The Icosian game made the vertex-visit puzzle tangible. The report of its 1857 presentation describes a plane projection of a regular dodecahedron, with holes at the vertices for ivory pins. The Trinity College Dublin archive connects this game with the search for closed routes visiting each vertex exactly once. This puzzle supplies a historical illustration of the Hamilton cycle requirement.^[ref]

A weighted Hamilton cycle is a Hamilton cycle in a weighted graph, with cost equal to the sum of its traversed edge weights. Here weights represent symmetric nonnegative costs, rather than conductances. Asking whether a Hamilton cycle exists is a feasibility question; asking for the cheapest one is an instance of the travelling salesman problem. A complete graph supplies every possible connection between distinct vertices, making feasibility automatic for $n\ge 3$, while choosing the cheapest visiting order remains a separate mathematical optimization problem.

Assign the complete graph $K_4$ the symmetric edge costs $w_{AB}=2$, $w_{BC}=3$, $w_{CD}=4$, $w_{AD}=5$, $w_{AC}=6$ and $w_{BD}=7$, in arbitrary units. Fix the start at $A$ and identify each Hamilton cycle with its reverse. The three possibilities areThese exhaust the possibilities: the permutations of the other three vertices give $3!=6$ orders, paired into three reversals. Hence the perimeter tour really is optimal for this toy instance. The diagram's geometric lengths are not costs, and no triangle inequality is assumed.

The usefulness of this model is to distinguish visiting required sites from covering required connections. If tasks are assigned to vertices and the route must return after one visit to each, a weighted Hamilton cycle captures the ordering requirement. If the job instead requires each edge once, the relevant object is an Euler circuit. The shortest path problem asks only for a cheapest route between specified endpoints. A 1954 primary research paper gave a concrete motivation for the travelling salesman problem: certify that a particular tour through 49 cities has minimum road-distance cost.^[ref]

These are abstract route models. If a cost stands for a whole graph path through an underlying network, travelling between consecutive task vertices can pass through intermediate network vertices more than once. One visit per task therefore does not imply one visit per physical junction. One-way connections require a directed graph; changing costs or extra timing constraints require a richer model than the fixed symmetric toy instance.

A Hamiltonian graph must be a connected graph and have minimum degree $\delta (G)\ge 2$: each vertex has two distinct neighbours on its Hamilton cycle. It also has no cut vertex. Indeed, deleting any vertex from the Hamilton cycle leaves a graph path through every remaining vertex, so the remaining graph is connected. These are quick ways to rule out a Hamilton cycle.^[ref]

Even these conditions together are not sufficient: the complete bipartite graph $K_{2,3}$ is connected, has minimum degree two and no cut vertex, but has no Hamilton cycle. Every graph cycle in a bipartite graph alternates between its two parts, using equally many vertices from each; it cannot use all two and all three.

In the other direction, Dirac's theorem gives a sufficient condition: a finite simple graph with $n\ge 3$ and $\delta (G)\ge n/2$ has a Hamilton cycle. Failure of this inequality proves nothing by itself: the cycle graph $C_5$ has a Hamilton cycle despite $\delta (C_5)=2<5/2$.^[ref]

A Hamiltonian graph is a graph containing a Hamilton cycle. This describes existence of a spanning graph cycle, rather than requiring every edge of the graph to belong to that graph cycle. In the finite simple graph convention, $K_n$ is a Hamiltonian graph for $n\ge 3$, because any ordering of its vertices closes using available edges.

The Hall coefficient relates a transverse electric field to longitudinal current density and transverse magnetic field, with oriented axes as in the displayed formula. Its SI unit is ${\mathrm{m}}^3/C$. For one isotropic Drude model species it is $1/(nq)$; multiple charge carriers need not follow that formula.

For a finite nonempty integer set $A$, put $D=2A-2A$. If an integer $N>|D|$, some $A^{\prime }\subseteq A$ of size at least $|A|/2$ has a Freiman 2-isomorphism into $\mathbb{Z}/N\mathbb{Z}$. Choose an auxiliary prime $P>N$ so large that reduction is injective on $A$ and every nonzero element of $D$ remains nonzero. Dilate modulo $P$ by a uniformly chosen nonzero multiplier. The forbidden residues are those represented by nonzero multiples of $N$ of absolute value less than $P$, at most $2\lfloor (P-1)/N\rfloor$ residues. The nonzero elements of $D$ occur in opposite pairs; the two elements in each pair have the same bad event. The union bound therefore gives failure probability at most $(|D|-1)/N<1$. Choose a multiplier avoiding all these events. Select the more populated of two half-intervals of residues modulo $P$. Differences of two pair sums of representatives then have absolute value less than $P$. An original additive relation must therefore lift to equality in the integers; conversely an equality modulo $N$ cannot have a nonzero lifted difference, by the forbidden-residue choice. Reduction modulo $N$ is the required isomorphism.

Let $h_i$ represent a nonzero two-torsion point of a complex torus, and write $e_i=\wp (h_i)$. The product $(\wp (z+h_i)-e_i)(\wp (z)-e_i)$ is an elliptic function whose apparent poles cancel against double zeros. It is therefore constant by Liouville theorem. Its value at $z=0$ is ${\wp }^{\prime \prime }(h_i)/2$. The Weierstrass elliptic differential equation gives ${\wp }^{\prime \prime }(h_i)/2=3e_i^2-g_2/4=(e_i-e_j)(e_i-e_l)$, proving the formula.

Let $A_j$ be the weights of the cubic Stokes dispersion symmetry and set $F(k)={\sum }_j{A}_j(k){\hat{q}}_0({\nu }_j(k))$. Orient $\partial D_{+}$ with $D_{+}$ on its left. Evaluating the global relation at its two lower-half-plane roots and interpolating the linear expression $ik{G}_1+G_2$ removes both unknown boundary traces. The resulting Fokas method formula isThe discarded transform of the solution is holomorphic in $D_{+}$ and has zero integral after closing its contour for $x>0$. Initial compatibility and sufficient smoothness and spatial decay are required. At the boundary, use a fixed time horizon later than $t$ before Fourier inversion, to avoid an endpoint half-value.

For $q_t=q_{xx}+\alpha q_x$ with $\alpha >0$, the dispersion relation is $\omega =k^2-i\alpha k$ and its symmetry is $\nu =i\alpha -k$. On the upper domain $\mathrm{Re}\omega <0$, $\mathrm{Im}k>\alpha$ and therefore $\mathrm{Im}\nu <0$. After removing the unknown trace through the global relation, deform the boundary contour to the displayed two rays, with $0<\theta <\pi /4$, oriented from left infinity to $i\alpha$ to right infinity. The factor $e^{ikx}$ decays like $e^{-rx\sin \theta }$, while $e^{-\omega t}$ has Gaussian decay for large $r$. Apparent poles introduced by splitting a finite-time forcing transform must be treated as removable singularities of the unsplit expression.

The baryon octet has rows $Y=1/2,0,-1/2$ with isospins $1/2$, $1\oplus 0$, $1/2$. The central weight has two states, ${\Sigma }^0$ and ${\Lambda }^0$. The baryon decuplet has rows $Y=1/2,0,-1/2,-1$ with isospins $3/2,1,1/2,0$. The three-quark flavour singlet has one state at the origin. Both abstract octet copies in $3^{\otimes 3}$ share the same weights.

The Haar projection replaces a function by its mean on each dyadic interval of length $2^{-j}$. For a function satisfying Lipschitz continuity with constant $L$, every value in a cell differs from its cell mean by at most $L{2}^{-j}$. This bound holds for all points with a consistent endpoint convention, and requires no global L2 norm assumption when the projection is defined by local averages.

For a pure scalar growing density contrast with time-independent boundary conditions, ${\nabla }^2\psi ={\delta }_{\ast }$ fixes the scaled peculiar gravitational potential up to an irrelevant constant. The linearized continuity equation fixes the longitudinal scaled velocity to $-\nabla \psi$. The remaining homogeneous velocity is ${\mathbf{u}}_h=\mathbf{C}(\mathbf{x})/(a^2\dot{b})$; a divergence-free such field does not change the growing density at linear order. Excluding this decaying velocity component is necessary for time-independent $\mathbf{u}$. In an Einstein-de Sitter universe, $b=a$ and ${\mathbf{u}}_h\propto a^{-3/2}$, furnishing a counterexample if the pure growing-velocity hypothesis is omitted.

For each treatment year let $C$ be cycles, $S$ singleton deliveries, $T$ twin deliveries and $H$ deliveries of three or more babies. A singleton-per-cycle model uses $S\sim \mathrm{Bin}(C,p)$. A multiple-delivery model conditional on a live delivery instead uses $T+H\sim \mathrm{Bin}(S+T+H,q)$. The latter denominator counts deliveries, not babies, and it excludes failed cycles. Confusing these denominators changes the statistical question.

For independent $Y_{ij}\sim \mathrm{Bin}(m,p_i)$ with $k$ replicates per group and $\mathrm{logit}(p_i)=\mu +{\alpha }_i$, ${\alpha }_1=0$, let $S_i={\sum }_j{Y}_{ij}$ and $N=km$. The likelihood equations give ${\hat{p}}_i=S_i/N$, $\hat{\mu }=\log (S_1/(N-S_1))$, and ${\hat{\alpha }}_i=\mathrm{logit}({\hat{p}}_i)-\hat{\mu }$ when $0<S_i<N$. Boundary totals give infinite logits. With $0<p_1<1$ fixed and $N\to \infty$, the central limit theorem and delta method giveThe other groups do not increase information about the baseline logit when they have free offsets. For two groups, the information matrix for $(\mu ,{\alpha }_2)$ is $\left(\begin{array}{cc}{w}_1+w_2& w_2\\ w_2& w_2\end{array}\right)$ with $w_i=N{p}_i(1-p_i)$, whose inverse has first diagonal entry $1/w_1$.

This subgroup uses Hamiltonians whose support in $[0,1]\times M$ is compact. Their flows are the identity outside a fixed compact set. On closed manifolds this support condition imposes no extra restriction. On an open symplectic manifold, it excludes nontrivial whole-fiber cotangent fiber translations, even when their generating one-form is exact.

If the closed span of the Rademacher functions were a complemented subspace of $L^1(0,1)$, compose a projection with the inverse of its l2 sequence space parametrization. The Grothendieck theorem for L1 to Hilbert operators would make this map an absolutely summing operator. Its composition with the parametrization would make the identity on the infinite-dimensional l2 sequence space absolutely summing, contradicting its coordinate-vector test.

A local two-body and three-body mean-field energy density $V_0{n}^2/2+W_0{n}^3/3$ generates cubic and quintic terms in the Gross–Pitaevskii equation. Variation with respect to ${\psi }^{\ast }$ gives $(V_{0}n+W_0{n}^2)\psi$, with $n=|\psi {|}^2$. A uniform stationary state's chemical potential is $\mu =V_0{n}_0+W_0{n}_0^2$. Its positive-compressibility condition is $V_0+2W_0{n}_0>0$. If $\mu >0$, scaling amplitude by $\sqrt{n_0}$ and distance by $\hslash /\sqrt{2m\mu }$ gives coefficients $\alpha =V_0{n}_0/\mu$, $\beta =W_0{n}_0^2/\mu$ and $\alpha +\beta =1$. Negative chemical potential requires a different sign in the dimensionless linear term.

Factorizing a normalized condensate field as ${\Psi }_{3D}(x,y,z)=\psi (x,y)\chi (z)$ with $\int |\chi {|}^{2}dz=1$ reduces its contact interaction energy to $(g_{2D}/2)\int |\psi {|}^{4}dxdy$, with the displayed coupling. Thus $g_{2D}$ has units of energy times area, whereas $g_{3D}=4\pi {\hslash }^{2}a/m$ has units of energy times volume for a physical scattering length $a$. For a normalized axial Gaussian $\chi (z)={\pi }^{-1/4}{a}_z^{-1/2}{e}^{-z^2/(2a_z^2)}$, direct Gaussian integration gives $g_{2D}=g_{3D}/(\sqrt{2\pi }{a}_z)$. A model with density normalized per unit axial length can instead retain $g_{3D}$ with a line-density normalization.

Let $D_n={\partial }_t+{\mathbf{v}}_n\cdot \nabla$ be the material derivative of the normal fluid, and $\zeta >0$. The displayed phenomenological density damping preserves the condensate continuity equation because its coefficient is real. Under a Galilean boost, both velocities shift by the boost velocity and the field acquires the Schrödinger phase; $D_n|\psi {|}^2$ is invariant. Thus the normal-fluid derivative, rather than a derivative along the condensate flow, is essential to the model.

Apply Green theorem with $P=-y/2$ and $Q=x/2$ to a positively oriented regular planar region. Since $Q_x-P_y=1$, its area is the displayed line integral. Clockwise orientation changes the sign; holes have the opposite boundary orientation to the outer curve.