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comment How Graph Isomorphism is used to determine Graph Automorphism?
"Now consider a transposition $\pi$ that moves $(i+1)$ th vertex to $j'$ th position where $i <j'\leq n$." Is it really sufficient to only look at transpositions $\pi$? What would happen, if you used this algorithm to compute the automorphism group of a directed circle (whose automorphism group doesn't contain any transposition)?
Jan
25
revised Bound on the size of Permutation Set for Isomorphism
question has been modified, wrote an extended comment on my attempts to help Jim!
Jan
19
accepted Graphs embeddable into tree like simplicial 2-complexes
Jan
18
answered Bound on the size of Permutation Set for Isomorphism
Jan
4
comment Isomorphism of Non-Symmetric Matrices
Your algorithm doesn't work as claimed. You basically only describe the divide phase of the algorithm, but use handwaving instead of a description for how to combine the results from the subproblems to form a global solution. You also rely on handwaving for your mathematical notation: $2^{\sum log_2(m)}$ is unclear (sum over what?), $E_2= \min(K_i,J_i)$ is unclear (minimum of two matrices?), $J_i < m_1/2$ is unclear (a matrix is smaller than some number?), ... I admit that I can guess what you want to say, but how can I help you understand where your algorithm fails, with such handwaving?
Jan
4
answered Isomorphism of Non-Symmetric Matrices
Dec
27
awarded  Popular Question
Dec
18
asked Graphs embeddable into tree like simplicial 2-complexes
Dec
16
reviewed Leave Open Can we always multiply some function that is not differentiable everywhere with function that is to obtain differentiable product?
Dec
16
reviewed Leave Open Solution about a Michael Aschbacher exercise [edit]
Dec
16
reviewed Leave Open Is there a non-trivial cyclic quotient group of a non-cyclic group?
Dec
15
comment What does it mean if a free algebra has an unsolvable word problem?
@MattSamuel In the context of the word problem for groups, one talks about presentation of a group which means that a group is given by a set of generators and a set of relations among those generators. In this context, a relation is not allowed to contain free variables. So there might be a polynomial time algorithm for deciding the word problem for lattices (including relations), even so the word problem for free modular lattices is unsolvable.
Dec
15
revised What does it mean if a free algebra has an unsolvable word problem?
added links to definitions and the "terminology" tag
Dec
15
comment What does it mean if a free algebra has an unsolvable word problem?
@MattSamuel I thought about it some more, and came to the conclusion that equational axioms and relations are completely different things. If $a,b,c$ are your generators, then a relation is an equation between $a,b,c$ containing no free variable (=universally quantified variable) at all. An equational axiom on the other hand is an equation between free variables, containing no generators at all.
Dec
15
comment What does it mean if a free algebra has an unsolvable word problem?
@MattSamuel Why do you mean that a free modular lattice is very far from free algebraically? It seems to fit exactly the definition of a free algebra. No sets of relations in addition to the equational axioms seem to be allowed, so it seems significantly more restricted than an algebra presented by a set of generators and relations. (In which case the relations would be allowed to explicitly name a specific generator, while the equational axioms only contain universal quantified variables without any reference to the generators).
Dec
15
asked What does it mean if a free algebra has an unsolvable word problem?
Dec
7
revised Zero divided by zero must be equal to zero
corrected typo, and added more context and applications to the answer
Dec
5
revised Zero divided by zero must be equal to zero
wrote down one symbolic expression for a/b+c/d, just to show that one can still compute
Dec
5
answered Zero divided by zero must be equal to zero
Nov
13
accepted Are pseudoheaps and heaps the same thing?