Analysis Population Size of Optimal Mixing

許泓崴, 廖翊雲
 

指導教授：于天立 教授

2017.06.27

Outline

  • Recap : Pasting Problem
  • Again : Pasting Problem
\text{General Lower Bound }\mathcal{O}\left(\log{L}\right)
General Lower Bound O(logL)\text{General Lower Bound }\mathcal{O}\left(\log{L}\right)
\text{Experiment curve }\approx\mathcal{O}\left(\log{L}\right)
Experiment curve ≈O(logL)\text{Experiment curve }\approx\mathcal{O}\left(\log{L}\right)
\text{Population Size}
Population Size\text{Population Size}
\text{Upper bound of lower bound for chain structure }\mathcal{O}\left(L^{2}\right)
Upper bound of lower bound for chain structure O(L2)\text{Upper bound of lower bound for chain structure }\mathcal{O}\left(L^{2}\right)
\text{Upper bound of lower bound for loop structure }\mathcal{O}\left(L^{3}\right)
Upper bound of lower bound for loop structure O(L3)\text{Upper bound of lower bound for loop structure }\mathcal{O}\left(L^{3}\right)

Pasting Problem

1
11
1
11
1
11
1
11
0
00
0
00
0
00
1
11
0
00
1
11
1
11
0
00

Optimal solution

Target solution

\text{Paste }\,\mathcal{O}{(logn)} \Rightarrow \text{Supply} \,\,2^{c\log{n}}=n^{c}
Paste O(logn)⇒Supply2clogn=nc\text{Paste }\,\mathcal{O}{(logn)} \Rightarrow \text{Supply} \,\,2^{c\log{n}}=n^{c}
\text{Expected cost }\,\mathbb{E}{\left[ 2^{len} \right]} =\mathcal{O}{(n^c)}
Expected cost E[2len]=O(nc)\text{Expected cost }\,\mathbb{E}{\left[ 2^{len} \right]} =\mathcal{O}{(n^c)}
\text{However,}\quad\mathbb{E}{\left[ 2^{len} \right]} \geq2^{\mathbb{E}{\left[ len\right]}}
However,E[2len]≥2E[len]\text{However,}\quad\mathbb{E}{\left[ 2^{len} \right]} \geq2^{\mathbb{E}{\left[ len\right]}}

Properties of Good Topologies

  • Tree structure / Ring structure

  • Random instances
  • Identical subfunction

We have discussed on problems with following properties

Counter Example

We have found counter example in 2d spin glass problem

\frac{n}{4}
n4\frac{n}{4}
\sqrt{n}
n\sqrt{n}

Flip from outer region

Flip from corner

Last semester

This semester

Outer Region

Fail

Success

Success

Fail

Degree 2, 3 nodes are important

topology dependent

\Rightarrow
⇒\Rightarrow

Back to Ring Structure

We want to solve NK landscape

Identical sub function
Spin-glass like sub function
value = {0, 1, 2, 3}
sub function
\mathcal{O}(1)
O(1)\mathcal{O}(1)
\mathcal{O}(1)
O(1)\mathcal{O}(1)
\mathcal{O}(1)
O(1)\mathcal{O}(1)

Longest differnet segment length

proportion of longest  different segment k is exponential to k

First flip method

Concept

From head, flip first met segment

P\left[\textrm{length}=l\right] = p^{-\left(l+1\right)}
P[length=l]=p−(l+1)P\left[\textrm{length}=l\right] = p^{-\left(l+1\right)}

Result

chain
ring
comb
p=\frac{1}{2}
p=12p=\frac{1}{2}
p<\frac{1}{2}
p<12p<\frac{1}{2}
\mathcal{O}(l^2)
O(l2)\mathcal{O}(l^2)
\mathcal{O}(l^3)
O(l3)\mathcal{O}(l^3)
\mathcal{O}(l)
O(l)\mathcal{O}(l)
\mathcal{O}(l)
O(l)\mathcal{O}(l)
\mathcal{O}(l^{m+1})
O(lm+1)\mathcal{O}(l^{m+1})
\mathcal{O}(lc^{m})
O(lcm)\mathcal{O}(lc^{m})

Problem

  • Bound direction
  • Subfunction size

Bound Direction

Upper of lower = ??

Subfunction Size

\mathbb{E}\left[2^{k\times\textrm{seg length}}\right] \gg \left( \mathbb{E}\left[2^\textrm{seg length}\right]\right)^{k}
E[2k×seg length]≫(E[2seg length])k\mathbb{E}\left[2^{k\times\textrm{seg length}}\right] \gg \left( \mathbb{E}\left[2^\textrm{seg length}\right]\right)^{k}

Tighter upper : DP

Upper of lower

\mathcal{O}(l\ \log l)
O(l logl)\mathcal{O}(l\ \log l)

Lower of Lower bound

C(l)=1+\sum\limits_{k=0}^{l-1}2^{-l}{l \choose k}C(l-k)
C(l)=1+∑k=0l−12−l(lk)C(l−k)C(l)=1+\sum\limits_{k=0}^{l-1}2^{-l}{l \choose k}C(l-k)
C(l)\leq 1+\frac{1}{2}C(\frac{l}{2}) + \frac{1}{2}C(l)
C(l)≤1+12C(l2)+12C(l)C(l)\leq 1+\frac{1}{2}C(\frac{l}{2}) + \frac{1}{2}C(l)
C(l)\leq 2 + 2C(\frac{l}{2}) \leq2\lceil\log_{2}{l}\rceil
C(l)≤2+2C(l2)≤2⌈log2l⌉C(l)\leq 2 + 2C(\frac{l}{2}) \leq2\lceil\log_{2}{l}\rceil
C(l)=\Sigma_{i=2}^{\infty}{i\left(\left(1-2^{-(i-1)}\right)^l-\left(1-2^{-(i-2)}\right)^l\right)}
C(l)=Σi=2∞i((1−2−(i−1))l−(1−2−(i−2))l)C(l)=\Sigma_{i=2}^{\infty}{i\left(\left(1-2^{-(i-1)}\right)^l-\left(1-2^{-(i-2)}\right)^l\right)}
C(l)\ge-\Sigma_{i=2}^{k-1}{\left(1-2^{-(i-1)}\right)^l}+k+\left(1-2^{-(k-1)}\right)^l
C(l)≥−Σi=2k−1(1−2−(i−1))l+k+(1−2−(k−1))lC(l)\ge-\Sigma_{i=2}^{k-1}{\left(1-2^{-(i-1)}\right)^l}+k+\left(1-2^{-(k-1)}\right)^l
C(l)\ge (1-\frac{1}{e})\log_2 l -2
C(l)≥(1−1e)log2l−2C(l)\ge (1-\frac{1}{e})\log_2 l -2

experiment, lower of lower

upper of lower bound

l\log l
llogll\log l
\log l
logl\log l

End

Analysis Population Size of Optimal Mixing

By 許泓崴

Made with Slides.com

Analysis Population Size of Optimal Mixing

Individual Study of DSMGA2 by Sammy

  • 325

More from 許泓崴