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Dec
8
comment Algorithm to check whether a graph has no cycles
Considering correctness, what do you suggest for a better approch with the cycle $C$?
Dec
8
comment Algorithm to check whether a graph has no cycles
Please have a look at my updated question. Is this the kind of proof for correctness you recommended me?
Dec
8
revised Algorithm to check whether a graph has no cycles
added a second attempt to solve this problem
Dec
8
comment Algorithm to check whether a graph has no cycles
Furthermore to reduce redundancy as $\mu$ and $V'$ would hold visited nodes, an option would be to use $\mu$ for a topsort.
Dec
8
comment Algorithm to check whether a graph has no cycles
How about the following idea: Each time we visit a node we could set µ to 0 and add the visited node to a new set $V'$. At the end of the while loop we could check for emptiness of $V\setminus V'=\emptyset$. If this is true we can return false because there are no more components left; otherwise we pick an arbitrary $v'\in V\setminus V'$ and then repeat everything with $U=\{v'\}$. Do you think that would solve my problem with the components?
Dec
8
revised Algorithm to check whether a graph has no cycles
added note that we have to use DFS/BFS
Dec
8
asked Algorithm to check whether a graph has no cycles
Dec
5
accepted Derivatives and their domains
Dec
4
awarded  Enthusiast
Dec
3
answered Derivatives and their domains
Dec
3
comment Derivatives and their domains
@Siminore: Do you have some other clues about the domains of (3)-(5)?
Dec
3
revised Derivatives and their domains
fixed a - sign which should be a + in (1)
Dec
3
comment Derivatives and their domains
@Siminore: Yet one can argument that e.g. $\sqrt{\sqrt{-8}}$ leads to problems if we don't care about complex numbers (which are excluded in this excercise - I did forgot to mention this), or am I wrong here?
Dec
3
revised Derivatives and their domains
added the notice that the domains should be based on R and not C
Dec
3
asked Derivatives and their domains
Nov
26
revised Determining limit points and proving there are no more of them
fixed indexes n and 2k and 2k+1
Nov
25
accepted Determining limit points and proving there are no more of them
Nov
25
comment Determining limit points and proving there are no more of them
@TonyK: Oh I see the mistake - using the absolute value of $b_n$ always yields positive results which is no reason why $b_n$ should be bounded.
Nov
25
revised Determining limit points and proving there are no more of them
added 89 characters in body
Nov
25
comment Determining limit points and proving there are no more of them
@TonyK: Could you give an example why this isn't true?