/// <summary>
/// Calculates the distance of every connection between a genre or tag from m1 and a genre or tag of m2.
/// </summary>
/// <param name="m1">Media 1</param>
/// <param name="m2">Media 2</param>
/// <returns>A vector with the distance of each connection in a component.</returns>
/// <remarks>
/// 1. Differs from arrow in that the direction is irrelevant, though if an arrow appears in both directions the distance is
/// halved (and round up).
/// 2. Theorem: Arrow A_1 -> B_2 exists explicitly in the collection of reversed arrows, that is arrows of the form C_2 -> D_1.
/// <=> Arrows A_1 -> B_2 and B_1 -> A_2 both exist explicitly.
/// Notational remark: subscript stands for the media and the symbol in front for the tag or genre name. So A_1 means tag
/// or genre A from media 1.
///
/// Proof: =>: By assumption A_1 -> B_2 exists explicitly in the collection of reversed arrows, so there exists an arrow
/// A_2 -> B_1. But this implies B is a genre or tag of media 1 and A is a genre or tag of media 2 so the arrow
/// B_1 -> A_2 exists in the collection or arrows from media 1 to media 2.
/// <=: Assume Arrows A_1 -> B_2 and B_1 -> A_2 both exist explicitly. Then A and B are both genre or tags of media 1
/// and media 2. Hence the arrow A_2 -> B_1 exists in the collectrion of reversed arrows.
/// 3. Since the arrow in the other direction exists explicitly if the reversed arrow exists (by the above theorem)
/// it needs to be found and removed anyway.
/// 4. Moreover since the existence of the reversed arrow in the initial direction implies the existence of the reversed arrow
/// in the collection of reversed arrows it is not necessary to work with the collection of reversed arrows. In fact it's less
/// efficient since there would be a list traversal to find the arrow in the collection of reversed arrows and there would need
/// to be a traversal to remove the reversed arrow, whoms existence is assured by the theorem above, from the initial collection
/// of arrows. One would rather look for the reversed arrow in the initial collection of arrows, remove it and halven the
/// distance of the remaining connection!
/// </remarks>
private List <GenreTagConnection> CompleteDifferenceVectorConnection(Media m1, Media m2)
{
List <GenreTagConnection> cdv = CompleteDifferenceVectorArrow(m1, m2);
// New list created, since rather not mend the list that's being looped over.
List <GenreTagConnection> cdvConnection = new List <GenreTagConnection>(cdv.Count);
foreach (GenreTagConnection gtc in cdv)
{
GenreTagConnection alreadyExistingSameConnection = cdvConnection.Find(cdvConnectionGtc => cdvConnectionGtc.SameConnection(gtc));
// The connection does already exist.
if (alreadyExistingSameConnection != null)
{
alreadyExistingSameConnection.HalvenDistance();
}
// The connection does not already exist.
else
{
cdvConnection.Add(gtc);
}
}
return(cdvConnection);
}
private BigInteger InnerProductWithBaseConnection(List <GenreTagConnection> dv)
{
BigInteger sum = 0;
foreach (GenreTagConnection baseGtc in baseDifferenceVector)
{
GenreTagConnection cBaseGtcConnectionInDV = dv.Find(cdvGtc => cdvGtc.SameConnection(baseGtc));
// The current base connection being considered exists in dv.
if (cBaseGtcConnectionInDV != null)
{
sum += cBaseGtcConnectionInDV.Distance * baseGtc.Distance;
}
// The current base connection being considered does not exist in dv.
else
{
continue;
}
}
return(sum);
}
