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4h
comment How $a_{13}=0$ in $\begin{bmatrix} {2}&{1}&{0}\\ {1}&{3}&{5} \end{bmatrix}$?
Yes, but wouldn't you deduce it from that figure?
4h
comment How $a_{13}=0$ in $\begin{bmatrix} {2}&{1}&{0}\\ {1}&{3}&{5} \end{bmatrix}$?
Yes, that's why I asked if it spins. I showed the reason why I thought that.
5h
comment How $a_{13}=0$ in $\begin{bmatrix} {2}&{1}&{0}\\ {1}&{3}&{5} \end{bmatrix}$?
@Ilya Here.
5h
revised How $a_{13}=0$ in $\begin{bmatrix} {2}&{1}&{0}\\ {1}&{3}&{5} \end{bmatrix}$?
added 4 characters in body
5h
comment How $a_{13}=0$ in $\begin{bmatrix} {2}&{1}&{0}\\ {1}&{3}&{5} \end{bmatrix}$?
Yes, but I am confronting that information with another information given in the book.
5h
asked How $a_{13}=0$ in $\begin{bmatrix} {2}&{1}&{0}\\ {1}&{3}&{5} \end{bmatrix}$?
6h
comment Why $f^{+}(v)-f^-(v) =val(f)$ if $v$ is the source?
@Casteels I guess that I thought something I'm unable to explain. But as a pointed in the text, I guess I understood it right now.
6h
comment Why $f^{+}(v)-f^-(v) =val(f)$ if $v$ is the source?
@Casteels Reverse? How? I thought that I had to think about arrows incoming to the source, but then I guessed that it's not possible to have arrows incoming to the source (or at least, in most of the cases, that doesn't happen).
7h
revised Why $f^{+}(v)-f^-(v) =val(f)$ if $v$ is the source?
added 213 characters in body
7h
answered Why $f^{+}(v)-f^-(v) =val(f)$ if $v$ is the source?
7h
asked Why $f^{+}(v)-f^-(v) =val(f)$ if $v$ is the source?
1d
answered Easy to read books on Graph Theory
1d
comment How many words can be written with $aabbbccdd$ such that no two equal letters are adjacent?
@jip I guess that's what I'm doing, no? (Sorry, my head is a little messy)
1d
revised How many words can be written with $aabbbccdd$ such that no two equal letters are adjacent?
deleted 1 character in body
1d
asked How many words can be written with $aabbbccdd$ such that no two equal letters are adjacent?
1d
comment Is my idea of incoming/outgoing arcs correct?
Is it okay to have arcs with $s$ as their head? Until now, I have a little impression that such thing is not not possible (given the examples I've seen in my lectures).
1d
accepted Is my idea of incoming/outgoing arcs correct?
2d
accepted Show that if $a,k\in \mathbb{Z}$ with $0\leq k \leq a$, then $\binom ak=\frac{a!}{k!(a-k)!}=\binom {a}{a-k}$.
2d
asked Is my idea of incoming/outgoing arcs correct?
2d
awarded  Good Question