QIP 2009 Day 5 Liveblogging

To make an mathematical expression for your blog, go to the Wikipedia sandbox (http://en.wikipedia.org/wiki/Wikipedia:Sandbox). Go to the Edit page and follow these directions: http://en.wikipedia.org/wiki/Help:Formula save the page, copy-and-paste the result to your blog.

Quantum eavesdropping à la Merkle (Gilles Brassard): As there doesn't seem to be an arxiv paper yet, let me write some comments from the talk.

Merkle's idea for public-key cryptography: Given a one-way function f: [N] -> [N], Alice and Bob each choose sqrt{N} inputs and publish f applied to those inputs. With good probability, their lists have overlap. They use the preimage of that element as their secret key. Note that their effort is sqrt{N}, but an adversary needs to use N time to invert the function to figure out the key. Thus there is a quadratic effort gap. (Not as good as the suppose exponential effort gap for, e.g., Diffie-Hellman, but has a weaker assumption and is perhaps still useful.)

Quantumly, this doesn't work, since the eavesdropper can apply Grover search to invert the permutation in time sqrt{N}, so there is no effort gap. However, if we also allow Bob to be quantum, then, when Alice publishes (classically) a list of t=N^{1/3} encodings, Bob can find a preimage in time sqrt{N/t} = N^{1/3}. The adversary's required effort is still sqrt{N}, so there is a (3/2)-power effort gap.

A similar algorithm works for oblivious transfer (recall: Bob should get exactly one of Alice's strings a or b, and Alice shouldn't know which). Alice now publishes two columns with N^{1/3} rows. Roughly, one column holds a and the other b; each row can be hidden by the same random mask; and then f is applied. (I missed the details of the construction but this seems to work.) Bob chooses which string/column he wants to learn and searches for the preimage of some entry in that column, in time N^{1/3}. He then asks Alice for the mask for that row to get the output bit.

This construction has a (3/2)-power gap classically, with an appropriate choice of t. It seems like the best attack Bob can make to learn both preimages for some row takes time N^{4/9} (?), giving a (4/3)-power gap. But this gap has not been proved.

More open questions: The classical quadratic effort gap is tight (just proved in 2008 by Barak and Mahmoody-Ghidary). Is 3/2 tight quantumly? Are there fully classical protocols secure against quantum attack. He said that they have a candidate.

Einstein's Nemesis: DI Her Eclipsing Binary Stars Solution
The problem that the 100,000 PHD Physicists could not solve

This is the solution to the "Quarter of a century" Smithsonian-NASA Posted motion puzzle that Einstein and the 100,000 space-time physicists including 109 years of Nobel prize winner physics and physicists and 400 years of astronomy and Astrophysicists could not solve and solved here and dedicated to Drs Edward Guinan and Frank Maloney
Of Villanova University Pennsylvania who posted this motion puzzle and started the search collections of stars with motion that can not be explained by any published physics
For 350 years Physicists Astrophysicists and Mathematicians and all others including Newton and Kepler themselves missed the time-dependent Newton's equation and time dependent Kepler's equation that accounts for Quantum - relativistic effects and it explains these effects as visual effects. Here it is

Universal- Mechanics

All there is in the Universe is objects of mass m moving in space (x, y, z) at a location
r = r (x, y, z). The state of any object in the Universe can be expressed as the product

S = m r; State = mass x location

P = d S/d t = m (d r/dt) + (dm/dt) r = Total moment

= change of location + change of mass

= m v + m' r; v = velocity = d r/d t; m' = mass change rate

F = d P/d t = d²S/dt² = Force = m (d²r/dt²) +2(dm/d t) (d r/d t) + (d²m/dt²) r

= m γ + 2m'v +m"r; γ = acceleration; m'' = mass acceleration rate

In polar coordinates system

r = r r(1) ;v = r' r(1) + r θ' θ(1) ; γ = (r" - rθ'²)r(1) + (2r'θ' + rθ")θ(1)

F = m[(r"-rθ'²)r(1) + (2r'θ' + rθ")θ(1)] + 2m'[r'r(1) + rθ'θ(1)] + (m"r) r(1)

F = [d²(m r)/dt² - (m r)θ'²]r(1) + (1/mr)[d(m²r²θ')/d t]θ(1) = [-GmM/r²]r(1)

d² (m r)/dt² - (m r) θ'² = -GmM/r²; d (m²r²θ')/d t = 0

Let m =constant: M=constant

d²r/dt² - r θ'²=-GM/r² ------ I

d(r²θ')/d t = 0 -----------------II

r²θ'=h = constant -------------- II
r = 1/u; r' = -u'/u² = - r²u' = - r²θ'(d u/d θ) = -h (d u/d θ)
d (r²θ')/d t = 2rr'θ' + r²θ" = 0 r" = - h d/d t (du/d θ) = - h θ'(d²u/d θ²) = - (h²/r²)(d²u/dθ²)
[- (h²/r²) (d²u/dθ²)] - r [(h/r²)²] = -GM/r²
2(r'/r) = - (θ"/θ') = 2[λ + á» Ï (t)] - h²u² (d²u/dθ²) - h²u³ = -GMu²
d²u/dθ² + u = GM/h²
r(θ, t) = r (θ, 0) Exp [λ + á» Ï (t)] u(θ,0) = GM/h² + Acosθ; r (θ, 0) = 1/(GM/h² + Acosθ)
r ( θ, 0) = h²/GM/[1 + (Ah²/Gm)cosθ]
r(θ,0) = a(1-ε²)/(1+εcosθ) ; h²/GM = a(1-ε²); ε = Ah²/GM

r(0,t)= Exp[λ(r) + á» Ï (r)]t; Exp = Exponential

r = r(θ , t)=r(θ,0)r(0,t)=[a(1-ε²)/(1+εcosθ)]{Exp[λ(r) + ì Ï(r)]t} Nahhas' Solution

If λ(r) â 0; then:

r (θ, t) = [(1-ε²)/(1+εcosθ)]{Exp[á» Ï(r)t]

θ'(r, t) = θ'[r(θ,0), 0] Exp{-2á»[Ï(r)t]}

h = 2Ï a b/T; b=aâ (1-ε²); a = mean distance value; ε = eccentricity
h = 2Ïa²â (1-ε²); r (0, 0) = a (1-ε)

θ' (0,0) = h/r²(0,0) = 2Ï[â(1-ε²)]/T(1-ε)²
θ' (0,t) = θ'(0,0)Exp(-2á»wt)={2Ï[â(1-ε²)]/T(1-ε)²} Exp (-2iwt)

θ'(0,t) = θ'(0,0) [cosine 2(wt) - Ỡsine 2(wt)] = θ'(0,0) [1- 2sine² (wt) - Ỡsin 2(wt)]
θ'(0,t) = θ'(0,t)(x) + θ'(0,t)(y); θ'(0,t)(x) = θ'(0,0)[ 1- 2sine² (wt)]
θ'(0,t)(x) â θ'(0,0) = - 2θ'(0,0)sine²(wt) = - 2θ'(0,0)(v/c)² v/c=sine wt; c=light speed

Πθ' = [θ'(0, t) - θ'(0, 0)] = -4Ï {[â (1-ε) ²]/T (1-ε) ²} (v/c) ²} radians/second
{(180/Ï=degrees) x (36526=century)

Πθ' = [-720x36526/ T (days)] {[â (1-ε) ²]/ (1-ε) ²}(v/c) = 1.04°/century

This is the T-Rex equation that is going to demolished Einstein's space-jail of time

The circumference of an ellipse: 2Ïa (1 - ε²/4 + 3/16(ε²)²---) â 2Ïa (1-ε²/4); R =a (1-ε²/4)
v (m) = â [GM²/ (m + M) a (1-ε²/4)] â â [GM/a (1-ε²/4)]; m<joenahhas1958@yahoo.com