Networks of Noisy Gates (Part 2)
September 20, 2011, 3:28 pm
Filed under: Uncategorized | Tags: , , ,

In today’s post, I want to focus on some of the highlights of the lower bound that I mentioned last time on the “Networks of Noisy Gates“ problem. Recall that, in this problem, you are given a function {f} that can be computed with a circuit of {n} regular gates. You want to construct a new circuit with the smallest number of noisy gates that computes {f} correctly with probability greater than {1/2}. These new noisy gates each fail independently with some small probability {\epsilon} and when they fail, they produce the complement of their correct output.

The result I want to talk about today is the lower bound by Pippenger, Stamoulis and Tsitsiklis which shows that there are some functions that can be computed with {n} gates without noise, but which require {\Omega(n \log n)} noisy gates. In particular, they show that this is true if {f} is the exclusive or function.

The first part of their lower bound is a result that shows a new model of failure where wires can also fail can be made equivalent to the old model of failure where only gates can fail. In particular, in the new model, each wire fails with some probability {\delta}, independently of all other wires and gates, and failure of a wire results in transmitting the complement of the bit that is sent down that wire.

Lemma 3.1 of the paper “Lower Bound for the Redundancy of Self-correcting Arrangements of Unreliable Functional Elements” by Dobrushin and Ortyukov shows the following. Let {\Delta} be the maximum number of input wires for any gate. Then for any {\delta \in [0, \epsilon/\Delta]} and for any set of inputs to a gate, there exists a vector of malfunction probabilities on the gate such that the probability that the gate fails to produce the correct output is exactly {\epsilon}. Essentially this shows that, the new model where we have a small probability of failure on both wires and gates is stochastically equivalent to the old model where we have a larger probability of error just on the gates.

Now we assume we have {n} inputs to the to our exclusive-or function {f}, and we let {X_{1}, X_{2}, \ldots, X_{n}} be these inputs. For the purposes of the lower bound, we’ll assume that each {X_{i}} is an independent random variable that is {1} with probability {1/2} and {0} otherwise. Now for this lower bound, we’re going to be very generous to the noisy circuit. We’re going to assume that 1) no gate ever fails and 2) the only wires that can fail are those that are directly connected to some input bit ( each of these wires fails independently with probability {\delta}).

For the {i}-th input bit, we’ll let {m_{i}} be the number of wires that are connected directly to that input bit, and hence carry the value {X_{i}} if they are not faulty. Now in the backend of the circuit, which is computing the exclusive-or of the input bits, there needs to be some estimate for each value of {X_{i}}. Not surprisingly, it is possible to prove (as the paper does) that Maximum Likelihood way to estimate {X_{i}} is to take the majority bit over all the {m_{i}} wires that are connected to input {X_{i}}. We’ll let {Z_{i}} be an indicator r.v. that is {1} iff the Maximum Likelihood estimate for {X_{i}} is wrong. Note that {Pr(Z_{i} = 1) \geq \delta^{m_{i}}}, since if every wire that carries that value {X_{i}} is faulty, then clearly the circuit will use the wrong value.

Next, note that the probability that the noisy circuit fails is just {Pr(Z_{1} \oplus \cdots \oplus Z_{n} = 1)}. So now we have a cute probability problem: assume you’re give {n} independent indicator random variables, you know the probability that each of them is {1}, and you want to compute the probability that their exclusive-or is {1}. The next lemma solves this problem with the help of generating functions.

Lemma 1

\displaystyle 1 - 2 Pr(Z_{1} \oplus \cdots \oplus Z_{n} = 1) = \prod_{i=1}^{n} (1 - 2 Pr(Z_{i} = 1))

Proof: Let {\phi_{i}(t) = E(t^{Z_{i}})} be the generating function for {Z_{i}}. Note that {\phi_{i}(t) = 1- Pr(Z_{i}= 1) + t Pr(Z_{i} = 1)}. Now let {Z = \sum_{i=1} ^{n} Z_{i}} and let {\psi(t) = E(t^{Z})}. Now comes the clever part: note that {E((1^{Z} - (-1)^{Z})/2) = Pr(Z_{1} \oplus \cdots \oplus Z_{n} = 1)}. This is true since {1^{Z}} is always {1} and {-1^{Z}} is {1} if {Z} is even and {-1} if {Z} is odd.

But by linearity of expectation, {Pr(Z_{1} \oplus \cdots \oplus Z_{n} = 1) = E((1^{Z} - (-1)^{Z})/2) = (\psi(1) - \psi(-1)) / 2}. Next, note that {\psi(t) = \prod_{i=1}^{n} \phi_{i}(t)} since the {Z_{i}} are independent. Thus, {\psi(1) = 1} and {\psi(-1) = 1-2 Pr(Z_{i} = 1)}. This completes the proof. \Box

The remainder of the proof for the lower bound is just algebra. We’ll make use of the inequality between arithmetic and geometric means, which states that for any set of {n} numbers, {x_{1}, x_{2}, \ldots x_{n}}, it holds that {(1/n) \sum_{i=1}^{n} x_{i} \geq \sqrt[n] {\prod_{i=1}^{n} x_{i}}}.

Lemma 2 { \sum_{i=1}^{n} m_{i}} is {\Omega(n \log n) }

Proof: Let {p<1/2} be the probability of error for the noisy circuit. By assumption, {Pr(Z_{1} \oplus \cdots \oplus Z_{n} = 1) < p}. Using Lemma~1 and our bound on {Pr(Z_{i} = 1)}, we have that

\displaystyle 1-2p \leq \prod_{i=1}^{n} (1 - 2 \delta^{m_{i}})

Now using the inequality between arithmetic and geometric means, we have that

\displaystyle 1-2p \leq \left(1/n \sum_{i=1}^{n} (1 - 2 \delta^{m_{i}})\right)^{n} = \left(1 - 2/n \sum_{i=1}^{n} \delta^{m_{i}}\right)^{n}

Again using the inequality between arithmetic and geometric means (now on the term that is being subtracted) and the fact that {1-x \leq e^{-x}}, we get:

\displaystyle 1-2p \leq \left( 1-2\delta^{1/n \sum_{i=1}^{n} m_{i}} \right)^{n} \leq \textrm{exp}(-2n\delta^{1/n \sum_{i=1}^{n} m_{i}})

Isolating the {\sum_{i=1}^{n} m_{i}} term in the above inequality gives us that

\displaystyle \sum_{i=1}^{n} m_{i} \geq n \frac{\ln (2n) - \ln \ln (1/(1-2p))}{\ln (1/\delta)}

Since {\delta} is a constant depending only on {\epsilon} and since {p<1/2}, the above inequality completes the proof. \Box

Finally, we note that since each gate is assumed to have constant fan in, if the number of wires is {\Omega(n \log n)}, it follows that the number of gates in the noisy circuit must also be {\Omega(n \log n)}. That basically completes the proof of the lower bound. The interesting thing is that the only difficulty we exploited is the difficulty of estimating the input values in the situation where the wires may be faulty!

2 Comments so far
Leave a comment

in lemma 1 it is sufficient to assume that just the pairs(i > 0)
(x_1+…x_i, x_{i+1}) are independent. In this case one can use
markov chains to compute the probabilites PR(x_1+…x_i = 0).
Bolis down to compute sum of entries of the matrix product
((1-p_1)I + p_1J)…((1-p_i)I + p_i J)
where J is 2 \times 2 cyclic shift. J is easily diagonalizable,
with eigenvalues (1, -1).

Nice post, best, Leonid.

Comment by Leonid Gurvits


Interesting point. Wonder if this has any implications for strengthening the lower bound. I guess at least it’s clear that pairwise independent failure of the ML estimates wouldn’t help…


Comment by Jared

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: