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### Counting Problems

Two big pieces:

1. Equivalence of counting and generating via self reducibility
2. Generating via Markov chains

### Application: Permanent

Counting perfect matchings

• Choose random $n$-edge set
• check if matching
• problem: rare event
• to solve, need sample space where matchings are dense

Idea: self reducibility by adding an edge (till reach complete graph)

• problem: don't know how to generate random matching

Different idea: ratio of $k$-edge to $k-1$-edge matchings

• telescope down to 1-edge matchings (self reduction)
• in dense graphs (degree $n/2$), ratio is at most $m^3$ both ways.
• map each $k$ edge matching by removing an edge: $n^2$ to 1
• map each $k-1$ edge matching to $k$-edge matching by augmenting path of length at most 3.
• take unmatched $u$ and $v$
• if unmatched neighbor of $u$ or $v$, done
• by $u$ and $v$ have $n/2$ neighbors, so if all matched, some neighbor $b$ of $u$ matched to some neighbor $a$ of $v$.
• so each size $k$ matching “receives” at most $m^3$ size $k-1$ matchings.

Generate via random walk

• based on using uniform generation to do sampling.
• applies to minimum degree $n/2$
• Let $M_k$ be $k$-edge matchings, $\|M_k\|=m_k$
• algorithm estimates all ratios $m_k/m_{k-1}$, multiplies
• claim: ratio $m_{k+1}/m_k$ polynomially bounded (dense).
• deduce sufficient to generate randomly from $M_k\cup M_{k-1}$, test frequency of $m_k$
• do so by random walk of local moves:
• with probability $1/2$. stay still
• else Pick random edge $e$
• if in $M_k$ and $e$ matched, remove
• if in $M_{k-1}$ end $e$ can be added, add.
• if in $M_k$, $e=(u,v)$, $u$ matched to $w$ and $v$ unmatched, then match $u$ to $w$.
• else do nothing
• Note that exactly one applies
• Matrix is symmetric (undirected), so double stochastic, so stationary distribution is uniform as desired.
• In text, prove $\lambda_2 =1-1/n^{O(1)}$ on an $n$ vertex graph (by proving expansion property)
• so within $n^{O(1)}$ steps, rpd is polynomially small
• so can pretend stationary

Need to prove rapid mixing

• Markov chain is an unweighted undirected graph
• Show markov chain is an expander
• but not quite: count edges leaving rather than neighboring nodes
• same argument implies rapid mixing

Canonical paths argument

• Suppose chain has $N$ vertices
• Want to prove every set $S$ has $\Omega(|S|)$ edges leaving
• Pick a path between every pair of verties
• Show no edge on more than $3n^4N$ paths
• Know $S(N-S)\ge SN/2$ paths leave $S$
• So at least that$/3n^4N=S/6n^4$ edges leave
• i.e, a polynomial fraction of $S$
• Suffices to prove mixing time polynomial (ie as if expansion was $1+1/n^4$ so mixing time like $n^4$

Canonical path construction:

• path each $M_{n-1}$ to an $M_n$ by a length 3 path
• so each $M_k$ only gets $m^3$ of the matching
• make a canonical path for each $M_n$ pair
• corresponds to $m^6$ canonical paths including $M_{n-1}$ “children”
• for two perfect matchings:
• XOR them
• get a collection of even cycles
• so arbitrarily number (order) all even cycles
• canonical path “unwinds” these cycles: remove first matching edge, add second matching edge, etc.
• do so in canonical order, starting with smallest numbered vertex in smallest numbered cycle and going up
• claim only $N$ canonical paths use an edge
• prove by fixing edge, and creating a bijection between canonical paths that use it and the $N$ perfect matchings
• given matchings $s$ and $t$, XOR is collection of cycles
• suppose $M$ is matching on transition edge
• XORing cycles with $M$ gives a different matching $M'$
• $M'$ is the representative of the $s$-$t$ path
• Show this is a bijection
• map from representative to $s$-$t$ pair
• from representative $M'$, XOR with $M$ yields cycles unwound on canonical path
• edges of $M$ not in any cycle are in both $s$ and $t$ (don't get unwound)
• edges of $M$ in a cycle are in exactly one of $s$ and $t$
• edges that precede edge that is unwound in this transition have already been unwound
• i.e. those in $M$ are in $t$, those not in $M$ are in $s$
• edges that follow edge unwound in this transition have not yet been unwound
• i.e. those in $M$ are in $s$, those not in $M$ are in $t$.

Recently, extended to non-dense case.

### Volume

Outline:

• Describe problem. Membership oracle
• $\sharp P$ hard to volume intersection of half spaces in $n$ dimensions
• In low dimensions, integral.
• even for convex bodies, can't do better than $(n/\log n))^n$ ratio
• what about FPRAS?

Estimating $\pi$:

• pick random in unit square
• check if in circle
• gives ratio of square to circle
• Extends to arbitrary shape with “membership oracle”
• Problem: rare events.
• Circle has good easy outer box

Problem: rare events:

• In 2d, long skinny shapes
• In high $d$, even round shape has exponentially larger bounding box

Solution: “creep up” on volume

• modify $P$ to contain unit sphere $B_1$, contined in larger $B_2$ of radius $r$ with $r/r_1$ polynomial
• choose $\rho=1-1/n$.
• Consider sequence of bodies $\rho^i r P \cap B_2$
• note for large $i$, get $P$
• but for $i=0$, body contains $B_2$
• so volume known
• so just need ratios
• At each step, need to random sample from $\rho^i r P \cap B_2$

Sample method: random walk forbidden to leave

• MC irreducible since body connected
• ensure aperiodic by staying put with prob. 1/2
• markov chain is “regular graph” so uniform stationary distribution
• eigenvalues show rapid mixing: after $t$ steps, r.p.d at most $(1-\frac{1}{10^17n^19})^t$
• eigenvalues small because body convex: no bottlenecks.

Observations:

• Key idea of self reducibility: compare size of sequence of “related” shapes, then telescope ratios.
• Sizes compared by sampling
• Sample by markov chain
• wait: markov chain not exact?
• doesn't matter: just get accurate to within $(1-1/poly)$ in each step, product of errors still tiny.

### 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: $$\begin{eqnarray*} \Pr[X_t=j] &= &\Pr[X_t=j \mid X_t=Y_t]\Pr[X_t=Y_t]+\Pr[X_t=j \mid X_t\ne Y_t]\Pr[X_t\ne Y_t]\\ &= &\Pr[Y_t=j]\Pr[X_t=Y_t]+\epsilon\Pr[X_t=j\mid X_t\ne Y_t] \end{eqnarray*}$$ 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 $k^n$ colorings of empty graph
• Bounding the ratio:
• note $G-e$ colorings outnumber $G$ colorings
• By how much? Let $L$ colorings in difference ($u$ and $v$ same color)
• to make an $L$ coloring a $G$ coloring, change $u$ to one of $k-\Delta=\Delta+1$ legal colors
• Each $G$-coloring arises at most one way from this
• So each $L$ coloring has at least $\Delta+1$ neighbors unique to them
• So $L$ is $1/(\Delta+1)$ fraction of $G$.
• So can estimate ratio with few samples
• The chain:
• Pick random vertex, random color, try to recolor
• loops, so aperiodic
• Chain is time-reversible, so uniform distribution.
• Coupling:
• choose random vertex $v$ (same for both)
• based on $X_t$ and $Y_t$, choose bijection of colors
• choose random color $c$
• apply $c$ to $v$ in $X_t$ (if can), $g(c)$ to $v$ in $Y_t$ (if can).
• 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:
• 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 $0,\pm1$
• 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 $\sum_{v \in A} \frac1n \cdot \frac{d'(v)}{k} = \frac{m'}{kn}$
• When does it decrease?
• must have $v \in D$, both move (since move to same color)
• 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. $\sum_{v \in D}\frac1n \cdot \frac{k-(2\Delta-d'(v))}{k} = \frac{k-2\Delta}{kn}\delta+\frac{m'}{kn}$
• 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 $(1-a)^t n$.
• But now note $a > 1/n^2$, so $n^2\log n$ 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: