Editing Linking number (section)

==Computing the linking number==
[[Image:Linking Number Example.svg|thumb|With six positive crossings and two negative crossings, these curves have linking number two.]]
There is an [[algorithm]] to compute the linking number of two curves from a link [[knot theory#Knot diagrams|diagram]].  Label each crossing as ''positive'' or ''negative'', according to the following rule:<ref>This is the same labeling used to compute the [[writhe]] of a [[knot (mathematics)|knot]], though in this case we only label crossings that involve both curves of the link.</ref>
[[File:Link Crossings.svg|center|350px]]

The total number of positive crossings minus the total number of negative crossings is equal to ''twice'' the linking number.  That is:
:<math>\text{linking number}=\frac{n_1 + n_2 - n_3 - n_4}{2}</math>
where ''n''<sub>1</sub>, ''n''<sub>2</sub>, ''n''<sub>3</sub>, ''n''<sub>4</sub> represent the number of crossings of each of the four types. The two sums <math>n_1 + n_3\,\!</math> and <math>n_2 + n_4\,\!</math> are always equal,<ref>This follows from the [[Jordan curve theorem]] if either curve is simple.  For example, if the blue curve is simple, then  ''n''<sub>1</sub>&nbsp;+&nbsp;''n''<sub>3</sub> and ''n''<sub>2</sub>&nbsp;+&nbsp;''n''<sub>4</sub> represent the number of times that the red curve crosses in and out of the region bounded by the blue curve.</ref> which leads to the following alternative formula
:<math>\text{linking number} \,=\, n_1 - n_4 \,=\, n_2 - n_3.</math>
The formula <math>n_1-n_4</math> involves only the undercrossings of the blue curve by the red, while <math>n_2-n_3</math> involves only the overcrossings.