Polling

Outline

• Set has size u, contains n ``special'' elements
• goal: count number of special elements
• sample with probability p = c(log n)/$\epsilon^{2}_{}$n
• with high probability, (1¬±$\epsilon$)np special elements
• if observe k elements, deduce n $\in$(1¬±$\epsilon$)k.
• Problem: what is p?

Related idea: Monte Carlo simulation

• Probability space, event A
• easy to test for A
• goal: estimate p = Pr[A].
• Perform n trials (sampling with replacement).
• expected outcome pn.
• estimator ${\frac{1}{n}}$$\sum$I_i
• prob outside $\epsilon$< exp(- $\epsilon^{2}_{}$np/3) ( $\epsilon$< 1)
• for prob. $\delta$, need

n = O$$\left(\vphantom{\frac{\log 1/\delta}{\epsilon^2 p}}\right.$$$${\frac{\log 1/\delta}{\epsilon^2 p}}$$$$\left.\vphantom{\frac{\log 1/\delta}{\epsilon^2 p}}\right)$$

• what if p unknown?
• What if p is small?

Handling unknown p

• Sample n times till get $\mu_{\epsilon,\delta}^{}$= O(log$\delta^{-1}_{}$/$\epsilon^{2}_{}$) hits
• w.h.p, p $\in$(1¬±$\epsilon$)$\mu_{\epsilon,\delta}^{}$n

Minimum Cut

Min-cut

• saw RCA, $\Olog$(n²) 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) hss pm edges, min-cut pc
• So improve Gabow runtime by p² 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(n²e^{- $\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)
• For midterm, you had to prove bound on number of $\alpha$-minimum cuts.
• I assume you all did that
• well, maybe not, but proof will be in solutions
• 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 n² smallest cuts. All have expectation at least $\mu$, so prob any deviates is e^{- $\epsilon^{2}$$\mu$/4} = 1/n² by choice of $\mu$
• Write larger cut values in increasing order c₁,...
• Then c_n^2$\alpha$ > $\alpha$c
• write k = n^2$\alpha$, means $\alpha_{k}^{}$= log k/log n²
• What prob c_k deviates? e^{- $\epsilon^{2}$pc}_k^/4 = e^{- $\epsilon^{2}$$\alpha_{k}$$\mu$/4}
• By choice of $\mu$, this is k^-2
• sum over k > n², get O(1/n)

Transitive closure

Problem outline

• databases want size
• matrix multiply time
• compute reachibility set of each vertex, add

Sampling algorithm

• generate vertex samples until $\mu_{\epsilon}^{}$$\delta$reachable from v
• deduce size of v's reachibility set.
• reachability test: O(m).
• number of sample: n/size.
• O(mn) per vertex--ouch!

Pipeline for all vertices simultaneously

• increase mean to O(log n/$\epsilon^{2}_{}$),
• so 1/n² failure
• O(mn) for all vertices (still ouch).

Avoid wasting work

• after O(n log n) samples, every vertex has log n hits. No more needed.
• Send at most log n samples over an edge: $\Olog$(m)