Last wins - DPChallenge Forums

Nikon AF Nikkor 50mm f/1.4D

01/24/2016 05:31:52 AM · #27401

my jimmies have a nice pattern, will they do ?

a win for the pair of them

raish

Olympus C-2500L

01/24/2016 05:51:33 PM · #27402

bit samey

Bear_Music

Samsung Galaxy S24 Ultra

Sony 24-70mm F4 Vario-Tessar T* FE OSS

01/24/2016 06:06:42 PM · #27403

and gamy...

Nikon AF Nikkor 50mm f/1.4D

01/26/2016 09:48:18 AM · #27404

Its about that time of the day again...

raish

Olympus C-2500L

01/26/2016 04:51:37 PM · #27405

daylight come and I want go home

Canon EF 200mm f/2.8 L II USM Lens

01/26/2016 05:04:49 PM · #27406

good bye till next we meet

skewsme

Canon EOS-6D

01/27/2016 06:14:10 AM · #27407

and i say hello

Canon EF 200mm f/2.8 L II USM Lens

01/29/2016 08:05:34 AM · #27408

A win of the highest order is mine.

skewsme

Canon EOS-6D

01/29/2016 08:21:26 AM · #27409

That is the highest odor. You really rank.

01/29/2016 09:14:28 AM · #27410

thanks

Sony 24-70mm F4 Vario-Tessar T* FE OSS

01/29/2016 12:18:15 PM · #27411

Your Welcome....

Bear_Music

Samsung Galaxy S24 Ultra

01/29/2016 12:27:55 PM · #27412

Tiny's RANK! Or does he "rank"?
What's the difference, people?
Even when he's slumped atop
the church, he ain't no steeple.

01/29/2016 02:01:54 PM · #27413

Its kind of like putting lipstick on a pig.... At the end of the day its still a pig.... Or is it????

Canon EF 200mm f/2.8 L II USM Lens

01/29/2016 05:29:13 PM · #27414

Another winning morning.

skewsme

Canon EOS-6D

01/29/2016 11:40:48 PM · #27415

An utter whining moaning. Whey cup.

01/30/2016 09:18:52 AM · #27416

Yes but " what's university "

I love to win on

01/31/2016 12:50:37 AM · #27417

Let T be a tree of order k, and let G be a graph with d(G) >= k - 1 (where d(G) signifies the minimum degree of any of the vertices of G). I would like to show that G contains a subgraph isomorphic to T.

02/01/2016 02:25:40 PM · #27418

No simple graph exists. When we consider the degree sequence and the removal of vertices,
we would have (4, 4, 4, 2, 2) → (3, 3, 1, 1) → (2, 0, 0). This would leave a vertex of degree 2,
but no vertices for it to be adjacent to.
We could have a multigraph, however. If we suppose that each vertex of degree 4 had a loop,
then the removal of those loops would leave us with a degree sequence (2, 2, 2, 2, 2), which is
the sequence for C5.

02/01/2016 03:36:34 PM · #27419

You can't have a simple graph of order 5 with those degree sequences. K5 with any one edge removed would have degree sequences (4, 4, 4, 3, 3). You can't remove any more edges without decrementing the degree of one of the vertices of order 4 (multigraphs notwithstanding).

02/04/2016 09:38:29 AM · #27420

Originally posted by bvy:

For k = 1, and k = 2, the trees K1 and K2 are subgraphs of every non-empty graph. For k > 2, let T' be a tree of order k - 1, and G' a graph with d(G') >= k - 2, and suppose T' is a subgraph of G'. We proceed by induction on k.

Let T be a tree of order k, and G be a graph with d(G) >= k - 1. For some edge uv in T, such that v is a leaf node, consider T - v. T - v has order k - 1, and so it is the subgraph of some graph G with d(G) >= k - 1 >= k -2. The vertex u in T then corresponds to some u' in G. deg(u) <= k - 2 in T , and deg(u') >= k -1 in G, therefore u' must be adjacent to some v' in G not in T. As such, T is a subgraph of G.

Canon EF 17-40mm f/4.0L USM

02/04/2016 08:36:22 PM · #27421

Tomorrow I would like to show that every subgraph of a connected tree is also an induced subgraph.

kawesttex

Apple iPhone 11 Pro

02/04/2016 09:02:29 PM · #27422

02/04/2016 09:11:07 PM · #27423

Originally posted by bvy:

Tomorrow I would like to show that every subgraph of a connected tree is also an induced subgraph.

I should qualify: every connected subgraph...

02/05/2016 08:57:42 AM · #27424

Originally posted by bvy:

Originally posted by bvy:
Tomorrow I would like to show that every subgraph of a connected tree is also an induced subgraph.

I should qualify: every connected subgraph...

This is a fairly straightforward result. Let T' be a connected subgraph of some tree T, and suppose T' is not an induced subgraph. Then for some vertices u and v in T', the edge uv exists in T but not in T'. However if we add the edge uv to T', we've created a cycle in T' implying there is a cycle in T, which contradicts the supposition that T is a tree. Therefore every connected subgraph of a tree is also an induced subgraph.