As a default, hashCode/equality of edges is determined by their ends. We say that edge ends are part of the edge key.
Whenever your custom edge needs to be a multi-edge, meaning that your graph is allowed to connect some nodes by an instance of this edge even if those nodes are already connected, the edge key needs be extended by adding at least one more class member, constant or whatsoever to the edge key.
For example edges representing flight connections between airports need be multi-edges to allow for different flights between the same airports.
Each value, such as a class member, you specify in this non-empty sequence will be added to the edge "key" which contains the edge ends, in terms of a directed edge the source and the target, by default. The term "key" refers to how equality of edges is controlled.
Each value, such as a class member, you specify in this non-empty sequence will be added to the edge "key" which contains the edge ends, in terms of a directed edge the source and the target, by default. The term "key" refers to how equality of edges is controlled.
It is reflexive: for any instance x of type Any, x.equals(x) should return true.
It is symmetric: for any instances x and y of type Any, x.equals(y) should return true if and only if y.equals(x) returns true.
It is transitive: for any instances x, y, and z of type Any if x.equals(y) returns true and y.equals(z) returns true, then x.equals(z) should return true.
If you override this method, you should verify that your implementation remains an equivalence relation. Additionally, when overriding this method it is usually necessary to override hashCode to ensure that objects which are "equal" (o1.equals(o2) returns true) hash to the same scala.Int. (o1.hashCode.equals(o2.hashCode)).
Value parameters
that
the object to compare against this object for equality.
Attributes
Returns
true if the receiver object is equivalent to the argument; false otherwise.
The default hashing algorithm is platform dependent.
Note that it is allowed for two objects to have identical hash codes (o1.hashCode.equals(o2.hashCode)) yet not be equal (o1.equals(o2) returns false). A degenerate implementation could always return 0. However, it is required that if two objects are equal (o1.equals(o2) returns true) that they have identical hash codes (o1.hashCode.equals(o2.hashCode)). Therefore, when overriding this method, be sure to verify that the behavior is consistent with the equals method.