SL = L. (Solving Big Mazes Using Small Space) Undirected ST Connectivity in Log-Space Reingold ECCC-TR04-94, Nov 11,2004
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1 SL = L (Solving Big Mazes Using Small Space) Undirected ST Connectivity in Log-Space Reingold ECCC-TR04-94, Nov 11,2004 Entropy Waves, the Zig-Zag Products and New Constant Degree Expanders Reingold-Vadhan-Wigderson AM, 155(1), 2001
2 Maze Problems and Graphs Theseus Ariadne Crete 1000 B.C
3 Graph Connectivity Problem : Given a graph G = (V, E) and vertices s and t. Is t reachable from s? Representation : 1. Adjacency matrix, A[u, v] = 1 if (u, v) E. 2. Rotation maps, labels all ends of edges. Rot G (u, i) = (v, j) if i th outgoing edge of u is the j th incoming edge of v. From Adjacency matrix to rotation map in log-space.
4 LOGSPACE Arena NL - Non-deterministic TMs, space O(log n) Complete : ST-connectivity in directed graphs. L - Deterministic TMs, space O(log n) Complete : ST-connectivity In directed outdegree 1 graphs. SL - Non-deterministic Symmetric TMs, space O(log n) Complete : ST-connectivity in undirected graphs. RLP : Onesided error Probabilistic TMs, time poly, space O(log n). ZPLP : Zero-error Probablistic TMs, time poly, space O(log n). Notation : L k for DSPACE(log k n). L SL ZPL RL NL L 2
5 It started way back... NL L 2 [S 1970] SL RLP [AKLLR 1979] NL = Co-NL [IS 1988] SL ZPLP [BCDRT 1989]
6 It started way back... NL L 2 [S 1970] SL RLP [AKLLR 1979] NL = Co-NL [IS 1988] SL ZPLP [BCDRT 1989] SL L 2 but poly time. [N 1990] SL L 3 2 [NSW 1992] SL = Co-SL [NT 1995] RL L 3 2 [SZ 1995] SL L 4 3 [ATWZ 1997] SL = L [R 2004]
7 SL is almost L Universal Traversal Sequences (UTS) : A sequence S of vertices is n-universal if for every graph on G on n vertices, and for every vertex v of G, WALK(G, v, S) visits all vertices in v s connected component. 1. A random walk of length n 4 visits all vertices of a connected n-vertex graph.[akllr 79] Most sequences of length n 6 are n-universal. 2. There is a pseudo-random generator for RL which uses only O(log 2 n) random bits and space.[nis 89] Computing UTS is in L Nisan s PRG can be utilised for SL L 1.5 [NSW 92]. 4. SL=L except on rare inputs. [using space efficient extractors].
8 Highly connected graphs Expanders... Are highly connected but sparse graphs. Every cut has many edges crossing it. Every small set of vertices have large neighbourhood. Are almost random graphs. Random walks quickly converges to uniform distribution. Formally, An undirected d-regular graph G = (V, E) is a c-expander for 1 c 0, if S V such that S V 2 S Γ(S) (1 + c) S
9 Algebraic properties of Adjacency matrix Let A be the normalized adjacency matrix of a d-regular undirected graph G on n vertices and m edges. λ 0 λ 1... λ k be the eigen values of A. λ 0 λ 1 = g is the Spectral Gap of G. λ 0 = 1. G is connected λ 0 < λ 1 g 0. G is bipartite λ i = λ k i. G is non-bipartite and connected = spectral gap 1 dn 2 [AS 2000]
10 Intuitions formalised.. For a d-regular graphs G = (V, E), g = 1 λ. G is a g 2 -expander. [A 1986] For any partition of vertices into B, C, E(B, C) d.g n B C For initial distribution p, A t. p u nλ t. Expander Mixing Lemma: For any B, C V, E(B, C) d B C λ B C n Notation : (n, d, λ)-graph - d-regular graph on n vertices with second largest eigen value as λ.
11 Intuitions formalised.. For a d-regular graphs G = (V, E), g = 1 λ. G is a g 2 -expander. [A 1986] For any partition of vertices into B, C, E(B, C) d.g n B C For initial distribution p, A t. p u nλ t. Expander Mixing Lemma: For any B, C V, E(B, C) d B C λ B C n Constructing Constant Degree Expanders M 73, GG 81, AM 85, AGM 87, JM 87, LPS 88, M 88, M 94, RVW 2001, RSW 2004
12 USTCONN L for Expanders Claim 1: Expander graphs with constant expansion coefficient have logarithmic diameter. Claim 2: USTCONN L for constant degree expanders with constant expansion coefficient. Given a c-expander graph G on N vertices. Starting in vertex s, using log N space enumerate all k log n length paths. Check whether each of them hits t or not. Then ACCEPT else REJECT.
13 USTCONN L for Expanders Claim 1: Expander graphs with constant expansion coefficient have logarithmic diameter. Claim 2: USTCONN L for constant degree expanders with constant expansion coefficient. Given a c-expander graph G on N vertices. Starting in vertex s, using log N space enumerate all k log n length paths. Check whether each of them hits t or not. Then ACCEPT else REJECT. Reigngold gives a logspace many-one reduction from USTCONN in arbitrary graphs into USTCONN on constant degree expanders. Intuition : Improve the spectral gap. Watch outs : Correctness, Size of the graph, Space used by the reduction
14 Graph Powering : To Improve Spectral Gap Given d-regular graph G, the product graph G k is a d k -regular. Rot G k (v 0, (a 1, a 2... a t )) = (v t, (b t, b t 1... b 1 )) where, Rot G (v i 1, a i ) = (v i, b i ) Claim : G k is a an (n, d k, λ k )-graph.
15 Graph Powering : To Improve Spectral Gap Given d-regular graph G, the product graph G k is a d k -regular. Rot G k (v 0, (a 1, a 2... a t )) = (v t, (b t, b t 1... b 1 )) where, Rot G (v i 1, a i ) = (v i, b i ) Claim : G k is a an (n, d k, λ k )-graph. Mind Block : Our λ to start with is depending on n in arbitrary way. Degree is boosted too much.
16 Graph Powering : To Improve Spectral Gap Given d-regular graph G, the product graph G k is a d k -regular. Rot G k (v 0, (a 1, a 2... a t )) = (v t, (b t, b t 1... b 1 )) where, Rot G (v i 1, a i ) = (v i, b i ) Claim : G k is a an (n, d k, λ k )-graph. Mind Block : Our λ to start with is depending on n in arbitrary way. Degree is boosted too much. RVW s Solution: Degree reduction without affecting the expansion properties.
17 From Arbitrary Graphs to Expanders Input Graph G Replacement Product Non-bipartite Graph G D-regular Spectral Gap 1 Dn 2 Expander on D nodes Iterated Zig-Zag Product and Powering Expander Graph G D-regular Spectral Gap 1 2
18 Cycle Replacement Products Given a graph G = (V, E), replace each vertex u by an n-cycle C u. For an edge (u, v) E, connect copy of v in C u and u in C v. Make it D-regular by selfloops. Observations The rotation maps can be computed in log-space. Reachability is preserved. Spectral Gap 1 Dn 2.
19 Zig-Zag Products - The Intuition Random Walks
20 Zig-Zag Products - The Intuition Random Walks
21 Zig-Zag Products - The Intuition Random Walks H
22 Zig-Zag Products - The Intuition Random Walks u i cloud v (i,j) cloud v[k] k j k
23 Zig-Zag Products - Formal Definition G is a (N, D, λ) graph. H is a (D, d, α) graph. ZIGZAG(G, H) is a d 2 -regular graph on ND vertices. (i, j) th edge of (v, u) connects to (v[u[i]], u[i][j]) if 1. (v, u) (v, u[i]) 2. (v, u[i]) (v[u[i]], u[i]) 3. (v[u[i]], u[i]) (v[u[i]], u[i][j]). Claim : If ZIGZAG(G, H) is an (ND, d 2, f (λ, α)) where f (λ, α) (1 α2 )(1 λ).
24 Zig-Zag products - Why does it work? u i cloud v (i,j) cloud v[k] k j k Case 1: cloud is not uniform
25 Zig-Zag products - Why does it work? u i cloud v (i,j) cloud v[k] k j k Case 1: cloud is not uniform
26 Zig-Zag products - Why does it work? u i cloud v (i,j) cloud v[k] k j k Case 1: cloud is not uniform
27 Zig-Zag products - Why does it work? u i cloud v (i,j) cloud v[k] k j k Case 1: cloud is not uniform
28 Zig-Zag products - Why does it work? u i cloud v (i,j) cloud v[k] k j k Case 2: cloud is uniform
29 Zig-Zag products - Why does it work? u i cloud v (i,j) cloud v[k] k j k Case 2: cloud is uniform
30 Zig-Zag products - Why does it work? u i cloud v (i,j) cloud v[k] k j k Case 2: cloud is uniform
31 Zig-Zag products - Why does it work? u i cloud v (i,j) cloud v[k] k j k Case 2: cloud is uniform
32 From Non-bipartite to an Expander Input : (N, D, λ) graph G and (D, d, α) graph H, given by their rotation maps. Output : Rotation map of the graph G r. Choose r such that (1 1 DN 2 ) 2r 1 2. G 0 = G Output G r. G i = [ZIGZAG(G i 1, H)] 8 Claim 1: G i is a (N.D i, d 16, λ i ) graph. Claim 2: α 1 2 λ r 1 2.
33 In Log Space? Given vertex v N D i and edge label a i. For any vertex w, the address is a tuple w = (w, a 0,... a i 1 ) where a j D - vertex in the graph replaced at the j th level w vertex in the original graph from which v was generated. Each a i is a 16-tuple b i 1 bi 2... bi 16.
34 In Log Space? Given vertex v N D i and edge label a i. For any vertex w, the address is a tuple w = (w, a 0,... a i 1 ) where a j D - vertex in the graph replaced at the j th level w vertex in the original graph from which v was generated. Each a i is a 16-tuple b i 1 bi 2... bi 16. For j = 1 to 16, (a i 1, b i j ) = Rot H(a i 1, b i j ) For odd j simulate jump into the next cloud. (v, a 0... a i 1 ) = Rot Gi 1 ((v, a 0... a i 2 ), a i 2 ). Reverse a i.
35 Final Algorithm Input : Undirected graph G = (V, E) with V = n and two vertices s and t. Let d be a constant. Convert the graph into a non-bipartite one with a cycle replacement product. Exhaustively search for a (d 16, d, λ) graph H, with λ 1 2. Choose constant r such that (1 1 d 16 ) 2r 1 2. Invoke the reachability algorithm on expander graph on the graph V [d 16 ] r, and vertices (s, 1 r ) and (t, 1 r ). Rotation maps given by the logspace algorithm given previously.
36 SL = L... Martyrs? Nonbipartiteness checking, Comparing no. of connected components. 2-SAT where each clause has only parity gates. More on ECCC-TR Recent ones, Planarity Testing is hard for L, and lies in SL [AM 2000]. Colored Graph Isomorphism Problem for color multiplicities 2 and 3 are complete for SL [KT 2002]. Some relaxed labelling constraints for construction of UTSs.
37 What Next?
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