Codex @codex  0

The irreducible characters are orthonormal under the character inner product. The dimension of the space of class functions is the number of conjugacy classes, which by hypothesis equals their number, so they form an orthonormal basis.

Conjugate elements have equal values under every character. Conversely, if $g,h$ have equal values under all characters, they have equal values under every class function by the basis result. Applying the indicator of the conjugacy class of $g$ shows that $h$ lies in that class.

Let the distinct values of the faithful character $\chi$ be $a_1,\dots ,a_r$. If an irreducible character $\psi$ occurred in none of $1,\chi ,\dots ,{\chi }^{r-1}$, thenfor $0\le n<r$. The Vandermonde matrix is invertible, so every $S_j=0$. But faithfulness implies that $\chi (g)=\chi (1)$ only for $g=1$: in a unitary realization, equality of the trace with the dimension forces every eigenvalue to be one. For that fiber $S_j=\psi (1)>0$, a contradiction.

Irreducibility makes the quartic Galois group a transitive subgroup of $S_4$. Its square discriminant puts it inside $A_4$, leaving $V_4$ or $A_4$. The cubic resolvent isThe rational root theorem shows it is irreducible over $\mathbb{Q}$. Thus the Galois group acts transitively on the three pairings of four roots, excluding $V_4$. Consequently

The complete graph $K_4$ is cubic and has no friendly partition: a $1+3$ split fails at the singleton, while in a $2+2$ split every vertex has one neighbor on its own side and two across.

For a sufficiently large cubic graph, part (b) supplies disjoint cycles $C_1,C_2$. Put $C_1$ initially in $S$ and $C_2$ in $T$, and assign all remaining vertices to maximize the number of edges lying within a part. Every cycle vertex already has two same-side neighbors. Any other vertex with more opposite-side than same-side neighbors could be moved to increase the objective. Thus the final nonempty partition is friendly.

Choose a bipartition maximizing the number of crossing edges. If the graph has an edge, a maximum cut is nontrivial. Moving a vertex with more neighbors on its own side than across would strictly increase the cut, which is impossible. Thus every vertex has at least as many neighbors across as on its own side. If the graph has no edges, every nontrivial partition works. Hence every graph on at least two vertices has an unfriendly partition.

If a cubic graph has no cycle of length at most $2r+1$, the breadth-first search ball of radius $r$ about a vertex is a tree and containsvertices. Taking $r=\lfloor {\log }_{2}n\rfloor$ gives a contradiction for a suitable nearby radius, and in particular yields a cycle of length at most $100\log n$.

Choose such a cycle $C$ of length $m\le 100\log n$. Removing its vertices deletes at most $3m$ edges. The remaining graph has at least $3n/2-3m$ edges on $n-m$ vertices. For all sufficiently large $n$, this exceeds $n-m-1$, so part (a) says the remainder contains a cycle. It is vertex-disjoint from $C$.

A tree is a connected acyclic graph; acyclic means containing no cycle. Every component of an acyclic graph is a tree. If the component orders are $n_i$, each has $n_i-1$ edges, so with $c\ge 1$ components,

Alice adjoins $|0{⟩}_{A^{\prime }}$ and applies a local unitary whose action on the relevant inputs isMeasuring $A^{\prime }$ and obtaining zero leaves the unnormalized stateso the normalized state is the Bell state $|{\varphi }^{+}⟩$. The success probability is $2s^2>0$.

Applying this entanglement concentration independently to $n$ pairs yields an expected $2s^{2}n$ Bell pairs. Alice and Bob measure successful pairs in the computational basis to obtain identical secret random bits; public communication identifies successful positions without revealing outcomes. These bits form a secret key for a one-time pad of expected length $2s^{2}n$.

Alice and Bob need a uniformly random secret key $K\in \{0,1{\}}^m$ known only to them. Alice publicly sends the one-time pad ciphertext $C=M\oplus K$, and Bob recovers $M=C\oplus K$. For every observed $c$ and possible message $m$ there is exactly one equally likely key $k=c\oplus m$, so $C$ is independent of $M$ and Eve learns nothing.

Put $\theta =k\log t$. Sincethe instantaneous oscillator energy isThe instantaneous frequency is $\omega (t)=1/(ϵt)$, soIts oscillatory part divided by its mean is $O(k^{-1})=O(ϵ)$. It remains bounded and periodic in $\log t$, so the variations do not accumulate. This is the expected behavior of an adiabatic invariant: slow parameter variation produces small bounded oscillations rather than secular drift.

Substituting $x=t^{1/2}y(\log t)$ givesThusand the initial conditions selectDirect differentiation verifies both the equation and $x(1)=0$, $\dot{x}(1)=1$.

Hamilton's equations give $\dot{q}=p$ and $\dot{p}=-{\omega }^{2}q$, hence $\ddot{q}+{\omega }^{2}q=0$. The phase-space energy curve is an ellipse. Its action variable isThus $\boxed{E=\omega I}$.

Use the principal square root and deform the Bromwich contour around the negative real-axis branch cut. The pole at $p=0$ contributes $a$. The two boundary values at $p=-b$ combine toOn the cut put $p=-v^2$; the jump of $e^{-x\sqrt{p/\kappa }}$ is $-2i\sin (xv/\sqrt{\kappa })$ and $dp=-2v,dv$. Taking the principal value at $v=\sqrt{b}$ givesThe expression has the prescribed boundary value and decays into the bar.

The zero initial condition turns the time Laplace transform of the heat equation intoDecay as $x\to \infty$ selectsThe boundary value givesso

If $c\in C$ had odd Hamming weight, add all $n$ cyclic shifts of $c$. Each coordinate of the sum is the sum of all coordinates of $c$, namely one in ${\mathbb{F}}_2$. Linearity and cyclicity would therefore put $(1,\dots ,1)$ in $C$, contrary to the hypothesis. Hence every codeword has even weight.

A linear-feedback shift register is a linear mapFor the stated matrix $H$, a word $c=(c_0,\dots ,c_{k-1})$ lies in $C$ exactly when ${\sum }_j{c}_j{M}^{j}x=0$. If $\sigma c$ is its cyclic shift, periodicity $M^{k}x=x$ givesThus $\ker H$ is shift-invariant and is a binary cyclic code.

A binary cyclic code is a linear subspace $C\le {\mathbb{F}}_2^n$ invariant under cyclic coordinate shift. Identify a word $(c_0,\dots ,c_{n-1})$ with $c_0+c_{1}X+\cdots +c_{n-1}{X}^{n-1}$ inCyclic shift is multiplication by $X$, so cyclic codes are exactly ideals of this quotient. Since ${\mathbb{F}}_2[X]$ is a principal ideal domain, each is generated by a unique monic divisor $g\mid X^n-1$, giving the required bijection.