup$
– Mike Earnest
Mar 19 at 3:48
$begingroup$
From what I gather after this, the largest collection of unrelated strings will be the strings with all even parity, which is exactly $frac{n}{2}$. Do you think this heading in the right direction? Thanks for the tip.
$endgroup$
– EJJJJ
Mar 19 at 3:50
1
$begingroup$
Yes, though I think you mean $2^n/2$. Half the strings in $Q_n$ have even parity. Now you just need to partition $Q_n$ into that many chains. I think with this new info you should go back to drawing pictures; if you find the chain partition by hand for $n=1,2,3,4$, you will likely see a pattern.
$endgroup$
– Mike Earnest
Mar 19 at 3:56
$begingroup$
You're totally right, I did mean $frac{2^n}{2}$, thanks for catching that. I really appreciate it!
$endgroup$
– EJJJJ
Mar 19 at 3:56
1
1
$begingroup$
If $x<y$, you must have $x$ has an odd number of ones and $y$ has an even number of ones. Two strings are guaranteed to be unrelated if they have the same parity...does this help you find a large antichain?
$endgroup$
– Mike Earnest
Mar 19 at 3:48
$begingroup$
If $x<y$, you must have $x$ has an odd number of ones and $y$ has an even number of ones. Two strings are guaranteed to be unrelated if they have the same parity...does this help you find a large antichain?
$endgroup$
– Mike Earnest
Mar 19 at 3:48
$begingroup$
From what I gather after this, the largest collection of unrelated strings will be the strings with all even parity, which is exactly $frac{n}{2}$. Do you think this heading in the right direction? Thanks for the tip.
$endgroup$
– EJJJJ
Mar 19 at 3:50
$begingroup$
From what I gather after this, the largest collection of unrelated strings will be the strings with all even parity, which is exactly $frac{n}{2}$. Do you think this heading in the right direction? Thanks for the tip.
$endgroup$
– EJJJJ
Mar 19 at 3:50
1
1
$begingroup$
Yes, though I think you mean $2^n/2$. Half the strings in $Q_n$ have even parity. Now you just need to partition $Q_n$ into that many chains. I think with this new info you should go back to drawing pictures; if you find the chain partition by hand for $n=1,2,3,4$, you will likely see a pattern.
$endgroup$
– Mike Earnest
Mar 19 at 3:56
$begingroup$
Yes, though I think you mean $2^n/2$. Half the strings in $Q_n$ have even parity. Now you just need to partition $Q_n$ into that many chains. I think with this new info you should go back to drawing pictures; if you find the chain partition by hand for $n=1,2,3,4$, you will likely see a pattern.
$endgroup$
– Mike Earnest
Mar 19 at 3:56
$begingroup$
You're totally right, I did mean $frac{2^n}{2}$, thanks for catching that. I really appreciate it!
$endgroup$
– EJJJJ
Mar 19 at 3:56
$begingroup$
You're totally right, I did mean $frac{2^n}{2}$, thanks for catching that. I really appreciate it!
$endgroup$
– EJJJJ
Mar 19 at 3:56
add a comment |
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$begingroup$
If $x<y$, you must have $x$ has an odd number of ones and $y$ has an even number of ones. Two strings are guaranteed to be unrelated if they have the same parity...does this help you find a large antichain?
$endgroup$
– Mike Earnest
Mar 19 at 3:48
$begingroup$
From what I gather after this, the largest collection of unrelated strings will be the strings with all even parity, which is exactly $frac{n}{2}$. Do you think this heading in the right direction? Thanks for the tip.
$endgroup$
– EJJJJ
Mar 19 at 3:50
1
$begingroup$
Yes, though I think you mean $2^n/2$. Half the strings in $Q_n$ have even parity. Now you just need to partition $Q_n$ into that many chains. I think with this new info you should go back to drawing pictures; if you find the chain partition by hand for $n=1,2,3,4$, you will likely see a pattern.
$endgroup$
– Mike Earnest
Mar 19 at 3:56
$begingroup$
You're totally right, I did mean $frac{2^n}{2}$, thanks for catching that. I really appreciate it!
$endgroup$
– EJJJJ
Mar 19 at 3:56