- p8 (Definition of the orders $\prec_i$): Replace $e\prec_i e'$ by $e'\prec_i e$
in the four instances.

- p11 (Lemma 1.15): $RS(G)$ instead of $|RS(G)|$.

**Note:**Rom Pinchasi has a new and very nice proof for Theorem 1.13.

The idea is to consider the convex set*K*spanned by the points with label_{i}*i*on the circle*C*. With Lemma 1.15 it follows that the*K*are pairwise disjoint. Take two copies_{i}*C*,_{1}*C*of the disk bounded by_{2}*C*and glue them along the cycles such that points*p*on*C*and_{1}*q*on*C*get matched whenever_{2}*p*and*q*are the different intersection points of a line*l*supporting an edge_{e}*e*. Blow air between the disks to make it a sphere. The intersection graph of the sets*K*and_{i}*K'*on the sphere is planar and bipartite with_{i}*2n*vertices. Hence, it has at most*2(2n) -4*edges. These edges are a double count of the edges of*G*.

Q.E.D.

- p15
**Note:**A simple proof for the linearity of edges of a quasi-planar graph was found by Eyal Ackerman and Gabor Tardos.

- p44 (last line): The upper bound of the sum ought to be $\lfloor\frac{m}{r}\rfloor -1$.

- p46 (between Thm 3.5 and proof): Let $G_3=G$ and for $i=2,1,0$
define a subgraph $G_i$ of $G_{i+1}$ by removing edges which have
$i+1$ crossings until there are no such edges left. With this
modified definition of $G_i$ it is true that an edge from
$E(G_{i+1}) \ E(G_i)$ has $i+1$ crossings with edges from $G_i$.

- p48 (Figure 3.4): The edges added in the outer faces are missing.
corrected figure

- p48 The constant 3.3 in Theorem 3.6 can be replaced by 2.7, in the
proof replace m^2 > cr(G) by m^2/2 > cr(G).

- p50 (line 3): $m(G') \leq 6n$.

- p51 (line 9): The new constant is 0.031.

- p51
**Note:**A simple proof Tutte's result stating that graphs with odd-crossing number zero are planar was obtained by Michael J. Pelsmajer, Marcus Schaefer and Daniel Stefankovic: Removing Even Crossings.

- p59 (two lines before Fact 1.) $(k+d)$-element subsets
- p59 (proof of Fact 1.) $s^k = \sum_j \binom{j}{k}\hh_j$

- p72 (bottom): The statement
*Every arrangement ......... a point contained in exactly two of the lines.*is stronger than the dual of Proposition 5.2. Upon dualization an ordinary line may end up as a pair of parallel lines.

Still the given statement is correct and admits a proof via a distance argument in the spirit of Kelly's proof.

This has been pointed out by Jonathan Lenchner. He is author of note discussing the dual of Sylvester's Theorem [pdf].

- p72 (line -6)
$px$ for $T_2$ and $qx$ for $T_1$

- p80 (line -3)
has at least $\lceil 2n/3 \rceil$ triangles

- p97 (last line)
lines to $\BB(n)$, i.e., $B_n \leq |\BB(n)|$.

- p 110(line 9)
define $\tau_t:\binom{[n]}{r-1}\to\{+,-\}$ by

- p 137-138
**Note:**There is a simpler proof for the implication $(3\rightarrow1)$ from Theorem 8.9. The new proof widely follow the approach taken by A.Frank and L.Szegö in [93]. New proof

Februar 2006