Lagrange multipliers Part Two

My own question on last week’s BMO2 notwithstanding, inequalities seem out of fashion at the moment among mainstream international olympiads. Such problems often involve minimising some function subject to a constraint, and word has, over the years, filtered down to students interested in such things, that there’s a general method for achieving this via Lagrange multipliers. The motivation for my talk in Hungary, summarised by the previous post and this one, is to dispute the claim made by some of the UK students that these are hard to justify rigorously. I dispute this because I don’t think it’s qualitatively much harder to justify Lagrange multipliers rigorously than an unconstrained optimisation problem, whereas I would claim instead that Lagrange multipliers are merely unlikely to work at a computational level on the majority of olympiad problems.

Unconstrained optimisation in two variables

Before we can possibly discuss constrained optimisation, we should discuss unconstrained optimisation. That is, finding minima of a function of several variables. We don’t lose too much by assuming that our function f(x,y) depends on two variables.

Recall that our method in the previous post for justifying the A-level approach to minima was to find a necessary condition to be a local minimum, and also a general reason why there should be a global minimum. That way, if there are finitely many points satisfying the condition, we just check all of them, and the one with the smallest value of f is the global minimum. We’ll discuss the existence of the global minimum later.

If we hold one coordinate fixed, the local variation of a function is equivalent to the one-dimensional case.

$f(x+h,y)-f(x,y)= h\frac{\partial f}{\partial x}(x,y)+O(h^2).$

In general we want to vary both variables, which is fine since

$f(x+h,y+\ell)-f(x+h,y)=\ell \frac{\partial f}{\partial y}(x+h,y) + O(h^2).$

But since we really want everything to be determined by the function at (x,y), we really want

$\frac{\partial f}{\partial y}(x+h,y) \approx \frac{\partial f}{\partial y}(x,y),$

and so we be mindful that we may have to assume that both partial derivatives are continuous everywhere they exist. Once we have this though, we can rewrite as

$f(x+h,y+\ell) - f(x,y)= h\frac{\partial f}{\partial x}(x,y) + \ell \frac{\partial f}{\partial y}(x,y) + O(h^2)$

$= (\frac{\partial f}{\partial x}(x,y) ,\frac{\partial f}{\partial y}(x,y) )\cdot (h,\ell) + O(h\vee \ell^2).$

In particular, if we define grad of f to be $\nabla f(x,y)=(\frac{\partial f}{\partial x}(x,y), \frac{\partial f}{\partial y}(x,y) )$ and apply a similar argument to that which we used in the original setting. If $\nabla f(x,y)\ne 0$, then we can choose some small $(h,\ell)$, such that $f(x+h,y+\ell). Thus a necessary condition for (x,y) to be a local minimum for f is that $\nabla f=0$.

Lagrange multipliers

This is natural time to discuss where Lagrange multipliers emerge. The setting now is that we still want to minimise some function $f(\mathbf{x})$, but only across those values of $\mathbf{x}$ which satisfy the constraint $g(\mathbf{x})=0$.

But our approach is exactly the same, namely we find a necessary condition to be a local minimum subject to the condition. As before, we have

$f(\mathbf{x}+\mathbf{h}) - f(\mathbf{x}) = \mathbf{h}\cdot \nabla f + O(|\mathbf{h}|^2),$

but we are only interested in those small vectors $\mathbf{h}$ for which $\mathbf{x}+\mathbf{h}$ actually satisfies the constraint, namely $g(\mathbf{x}+\mathbf{h})=0$. But then

$0 = g(\mathbf{x}+\mathbf{h})- g(\mathbf{x})=\mathbf{h}\cdot \nabla g + O(|\mathbf{h}|^2).$

From this, we conclude that the set of small relevant $\mathbf{h}$ is described by $\mathbf{h}\nabla g=O(|\mathbf{h}|^2)$. And now we really can revert to the original argument. If there’s a small $\mathbf{h}$ such that $\mathbf{h}\cdot \nabla g=0$ but $\mathbf{h}\cdot \nabla f \ne 0$, then we can find some $\mathbf{h'_+},\mathbf{h'_-}=\pm \mathbf{h}+O(|\mathbf{h}|^2)$ such that at least one of $f(\mathbf{h'_+}),f(\mathbf{h'_-}).

So a necessary condition to be a constrained local minimum is that every vector which is perpendicular to $\nabla g$ must also be perpendicular to $\nabla f$. From which it follows that these two vectors must be parallel, that is $\nabla f(\mathbf{x})=\lambda \nabla g(\mathbf{x})$, where $\lambda$ is the so-called Lagrange multiplier. Of course we must also have that $g(\mathbf{x})=0$, and so we have a complete characterisation for a necessary condition that the constrained optimisation has a local minimum at $\mathbf{x}$, assuming that all the derivatives of both f and g exist with suitable regularity near $\mathbf{x}$.

Technicalities

The point of setting up the one-variable case in the unusual way in the previous post was to allow me to say at this stage: “it’s exactly the same”. Well, we’ve already seen an extra differentiability condition we might require, but apart from that, the same approach holds. Multi-variate continuous functions also attain their bounds when the domain is finite and includes its boundary.

Checking the boundary might be more complicated in this setting. If the underlying domain is $\{x,y,z\ge 0\}$, then one will have to produce a separate argument for why the behaviour when at least one of the variables is zero fits what you are looking for. Especially in the constrained case, it’s possible that the surface corresponding to the constraint doesn’t actually have a boundary, for example if it is the surface of a sphere. Similarly, checking that the objective function gets large as the variables diverge to infinity can be annoying, as there are many ‘directions’ down which to diverge to infinity.

Motivating Cauchy-Schwarz

In practise, you probably want to have such methods in hand as a last resort on olympiad problems. It’s always possible that something will slip through the net, but typically problem-setters are going to trying to ensure that their problems are not amenable to mindless application of non-elementary methods. And even then, one runs the risk of accusations of non-rigour if you don’t state exact, precise results which justify everything which I’ve presented above.

One thing that can be useful, on the other hand, is to observe that the Lagrange multiplier condition looks a lot like the equality condition for Cauchy-Schwarz. So, even if you can’t solve the family of Lagrange multiplier ‘equations’, this does suggest that applying Cauchy-Schwarz to the vectors involved might give you some insight into the problem.

The following inequality from the IMO 2007 shortlist is a good example.

Suppose $a_1,\ldots,a_{100}\ge 0$ satisfy $a_1^2+\ldots+a_{100}^2=1$. Prove that

$a_1^2a_2+a_2^2a_3+\ldots + a_{100}^2 a_1<\frac{12}{25}.$

We shouldn’t be perturbed by the strictness. Maybe we’ll end up showing a true bound in terms of surds that is less neat to write down…

Anyway, applying Lagrange multipliers would require us to solve

$a_{k-1}^2 + 2a_ka_{k+1}=2\lambda a_k,\quad k=1,\ldots,100,$

with indices taken modulo 100. As so often with these cyclic but non-symmetric expressions, this looks quite hard to solve. However, it turns out that by applying Cauchy-Schwarz to the vectors $(a_k),(a_{k-1}^2+2a_ka_{k+1})$ gets us a long way into the problem by classical means. Working all the way through is probably best left as an exercise.

The Configuration Model

In the past, I’ve talked about limitations of the Erdos-Renyi model of homogeneous random graphs for applications in real-world networks. In a previous post, I’ve discussed a dynamic model, the Preferential Attachment mechanism, that ‘grows’ a graph dynamically by adding edges from new vertices preferentially to existing vertices with high degree. The purpose of this adjustment is to ensure that the distribution of the degrees is not concentrated around some fixed value (which would be c in G(n,c/n) ) but rather exhibits a power-law tail such as observed in many genuine examples.

In this post, we introduce some aspects of the configuration model, which achieves this property more directly. This idea probably first arose in the guise of regular graphs. Recall a regular graph has all degrees equal. How would we construct a random d-regular graph on a large number of vertices?

What we probably want to do is to choose uniformly at random from the set of such graphs, but it is not clear even how large this set is, let alone how one would order its elements to make it possible to make this uniform choice. Instead, we try the following. Assign to each vertex d so-called stubs, which will end up being ‘half-edges’. We then choose two stubs uniformly at random, and glue them together. More formally, we construct an edge between the host vertices, and then delete the chosen stubs. We then continue.

The construction makes no reference to the distribution of stubs, so we are free to choose this as we please. We could for example specify some sequence of degrees which approximates a power-law, so we could sample a random sequence of degrees in some way. So long as we have a sequence of stub set sizes before we start building the edges of the graph we will be able to use the above algorithm.

So what might go wrong? There seem to me to be three potential problems that might arise with this construction.

Firstly, there might be a stub left over, if the sum of the stub set sizes is odd. Recall that in a graph the sum of the degrees is twice the sum of the number of edges, and so in particular the sum of the degrees should be even. But this is a small problem. When the degree sequence is deterministic we can demand that it have even sum, and if it is random, we will typically be working in a large N regime, and so deleting the solitary stub, if such a thing exists, will not affect the sort of properties of the graph we are likely to be interested in.

The second and third objections are perhaps more serious. If we glue together stubs naively, we might end up with loops, that is, edges that ‘begin’ and ‘end’ at the same vertex. These are not allowed in the standard definition of a graph. Alternatively, we might end up with more than one edge between the same pair of vertices.

Our overall aim is that this mechanism gives a convenient way of simulating the uniform distribution on simple graphs with a given degree sequence. At present we have the uniform distribution on potential multigraphs, with a weighting of 1/k! for every multi-edge with multiplicity k, and a weighting of 1/2 for every loop. The latter can be seen because there is an initial probability proportional to $d(v_i)d(v_j)$ that vertices v_i and v_j will be joined, whereas a probability proportional (with the same constant) to $d(v_i)^2$ that v_i will receive a loop. The multi-edge weighting justification is similar.

However, conditional on getting a simple graph, the distribution is uniform on the set of simple graphs with that degree sequence. So it remains to investigate the probability that a graph generated in this way is simple. So long as this probability does not tend to 0 as n grows, we will probably be happy.

The strongest results on this topic are due to Janson. First observe that if the sum of the degrees grows faster than the number of vertices n, we fail to get a graph without loops with high probability. Heuristically, note that on the first pass, we are taking two picks from the set of vertices, biased by the number of stubs. By Cauchy-Schwarz, Rearrangement Inequality or just intuition, the probability of getting the same vertex is greater than if we picked uniformly from the set of vertices without biasing. So the probability of getting no loop on the first pass is $\le (1-\frac{1}{n})$. Take some function a(n) that grows faster than n, but slower than the sum of the degrees. Then after a(n) passes, the degree distribution is still roughly the same. In particular, the sum of the degrees is still an order of magnitude greater than n. So we obtain:

$\mathbb{P}(\text{no loops})\leq (1-\frac{1}{n})^{a(n)}\approx e^{-\frac{a(n)}{n}}\rightarrow 0.$

So, since isolated vertices have no effect on the simplicity or otherwise, we assume the sum of the degrees is $\Theta(n)$. Then, Janson shows that the further condition

$\sum_{i=1}^n d_i^2=O(n),$

is essentially necessary and sufficient for simplicity. We can see why this might be true by looking at the probability that the first edge added is a loop, which is roughly

$\frac{d_1^2+d_2^2+\ldots+d_n^2}{2(\sum d_i)^2}.$

We have to consider $O(\sum d_i)$ edges, so if the above expression is much larger than this, we can perform a similar exponential estimate to show that the probability there are no loops is o(1). The technical part is showing that this probability doesn’t change dramatically as the first few stubs disappear.

Note that in both cases, considering only loops is sufficient for simplicity. Although it looks like loop appearance is weaker than multiplicity of edges, in fact they have the same threshold. It should also be pointed out that, like the uniform random forests, an alternative approach is simply to count the number of simple graphs and multigraphs with a given degree sequence. Good asymptotics can then be found for the probability of simplicity.

In the case of G(n,c/n), we were particularly interested in the emergence of the giant component at time c=1. While first-moment methods can be very effective in demonstrating such results, a branching process local limit representation is probably easiest heuristic for this phase transition.

So long as the degree sequences converge in a natural way, we can apply a similar approach to this configuration model. Concretely, we assume that the proportion of vertices with degree i is $\lambda_i$ in the limit. Although the algebra might push through, we should be aware that this means we are not explicitly specifying how many vertices have degree, eg $\Theta(n^{1/2})$. For now assume the $\lambda_i$s sum to 1, so specify a probability distribution for degree induced by choosing a vertex uniformly at random.

So we start at a vertex, and look at its neighbours. The expected number of neighbours of this root vertex is $\sum i\lambda i$. Thereafter, when we consider a child vertex, based on how the stubs are paired up (and in particular the fact that the order of the operations does not matter – the choice of partner of a given stub is chosen uniformly at random), we are really choosing a stub uniformly at random. This corresponds to choosing a vertex at random, biased by the number of stubs available. The quantity of interest is how many additional stubs (other than the one that led to the vertex) are attached to this vertex. We assume we don’t need to worry too much about repeating vertices, in a similar way to G(n,c/n). So the expected number of additional stubs is

$\frac{1}{\sum i\lambda_i}\sum i\lambda_i(i-1).$

For an infinite component, we required the expectation to be > 1, which is equivalent to

$\sum \lambda_i i(i-2)>0.$

This was proven by Molloy and Reed (95), then with fewer conditions by Janson (07). The latter also shows how to use this construction to derive the giant component for G(n,c/n) result.

REFERENCES

Janson – A New Approach to the Giant Component Problem

Molloy, Reed – A Critical Point for Random Graphs with a Given Degree Sequence

Janson – The Probability that  Random Multigraph is Simple