### Minimum Cut

deterministic algorithms

- Max-flow
- Gabow

### Recursive Contraction Algorithm

Recall contraction algorithm

- success probability $1/n^2$
- why?
- high failure probability when graph is small
- how fix? repetition
- how do better? repeat only on small problems!

- subtracting $p_{k-1}/4$ fraction of $p_{k-1}$ each time
- so requires about $4/p_{k-1}$ levels to cut $p_{k-1}$ in half.
- if $q_k=1/p_k$, requires about $q_k$ steps to double $q_k$
- i.e. $q_{2k}=2q_k$
- so $q_k=k$
- and $p_k=1/k$

### Sampling

Min-cut

- saw RCA, $\Olog(n^2)$ time
- Another candidate: Gabow's algorithm: $\Olog(mc)$ time on $m$-edge graph with min-cut $c$
- nice algorithm, if $m$ and $c$ small. But how could we make that happen?
- Similarly, for those who know about it, augmenting paths gives $O(mv)$ for max flow. Good if $m$, $v$ small. How make happen?
- Sampling! What's a good sample? (take suggestions, think about them.
- Define $G(p)$---pick each edge with probability $p$

Intuition:

- $G$ has $m$ edges, min-cut $c$
- $G(p)$ has $pm$ edges, min-cut $pc$
- So improve Gabow runtime by $p^2$ factor!

What goes wrong? (pause for discussion)

- expectation isn't enough
- so what, use chernoff?
- min-cut has $c$ edges
- expect to sample $\mu=pc$ of them
- chernoff says prob. off by $\epsilon$ is at most $2e^{-\epsilon^2 \mu/4}$
- so set $pc=8\log n$ or so, deduce with high probability, no min-cut deviates.

- (pause for objections)
- yes, a problem: exponentially many cuts.
- so even though Chernoff gives “exponentially small” bound, accumulation of union bound means can't bound probability of small deviation over all cuts.

Surprise! It works anyway.

- Theorem: if min cut $c$ and build $G(p)$,
then “min expected cut” is $\mu=pc$. Probability any cut deviates
by more than $\epsilon$ is $O(n^2 e^{-\epsilon^2 \mu/3})$.
- So, if get $\mu$ around $12(\log n)/\epsilon^2$, all cuts within $\epsilon$ of expectation with high probability.
- Do so by setting $p=12(\log n)/c$

- Application: min-cut approximation.
- Theorem says a min-cut will get value at most $(1+\epsilon)\mu$ whp
- Also says that any cut of original value $(1+\epsilon)c/(1-\epsilon)$ will get value at most $(1+\epsilon)\mu$
- So, sampled graph has min-cut at most $(1+\epsilon)\mu$, and whatever cut is minimum has value at most $(1+\epsilon)c/(1-\epsilon) \approx (1+2\epsilon)c$ in original graph.
- How find min-cut in sample? Gabow's algorithm
- in sample, min-cut $O((\log n)/\epsilon^2)$ whp, while number of edges is $O(m(\log n)/\epsilon^2 c)$
- So, Gabow runtime $\Olog(m/\epsilon^2 c)$
- constant factor approx in near linear time.

Proof of Theorem

- Suppose min-cut $c$ and build $G(p)$
- Lemma: bound on number of
$\alpha$-minimum cuts is $n^{2\alpha}$.
- Base on contraction algorithm

- So we take as given: number of cuts of value less than $\alpha c$ is at most $n^{2\alpha}$ (this is true, though probably slightly stronger than what you proved. If use $O(n^{2\alpha})$, get same result but messier.
- First consider $n^2$ smallest cuts. All have expectation at least $\mu$, so prob any deviates is $e^{-\epsilon^2 \mu/4} = 1/n^2$ by choice of $\mu$
- Write larger cut values in increasing order $c_1,\ldots$
- Then $c_{n^{2\alpha}} > \alpha c$
- write $k=n^{2\alpha}$, means $\alpha_k = \log k/\log n^2$
- What prob $c_k$ deviates? $e^{-\epsilon^2 pc_k/4} = e^{-\epsilon^2 \alpha_k \mu/4}$
- By choice of $\mu$, this is $k^{-2}$
- sum over $k>n^2$, get $O(1/n)$

Issue: need to estimate $c$.

Las Vegas:

- Tree pack sample? Note good enough: too few trees
- Partition into pieces, pack each one
- get $(1-\epsilon)\mu$ trees in each piece, so $(1-\epsilon)c$ total.
- So, run sampling alg till cut found and trees packed are within $\epsilon$.
- happens whp first time, repeat if not. Expected number of iterations less that 2, so poly expected time.

Idea for exact:

- Las vegas algorithm gave approximately maximum packing
- how turn maximum? Gabow augmentations.
- Idea: run approx alg for some $\epsilon$, then augment to optimum
- Gives faster algorithm.
- wait, faster algorithm could be used instead of Gabow's in approximation algorithm to get faster approximation algorithm
- then could use faster approx alg (followed by augmentations) to get faster exact algorithm
- each algorithms seems to imply a faster one
- What is limit? Recursive algorithm

DAUG:

- describe alg
- give recurrence: $T(m,c) = 2T(m/2,c/2)+\Olog(m\sqrt{c}) = \Olog(m\sqrt{c})$
- Are we done? (wait for comment)
- No! Subproblem sizes are random variables
- Wait, have used this trick in recurrences before
- But that was because recurrences were linear, could use linearity of expectation
- Here, recurrence nonlinear. Dead.

Recursion Tree:

- expand all nodes
- depth of tree $O(\log m)$ since unlikely to get 2 edges at same leaf
- wlog keep $n \le m$ by discarding isolated vertices
- successfull and unsuccessful augmentations
- telescoping of successful augmentations
- analyze “high nodes” where cut value near expectation
- analyze “low nodes” where cut values small (a fortiori) but rely mainly on having few edges.

Later work. Just talk about where this went.

- Linear time cut by tree packing
- applications to max-flow
- current state of affairs, open problems.

## Reliability

Idea

- only small cuts matter
- enumerate them
- DNF count

Monte Carlo

- suppose $u_G(p)$ large
- Monte Carlo can estimate
- so assume $u_G(p)$ small

Small cuts

- $u_G(p) \ge p^c$
- so assume $p^c \le n^{-3}$
- show large cuts contribute much less
- before, we bounded $\alpha$-mincuts by $2\alpha$
- so in particulat, $n^{2\alpha+2}$ cuts between $\alpha c$ and $(\alpha+1)c$
- contribution to $u_G(p)$ is at most $$ \begin{align*} n^{2\alpha+2}p^{\alpha c} &=n^{2\alpha+2}p^{3(\alpha-1)c}p^c\\ &=n^{2\alpha+2}n^{-3(\alpha-1)}p^c\\ &=n^{5-\alpha}p^c \end{align*} $$
- geometric sum over $\alpha$
- so dominated by first term
- if $\alpha > 6$, this is negligible.

Application

- enumerate small cuts using CA
- their failure probability is near $u_G(p)$
- pass to DNF counting algorithm
- get approximation to that

Alternative: coupling.

- which graph is least reliable, most likely to fail given $n$?
- cycle
- coupling argument