Triangle, Parallelogram, and Trapezoid Orders Kenneth P. Bogart ∗ Department of Mathematics 27 N. Main Street 6188 Kemeny Hall Hanover, NH 03755

Joshua D. Laison Mathematics Department Willamette University Salem, OR 97301 [email protected]

Stephen P. Ryan † Department of Mathematics 27 N. Main Street 6188 Kemeny Hall Hanover, NH 03755 [email protected] December 21, 2009

Abstract In his 1998 paper, Ryan classified the sets of unit, proper, and plain trapezoid and parallelogram orders. We extend this classification to include unit, proper, and plain triangle orders. We prove that there are 20 combinations of these properties that give rise to distinct classes of ordered sets, and order these classes by containment.

Mathematics Subject Classifications (2000) 06A06, 05C62 Keywords: trapezoid order, triangle order, parallelogram order, trapezoid graph ∗

deceased Supported by ONR contract N0014-94-1-0950 and NSF grant DMS-9022140. Current affiliation Dartware, LLC, 66-7 Benning Street, West Lebanon, NH 03784 †

1

1

Introduction

Suppose that B1 and B2 are two horizontal lines in the plane, with B1 below B2 , called baselines, and suppose that R is a finite set of trapezoids, each of which has one base on each of the baselines. We allow trapezoids to have one or two bases of length zero. If x and y are trapezoids in R, we define x < y if and only if x and y are disjoint, and the base of x on B1 is to the left of the base of y on B1 . An ordered set P is called a trapezoid order if there exists such an ordered set of trapezoids R such that P ∼ = R. This set of trapezoids R is called a trapezoid representation of P . Following [1] and [4], if x is a trapezoid in a representation, we denote its left and right vertices on B1 by l(x) and r(x), respectively, and its left and right vertices on B2 by L(x) and R(x), respectively. If the line segments l(x)L(x) and r(x)R(x) are parallel for every element of R, then R is called a parallelogram representation of P and P a parallelogram order. If either l(x) = r(x) or L(x) = R(x) for every element x ∈ R, then R is called a triangle representation of P and P a triangle order. If L(x) = R(x) for every element x ∈ R, then we call R an up-triangle representation of P and P an up-triangle order. We discuss up-triangle orders in Section 4. If all of the geometric objects in the set R have unit area, then we say that R is a unit representation of P and P is a unit order. If no object in R is properly contained in any other, then we say that R is a proper representation of P and P is a proper order. When R might not be unit or proper, we emphasize this fact by calling R a plain representation and P a plain order. The terms unit, proper, and plain will always be thought of as modifying one of the terms trapezoid, parallelogram, or triangle. All ordered sets in this paper are trapezoid orders. In particular, any ordered set which has a triangle or parallelogram representation is a trapezoid order. The goal of this paper is to compare the different classes of unit, proper, and plain trapezoid, parallelogram, and triangle orders. In Section 2 we determine that there are 20 distinct classes of ordered sets defined in this way. In Section 3 we complete the classification of these 20 classes of ordered sets. Figures 2 and 3 summarize our results.

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2

Related Classes of Trapezoid Orders

We begin by discussing the various combinations of these definitions which are permissable. First, a unit order is always a proper order, since one polygon cannot be properly contained in another with the same area. Also, a plain triangle order is always a proper triangle order, as the following lemma shows. Lemma 1. Every (finite) triangle order is a proper triangle order. Proof. Suppose P is a triangle order. Then there exists a triangle representation R of P . If R is not a proper representation, then there exist triangles x and y such that x is properly contained in y, as in Figure 1. The triangles x and y must both have a single vertex on the same baseline, which we may take to be B2 . Since R is finite, the nearest vertex to L(x) = L(y) is some positive distance ² away. We can therefore move the vertex L(x) to the right by 2² , and the new set of triangles R is still a triangle representation of P . In this way, we may eliminate all proper containment of triangles in R.

Figure 1: The elimination of one proper containment of triangles in a triangle representation. Any parallelogram representation is also a trapezoid representation, so every parallelogram order is also a trapezoid order. If the representation is unit or proper, then it will still be unit or proper as a trapezoid representation. Therefore every proper parallelogram (resp. triangle) order is a proper trapezoid order, and every unit parallelogram (resp. triangle) order is a unit trapezoid order.

3

Parallelogram

Trapezoid orders

8

Proper Proper Parallelogram

Unit Unit parallelogram Triangle (proper triangle)

5

3 Unit Triangle

2

4

6

1 7

Figure 2: The Venn diagram of classes of trapezoid orders.

Non-P

8

Non-Proper Parallelogram

2

NonParallelogram

6 4 Proper Non-Unit Parallelogram

Unit Parallelogram

ngle n-Tria o N t i 7 -Un r Non it Prope le Non-Un 5 Triang -Triangle on Unit N t Both 3 nit, no U r o e l e Triang riangl 1 Unit T

roper

Figure 3: The ordered set of classes of trapezoid orders.

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3

Separating Examples

Taking all combinations of unit, proper, plain, or non-triangle, unit, proper, plain, or non-parallelogram, and unit, proper, or plain trapezoid orders yields 48 possible classes of ordered sets. After applying the relations in the above section, we are left with 20 possible classes. In this section, we show that all 20 of these classes are in fact distinct. The 20 distinct classes of trapezoid orders are shown in Figure 2 in a Venn diagram, and in Figure 3 in an order diagram. In fact, the ordered set shown in Figure 3 is a concept lattice [2, 3]. The eight join-irreducible elements in the lattice are labeled 1 through 8, and examples of ordered sets in each of these eight classes are given below. If P and Q are elements of the concept lattice in Figure 3, with ordered sets P ∈ P and Q ∈ Q, then the sum P + Q is an ordered set in the class P ∨ Q. Note also that P + Q ∈ P if and only if Q ⊆ P. In this way we obtain an example in every class of ordered sets in Figure 3, but not in any proper subclass shown in the figure, by taking sums of the eight examples below. We denote the ordered set in Example i by Pi . For each of the eight examples, we display an order diagram and a proof that the given ordered set has the claimed properties. Some of these examples are modified from [1] and [4]. Example 1. Unit triangle and unit parallelogram G

3

D

F

2

C

E

1

B

Figure 4: The ordered set P1 , also called the jaw order. Although every linear order is both a unit triangle and a unit parallelogram order, it will be useful to have the following example of a unit triangle, unit parallelogram order for later use. The order P1 shown in Figure 4 is also called the jaw order [1]. The proof that the jaw order is a unit parallelo5

gram order appears in [1]. Since P1 ∼ = P2 − x, a unit triangle representation of P1 can be obtained from Figure 7 by deleting the triangle x. The key property of the ordered set P1 is given in the following lemma, which appears in [1]. Lemma 2. In any trapezoid representation of P1 , we must have either r(E) < l(2) ≤ r(2) < l(D) and R(B) < L(C) ≤ R(1) < L(2) ≤ R(E) < L(D) ≤ R(2) < L(3) ≤ R(F ) < L(G) or R(E) < L(2) ≤ R(2) < L(D) and r(B) < l(C) ≤ r(1) < l(2) ≤ r(E) < l(D) ≤ r(2) < l(3) ≤ r(F ) < l(G). Figure 5 shows the relations l(2) ≤ r(E) < l(D) ≤ r(2) given in the second of the two possibilities of Lemma 2. Note that in this case any trapezoid X greater than E and less than D must have its bottom base properly contained in the bottom base of the trapezoid 2. For this reason, the elements E and D are called the teeth of the jaw [1].

E

2 l(2)

r(E)

D l(D)

r(2)

Figure 5: A visual representation of Lemma 2. Example 2. Unit triangle, and proper parallelogram, but not unit parallelogram The ordered set P2 is shown in Figure 6. We consider a parallelogram representation S of P2 , and show that S is not unit. Since P2 is obtained from P1 by adding the element x, by Lemma 2, the parallelogram x in S must have one of its bases properly contained in a base of the parallelogram 2. Since opposite bases of parallelograms have equal lengths, this implies that the area of x must be smaller than the area of 2. Therefore S is not a unit parallelogram representation of P2 . 6

G x

F E

3

D

2

C

1

B

Figure 6: The ordered set P2 . x

1 3 G

2

B

C

D

E

F

Figure 7: A unit triangle representation of P2 .

2 3 1

x B

C

F D

E

G

Figure 8: A proper parallelogram representation of P2 . Figure 7 shows a unit triangle representation of P2 , and Figure 8 shows a proper parallelogram representation of P2 . Example 3. Triangle, and unit parallelogram, but not unit triangle We display a triangle representation of P3 in Figure 10 and a unit parallelogram representation of P3 in Figure 11. It remains to show that no triangle representation of P3 is unit. To prove this, we first note that the two suborders of P3 induced on the subsets {1, 3, 5, b, c, d, e, f, g} and {1, X, 5, b, c, d, e, f, g} are both isomorphic to P1 . 7

5 h

4

d

g

3

c

f

2

b

e

1

a

X

Figure 9: The ordered set P3 .

e

g

f

h X 5

1 3

2 a

c

b

4

d

Figure 10: A triangle representation of P3 .

f

e

g 3

X

1 2

4 5

h

d a

b

c

Figure 11: A unit parallelogram representation of P3 . Therefore by Lemma 2, l(3) ≤ r(e) < l(d) ≤ r(3) or L(3) ≤ R(e) < L(d) ≤ R(3), and l(X) ≤ r(e) < l(d) ≤ r(X) or L(X) ≤ R(e) < L(d) ≤ R(X) 8

By Theorem 2 of [1], we cannot have the above inequalities occur on different baselines. Therefore without loss of generality, l(3) ≤ r(e) < l(d) ≤ r(3) and l(X) ≤ r(e) < l(d) ≤ r(X). Therefore Lemma 2 also gives us l(b) < r(1), and since a < b and 1 < X in P3 , we must have L(X) ≤ R(a) for the triangles representing X and a to intersect. Since a < 4 in P3 , R(a) < L(4), so L(X) < L(4). By the same argument, R(2) < R(X), and since X has its base on B1 , R(X) = L(X), so the apex of X is strictly between 2 and 4 on B2 . Since X||2 and X||4 in P3 , we must therefore have r(X) > l(4) and l(X) < r(2). Since 2 < 3 < 4 in P3 , r(2) < l(3) < r(3) < l(4). Along with the statement above, this implies that l(X) < r(2) < l(3) < r(3) < l(4) < r(X). Since the base of 3 is contained in the base of X, 3 must have a smaller area than X in any triangle representation of P3 , and so P3 is not a unit triangle order. Example 4. Unit triangle, and parallelogram, but not proper parallelogram d2

3 g

d1 x2

f

c

2

e2

b x1

e1

a

1

Figure 12: The ordered set P4 . Figure 13 shows a parallelogram representation of P4 , and Figure 14 shows a unit triangle representation of P4 . Theorem 3. Every parallelogram representation R of P4 is non-proper. Proof. We refer to parallelograms in R by their associated elements in P4 . Step 1. R(e2) < L(2) < L(d1). The ordered set P4 has two suborders isomorphic to the jaw order, {b, c, d1, e1, f, g, 1, 2, 3} and {b, c, d2, e2, f, g, 1, 2, 3}. Therefore by Lemma 2, 9

x1

3

1 e1

a

b

2

e2

x2

d1 d2 f

c

g

Figure 13: A parallelogram representation of P4 .

a

x1

e2

e1

b

c

x2

d1

d2

2 3

1

f

g

Figure 14: A unit triangle representation of P4 .

r(e1) < l(2) ≤ r(2) < l(d1) and R(b) < L(c) ≤ R(1) < L(2) ≤ R(e1) < L(d1) ≤ R(2) < L(3) ≤ R(f ) < L(g)

(1)

or R(e1) < L(2) ≤ R(2) < L(d1) and r(b) < l(c) ≤ r(1) < l(2) ≤ r(e1) < l(d1) ≤ r(2) < l(3) ≤ r(f ) < l(g) and

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(2)

r(e2) < l(2) ≤ r(2) < l(d2) and R(b) < L(c) ≤ R(1) < L(2) ≤ R(e2) < L(d2) ≤ R(2) < L(3) ≤ R(f ) < L(g).

(3)

or R(e2) < L(2) ≤ R(2) < L(d2) and r(b) < l(c) ≤ r(1) < l(2) ≤ r(e2) < l(d2) ≤ r(2) < l(3) ≤ r(f ) < l(g)

(4)

The combination of equations (2) and (3) above results in r(e2) < l(2) ≤ r(e1), which is a contradiction, since e1 < e2 in P4 . Similarly, the combination of equations (1) and (4) results in the contradiction R(e2) < R(e1). The remaining two possibilities are symmetric; without loss of generality we conclude that equations (2) and (4) are true. This gives us the equations R(e1) < R(e2) < L(2) ≤ R(2) < L(d1) < L(d2), and r(b) < l(c) ≤ r(1) < l(2) ≤ {r(e1), r(e2), l(d1), l(d2)} ≤ r(2) < l(3) ≤ r(f ) < l(g), since e1 < e2 and d1 < d2 in P4 . Figure 13 shows a parallelogram representation of P4 with these inequalities satisfied. In particular, R(e2) < L(2) < L(d1). Step 2. L(x2) < R(x1). Since e2||d1 in P4 , we must have l(d1) < r(e2). Since x1 < d1 and e2 < x2 in P4 , the bottom interval of x1 occurs entirely to the left of the bottom interval of x2. But x1||x2 in P4 , so we must have L(x2) < R(x1). Step 3. L(x1) < R(x2), and the top intervals of x1 and x2 intersect. Since x1||1 in P4 , we know that L(x1) < R(1). However, x2 < a in P4 , which overlaps 2 on B2 , and so L(x1) ≤ R(1) < L(2) ≤ R(a) < L(x2) < R(x2). Step 4. x2 is properly contained in 2. Since the bottom bases of both x1 and x2 are contained in the bottom base of 2 and are disjoint, the 11

sum of the lengths of the bottom bases of x1 and x2 must be less than the length of the bottom base of 2. Since they are parallelograms, this must be true for the top bases of x1, x2, and 2 as well. Therefore since L(x1) < L(2) and L(x2) < R(x1), we must also have R(x2) < R(2). Therefore, x2 is completely contained in 2, and thus the parallelogram representation R of P4 is not proper. Example 5. Unit parallelogram, but not triangle d

z y

3

c

x

2

b

w

1

a

Figure 15: The ordered set P5 .

w

y

x

z

1

a

b

c

3

2

d

Figure 16: A unit parallelogram representation of P5 Theorem 2 in [1] proved that in the ordered set P8 , shown in Figure 21, l(2) < r(2) and L(2) < R(2). Therefore the element 2 in P8 cannot be represented as a triangle. This conclusion still holds without the element N , so that P5 is also not a triangle order. Figure 21 shows a unit parallelogram representation of P5 . Example 6. Unit triangle, not parallelogram 12

3

G

2

D x

F

C 1

E

B

Figure 17: The ordered set P6 . x

1 3

B

G

2 C

D

E

F

Figure 18: A unit triangle representation of P6 .

The proof that the ordered set P6 is not a parallelogram order appears in [4]. Figure 18 shows a unit triangle representation of P6 . Example 7. Triangle, and proper parallelogram, but not unit trapezoid h g

z

4

d

3

c

2

b

1

a

y f

x

e

Figure 19: The ordered set P7 . The ordered set P7 is not a unit trapezoid order [1], and therefore not a unit triangle order or unit parallelogram order. Figure 20 shows a proper triangle representation of P7 . 13

a

h 4

3 1

d

2 e

bc

x y z

fg

Figure 20: A triangle representation of P7 .

Example 8. Parallelogram, but not triangle, and not proper trapezoid d

z 3

c

x

2

b

w

1

a

y N

Figure 21: The ordered set P8 . The ordered set P5 is a suborder of P8 , and therefore P8 is not a triangle order. The proof that P8 is a parallelogram order and not a proper trapezoid order again appears in [1].

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Up-Triangle Orders

We might also consider triangle representations in which the base of every triangle is on the bottom baseline, called up-triangle orders. The ordered sets representable this way are exactly the ordered sets which are the intersection of a linear order and an interval order. Since the following proposition shows that every such ordered set is also a unit trapezoid order, P7 is also an example of a triangle order that is not an up-triangle order. Proposition 4. Every up-triangle order is a unit trapezoid order. 14

Proof. Let P be an up-triangle order, and R an up-triangle representation of P . Let B be the longest base of all triangles in R, and suppose that B has length β. For each triangle T in R, if its base has length τ , we stretch its top vertex to an interval of length β − τ . We shift the intervals to the right of T on B2 further to the right as this stretch is performed, so that T has the same intersections with other polygons in R that it did previously. In the resulting trapezoid representation R0 , the sum of the bases of each trapezoid is β, by construction. Thus, R0 is a unit trapezoid representation, and P is a unit trapezoid order. By Proposition 4, the class of up-triangle orders appears below the join of the classes labeled 3 and 6 in Figure 3. We leave the classification of up-triangle orders within this sub-lattice as an open problem.

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Acknowledgements

Authors Laison and Ryan are deeply indebted to Kenneth P. Bogart, their advisor, co-author, and friend, without whom this paper and much more would not have been possible.

References [1] Kenneth P. Bogart, Rolf H. M¨ohring, and Stephen P. Ryan. Proper and unit trapezoid orders and graphs. Order, 15:325–340, 1998. [2] B.A. Davey and H.A. Priestley. Introduction to Lattices and Order, 2E. Cambridge University Press, 2002. [3] Bernhard Ganter and Rudolf Wille. Formal concept analysis. SpringerVerlag, Berlin, 1999. Mathematical foundations, Translated from the 1996 German original by Cornelia Franzke. [4] Stephen P. Ryan. Trapezoid order classification. Order, 15:341–354, 1998.

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