#### Note on the Pair-crossing Number and the Odd-crossing Number

Discrete Comput Geom
Note on the Pair-crossing Number and the Odd-crossing Number
Géza Tóth 0
0 G. Tóth ( ) Rényi Institute, Hungarian Academy of Sciences , Budapest , Hungary
The crossing number CR(G) of a graph G is the minimum possible number of edge-crossings in a drawing of G, the pair-crossing number PAIR-CR(G) is the minimum possible number of crossing pairs of edges in a drawing of G, and the oddcrossing number ODD-CR(G) is the minimum number of pairs of edges that cross an odd number of times. Clearly, ODD-CR(G) ≤ PAIR-CR(G) ≤ CR(G). We construct graphs with 0.855 · PAIR-CR(G) ≥ ODD-CR(G). This improves the bound of Pelsmajer, Schaefer and Štefankovicˇ. Our construction also answers an old question of Tutte. Slightly improving the bound of Valtr, we also show that if the pair-crossing number of G is k, then its crossing number is at most O(k2/ log2 k).
Drawings of a graph; Crossing number
1 Introduction
In a drawing of a graph G vertices are represented by points and edges are
represented by Jordan curves connecting the corresponding points. If it does not lead to
confusion, we do not make any notational distinction between vertices (resp. edges)
and points (resp. curves) representing them. We assume that the edges do not pass
through vertices, any two edges have finitely many common points and each of them
is either a common endpoint, or a proper crossing. We also assume that no three edges
cross at the same point.
The crossing number CR(G) is the minimum number of edge-crossings (i.e.
crossing points) over all drawings of G. The pair-crossing number PAIR-CR(G) is the
G. Tóth’s research was supported by the Hungarian Research Fund grant OTKA-K-60427 and
the Research Foundation of the City University of New York.
minimum number of crossing pairs of edges over all drawings of G, and the
oddcrossing number ODD-CR(G) is the minimum number of pairs of edges that cross an
odd number of times over all drawings of G.
Clearly, for any graph G we have
ODD-CR(G) ≤ PAIR-CR(G) ≤ CR(G).
Pach and Tóth [
1
] proved that CR(G) cannot be arbitrarily large if ODD-CR(G) is
bounded, namely, for any G, if ODD-CR(G) = k, then CR(G) ≤ 2k2 and this is the
best known bound. Obviously it follows that PAIR-CR(G) ≤ 2k2 as well and this is
also the best known bound. On the other hand, Pelsmajer, Schaefer and Štefankovicˇ
[
4
] proved that ODD-CR(G) and PAIR-CR(G) are not necessarily equal, they
constructed a series of graphs with
ODD-CR(G) <
+ o(1) · PAIR-CR(G).
ODD-CR(G) <
+ o(1) · ALG-CR(G).
We show that ALG-CR(G) and PAIR-CR(G) are not necessarity equal either.
We slightly improve their bound with a completely different construction.
Theorem 1 There is a series of graphs G with
√3
2
√3
2
ODD-CR(G) <
3√5
2
5
− 2 + o(1) · PAIR-CR(G).
Note that √23 ≈ 0.866 and 3√2 5 − 25 ≈ 0.855. There are many other versions of
the crossing number (see e.g. [
2, 3
]). Tutte [6] defined the following version which
we call independent algebraic crossing number, IALG-CR(G), and we also define its
close relative the algebraic crossing number, ALG-CR(G).
Orient the edges of G arbitrarily. For any drawing D of G, and any two edges e
and f , let c+ (resp. c−) be the number of e–f crossings where e crosses f from left
to right (resp. from right to left). Let c(e, f ) = |c+ − c−|, and let c(D) = c(e, f )
where the summation is for all pairs of independent edges. Similarly, let c (D) =
c(e, f ) where the summation is for all pairs of edges. Finally, let IALG-CR(G) be
the minimum of c(D) for all drawings D of G, and let ALG-CR(G) be the minimum
of c (D) for all drawings D of G.
It is easy to see that for any graph G we have IALG-CR(G) ≤ ALG-CR(G) and
ODD-CR(G) ≤ ALG-CR(G) ≤ CR(G).
In the construction of Pelsmajer, Schaefer and Štefankovicˇ [
4
] for each of the
graphs the pair-crossing number and the algebraic crossing number are equal.
Therefore, for their series of graphs
Theorem 2 There is a series of graphs G with
3√5
2
5
− 2 + o(1) · PAIR-CR(G).
Since ODD-CR(G) ≤ ALG-CR(G) for every graph G, Theorem 1 is an immediate
consequence of Theorem 2. Tutte [
6
] asked if IALG-CR(G) = CR(G) holds for every
graph G. Since IALG-CR(G) ≤ ALG-CR(G), Theorem 2 gives a negative answer for
this question. Finally, since PAIR-CR(G) ≤ CR(G), Theorems 1 and 2 hold also for
CR(G) instead of PAIR-CR(G). Moreover, the whole argument works, without any
change.
It is still a challenging open question whether CR(G) = PAIR-CR(G) holds for
all graphs G. Pach and Tóth [
1
] proved that for any G, if PAIR-CR(G) = k, then
CR(G) ≤ 2k2. Valtr [
7
] managed to improve this bound to CR(G) ≤ 2k2/ log k. Based
on the ideas of Valtr, in this note we give a further little improvement.
Theorem 3 For any graph G, if PAIR-CR(G) = k, then CR(G) ≤ 9k2/ log2 k.
2 Proof of Theorem 2
The Idea and Sketch of the Construction For simplicity, we write alg-crossing num
ber for the algebraic crossing number. In the description we use weights on the edges
of the graph. If we substitute each weighted edge by an appropriate number of
parallel paths, say, each of length two, we can obtain an unweighted simple graph whose
ratio of the pair-crossing and alg-crossing numbers is arbitrarily close to that of the
weighted construction.
First of all, take a “frame”F , which is a cycle K with very heavy edges, together
with a vertex V connected to all vertices of the cycle, also with very heavy edges.
In the optimal drawings the edges of F do not participate in any crossing, and we
can assume that V is drawn outside the cycle K . Therefore, all additional edges and
vertices of the graph will be inside K .
We have four further vertices, each connected to three different vertices of the
frame-cycle K . These three edges have weights 1, 1, w respectively, with some
1 < w < 2. Each one of these four vertices, together with the adjacent three edges,
and the frame F , is called a component of the construction.
If we take any two of the components, it is easy to see how to draw them optimally,
both in the alg-crossing and pair-crossing sense. See Fig. 1. The point is that if we
take all four components, we can still draw them such that each of the six pairs are
drawn optimally, in the alg-crossing sense. See Fig. 2. On the other hand, it is easy
to see that it is impossible to draw all six pairs optimally in the pair-crossing sense,
some pairs will not have their best drawing. See Fig. 3. Note that we did not indicate
vertex V of the frame. √
We get the best result with w = 52+1 . Actually, we will see that among any three
components there is a pair which is not drawn optimally in the pair-crossing sense.
So, we could take the union of just three components, but that gives a weaker bound.
Proof of Theorem 2 A weighted graph G is a graph with positive weights on its
edges. For any edge e let w(e) denote its weight. For any fixed drawing G of G, the
pair-crossing value PAIR-CR(G) = w(e)w(f ) where the sum goes over all
crossing pairs of edges e, f . For the alg-crossing value ALG-CR(G), orient the edges of G
arbitrarily, let c+ (resp. c−) be the number of e–f crossings where e crosses f from
left to right (resp. from right to left), let c(e, f ) = |c+ − c−|. The alg-crossing value
ALG-CR(G) = w(e)w(f )c(e, f ) where the sum goes over all pairs of edges e, f .
The pair-crossing number (resp. alg-crossing number) is the minimum of the
paircrossing value (resp. alg-crossing value) over all drawings. That is,
PAIR-CR(G) = over all drawings
min
ALG-CR(G) = over all drawings for all pairs
min
of edges e,f
for all crossing pairs
of edges e,f
w(e)w(f ),
w(e)w(f )c(e, f ).
Theorem 4 There exists a weighted graph G with
Proof First we define the weighted graph G. Take nine vertices, A1, B3, A2, C1, D3,
C2, B1, A3, B2, D1, C3, D2 which form cycle K in this order. Vertex V is connected
to all of the nine vertices of K . These vertices and edges form the “frame” F . All
edges of F have extremely large weights, therefore, they do not participate in any
crossing in an optimal drawing. We can assume without loss of generality that V is
drawn outside the cycle K , so all further edges and vertices of G will be inside K .
There are four more vertices, A0, B0, C0, D0, and for X = A, B, C, D, X0 is
connected to X1, X2, and X3. The weight w(X0X1) = w(X0X2) = 1 and w(X0X3) =
√
w = 52+1 . Graph X is a subgraph of G, induced by the frame and X0. See Fig. 1.
Finally, for any X, Y = A, B, C, D, X = Y , let PAIR-CR(X, Y ) = PAIR-CR(X ∪ Y ),
and ALG-CR(X, Y ) = ALG-CR(X ∪ Y ).
First we find all these crossing numbers. Moreover, we also find the second
smallest pair-crossing values.
Start with A ∪ C. Since the path A1B3A2 is not intersected by any edge in an
optimal drawing, we can contract it to one vertex, without changing the pair-crossing
number, so now A1 = A2. Consider the edges e1 = A1A0 and e2 = A2A0. Now they
connect the same vertices. Suppose that they do not go parallel in an optimal drawing.
Let w∗(e1) (resp. v∗(e2)) be the sum of the weights of the edges crossing e1 (resp. e2)
and assume without loss of generality that w∗(e1) ≤ w∗(e2). Then draw e2 parallel
with e1, the drawing obtained is at least as good as the original drawing was, so it is
optimal as well. Therefore, we can assume without loss of generality that e1 and e2
go parallel in an optimal drawing, so we can substitute them by one edge of weight 2.
Similarly, we can contract the path C1D3C2 and substitute the edges C1C0 and C2C0
by one edge of weight 2. Now we have a very simple graph, whose pair-crossing
number is immediate, that is, we have two paths C1C0C3 and A1A0A3, which have
to cross each other, and on both paths one edge has weight w, the other one has
weight 2. Since w < 2, in the optimal drawing the edges A0A3 and C0C3 will cross
each other and no other edges cross so we have PAIR-CR(A, C) = w2. Moreover, it
is also clear that the second smallest pair-crossing value is 2w.
The same argument holds for ALG-CR(A, C), moreover, by symmetry, we can
argue exactly the same way for the pairs (A, D), (B, C), and (B, D).
Now we determine PAIR-CR(A, B) and the second smallest pair-crossing value.
The edges a1 = A0A1, a2 = A0A2, a3 = A0A3 divide the interior of F into three
regions R1, R2 and R3. Number them in such a way that for i = 1, 2, 3, ai is outside
Ri . See Fig. 1. Once we place B0 into one of these regions, it is clear how to draw
the edges b1 = B0BB1, b2 = B0BB2, b3 = B0BB3 to get the best of the possible
drawings. If B0 is in R1 or in R2, we get the pair-crossing value 2w, but if we place
B0 in R3, then we get 2. Again, the same argument holds for ALG-CR(A, B), and by
symmetry, the situation is the same with the pair (C, D). See Fig. 1.
Lemma 1 ALG-CR(G) = 4w2 + 4.
Proof We have ALG-CR(G) ≥ ALG-CR(A, B) + ALG-CR(A, C) + ALG-CR(A, D) +
ALG-CR(B, C) + ALG-CR(B, D) + ALG-CR(C, D) = 4w2 + 4, and there is a drawing
(see Fig. 2) with exactly this alg-crossing value.
Lemma 2 PAIR-CR(G) = 4w2 + 4w.
Proof The argument, except for the exact calculation, should be clear from the
figures. While we have a drawing which is optimal for all six pairs in the alg-crossing
sense (see Fig. 2), in the pair-crossing sense some of the pairs will not be optimal,
they have to take at least the second smallest pair-crossing value. We start with an
observation that in any triple at least one pair is not optimal. Then we will distinguish
three cases.
Take a drawing G of G. Suppose that we have a drawing G of G where the pairs
(A, C) and (A, D) are drawn optimally, that is, PAIR-CR(A, C) = PAIR-CR(A, D) =
w2. Recall that the edges a1 = A0A1, a2 = A0A2, a3 = A0A3 divide the interior of F
into three regions R1, R2 and R3. It follows from the above argument that C0 ∈ R1,
D0 ∈ R2. But then the pair (C, D) is not drawn optimally, that is, PAIR-CR(C, D) > 2,
so we have PAIR-CR(C, D) ≥ 2w. In other words, it is impossible that all three pairs
(A, C), (A, D), (C, D) are drawn optimally at the same time. By symmetry, this
observation holds for any triple of A, B, C, D.
We have to distinguish three cases.
Case 1 Neither (A, B), nor (C, D) are drawn optimally. In this case,
PAIR-CR(A, B) > 2 so by the above argument we have PAIR-CR(A, B) ≥ 2w,
and similarly PAIR-CR(C, D) ≥ 2w. For all other pairs we have pair-crossing
value at least w2, therefore, PAIR-CR(G) = PAIR-CR(A, B) + PAIR-CR(A, C) +
PAIR-CR(A, D) + PAIR-CR(B, C) + PAIR-CR(B, D) + PAIR-CR(C, D) ≥ 4w2 +
4w.
Case 2 (A, B) is drawn optimally, (C, D) is not. Since (A, B) is drawn optimally,
one of the pairs (A, C) and (B, C) and one of the pairs (A, D) and (B, D)
is not drawn optimally so we have PAIR-CR(A, C) + PAIR-CR(B, C) ≥ w2 +
2w and analogously PAIR-CR(A, D) + PAIR-CR(B, D) ≥ w2 + 2w therefore,
PAIR-CR(G) = PAIR-CR(A, B) + PAIR-CR(A, C) + PAIR-CR(A, D) +
PAIR-CR(B, C) + PAIR-CR(B, D) + PAIR-CR(C, D) ≥ 2w2 + 6w + 2 = 4w2 + 4w.
The last equality can be verified by solving the quadratic equation.
Case 3 Both (A, B) and (C, D) are drawn optimally. If none of the other four
pairs is optimal, then we have PAIR-CR(G) = PAIR-CR(A, B) + PAIR-CR(A, C) +
PAIR-CR(A, D)+ PAIR-CR(B, C)+ PAIR-CR(B, D)+ PAIR-CR(C, D) ≥ 8w +4 =
4w2 + 4w. So we can assume that one of them, say (A, C) is drawn optimally,
that is, PAIR-CR(A, C) = w2. Since in any triple we have at least one
nonoptimal pair, we have PAIR-CR(B, C) ≥ 2w and PAIR-CR(A, D) ≥ 2w. We
estimate PAIR-CR(B, D) now.
Again, the edges a1 = A0A1, a2 = A0A2, a3 = A0A3 of A divide the
interior of F into three regions R1, R2 and R3 with Ri is the one to the
opposite of ai . Similarly define the regions Q1, Q2, Q3 for C. Since (A, C) is
drawn optimally, R3 and Q3 are disjoint. Since (A, B) is drawn optimally,
B0 ∈ R3, and since (C, D) is also drawn optimally, D0 ∈ Q3. See Fig. 3.
Now it is not hard to see that the edge D0D1 either crosses A0A1, A0A2,
and B0B3, or B0B1, B0B2, and A0A3. The same holds for the edge D0D1,
so PAIR-CR(A, D) + PAIR-CR(B, D) ≥ 2w + 4. So we have PAIR-CR(G) =
PAIR-CR(A, B) + PAIR-CR(A, C) + PAIR-CR(A, D) + PAIR-CR(B, C) +
PAIR-CR(B, D) + PAIR-CR(C, D) ≥ w2 + 4w + 8 > 4w2 + 4w. This concludes
the proof of Lemma 2.
Now we have
ALG-CR(G) 4w2 + 4 −5 3√5
PAIR-CR(G) = 4w2 + 4w = 2 + 2
,
and Theorem 4 follows immediately.
Proof of Theorem 2 Let ε > 0 an arbitrary small number. Let p and q be positive
integers with the property that w(1 + 1ε0 ) > pq > w(1 − 1ε0 ). Let Gε be the following
graph. In the weighted graph G of Theorem 4, (i) substitute each edge e = XY of
weight 1 with q paths between X and Y , each of length 2, (ii) substitute each edge
e = XY of weight w with p paths between X and Y , each of length 2, and (iii)
substitute each edge e = XY of the frame F with a huge number of paths between X
and Y , each of length 2. Then
ALG-CR(Gε)
PAIR-CR(Gε)
<
ALG-CR(G)
PAIR-CR(G)
3√5
(1 + ε) < −25 + 2
+ ε.
3 Proof of Theorem 3
Let G be a graph, PAIR-CR(G) = k and take a drawing of G which has exactly k
crossing pairs of edges. Let t be a parameter, to be defined later. We distinguish three
types of edges. An edge e is
good if it is not crossed by any other edge;
light if it is crossed by at least one and at most t other edges;
heavy if it is crossed by more than t other edges.
We will apply the following result of Schaefer and Štefankovicˇ [
5
].
Fig. 4 Switch the uv segment
of e and f
Lemma (Schaefer and Štefankovicˇ [5]) Suppose that a graph is drawn in the plane,
and edge e is crossed by m other edges. If there are at least 2m crossings on e, then
the drawing can be modified such that (i) the number of crossings between any two
edges does not increase, and (ii) the number of crossings on e decreases.
Return to the proof of Theorem 3. Suppose that there is a light edge that has at
least 2t crossings. Then we can modify the drawing according to the lemma. This
modification does not increase the number of crossings on any edge and does not
introduce new pairs of crossing edges. On the other hand, it decreases the total
number of crossings, so after finitely many applications, all light edges have less than 2t
crossings.
Now we apply two other types of redrawing steps.
Suppose that in our drawing two heavy edges e and f cross at least twice and
let u and v be two crossings. Then switch the uv segment of e and f . This way (i)
we reduced the number of crossings between e and f and (ii) the total number of
crossings on any other edge remains the same.
Observe that this way we could have introduced self-crossings, in this case remove
the loop formed by the self-crossing edge. This way (i) the number of crossings on
any edge does not increase, and (ii) the total number of crossings decreases.
Apply the above redrawing steps as long as there are two heavy edges that cross
more than once or there is a self-crossing edge. Since the total number of crossings
decreases in each step, after finitely many applications any two heavy edges will cross
at most once and no edge crosses itself.
Now count the number of crossings for the drawing obtained. Originally there
were k pairs of crossing edges. A heavy edge crosses more than t other edges, so
there are less than 2k/t heavy edges. The total number of light edges is at most 2k.
Each light edge has less than 2t crossings, so the total number of crossings on the
light edges is less than 2k2t . On the other hand, since any two heavy edges cross at
most once, we have less than
of crossings C we have
Set t = (log k)/2, we obtain CR(G) < 9k2/ log2 k.
Acknowledgement
We are very grateful to Daniel Štefankovicˇ for his comments.
heavy-heavy crossings. So, for the total number
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