Testing a Quasi-Isometric Quantized Embedding with Percentile Analysis


$$ \newcommand{\bs}{\boldsymbol} \newcommand{\cl}{\mathcal} \newcommand{\bb}{\mathbb} $$ This simple python notebook aims at estimating the validity of some of the results explained in the paper

A Quantized Johnson Lindenstrauss Lemma: The Finding of Buffon’s Needle"
by Laurent Jacques (arXiv)

“In 1733, Georges-Louis Leclerc, Comte de Buffon in France, set the ground of geometric probability theory by defining an enlightening problem: What is the probability that a needle thrown randomly on a ground made of equispaced parallel strips lies on two of them? In this work, we show that the solution to this problem, and its generalization to N dimensions, allows us to discover a quantized form of the Johnson-Lindenstrauss (JL) Lemma, i.e., one that combines a linear dimensionality reduction procedure with a uniform quantization of precision $\delta>0$. In particular, given a finite set $\cl S \in \bb R^N$ of $S$ points and a distortion level $\epsilon>0$, as soon as $M > M_0 = O(\log |S|/\epsilon^2)$, we can (randomly) construct a mapping from $(\cl S, \ell_2)$ to $((\delta \bb Z)^M, \ell_1)$ that approximately preserves the pairwise distances between the points of $\cl S$. Interestingly, compared to the common JL Lemma, the mapping is quasi-isometric and we observe both an additive and a multiplicative distortions on the embedded distances. These two distortions, however, decay as $O(\sqrt{\log |S|/M})$ when $M$ increases. Moreover, for coarse quantization, i.e., for high $\delta$ compared to the set radius, the distortion is mainly additive, while for small $\delta$ we tend to a Lipschitz isometric embedding. Finally, we show that there exists “almost” a quasi-isometric embedding of $(\cl S, \ell_2)$ in $((\delta \bb Z)^M, \ell_2)$. This one involves a non-linear distortion of the $\ell_2$-distance in $\cl S$ that vanishes for distant points in this set. Noticeably, the additive distortion in this case decays slower as $O((\log S/M)^{1/4})$.”

In particular, our objective is to study the behavior of a nonlinear mapping from $\bb R^N$ to $\bb R^M$, for two specific dimensions $M,N \in \bb N$. This mapping is defined as follows.

Let a random (sensing) matrix $\bs \Phi \in \bb R^{M \times N}$ be such that each component $\Phi_{ij} \sim_{\rm iid} \cl N(0,1)$, or more shorthly $\bs \Phi \sim \cl N^{M\times N}(0,1)$.

We want to combine the common random projection realized by $\bs \Phi$ with the following non-linear operation, i.e., a uniform scalar quantizer of resolution $\delta > 0$:

$$ \qquad\cl Q_\delta[\lambda] = \delta \lfloor \lambda / \delta \rfloor. $$
(remark: we could define instead a midrise quantizer by adding $\delta/2$ to the definition above, and the rest of the development would be unchanged)

This combination reads

$$ \qquad\bs \psi_\delta(\bs x) := \cl Q_\delta(\bs\Phi \bs x + \bs \xi) $$
with $\bs \xi \sim \cl U^M([0, \delta])$ is a dithering vector, i.e., a random uniform vector in $\bb R^M$ such that $u_i \sim_{\rm iid} \cl U([0, \delta])$, and $\bs \psi_\delta:\bb R^N \to (\delta \bb Z)^M$ $ is our mapping of interest.

In Proposition 13 (page 16) of the paper above, it shown that

Proposition 13: Fix $\epsilon_0>0$, $0< \epsilon\leq \epsilon_0$ and $\delta>0$. There exist two values $c,c’>0$ only depending on $\epsilon_0$ such that, for $\bs\Phi \sim \cl N^{M\times N}(0,1)$ and $\bs \xi\sim \cl U^M([0, \delta])$, both determining the mapping $\bs\psi_\delta$ above, and for $\bs u, \bs v\in\bb R^N$,

$$\qquad(1 - c\epsilon) |\bs u - \bs v| - c’\epsilon\delta\ \leq\ \tfrac {\sqrt\pi}{\sqrt 2 M} |\bs\psi_\delta(\bs u) - \bs\psi_\delta(\bs v)|_1\ \leq\ (1 + c\epsilon)|\bs u - \bs v| + c’\epsilon\delta.\tag{1} $$
with a probability higher than $1 - 2 e^{-\epsilon^2M}$.

This proposition shows that the random quantity $c_\psi M^{-1} |\bs\psi_\delta(\bs u) - \bs\psi_\delta(\bs v)|1$ with $c\psi = \sqrt{\pi/2}$ concentrates around its mean $|\bs u - \bs v|$. However, conversely to linear random projections, this concentration phenomenon displays two kind of deviation: a standard multiplicative one (the $1\pm c\epsilon$ factor) and an additive one in $\pm c’\epsilon\delta$.

Moreover, forcing (1) to be valid at constant probability, we deduce also that $\epsilon = O(1/\sqrt M)$, i.e., showing that both the multiplicative and the additive distortion decay as $O(1/\sqrt M)$ with respect to $M$.

Remark: From a standard union bound argument (see the paper), this result allows one to show the existence of a non-linear embedding from a set $\cl S \subset \bb R^N$ to $(\delta \bb Z)^M$, as soon as $M \geq M_0 = O( \epsilon^{-2},\log |\cl S|)$. Proposition 13 is thus central to reach this property.

This notebook proposes to test the behavior of the two distortions displayed in (1).

For that, arbitrarily fixing $|\bs u - \bs v|=1$ by normalizing conveniently $\bs u$ and $\bs v$ (e.g., after their random selection in $\bb R^N$), we study the experimental distribution of the random variable

$$ \qquad V_\psi = \tfrac{\sqrt\pi}{\sqrt 2 M} |\bs\psi_\delta(\bs u) - \bs\psi_\delta(\bs v)|_1. $$

Remark: the concentration (1) is invariant under the coordinate change $\bs u \to \lambda \bs u$, $\bs v \to \lambda \bs v$ and $\delta \to |\lambda| \delta$ for any $\lambda \neq 0$, which allows us to set $|\bs u - \bs v|=1$ without any loss of generality (providing we study the variability of (1) in $\delta$).

Assuming $N$ fixed, we study the behavior of this random variable in function of $\delta$ and $M$, i.e., $V_\psi = V_\psi(\delta, M)$.

Although $V_\psi$ displays a non-trivial distribution1, we can estimate numerically the failure curve given a fixed failure probability $0 < p_{\rm fail} < 1$:

$$ \quad{\rm cdf}^{-1}\psi(\delta, M;, p{\rm fail}) := {t: \bb P[V_\psi(\delta, M) - 1 > t] = p_{\rm fail}}, $$

where we use the fact that $\bb E V_\psi = |\bs u - \bs v| = 1$

We expect that, if Prop. 13 above is true,

$$ \qquad t_\psi(\delta,M) := {\rm cdf}^{-1}\psi(\delta, M; p{\rm fail}) \quad \simeq\quad a\ \epsilon(M)\ +\ b\ \epsilon(M)\ \delta,\quad \text{for some }a,b > 0, $$

and possibly find the values $a, b>0$, or their product with $\epsilon(M)$ by a simple linear fit at specific values of $M$.

Moreover, if $p_{\rm fail} \simeq e^{-\epsilon^2M}$ holds (as in Prop. 13 above), then, once this linear fit will be obtained at several $M$s, we should also observe that $\epsilon \simeq 1/\sqrt{M}$ since $p_{\rm fail}$ is set to a constant value.

Well, let’s now test the validity of all these phenomena!

1: Actually, a mixture of *Buffon* random variable and of a Chi distribution with $N$ degrees of freedom.

The simulations:

Remark: launch ipython notebook with the pylab option, i.e.,
ipython notebook –pylab=inline

import numpy as np
from scipy import stats

Defining the $\bs \psi_\delta:\bb R^N \to (\delta \bb Z)^M$ with two functions:

def quant(val, d): # the uniform quantizer of resolution 'd'
    return d * np.floor(val/d)
def psi(vec, mat, dith, d): # the mapping (matrix A, dithering u, quantization d)
    return quant(mat.dot(vec) + dith, d)

The following function returns the inverse cumulative density function, i.e., given a pdf $f$ associated to the random variable $X$, this function is defined as

$$ \quad{\rm cdf}^{-1}(p; f) = {t: P[X > t] = \textstyle\int_{t}^{+\infty} f(u)\ {\rm d}u = p} $$

We use the python function

g(q) = stats.scoreatpercentile(samples, q)

such that

$$ \quad g(q) = {t: P[X < t] = \textstyle\int_{0}^{t} f(u)\ {\rm d}u = q} $$

where $f$ is estimated experimentally from the samples.

By definiton, $P[X < g(q)] = q$ so that

$$ \quad q = P[X < g(q)] = 1 - P[X > {\rm cdf}^{-1}(p; f)] = 1 - p $$

so that, for $q = 1- p$, $g(1-p) = {\rm cdf}^{-1}(p; f)$.

We can thus define this function:

def inv_cdf(samples, p): # returns {t : P[X > t] = p}
    return stats.scoreatpercentile(samples, 100*(1 - p))

Entering into the simulations: (time for a coffee/tea break)

Experimental setup: The standard deviation of $V_\psi$ is evaluated for several number of measurements $M$ and for several values of $\delta$. For each of them a certain number of trials are generated. In the configuration, there are 10 000 trials and the sensing matrix $\bs \Phi$ and the dithering $\bs \xi$ are regenerated every 100 trials.

# space dimension
N = 256 

# Normalising constant for the quantized embedding (see above)
c_psi = np.sqrt(np.pi/2) 

# Number of trials
nb_trials_phi = 100 # number of regeneration of Phi/dithering
nb_trials_pt = 100 # number of trials for points (i.e., per phi/dithering)
nb_trials = nb_trials_pt * nb_trials_phi 

# probability of failure for drawing the deviation curve 
#   f(t) = {t : P[ V_psi > 1 + t ] = p_fail}
p_fail = 0.05

# variable number of dimensions
M_v = np.array([64, 128, 256, 512, 1024])
nb_M = M_v.size

nb_delta = 8

# The fixed value for ||u - v|| = 1
dist_l2 = 1 

# Testing several delta, a logspace arrangement seems logical, here from 10^-3 to 10
delta_v = np.linspace(0.1,4,nb_delta) 

# Curve at precentile p_fail, one per measurement
V_psi_per = np.zeros([nb_M,nb_delta])

# Checking if the mean is around 1
V_psi_mean = np.zeros([nb_M,nb_delta])

# Only the standard deviation of V_psi matters here, 
# but interested readers could evaluate its mean too to see that it matches ||u - v||. 
# In this case, just uncomment this line and the one 
# V_psi_avg = np.zeros([nb_M,nb_delta])

for j in range(nb_M):

    M = M_v[j]
    V_psi = np.zeros(nb_trials) # temporary store distances in the embedded space
    for k in range(nb_delta):
        for i in range(nb_trials):
            # local variables for the loop
            delta = delta_v[k]
            # Regenerating the sensing matrix/dithering every 'nb_trials_pt' trials
            if (np.mod(i, nb_trials_pt) == 0):
                # Defining the sensing matrix and dithering
                Phi = np.random.randn(M,N)
                # and dithering, i.e., uniform random vector in [0, \delta]
                xi = np.random.rand(M,1)*delta 
            # Generating two points in R^N and normalizing their distance to 1 
            tmp_u = np.random.randn(N,1)
            tmp_v = np.random.randn(N,1)
            tmp_dist_uv = np.linalg.norm(tmp_u - tmp_v)
            u = tmp_u * dist_l2 / tmp_dist_uv;
            v = tmp_v * dist_l2 / tmp_dist_uv;
            # projecting the two points according to psi
            u_proj = psi(u, Phi, xi, delta);
            v_proj = psi(v, Phi, xi, delta);
            # storing one sample of the normalized l1 distance
            V_psi[i] = c_psi*np.linalg.norm(u_proj - v_proj,1)/M
        # Finding the percentile curve
        V_psi_per[j,k] = inv_cdf(V_psi - 1, p_fail)
        # For mean evaluation, you could estimate too: 
        V_psi_mean[j,k] = np.mean(V_psi)
    print j, # observing the loop progression (sober progress bar ;-)
 0 1 2 3 4

Let’s observe now the diferent evolutions of $V_{\psi}$ vs. $\delta$ for the different number of measurements $M\in {64, \cdots, 1024}$ selected above:

# Mean plot
fig, ax = plt.subplots()

for k in range(M_v.size):
    ax.plot(delta_v, V_psi_mean[k], '.-', label = "M = %i" % M_v[k])
ax.set_title(r'${\rm E}[V_\psi]$ vs $\delta$')
ax.set_ylabel(r'${\rm E}[V_\psi]$ vs $\delta$')
ax.legend(bbox_to_anchor=(1.40, 1))

# Percentile plot
fig, ax = plt.subplots()

for k in range(M_v.size):
    ax.plot(delta_v, V_psi_per[k], '.-', label = "M = %i" % M_v[k])
ax.set_title(r'$t_\psi$ vs $\delta$   (zoom on x-axis)')
ax.legend(bbox_to_anchor=(1.40, 1))
<matplotlib.legend.Legend at 0x10ebcd990>



From this last plot, we conclude that:

  • for $\delta \gtrsim 0.2$, $t_\psi(\delta,M)$ displays roughly a linear progression in $\delta$;
  • for different value of $M$, this linear trend has an offset as $\delta \simeq 0$;
  • the slope and the offset of the linear progresison clearly depend on $M$.

Let’s go one step further in our analysis.

From this plot, we perform a linear fit to estimate $t_\psi(\delta,M)$ as

$$ \quad t_\psi(\delta,M)\quad \simeq\quad v_\alpha +\ v_\beta\ \delta,\qquad \text{for some }v_\alpha, v_\beta > 0. $$

Actually, we have to estimate one couple of $(v_\alpha,v_\beta)$ for each value $M$, since, from what we explained above, we should have

$$\qquad v_\alpha\ =\ a\ \epsilon$$


$$\qquad v_\beta\ =\ b\ \epsilon$$

for some constants $a$ and $b$ and $\epsilon = O(1/\sqrt{M})$ depending on $M$ (at fixed failure probability for inequality (1) above).

Let us see if this is what we get. We first do the linear fit:

# initialization
v_alpha = np.zeros(M_v.size)
v_beta = np.zeros(M_v.size)

# For each M, use np.polyfit at degree 1
for k in range(M_v.size):
    v_beta[k], v_alpha[k] = np.polyfit(delta_v, V_psi_per[k], 1)

If $v_\beta = b \epsilon$, where we recall that $v_\beta$ is the slope of each curves in the plots above, we should see a good match between $v_\beta(M)$ and

$$ \qquad v^{\rm th}\beta(M)\ :=\ \sqrt{\tfrac{M_0}{M}}\ v\beta(M_0),\qquad \text{with $v^{\rm th}\beta(M_0)=v\beta(M_0)$} $$

for one fixed value $M_0$.

Is it what we have?

plt.semilogx(M_v, v_beta, 'o-', label=r'$v_\beta(M)$')
plt.semilogx(M_v, v_beta[0]*np.sqrt(M_v[0])/np.sqrt(M_v), 'o-', label=r'$v^{\rm th}_\beta(M)$') # M_0 is the first M of M_v
plt.ylim(0, v_beta[0])


The match sounds good enough!

Finally, if we really have $v_\alpha = a \epsilon$ and $v_\beta = b \epsilon$, $v_\beta$ should be a linear function of $v_\alpha$, i.e., $v_\alpha/v_\beta \simeq a/b$.

Let’s test this fact.

fig, ax = plt.subplots()
ax.plot(v_alpha, v_beta, '.-', label = r'$v_\beta$ vs. $v_\alpha$')
ax.plot(np.linspace(0,v_alpha.max(),10), np.linspace(0,v_alpha.max(),10)*v_beta.max()/v_alpha.max(), label = 'linear trend')
ax.set_xlim(0, v_alpha.max())
ax.set_ylim(0, v_beta.max())
ax.legend(bbox_to_anchor=(.5, 1))

ratio, offset = np.polyfit(v_alpha,v_beta,1)
print "We approximatley have v_beta = %f v_alpha + %f" % (ratio, offset)
We approximatley have v_beta = 0.574426 v_alpha + -0.002124


Laurent Jacques, November 2013

Laurent Jacques
Laurent Jacques
FNRS Senior Research Associate and Professor