@Ryan The Hungarian algorithm yields a selection of elements in $S$ such that their sum is maximal. (It does not return an integer.) Since each row/column in $S$ corresponds to a vertex in the first/second graph to compare, we can consider an entry in $S$ to provide a map from the first graph to the second graph. The row/columns of $S$ that are selected then generates a mapping between the two graphs. Does that help, or am I misunderstanding your question?

Some context for the question, in case people were wondering: academia.stackexchange.com/questions/31410/…

See page 13 of this slide deck. Per my understanding, the entries in $S$ need not be integral for the algorithm to still work.

Ok, thanks! That answers all my questions! :)

(I'm marking this as accepted because it does answer my question and I don't want to forget to mark it as such. That said, I'd still like to know the answers to the clarifications I requested above. :) Thank you very much for your answer.)

Thank you very much! Just to confirm: I have it backwards because people don't typically care about the size of $N$ unless they are talking about complexity classes? (I want to make sure exactly what I have backwards. ;)) And, also to confirm, you're using $\langle N\rangle$ to denote the size of $N$? (I haven't seen those symbols before.)

Hmmm... I guess I concede--I see your point now (the sentence "prints the number on screen..." was what I needed to read to see what you were saying.) Thanks for your patience with an obstinate person... :)

This does pass the number through base 10 representation, as the fromDigits returns the number in base 10.