What does Gaussian efficiency mean? In case of robust estimators, What does Gaussian efficiency means? For example $Q_{_n}$ has 82% Gaussian efficiency and 50% breakdown point.
The reference is: Rousseeuw P.J., and Croux, C. (1993). “Alternatives to median absolute deviation.” J. American Statistical Assoc., 88, 1273-1283
 A: I agree with @cardinal and @whuber that this is about asymptotic relative efficiency and here I write an answer if other are interested in the concept.
In statistics, it is very common to measure how efficient an estimator is using its asymptotic variance. Then, in this context, the asymptotic relative efficiency of $d_n$ relative to $s_n$ is defined by
$$ARE=\lim_{n \to \infty}\frac{var(s_n)/E(S_n)^2}{var(d_n)/E(d_n)^2}.$$
In robust statistics, the law with respect to which we take the expectation in ARE is often a corrupted distribution.
For instance, I present an example from the book Robust Statistics by Huber and Ronchetti (see page 3). Suppose $X_1,\dots, X_n$ are from a corrupted gaussian distribution
$$F(x)=(1-\varepsilon)\Phi(x)+\varepsilon \Phi(\frac{x}{3})  $$
so that with high proba ($1-\varepsilon$) the data are standard normal and with small proba ($\varepsilon$) the data are normal with a higher variance. Then, the ARE between
$$d_n=\frac{1}{n}\sum_{i=1}^n |X_i-\overline{X}|\text{ and }s_n=\left(\frac{1}{n}\sum_{i=1}^n (X_i-\overline{X})^2\right)^{1/2}$$ is $\simeq 0.876$ when $\varepsilon = 0$ but as soon as $\varepsilon>0.005$, we have $ARE>1$ and for instance for $\varepsilon=0.01$, we have $ARE \simeq 1.44$. We conclude that the mean absolute deviation is more efficient than the std when the data are corrupted.
A: I guess Gaussian efficiency is something related to computation cost.
The efficiency of Gaussian adaptation relies on the theory of information due to Claude E. Shannon. When an event occurs with probability P, then the information −log(P) may be achieved. For instance, if the mean fitness is P, the information gained for each individual selected for survival will be −log(P) – on the average - and the work/time needed to get the information is proportional to 1/P. Thus, if efficiency, E, is defined as information divided by the work/time needed to get it we have:
E = −P log(P).
This function attains its maximum when P = 1/e = 0.37. The same result has been obtained by Gaines with a different method.
I may simply conclude that the higher the Gaussian Efficiency is, less resources (RAM) is needed for computing something like a robust scale estimator of a large sample. Since CPUs are much faster than the rest of computer we prefer to run a trial/error algorithm for times rather doing it at once with saying 128GB of RAM. when the Gaussian Efficiency is high the job will be done in a shorter time. 
