That looks correct. Indeed, the lexiographic ordering establishes a total order on R^n which would allow the use of set. Thanks.
As a related question, sets in STL are implemented as trees. ...
Type: Posts; User: Steff
That looks correct. Indeed, the lexiographic ordering establishes a total order on R^n which would allow the use of set. Thanks.
As a related question, sets in STL are implemented as trees. ...
That's the point. There is no total ordering on vectors in R^n. In other words, while [2,2]>[1,1], we do not have that [1,2] > [2,1] nor [2,1]>[1,2]. We do have a partial ordering, but sets in STL...
Hi,
Sets in STL require a comparison operation. However, I have a situation where I'd like to create a set of vectors of numbers. So, I have a partial order, but no strict ordering. What's...