Date: Feb 6, 2013 3:44 PM
Author:
Subject: Proof leaves burn in marks from local content online
On Feb 6, 12:08 pm, quasi <qu...@null.set> wrote:
> Butch Malahide wrote:
> >quasi wrote:
>
> >> Here's a finite version of the problem, cast as a game.
>
> >> Perhaps the game is well known, I'm not sure, however the
> >> resolution of the game is nice, and not immediately obvious,
> >> so I'll pose it here as a challenge.
>
> >> Two players play a game, win or lose, for 1 dollar, based on a
> >> fixed positive integer n > 1, known in advance to both players.
>
> >> Player 1 chooses two distinct integers from 1 to n inclusive,
> >> writes them on separate index cards, and places them face down
> >> on the table.
>
> >> Player 2 then selects one of the cards and turns it face up,
> >> exposing the hidden value. Player 2 can then either "stay",
> >> yielding the value on the chosen card, or "switch", yielding the
> >> value on the other card instead. Player 2 wins (and player 1
> >> loses) if player 2's final value is the higher of the 2 values,
> >> otherwise player 2 loses (and player 1 wins).
>
> >> In terms of n, find the value of the game for player 2, and
> >> specify optimal strategies for both players.
>
> >I haven't tried to prove anything, but I'll guess that the
> >obvious strategies are optimal.
>
> Obvious to you perhaps, but the solution was not immediately
> obvious to me.
>
> >Let S = {3/2, 5/2, . . ., (2n1)/2}.
>
> >Player 1's strategy: choose random x in S (uniform distribution)
> >and play {x  1/2, x + 1/2}.
>
> >Player 2's strategy: choose random y in S (uniform distribution),
> >stay n values > y, switch on values < y; otherwise put (in the
> >form of a behavior strategy), if the observed value is k, stay
> >with probability (k1)/(n1).
>
> >The value of the game for player 2 is $1/(n1): player 1 wins
> >a dollar if x = y, otherwise the chances are even.
>
> I waited to give others a chance before responding, but yes,
> your proposed solution is correct.
>
> An equivalent formulation of the answer is this:
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?On the Nature of Opt
imality: a Symbolic and Literal.
?
Martin M. Musatov, 2/3/2009NP
P equals a tightly bound 4
th
Dimensional Transition
. ?Of the sum of two edges
,
one always faces up.?
For the Ricci Flow equation there isan evolution equation for
aRiemannian metric , where isthe Ricci curvature tensor.Hamilton(1982)
demonstrated a uniquesolution to this equation for anarbitrary smooth
metric on a closedmanifold over a sufficiently short amount of time.
Hamilton (1986)also showed that Ricci flowpreserves positivity of the
Riccicurvature tensor in threedimensions and the curvatureoperator has
been explicitly provenin all dimensions (Perelman 2002).Given latency
between ellipticalcurves and linear expansion, inwarring disciplines
to approach anequilibrium solution, a linearprogression of an
evolutionequation is popularly distorted indistortion to its rate of
growth. 1
?
??????????
?
??????
??
?On the Nature of Opt
imality: a Symbolic and Literal.
?
Martin M. Musatov, 2/3/2009NP
P equals a tightly bound 4
th
Dimensional Transition
. ?Of the sum of two edges
,
one always faces up.?
1.
Statement of the Solution
P =
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N
???????
P
A solution to
P
versus
NP
must necessarily explain how every language accepted by some non
deterministic algorithm inpolynomial time has been or will be accepted
by some (deterministic) algorithm in polynomial time.To meet this
requirement it is essential to observe the complete model of a
computer,
or Turing machine and process informationin
?
realtime
?
as it is received as a
computable function
or linear stream.In concordance, the Encarta Dictionary: English
(North America):
Formally, the class P contains the ?in

decision? problems.
a(1)(noun) = a [ay]
(a?s)(A?s, As)
1.
1
st
letter of English alphabet=the first letter of the English alphabet,
representing a vowel sound2.
Letter ?a? written=a written representation of the letter ?a?
a (2)(symbol)=a1.
acceleration=acceleration=ac·cel·er·a·tionPHYSICS
?
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?
??????
??
?On the Nature of Opt
imality: a Symbolic and Literal.
?
Martin M. Musatov, 2/3/2009NP
P equals a tightly bound 4
th
Dimensional Transition
. ?Of the sum of two edges
,
one always faces up.?
REFERENCES:
Collins, G. P. "The Shapes of Space."
Sci. Amer.
291
, 94103, July 2004.Hamilton, R. S. "Three Manifolds with Positive
Ricci Curvature."
J. Diff. Geom.
17
, 255306, 19
82.Hamilton, R. S. "Four Manifolds with Positive Curvature Operator."
J. Diff. Geom.
24
, 153179, 1986.Kleiner, B. and Lott, J. "Notes and Commentary on
Perelman's Ricci Flow Papers."http://www.math.lsa.umich.edu/research/
ricciflow/perelman.html. Perelman, G. "The Entropy Formula for the
Ricci Flow and Its Geometric Application" 11 Nov 2002.http://arxiv.org/
abs/math.DG/0211159/ .Robinson, S. "Russian Reports He Has Solved a
Celebrated Math Problem."
The New York Times
, p. D3, April 15, 2003.Clay Mathematics Institute. "P vs. NP."
http://www.claymath.org/millennium/P_vs_NP/.
>
> For player 1, an optimal strategy is to choose one of the
> n1 pairs (k,k+1) with equal probability, ignoring the other
> (n choose 2)  (n1) pairs.
>
> For player 2, an optimal strategy is to stay on value k
> with probability (k1)/(n1), and switch with probability
> 1  (k1)/(n1).
>
> As you indicated, the value of the game for player 2, in
> dollars, is 1/(n1).
>
> quasi
musatov.M