Talk:Linkless embedding

I am confused by the definition here, carried over from an article of Robertson, Seymour and Thomas:

"The version of the linking number used for defining linkless embeddings of graphs is found by projecting the embedding onto the plane and counting the number of crossings of the projected embedding in which the first curve passes over the second one, modulo 2".

In an example from the page about the linking number (https://en.wikipedia.org/wiki/Linking_number#Definition), the first example from the left has linking number -2. But by the above definition, the blue curve passes over the red one two times, so the linking number is zero. Is mod 2 a mistake here? — Preceding unsigned comment added by 158.129.140.71 (talk) 08:15, 18 January 2016 (UTC)

English is not may native language, and for me sentence Although linkless and flat embeddings are not the same, the graphs that have linkless embeddings turn out to be the same as the graphs that have flat embeddings is unclear, perhaps because of idiomatic expression turn out to be, that hard to tramslate in Russian. May you explain it in another words? Jumpow (talk) 21:59, 1 April 2016 (UTC)

:Replacing "turn out to be" with "are" preserves the meaning and is simpler. —David Eppstein (talk) 22:39, 1 April 2016 (UTC)

:: Thank you. Now I understood sentence. Jumpow (talk) 15:34, 2 April 2016 (UTC)

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Does anyone have {{harvtxt|Kawarabayashi|Kreutzer|Mohar|2010}}? From the abstract it seems to be about O(n^2) time for constructing a flat embedding, not in linear time as it appears in the text. SyP (talk) 14:32, 26 December 2018 (UTC)

:It appears to be freely available online at http://logic.las.tu-berlin.de/Members/Kreutzer/Publications/12-dcg.pdf and it does indeed appear to be O(n^2) not O(n). —David Eppstein (talk) 17:52, 26 December 2018 (UTC)