Sometimes randomness helps in computation. · 𝐑𝐋Algorithm: Start a random walker at and use...

randomized computation

Sometimes randomness helps in computation.


Augment our usual Turing machines with a read-only tape of random bits.

𝑞0

𝑞1

𝑞start

𝑞accept

𝑞2

𝑎𝑎 ## 𝑏 𝑎

00 10 1 0 1 1


The notion of acceptance changes.

One-sided error [two variants]:

Two-sided error:

𝑥 ∈ 𝐿 ⇒ Pr 𝑀 accepts 𝑥 = 1

𝑥 ∉ 𝐿 ⇒ Pr 𝑀 accepts 𝑥 <9

10

What does it mean for a TM 𝑀 with access to random bits to decide a language 𝐿?

𝑥 ∈ 𝐿 ⇒ Pr 𝑀 accepts 𝑥 >2

3

𝑥 ∉ 𝐿 ⇒ Pr 𝑀 accepts 𝑥 <1

3

𝑥 ∈ 𝐿 ⇒ Pr 𝑀 accepts 𝑥 >1

10

𝑥 ∉ 𝐿 ⇒ Pr 𝑀 accepts 𝑥 = 0

In both cases, we can make the error much smaller by repeating a few times.


For instance, the two-sided error variant of 𝐏 is called 𝐁𝐏𝐏.𝐁𝐏𝐏 stands for “bounded-error probabilistic polynomial time”

Derandomization problem:

Can every efficient randomized algorithm be replaced by a deterministic one?

We don’t know the answer, though it is expected to be “yes.”This is the question of 𝐏 vs. 𝐁𝐏𝐏.


Recall the complexity class 𝐋 = SPACE log 𝑛

The randomized one-sided error variant of 𝐋 is called 𝐑𝐋.

𝑥 ∈ 𝐴 ⇒ Pr 𝑀 accepts 𝑥 ≥1

2

𝑥 ∉ 𝐴 ⇒ Pr 𝑀 accepts 𝑥 = 0

A language 𝐴 is in 𝐑𝐋 if there is a randomized 𝑂 log𝑛 space Turing machine such that [one way access to random tape]:

The question of whether 𝐋 = 𝐑𝐋 is also open.

We know that 𝐑𝐋 ⊆ 𝐍𝐋 ⊆ SPACE log 𝑛 2

Nisan showed how to do this using a pseudo-random number generator for space-bounded machines.

undirected connectivity

Recall the language:DIRPATH= 𝐺, 𝑠, 𝑡 ∶ 𝐺 is a directed graph with a directed 𝑠 → 𝑡 path

We saw that DIRPATH is 𝐍𝐋-complete.

Thus 𝐋 = 𝐍𝐋 if and only if DIRPATH ∈ 𝑳.

What about the language?PATH = 𝐺, 𝑠, 𝑡 ∶ 𝐺 is an 𝐮𝐧𝐝𝐢𝐫𝐞𝐜𝐭𝐞𝐝 graph with an 𝑠 ↔ 𝑡 path

undirected connectivity

What about the language?PATH = 𝐺, 𝑠, 𝑡 ∶ 𝐺 is an 𝐮𝐧𝐝𝐢𝐫𝐞𝐜𝐭𝐞𝐝 graph with an 𝑠 ↔ 𝑡 path

Is PATH ∈ 𝐋?

random walks

Let’s show that PATH ∈ 𝐑𝐋.

One step of random walk: Move to a uniformly random neighbor.

random walks

What we will show: If 𝐺 = (𝑉, 𝐸) is a graph with 𝑛 vertices and 𝑚 edges, then a random walker started at 𝑠 ∈ 𝑉 who takes 4𝑚𝑛 steps will visit everyvertex in the connected component of 𝑠 with probability at least ½.In particular, if there is an 𝑠 ↔ 𝑡 path in 𝐺, she will visit 𝑡 with probability at least ½.

random walks

What we will show: If 𝐺 = (𝑉, 𝐸) is a graph with 𝑛 vertices and 𝑚 edges, then a random walker started at 𝑠 ∈ 𝑉 who takes 4𝑚𝑛 steps will visit everyvertex in the connected component of 𝑠 with probability at least ½.In particular, if there is an 𝑠 ↔ 𝑡 path in 𝐺, she will visit 𝑡 with probability at least ½.

𝐑𝐋 Algorithm: Start a random walker at 𝑠 and use the random bits to simulate a random walk for 4𝑚𝑛 steps (can count this high in 𝑂 log𝑛space). If we visit 𝑡 at any point, accept. Otherwise, reject.

If there is an 𝑠 ↔ 𝑡 path in 𝐺: Pr accept 𝐺, 𝑠, 𝑡 ≥ 12

If no 𝑠 ↔ 𝑡 path in 𝐺: Pr accept 𝐺, 𝑠, 𝑡 = 0

cover time of a graph

Cover time of a graph = expected # of steps before all vertices are visited.

cover time of a graph

We will show that if 𝐺 has 𝑛 vertices and 𝑚 then the expected time before the random walk covers 𝐺 (from any starting point) is 2𝑚𝑛.

Then it must be that with probability at least ½, we cover 𝐺 after 4𝑚𝑛 steps.(This fact is called “Markov’s inequality.”)

heat dispersion on a graph

spectral embedding

𝑣2

hitting times and commute times

For vertices 𝑢, 𝑣 ∈ 𝑉, the hitting time is𝐻 𝑢, 𝑣 = expected time for randomwalk to hit 𝑣 starting at 𝑢

𝑢

𝑣

The commute time between 𝑢 and 𝑣 is𝐶 𝑢, 𝑣 = 𝐻 𝑢, 𝑣 + 𝐻(𝑣, 𝑢)


For vertices 𝑢, 𝑣 ∈ 𝑉, the hitting time is𝐻 𝑢, 𝑣 = expected time for randomwalk to hit 𝑣 starting at 𝑢

𝑢𝑣

The commute time between 𝑢 and 𝑣 is𝐶 𝑢, 𝑣 = 𝐻 𝑢, 𝑣 + 𝐻(𝑣, 𝑢)

Claim: If {𝑢, 𝑣} is an edge of 𝐺, then 𝐶 𝑢, 𝑣 ≤ 2𝑚.


𝑠


Let’s use the claim to show that the cover time of 𝐺 is at most 2𝑚(𝑛 − 1).


𝑠



Choose an arbitrary spanning tree of 𝐺.


𝑠




Consider the “tour around the spanning tree”

This is a path 𝑠 = 𝑢0, 𝑢1, 𝑢2, … , 𝑢2𝑛 = 𝑠Where 𝑢𝑖 , 𝑢𝑖+1 is an edge for each 𝑖, and each edge appears exactly twice.


𝑠





Claim: Cover time of 𝐺 is at most𝐶 𝑢0, 𝑢1 + 𝐶 𝑢1, 𝑢2 + 𝐶 𝑢2, 𝑢3 +⋯+ 𝐶 𝑢2𝑛−1, 𝑢2𝑛


𝑠





Claim: Cover time of 𝐺 is at most

𝑢,𝑣 ∈𝑇

𝐶(𝑢, 𝑣) ≤ 2𝑚(𝑛 − 1)

high school physics


high school physics


View the graph 𝐺 as an electrical network with unit resistances on the edges.

When a potential difference is applied between the nodes from an external source, current flows in the network in accordance with Kirchoff’s and Ohm’s laws.

K1: The total current flowing into a vertex = total current flowing outK2: The sum of potential differences around any cycle is zero.Ohm: The current flowing along any edge {𝑢, 𝑣} is equal to

(potential 𝑢 − potential 𝑣 )/resistance(𝑢, 𝑣)

The effective resistance between 𝑢 and 𝑣 is defined as the potential difference required to send one unit of current from 𝑢 to 𝑣.

𝑢 𝑣

presentation agenda

Error-correcting codes:

Simple uses of randomness in computation:

1. Intro [Amr]2. Reed-Solmon codes [Jie]3. Expander codes [Ali]

1. Checking matrix multiplication [Mahmoud]2. Karp-Rabin matching algorithm [Stanislav]

random

pseudo-random

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