# Random Walk on Graphs and its Algorithmic Shengyu Random Walk on Graphs and its Algorithmic Shengyu Zhang Applications Winter School, [email protected], 2009 Random walk on graphs On an undirected graph G: Starting from vertex v 0 Repeat for a number of steps: Go to a random neighbor. Simple but powerful. Road map: Random walk

Parameters /Properties Algorithms Hitting time k-SAT st-connectivity Mixing time PageRank Approximate counting Error-reduction Road map: Quantum walk Short intro to math model of quantum mechanics Types Algorithms

Discrete QW Element Distinctness Continuous QW Formula Evaluation PART I. Random Walk Key parameter 1: Hitting time Hitting time Recall the process of random walk on a graph G. Starting vertex v0 Repeat for a number of steps:

j Go to a random neighbor. i Hitting time: H(i,j) = expected time to visit j (for the first time), starting at i Undirected graphs Complete graph Line:

H(i,j) = n-1 (ij) 1 2 n-1 H(i,j) = j2-i2 (i (n/2)-complete graph Algorithm 1: 2-SAT k-SAT: satisfiability of k-CNF formula n variables x1, , xn{0,1} m clauses, each being OR of k literals

Literal: xi or xi e.g. (k=3): (x1) x5 x7 3-CNF formula: AND of these m clauses e.g. ((x1) x5 x7) (x2 (x5) (x7)) (x1 x7 x8) 3-SAT Problem: Given a 3CNF formula, decide whether there is an assignment of variables s.t. the formula evaluates to 1. For the above example, Yes: x5=1, x7=0, x1=1 P vs. NP P: problems that can be easily solved

easily: in polynomial time NP: problems that can be easily verified. Formally: a polynomial time verifier V, s.t. for any input x, If the answer is YES, then y s.t. V(x,y) = 1 If the answer is NO, then y, V(x,y) = 1 The question of TCS: Is P = NP? Intuitively, no. NP should be much larger.

Its much easier to verify (with help) than to solve (by yourself) mathematical proof, appreciation of good music/food, Formal proof? We dont know yet. One of the 7 Millennium Problems by CMI. http://www.claymath.org/millennium/P_vs_NP/ NP-completeness k-SAT is NP-complete, for any k 3. NP-complete:

NP-complete problems are the hardest ones in NP. 3-SAT is in NP: In NP All other problems in NP can be reduced to it in poly. time witness --- satisfying assignment A Verification: evaluate formula with variables assigned by A [Cook-Levin] 3-SAT is NP-complete. How about 2-SAT? While 3-SAT is the hardest in NP, 2-SAT is solvable in polynomial time.

Here we present a very simple randomized algorithm, which has polynomial expected running time. Algorithm for 2-SAT 2SAT: each clause has two variables/negations (x1xx2)(x(x2xx3) (x(x4xx3) (x(x5xx1) Alg [Papadimitriou]: Pick any assignment Repeat O(n2) time

If all satisfied, done Else x1, x2, x3, x4, x5 0, 1, 0, 1, 0 1 Pick any unsatisfied clause Pick one of the two literals each with probability, and flip the assignment on that variable Analysis (x1xx2)(x(x2xx3) (x(x4xx3) (x(x5xx1) x1, x2, x3, x4, x5

0, 1, 0, 1, 0 If unsatisfiable: never find an satisfying assignment If satisfiable, there exists a satisfying assignment x If our initially picked assignment x is satisfying, then done. Otherwise, for any unsatisfied clause, at least one of the two variables is assigned a value different than that in x Randomly picking one of the two variables and flipping its value increases # correct assignments by 1 w.p. Analysis (continued) Consider a line of n+1 points, where k represents weve assigned k variables correctly

Last slide: Randomly picking one of the two variables and flipping its value increases # correct assignments by 1 w.p. Thus the algorithm is actually a random walk on the line of n+1 points, with Pr[going right] . correctly: the same way as x Hitting time (i n): O(n2) So by repeating this flipping process O(n2) steps, well reach x with high probability. Algorithm 2: stconnectivity

st-Connectivity Problem: Given a graph G and two vertices s and t on it, decide whether they are connected. BFS/DFS (starting at s) solves the problem in linear time. But uses linear space as well. Question: Can we use much less space? Simple random walk algorithm A randomized algorithm takes O(log n)

space. Starting at s, perform the random walk for O(n3) steps. If ever see t, output YES and stop. output NO. Why it works? H(s,t) = O(n3). If s can reach t, then we should see it within O(n3) time. Key parameter 2: Mixing time Convergence

Now lets study the probability distribution of v the particle after a long time. u w What do we observe? The distribution converges to uniform wherever it starts. Q: Does this hold in general? t 0

1 0 1 1/2 0 1/2 2 1/4 2/4 1/4 3 3/8

2/8 3/8 4 5/16 6/16 5/16 5 11/32 10/32 11/32 6

21/64 22/64 21/64 Convergence v1 Well, Yes and No. Consider the following cases on undirected graphs. v2

v3 v4 v5 v1 v2 3 4 5 6 Case 1: The graph is unconnected. v v v v

Case 2: The graph is bipartite. [Thm] For any connected non-bipartite graph, and any starting point, the random walk converges. Converges to what? v u w In the previous triangle example: uniform. In general? As a result of the convergence, the distribution doesnt change by the matrix If the particle is on the graph according to the

distribution, then further random walk will result in the same distribution. We call it the stationary distribution. [Fact] Its unique. stationary distribution [Fact] Its the following distribution: (v) = d(v)/2m d(v) = degree of v, i.e. # of neighbors. m: |E|, i.e. # of edges.

[Proof] Consider one step of walk: (v) = u: (u,v)E p(u)[1/d(u)] = u: (u,v)E [d(u)/2m][1/d(u)] = u: (u,v)E 1/2m = d(v)/2m = (v) So is the stationary distribution. For regular graphs, is the uniform distribution. Proportional to degree Note: the stationary distribution (v) = d(v)/2m is proportional to the degree of v. Whats the intuition? The more neighbors you have, the more chance youll be reached. We will see another natural interpretation

shortly. Speed of the convergence Weve seen that random walk converges to the stationary distribution. Next question: how fast is the convergence? Lets define the mixing time as min {T: pi(T) - } where pi(T): the distribution in time T, starting at i. : some norm. Well see the reason of mixing later. Now lets first see an example you run into everyday. Algorithm 3: PageRank

PageRank Google gives each webpage a number for its importance. [IT] Google: 10, Microsoft: 9, Apple: 9 [Media] NYTimes: 9, CNN: 10, sohu: 8, newsmth: 7 [Sports] NBA: 7, CBA: 7, CFA: 7 [University] MIT: 9, CUHK: 8, Tsinghua, Pku, Fudan, IIT(B): 9 When you search for something by making a query, a large number of related webpages are retrieved. What webpages to retrieve? Information Retrieval. Thats

an orthogonal issue. Ranking How to give this big corpus to you? Search engines rank them based on the importance, and give them in descending order. Thus presumably the first page contains the 10 most important webpages related to your query. Question: How to rank? PageRank: Use the vast link structure as in indicator of an individual pages value. Reference system

Webpage A has a link to webpage B A thinks B is useful. So a webpage with a lot of other pages pointing to it is probably important. A guy getting a lot of letters is strong. Further, pointers from pages that are themselves important bear more weight. Think of it as A writing a recommendation letter for B.

Letters by Noga, Lszl, Sasha, Avi, Andy, mean a lot. But the importance of those pages also need to be calculated we have a recurrence equation. Furthermore If page A has a lot of links, then each link means less. Ok, you get Einsteins letter, but you know what, last year everyone on the market got his letter. So lets assume that page As reference weight is divided evenly to all pages B that A links to.

Recurrence equation: R(A) = B: BA R(B)/d(B) where d(B) = # pages C that B links to Sink issue R(A) = B: BA R(B)/d(B) has a problem. There are some sink pages that contain no links to other pages. Sinks accumulate weights without giving out. The recurrence equation only has solutions with weight on sinks, losing the original intension of indicating the importance of all pages. To handle this, we modify the recursion: R(A) = (1-)/N + B: BA R(B)/d(B)

Force each page to have a (1- )-fraction of weights (evenly) going to all pages. So each A also receives a weight of (1- )/N from all pages : around 0.85 Random Walk view Note that the recursion R(A) = (1-)/N + B: BA R(B)/d(B) is exactly the random walk on the graph, where at each point A, we w/ prob. , follow a random link;

w/ prob. (1-), go to a random page. Question: How to solve this recurrence equation? # webpages: ~ 30 billion (and counting) Algorithm Recall: the random walk converges to the stationary distribution! Algorithm: start from any distribution, run a few iterations of random walk, and output the result. Its a bit different since its random walk on directed graphs,

but this PageRank matrix has all good properties we need so that the random walk also converges to the solution. Google: 50-100 iterations, need a few days. That should be close to the stationary distribution, which serves as indictor of the importance of pages. Next Well see a bit math behind the mixing story. Mathematics behind the mixing [Eigenvalue decomposition] A symmetric matrix MNN can be written as M = i=1,,n iiiT

The eigenvectors are orthonormal. i: eigenvalue. Order them: |1| |2| |3| |n| i: (column) eigenvector, i.e. Mi = ii i2=1 i, j = 0, for all ij M2 = (i=1,,n iiiT)(j=1,,n jjjT) = i,j iiiT jjjT = i,j i j i,i jjT = i i2 jiT Mt = i it jiT Random walk in matrix form

A: adjacency matrix. (Aij = 1 if (i,j) E; 0 o.w.) P: probability transition matrix. N = D-1/2AD-1/2 = D1/2PD-1/2 Pij = 1/di if (i,j) E; 0 o.w. P = D-1A, where D = diag(d1, , dn). (di: deg(i)) For any distribution p, PTp is the distribution after one step of random walk Nij = 1/(didj)1/2 if (i,j) E; 0 o.w. N is symmetric, so N can be written as N = i=1,,n iiiT Random Walk for t steps:

Pt = (D-1/2ND1/2)t = D-1/2NtD1/2 = D-1/2 (iitiiT)D1/2 = iitD-1/2iiTD1/2 Why mixing? Why to ? [Thm] For connected non-bipartite graph, N has 1 = 1 and 1 = 1/2 = ((d1/2m)1/2, , (dn/2m)1/2) is the square root version of the stationary distribution (so that the l2 norm is 1). |2| < 1. (So is all other is.) Random Walk for t steps: Pt = i itD-1/2iiTD1/2 First item: 1tD-1/211TD1/2 = D-1/2(1/2m)[(didj)1/2]ijD1/2= (1/2m)[dj]ij For any starting distribution p, (1/2m)[dj]ijT p = . Speed of the convergence All other items: since |i|<1, itD-1/2iiTD1/20! Speed of convergence depends on how close |2| is to 1. PageRank matrix: |2| = ( 0.85) Expander: 1-|2| = (1) The rest of the talk Only ideas are given. Many details are omitted. I may cheat a bit to illustrate the main steps. Algorithm 4: Approximately counting Algorithm 4: Approximately counting Task: estimate the size of an exponentially large set V

Approach: find a chain V0VV1VVVm=V |V0| is easy to compute; m = poly(n) layers; Each ratio |Vi+1|/|Vi| = poly(n) is easy to estimate. Example: Given G, count # of perfect matchings. Then |V| = |V0| (|V1|/|V0|) (|V2|/|V1|) (|Vn|/|Vn-1|) can

be estimated. Question: How to estimate |Vi+1|/|Vi|? Estimate the ratio Estimate by random sampling: Generate a random element uniformly distributed in Vi+1, see how often it hits Vi. Question: How to uniformly sample Vi+1? By random walk

We construct a regular graph with vertex set Vi+1 The algorithm runs efficiently (i.e. in poly(n) time) if the walk converges to uniform rapidly (i.e. in poly(n) time). Its the case in perfect matching counting problem. Algorithm 5: Error Reduction with efficient randomness Error reduction Task: Reduce the error of a randomized algorithm A. Error: Prr{0,1}^{0,1}^m[r is bad for A] = 1/3

? Nave approach: Draw k random strings r1, , rk, Run algorithm A using r1, , rk and get k answers Output the majority of the answers B {0,1}m [Fact] the error drops to 2-(k) . --- Chernoffs bound Randomness complexity: mk. Question: Can we reduce the error prob w/o using too many additional random bits?

Expander graph Expander: Eigenvalue gap is large --- (1) Random walk converges very fast --- O(log n). Highly connected --- diameter O(log n) Large boundary --- any subset has lots of edges going out Could be sparse --- constant degree expanders. We know how to construct them explicitly. Algorithm for error reduction

New Algorithm A: B {0,1}m Perform a random walk (r1, r2,, rk) of length k Random algorithm A using r1, r2,, rk and get k answers Output the majority of the answers

Thm: the error prob of A is also 2-(k) How many random bits used? m + O(k) Construct an expander with V = {0,1}m Start from a random vertex r1 The expander is of constant degree. Highly dependent! Why it works? Expander: large boundary For simplicity, consider the one-side error case:

Algorithm A wrong Whole walk in B. Expander: Every set has a large boundary. At every step, it walks outside B w.p. (1). Pr[k steps never out] = 2-(k) Non-expander {0,1}m: all random strings B Boundary PART II. Quantum Walk

Quantum mechanics in one slide |1 Physics Math |0+|1 Physical System Unit Vector (||2+||2=1) ,: amplitudes Evolution Unitary Matrix

Measurement Projection Composition Tensor Product 1 Classical: 0 |1 |1 Quantum: |0 A classical bit A quantum bit (qubit) Measure by |0 and |1: - get

0 w.p. ||2; system |0; State space for 2 bits: 2 - get 1{00, w.p. ||10, ; system |1. all combinations 01, 11} 1 0 |0 |0 State space for 2 qubits: the space span{|00,|01,|10,|11}

Quantum walk Many things become quite tricky. Even the definition of quantum walk. Well ignore the formal definitions here, but only present some results. Type 1: Discrete Quantum Walk Coin register The most natural quantum counterpart of random

walk is the following: each vertex v gives its neighbors (equally) some mass so that the l2 norm of the total mass is still 1. However, simply accumulating all mass from neighbors does not work The operator is not unitary. To overcome this, we need a register to remember where the mass is from. It amounts to add a coin register. 1 1 1 p 2

Flip the quantum coin: |v|cc |vH|cc, where H = Walk accordingly: |v|cc |nc(v)|cc, where nc(v) is the cth neighbor of v. 1 1 General graph Given any random walk on an undirected graph G, Szegedy proposed a method to transform it into a quantum walk Set hitting time: Classical set hitting time = O(()-1) : density of the target set, i.e. |T|/|V|

A a target set TV V is hidden in the graph Starting from the uniform distribution, whats the expected time to hit some point in T? : eigenvalue gap of matrix P [Szegedy] Quantum set hitting time = O(Classical set hitting time) = O(()-1/2) V T Quantum algorithm for Element Distinctness Element Distinctness: To decide whether all input

integers are distinct. [Ambainis] Discrete quantum walk on the following graph: (r = n2/3) V = {S[n]: |S| = r}; E = {(S,T): |ST| = r-2} Eigenvalue gap: = 1/r Target set density: = r2/n2 Quantum running time = setup time + hitting time = O(r) + O(()-1/2) = O(r + n/r) = O(n2/3). Type 2: Continuous Quantum Walk Continuous quantum walk on a graph

Given an undirected graph G, and a Hermitian matrix H that respects the graph structure Hermitian: (HT)* = H. Respect G: Hij = 0 if (i,j)E. The dynamics (t) = j i(t)|i for H is governed by the Schrodinger equation i(d/dt) i(t) = j:(i,j)EHijj(t), which has solution |c(t) = eiHt |c(0). Formula Evaluation Grover Search: find a marked point in an n-point set in time O(n1/2).

Equivalently: evaluate OR in time O(n1/2). Generalizations: (x x (x x x (Balanced) AND-OR Tree (x General AND-OR-NOT Formula

(General Game Tree) Formula evaluation Motivation: a natural generalization of OR --- Grover search well studied subject in TCS matching lower bound known long ago game: min-max tree. Same up to log(n) max min min

Breakthroughs [FGG, 07] An O(n1/2) algorithm for Balanced ANDOR Tree. Method: scattering theory Not really understandable for computer scientists [CRSZ, A, FOCS07] An O(n1/2+o(1)) algorithm for General ANDORNOT Formula. Method: phase estimation + quantum walk. Simpler algorithm, simpler proof. For special cases like Balanced ANDOR Tree: O(n1/2)

Classical implications [OS, STOC03] Any formula f of size n has polynomial threshold function thr(f) = O(n1/2)? Our result: ~ Q( f ) thrn(1/2+o(1) f ) deg( f ) [KlivansServedio, KOS, FOCS02] DefinitionSTOC01; of thr [BBCMdW, Class C of Boolean functions has thr(f) r for all fC, C can JACM01]

be learned in time nO(r) (in both PAC or adversarially generated examples) n1/2+o(1) [Implication] Formulas are learnable in time 2 This is very interesting because studies of quantum algorithms solve a purely classical open problem! This is not uncommon in quantum computing! Sketch of the algorithm {AND,OR,NOT}{NAND} For the formula f and an input x, construct a graph Gx x1:0 x2

:1 x3:0 x4:1 x5 :1 Let A be the adjacency matrix of Gx; consider As spectrum. Key observation Red: function value at a vertex green: amplitude of 0-eigenvector Function values and amplitudes of 0eigenvector propagate in the same way. 0 1 (x 0 0

(x 0 1 1 0 (x 0 0 If a function evaluates to 0, then an 0-eigenvector (x (x =0 (x 1 =0 1 =0 If a function evaluates to 1,

then 0-eigenvector has no support on the root. Phase Estimation Thus qualitatively, its enough to know whether there is a 0-eigenvector with support on root. Phase estimation: Given an unitary operator U (to use) and an eigenvector |, find the phase in the corresponding eigenvalue ei. Unitary matrix: all eigenvalues have l2-norm 1 For eiHt, the phase is nothing but the corresponding eigenvalue of H.

The quantum walk is here! Thus use phase estimation, we can find an eigenvalue of H. Depending on whether its 0, output the answer to f(x). A highlevel structure of algorithm We need to carefully choose weight hij on edges (i,j) s.t. the above qualitative connection also works quantitatively. For f(x) = 0, For f(x) = 1,

an 0-eigenvector with (1) support on root. eigenvector with support on root has eigenvalue (n-1/2) Algorithm: Starting from the root, use quantum walk to carry out the Phase Estimation, output the answer depending on whether the phase is 0. Summary Random walk is a simple but powerful tool. There will sure be more important algorithmic applications to be found.

Theories of quantum walk have been rapidly developed in the past couple of years. A lot of fundamental issues yet to be better understood. Significant breakthrough algorithms using quantum walk are ahead! Reference For random walk on graphs: Lovsz, Random Walks on Graphs: A Survey, Combinatorics, 353-398, 1996. Chung, Spectral Graph Theory, American Mathematical Society, 1997. For quantum walk:

Quantum computing in general: Textbook by Nielsen and Chuang Lecture notes by Vazirani, by Ambainis, and by Childs Quantum walk: Surveys by Kempe and by Santha. All above lectures have nice chapters for quantum walk. Thanks!