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0 $begingroup$ I tried to think about some characteristics of the trees, for example, they have a number of edges equal to $|text{Vertices}|-1$ and of course the handshaking lemma but I couldn't find properly logical reasoning. Any help? discrete-mathematics graph-theory trees share | cite | improve this question edited Dec 24 '18 at 18:01 EdOverflow 251 1 9 asked Dec 24 '18 at 17:44 PCNF PCNF 133 8 ...

0 $begingroup$ I'm trying to create a 3-rank tensor in numpy, with python3.x. I need to create this 3-rank tensor A in a very particular way. If I have 3 matrices, let's say that all of them are some $Y$ matrix (which is dimensionally $4x4$ ), then it must be of the following mathematical form: $A((in),(jk),(lm))$ from some type of multiplication roughly like: $"Y_{4}(i,j) otimes Y_{4}(k,l) otimes Y_{4}(m,n)"$ . So, in a sense, I must connect the "input" of a matrix to the "output" of another identical matrix, for three identical matrices. The final 3-rank tensor $A$ should have dimensions 16x16x16. Mathematically, I'm not sure if this corresponds to a tensor product or a kronecker product. Obviously this makes a big difference, as in numpy for python, the funct...

2 $begingroup$ I'm concerning myself with factoring semi-primes and believe that if given a large semi-prime ( $N$ ) one finds a non-trivial sum of squares representation: $$ x^2 + y^2 = N$$ Then one can efficiently retrieve a factorization of $N$ . My intuition for this belief is that I think the "factoring" problem over the gaussian integers is difficult and note that such a representation can be in linear time converted into a factorization $(x + yi) ( x- yi)$ when considering gaussian integers. But it's not clear to me how to leverage this information for regular integers. More Notes: So I realized after @Jack D'Aruzio's comment that this information could be exploited by computing $i$ mod $N$ That is finding some constant $c$ such that $c^2 = -1 mod N$ . Then we have...