Transcript Lecture 5 notes, ppt file

Lecture 5. The Incompressibility
Method
 A key problem in computer science: analyze the
average case performance of a program.
 Using the Incompressibility Method:
 Give the program a random input (with high
Kolmogorov complexity)
 Analyze the program with respect to this single and
fixed input. This is usually easier using the fact this
input is incompressible.
 The running time for this single input is the average
case running time of all inputs!
Formal language theory
 Example: Show L={0k1k | k>0} not regular. By
contradiction, assume that DFA M accepts L.
Choose k so that C(k) >> 2|M|. Simulate M:
k
k
000 … 0 111 … 1
stop here
M q
C(k) < |M| + q + O(1) < 2|M|. Contradiction.
 Remark. Generalize to iff condition: more powerful
& easier to use than “pumping lemmas”.
Combinatorics
 Example There is a tournament (complete directed graph) T of n
players that contains no large transitive subtournaments (>1 + 2 log n).
Proof by Picture: Choose a random T.


One bit codes an edge. C(T) n(n-1)/2.
If there is a large transitive subtournament, then a large number of
edges are given for free!
C(T)< n(n - 1)/2 - subgraph-edges +overhead
T
Linearly ordered
subgraph.
Easy to describe
Fast adder
 Example. Fast addition on average.
 Ripple-carry adder: n steps adding n-bit numbers.
 Carry-lookahead adder: 2 log n steps.
 Burks-Goldstine-von Neumann (1946): logn expected steps.
S= xy; C= carry sequence;
while (C≠0) {
S= SC;
C= new carry sequence; }
Average case analysis: Fix x, take random y s.t. C(y|x)≥|y|
x = … u1 …
(Max such u is carry length)
y = … û1 …, û is complement of u
If |u| > log n, then C(y|x)<|y|. Average over all y, get logn. QED
Sorting
 Given n elements (in an array). Sort them into
ascending order.
 This is the most studied fundamental problem in
computer science.
 Shellsort (1959): P passes. In each pass, move the
elements “in some stepwise fashion” (Bubblesort)
 Open for over 40 years: a nontrivial general average
case complexity lower bound of Shellsort?
Shellsort Algorithm
 Using p increments h1, … , hp, with hp=1
 At k-th pass, the array is divided in hk
separate sublists of length n/hk (taking every
hk-th element).
 Each sublist is sorted by insertion/bubble sort.
------------ Application: Sorting networks --- nlog2 n
comparators.
Shellsort history
 Invented by D.L. Shell [1959], using pk= n/2k for step


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k. It is a Θ(n2) time algorithm
Papernow&Stasevitch [1965]: O(n3/2) time.
Pratt [1972]: O(nlog2n) time.
Incerpi-Sedgewick, Chazelle, Plaxton, Poonen, Suel
(1980’s) – worst case, roughly,Θ(nlog2n / (log logn)2).
Average case:


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Knuth [1970’s]: Θ(n5/3) for p=2
Yao [1980]: p=3
Janson-Knuth [1997]: Ω(n23/15) for p=3.
Jiang-Li-Vitanyi [J.ACM, 2000]: Ω(pn1+1/p) for any p.
Shellsort Average Case Lower bound
Theorem. p-pass Shellsort average case T(n) ≥ pn1+1/p
Proof. Fix a random permutation Π with Kolmogorov complexity
nlogn. I.e. C(Π)≥ nlogn. Use Π as input.
For pass i, let mi,k be the number of steps the kth element
moves. Then T(n) = Σi,k mi,k
From these mi,k's, one can reconstruct the input Π, hence
Σ log mi,k ≥ C(Π) ≥ n logn
Maximizing the left, all mi,k must be the same. Call it m.
Σ log m = pn log m ≥ Σ log mi,k ≥ nlogn  mp ≥ n.
So T(n) = pnm > pn1+1/p.
■
Corollary: p=1: Bubblesort Ω(n2) average case lower bound.
p=2: n1.5 lower bound. p=3, n4/3 lower bound
Heapsort
 1964, JWJ Williams [CACM 7(1964), 347-
348] first published Heapsort algorithm
 Immediately it was improved by RW Floyd.
 Worst case O(nlogn).
 Open for 40 years: Which is better in average
case: Williams or Floyd?
 R. Schaffer & Sedgewick (1996). Ian Munro
provided the solution here.
Heapsort average analysis (I. Munro)
 Average-case analysis of Heapsort.
Heapsort: (1) Make Heap. O(n) time.
(2) Deletemin, restore heap, repeat.
Williams
Floyd
log n
d
2 log n - 2d
d
log n + d comparisons/round
Fix random heap H, C(H) > n log n. Simulate Step (2). Each round,
encode the red path in log n -d bits. The n paths describe the heap!
Hence, total n paths, length n log n, d must be a constant.
Floyd takes n log n comparisons, and Williams takes 2n log n.
A selected list of results proved by the
incompressibility method
 Ω(n2) for simulating 2 tapes by 1 (20 years)
 k heads > k-1 heads for PDAs (15 years)
 k one-ways heads can’t do string matching (13 yrs)
 2 heads are better than 2 tapes (10 years)
 Average case analysis for heapsort (30 years)
 k tapes are better than k-1 tapes. (20 years)
 Many theorems in combinatorics, formal
language/automata, parallel computing, VLSI
 Simplify old proofs (Hastad Lemma).
 Shellsort average case lower bound (40 years)
 Lovazs Local Lemma short proof.
More on formal language theory
Lemma (Li-Vitanyi) Let L  V*, and Lx={y: xy 
L}. Then L is regular implies there is c for all
x,y,n, let y be the n-th element in Lx, we have
C(y) ≤ C(n)+c.
Proof. Like example. QED.
Example 2. {1p : p is prime} is not regular.
Proof. Let pi, i=1,2 …, be the list of primes.
Then pk+1 is the first element in LPk, hence by
Lemma, C(pk+1)≤O(1). Impossible. QED
Characterizing regular sets
 For any enumeration of *={y1,y2, …}, define
characteristic sequence X of Lx={yi : xyi L}
by
Xi = 1 iff xyi L
Theorem. L is regular iff there is a c for all x,n,
C(Xn|n) < c
Robin Moser
Low probability events -- LLL
 So far, we have used Kolmogorov complexity for
“very high probability events” hence we obtained
“average case” complexity of our problems.
 But there are times when things happen with low
probabilities (and do not hold for average case), can
the incompressibility method still be applied?
 A typical such case is captured by the Lovasz Local
Lemma.
 Moser gave such a simple (and contructive –
traditional probabilistic proof is not constructive)
proof at STOC’2009. We follow Lance Fortnow’s
proof here.
The Lovasz Local Lemma
(Symmetric version, all events with same probability)
Consider a k-CNF formula Ф with n variable and
n clauses. Assume each clause shares a
variable with at most r other clauses. Then
there is a constant d such that if r < 2k-d , then
Ф is satisfiable. Moreover, we can find that
assignment in time polynomial in m and n.
Interpretation: if we have n independent events, each
with probability <1, then there is a positive probability
that none of the events will occur. LLL relaxes this to
“almost independent”
Lance Fortnow
Incompressibility Proof of the LLL
Proof. Fix a Kolmogorov random string x of length n+sk (s determined later) such that
C(x | Ф, k, s, r, m, n) ≥ |x| = n+sk.
Algorithm Solve(Ф)
 1. Use the first n bits as assignment for Ф
 2. While there is an unsatisfied clause C
 3.
Fix(C);
Algorithm Fix(C)
 4. Replace the variables of C with next k bits in x;
 5. While there is a unsatisfied clause D that shares a variable with C
 6.
Fix(D);
Suppose the algorithm makes s Fix calls. If we know which clause is being fixed, we know the
clause is not satisfied, hence we know all k bits corresponding to this clause. Hence we can
describe x by its first n bits, plus the description of the clauses being fixed.
For each of the m Fix calls from Solve, it takes log m bits to indicate the clause. For other
(recursive) Fix calls from Fix, it takes log r + O(1)$ bits since there are only r other clauses
that share a variable with the current clause. Thus, we must have
m log m + s(log r + O(1)) + n ≥ n + sk.
We must have r ≤ 2k-d, and letting s ≥ m log m finishes the proof.