dear sir, give me a little time (2 days or 3 days) to give a more
detailed demonstration.

but I must know two things if you enjoy:
1) are that I must show that the distance between A and B equals the
distance between f (A) and f (B)?

2) are that the algorithm is to explain or not?

and thank you

On 18 juil, 16:29, Miroslav Balaz <[email protected]> wrote:
> Now it is more clear to me.
> So labeling of vertex is set of n fourtuples.
>
> Still i do not understand the proof( first implication trivialy holds,
> because such labeling is in variant to permutation)
>
> 2) A and B have the same labeling.
> If the labeling of A in the pseudo-tree x is nhmb, labeling B in the
> pseudo-tree is also: n hmb because it af (x) = y. with the same idea
> (the automorphism keeps distance), we find that f (A) = B.
>
> try to explain it in more details. this is not proof. Try to be more formal
> at high level. If you cant prove it, than it is not worth anything.
> If you say that something is trivial and i don't understand, it means, that
> it is not trivial and you can't prove it. If it were trivial then i would
> understand it.
>
> you were may thinking about graph isomorphism more than me, so from your
> point of view it can be trivial because you are on higher level.
>
> 2009/7/17 mimouni <[email protected]>
>
>
>
>
>
> > On 16 juil, 18:19, Miroslav Balaz <[email protected]> wrote:
> > > I think that there is logical error, in the proof what do you think about
> > > it?
> > > f(A)=B iff A and B have the same labelig, but what if there are 3
> > vertices
> > > with the same labeling? say A,B,C
> > > then F(A)=B and F(A)=C
>
> > > you forget to quantify the f. I think everyone stops reading it if you
> > will
> > > have such errors there.
>
> > > 2009/7/15 mimouni <[email protected]>
>
> > > > you can consult in:http://www.wbabin.net/science/mimouni2e.pdf
> > > > and I finished on implimentation schedule a php (to find the labels
> > > > for a graph exceeds 5000 vertices).
>
> > > > On 14 juil, 19:25, Miroslav Balaz <[email protected]> wrote:
> > > > > Graph isomorphism is not very good problem, because for human
> > generated
> > > > > graphs the algorithhm for tree-isomprphism wlll work.
> > > > > But that is only my personal opinion.
>
> > > > > But it is hard to understand your algorithm.
> > > > > Mainly because i do not understand the words you are using
> > > > > peak-?
> > > > > summit-?
> > > > > pseudo tree-?
> > > > > stoppage-?
> > > > > nhbm-?
> > > > > Also you have there a lot of errors( i do not mean englis erros)
> > > > > You should rework that, i was rewriting my master's thesis proofs at
> > > > least 3
> > > > > times each.
>
> > > > > 2009/7/14 mimouni <[email protected]>
>
> > > > > > Hello, I found a new labeling vertex, which can make the deference
> > > > > > between the peaks of a graph, and thus resolve the automorphism and
> > > > > > isomorphism. Its complexity is estimated to O (n^3).
> > > > > > And here is the procedure:
> > > > > > To build a pseudo tree this way:
> > > > > > 1. Put a single vertices (example: A) in the Level 1.
> > > > > > 2. Putting all the peaks surrounding the vertices An in Level 2.
> > And
> > > > > > not forgetting the edges.
> > > > > > 3. Putting all the peaks adjacent to each vertex exists in the
> > > > > > nouveau2, and without duplication and without forgetting the edges.
> > > > > > In
> > > > > > the level 3.
> > > > > > 4. Repeat Step 3 until more vertices.
> > > > > > Labeling the vertices; is therefore in this way:
> > > > > > 1. in built all the pseudo trees.
> > > > > > 2. In seeking pseudo tree that’s a vertices lies in the level x.
> > > > > > 3. Labeling the vertices in the A-level x is composed of four
> > parts:
> > > > > > the number of times or A lies in the level x, the total number of
> > > > > > stoppages A up, the total number of stoppages in the same A level,
> > > > > > and
> > > > > > finally the total number of stoppages A down.
> > > > > > 4. And labeling a vertex is the labeling on all levels.
> > > > > > Making the deference between A and B.
> > > > > > the two vertices A and B are isomorphism between waters if they
> > both
> > > > > > have the same labeling.
> > > > > > If the labeling of A in a level x is deferential to the labeling of
> > B
> > > > > > at the same level, then A and B are deferens.
> > > > > > ========================
> > > > > > Validity of the algorithm
> > > > > > The demonstration validation of this algorithm is trivial!
> > > > > > Theorem Let A and B, two peaks in a graph G. function of the
> > > > > > automorphism of G to G is noted f.
> > > > > > f (A) = B if and only if, A and B have the same labeling.
> > > > > > Proof 1) f (A) = B.
> > > > > > Here we will show that A and B on the same labeling. Let x and two
> > > > > > other top graph G, such that f (x) = y. Labeling is based on
> > pseudo-
> > > > > > tree, so if the tree with pseudo-header as x, A is in the p, and B
> > is
> > > > > > in the level q. then the automorphism keeps the distance, then:
> > > > > > For the pseudo-tree with it as header, B is in the p, and A is in
> > the
> > > > > > level q.
> > > > > > With the same idea was for the pseudo-tree x, adjacent to A summits
> > > > > > are divided into three parts (top, at the same level as A, and
> > > > > > bottom), then the pseudo-tree there, the adjacent peaks B are also
> > > > > > divided into three parts (top, at the same level as B, and bottom).
> > > > > > So the two summits: A and B have the same labeling
> > > > > > 2) A and B have the same labeling.
> > > > > > If the labeling of A in the pseudo-tree x is nhmb, labeling B in
> > the
> > > > > > pseudo-tree is also: n hmb because it af (x) = y. with the same
> > idea
> > > > > > (the automorphism keeps distance), we find that f (A) = B.
> > > > > > So: f (A) = B if and only if, A and B have the same labeling.
> > > > > > Complexity of the algorithm
> > > > > > the complexity of a pseudo-tree is O(n²).
> > > > > > the complexity of all pseudo is so O(n³).
> > > > > > the complexity of labeling a summit from a pseudo-tree is O(n).
> > > > > > the complexity of the labeling is a summit O(n²).
> > > > > > So the algorithm is polynomial
> > > > > > =======
> > > > > > implementation
> > > > > > an application in beta (for small graphs) in php is available on:
> > > > > >http://mohamed.mimouni1.free.fr/
> > > > > > and for big graphs is avaibles on:
> > > > > >http://sites.google.com/site/isomorphismproject/

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