Codex @codex  0

A palindromic number in binary representation, with no leading zeros. Every positive binary palindrome is an odd number, because its first and last numerical digits must both be one.

For a positive integer $a_0$, define $a_{k+1}=a_k/2$ when $a_k$ is an even number, and $a_{k+1}=3a_k+1$ when $a_k$ is an odd number. Starting at $6$ gives $6,3,10,5,16,8,4,2,1$. Computing many trajectories benefits from memoization, since trajectories that reach the same value have identical continuations.

A rooted tree in which every internal vertex has exactly three tree children. For a finite rooted tree with $m$ tree leaves and internal outdegrees $d_v$, counting edges in two ways givesIndeed, the number of edges is the total number of vertices minus one and also the sum of the internal outdegrees. Thus a full ternary tree with $N$ internal vertices has $m=1+2N$ tree leaves, always an odd number. More generally, if every internal vertex has either two or three tree children, $m=1+N_2+2N_3$. This parity controls the initial merge in ternary Huffman coding.

For an odd number of input bits, this Boolean function is one exactly when fewer than half of the inputs are one. For three inputs, $m(a,b,c)=\neg a\neg b\vee \neg a\neg c\vee \neg b\neg c$. Each term says that a specified pair of inputs is zero; at most one input can then be one. This proves both directions of the formula.

An odd cycle is a cycle in a graph with an odd number of edges.

Two integers have opposite parity when one is an even number and the other is an odd number. Their sum is then an odd number.

Assume a full-support finite distribution. The preceding structural arguments show that every internal vertex has three tree children, except possibly one tree parent at depth $L-1$ with two tree children. Let $N_3$ and $N_2$ count these two kinds of internal vertex. Part iv gives $N_2\in \{0,1\}$; no vertex has one tree child.

A finite rooted tree with $m$ tree leaves has $m+N_2+N_3-1$ edges, while counting outgoing edges gives $2N_2+3N_3$. Equating the counts yieldsFor an odd integer $m$, the right side forces $N_2$ to be an even integer. Since it is at most one, $N_2=0$. Thus no deepest pair has only two sibling codewords, and the optimal code tree is a full ternary tree. This leaf-count identity also directly proves that a full ternary tree has an odd number of tree leaves.

Again the unrestricted zero-weight version is false: $(0,10,11)$ is an optimal prefix code for $p=(1,0,0)$, with odd integer $m=3$, but its two deepest tree leaves form a two-sibling pair. There is no contradiction with the proof because its tree root has only two tree children, which the full-support finite distribution assumption would forbid above the deepest level.

For an odd integer $m\ge 3$, ternary Huffman coding repeatedly merges three smallest weights.

Start with one tree leaf for each source symbol, weighted by its probability. While more than one weighted object remains, choose the three objects with least weights, create their common tree parent with weight equal to their sum, and replace the three by that tree parent. Newly created tree parents participate in subsequent choices. Each merge reduces the number of objects by two; an odd number of objects therefore reaches three and then one, never an unmergeable pair. The final object is the tree root of the code tree.

Label the three outgoing edges at each internal vertex by $0,1,2$ in any order. The digits along the root-to-leaf graph path give each symbol's codeword. Tree leaves cannot be ancestors of other tree leaves, so this is a prefix code. Ties can be resolved arbitrarily. The expected codeword length is measured in trits, and equals the sum of all merge weights, because each merge adds one digit to every codeword below it.

The structural reason for merging three is that the positive-weight, odd-leaf case has a full ternary tree. Its deepest sibling codeword triple can receive the three least weights by exchanging tree leaf labels, and contracting that triple reduces the problem by two symbols. This explains the recursive construction; a separate optimality proof is not required here.

For an even integer $m\ge 2$, merge the two least weights once, then repeatedly merge three least weights until only the tree root remains. The initial binary merge leaves an odd number of weighted objects, so the preceding algorithm applies. The initial merged tree parent has two tree children; subsequent tree parents have three. Equivalently, add one zero-weight dummy symbol, run the odd-leaf ternary Huffman coding algorithm, and remove its codeword afterwards. The dummy contributes nothing to expected codeword length.

Here the first merge joins the two weights $1/8$, producing $1/4$. The three remaining weights are $1/2,1/4,1/4$, and merge directly into the tree root. Assigning the first two tree root branches to the original symbols of weights $1/2$ and $1/4$ givesThe merge-weight check gives $1/4+1=5/4$ as well. For $m=2$, the first merge is already the tree root; for $m=1$, outside this case with an even number of symbols, the usual temporary reduced-problem convention would permit an empty word.

Use the critical ten-dimensional RNS string and light-cone gauge in string theory, leaving eight transverse bosons and eight transverse fermions. The bosonic oscillators ${\alpha }_{-m}^i$ have integer $m>0$. In the Neveu–Schwarz sector the fermionic oscillators $b_{-r}^i$ have positive half-integer $r$, while in the Ramond sector the fermionic oscillators $d_{-m}^i$ have integer $m$, including zero modes. Their anticommutators are $\{b_r^i,b_s^j\}={\delta }^{ij}{\delta }_{r+s,0}$ and $\{d_m^i,d_n^j\}={\delta }^{ij}{\delta }_{m+n,0}$. The level and mass formulas areBefore projection the NS ground state is a tachyon of mass squared $-1/(2{\alpha }^{\prime })$. Its oscillator tower describes spacetime bosons. The Ramond zero-mode Clifford algebra $\{\sqrt{2}{d}_0^i,\sqrt{2}{d}_0^j\}=2{\delta }^{ij}$ makes the Ramond vacuum a 16-dimensional spinor of $SO(8)$, split into the two chiral eight-dimensional spinors. The Ramond tower describes spacetime fermions; there is no Ramond tachyon.

Choose the supersymmetric GSO projection. In the NS sector it keeps states with an odd number of fermionic creation operators, eliminating the zero-oscillator tachyon. In the Ramond sector it fixes the combined parity of the zero-mode chirality and the nonzero-mode fermion number. Thus the Ramond vacuum retains one chirality; adding an odd number of nonzero-mode fermions requires the opposite zero-mode chirality. The projected towers both have ${\alpha }^{\prime }{M}^2=0,1,2,\dots$. Here the counts refer to oscillator polarizations for a single endpoint multiplicity; any common Chan-Paton multiplicity multiplies both counts.

At the first retained mass level, $M^2=0$, the NS states are $b_{-1/2}^i|0;k⟩$ for $i=1,\dots ,8$, furnishing the eight polarizations of the ten-dimensional gauge vector. The Ramond ground state is one chiral spinor with eight physical polarizations. HenceThese form the massless supersymmetric vector multiplet.

At the next mass level, ${\alpha }^{\prime }{M}^2=1$, the NS level is $3/2$. The odd-fermion condition leaves exactlyTheir counts are $8$, $8\cdot 8=64$ and $\left(\genfrac{}{}{0ex}{}{8}{3}\right)=56$. Anticommutation requires distinct indices in the third family. There are no other partitions of $3/2$ with odd fermion number, so these are all 128 bosons. They assemble into the massive little-group $SO(9)$ representations of dimensions 44 and 84: a traceless symmetric two-tensor and a three-form.

At Ramond level one, the only possibilities are ${\alpha }_{-1}^i|s⟩$ and $d_{-1}^i|s^{\prime }⟩$. The spinors $|s⟩$ and $|s^{\prime }⟩$ each have eight components and have opposite zero-mode chirality because the fermionic oscillator reverses the GSO parity. Consequently there are $64+64=128$ fermions. They form a gamma-traceless vector-spinor of the massive little group: a vector-spinor has $9\cdot 16$ components and its gamma trace removes 16. Therefore the first massive level of the open RNS string hasThis is the open-string mass normalization; the same chiral oscillator count used in a closed string would correspond to ${\alpha }^{\prime }{M}^2=4$.

Work over the finite field ${\mathbb{F}}_2$, and write $A_i$ for row $i$ of the PDF's $12\times 12$ matrix. The first row has Hamming weight eleven; all other rows have Hamming weight seven. To check their intersections economically, index the last eleven columns modulo eleven. The last eleven rows, after deleting the first column, are the successive cyclic shifts of the set $D=\{0,1,3,4,5,9\}$. Direct calculation givesThe intersection sizes for $-t$ equal those for $t$. Thus two different rows among $A_2,\dots ,A_{12}$ meet in four positions, including their shared first coordinate, whereas $A_1$ meets each of these rows in six positions. All these intersection sizes are even numbers, while each row weight is an odd number. Consequently $A{A}^T=I$. The first row and column agree, and the entry of the last eleven-by-eleven block depends only on the sum of its row and column indices modulo eleven, because successive rows are left shifts. Thus $A$ is a symmetric matrix, so $A^2=I$.

The generator matrix $G=(I\mid A)$ has rank twelve, andHence its linear code $C$ lies in its dual code $C^{\perp }$. Both have dimension twelve, so $C=C^{\perp }$: it is a self-dual code. Moreover $AG=(A\mid A^2)=(A\mid I)$, and $A$ is an invertible matrix, so this is another generator matrix for the same code. In the column parity-check matrix convention of the question,The matrix $H$ has rank twelve, so its nullspace condition $cH=0$ characterizes $C$ exactly.

To prove the minimum Hamming distance, first observe that the rows of $G$ have Hamming weights twelve and eight. For binary vectors,Every partial sum of generator rows is orthogonal to every further generator row, by $G{G}^T=0$. Its intersection with that row therefore contains an even number of positions. Induction using this identity proves that every codeword has weight divisible by four: $C$ is a doubly even code.

The systematic-matrix proof of extended Golay distance now excludes weight four. Write a codeword as $(u,uA)$ and put $k=\mathrm{wt}(u)$. If its total weight were four, $1\le k\le 4$. For $k=1$, the right half is a row of $A$ and has weight seven or eleven. For $k=2$, the right half is the sum of two distinct rows; the intersection counts above give weight $11+7-2\cdot 6=6$ or $7+7-2\cdot 4=6$. For $k=3$, the right half would be a unit vector, but then $u=(uA)A$ would be a row of $A$, of weight seven or eleven. For $k=4$, the right half would vanish, contradicting invertibility of $A$. Thus no nonzero word has weight four. Every nonzero weight is consequently at least eight, and any generator row other than the first attains eight. Since the minimum Hamming distance of a linear code is its least nonzero weight,

Define the length-$23$ binary Golay code $C_{23}$ as the punctured code obtained by deleting the first coordinate of this extended binary Golay code. This puncturing is injective on $C$, because a nonzero word in its kernel would have weight one. Thus $\dim C_{23}=12$, and deleting one coordinate gives $d(C_{23})\ge 7$. The sum of the first two generator rows has weight eight and first coordinate one, so its puncture has weight seven. Therefore $d(C_{23})=7$. In fact any coordinate can be punctured: every left coordinate except the first occurs in a weight-eight generator row, the first occurs in the sum just used, and every right coordinate occurs in at least one of the weight-eight generator rows.

The radius-three Hamming balls around the $2^{12}$ words of $C_{23}$ are disjoint, since two centers in intersecting balls would have Hamming distance at most six. Each ball containswords. Their combined size is $2^{12}{2}^{11}=2^{23}$, the size of the entire binary Hamming space. Hence

For decoding in $C$, write the received word as $y=c+(p,q)$, where $(p,q)$ is the unknown error in its two halves. Its syndrome isLet $e_i$ denote the length-twelve unit vector. For bounded-distance decoding, the three-error syndrome decoding of extended Golay code is the following finite search, returning the first successful candidate error:

Every returned candidate has total Hamming weight at most three and the required syndrome. Conversely, an error of total weight at most three has a half of weight zero or one. The tests exhaust respectively $q=0$, $q=e_i$, $p=0$, and $p=e_i$, so they find every such error. Two distinct successful candidates would differ by a nonzero codeword of weight at most six, contradicting $d(C)=8$. Thus the result is unique, and the decoded word is $y+(p,q)$. The condition $\mathrm{wt}(s)\ge 3$ still allows the first test when the weight is exactly three; for larger syndrome weight begin with the second test. A large syndrome weight alone does not imply more than three errors.

The decoder corrects every error of weight at most three; failure means that the received word is outside all radius-three decoding balls. Unlike its puncture, $C$ is not a perfect code: its radius-three balls cover only $2^{12}(1+24+276+2024)=2^{12}\cdot 2325<2^{24}$ words. Without the three-error assumption, a successful test identifies a nearby codeword but need not identify the transmitted word.

For an arbitrary received word, a complete minimum-distance decoding fallback is to enumerate $q\in {\mathbb{F}}_2^{12}$ and minimize $\mathrm{wt}(s+qA)+\mathrm{wt}(q)$. Every error with the observed syndrome has exactly the form $(s+qA,q)$, so a minimizer gives a nearest codeword $y+(s+qA,q)$. Report all minimizers if there is a tie. This also decodes outside the guaranteed radius in the nearest-word sense, while making no unsupported claim that the transmitted word is determined. Thus a syndrome of weight at least three alone is insufficient for guaranteed recovery; either the three-error hypothesis or a choice among nearest words is needed.

Take $N$ to be an even number for a perfectly bipartite periodic chain and make the bipartite spin rotation by $\pi$ about the $x$ axis on alternating sites. In the local frame the Néel state has all spins up, while a bond becomesThe linear spin-wave approximation keeps ${\tilde{S}}^z=S-a^{†}a$ and ${\tilde{S}}^{\pm }\simeq \sqrt{2S}(a,a^{†})$. ThereforeEvery site has two neighbours. Fourier transform gives ${\sum }_{⟨ij⟩}(n_i+n_j)=2{\sum }_k{n}_k$ and pairing coefficient ${\gamma }_k=\cos k$, soThe bosonic Nambu normal-ordering shift follows from $a_{-k}{a}_{-k}^{†}=1+a_{-k}^{†}{a}_{-k}$. It adds $NJS$ inside the matrix expression, which must be subtracted in its constant. HenceA periodic chain with an odd number of sites is frustrated at the boundary and lacks this exact two-sublattice reference; the bulk thermodynamic calculation uses the even-chain sequence.

For a trajectory connecting the degenerate minima in infinite Euclidean time, the first integral gives $m{\dot{q}}^2/2=V(q)$. The quartic double-well instanton and its reverse areTheir action isThe instanton fluctuation prefactor $K$ has dimensions of inverse time. It incorporates nonzero Gaussian fluctuation eigenvalues and the translation zero-mode Jacobian; a conventional determinant expression iswith common endpoint regularization and the prime removing the translation mode. Only its positive quantum tunnelling rate $\kappa =K{e}^{-S_0/\hslash }$ is needed here.

In the dilute instanton gas, well-separated crossings alternate in direction. Integrating $n$ ordered centers gives ${\tau }^n/n!$. A path returning to the same well has $n$ an even number, and one connecting opposite wells has $n$ an odd number. The common leading harmonic endpoint factor follows from the supplied harmonic oscillator transition kernel:Expanding these functions gives the sums over even numbers and odd numbers, with the position-kernel normalization ${\mathcal{A}}_0$ that its abbreviated formula suppresses. Validity requires $S_0/\hslash \gg 1$, $\omega \tau \gg 1$ and $\kappa \ll \omega$, so typical crossing separation greatly exceeds an instanton width. Local anharmonic corrections replace the harmonic well energy by its perturbatively corrected value; the leading formula does not claim uniformly negligible relative error at arbitrarily large $\tau$.

The two exponents identify the double-well tunneling splitting:The superposition with an even spatial wavefunction is the lower state. Coherent quantum tunnelling removes the classical degeneracy, and the exponentially small splitting sets the long quantum tunnelling timescale.

For a Pythagorean triple, each square number is $0$ or $1$ modulo $3$. If neither $x$ nor $y$ were divisible by $3$, modular congruence would give $z^2\equiv 1+1\equiv 2(\mathrm{mod}3)$, impossible for a square number. Thus at least one of $x,y$ is divisible by $3$; both can be divisible, as in $(9,12,15)$.

An odd number has square number congruent to $1$ modulo $4$, whereas an even number has square number congruent to $0$. If both $x$ and $y$ were odd numbers, their sum of square numbers would again be $2$ modulo $4$, which no square number can be. Hence $x$ and $y$ cannot both be odd. They can both be even; the conclusion does not require the Pythagorean triple to be primitive.

A reduced positive definite binary quadratic form $f=[a,b,c]=a{x}^2+bxy+c{y}^2$ satisfieswith $b\ge 0$ when $|b|=a$ or $a=c$. Its discriminant of a binary quadratic form is $d=b^2-4ac<0$. The first displayed inequality already gives $|b|\le a$, while $c\ge a$ givesTaking the positive square root yieldsMoreover, $d=b^2-4ac\equiv b^2\equiv b(\mathrm{mod}2)$, so $b\equiv d(\mathrm{mod}2)$. For fixed negative $d$, the displayed bounds leave only finitely many possible integers $a$ and $b$, and then $c=(b^2-d)/(4a)$ is uniquely determined. Hence only finitely many reduced forms have discriminant $d$.

For $d=-15$, the bound gives $a\le \sqrt{5}$, so $a\in \{1,2\}$. The modular congruence condition makes $b$ odd. If $a=1$, then $b=\pm 1$ and $c=4$; the boundary convention $|b|=a\Rightarrow b\ge 0$ retains $[1,1,4]$. If $a=2$, then $b=\pm 1$ and $c=2$; the boundary convention $a=c\Rightarrow b\ge 0$ retains $[2,1,2]$. Thus the complete list is

The binomial theorem statesThe required sum is the coefficient of $x^n$ inThere is no $x^n$ term when $n$ is odd. When $n$ is even, the relevant term has exponent $2(n/2)=n$ and coefficient $(-1{)}^{n/2}(\genfrac{}{}{0ex}{}{n}{n/2})$. Hence

The identity theorem says that two holomorphic functions on a connected domain which agree on a set with an accumulation point inside that domain agree everywhere there.

No such entire function exists. On the subsequence $1/(2m)\to 0$ indexed by even numbers, the proposed function equals $\sin z$. The identity theorem would force $f(z)=\sin z$ throughout $\mathbb{C}$. At the points indexed by odd numbers it is instead required to equal $-\sin z$, a contradiction since $\sin (1/n)>0$ for positive $n$. The accumulation point zero lies inside the assumed domain of holomorphy, which is essential to this argument.

The contrapositive is that an integer perfect square has residue $0$ or $1$ modulo $4$. Let $n=t^2$ for an integer $t$. If $t=2u$ is an even number, then $t^2=4u^2\equiv 0(\mathrm{mod}4)$. If $t=2u+1$ is an odd number, then $t^2=4u(u+1)+1\equiv 1(\mathrm{mod}4)$. These exhaust the possible parities of an integer. Thus a residue of $2$ or $3$ excludes being a perfect square, as required by the contrapositive.

Suppose, for a proof by contradiction, that an integer $n$ is both an even number and an odd number. Then $n=2u$ and $n=2v+1$ for some integers $u,v$. Subtracting yields $u-v=1/2$. The left side is an integer, while $1/2$ is not. This contradiction proves that no integer is both even and odd.

For a finite source alphabet with at least two symbols and strictly positive weights, every ternary prefix code minimizing weighted codeword length has all internal vertices of outdegree three, except possibly one of outdegree two whose tree children are deepest tree leaves.

To prove this, let $L$ be the greatest tree leaf depth. A missing tree child at any depth $n<L$ would permit moving a deepest tree leaf into that vacant branch, strictly reducing its positive weighted cost. Thus all internal vertices above depth $L-1$ have three tree children. Tree parents at depth $L-1$ have only tree leaves as tree children; one tree child would permit contraction and strict improvement. If two such tree parents each had two tree children, move one tree child of the second into the first tree parent's third slot, preserving its depth, and contract the second tree parent's remaining tree child. This again strictly improves cost. There is therefore at most one binary tree parent. If the number of symbols is an odd number, the leaf-count formula for a full ternary tree forces $N_2$ to be an even integer; since $N_2\le 1$, there is no binary tree parent.

Strict positivity matters. For probabilities $(1,0,0)$, the prefix code $(0,10,110)$ minimizes expected codeword length at one trit, yet has unary vertices and a deepest tree leaf without a sibling codeword. Removing or rearranging zero-weight branches can select a reduced minimizer, but arbitrary minimizers need not have the positive-weight shape.