Study of Optimal Mixing

Hsu, Hung-Wei

Juang, Yi-Lin

 

Prof. Yu, Tian-Li

2016.06.28

Outline

  • Pasting Problem
  • Trials of improving DSMGA2

Pasting Problem

Properties of Good Topologies

  • Tree structure

  • Random instances

We first discuss on problems with following properties

Simple Case

  • Spin-glass like subfunctions

  • Graph degree = 4 (same as spin-glass)

  • Not compatible with high degree

Properties of simple case

Proof of O(1) Solution In Simple Case

E \leq 1 + \frac{\binom{4}{1}}{2^{4}}2E + \frac{\binom{4}{0}}{2^{4}}4E \leq 1 + \frac{3}{4}E
E≤1+(41)242E+(40)244E≤1+34EE \leq 1 + \frac{\binom{4}{1}}{2^{4}}2E + \frac{\binom{4}{0}}{2^{4}}4E \leq 1 + \frac{3}{4}E
E \leq 4
E≤4E \leq 4

High degrees doesn’t converge

E \leq 1 + \frac{\binom{7}{3}}{2^{7}}E +\frac{\binom{7}{2}}{2^{7}}3E+\frac{\binom{7}{1}}{2^{7}}5E+\frac{\binom{7}{0}}{2^{7}}7E
E≤1+(73)27E+(72)273E+(71)275E+(70)277EE \leq 1 + \frac{\binom{7}{3}}{2^{7}}E +\frac{\binom{7}{2}}{2^{7}}3E+\frac{\binom{7}{1}}{2^{7}}5E+\frac{\binom{7}{0}}{2^{7}}7E
\leq 1 +\frac{140}{128}E
≤1+140128E\leq 1 +\frac{140}{128}E
E\leq 1 + \frac{\binom{8}{3}}{2^{8}}2E+ \frac{\binom{8}{2}}{2^{8}}4E+ \frac{\binom{8}{1}}{2^{8}}6E+ \frac{\binom{8}{0}}{2^{8}}8E
E≤1+(83)282E+(82)284E+(81)286E+(80)288EE\leq 1 + \frac{\binom{8}{3}}{2^{8}}2E+ \frac{\binom{8}{2}}{2^{8}}4E+ \frac{\binom{8}{1}}{2^{8}}6E+ \frac{\binom{8}{0}}{2^{8}}8E
\leq 1 + \frac{280}{256}E
≤1+280256E\leq 1 + \frac{280}{256}E

Experiment results

metric mean std mean std
original 51k 25k 545k 251k
simple 56k 23k 708k 316k
kai 48k 19k 556k 241k
mi3 85k 46k - -
mi3ex 70k 32k - -
dRank 51k 23k 567k 219k

Instance with One Loop

  • Best solution must match all edges

  • Case 1 : all edges on loop matched.

  • Case 2 : some edges on loop mismatched.

Properties of instance with one loop

Case 1: All edges matched

Case 2: Some edges mismatched

  • Break that edge to form a tree structure

  • Fix this intance with previous method

Case 2: Some edges mismatched

How about 2D Spin Glass

  • Hard to decide loop numbers

  • Break mismatch edges

  • Estimate average loop number

Procedure:

A Wrong Approach

  • V = N, E = 2N

  • N mismatched edges  in average

  • V = N, E = N implies one loop in average

Obstacles

  • Several connected components
  • All edges matched after removing edges
  • This analysis may only work for spin-glass

General analysis

The model we try :

  • One-bit overlapping between BBs
  • Quantize the value of each subfunctions

Expect that we can derive some properties

1

General analysis

0

0

0

1

1

0

1

1

We quantize the value of each
subfunctions into four cases :

 1, 2, 3 and 4
 

By this method, we derive that optimal method must have 2.5 in average 

Obstacles

  • Few properties derived
  • Cannot handle tree structure easily
  • This analysis is a brute force approach
  • No graph theory is applied

An elegant approach

An elegant approach

  • Find the area different to optimal solution
  • Estimate the distribution of these areas size
  • Bound the value by integration

Expected result

We first try exponential distribution : 

\int_{0}^{N}2^{n}\,pdf\left(n\right)\,dn = \mathcal{O}\left(N^{2}\right)
∫0N2npdf(n)dn=O(N2)\int_{0}^{N}2^{n}\,pdf\left(n\right)\,dn = \mathcal{O}\left(N^{2}\right)
pdf \left( n \right) = \lambda \, e^{-\lambda n}
pdf(n)=λe−λnpdf \left( n \right) = \lambda \, e^{-\lambda n}
\lambda \geq \ln{2}
λ≥ln2\lambda \geq \ln{2}

We need : 

pdf \left( n \right) = n^{2}\,2^{-n}
pdf(n)=n22−npdf \left( n \right) = n^{2}\,2^{-n}

A Smarter Way

  • Ability to cut large areas into small ones

  • Apply a recusive method

  • Probability model

Basic Concept

+4

+3

-2

Basic Concept (Worst)

+1

+3

-4

Basic Concept (Illegal)

+1

+3

+2

Model consideration

  • Invariants

  • Constraints

  • Approximations

Model We Chose

  • Energy function

  • Proper assumption to preserve some properties

  • Estimated distribution to random variable

Order Assumption

+1

+3

-2

Order Assumption (recursive)

+5

+4

+1

+3

+0

Other Constraints

We want to know more about I to determine cuts

\sum_{c \in C }f\left(c\right) = I
∑c∈Cf(c)=I\sum_{c \in C }f\left(c\right) = I

Build Probability Model from Experiment

  1. Calculate the value of optimal solution

  2. Use half binomial distribution to bound

  3. Use parameter p in model to determine cuts

  4. Describe that we can usually cut the instance

Future Prospectives 

  • Better Probability Model

End

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