Resume

Research

Learning

Blog

Teaching

Jokes

Kernel Papers

Oftentimes, we have some set with some notion of distances between elements i.e. a metric space, and we want to embed that metric space inside another, either for easier analyses or visualization. For example, we may want to convert a graph metric space into an \(\ell_p\) metric space.

Given 2 metric spaces \((X, d_X), (Y, d_Y)\), a function \(f: X \rightarrow Y\) is an isometric embedding iff \(\forall x, x' \in X\):

\[d_Y(f(x), f(x')) = d_X(x, x')\]Any finite metric space can always bee isometrically embedded into \(\ell_{\infty}\). The embedding is as follows: Given \((X, d)\), with \(|X| = n\), define \(f: X \rightarrow \mathbb{R}^n\) by

\[f(x_i) := [d(x_1, x_i), d(x_2, x_i), ..., d(x_n, x_i)\]This is called a Frechet embedding, and has the property that \(d(x_, x_j) = d_{\infty}(f(x_i), f(x_j))\). However, an isometric embedding to a target metric space doesn’t always exist. For instance, a square graph (4 vertices) with edge weights 1 cannot be embedded into \(\ell_2\). This raises the prospect of ““low distortion” embeddings.

Given 2 metric spaces \((X, d_X), (Y, d_Y)\), a function \(f: X \rightarrow Y\) is an embedding of \(X into\)Y \(with distortion\)\alpha\(and scaling\)\beta\(if\)\forall x, x’ \in X$$:

\[\beta d_X(x, x') \leq d_Y(f(x), f(x')) \leq \alpha \beta d_X(x, x')\]Given any finite metric space \((X, d)\) with \(|X| = n\), there exists an embedding of \((X, d)\) into \((\mathbb{R}^k, d_1)\) where \(k = O(\log^2 n)\) (also works for \(\ell_p\)) with distortion \(O(\log n)\).

- For \(i = 1, ..., \log n\):
- For \(j = 1, ..., c \log n\):
- Choose \(S_{i, j} \subseteq X\) at random such that every \(x \in XX\) is contained in \(S_{i,j}\) with probability \(2^{i}\)

- For \(j = 1, ..., c \log n\):

Define \(f(x) = [d(x, S_{11}), d(x, S_{12}),..., d(x, S_{\log n, c\log n}),]\) where \(d(x, S) = \min_{s \in S} d(x, s)\)

In order to show that an embedding is low-distortion, we need to show both that distances don’t expand or contract by too much. Let’s first consider expansion. Consider \(f(x) := d_S(x, S)\) i.e. the distance from \(x\) to the point in set \(S\) nearest \(x\). This distance has the property of being non-expanding:

\[\forall x, y \in X, ||f_S(x) - f_S(y)|| \leq d(x, y)\]Thus, Bourgain’s Embedding is non-expanding. Next, we argue that Bourgain’s Embedding isn’t too “shrinking” i.e. that \(|f_S(x) - f_S(y)| \geq \Delta\). The argument is going to consist of two parts: First, we’re going to assume a particular picture, and argue that under this particular picture, the embedding isn’t too shrinking. Second, we’re going to argue that this particular picture is highly likely.

First, picture a ball around \(x\) and another ball around \(y\) with slightly different radii. Call the balls \(B_x := \{z \in X: d(z, x) \leq \delta \}\) and \(B_y := \{z \in X: d(z, x) \leq \delta + \Delta\}\). Imagine that the set \(S\) intersects with \(B_x\) but not with \(B_y\). In this picture, we know that \(d(x, S) \leq \delta\), since \(S\) intersects \(B_x\), and we know that \(d(y, S) > \delta + \Delta\), since \(B_y\) does not intersect \(S\). Consequently:

\[d(y, S)- d(x, S)\geq \delta + \Delta - \delta = \Delta\]This means that the original distances don’t shrink by more than \(\Delta\). By constructing Bourgain’s Embedding in the above manner, this picture will turn out to be likely with high probability. The inner loop over \(j\) tries to find this good picture, while the outer loop tries to find the right density for \(S\), which depends on \(d(x,y)\) in the data.

**Bourgain’s Theorem**: Let \(k = c \log^2 n\) be the number of sets \(S_{ij}\). Let \(f\) be Bourgain’s embedding.
There exists a constant \(b\) such that \(\forall x, y \in X\):

Proof: Starting with the upper bound, it’s enough to show that for every \(S \subset X\):

\[|d(x, S) - d(y, S)| \leq d(x, y)\]If that were true, then:

\[||f(x) - f(y) ||_1 = \sum_{ij} |d(x, S_{ij}) - d(y, S_{ij}) | \leq k d(x, y)\]The proof is via the triangle inequality. Because the metric \(d\) must obey the triangle inequality, the distance from \(x\) to the nearest element of \(S\) must be less than the distance from \(x\) to \(y\), then from \(y\) to the point in \(S\) nearest to \(x\). Turning next to the lower bound, we want to show:

\[\frac{k}{b \log n} d(x, y) \leq ||f(x) - f(y)||_1\]Recall, in our good situation, \(|d(x, S) - d(y, S)| \geq \Delta\). We now choose a bunch of different \(\delta\) and show that the good situation occurs with decent probability. Fix \(x, y \in X\). Choose \(0 < \delta_1 < \delta_2 < ...\) such that \(\delta_i\) is the smallest value such that \(B(x; \delta_i) := \{ z \in X : d(x, z) \leq \delta \}\) and \(B(y; \delta_i)\) both have \(2^i\) points in them. We keep choosing \(\delta_i\) until \(\delta_i < d(x, y) / 3\). Let \(\delta_{i+1} = d(x, y) / 3\). There are two subclaims: \(|d(x, S) - d(y, S)| \geq \delta_{i+1} - \delta_i\), and that this is decently likely to produce the nice picture above.

Why is the nice situation decently likely? By definition of \(\delta_{i+1}\), either \(<2^{i+1}\) points are in \(B(x, \delta_{i+1})\) or the same for \(B(y, \delta_{i+1})\). WLOG, assume this occurs for \(y\). By definition of \(\delta_i\), \(\geq 2^{i}\) points live in \(B(x, \delta_i)\). Since \(\mathbb{P}[x \in S_{ij}] = 2^{-i} \forall x \in X\) the good situation can be decomposed into the product of the probability that \(B(x, \delta_i)\) intersects with \(S\) and the probability that \(B(y, \delta_i)\) does not intersect with \(S\):

\[\mathbb{P}[\text{good picture}] = \mathbb{P}[B(x) \cap S] \mathbb{P}[B(y) \not \cap S] \geq (1 - (1 - 2^{-i}))^{2_i} (1 - 2^{-i})^{2i+1} \geq 2^{-5}\]Fix \(i\). The probability that at least \(2^{-6}\) of the \(S_{ij}\) have the “nice” situation is

\[1 - \mathbb{P}[\sum_j^{c \log n} \mathbb{I}[nice] \leq \frac{1}{2} \frac{c \log n}{2^5}] \geq 1 - \exp(-\frac{c log n}{8 2^5}) \geq 1 - n^{-3}\]if we choose \(c \geq 3 * 2^8\). Putting it together, we showed that when our nice situation occurs, \(|d(x, S) - d(y, S)| \geq \delta_{i+1} - \delta_i\), and that for each \(i\), the probability that at least \(d \log n / 2^6\) of the \(S_{ij}\) have this nice situation is at least \(1 - n^{-3}\). Then, by a union bound over all \(n^2\) pairs \((x, y)\) and \(\leq \log n\) choices of \(i\), with high probability, \(\forall i, x, y\), \(\geq c \log n / 2^6\) of the \(S_{ij}\) have \(|d(x, S) - d(y, S)| \geq \delta_{i+1} - \delta_i\), meaning:

\[\begin{align*} ||f(x) - f(y) ||_1 \geq \sum_{ij} |d(x, S_{ij}) - d(y, S_{ij})| \geq \sum_{ij} (\frac{c \log n}{2^6}) (\delta_{i+1} - \delta_i)\\ &= \frac{c \log n}{2^6} \delta_{t+1}\\ &= \frac{c \log n}{3 2^6} d(x, y)\\ \end{align*}\]Recalling that \(k = c \log^2 n\), we have

\[||f(x) - f(y) ||_1 \geq \frac{k}{3 2^t \log n} d(x, y)\]Bourgain’s Embedding says we can embed *any* metric space into \(\ell_1\) with distortion \(O(\log n)\).
One naturally asks next whether a better embedding is possible. In general, the answer is no, but
for more structured metric spaces, the answer is yes. The Johnson-Lindenstrauss (JL) Lemma tells us that if we have \(n\)
points that live in an \(\ell_2\) space of dimension \(d\), then we can embed those points into dimension
\(O(\frac{\log n}{\epsilon^2})\) with distortion \(1 + \epsilon\). Formally:

**Theorem:** Given \(\epsilon \in (0, 1)\) and a set \(X \subset \mathbb{R}^d\) with \(|X| = n\), there exist a linear
map \(f: \mathbb{R}^d \rightarrow \mathbb{R}^m\) where \(m = O(\frac{\log n}{\epsilon^2})\) that embeds
\((X, d_2)\) into \((\mathbb{R}^m, d_2)\) with distortion \(\leq 1 + \epsilon\).

Recall, distortion means that if the distance between any two points in the original space is \(\delta\), then in the new space, the distance is \((1 \pm \epsilon)\delta\).

Proof: Let \(A \in \mathbb{R}^{m \times d}\) be a matrix with entries chosen i.i.i. \(\mathcal{N}(0, 1/m)\). Define \(f(x)= Ax\). WLOG, because an isotropic Gaussian is rotationally symmetric, for any two \(x, y\), we can think of \(x - y = ||x - y||_2 e_1\), where \(e_1\) is the first canonical basis vector. Our goal is to show \(||A (x - y)||_2 = (1 \pm \epsilon) ||x - y|| \Leftrightarrow ||A e_1|| = (1 \pm \epsilon)\).

Now, we just need to consider the first column of \(A\) in our rotated system. We can use a Chernoff-style claim: If \(Z_1, ..., Z_m \sim N(0, 1)\), then \(\mathbb{P}[\sum Z_i^2 > (1+\epislon)m] \leq \exp(- m \epsilon^2 / 8)\). Thus, for any pair \(x, y\), the probability \(A(x -y)\) is distorted by \(\epsilon\) decays exponentially in \(\epsilon\). By a union bound, we have:

\[\mathbb{P}[\exists x, y \in X s.t. ||A (x-y)||_2 \geq (1+\epsilon) ||x - y||_2] \leq n^2 \exp(-m\epsilon^2/8)\]Choose \(m = O(\log(n)/\epsilon^2)\) and we’re good. A similar argument holds for the lower bound \(1-\epsilon\).