reachability Is NL-Complete

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Logarithmic Space

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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.

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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

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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).

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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

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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 ].

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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 ].

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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 ]

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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.

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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.

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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.

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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.

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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.

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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 |.

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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.

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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 ].

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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”;

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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

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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.

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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).

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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).

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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).

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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.

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Finis

Figure

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