Yes—your congruence is an immediate consequence of the classical congruence for Dedekind sums 12 k s(h,k) ≡ h + h^{-1} (mod k), valid for coprime h,k, which itself follows directly from the reciprocity law. You can find this (with your convention for s) in Rademacher–Grosswald, Dedekind Sums, or derive it in two lines from the reciprocity formula s(h,k) + s(k,h) = −1/4 + (1/12)(h/k + k/h + 1/(hk)) by reducing modulo k.

Here is a short derivation of your n=3 statement from that congruence.

Set ...

Short answer: Yes, the idea that living creatures can be found “inside” inanimate matter (trees, earth, even stone) is old in India, and it is explicitly used in philosophical argument long before early modern Europe. However, earlier Indian texts do not, as far as we can securely document, single out the precise marvel “a live toad sealed in the heart of a solid rock.” What we do find well before the 13th century are (a) the general doctrine that some beings arise from moisture or from the eart...

No. Even for a very simple first‑order (indeed, positive existential) definable set, the density need not exist and can oscillate wildly along the chain.

Here is a concrete counterexample.

• Let G be the (locally finite) direct sum of countably many copies of a cyclic group of order 4: G = ⊕m≥1 C4. In this group, the set of squares is exactly 2G (coordinatewise squaring). In particular, an element is a square in G iff each of its C4 coordinates is either 0 or 2. The set of squares is first...

Short answer: Nobody knows (as of 2025).

Details:

• The opening move 16 in Sylver Coinage is a famous open problem. It has been studied since Conway introduced the game, and it is still not known whether the first player can force a win after opening with 16, assuming optimal play by both players.
• It is widely believed (but unproven) that 16 is a first-player win.
• A standard reduction of the question is this: after 16, the only plausible reply for the second player that is not known to lose...

Short answer: As far as we know today, both questions are open for unrestricted (i.e., arbitrary-depth) circuits. We do not know any explicit monotone function family in P/poly for which every monotone circuit that achieves even constant-bounded error under the uniform distribution (or at its critical p) must have superpolynomial size; conversely, we also do not have a general positive theorem showing that such monotone approximators always exist.

Some context and related results:

• Exact lowe...

Short answer The first state-organized, maintained overland route that let you travel by land from “Sweden proper” into “Swedish Finland” ran at the head of the Gulf of Bothnia via the Torneå/Tornio crossing. It came together in the early 17th century and was formalized as a postal road in the 1630s:

• On the Swedish side: the Norrland coastal road (Kustlandsvägen/Norrstigen) was extended north and improved in connection with the crown’s 17th‑century town foundations (Umeå 1622; Piteå and Luleå ...

Yes. In fact, for any diffuse abelian von Neumann algebra A (in particular A = L∞[0,1]), if M and N are factors and A ⊗ M ≅ A ⊗ N, then M ≅ N. No separability assumption on M or N is needed.

Here is a short proof.

• Let A = L∞[0,1]. Since M and N are factors, the center of A ⊗ M (resp. A ⊗ N) is A ⊗ 1. Thus any ∗-isomorphism θ: A ⊗ M → A ⊗ N maps the center onto the center and hence induces an automorphism α of A. Composing θ with the automorphism α−1 ⊗ idN of A ⊗ N, we may and do assume θ fix...

Short answer Yes. Shor’s algorithm admits substantial time–memory (depth–qubit) tradeoffs. There isn’t a single hard qubit threshold beyond which you suddenly can factor and below which you can’t. If you are “one qubit short” of a particular circuit layout, you can often recompile to save that qubit at the cost of longer runtime. However, there is a fundamental floor: you need on the order of n logical qubits to factor an n‑bit modulus at all. If you have fewer than about n + constant logical qu...

Short answer: I’m not aware of any peer‑reviewed study that documents a nonhuman group systematically changing their communication specifically for a chronically deaf conspecific (e.g., consistently calling louder/closer to that individual or routinely switching to tactile signals only with that individual). Such cases are probably rare for two practical reasons: (1) confirmed deafness is hard to diagnose in free‑ranging animals, and (2) in taxa where hearing is crucial for survival, fully deaf ...

Short answer

• If the two moduli come from the same prime and one field is a subfield of the other (GF(p^m) ⊆ GF(p^k) with m | k), then conversion is trivial and non‑interactive: just re‑interpret the shares in the larger field.

• In all other cases (different primes, no subfield relation, or rings Zp vs Zq), there is no purely local, linear, black‑box way to convert an LSSS sharing over Zp (or GF(p^m)) into an LSSS sharing over Zq that preserves the secret for all inputs. You need interaction...

No in general (in dimensions ≥ 3).

There are plenty of continuous (topological) finite group actions on smooth manifolds that are not topologically conjugate to smooth actions. The classical source of such examples are wild actions, whose fixed point sets are wildly embedded and hence cannot occur for smooth actions (where fixed point sets are smoothly embedded submanifolds).

Concrete mechanism:

• In dimension 3, there is Bing’s involution of S^3: a Z/2-action by homeomorphisms whose fixed po...

Short answer: This is open.

More detail:

• Model. A “sorting network” here means a fixed (oblivious) comparator network: the sequence of compare–exchange operations does not depend on the input. One can specialize such a network to compute only the median (or, more generally, the k-th order statistic), often called a selection network.

• Upper bounds. The best general deterministic constructions for exact selection in comparator networks have size O(n log k), where k is the number of selected ...

Yes. If you allow “explicit” to mean “effectively presented with decidable equality” (but without requiring the field operations to be efficient), then you can make the irreducibility problem as hard as you like, up to any decidable complexity class. In particular you can get EXPTIME-hardness (hence, under P ≠ EXPTIME, no polynomial-time algorithm), and you also recover the classical undecidability examples by the same template.

A simple construction

• Start with the purely transcendental fiel...

Short answer: This is open in general. There is no known theorem that every number field K embeds in a number field L whose ring of integers OL is generated by one element over Z; nor is there a known counterexample.

Remarks and partial results:

• Positive cases:
  • If K is abelian over Q, then by Kronecker–Weber there exists m with K ⊂ Q(ζm), and cyclotomic fields are monogenic: OQ(ζm) = Z[ζm]. Thus the answer is yes for abelian K (take L = Q(ζm)).
  • Quadratic fields are themselves monogeni...

Short answer:

• For the Cartesian power Cn(G), the chromatic number is trivial: χ(Cn(G)) = χ(G) for all n (Sabidussi, 1957/58). For certain G (e.g., complete graphs Kq and K2 in particular), α(Cn(G)) also has a closed form and can be computed in time polynomial in n.
• For the strong power Sn(G), computing α(Sn(G)) is exactly the “finite-n” version of the Shannon-capacity problem. Even for very small graphs this is notoriously hard; there is no general efficient algorithm known, nor are there pr...

Short answer: No. There are “easy” (in the Lovász–Local-Lemma sense) satisfiable formulas on which DPLL with random variable splits (with unit-propagation and pure-literal rules) takes exponential expected time.

Why the LLL condition does not help DPLL

• The LLL condition (each t-clause conflicts with fewer than 2^t/e other clauses) guarantees existence of a satisfying assignment and, via the Moser–Tardos resampling algorithm, a polynomial-time randomized algorithm to find one. But this algorit...

Short answer There was no fixed or uniform price. On the late‑19th‑century frontier, criminal defense fees were privately negotiated and varied with the seriousness of the charge, the standing of the lawyer, the amount of travel and preparation involved, and whether there would be separate trials for multiple defendants. Contemporary western “minimum fee schedules” adopted by local bar associations suggest that, circa 1880s–1890s, a privately retained lawyer’s fee for a felony murder case in a t...

Short answer Among deterministic context–free languages (DCFLs), those that occur as the word problem of a finitely generated group are exactly the word problems of virtually free groups. This is a very small subclass of DCFLs, constrained by strong algebraic invariances (conjugation, inversion, free reduction). There is no known purely “language–theoretic” intrinsic characterization beyond these structural invariances; the sharpest description comes from group theory (Bass–Serre theory) or from...

Short answer: No analogue of Conway’s “Mathieu groupoid” for M24 (on 31 holes) exists in the same canonical sense as M13. This was looked at quite seriously (by Conway and others), and there are clean obstructions.

What does work, and why one might hope for it

• The “right” ambient set is indeed V = F2^5 \ {0}, with |V| = 31.
• A “hole” should be a 3-dimensional subspace U ≤ V (a plane in PG(4,2)), which contributes 7 excluded points, leaving 24 points to permute.
• As you observed, this matche...

Short answer: there is no new, accepted “central insight” that suddenly turns the Wang–Xu 61‑stem computation into a proof that a Kervaire invariant one class exists in dimension 126. The existence in 126 is still open. The 8‑page preprint you cite does not contain a vetted argument; it relies on a step that is exactly the piece of 2‑primary information Wang and Xu left unresolved because it is equivalent to deciding the 126‑dimensional Kervaire problem.

What experts have known for a long time ...

Short answer: not as a single sheaf. The right higher‑codimension replacement of “Cartier divisors” is no longer captured by global sections of one sheaf; instead one needs a (Gersten) complex of sheaves coming from K‑theory whose last term is the sheaf of codimension‑p cycles.

More precisely (assume X is regular, noetherian; e.g. a smooth variety), set Z^p_X := ⊕{x∈X^{(p)}} i{x,*} Z, the sheaf whose sections over U are the free abelian group on codimension‑p points of U (so Γ(X, Z^p_X)=Z^p(X...

Short answer.

• For ordinary tangential structures (X, ζ) the cobordism hypothesis does not give anything “new”: one has a canonical identification |Bordn(X,ζ)| ≃ Ω∞−n Mζ, i.e. the geometric realization of the n-dimensional (∞,n)-bordism category is the (∞−n)-fold loop space of the Thom spectrum Mζ. This is the same identification one gets from Pontryagin–Thom/“GMTW” results; it is not a different universal characterization.

• For manifolds with Baas–Sullivan singularities, Lurie’s universa...

Short answer: Both conjectures are true (in the finite-dimensional setting you state) for the Petz-type Rényi conditional mutual information you defined, for the whole range α ∈ [0,2]. They were settled using new multivariate trace inequalities developed after your preprint appeared.

More detail and references

• Monotonicity under local operations on A (Conjecture 1). Sutter, Berta, and Tomamichel proved a multivariate Golden–Thompson/Araki–Lieb–Thirring type inequality via complex interpolati...

Short answer: you can’t make a diagnostic pragma inside code that’s compiled into a Clang module suppress a warning that’s emitted later when that code is instantiated in another TU. This is a known limitation of Clang modules: the diagnostic state from #pragma clang diagnostic … is not preserved/rehonored across module boundaries for things like template instantiations that happen in the importing TU. That’s why your pragma works with textual inclusion but not after the header is compiled into ...

No.

A convenient way to see this is via the subadditive (almost additive) thermodynamic formalism. There are natural quasi-Bernoulli equilibrium measures that come from almost additive potentials, and in general these cannot be realised as Gibbs measures for any single continuous (additive) potential.

Outline.

1. Almost additive potentials and their equilibrium states.
   On the full shift X = {0,1}^N, let Φ = {φn}n≥1 be an almost additive potential, i.e. there exists K ≥ 0 such that for all x...

Short answer: No. Cohomology with coefficients in a ZG–module (i.e. Borel-equivariant cohomology with local coefficients pulled back from BG) is not a special case of “genuine,” RO(G)-graded equivariant cohomology. It is a different flavor of equivariant theory. It is representable, but not by an RO(G)-graded Eilenberg–MacLane G-spectrum of the sort used in May’s book; rather it is represented either in the naive (integer-graded) equivariant stable category or, more naturally, in the parametrize...

Short answer: No. For any choice of the inertia M and the fixed rigid thruster T, a single nonreversible input f(t) ≥ 0 cannot make the endpoint map locally surjective at 0, hence (by your own time/scale symmetries) global controllability between arbitrary angular velocities is impossible.

A simple quantitative obstruction

It is convenient to work with the angular momentum L = {M,Ω} ∈ so(4), for which Euler’s equation becomes the control-affine system L' = −[K L, L] + f(t) T, L(0) = 0, wher...

Yes. The analogue of Uhlenbeck’s result holds if you restrict to Kähler metrics, even within a fixed Kähler class: for a residual (comeagre) set of Kähler metrics in a given class, the scalar Laplacian has simple spectrum.

Why the same argument goes through

• Parametrization of the space: Fix a background Kähler form ω0 in the given Kähler class. Every other Kähler form in the class can be written as ωφ = ω0 + i∂∂̄φ, where φ is a smooth real-valued potential and ωφ > 0. For k large, the set of ...

No. If X × Y has the homotopy type of a finite-dimensional CW complex, then X and Y each do as well.

Here is a clean argument.

• Let Z ≃ X × Y be homotopy equivalent to an m-dimensional CW complex. Then for every local coefficient system L on Z we have H^i(Z; L) = 0 for i > m, since cohomology with local coefficients is computed by a length-m cochain complex.

• Fix any local coefficient system A on X. Consider the pullback local system p_X^*A on X × Y (where p_X: X × Y → X is the projection)....

Short answer: Yes in principle, but in practice no one seems to have published a clean, usable value of the “sufficiently large” threshold for the best-known exponents for π; extracting it from the proofs is possible but leads to astronomically large numbers, making it of no practical use. The only widely available completely explicit inequality that holds for all q is still Mahler’s old result (with a very poor exponent).

Details and references:

• Effectivity of the best results
  • Hata’s 19...

Yes – there is a clean answer that simultaneously addresses your three questions.

Set-up and a standard description of Aut(Gn) For any group G, write Zi = Z(G) for its center. A standard and very useful fact is that Aut(Gn) is generated by:

• permutations of the n coordinates,
• componentwise automorphisms (α1,…,αn) with each αi ∈ Aut(G), and
• “elementary transvections” tij(φ) (i ≠ j) defined by tij(φ)(g1,…,gn) = (g1,…,gi φ(gj),…,gn), where φ: G → Z(G) is a homomorphism. This is an automor...

Short answer.

• For a Lefschetz Landau–Ginzburg model (Y, ω, W) with only Morse critical points (the setting in which the Fukaya–Seidel category FS(Y, W) is defined), the “closed string” A-model state space is the relative cohomology H*(Y, Y−∞; C), and there are natural open–closed and closed–open maps OC: HH∗(FS(Y, W)) → H∗+n(Y, Y−∞), CO: H∗(Y, Y−∞) → HH∗(FS(Y, W)), where n = dimC Y.
• The open–closed map is an isomorphism under the standard generation hypotheses (vanishing cycles generat...

Short answer for the simplicial case: yes.

Let X• be a simplicial object in finite CW complexes. Then there is a simplicial object K• in finite simplicial sets and a zig–zag of levelwise weak equivalences of simplicial objects |K•| ~→ |Sing X•| ~→ X• so that, in particular, each |Kn| has the homotopy type of Xn and the maps respect the simplicial structure (strictly) in the middle object.

Here is a clean way to see this.

1. Replace Top by sSet. The Quillen adjunction |–| ⊣ Sing between si...

Yes. In fact you don’t need the hypothesis “every atlas is a scheme”; the existence of one étale atlas by a scheme already forces the coarse moduli space to be a scheme.

Here is a convenient way to see it.

• Let π: X → Y be the coarse moduli space of a separated DM stack X with finite inertia (so Y is an algebraic space).
• Pick an étale atlas p: U → X with U a scheme (such an atlas always exists for a DM stack).
• By the basic properties of coarse moduli spaces (Keel–Mori), the composite f: U...

Short answer

• (a) is classical.
• (c)–(d) are standard “scaled” variants of (a) and can be found in tables.
• (e)–(h) follow from Jonquière’s inversion formula for polylogarithms together with the Mellin–Beta evaluation used in (a) and the classical integrals ∫ ln^m(x)/(1+x^2). They are not usually tabulated in exactly your binomial/Euler-number form, but they are direct consequences of well-known general identities.
• (i)–(j) are depth‑1 alternating Euler sums with odd denominators (level 4). ...

Short answer: Flex is far smaller than the class of primitive recursive (PR) functions. In particular, every Flex-function has a fixed, finite “exponential height,” so a standard PR function such as tetration n ↦ 2↑↑n (defined by g(0)=1, g(n+1)=2^{g(n)}) is not in Flex. Exponentiation is also not superfluous: without it, Flex collapses to at most polynomial growth and cannot define, e.g., n!.

What “can be made from +, −, ×, ÷, ^, ⌊·⌋” precisely means

• Interpret Flex as the set of functions f: ...

Short answer: there is no known method to determine which digit occurs most often in the decimal expansion of 4↑↑↑↑4 without essentially knowing (i.e., computing) all of its digits.

Some context and what we can say:

• 4↑↑↑↑4 is a power of 2. In fact, for any number of arrows ≥2, starting with base 4, the result is an integer power of 2. Thus 4↑↑↑↑4 = 2^M for some astronomically large integer M.

• Moreover, one can show M is divisible by 4 (indeed, from the tetration stage onward the exponents...

Short answer As stated, the identity need not hold in general: the operator S need not be of trace class. A counterexample and a correction are given by Lyons in the erratum to his paper. The identity does hold (and S is trace class) under additional structure on the marginal metric spaces, namely when both metrics admit an isometric embedding into a Hilbert space (i.e., are of negative type so that the metric equals a squared Hilbert distance). Under those assumptions one can factor S and compu...

Short answer: You can’t safely do this any more. NDSolveStateData is now an opaque object that contains internal handles (e.g. InternalBag) whose identity has to be preserved. Any attempt to “crack it open” and rebuild it (including via the loopback-link Hold/ReleaseHold trick) will create new internal objects that no longer match the other references inside the state. That breaks NDSolve’s invariants and can crash the kernel. The right way to change things like the finite-difference order is ...

Write the superpotential so that the only combination of X and Φ2 that talks to Φ1 is manifest: W = 1/2 Φ1^2 (h1 X + h2 Φ2) + f X.

Make a linear, unitary change of variables in the X–Φ2 plane:

• S ≡ (h1 X + h2 Φ2)/h,
• T ≡ (−h2 X + h1 Φ2)/h, with h ≡ √(|h1|^2 + |h2|^2).

In this basis, W = 1/2 h S Φ1^2 + f (h1 S − h2 T)/h.

Two immediate consequences follow:

• SUSY breaking (tree level): Along the classical pseudomoduli space Φ1 = 0, one has F_S = f h1/h and F_T = f h2/h, so V = |f|^2 and SUSY...

Short answer: The coefficient 4 β^3 β,φ in your φ–equation is an artifact of treating the matter/radiation piece as S_m ∼ ∫√−g β^4(φ) ρ and varying it as if ρ were an external constant. That is not correct in GR. The matter action depends on the metric seen by matter, ĝμν = β^2(φ) gμν, and when you vary the action properly you find that φ couples to the trace of the matter stress–energy, not to ρ with a factor 4. In particular, radiation does not source φ (its stress–energy is traceless), while ...

Short answer

• In current glibc (ptmalloc2), free() does not call the public malloc_trim(). It never has in the multi-arena ptmalloc2 code used by glibc.
• What free() does do “automatically” is a per-arena top-of-heap trim via an internal helper (systrim/sys_trim/heap_trim) when the top chunk becomes large enough. That is the behavior the old comments and the man page sentence allude to.
• The public malloc_trim() is only run when your program calls it (or in glibc’s tests). It is not invoked i...

Yes. The profunctors package gives you exactly the “continuation/CPS for arrows” you’re looking for: the Cayley construction for profunctors.

• Type: newtype Cayley p a b = Cayley (forall r. p b r -> p a r)
• Module: Data.Profunctor.Cayley (in the profunctors package)

Intuition

• Arrows are monoids in the monoidal category of profunctors (Paterson: “Arrows are Monads?” / “Arrows as Monoids”).
• The Cayley construction is the same trick that turns any monoid M into a submonoid of its endomorphi...

Short answer: No. There is no universal, state‑independent “constant collision term” that should be added to the action of a classical gas. Collisions are already fully accounted for by the interparticle potential in the standard Lagrangian L = T − U, and any additive constant to L has no physical effect on the dynamics or thermodynamics.

Why a constant “collision action” does not arise

• Proper Lagrangian for a gas: For N classical particles with pair potential V(rij), the Lagrangian is L =...

Short answer: Your statements A and B are false. You can construct sequences where the crossing happens in fewer than n steps but the log-return is negative, and also sequences where the crossing takes at least n steps but the log-return is positive. So there’s no such deterministic relation in general.

First, a quick clarification: you wrote winnum = log(tstarstar) − log(tstar). If tstarstar > tstar (the game ends after it starts), that’s always positive, contradicting your note that winnum co...

Short answer

Not in the purely syntactic way you are doing it. If you insist on (i) local, constant‑time “incremental” copying that can pull a λ out across the duplicator and (ii) the usual lexical notion of scope (a binder’s variable never occurs outside its body), then the two requirements conflict. To keep closed scopes, crossing a λ must bring along its whole body at once; to keep copying incremental, you let pieces of the body remain behind and then some bound occurrences must temporarily ...

Short answer: No. Replacing the max by “sampling with probability proportional to Q” does not give a non-expansive map, hence the corresponding Bellman operator is not a contraction in general (except in very special parameter ranges such as very small discount).

Why your inequality does not apply

• Your inequality |Eρ[f] − Eρ[g]| ≤ Eρ[|f − g|] holds only when the sampling distribution ρ is the same for both expectations.
• In your proposed update, the sampling distribution itself depends on th...

Short answer: Yes—adding “bisimilarity implies Leibniz equality” for (non-indexed, strictly positive) coinductive datatypes such as streams is consistent with Coq’s logic. It is not provable in intensional Coq in general, but it holds in standard set-theoretic models and therefore is consistent relative to Coq.

Why it is reasonable

• For a coinductive datatype given as the greatest fixed point νF of a functor F (e.g., streams for F(X) = A × X), the canonical notion of bisimulation is the larges...

Short answer

• A uniform, easy-to-check sufficient condition is: α must be part of a definable retraction. That is, there is a pure term β0 with β0 ∘ α ≡βη id. Then for every g and f = M(α) g we can take β = β0 and get g = M(β) f. Your two positive examples are exactly of this kind: one is an isomorphism (argument swap) and the other is a section with a definable left-inverse (a ↦ eval-at-a, left-inverse y ↦ y I).
• In the absence of such a retraction, there is in general no way to “invert” α by...

Come and Go Mad by Fredric Brown (first published in Thrilling Wonder Stories, 1949)

• It’s set in a mental hospital, where the narrator/patient tells how he learned a simple occult technique for exchanging minds.
• The “ritual” requires a burnt offering (even a cigarette will do) and saying the exact words.
• While recounting the story, he asks the attendant/doctor for a cigarette; by the end of the tale he has already performed the swap, now speaking from the corridor as the staff member.
• H...

Short version: There doesn’t seem to be a published SF/F source for this passage. All signs point to it being an original piece of creative writing (likely from a class, zine, or personal journal) rather than something from a commercially published book, story, or magazine.

Why:

• No textual matches: Key, distinctive strings such as “land of Narda” and “pockmarked by the cosmic acne of space” produce no hits in major indexed corpora (books, magazines, newspapers) or common quotation databases. ...

Short answer Your problem is exactly the weighted correlation clustering objective on a (not necessarily complete) graph: you want to find a partition of the vertices so as to maximize agreements, where an agreement is “put endpoints together” on a negative-weight edge and “put endpoints apart” on a positive-weight edge. Formally, if we let A be the total weight of agreements A = sum_{e=(u,v), w(e)>0} w(e)·[u and v in different parts] + sum_{e=(u,v), w(e)<0} |w(e)|·[u and v in the same part], th...

The Time Mercenaries by Philip E. High (first published 1968)

Why it matches:

• Protagonist: A 20th‑century military airman (from WWII) is brought forward into the future and “upgraded” to serve as a commando. High’s text specifically mentions physiological enhancements, including greatly increased strength and a blood/oxygenation modification that lets him hold his breath for extended periods—exactly the ability later used underwater.
• Mission: He’s placed with a special-ops team sent by spac...

Counterexample (valid for every n ≥ 4)

Fix n ≥ 4 and define a probability space with n+1 atoms ω0, ω1, …, ωn having probabilities

• P(ω0) = s := 1/(n−1)^2,
• P(ωi) = r := (n−2)/(n−1)^2 for i = 1, 2, …, n. Note that s + n r = 1.

Define Bernoulli variables X1, …, Xn by

• Xi(ω0) = 1 for every i,
• Xi(ωi) = 1 and Xj(ωi) = 0 for j ≠ i.

Equivalently, the random vector X = (X1,…,Xn) takes the value 1^n with probability s and the value ei (the i-th standard basis vector) with probability r for i=1,…,...

Short answer

• There is no user-selectable “encryption type” for Windows Hello biometrics, and it is not comparable to BitLocker’s XTS-AES options.
• Windows Hello does not store a fingerprint image or a hash (so SHA-1 vs SHA-2 is not applicable). It stores a biometric template that must be compared with some tolerance, which cannot be done with a one‑way hash.
• The template is stored only on the device and is encrypted/protected by the platform. On modern hardware it is often stored and matche...

Short answer Yes—if by “optimal” you mean locally (asymptotically) most powerful for contiguous alternatives within the natural semiparametric model—then the Wilcoxon–Mann–Whitney (WMW) test is optimal for proportional-odds (PO) alternatives. In the two-sample problem with a single binary covariate and a continuous outcome (no ties), the WMW statistic is exactly the efficient score test for β=0 in the semiparametric proportional-odds cumulative logit model. Consequently, when PO holds, the WMW t...

Short answer

1. For an i.i.d. scale model X1,…,Xn with distribution F(x/σ), the “ratio” characterization is not only sufficient but (modulo null sets and under standard regularity) also essentially necessary: an exact ancillary is, almost surely, a function of a maximal invariant under the scale group. Vectors such as (X1/Xn,…,Xn−1/Xn), or any equivalent maximal invariant (e.g. X/∥X∥ for a chosen norm, ratios of order statistics), generate the ancillary σ-field. Thus, S is ancillary if and only...

Let X = D M, where M is N×k with i.i.d. N(0,1) entries and D = diag(λ1,…,λN) is fixed. Then X has a matrix normal distribution X ∼ MNN×k(0, Σ, Ik) with row covariance Σ = D^2 and column covariance Ik: vec(X) ∼ N(0, Ik ⊗ Σ).

Write the QR (more conveniently, polar) decomposition of X as X = Q R, with Q ∈ VNk (the Stiefel manifold of N×k matrices with orthonormal columns) and R upper triangular with positive diagonal. Equivalently, let T = R R′ ∈ Sym+ k and write X = Q T1/2.

Distribution of Q (ge...

Short answer

1. It is exactly 0. Because E[Xi^d]=0, we have E[Sn^d]=(1/√n)∑i E[Xi^d]=0. You also assumed Y^d has mean 0. Hence |E[Sn^d]−E[Y^d]|=0 for all n,d.

2. As written, there is no general useful upper bound. The two conditional expectations are random variables defined on different probability spaces; if you interpret them as conditional expectation functions mS(x)=E[Sn(i)|Sn(1:i−1)=x] and mY(x)=E[Y(i)|Y(1:i−1)=x], then mY(x) is linear in x (Gaussian regression), but mS(x) can be highly ...

Это не баг и не про лимит 3 rps. Для поисковых методов (в первую очередь users.search, но также groups.search и т.п.) у ВК есть отдельные антискрейпинг‑ограничения. При их превышении платформа «тихо» режет поиск: метод начинает возвращать пустую выдачу (count: 0, items: []), и то же самое вы видите в веб‑поиске из того же аккаунта. Ошибки flood control (code 6) при этом может не быть.

Что известно про эти ограничения

• Они завязаны на аккаунт (access token) и часто на IP. execute не помогает: в...

You’re almost certainly thinking of VDM — the Vienna Development Method — and, more specifically, the VDM specification notation used inside IBM in that period (often called Meta‑IV at the time, later standardized as VDM‑SL).

Why this matches your recollection

• Origin in IBM: VDM grew out of work at IBM’s Vienna Laboratory in the 1970s (initially “Vienna Definition Language” for formally defining PL/I), and by the 1980s IBM groups were using the associated specification notation for software s...

Short answer There isn’t a single, meaningful “qubit fidelity” number for a D‑Wave annealer, and randomized benchmarking (a gate‑model tool) does not apply. D‑Wave devices are analog quantum annealers; the relevant question is how faithfully the hardware realizes the programmed Ising Hamiltonian over the whole chip. That faithfulness is characterized by control errors and noise rather than gate fidelities, and D‑Wave does not publish per‑qubit calibration/fidelity tables. They do, however, routi...

Short answer

• Proliferation is the net rate at which tumor cells are created by division (mitosis) minus the rate they are lost (death). It increases total cell number and, under normal biology, total tumor mass.
• Diffusion (in this context) is a mathematical way to describe random cell motility that causes cells to spread from regions of high to low density. It redistributes cells in space but does not create or destroy them.

Details and how they relate to “tumor growth”

• Proliferation
  • ...

Short answer

• Metallic Ni or Co do not appreciably “react” (form bulk compounds) with Rb at room temperature in UHV; Rb atoms simply chemisorb on clean Ni/Co surfaces and can desorb upon heating. Bulk alloying or intermetallic formation requires much higher temperatures.
• What Rb does react with very readily is the surface oxide that is essential for a glass-to-metal (Kovar–glass) seal. Rb reduces Fe/Ni/Co oxides to the metals, forming Rb2O. That chemical reduction can undercut or embrittle th...

Short answer: Not with the known proofs, and very likely not. All existing BQP–hardness proofs for the quantum linear systems problem (QLSP/HHL) fundamentally use negative (or complex) matrix elements to encode quantum phases. If you forbid those and insist that every entry of A and of b is in {0,1}, the standard reduction to a generic quantum computation breaks down. Moreover, under natural extra promises that keep the problem well-posed, the resulting linear-systems task lies in much weaker cl...

Short answer: it’s not CPU-dependent. What you’re seeing is a JIT bug in certain .NET/runtime versions, not a fundamental precision/accuracy issue with integer vectorization or division.

What’s going wrong

• Your alpha is a ushort (0x8000 = 32768). When it’s widened to 32-bit and then broadcast into a Vector128<int>/Vector128<uint>, the JIT sometimes sign-extends the 16-bit value instead of zero-extending it. So 0x8000 becomes 0xFFFF8000 (-32768) instead of 0x00008000 (32768).
• Once that “nega...

Short answer: no. Both the word and the modern separation of “isomorphism” vs “homomorphism” are older than the examples you list and were in circulation already in the 1890s, first in the algebra-of-logic/universal-algebra literature and, on the group-theory side, in German algebra.

What changed

• In Jordan (1870) and much of the 19th‑century group literature, “isomorphism” covered what we would now describe as a structure‑preserving correspondence that need not be bijective; a bijective one w...

The Last of the Really Great Whangdoodles by Julie Andrews Edwards (1974).

Why it matches your memories:

• Three children travel with a professor into a whimsical but perilous fantasy land. The chief adversary they keep running up against there is the Prock (the Whangdoodle’s gatekeeper), which can come across as an “evil ruler/prince” figure in memory.
• There’s a set piece where the kids ride a big inchworm-like creature with saddles; because of the way it moves (looping up in the middle), th...

Short answer

• In general, (X ∩ Y)/F ⊆ X/F ∩ Y/F always holds, but the reverse inclusion need not hold.
• For most interesting pairs of classes (in particular X = NP and Y = coNP with F = poly), it is still open whether equality holds. In particular, it remains open whether NP/poly ∩ coNP/poly = (NP ∩ coNP)/poly.
• There are no known unconditional separations for natural classes like NP and coNP; on the other hand, relativized separations are known (i.e., there are oracles relative to which th...

Short answer: No. At present there are no results in the Langlands program (or comparable long‑standing problems outside logic/topos theory) whose proofs essentially use Caramello’s “toposes as bridges” methodology. Lafforgue’s remark is aspirational: the Langlands correspondence “looks like” a Morita‑type equivalence between two descriptions of the same structure, and so it is tempting to hope it could someday be expressed/understood via classifying toposes. But to date there is no such realiza...

Short answer: This is not known. With the operations you allow (starting from N or Q and using only finitely many applications of +, −, ·, ÷, and exponentiation), it is an open problem whether one can obtain π, and it is also open whether π is a root of any nontrivial equation built from these same operations.

More context

• The set of numbers you can produce this way is countable (there are only countably many finite expressions over a countable alphabet). That by itself doesn’t tell you whet...

Short answer: This is open in general. It would indeed follow from the (strictly stronger) statement that a graph on n vertices is determined by the set (not multiset) of isomorphism types of its edge-deleted subgraphs, but that stronger statement is false, so a different approach would be needed.

A few relevant points.

• Automorphisms preserve rank. The poset is graded by the number of edges, with unique minimum (empty graph) and maximum (complete graph), so any poset automorphism fixes those...

Short answer: it’s not a new propagator or a magical “AI” breakthrough; it’s a modern, commercial SSA service whose main differentiator is data fusion, provenance, and operator data‑sharing. For many operators that had been working off public TLEs and emailed CDMs, that is indeed “better.” For operators already using premium SSA providers (e.g., COMSPOC/Ansys, LeoLabs, ExoAnalytic, Kayhan Space, Numerica, EU SST, SDA), Crow’s Nest is more of an alternative than a step change in the physics or ra...

Short answer

• The Büchi–Elgot–Trakhtenbrot equivalence (MSO on finite words = WS1S on N with finite-set quantification) is robust under adding any relations that are already MSO-definable in the basic signature; in particular, all “regular” and modular-counting predicates preserve both equivalence and decidability.
• A much wider class also works: adding Presburger-style cardinality constraints on monadic variables (e.g., |X|<|Y|, |X|≡r mod m, and more general linear constraints on sizes of MSO...

Short answer. Yes: for n ≥ 3 the mapping class group of CP^n (in each of the topological, PL and smooth categories) is Z/2, generated by complex conjugation. Equivalently:

• π0(Homeo(CP^n)) ≅ π0(PL(CP^n)) ≅ π0(Diff(CP^n)) ≅ Z/2.
• The nontrivial class is represented by complex conjugation [z0:…:zn] ↦ [z̄0:…:z̄n].
• Orientation: conjugation reverses orientation when n is odd and preserves it when n is even (since it multiplies the generator x ∈ H^2 by −1, hence x^n by (−1)^n). Thus:
  • π0(Diff^+...

Short answer: as stated, Ω is not a linear order, and it is far too complicated to admit a nice “universal” combinatorial description. In fact, it already contains, as a suborder, the very complicated poset (P(ω), ⊆*) of subsets of the natural numbers ordered by inclusion modulo finite. So any hope for a simple colimit description in the spirit of your Θn- or Φ-examples cannot work without imposing strong regularity restrictions on the functions.

Details:

1. Your relation is just Big-O. For re...

Let V = C^m and G = span{G1,…,Gn} ⊆ End(V). For a subspace I ⊆ V and a linear subspace S ⊆ G, “I is invariant under every S ∈ S” means S(I) ⊆ I for all S ∈ S.

Key observation (reduction to a rank condition)

• Fix k with 1 ≤ k ≤ m−1 and a k-dimensional subspace I ⊆ V.
• Let i_I: I → V be the inclusion and q_I: V → V/I the quotient map.
• Define the linear map Φ_I: G → Hom(I, V/I) by Φ_I(A) = q_I ∘ A ∘ i_I.

Then:

• I is invariant under A ∈ G iff Φ_I(A) = 0.
• The largest subspace of G whose ever...

Think of your matrix as the evaluation matrix of the bivariate polynomial space Π = span{x^b y^a : 0 ≤ a ≤ m−1, 0 ≤ b ≤ m−1}, which has dimension m^2. Your matrix V has one column per point (x_i, y_i) and one row per monomial in Π, so V ∈ R^{m^2 × (m^2+1)}.

1. Dimension of the kernel

• For “generic” points (x_i, y_i) (i.e., away from an algebraic exceptional set), rank(V) = m^2, hence dim ker(V) = (m^2+1) − m^2 = 1.
• Rank(V) < m^2 (so ker dimension > 1) if and only if every m^2 × m^2 minor van...

Two very different behaviors show up depending on whether you invert 6 or not. Over Z[1/6] (and hence over Z(p) for p>3) the situation is completely rigid and all vector bundles split as sums of powers of the Hodge line bundle ω. At p=3 (and similarly at p=2) there are many nontrivial extensions and indeed indecomposables of all ranks.

Here are precise statements and proofs/sketches that address your two strategies.

1. Over Z[1/6] (in particular over Z(p) for p>3): every vector bundle splits a...

Let X ∼ tα be a standard (mean 0, scale 1) Student-t with ν=α degrees of freedom and let Y=|X|. Then Y has the half‑t (a.k.a. folded t) density you wrote: g(y) = 2 Γ((α+1)/2) / (√(απ) Γ(α/2)) (1 + y^2/α)^{-(α+1)/2}, y ≥ 0.

The sum Sn = ∑i=1..n Yi is the n-fold convolution of this half‑t density.

1. Closed form? In general, no

• The family of half‑t distributions is not closed under convolution. For n ≥ 2 there is no simple closed-form pdf/cdf for Sn in terms of elementary functions (this is tr...

Short answer There is no “Shadu‑kām” in the classical Persian mythological corpus (e.g., Ferdowsi’s Shahnameh, Asadi Ṭusi’s Garshāsp‑nāmeh, Nezami’s Khamsa). In Iranian tradition the fairies’ realm is Paristān (and, more generally, Jinnistān for jinn) beyond Mount Qāf; a specific province called Shād‑o‑kām does not occur in the standard primary sources. The name Shād‑o‑kām (Persian shād o kām = “joy and desire/pleasure,” hence “bliss”) is an ordinary Persian phrase, and when it appears as a plac...