diff b34550f..b1f9bc4 uniform_trace_reconstruction_v2.txt
diff --git a/uniform_trace_reconstruction_v2.txt b/uniform_trace_reconstruction_v2.txt
index 584cde7..5744693 100644
--- a/uniform_trace_reconstruction_v2.txt
+++ b/uniform_trace_reconstruction_v2.txt
@@ -10,7 +10,7 @@
-\usetikzlibrary{arrows.meta,positioning,calc,decorations.pathreplacing}
+usetikzlibrary{arrows.meta,positioning,calc,decorations.pathreplacing}
@@ -30,7 +30,7 @@
-Fix a binary string $x=x_1... x_n\in\{0,1\}^n$. A deletion trace
+Fix a binary string $x=x_1... x_nin{0,1}^n$. A deletion trace
is the random string produced by the following experiment: walk through
the positions $1,2,...,n$; keep position $i$ with probability $p$
and drop it with probability $q=1-p$, independently at each $i$; the
@@ -54,41 +54,41 @@ a family of carefully chosen nonlocal statistics (products of
bits at ranks that may be far apart) together with a way to test, bit
by bit, whether a partial guess is consistent with the data.
-Let $n\ge 2$ and let $p=a/b\in(0,1]$ be a rational in lowest terms.
+Let $nge 2$ and let $p=a/bin(0,1]$ be a rational in lowest terms.
Write $L$ for the total number of bits in $a$ and $b$ when written in
binary, so $L$ is the length of the honest input to an algorithm.
Write $q=1-p$. All logarithms are natural unless a base is shown.
Define the budget proxy
- \mathsf X(n,p) &=
+ mathsf X(n,p)sfX(n,p) &=
- 1+(\log n)/(1+\log(1/q)), & q>0,\\[6pt]
+ 1+(log n)/(1+log(1/q)), & q>0,[6pt]
1, & q=0,
- \\[4pt]
- B(n,p) &= p^{-1}\,(\log n)\,\mathsf X(n,p)^2\,
- (1+\log \mathsf X(n,p))^6 .
+ [4pt]
+ B(n,p) &= p^{-1},(log n),mathsf X(n,p)^2sfX(n,p)^2,
+ (1+log mathsf X(n,p))^6sfX(n,p))^6 .
[Uniform trace reconstruction]
-There are absolute constants $C,c\ge 1$ and one randomized multitape
-Turing machine $\mathcal A$ with the following property. Take any
-$n\ge 2$, any rational $p\in(0,1]$, and any $x\in\{0,1\}^n$. Given
+There are absolute constants $C,cge 1$ and one randomized multitape
+Turing machine $mathcal A$calA$ with the following property. Take any
+$nge 2$, any rational $pin(0,1]$, and any $xin{0,1}^n$. Given
$n$ and $p$ in binary and access to independent deletion traces of $x$
with retention probability $p$, the machine
-[label=(\roman*)]
-\item reads at most $M_C(n,p)=\bigl\lceil \exp(C\,B(n,p))\bigr\rceil$
+[label=(roman*)]
+item reads at most $M_C(n,p)=bigllceil exp(C,B(n,p))bigrrceil$
traces,
-\item outputs a string in $\{0,1\}^n$, and
-\item returns $x$ with probability at least $2/3$.
+item outputs a string in ${0,1}^n$, and
+item returns $x$ with probability at least $2/3$.
Moreover, for every internal sequence of random bits and
every stream of well-formed traces of strings of length at most
$n$, the machine halts within
-\[
+[
(n+L+M_C(n,p))^c
-\]
+]
bit operations. This count already includes reading every trace bit
and every delimiter, all preprocessing, all finite-precision arithmetic,
and writing the output. The only thing not counted is the external
@@ -96,16 +96,16 @@ generation of the traces.
For a fixed $p$, plugging $q>0$ constant into (eq:budget) gives
-$\mathsf X(n,p)=O(\log n)$ and hence
-$B(n,p)=O_p((\log n)^3 (1+\log\log n)^6)$. Since
-$M_C=\exp(CB)$, the number of traces is
-$\exp(O_p((\log n)^3(1+\log\log n)^6))$, which is
+$mathsf X(n,p)=O(sfX(n,p)=O(log n)$ and hence
+$B(n,p)=O_p((log n)^3 (1+loglog n)^6)$. Since
+$M_C=exp(CB)$, the number of traces is
+$exp(O_p((log n)^3(1+loglog n)^6))$, which is
quasipolynomial in $n$. The running time bound
$(n+L+M_C)^c$ is then also quasipolynomial. So both resources are
-quasipolynomial for fixed $p$. When instead $q\le n^{-\kappa}$ for a
-fixed $\kappa>0$, one has $1+\log(1/q)\ge \kappa\log n$, so
-$\mathsf X(n,p)\le 1+1/\kappa$ is bounded and $p^{-1}$ is bounded by a
-constant depending only on $\kappa$; then $B(n,p)=O_\kappa(\log n)$
+quasipolynomial for fixed $p$. When instead $qle n^{-kappa}$ for a
+fixed $kappa>0$, one has $1+log(1/q)ge kappalog n$, so
+$mathsf X(n,p)sfX(n,p)le 1+1/kappa$ is bounded and $p^{-1}$ is bounded by a
+constant depending only on $kappa$; then $B(n,p)=O_kappa(log n)$
and everything is polynomial in $n$.
The proof is one long chain. The decoder is tested against a family of
@@ -132,13 +132,13 @@ many input positions.
doi.org/10.1109/18.904499
For a fixed deletion probability, Holenstein, Mitzenmacher, Panigrahy,
-and Wieder proved a worst-case sample bound of $\exp(\widetilde
-O(\sqrt n))$. In independent STOC 2017 papers, De, O'Donnell, and
+and Wieder proved a worst-case sample bound of $exp(widetilde
+O(sqrt n))$. In independent STOC 2017 papers, De, O'Donnell, and
Servedio, and separately Nazarov and Peres, proved
-$\exp(O(n^{1/3}))$ bounds together with matching limitations for
+$exp(O(n^{1/3}))$ bounds together with matching limitations for
methods that use only trace means. Their complex-analytic approach ties
deletion statistics to bounded-coefficient polynomials, with analytic
-inequalities of Borwein and Erd\'elyi in the background. That
+inequalities of Borwein and Erd'elyi in the background. That
mean-based lower bound restricts a method, not arbitrary trace
algorithms.
@@ -148,11 +148,11 @@ doi.org/10.1145/3055399.3055494
doi.org/10.1512/iumj.1997.46.1435
Chase improved the general sample upper bound to
-$\exp(O(n^{1/5}\log^5 n))$. On the lower-bound side, Holden and Lyons
+$exp(O(n^{1/5}log^5 n))$. On the lower-bound side, Holden and Lyons
proved an almost-$n^{5/4}$ bound and Chase strengthened it to
-$Ω(n^{3/2}/\log^7 n)$ for fixed deletion probability. Burudgunte,
+$Ω(n^{3/2}/log^7 n)$ for fixed deletion probability. Burudgunte,
Valiant, and Wang then proved the sample bound
-$\exp(p^{-7/3}(\log_2 n)^{c_0})$ for an absolute $c_0$, using
+$exp(p^{-7/3}(log_2 n)^{c_0})$ for an absolute $c_0$, using
nonlocal low-order statistics and a multiscale propagation argument;
their bound is quasipolynomial when $p^{-1}$ is polylogarithmic in $n$.
The uniform reconstruction algorithm developed later builds on this
@@ -212,26 +212,26 @@ semidefinite (positivity) condition, and solves for a completion
of a partial guess. That relaxation is set up first, along with the
separation theorem that makes it useful.
-Throughout, $[n]=\{1,...,n\}$ and $X_1,...,X_n$ are formal Boolean
+Throughout, $[n]={1,...,n}$ and $X_1,...,X_n$ are formal Boolean
variables standing for the unknown bits. Products are always reduced by
$X_i^2=X_i$ (a bit times itself is itself), so every polynomial has
degree at most the number of distinct variables it mentions.
-Fix a nonempty set of ranks $I=\{\ell_1<...<\ell_k\}⊆[n]$. The
+Fix a nonempty set of ranks $I={ell_1<...<ell_k}⊆[n]$. The
corresponding trace observable is the product of the bits that sit
-at those ranks in a single trace; if the trace is shorter than $\ell_k$,
+at those ranks in a single trace; if the trace is shorter than $ell_k$,
the observable is $0$. Let $T_I(X)$ be its expectation over the
deletion channel, viewed as a polynomial in the input bits:
- T_I(X)=\sum_{i_1<...<i_k} c_{I;\,i_1,...,i_k}\,X_{i_1}... X_{i_k}.
+ T_I(X)=sum_{i_1<...<i_k} c_{I;,i_1,...,i_k},X_{i_1}... X_{i_k}.
-The coefficient $c_{I;\,i_1,...,i_k}$ is the probability that the
+The coefficient $c_{I;,i_1,...,i_k}$ is the probability that the
input positions $i_1,...,i_k$ land at trace ranks
-$\ell_1,...,\ell_k$. As the input tuple varies these events are
+$ell_1,...,ell_k$. As the input tuple varies these events are
disjoint, so
- c_{I;\,i_1,...,i_k}\ge 0,\qquad
- \sum_{i_1<...<i_k} c_{I;\,i_1,...,i_k}\le 1.
+ c_{I;,i_1,...,i_k}ge 0,qquad
+ sum_{i_1<...<i_k} c_{I;,i_1,...,i_k}le 1.
Thus $T_I$ has degree at most $k$, nonnegative coefficients, and
coefficient sum at most one; its coefficients are rational when $p$ is
@@ -244,37 +244,37 @@ semidefinite moment and sum-of-squares frameworks of
Lasserre and Parrilo. Only the defining positivity condition and its
elementary consequences are used, all proved below.
-Fix a proposed prefix $u\in\{0,1\}^d$ and a degree bound $D\ge 1$. On the
-remaining positions $U=\{d+1,...,n\}$ introduce sign variables
+Fix a proposed prefix $uin{0,1}^d$ and a degree bound $Dge 1$. On the
+remaining positions $U={d+1,...,n}$ introduce sign variables
$Y_i=2X_i-1$, so that $Y_i^2=1$ and $X_i=(1+Y_i)/2$. A moment
array assigns real numbers $m_J$ to subsets $J⊆ U$ with
-$|J|\le 2D$, with $m_∅=1$, subject to the single
+$|J|le 2D$, with $m_∅=1$, subject to the single
positive-semidefiniteness condition
- \mathcal M_D(m)=(m_{I△ J})_{\substack{I,J⊆ U\\
- |I|,|J|\le D}}\succeq 0,
+ mathcal M_D(m)=(m_calM_D(m)=(m_{I△ J})_{substack{I,J⊆ U
+ |I|,|J|le D}}succeq 0,
where $△$ is symmetric difference. Only subsets that actually
exist are indexed, so the definition still makes sense when $D>|U|$ or
$U$ is empty.
-The array defines a functional $\EE$ on polynomials of degree at most
-$2D$ as follows: substitute $X_i=u_i$ for $i\le d$; expand each remaining
+The array defines a functional $EE$ on polynomials of degree at most
+$2D$ as follows: substitute $X_i=u_i$ for $ile d$; expand each remaining
bit as $(1+Y_i)/2$; reduce $Y_i^2=1$; then set
-$\EE[\prod_{i\in J}Y_i]=m_J$ and extend complex-linearly. Call
-$u\mapsto\EE$ a positive moment functional of order $2D$ fixing
+$EE[prod_{iin J}Y_i]=m_J$ and extend complex-linearly. Call
+$umapstoEE$ a positive moment functional of order $2D$ fixing
$u$. The word ``positive'' asserts (eq:moment-matrix), not the
existence of an actual probability law realising all these moments; any
feasible array, even a fictitious one, is allowed in the test.
[Elementary moment bounds]
For every array obeying (eq:moment-matrix):
-[label=(\alph*)]
-\item $\abs{m_J}\le 1$ for every $|J|\le 2D$;
-\item under $\EE$, the value of any product of at most $2D$ distinct
+[label=(alph*)]
+item $abs{m_J}le 1$ for every $|J|le 2D$;
+item under $EE$, the value of any product of at most $2D$ distinct
input bits has magnitude at most one;
-\item on polynomials of degree at most $D$, the form
- $< P,R>_{\EE}=\EE[P\,\overline R]$ is a positive
+item on polynomials of degree at most $D$, the form
+ $< P,R>_{EE}=EE[P,overlineoverline R]$R]$ is a positive
semidefinite Hermitian form, linear in its first argument; it
satisfies Cauchy--Schwarz and becomes an inner product after
quotienting by its zero-norm subspace.
@@ -282,68 +282,68 @@ For every array obeying (eq:moment-matrix):
Proof.
-The empty moment equals one. Any nonempty $J$ with $|J|\le 2D$ splits as
-$J=I\sqcup K$ with $|I|,|K|\le D$. In (eq:moment-matrix) the
+The empty moment equals one. Any nonempty $J$ with $|J|le 2D$ splits as
+$J=Isqcup K$ with $|I|,|K|le D$. In (eq:moment-matrix) the
$I,K$ principal submatrix has diagonal entries one and off-diagonal
entry $m_J$; positive semidefiniteness of a real symmetric matrix with
-unit diagonal forces $|m_J|\le 1$. This is (a). For (b), write each
+unit diagonal forces $|m_J|le 1$. This is (a). For (b), write each
unfixed bit as $(1+Y_i)/2$; a product of $t$ unfixed bits becomes
-$2^{-t}\prod(1+Y_i)$, a convex combination of sign monomials of the
+$2^{-t}prod(1+Y_i)$, a convex combination of sign monomials of the
$Y_i$, each of modulus at most one by (a); fixed bits simply remove
factors or send the product to zero. For (c), the Gram-form reading of
the real symmetric positive semidefinite matrix (eq:moment-matrix),
extended to complex coefficient vectors, gives the stated form;
-Cauchy--Schwarz follows by applying positivity to $P+zR$ for $z\in\CC$,
+Cauchy--Schwarz follows by applying positivity to $P+zR$ for $zinCC$,
and it shows every zero-norm vector is orthogonal to every vector, so
the quotient is well defined.
-Now fix the true input $x\in\{0,1\}^n$ and a prefix $u$ of length $d$
+Now fix the true input $xin{0,1}^n$ and a prefix $u$ of length $d$
that is wrong exactly at its last position,
- u_i=x_i\ \ (i<d),\qquad u_d=1-x_d.
+ u_i=x_i (i<d),qquad u_d=1-x_d.
For any positive moment functional fixing $u$, define the discrepancy
- \Delta[P]=\EE[P]-P(x).
+ Delta[P]=EE[P]-P(x).
-Because $u$ agrees with $x$ before $d$, one has $\Delta[X_d]=\pm 1$;
+Because $u$ agrees with $x$ before $d$, one has $Delta[X_d]=pm 1$;
because every monomial using only positions strictly before $d$ agrees
-with $x$ there, $\Delta[P]=0$ whenever $P$ mentions only positions
-$<d$. If $P=\sum_A c_A\prod_{i\in A}X_i$ has degree at most $2D$, then
+with $x$ there, $Delta[P]=0$ whenever $P$ mentions only positions
+$<d$. If $P=sum_A c_Aprod_{iin A}X_i$ has degree at most $2D$, then
point (b) of Lemma~lem:moment-bounds gives the useful bound
- \abs{\Delta[P]}\le 2\sum_A \abs{c_A}.
+ abs{Delta[P]}le 2sum_A abs{c_A}.
For $p<1$ introduce the bookkeeping quantities
- q_*=\max(q,\,1/n),\qquad
- Q=1+\log(1/q_*),\qquad
- Y=1+(\log n)/(Q).
+ q_*=max(q,,1/n),qquad
+ Q=1+log(1/q_*),qquad
+ Y=1+(log n)/(Q).
-When $q\ge 1/n$ these give $Y=\mathsf X(n,p)$ exactly. When $q<1/n$
-they give $1\le \mathsf X(n,p)\le Y<2$. In all cases
+When $qge 1/n$ these give $Y=mathsf X(n,p)$sfX(n,p)$ exactly. When $q<1/n$
+they give $1le mathsf X(n,p)sfX(n,p)le Y<2$. In all cases
- \mathsf X(n,p)\le Y\le 2\,\mathsf X(n,p),\qquad
- QY=Q+\log n\le C\log n
+ mathsf X(n,p)sfX(n,p)le Yle 2,mathsf X(n,p),sfX(n,p),qquad
+ QY=Q+log nle Clog n
-for an absolute constant $C$, since $n\ge 2$.
+for an absolute constant $C$, since $nge 2$.
[Separation from a wrong prefix]
-There exist absolute constants $C_{sep}\ge 1$ and an integer
-$H\ge 1$ with the following property. For $n\ge 2$ and $0<p<1$ define
-\[
- D_*=4^{\lceil \log_2 Y\rceil+H}\le 4^{H+1}Y^2
-\]
-using (eq:aux-parameters). Take any $x\in\{0,1\}^n$, any
-$d\in[n]$, and any positive moment functional of order $2D$ fixing
-$x_1,...,x_{d-1},1-x_d$ with $D\ge D_*$. Then some nonempty
-$I⊆[n]$ with $|I|\le D_*$ satisfies
-\[
- |\EE[T_I]-T_I(x)|
- > \exp(-C_{sep}\,B(n,p)).
-\]
+There exist absolute constants $C_{sep}ge 1$ and an integer
+$Hge 1$ with the following property. For $nge 2$ and $0<p<1$ define
+[
+ D_*=4^{lceil log_2 Yrceil+H}le 4^{H+1}Y^2
+]
+using (eq:aux-parameters). Take any $xin{0,1}^n$, any
+$din[n]$, and any positive moment functional of order $2D$ fixing
+$x_1,...,x_{d-1},1-x_d$ with $Dge D_*$. Then some nonempty
+$I⊆[n]$ with $|I|le D_*$ satisfies
+[
+ |EE[T_I]-T_I(x)|
+ > exp(-C_{sep},B(n,p)).
+]
The theorem holds for every feasible array, including arrays that
@@ -356,39 +356,39 @@ such a statistic. A bounded rational computation then finds the feasible
extension.
The separation proof is analytic, so the objects it uses come first.
-Fix $x,d,\EE$ as above. For $1\le k\le 2D$ an order-$k$ feature is
+Fix $x,d,EE$ as above. For $1le kle 2D$ an order-$k$ feature is
a family of polynomials indexed by the last position $j$,
- f_j=\sum_{i_1<...<i_k=j}
+ f_j=sum_{i_1<...<i_k=j}
X_{i_1}... X_{i_k}
- \prod_{h=1}^{k-1} z_h^{\,i_{h+1}-i_h},
- \qquad j\in[n],
+ prod_{h=1}^{k-1} z_h^{,i_{h+1}-i_h},
+ qquad jin[n],
-where $z_h=\exp(-a_h+\ii\theta_h)$ with $a_h>0$ and
-$\theta_h\in\RR$. For $k=1$ this is just $f_j=X_j$. Each $f_j$ sums over
+where $z_h=exp(-a_h+iitheta_h)$ with $a_h>0$ and
+$theta_hinRR$. For $k=1$ this is just $f_j=X_j$. Each $f_j$ sums over
increasing tuples of input positions with exponential weights on the
gaps. Put
- H_f=\prod_{h=1}^{k-1}(1+\frac1{a_h}).
+ H_f=prod_{h=1}^{k-1}(1+frac1{a_h}).
Summing geometric series over the positive gaps and using
-$(1-e^{-a})^{-1}\le 1+1/a$ shows that $H_f\ge 1$ bounds the sum of the
+$(1-e^{-a})^{-1}le 1+1/a$ shows that $H_fge 1$ bounds the sum of the
absolute coefficients of every $f_j$.
-For $s>0$ and $\omega\in\RR$, the signal of the feature is the
+For $s>0$ and $omegainRR$, the signal of the feature is the
exponentially damped Fourier sum
- \mathcal S_f(s,\omega)
- =\sum_{j=1}^n \Delta[f_j]\,e^{(-s+\ii\omega)(j-d)}.
+ mathcal S_f(s,calS_f(s,omega)
+ =sum_{j=1}^n Delta[f_j],e^{(-s+iiomega)(j-d)}.
Every monomial in $f_j$ uses positions at most $j$, so
-$\Delta[f_j]=0$ for $j<d$; hence, despite the exponent being centred at
-$d$, only $j\ge d$ contribute. For the base case $k=1$, where $f_j=X_j$,
-this reads $\mathcal S_f(s,0)=\sum_{j\ge d}\Delta[X_j]\,e^{-s(j-d)}$, whose
-first term is $\Delta[X_d]=\pm1$; that single term is what makes the
+$Delta[f_j]=0$ for $j<d$; hence, despite the exponent being centred at
+$d$, only $jge d$ contribute. For the base case $k=1$, where $f_j=X_j$,
+this reads $mathcal S_f(s,0)=calS_f(s,0)=sum_{jge d}Delta[X_j],e^{-s(j-d)}$, whose
+first term is $Delta[X_d]=pm1$; that single term is what makes the
starting signal about $1/2$ at decay $s=4$, as made precise in
(eq:initial-signal). The whole multiscale argument has one goal: to preserve a detectable signal
-$\mathcal S_f(s,\omega)$ while shrinking both $s$ and $|\omega|$ to
+$mathcal S_f(s,calS_f(s,omega)$ while shrinking both $s$ and $|omega|$ to
$O(1/n)$, at the cost of raising the order $k$ by at most a factor of
four per step.
@@ -405,36 +405,36 @@ positive moment functional, and the scale step uses Gram matrices of
polynomial vectors.
[One multiscale step]
-There are absolute constants \(0<\beta,\eta_0<1\) and \(C_0,C_1\ge1\)
-with this property. Let \(f\) be an order-\(k\) feature from
-(eq:feature), and let \(\Delta\) come from a moment functional of
-order \(2D\) satisfying the prefix conditions stated earlier.
+There are absolute constants (0<beta,eta_0<1) and (C_0,C_1ge1)
+with this property. Let (f) be an order-(k) feature from
+(eq:feature), and let (Delta) come from a moment functional of
+order (2D) satisfying the prefix conditions stated earlier.
Suppose
-\[
- 0<S\le4,\qquad 0<S'\le\min\{1,\eta_0S\},\qquad 4k\le D,
-\]
-and, for some \(\beta S\le s\le S\) and \(\abs{\omega}\le2.002s\),
+[
+ 0<Sle4,qquad 0<S'lemin{1,eta_0S},qquad 4kle D,
+]
+and, for some (beta Sle sle S) and (abs{omega}le2.002s),
- \abs{\mathcal S_f(s,\omega)}\ge e^{-E},\qquad
- E\ge C_0(1+k\log(2k)+\log H_f+\log\frac1{S'}).
+ abs{mathcal S_f(s,calS_f(s,omega)}ge e^{-E},qquad
+ Ege C_0(1+klog(2k)+log H_f+logfrac1{S'}).
-Then there is an order-\(k'\) feature \(f'\), with \(1\le k'\le4k\),
+Then there is an order-(k') feature (f'), with (1le k'le4k),
satisfying
- \abs{\mathcal S_{f'}(s',\omega')}\ge e^{-7.98E},\qquad
- \beta S'\le s'\le S',\qquad \abs{\omega'}\le2.002s',
+ abs{mathcal S_calS_{f'}(s',omega')}ge e^{-7.98E},qquad
+ beta S'le s'le S',qquad abs{omega'}le2.002s',
and
- H_{f'}\le H_f^4(1+\frac1{\beta S'})^3 .
+ H_{f'}le H_f^4(1+frac1{beta S'})^3 .
-Moreover there are \(1\le m\le4\) lists, each equal to the old parameter
-list \((z_1,...,z_{k-1})\) or its complex conjugate, and at most
-\(m-1\) added parameters, so that: each new inner-gap parameter is a
+Moreover there are (1le mle4) lists, each equal to the old parameter
+list ((z_1,...,z_{k-1})) or its complex conjugate, and at most
+(m-1) added parameters, so that: each new inner-gap parameter is a
nonempty product using at most one entry from each list and at most one
-added parameter; and every added parameter \(e^{-a+\ii\theta}\) obeys
+added parameter; and every added parameter (e^{-a+iitheta}) obeys
- a\ge\beta S',\qquad \abs{\theta}\le C_1S.
+ agebeta S',qquad abs{theta}le C_1S.
@@ -447,209 +447,209 @@ parameters, built later.
Throughout the proof of Proposition~prop:multiscale set
- \sigma=(K\sqrt E)/(S'),\qquad
- W(l)=\exp(-(l^2)/(2\sigma^2)),\qquad b=\beta S',
+ sigma=(Ksqrt E)/(S'),qquad
+ W(l)=exp(-(l^2)/(2sigma^2)),qquad b=beta S',
with $K$ a sufficiently large absolute constant. Constants are chosen in
-the order: $K$ first, then $\beta$ small enough, then $\eta_0$ small
-enough, and finally $C_0$ large enough. The choices of $K$ and $\beta$
+the order: $K$ first, then $beta$ small enough, then $eta_0$ small
+enough, and finally $C_0$ large enough. The choices of $K$ and $beta$
also force the contour estimates later.
Repeated use will be made of this consequence of
(eq:multiscale-cost). For any fixed product of powers of
-$H_f,1+\sigma,1/S',1/b$, together with a factor $\exp(O(k\log(2k)))$,
+$H_f,1+sigma,1/S',1/b$, together with a factor $exp(O(klog(2k)))$,
the logarithm is at most
-\[
- C(1+k\log(2k)+\log H_f+\log\frac1{S'})+C\log E,
-\]
+[
+ C(1+klog(2k)+log H_f+logfrac1{S'})+Clog E,
+]
with $C$ depending only on constants already fixed. Enlarging $C_0$
makes this at most any chosen positive multiple of $E$. Only finitely
many such estimates occur, so one final choice of $C_0$ handles them all.
For real $t$ define the polynomial and its value at $x$,
- A(t)=\sum_{j=1}^n f_j\,W(j-d)\,e^{\ii t(j-d)},\qquad
+ A(t)=sum_{j=1}^n f_j,W(j-d),e^{ii t(j-d)},qquad
B_x(t)=A(t)(x).
The coefficient bound on $f_j$ and the Gaussian sum give
- \abs{B_x(t)},\quad \abs{\EE[A(t)]},\quad \abs{\Delta[A(t)]}
- \le U:=C H_f(1+\sigma),
+ abs{B_x(t)},quad abs{EE[A(t)]},quad abs{Delta[A(t)]}
+ le U:=C H_f(1+sigma),
-with $C$ absolute and $U\ge1$. By prefix cancellation,
-\[
- P(z):=\sum_{j=d}^n \Delta[f_j]\,W(j-d)\,z^{j-d}
- \quadsatisfies\quad P(e^{\ii t})=\Delta[A(t)].
-\]
+with $C$ absolute and $Uge1$. By prefix cancellation,
+[
+ P(z):=sum_{j=d}^n Delta[f_j],W(j-d),z^{j-d}
+ quadsatisfiesquad P(e^{ii t})=Delta[A(t)].
+]
First, $P$ is shown to keep a signal at almost the original decay:
- \abs{P(e^{-s_1+\ii\omega})}\ge\tfrac12 e^{-E}
- \quadfor some\quad \abs{s_1/s-1}\le10^{-4}.
-
-
-Let $V$ be a centred real normal variable with variance $\sigma^{-2}$.
-Since $W(l)\,\mathbf E\,e^{Vl}=1$,
-\[
- \mathcal S_f(s,\omega)=\mathbf E\,P(e^{-s+V+\ii\omega}).
-\]
-Write $\delta=10^{-4}$. Exponential tilting gives
-\[
- W(l)\,\mathbf E[e^{Vl}\mathbf 1_{\{\abs V>\delta s\}}]
- =\Pr\!(\abs{N(l/\sigma^2,\sigma^{-2})}>\delta s).
-\]
-For $0\le l\le\delta s\sigma^2/2$ this probability is at most
-$2e^{-c\delta^2s^2\sigma^2}$; for larger $l$ use the bound one and sum
-$e^{-sl}$. Because $\abs{\Delta[f_j]}\le2H_f$, the contribution of
-$\abs V>\delta s$ is at most
-
- 2H_f(1+1/s)(2e^{-c\delta^2s^2\sigma^2}+e^{-\delta s^2\sigma^2/2}).
-
-Here $1/s\le1/(\beta S')$ and
-$s^2\sigma^2\ge\beta^2K^2E/\eta_0^2$. Choose $\eta_0$ small and then
+ abs{P(e^{-s_1+iiomega})}getfrac12 e^{-E}
+ quadfor somequad abs{s_1/s-1}le10^{-4}.
+
+
+Let $V$ be a centred real normal variable with variance $sigma^{-2}$.
+Since $W(l),mathbf EbfE,e^{Vl}=1$,
+[
+ mathcal S_f(s,calS_f(s,omega)=mathbf EbfE,P(e^{-s+V+iiomega}).
+]
+Write $delta=10^{-4}$. Exponential tilting gives
+[
+ W(l),mathbf E[e^bfE[e^{Vl}mathbf 1_bf1_{{abs V>delta s}}]
+ =Pr!(abs{N(l/sigma^2,sigma^{-2})}>delta s).
+]
+For $0le lledelta ssigma^2/2$ this probability is at most
+$2e^{-cdelta^2s^2sigma^2}$; for larger $l$ use the bound one and sum
+$e^{-sl}$. Because $abs{Delta[f_j]}le2H_f$, the contribution of
+$abs V>delta s$ is at most
+
+ 2H_f(1+1/s)(2e^{-cdelta^2s^2sigma^2}+e^{-delta s^2sigma^2/2}).
+
+Here $1/sle1/(beta S')$ and
+$s^2sigma^2gebeta^2K^2E/eta_0^2$. Choose $eta_0$ small and then
$C_0$ large to make (eq:window-tail) less than $e^{-2E}$. Then the
-integral over $\abs V\le\delta s$ has magnitude at least
-$e^{-E}-e^{-2E}\ge e^{-E}/2$, so its integrand attains magnitude at
+integral over $abs Vledelta s$ has magnitude at least
+$e^{-E}-e^{-2E}ge e^{-E}/2$, so its integrand attains magnitude at
least that large. This proves (eq:window-interior-signal).
Set
-\[
- Λ(t)=\log\frac{U}{\abs{P(e^{\ii t})}}\ge0,
-\]
+[
+ Λ(t)=logfrac{U}{abs{P(e^{ii t})}}ge0,
+]
with value $+∞$ at a boundary zero. The key estimate is
- \int_{\RR}Λ(t)\,(s_1\,dt)/(\pi(s_1^2+(t-\omega)^2))\le1.001E.
+ int_{RR}Λ(t),(s_1,dt)/(pi(s_1^2+(t-omega)^2))le1.001E.
For completeness, the Cauchy Fourier transform is
- \int_{\RR}\frac{e^{\ii\xi v}}{\pi(1+v^2)}\,dv=e^{-\abs\xi}.
+ int_{RR}frac{e^{iixi v}}{pi(1+v^2)},dv=e^{-absxi}.
-For $\xi>0$ close the contour in the upper half-plane and take the residue
-at $\ii$; for $\xi<0$ use the lower half-plane; $\xi=0$ is direct. Thus
+For $xi>0$ close the contour in the upper half-plane and take the residue
+at $ii$; for $xi<0$ use the lower half-plane; $xi=0$ is direct. Thus
periodizing the density in (eq:poisson-log-bound) gives the circle
-Poisson measure at $e^{-s_1+\ii\omega}$.
+Poisson measure at $e^{-s_1+iiomega}$.
To apply it, factor the nonzero polynomial $P$ into linear factors. A
factor with a root outside the unit disk has harmonic logarithmic
magnitude inside. For a root $a$ inside the disk, the identity
-$\abs{e^{\ii t}-a}=\abs{1-\overline a e^{\ii t}}$ and the inequality
-\[
- \abs{1-\overline a z}^2-\abs{z-a}^2=(1-\abs a^2)(1-\abs z^2)\ge0
-\]
+$abs{e^{ii t}-a}=abs{1-overlineoverline aa e^{ii t}}$ and the inequality
+[
+ abs{1-overlineoverline aa z}^2-abs{z-a}^2=(1-abs a^2)(1-abs z^2)ge0
+]
bound its contribution by its Poisson integral. Roots on the circle
follow by a radial limit; their logarithmic singularities are integrable.
Summing over factors gives
-\[
- \int_{\RR}Λ(t)(s_1\,dt)/(\pi(s_1^2+(t-\omega)^2))
- \le\log U-\log\abs{P(e^{-s_1+\ii\omega})}
- \le E+\log(2U).
-\]
+[
+ int_{RR}Λ(t)(s_1,dt)/(pi(s_1^2+(t-omega)^2))
+ lelog U-logabs{P(e^{-s_1+iiomega})}
+ le E+log(2U).
+]
The last term is at most $1.001E$ by the choice of $C_0$, proving the
claim.
-The goal is a frequency $\lambda\in[-2s_1,2s_1]$ with a large signal, and,
+The goal is a frequency $lambdain[-2s_1,2s_1]$ with a large signal, and,
for every prescribed difference in that interval, pairs of frequencies
with that difference. The product $A(t)A(t')bar$ carries signed
frequency $t-t'$. After normalizing by the evaluations at $x$, these
products will be the interior vectors of a chain from $1$ to a
-normalized copy of $A(\lambda)$, with consecutive frequency labels
+normalized copy of $A(lambda)$, with consecutive frequency labels
differing by at most $b$. Each neighbouring inner product then has signed
-frequency of magnitude at most $b$, even though $\abs\lambda$ may be
+frequency of magnitude at most $b$, even though $abslambda$ may be
much larger. The logarithmic estimate turns the existence of enough
pairs into a question about pairing Cauchy mass.
[Cauchy matching]
-Let $w(v)=1/(\pi(1+v^2))$. For every $r\ge0$ there is a finite positive
-measure $\mu_r$ on $\RR^2$, supported on $v-v'=r$, whose two marginals
-sum to at most $w(v)\,dv$. For $r>0$ its total mass is
+Let $w(v)=1/(pi(1+v^2))$. For every $rge0$ there is a finite positive
+measure $mu_r$ on $RR^2$, supported on $v-v'=r$, whose two marginals
+sum to at most $w(v),dv$. For $r>0$ its total mass is
- 2\mu_r(\RR^2)=1-\frac4\piarctan(e^{-\pi/r});
+ 2mu_r(RR^2)=1-frac4piarctan(e^{-pi/r});
for $r=0$ its total mass is $1/2$. In particular
-(eq:cauchy-matching-mass) exceeds $0.737$ for $0<r\le2$.
+(eq:cauchy-matching-mass) exceeds $0.737$ for $0<rle2$.
Proof.
-For $r=0$ put half of $w(v)\,dv$ on the diagonal. Suppose $r>0$ and
-partition the line into chains $v+jr$, $j\in\mathbb Z$, with
-$-r/2\le v<r/2$. Write $w_j=w(v+jr)$. On the edge from $j$ to $j-1$, for
-$j\ge1$, put weight
-\[
- e_j=\sum_{m\ge0}(-1)^m w_{j+m}.
-\]
+For $r=0$ put half of $w(v),dv$ on the diagonal. Suppose $r>0$ and
+partition the line into chains $v+jr$, $jinmathbb Z$,bbZ$, with
+$-r/2le v<r/2$. Write $w_j=w(v+jr)$. On the edge from $j$ to $j-1$, for
+$jge1$, put weight
+[
+ e_j=sum_{mge0}(-1)^m w_{j+m}.
+]
These weights are nonnegative because the summands decrease along the
tail, and $e_j+e_{j+1}=w_j$, so the edges exhaust the weight at every
positive vertex. Build the negative-side edges the same way, from the
negative tail toward zero. The only possibly unused weight sits at the
central vertex, with value
-\[
- R(v)=\sum_{j\in\mathbb Z}(-1)^j w(v+jr).
-\]
+[
+ R(v)=sum_{jinmathbb Z}bbZ(-1)^j w(v+jr).
+]
To check no weight is overused, periodize $w$ at period $2r$ and
subtract its translate by $r$. By (eq:cauchy-transform), with
-$\tau=e^{-\pi/r}$, this gives the absolutely convergent Fourier series
-\[
- R(v)=\frac2r\sum_{m\ge0}\tau^{2m+1}\cos((2m+1)\pi v)/(r).
-\]
-Writing $a=\pi v/r$, the sum equals
-\[
- \re\frac{\tau e^{\ii a}}{1-\tau^2e^{2\ii a}}
- =(\tau(1-\tau^2)\cos a)/(1-2\tau^2\cos(2a)+\tau^4)\ge0,
-\]
-since $\abs a\le\pi/2$. Orient each edge with its larger endpoint first
+$tau=e^{-pi/r}$, this gives the absolutely convergent Fourier series
+[
+ R(v)=frac2rsum_{mge0}tau^{2m+1}cos((2m+1)pi v)/(r).
+]
+Writing $a=pi v/r$, the sum equals
+[
+ refrac{tau e^{ii a}}{1-tau^2e^{2ii a}}
+ =(tau(1-tau^2)cos a)/(1-2tau^2cos(2a)+tau^4)ge0,
+]
+since $abs alepi/2$. Orient each edge with its larger endpoint first
and integrate the edge weights against $dv$ over $[-r/2,r/2)$. The
resulting measure has the claimed marginal domination. Its uncovered
mass is
-\[
- \int_{-r/2}^{r/2}R(v)\,dv
- =\frac4\pi\sum_{m\ge0}\frac{(-1)^m\tau^{2m+1}}{2m+1}
- =\frac4\piarctan\tau,
-\]
+[
+ int_{-r/2}^{r/2}R(v),dv
+ =frac4pisum_{mge0}frac{(-1)^mtau^{2m+1}}{2m+1}
+ =frac4piarctantau,
+]
proving (eq:cauchy-matching-mass). The expression decreases with
$r$ and at $r=2$ exceeds $0.737$; for instance
-$e^{-\pi/2}<0.209$ together with $arctan u\le u-u^3/3+u^5/5$ for
-$0\le u\le1$ gives this numeric bound.
+$e^{-pi/2}<0.209$ together with $arctan ule u-u^3/3+u^5/5$ for
+$0le ule1$ gives this numeric bound.
[Frequencies from the logarithmic bound]
-Let $s_1,E>0$, let $\abs\omega\le2.003s_1$, and let
-$Λ:\RR\to[0,∞]$ be measurable satisfying
-(eq:poisson-log-bound). There is a real $\lambda$ with
+Let $s_1,E>0$, let $absomegale2.003s_1$, and let
+$Λ:RRto[0,∞]$ be measurable satisfying
+(eq:poisson-log-bound). There is a real $lambda$ with
- \abs\lambda\le2s_1,\qquad Λ(\lambda)\le2.40E.
+ abslambdale2s_1,qquad Λ(lambda)le2.40E.
-There is also an absolute $C_2$ such that, for every $\abs y\le2s_1$,
+There is also an absolute $C_2$ such that, for every $abs yle2s_1$,
some real $t,t'$ satisfy
- t-t'=y,\qquad \abs t,\abs{t'}\le C_2s_1,\qquad
- Λ(t)+Λ(t')\le2.74E.
+ t-t'=y,qquad abs t,abs{t'}le C_2s_1,qquad
+ Λ(t)+Λ(t')le2.74E.
Proof.
The Cauchy mass of $[-2s_1,2s_1]$ is at least
-\[
- \frac{arctan(2-2.003)+arctan(2+2.003)}\pi>0.418.
-\]
+[
+ frac{arctan(2-2.003)+arctan(2+2.003)}pi>0.418.
+]
Because $1.001/0.418<2.40$, averaging proves (eq:single-frequency).
For (eq:paired-frequencies), apply Lemma~lem:cauchy-matching
-with $r=\abs y/s_1\le2$ and substitute $t=\omega+s_1v$, reversing the
+with $r=abs y/s_1le2$ and substitute $t=omega+s_1v$, reversing the
orientation if needed. The two marginals sum to at most the Cauchy
measure in (eq:poisson-log-bound), while twice the pair mass
exceeds $0.737$. Discard pairs with an endpoint outside
$[-C_2s_1,C_2s_1]$; twice the discarded mass is at most twice the Cauchy
tail outside that interval, which tends uniformly to zero as
-$C_2\to∞$ since $\abs\omega/s_1\le2.003$. Choose absolute $C_2$ so
+$C_2to∞$ since $absomega/s_1le2.003$. Choose absolute $C_2$ so
the retained mass exceeds $0.733/2$. Its integral of
$Λ(t)+Λ(t')$ is at most $1.001E$, so some retained pair has
sum at most
-\[
+[
(2(1.001))/(0.733)E<2.74E,
-\]
+]
as required. The diagonal construction gives the same when $y=0$.
-The window parameters satisfy $\abs\omega/s_1\le2.002/0.9999<2.003$, so
+The window parameters satisfy $absomega/s_1le2.002/0.9999<2.003$, so
Lemma~lem:good-frequencies applies to this $Λ$. Individual
frequencies in (eq:paired-frequencies) need not be small, but their
differences can be prescribed. These pairs build a chain of
@@ -658,108 +658,108 @@ frequency. The following Hilbert-space fact explains why such a chain
keeps the needed exponential accuracy.
[Gram chaining]
-Let $v_0,...,v_J$, $J\ge1$, lie in a complex Hilbert space whose inner
-product is linear in its first argument. Suppose $\norm{v_0}=1$ and, for
-some $\xi\ge0$,
-\[
- \abs{\norm{v_i}^2-1}\le\xi\ (0\le i\le J),\qquad
- \abs{< v_i,v_{i-1}>-1}\le\xi\ (1\le i\le J).
-\]
-Then $\abs{< v_J,v_0>-1}\le2J^2\xi$.
+Let $v_0,...,v_J$, $Jge1$, lie in a complex Hilbert space whose inner
+product is linear in its first argument. Suppose $norm{v_0}=1$ and, for
+some $xige0$,
+[
+ abs{norm{v_i}^2-1}lexi (0le ile J),qquad
+ abs{< v_i,v_{i-1}>-1}lexi (1le ile J).
+]
+Then $abs{< v_J,v_0>-1}le2J^2xi$.
Proof.
Put $h_i=v_i-v_{i-1}$. The hypotheses give
-\[
- \norm{h_i}\le2\sqrt\xi,\qquad
- \abs{< h_i,v_{i-1}>}\le2\xi,\qquad
- \norm{v_{i-1}-v_0}\le2(i-1)\sqrt\xi.
-\]
+[
+ norm{h_i}le2sqrtxi,qquad
+ abs{< h_i,v_{i-1}>}le2xi,qquad
+ norm{v_{i-1}-v_0}le2(i-1)sqrtxi.
+]
So Cauchy--Schwarz gives
-$\abs{< h_i,v_0>}\le2\xi+4(i-1)\xi$. Sum over $i$ and use
+$abs{< h_i,v_0>}le2xi+4(i-1)xi$. Sum over $i$ and use
$< v_0,v_0>=1$.
Choose an absolute $C_3$ large enough that all frequencies supplied by
Lemma~lem:good-frequencies have magnitude at most $C_3S$. A product
of one to four factors, each of the form $A(t)$ or $A(t)bar$ with
-$\abs t\le C_3S$, is called balanced if the sum of its signed
+$abs tle C_3S$, is called balanced if the sum of its signed
frequencies has magnitude at most $b$; conjugation flips the sign of a
frequency.
[A balanced product keeps a signal]
Some balanced product $G$ satisfies
- \abs{\Delta[G]}\ge e^{-7.94E}.
+ abs{Delta[G]}ge e^{-7.94E}.
Proof.
Suppose not: every balanced product has signal below
-$ε=e^{-7.94E}$. The squared modulus $\abs{A(t)}^2$ is one such
-product. Positivity of $\EE$ on $A(t)$ and $1$ gives
-\[
- \abs{\EE[A(t)]}^2\le\EE[\abs{A(t)}^2]\le\abs{B_x(t)}^2+ε,
-\]
+$ε=e^{-7.94E}$. The squared modulus $abs{A(t)}^2$ is one such
+product. Positivity of $EE$ on $A(t)$ and $1$ gives
+[
+ abs{EE[A(t)]}^2leEE[abs{A(t)}^2]leabs{B_x(t)}^2+ε,
+]
hence
- \abs{\Delta[A(t)]}\le2\abs{B_x(t)}+\sqrtε.
+ abs{Delta[A(t)]}le2abs{B_x(t)}+sqrtε.
-At $\lambda$ from (eq:single-frequency) this implies
+At $lambda$ from (eq:single-frequency) this implies
- \abs{B_x(\lambda)}\ge e^{-2.41E}.
+ abs{B_x(lambda)}ge e^{-2.41E}.
-Indeed $\abs{\Delta[A(\lambda)]}\ge e^{-2.40E}$ whereas
-$\sqrtε=e^{-3.97E}$; fixed factors are absorbed by the lower bound
+Indeed $abs{Delta[A(lambda)]}ge e^{-2.40E}$ whereas
+$sqrtε=e^{-3.97E}$; fixed factors are absorbed by the lower bound
on $E$ in (eq:multiscale-cost). For a pair in
(eq:paired-frequencies), nonnegativity of $Λ$ gives both
signals at least $e^{-2.74E}$, and
-\[
- \abs{\Delta[A(t)]\Delta[A(t')]}=U^2e^{-Λ(t)-Λ(t')}
- \ge e^{-2.74E}.
-\]
+[
+ abs{Delta[A(t)]Delta[A(t')]}=U^2e^{-Λ(t)-Λ(t')}
+ ge e^{-2.74E}.
+]
Applying (eq:evaluation-lower-bound) to each gives
- \abs{B_x(t)B_x(t')bar}\ge e^{-2.75E}.
+ abs{B_x(t)B_x(t')bar}ge e^{-2.75E}.
Each evaluation is at least a third of its signal once $E$ is large; the
factor $1/9$ is absorbed in $e^{-0.01E}$.
-Take a monotone chain $0=y_0,y_1,...,y_J=\lambda$ with consecutive
+Take a monotone chain $0=y_0,y_1,...,y_J=lambda$ with consecutive
distances at most $b$ and
-\[
- 1\le J\le C(1+S/b).
-\]
-If $\lambda=0$, use two endpoints and $J=1$. For each interior $y_i$
+[
+ 1le Jle C(1+S/b).
+]
+If $lambda=0$, use two endpoints and $J=1$. For each interior $y_i$
choose a pair from (eq:paired-frequencies) and set
-\[
- V_0=1,\qquad
- V_i=\frac{A(t)A(t')bar}{B_x(t)B_x(t')bar}\ (0<i<J),
- \qquad V_J=(A(\lambda))/(B_x(\lambda)).
-\]
+[
+ V_0=1,qquad
+ V_i=frac{A(t)A(t')bar}{B_x(t)B_x(t')bar} (0<i<J),
+ qquad V_J=(A(lambda))/(B_x(lambda)).
+]
The denominators are nonzero by
(eq:endpoint-normalization)--(eq:pair-normalization), and
every $V_i(x)=1$. View these polynomials as vectors $v_i$ in the Hilbert
-space of $\EE$; they have degree at most $2k\le D$.
+space of $EE$; they have degree at most $2kle D$.
-For $1\le i\le J$ the norm square of $v_i$ and its inner product with
+For $1le ile J$ the norm square of $v_i$ and its inner product with
$v_{i-1}$ are expectations of normalized balanced products, whose signed
frequencies are respectively zero and $y_i-y_{i-1}$; each product has at
most four factors and its normalizing denominator has magnitude at least
$e^{-5.50E}$. The contradictory assumption therefore gives
-\[
- \abs{\norm{v_i}^2-1}\le\xi,\qquad
- \abs{< v_i,v_{i-1}>-1}\le\xi,\qquad
- \xi=ε e^{5.50E}=e^{-2.44E}.
-\]
+[
+ abs{norm{v_i}^2-1}lexi,qquad
+ abs{< v_i,v_{i-1}>-1}lexi,qquad
+ xi=ε e^{5.50E}=e^{-2.44E}.
+]
Since $v_0=1$ has norm one, Lemma~lem:gram-chain yields
-\[
- \abs{\Delta[A(\lambda)]}
- =\abs{B_x(\lambda)}\,\abs{< v_J,v_0>-1}
- \le2UJ^2e^{-2.44E}<e^{-2.40E}.
-\]
+[
+ abs{Delta[A(lambda)]}
+ =abs{B_x(lambda)},abs{< v_J,v_0>-1}
+ le2UJ^2e^{-2.44E}<e^{-2.40E}.
+]
The last strict inequality holds after enlarging $C_0$, because
-$\log(2UJ^2)<0.04E$ by (eq:multiscale-cost). This contradicts
+$log(2UJ^2)<0.04E$ by (eq:multiscale-cost). This contradicts
(eq:single-frequency) and proves the lemma.
@@ -776,147 +776,147 @@ Proposition~prop:multiscale requires; what remains is to bound
the damping on the inner gaps.
Fix a product from Lemma~lem:balanced-product and let
-$1\le m\le4$ be its number of factors. Expand each factor into the
+$1le mle4$ be its number of factors. Expand each factor into the
tuples that define $f$. An interleaving is an ordering, ties
included, of the $mk$ labeled positions in these tuples. There are at
-most $(mk)^{mk}=\exp(O(k\log(2k)))$ interleavings: assign each labeled
+most $(mk)^{mk}=exp(O(klog(2k)))$ interleavings: assign each labeled
position its rank among the distinct positions. Keep only interleavings
that increase strictly inside each old tuple. Fix one interleaving and
-relabel its copies so $j_1\le...\le j_m$. Ties between endpoints may
+relabel its copies so $j_1le...le j_m$. Ties between endpoints may
be ordered arbitrarily. Absorbing conjugation into the signs of the
frequencies, the window weight is
- \prod_{r=1}^m W(j_r-d)\exp(\ii t_r(j_r-d)),
- \qquad |t_r|\le C_3S,\qquad
- |\sum_{r=1}^m t_r|\le b.
+ prod_{r=1}^m W(j_r-d)exp(ii t_r(j_r-d)),
+ qquad |t_r|le C_3S,qquad
+ |sum_{r=1}^m t_r|le b.
Conjugation also conjugates the old inner-gap parameters, leaving their
dampings unchanged.
-{\itshape A Gaussian contour with controlled total frequency.}
-Recall $\sigma=K\sqrt E/S'$ and $b=\beta S'$. For every real $l$,
+{itshape A Gaussian contour with controlled total frequency.}
+Recall $sigma=Ksqrt E/S'$ and $b=beta S'$. For every real $l$,
- W(l)=\frac{\sigma}{\ii√(2\pi)}
- \int_{-\ii∞}^{\ii∞}
- \exp((\sigma^2v^2)/(2)+vl)\,dv.
+ W(l)=frac{sigma}{ii√(2pi)}
+ int_{-ii∞}^{ii∞}
+ exp((sigma^2v^2)/(2)+vl),dv.
-Indeed, $v=\ii u$ gives the Gaussian Fourier transform; its value at
+Indeed, $v=ii u$ gives the Gaussian Fourier transform; its value at
$l=0$ is one, and integration by parts shows its $l$-derivative is
-$-l/\sigma^2$ times itself, fixing the formula.
+$-l/sigma^2$ times itself, fixing the formula.
-Set $\rho=1/2.002$. For $u=(u_1,...,u_m)\in\RR^m$ define
+Set $rho=1/2.002$. For $u=(u_1,...,u_m)inRR^m$ define
- Ω=\sum_{r=1}^m(u_r+t_r),\qquad
- v_r=b+\ii u_r\ (r<m),\qquad
- v_m=-mb-\rho|Ω|+\ii u_m.
+ Ω=sum_{r=1}^m(u_r+t_r),qquad
+ v_r=b+ii u_r (r<m),qquad
+ v_m=-mb-rho|Ω|+ii u_m.
These real parts have partial sums $rb$ for $r<m$ and total
-$-b-\rho|Ω|$. Centred at $j_m$, the total sum gives the real part
+$-b-rho|Ω|$. Centred at $j_m$, the total sum gives the real part
of the outer coefficient and the negatives of the partial sums give the
real parts on consecutive endpoint gaps. So the contour is built to
-produce outer decay $b+\rho|Ω|$ and positive inner dampings $rb$;
+produce outer decay $b+rho|Ω|$ and positive inner dampings $rb$;
the exact rewriting is checked below.
For each fixed tuple of positions, first shift the first $m-1$ contours
-of (eq:gaussian-contour) to the vertical lines $\re v_r=b$,
+of (eq:gaussian-contour) to the vertical lines $re v_r=b$,
leaving the last on the imaginary axis. The linking horizontal
integrals vanish by Gaussian decay. Next fix $(u_1,...,u_{m-1})$ and
deform the last contour to the piecewise-linear graph in
-(eq:bent-contour), whose slopes have absolute value $\rho<1$. Along
+(eq:bent-contour), whose slopes have absolute value $rho<1$. Along
horizontal segments that join its truncations to the imaginary axis at
-height $\pm T$, the Gaussian exponent has real part at most $-cT^2+O(T)$
-for some $c>0$, while $\exp(v_m(j_m-d))$ adds only a linear term in $T$,
+height $pm T$, the Gaussian exponent has real part at most $-cT^2+O(T)$
+for some $c>0$, while $exp(v_m(j_m-d))$ adds only a linear term in $T$,
so those linking integrals also vanish. The integrands are entire, so no
residues occur. This proves the identities as iterated integrals; joint
absolute convergence follows from the bound below.
-Let $\nu$ be the resulting complex measure on $\RR^m$ after removing the
+Let $nu$ be the resulting complex measure on $RR^m$ after removing the
position-dependent factors. Outside the measure-zero set $Ω=0$,
- d\nu(u)=
- (\frac{\sigma}{√(2\pi)})^m
- \exp((\sigma^2)/(2)\sum_{r=1}^m v_r^2)
- (1+\ii\rhosgnΩ)\,du.
+ dnu(u)=
+ (frac{sigma}{√(2pi)})^m
+ exp((sigma^2)/(2)sum_{r=1}^m v_r^2)
+ (1+iirhosgnΩ),du.
The last factor is the differential of the bent contour divided by
-$\ii\,du_m$. This quantity is bounded uniformly over the frequencies in
+$ii,du_m$. This quantity is bounded uniformly over the frequencies in
(eq:endpoint-window); the bounds also give absolute convergence.
-Put $\delta_0=1-4\rho^2>0$ and $c_1=\delta_0/2$. Since $m\le4$ and
-$|\sum t_r|\le b$, Young's inequality gives an absolute $C_4$ with
+Put $delta_0=1-4rho^2>0$ and $c_1=delta_0/2$. Since $mle4$ and
+$|sum t_r|le b$, Young's inequality gives an absolute $C_4$ with
- \sum_{r=1}^m(\re v_r)^2
- &=(m-1)b^2+(mb+\rho|Ω|)^2 \notag\\
- &\le 3b^2+(2\rho√(\textstyle\sum_r u_r^2)+(4+\rho)b)^2
- \notag\\
- &\le (1-c_1)\sum_r u_r^2+C_4b^2.
+ sum_{r=1}^m(re v_r)^2
+ &=(m-1)b^2+(mb+rho|Ω|)^2 notag
+ &le 3b^2+(2rho√(textstylesum_r u_r^2)+(4+rho)b)^2
+ notag
+ &le (1-c_1)sum_r u_r^2+C_4b^2.
-The strict inequality $4\rho^2<1$ is essential: summing at most four
+The strict inequality $4rho^2<1$ is essential: summing at most four
imaginary coordinates can multiply their squared norm by four, and the
sector constant $2.002$ leaves a positive Gaussian margin. For example
-$C_4=3+(4+\rho)^2(1+8\rho^2/\delta_0)$ works. With
+$C_4=3+(4+rho)^2(1+8rho^2/delta_0)$ works. With
(eq:contour-measure) this gives
-\[
- |d\nu(u)|\le C\sigma^m
- \exp((C_4\sigma^2b^2)/(2)
- -(c_1\sigma^2)/(2)\sum_r u_r^2)du.
-\]
+[
+ |dnu(u)|le Csigma^m
+ exp((C_4sigma^2b^2)/(2)
+ -(c_1sigma^2)/(2)sum_r u_r^2)du.
+]
For any fixed tuple the remaining exponential factor has modulus at most
-$\exp(C_n(1+\sum_r|u_r|))$ for a finite $C_n$, which is integrable
+$exp(C_n(1+sum_r|u_r|))$ for a finite $C_n$, which is integrable
against the Gaussian bound. So the iterated deformation is a jointly
absolutely convergent integral, and finite tuple sums may be exchanged
with it.
-Write $\mathcal U=\{u:\max_r|u_r|\le S'/16\}$. Integrating the last bound
+Write $mathcal U=calU={u:max_r|u_r|le S'/16}$. Integrating the last bound
and splitting the negative quadratic term into two equal halves on
-$\mathcal U^c$ gives absolute constants $C,C'$ with
+$mathcal U^c$calU^c$ gives absolute constants $C,C'$ with
- \|\nu\|_{abs}
- &\le C\exp(C_4K^2\beta^2E/2),\\
- \int_{\mathcal U^c}|d\nu|
- &\le C'\exp\Bigl[((C_4K^2\beta^2)/(2)
- -(c_1K^2)/(1024))E\Bigr].
+ |nu|_{abs}
+ &le Cexp(C_4K^2beta^2E/2),
+ int_{mathcal U^c}calU^c|dnu|
+ &le C'expBigl[((C_4K^2beta^2)/(2)
+ -(c_1K^2)/(1024))EBigr].
-Here $\|\nu\|_{abs}$ is total variation; the powers $\sigma^m$
-cancel under Gaussian rescaling. Choose $K$ with $c_1K^2/1024\ge30$,
-then $\beta\le1/4$ with $C_4K^2\beta^2/2\le0.001$. Enlarging $C_0$ to
+Here $|nu|_{abs}$ is total variation; the powers $sigma^m$
+cancel under Gaussian rescaling. Choose $K$ with $c_1K^2/1024ge30$,
+then $betale1/4$ with $C_4K^2beta^2/2le0.001$. Enlarging $C_0$ to
absorb fixed prefactors gives
- \|\nu\|_{abs}\le e^{0.005E},\qquad
- \int_{\mathcal U^c}|d\nu|\le e^{-20E}.
+ |nu|_{abs}le e^{0.005E},qquad
+ int_{mathcal U^c}calU^c|dnu|le e^{-20E}.
-This respects the constant order fixed earlier: $K$ first, $\beta$
-second, then $\eta_0$ small to meet the window estimate, and only
+This respects the constant order fixed earlier: $K$ first, $beta$
+second, then $eta_0$ small to meet the window estimate, and only
afterward $C_0$ enlarged.
-{\itshape Outer parameters and endpoint gaps.}
-For $c_r=v_r+\ii t_r$, telescoping the endpoint differences gives
+{itshape Outer parameters and endpoint gaps.}
+For $c_r=v_r+ii t_r$, telescoping the endpoint differences gives
- \sum_{r=1}^m c_r(j_r-d)
- =(\sum_{r=1}^m c_r)(j_m-d)
- -\sum_{r=1}^{m-1}(\sum_{l=1}^r c_l)(j_{r+1}-j_r).
+ sum_{r=1}^m c_r(j_r-d)
+ =(sum_{r=1}^m c_r)(j_m-d)
+ -sum_{r=1}^{m-1}(sum_{l=1}^r c_l)(j_{r+1}-j_r).
-So the largest endpoint has coefficient $-s'+\iiΩ$ with
+So the largest endpoint has coefficient $-s'+iiΩ$ with
- s'=b+\rho|Ω|.
+ s'=b+rho|Ω|.
The link from $j_r$ to $j_{r+1}$ receives the parameter
- \exp(-rb-\ii\sum_{l=1}^r(u_l+t_l)),\qquad 1\le r<m.
+ exp(-rb-iisum_{l=1}^r(u_l+t_l)),qquad 1le r<m.
Thus every added damping is at least $b$, uniformly in $u$. On
-$\mathcal U$,
-\[
- |Ω|\le(\beta+1/4)S',\qquad
- \beta S'\le s'\le[\beta+\rho(\beta+1/4)]S'\le S',
- \qquad |Ω|\le2.002s'.
-\]
+$mathcal U$,calU$,
+[
+ |Ω|le(beta+1/4)S',qquad
+ beta S'le s'le[beta+rho(beta+1/4)]S'le S',
+ qquad |Ω|le2.002s'.
+]
The angle in (eq:endpoint-link) has absolute value at most
-$3(C_3S+S'/16)\le C_1S$ for a fixed absolute $C_1$. Hence the outer
+$3(C_3S+S'/16)le C_1S$ for a fixed absolute $C_1$. Hence the outer
parameters satisfy (eq:multiscale-output) and the endpoint links
satisfy (eq:added-parameters). The choice of $C_1$ uses only $C_3$
-and $S'\le S$, imposing no new condition on $K$ or $\beta$.
+and $S'le S$, imposing no new condition on $K$ or $beta$.
The outer weights are at the right scale. To finish, each interleaving
must have exactly the form of a feature with controlled coefficient sum.
@@ -925,7 +925,7 @@ charging its damping at every crossing would overcount. A distinct-edge
assignment prevents that loss.
For the fixed interleaving write its distinct positions as
-$i_1<...<i_v$ with $1\le v\le mk$. Form a multigraph on these
+$i_1<...<i_v$ with $1le vle mk$. Form a multigraph on these
vertices: include the inner edges of each old tuple and the links
between consecutive endpoints in (eq:endpoint-link), each edge
carrying its exponential parameter. It is connected, since each old
@@ -942,13 +942,13 @@ follows that each new parameter uses at most one old parameter from each
copy and at most one added endpoint parameter. This is exactly the
factor-count assertion of Proposition~prop:multiscale.
-Ties merge repeated bit variables by $X_i^r=X_i$ for $r\ge1$. Once the
+Ties merge repeated bit variables by $X_i^r=X_i$ for $rge1$. Once the
interleaving is fixed, choosing arbitrary strictly increasing values for
its $v$ distinct positions specifies exactly one assignment to its
labeled tuples, so no further restriction remains on the new tuple. Its
largest position is $i_v=j_m$. Summing all these assignments gives an
-order-$v$ feature $f^{\mathcal I,u}$ with outer weight
-$\exp((-s'+\iiΩ)(i_v-d))$.
+order-$v$ feature $f^{mathcal I,u}calI,u$ with outer weight
+$exp((-s'+iiΩ)(i_v-d))$.
The needed combinatorial fact controls its height.
@@ -962,7 +962,7 @@ the assigned edge crosses that gap.
Proof.
For $v=1$ there is nothing to prove. Otherwise keep a spanning tree and
orient each edge from its smaller to its larger endpoint. Let $M$ be the
-$(v-1)\times(v-1)$ matrix whose $(e,a)$ entry is one if tree edge $e$
+$(v-1)times(v-1)$ matrix whose $(e,a)$ entry is one if tree edge $e$
crosses gap $a$ and zero otherwise. It is invertible: for $g$ in its
kernel set $y_1=0$ and $y_{a+1}=y_a+g_a$; then $Mg=0$ says $y$ is equal
at the endpoints of every tree edge, and connectivity forces all values
@@ -975,35 +975,35 @@ To see why distinctness matters, look at Figure~fig:edge-gap-assignment:
one edge may contribute to several new parameters, but it need pay for
only one of their upper bounds. Concretely, take two old tuples
$(i_1,i_3,i_5)$ and $(i_2,i_4,i_6)$ joined by one dashed endpoint
-link-edge, with gaps $g_a=(i_a,i_{a+1})$ for $1\le a\le5$. The assignment
-$g_a\mapsto e_a$, $1\le a\le5$, gives every gap a distinct crossing edge:
+link-edge, with gaps $g_a=(i_a,i_{a+1})$ for $1le ale5$. The assignment
+$g_amapsto e_a$, $1le ale5$, gives every gap a distinct crossing edge:
$e_1$ crosses both $g_1$ and $g_2$ but is charged only to $g_1$, and $e_2$
supplies the bound for $g_2$.
[htbp]
-\centering
+centering
[x=1.28cm,y=0.9cm,>=stealth,
olda/.style={draw=blue!70!black,thick},
oldb/.style={draw=orange!85!black,thick},
added/.style={draw=black,thick,dashed}]
- \draw[gray] (0,0)--(5,0);
- \foreach \a in {1,...,6} {
- \fill (\a-1,0) circle (1.8pt);
- \node[below=3pt] at (\a-1,0) {$i_{\a}$};
+ draw[gray] (0,0)--(5,0);
+ foreach a in {1,...,6} {
+ fill (a-1,0) circle (1.8pt);
+ node[below=3pt] at (a-1,0) {$i_{a}$};
}
- \draw[olda] (0,0) to[bend left=30] node[above,pos=.5] {$e_1$} (2,0);
- \draw[olda] (2,0) to[bend left=30] node[above,pos=.5] {$e_3$} (4,0);
- \draw[oldb] (1,0) to[bend left=65] node[above,pos=.5] {$e_2$} (3,0);
- \draw[oldb] (3,0) to[bend left=65] node[above,pos=.5] {$e_4$} (5,0);
- \draw[added] (4,0) to[bend right=60] node[below=2pt,pos=.5] {$e_5$} (5,0);
- \foreach \a in {1,...,5} {
- \node at (\a-.5,-1.05) {$g_{\a}$};
+ draw[olda] (0,0) to[bend left=30] node[above,pos=.5] {$e_1$} (2,0);
+ draw[olda] (2,0) to[bend left=30] node[above,pos=.5] {$e_3$} (4,0);
+ draw[oldb] (1,0) to[bend left=65] node[above,pos=.5] {$e_2$} (3,0);
+ draw[oldb] (3,0) to[bend left=65] node[above,pos=.5] {$e_4$} (5,0);
+ draw[added] (4,0) to[bend right=60] node[below=2pt,pos=.5] {$e_5$} (5,0);
+ foreach a in {1,...,5} {
+ node at (a-.5,-1.05) {$g_{a}$};
}
-\caption{A schematic interleaving of the two old tuples $(i_1,i_3,i_5)$
+caption{A schematic interleaving of the two old tuples $(i_1,i_3,i_5)$
and $(i_2,i_4,i_6)$, joined by the dashed endpoint link $e_5$. Here
-$g_a$ is the gap $(i_a,i_{a+1})$. The assignment $g_a\mapsto e_a$ for
-$1\le a\le5$ uses a distinct crossing edge for every gap: $e_1$ crosses
+$g_a$ is the gap $(i_a,i_{a+1})$. The assignment $g_amapsto e_a$ for
+$1le ale5$ uses a distinct crossing edge for every gap: $e_1$ crosses
both $g_1$ and $g_2$ but is charged only to $g_1$, while $e_2$ supplies
the bound for $g_2$.}
@@ -1014,58 +1014,58 @@ is the sum of $a_e$ over edges crossing it, hence at least that of its
assigned edge. Since $1+1/a$ decreases with $a$,
Lemma~lem:edge-gap-assignment gives
- H_{f^{\mathcal I,u}}
- &\le\prod_{e\ assigned}(1+1/a_e)
- \le\prod_{e\ present}(1+1/a_e) \notag\\
- &\le H_f^m(1+1/b)^{m-1}\le H_f^4(1+1/b)^3.
+ H_{f^{mathcal I,u}calI,u}
+ &leprod_{e assigned}(1+1/a_e)
+ leprod_{e present}(1+1/a_e) notag
+ &le H_f^m(1+1/b)^{m-1}le H_f^4(1+1/b)^3.
The second inequality uses that all factors are at least one; the
-endpoint-link dampings are $rb\ge b$. This covers all overlaps between
+endpoint-link dampings are $rbge b$. This covers all overlaps between
old tuples and also the single-distinct-position case, where the height
is the empty product one.
-{\itshape Completing the multiscale estimate.}
-Apply $\Delta$ to the expanded product and use the integral
-representation for each interleaving. With $\nu_{\mathcal I}$ the
+{itshape Completing the multiscale estimate.}
+Apply $Delta$ to the expanded product and use the integral
+representation for each interleaving. With $nu_{mathcal I}calI$ the
corresponding measure,
- \Delta[product]
- =\sum_{\mathcal I}\int_{\RR^m}
- \mathcal S_{f^{\mathcal I,u}}(s',Ω)\,d\nu_{\mathcal I}(u).
+ Delta[product]
+ =sum_{mathcal I}calIint_{RR^m}
+ mathcal S_calS_{f^{mathcal I,u}calI,u}(s',Ω),dnu_{mathcal I}calI(u).
All estimates are uniform over interleavings, and the integrands are
-uniformly bounded even outside $\mathcal U$. Indeed, for each fixed
-endpoint $j$ the coefficients of $f_j^{\mathcal I,u}$ have absolute sum at
+uniformly bounded even outside $mathcal U$.calU$. Indeed, for each fixed
+endpoint $j$ the coefficients of $f_j^{mathcal I,u}calI,u$ have absolute sum at
most the height in (eq:interleaving-height), terms with $j<d$
-vanish under $\Delta$ by prefix agreement, and $s'\ge b$ gives
+vanish under $Delta$ by prefix agreement, and $s'ge b$ gives
- |\mathcal S_{f^{\mathcal I,u}}(s',Ω)|
- \le 2H_f^4(1+1/b)^3\sum_{l\ge0}e^{-bl}
- \le 2H_f^4(1+1/b)^4.
+ |mathcal S_calS_{f^{mathcal I,u}calI,u}(s',Ω)|
+ le 2H_f^4(1+1/b)^3sum_{lge0}e^{-bl}
+ le 2H_f^4(1+1/b)^4.
No exponential factor from negative distances appears.
Let $N_{int}$ be the number of interleavings and
$V=2H_f^4(1+1/b)^4$. Hypothesis~(eq:multiscale-cost) with large
enough absolute $C_0$ gives
-\[
- E\ge100,\qquad \log N_{int}\le0.01E,\qquad \log V\le0.01E.
-\]
-Here $\log(1+1/b)$ is at most an absolute constant plus $\log(1/S')$,
-because $b=\beta S'$, $S'\le1$, and $\beta$ is already fixed. If every
-signal in (eq:product-feature-integral) with $u\in\mathcal U$ had
+[
+ Ege100,qquad log N_{int}le0.01E,qquad log Vle0.01E.
+]
+Here $log(1+1/b)$ is at most an absolute constant plus $log(1/S')$,
+because $b=beta S'$, $S'le1$, and $beta$ is already fixed. If every
+signal in (eq:product-feature-integral) with $uinmathcal U$calU$ had
magnitude below $e^{-7.98E}$, then (eq:contour-mass) and
(eq:contour-uniform-signal) would give
-\[
- |\Delta[product]|
- \le N_{int}(e^{(0.005-7.98)E}+Ve^{-20E})
- \le e^{-7.965E}+e^{-19.98E}<e^{-7.94E}.
-\]
+[
+ |Delta[product]|
+ le N_{int}(e^{(0.005-7.98)E}+Ve^{-20E})
+ le e^{-7.965E}+e^{-19.98E}<e^{-7.94E}.
+]
This contradicts Lemma~lem:balanced-product. Some interleaving and
-some $u\in\mathcal U$ therefore give a feature with signal at least
-$e^{-7.98E}$. Its order is at most $mk\le4k$; (eq:new-outer-decay)
-and the bounds on $\mathcal U$ give the outer parameters;
+some $uinmathcal U$calU$ therefore give a feature with signal at least
+$e^{-7.98E}$. Its order is at most $mkle4k$; (eq:new-outer-decay)
+and the bounds on $mathcal U$calU$ give the outer parameters;
(eq:interleaving-height) gives (eq:multiscale-height); and
the crossing description gives the parameter factors and
(eq:added-parameters). This completes the proof of
@@ -1079,125 +1079,125 @@ control the signal, so a single one-variable continuation bridges the gap.
Throughout, $0<p<1$ and $q_*,Q,Y$ are as in
(eq:aux-parameters). Recall from (eq:aux-comparisons) that
- 1\le\mathsf X(n,p)\le Y\le2\mathsf X(n,p),\qquad
- QY=Q+\log n\le C\log n.
+ 1lemathsf X(n,p)sfX(n,p)le Yle2mathsf X(n,p),sfX(n,p),qquad
+ QY=Q+log nle Clog n.
Set
- \mu=7.98,\qquad \alpha=0.001,\qquad
- h=\lceil\log_2Y\rceil+H,\qquad D_*=4^h,
+ mu=7.98,qquad alpha=0.001,qquad
+ h=lceillog_2Yrceil+H,qquad D_*=4^h,
where the absolute integer $H$ is fixed below; in particular
-$D_*\le4^{H+1}Y^2$. Fix any positive moment functional of order $2D$ with
-$D\ge D_*$ and first wrong position $d$, and let $\Delta$ be its
+$D_*le4^{H+1}Y^2$. Fix any positive moment functional of order $2D$ with
+$Dge D_*$ and first wrong position $d$, and let $Delta$ be its
difference from the true word, as set out earlier. Every
constant chosen below is independent of this system, of $n$, and of $p$.
-For an absolute constant $\eta>0$ define
+For an absolute constant $eta>0$ define
- S_0=4,\qquad
- S_j=\eta\,q_*\,Y^{-\alpha}\,S_{j-1}\,
- \min\{1,D_*S_{j-1}\}\quad(1\le j\le h).
+ S_0=4,qquad
+ S_j=eta,q_*,Y^{-alpha},S_{j-1},
+ min{1,D_*S_{j-1}}quad(1le jle h).
The linear regime of this recurrence pushes $D_*S_j$ below one; its
quadratic regime then makes $S_j$ shrink fast. Both regimes also control
the arguments of the final gap parameters.
-Choose an integer $J_*$ with $\alpha J_*\ge3$ and set $H=J_*+3$. To check
-the decrease, write $θ=\eta q_*Y^{-\alpha}$ and $t_j=D_*S_j$, so
-$t_j=θ\,t_{j-1}\min\{1,t_{j-1}\}$. Then, whenever the quadratic
+Choose an integer $J_*$ with $alpha J_*ge3$ and set $H=J_*+3$. To check
+the decrease, write $θ=eta q_*Y^{-alpha}$ and $t_j=D_*S_j$, so
+$t_j=θ,t_{j-1}min{1,t_{j-1}}$. Then, whenever the quadratic
regime begins,
-\[
- t_{J_*}\leθ^{J_*}t_0
- \le4^{H+2}\eta^{J_*}Y^{2-\alpha J_*}\le1
-\]
-once $\eta$ is small enough. Require also
-$\eta\le\min\{\eta_0,1/4,e^{-1}\}$, with $\eta_0$ from
-Proposition~prop:multiscale. Now $\log(1/θ)\ge Q$ and
-$t_j\le1$ for $j\ge J_*$. Iterating $t_j=θ t_{j-1}^2$ in that
+[
+ t_{J_*}leθ^{J_*}t_0
+ le4^{H+2}eta^{J_*}Y^{2-alpha J_*}le1
+]
+once $eta$ is small enough. Require also
+$etalemin{eta_0,1/4,e^{-1}}$, with $eta_0$ from
+Proposition~prop:multiscale. Now $log(1/θ)ge Q$ and
+$t_jle1$ for $jge J_*$. Iterating $t_j=θ t_{j-1}^2$ in that
range gives
- t_h\leθ^{\,2^{h-J_*}-1}
- \le\exp\{-Q(2^{h-J_*}-1)\}\le n^{-1},\qquad S_h\le n^{-1}.
+ t_hleθ^{,2^{h-J_*}-1}
+ leexp{-Q(2^{h-J_*}-1)}le n^{-1},qquad S_hle n^{-1}.
-The last inequality uses $2^{h-J_*}\ge8Y$ and $Q(Y-1)=\log n$. At each
-transition $0<S_j\le\min\{1,\eta_0S_{j-1}\}$, as the proposition requires.
+The last inequality uses $2^{h-J_*}ge8Y$ and $Q(Y-1)=log n$. At each
+transition $0<S_jlemin{1,eta_0S_{j-1}}$, as the proposition requires.
-The cost of these small scales is also needed. Put $A_0=Q+\log Y$. With
-$\ell_j=\max\{0,\log(1/S_j)\}$ the recurrence gives
-\[
- \ell_j\le2\ell_{j-1}+\log(1/θ),\quad\ell_0=0,\quad
- \log(1/θ)\le CA_0,
-\]
+The cost of these small scales is also needed. Put $A_0=Q+log Y$. With
+$ell_j=max{0,log(1/S_j)}$ the recurrence gives
+[
+ ell_jle2ell_{j-1}+log(1/θ),quadell_0=0,quad
+ log(1/θ)le CA_0,
+]
hence
- \log(1/S_j)\le C2^jA_0\qquad(0\le j\le h).
+ log(1/S_j)le C2^jA_0qquad(0le jle h).
Now iterate Proposition~prop:multiscale. Start with the order-one
feature $f_i^{(0)}=X_i$ at decay $s_0=4$ and frequency zero. Prefix
-agreement gives $\Delta[X_i]=0$ for $i<d$ while $\abs{\Delta[X_d]}=1$;
-since $\abs{\Delta[X_i]}\le2$ for all $i$,
+agreement gives $Delta[X_i]=0$ for $i<d$ while $abs{Delta[X_d]}=1$;
+since $abs{Delta[X_i]}le2$ for all $i$,
- \abs{\mathcal S_{f^{(0)}}(4,0)}
- \ge1-2\sum_{r\ge1}e^{-4r}
- =1-(2)/(e^4-1)>\frac12 .
+ abs{mathcal S_calS_{f^{(0)}}(4,0)}
+ ge1-2sum_{rge1}e^{-4r}
+ =1-(2)/(e^4-1)>frac12 .
Use error exponents
- E_j=C_5A_0\mu^j\qquad(0\le j\le h),
+ E_j=C_5A_0mu^jqquad(0le jle h),
-with absolute $C_5$ chosen after $H$ and $\eta$.
+with absolute $C_5$ chosen after $H$ and $eta$.
The induction maintains this statement. At stage $j$ there is a feature
-$f^{(j)}$ of order $k_j\le4^j$ with
+$f^{(j)}$ of order $k_jle4^j$ with
- \beta S_j\le s_j\le S_j,\quad
- \abs{\omega_j}\le2.002s_j,\quad
- \abs{\mathcal S_{f^{(j)}}(s_j,\omega_j)}\ge e^{-E_j},\\
- \log H_{f^{(j)}}\le C4^jA_0.
+ beta S_jle s_jle S_j,quad
+ abs{omega_j}le2.002s_j,quad
+ abs{mathcal S_calS_{f^{(j)}}(s_j,omega_j)}ge e^{-E_j},
+ log H_{f^{(j)}}le C4^jA_0.
The base case follows from (eq:initial-signal) and $H_{f^{(0)}}=1$
-on increasing $C_5$. At the transition $j-1\to j$ the degree requirement
-is $4k_{j-1}\le4^j\le D_*\le D$. The proposition's height conclusion and
+on increasing $C_5$. At the transition $j-1to j$ the degree requirement
+is $4k_{j-1}le4^jle D_*le D$. The proposition's height conclusion and
(eq:scale-cost) give
-\[
- \log H_{f^{(j)}}\le4\log H_{f^{(j-1)}}+3\log(1+1/(\beta S_j))
- \le4\log H_{f^{(j-1)}}+C2^jA_0,
-\]
+[
+ log H_{f^{(j)}}le4log H_{f^{(j-1)}}+3log(1+1/(beta S_j))
+ le4log H_{f^{(j-1)}}+C2^jA_0,
+]
and summing this proves the height bound in (eq:scale-induction).
For the exponent hypothesis, the proposition's bracket is at most
-\[
+[
C(1+j4^{j-1}+4^{j-1}A_0+2^jA_0).
-\]
-Because $\mu>4$, the sequences $j(4/\mu)^{j-1}$ and $2^j/\mu^{j-1}$ are
+]
+Because $mu>4$, the sequences $j(4/mu)^{j-1}$ and $2^j/mu^{j-1}$ are
bounded. Thus one choice of $C_5$ makes this bracket times $C_0$ at most
$E_{j-1}$ for every $j$. All hypotheses of
Proposition~prop:multiscale hold, and its signal bound is
-$e^{-\mu E_{j-1}}=e^{-E_j}$.
+$e^{-mu E_{j-1}}=e^{-E_j}$.
-Write $f=f^{(h)}$, $k=k_h$, $s=s_h$, $\omega=\omega_h$. There is a signal
-of magnitude at least $e^{-E_h}$ with $k\le D_*$ and $s\le1/n$. The next
+Write $f=f^{(h)}$, $k=k_h$, $s=s_h$, $omega=omega_h$. There is a signal
+of magnitude at least $e^{-E_h}$ with $kle D_*$ and $sle1/n$. The next
step records where its parameters sit inside the unit disk; this controls
the cost of continuing the signal to the trace domain.
Call the parameters introduced between endpoints in any application of
Proposition~prop:multiscale elementary parameters, and
-include the final outer parameter $z_0=e^{-s+\ii\omega}$. An elementary
-parameter $e^{-a+\ii\theta}$ introduced at stage $j$ has $a\ge\beta S_j$
-and $\abs{\theta}\le C_1S_{j-1}$. Using (eq:scale-schedule), pick a
+include the final outer parameter $z_0=e^{-s+iiomega}$. An elementary
+parameter $e^{-a+iitheta}$ introduced at stage $j$ has $agebeta S_j$
+and $abs{theta}le C_1S_{j-1}$. Using (eq:scale-schedule), pick a
small absolute $c_2>0$ so that every elementary parameter obeys
- a\ge\gamma\min\{\abs{\theta},D_*\theta^2\},\qquad
- \gamma=c_2q_*Y^{-\alpha}\le1.
+ agegammamin{abs{theta},D_*theta^2},qquad
+ gamma=c_2q_*Y^{-alpha}le1.
-Indeed, the increasing map $u\mapsto u\min\{1,D_*u\}$ changes by at most
-a factor $\max\{C_1,C_1^2\}$ when $u$ becomes $C_1u$, so
-$c_2\le\beta\eta/\max\{C_1,C_1^2\}$ suffices; shrinking $c_2$ further
-covers $z_0$ since $\abs\omega\le2.002s$. Conjugating an elementary
+Indeed, the increasing map $umapsto umin{1,D_*u}$ changes by at most
+a factor $max{C_1,C_1^2}$ when $u$ becomes $C_1u$, so
+$c_2lebetaeta/max{C_1,C_1^2}$ suffices; shrinking $c_2$ further
+covers $z_0$ since $absomegale2.002s$. Conjugating an elementary
parameter preserves (eq:elementary-damping).
Each new inner gap holds at most one parameter from each of at most four
@@ -1205,84 +1205,84 @@ copies of the preceding feature, and at most one new elementary parameter;
count repeated factors with multiplicity, and conjugation does not change
the count. Starting with no inner parameters, the maximum number $N_j$ of
elementary factors in a stage-$j$ inner gap satisfies
-\[
- N_0=0,\qquad N_j\le4N_{j-1}+1,\qquad N_j\le(4^j-1)/(3).
-\]
+[
+ N_0=0,qquad N_jle4N_{j-1}+1,qquad N_jle(4^j-1)/(3).
+]
In particular an inner gap of $f$, even after multiplying by $z_0$,
-contains at most $(4^h+2)/3\le D_*$ elementary factors.
+contains at most $(4^h+2)/3le D_*$ elementary factors.
The next lemma explains why this count matters. The Cayley map
-$z\mapsto(1+z)/(1-z)$ sends the unit disk to the right half-plane, and a
+$zmapsto(1+z)/(1-z)$ sends the unit disk to the right half-plane, and a
quantitative distance from its boundary is needed.
[Products with controlled damping]
-Let $N\ge1$ be an integer and $0<\gamma\le1$. Suppose $z$ is a nonempty
-product of at most $N$ numbers $e^{-a_j+\ii\theta_j}$ with $a_j>0$,
-$\theta_j\in\RR$, and $a_j\ge\gamma\min\{\abs{\theta_j},N\theta_j^2\}$.
+Let $Nge1$ be an integer and $0<gammale1$. Suppose $z$ is a nonempty
+product of at most $N$ numbers $e^{-a_j+iitheta_j}$ with $a_j>0$,
+$theta_jinRR$, and $a_jgegammamin{abs{theta_j},Ntheta_j^2}$.
Then for some absolute $c_3>0$,
-\[
- \re(1+z)/(1-z)\ge c_3\gamma.
-\]
+[
+ re(1+z)/(1-z)ge c_3gamma.
+]
Proof.
-Write $z=e^{-T+\ii\psi}$ with $T=\sum_j a_j>0$ and
-$\psi=\sum_j\theta_j$. Let
-\[
- U=\sum_{\abs{\theta_j}\ge1/N}\abs{\theta_j},\qquad
- V=\sum_{\abs{\theta_j}<1/N}\abs{\theta_j}.
-\]
+Write $z=e^{-T+iipsi}$ with $T=sum_j a_j>0$ and
+$psi=sum_jtheta_j$. Let
+[
+ U=sum_{abs{theta_j}ge1/N}abs{theta_j},qquad
+ V=sum_{abs{theta_j}<1/N}abs{theta_j}.
+]
Summing the hypotheses and applying Cauchy--Schwarz to at most $N$ small
angles gives
-\[
- T\ge\gamma(U+N\!\!\sum_{\abs{\theta_j}<1/N}\theta_j^2)
- \ge\gamma(U+V^2).
-\]
-If $U\ge1$ or $V\ge1$ the bracket is at least one; otherwise
-$U+V^2\ge U^2+V^2\ge(U+V)^2/2$. Since $\abs\psi\le U+V$, in all cases
-\[
- T\ge\tfrac12\gamma\min\{1,\psi^2\}.
-\]
-For $0<T\le1$ this gives
-\[
- 1-\abs z^2\ge cT,\qquad
- \abs{1-z}^2\le C(T^2+\min\{1,\psi^2\})\le C'T/\gamma.
-\]
-Their ratio is $\re((1+z)/(1-z))$. If $T>1$, that ratio is at least
-$(1-e^{-2})/4$, which also suffices because $\gamma\le1$.
+[
+ Tgegamma(U+N!!sum_{abs{theta_j}<1/N}theta_j^2)
+ gegamma(U+V^2).
+]
+If $Uge1$ or $Vge1$ the bracket is at least one; otherwise
+$U+V^2ge U^2+V^2ge(U+V)^2/2$. Since $abspsile U+V$, in all cases
+[
+ Tgetfrac12gammamin{1,psi^2}.
+]
+For $0<Tle1$ this gives
+[
+ 1-abs z^2ge cT,qquad
+ abs{1-z}^2le C(T^2+min{1,psi^2})le C'T/gamma.
+]
+Their ratio is $re((1+z)/(1-z))$. If $T>1$, that ratio is at least
+$(1-e^{-2})/4$, which also suffices because $gammale1$.
Let $z_1,...,z_{k-1}$ be the inner gap parameters of the final feature
$f$. Define the polynomial in $k$ complex variables
F(Z_0,...,Z_{k-1})
- =\sum_{1\le i_1<...<i_k\le n}
- \Delta\Bigl[\prod_{r=1}^kX_{i_r}\Bigr]
- Z_0^{i_1}\prod_{r=1}^{k-1}Z_r^{i_{r+1}-i_r}.
+ =sum_{1le i_1<...<i_kle n}
+ DeltaBigl[prod_{r=1}^kX_{i_r}Bigr]
+ Z_0^{i_1}prod_{r=1}^{k-1}Z_r^{i_{r+1}-i_r}.
-At the point $Z_0^*=z_0$, $Z_r^*=z_rz_0$ $(1\le r<k)$, the powers of
+At the point $Z_0^*=z_0$, $Z_r^*=z_rz_0$ $(1le r<k)$, the powers of
$z_0$ telescope to $i_k$, so
- F(Z^*)=z_0^d\,\mathcal S_f(s,\omega),\qquad
- \abs{F(Z^*)}\ge e^{-E_h-1},
+ F(Z^*)=z_0^d,mathcal S_f(s,calS_f(s,omega),qquad
+ abs{F(Z^*)}ge e^{-E_h-1},
-where $sd\le nS_h\le1$. Each coordinate of $Z^*$ is a nonempty product of
+where $sdle nS_hle1$. Each coordinate of $Z^*$ is a nonempty product of
at most $D_*$ elementary parameters, so Lemma~lem:product-cayley
gives
- \re(1+Z_r^*)/(1-Z_r^*)\ge c_3\gamma\qquad(0\le r<k).
+ re(1+Z_r^*)/(1-Z_r^*)ge c_3gammaqquad(0le r<k).
A bound on the coordinates' collective closeness to the unit circle is also needed. If
-$z_r=e^{-a_r+\ii\theta_r}$, then
-\[
- \frac1{1-\abs{Z_r^*}}\le1+\frac1{a_r+s}\le1+\frac1{a_r}\ (r\ge1),
- \qquad \frac1{1-\abs{Z_0^*}}\le1+\frac1s,
-\]
-using $e^u\ge1+u$. The height bound in (eq:scale-induction) and
+$z_r=e^{-a_r+iitheta_r}$, then
+[
+ frac1{1-abs{Z_r^*}}le1+frac1{a_r+s}le1+frac1{a_r} (rge1),
+ qquad frac1{1-abs{Z_0^*}}le1+frac1s,
+]
+using $e^uge1+u$. The height bound in (eq:scale-induction) and
(eq:scale-cost) therefore give
- W_*:=\log\prod_{r=0}^{k-1}\frac1{1-\abs{Z_r^*}}
- \le\log H_f+\log(1+1/(\beta S_h))\le C4^hA_0.
+ W_*:=logprod_{r=0}^{k-1}frac1{1-abs{Z_r^*}}
+ lelog H_f+log(1+1/(beta S_h))le C4^hA_0.
Now express the same polynomial through observable trace moments on a
@@ -1290,17 +1290,17 @@ different domain. This is a multivariate trace generating identity from the smoo
literature on higher-order trace reconstruction, and the same counting fact
underlies the consecutive-block statistical-query work. The counting
argument is given directly. For a rank tuple
-$\ell=(\ell_1,...,\ell_k)$ with $1\le\ell_1<...<\ell_k\le n$ write
-$T_\ell=T_{\{\ell_1,...,\ell_k\}}$. For $u=(u_0,...,u_{k-1})\in\CC^k$
+$ell=(ell_1,...,ell_k)$ with $1leell_1<...<ell_kle n$ write
+$T_ell=T_{{ell_1,...,ell_k}}$. For $u=(u_0,...,u_{k-1})inCC^k$
set
-\[
- G(u)=\sum_{1\le\ell_1<...<\ell_k\le n}
- \Delta[T_\ell]\,
- u_0^{\ell_1-1}\prod_{r=1}^{k-1}u_r^{\ell_{r+1}-\ell_r-1}.
-\]
+[
+ G(u)=sum_{1leell_1<...<ell_kle n}
+ Delta[T_ell],
+ u_0^{ell_1-1}prod_{r=1}^{k-1}u_r^{ell_{r+1}-ell_r-1}.
+]
Then, with $Z_r=q+pu_r$, the exact identity is
- (\prod_{r=0}^{k-1}Z_r)G(u)=p^kF(Z).
+ (prod_{r=0}^{k-1}Z_r)G(u)=p^kF(Z).
To prove it, fix input positions $i_1<...<i_k$ that supply the
specified retained bits. Their own retention contributes $p^k$. Every input
@@ -1308,205 +1308,205 @@ position before $i_1$ contributes $q+pu_0$ according to whether it is
deleted or retained; there are $i_1-1$ such positions. Similarly, the
$i_{r+1}-i_r-1$ positions between $i_r$ and $i_{r+1}$ each contribute
$q+pu_r$, and positions after $i_k$ contribute a factor of one. The
-retained counts in these intervals are exactly $\ell_1-1$ and
-$\ell_{r+1}-\ell_r-1$. Summing over all input tuples and applying $\Delta$
+retained counts in these intervals are exactly $ell_1-1$ and
+$ell_{r+1}-ell_r-1$. Summing over all input tuples and applying $Delta$
gives $G(u)$ as $p^k$ times the polynomial of (eq:input-polynomial)
with every displayed gap exponent reduced by one; multiplying by
-$\prod_rZ_r$ proves (eq:trace-generating-identity), including points
+$prod_rZ_r$ proves (eq:trace-generating-identity), including points
where some $Z_r=0$.
Suppose for the moment that
- \abs{\Delta[T_I]}\leε
- \quadfor every I⊆\{1,...,n\} with 1\le\abs I\le D_*.
+ abs{Delta[T_I]}leε
+ quadfor every I⊆{1,...,n} with 1leabs Ile D_*.
-Since $k\le D_*$ the rank sum has at most $n^k$ terms, so
-$\abs{G(u)}\le n^kε$ when all $\abs{u_r}\le1$; on that polydisk
-$\abs{Z_r}\le q+p=1$. Thus (eq:trace-generating-identity) gives
+Since $kle D_*$ the rank sum has at most $n^k$ terms, so
+$abs{G(u)}le n^kε$ when all $abs{u_r}le1$; on that polydisk
+$abs{Z_r}le q+p=1$. Thus (eq:trace-generating-identity) gives
- \abs{F(Z)}\le(n/p)^kε
- \qquadwhenever \abs{Z_r-q}\le p\ (0\le r<k).
+ abs{F(Z)}le(n/p)^kε
+ qquadwhenever abs{Z_r-q}le p (0le r<k).
This uses only ordinary trace ranks and the convention that a rank
product is zero when the trace is too short.
The lower bound (eq:input-polynomial-signal) and the upper bound
(eq:trace-polydisk-bound) live at different points. In Cayley
-coordinates the disk $\abs{Z-q}\le p$ becomes a half-plane:
+coordinates the disk $abs{Z-q}le p$ becomes a half-plane:
- \abs{Z-q}\le p
- \iff \re(1+Z)/(1-Z)\ge\frac qp\qquad(\abs Z<1).
+ abs{Z-q}le p
+ iff re(1+Z)/(1-Z)gefrac qpqquad(abs Z<1).
Indeed, writing $w=(1+Z)/(1-Z)$ and expanding
-$\abs{pw-(1+q)}^2\le p^2\abs{w+1}^2$ reduces it to $4q\le4p\re w$. All
+$abs{pw-(1+q)}^2le p^2abs{w+1}^2$ reduces it to $4qle4pre w$. All
coordinates are moved into this half-plane with one complex variable. A single
strip interpolation avoids paying a separate continuation loss per
coordinate.
Write
-\[
- w_r=(1+Z_r^*)/(1-Z_r^*)=\xi_r+\ii\zeta_r,\qquad \xi_r\ge c_3\gamma,
-\]
-with $\xi_r,\zeta_r\in\RR$. For $v\in\CC$ set
+[
+ w_r=(1+Z_r^*)/(1-Z_r^*)=xi_r+iizeta_r,qquad xi_rge c_3gamma,
+]
+with $xi_r,zeta_rinRR$. For $vinCC$ set
- w_r(v)=(1+v)\xi_r+\ii\zeta_r,\quad
- Z_r(v)=(w_r(v)-1)/(w_r(v)+1),\quad
- R=1+(q)/(pc_3\gamma).
+ w_r(v)=(1+v)xi_r+iizeta_r,quad
+ Z_r(v)=(w_r(v)-1)/(w_r(v)+1),quad
+ R=1+(q)/(pc_3gamma).
A common real value of $v$ moves every starting coordinate into the trace
half-plane at once. Concretely, in the $Z$-plane the region
-$\abs{Z-q}\le p$ is a disk; after the Cayley map
-$w=(1+Z)/(1-Z)$ it becomes the half-plane $\re w\ge q/p$. Each path
-$w_r(v)=(1+v)\re w_r+\ii\,{\rm Im}\,w_r$ is a horizontal segment starting
-at $w_r$ and ending at $w_r(R)$ with $\re w_r(R)=(1+R)\xi_r\ge q/p$. All
-coordinates use the same parameter $0\le v\le R$; the proof interpolates
+$abs{Z-q}le p$ is a disk; after the Cayley map
+$w=(1+Z)/(1-Z)$ it becomes the half-plane $re wge q/p$. Each path
+$w_r(v)=(1+v)re w_r+ii,{rm Im},w_r$ is a horizontal segment starting
+at $w_r$ and ending at $w_r(R)$ with $re w_r(R)=(1+R)xi_rge q/p$. All
+coordinates use the same parameter $0le vle R$; the proof interpolates
$F(Z_0(v),...,Z_{k-1}(v))$ on one complex strip, simultaneously for
every coordinate (Figure~fig:simultaneous-cayley).
-\suppressfloats[t]
+suppressfloats[t]
[htbp]
-\centering
-[x=1.2cm,y=1.2cm,>=Stealth,font=\small]
+centering
+[x=1.2cm,y=1.2cm,>=Stealth,font=small]
- \fill[blue!4] (0,0) circle (1);
- \fill[teal!17] (0.4,0) circle (0.6);
- \draw[black!35,->] (-1.2,0)--(1.32,0) node[right] {$\re Z$};
- \draw[black!35,->] (0,-1.16)--(0,1.2);
- \draw[black!65] (0,0) circle (1);
- \draw[teal!65!black] (0.4,0) circle (0.6);
- \fill (0.4,0) circle (0.022) node[below right] {$q$};
- \node[below right] at (1,0) {$1$};
- \fill[blue!70!black] (-0.2027,0.5791) circle (0.027)
+ fill[blue!4] (0,0) circle (1);
+ fill[teal!17] (0.4,0) circle (0.6);
+ draw[black!35,->] (-1.2,0)--(1.32,0) node[right] {$re Z$};
+ draw[black!35,->] (0,-1.16)--(0,1.2);
+ draw[black!65] (0,0) circle (1);
+ draw[teal!65!black] (0.4,0) circle (0.6);
+ fill (0.4,0) circle (0.022) node[below right] {$q$};
+ node[below right] at (1,0) {$1$};
+ fill[blue!70!black] (-0.2027,0.5791) circle (0.027)
node[above left] {$Z_0^*$};
- \fill[orange!80!black] (-0.1753,-0.4310) circle (0.027)
+ fill[orange!80!black] (-0.1753,-0.4310) circle (0.027)
node[below left] {$Z_1^*$};
- \node at (0,1.5) {The $Z$-plane};
- \node at (0,-1.42) {$\abs{Z-q}\le p$ is shaded};
+ node at (0,1.5) {The $Z$-plane};
+ node at (0,-1.42) {$abs{Z-q}le p$ is shaded};
- \draw[->,thick] (1.68,0.22)--(3.78,0.22)
+ draw[->,thick] (1.68,0.22)--(3.78,0.22)
node[midway,above=5pt] {$w=(1+Z)/(1-Z)$};
[xshift=6cm]
- \fill[blue!4] (0,-1.1) rectangle (2.85,1.1);
- \fill[teal!17] (0.6667,-1.1) rectangle (2.85,1.1);
- \draw[black!35,->] (-0.15,0)--(3,0) node[right] {$\re w$};
- \draw[black!65,->] (0,-1.16)--(0,1.22);
- \draw[dashed,teal!65!black] (0.6667,-1.1)--(0.6667,1.1);
- \node[below right] at (0.6667,0) {$q/p$};
- \draw[->,thick,blue!70!black] (0.35,0.65)--(1.6333,0.65);
- \fill[blue!70!black] (0.35,0.65) circle (0.027)
+ fill[blue!4] (0,-1.1) rectangle (2.85,1.1);
+ fill[teal!17] (0.6667,-1.1) rectangle (2.85,1.1);
+ draw[black!35,->] (-0.15,0)--(3,0) node[right] {$re w$};
+ draw[black!65,->] (0,-1.16)--(0,1.22);
+ draw[dashed,teal!65!black] (0.6667,-1.1)--(0.6667,1.1);
+ node[below right] at (0.6667,0) {$q/p$};
+ draw[->,thick,blue!70!black] (0.35,0.65)--(1.6333,0.65);
+ fill[blue!70!black] (0.35,0.65) circle (0.027)
node[above left] {$w_0$};
- \fill[blue!70!black] (1.6333,0.65) circle (0.027)
+ fill[blue!70!black] (1.6333,0.65) circle (0.027)
node[above] {$w_0(R)$};
- \draw[->,thick,orange!80!black] (0.5,-0.55)--(2.3333,-0.55);
- \fill[orange!80!black] (0.5,-0.55) circle (0.027)
+ draw[->,thick,orange!80!black] (0.5,-0.55)--(2.3333,-0.55);
+ fill[orange!80!black] (0.5,-0.55) circle (0.027)
node[below left] {$w_1$};
- \fill[orange!80!black] (2.3333,-0.55) circle (0.027)
+ fill[orange!80!black] (2.3333,-0.55) circle (0.027)
node[below] {$w_1(R)$};
- \node at (1.4,1.5) {The $w$-plane};
- \node at (1.4,-1.42) {Both paths use the same $0\le v\le R$};
+ node at (1.4,1.5) {The $w$-plane};
+ node at (1.4,-1.42) {Both paths use the same $0le vle R$};
-\caption{Geometry of the simultaneous continuation. The Cayley map turns
-the deletion disk into $\re w\ge q/p$. Two possible starting coordinates
+caption{Geometry of the simultaneous continuation. The Cayley map turns
+the deletion disk into $re wge q/p$. Two possible starting coordinates
are shown; the horizontal arrows have different lengths but share the
-same parameter: $w_r(v)=(1+v)\re w_r+\iiImw_r$ for
-$0\le v\le R$. The proof interpolates
+same parameter: $w_r(v)=(1+v)re w_r+iiImw_r$ for
+$0le vle R$. The proof interpolates
$F(Z_0(v),...,Z_{k-1}(v))$ on one complex strip, simultaneously for
every coordinate.}
-On the closed strip $-1/2\le\re v\le R$ there holds
-$\re w_r(v)=(1+\re v)\xi_r>0$, so all $Z_r(v)$ lie in the unit disk and
+On the closed strip $-1/2lere vle R$ there holds
+$re w_r(v)=(1+re v)xi_r>0$, so all $Z_r(v)$ lie in the unit disk and
depend analytically on $v$; at $v=0$ they equal $Z_r^*$. On the right
-boundary $\re w_r(v)=(1+R)\xi_r\ge q/p$, so (eq:trace-polydisk-bound)
+boundary $re w_r(v)=(1+R)xi_rge q/p$, so (eq:trace-polydisk-bound)
applies. Moreover
- 1+R\le C(Y^\alpha)/(p),
+ 1+Rle C(Y^alpha)/(p),
-because $\gamma=c_2q_*Y^{-\alpha}$ and $q\le q_*$.
+because $gamma=c_2q_*Y^{-alpha}$ and $qle q_*$.
A left-boundary bound free of $n^k$ is needed. The identity
-$1-\abs Z^2=4\re w/\abs{w+1}^2$ gives, at $\re v=-1/2$,
+$1-abs Z^2=4re w/abs{w+1}^2$ gives, at $re v=-1/2$,
- 1-\abs{Z_r(v)}\ge\frac{1-\abs{Z_r^*}}{4(1+\abs v)^2}.
+ 1-abs{Z_r(v)}gefrac{1-abs{Z_r^*}}{4(1+abs v)^2}.
To check the constant, first use
-$1-\abs{Z_r(v)}\ge\xi_r/\abs{w_r(v)+1}^2$ and
-$1-\abs{Z_r^*}\le4\xi_r/\abs{w_r+1}^2$, then
-$\abs{w_r(v)+1}\le\abs{w_r+1}+\xi_r\abs v\le(1+\abs v)\abs{w_r+1}$.
+$1-abs{Z_r(v)}gexi_r/abs{w_r(v)+1}^2$ and
+$1-abs{Z_r^*}le4xi_r/abs{w_r+1}^2$, then
+$abs{w_r(v)+1}leabs{w_r+1}+xi_rabs vle(1+abs v)abs{w_r+1}$.
The coefficients of $F$ have magnitude at most two, so summing independent
geometric series over the positive prefix and gap lengths gives, in the
open unit polydisk,
-\[
- \abs{F(Z)}\le2\prod_{r=0}^{k-1}\frac1{1-\abs{Z_r}}.
-\]
+[
+ abs{F(Z)}le2prod_{r=0}^{k-1}frac1{1-abs{Z_r}}.
+]
It follows from (eq:left-boundary-distance) that
-\[
- g(v):=\frac{F(Z_0(v),...,Z_{k-1}(v))}{(2+v)^{2k}}
-\]
+[
+ g(v):=frac{F(Z_0(v),...,Z_{k-1}(v))}{(2+v)^{2k}}
+]
has the boundary bounds
- \log\abs{g(v)}&\le W_*+Ck+C &&(\re v=-1/2),\\
- \log\abs{g(v)}&\le k\log(n/p)+\logε &&(\re v=R).
+ logabs{g(v)}&le W_*+Ck+C &&(re v=-1/2),
+ logabs{g(v)}&le klog(n/p)+logε &&(re v=R).
-On the left $(1+\abs v)/\abs{2+v}$ is bounded absolutely; on the right
-$\abs{2+v}\ge2+R>1$. The function $g$ is analytic and bounded on the strip,
-since $F$ is a finite polynomial on the unit polydisk and $\abs{2+v}\ge3/2$.
+On the left $(1+abs v)/abs{2+v}$ is bounded absolutely; on the right
+$abs{2+v}ge2+R>1$. The function $g$ is analytic and bounded on the strip,
+since $F$ is a finite polynomial on the unit polydisk and $abs{2+v}ge3/2$.
The strip estimate used here is a consequence of the maximum principle. If a
-bounded analytic function on $a\le\re v\le b$ has logarithmic boundary
-bounds $L_-,L_+$, then at real $t\in[a,b]$ its logarithm is at most
+bounded analytic function on $alere vle b$ has logarithmic boundary
+bounds $L_-,L_+$, then at real $tin[a,b]$ its logarithm is at most
$((b-t)L_-+(t-a)L_+)/(b-a)$. Multiply by the exponential of the negative
affine interpolant to make both sides one; on rectangles of height tending
to infinity apply the maximum principle after multiplying by
-$\exp\{\tau(v^2-\max\{a^2,b^2\})\}$ for $\tau>0$, which has modulus at most
+$exp{tau(v^2-max{a^2,b^2})}$ for $tau>0$, which has modulus at most
one on the vertical sides and tends to zero on the horizontal sides. Let
-the height tend to infinity, then $\tau\to0$; this proves the estimate,
+the height tend to infinity, then $tauto0$; this proves the estimate,
including zeros of the function by continuity.
At $t=0$ the right boundary has weight $1/(2R+1)$. By
-(eq:input-polynomial-signal), $\log\abs{g(0)}\ge-E_h-1-2k\log2$.
+(eq:input-polynomial-signal), $logabs{g(0)}ge-E_h-1-2klog2$.
Substituting (eq:strip-side-bounds) gives
- \log(1/ε)
- &\le k\log(n/p)+(2R+1)(E_h+1+2k\log2)+2R(W_*+Ck+C)\notag\\
- &\le k\log(n/p)+C(1+R)(E_h+4^hA_0).
+ log(1/ε)
+ &le klog(n/p)+(2R+1)(E_h+1+2klog2)+2R(W_*+Ck+C)notag
+ &le klog(n/p)+C(1+R)(E_h+4^hA_0).
This is the quantitative transfer from input coordinates to trace
coordinates.
It remains to compare this exponent with the budget (eq:budget). Put
-$t=\log_2\mu$. The choices in (eq:separation-cutoff) give
-$t+\alpha<3$, and
-\[
- 4^h\le CY^2,\qquad E_h\le CA_0Y^t,\qquad
- A_0\le Q(1+\log Y),\qquad QY\le C\log n.
-\]
+$t=log_2mu$. The choices in (eq:separation-cutoff) give
+$t+alpha<3$, and
+[
+ 4^hle CY^2,qquad E_hle CA_0Y^t,qquad
+ A_0le Q(1+log Y),qquad QYle Clog n.
+]
Thus (eq:strip-width) yields
(1+R)E_h
- &\le\frac Cp(\log n)Y^{t+\alpha-1}(1+\log Y)
- \le\frac Cp(\log n)Y^2(1+\log Y),\\
+ &lefrac Cp(log n)Y^{t+alpha-1}(1+log Y)
+ lefrac Cp(log n)Y^2(1+log Y),
(1+R)4^hA_0
- &\le\frac Cp(\log n)Y^{1+\alpha}(1+\log Y)
- \le\frac Cp(\log n)Y^2(1+\log Y).
-
-Also, since $n\ge2$ and $\log(1/p)\le1/p$,
-\[
- k\log(n/p)\le CY^2(\log n+\log(1/p))
- \le\frac Cp(\log n)Y^2.
-\]
+ &lefrac Cp(log n)Y^{1+alpha}(1+log Y)
+ lefrac Cp(log n)Y^2(1+log Y).
+
+Also, since $nge2$ and $log(1/p)le1/p$,
+[
+ klog(n/p)le CY^2(log n+log(1/p))
+ lefrac Cp(log n)Y^2.
+]
Using (eq:aux-comparisons-2), the right-hand side of
(eq:separation-exponent) is at most
- \frac Cp(\log n)Y^2(1+\log Y)
- \le C_6\,p^{-1}(\log n)\mathsf X(n,p)^2
- (1+\log\mathsf X(n,p))^6
- =C_6\,B(n,p).
+ frac Cp(log n)Y^2(1+log Y)
+ le C_6,p^{-1}(log n)mathsf X(n,p)^2sfX(n,p)^2
+ (1+logmathsf X(n,p))^6sfX(n,p))^6
+ =C_6,B(n,p).
Taking $ε=e^{-C_{sep}B(n,p)}$ with
$C_{sep}>C_6$ contradicts (eq:separation-exponent), so
@@ -1514,10 +1514,10 @@ $C_{sep}>C_6$ contradicts (eq:separation-exponent), so
Theorem~thm:separation with the cutoff (eq:separation-cutoff).
All constants were fixed in this order: those of
-Proposition~prop:multiscale; then $J_*,H,\eta$; then $C_5,c_2$; and
+Proposition~prop:multiscale; then $J_*,H,eta$; then $C_5,c_2$; and
finally $C_{sep}$. Nothing depends on the denominator of $p$ or
on how close $p$ is to an endpoint. In particular, when $q<1/n$ the cutoff
-$q_*=1/n$ keeps $Y<2$ and $Q=1+\log n$, while $q/q_*\le1$ keeps the strip
+$q_*=1/n$ keeps $Y<2$ and $Q=1+log n$, while $q/q_*le1$ keeps the strip
estimate uniform; when $p$ is small the strip width is controlled by the
single factor $p^{-1}$ already in the budget. The case $p=1$ is handled
directly in the decoding argument.
@@ -1532,75 +1532,75 @@ condition. That slack is exhibited first, then a rational procedure is given
that is guaranteed to find a feasible system whenever slack exists.
Assume for now $p<1$. Choose an integer $D$ and a positive dyadic number
-$\zeta=2^{-b_\zeta}$ so that
+$zeta=2^{-b_zeta}$ so that
- D_*\le D\le CY^2,\quad
- \zeta\le\tfrac14,\quad
- 4\zeta<e^{-C_{sep}B(n,p)},\quad
- b_\zeta=O(B(n,p)),
+ D_*le Dle CY^2,quad
+ zetaletfrac14,quad
+ 4zeta<e^{-C_{sep}B(n,p)},quad
+ b_zeta=O(B(n,p)),
with $C_{sep}$ from Theorem~thm:separation. Integer
formulas for these choices come later. Let
-\[
- \mathcal I_D=\{I⊆\{1,...,n\}:1\le|I|\le D\},\quad
- N=|\mathcal I_D|,\quad
- T=512(1+N)\zeta^{-2},
-\]
+[
+ mathcal I_D=calI_D={I⊆{1,...,n}:1le|I|le D},quad
+ N=|mathcal I_D|,calI_D|,quad
+ T=512(1+N)zeta^{-2},
+]
which is an integer. From $T$ independent traces form the empirical mean
-$\widehat t_I$ of the rank observable defining $T_I$, for every
-$I\in\mathcal I_D$. These are ordinary trace observables: a requested
+$widehatwidehat t_I$t_I$ of the rank observable defining $T_I$, for every
+$Iinmathcal I_D$.calI_D$. These are ordinary trace observables: a requested
rank that is absent contributes zero.
-For each fixed input $x$ let $\mathcal G$ be the event
+For each fixed input $x$ let $mathcal G$calG$ be the event
- |\widehat t_I-T_I(x)|\le\zeta/8\quadfor every I\in\mathcal I_D.
+ |widehatwidehat t_I-T_I(x)|t_I-T_I(x)|lezeta/8quadfor every Iinmathcal I_D.calI_D.
-Each observable takes values in $\{0,1\}$, so its empirical mean has
+Each observable takes values in ${0,1}$, so its empirical mean has
variance at most $1/(4T)$. Chebyshev and a union bound give
- \mathbb P_x(\mathcal G^c)\le(16N)/(T\zeta^2)\le\frac1{32}.
+ mathbb P_x(bbP_x(mathcal G^c)calG^c)le(16N)/(Tzeta^2)lefrac1{32}.
The same event controls all the prefix tests, even though the prefixes are
chosen using the data.
For a proposed prefix, use the remaining sign variables and their moment
-matrix $M(m)=\mathcal M_D(m)$ from (eq:moment-matrix); its nonempty
+matrix $M(m)=mathcal M_D(m)$calM_D(m)$ from (eq:moment-matrix); its nonempty
moment coordinates form a real vector $m$. Consider the rational
constraints
- m\in[-1,1]^l,\quad M(m)\succeq0,\quad
- |\EE_m[T_I]-\widehat t_I|\le\zeta\quad(I\in\mathcal I_D),
+ min[-1,1]^l,quad M(m)succeq0,quad
+ |EE_m[T_I]-widehatwidehat t_I|t_I|lezetaquad(Iinmathcal I_D),calI_D),
where $l$ is the number of moment coordinates. Substituting the prefix and
writing each remaining bit as one half of one plus its sign makes
-$\EE_m[T_I]$ an affine function of $m$; its nonconstant coefficients have
+$EE_m[T_I]$ an affine function of $m$; its nonconstant coefficients have
absolute sum at most one by (eq:coeff-bounds).
-On $\mathcal G$, a prefix whose first incorrect bit is at its last
+On $mathcal G$,calG$, a prefix whose first incorrect bit is at its last
position has no solution to (eq:prefix-feasibility). Every correct
-prefix has a moment vector $m^*\in[-1,1]^l$ such that, with $G$ the
+prefix has a moment vector $m^*in[-1,1]^l$ such that, with $G$ the
dimension of its moment matrix,
- M(m^*)\succeq(\zeta/8)Id_G,\qquad
- |\EE_{m^*}[T_I]-\widehat t_I|\le\zeta/2\quad(I\in\mathcal I_D).
+ M(m^*)succeq(zeta/8)Id_G,qquad
+ |EE_{m^*}[T_I]-widehatwidehat t_I|t_I|lezeta/2quad(Iinmathcal I_D).calI_D).
Proof.
If a wrong prefix had a feasible moment vector, then
(eq:empirical-good) and (eq:prefix-feasibility) would give
-$|\EE_m[T_I]-T_I(x)|\le9\zeta/8$ for all $I\in\mathcal I_D$, contradicting
+$|EE_m[T_I]-T_I(x)|le9zeta/8$ for all $Iinmathcal I_D$,calI_D$, contradicting
Theorem~thm:separation and (eq:decoder-parameters).
For a correct prefix, mix the moments of the true completion with weight
-$1-\zeta/8$ and the moments of independent uniform signs on the remaining
-positions with weight $\zeta/8$. The latter moment matrix is the identity:
+$1-zeta/8$ and the moments of independent uniform signs on the remaining
+positions with weight $zeta/8$. The latter moment matrix is the identity:
its $(I,J)$ entry is zero unless $I=J$. The mixture has the stated matrix
lower bound and coordinates in $[-1,1]$. Both constituent distributions
give channel expectations in $[0,1]$, so mixing changes each $T_I(x)$ by
-at most $\zeta/8$; (eq:empirical-good) then bounds the discrepancy
-from $\widehat t_I$ by $\zeta/4$, stronger than required.
+at most $zeta/8$; (eq:empirical-good) then bounds the discrepancy
+from $widehatwidehat t_I$t_I$ by $zeta/4$, stronger than required.
The next lemma gives the bounded procedure for
@@ -1612,31 +1612,31 @@ Motzkin--Schoenberg; the iteration and its rounding analysis are given
explicitly.
[Rational feasibility with slack]
-Let $r\ge0$ and $D\ge1$ be integers, and let $m$ consist of the
-\[
- l=\sum_{j=1}^{\min(2D,r)}\binom rj
-\]
+Let $rge0$ and $Dge1$ be integers, and let $m$ consist of the
+[
+ l=sum_{j=1}^{min(2D,r)}binom rj
+]
nonempty sign moments. Let $M(m)$ be their moment matrix, of dimension
-\[
- G=\sum_{j=0}^{\min(D,r)}\binom rj.
-\]
-Suppose a finite list of rational affine inequalities $\ell_t(m)\le b_t$
-is given, each with nonconstant coefficient vector of $\ell^1$-norm at
-most one, together with a dyadic $0<\zeta\le1/4$. Then a deterministic
+[
+ G=sum_{j=0}^{min(D,r)}binom rj.
+]
+Suppose a finite list of rational affine inequalities $ell_t(m)le b_t$
+is given, each with nonconstant coefficient vector of $ell^1$-norm at
+most one, together with a dyadic $0<zetale1/4$. Then a deterministic
rational procedure either returns a vector with
-\[
- m\in[-1,1]^l,\quad M(m)\succeq0,\quad \ell_t(m)\le b_t\ for all t,
-\]
-or reports failure. It returns a vector whenever some $m^*\in[-1,1]^l$
+[
+ min[-1,1]^l,quad M(m)succeq0,quad ell_t(m)le b_t for all t,
+]
+or reports failure. It returns a vector whenever some $m^*in[-1,1]^l$
satisfies
-\[
- M(m^*)\succeq(\zeta/8)Id_G,\quad
- \ell_t(m^*)\le b_t-\zeta/2\ for all t.
-\]
-Put $\lambda=\zeta/(16G)$. The procedure performs at most
-$\lceil10(1+l)/\lambda^2\rceil$ feasibility checks, each taking polynomially
+[
+ M(m^*)succeq(zeta/8)Id_G,quad
+ ell_t(m^*)le b_t-zeta/2 for all t.
+]
+Put $lambda=zeta/(16G)$. The procedure performs at most
+$lceil10(1+l)/lambda^2rceil$ feasibility checks, each taking polynomially
many bit operations in the list dimensions, the input bit lengths, and
-$\log(1/\zeta)$. These time bounds hold without the slack assumption.
+$log(1/zeta)$. These time bounds hold without the slack assumption.
Proof.
@@ -1644,85 +1644,85 @@ If $l=0$ the matrix is $[1]$ and all inequalities are constant, so direct
rational comparisons suffice. Suppose $l>0$. Exact separating
directions, then an iteration with bounded precision.
-{\itshape Directions from violated constraints.}
-At a current point $m^{cur}\in[-1,1]^l$ first check the affine
+{itshape Directions from violated constraints.}
+At a current point $m^{cur}in[-1,1]^l$ first check the affine
inequalities and the matrix condition exactly. If all hold, return the
-current vector. A violated inequality $\ell_t(m^{cur})>b_t$ gives
+current vector. A violated inequality $ell_t(m^{cur})>b_t$ gives
the direction $a$ equal to minus its nonconstant coefficient vector. Then
-$\norm a_2\le1$ and, under the slack assumption,
-$< m^*-m^{cur},a>\ge\zeta/2\ge\lambda$.
+$norm a_2le1$ and, under the slack assumption,
+$< m^*-m^{cur},a>gezeta/2gelambda$.
If instead the matrix condition fails, find a nonzero rational vector $v$
-with $v^\top M(m^{cur})v<0$ (constructed just below). Write the
-associated affine form as $v^\top M(m)v=c+h\cdot m$ and set
-$a=h/(G\,v^\top v)$. Each matrix entry is the constant $1$ or one moment
-coordinate, so $\norm h_1\le\norm v_1^2\le G\norm v_2^2$ and
-$\norm a_2\le1$. The slack matrix bound gives
-\[
+with $v^top M(m^{cur})v<0$ (constructed just below). Write the
+associated affine form as $v^top M(m)v=c+hcdot m$ and set
+$a=h/(G,v^top v)$. Each matrix entry is the constant $1$ or one moment
+coordinate, so $norm h_1lenorm v_1^2le Gnorm v_2^2$ and
+$norm a_2le1$. The slack matrix bound gives
+[
< m^*-m^{cur},a>
- =\frac{v^\top(M(m^*)-M(m^{cur}))v}{G\,v^\top v}
- \ge(\zeta)/(8G)\ge\lambda.
-\]
+ =frac{v^top(M(m^*)-M(m^{cur}))v}{G,v^top v}
+ ge(zeta)/(8G)gelambda.
+]
Thus either violated condition yields a rational direction with
- \norm a_2\le1,\qquad < m^*-m^{cur},a>\ge\lambda
+ norm a_2le1,qquad < m^*-m^{cur},a>gelambda
whenever a slack solution exists. A violated constant inequality has zero
direction; that cannot happen in the slack case, and otherwise the
iteration is still bounded.
-{\itshape An exact matrix check.}
+{itshape An exact matrix check.}
For a rational symmetric matrix, eliminate a positive diagonal pivot via
its Schur complement. Congruence by an invertible triangular matrix shows
the original is positive semidefinite exactly when the positive pivot and
this Schur complement are. A negative diagonal gives a negative unit
vector. If a remaining diagonal at $i$ is zero and its entry $b$ in column
-$j$ is nonzero, let $c$ be the diagonal at $j$; assume $c\ge0$,
+$j$ is nonzero, let $c$ be the diagonal at $j$; assume $cge0$,
else a negative diagonal already exists. Then
-\[
+[
y=e_i-(b)/(1+c+|b|)e_j
-\]
+]
has quadratic form
$(b^2)/((1+c+|b|)^2)[-2(1+c+|b|)+c]<0$. Otherwise continue
with a positive diagonal pivot; when none remain, all entries are zero and
the remainder is positive semidefinite. A negative vector of a Schur
complement lifts to one of the original matrix using only rational
arithmetic: with $P$ the eliminated block and $B$ its block against the
-remaining indices, the lift of $y$ is $v=(-P^{-1}By,\,y)$, whose quadratic
+remaining indices, the lift of $y$ is $v=(-P^{-1}By,,y)$, whose quadratic
form equals that of $y$ for the Schur complement.
These operations have polynomial bit complexity. Clear denominators in a
rational matrix with $t$-bit entries; the product of all denominators has
at most $G^2t$ bits. If $H$ is the resulting integer matrix with $s$-bit
entries, the determinant expansion bounds every $j$-row minor by
-$j!\,2^{js}$, of bit length $O(G(s+\log G))$. Schur-complement entries are
+$j!,2^{js}$, of bit length $O(G(s+log G))$. Schur-complement entries are
ratios of minors, and the inverse formula for $P$ gives the same bound for
the lift. Fraction-reduced elimination with rational comparisons computes
the check and the negative vector in polynomially many bit operations,
however small a positive pivot or negative quadratic form may be.
-{\itshape Iteration and rounding.}
+{itshape Iteration and rounding.}
Start with $m^{cur}=0$. Choose the dyadic grid with mesh
-\[
- \eta=2^{-g},\qquad
- g=\Bigl\lceil\log_2(100(1+l))/(\lambda^2)\Bigr\rceil.
-\]
+[
+ eta=2^{-g},qquad
+ g=Bigllceillog_2(100(1+l))/(lambda^2)Bigrrceil.
+]
The integer $g$ is computed by rational comparisons with powers of two, so
-$\eta\le\lambda^2/(100(1+l))$, with $O(1+\log((1+l)/\lambda))$ fractional
-bits. After an unsuccessful check, add $\lambda a$, clip every coordinate
+$etalelambda^2/(100(1+l))$, with $O(1+log((1+l)/lambda))$ fractional
+bits. After an unsuccessful check, add $lambda a$, clip every coordinate
to $[-1,1]$, and round to the nearest grid point in the box (both box
endpoints lie on the grid). Repeat up to the stated number of checks,
reporting failure if none accepts.
Under the slack assumption (eq:feasibility-direction) gives
-\[
- \norm{m^{cur}+\lambda a-m^*}_2^2
- \le\norm{m^{cur}-m^*}_2^2-\lambda^2.
-\]
+[
+ norm{m^{cur}+lambda a-m^*}_2^2
+ lenorm{m^{cur}-m^*}_2^2-lambda^2.
+]
Clipping cannot increase distance to $m^*$; since both the clipped vector
and $m^*$ lie in the box, rounding changes the squared distance by at most
-$4l\eta+l\eta^2<\lambda^2/2$. Each failed check thus decreases squared
-distance by at least $\lambda^2/2$, whose initial value is at most $l$, so
+$4leta+leta^2<lambda^2/2$. Each failed check thus decreases squared
+distance by at least $lambda^2/2$, whose initial value is at most $l$, so
acceptance occurs before the cap. Finally, rounding keeps every iterate on
a single grid of the indicated bit length; the exact matrix check, the
affine comparisons, forming and normalizing $h$, and one update involve
@@ -1749,11 +1749,11 @@ The choices in (eq:decoder-parameters) are now realised using only the
binary input. Continue to assume $p<1$ and use $q_*,Q,Y$ from
(eq:aux-parameters). Define the positive integers
- u&=1+\lceil\log_2 n\rceil, &
- v&=1+\lfloor\log_2(1/q_*)\rfloor,\\
- y&=1+\lceil u/v\rceil, &
- w&=1+\lceil\log_2 y\rceil, &
- a_p&=\lceil1/p\rceil.
+ u&=1+lceillog_2 nrceil, &
+ v&=1+lfloorlog_2(1/q_*)rfloor,
+ y&=1+lceil u/vrceil, &
+ w&=1+lceillog_2 yrceil, &
+ a_p&=lceil1/prceil.
All logarithms here mean integer length comparisons. In particular the
integer part of a logarithm of a rational number is found by comparing its
@@ -1761,32 +1761,32 @@ numerator with power-of-two multiples of its denominator, and the rational
maximum defining $q_*$ is found by an integer comparison.
The elementary comparisons
-\[
- u\asymp\log n,\quad v\asymp Q,\quad Y\le y\le C Y,\quad
- w\asymp1+\log Y,\quad p^{-1}\le a_p\le2p^{-1}
-\]
-have absolute constants. For the bound on $y$ use $u\ge(\log n)/\log2$
-and $v\le Q/\log2$. With $Y\asymp\mathsf X(n,p)$ these give
+[
+ uasymplog n,quad vasymp Q,quad Yle yle C Y,quad
+ wasymp1+log Y,quad p^{-1}le a_ple2p^{-1}
+]
+have absolute constants. For the bound on $y$ use $uge(log n)/log2$
+and $vle Q/log2$. With $Yasympmathsf X(n,p)$sfX(n,p)$ these give
- B(n,p)\le a_p\,u\,y^2\,w^6\le C\,B(n,p).
+ B(n,p)le a_p,u,y^2,w^6le C,B(n,p).
Choose fixed positive integers $K_D,K_m$ large enough for the constants in
Theorem~thm:separation and set
- D=K_Dy^2,\qquad b_\zeta=K_m\,a_p\,u\,y^2\,w^6,\qquad
- \zeta=2^{-b_\zeta}.
+ D=K_Dy^2,qquad b_zeta=K_m,a_p,u,y^2,w^6,qquad
+ zeta=2^{-b_zeta}.
-Since $D_*\le C Y^2$ and $B(n,p)\ge\log2$, these ensure all of
-(eq:decoder-parameters). The integers $D,b_\zeta$ in binary are
+Since $D_*le C Y^2$ and $B(n,p)gelog2$, these ensure all of
+(eq:decoder-parameters). The integers $D,b_zeta$ in binary are
computable in time polynomial in the input bit lengths; writing the
-denominator $2^{b_\zeta}$ then costs $O(b_\zeta)$ bits, included below.
+denominator $2^{b_zeta}$ then costs $O(b_zeta)$ bits, included below.
-Next form the rational constraints. For $I=\{l_1<...<l_k\}$ and
+Next form the rational constraints. For $I={l_1<...<l_k}$ and
$i_1<...<i_k$ the coefficient of $X_{i_1}... X_{i_k}$ in $T_I$ is
- \binom{i_1-1}{l_1-1}
- \prod_{j=1}^{k-1}
- \binom{i_{j+1}-i_j-1}{l_{j+1}-l_j-1}\,
+ binom{i_1-1}{l_1-1}
+ prod_{j=1}^{k-1}
+ binom{i_{j+1}-i_j-1}{l_{j+1}-l_j-1},
p^{l_k}q^{i_k-l_k},
when all gap counts are possible and zero otherwise. Indeed the selected
@@ -1799,33 +1799,33 @@ If the input representation is $p=a/b$, every denominator in
(eq:explicit-channel-coefficient) divides $b^n$. Binomial
coefficients have $O(n)$ bits. Sign expansion adds a denominator dividing
$2^D$, and empirical means have denominator $T$. Hence the constraint
-coefficients have polynomial bit length in $n+L+D+\log T$. Enumerating the
+coefficients have polynomial bit length in $n+L+D+log T$. Enumerating the
rank and input tuples, expanding the resulting bit monomials, and
substituting the prefix constructs them directly; no list of complete
trace masks is needed.
These enumerations fit the resource scale. From $D=O(Y^2)$,
-$Y\asymp\mathsf X(n,p)$, and (eq:budget),
+$Yasympmathsf X(n,p)$,sfX(n,p)$, and (eq:budget),
- D\log n=O(B(n,p)).
+ Dlog n=O(B(n,p)).
The rank list, every matrix basis, and every moment list have size at most
-$(n+1)^{2D}=\exp(O(B(n,p)))$, and so do input tuple lists. A sign expansion
+$(n+1)^{2D}=exp(O(B(n,p)))$, and so do input tuple lists. A sign expansion
has at most $2^D$ terms; products of these enumerations remain
-$\exp(O(B(n,p)))$. Subsets larger than the available positions are simply
+$exp(O(B(n,p)))$. Subsets larger than the available positions are simply
absent, so this includes $D>n$ and prefixes with very few remaining
positions.
-By (eq:decoder-parameters), $T$ is $\exp(O(B(n,p)))$. For every
+By (eq:decoder-parameters), $T$ is $exp(O(B(n,p)))$. For every
prefix the iteration cap of Lemma~lem:rational-feasibility is at most
-\[
- 2560(1+l)G^2\zeta^{-2}+1=\exp(O(B(n,p))),
-\]
-and the bit length of its grid is $O(b_\zeta+\log G+\log(1+l))=O(B(n,p))$.
+[
+ 2560(1+l)G^2zeta^{-2}+1=exp(O(B(n,p))),
+]
+and the bit length of its grid is $O(b_zeta+log G+log(1+l))=O(B(n,p))$.
The coefficient bounds and that lemma bound all computation by a fixed
polynomial in
- n+L+\exp(O(B(n,p))),
+ n+L+exp(O(B(n,p))),
including parameter and constraint construction, exact rational
arithmetic, and the at most $2n$ prefix tests.
@@ -1840,15 +1840,15 @@ For $p<1$, use (eq:integer-decoder-parameters), request the $T$
traces, and compute the empirical statistics. Run the prefix rule. This
defines the machine on every stream of well-formed traces.
-On $\mathcal G$, induction on prefix length shows precisely the correct
+On $mathcal G$,calG$, induction on prefix length shows precisely the correct
extension accepts: the incorrect extension is infeasible by
Lemma~lem:prefix-slack, and the correct one has the slack that
-guarantees acceptance. Thus the output is $x$ on $\mathcal G$, whose
+guarantees acceptance. Thus the output is $x$ on $mathcal G$,calG$, whose
probability is at least $31/32>2/3$ for each fixed $x$.
Each trace has at most $n$ bits and one delimiter. Reading the traces and
computing each of the finitely many rank observables costs polynomially
-many bit operations in $n+T+N$; the empirical counters have $O(\log T)$
+many bit operations in $n+T+N$; the empirical counters have $O(log T)$
bits. All loops, including all feasibility tests, have the input-dependent
caps already given. Thus (eq:decoder-total-resource) bounds every
well-formed trace stream, independently of the success event, and the
@@ -1859,14 +1859,14 @@ on tapes; replacing an indexed lookup by a scan costs at most a polynomial
factor in their total encoded length. Integer addition, multiplication,
division, and fraction reduction have elementary polynomial-time multitape
implementations. The bit-length and loop bounds thus give absolute
-constants $K,d$ with trace count at most $\exp(KB(n,p))$ and total bit cost
-at most $(n+L+\exp(KB(n,p)))^d$, after enlarging $K,d$ to absorb fixed
-factors. This enlargement is uniform because $B(n,p)\ge\log2$. Choose
-absolute $C,c$ with $C\ge K$ and $c\ge d$; both the sample bound $M_C(n,p)$
+constants $K,d$ with trace count at most $exp(KB(n,p))$ and total bit cost
+at most $(n+L+exp(KB(n,p)))^d$, after enlarging $K,d$ to absorb fixed
+factors. This enlargement is uniform because $B(n,p)gelog2$. Choose
+absolute $C,c$ with $Cge K$ and $cge d$; both the sample bound $M_C(n,p)$
and the runtime $(n+L+M_C(n,p))^c$ follow. All constants in parameter
selection are fixed integers in one finite program.
-\par
+par