Networks of Noisy Gates (Part 4)
October 21, 2011, 5:40 pm
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In this post, I want to finish up the result that for any function of n inputs, there is a network of O(2^n/n) noisy gates that will compute that function with probability better than 1/2.  Since most functions require O(2^n/n) noise-free gates, this will show that, for most functions, there is at most a constant blowup in gates when all gates are noisy.

Recall from last time that we can build a MUX over 2^x signals with O(2^x) noisy gates.  Using a similar technique, for any set of x inputs, we can build an “All Functions” circuit that computes all 2^(2^x) possible functions of these inputs with O(2^(2^x)) gates. Our result today will make use of both 1) an “All Functions” circuit and 2) a MUX.

The technique is illustrated in the figure below.  Assume we’re trying to compute a function f of n inputs.  Let a = log (n-log n) Basically, we send a of the inputs the “All Functions” circuit and we send n-a of the inputs to the MUX.  The MUX selects, based on the first n-a inputs, the exact function of the remaining a inputs that determines the function f over all n inputs.  The MUX selects 1 of 2^(n-a) such “completion functions”.

The beautiful thing here is that both the MUX and the “All Functions” circuit require only O(2^n/n) gates.  Thus, this clever combination of both types of circuits has saved us a factor of n in the total number of noisy gates required.



Networks of Noisy Gates (Part 3)
October 15, 2011, 3:06 am
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Today I want to talk about an interesting result in Pippenger’s paper “Networks of Noisy Gates” for the Von Neumann noisy gates model.  Recall that in this model,  each noisy gate fails independently with some fixed probability \epsilon.  We are given a circuit to compute a function with m regular gates and our goal is to compute the same function with probability great than 1/2 with as few noisy gates as possible.

In previous posts, we showed that n \log n gates are always sufficient and we showed that for the exclusive-or function, n \log n gates are necessary.  This implies that in the worst case, there is a log n blowup in the number of gates required.

There is another result by Pippenger that shows that for *most* boolean functions, only a *constant* blowup is required.  Today I want to focus on the first part of this result, which is to show that a multiplexer (MUX) over 2^r signals can be computed with only O(2^r) noisy gates.  Recall that a MUX over 2^r signals is a circuit that takes as input both 1) 2^r possible signals; and 2) an r bit selector.  The MUX outputs exactly one of the 2^r possible signals, specifically the one that is specified by the r bit selector.

For example, a MUX for r=1 has as input 1) 2 signal bits; and 2) a single selector bit.  If the selector bit is 0, the MUX outputs the first signal bit and if the selector bit is 1, the MUX outputs the second selector bit.  Let G be a gate that computes the MUX for r=1.  Then it’s easy to create a MUX for any r by wiring up O(2^r) copies of G (try it).  In particular, a MUX for r requires at least 2^r noise free gates.

Pippenger shows that you can create a MUX for any r using only O(2^r) noisy gates.  Clearly this is just a constant blowup over the number of noiseless gates needed to create such a MUX.  The construction used to prove this result is sketched in the figure below. See also this pdf: pipp-mux

Note that any boolean function over x inputs can be computed using a MUX, where r=x (the computation is done essentially by table lookup).  This means that we can now compute any boolean function over x inputs with O(2^x) noisy gates.  It turns out that most boolean functions over x variables require 2^x/x noise-free gates.   In my next post, I’ll show how we can shave off this factor of x in the noisy gate world.