Abstract
In this paper we combine ideas from tolerance orders with recent work on OC interval orders. We consider representations of posets by unit intervals Iv in which the interval endpoints (L(v) and R(v)) may be open or closed as well as the center point (c(v)). This yields four types of intervals: A (endpoints and center points closed), B (endpoints and center points open), C (endpoints closed, center points open), and D (endpoints open, center points closed). For any non-empty subset S of {A, B, C, D}, we define an S-order as a poset P that has a representation as follows: each element v of P is assigned a unit interval Iv of type belonging to S, and x ≺ y if and only if either (i) R(x) < c(y) or (ii) R(x) = c(y) and at least one of R(x), c(y) is open and at least one of L(y), c(x) is open. We characterize several of the classes of S-orders and provide separating examples between unequal classes. In addition, for each S ⊆{A, B, C, D} we present a polynomial-time algorithm that recognizes S-orders, providing a representation when one exists and otherwise providing a certificate showing it is not an S-order.
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We appreciate the valuable suggestions made by the referees that improved the algorithm and the clarity of the paper.
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The third author’s work was supported by a Simons Foundation grant (#426725, Ann Trenk).
Appendix
Appendix
In Example 3.5 we showed that the posets 2 + 2, 3 + 1, 4 + 1, V, and Z are positioned correctly in the Venn diagrams of Figs. 4 and 5. In this appendix, we provide proofs for the remaining posets that appear in those figures.
The poset 3 + 1 + 1.
This poset has the forcing cycle \({\mathcal C}: x \prec y \prec z \parallel u \parallel x \prec y \prec z \parallel v \parallel x\) with \(val({\mathcal C}) = 0\). Suppose 3 + 1 + 1 were an S-order and without loss of generality fix an S-representation \(\mathcal I\) of it in which all intervals have length 2 and c(x) = 0. Now Theorem 3.3 implies that c(y) = 1, c(z) = 2, c(u) = 1, c(v) = 1. There are three unit intervals with center at 1, thus if |S| ≤ 2, two of these intervals must be identical. Identical intervals result in twins, so the poset 3 + 1 + 1 can not be induced in any twin-free S-order when |S| ≤ 2.
We also consider cases where |S| = 3. A representation is possible if S = {A, C, D}, namely by making Ix, Iy, Iz of type C, Iu of type A and Iv of type D, and if S = {A, B, C}, namely by making Ix, Iu, Iz of type A, Iy of type B and Iv of type C. However, an S-representation is not possible for S = {B, C, D}, as we now show. Since u ∥ z and v ∥ z, by Remark 2.4, we know neither Iu nor Iv can be of type B, so without loss of generality, Iu is type C and Iv is type D. Now if Iz is type B or C we get v ≺ z, a contradiction, and if Iz is type D we get u ≺ z, a contradiction.
The poset H.
This poset has the forcing cycle \({\mathcal C}: x \prec y \prec z \prec w \parallel u \parallel v \parallel x\) with \(val({\mathcal C}) = 0\). Suppose H were an S-order for some S. Without loss of generality, fix an S-representation \(\mathcal I\) of it which all intervals have length 2 and c(x) = 0. Now Theorem 3.3 implies that c(y) = 1, c(z) = 2, c(w) = 3, c(u) = 2, and c(v) = 1. By choosing Iy and Iz to be type B and the remaining intervals to be type A, we get an AB-representation of H.
By Remark 2.4, intervals Ix, Iw, Iu, Iv cannot be of type B. If no intervals are type B, then without loss of generality, Ix, Iy, Iz, Iw are all of type C and because y ≺ u and v ≺ z, this forces Iu, Iv also to be of type C, a contradiction. If no interval is of type A, without loss of generality, Ix is of type C, forcing Iy, Iz, Iw to also be of type C, and then it is not possible to assign a type to Iu because y ≺ u and u ∥ w. Thus H has an S-representation if and only if {A, B}⊆ S.
The poset Y.
We show that poset Y is an S-order for S = {A, B, C} but not when S = {A, C, D} or {B, C, D}, nor for any S with |S| ≤ 2. The same is true of its dual.
Poset Y has the forcing cycle \({\mathcal C}: x \prec y \prec z \parallel v \parallel x \prec y \prec w \parallel v \parallel x\) with \(val({\mathcal C}) = 0\). Suppose Y were an S-order for some S. Without loss of generality, fix an S-representation \(\mathcal I\) of it which all intervals have length 2 and c(x) = 0. Now Theorem 3.3 implies that c(y) = 1, c(z) = 2, c(w) = 2, c(v) = 1.
A representation is possible if S = {A, B, C}, namely by making Ix, Iz, Iv, of type A, Iy of type B, and Iw of type C. If \(\mathcal I\) contains only intervals of type A and B, then Iz and Iw must be of type A by Remark 2.4, resulting in z and w being twins.
Now consider S = {A, C, D} and for a contradiction, assume an S-representation is possible. By Remark 2.4, intervals Iy, Iz, Iw must all be of type C or all of type D and each of these cases leads to Iz, Iw getting identical intervals. Similarly, if S = {B, C, D}, intervals Iv, Iw, Iz must be type C or D by Remark 2.4, but z and w are both incomparable to v, so they are forced to get identical intervals, a contradiction.
Finally, Y is not induced in a twin-free S order for |S| ≤ 2, since we have shown this to be true for S = {A, B} and each other such S is a subset of {B, C, D} or {A, C, D}.
The same is true of its dual.
The poset X 1.
This poset has the forcing cycle \({\mathcal C}: x \prec t \prec w \parallel y \parallel u \prec t \prec z \parallel v \parallel x\) with \(val({\mathcal C}) = 0\). Suppose X1 were an S-order for some S. Without loss of generality, fix an S-representation \(\mathcal I\) of it in which all intervals have length 2 and c(x) = 0. Now Theorem 3.3 implies that c(y) = 1, c(z) = 2, c(u) = 0, c(v) = 1, c(w) = 2, c(t) = 1.
An S-representation is possible for S = {B, C, D}, namely by making Ix, Iy, Iz of type C, Iu, Iv, Iw of type D, and It of type B. We show that X1 is not an S-order for any other S, |S| ≤ 3.
Note that the elements x, y, u, v induce a 2 + 2 in X1. As seen in Example 3.5, without loss of generality Ix, Iy are type C and Iu, Iv are type D. Since u ≺ t and v ≺ t, interval It must be type B.
The poset X 2.
This poset has the forcing cycle \({\mathcal C}: x \prec y \prec z \parallel t \parallel u \prec v \prec w \parallel t \parallel x\) with \(val({\mathcal C}) = 0\). Suppose X2 were an S-order for some S. Without loss of generality, fix an S-representation \(\mathcal I\) of it in which all intervals have length 2 and c(x) = 0. Now Theorem 3.3 implies that c(y) = 1, c(z) = 2, c(u) = 0, c(v) = 1, c(w) = 2, c(t) = 1.
An S-representation is possible for S = {A, C, D}, namely by making Ix, Iy, Iz type C, Iu, Iv, Iw type D, and It type A. We show that X2 is not an S-order for any other S, |S| ≤ 3.
As in the case of poset X1, the elements x, y, u, v induce a 2 + 2 in X2. Again, without loss of generality Ix, Iy are type C and Iu, Iv are type D. Since u ∥ t and x ∥ t, interval It must be type A.
The poset X 3.
The poset X3 contains both posets X1 and X2 and hence is not an S-order for |S| ≤ 3. A representation is possible for S = {A, B, C, D} by starting with the BCD-representation for X1 given above and introducing an additional interval, of type A, centered at 1.
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Shuchat, A., Shull, R. & Trenk, A.N. Tolerance Orders of Open and Closed Unit Intervals. Order 36, 313–333 (2019). https://doi.org/10.1007/s11083-018-9468-1
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DOI: https://doi.org/10.1007/s11083-018-9468-1