Maximal independent set

trivial sequential algorithm

• inherently sequential
• from node point of view: each thinks can join MIS if others stay out
• randomization breaks this symmetry

Randomized idea

• each node joins with some probability
• all neighbors excluded
• many nodes join
• few phases needed

Algorithm:

• all degree 0 nodes join
• node v joins with probability 1/2d (v)
• if edge (u, v) has both ends marked, unmark lower degree vertex
• put all marked nodes in IS
• delete all neighbors

Intuition: d-regular graph

• vertex vanishes if it or neighbor gets chosen
• mark with probability 1/2d
• prob (no neighbor marked) is (1 - 1/2d )^d, constant
• so const prob. of neighbor of v marked--destroys v
• const fraction of neighbors vanish: O(log n) iters
• what about unmarking?
• prob(unmarking forced) only constant.
• So just changes constants

Implementing a phase trivial in O(log n).

Prob chosen for IS, given marked, exceeds 1/2

• suppose w marked. only unmarked if higher degree neighbor marked
• higher degree neighbor marked with prob. $\le$1/2d (w)
• only d (w) neighbors
• prob. any marked at most 1/2.
• deduce prob. good vertex killed exceeds (1 - e^-1/6)/2

Good vertices

• good: at least 1/3 neighbors have lower degree
• prob. no neighbor of good marked $\le$(1 - 1/2d (v))^d(v)/3$\le$e^-1/6.

Good edges

• any edge with a good neighbor
• has const prob. to vanish
• show half edges good
• deduce O(log n) iterations.

Proof

• Let V_B be bad vertices; we count edges with both ends in V_B.
• direct edges from lower to higher degree d_i is indegree, d_o outdegree
• if v bad, then d_i(v)$\le$d (v)/3
• deduce

$$\sum_{V_B}^{}$$d_i(v)$$\le$$$${\textstyle\frac{1}{3}}$$$$\sum_{V_B}^{}$$d (v) = $${\textstyle\frac{1}{3}}$$$$\sum_{V_B}^{}$$(d_i(v) + d_o(v))

• so $\sum_{V_B}^{}$d_i(v)$\le$${\frac{1}{2}}$$\sum_{V_B}^{}$d_o(v)
• which means indegree can only ``catch'' half of outdegree; other half must go to good vertices.
• more carefully,
• d_o(v) - d_i(v)$\ge$${\frac{1}{3}}$(d (v)) = ${\frac{1}{3}}$(d_o(v) + d_i(v)).
• Let V_G, V_B be good, bad vertices
• degree of bad vertices is

$$\sum_{v\in V_B}^{}$$

$$\le \sum$$

$$\le$$

• 
  
• Deduce e(V_B, V_B)$\le$e(V_B, V_G) + e(V_G, V_B). result follows.

Derandomization:

• Analysis focuses on edges,
• so unsurprisingly, pairwise independence sufficient
• not immediately obvious, but again consider d-uniform case
• prob vertex marked 1/2d
• neighbors 1,..., d in increasing degree order
• Let E_i be event that i is marked.
• Let E'_i be E_i but no E_j for j < i
• A_i event no neighbor of i chosen
• Then prob eliminate v at least

$$\sum \cap \sum$$

$$\ge \sum$$

• 
  
• Wait: show Pr[A_i | E'_i]$\ge$Pr[A_i]
• true if independent
• measure Pr[¬¨A_i | E'_i]$\le$$\sum$Pr[E_w | E'_i]
• measure

$${\frac{\Pr[E_w \cap E']}{\Pr[E'_i]}}$$

$${\frac{\Pr[E_w \cap \neg E_1 \cap \cdots \mid E_i]}{\Pr[\neg E_1 \cap \cdots \mid E_i]}}$$

$$\le {\frac{\Pr[E_w \mid E_j]}{1-\sum\Pr[E_j \mid E_i]}}$$

$$\Theta$$

• 
  
• But expected marked neighbors 1/2, so by Markov Pr[A_i] > 1/2
• so prob eliminate v exceeds $\sum$Pr[E'_i] = Pr[ $\cup$E_i]
• lower bound as $\sum$Pr[E_i] - $\sum$Pr[E_i $\cap$E_j] = 1/2 - d (d - 1)/8d² > 1/4
• so 1/2d prob. v marked but no neighbor marked, so v chosen
• Generate pairwise independent with O(log n) bits
• try all polynomial seeds in parallel
• one works
• gives deterministic NC algorithm

with care, O(m) processors and O(log n) time (randomized)

LFMIS P-complete.

Perfect Matching

We focus on bipartite; book does general case.

Detecting one easy in $\NC$:

• Tutte matrix
• Determinant nonzero iff PM
• Replace vars with values 1,..., 2^m, same holds
• Matrixu Mul, Determinant in $\NC$
• Wait: big numbers?
• Who cares: poly bits, NC to multiply etc

How about finding one?

• If unique, no problem
• Remove each edge, see if still PM in parallel
• multiplies processors by m
• still NC
• generalizes to polynomial number of matchings

Idea:

• make unique minimum weight perfect matching
• find it

Isolating lemma:

• Family of distinct sets over x₁,..., x_m
• assign random weights in 1,..., 2m
• Pr(unique min-weight set)$\ge$1/2
• Odd: no dependence on number of sets!
• (of course < 2^m)

Proof:

• Fix item x_i
• Y is min-sets containing x_i
• N is min-sets no containing x_i
• true min-sets are either those in Y or in N
• how decide? Value of x_i
• For x_i = - $\infty$, min-sets are Y
• For x_i = + $\infty$, min-sets are N
• As increase from - $\infty$to $\infty$, single transition value when both X and Y are min-weight
• If only Y min-weight, then x_i in every min-set
• If only X min-weight, then x_i in no min-set
• If both min-weight, x_i is ambiguous
• Suppose no x_i ambiguous. Then min-weight set unique!
• Exactly one value for x_i makes it ambiguous given remainder
• So pr(ambiguous)1/2m
• So pr(any ambiguous)< m/2m = 1/2

Usage:

• Consider tutte matrix A
• Assign random value 2^w_i to x_i, with w_i $\in$1,..., 2m
• Weight of matching is 2^$\sum$w_i
• Let W be minimum sum
• Unique w/pr 1/2
• If so, determinant is odd multiple of 2^W
• Try removing edges one at a time
• Edge in PM iff new determinant/2^W is odd.

NC algorithm open.

For exact matching, P algorithm open.