Study of Optimal Mixing

Hsu, Hung-Wei

Juang, Yi-Lin

 

Prof. Yu, Tian-Li

2016.06.28

Outline

  • Pasting Problem
  • Tree Structure
  • Generalize
  • Area Distribution
  • Cutting Method

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)

 

A solution 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

Degree Avg Time Max Time
2 1.34 12
4 1.41 12
6 1.58 16
8 1.74 21
10 1.87 35
15 2.16 29

Instance with One Loop

  • Best solution is simple

  • 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

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General analysis

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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^{-n}
pdf(n)=n 2−npdf \left( n \right) = n\ 2^{-n}
\lambda \approx 0.14
λ≈0.14\lambda \approx 0.14

A Smarter Way

  • Ability to cut large areas into small ones

  • Apply a recusive method

  • Probability model

Basic Concept

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Basic Concept (Worst)

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Basic Concept (Illegal)

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Model consideration

  • Invariants

  • Constraints

  • Approximations

Model We Chose

  • Energy function

  • Proper assumption to preserve some properties

  • Estimated distribution to random variable

Order Assumption

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Order Assumption (recursive)

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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

Conclusion

End

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