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Dilworth's Theorem for Interval Orders
As remarked previously, we do not yet have an efficient process for determining the width of a poset and a minimum partition into chains. For interval orders, there is indeed a simple way to find both.
Dilworth's Theorem for Interval Orders
As remarked previously, we do not yet have an efficient process for determining the width of a poset and a minimum partition into chains. For interval orders, there is indeed a simple way to find both. The explanation is just to establish a connection with coloring of interval graphs as discussed in .
Let \(\PXP\) be an interval order and let \(\{[a_x,b_x]:x\in X\}\) be intervals of the real line so that \(x\lt y\) in \(\bfP\) if and only \(b_x\lt a_y\). Then let \(\bfG\) be the interval graph determined by this family of intervals. Note that if \(x\) and \(y\) are distinct elements of \(X\), then \(x\) and \(y\) are incomparable in \(\bfP\) if and only if \(xy\) is an edge in \(\bfG\). In other words, \(\bfG\) is just the incomparability graph of \(\bfP\).
Recall from that interval graphs are perfect, , \(\chi(\bfG)=\omega(\bfG)\) for every interval graph \(\bfG\). Furthermore, you can find an optimal coloring of an interval graph by applying first fit to the vertices in a linear order that respects left end points. Such a coloring concurrently determines a partition of \(\bfP\) into chains.
In fact, if you want to skip the part about interval representations, take any linear ordering of the elements as \(x_1\), \(x_2,\dots,x_n\) so that \(i\lt j\) whenever \(D(x)\) is a proper subset of \(D(y)\). Then apply First Fit with respect to chains. For example, using the \(10\) point interval order illustrated in , here is such a labeling: \[\begin{aligned}x_1 \amp=g \amp x_2 \amp=f \amp x_3 \amp=c \amp x_4 \amp=d \amp x_5 \amp=h \\ x_6 \amp=a \amp x_7 \amp=j \amp x_8 \amp=b \amp x_9 \amp=i \amp x_{10} \amp=e\end{aligned}\]
Now apply the First Fit algorithm to the points of \(\bfP\), in this order, to assign them to chains \(C_1\), \(C_2,\dots\). In other words, assign \(x_1\) to chain\(C_1\). Thereafter if you have assigned points \(x_1\), \(x_2,\dots,x_i\) to chains, then assign \(x_{i+1}\) to chain \(C_j\) where \(j\) is the least positive integer for which \(x_{i+1}\) is comparable to \(x_k\) whenever \(1\le k\le i\) and \(x_k\) has already been assigned to \(C_j\). For example, this rule results in the following chains for the interval order \(\bfP\) \[\begin{aligned}C_1 \amp = \{g,h,b\} \\ C_2 \amp = \{f,a,e\} \\ C_3 \amp = \{c,d\} \\ C_4 \amp = \{j\} \\ C_5 \amp = \{i\}\end{aligned}\] In this case, it is easy to see that the chain partition is optimal since the width of \(\bfP\) is \(5\) and \(A=\{a,b,d,i,j\}\) is a \(5\)-element antichain.
Condensed: the full section is in Keller & Trotter, Applied Combinatorics.
Now you Không máy tính nào có thể tính toán được nó, nhưng các mảnh của nó có thể tính toán được. Thử một trong những cách dưới đây, hoặc gõ theo ý bạn.
Luyện tập (2)
Đầu tiên thử mỗi câu trên giấy. Cho thấy câu trả lời để kiểm tra; những câu trả lời đã kiểm tra có thể mở trong trình giải cho mỗi bước.
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Start applying the First Fit algorithm for chain partitioning to the interval order shown below, ordering the intervals by left endpoint (and then right endpoint if you have a tie to break). Only apply the algorithm until you assign interval \(q\) to a chain! What chain is interval \(q\) assigned to? Why?
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Consider the interval order with representation shown below. Can you see how to visually inspect the interval representation to find its width without needing to apply the First Fit algorithm? If so, what is the width of this interval order and explain how you came up with this. If you're not able to see how to do this, briefly describe some things you've thought about and ask a question that might help get you unstuck.
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Nhấn vào bất kỳ ký hiệu nào để xem định nghĩa đầy đủ, một bức ảnh, và ý nghĩa của từng chữ cái trong nó.
Câu hỏi người ta hỏi
Permutation or combination?
Ask whether order matters. A lock code is a permutation (order matters); a hand of cards is a combination (it does not).
What is a graph in this sense?
Dots (vertices) joined by lines (edges), not a plot. Road maps, social networks and molecules are graphs; questions like "is there a route" and "how few colours" are graph theory.
Một số phần của trang này được chuyển từ Keller & Trotter, Applied Combinatorics (CC BY-SA 4.0). Được nén lại và giải thích lại ở đây; lỗi là của chúng tôi.
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