# Homework 3

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### Tom Hayes

May 7, 2010, 4:44:29 AM5/7/10
to CS 510 Randomized Algorithms
Hi class,

Sorry for the delay. Here is the full list of homework questions.
These are challenging, so I encourage you to work together, and
consult me if you get stuck. You can also post discussion in this
google group, which I will be checking regularly. I need your
solution
sets by Thursday midnight at the latest.

-Tom

There are 3 problems (in random order):

1) As described in class, do the exploratory assignment in Section 5.8
of the textbook. Moreover, although the book doesn't say to do so, I
want you to try to prove the tightest upper AND lower bounds you can
for each of the 3 expectations.

2) Recall from class the Metropolis Glauber dynamics on independent
sets:
Let lambda < 1 be given, and a graph G = (V,E).
Given a set X_{t-1}, choose a random vertex v_t, and
(a) if v_t is in X_{t-1}, then X_t := X_{t-1} minus {v_t}
(b) if v_t is not in X_{t-1} and no neighbor of v_t is in X_{t-1},
then
with probability lambda, X_t := X_{t-1} union {v_t}.
(c) Otherwise, X_t := X_{t-1}.

We proved in class that, when lambda < 1/Delta, where Delta is the
maximum degree of G, the mixing time is O(n log(n)/epsilon).

"slide" rule, thus:
Let lambda < 1 be given, and a graph G = (V,E).
Given a set X_{t-1}, choose a random vertex v_t, and
(a) if v_t is in X_{t-1}, then X_t := X_{t-1} minus {v_t}
(b) if v_t is not in X_{t-1} and no neighbor of v_t is in X_{t-1},
then
with probability lambda, X_t := X_{t-1} union {v_t}. Otherwise,
X_t := X_{t-1}.
(c) if v_t is not in X_{t-1} and exactly one neighbor, w, of v_t is in
X_{t-1}, then,
with probability alpha, let X_t = X_{t-1} union {v_t} minus {w}.
Informally, we
"slide" one element of the set along the edge {w, v_t}.
(d) Otherwise, X_t := X_{t-1}.

Using path coupling, analyze the worst-case expected change in
Hamming
distance after one step of this Markov chain. Set alpha to get the
best upper
bound on mixing time. Show that, for this chain, the mixing time is
O(n log(n))
for twice the range of lambda values.

3) Let G be a connected regular graph on n vertices, where the degree
of
every vertex equals d. Prove that the mixing time for lazy random
walk
on G is, in general, at most polynomial in n and d. Try to pin down
the
best such general polynomial with the best upper and lower bounds you
can.

### Amitabh Trehan

May 8, 2010, 12:50:57 AM5/8/10
Tom,
Vamsi and I are working on the first question. Our conjecture is that you need to sample at least a number of nodes which is of the form 1/2(a - \sqrt(a)), and (as if in part 3 of the question) if you don't pick any marked nodes, you will not need more than a/2 samples.
So, we think the upper and lower bounds for the three cases can be worked out using this. In the  first case we apply coupons collector?

Amitabh
--

“I don't know the key to success, but the key to failure is trying to please everybody.”
“Is the glass half full, or half empty? It depends on whether you're pouring, or drinking.”

### Tom Hayes

May 8, 2010, 1:19:57 PM5/8/10
to CS 510 Randomized Algorithms
Hi,

Well, every positive number is of the form 1/2(a - sqrt(a)) for some
a,
so your conjecture is going to be true in that sense. But you would
need to say something about what a is.

Yes, the first case can be reduced to coupon collector (for both the
upper and lower bounds).

Keep at it!

Tom

### Amitabh Trehan

May 8, 2010, 1:42:11 PM5/8/10
Yep, of course, I guess we were trying to reveal the least information ;) - but yes, a is a linear function of n (almost n actually). If a were n, then wouldn't this be like mean - variance (\sqrt(n)). We have some more explicit numbers - we were wondering if we should put them on the list. In asymptotic terms, we think:
Case i: lower and upper: O(n log n)
Case ii: lower and upper: O(n)
Case iii: lower and upper: O(n)

Any hints on how to approach the expected values?

Amitabh

### Tom Hayes

May 9, 2010, 12:31:06 AM5/9/10
to CS 510 Randomized Algorithms
Hi again,

Try for the following:
Case 1: Theta(n log n).
Case 2: n - Theta(sqrt(n)).
Case 3: I don't want to spoil this yet, but doing
some computer simulation should put you on the right track.

Tom

### Vamsi K. Potluru

May 11, 2010, 7:01:30 PM5/11/10
Hello Tom,

I think we got the following:

Lower bound Upper bound
1. Theta(n log n) Theta(n log n)
2. (n+1)/2-|_sqrt(n+1)/2 _| n-2
3. (n+1)/2 - |_sqrt(n+1)/2_| (n+1)/2

where |_ _| is the floor function.
Let me know if they are right.

Thanks,

~Vamsi.

### Viktor Chekh

May 11, 2010, 8:27:50 PM5/11/10
I believe that the case (3) is very, very different comparing to other two ones and its lower bound is very, very,...,very close to the upper bound :-)

### Vamsi K. Potluru

May 12, 2010, 1:37:43 PM5/12/10
Hello Tom,

For the third problem, is the mixing time equivalent to
calculating the second eigenvalue of the transition matrix?

Thanks,

~Vamsi.

### Thomas Hayes

May 12, 2010, 2:43:14 PM5/12/10
Hi Vamsi,

Not quite equivalent, but if we call "gap" the difference
between the first and second eigenvalues of the transition
matrix, and you can prove (1/gap) <= poly(n,d), then that
will imply that the mixing time is also poly(n,d).
So, basically, yes.

See, for instance, Theorem 12.3 on p. 155 of this book:
(This is an excellent book, by the way, although it can be a bit dry.)

Tom

### Viktor Chekh

May 13, 2010, 5:30:39 PM5/13/10
Dear Professor Hayes,

For the 3-rd question:
is it enough to show that t = log(n/eps)/delta ? where delta = 1-lamda_1 (and depends on n and d) and eps is given distance?

Thanks,
V.

### Tom Hayes

May 13, 2010, 10:20:23 PM5/13/10
to CS 510 Randomized Algorithms
It is enough to prove that what you are calling delta is at least
1/poly(n,d). Probably the easiest way to do this is by proving a
the same book I gave the link to earlier.

A more direct approach would be to show that, if mu_t is the
distribution at time t, then the 1-(or 2)-norm of (mu_t - pi) is at
most
(1 - 1/poly(n,d)) times (mu_{t-1} - pi). Then you can induct on
t to bound (mu_{t} - pi) explicitly. For this, you would of course
make use of the fact that, since the graph is d-regular, pi is
uniform.

Hope these suggestions help!

Tom