Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 339 (Topics in Convex Optimisation)

Suppose $y$ in the capped simplex has a coordinate strictly between zero and one. Since ${\sum }_i{y}_i=k$ is an integer, it cannot have exactly one such coordinate: the other coordinates would contribute an integer sum. Hence there are distinct $i,j$ with $0<y_i,y_j<1$.

Choose $0<\epsilon <\min \{y_i,1-y_i,y_j,1-y_j\}$ and set $y^{\pm }=y\pm \epsilon (e_i-e_j)$, where $e_i,e_j$ are standard basis vectors. Both perturbed vectors satisfy the coordinate bounds and have the same coordinate sum. They are distinct and $y=(y^{+}+y^{-})/2$, so $y$ is not an extreme point.

Conversely, every zero-one vector in this convex polytope is an extreme point. If it were a nontrivial convex combination of two feasible vectors, each coordinate equal to zero would force both corresponding coordinates to be zero, and each coordinate equal to one would force both to be one. Thus the two vectors would equal the original one. The extreme points are therefore exactly the indicators of $k$-element subsets.

The required linear program isThere are $2n$ inequality constraints and one equality constraint, all independent of $x$.

The feasible set is the capped simplex, a nonempty compact set. A linear function attains its maximum at an extreme point; the previous part identifies these as the indicators of $k$-element subsets. At such a vector, the objective is the sum of the selected coordinates. Maximizing selects the $k$ largest coordinates, including when some are negative or tied. Equivalently, $f_k$ is the support function called the sum of the largest components.

Keep $y\ge 0$ as the domain restriction, introduce nonnegative Lagrange multipliers $s_i$ for $y_i\le 1$, and introduce a free multiplier $t\in \mathbb{R}$ for ${\sum }_i{y}_i=k$. The maximizing Lagrangian isIts supremum over $y\ge 0$ is finite exactly when $s+t1\ge x$, and then equals $kt+{\sum }_i{s}_i$. By linear programming duality, since the primal is feasible and bounded,Minimizing over $s$ for fixed $t$ gives $s_i=(x_i-t{)}_{+}$, the positive part of a real-valued function. Thus the threshold formula for the sum of the largest components isFor a direct attainment check, order the coordinates $x_{[1]}\ge \cdots \ge x_{[n]}$. Any $t\in [x_{[k+1]},x_{[k]}]$ is optimal for $k<n$, while any $t\le x_{[n]}$ is optimal for $k=n$.

Substitute the dual linear program from the previous part and optimize jointly over $x,t,s$. The result isThe objective is linear, and the constraints are equalities or inequalities between affine functions, so this is a linear program. For each fixed feasible $x$, minimizing over $t,s$ gives exactly $f_k(x)$ by the threshold formula for the sum of the largest components; hence the reformulation preserves the optimal value and optimal $x$ whenever an optimum exists.

Replace coordinate inequalities by the Loewner order and coordinate sums by matrix trace. The maximizing semidefinite program isIts feasible set is the fantope. In an orthonormal eigenbasis of $X$, its objective is ${\sum }_i{\lambda }_i(X)Y_{ii}$, and the diagonal entries of $Y$ belong to the capped simplex. Choosing $Y$ as the orthogonal projection matrix onto the largest $k$ eigenvectors attains the sum of the largest eigenvalues, as in the Ky Fan maximum principle.

The minimizing semidefinite program isFor example, its Lagrangian arises by assigning $S⪰0$ to $I-Y⪰0$, keeping $Y⪰0$ in the domain, and using free $t$ for $\mathrm{tr}Y=k$. Finiteness of the supremum over $Y$ requires $S+tI-X⪰0$.

Equality follows directly by choosing $S$ to have eigenvalues $({\lambda }_i(X)-t{)}_{+}$ in the same orthonormal eigenbasis, with $t$ between the $k$th and $(k+1)$st eigenvalues, or below the smallest one for $k=n$. This is the threshold semidefinite program for the largest eigenvalues, and it also verifies the endpoint case $k=n$.

Let $J=11^T$ be the all-ones matrix. Every feasible block matrix in the first formulation has $t\ge 1$, because its principal $2\times 2$ submatrix on indices $0,i$ is $\left(\begin{array}{cc}t& 1\\ 1& 1\end{array}\right)$ and is a positive semidefinite matrix. In particular, $t>0$.

By the Schur complement,Set $U=tZ-J$. Then $U⪰0$, $U_{ii}=t-1$, and $U_{ij}=-1$ on every edge. Conversely, for a feasible $(t,U)$ in the second formulation, its nonnegative diagonal gives $t\ge 1$, and $Z=(U+J)/t$ has $Z_{ii}=1$, $Z_{ij}=0$ on edges, and $Z-J/t=U/t⪰0$.

Both transformations preserve the objective $t$. Hence the two semidefinite programs defining the complement theta number have the same feasible objective values and the same minimum.

For a proper three-graph colouring $c$, associate the colors with the three roots of unity $d(i)=\exp (2\pi i(c(i)-1)/3)$, viewed as unit vectors in ${\mathbb{R}}^2$. Define $U_{ij}=2⟨d(i),d(j)⟩$ using the real inner product.

This is twice a Gram matrix, so it is a positive semidefinite matrix. Its diagonal is $2$. On every edge, the endpoint colors differ, so their vectors make angle $2\pi /3$ or $4\pi /3$ and $U_{ij}=2\cos (2\pi /3)=-1$. Thus $t=3$ and this $U$ are feasible in the second semidefinite program, provingThis is the regular simplex construction of a strict vector coloring.

For an edge $ij$, put $\rho =⟨v_i,v_j⟩$ and $\theta =\arccos \rho$. Since $\rho \le -1/2$, we have $\theta \ge 2\pi /3$.

The random normal $a$ has the isotropic Gaussian distribution $N(0,I_p)$. Its distribution is invariant under orthogonal transformations; if $v_i,v_j$ are linearly independent, its projection onto their plane has a uniformly distributed direction. The signs of its inner products with $v_i,v_j$ differ in two sectors of total angle $2\theta$, out of $2\pi$. Thus random hyperplane rounding givesIf $v_j=-v_i$, the signs differ with probability one and the same formula holds with $\theta =\pi$. Zero inner products have probability zero, since each $v_i$ is a unit vector.

If the graph has an edge, its corresponding principal $2\times 2$ block of $U$ forces $t\ge 2$, so the division by $t-1$ used to obtain the Gram matrix is valid. An edgeless graph can instead be colored with one color directly.

A fixed edge is monochromatic exactly when none of the $r$ hyperplanes separates its endpoints. Each hyperplane fails to cut it with probability at most $1/3$, and the hyperplanes are independent. ThereforeLet $B={\sum }_{ij\in E}{1}_{\{c(i)=c(j)\}}$ count monochromatic edges. Applying linearity of expectation to these indicator random variables givesIndependence between different edges is unnecessary; only the independently sampled hyperplanes are used. The $r$ signs provide at most $2^r$ colors, whether or not every sign region is nonempty.

If $m=0$, use a single color. Otherwise chooseThe previous bound gives $\mathbb{E}B\le n/4$, including the case $r=0$. By Markov inequality, $\mathbb{P}(B>n/2)\le 1/2$. Sample the hyperplanes, count the monochromatic edges, and repeat if $B>n/2$. The success probability is at least $1/2$, so at most two draws are needed on average.

For a successful draw, select one endpoint of each monochromatic edge and let $D$ be the union of the selected vertices. Then $|D|\le B\le n/2$. Every monochromatic edge meets $D$, so the original colors give a proper graph colouring on the induced subgraph with vertex set $W=V\setminus D$. Since $|W|\ge n/2$, the coloring of all vertices is a semicoloring. This is a random alteration method; no additional colors for $D$ are required.

The number of available colors satisfies $2^r\le 2\max \{1,(4m/n{)}^{{\log }_{3}2}\}$. A simple graph has $m\le n(n-1)/2$, hence semicoloring by independent hyperplanes gives

For $z\in {\mathbb{R}}^n$, define $x_i=z_i^2$. Then $x$ lies in the nonnegative orthant andIf $A$ is a copositive matrix, this is nonnegative for every $z$, so $p$ is a globally nonnegative polynomial. Conversely, every $x\ge 0$ has the form $x_i=z_i^2$ by taking $z_i=\sqrt{x_i}$. If $p$ is globally nonnegative, then $x^{T}Ax=p(z)\ge 0$ for every such $x$, proving that $A$ is a copositive matrix. The map $z\mapsto (z_i^2{)}_i$ covers the entire nonnegative orthant, which is the key to both directions.

First suppose $A=P+N$, with $P$ a real positive semidefinite matrix and $N$ a symmetric nonnegative matrix. Factor $P=B^{T}B$, and write $v(z)=(z_1^2,\dots ,z_n^2{)}^T$. ThenEach term is a square multiplied by a nonnegative scalar, so $p$ is a sum of squares polynomial.

Conversely, suppose $p={\sum }_{\ell }{q}_{\ell }^2$. Because $p$ is a degree-four homogeneous polynomial, the homogeneous sum of squares representation allows every $q_{\ell }$ to be quadratic and homogeneous. Explicitly, higher-degree parts cannot cancel in a sum of squares; constant parts vanish because $p(0)=0$, and the degree-two part ${\sum }_{\ell }(q_{\ell }^{(1)}{)}^2$ forces all linear parts to vanish. WriteSince $p$ is invariant under every coordinate sign change, sign averaging of a sum of squares over independent Rademacher random variables givesThe cross terms vanish because their sign products contain an odd power of at least one independent sign. SetThen $P$ is a positive semidefinite matrix and $N$ is a symmetric nonnegative matrix. Comparing the coefficients of $z_i^4$ and $z_i^2{z}_j^2$ gives $A=P+N$. This proves the sum of squares criterion for a biquadratic form.

The Horn copositive matrix is invariant under cyclic permutation of its five coordinates: its negative entries correspond precisely to neighboring indices on the five-cycle. For any $x\ge 0$, cyclically relabel the coordinates so that $x_5$ is a smallest coordinate. In particular, $x_4-x_5\ge 0$.

Using the stated identity in this coordinate order,The square is nonnegative, and both remaining terms are nonnegative since $x_i\ge 0$ and $x_4\ge x_5$. Cyclic invariance means the relabeling has not changed the quadratic form. Thus $H$ is a copositive matrix for every original ordering of $x$.

Let $w=(1,2,1,0,0{)}^T$. Direct multiplication gives $Hw=(0,0,0,2,2{)}^T$, so $w^{T}Hw=0$.

Suppose $H=P+N$, where $P$ is a positive semidefinite matrix and $N$ is a symmetric nonnegative matrix. Since $w\ge 0$, both $w^{T}Pw$ and $w^{T}Nw$ are nonnegative. Their sum is zero, so both vanish. In particular,Every coefficient $w_i{w}_j$ in this sum is strictly positive and every $N_{ij}$ is nonnegative, forcing $N_{ij}=0$ for $1\le i,j\le 3$.

Now apply the same argument to all five cyclic shifts of $w$. The Horn copositive matrix is cyclically invariant, so every shifted vector also has zero quadratic form. It follows that the entries of $N$ vanish on every cyclic block of three consecutive indices. Every pair of indices on a five-cycle lies in such a block, hence $N=0$.

This would imply $P=H$. But the zero quadratic form of a positive semidefinite matrix would then force $Hw=0$, contradicting $Hw=(0,0,0,2,2{)}^T$. Therefore $H$ lies outside the positive-semidefinite-plus-nonnegative cone. By the previous equivalence, its quartic form is a nonnegative polynomial that is not a sum of squares polynomial.