Tolerance Orders of Open and Closed Unit Intervals

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 I_v 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 I_v 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.

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 I_x, I_y, I_z of type C, I_u of type A and I_v of type D, and if S = {A, B, C}, namely by making I_x, I_u, I_z of type A, I_y of type B and I_v 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 I_u nor I_v can be of type B, so without loss of generality, I_u is type C and I_v is type D. Now if I_z is type B or C we get v ≺ z, a contradiction, and if I_z 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 I_y and I_z to be type B and the remaining intervals to be type A, we get an AB-representation of H.

By Remark 2.4, intervals I_x, I_w, I_u, I_v cannot be of type B. If no intervals are type B, then without loss of generality, I_x, I_y, I_z, I_w are all of type C and because y ≺ u and v ≺ z, this forces I_u, I_v also to be of type C, a contradiction. If no interval is of type A, without loss of generality, I_x is of type C, forcing I_y, I_z, I_w to also be of type C, and then it is not possible to assign a type to I_u 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 I_x, I_z, I_v, of type A, I_y of type B, and I_w of type C. If $\mathcal I$ contains only intervals of type A and B, then I_z and I_w 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 I_y, I_z, I_w must all be of type C or all of type D and each of these cases leads to I_z, I_w getting identical intervals. Similarly, if S = {B, C, D}, intervals I_v, I_w, I_z 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 I_x, I_y, I_z of type C, I_u, I_v, I_w of type D, and I_t 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 I_x, I_y are type C and I_u, I_v are type D. Since u ≺ t and v ≺ t, interval I_t 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 I_x, I_y, I_z type C, I_u, I_v, I_w type D, and I_t 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 I_x, I_y are type C and I_u, I_v are type D. Since u ∥ t and x ∥ t, interval I_t 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.