Coupling:

Method

• Run two copies of Markov chain X_t, Y_t
• Each considered in isolation is a copy of MC (that is, both have MC distribution)
• but they are not independent: they make dependent choices at each step
• in fact, after a while they are almost certainly the same
• Start Y_t in stationary distribution, X_t anywhere
• Coupling argument:

$$\ne \ne$$

$$\epsilon \ne$$

• 
  
• So just need to make $\epsilon$(which is r.p.d.) small enough.

n-bit Hypercube walk: at each step, flip random bit to random value

• At step t, pick a random bit b, random value v
• both chains set but b to value v
• after O(n log n) steps, probably all bits matched.

Counting k colorings when k > 2$\Delta$ + 1

• The reduction from (approximate) uniform generation
• compute ratio of coloring of G to coloring of G - e
• Recurse counting G - e colorings
• Base case kⁿ colorings of empty graph

·       Bounding the ratio:

o      note G - e colorings outnumber G colorings

o      By how much? Let L colorings in difference (u and v same color)

o      to make an L coloring a G coloring, change u to one of k - $\Delta$= $\Delta$+ 1 legal colors

o      Each G-coloring arises at most one way from this

o      So each L coloring has at least $\Delta$+ 1 neighbors unique to them

o      So L is 1/($\Delta$ + 1) fraction of G.

o      So can estimate ratio with few samples

·       The chain:

o      Pick random vertex, random color, try to recolor

o      loops, so aperiodic

o      Chain is time-reversible, so uniform distribution.

·       Coupling:

o      choose random vertex v (same for both)

o      based on X_t and Y_t, choose bijection of colors

o      choose random color c

o      apply c to v in X_t (if can), g(c) to v in Y_t (if can).

o      What bijection?

§       Let A be vertices that agree in color, D that disagree.

§       if v $\in$D, let g be identity

§       if v $\in$A, let N be neighbors of v

§       let C_X be colors that N has in X but not Y (X can't use them at v)

§       let C_Y similar, wlog larger than C_X

§       g should swap each C_X with some C_Y, leave other colors fixed. Result: if X doesn't change, Y doesn't

·       Convergence:

o      Let d'(v) be number of neighbors of v in opposite set, so

$$\sum_{v \in A}^{}$$d'(v) = $$\sum_{v \in D}^{}$$d'(v) = m'

• Let $\delta$= | D|
• Note at each step, $\delta$changes by 0,±1
• When does it increase?
• v must be in A, but move to D
• happens if only one MC accepts new color
• If c not in C_X or C_Y, then g(c) = c and both change
• If c $\in$C_X, then g(c) $\in$C_Y so neither moves
• So must have c $\in$C_Y
• But | C_Y|$\le$d'(v), so probability this happens is

$$\sum_{v \in A}^{}$$$${\frac{1}{n}}$$^. $${\frac{d'(v)}{k}}$$= $${\frac{m'}{kn}}$$

• When does it decrease?
• must have v $\in$D, only one moves
• sufficient that pick color not in either neighborhood of v,
• total neighborhood size 2$\Delta$, but that counts the d'(v) elements of A twice.
• so Prob.

$$\sum_{v \in D}^{}$$$${\frac{1}{n}}$$^. $${\frac{k-(2\Delta-d'(v))}{k}}$$= $${\frac{k-2\Delta}{kn}}$$$$\delta$$+ $${\frac{m'}{kn}}$$

• Deduce that expected change in $\delta$is difference of above, namely

- $${\frac{k-2\Delta}{kn}}$$$$\delta$$= - a$$\delta$$.

• So after t steps, E[$\delta_{t}^{}$]$\le$(1 - a)^t$\delta_{0}^{}$$\le$(1 - a)^tn.
• Thus, probability $\delta$> 0 at most (1 - a)^tn.
• But now note a > 1/n², so n²log n steps reduce to one over polynomial chance.

Note: couple depends on state, but who cares

• From worm's eye view, each chain is random walk
• so, all arguments hold

Counting vs. generating:

• we showed that by generating, can count
• by counting, can generate:

Parallel Algorithms

PRAM

• P processors, each with a RAM, local registers
• global memory of M locations
• each processor can in one step do a RAM op or read/write to one global memory location
• synchronous parallel steps
• various conflict resolutions (CREW, EREW, CRCW)
• not realistic, but explores ``degree of parallelism''

Randomization in parallel:

• load balancing
• symmetry breaking
• isolating solutions

Classes:

• NC: poly processor, polylog steps
• RNC: with randomization. polylog runtime, monte carlo
• ZNC: las vegas NC
• immune to choice of conflict resolution

Practical observations:

• very little can be done in o(log n) with poly processors
• lots can be done in $\Theta$(log n)
• often concerned about work which is processors times time
• algorithm is ``optimal'' if work equals best sequential

Basic operations

• and, or
• counting ones

Sorting

Quicksort in parallel:

• n processors
• each takes one item, compares to splitter
• count number of predecessors less than splitter
• determines location of item in split
• total time O(log n)
• combine: O(log n) per layer with n processors
• problem: $\Omega$(log²n) time bound
• problem: n log²n work

Parallel recursion:

• paradigm: reduce problem size from n to $\sqrt{n}$in O(log n) time.
• total time O(log n + log$\sqrt{n}$ + ^... ) = O(log n)

More processors:

• n² processors
• do all comparisons
• count number of items smaller than me: O(log n)
• put into place
• result: O(log n) time with n² processors
• or, O(n) time with n processors

BoxSort:

• n processors
• Choose $\sqrt{n}$random splitters
• sort in O(log n) time
• insert items in splitters: O(log n) time
• solve each piece separately, recursively

Intuition:

• expected subproblem size O($\sqrt{n}$)
• so expected time spent on a branch is O(log n) as above
• problem: many branches: need high probability result.
• solution: analyze each path, show O(log n) time whp
• thus max path is O(log n)

High probability:

• consider item x
• claim splitter within $\alpha$$\sqrt{n}$on each side
• since prob. not at most (1 - $\alpha$$\sqrt{n}$/n)^$\sqrt{n}$$\le$e^{- $\alpha$}
• fix $\gamma$, d < 1/$\gamma$
• define $\tau_{k}^{}$= d^k
• define $\rho_{k}^{}$= n^{$\gamma^{k}$}
• note size $\rho_{k}^{}$problem takes $\gamma^{k}_{}$log n time
• argue at most d^k size-$\rho_{k}^{}$ problems whp
• deduce runtime $\sum$d^k$\gamma_{k}^{}$ = $\sum$(d$\gamma$)^klog n = O(log n)
• note: as problem shrinks, allowing more divergence in quantity for whp result
• minor detail: ``whp'' dies for small problems
• OK: if problem size log n, finish in log n time with log n processors