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(1)

Logarithmic Space

(2)

reachability Is NL-Complete

• reachability ∈ NL (p. 95).

• Suppose L is decided by the log n space-bounded TM N .

• Given input x, construct in logarithmic space the

polynomial-sized configuration graph G of N on input x (see Theorem 21 on p. 176).

• G has a single initial node, call it 1.

• Assume G has a single accepting node n.

• x ∈ L if and only if the instance of reachability has a

“yes” answer.

(3)

2sat Is NL-Complete

• 2sat ∈ NL (p. 265).

• As NL = coNL (p. 191), it suffices to reduce the coNL-complete unreachability to 2sat.

• Start without loss of generality an acyclic graph G.

• Identify each edge (x, y) with clause ¬x ∨ y.

• Add clauses (s) and (¬t) for the start and target nodes s and t.

• The resulting 2sat instance is satisfiable if and only if

(4)

The Class RL

• reachability is for directed graphs.

• It is not known if undirected reachability is in L.

• But it is in randomized logarithmic space, called RL.

• RL is RP in which the space bound is logarithmic.

• We shall prove that undirected reachability ∈ RL.a

• As a note, undirected reachability ∈ coRL.b

aAleliunas, Karp, Lipton, Lov´asz, and Rackoff (1979).

bBorodin, Cook, Dymond, Ruzzo, and Tompa (1989).

(5)

Random Walks

• Let G = (V, E) be an undirected graph with 1, n ∈ V .

• Add self-loops { i, i } at each node i.

• The randomized algorithm for testing if there is a path from 1 to n is a random walk.

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The Random Walk Framework

1: x := 1;

2: while x 6= n do

3: Pick y uniformly from x’s neighbors (including x);

4: x := y;

5: end while

(7)

Some Terminology

• vt is the node visited by the random walk at time t.

• In particular, v0 = 1.

• di denotes the degree of i (including the self-loops).

• Let pt[ i ] = prob[ vt = i ].

(8)

A Convergence Result

Lemma 102 If G = (V, E) is connected, then limt→∞ pt[ i ] = 2·| E |di for all nodes i.

• Here is the intuition.

• The random walk algorithm picks the edges uniformly randomly.

• In the limit, the algorithm will be well “mixed” and forgets about the initial node.

• Then the probability of each node being visited is proportional to its number of incident edges.

• Finally, observe that Pn

i=1 di = 2 · | E |.

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Proof of Lemma 102

• Let δt[ i ] = pt[ i ] − 2·| E |di , the deviation.

• Define ∆t = P

i∈Vt[ i ]|, the total absolute deviation.

• Now we calculate the pt+1[ i ]’s from the pt[ i ]’s.

• Each node divides its pi[ t ] into di equal parts and distributes them to its neighbors.

• Each node adds those portions from its neighbors (including itself) to form pi[ t + 1 ].

(10)

The Flows

pt+ 1[ xi ]/dix

pt+ 1[ x2]/ dx2 pt+ 1[ x1]/ dx1

pt+ 1[ i ] pt[ i ]/ di

pt[ i ]/di pt[ i ]/di

pt[ i ]

(11)

Proof of Lemma 102 (continued)

• pt[ i ] = δt[ i ] + 2·| E |di by definition.

• Splitting and giving the 2·| E |di part does not affect pt+1[ i ] because the same 2·| E |1 is exchanged between any two neighbors.

• So we only consider the splitting of the δt[ i ] part.

• The δt[ i ]’s are exchanged between adjacent nodes.

(12)

Proof of Lemma 102 (continued)

• Clearly P

i δt+1[ i ] = P

i δt[ i ] because of conservation.

• But ∆t+1 = P

it+1[ i ]| ≤ P

it[ i ]| = ∆t. – If δt[ i ]’s are all of the same sign, then

t+1 = P

it+1[ i ]| = P

it[ i ]| = ∆t.

– When δt[ i ]’s of opposite signs meet at a node, that will reduce P

it+1[ i ]|.

• We next quantify the decrease ∆t − ∆t+1.

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Proof of Lemma 102 (continued)

• There is a node i+ with δt[ i+ ] ≥ 2·| V |t , and there is a node i with δt[ i ] ≤ −2·| V |t .

– Recall that P

i δt[ i ] = 0 and P

i∈Vt[ i ]| = ∆t. – So the sum of all δt[ i ] ≥ 0 equals ∆t/2.

– As there are at most | V | such δt[ i ], there must be one with magnitude at least (∆t/2)/| V |.

– Similarly for δt[ i ] ≤ 0.

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Proof of Lemma 102 (continued)

• There is a path [ i0 = i+, i1, i2, . . . , i2m = i ] with an even number of edges between i+ and i.

– Add self-loops to make it true.

• The positive deviation δt[ i+ ] from i+ will travel along this path for m steps, always subdivided by the degree of the current node.

• Similarly for the negative deviation δt[ i ] from i.

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Proof of Lemma 102 (continued)

• At least a positive deviation equal to |V |1m of the original amount will arrive at the middle node im.

• Similarly for a negative deviation from the opposite direction.

• So after m ≤ n steps, a positive deviation of at least

t

2·|V |n will cancel an equal amount of negative deviation.

• We do not need to care about cases where numbers of the same sign meet at a node; they will not change ∆t.

(16)

Proof of Lemma 102 (concluded)

• So in n steps the total absolute deviation decreases from

t to at most ∆t(1 − |V |1n).

• But we already knew that ∆t will never increase.a

• So in the limit, ∆t → 0 (but exponentially slow).

aContributed by Mr. Chih-Duo Hong (R95922079) on January 11, 2007.

(17)

First Return Times

• Lemma 102 (p. 783) and theory of Markov chainsa imply that the walk returns to i every 2 · | E |/di steps,

asymptotically and on the average.

• Equivalently, if vt = i, then the expected time until the walk comes back to i for the first time after t is

2 · | E |/di, asymptotically.

– This is called the mean recurrence time.

aParticularly, theory of homogeneous Markov chains on first passage time.

(18)

First Return Times (concluded)

• Although the above is an asymptotic statement, the said expected return time is the same for any t—including the beginning t = 0.

• So from the beginning and onwards, the expected time between two successive visits to node i is exactly

2 · | E |/di.

(19)

Average Time To Reach Target Node n

• Assume there is a path [ 1, i1, . . . , im = n ] from 1 to n.

– If there is none, we are done because the algorithm then returns no false positives.

• Starting from 1, we will return to 1 every expected 2 · | E |/d1 steps.

• Every cycle of leaving and returning uses at least two edges of 1.

– They may be identical.

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Average Time To Reach Target Node n (continued)

• So after an expected d21 of such returns, the walk will head to i1.

– There are d21 pairs of edges incident on node 1 used for the cycles.

– Among them, d1 of them leave node 1 by way of i1 and d1 of them return by way of i1.

• The expected number of steps is d1

2

2 · | E |

d1 = | E |.

(21)

Average Time To Reach Target Node n (concluded)

• Repeat the above argument from i1, i2, . . .

• After an expected number of ≤ n · |E| steps, we will have arrived at node n.

• Markov’s inequality (p. 410) suggests that we run the algorithm for 2n · | E | steps to obtain the desired

probability of success, 0.5.

(22)

Probability To Visit All Nodes

Corollary 103 With probability at least 0.5, the random walk algorithm visits all nodes in 2n · | E | steps.

• Repeat the above arguments for this particular path:

[ 1, 2, . . . , n ].

(23)

The Complete Algorithm

1: x := 1;

2: c := 0;

3: while x 6= n and c < 2n · | E | do

4: Pick y uniformly from x’s neighbors (including x);

5: x := y;

6: c := c + 1;

7: end while

8: if x = n then

9: “yes”;

10: else

11: “no”;

(24)

Some Graph-Theoretic Notions

• A d-regular (undirected) graph has degree d for each node.

• Let G be d-regular.

• Each node’s incident edge is labeled from 1 to d.

– An edge is labeled at both ends.

1 2 3

1 2

3

1 2

3

1 2

3

1 2

3 1

2 3

1 2

3

1 2

3 2

4

3

5

6

7 1

8

(25)

Universal Sequences

a

• A sequence of numbers between 1 and d results in a walk on the graph if given the starting node.

– E.g., (1, 3, 2, 2, 1, 3) from node 1.

• A sequence of numbers between 1 and d is called universal for d-regular graphs with n nodes if:

– For any labeling of any n-node d-regular graph G, and for any starting node, all nodes of G are visited.

– A node may be visited more than once.

• Useful for museum visitors, security guards, etc.

(26)

Existence of Universal Sequences

Theorem 104 For any n, a universal sequence exists for the set of d-regular connected undirected n-node graphs.

• Enumerate all the different labelings of d-regular n-node connected graphs and all starting nodes.

• Call them (G1, v1), (G2, v2), . . . (finitely many).

• S1 is a sequence that traverses G1, starting from v1. – A spanning tree will accomplish this.

• S2 is a sequence that traverses G2, starting from the node at which S1 ends when applied to (G2, v2).

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The Proof (concluded)

• S3 is a sequence that traverses G3, starting from the node at which S1S2 ends when applied to (G3, v3), etc.

• The sequence S ≡ S1S2S3 · · · is universal.

– Suppose S starts from node v of a labeled d-regular n-node graph G.

– Let (G, v) = (Gk, nk), the kth enumerated pair.

– By construction, Sk will traverse G (if not earlier).

(28)

A O(n

3

log n) Bound on Universal Sequences

Theorem 105 For any n and d, a universal sequence of length O(n3 log n) for d-regular n-node connected graphs exists.

• Fix a d-regular labeled n-node graph G.

• A random walk of length 2n · | E | = n2d = O(n2) fails to traverse G with probability at most 1/2.

– By Corollary 103 (p. 797).

– This holds wherever the walk starts.

• The failure probability for G drops to 2−Θ(n log n) if the random walk has length Θ(n3 log n).

(29)

The Proof (continued)

• There are 2O(n log n) d-regular labeled n-node graphs.

– Each node has ≤ nd choices of neighbors.

– So there are ≤ nd+1 d-regular graphs on nodes { 1, 2, . . . , n }.

– Each node’s d edges are labeled with unique integers between 1 and d.

– Hence the count is

≤ nd+1(d!)n = nO(n) = 2O(n log n).

(30)

The Proof (concluded)

• The probability that there exists a d-regular labeled

n-node graph that the random walk fails to traverse can be made at most 1/2.

– Lengthen the length of the walk suitably.

• Because the probability is less than one, there exists a walk that traverses all labeled d-regular graphs.

(31)

Finis

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