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| Maury Barbato... |
 Posted: Tue Oct 27, 2009 11:57 pm |
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Guest
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Hello,
let G and G' two isomorphic finite planar graphs
embedded in R^2. Then, I suppose that there exists a
homeomorphism f:R^2 -> R^2 such that:
(I) if v is a vertex of G, then f(v) is a vertex of G',
(II) if e is an edge of G, then f(e) is an edge of G'.
Maybe, my question is trivial, but I have a very
poor knowledge of topology, so ...
Thank you very much for your attention.
My Best Regards,
Maury Barbato |
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| Peter Webb... |
| Posted: Wed Oct 28, 2009 5:19 am |
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Guest
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"Maury Barbato" <mauriziobarbato at (no spam) aruba.it> wrote in message
news:1414614252.125017.1256723881279.JavaMail.root at (no spam) gallium.mathforum.org...
[quote]Hello,
let G and G' two isomorphic finite planar graphs
embedded in R^2. Then, I suppose that there exists a
homeomorphism f:R^2 -> R^2 such that:
(I) if v is a vertex of G, then f(v) is a vertex of G',
(II) if e is an edge of G, then f(e) is an edge of G'.
[/quote]
It seems like you are trying to give a definition for two graphs being
isomorphic. Assuming we are talking about the simplest types of graphs with
just vertexes and edges, you are on the right track but not quite there. You
have "injected" G into G', but you need to form a bijection. The way you
have defined it, G' could be a superset of G, or even a subset of G, as
multiple points and edges could map to a single point and edge.
[quote]Maybe, my question is trivial, but I have a very
poor knowledge of topology, so ...
Thank you very much for your attention.
My Best Regards,
Maury Barbato[/quote] |
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| David C. Ullrich... |
| Posted: Wed Oct 28, 2009 5:34 am |
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Guest
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On Wed, 28 Oct 2009 05:57:31 EDT, Maury Barbato
<mauriziobarbato at (no spam) aruba.it> wrote:
[quote]Hello,
let G and G' two isomorphic finite planar graphs
embedded in R^2. Then, I suppose that there exists a
homeomorphism f:R^2 -> R^2 such that:
(I) if v is a vertex of G, then f(v) is a vertex of G',
(II) if e is an edge of G, then f(e) is an edge of G'.
[/quote]
Not true. Let the vertices of G be the vertices of a
square, and let the edges of G be the edges of that
square plus one of the diagonals. Let G' be the
same as G except the diagonal edge is replaced
by an edge looping outside the square, connecting
the same two vertices.
Then the complement of G' has a compact component
bounded by four edges, which is not true of G.
That example wouldn't work with S^2 in place
of R^2, but I think it's clear that a more elaborate
version of the same idea would work in S^2 as well.
[quote]Maybe, my question is trivial, but I have a very
poor knowledge of topology, so ...
Thank you very much for your attention.
My Best Regards,
Maury Barbato
[/quote]
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.) |
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| Butch Malahide... |
| Posted: Wed Oct 28, 2009 7:23 pm |
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Guest
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On Oct 29, 12:10 am, Gerry Myerson <ge... at (no spam) maths.mq.edi.ai.i2u4email>
wrote:
[quote]In article <61bge5p81jc5nct8itjugckl6ba7q68... at (no spam) 4ax.com>,
David C. Ullrich <dullr... at (no spam) sprynet.com> wrote:
On Wed, 28 Oct 2009 05:57:31 EDT, Maury Barbato
mauriziobarb... at (no spam) aruba.it> wrote:
Hello,
let G and G' two isomorphic finite planar graphs
embedded in R^2. Then, I suppose that there exists a
homeomorphism f:R^2 -> R^2 such that:
(I) if v is a vertex of G, then f(v) is a vertex of G',
(II) if e is an edge of G, then f(e) is an edge of G'.
Not true. Let the vertices of G be the vertices of a
square, and let the edges of G be the edges of that
square plus one of the diagonals. Let G' be the
same as G except the diagonal edge is replaced
by an edge looping outside the square, connecting
the same two vertices.
Then the complement of G' has a compact component
bounded by four edges, which is not true of G.
That example wouldn't work with S^2 in place
of R^2, but I think it's clear that a more elaborate
version of the same idea would work in S^2 as well.
I wonder if this works for S^2.
Here are two graphs, each with 6 vertices and 6 edges.
Each has a K_3, an edge joining the 4th vertex to one of
the vertices of the K_3, and edges joining the other two
vertices to one of the other vertices of the K_3. They
differ in which of the non-K_3 edges is interior to the K_3.
The graphs are isomorphic, and I think you can set it up
so there's no homeomorphism of S^2 carrying one to
the other.
[/quote]
Doesn't this work, with just 5 vertices and 5 edges? Let G_1 be a K_3
plus two pendant vertices, each joined to a different vertex of the
K_3; i.e., the self-complementary graph of order 5; and embed it in
the sphere by putting the K_3 on the equator, one pendant vertex in
the northern hemisphere, the other in the southern hemisphere. Let G_2
be the same except that both of the pendant vertices are in the
northern hemisphere. |
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| Gerry Myerson... |
| Posted: Wed Oct 28, 2009 11:10 pm |
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Guest
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In article <61bge5p81jc5nct8itjugckl6ba7q68cd1 at (no spam) 4ax.com>,
David C. Ullrich <dullrich at (no spam) sprynet.com> wrote:
[quote]On Wed, 28 Oct 2009 05:57:31 EDT, Maury Barbato
mauriziobarbato at (no spam) aruba.it> wrote:
Hello,
let G and G' two isomorphic finite planar graphs
embedded in R^2. Then, I suppose that there exists a
homeomorphism f:R^2 -> R^2 such that:
(I) if v is a vertex of G, then f(v) is a vertex of G',
(II) if e is an edge of G, then f(e) is an edge of G'.
Not true. Let the vertices of G be the vertices of a
square, and let the edges of G be the edges of that
square plus one of the diagonals. Let G' be the
same as G except the diagonal edge is replaced
by an edge looping outside the square, connecting
the same two vertices.
Then the complement of G' has a compact component
bounded by four edges, which is not true of G.
That example wouldn't work with S^2 in place
of R^2, but I think it's clear that a more elaborate
version of the same idea would work in S^2 as well.
[/quote]
I wonder if this works for S^2.
Here are two graphs, each with 6 vertices and 6 edges.
Each has a K_3, an edge joining the 4th vertex to one of
the vertices of the K_3, and edges joining the other two
vertices to one of the other vertices of the K_3. They
differ in which of the non-K_3 edges is interior to the K_3.
The graphs are isomorphic, and I think you can set it up
so there's no homeomorphism of S^2 carrying one to
the other.
--
Gerry Myerson (gerry at (no spam) maths.mq.edi.ai) (i -> u for email) |
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| David C. Ullrich... |
| Posted: Thu Oct 29, 2009 7:43 am |
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Guest
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On Wed, 28 Oct 2009 22:19:40 +1100, "Peter Webb"
<webbfamily at (no spam) DIESPAMDIEoptusnet.com.au> wrote:
[quote]
"Maury Barbato" <mauriziobarbato at (no spam) aruba.it> wrote in message
news:1414614252.125017.1256723881279.JavaMail.root at (no spam) gallium.mathforum.org...
Hello,
let G and G' two isomorphic finite planar graphs
embedded in R^2. Then, I suppose that there exists a
homeomorphism f:R^2 -> R^2 such that:
(I) if v is a vertex of G, then f(v) is a vertex of G',
(II) if e is an edge of G, then f(e) is an edge of G'.
It seems like you are trying to give a definition for two graphs being
isomorphic.
[/quote]
Huh? It doesn't seem that way to me - it seems like he _knows_
the (perfectly standard) definition, and he's asking whether it
implies something.
[quote]Assuming we are talking about the simplest types of graphs with
just vertexes and edges, you are on the right track but not quite there. You
have "injected" G into G', but you need to form a bijection. The way you
have defined it,
[/quote]
What "definition"?
Suppose I asked whether it was true that every even integer was
the sum of two primes. Would it seem to you like I was trying
to _define_ the word "even"?
[quote]G' could be a superset of G, or even a subset of G, as
multiple points and edges could map to a single point and edge.
Maybe, my question is trivial, but I have a very
poor knowledge of topology, so ...
Thank you very much for your attention.
My Best Regards,
Maury Barbato
[/quote]
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.) |
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| Gerry Myerson... |
| Posted: Thu Oct 29, 2009 6:10 pm |
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Guest
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In article
<d0fd6b00-a53f-47f0-8875-19af0d66f3a0 at (no spam) j19g2000yqk.googlegroups.com>,
Butch Malahide <fred.galvin at (no spam) gmail.com> wrote:
[quote]On Oct 29, 12:10 am, Gerry Myerson <ge... at (no spam) maths.mq.edi.ai.i2u4email
wrote:
In article <61bge5p81jc5nct8itjugckl6ba7q68... at (no spam) 4ax.com>,
David C. Ullrich <dullr... at (no spam) sprynet.com> wrote:
On Wed, 28 Oct 2009 05:57:31 EDT, Maury Barbato
mauriziobarb... at (no spam) aruba.it> wrote:
Hello,
let G and G' two isomorphic finite planar graphs
embedded in R^2. Then, I suppose that there exists a
homeomorphism f:R^2 -> R^2 such that:
(I) if v is a vertex of G, then f(v) is a vertex of G',
(II) if e is an edge of G, then f(e) is an edge of G'.
Not true. Let the vertices of G be the vertices of a
square, and let the edges of G be the edges of that
square plus one of the diagonals. Let G' be the
same as G except the diagonal edge is replaced
by an edge looping outside the square, connecting
the same two vertices.
Then the complement of G' has a compact component
bounded by four edges, which is not true of G.
That example wouldn't work with S^2 in place
of R^2, but I think it's clear that a more elaborate
version of the same idea would work in S^2 as well.
I wonder if this works for S^2.
Here are two graphs, each with 6 vertices and 6 edges.
Each has a K 3, an edge joining the 4th vertex to one of
the vertices of the K 3, and edges joining the other two
vertices to one of the other vertices of the K 3. They
differ in which of the non-K 3 edges is interior to the K 3.
The graphs are isomorphic, and I think you can set it up
so there's no homeomorphism of S^2 carrying one to
the other.
Doesn't this work, with just 5 vertices and 5 edges? Let G 1 be a K 3
plus two pendant vertices, each joined to a different vertex of the
K 3; i.e., the self-complementary graph of order 5; and embed it in
the sphere by putting the K 3 on the equator, one pendant vertex in
the northern hemisphere, the other in the southern hemisphere. Let G 2
be the same except that both of the pendant vertices are in the
northern hemisphere.
[/quote]
Looks good to me.
--
Gerry Myerson (gerry at (no spam) maths.mq.edi.ai) (i -> u for email) |
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| }niaM olleH dlroW ediW beW... |
| Posted: Thu Oct 29, 2009 8:12 pm |
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Guest
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On Oct 29, 8:10Â pm, Gerry Myerson <ge... at (no spam) maths.mq.edi.ai.i2u4email>
wrote:
[quote]In article
d0fd6b00-a53f-47f0-8875-19af0d66f... at (no spam) j19g2000yqk.googlegroups.com>,
 Butch Malahide <fred.gal... at (no spam) gmail.com> wrote:
On Oct 29, 12:10Â am, Gerry Myerson <ge... at (no spam) maths.mq.edi.ai.i2u4email
wrote:
In article <61bge5p81jc5nct8itjugckl6ba7q68... at (no spam) 4ax.com>,
 David C. Ullrich <dullr... at (no spam) sprynet.com> wrote:
On Wed, 28 Oct 2009 05:57:31 EDT, Maury Barbato
mauriziobarb... at (no spam) aruba.it> wrote:
Hello,
let G and G' two isomorphic finite planar graphs
embedded in R^2. Then, I suppose that there exists a
homeomorphism f:R^2 -> R^2 such that:
(I) if v is a vertex of G, then f(v) is a vertex of G',
(II) if e is an edge of G, then f(e) is an edge of G'.
Not true. Let the vertices of G be the vertices of a
square, and let the edges of G be the edges of that
square plus one of the diagonals. Let G' be the
same as G except the diagonal edge is replaced
by an edge looping outside the square, connecting
the same two vertices.
Then the complement of G' has a compact component
bounded by four edges, which is not true of G.
That example wouldn't work with S^2 in place
of R^2, but I think it's clear that a more elaborate
version of the same idea would work in S^2 as well.
I wonder if this works for S^2.
Here are two graphs, each with 6 vertices and 6 edges.
Each has a K 3, an edge joining the 4th vertex to one of
the vertices of the K 3, and edges joining the other two
vertices to one of the other vertices of the K 3. They
differ in which of the non-K 3 edges is interior to the K 3.
The graphs are isomorphic, and I think you can set it up
so there's no homeomorphism of S^2 carrying one to
the other.
Doesn't this work, with just 5 vertices and 5 edges? Let G 1 be a K 3
plus two pendant vertices, each joined to a different vertex of the
K 3; i.e., the self-complementary graph of order 5; and embed it in
the sphere by putting the K 3 on the equator, one pendant vertex in
the northern hemisphere, the other in the southern hemisphere. Let G 2
be the same except that both of the pendant vertices are in the
northern hemisphere.
Looks good to me.
--
Gerry Myerson (ge... at (no spam) maths.mq.edi.ai) (i -> u for email)
[/quote]
Results 1 - 10 for {> Doesn't this work, with just 5 vertices and 5
edges? Let G 1 be a K 3> plus two pendant vertices, each joined to a
different vertex of the> K 3; i.e., the self-complementary graph of
order 5; and embed it in> the sphere by putting the K 3 on the
equator, one pendant vertex in> the northern hemisphere, the other in
the southern hemisphere. Let G 2> be the same except that both of the
pendant vertices are in the> northern hemisphere.Looks good to me.--
Gerry Myerson (ge... at (no spam) maths.mq.edi.ai) (i -> u for email)}. (0.27
seconds)
[PDF] Graphs and AlgorithmsFile Format: PDF/Adobe Acrobat - View as
HTML
vertices in a graph G and let k be a positive integer. ..... 1.1.5,
that there exist two vertex disjoint paths each with one end in {u, v}
and ...... pendant edge, that is adding a new vertex and a new edge
from the new vertex ... Indeed, if G∗ is the dual of a plane embedding
of G, then, by (3.4.3), the max- ...
homepages.cwi.nl/~bgerards/personal/ga.pdf
[PDF] CanaDAM 2009File Format: PDF/Adobe Acrobat - View as HTML
Les sujets abordés pendant le congrès proviendront de tous les
domaines des ..... number of vertices or edges) satisfy cn ∼ cgnn
−5/2 ..... itive integers n and k, n<k, a self-complementary ...... k
such that each vertex is reachable under every distri- ..... and K∆
−1,∆, any bipartite graph G with ∆ ≥ 3 has ....
www.crm.umontreal.ca/CanaDAM2009/canadam09.pdf
The "I need Help Choosing a Name" Thread! - Page 5 - Graffiti ...oh
for fuck sake put a fucking k in it instead, or just put any fucking
letters you like ... Activity, Longevity. 1/20, 19/20. Today, Posts.
0/5, s 2 3 1 4 ...
www.puregraffiti.com/graffiti-space/showthread.php?t=787&page=5
Result for query "keyword(s)=in author= title="We now turn our
attention to the case $k=3$. In turns out that in this ...... In \cite
[Theorem~0.2] oh, it is shown; $G (1, \bold P^5)$ embedded in $ \bold
P^ ...... Each; edge of an elementary disk is a straight line in a 2-
simplex of ...... Let $A_1, \dotsc, A_t$ be the; vertices in $\Gamma$
that are adjacent to ...
nyjm.albany.edu:8000/cgi-bin/aglimpse/19/nyjm/Http/.../j?...in
[PDF] 8 Cologne-Twente Workshop on Graphs and Combinatorial
Optimization ...File Format: PDF/Adobe Acrobat
Suppose G1 and G2 are graphs with disjoint vertex-sets and k ≥ 0 is an
integer. ... (Fomin and Thilikos [3]) Let G1 and G2 be graphs with one
edge or ..... if G has at least two vertices then each internal node
of the tree has at least ...... 5: Check if there are l(j + 1)
vertices with pairwisely different colors ...
www.lix.polytechnique.fr/ctw09/ctw09-proceedings.pdf
Result for query "keyword(s)=in author= title="For each $a \in X$ we
can identify; For two distinct vertices $a, c$ in $X$ we write $X
(c, ...... K1 of separative exchange rings and C-algebras with real
rank zero ...... Embeddings of Z2-homology 3-spheres in R up to
regular homotopy ..... In a graph $G$, the volume of a subset $X$ of
the vertex set $V$, ...
nyjm.albany.edu:8000/cgi-bin/aglimpse/19/nyjm/Http/.../j?...in
PSX MASTER LIST - Page 2Aug 19, 2009 ... SLPM-86812 : Jet de Go! -
Let's Go By Airliner [Playstation The Best] SCUS-94309 : Jet Moto ....
SLPM-86763 : K-1 World Grand Prix 2001 - Kaimakuban by Xing .....
SLPM-86933 : Kuma no Pooh-San - Mori no Nakamato 1 2 3 ......
SCUS-94212 : Official U.S. PlayStation Magazine Demo Disc 5 ...
cybertopic.youneed.us/ps1-and-ps2-f14/psx-master-list-t65-15.htm
Source for "MobWars Auto-Searcher" – Userscripts.orgBy mobwars — Last
update Sep 3, 2009 — Installed 2172 times. ......
","vertebra","vertebrae","vertebral","vertebrate","vertex","vertical","vertically","vertices","vertiginous" .....
5); var searchBeginning; var useQuestionMark = 1; ..... for Mob Wars
Auto-Searcher\nWould you like to go to the install page now? ...
userscripts.org/scripts/review/54848
Godel Escher Bach by Douglas R. HofstadterIn order for a theme to work
as a canon theme, each of its notes must be able to .... Bach and
Escher are playing one single theme in two different "keys": music and
art. ...... The string --p---q--- is a theorem because 2 plus 3 equals
5. ...... And besides, metawishes are my favorite k of wish. Let me
just see if ...
www.scribd.com/doc/.../Godel-Escher-Bach-by-Douglas-R-Hofstadter-
Result for query "keyword(s)=and author= title="In Section~ 5 we show
that Theorems~\thmref 1.2 and~ 1.3 remain valid ... This result was
investigated by D. Hoff and K. Zumbrun \cite 2, 3; fluid flow, .....
Let us remark that if two increasing sequences of integers $(n_k)$ and
$(m_k)$ ...... One of the purposes of; polygon with $n+2$ vertices on
the geometric ...
nyjm.albany.edu:8000/cgi-bin/aglimpse/19/nyjm/Http/.../j?...and
1 2 3 4 5 6 7 8 9 Next |
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| }niaM olleH dlroW ediW beW... |
| Posted: Thu Oct 29, 2009 8:13 pm |
|
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Guest
|
On Oct 29, 8:10Â pm, Gerry Myerson <ge... at (no spam) maths.mq.edi.ai.i2u4email>
wrote:
[quote]In article
d0fd6b00-a53f-47f0-8875-19af0d66f... at (no spam) j19g2000yqk.googlegroups.com>,
 Butch Malahide <fred.gal... at (no spam) gmail.com> wrote:
On Oct 29, 12:10Â am, Gerry Myerson <ge... at (no spam) maths.mq.edi.ai.i2u4email
wrote:
In article <61bge5p81jc5nct8itjugckl6ba7q68... at (no spam) 4ax.com>,
 David C. Ullrich <dullr... at (no spam) sprynet.com> wrote:
On Wed, 28 Oct 2009 05:57:31 EDT, Maury Barbato
mauriziobarb... at (no spam) aruba.it> wrote:
Hello,
let G and G' two isomorphic finite planar graphs
embedded in R^2. Then, I suppose that there exists a
homeomorphism f:R^2 -> R^2 such that:
(I) if v is a vertex of G, then f(v) is a vertex of G',
(II) if e is an edge of G, then f(e) is an edge of G'.
Not true. Let the vertices of G be the vertices of a
square, and let the edges of G be the edges of that
square plus one of the diagonals. Let G' be the
same as G except the diagonal edge is replaced
by an edge looping outside the square, connecting
the same two vertices.
Then the complement of G' has a compact component
bounded by four edges, which is not true of G.
That example wouldn't work with S^2 in place
of R^2, but I think it's clear that a more elaborate
version of the same idea would work in S^2 as well.
I wonder if this works for S^2.
Here are two graphs, each with 6 vertices and 6 edges.
Each has a K 3, an edge joining the 4th vertex to one of
the vertices of the K 3, and edges joining the other two
vertices to one of the other vertices of the K 3. They
differ in which of the non-K 3 edges is interior to the K 3.
The graphs are isomorphic, and I think you can set it up
so there's no homeomorphism of S^2 carrying one to
the other.
Doesn't this work, with just 5 vertices and 5 edges? Let G 1 be a K 3
plus two pendant vertices, each joined to a different vertex of the
K 3; i.e., the self-complementary graph of order 5; and embed it in
the sphere by putting the K 3 on the equator, one pendant vertex in
the northern hemisphere, the other in the southern hemisphere. Let G 2
be the same except that both of the pendant vertices are in the
northern hemisphere.
Looks good to me.
--
Gerry Myerson (ge... at (no spam) maths.mq.edi.ai) (i -> u for email)
[/quote]
Results 1 - 10 for {> Doesn't this work, with just 5 vertices and 5
edges? Let G 1 be a K 3> plus two pendant vertices, each joined to a
different vertex of the> K 3; i.e., the self-complementary graph of
order 5; and embed it in> the sphere by putting the K 3 on the
equator, one pendant vertex in> the northern hemisphere, the other in
the southern hemisphere. Let G 2> be the same except that both of the
pendant vertices are in the> northern hemisphere.Looks good to me.--
Gerry Myerson (ge... at (no spam) maths.mq.edi.ai) (i -> u for email)}. (0.27
seconds)
[PDF] Graphs and AlgorithmsFile Format: PDF/Adobe Acrobat - View as
HTML
vertices in a graph G and let k be a positive integer. ..... 1.1.5,
that there exist two vertex disjoint paths each with one end in {u, v}
and ...... pendant edge, that is adding a new vertex and a new edge
from the new vertex ... Indeed, if G∗ is the dual of a plane embedding
of G, then, by (3.4.3), the max- ...
homepages.cwi.nl/~bgerards/personal/ga.pdf
[PDF] CanaDAM 2009File Format: PDF/Adobe Acrobat - View as HTML
Les sujets abordés pendant le congrès proviendront de tous les
domaines des ..... number of vertices or edges) satisfy cn ∼ cgnn
−5/2 ..... itive integers n and k, n<k, a self-complementary ...... k
such that each vertex is reachable under every distri- ..... and K∆
−1,∆, any bipartite graph G with ∆ ≥ 3 has ....
www.crm.umontreal.ca/CanaDAM2009/canadam09.pdf
The "I need Help Choosing a Name" Thread! - Page 5 - Graffiti ...oh
for fuck sake put a fucking k in it instead, or just put any fucking
letters you like ... Activity, Longevity. 1/20, 19/20. Today, Posts.
0/5, s 2 3 1 4 ...
www.puregraffiti.com/graffiti-space/showthread.php?t=787&page=5
Result for query "keyword(s)=in author= title="We now turn our
attention to the case $k=3$. In turns out that in this ...... In \cite
[Theorem~0.2] oh, it is shown; $G (1, \bold P^5)$ embedded in $ \bold
P^ ...... Each; edge of an elementary disk is a straight line in a 2-
simplex of ...... Let $A_1, \dotsc, A_t$ be the; vertices in $\Gamma$
that are adjacent to ...
nyjm.albany.edu:8000/cgi-bin/aglimpse/19/nyjm/Http/.../j?...in
[PDF] 8 Cologne-Twente Workshop on Graphs and Combinatorial
Optimization ...File Format: PDF/Adobe Acrobat
Suppose G1 and G2 are graphs with disjoint vertex-sets and k ≥ 0 is an
integer. ... (Fomin and Thilikos [3]) Let G1 and G2 be graphs with one
edge or ..... if G has at least two vertices then each internal node
of the tree has at least ...... 5: Check if there are l(j + 1)
vertices with pairwisely different colors ...
www.lix.polytechnique.fr/ctw09/ctw09-proceedings.pdf
Result for query "keyword(s)=in author= title="For each $a \in X$ we
can identify; For two distinct vertices $a, c$ in $X$ we write $X
(c, ...... K1 of separative exchange rings and C-algebras with real
rank zero ...... Embeddings of Z2-homology 3-spheres in R up to
regular homotopy ..... In a graph $G$, the volume of a subset $X$ of
the vertex set $V$, ...
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PSX MASTER LIST - Page 2Aug 19, 2009 ... SLPM-86812 : Jet de Go! -
Let's Go By Airliner [Playstation The Best] SCUS-94309 : Jet Moto ....
SLPM-86763 : K-1 World Grand Prix 2001 - Kaimakuban by Xing .....
SLPM-86933 : Kuma no Pooh-San - Mori no Nakamato 1 2 3 ......
SCUS-94212 : Official U.S. PlayStation Magazine Demo Disc 5 ...
cybertopic.youneed.us/ps1-and-ps2-f14/psx-master-list-t65-15.htm
Source for "MobWars Auto-Searcher" – Userscripts.orgBy mobwars — Last
update Sep 3, 2009 — Installed 2172 times. ......
","vertebra","vertebrae","vertebral","vertebrate","vertex","vertical","vertically","vertices","vertiginous" .....
5); var searchBeginning; var useQuestionMark = 1; ..... for Mob Wars
Auto-Searcher\nWould you like to go to the install page now? ...
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Godel Escher Bach by Douglas R. HofstadterIn order for a theme to work
as a canon theme, each of its notes must be able to .... Bach and
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Result for query "keyword(s)=and author= title="In Section~ 5 we show
that Theorems~\thmref 1.2 and~ 1.3 remain valid ... This result was
investigated by D. Hoff and K. Zumbrun \cite 2, 3; fluid flow, .....
Let us remark that if two increasing sequences of integers $(n_k)$ and
$(m_k)$ ...... One of the purposes of; polygon with $n+2$ vertices on
the geometric ...
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| Maury Barbato... |
| Posted: Fri Oct 30, 2009 1:25 am |
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Guest
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Maury Barbato wrote:
[quote]On Oct 29, 12:10Â am, Gerry Myerson
ge... at (no spam) maths.mq.edi.ai.i2u4email
wrote:
In article
61bge5p81jc5nct8itjugckl6ba7q68... at (no spam) 4ax.com>,
 David C. Ullrich <dullr... at (no spam) sprynet.com> wrote:
On Wed, 28 Oct 2009 05:57:31 EDT, Maury Barbato
mauriziobarb... at (no spam) aruba.it> wrote:
Hello,
let G and G' two isomorphic finite planar graphs
embedded in R^2. Then, I suppose that there
exists a
homeomorphism f:R^2 -> R^2 such that:
(I) if v is a vertex of G, then f(v) is a vertex
of G',
(II) if e is an edge of G, then f(e) is an edge
of G'.
Not true. Let the vertices of G be the vertices
of a
square, and let the edges of G be the edges of
that
square plus one of the diagonals. Let G' be the
same as G except the diagonal edge is replaced
by an edge looping outside the square, connecting
the same two vertices.
Then the complement of G' has a compact component
bounded by four edges, which is not true of G.
That example wouldn't work with S^2 in place
of R^2, but I think it's clear that a more
elaborate
version of the same idea would work in S^2 as
well.
I wonder if this works for S^2.
Here are two graphs, each with 6 vertices and 6
edges.
Each has a K_3, an edge joining the 4th vertex to
one of
the vertices of the K_3, and edges joining the
other two
vertices to one of the other vertices of the K_3.
They
differ in which of the non-K_3 edges is interior to
the K_3.
The graphs are isomorphic, and I think you can set
it up
so there's no homeomorphism of S^2 carrying one to
the other.
Doesn't this work, with just 5 vertices and 5 edges?
Let G_1 be a K_3
plus two pendant vertices, each joined to a different
vertex of the
K_3; i.e., the self-complementary graph of order 5;
and embed it in
the sphere by putting the K_3 on the equator, one
pendant vertex in
the northern hemisphere, the other in the southern
hemisphere. Let G_2
be the same except that both of the pendant vertices
are in the
northern hemisphere.
[/quote]
I haven't understood what you mean by saying "putting
K_3 on the equator" ... Maybe, your idea is: embed K_3
on S^2, and let C_1 and C_2 the two connected components
of the complement of K_3. Then build G_1 joining two points of C_1 to two distinct vertices of K_3, while G_2
is built up joining two points, one in C_1 and one in
C_2, to two distinct vertices of K_3.
If it's waht you mean, it works.
Maury Barbato |
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| Maury Barbato... |
| Posted: Sun Nov 01, 2009 7:18 am |
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Guest
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I wrote:
[quote]Maury Barbato wrote:
On Oct 29, 12:10Â am, Gerry Myerson
ge... at (no spam) maths.mq.edi.ai.i2u4email
wrote:
In article
61bge5p81jc5nct8itjugckl6ba7q68... at (no spam) 4ax.com>,
 David C. Ullrich <dullr... at (no spam) sprynet.com> wrote:
On Wed, 28 Oct 2009 05:57:31 EDT, Maury
Barbato
mauriziobarb... at (no spam) aruba.it> wrote:
Hello,
let G and G' two isomorphic finite planar
graphs
embedded in R^2. Then, I suppose that there
exists a
homeomorphism f:R^2 -> R^2 such that:
(I) if v is a vertex of G, then f(v) is a
vertex
of G',
(II) if e is an edge of G, then f(e) is an
edge
of G'.
Not true. Let the vertices of G be the
vertices
of a
square, and let the edges of G be the edges of
that
square plus one of the diagonals. Let G' be
the
same as G except the diagonal edge is replaced
by an edge looping outside the square,
connecting
the same two vertices.
Then the complement of G' has a compact
component
bounded by four edges, which is not true of G.
That example wouldn't work with S^2 in place
of R^2, but I think it's clear that a more
elaborate
version of the same idea would work in S^2 as
well.
I wonder if this works for S^2.
Here are two graphs, each with 6 vertices and 6
edges.
Each has a K_3, an edge joining the 4th vertex
to
one of
the vertices of the K_3, and edges joining the
other two
vertices to one of the other vertices of the
K_3.
They
differ in which of the non-K_3 edges is interior
to
the K_3.
The graphs are isomorphic, and I think you can
set
it up
so there's no homeomorphism of S^2 carrying one
to
the other.
Doesn't this work, with just 5 vertices and 5
edges?
Let G_1 be a K_3
plus two pendant vertices, each joined to a
different
vertex of the
K_3; i.e., the self-complementary graph of order
5;
and embed it in
the sphere by putting the K_3 on the equator, one
pendant vertex in
the northern hemisphere, the other in the southern
hemisphere. Let G_2
be the same except that both of the pendant
vertices
are in the
northern hemisphere.
I haven't understood what you mean by saying
"putting
K_3 on the equator" ... Maybe, your idea is: embed
K_3
on S^2, and let C_1 and C_2 the two connected
components
of the complement of K_3. Then build G_1 joining two
points of C_1 to two distinct vertices of K_3, while
G_2
is built up joining two points, one in C_1 and one
in
C_2, to two distinct vertices of K_3.
If it's waht you mean, it works.
Maury Barbato
[/quote]
Ok, I have understood: you mean "putting K_3 on a great
circle of the sphere. Ok!
My Best Regards,
Maury Barbato |
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