1 Introduction
For the purpose of this paper, an arrangement is a (finite) collection of curves such as lines or circles in the plane. The study of arrangements has a long history; for example, Grünbaum [14] studied arrangements of lines in the projective plane. Arrangements of circles and other closed curves have also been studied extensively [1, 2, 12, 18, 21]. An arrangement is simple if no point of the plane belongs to more than two curves and every two curves intersect. A face of an arrangement in the projective or Euclidean plane is a connected component of the subdivision induced by the curves in , that is, a face is a component of .
For a given type of curves, people have investigated the maximum number of faces that an arrangement of such curves can form. Already in 1826, Steiner [22] showed that a simple arrangement of straight lines can have at most faces while an arrangement of circles can have at most faces. Alon et al. [2] studied the number of faces of degree 2, that is, faces that are bounded by two edges, for various kinds of arrangements of circles: arrangements of unit circles, unit circles where every two circles intersect (see also [21]), arbitrary circles with sufficiently many intersecting pairs, or arbitrary circles where every two pairs intersect. For example, arrangements of unit circles can have at most faces of degree 2, where is the number of circles in the arrangement [2]. Pinchasi [21] showed that the bound is for unit circle arrangements in which every pair of circles intersects whereas Alon et al. [2] showed that arrangements of circles with arbitrary radii have at most faces of degree 2 if every pair of circles intersect.
The same arrangements can, however, have quadratically many triangular faces. A lower bound example can be constructed from a simple arrangement of lines with quadratically many triangular faces by projecting it on a sphere (disjoint from the plane containing ) and having each line become a great circle. This is always possible since the line arrangement is simple; for more details see [11, Section 5.1]. In this process we obtain triangular faces, where is the number of triangular faces in the line arrangement. The great circles on the sphere can then be transformed into a circle arrangement in a different plane using the stereographic projection. This gives rise to an arrangement of circles with triangular faces in this plane. Füredi and Palásti [13] provided simple line arrangements with triangular faces. With the argument above, this immediately yields a lower bound of on the number of triangular faces of arrangements of circles. Felsner and Scheucher [12] showed that this lower bound is tight by proving that an arrangement of pseudocircles (that is, closed curves that can intersect at most twice and no point belongs to more than two curves) can have at most triangular faces.
One can also specialize circle arrangements by fixing an angle at which each pair of intersecting circles intersect^{1}^{1}1i.e., measured as the angle between the two tangents at either intersection point.; this was recently discussed by Eppstein [9]. In this paper, we consider arrangements of circles with the restriction that each pair of circles must intersect at a right angle. Two circles intersecting at a right angle are called orthogonal. We make the following simple observation regarding orthogonal circles; see Fig. 2.
Observation 1
Let and be two circles with centers , and radii , , respectively. Then and are orthogonal if and only if .
We discuss further basic properties of orthogonal circles in Section 2. In particular, in an arrangement of orthogonal circles no two circles can touch and no three circles can intersect at the same point.
The main result of our paper is that arrangements of orthogonal circles have at most intersection points and at most faces; see Theorem 3.1 (in Section 3). This is especially interesting when contrasted with the fact that arrangements of orthogonal circular arcs can have quadratically many quadrangular faces; see the construction in Fig. 2. In addition, we consider faces of different types, that is, bounded by two, three, and four edges and provide bounds on the number of such faces in any arrangement of orthogonal circles.
Given a set of geometric objects, their intersection graph is a graph whose vertices correspond to the objects and whose edges correspond to the pairs of intersecting objects. Restricting the geometric objects to a certain shape restricts the class of graphs that admit a representation with respect to this shape. For example, graphs represented by disks in the Euclidean plane are called disk intersection graphs. The special case of unit disk graphs—intersection graphs of unit disks—has been studied extensively. Recognition of such graphs as well as many combinatorial problems restricted to these graphs such as coloring, independent set, and domination are all NPhard [5]; see also the survey of Hliněný and Kratochvíl [16]. Instead of restricting the radii of the disks, people have also studied restrictions of the type of intersection. If the disks are only allowed to touch, the corresponding graphs are called coin graphs. Koebe’s classical result says that the coin graphs are exactly the planar graphs. If all coins have the same size, the represented graphs are called penny graphs. These graphs have also been studied extensively [7, 4, 10], e.g., they are NPhard to recognize [3] [6, Section 11.2.3].
As with the arrangements above, we again consider a restriction on the intersection angle. We define the orthogonal circle intersection graphs as the intersection graphs of arrangements of orthogonal circles. In Section 4, we investigate properties of these graphs. For example, similar to the proof of our linear bound on the number of intersection points for arrangements of orthogonal circles (Theorem 3.1), we observe that such graphs have only a linear number of edges.
We also consider orthogonal unit circle intersection graphs, that is, orthogonal circle intersection graphs with a representation that consists only of unit circles. We show that these graphs are a proper subclass of penny graphs. As it turns out, the NPhardness proof for the recognition of penny graphs using the logic engine [8] detailed in the book of Di Battista et al. [6, Section 11.2.3] can be modified to yield the NPhardness for recognizing orthogonal unit circle intersection graphs (Theorem 4.2).
2 Preliminaries
In this section, we collect a few useful observations concerning orthogonal circles. We first give easy observations that follow from Observation 1. Afterwards, we discuss a classic arrangement of orthogonal circles and its consequences.
Observation 2
If and are orthogonal circles with centers and , respectively, then is not contained in and is not contained in .
Observation 3
If and are orthogonal circles with centers and , radii and , and intersection points and , then ; see Fig. 2.
Apollonian circles are two families of circles such that every circle in the first family is orthogonal to every circle in the second family. One family is called the coaxal intersecting circles and consists of circles that pass through two same points and ; the other family is called the coaxal nonintersecting circles and consists of nonintersecting circles that form two nested structures of circles, one containing , the other one containing in its interior; see Fig. 2. For a precise definition of Apollonian circles, see Ogilvy’s geometry textbook [19, Section 2].
Let be an arrangement of circles and be a face of . If is a triangular or quadrangular face, then its sides are formed by at least three circles. Thus, there are two circles that are orthogonal to either one or two other circles. So we can partition these circles into two families where each circle of one family is orthogonal to each circle of the other family, that is, they are Apollonian circles. We highlight this fact as Observation 4. In addition, we state and justify some further observations that directly follow from the structure of Apollonian circles.
Observation 4
Observation 5
Orthogonal circle intersection graphs do not contain any . In other words, arrangements of orthogonal circles do not contain any four pairwise orthogonal circles.
This is due to the fact that we can partition the circles of a quadruple of pairwise orthogonal circles into two pairs of circles such that each circle of one pair is orthogonal to each circle of the other pair, so they are Apollonian circles, that is, one pair must consist of coaxal nonintersecting circles and the other pair must consist of coaxal interesting circles. But for each pair its circles intersect—a contradiction. Similarly, we can show the following.
Observation 6
Orthogonal circle intersection graphs do not contain any induced . In other words, in an arrangement of orthogonal circles there cannot be two pairs of orthogonal circles such that each circle of one pair is orthogonal to each circle of the other pair and the circles within the pairs are not orthogonal.
3 Arrangements of Orthogonal Circles
In this section we study the number of faces of an arrangement of orthogonal circles. In Section 3.1, we give a bound on the total number of faces. In Section 3.2, we separately bound the number of faces formed by two, three, and four edges.
Let be an arrangement of orthogonal circles in the plane. By a slight abuse of notation, we will say that a circle contains a geometric object and mean that the disk bounded by contains . We say that a circle is nested in a circle if is contained in . We say that a circle is nested consecutively in a circle if is nested in and there is no other circle such that is nested in and is nested in . Consider a subset of maximum cardinality such that for each pair of circles one is nested in the other. The innermost circle in is called a deepest circle in ; see Fig. 4.
Lemma 1
Let be a circle of radius , and let be a set of circles orthogonal to . If does not contain nested circles and each circle in has radius at least , then . Moreover, if , then all circles in have radius and is contained in the union of the circles in .
Proof
Let be the center of . Consider any two circles and in with centers and and with radii and , respectively. Since and , the edge is the longest edge of the triangle ; see Fig. 4. So the angle is at least . Thus, .
Moreover, if then, for each pair of circles and in that are consecutive in the circular ordering of the circle centers around , it holds that . This is only possible if . Thus, all the circles in have radius . This implies that is contained in the union of the circles in ; see Fig. 4(b).
3.1 Counting Faces
Theorem 3.1 ()
Every arrangement of orthogonal circles has at most intersection points and faces.
The above theorem (whose formal proof is at the end of the section) follows from the fact that any arrangement of orthogonal circles contains a circle with at most seven neighbors (that is, circles that are orthogonal to ).
Lemma 2
Every arrangement of orthogonal circles has a circle that is orthogonal to at most seven other circles.
Proof
If no circle is nested within any other, Lemma 1 implies that the smallest circle has at most six neighbors, and we are done.
So, among the deepest circles in , consider a circle with the smallest radius. Let be the radius of . Note that is nested in at least one circle. Let be a circle such that and are consecutively nested. Denote the set of all circles in that are orthogonal to but not to by . All circles in are nested in . Since is deepest, contains no nested circles; see Fig. 4(a). Since the radius of every circle in is at least , Lemma 1 ensures that contains at most six circles. Given the structure of Apollonian circles, there can be at most two circles that intersect both and . This together with Lemma 1 immediately implies that cannot be orthogonal to more than eight circles. In the following we show that there can be at most seven such circles.
If there is only one circle intersecting both and , then is orthogonal to at most seven circles in total, and we are done.
Otherwise, there are two circles orthogonal to both and . Let these circles be and . We assume that contains exactly six circles. Hence, by Lemma 1, all circles in have radius . Let be ordered clockwise around so that every two circles and with are orthogonal.
Let and be the intersection points of and ; see Fig. 4(a). Note that, by the structure of Apollonian circles, one of the intersection points, say , must be contained inside , whereas the other intersection point must lie in the exterior of . Since the circles in are contained in , none of them contains . Further, no circle in contains , as otherwise the circles , , , and would be pairwise orthogonal, contradicting Observation 5. Recall that, by Lemma 1, is contained in the union of the circles in . Since is not contained in this union, intersects two different circles and , and intersects two different circles and . Note that and cannot intersect the same circle in , because , , , and would be pairwise orthogonal, contradicting Observation 5. Therefore, the indices , , , and are pairwise different.
We now consider possible values of the indices , , , and , and show that in any case we get a contradiction to Observations 5 or 6. If , then , , , and would be pairwise orthogonal, contradicting Observation 5; see Fig. 4(b). If , then , , , and would form an induced in the intersection graph; see Fig. 4(c). This would contradict Observation 6. If and , then either or ; see Fig. 4(d). W.l.o.g., assume the latter and observe that then , , , would form an induced , again contradicting Observation 6.
We conclude that contains at most five circles. Together with and , at most seven circles are orthogonal to .
Using the lemma above and Euler’s formula, we now can easily prove Theorem 3.1.
Proof (of Theorem 3.1)
Let be an arrangement of orthogonal circles. By Lemma 2, contains a circle orthogonal to at most seven circles. The circle yields at most intersection points. By induction, the whole arrangement has at most intersection points.
Consider the planarization of , and let , , , and denote the numbers of vertices, edges, faces, and connected components of , respectively. Since every vertex in the planarization corresponds to an intersection, the resulting graph is regular and therefore . By Euler’s formula, we obtain . This yields since and .
3.2 Counting Small Faces
In the following we study the number of faces of each type in particular, that is, the number of digonal, triangular, and quadrangular faces. We begin with some notation. Let be an arrangement of orthogonal circles in the plane. Let be some subset of circles of . A face in is called a region in formed by ; see for instance Fig. 9. Note that a face is also a region. Let be a digonal or triangular region with a convex side formed by a circle and let denote the convex arc of bounding . The arc subtends a certain angle in . We say in this case that subtends an angle of in . Note that if forms a concave side of , then . If is a digonal region, that is, it is formed by only two circles and , then we simply say that subtends an angle of in to mean .
For a digonal region formed by circles and or a triangular region formed by circles , , and we say that subtends a total angle of , for or if is digonal or triangular region respectively.
Counting Digonal and Triangular Faces.
In this section we give an upper bound on the number of digonal and triangular faces in an arrangement of orthogonal circles. Because faces do not overlap, each digonal or triangular face uses a unique convex arc of a circle bounding this face. Therefore, the sum of angles subtended by digonal or triangular faces formed by the same circle must be at most . Analogously, the sum of total angles over all digonal or triangular faces cannot exceed . First, we show that each digonal face subtends a total angle of size at least ; this gives an upper bound of on the number of digonal faces. Then we show that each triangular face subtends a total angle of size at least ; this gives an upper bound of on the number of triangular faces.
We start with digonal faces. As in Fig. 9, a digonal face is formed by either:

[label=(D0), ref=D0,align=left,leftmargin = 7.5ex]

two convex arcs, or

a convex and a concave arc.
Lemma 3
Proof
Let be a digonal face formed by a pair of orthogonal circles and . If is of type 1, the statement of the lemma follows from Observation 3. Let be a digonal face of type 2, assume wlog. that is contained in the disk of . Since and are orthogonal, the intersection of the disks of and forms a digonal region of type 1. Hence, by Observation 3, . Therefore, since , it follows that .
Lemma 4
Every arrangement of orthogonal circles has at most digonal faces.
Proof
The lemma follows directly from Lemma 3.
Now we turn to triangular faces. According to Observation 4
a triangular face must be formed by Apollonian circles. For this reason there is a topologically unique arrangement of three orthogonal circles that can form a triangular face. Therefore, we can classify triangular faces into three types (see Fig.
9):
[label=(T0), ref=T0,align=left,leftmargin = 7.5ex]

those with exactly one convex side;

those with exactly two convex sides;

those with exactly three convex sides.
Lemma 5
A triangular region of type 3 subtends a total angle of exactly .
Proof
Let denote the triangular region of type 3 bounded by , and , that is ; see Fig. 9. Observe that is the intersection of three digonal regions (the intersection of the disks of and ), (the intersection of the disks of and ), and (the intersection of the disks of and ).
Consider the hexagon . Obviously, it holds that
(1) 
and similarly,
(2) 
and
(3) 
Since , it holds, that the sum of angles
Lemma 6
A triangular region of type 2 subtends a total angle larger than .
Proof
Wlog., as in Fig. 9, let denote the triangular region of type 2 whose convex sides are formed by and , that is, . Note that is a subregion of the digonal region (that is, the intersection of the disks of and ). Note that, in addition to , the triangular region of type 3 is also formed by , and . Due to Lemma 5, we know that . Since and , Lemma 3 implies that . Hence, as required.
Lemma 7
A triangular region of type 1 subtends a total angle of at least .
Proof
Wlog., as in Fig. 9, let denote the triangular region of type 1 whose convex side is formed by , that is . Observe that is located inside . Since is a crossing, there exist two perpendicular rays pointing inside which follow the tangents of and at . These tangents lie entirely in the cone spanned by angle which therefore is at least ; see the grayshaded cone in Fig. 9. By Observation 2, is located in this cone as well. Hence subtends an angle in at least as large as , that is, .
Theorem 3.2
Every arrangement of orthogonal circles has at most triangular faces.
Counting Quadrangular Faces.
According to Observation 4, a quadrangular face must be formed by Apollonian circles. We provide a complete classification of such faces below. In particular, we group quadrangular faces into those formed by two nested circles and those formed by two disjoint circles.
A quadrangular face of an arrangement of orthogonal circles is of one of the following types:

[label=(Q0),ref=Q0,align=left,leftmargin=*]

with two sides formed by two circles and whose disks are disjoint and

[label=(Q2.0),ref=Q2.0,align=left,leftmargin=*]

the two other sides formed by one more circle ; see Fig. 9(c).

the two other sides formed by two intersecting circles and and the disks of both of them contain ; see Fig. 9(d).

the two other sides formed by two intersecting circles and and the disk of only one of them contains ; see Fig. 9(d).

We will bound the number of quadrangular faces of each type. First, we consider quadrangular faces of type 2a and 2c.
Observation 7
This observation implies the following.
Next let us consider quadrangular faces of type 1. Recall that they have two sides formed by a pair of consecutively nested circles.
Lemma 9
Every arrangement of orthogonal circles has at most quadrangular faces of type 1.
Proof
Consider a pair of consecutively nested circles and . The other circles forming quadrangular faces together with and must intersect both and and so they must be coaxal intersecting circles, which all pass through the same two points. Therefore, there can only be two of them. And so, and can form at most four quadrangular faces; see Fig. 10.
We assign each quadrangular face of type 1 to the inner circle of the pair of consecutively nested circles forming this face.
Call a pair of consecutively nested circles such that its inner circle is unique to this pair a single pair and let be the number of single pairs. And call a pair of consecutively nested circles such that the inner circle of this pair is also the inner circle of at least another pair a sharing pair and let be the number of sharing pairs.
As discussed above any pair of consecutively nested circles can form at most four quadrangular faces, therefore, inner circles of single pairs get at most quadrangular faces in total. A sharing pair can form at most three quadrangular faces (see Fig. 11) so inner circles of sharing pairs get at most quadrangular faces in total.
Because a circle cannot be the inner circle of a single pair and a sharing pair at the same time, quadrangular faces assigned to inner circles of single pairs are different from those assigned to inner circles of shared pairs. Therefore, there are quadrangular faces in total.
Since there are single pairs, there are different circles such that each of them is inner in some single pair. So there can be at most circles such that each is inner in some sharing pair. Due to Observation 5 a circle can be the inner circle in at most three sharing pairs, so . Thus, the total number of quadrangular faces is and the left side of the inequality is the largest, namely , if , that is, there are only sharing pairs and no single pairs.
We now turn to type 2b.
Lemma 10
Every arrangement of orthogonal circles has at most quadrangular faces of type 2b.
Proof
Consider an arrangement of orthogonal circles. For each quadrangular face of type 2b we assign it to a circle of the smallest radius forming one of its convex sides. Consider some circle and the subset of circles that form the other convex sides of the quadrangular faces assigned to . Note that each circle in has larger radius than that of .
Let us first show that does not contain nested circles. Assume that there are two circles that are nested, wlog. is contained in . Consider the quadrangular face whose convex sides are formed by and and the concave sides by two circles and ; see Fig. 12. Consider the region formed by the intersection of disks of and . Since is orthogonal to and is contained in , it intersects . Note that neither nor can intersect in , as otherwise would not be a quadrangular face. Therefore, the face divides the region into two triangular regions of type 3 and intersects in one of these regions. Let be this region. Therefore, . Assume wlog. that is formed by the circles , , and .
By Lemma 5, it holds that whereas, due to Observation 3, , because the radius of is smaller than or equal to that of ; contradiction.
By Lemma 1 cannot contain more than six circles.
Therefore, our assignment maps at most six quadrangular faces of type 2b to each circle and the whole arrangement can have at most of such faces.
We conclude this section by noting that the above bounds on the number of faces of degree at most 4 also provide a linear bound on the total number of faces in an arrangement of orthogonal circles; however, this bound is weaker than the one in Theorem 3.1. Namely, since the average degree of a face in an arrangement of orthogonal circles is 4, a bound on the number of faces of degree at most 4 gives a bound on the number of all faces in the arrangement (via Euler’s formula). The bound one can achieve in this way is roughly .
4 Intersection Graphs of Orthogonal Circles
Given an arrangement of orthogonal circles, consider its intersection graph, which is the graph with vertex set that has an edge between any pair of intersecting circles in . Observations 5 and 6 imply that such a graph does not contain any and any induced . First, we show that such graphs may be nonplanar.
Lemma 11
For every , there is an intersection graph of orthogonal circles that contains as a minor. The representation uses circles of three different radii.
Proof
Let a chain be an arrangement of orthogonal circles whose intersection graph is a path. We say that two chains and cross if two disjoint circles and of one chain, say , are orthogonal to the same circle of the other chain ; see Fig. 12(a) (left). If two chains cross, their paths in the intersection graph are connected by two edges; see the dashed edges in Fig. 12(a) (right).
Consider an arrangement of rectilinear paths embedded on a grid where each pair of curves intersect exactly once; see the inset in Fig. 12(b). We convert the arrangement of paths into an arrangement of chains such that each pair of chains crosses; see Fig. 12(b). Now consider the intersection graph of the orthogonal circles in the arrangement of chains. If we contract each path in the intersection graph that corresponds to a chain, we obtain .
Next, we discuss the density of orthogonal circle intersection graphs. Gyárfás et al. [15] have shown that any free graph on vertices with average degree at least has clique number at least . Due to Observation 5, we know that orthogonal circle intersection graphs have clique number at most 3. Thus, their average degree is bounded from above by . Lemma 2, however, implies that the average degree is seven.
Theorem 4.1
The intersection graph of a set of orthogonal circles has at most edges.
Proof
The geometric representation of an orthogonal circle intersection graph is an arrangement of orthogonal circles. By Lemma 2, an arrangement of orthogonal circles always has a circle orthogonal to at most seven circles. Therefore, the corresponding intersection graph always has a vertex of degree at most seven. Thus, it has at most edges.
For the remainder of this section, we focus on a natural subclass of orthogonal circle intersection graphs, namely orthogonal unit circle intersection graphs. Recall that these are orthogonal circle intersection graphs with a representation that consists of unit circles only. As Fig. 13(a) shows, every representation of an orthogonal unit circle intersection graph can be transformed (by scaling each circle by a factor of ) into a representation of a penny graph, that is, a contact graph of equalsize disks. Hence, every orthogonal unit circle intersection graph is a penny graph – whereas the converse is not true. For example, or the 5star are penny graphs but not orthogonal unit circle intersection graphs (see Fig. 13(b)).
Orthogonal unit circle intersection graphs being penny graphs implies that they inherit the properties of penny graphs, e.g., their maximum degree is at most six and their edge density is at most , where is the number of vertices [20, Theorem 13.12, p. 211]. Because triangular grids are orthogonal unit circle intersection graphs, this upper bound is tight.
As it turns out, orthogonal unit circle intersection graphs share another feature with penny graphs: their recognition is NPhard. The hardness of pennygraph recognition can be shown using the logic engine [6, Section 11.2], which simulates an instance of the NotAllEqual3Sat (NAE3SAT) problem. We establish a similar reduction for the recognition of orthogonal unit circle intersection graphs; for details, see Appendix 0.A.
Theorem 4.2 ()
It is NPhard to recognize orthogonal unit circle intersection graphs.
5 Discussions and Open Problems
In Section 3 we provided upper bounds for the number of faces of an orthogonal circle arrangement. As for lower bounds on the number of faces, we found only very simple arrangements containing digonal, triangular, or quadrangular faces; see Fig. 14(a), 14(b), and 14(c), respectively. Can we construct better lower bound examples or improve the upper bounds?
What’s the complexity of recognizing general orthogonal circle intersection graphs? Recognizing unit disk intersection graphs is complete [17]. Is the same true for recognizing orthogonal (unit) circle intersection graphs?
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Appendix 0.A Recognizing Orthogonal Unit Circle Intersection Graphs
In this section, we show how to realize the logic engine with orthogonal unit circle intersection graphs. The logic engine simulates the NotAllEqual3Sat (NAE3SAT) problem where a set of clauses each containing three literals from a set of boolean variables is given and the question is to find a truth assignment to the variables so that each clause contains at least one true literal and at least one false literal.
See 4.2
Proof
We closely follow the description from [6, Section 11.2] and use their notations and definitions. The logic engine consists of the following parts (we will mostly refer to Figs. 16 and 18 to explain how the parts of the logic engine are connected). The frame and armatures (drawn blue and black respectively in Fig. 16) for the logic graph are built of hexagonal blocks, as shown in Fig. 16(a) whose orthogonal unit circle intersection representation is shown in Fig. 16(b). It is easy to see that they are uniquely drawable (up to rotation, reflection, and translation) since has a unique orthogonal unit circle intersection representation. Each armature corresponds to a variable in .
A chain graph (represented by gray circles in Fig. 16) is a sequence of links, as shown in Fig. 16(e) whose orthogonal unit circle intersection representation is shown in Fig. 16(f). The number of links in a chain corresponds to the number of clauses in . The shaft (green in Fig. 16) is a simple path and serves as an axle for the armatures, that is, the armatures can be flipped around the shaft. Each armature corresponding to a variable has two chains and each suspended between one of the ends of the armature and the shaft. For that reason in an orthogonal unit circle intersection representation each chain is taut.
So far we have described the universal part of the logic engine, that is, the part that only depends on the number of clauses in and the number of variables in ; it is illustrated in Fig. 16. The frame, armatures, and chain graphs have a unique orthogonal unit circle intersection representation up to flipping armatures (see Fig. 16), since they are built up of hexagonal blocks which are uniquely drawable. We still need to show that the shaft is taut. This is enforced by the bottom part of the frame. Consider the middle horizontal sequence of circles in the bottom part of the frame that spans the frame from the left side to the right; in light blue in Fig. 16. It is easy to see that the shaft must be drawn as this sequence, because it consists of the same number of circles and must also span the frame from the left side to the right. Since the sequence is taut, the shaft is also taut. Notice that there is still the freedom of flipping each armature together with its chains around the shaft, that is, it can take two possible positions where one part of the armature is either above or below the shaft. This is the flexibility that allows our logic engine to encode a solution to a NAE3SAT problem.
Now let us show how to customize the logic engine according to an instance of NAE3SAT. A chain link graph can be extended to a flagged link by the addition of three new vertices as shown in Fig. 16(c) whose orthogonal unit circle representation is shown in Fig. 16(d). Note that it also has a unique drawing. To simulate the given NAE3SAT instance we replace link graphs with flagged link graphs according to the incidence between literals and clauses. If the literal appears in clause , then link of chain is unflagged. If the literal appears in clause , then link of of chain is unflagged. For an example see Fig. 18.
It is easy to see that by adjusting the sizes of the frame and the armatures we can ensure that in an orthogonal unit circle intersection representation of the logic engine two flagged links which lie in the same row and are attached to chains of adjacent armatures collide if and only if they are flipped so that they point towards each other; see Fig. 19.
Similarly we can ensure that any flag attached to the chain of the outermost armature collides with the frame if it points toward the front edge of the frame, and any flag attached to the chain of the innermost armature collides with that armature if it points toward the rear. Therefore, we can use [6, Theorem 11.2] to show that the corresponding customized logic engine has an orthogonal unit circle representation if and only if the corresponding instance of NAE3SAT is a yesinstance.