Over ${\mathbb{F}}_2$, $X^5-1=(X+1)(X^4+X^3+X^2+X+1)$ and the degree-four factor is irreducible: it has no linear root and is not divisible by the sole irreducible quadratic $X^2+X+1$. The four monic divisors give precisely four cyclic codes: the full five-dimensional space, the four-dimensional even-weight code, the one-dimensional repetition code, and the zero code. Their generator polynomials of a cyclic code are respectively $1$, $X+1$, $X^4+X^3+X^2+X+1$, and $X^5-1$.

For a cyclic code with monic generator $g\mid X^N-1$, the canonical check polynomial is $h=(X^N-1)/g$. A word $a$ belongs to the code exactly when $ah$ is zero modulo $X^N-1$, since cancellation in the polynomial ring turns this condition into $g\mid a$. The kernel condition alone allows additional polynomials: it is equivalent to $\gcd (\tilde{h},X^N-1)=h$. Requiring the monic divisor of the modulus singles out the canonical choice.

// Target: algebra.bigb

A repeated-root cyclic code has length divisible by the characteristic of its field, so the polynomial $X^N-1$ has repeated factors. Its generator can involve powers of those factors. In characteristic two, length $2^t$ gives $X^{2^t}+1=(X+1{)}^{2^t}$.

// Target: algebra.bigb

A zero of a cyclic code is a polynomial root of its generator polynomial, taken in a splitting field. Every codeword, represented as a polynomial, vanishes there. For a repeated-root cyclic code, the multiplicity of a root must also be retained: ordinary evaluations alone need not determine the code.

A linear code is negacyclic if $(c_0,\dots ,c_{n-1})$ being a codeword implies that $(-c_{n-1},c_0,\dots ,c_{n-2})$ is a codeword. Polynomial identification makes it an ideal modulo $X^n+1$. For odd $n$, the substitution $X\mapsto -X$ gives a monomial equivalence of linear codes between negacyclic and cyclic codes.

A topological graph is a CW complex with only zero-cells and one-cells, called vertices and edges. Edge endpoints attach to vertices; loops, multiple edges and infinitely many edges are allowed. A subgraph in this setting is a CW subcomplex. A connected acyclic graph is a tree and is a contractible space.

An open $q$-cell of a CW complex is a copy of an open $q$-dimensional ball. Its characteristic map is a map from a closed ball whose interior parametrizes that cell and whose boundary lands in the preceding skeleton. The characteristic map need not be injective on the boundary. Each oriented cell gives one generator of the cellular chain complex.

The $q$-skeleton of a CW complex is the subcomplex consisting of its topological cells of dimension at most $q$. A cellular map preserves these skeleta. The cellular approximation theorem lets continuous maps respect this filtration up to homotopy.

A compact subset of a CW complex is contained in a finite CW complex that is a subcomplex of the original complex. Otherwise choose one point in each of infinitely many cells met by the subset. The closure-finiteness axiom makes each closed cell meet the chosen set in finitely many points. The weak topology axiom then makes the chosen set, and each of its subsets, closed. It is consequently an infinite closed discrete subspace of the compact subset, a contradiction. Only finitely many cells are met; their closures form a finite subcomplex containing the subset.

For a finite simple graph that is an undirected graph, the DFS lowpoint of $v$ is the smallest DFS discovery time reachable by following zero or more discovery edges down its rooted tree, then at most one non-discovery edge to an ancestor. Staying at $v$ is allowed. Ignore the reverse traversal of the discovery edge to the parent in a rooted tree: it is the same edge, not an escape route. Start with $\mathrm{low}(v)=\mathrm{disc}(v)$. On returning from a child in a rooted tree $w$, replace $\mathrm{low}(v)$ by $\min \{\mathrm{low}(v),\mathrm{low}(w)\}$. For a non-discovery edge from $v$ to an earlier ancestor $a$, replace it by $\min \{\mathrm{low}(v),\mathrm{disc}(a)\}$. This summarizes how far upward the explored branch can reconnect. ^[ref]

The cut vertex criterion has a separate case for each root of a tree in the discovery forest:

Why does equality matter? If $\mathrm{low}(w)=\mathrm{disc}(u)$, the branch below $w$ can reconnect to $u$, so the discovery edge $uw$ lies on a graph cycle and is not a bridge in a graph. Deleting the vertex $u$ destroys that reconnection too. If $u$ has a parent in a rooted tree, the branch is then separated from that parent in a rooted tree, explaining the non-strict cut vertex test. For a root of a tree, there is no earlier side to separate from; distinct discovery branches cannot have an edge between them, since depth-first search would have explored them as one branch. This gives the two-children in a rooted tree rule. These are proof outlines of the criteria. ^[ref][ref]

In the joined-$CD$ example with DFS discovery times $1,2,3,4,5,6$, the DFS lowpoints in label order are $1,1,1,4,4,4$. For $C$ and its child in a rooted tree $D$, $4>3$: $CD$ is a bridge in a graph, and $C$ is a cut vertex. For $D$ and its child in a rooted tree $E$, $4=4$: $D$ is a cut vertex, although $DE$ is not a bridge in a graph. The root of a tree $A$ has only one child in a rooted tree, $B$, so it is not a cut vertex. In the shared-$A$ example, the root of a tree $A$ has two children in a rooted tree, $B$ and $D$, and is a cut vertex, while every DFS lowpoint is $1$ and no edge passes the strict bridge in a graph test.

Restart depth-first search at every undiscovered vertex to cover a disconnected graph, applying the root of a tree rule separately in each connected component of a graph. With adjacency lists and constant-cost label operations, computing all DFS lowpoints, cut vertices and bridges in a graph has time complexity $O(|V|+|E|)$ in Big O notation, because each vertex is discovered once and each edge is inspected twice. The simple parent-skipping rule assumes a simple graph. In a multigraph, skip only the actual parent edge, identified individually: a parallel edge to the same parent in a rooted tree is a valid reconnection and can prevent the first edge from being a bridge in a graph. The criteria here concern undirected graphs; a directed graph requires a different connectivity analysis.

The DFS discovery time of a vertex is its first-visit number in depth-first search, assigned by an increasing counter. Discovery edges form a rooted tree in each connected component of a graph. For the joined-$CD$ example, starting at $A$ and visiting graph neighbours in lexicographic order gives first visits $A,B,C,D,E,F$, numbered $1,2,3,4,5,6$; its discovery edges are $AB,BC,CD,DE,EF$. These numbers encode the chosen traversal, not an invariant of the graph.

A biconnected component is a maximal connected subgraph with no cut vertex of its own. We use the block convention that also includes each bridge in a graph with its two endpoint vertices, and each isolated vertex, as a biconnected component. A biconnected component with at least three vertices stays connected after any single vertex deletion. Distinct biconnected components share at most one vertex; the vertices belonging to more than one biconnected component are exactly the cut vertices. The shared-$A$ example has two biconnected components, its two triangles in a graph. The joined-$CD$ example has three: $ABC$, the bridge in a graph $CD$ with its endpoints, and $DEF$. The edges partition into these pieces, while the shared cut vertices occur in several pieces. ^[ref]

An unfriendly partition $V=A\bigsqcup B$ puts at least as many neighbours of every vertex in the opposite part as in its own part.

A parametric curve is described by a vector-valued map $t\mapsto Q(t)$. Its derivative gives a tangent vector wherever it is nonzero.

// Target: geometry-and-topology.bigb

An asymptote of an unbounded curve is a straight line whose distance from the curve tends to zero along a specified end. For a graph, a sloping asymptote $y=mx+b$ satisfies $y(x)-(mx+b)\to 0$ as $x$ tends to the relevant infinity. Factorization of the leading quadratic part locates the two asymptotes of a nondegenerate hyperbola.

A curve $\gamma :[0,1]\to X$ whose endpoints agree, $\gamma (0)=\gamma (1)$. This is the same unbased notion as a loop; self-intersections are allowed.

For a unit-speed curve with nonzero curvature, torsion is $\tau =-B^{\prime }\cdot N$, measuring rotation of the osculating plane.

Fix the standard orientation of a vector space on the plane. For an oriented regular curve in that plane, let $T=(T_x,T_y)$ be its unit tangent vector and let $N_L=(-T_y,T_x)$ be its left unit normal, obtained by an anticlockwise rotation through $\pi /2$. Its signed curvature $k$, also commonly written ${\kappa }_s$, is defined byHere $s$ increases with the chosen traversal and $\det$ is the determinant. Positive signed curvature means turning left; negative signed curvature means turning right. The left unit normal exists even when $k=0$; when $k<0$ it points opposite to the principal normal vector. See Ghomi's university lecture notes, section 1.10.

For an arbitrary regular parametric curve $r(t)=(x(t),y(t))$ traversed with increasing $t$, the signed curvature isFor the anticlockwise circle $r(t)=(R\cos t,R\sin t)$, the numerator is $R^2$ and the denominator is $R^3$, so $k=1/R$. For the clockwise traversal $q(t)=(R\cos t,-R\sin t)$, the numerator is $-R^2$, so $k=-1/R$, while unsigned curvature is $1/R$ in both cases. On this circle $N_L$ points inward for anticlockwise traversal and outward for clockwise traversal. For any simple closed regular curve traversed anticlockwise, $N_L$ is the inward unit normal, but signed curvature can still be negative along nonconvex portions.

To see why the formula uses the cube of the speed, write $r^{\prime }=vT$ and $r^{\prime \prime }=v^{\prime }T+v^2(dT/ds)$. The term parallel to $T$ contributes zero to the determinant, giving $\det (r^{\prime },r^{\prime \prime })=v^3\det (T,dT/ds)=v^{3}k$. Under a reparametrization $q(u)=r(\varphi (u))$ with ${\varphi }^{\prime }\ne 0$, the numerator gains a factor $({\varphi }^{\prime }{)}^3$ and the denominator gains $|{\varphi }^{\prime }{|}^3$. Thus orientation-preserving reparametrization preserves signed curvature, while reversing traversal changes its sign at corresponding points. Unsigned curvature is preserved in either case.

Locally write the unit tangent vector as $T=(\cos \theta ,\sin \theta )$. Differentiating with respect to arc length gives $k=d\theta /ds$: signed curvature records the direction as well as the amount of turning. Its integral therefore measures net turning, as in the turning tangent theorem; the integral of $|k|$ is total curvature and counts both left and right turns positively.

For a uniform-mass density cylinder with surface displacement $\eta e^{im\varphi }$, the interior and exterior potential perturbations solve Laplace's equation and are proportional to $R^m$ and $R^{-m}$. Continuity of potential and the Poisson derivative jump ${\Phi }_{R,\mathrm{out}}^{\prime }-{\Phi }_{R,\mathrm{in}}^{\prime }=4\pi G{\rho }_0\eta$ give the displayed surface amplitude. The surface mass density term must be retained even though bulk incompressibility makes the interior mass density perturbation zero.

The regular $m=1$ velocity family satisfies the displayed formulas. The surface pressure condition and gravitational matching give $W_s=-{\Omega }_s^2{R}_0\eta$, while the kinematic condition is $i(\omega +{\Omega }_s)\eta =u_R(R_0)$. If the material radial displacement $u_R/[i(\omega +kR)]$ is continued regularly to the boundary, then $\eta =-ia$ and ${\omega }^2=0$: translation of an isolated cylinder has no restoring force. Using only the Eulerian velocity conditions instead leaves ${\omega }^2(\omega +{\Omega }_s)=0$. The additional surface-corotation case has $\eta =0$, ${\Phi }^{\prime }=0$, $u_R=a(kR-{\Omega }_s)$ and $W=iaR[(kR{)}^2-{\Omega }_s^2]$; it satisfies the velocity equations and zero surface pressure. Excluding that degenerate branch is an extra admissibility assumption, not a valid division by a nonzero frequency at every surface.