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Showing posts 26726 - 26750 of 30665, (reverse)
AuthorThread
02/08/2015 04:09:32 AM · #26726
winning is me
02/08/2015 12:59:07 PM · #26727
Originally posted by the99:

So, the sun orbits the earth.

Of course it does! You can watch it happen! Are we supposed to believe some sort of scientific *theory* over the evidence of our own, God-given eyes?
02/08/2015 02:14:11 PM · #26728
I'm back, and I win!
02/08/2015 04:16:59 PM · #26729
Yes, God is the problem. Spork is the heretic with the answer.

Spork wins

Originally posted by Bear_Music:

Originally posted by the99:

So, the sun orbits the earth.

Of course it does! You can watch it happen! Are we supposed to believe some sort of scientific *theory* over the evidence of our own, God-given eyes?
02/08/2015 05:23:37 PM · #26730
Who,s that girl, she looks like fun.

but i still win.
02/08/2015 05:25:06 PM · #26731
There was someone like that here long long ago.
02/08/2015 07:51:44 PM · #26732
beaty!
02/08/2015 10:15:25 PM · #26733
Meaty, Beaty, Big and Bouncy!

Spork wins
02/08/2015 11:02:41 PM · #26734
On edge colorings of connected graphs. Vizing's Theorem tells us that a graph's edges can be colored with D(G) or D(G) + 1 colors, where D(G) is the maximum degree of any vertex of G. A proper edge coloring is one requiring as few colors as possible such that no vertex is adjacent to two edges of the same color. A graph that requires D(G) colors is said to be of class 1. Every line graph is class 1. A graph that required D(G) + 1 colors is said to be of class 2. K3 is class 2.

With this in mind, over the coming days I'd like to share some elementary results of edge colorings of graphs.
02/08/2015 11:06:25 PM · #26735
only if you sing
02/09/2015 03:21:24 PM · #26736
Okay, good, we have some encouraging feedback thus far. Let B(G) denote the edge independence number of G, the maximum cardinality of any set of non-adjacent edges of G. Let m be the size of G. If m > D(G) B(G), then G is of class 2.

Proof: We show that the contrapositive is true. Suppose G is of class 1. Then there are exactly X(G) = D(G) independent edge sets of G, and since each set has at most B(G) edges, m <= D(G) B(G).

There is a corollary to this. We call a graph G of order n overfull if m >= D(G) floor(n/2). The simplest example of this is the complete graph on three vertices, K3. In this case, D(G) = 2 and floor(n/2) = floor(3/2) = 1, and m = 3 > 2. Since B(G) <= floor(n/2), we can state the following:

Every overfull graph is of class 2.

This is useful in that we don't have to consider the edge independence number of the graph to apply this test; the edge independence number will never exceed n/2.
02/09/2015 03:28:09 PM · #26737
but what is a song without words?
02/09/2015 04:09:30 PM · #26738
Like a graph without edges?
02/09/2015 04:53:13 PM · #26739
mere shrubbery
02/09/2015 05:08:53 PM · #26740
A word without a song would be
a leaf without a tree.
"A graph without an edge," she said,
"is merely shrubbery."
Oh, bail me out! Oh, set me free!
There's more to life than this!
The Jail of Reality
is a man who needs to piss.
Always stand upwind of him
and you'll usually smell just fine,
but I'm upwind of YOU, you see,
so the victory is MINE!
02/09/2015 07:28:51 PM · #26741
alas, this lass
will always outepistle
Capecodians
02/09/2015 10:25:25 PM · #26742
Originally posted by bvy:

Okay, good, we have some encouraging feedback thus far. Let B(G) denote the edge independence number of G, the maximum cardinality of any set of non-adjacent edges of G. Let m be the size of G. If m > D(G) B(G), then G is of class 2.

Proof: We show that the contrapositive is true. Suppose G is of class 1. Then there are exactly X(G) = D(G) independent edge sets of G, and since each set has at most B(G) edges, m <= D(G) B(G).

There is a corollary to this. We call a graph G of order n overfull if m >= D(G) floor(n/2). The simplest example of this is the complete graph on three vertices, K3. In this case, D(G) = 2 and floor(n/2) = floor(3/2) = 1, and m = 3 > 2. Since B(G) <= floor(n/2), we can state the following:

Every overfull graph is of class 2.

This is useful in that we don't have to consider the edge independence number of the graph to apply this test; the edge independence number will never exceed n/2.


It's important to note that the converse of our above theorem is not necessarily true -- that is, if m <= D(G) B(G), then G is of class 1. A notable example of this is the overfull Petersen graph, with m = 15, D(G) = 3, and floor(n/2) = floor(10/2) = 5. 15 <= 3 * 5 = 15, and yet the Petersen Graph is of class 2.

Also, no one caught the error in my earlier post. A graph is overfull if m is strictly greater than the product given; the correct definition of an overfull graph is one with size m > D(G) floor(n/2).
02/09/2015 10:30:08 PM · #26743
With this in mind, some exciting results will follow. Stick with me...
02/09/2015 11:10:33 PM · #26744
my graph runneth over.
02/09/2015 11:40:14 PM · #26745
When thy graph runneth over
thy numbers sink
without a trace
into the muck
adieu
02/09/2015 11:44:11 PM · #26746
stick in the muck?
bear with me?
02/10/2015 12:08:00 AM · #26747
The piney watchers watch.
A ripple wakes the pond.
Those stagnant waters lie
deeper than eye can reach,
as deep as mud can sift.
A tree breaks from its leaves.
Nothing that lives, but grieves.
02/10/2015 08:31:29 AM · #26748
Rubbish.

Another win for us.
02/10/2015 08:31:31 AM · #26749
Rubbish.

Another win for us.
02/10/2015 08:32:33 AM · #26750
So good they made it twice
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