Is there any work on partition a partial order set into minimum number total order subsets?
The problem is what's the minimum number of total order subsets can a partial order set partition into?
For example, (1,2) and (3,4) are comparable i.e. (1,2) < (3,4), and (1,2) and (2,1) are incomparable. For the partial order set { (1,2), (3,4), (2,1)} we can partition into two subsets {(1,2), (3,4)} and {(2,1)}, which both are total order sets.
I searched on the web and found little related documents. The question is that whether any works or solutions on this problem exist?
discrete-mathematics order-theory set-partition
edited Jan 4 at 8:58
Asaf Karagila♦
302k32427757
asked Jan 4 at 8:23
wang zhihaowang zhihao
1166
add a comment |
The problem is what's the minimum number of total order subsets can a partial order set partition into?
For example, (1,2) and (3,4) are comparable i.e. (1,2) < (3,4), and (1,2) and (2,1) are incomparable. For the partial order set { (1,2), (3,4), (2,1)} we can partition into two subsets {(1,2), (3,4)} and {(2,1)}, which both are total order sets.
I searched on the web and found little related documents. The question is that whether any works or solutions on this problem exist?
discrete-mathematics order-theory set-partition
edited Jan 4 at 8:58
Asaf Karagila♦
302k32427757
asked Jan 4 at 8:23
wang zhihaowang zhihao
1166
4
Consider Dilworth's theorem.
– William Elliot
Jan 4 at 9:10
1
@WilliamElliot Why not make that an answer?
– Arthur
Jan 4 at 9:33
@Arthur. Too short.
– William Elliot
Jan 4 at 11:31
@WilliamElliot Not at all.
– Arthur
Jan 4 at 11:36
add a comment |
The problem is what's the minimum number of total order subsets can a partial order set partition into?
For example, (1,2) and (3,4) are comparable i.e. (1,2) < (3,4), and (1,2) and (2,1) are incomparable. For the partial order set { (1,2), (3,4), (2,1)} we can partition into two subsets {(1,2), (3,4)} and {(2,1)}, which both are total order sets.
I searched on the web and found little related documents. The question is that whether any works or solutions on this problem exist?
discrete-mathematics order-theory set-partition
edited Jan 4 at 8:58
Asaf Karagila♦
302k32427757
asked Jan 4 at 8:23
wang zhihaowang zhihao
1166
The problem is what's the minimum number of total order subsets can a partial order set partition into?
For example, (1,2) and (3,4) are comparable i.e. (1,2) < (3,4), and (1,2) and (2,1) are incomparable. For the partial order set { (1,2), (3,4), (2,1)} we can partition into two subsets {(1,2), (3,4)} and {(2,1)}, which both are total order sets.
I searched on the web and found little related documents. The question is that whether any works or solutions on this problem exist?
discrete-mathematics order-theory set-partition
discrete-mathematics order-theory set-partition
edited Jan 4 at 8:58
Asaf Karagila♦
302k32427757
asked Jan 4 at 8:23
wang zhihaowang zhihao
1166
edited Jan 4 at 8:58
Asaf Karagila♦
302k32427757
asked Jan 4 at 8:23
wang zhihaowang zhihao
1166
edited Jan 4 at 8:58
Asaf Karagila♦
302k32427757
edited Jan 4 at 8:58
Asaf Karagila♦
302k32427757
edited Jan 4 at 8:58
Asaf Karagila♦
302k32427757
302k32427757
asked Jan 4 at 8:23
wang zhihaowang zhihao
1166
asked Jan 4 at 8:23
wang zhihaowang zhihao
1166
asked Jan 4 at 8:23
wang zhihaowang zhihao
1166
1166
4
Consider Dilworth's theorem.
– William Elliot
Jan 4 at 9:10
1
@WilliamElliot Why not make that an answer?
– Arthur
Jan 4 at 9:33
@Arthur. Too short.
– William Elliot
Jan 4 at 11:31
@WilliamElliot Not at all.
– Arthur
Jan 4 at 11:36
add a comment |
4
Consider Dilworth's theorem.
– William Elliot
Jan 4 at 9:10
1
@WilliamElliot Why not make that an answer?
– Arthur
Jan 4 at 9:33
@Arthur. Too short.
– William Elliot
Jan 4 at 11:31
@WilliamElliot Not at all.
– Arthur
Jan 4 at 11:36
4
4
Consider Dilworth's theorem.
– William Elliot
Jan 4 at 9:10
Consider Dilworth's theorem.
– William Elliot
Jan 4 at 9:10
1
1
@WilliamElliot Why not make that an answer?
– Arthur
Jan 4 at 9:33
@WilliamElliot Why not make that an answer?
– Arthur
Jan 4 at 9:33
@Arthur. Too short.
– William Elliot
Jan 4 at 11:31
@Arthur. Too short.
– William Elliot
Jan 4 at 11:31
@WilliamElliot Not at all.
– Arthur
Jan 4 at 11:36
@WilliamElliot Not at all.
– Arthur
Jan 4 at 11:36
add a comment |
1 Answer
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Given a finite partially ordered set, Dilworth's theorem says that the minimal number of totally ordered subsets needed to cover the entire set is equal to the size of the largest antichain (also known as the width of the set). An antichain is a subset where no element is comparable to any of the other elements in the antichain.
answered Jan 4 at 11:36
ArthurArthur
111k7105186
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Given a finite partially ordered set, Dilworth's theorem says that the minimal number of totally ordered subsets needed to cover the entire set is equal to the size of the largest antichain (also known as the width of the set). An antichain is a subset where no element is comparable to any of the other elements in the antichain.
answered Jan 4 at 11:36
ArthurArthur
111k7105186
add a comment |
Given a finite partially ordered set, Dilworth's theorem says that the minimal number of totally ordered subsets needed to cover the entire set is equal to the size of the largest antichain (also known as the width of the set). An antichain is a subset where no element is comparable to any of the other elements in the antichain.
answered Jan 4 at 11:36
ArthurArthur
111k7105186
add a comment |
Given a finite partially ordered set, Dilworth's theorem says that the minimal number of totally ordered subsets needed to cover the entire set is equal to the size of the largest antichain (also known as the width of the set). An antichain is a subset where no element is comparable to any of the other elements in the antichain.
answered Jan 4 at 11:36
ArthurArthur
111k7105186
Given a finite partially ordered set, Dilworth's theorem says that the minimal number of totally ordered subsets needed to cover the entire set is equal to the size of the largest antichain (also known as the width of the set). An antichain is a subset where no element is comparable to any of the other elements in the antichain.
answered Jan 4 at 11:36
ArthurArthur
111k7105186
edited Jan 4 at 12:32
answered Jan 4 at 11:36
ArthurArthur
111k7105186
answered Jan 4 at 11:36
ArthurArthur
111k7105186
answered Jan 4 at 11:36
ArthurArthur
111k7105186
111k7105186
add a comment |
add a comment |
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4
Consider Dilworth's theorem.
– William Elliot
Jan 4 at 9:10
1
@WilliamElliot Why not make that an answer?
– Arthur
Jan 4 at 9:33
@Arthur. Too short.
– William Elliot
Jan 4 at 11:31
@WilliamElliot Not at all.
– Arthur
Jan 4 at 11:36