At $\sigma =1$, complete the square:For fixed $y$, all factors not involving $x$ enter its conditional normalizing constant. Interchanging the roles of the coordinates givesThese are the full conditional distributions of the standard bivariate normal distribution with correlation $\rho$.

A systematic Gibbs sampler starts at a chosen $(x_0,y_0)$ and, at each sweep, draws independent standard normals $Z_{1,n},Z_{2,n}$ and setsThe second update uses the newly drawn $x_{n+1}$. Both coordinate updates preserve the target by part (a), hence so does the sweep. The Gaussian two-coordinate Gibbs recursion is explicitlyan AR(1) recursion with innovation variance $1-{\rho }^4$. Its stationary variance is one, and its correlation at sweep lag $k$ is ${\rho }^{2k}$, displaying the dependence of the sample.

The printed general-$\sigma$ expression has a separate normalization defect: for $\sigma >0$ its integral is $\sigma$, not one, since the exponent has covariance ${\sigma }^2\left(\begin{array}{cc}1& \rho \\ \rho & 1\end{array}\right)$, whose normalizing denominator is $2\pi {\sigma }^2\sqrt{1-{\rho }^2}$. Retaining that exponent requires ${\sigma }^2$ in the prefactor. The requested $\sigma =1$ target is already normalized, so this defect does not alter the conditional distributions above.

The supplied hint interchanges its item and bidder indices in the first-value identity. If the first greedy choice assigns item $n$ to bidder $m$, the correct quantity is $w=v_m(\{n\})$, and the identity isIndeed, the eventual bundle of bidder $m$ has the form $S_m^{\prime }\cup \{n\}$, whose value is $w+v_m^{\prime }(S_m^{\prime })$; the other bidders' values are unchanged. Every later marginal contribution for bidder $m$ is exactly its marginal in the residual problem, and this is also true for every other bidder. Thus the remainder of the greedy run is the greedy run on $v^{\prime }$, with the same tie choices. The residual values are nonnegative and increasing; they remain submodular set functions as allowed in the question.

We prove the greedy half-approximation for submodular welfare by induction on the number of items. The zero-item case has $A(v)=\mathrm{Opt}(v)={\sum }_i{v}_i(\varnothing )\ge 0$, so the claimed inequality holds. This base case also handles the fact that the question does not explicitly normalize empty-bundle values to zero.

For the induction step, letbe the first greedy marginal. Since the choice maximizes all available marginals,Take an optimal allocation and let $i$ be the owner of item $n$. If $i\ne m$, remove that item from $i$'s bundle and give it to $m$. By submodularity, the loss to its old owner isMonotonicity makes the recipient's gain nonnegative. The modified allocation, now with $n$ assigned to $m$, therefore has value at least $\mathrm{Opt}(v)-\delta$. If $i=m$, no modification is necessary and this bound still holds. Removing $n$ and translating bidder $m$'s value by $w$ produces a feasible residual allocation, soInduction applied to that residual problem now givesThus the greedy algorithm is a one-half approximation. Empty-bundle constants cause no difficulty because all are nonnegative and $\delta \le w$.

At each of $n$ stages evaluate the marginal for at most $m$ bidders and $n$ remaining items. This requires $O(m{n}^2)$ value queries, and polynomial additional work with bundle bit-vectors and cached current values. Hence it is a polynomial time approximation algorithm in the value-oracle model, assuming polynomial time bid evaluations. If every bundle bid is explicitly listed, this query bound is polynomial in that larger explicit input as well. The exponentially many possible bundles do not require enumerating allocations.

For the unknot the three-manifolds are $S^3$, $U_0=S^1\times S^2$, and $U_n=L_n$, where $L_n$ denotes positive $n$-surgery with its surgery orientation. This fixes the orientation independently of competing conventions for the name $L(n,1)$.

Use the grading convention $\mathrm{gr}{\Theta }_{S^3}^{+}=0$ and $\mathrm{gr}{\Theta }_{S^3}^{-}=-2$. Let $R=\mathbb{Z}[U]$ and $\hat{R}=\mathbb{Z}[[U]]$. The ordinary minus groups arewhere $0\le k<n$ andThe lens-space absolute grading can be calculated from the standard torus surgery triple. For $0\le k<n$, the reverse trace structure with evaluation $2k-n$ has a small embedded holomorphic triangle with basepoint multiplicity zero and signed count $\pm 1$, taking $y_k$ to ${\Theta }^{-}$. Its degree shift is ${\Delta }_k=(1-(2k-n{)}^2/n)/4$. Thus $(d_k-2)+{\Delta }_k=-2$, which gives the boxed formula for $d_k$. Additional triangles wind around the surgery region and give the higher powers of $U$ calculated below.

Their infinity groups are obtained by inverting $U$: one Laurent tower for $S^3$ and each lens-space structure, two for the torsion structure on $S^1\times S^2$, and $\mathbb{Z}[U,U^{-1}]/(1-U^{|i|})$ in its non-torsion structures. Their plus groups are respectively ${\mathcal{T}}_0^{+}$, ${\mathcal{T}}_{1/2}^{+}\oplus {\mathcal{T}}_{-1/2}^{+}$, and ${\mathcal{T}}_{d_k}^{+}$. The hat groups are $\mathbb{Z}$ in degree zero, $\mathbb{Z}$ in each degree $\pm 1/2$, and $\mathbb{Z}$ in degree $d_k$. These follow directly by straightening the standard genus-one Heegaard diagrams: the lens space has one generator in each structure and no chain differential, while the two bigons in the torsion $S^1\times S^2$ diagram cancel with coherent signs. The sphere adjunction condition eliminates its non-torsion plus and hat groups, but not its ordinary minus group. The Non-torsion Heegaard Floer homology of the sphere-circle product calculation comes from a two-generator chain complex with chain differential $1-U^{|i|}$; its cokernel is the displayed quotient. For the triangle, complete the free minus towers by replacing $R$ with $\hat{R}$. The non-torsion summands disappear after completion, because $1-U^{|i|}$ has an inverse $1+U^{|i|}+U^{2|i|}+\cdots$ in $\hat{R}$.

For $k=0$, choose signs so that the exact triangle becomesThe zero-surgery trace has $\chi =1$, $\sigma =0$, and $c_1^2=0$ in the torsion structure; hence its absolute degree shift is $-1/2$, taking ${\Theta }^{-}$ to $x_{-}$. The auxiliary-generator map takes $x_{+}$ to $y_0$, so its effective grading shift isFor the underlying cobordism $Z_n$ in its torsion structure, $c_1^2=0$, $\chi =1$, $\sigma =0$, and its actual cobordism shift is $-1/2$. Tensoring with the canonical lens generator adds $d_0=(n-1)/4$ under the connected-sum grading convention, explaining the effective shift. For $n=1$ the auxiliary lens space is $S^3$, and this is simply the other degree-$-1/2$ surgery trace map.

The reverse trace components are explicit in every $k$. For $j=k+n\ell$, put $m_j=2j-n$. Its intersection form is $(-n)$, soIndeed $c_1^2=-m_j^2/n$, $\chi =1$, $\sigma =-1$, and matching source and target degrees gives precisely the displayed exponent. The exponent is nonnegative, and the coefficient is a unit: the infinity trace map is an isomorphism, and restriction to the free minus towers leaves the unique grading-compatible monomial.

With a compatible choice of Coherent orientations in Heegaard Floer homology, the sum $C_k$ is multiplication byFor $k=0$, pair $\ell$ with $1-\ell$: their exponents agree and their signs are opposite, so $P_0=0$, in agreement with the first triangle. For $0<k<n$ the term $\ell =0$ is the unique constant term, and every other exponent is positive. Thus $P_k=1+U(\cdots )$ is a unit in $\hat{R}$. The zero-surgery summand for such a $k$ vanishes, and the exact triangle isIt is isomorphic, as an ungraded completed-module triangle, to one with the final map the identity. The sum generally has no single absolute Heegaard Floer grading shift; each individual trace component has shift ${\Delta }_j$. This distinction and the power series explain why replacing the completed triangle by a polynomial-module triangle is not innocuous.

In Cartesian coordinates on Euclidean space, defineThis is an affine connection by the ordinary product rule. Its connection coefficients in this coordinate frame are all zero. The component formula for the Lie bracket of vector fields gives ${\nabla }_{X}Y-{\nabla }_{Y}X=[X,Y]$, so it is torsion-free. Also, since the Euclidean metric has constant coefficients,This is exactly metric compatibility. By uniqueness it is the Euclidean Levi-Civita connection. In particular constant vector fields are parallel, while the derivative of a general field is given by the boxed formula.

The space is the mapping torus of the antipodal map $A:S^n\to S^n$. For $n\ge 1$, its mapping degree is $\delta =(-1{)}^{n+1}$: $A$ is the restriction of $-I$ on ${\mathbb{R}}^{n+1}$, whose determinant is $(-1{)}^{n+1}$. The Wang sequence givesThe kernel terms are free abelian, so these short exact sequences split as groups. On $H_0(S^n)$ the map is the identity, and on $H_n(S^n)$ it is multiplication by $\delta$. For $n\ge 2$ this givesAt $n=1$, both contributions to degree one are present, giving $H_0=H_2=\mathbb{Z}$ and $H_1={\mathbb{Z}}^2$, with all other groups zero. At $n=0$, the antipodal map exchanges the two points of $S^0$, and the two cylinder intervals join into a single circle; thus $H_0=H_1=\mathbb{Z}$ and higher groups vanish. These are the homology of the antipodal sphere mapping torus cases.

Let $a$ be the projective-line loop and $b$ the core loop of the Möbius band. The boundary of the Möbius band goes twice around its core. Its boundary inclusion is a cofibration, so replacing the gluing map by a homotopic cellular degree-$n$ map preserves the homotopy type. More precisely, the band is a mapping cylinder of the boundary double covering onto the core circle. After attaching its other end and collapsing the connecting tree, the resulting space has a homotopy-equivalent CW complex with one vertex, edges $a,b$, and two two-cells with attaching wordsThe first is the projective-plane face; the second records equality of the attached boundary loop $a^n$ and the band boundary loop $b^2$. Taking exponent sums gives the cellular chain complexSince $\det D=4$, its kernel is zero. The first invariant factor in the Smith normal form is the greatest common divisor of its entries, $d=\gcd (2,n)$, and the second is $4/d$. HenceThis includes $n=0$ and negative degrees. The fundamental group after attaching a Möbius band is $⟨a,b\mid a^2=1,b^2=a^n⟩$, consistent with the computed abelianization. The calculation is an example of homology after attaching a Möbius band to the real projective plane.

For any fixed $t>0$, the characteristic map $X_t(\xi )=\xi +t{u}_I(\xi )$ has derivative $1+t{u}_I^{\prime }(\xi )>1$. It is strictly increasing and onto: for $\xi \ge 0$ it is at least $\xi +t{u}_I(0)$, and for $\xi \le 0$ at most that quantity. Hence the inverse function theorem supplies a global $C^1$ inverse. Define $u(x,t)=u_I(X_t^{-1}(x))$. Differentiating givesThe derivatives are continuous, and $u_x\le 1/t$ for positive times. Thus a classical solution exists for all $t>0$, without any characteristic intersection or shock formation.

The integrand has simple poles at $z=\pm 1/2$, both inside the positively oriented unit circle. At either pole $z_0$, the residue is $z_0/(-8z_0)=-1/8$. Their sum is $-1/4$, so the residue theorem gives

Choose $0\ne v\in A_3$ and write $v=a_1+a_2$ using the direct sum of vector spaces $V=A_1\oplus A_2$. Neither component is zero: otherwise $v$ would lie in one of the zero intersections $A_3\cap A_1$ or $A_3\cap A_2$. They are linearly independent, so $W=\mathrm{span}\{a_1,a_2\}$ is a two-dimensional vector subspace. Uniqueness of the direct-sum decomposition gives $W\cap A_1=\mathrm{span}\{a_1\}$ and $W\cap A_2=\mathrm{span}\{a_2\}$. Also $v\in W\cap A_3$, but $W$ cannot be contained in $A_3$ because $a_1\in A_1\setminus \{0\}$. Hence $W\cap A_3$ is one-dimensional. This constructs the required common plane for three pairwise complementary subspaces.

Let $r=\dim \mathrm{im}T$. Since the kernel of a linear map and its image agree, the rank-nullity theorem gives $\dim V=r+r$. Thus $\dim V=2r$ is even. Every $Tv$ lies in $\ker T$, so $T^2=0$. The minimal polynomial therefore divides $X^2$. It cannot be $1$, and it cannot be $X$ since $T\ne 0$; hence the minimal polynomial is $X^2$.

Use the universe $\Omega =\{B\subseteq A:|B|=k\}$ and define the exact-occupancy events $E_i=\{B\in \Omega :|B\cap A_i|=3\}$. The desired collection is ${\bigcup }_{i=1}^k{E}_i$, so apply inclusion-exclusion to these events.

Fix $r$ distinct block indices. For $B$ to belong to the intersection of their events, it must contain exactly three elements in each selected block. There are ${\left(\genfrac{}{}{0ex}{}{m}{3}\right)}^r$ ways to make those choices. None of the other elements in these selected blocks may be included; the remaining $k-3r$ elements must come from the $n-rm$ elements outside the selected blocks. ThusThere are $\left(\genfrac{}{}{0ex}{}{k}{r}\right)$ choices of block indices, and the intersection is empty when $3r>k$. Substitution into inclusion-exclusion givesHere binomial coefficients with a lower argument exceeding a nonnegative upper argument are zero. In particular, when $m<3$ every event is empty, and when $k<3$ the sum is empty; both give the correct answer zero. The subtraction of entire selected blocks, not merely the $3r$ chosen elements, is what enforces exact rather than at-least-three occupancy. This is inclusion-exclusion for exact block occupancy with the number of blocks and subset size both equal to $k$.

False. On the additive topological group $\mathbb{R}$, put $d(x,y)=|x-y|$ and $d^{\prime }(x,y)=\sqrt{|x-y|}$. Both are invariant under left and right translations and induce the usual topology. The triangle inequality for $d^{\prime }$ follows from $\sqrt{a+b}\le \sqrt{a}+\sqrt{b}$ for $a,b\ge 0$. Butcan be arbitrarily close to zero and arbitrarily large. Consequently no $K>0$ can satisfy both proposed inequalities. Compatibility of the two invariant metrics is topological equivalence, which need not be bilipschitz equivalence.

True. Given a compatible left-invariant group metric $d$, definePullback by the bijection $x\mapsto x^{-1}$ preserves all metric axioms, and because inversion is a homeomorphism, $d_R$ induces the same topology. For right translation by $a$, left invariance givesThus $d_R$ is the required compatible right-invariant group metric. No metrizability theorem is needed.

False. Give the two-element group $C_2$ the indiscrete topology and take $G=\mathbb{R}\times C_2$ with the product topology. The indiscrete factor is a topological group, since every map into an indiscrete space is continuous, and the product group operations are continuous. Points $(0,e)$ and $(0,a)$, $a\ne e$, have identical neighbourhoods, so $G$ is not a Hausdorff space. However, its continuous projection onto $\mathbb{R}$ is surjective. If $G$ were compact, its image $\mathbb{R}$ would be compact, which it is not.

Let $G_1=\varphi (\mathbf{r}|{\mathbf{r}}_1)$ and $G_2=\varphi (\mathbf{r}|{\mathbf{r}}_2)$. Green's second identity givesThe common Robin boundary condition makes the surface integrand vanish. Where $\beta \ne 0$, substitute ${\partial }_n{G}_i=-\alpha G_i/\beta$; a Dirichlet portion also contributes zero. The delta sources therefore give the Helmholtz Green-function reciprocityThis is a bilinear identity, with no complex conjugation. Assume the boundary-value problem has a unique Green kernel; in an exterior problem use the same outgoing radiation condition for both kernels.

For the weighted operator, assume smooth positive $f$ and set $\mathbf{b}=\nabla \log f$. ThenIts unweighted transpose, obtained by integration by parts, isThe associated boundary identity isTo cancel it when $v$ satisfies the given Robin condition, the required adjoint kernel satisfiesUsing these two delta sources in the boundary identity proves $\psi ({\mathbf{r}}_2|{\mathbf{r}}_1)=\tilde{\psi }({\mathbf{r}}_1|{\mathbf{r}}_2)$, the adjoint reciprocity for a weighted Helmholtz operator. As a check, $L^{\mathrm{T}}(fv)=fLv$, soThis is equivalent to weighted acoustic Green-function reciprocity, $f({\mathbf{r}}_2)\psi ({\mathbf{r}}_2|{\mathbf{r}}_1)=f({\mathbf{r}}_1)\psi ({\mathbf{r}}_1|{\mathbf{r}}_2)$. Unit delta normalization is essential here. A Hermitian adjoint would instead conjugate coefficients and change the radiation convention; the requested equality uses the transpose.

Let $z_f(t)$ be the filling-box first front, with $z_f(0)=H$. Below it the original fluid has not yet been through the plume. In the filling box model, the turbulent plume adjusts rapidly and occupies a negligible fraction of the room's area. A point source with negligible source volume flux supplies $Q(z)=C{F}_s^{1/3}{z}^{5/3}$. All its volume crossing the front grows the processed region, so volume conservation givesPut $a_p=\pi C{F}_s^{1/3}/A_c$. Integrating $d(z_f^{-2/3})/dt=2a_p/3$ yieldsThe first front approaches the floor only as $t\to \infty$. A nonzero source volume flux would require an equal exhaust to keep the room volume fixed; it is treated in the following parts.

Since $2p-q=1$, the leading turbulent viscosity is linear:Using $U_0=C_{\mu }p\sqrt{k_0}/{\sigma }_k$ and $1/p=2-\sigma$ gives

Let $p$ be the best uniform approximation to the exponential function from the degree-at-most-$n-1$ polynomial space. Suppose that $E_{n-1}(e^x)=E_n(e^x)$. Then the same $p$ is best in the larger space. Its error $g(x)=e^x-p(x)$ is nonzero, since the exponential is not a polynomial. The Chebyshev alternation theorem supplies $n+2$ ordered points where $g$ has alternating signs. By the intermediate value theorem, $g$ has at least one zero in each of the $n+1$ intervening intervals, hence at least $n+1$ distinct zeros.

Applying Rolle's theorem successively $n$ times forces a zero of $g^{(n)}$. But $\mathrm{deg}p\le n-1$, soeverywhere. This is a contradiction. Together with the automatic non-increase of best errors, it provesThe argument is an instance of strict polynomial error decrease from nonvanishing derivatives.