Quick Math Intuitions https://quickmathintuitions.org Sharing quick intuitions for math ideas Mon, 07 Dec 2020 15:33:15 +0000 en-US hourly 1 https://wordpress.org/?v=5.4.11 Projection methods in linear algebra numerics https://quickmathintuitions.org/projection-methods-linear-algebra-numerics/ https://quickmathintuitions.org/projection-methods-linear-algebra-numerics/#respond Mon, 07 Dec 2020 15:33:15 +0000 https://quickmathintuitions.org/?p=409 Linear algebra classes often jump straight to the definition of a projector (as a matrix) when talking about orthogonal projections in linear spaces. As often as it happens, it is…

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Linear algebra classes often jump straight to the definition of a projector (as a matrix) when talking about orthogonal projections in linear spaces. As often as it happens, it is not clear how that definition arises. This is what is covered in this post.

Orthogonal projection: how to build a projector

Case 1 – 2D projection over (1,0)

It is quite straightforward to understand that orthogonal projection over (1,0) can be practically achieved by zeroing out the second component of any 2D vector, at last if the vector is expressed with respect to the canonical basis \{ e_1, e_2 \}. Albeit an idiotic statement, it is worth restating: the orthogonal projection of a 2D vector amounts to its first component alone.

How can this be put math-wise? Since we know that the dot product evaluates the similarity between two vectors, we can use that to extract the first component of a vector v. Once we have the magnitude of the first component, we only need to multiply that by e_1 itself, to know how much in the direction of e_1 we need to go. For example, starting from v = (5,6), first we get the first component as v \cdot e_1 = (5,6) \cdot (1,0) = 5; then we multiply this value by e_1 itself: 5e_1 = (5,0). This is in fact the orthogonal projection of the original vector. Writing down the operations we did in sequence, with proper transposing, we get

    \[e_1^T (e_1 v^T) = \begin{bmatrix} 1 \\ 0 \end{bmatrix} ([1, 0] \begin{bmatrix} 5 \\ 6 \end{bmatrix}) .\]

One simple and yet useful fact is that when we project a vector, its norm must not increase. This should be intuitive: the projection process either takes information away from a vector (as in the case above), or rephrases what is already there. In any way, it certainly does not add any. We may rephrase our opening fact with the following proposition:

PROP 1: ||v|| \geq ||Projection(v)||.

This is can easily be seen through the pitagorean theorem (and in fact only holds for orthogonal projection, not oblique):

    \[||v||^2 = ||proj_u(v)||^2 + ||v - proj_u(v)||^2 \geq ||proj_u(v)||^2\]

Case 2 – 2D projection over (1,1)

Attempt to apply the same technique with a random projection target, however, does not seem to work. Suppose we want to project over (1,1). Repeating what we did above for a test vector [3,0], we would get

    \[\begin{bmatrix} 1 \\ 1 \end{bmatrix} ([3, 0] \begin{bmatrix} 1 \\ 1 \end{bmatrix}) =  [3,3].\]

This violates the previously discovered fact the norm of the projection should be \leq than the original norm, so it must be wrong. In fact, visual inspection reveals that the correct orthogonal projection of [3,0] is [\frac{3}{2}, \frac{3}{2}].

The caveat here is that the vector onto which we project must have norm 1. This is vital every time we care about the direction of something, but not its magnitude, such as in this case. Normalizing [1,1] yields [\frac{1}{\sqrt 2}, \frac{1}{\sqrt 2}]. Projecting [3,0] over [\frac{1}{\sqrt 2}, \frac{1}{\sqrt 2}] is obtained through

    \[\begin{bmatrix} \frac{1}{\sqrt 2} \\ \frac{1}{\sqrt 2} \end{bmatrix} ([3, 0] \begin{bmatrix} \frac{1}{\sqrt 2} \\ \frac{1}{\sqrt 2} \end{bmatrix}) =  [\frac{3}{2}, \frac{3}{2}],\]

which now is indeed correct!

PROP 2: The vector on which we project must be a unit vector (i.e. a norm 1 vector).

Case3 – 3D projection on a plane

A good thing to think about is what happens when we want to project on more than one vector. For example, what happens if we project a point in 3D space onto a plane? The ideas is pretty much the same, and the technicalities amount to stacking in a matrix the vectors that span the place onto which to project.

Suppose we want to project the vector v = [5,7,9] onto the place spanned by \{ [1,0,0], [0,1,0] \}. The steps are the same: we still need to know how much similar v is with respect to the other two individual vectors, and then to magnify those similarities in the respective directions.

    \[\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 5 \\ 7 \\ 9 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 5 \\ 7 \end{bmatrix} = 5 \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + 7 \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 5 \\ 7 \\ 0 \end{bmatrix}\]

The only difference with the previous cases being that vectors onto which to project are put together in matrix form, in a shape in which the operations we end up making are the same as we did for the single vector cases.

The rise of the projector

As we have seen, the projection of a vector v over a set of orthonormal vectors Z is obtained as

    \[Projection_Z(v) = Z^T Z v^T .\]

And up to now, we have always done first the last product Z v^T, taking advantage of associativity. It should come as no surprise that we can also do it the other way around: first Z^T Z and then afterwards multiply the result by v^T. This Z^T Z makes up the projection matrix. However, the idea is much more understandable when written in this expanded form, as it shows the process which leads to the projector.

THOREM 1: The projection of v over an orthonormal basis Z is

    \[Projection_Z(v) = Z^T Z v^T = \underbrace{P}_{Projector} v^T .\]

So here it is: take any basis of whatever linear space, make it orthonormal, stack it in a matrix, multiply it by itself transposed, and you get a matrix whose action will be to drop any vector from any higher dimensional space onto itself. Neat.

Projector matrix properties

  • The norm of the projected vector is less than or equal to the norm of the original vector.
  • A projection matrix is idempotent: once projected, further projections don’t do anything else. This, in fact, is the only requirement that defined a projector. The other fundamental property we had asked during the previous example, i.e. that the projection basis is orthonormal, is a consequence of this. This is the definition you find in textbooks: that P^2 = P. However, if the projection is orthogonal, as we have assumed up to now, then we must also have P = P^T.
  • The eigenvalues of a projector are only 1 and 0. For an eigenvalue \lambda,

        \[\lambda v = Pv = P^2v = \lambda Pv = \lambda^2 v \Rightarrow \lambda = \lambda^2 \Rightarrow \lambda = \{0,1\}\]

  • It exists a basis X of \mathbb{R}^n such that it is possible to write P as P = [I_k \ 0_{n-k}], with k being the rank of P. If we further decompose X = [X_1, X_2], with X_1 being N \times k and X_2 being N \times N-k, the existence of the basis X shows that P really sends points from \mathbb{R}^N into Im(X_1) = Im(P) and points from \mathbb{R}^N - P(\mathbb{R}^N) into Ker(P). It also shows that \mathbb{R}^N = Im(P) + Ker(P).

Model Order Reduction

Is there any application of projection matrices to applied math? Indeed.

It is often the case (or, at least, the hope) that the solution to a differential problem lies in a low-dimensional subspace of the full solution space. If some \textbf{w}(t) \in \mathbb{R}^N is the solution to the Ordinary Differential Equation

    \begin{equation*} \frac{d\textbf{w}(t)}{dt} = \textbf{f}(\textbf{w}(t), t) \end{equation*}

then there is hope that there exists some subspace \mathcal{S} \subset \mathbb{R}^, s.t. dim(\mathcal{S}) < N in which the solution lives. If that is the case, we may rewrite it as

    \[\textbf{w}(t) = \textbf{V}_\mathcal{S}\textbf{q}(t)\]

for some appropriate coefficients (q_i(t)), which are the components of \textbf{w}(t) over the basis \textbf{V}_\mathcal{S}.

Assuming that the base \textbf{V} itself is time-invariant, and that in general \textbf{Vq(t)} will be a good but not perfect approximation of the real solution, the original differential problem can be rewritten as:

    \begin{equation*} \begin{split} \frac{d}{dt}\textbf{Vq(t)} =  \textbf{f}(Vq(t), t) + \textbf{r}(t) \\ \textbf{V}\frac{d}{dt}\textbf{q(t)} =  \textbf{f}(Vq(t), t) + \textbf{r}(t) \\ \end{split} \end{equation*}

where \textbf{r(t)} is an error.

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Reproducing a transport instability in convection-diffusion equation https://quickmathintuitions.org/reproducing-transport-instability-convection-diffusion-equation/ https://quickmathintuitions.org/reproducing-transport-instability-convection-diffusion-equation/#respond Tue, 10 Nov 2020 14:39:47 +0000 https://quickmathintuitions.org/?p=411 Drawing from Larson-Bengzon FEM book, I wanted to experiment with transport instabilities. It looks there might be an instability in my ocean-ice model but before being able to address that,…

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Drawing from Larson-Bengzon FEM book, I wanted to experiment with transport instabilities. It looks there might be an instability in my ocean-ice model but before being able to address that, I wanted to wrap my head around the 1D simplest example one could find. And that is the Convection-Diffusion Equation:

(1)   \begin{equation*} \begin{split} - \epsilon \Delta u + b \cdot \nabla u &= f  \ \ in \ \Omega \\ u &= 0 \ \ on \ \partial \Omega \end{split} \end{equation*}

The first term -\epsilon \Delta u is responsible of the smearing of the velocity field u proportionally to \epsilon, and is thus called the diffusion term. Intuitively, it controls how much the neighbors of a given point x are influenced by the behavior of x; how much velocities (or temperatures: think of whatever you want!) diffuse in the domain.

The second term b \cdot \nabla u controls how much the velocities u are transported in the direction of the vector field b, and is thus called the convective term. A requirement for this problem to be well-posed is that \nabla \cdot b = 0 — otherwise it would mean that we allow velocities to vanish or to come into existence.

The instability is particularly likely to happen close to a boundary where a Dirichlet boundary condition is enforced. The problem is not localized only close to the boundary though, as fluctuations can travel throughout the whole domain and lead the whole solution astray.

Transport instability in 1D

The simplest convection-diffusion equation in 1D has the following form:

(2)   \begin{equation*} \epsilon u_{xx} + u_x = 1 \ \ in \ (0,1), \ \ u(0) = u(1) = 0 \end{equation*}

whose solution, for small \epsilon is approximately just u = x. This goes well with the boundary condition at 0, but not with the condition at 1, where the solution needs to go down to 0 quite abruptly to satisfy the boundary condition.

It’s easy to simulate the scenario with FEniCS and get this result (with \epsilon = 0.01 and the unit interval divided in 10 nodes):

transport-instability

in which we can infer two different trends: one with odd points and one with even points! In fact, if we discretize the 1D equation with finite elements we obtain:

(3)   \begin{equation*} \epsilon \frac{u_{i+1}-2u_i-u_{i-1}}{h^2} + \frac{u_{i+1}-u_{i-1}}{2h} = 1$ \end{equation*}

aha! From this we see that, if \epsilon is not of the same order of magnitude of h (i.e. if \epsilon << h), then the first factor becomes negligible. The problem then is that the second term contains only u_{i-1} and u_{i+1}. but not u_i. This will make it such that each node only talks to its second closest neighbor, explaining the behavior we saw in the plot before. It’s like even nodes make one solution and odd nodes a separate solution!

If \frac{\epsilon}{h} \approx 1, the solution that comes out is quite different:

transport-instability-solved

As we expected: a linear solution rapidly decaying towards the right. This is because in the above plot we had \epsilon = 0.01, h = 0.01, i.e. unit interval divided in 100 nodes.
Also notice how the problem does not pop up if the boundary conditions agree with the ideal solution (i.e. if the BC on the right is 1 instead of 0).

Solving the transport instability with a stabilization coefficient

The easiest dynamic way to fix the issue is to introduce an artificial component to \epsilon to make sure that the first term of the transport equation is never neglected, regardless of the relationship between mesh size h and \epsilon. This is a stabilization parameter:

(4)   \begin{equation*} (\epsilon + \beta h_{min}) \ u_{xx} + u_x = 1 \ \ in \ (0,1), \ \ u(0) = u(1) = 0 \end{equation*}

where h_{min} is the mesh smallest diameter. There is no single correct value for \beta: it quite depends on the other values (although it must be 0 \leq \beta < 1). Anyway, a good starting point is \beta = 0.5, which can then be tweaked according to results. This way feels a bit hacky though: “if we can’t solve it for \epsilon, let’s bump it up a bit” is pretty much the idea behind it.

With this formulation it’s also possible to derive what mesh size is needed to actually use a particular value for \epsilon. For example, if we’d like the second derivative term to have a 10^{-4} coefficient, then we need a mesh size h_{min} = \frac{10^{-4}}{\beta} \approx 10^{-3}, achieved with a 300×300 mesh, for example (which you can find out with
m=300; mesh=fenics.UnitSquareMesh(m,m); mesh.hmin()
). A uniformly fine mesh might not be needed though: it is often enough to have a coarse mesh in points where not much is happening, and very fine at problematic regions (such as boundaries, for this example).

Code — Convection-diffusion equation 1D

from fenics import *
import matplotlib.pyplot as plt

mesh = UnitIntervalMesh(100)
V = FunctionSpace(mesh, 'P', 1)

bcu = [
DirichletBC(V, Constant(0), 'near(x[0], 0)'),
DirichletBC(V, Constant(0), 'near(x[0], 1)'),
]
u = TrialFunction(V)
v = TestFunction(V)
u_ = Function(V)
f = Constant(1)
epsilon = Constant(0.01)
beta = Constant(0.5)
hmin = mesh.hmin()

a = (epsilon+beta*hmin)*dot(u.dx(0), v.dx(0))*dx + u.dx(0)*v*dx
L = v*dx

solve(a == L, u_, bcs=bcu)

print("||u|| = %s, ||u||_8 = %s" % ( \
round(norm(u_, 'L2'), 2), round(norm(u_.vector(), 'linf'), 3)
))

fig2 = plt.scatter(mesh.coordinates(), u_.compute_vertex_values())
plt.savefig('velxy.png', dpi = 300)
plt.close()

#plot(mesh)
#plt.show()

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What is the Rossby number? https://quickmathintuitions.org/what-is-the-rossby-number/ https://quickmathintuitions.org/what-is-the-rossby-number/#respond Wed, 21 Oct 2020 09:50:36 +0000 https://quickmathintuitions.org/?p=400 The Rossby number is used to describe whether a phenomenon is large-scale, i.e. if it is affected by earth’s rotation. But do we actually quantify if a fluid flow is…

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The Rossby number is used to describe whether a phenomenon is large-scale, i.e. if it is affected by earth’s rotation. But do we actually quantify if a fluid flow is affected by earth’s rotation?

Consider two quantities L and U, with L being a characteristic scale-length of the phenomenon (ex. distance between two peaks, distance between two isobars, length of simulation domain) and U the horizontal velocity scale of the motion. The ratio \frac{L}{U} is the time it takes to the motion to cover a distance L with velocity U. If this time is bigger than the period of earth’s rotation, then the phenomenon IS affected by the rotation.

So if \frac{L}{U} \geq \frac{1}{\Omega}, then the phenomenon IS a large-scale one. Thus we can define \epsilon = \frac{U}{2L \Omega} and say that for \epsilon \leq 1 a phenomenon is large scale. Phenomena with small Rossby number are dominated by Coriolis force behavior, while those with large Rossby number are dominated by inertial forces (ex: a tornado). However, rotational effects are more evident for low latitudes (i.e. near the equator), so the Rossby number can be different depending on where on earth we are.

(Notice that \Omega is in theory equal to 2 \Omega \sin(\phi), with \Omega being the earth rotational velocity and \phi the angle between the axis of rotation and the direction of fluid movement. In the geophysical context, flows are mostly horizontal (also due to density stratification in both atmosphere and ocean), so \sin(\phi) can be approximated with 1. There is a bunch of different notation, but this \Omega is also referred to as f, called the Coriolis frequency.)

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How do Dirichlet and Neumann boundary conditions affect Finite Element Methods variational formulations? https://quickmathintuitions.org/dirichlet-neumann-boundary-conditions-affect-finite-element-methods-variational-formulations/ https://quickmathintuitions.org/dirichlet-neumann-boundary-conditions-affect-finite-element-methods-variational-formulations/#respond Wed, 21 Oct 2020 09:43:09 +0000 https://quickmathintuitions.org/?p=382 To solve a classical second-order differential problem     with FEM, we first need to derive its weak formulation. This is achieved by multiplying the equation by a test function…

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To solve a classical second-order differential problem

    \begin{equation*} -(au')' + bu' + cu = f \ in \ \Omega \end{equation*}

with FEM, we first need to derive its weak formulation. This is achieved by multiplying the equation by a test function \phi and then integrating by parts to get rid of second order derivatives:

(1)   \begin{equation*} \begin{split} 0 &= \int_\Omega ((-a u')' + b u' + c u - f) \phi dx \\ &= \underbrace{\int_\Omega ((a u' \phi' + b u' \phi + c u \phi) dx}_{a(u, \phi)} \underbrace{- \int_\Omega f \phi dx - (a u' \phi)|_{\partial\Omega}}_{L(\phi)} \end{split} \end{equation*}

A typical FEM problem then reads like:

    \begin{equation*} \begin{split} \text{Find } u \in H_0'(\Omega) \ s.t. \ a(u, \phi) + L(\phi) = 0 \ \ \forall \phi \in H_0'(\Omega), \\ \text{where } H_0'(\Omega) = \{ v: \Omega \rightarrow \mathbb{R} : \int_0^1v^2(x) + v'(x)^2 dx < \infty \}. \end{split} \end{equation*}

What is the difference between imposing Dirichlet boundary conditions (ex. u(\partial \Omega) = k) and Neumann ones (u'(\partial \Omega) = k(x)) from a math perspective? Dirichlet conditions go into the definition of the space H_0', while Neumann conditions do not. Neumann conditions only affect the variational problem formulation straight away.

For example, in one dimension, adding the Dirichlet condition v(0) = v(1) = 0 results in the function space change H_0'(\Omega) = \{ v \in \Omega_0' : v(0)=v(1)=0 \}. With this condition, the boundary term (a u' \phi)|_{\partial\Omega} would also zero out in the variational problem. because the test function \phi belongs to H_0'.

On the other hand, by adding the Neumann condition u'(0) = u'(1) = 0, the space H_0' does not change, even though the boundary term vanishes from the variational problem in the same way as the for the Dirichlet condition. However, that term goes to zero not because of the test function anymore, but because of the value of the derivative u'. If the Neumann condition had specified a different value, such as u'(0) = u'(1) = 5, then the boundary term would not zero out!

In other words, Dirichlet conditions have the effect of further constraining the solution function space, while Neumann conditions only affect the equations.

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A gentle (and short) introduction to Gröbner Bases https://quickmathintuitions.org/gentle-introduction-grobner-bases/ https://quickmathintuitions.org/gentle-introduction-grobner-bases/#respond Wed, 03 Jun 2020 08:19:27 +0000 https://quickmathintuitions.org/?p=372 Taken from my report for a Computer Algebra course. Motivation We know there are plenty of methods to solve a system of linear equations (to name a few: Gauss elimination,…

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Taken from my report for a Computer Algebra course.

Motivation

We know there are plenty of methods to solve a system of linear equations (to name a few: Gauss elimination, QR or LU factorization). In fact, it is straightforward to check whether a linear system has any solutions, and if it does, how many of them there are. But what if the system is made of non-linear equations? The invention of Groebner bases and the field of computational algebra came up to answer these questions.

In this text we will recap the theory behind single-variable polynomials and extend it to multiple-variable ones, ultimately getting to the definition of Groebner bases.

In some cases, the transition from one to multiple variables is smooth and pretty much an extension of the simple case (for example for the Greatest Common Divisor algorithm). In other cases, however, there are conceptual jumps to be made. To give an example, single variable polynomials have always a finite number of roots, while this does not hold for multivariable polynomials. Intuitively, the reason is that a polynomial in one variable describes a curve in the plane, which can only intersect the x-axis a discrete and finite number of times. On the other hand, a multivariate polynomial describes a surface in space, which will always intersect the 0-plane in a continuum of points.

Preliminaries

All throughout these notes, it will be important to have in mind some basic algebra definitions.

To begin with, we ask what is the most basic (but useful) structure we can put on a set. We ask, for example, given the set of natural numbers, what do we need to do to allow basic manipulation (i.e. summation)? This leads us to the definition of group.

DEF 1: A group is made of a set \mathcal{G} with one binary operation + such that:

  • The operation is closed: a+b \in \mathcal{G} \ \forall a,b \in \mathcal{G}
  • The operation is associative: a+(b+c)=(a+b)+c \ \forall a,b,c \in \mathcal{G}
  • The operation + has an identity element 0 s.t. g+0 = g \ \forall g \mathcal{G}
  • Each element has an inverse element: \forall g \in \mathcal{G}, \exists h \in \mathcal{G} : g+h=0

A group is usually denoted with (\mathcal{G}, +).
Notice that we did not ask anything about commutativity!

Then, the notion of group can be made richer and more complex: first into that of ring, then into that of field.

DEF 2: A ring is a group with an extra operation (\mathcal{G}, +, *) which sastisfies the following properties:

  • The operation + is commutative: a+b=b+a \ \forall a,b \in \mathcal{G}
  • The operation * is closed: a*b \in \mathcal{G} \ \forall a,b \in \mathcal{G}
  • The operation * has an identity element 1 s.t. g*1 = g \ \forall g
  • The operation * is associative: a*(b*c)=(a*b)*c \ \forall a,b,c \in \mathcal{G}
  • The operation * is distributive with respect to +

DEF. 3: A field \mathcal{K} is a ring in which all elements have an inverse with respect to the operation *.

All throughout these notes, the symbol \mathcal{K} will denote a field.

DEF 4: A monomial is a product x_1^{\alpha_1} \cdots x_n^{\alpha_n}, with \alpha_i \in \mathbb{N}. Its degree is the sum of the exponents.

DEF 5: A polynomial is a linear combinations of monomials.

We conclude by noting that the space of polynomials with coefficients taken from a field \mathcal{K} makes a ring, denoted with \mathcal{K}[x_1, \cdots, x_n].

Affine varieties and ideals

Our first step towards formalizing the theory for non-linear systems is to understand what the space of solutions looks like. As much as we know that linear spaces are the solutions spaces for linear systems, there is something analogous for non-linear systems, and that is affine varieties.

DEF 6: Given f_1, \cdots, f_s polynomials in \mathcal{K}[x_1, \cdots, x_n], the affine variety over them is the set of their common roots:

    \[V(f_1, \cdots, f_s) = \{ (a_1, \cdots, a_n) \in \mathcal{K}^n : f_i(a_1, \cdots, a_n) = 0 \ \forall i = 1, \cdots, s\}\]

EX 1: V(x_1+x_2-1, x_2+1) = \{ (2, -1) \}

When working with rings, as it is our case, the notion of ideal is important. The reason for its importance is that ideals turn out to be kernels of ring homomorphisms — or, in other words, that they are the “good sets” that can be used to take ring quotients.

DEF 7: An ideal is a subset I \subset \mathcal{K}[x_1, \cdots, x_n] such that:

  • 0 \in I
  • it is closed w.r.t +: f+g \in I \ \forall f,g \in I
  • it is closed w.r.t * for elements in the ring: f*g \in I \ \forall f \in I, g \in \mathcal{K}[x_1, \cdots, x_n]

Given some elements of a ring, we might wonder what is the way to build an ideal (the smallest) that would contain them.

DEF 8: Given f_1, \cdots, f_s polynomials, the ideal generated by them is the set of combinations with coefficients taken from the ring:

    \[<f_1, \cdots, f_s> = \{ \sum_i^s h_i f_i, \ \ h_i \in \mathcal{K}[x_1, \cdots, x_n] \}\]

Having introduced ideals, we immediately find a result that is linked to our purpose of non-linear systems inspection: a simple way to check if a system has solutions or not.

THEO 1: If 1 \in I=<f_1, \cdots, f_s>, then V(I) = \emptyset.
PROOF: Since 1 \in I, it must be possible to write it as a combination of the form 1 = \sum h_i f_i. Now, if we suppose that V(I) is not empty, then one of its points a is a root of all the f_i. This would mean that \sum h_i f_i(a) = 0 \neq 1, which is absurd.

Groebner bases

Groebner bases give a computational method for solving non-linear systems of equations through an apt sequence of intersection of ideals. To state its definition, we first need to know what a monomial ordering is. Intuitively, we can think of such an ordering as a way to compare monomials — the technical definition does not add much more concept. Different orderings are possible.

Once we have a way of ordering monomials, it is also possible to define the leading monomial (denoted as LM) of a given polynomial. For single variable polynomials it is pretty straightforward, but for the multi-variate case we need to define an ordering first (some possible options are: lexicographic, graded lexicographic, graded reverse lexicographic).

DEF 9: Given a monomial ordering, a Groebner basis of an ideal I w.r.t the ordering is a finite subset G = \{ g_1, \cdots, g_s \} \subset I s.t. <LM(g_1), \cdots, LM(g_s)> = LM(I).

This basis is a generating set for the ideal, but notice how it depends on the ordering! Finally, it is possible to prove that every ideal has a Groebner basis (Hilbert’s basis theorem).

From here now, the rationale is that, given a system of polynomial equations, we can see the polynomials as generators of some ideal. That ideal will have a Groebner basis, and there is an algorithm to build one (Buchberger algorithm). From there, apt ideal operations will allow to solve the system by eliminating the variables.

We now describe this elimination algorithm with an example:

(1)   \begin{equation*}  \begin{cases} x^2+y+z=1 \\ x + y^2 +z=1 \\ x+y+z^2=1 \end{cases} \end{equation*}

Given the ideal

    \[I = <x^2+y+z-1, x + y^2 +z-1, x+y+z^2-1>,\]

then a Groebner basis with respect to the (lexicographical order) is

(2)   \begin{equation*} \begin{cases} g_1=x+y+z^2-1 \\ g_2=y^2-y-z^2+z \\ g_3=2yz^2+z^4-z^2\\ g_4=z^6-4z^4+4z^3-z^2 \end{cases} \end{equation*}

which can be used to compute the solutions of the initial system (1).

To do so, first consider the ideal I \cap \mathbb{C}[z], which practically corresponds to all polynomials in I where x,y are not present. In our case, we are left only with one element from the basis which only involve z: g_4=z^6-4z^4+4z^3-z^2. The roots of g_4 are 0,1,-1 \pm \sqrt{2}.

The values for z can then be used to find the possible values for y using polynomial g_3, g_2, which only involve y,z. Finally, once possible values for y,z are known, they can be used to find the corresponding values for x through g_1.

This example will yield the following solutions:

(3)   \begin{equation*} \begin{cases} (1, 0, 0), (0, 1, 0), (0, 0, 1), \\ (-1 + \sqrt{2}, -1 + \sqrt{2}, -1 + \sqrt{2}), \\ (-1 - \sqrt{2}, -1 - \sqrt{2}, -1 - \sqrt{2}) \end{cases} \end{equation*}

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What is the difference between Finite Differences and Finite Element Methods? https://quickmathintuitions.org/different-finite-differences-finite-element-methods/ https://quickmathintuitions.org/different-finite-differences-finite-element-methods/#respond Wed, 27 May 2020 08:46:23 +0000 https://quickmathintuitions.org/?p=368 With Finite Differences, we discretize space (i.e. we put  a grid on it) and we seek the values of the solution function at the mesh points. We still solve a…

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With Finite Differences, we discretize space (i.e. we put  a grid on it) and we seek the values of the solution function at the mesh points. We still solve a discretized differential problem.

With Finite Elements, we approximate the solution as a (finite) sum of functions defined on the discretized space. These functions make up a basis of the space, and the most commonly used are the hat functions. We end up with a linear system whose unknowns are the weights associated with each of the basis functions: i.e., how much does each basis function count for out particular solution to our particular problem?

Brutally, it is finding the value of the solution function at grid points (finite differences) vs the weight of the linear combinations of the hat functions (finite elements).

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The role of intuitions in mathematics https://quickmathintuitions.org/role-intuitions-mathematics/ https://quickmathintuitions.org/role-intuitions-mathematics/#respond Fri, 01 Nov 2019 21:21:09 +0000 http://quickmathintuitions.org/?p=365 Some thoughts and questions about the role of intuition in mathematics: Is intuition needed to really understand a topic? I would say yes, since in the end we reason through…

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Some thoughts and questions about the role of intuition in mathematics:

  • Is intuition needed to really understand a topic?
    I would say yes, since in the end we reason through ideas, of which we have an intuitive representation. Without intuitions, it is difficult to relate topics with each other as we lack in hooks, and we often lack a deep understanding as well.
  • Do you feel like you have understood something even if you do not have an intuitive representation of it?
  • How does formalism complete intuition?
    It shows whether and how an intuition is right. Sometimes intuition can be deceitful and/or tricky, especially in high dimensions or very abstract topics.
  • Can/Should intuitions be taught? Or are they only effective when discovered on one’s own?
    I side more with the latter. This is bordering with Maths Education, but I deem the process more important than the result – it is the tough digestion of some math material that ultimately leads to developing an intuition what really makes the intuition strong in one’s mind. If somebody else (like a teacher) does the work for us, then the result does not really stick, albeit nice it may be.
  • Can we say somebody with only intuitions (well understood and well reasoned) is a mathematician?
    I would say yes. I often find the intuitive side more important than the formal one.
  • Is it possible to develop intuitions for very abstract topics? If yes, what shape would they have, since there is rarely anything visual we can hook up to?
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A note on the hopes for Fully Homomorphic Signatures https://quickmathintuitions.org/note-hopes-fully-homomorphic-signatures/ https://quickmathintuitions.org/note-hopes-fully-homomorphic-signatures/#respond Thu, 29 Aug 2019 15:34:49 +0000 http://quickmathintuitions.org/?p=351 This is taken from my Master Thesis on Homomorphic Signatures over Lattices. What are homomorphic signatures Imagine that Alice owns a large data set, over which she would like to…

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This is taken from my Master Thesis on Homomorphic Signatures over Lattices.

What are homomorphic signatures

Imagine that Alice owns a large data set, over which she would like to perform some computation. In a homomorphic signature scheme, Alice signs the data set with her secret key and uploads the signed data to an untrusted server. The server then performs the computation modeled by the function g to obtain the result y = g(x) over the signed data.

Alongside the result y, the server also computes a signature \sigma_{g,y} certifying that y is the correct result for g(x). The signature should be short – at any rate, it must be independent of the size of x. Using Alice’s public verification key, anybody can verify the tuple (g,y,\sigma_{g,y}) without having to retrieve all the data set x nor to run the computation g(x) on their own again.

The signature \sigma_{g,y} is a homomorphic signature, where homomorphic has the same meaning as the mathematical definition: ‘mapping of a mathematical structure into another one in such a way that the result obtained by applying the operations to elements of the first structure is mapped onto the result obtained by applying the corresponding operations to their respective images in the second one‘. In our case, the operations are represented by the function f, and the mapping is from the matrices U_i \in \mathbb{Z}_q^{n \times n} to the matrices V_i \in \mathbb{Z}_q^{n \times m}.

homomorphic signatures

Notice how the very idea of homomorphic signatures challenges the basic security requirements of traditional digital signatures. In fact, for a traditional signatures scheme we require that it should be computationally infeasible to generate a valid signature for a party without knowing that party’s private key. Here, we need to be able to generate a valid signature on some data (i.e. results of computation, like g(x)) without knowing the secret key. What we require, though, is that it must be computationally infeasible to forge a valid signature \sigma' for a result y' \neq g(x). In other words, the security requirement is that it must not be possible to cheat on the signature of the result: if the provided result is validly signed, then it must be the correct result.

The next ideas stem from the analysis of the signature scheme devised by Gorbunov, Vaikuntanathan and Wichs. It relies on the Short Integer Solution hard problem on lattices. The scheme presents several limitations and possible improvements, but it is also the first homomorphic signature scheme able to evaluate arbitrary arithmetic circuits over signed data.

Def. – A signature scheme is said to be leveled homomorphic if it can only evaluate circuits of fixed depth d over the signed data, with d being function of the security parameter. In particular, each signature \sigma_i comes with a noise level \beta_i: if, combining the signatures into the result signature \sigma, the noise level grows to exceed a given threshold \beta^*, then the signature \sigma is no longer guaranteed to be correct.

Def. – A signature scheme is said to be fully homomorphic if it supports the evaluation of any arithmetic circuit (albeit possibly being of fixed size, i.e. leveled). In other words, there is no limitation on the richness” of the function to be evaluated, although there may be on its complexity.

Let us remark that, to date, no (non-leveled) fully homomorphic signature scheme has been devised yet. The state of the art still lies in leveled schemes. On the other hand, a great breakthrough was the invention of a fully homomorphic encryption scheme by Craig Gentry.

On the hopes for homomorphic signatures

The main limitation of the current construction (GVW15) is that verifying the correctness of the computation takes Alice roughly as much time as the computation of g(x) itself. However, what she gains is that she does not have to store the data set long term, but can do only with the signatures.

To us, this limitation makes intuitive sense, and it is worth comparing it with real life. In fact, if one wants to judge the work of someone else, they cannot just look at it without any preparatory work. Instead, they have to have spent (at least) a comparable amount of time studying/learning the content to be able to evaluate the work.

For example, a good musician is required to evaluate the performance of Beethoven’s Ninth Symphony by some orchestra. Notice how anybody with some musical knowledge could evaluate whether what is being played makes sense (for instance, whether it actually is the Ninth Symphony and not something else). On the other hand, evaluating the perfection of performance is something entirely different and requires years of study in the music field and in-depth knowledge of the particular symphony itself.

That is why it looks like hoping to devise a homomorphic scheme in which the verification time is significantly shorter than the computation time would be against what is rightful to hope. It may be easy to judge whether the result makes sense (for example, it is not a letter if we expected an integer), but is difficult if we want to evaluate perfect correctness.

However, there is one more caveat. If Alice has to verify the result of the same function g over two different data sets, then the verification cost is basically the same (amortized verification). Again, this makes sense: when one is skilled enough to evaluate the performance of the Ninth Symphony by the Berlin Philharmonic, they are also skilled enough to evaluate the performance of the same piece by the Vienna Philharmonic, without having to undergo any significant further work other than going and listening to the performance.

 

So, although it does not seem feasible to devise a scheme that guarantees the correctness of the result and in which the verification complexity is significantly less than the computation complexity, not all hope for improvements is lost. In fact, it may be possible to obtain a scheme in which verification is faster, but the correctness is only probabilistically guaranteed.

Back to our music analogy, we can imagine the evaluator listening to a handful of minutes of the Symphony and evaluate the whole performance from the little he has heard. However, the orchestra has no idea at what time the evaluator will show up, and for how long they will listen. Clearly, if the orchestra makes a mistake in those few minutes, the performance is not perfect; on the other hand, if what they hear is flawless, then there is some probability that the whole play is perfect.

Similarly, the scheme may be tweaked to only partially check the signature result, thus assigning a probabilistic measure of correctness. As a rough example, we may think of not computing the homomorphic transformations over the U_i matrices wholly, but only calculating a few, randomly-placed entries. Then, if those entries are all correct, it is very unlikely (and it quickly gets more so as the number of checked entries increases, of course) that the result is wrong. After all, to cheat, the third party would need to guess several numbers in \mathbb{Z}_q, each having 1/q likelihood of coming up!

Another idea would be for the music evaluator to delegate another person to check for the quality of the performance, by giving them some precise and detailed features to look for when hearing the play. In the homomorphic scheme, this may translate in looking for some specific features in the result, some characteristics we know a priori that must be in the result. For example, we may know that the result must be a prime number, or must satisfy some constraint, or a relation with something much easier to check. In other words, we may be able to reduce the correctness check to a few fundamental traits that are very easy to check, but also provide some guarantee of correctness. This method seems much harder to model, though.

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Probability as a measure of ignorance https://quickmathintuitions.org/probability-as-measure-of-ignorance/ https://quickmathintuitions.org/probability-as-measure-of-ignorance/#comments Sat, 11 May 2019 06:29:32 +0000 http://quickmathintuitions.org/?p=335 One of the most beautiful intuitions about probability measures came from Rovelli’s book, that took it in turn from Bruno de Finetti. What does a probability measure measure? Sure, the…

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One of the most beautiful intuitions about probability measures came from Rovelli’s book, that took it in turn from Bruno de Finetti.

What does a probability measure measure? Sure, the open sets of the \sigma-algebra that supports the measure space. But really, what? Thinking about it, it is very difficult to define probability without using the word probable or possible.

Well, probability measures our ignorance about something.

When we make some claim with 90% probability, what we are really saying is that the knowledge we have allows us to make a prediction that is that much accurate. And the main point here is that different people may assign different probabilities to the very same claim! If you have ever seen weather forecasts for the same day disagree, you know what I am talking about. Different data or different models can generate different knowledge, and thus different probability figures.

But we do not have to go that far to find reasonable examples. Let’s consider a very simple one. Imagine you found yourself on a train, and in front of you is sitting a girl with clothes branded Patagonia. What would be the odds that the girl has been to Patagonia? Not more than average, you would guess, because Patagonia is just a brand that makes warm clothes, and can be purchased in several stores all around the world, probably even more than in Patagonia itself! So you would probably say that is surely no more than 50% likely.

But now imagine a kid in the same scenario. If they see a girl with Patagonia clothes, they would immediately think that she had been to Patagonia (with probability 100% this time), because they are lacking a good amount of important information that you instead hold. And so the figure associated with \mathbb{P}(\text{The girl has been to Patagonia} | \text{The girl has a Patagonia jacket}) is pretty different depending on the observer, or rather on the knowledge (or lack of) they possess. In this sense probability is a measure of our ignorance.

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But WHY is the Lattices Bounded Distance Decoding Problem difficult? https://quickmathintuitions.org/why-lattices-bounded-distance-decoding-problem-difficult/ https://quickmathintuitions.org/why-lattices-bounded-distance-decoding-problem-difficult/#respond Wed, 08 May 2019 20:54:04 +0000 http://quickmathintuitions.org/?p=306 This is taken from my Master Thesis on Homomorphic Signatures over Lattices. Introduction to lattices and the Bounded Distance Decoding Problem A lattice is a discrete subgroup , where the…

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This is taken from my Master Thesis on Homomorphic Signatures over Lattices.

Introduction to lattices and the Bounded Distance Decoding Problem

A lattice is a discrete subgroup \mathcal{L} \subset \mathbb{R}^n, where the word discrete means that each x \in \mathcal{L} has a neighborhood in \mathbb{R}^n that, when intersected with \mathcal{L} results in x itself only. One can think of lattices as being grids, although the coordinates of the points need not be integer. Indeed, all lattices are isomorphic to \mathbb{Z}^n, but it may be a grid of points with non-integer coordinates.

Another very nice way to define a lattice is: given n independent vectors b_i \in \mathbb{R}^n, the lattice \mathcal{L} generated by that base is the set of all linear combinations of them with integer coefficients:

    \[\mathcal{L} = \{\sum\limits_{i=0}^{n} z_i b_i, \ b_i \in \mathbb{R}^n, z_i \in \mathbb{Z} \}\]

Then, we can go on to define the Bounded Distance Decoding problem (BDD), which is used in lattice-based cryptography (more specifically, for example in trapdoor homomorphic encryption) and believed to be hard in general.

Given an arbitrary basis of a lattice \mathcal{L}, and a point x \in \mathbb{R}^n not necessarily belonging to \mathcal{L}, find the point of \mathcal{L} that is closest to x. We are also guaranteed that x is very close to one of the lattice points. Notice how we are relying on an arbitrary basis – if we claim to be able to solve the problem, we should be able to do so with any basis.

Bounded Distance Problem example

Now, as the literature goes, this is a problem that is hard in general, but easy if the basis is nice enough. So, for example for encryption, the idea is that we can encode our secret message as a lattice point, and then add to it some small noise (i.e. a small element v \in \mathbb{R}^n). This basically generates an instance of the BDD problem, and then the decoding can only be done by someone who holds the good basis for the lattice, while those having a bad basis are going to have a hard time decrypting the ciphertext.

However, albeit of course there is no proof of this (it is a problem believed to be hard), I wanted to get at least some clue on why it should be easy with a nice basis and hard with a bad one (GGH is an example schema that employs techniques based on this).

So now to our real question. But WHY is the Bounded Distance Decoding problem hard (or easy)?

Why the Bounded Distance Decoding problem is easy with a nice basis

Let’s first say what a good basis is. A basis is good if it is made of nearly orthogonal short vectors. This is a pretty vague definition, so let’s make it a bit more specific (although tighter): we want a base in which each of its b_i is of the form (0, ..., 0, k, 0, ..., 0) for some k \in \mathbb{R}. One can imagine k being smaller than some random value, like 10. (This shortness is pretty vague and its role will be clearer later.) In other words, a nice basis is the canonical one, in which each vector has been re-scaled by an independent real factor.

To get a flavor of why the Bounded Distance Decoding problem is easy with a nice basis, let’s make an example. Consider \mathbb{R}^2, with b_0 = (\frac{1}{2}, 0), b_1 = (0, \frac{5}{4}) as basis vectors. Suppose we are given x = (\frac{3}{7}, \frac{9}{10}) as challenge point. It does not belong to the lattice generated by b_0, b_1, but it is only (\frac{1}{14}, \frac{9}{25}) away from the point (\frac{1}{2}, \frac{5}{4}), which does belong to the lattice.

Now, what does one have to do to solve this problem? Let’s get a graphical feeling for it and formalize it.

Buonded Distance Decoding problem example with good basis
Buonded Distance Decoding problem example with good basis

We are looking for the lattice point closest to x. So, sitting on x, we are looking for the linear combination with integer coefficients of the basis vectors that is closest to us. Breaking it component-wise, we are looking for \min y, z \in \mathbb{R} and k, j \in \mathbb{Z} such that they are solution of:

    \[\begin{cases} \frac{3}{7} + y = \frac{1}{2} k \\ \frac{9}{10} + z = \frac{5}{4} j \end{cases}\]

This may seem a difficult optimization problem, but in truth it is very simple! The reason is that each of the equations is independent, so we can solve them one by one – the individual minimum problems are easy and can be solved quickly. (One could also put boundaries on y, z with respect to the norm of the basis vectors, but it is not vital now.)

So the overall complexity of solving BDD with a good basis is \theta(\theta(\min)n), which is okay.

Why the Bounded Distance Decoding problem is hard with a bad basis

A bad basis is any basis that does not satisfy any of the two conditions of a nice basis: it may be poorly orthogonal, or may be made of long vectors. We will later try to understand what roles these differences play in solving the problem: for now, let’s just consider an example again.

Another basis for the lattice generated by the nice basis we picked before ((\frac{1}{2}, 0), (0, \frac{5}{4})) is b_0 = (\frac{9}{2}, \frac{5}{4}), b_1 = (5, \frac{10}{4}). This is a bad one.

Buonded Distance Decoding problem example with bad basis
Buonded Distance Decoding problem example with bad basis

Let’s write down the system of equations coordinate-wise as we did for the nice basis. We are looking for \min y, z \in \mathbb{R} and k, j \in \mathbb{Z} such that they are solution of:

    \[\begin{cases} \frac{3}{7} + y = \frac{9}{2} k + 5 j \\ \frac{9}{10} + z = \frac{5}{4} k + \frac{10}{4} j \end{cases}\]

Now look! This may look similar as before, but this time it really is a system, the equations are no longer independent: we have 3 unknowns and 2 equations. The system is under-determined! This already means that, in principle, there are infinite solutions. Moreover, we are also trying to find a solution that is constrained to be minimum. Especially with big n, solving this optimization problem can definitely be non-trivial!

On the differences between a good and a bad basis

So far so good: we have discovered why the Bounded Distance Decoding problem is easy with a good basis and difficult with a bad one. But still, what does a good basis have to make it easy? How do its properties related to easy of solution?

We enforced two conditions: orthogonality and shortness. Actually, we even required something stronger than orthogonality: that the good basis was basically a stretched version of the canonical one – i.e. had only one non-zero entry.

Let’s think for a second in terms of canonical basis \{e_i = (0, ..., 0, 1, 0, ... 0)\}. This is what makes the minimum problems independent and allows for easy resolution of the BDD problem. However, when dealing with cryptography matters, we cannot always use the same basis, we need some randomness. That is why we required to use a set of independent vectors each having only one non-zero coordinate: it is the main feature that makes the problem easy (at least for the party having the good basis).

We also asked for shortness. This does not give immediate advantage to who holds the good basis, but makes it harder to solve the problem for those holding the bad one. The idea is that, given a challenge point x \in \mathbb{R}^n, if we have short basis vectors, we can take small steps from it and look around us for nearby points. It may take some time to find the best one, but we are still not looking totally astray. Instead, if we have long vectors, every time we use one we have to make a big leap in one direction. In other words, who has the good basis knows the step size of the lattice, and thus can take steps of considerate size. slowly poking around; who has the bad basis takes huge jumps and may have a hard time pinpointing the right point.

It is true, though, that the features of a good basis usually only include shortness and orthogonality, and not the “rescaling of the canonical basis” we assumed in the first place. So, let’s consider a basis of that kind, like \{v_1 = (\frac{\sqrt{3}}{2}, \frac{1}{2}), v_2 = (\frac{1}{2}, \frac{\sqrt{3}}{2})\}. If we wrote down the minimum problem we would have to solve given a challenge point, it would be pretty similar to the one with the bad basis, with the equations not being independent. Looks like bad luck, uh?

However, not all hope is lost! In fact, we can look for the rotation matrix that will turn that basis into a stretching of the canonical one, finding v_1', v_2'! Then we can rotate the challenge point x as well, and solve the problem with respect to those new basis vectors. Of course that is not going to be the solution to the problem, but we can easily rotate it back to find the real solution!

However, given that using a basis of this kind does not make the opponent job any harder, but only increases the computational cost for the honest party, I do not see why this should ever be used. Instead, I guess the best choices for good basis are the stretched canonical ones.

(This may be obvious, but having a generic orthogonal basis is not enough for an opponent to break the problem. If it is orthogonal, but its vectors are long, bad luck!)

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