Median finding.

change from book. List L

• idea; random sampling
• median of sample looks like median of whole. neighborhood.
• Algorithm
• choose s samples with replacement
• take fences before and after sample median
• keep items between fences. sort.
• Analysis
• claim (i) median within fences and (ii) few items between fences.
• Without loss of generality, L contains 1,..., n.
• Samples s₁,..., s_m in sorted order.
• lemma: S_r near rn/s.
• Chernoff: $\forall$k, number elements before k is (1¬±$\epsilon$)ks/n, where $\epsilon$= $\sqrt{(6n\ln n)/ks}$.
• Thus, when k > n/4, error ks/n(1¬±$\sqrt{24\ln n/s}$) = ks/n(1¬±$\epsilon$).
• S_{(1 + $\epsilon$)ks/n} > k
• S_r > rn/s(1 + $\epsilon$)
• S_r < rn/s(1 - $\epsilon$).
• Let r₀ = ${\frac{s}{2}}$(1 - $\epsilon$)
• Then w.h.p., ${\frac{n}{2}}$(1 - $\epsilon$)/(1 + $\epsilon$) < S_r0 < n/2
• Let r₁ = ${\frac{s}{2}}$(1 - $\epsilon$)
• Then S_r1 > n/2
• But S_r1 - S_r0 = O($\epsilon$n)
• Number of elements to sort: s
• Set containing median: O($\epsilon$n) = O(n$\sqrt{(\log n)/s}$).
• balance: $\Olog$(n^2/3) in both steps.

Randomized is strictly better:

• Optimum deterministic: $\ge$(2 + $\epsilon$)n
• Optimum randomized: $\le$(3/2)n + o(n)

Routing

• synchronous message passing
• bidirectional links, one message per step
• queues on links
• permutation routing
• oblivious algorithms only consider self packet.
• Theorem Any deterministic oblivious permutation routing requires $\Omega$($\sqrt{N/d}$) steps on an N node degree d machine.
• reason: some edge has lots of paths through it.
• homework: special case
• Hypercube.
• N nodes, n = log₂N dimensions
• bit representation
• natural routing: bit fixing (left to right)
• paths of length n
• Nn edges for N length n paths
• lower bound n
• Routing algorithms:
• O(n) = O(log N) randomized
• beats $\Omega$($\sqrt{N/n}$) deterministic
• how? load balance paths.
• Random destination (not permutation!), bit correction
• Average case, but a good start.
• T(e_i) = number of paths using e_i
• by symmetry, all E[T(e_i)] equal
• expected path length n/2
• LOE: expected total path length Nn/2
• nN edges in hypercube
• E[T(e_i)] = 1/2
• Chernoff: every edge gets $\le$3n (prob 1 - 1/N)
• Naive usage:
• n phases, one per bit
• 3n time per phase
• O(n²) total
• From intermediate destination, route back!
• routes worst case permutation in O(n²).
• What if don't wait for next phase?
• FIFO queuing
• total time is length plus delay
• Expected delay $\le$E[$\sum$T(e_l)] = n/2.
• Chernoff bound? no. dependence of T(e_i).
• High prob. bound:
• consider paths sharing route (e₀,..., e_k)
• Suppose S packets intersect route (use at least one of e_i)
• claim delay $\le$| S|
• Suppose true: Let H_ij = 1 if j hits i's (fixed) route.

$$\sum$$

$$\le \sum$$

$$\le$$

• 
  
• Now Chernoff does apply (H_ij independent for fixed i-route).
• | S| = O(n) w.p. 1 - 2^-5n, so O(n) delay for all 2ⁿ paths.
• Lag argument
• Exercise: once packets separate, don't rejoin
• Route for i $\rho_{i}^{}$= (e₁,..., e_k)
• charge each delay to a departure of a packet from $\rho_{i}^{}$.
• Packet waiting to follow e_j at time t has: Lag t - j
• Delay of v_i is lag crossing e_k
• When v_i delay rises to l + 1, some packet from S has lag l (since crosses e_j instead of v_i).
• Consider last time t' where a lag-l packet exists
• some lag-l packet w crosses e_j' at t' (others increase to lag-(l + 1))
• w leaves at this point (if not, then l at e_{j' + 1} next time)
• charge one delay to w.