Last wins - DPChallenge Forums

06/26/2013 11:46:16 AM · #21753

ausfart

Canon EF 200mm f/2.8 L II USM Lens

06/26/2013 12:23:39 PM · #21754

Uh oh...Spork sharted.

Spork even wins with soiled underpants.

skewsme

Canon EOS-6D

06/26/2013 12:31:39 PM · #21755

spork i thought you wear a thong
to floss your tine-y prongs

06/26/2013 12:37:15 PM · #21756

Confession: I am really a spambot
(Don't tell Slippy)

06/26/2013 12:39:08 PM · #21757

Originally posted by Art Roflmao:

Confession: I am really a spambot
(Don't tell Slippy)

Spork challenges you to a fight...in the vacant lot after school.

Spork wins!

beatabg

Google Pixel

06/26/2013 12:58:30 PM · #21758

Sporks are fragile.

06/26/2013 01:11:10 PM · #21759

Originally posted by beatabg:

Sporks are fragile.

Sporks are sharp and make a mean shiv.

Canon EF 200mm f/2.8 L II USM Lens

06/26/2013 01:25:24 PM · #21760

Originally posted by Spork99:

Originally posted by Art Roflmao:
Confession: I am really a spambot
(Don't tell Slippy)

Spork challenges you to a fight...in the vacant lot after school.

Spork wins!

Hey now...watch out about all that violence. It leads to bad things

bvy

Sigma DP1x

06/26/2013 01:49:12 PM · #21761

To the Last Wins community, I sincerely apologize. Some time back, I promised to show the following: Given a k-chromatic graph G (k >= 2) with maximum degree D, that for any r >= D, there exists a k-chromatic, r-regular graph H that has G as an induced subgraph. I discovered this to be a simple extension of a more classic result -- identical to the one stated above but with no restrictions on the chromatic number of G or H.

I'm preparing a proof of the original statement now and will post it soon. I appreciate your patience.

skewsme

Canon EOS-6D

06/26/2013 02:01:09 PM · #21762

Invalid. You must include a null hypothesis.

06/26/2013 02:31:25 PM · #21763

Originally posted by cowboy221977:

Originally posted by Spork99:

Originally posted by Art Roflmao:
Confession: I am really a spambot
(Don't tell Slippy)

Spork challenges you to a fight...in the vacant lot after school.

Spork wins!

Hey now...watch out about all that violence. It leads to bad things

Spork on Spambot violence should only be encouraged.

Spork wins

06/26/2013 03:11:23 PM · #21764

Originally posted by bvy:

Given a k-chromatic graph G (k >= 2) with maximum degree D, that for any r >= D, there exists a k-chromatic, r-regular graph H that has G as an induced subgraph. I discovered this to be a simple extension of a more classic result -- identical to the one stated above but with no restrictions on the chromatic number of G or H.

06/26/2013 03:27:11 PM · #21765

Ha funny

06/26/2013 03:51:11 PM · #21766

Spork is thinking about starting a thread where different spambots can duke it out.

Kinda like UFC, but less blood.

Spork wins

06/26/2013 05:46:55 PM · #21767

Bookies are welcome

06/26/2013 06:19:11 PM · #21768

06/27/2013 08:01:39 AM · #21769

Spork wonders, "Does that violate the TOS?"

Spork does not give a (lump of feces), fighting robots are cool.

Spork wins

bvy

Sigma DP1x

06/27/2013 08:12:48 AM · #21770

Originally posted by bvy:

Well, I'm not a bot. Anyway...

I'd like to start things off with a simple prerequisite proof; I'll be referring to it later on. Let G1 be a k-chromatic graph with k >= 2, let G2 be a copy of G1, and let G be the graph which results by adding edges between corresponding vertices of G1 and G2. Then G is also k-chromatic. This is not difficult to visualize. Suppose G1 has colors {c1, c2, ..., ck}. We simply permute the colors of G2, such that it uses colors {c2, c3, ..., ck, c1}. So for any u in G1 and its corresponding vertex v in G2, if u is colored c(i) in G1 then it is colored c(i+1) in G2 (or if it's colored ck in G1 then it's colored c1 in G2). G1 and G2 are both k-chromatic, and the resulting construction of G is also k-chromatic since, as we've shown, we can add edges between G1 and G2 without connecting like-colored vertices.

06/27/2013 08:17:48 AM · #21771

agh the mathbot is back

06/27/2013 12:37:20 PM · #21772

Death to mathbot!
(Channeling my inner Slippy. ...and WINNING)

06/27/2013 12:51:50 PM · #21773

Spork challenges mathbot to a fight.

Spork wins.

06/27/2013 02:32:00 PM · #21774

Spork vs. mathbot
Friday at the Flagpole
Tickets on sale now

mathbot: "Spork's days are numbered. ...in fractions!"
Spork: *throws a pi in mathbot's face*