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


Min-cut Max-flow $Rightarrow$ Dilworth's theoremThe maximum size of an antichain in a posetFinding a Hamiltonian cycle for $Q_4$Finding the minimum number of chains of a given poset (partially ordered set)Is it always possible to get MC/DC coverage on an $n$-input Boolean function with $n + 1$ test cases?Number of ways of coloring projected faces of any hypercubeCounting certain digits - How to do this efficiently?What is the name of this class of (combinatorial?) problems?What is the Length of the maximal antichain in a Poset of cardinality n?Some doubt about minimal antichain cover of poset.













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$begingroup$


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.










share|cite|improve this question









$endgroup$








  • 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
















0












$begingroup$


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.










share|cite|improve this question









$endgroup$








  • 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














0












0








0





$begingroup$


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.










share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










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