A Gabor filter has a kernel formed by multiplying a Gaussian function by a sinusoid or a complex oscillation. In two-dimensional image processing, its orientation and spatial frequency select local image structure; evaluating the kernel's translated dot product with an image is spatial filtering. The real even example under signed spatial filter model of a simple cell uses a cosine.

There is experimental motivation for this family without making every simple cell an exact Gabor filter. Jones and Palmer fitted two-dimensional Gabor filters to measured spatial and spectral response profiles of simple cells in cat cortex and compared independently estimated parameters as a test of spatial linearity. Their reported residuals supported the approximation in that dataset. The chosen parameters and orientation tuning curve in this article are not extracted from their measurements. ^[ref]

Communication complexity studies how much information distributed parties must exchange to compute a function of their separately held inputs. Local computation is free; the resource being counted is communicated bits or qubits, with the allowed shared randomness, quantum entanglement or other correlations specified separately. For a Boolean function $f(x,y)$, a one-way exact protocol has the receiver output $f$ correctly for every input after a single message. One-bit computation using Popescu–Rohrlich boxes shows that this resource measure changes drastically if ideal superquantum boxes are available.

Compilation is the translation process performed by a compiler, which may include checking and transforming an abstract syntax tree. Translation can finish before execution, or happen during execution through just-in-time compilation. Reusing the translated representation can avoid repeated parsing and repeated constant folding when only the inputs change. This does not guarantee a speedup: translation itself costs work, and the choice of representation and execution mechanism matters.

Online text: old.goldbook.iupac.org/html/I/I03332.html

Cited passages: IUPAC Gold Book, isotopic abundance: relative number of atoms of an isotope among all atoms of that element.

For a completely regular Hausdorff space, the Stone-Čech compactification is the weak-star closure of the evaluation functionals in the dual unit ball of $C_b(X)$. The Banach-Alaoglu theorem gives compactness; complete regularity makes evaluation a topological embedding. Every bounded real continuous function extends by $\tilde{f}(p)=p(f)$, yielding an isometric isomorphism $C_b(X)\cong C(\beta X)$. For a continuous map into a compact Hausdorff space $K$, the adjoint of pullback sends this closure into the closed evaluation copy of $K$, yielding the unique continuous extension. Uniqueness follows from density. It is a universal compactification, not a homeomorphism between $X$ and the whole dual space.

Large-extra-dimension models lower the fundamental higher-dimensional gravitational scale $M_{\ast }$ while obtaining a large four-dimensional reduced Planck mass from compact volume $V_n$. Ordinary matter can be localized on a brane while gravity propagates in the bulk. The Planck mass from compactification volume relation explains weak effective four-dimensional gravity but leaves volume stabilization and the origin of a large internal volume as separate problems.

// Target: physics.bigb

A warped five-dimensional geometry redshifts masses of fields localized on the distant brane: $m_{\mathrm{IR}}=e^{-kL}{m}_0$. For the usual doubled orbifold interval, integration of the Einstein-Hilbert action gives $M_{\mathrm{Pl}}^2=M_5^3(1-e^{-2kL})/k$. Exponential redshift can generate a large separation of observable scales with a moderately sized dimensionless distance $kL$. Stabilizing the radion remains a dynamical requirement.

// Target: physics.bigb

A radion is a lower-dimensional scalar field describing the size or separation modulus of compact extra dimensions. In a two-brane Randall–Sundrum model, its expectation determines the inter-brane separation and the warp-generated hierarchy. A background size alone does not supply a radion potential or a stabilization mechanism.

// Target: supersymmetry.bigb

Kaluza-Klein theory obtains lower-dimensional fields from a higher-dimensional metric and other fields on compact dimensions. In the displayed fixed-radius circle ansatz, the off-diagonal metric supplies a vector. The coordinate change $y\mapsto y+\lambda (x)$ is compensated by $A\mapsto A-d\lambda$, producing an Abelian gauge symmetry. Allowing the radius to vary adds a scalar modulus. Kaluza-Klein modes also include nonzero internal momenta, giving a tower of massive fields.

// Target: quantum-field-theory.bigb

In a brane-world model some fields, typically ordinary matter and gauge fields, are localized on a lower-dimensional brane, while gravity propagates through the higher-dimensional bulk. Compact volume or a warped metric can then separate the observed gravitational scale from local matter scales. A model must specify field localization, boundary conditions and stabilization rather than identifying a brane with an arbitrary compactification.

A coloring necklace is an equivalence class of colorings of cyclically arranged positions under rotation. Burnside lemma counts necklaces by averaging the numbers of colorings fixed by each rotation.

Stars and bars counts tuples of nonnegative integers with a prescribed sum. For integers $N\ge 0$ and $k\ge 1$, its statement isSuch a tuple is a weak composition of an integer. Think of $k$ labelled bins: $x_1$ counts identical objects in the first bin, $x_2$ in the second, and so on. Bin names distinguish positions even when counts agree. Stars and bars counts these count tuples; assigning individually distinct objects requires additional information. ^[ref]

Computed example. With $N=5$ and $k=3$, the number of weak compositions of an integer is $\left(\genfrac{}{}{0ex}{}{7}{2}\right)=21$. The string ••|•|•• encodes $(2,1,2)$: two dots before the first bar, one between the bars, and two after the second. For a fixed first-bin count $x_1=a$, the second-bin count can be any integer from $0$ to $5-a$ and the third is forced. Thus the direct count is $6+5+4+3+2+1=21$.

Adjacent bars encode an empty middle bin: ••||••• gives $(2,0,3)$. A bar at the left end encodes an empty first bin, as in |••|••• for $(0,2,3)$; a bar at the right end encodes an empty third bin, as in ••|•••| for $(2,3,0)$. These are valid weak compositions of an integer. For $N=0$ every count is zero, and for $k=1$ the unique tuple is $(N)$, consistent with $\left(\genfrac{}{}{0ex}{}{N}{0}\right)=1$.

Proof outline. Replace each count $x_i$ by that many dots and separate successive groups with $k-1$ bars. Reading the group sizes reverses this construction, so it is a bijection between weak compositions of an integer and strings of $N$ dots and $k-1$ bars. Choose the bar positions among $N+k-1$ positions; the binomial coefficient counts the choices. The beauty is that empty bins need no separate cases: end bars and adjacent bars encode them automatically.

If every bin must be nonempty, the tuples are compositions of an integer, and their count is $\left(\genfrac{}{}{0ex}{}{N-1}{k-1}\right)$ for $N\ge k\ge 1$; for $N<k$ there are none. Subtract one from every count, setting $y_i=x_i-1$. This is a bijection to weak compositions of an integer of $N-k$, so stars and bars gives $\left(\genfrac{}{}{0ex}{}{(N-k)+k-1}{k-1}\right)=\left(\genfrac{}{}{0ex}{}{N-1}{k-1}\right)$. For five objects and three bins, this is $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$, with the six positive tuplesFor instance, $(2,1,2)$ shifts to $(1,0,1)$, whose sum is $2=5-3$. Sorting the six positive tuples collapses them to just two integer partitions, $3+1+1$ and $2+2+1$. Thus named bins and unordered summands answer different questions.

Stars and bars is useful for allocating identical resources or enumerating integer solutions without listing them individually. Prescribed lower bounds can be subtracted just like the nonempty-bin requirement. Upper bounds require removing forbidden tuples; the existing bounded weak compositions article applies the inclusion-exclusion principle to those violations.

Double counting establishes an identity by computing the cardinality of the same finite set in two ways. The useful step is to identify exactly what the members of that collection are; an algebraic equality then records two legitimate ways to organize them.

Take the ordered pairs $(S,x)$ where $S$ is a $k$-element subset of an $n$-element set and $x\in S$ is marked. Choosing $S$ first gives $k\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ pairs. Choosing $x$ first leaves $k-1$ members to choose from the other $n-1$, giving $n\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)$. HenceFor five people and two-person teams, both counts give $20$: ten teams with two choices of a marked member, or five choices of that member with four choices of a teammate. The identity is explained by the bijection between these descriptions, rather than merely verified through factorials.

A tree is a nonempty connected graph containing no graph cycle. Think of seven locations joined so that everyone can reach everyone else, but no connection supplies an alternative route. For a finite simple graph with $n\ge 1$ vertices, three descriptions are equivalent: being a tree; having exactly one simple graph path between every pair of distinct vertices; and being a connected graph with exactly $n-1$ edges. Here a simple graph path repeats no vertex. Connectedness in the last description is essential. ^[ref]

For a concrete tree, label the vertices $1,\dots ,7$ and take the edgeswhere $ij$ abbreviates the unordered pair $\{i,j\}$. There are seven vertices and six edges. The unique graph path from $4$ to $7$ is $4,2,1,3,6,7$, using five edges. The degree sequence in label order is $(2,3,2,1,1,2,1)$: its sum is $12=2\cdot 6$, as the degree-sum formula predicts. Each of $4,5,7$ is a leaf of a graph, having vertex degree one.

Deleting edge $13$ leaves precisely two connected components of a graph, on $\{1,2,4,5\}$ and $\{3,6,7\}$. Every edge of a tree is a bridge in a graph. Instead add edge $57$ to the original tree: it closes the unique existing graph path $5,2,1,3,6,7$ into exactly one graph cycle. Thus a tree provides economical connections with no redundancy: a single failed link separates locations. Every finite tree with at least two vertices has at least two instances of a leaf of a graph; the endpoints of a longest graph path supply them. ^[ref]

The number $n-1$ alone proves nothing about connectedness: a cycle graph on $1,\dots ,6$ together with an isolated vertex $7$ has seven vertices and six edges, yet is disconnected. Briefly, two different simple graph paths between the same vertices would contain a graph cycle. Repeatedly removing a leaf of a graph and its incident edge proves the tree count $n-1$ by mathematical induction, starting with one vertex and no edges.

The indegree $d^{-}(v)$ of a vertex $v$ is the number of directed edges entering it, counting repeated directed edges separately if permitted. In the toy directed acyclic graph, the initial indegrees of $A,B,C,D,E$ are $0,0,1,2,2$. In Kahn's algorithm, these counts are updated to describe only the remaining directed graph, so indegree zero means that no unfinished predecessor remains. ^[ref]

For finite sets, the number of ordered choices of one element from each is the product of the individual numbers of choices. More generally, if stage $j$ has exactly $n_j$ possible choices after every permitted history of earlier stages, the total number of full choice sequences is ${\prod }_j{n}_j$, by mathematical induction on the stages. This counts $b^a$ functions between sets of sizes $a,b$, and counts the injective functions by the falling factorial $b(b-1)\cdots (b-a+1)$ when $a\le b$.

With five people, choose a captain in five ways and a different deputy in four ways: $5\cdot 4=20$ ordered assignments. The second choice depends on the first because the captain is excluded, but its number of permitted options is always four. The rule of product therefore applies without any assumption about independent events. To count two-person teams instead, each subset has two possible role assignments, so division by $2!=2$ gives ten teams. This division works because every team was counted equally often.

A matroid is an independence system in which a smaller admissible selection can always grow using an element of a larger admissible selection. This exchange rule abstracts linear independence of vectors and the selection of edges forming a forest. It forces all matroid bases to have the same cardinality and makes the matroid greedy algorithm correct for maximizing their total weight. The useful surprise is that local choices become globally optimal because of a structural rule about admissible subsets, rather than a special numerical coincidence.

The weighted three-vector matroid below lets us inspect every subset and every possible size comparison. Its three vectors give three two-element matroid bases, yet their total weights differ. The graphic matroid of a graph triangle has exactly the same admissible labels. A pregeometry is a finitary closure-theoretic extension to possibly infinite sets.

A combinatorial structure is a discrete mathematical object specified by underlying elements and rules organizing them. A graph equips a set of vertices with edges; a permutation orders labels; a set partition groups them into nonempty subsets. Counting requires an explicit equality convention. Two labelled combinatorial structures may differ only by their labels, whereas an unlabelled combinatorial structure identifies objects related by an appropriate isomorphism. Thus there are three labelled two-element subsets of a three-element set, but only one abstract two-element set up to bijection.

Enumerative combinatorics counts families of finite combinatorial structures, usually as a function of size. Selecting $k$ labels from $n$ gives a binomial coefficient; arranging all labels gives a factorial; grouping labels gives a set partition. A formula is only one kind of answer: a bijection, a recurrence relation or a generating function can expose how the objects are assembled and make their counts computable. The point is to specify exactly which distinctions, such as ordering or labels, count as different objects.^[ref]

In a two-level ground system, balancing collisional upward excitation against radiative and collisional downward loss gives $n_2/n_1=n_e{q}_{12}/(A_{21}+n_e{q}_{21})$. Excited lines fed in different proportions by these levels have a density-sensitive ratio. At low density the population ratio is proportional to density; at high density it saturates at $q_{12}/q_{21}$. A reliable calibration uses a full multilevel model when upper-state branching and cascades matter.