New Thread v. ask me anything

[tapout]

1. You're reading my post.
2. Now you're saying/thinking this is stupid.
4. You didnt notice that I skipped 3.
5. You're checking it now.
6. You're smiling.
7. You're still reading my post.
8. You know all you have read is true.
10. You didnt notice that I skipped 9.
11. You're checking it now.
12. Dumbass

[bluecar]

Why do I keep getting logged out?

[wasted ink]

First post is win. You'll be raped.

Why the fuck aren't a crunk yet?

[tapout]

Why do I keep getting logged out?

Your computer is set up to not keep cookies long.

[tapout]

First post is win. You'll be raped.

Why the fuck aren't a crunk yet?

Iam hasnt said so. When he says "wasted is crunk" you shall be green.

[wasted ink]

Are you as buff as everyone says?












I feel like i'm talking to a magic eight ball.

[tapout]

Are you as buff as everyone says?












I feel like i'm talking to a magic eight ball.

yes, but not as cut as my older pictures. Ill never do that again.

[97gp]

I like those pictures of you in your skirt.

[tapout]

I like those pictures of you in your skirt.

http://i162.photobucket.com/albums/t278/tap0ut3/9729368634694ed09ee2cc.jpg

[Toasted1]

lol that would be a funny shirt to wear while kilting :laugh:

[bluecar]

Why did I fuckup the first day I was a mod?

[tapout]

Why did I fuckup the first day I was a mod?

you were looking for teh lols?

[97gp]

Why did I fuckup the first day I was a mod?

Atleast you had a day :fyi:

[bluecar]

you were looking for teh lols?

I was.

Atleast you had a day :fyi:

Indeed.


Perhaps on graduation day, I can redeem myself.

[darthvibrator]

1. You're reading my post.
2. Now you're saying/thinking this is stupid.
4. You didnt notice that I skipped 3.
5. You're checking it now.
6. You're smiling.
7. You're still reading my post.
8. You know all you have read is true.
10. You didnt notice that I skipped 9.
11. You're checking it now.
12. Dumbass

dude i got fucked at 9. i did see three. hers an idea, add in a part about you thinking you tricked me, but me knowing i saw 3, and then springing the number 9 line on me.

damn im baked, but i think thats a really fucking good change to the post. no?

[darthvibrator]

yes, but not as cut as my older pictures. Ill never do that again.

whats funny is, this nigga is HUGE lawl.




fuck.

[tapout]

dude i got fucked at 9. i did see three. hers an idea, add in a part about you thinking you tricked me, but me knowing i saw 3, and then springing the number 9 line on me.

damn im baked, but i think thats a really fucking good change to the post. no?

lolwut?

[darthvibrator]

lolwut?

fuck i cant explain that shit any better. im an American born foreigner. it means i sound like an American, but i think like a fucking idiot. the more i explain the worse its gonna get

[tapout]

you got a question?

[infamous_e46]

i would like to see pix wtf where

[darthvibrator]

i would like to see pix wtf where

what are you babblin about?

[IDSkot]

what are you babblin about?

He wants to see pictures of Tapout's delts

[darthvibrator]

He wants to see pictures of Tapout's delts

ahhh. niggas huge dude.

[infamous_e46]

huge like me ??

http://graysmatter.codivation.com/content/binary/Ronniecoleman.jpg

[tapout]

Yeah buddy.

Light weight.

[pl8er]

What is the answer please?

http://threesixty360.files.wordpress.com/2008/04/road-coloring.jpg

[infamous_e46]

me standing next to tapout

http://www.vojvodina.com/fitness/galery1/images/03_psd.jpg

[tapout]

What is the answer please?



13

[pl8er]

Proof. Two states from an F-clique could not belong to one F-maximal set
because this pair is not synchronizing. By Theorem 1 there exists a partition
of �� on F-maximal sets of weight w. So the partition consists from w(��)/w
F-maximal sets and to every F-maximal set belongs at most one state from
F-clique. Consequently, the size of any F-clique is not greater than w(��)/w.
Let ��s be an F-clique. The sum of the weights qs−1 for all q 2 ��s is the
weight of ��. So
w(��) =
X
q2��s
w(qs−1)
3

[pl8er]

ok ok...i am just kidding, that isn't the answer

[darthvibrator]

me standing next to tapout

http://www.vojvodina.com/fitness/galery1/images/03_psd.jpg
better keep drinking that elephants milk

[pl8er]

here is the actual answer

[pl8er]

alright fuck it, lets just call it 13 :P

[Pop Da Hatch]

why is my left testicle bigger than my right?

[undeadxsunshine]

The road coloring problem was formulated for AGW graphs [1] and only such
graphs are considered below. We exclude from the consideration also the primitive
cases of graphs with loops and of only one color [1], [13]. Let us formulate
two important results from [6] and [10] in the following form:
Theorem 1 [6] There exists a partition of 􀀀 on F-maximal sets (of the same
weight).
Some side conditions on the AGW graph stated in [6] were not used in the proof
of this statement.
Let us recall that a binary relation  on the set of the states of an automaton
is called congruence if  is equivalence and for any word u from p  q follows
pu  qu.
Theorem 2 [10] Let us consider a coloring of AGW graph 􀀀. Stability of
states is a binary relation on the set of states of the obtained automaton; denote
this relation by . Then  is a congruence relation, 􀀀/ presents an AGW graph
and synchronizing coloring of 􀀀/ implies synchronizing recoloring of 􀀀.
The last theorem shows that if every AGW graph has a coloring with a stable
pair, then every AGW graph has a synchronizing coloring.
Lemma 1 Let w be the weight of F-maximal set of the AGW graph 􀀀 via
some coloring. Then the size of every F-clique of the coloring is the same and
equal to w(􀀀)/w (the size of partition of 􀀀 on F-maximal sets).
Proof. Two states from an F-clique could not belong to one F-maximal set
because this pair is not synchronizing. By Theorem 1 there exists a partition
of 􀀀 on F-maximal sets of weight w. So the partition consists from w(􀀀)/w
F-maximal sets and to every F-maximal set belongs at most one state from
F-clique. Consequently, the size of any F-clique is not greater than w(􀀀)/w.
Let 􀀀s be an F-clique. The sum of the weights qs−1 for all q 2 􀀀s is the
weight of 􀀀. So
w(􀀀) =
X
q2􀀀s
w(qs−1)
3
The number of addends (the size of the F-clique) is not greater than w(􀀀)/w.
The weight of the set qs−1 for every q 2 􀀀s is not greater than w. Therefore
qs−1 is an F-maximal set of weight w for every q 2 􀀀s and the size of any Fclique
is w(􀀀)/w, the number of F-maximal sets in the corresponding partition
of 􀀀.
Lemma 2 Let F be F-clique via some coloring of AGW graph 􀀀. For any
word s the set Fs is also an F-clique and any state [vertex] p belongs to some
F-clique.
Proof. Any pair p, q from an F-clique F is a deadlock. To be deadlock is a
stable binary relation, therefore for any word s the pair ps, qs from Fs also is
a deadlock. So all pairs from Fs are deadlocks.
For the F-clique F there exists a word t such that 􀀀t = F. Thus 􀀀ts = Fs,
whence Fs is an F-clique.
For any r from a strongly connected graph 􀀀, there exists a word u such
that r = pu for p from the F-clique F, whence r belongs to the F-clique Fu.
Lemma 3 Let A and B (|A| > 1) be distinct F-cliques via some coloring
without stable pairs of the AGW graph 􀀀. Then |A|−|A\B| = |B|−|A\B| > 1.
Proof. Let us assume the contrary: |A| − |A \ B| = 1. By Lemma 1, |A| = |B|.
So |B| − |A \ B| = 1, too. The pair of states p 2 A \ B and q 2 B \ A is not
stable. Therefore for some word s the pair (ps, qs) is a deadlock. Any pair of
states from the F-clique A and from the F-clique B as well as from F-cliques As
and Bs is a deadlock. So any pair of states from the set (A[B)s is a deadlock.
One has |(A [ B)s| = |A| + 1 > |A|.
In view of Theorem 1, there exists a partition of size |A| (Lemma 1) of 􀀀
on F-maximal sets. To every F-maximal set belongs at most one state from
(A[B)s because every pair of states from this set is a deadlock and no deadlock
could belong to an F-maximal set. This contradicts the fact that the size of
(A [ B)s is greater than |A|.
Lemma 4 Let some vertex of AGW graph 􀀀 have two incoming bunches. Then
any coloring of 􀀀 has a stable couple.
Proof. If a vertex p has two incoming bunches from vertices q and r, then the
couple q, r is stable for any coloring because q = r = p for any letter (color)
 2 .
2 The spanning subgraph of cycles and trees
with maximal number of edges in the cycles
D´efinition 1 Let us call a subgraph S of the AGW graph 􀀀 a spanning subgraph
of 􀀀 if to S belong all vertices of 􀀀 and exactly one outgoing edge of every
vertex.
4
A maximal subtree of the spanning subgraph S with root on a cycle from S
and having no common edges with cycles from S is called a tree of S.
The length of path from a vertex p through the edges of the tree of the spanning
set S to the root of the tree is called the level of p in S.
Remark 1 Any spanning subgraph S consists of disjoint cycles and trees with
roots on cycles; any tree and cycle of S is defined identically, the level of the
vertex from cycle is zero, the vertices of trees except root have positive level, the
vertex of maximal positive level has no incoming edge from S.
Lemma 5 Let N be a set of vertices of level n from some tree of the spanning
subgraph S of AGW graph 􀀀. Then in a coloring of 􀀀 where all edges of S have
the same color , any F-clique F satisfies |F \ N|  1.
Proof. Some power of  synchronizes all states of given level of the tree and
maps them into the root. Any couple of states from an F-clique could not be
synchronized and therefore could not belong to N.
Lemma 6 Let AGW graph 􀀀 have a spanning subgraph R of only disjoint
cycles (without trees). Then 􀀀 also has another spanning subgraph with exactly
one vertex of maximal positive level.
Proof. The spanning subgraph R has only cycles and therefore the levels of all
vertices are equal to zero. In view of gcd =1 in the strongly connected graph 􀀀,
not all edges belong to a bunch. Therefore there exist two edges u = p ! q 62 R
and v = p ! s 2 R with common first vertex p but such that q 6= s. Let us
replace the edge v = p ! s from R by u. Then only the vertex s has maximal
level L > 0 in the new spanning subgraph.
Lemma 7 Let any vertex of an AGW graph 􀀀 have no two incoming bunches.
Then 􀀀 has a spanning subgraph such that all its vertices of maximal positive
level belong to one non-trivial tree.
Proof. Let us consider a spanning subgraph R with a maximal number of vertices
[edges] in its cycles. In view of Lemma 6, suppose that R has non-trivial trees
and let L > 0 be the maximal value of the level of a vertex.
Further consideration is necessary only if at least two vertices of level L
belong to distinct trees of R with distinct roots.
Let us consider a tree T from R with vertex p of maximal level L and edge
¯b
from vertex b to the tree root r 2 T on the path of length L from p. Let
the root r belong to the cycle H of R with the edge ¯c = c ! r 2 H. There
exists also an edge ¯a = a ! p that does not belong to R because 􀀀 is strongly
connected and p has no incoming edge from R.
c ?c c
c c
c
 
A

AK


c
p
r
a
d
c
-· · · - b
· · ·
· · · · · · · · · · · ·

¯a
¯ w
¯c
¯b
H
T

􀀀
􀀀
5
Let us consider the path from p to r of maximal length L in T. Our aim is
to extend the maximal level of the vertex on the extension of the tree T much
more than the maximal level of vertex of other trees from R. We plan to use
the following three changes:
1) replace the edge ¯ w from R with first vertex a by the edge ¯a = a ! p,
2) replace the edge¯b from R by some other outgoing edge of the vertex b,
3) replace the edge ¯c from R by some other outgoing edge of the vertex c.
If one of the ways does not succeed let us go to the next assuming the
situation in which the previous way fails and excluding the successfully studied
cases. So we diminish the considered domain. We can use sometimes two
changes together. Let us begin with
1) Suppose first a 62 H. If a belongs to a path in T from p to r then a new
cycle with part of the path and edge a ! p is added to R extending the number
of vertices in its cycles in spite of the choice of R. In opposite case the level of
a in the new spanning subgraph is L + 1 and the vertex r is a root of the new
tree containing all vertices of maximal level (in particular, the vertex a or its
ancestors in R).
So let us assume a 2 H and suppose ¯ w = a ! d 2 H. In this case the
vertices p, r and a belong to a cycle H1 with new edge ¯a of a new spanning
subgraph R1. So we have the cycle H1 2 R1 instead of H 2 R. If the length of
path from r to a in H is r1 then H1 has length L+r1 +1. A path to r from the
vertex d of the cycle H remains in R1. Suppose its length is r2. So the length
of the cycle H is r1 +r2 +1. The length of the cycle H1 is not greater than the
length of H because the spanning subgraph R has maximal number of edges in
its cycles. So r1 + r2 + 1  L + r1 + 1, whence r2  L. If r2 > L, then the
length r2 of the path from d to r in a tree of R1 (and the level of d) is greater
than L and the level of d (or of some other ancestor of r in a tree from R1) is
the desired unique maximal level.
So assume for further consideration L = r2 and a 2 H. Analogously, for
any vertex of maximal level L with root in the cycle H and incoming edge from
a vertex a1 the proof can be reduced to the case a1 2 H and L = r2 for the
corresponding new value of r2.
2) Suppose the set of outgoing edges of the vertex b is not a bunch. So one
can replace in R the edge ¯b from the vertex b by an edge ¯v from b to a vertex
v 6= r.
The vertex v could not belong to T because in this case a new cycle is added
to R and therefore a new spanning subgraph has a number of vertices in the
cycles greater than in R.
If the vertex v belongs to another tree of R but not to cycle, then T is a
part of a new tree T1 with a new root of a new spanning subgraph R1 and the
path from p to the new root is extended. So only the tree T1 has states of new
maximal level.
If v belongs to some cycle H2 6= H from R, then together with replacing ¯b
by ¯v, we replace also the edge ¯ w by ¯a. So we extend the path from p to the new
root v at least by the edge ¯a = a ! p and by almost all edges of H. Therefore
6
the new maximal level L1 > L has either the vertex d or its ancestors from the
old spanning subgraph R.
Now there remains only the case when v belongs to the cycle H. The vertex
p also has level L in new tree T1 with root v. The only difference between T
and T1 (just as between R and R1) is the root and the incoming edge of the
root. The new spanning subgraph R1 has also a maximal number of vertices in
cycles just as R. Let r3 be the length of the path from d to the new root v 2 H.
For the spanning subgraph R1, one can obtain L = r3 just as it was done on
the step 1) for R. From v 6= r follows r3 6= r2, though L = r3 and L = r2.
So for further consideration suppose that the set of outgoing edges of the
vertex b is a bunch to r.
3) The set of outgoing edges of the vertex c is not a bunch to r because r
has another bunch from b.
Let us replace in R the edge ¯c by an edge ¯u = c ! u such that u 6= r.
The vertex u could not belong to the tree T because in this case the cycle H is
replaced by a cycle with all vertices from H and some vertices of T whence its
length is greater than |H|. Therefore the new spanning subgraph has a number
of vertices in its cycles greater than in spanning subgraph R in spite of the
choice of R.
So remains the case u 62 T. Then the tree T is a part of a new tree with a
new root and the path from p to the new root is extended at least by a part of
H from the former root r. The new level of p therefore is maximal and greater
than the level of any vertex in some another tree.
Thus anyway there exists a spanning subgraph with vertices of maximal level
in one non-trivial tree.
Theorem 3 Any AGW graph 􀀀 has a coloring with stable couples.
Proof. By Lemma 4, in the case of vertex with two incoming bunches 􀀀 has a
coloring with stable couples. In opposite case, by Lemma 7, 􀀀 has a spanning
subgraph R such that the vertices of maximal positive level L belong to one
tree of R.
Let us give to the edges of R the color  and denote by C the set of all
vertices from the cycles of R. Then let us color the remaining edges of 􀀀 by
other colors arbitrarily.
By Lemma 2, in a strongly connected graph 􀀀 for every word s and F-clique
F of size |F| > 1, the set Fs also is an F-clique (of the same size by Lemma 1)
and for any state p there exists an F-clique F such that p 2 F.
In particular, some F has non-empty intersection with the set N of vertices
of maximal level L. The set N belongs to one tree, whence by Lemma 5 this
intersection has only one vertex. The word L−1 maps F on an F-clique F1 of
size |F|. One has |F1 \C| = 1 because the sequence of edges of color  from any
tree of R leads to the root of the tree, the root belongs to a cycle colored by 
from C and only for the set N with vertices of maximal level holds NL−1 6 C.
So |NL−1 \ F1| = |F1 \ C| = 1 and |C \ F1| = |F1| − 1.
Let the integer m be a common multiple of the lengths of all considered cycles
from C colored by . So for any p from C as well as from F1\C holds pm = p.
7
Therefore for an F-clique F2 = F1m holds F2  C and C \ F1 = F1 \ F2.
Thus two F-cliques F1 and F2 of size |F1| > 1 have |F1|−1 common vertices.
So |F1 \ (F1 \F2)| = 1. Consequently, in view of Lemma 3, there exists a stable
couple in the considered coloring.
Theorem 4 Every AGW graph 􀀀 has synchronizing coloring.
The proof follows from Theorems 3 and 2.

I think you just blew my mind... :wow:

[wasted ink]

WHERE'S THE FUCKING PITCHERS















































































http://www.villagekitchen.com/mfg/arc/luminarc/pitchers/pitcher_assortment/art/pitcher.jpg

[pl8er]

why is my left testicle bigger than my right?

I know tapout is supposed to answer, but it is because you have a natural tendency to lean one direction or the other while fapping. This has altered the size of one nut.

[wasted ink]

I know tapout is supposed to answer, but it is because you have a natural tendency to lean one direction or the other while fapping. This has altered the size of one nut.

So my balls work the same way the differential in my truck does?

[IDSkot]

So my balls work the same way the differential in my truck does?

Exactly.

[Pop Da Hatch]

why did wasted ink steal my avatar?

[IDSkot]

why did wasted ink steal my avatar?

Your ava has really fucked me up on more than one occasion. :fyi:

[wasted ink]

why did wasted ink steal my avatar?

:laugh:

[pl8er]

So my balls work the same way the differential in my truck does?

you must put that poor fella under a hell of a load...until you release it


and boom goes the dynamite.

[darthvibrator]

So my balls work the same way the differential in my truck does?

i has lsd

[tapout]

I know tapout is supposed to answer, but it is because you have a natural tendency to lean one direction or the other while fapping. This has altered the size of one nut.

this

[wasted ink]

you must put that poor fella under a hell of a load...until you release it


and boom goes the dynamite.

I meant the lean one way, one side gets more wear thing :boobies:

[pl8er]

I meant the lean one way, one side gets more wear thing :boobies:

lol...i did too, i was talking about the lean on your nut ;)

[wasted ink]

lol...i did too, i was talking about the lean on your nut ;)

Ohh. I did that the other day, sat right down on the left neighbor.

Edit: :fu:

[pl8er]

Why didn't I discover Windows Home Server sooner?

[darthvibrator]

because you were talking about hairy bubble gum sacks that hole testicles in them?

[wasted ink]

Bring on the pitchers, tappy. The fuck is this shit.

[tapout]

Bring on the pitchers, tappy. The fuck is this shit.

of what?

[tapout]

http://i162.photobucket.com/albums/t278/tap0ut3/hiphopzombupsidedownbackwardsgif.gif

[tapout]

http://i162.photobucket.com/albums/t278/tap0ut3/1236210177533.jpg

[IDSkot]

http://i162.photobucket.com/albums/t278/tap0ut3/1236210177533.jpg

WOW! Look at that wave.

EDIT:: Fuck it. I can't do it. Look at that fucking ass!

[tapout]

WOW! Look at that wave.

EDIT:: Fuck it. I can't do it. Look at that fucking ass!

never noticed it but now that you mention it, it is a nice wave.

[IDSkot]

never noticed it but now that you mention it, it is a nice wave.

I wish Miami had good waves. The best waves we get are like 1-2 foot waves.

[wasted ink]

of what?

Your delts adurrrr.

[tapout]

.

[infamous_e46]

http://i162.photobucket.com/albums/t278/tap0ut3/hiphopzombupsidedownbackwardsgif.gif

[tapout]

,

[infamous_e46]

http://www.ifbbpro.com/wp-content/uploads/image/halloffame/ArthurSaxon1.jpg

[IDSkot]

Kilts are manly.

[wasted ink]

The fuck on!

[Spider Monkey]

First post is win. You'll be raped.


This


Why the fuck aren't a crunk yet?

Because you have not passed all of the crunk missions yet.

Iam hasnt said so. When he says "wasted is crunk" you shall be green.

Bull shit. Iam has authority over Serious Business. I have authority over crunkness by verifying that people have passed enough crunk missions to be crunk.

[IamDeMan]

This



Because you have not passed all of the crunk missions yet.



Bull shit. Iam has authority over Serious Business. I have authority over crunkness by verifying that people have passed enough crunk missions to be crunk.
Serious business supersedes and may veto crunk decisions. This ultimately make me in control.

[Spider Monkey]

Serious business supersedes and may veto crunk decisions. This ultimately make me in control.

Lies. Crunk cannot be denied. It is a birth right.

[Spider Monkey]

:snicker:

[IamDeMan]

:snicker:

You really do forget one thing. You = me. Therefore you are only hurting yourself with this act.

[Spider Monkey]

You really do forget one thing. You = me. Therefore you are only hurting yourself with this act.

Shit. I just deleted one of my own poasts. :crap:

[IamDeMan]

Shit. I just deleted one of my own poasts. :crap:

that's what I'm sayin. On the same token, do you think I would ever veto one of your decisions?

Wouldn't make much sense.

[Spider Monkey]

that's what I'm sayin. On the same token, do you think I would ever veto one of your decisions?

Wouldn't make much sense.

I bet it might happen if we both showed up at the shrink's office at the same time. :fyi:

[IamDeMan]

I bet it might happen if we both showed up at the shrink's office at the same time. :fyi:

Paradox.

[Spider Monkey]

On second thought. One of us would have to be wrong for that to occur, and that would be like dividing by zero.

[IamDeMan]

On second thought. One of us would have to be wrong for that to occur, and that would be like dividing by zero.

Your head is hurting me just thinking about it.

[Spider Monkey]

Okay, I will get out.