text:
\section{\bf Introduction}\label{intro}

Let $I\subset S=K[\bx]=K[x_1,\dots,x_n]$ be a monomial ideal and let $M$ be the set of {\it standard monomials}, that is, the monomials $\bx^{\bk}=x_1^{k_1}\cdots x_n^{k_n}\in K[\bx]$ that are {\it not} in $I$.  Each coset in $S/I$ has exactly one representative as a {\it standard polynomial} in $\Span(M)$, and $S/I$ is isomorphic to $\Span(M)$ as a vector space over $K$, although the ring structure of $S/I$ is lost.  A {\it \bld} of $M$  (defined precisely in Section \ref{block}) is an expression for $M$ as a disjoint union of rectangular {\it blocks} of standard monomials, regarded as points in Newton space.  A block will be called a {\it Stanley block} if its span is a Stanley space; a \bld\ of $A$ into Stanley blocks is equivalent to a \sd\ of $S/I$.

In this paper and its sequel we study algorithms leading from $I$ to a \bld\ for $M$, emphasizing that different algorithms produce \blds\ with different algebraic and geometric properties.  Our long-range goal is to produce \blds\ that are as simple as possible, with respect to various criteria.  In this first paper we present two algorithms, producing what we call the {\it elementary} and {\it gnomon} \blds\ respectively, and show that these are associated with {\it subprime filtrations} of $I$, a natural generalization of prime filtrations.  In doing so, we generalize Soleyman-Jahan's criterion in \cite[Prop. 2.7]{AliPrime} for a \sd\ to be associated with a prime filtration, and simplify the proof.  The gnomon decomposition is shorter (contains fewer blocks) than the elementary decomposition.  In the second paper we present algorithms leading to what we call the {\it organized} and {\it stacked} \blds.   These are {\it incompressible}, in the sense that they cannot be shortened by combining blocks.  We have not yet found an algorithm that always produces a minimal \bld\ (one with the fewest blocks), but since any such \bld\ must be incompressible, these algorithms are a step in that direction.

From one point of view, \blds\ are a generalization of \sds, and are in general {\it coarser} than \sds.  From another point of view, each \bld\ has a unique minimal Stanley refinement, given by Algorithm \ref{SDfromBD} below, and since this algorithm is very easy, the \bld\ can be taken as a {\it compressed notation} for this \sd.  From this viewpoint, we are studying algorithms leading from $I$ to a \sd\ of $S/I$.  The usual algorithm of this type, \cite[Lemma 2.4]{SturmWh}, proceeds by induction on the variables $x_1,\dots,x_n$.  Our gnomon algorithm works by induction on the generators of $I$, so that if one has determined a \bld\ for $S/I$ and then adds a generator to $I$, it is not necessary to start over from the beginning.

In Section \ref{block} we define \blds\ and \sds\ and present Algorithm \ref{SDfromBD} that relates them. Sections \ref{elem} and \ref{gnomon} contain the algorithms for the elementary and gnomon block decompositions.  Section \ref{prime} contains the generalization of Soleyman-Jahan's theorem and the proof that elementary and gnomon decompositions are subprime.

Stanley decompositions are best known among algebraists in connection with the conjecture that the Stanley depth of a module over a polynomial algebra is an upper bound for the classical depth, but have a separate life in applied mathematics through their use in describing subalgebras.  This life is based entirely on the first two pages of \cite{SturmWh} and, for one  application, on \cite{SturmWhBracket}.  The motivation for our work comes from that direction.  Let $\CA$ be a subalgebra of $K[\by]=K[y_1,\dots,y_m]$.  It is desired to find a formula, containing finitely many arbitrary polynomial functions, that expresses each element of $\CA$ exactly once as the arbitrary functions are varied.  A set of polynomials $f_1,\dots,f_n\in\CA$ is a {\it \hb} for $\CA$ if the monomials $f_1^{k_1}\cdots f_n^{k_n}$  span $\CA$ as a vector space over $K$.  The expression for a given $f$ in terms of the \hb\ is usually not unique, because the monomials $f_1^{k_1}\cdots f_n^{k_n}$ may not be linearly independent.  There always exist linearly independent subsets of these monomials, which we call {\it preferred sets}, and the expression of any $f$ as a linear combination of these is unique.  But a preferred set of monomials is infinite, and may not have a finite description leading to an expression involving arbitrary polynomial functions.  Stanley decompositions in the manner of \cite{SturmWh} provide suitable preferred sets . Let $\Phi:K[\bx]\ra K[\by]$ be the unique algebra homomorphism such that $\Phi(x_i)=f_i$; then $J=\ker\Phi$ is the ideal of relations among the $f_i$.  Compute a \Gro\ basis for $J$ using an appropriate elimination order on $K[\bx,\by]$ (\cite[p. 81]{AdLou}), and obtain the monomial ideal $I$ of leading terms of $J$, which is generated by the leading terms of the \Gro\ basis.  The standard monomials of $I$ provide a preferred set of monomials in $f_1,\dots,f_n$ when each $x_i$ is replaced by $f_i$.  In fact $\CA$ is isomorphic as an algebra to $K[\bx]/J$, and isomorphic as a vector space to both $K[\bx]/I$ and $\Span(M)$.  Furthermore, a \sd\ for $K[\bx]/I$ provides the desired formula expressing each element of $\CA$ uniquely.  For instance, if $\CA$ has a \hb\ of four elements $f_1,\dots,f_4$ with the single relation $f_3^2-f_2^3-f_1^2f_4=0$, and $f_3^2$ is taken as the leading term, then $I=\langle x_3^2\rangle$ and the \sd\ given by the algorithm of \cite{SturmWh} is
\begin{equation}\label{N4}
K[x_1,x_2,x_3,x_4]/\langle x_2^3\rangle \cong K[x_1,x_2,x_4]\oplus K[x_1,x_2,x_4]x_3.
\end{equation}
It follows that every element of $\CA$ can be written uniquely in the form
\[
f=F_1(f_1,f_2,f_4)+F_2(f_1,f_2,f_4)f_3,
\]
where $F_1$ and $F_2$ are arbitrary polynomials in three variables.  (This example arises for the  classical seminvariants of the binary cubic form.)

Notice that the number of arbitrary polynomials is the same as the number of Stanley spaces.  The number of such polynomials can become large in larger problems, and it is natural to ask if this can be reduced.  In part this can be done by a good choice of the term order; for instance, if $f_2^3$ was taken as the leading term in the example, there would be three Stanley spaces instead of two. But even once $I$ is selected, there can be large differences in the number of Stanley terms in different \sds\ for $S/I$.  This suggests the problem:  {\it Find a minimal \sd\ for $S/I$ given a fixed monomial ideal $I$.}   A related, but different, problem is to {\it find a minimal \bld\ for $M$ given $I$}; here the minimality is with regard to the number of blocks, not the number of Stanley spaces.

It is not an accident that the example mentioned above comes from classical invariant theory.  Invariant theory is closely linked with normal form theory for systems of differential equations with nilpotent linear part.  There was a flurry of work on this topic in the late 1980s and early 1990s (\cite{CSNilp}, \cite{CSN4}, \cite{CSW}, \cite{SturmWhBracket}, \cite{CSSurv}, \cite{SturmWh}), and a second flurry more recently (\cite{MurdStruc}, \cite{MurdNTA}, \cite[Ch.12]{SanVerMurd}, \cite{SanJoint}, \cite{Mal}, \cite{Mal-2}, \cite{GachInv}, \cite{GachNF}).  Additional motivation for our questions comes from these papers.  The algebra of scalar fields that are invariant under the the group $\{e^{Mt}:t\in\R\}$, where $M$ is a nilpotent matrix, plays an important role.  These invariants, in turn, are the same as classical seminvariants of a binary form, or joint seminvariants of several forms.  Knowing the seminvariants of several forms, {\it transvectants} are used to find a \hb\ for the the joint seminvariants.  The {\it box product} method introduced in \cite{MurdNTA} lifts this procedure to the level of \sds.  The box product of two \sds\ for the seminvariants of two binary forms is a \sd\ for the joint seminvariants of the two forms; thus the uniqueness issue is taken care of automatically, without further \Gro\ basis work.  In \cite{MurdBox} I (J.M.) have reformulated the box product as a product of block decompositions, in which $\boxtimes$ distributes over disjoint union; the basic unit of computation is the box product of two blocks, for which algorithms are given, and the final result (expressed in \sds) is shorter than for the method in \cite{MurdNTA}.  It is best, then, to start with \blds\ with a minimal number of blocks.

This research began when I (J.M.) asked T.M., a software engineer, to write programs for me concerning \sds.  He created block decompositions as a computer-friendly shorthand for \sds, and the gnomon algorithm to provide an interactive program in which ideal generators could be added one at a time.  I provided the mathematical write-up, proofs, and examples, and the material on subprime filtrations.  (The word {\it subprime} was provided by the mortgage crisis.)

\section{\bf Block decompositions}\label{block}
Let $V\subset S=K[\bx]$ be a vector subspace; $V$ is a {\bf monomial space} if it is spanned by the set $M=\mon(V)$ of monomials in $V$, which is then also the unique monomial basis for $V$.   Under the one-to-one correspondence $V=\Span(M)$, $M=\mon(V)$ between monomial spaces and their monomial bases, a direct sum of two spaces corresponds to the disjoint union of their bases.  We often identify $\bx^{\bbm}$ with $\bbm=(m_1,\dots,m_n)$ in the {\bf Newton space} $\N^n$.

A {\bf Stanley space} is a monomial space of the form
\[
K[X]\p = \{f\p:f\in K[X]\},
\]
where $X$ is a subset of the set of variables $\{x_1,\dots,x_n\}$ and $\p$ is a monomial.  An expression exhibiting a monomial space $V$ as a direct sum of finitely many Stanley spaces is called a {\bf Stanley decomposition} of $V$.  (Because of this finiteness requirement, there are monomial spaces, such as $\Span\{1,x^2,x^4,x^6,\dots\}\subset K[x]$, that do not have Stanley decompositions.)

Let $I$ be a monomial ideal in $S$ and let $M=\mon(S)\smallsetminus \mon(I)$ be the set of {\bf standard monomials} for $I$.  Then $V=\Span(M)$ is the space of {\bf standard polynomials}, and as noted in the introduction, $S/I\cong V$ as vector spaces.  If $S=K[x,y]$ and $I=\langle xy^3, x^3y\rangle$, a Stanley decomposition for $S/I$ is
\begin{equation}\label{sd1}
K[x]\oplus K[y]y \oplus Kxy\oplus Kx^2y\oplus Kxy^2\oplus Kx^2y^2.
\end{equation}
Applying the correspondences above, we replace each Stanley space by its set of monomials and replace $\oplus$ by the disjoint union symbol $\sqcup$, obtaining
\[
\{1,x,x^2,\dots\}\sqcup\{y,y^2,y^3,\dots\}\sqcup\{xy, xy^2, x^2y, x^2y^2\}.
\]
In this way the \sd\ (\ref{sd1}) becomes the following {\it block decomposition} in Newton space:

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More generally, let $[a,b]$ denote an integer interval, with the understanding that $[a,\infty]$ does not include $\infty$ and that $[a,b]=\emptyset$ if $b<a$.  Define a {\bf block} $B$ in $\N^n$ to be a Cartesian product of $n$ intervals:
\begin{equation}\label{defblock}
B=\bm \bb\\ \ba\ebm = \bm b_1&b_2&\cdots&b_n\\a_1&a_2&\cdots&a_n\ebm = [a_1,b_1]\times\cdots\times [a_n,b_n].
\end{equation}
If any $b_i<a_i$, the block is empty.  The bottom row $\ba$ of a block is called its {\bf inner corner}.  The letter $B$ will be used ambiguously to mean the matrix as such, the set of integer vectors represented by the matrix, and the set of monomials having those integer vectors as exponents.  Let $M\subset\mon(S)$.  A {\bf block decomposition} of $M$ is an expression
\begin{equation}\label{bd}
M=B^1\sqcup B^2\sqcup \dots \sqcup B^s,
\end{equation}
where each $B^k$ is a block.
For example, the Stanley decomposition (\ref{sd1}) can be written as the block decomposition
\begin{equation}\label{bd1}
B^1\sqcup B^2\sqcup B^3 = \bm \infty&0\\0&0\ebm \sqcup \bm 0&\infty\\0&1\ebm \sqcup \bm 2&2\\1&1\ebm,
\end{equation}
corresponding to the diagram shown above in Newton space.

\brem After this research was mostly completed, we learned of the {\bf interval decompositions} defined in \cite{HerVladZh}.  They define the {\bf interval} $[\ba,\bb]$ to be the set of monomials lying between $\bx^{\ba}$ and $\bx^{\bb}$ in division partial order.  This is essentially the same as our block $B$ in (\ref{bd1}), except that they consider only monomials that divide a chosen $\bx^{\bc}$, so that, in particular, $\infty$ is not allowed in the top row.  This is sufficient since they only consider interval decompositions for $I/J$, where $I$ and $J$ are two monomial ideals in $S$.\erem

A block decomposition is called {\bf compressible} if the union of some subset of the blocks is itself a block.  In this case the decomposition can be simplified by performing the union.  However, performing one such union may prevent another one, so a given decomposition may be compressible in several ways to give distinct incompressible decompositions.  A decomposition is called {\bf minimal} if there is no decomposition (describing the same set of monomials) having fewer blocks.  Incompressibility and minimality are two possible notions of ``simplicity'' of a decomposition (as briefly discussed in the introduction).  An incompressible decomposition need not be minimal, and a minimal decomposition need not be unique.  The following example shows that incompressibility is a global property of a decomposition, and cannot be detected by examining the decomposition locally:  The disjoint union
\[
\bm 1&0\\0&0\ebm\sqcup\bm 0&2\\0&1\ebm\sqcup\bm1&1\\1&1\ebm\sqcup\bm2&1\\2&0\ebm\sqcup\bm2&2\\1&2\ebm
\]
is compressible to the single block $\bm 2&2\\0&0\ebm$, although no proper subset of the five blocks can be compressed.

A {\bf Stanley block} is a block (\ref{defblock}) in which each $b_i$ equals either $a_i$ or $\infty$.  The span of a Stanley block is a Stanley space, and any Stanley decomposition can be changed to a block decomposition by replacing Stanley spaces by Stanley blocks and $\oplus$ by $\sqcup$.  Such a block decomposition can almost always be compressed.  For example, $K[x_1,x_2,x_4]\oplus K[x_1,x_2,x_4]x_3$ becomes
\[
\bm \infty&\infty&0&\infty\\ 0&0&0&0\ebm \sqcup \bm \infty&\infty&1&\infty\\0&0&1&0\ebm=\bm \infty&\infty&1&\infty\\0&0&0&0\ebm.
\]

Any block $B$ can be written as a disjoint union of its {\bf bounded part} and its {\bf unbounded part}; the bounded part is obtained by replacing the unbounded columns with zeroes, the unbounded part by doing the same with the bounded columns.  For instance,
\[
B=\bm \infty&5&\infty&2\\1&3&0&1\ebm = \bm 0&5&0&2\\0&3&0&1\ebm \sqcup \bm \infty&0&\infty&0\\1&0&0&0\ebm.
\]
Let $X$ be the set of variables associated with unbounded columns of $B$ ($x_1$ and $x_3$ in the example).  Let $\psi$ be the monomial in the bottom row of the unbounded part ($x_2^2x_4$ in the example).  Let $\th_1,\dots,\th_s$ be the finite set of monomials in the bounded part ($x_2^3x_4$, $x_2^4x_4$, $x_2^5x_4$, $x_2^3x_4^2$, $x_2^4x_4^2$, $x_2^5x_4^2$, with $s=6$).  Let $\p_k=\th_k\psi$ for $k=1,\dots,s$.  Then the block $B$ is converted to its minimal \sd\ as follows.
\balg\label{SDfromBD}   The Stanley decomposition with the fewest Stanley spaces that describes the span of the monomials in $B$ is
\[
\Span(B)=K(X)\p_1\oplus\cdots\oplus K(X)\p_s.
\]
\ealg

\noindent Note that all of the coefficient algebras $K(X)$ in the minimal \sd\ of a block are equal, and that the number of Stanley spaces $s$ is the product of the numbers of elements in the intervals defined by the bounded columns of $B$.  When Algorithm \ref{SDfromBD} is applied to each block in a \bld, the coefficient rings may differ from block to block, but the total number of Stanley spaces is just the sum of the number for each block.

As an introduction to the next two sections, we now give examples of the {\it elementary} and {\it gnomon} block decompositions defined in those sections.  For
\begin{equation}\label{firstexample0}
I=\langle x^3y^9, x^7y^5\rangle\subset S=K[x,y],
\end{equation}
the {\it elementary} decomposition of $S/I$ is the disjoint union of the blocks
\begin{equation}\label{firstexample1}
\bm 2&4\\0&0\ebm, \bm 2&8\\0&5\ebm, \bm 2&\infty\\0&9\ebm,\bm 6&4\\3&0\ebm, \bm 6&8\\3&5\ebm, \bm \infty&4\\7&0\ebm.
\end{equation}
The {\it gnomon} decomposition comes in two forms,
\begin{equation}\label{firstexample2}
\bm 2&\infty\\0&0\ebm\sqcup \bm 6&8\\3&0\ebm\sqcup \bm \infty&4\\7&0\ebm
\end{equation}
and
\begin{equation}\label{firstexample3}
\bm \infty&4\\0&0\ebm\sqcup \bm6&\infty\\0&5\ebm\sqcup \bm 2&\infty\\0&9\ebm.
\end{equation}
The elementary decomposition is compressible to either of the gnomon decompositions, and (in this case, but not always) the gnomon decompositions are incompressible.

\section{The elementary block decomposition}\label{elem}

In Newton space, the {\bf division partial order} $\bx^{\bbm}\vert \bx^{\bbm'}$ will be written as $\bbm \preceq \bbm'$, meaning $m_i$ $\leq m'_i$ for $i=1,\dots,n$.  For variables in Newton space we use $\bmu=(\m_1,\dots,\m_n)$, reserving  $\bbm=(m_1,\dots,m_n)$ for constants, so that an equation such as $\m_2=m_2$ will represent the ``hyperplane'' through $\bbm$ perpendicular to the $\m_2$ axis in $\N^n$.  Let $S=K[\bx]$ and let $I=\langle \bx^{\bbm^1},\dots,\bx^{\bbm^r}\rangle \subset S$ be the monomial ideal with the indicated generators; we also write $I=\langle \bbm^1,\dots,\bbm^r\rangle$.  It is assumed that these generators are minimal, that is, no redundant generators (divisible by another generator) are included.  The {\bf elementary block decomposition} for $S/I$ is created by first gridding the Newton space $\N^n$ with the hyperplanes $\m_i=m^r_i$ for $i=1,\dots,n$ and $j=1,\dots,s$, passing through all points of the minimal generating set, and then discarding those of the resulting blocks that belong to $\mon(I)$ rather than $\mon(S)\smallsetminus \mon(I)$.  The following figure illustrates the elementary decomposition (\ref{firstexample1}) at the end of Section \ref{block}.  The solid dots are the generators at $(3,9)$ and $(7,5)$, the dark lines are the boundary of $\mon(I)$, and the blocks are numbered in the order that they appear in (\ref{firstexample1}).

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The elementary block decomposition is almost always compressible, and so is not very desirable in itself, but it shows that a natural block decomposition always exists.  The other decompositions (gnomon, organized, and stacked) in this paper and its sequel can be obtained by compressing the elementary decomposition, so we will never need to consider blocks smaller than the blocks in the elementary decomposition, or blocks that are not disjoint unions of elementary blocks.

The following is a precise algorithm to create the elementary decomposition.

\balg\label{ebd} Let $I=\langle \bbm^1,\dots,\bbm^r\rangle$ be presented by its minimal generating set.  The elementary block decomposition is generated as follows.
\blist
\item For $i=1,\dots,n$, create the list of exponents of $x_i$ in the set of generators, adding zero at the beginning of each list:
\begin{align*}
&L_1:  0, m^1_1,\dots, m^r_1;\\
&L_2:  0, m^1_2,\dots, m^r_2;\\
&\vdots\\
&L_n:  0, m^1_n,\dots,m^r_n.
\end{align*}
\item\label{reorder} Delete any repetitions in each list $L_i$, and re-order each list in increasing numerical order.
\item Create a preliminary list of inner corners as follows.  For $i=1,\dots,n$, choose an entry $a_i$ from $L_i$; then create an inner corner $\ba=(a_1,\dots,a_n)$.  Do this in all possible ways.
\item\label{refine} Refine the list of inner corners by discarding any $\ba\in I$, that is, any $\ba$ such that $\bbm^i\prec \ba$ for some $i$.
\item\label{createblock} Create a block $B$ for each inner corner $\ba$ in the refined list, by determining its outer corner $\bb$, as follows.  If $a_i$ has a successor $a_i'$ in $L_i$ (under the ordering from step \ref{reorder}), put $b_i=a_i'-1$.  If $a_i$ is the last element of $L_i$, but $b_i=\infty$.
\item\label{elemenum} Let $s$ be the number of resulting blocks, and enumerate these as
\[
B^k=\bm \bb^k\\ \ba^k \ebm
\]
for $k=1,\dots,s$.
\elist
\ealg

\brem\label{elemorder} The order of enumeration in Step \ref{elemenum} does not matter except in Section \ref{prime}.  For that section, enumerate the blocks in lexicographic order by their inner corners.\erem

For the example (\ref{firstexample0}), the lists from Step 2 are $L_1=\{0,3,7\}$, $L_2=\{0,5,9\}$.  The preliminary list of inner corners (in lexicographic order) is $\{(0,0), (0,5),(0,9),(3,0),(3,5)(3,9),(7,0),(7,5),(7,9)\}$.  From this we discard $(3,9)$, $(7,5)$, and $(7,9)$.  It is now easy to check that the outer corners in (\ref{firstexample1}) follow from step 5 above.

\brem\label{elemprop} The elementary decomposition has the following property, which will be used in Section \ref{prime}.  Let $\bc \notin I$ be a monomial.  Then there exists a unique inner corner $\ba^k$ of the elementary decomposition for $S/I$ that is maximal (under $\prec$) among all inner corners satisfying $\ba^k\preceq\bc$.  The inner corner $\ba^k$ satisfying this condition determines the block $B^k$ that contains $\bc$.  (The proof is trivial: The block containing $\bc$ has for the $i$-th component of its inner corner the largest element of $L_i$ that is $\leq c_i$.)\erem

\section{The gnomon decomposition}\label{gnomon}

Consider the following {\bf Block Subtraction Problem:}   Given a block (\ref{defblock}) and a principal monomial ideal $I=\langle \bbm \rangle$, find a block decomposition of $B\smallsetminus I$.  If $\bbm$ is within the block $B$, there are $n!$ distinct natural solutions.  For $n=2$, the following figure shows a block $B$ in $\N^2$, an ideal $I$ generated by the heavy dot, and the two block decompositions of $B\smallsetminus I$.  The set of points in $B\smallsetminus I$ is the sort of L-shaped region the Pythagoreans called a  {\bf gnomon} ($\g\n\o\m\o\n$).

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\noindent The first of these decompositions, expressed algebraically, takes the form
\begin{equation}\label{leftsolution}
\bm b_1&b_2\\a_1&a_2\ebm \smallsetminus \langle (m_1,m_2) \rangle = \bm m_1-1&b_2\\a_1&a_2\ebm \sqcup \bm b_1&m_2-1\\m_1&a_2\ebm.
\end{equation}
We will generalize this formula to arbitrary $n$ in such a way that it gives all $n!$ solutions, one for each permutation of the variables $x_1,\dots,x_n$.  First we focus on the natural order of variables (the trivial permutation), which for $n=2$ gives (\ref{leftsolution}).

The following solution for $n=2$ is valid whether or not $\bbm$ is in the interior of $B$, and reduces to (\ref{leftsolution}) when it is:
\begin{align*}
&\bm b_1&b_2\\a_1&a_2\ebm \smallsetminus \langle (m_1,m_2)\rangle =\\
&\qquad \bm \min\{b_1,m_1-1\}&b_2\\a_1&a_2\ebm \sqcup \bm b_1&\min\{b_2,m_2-1\}\\\max\{a_1,m_1\}&a_2\ebm.
\end{align*}
The first block consists of all monomials in $B$ that are not divisible by $x_1^{m_1}$, and is empty if $m_1\leq a_1$.  The second block consists of those monomials in $B$ that {\it are} divisible by $x_1^{m_1}$ but not by $x_2^{m_2}$.  But this solution has a defect: it sometimes divides $B$ unnecessarily into two nonempty blocks whose union is again $B$.  This happens whenever $a_1<m_1<b_1$ but $b_2<m_2$.  In fact, if $m_i>b_i$ for $i=$ 1 or 2 or both, then $B\cap I =\emptyset$ and $B\smallsetminus I=B$.  This observation also simplifies the formula in the remaining cases (where $m_i\leq b_i$ for all $i$), because then $\min\{b_i,m_i-1\}$ can be replaced by $m_i-1$.

Generalizing to higher dimensions, we define the {\bf gnomon} determined by $B$ and $I$ to be the set represented by the $3\times n$ matrix
\[
\bm \bb \\ \bbm \\ \ba \ebm = \bm b_1&\cdots&b_n\\m_1&\cdots&m_n\\a_1&\cdots&a_n\ebm
\]
as follows.  (Placing $\bbm$ between $\ba$ and $\bb$ is intended to suggest ``cutting $B$ by $\bbm$''.)
\blist
\item If there exists $i\in\{1,\dots,n\}$ such that $b_i<m_i$ (that is, if $\bbm \npreceq \bb$), then
\begin{equation}\label{gnomonformula1}
\bm \bb \\ \bbm \\ \ba \ebm = \bm \bb \\ \ba \ebm.
\end{equation}
\item If for all $i$, $m_i\leq b_i$ ($\bbm\preceq\bb$), let
\begin{equation}\label{defc}
c_i = \max\{a_i,m_i\}.
\end{equation}
Then
\begin{align}\label{gnomonformula2}
&\bm \bb \\ \bbm \\ \ba \ebm
= \bm m_1-1 & b_2&\cdots&b_n\\a_1&a_2&\cdots&a_n\ebm\\
&\qquad\sqcup \bm b_1&m_2-1&b_3&\cdots&b_n\\c_1&a_2&a_3&\cdots&a_n\ebm \notag\\
&\qquad\sqcup \bm b_1&b_2&m_3-1 &b_4&\cdots&b_n\\c_1&c_2&a_3&a_4&\cdots&a_n \ebm \notag\\
&\qquad\sqcup\cdots\sqcup \bm b_1&b_2&\cdots&b_{n-1}&m_n-1\\c_1&c_2&\cdots&c_n&a_n\ebm. \notag
\end{align}
\elist

The following lemma shows that the solution to the Block Subtraction Problem is a gnomon as defined above.

\blem\label{gnomonlem} For all nonnegative values of $a_i$, $b_i$, and $m_i$,
\begin{equation}\label{gnomonlem1}
\bm \bb \\ \ba \ebm \smallsetminus \langle \bbm\rangle
  = \bm \bb \\ \bbm \\ \ba \ebm.
\end{equation}
\elem

\bpf    Assume first that $b_i<m_i$ for some $i$. Then the ideal $\langle \bbm \rangle$ does not intersect the given block, and (\ref{gnomonlem1}) follows from (\ref{gnomonformula1}).

Next assume that $a_i<m_i\leq b_i$ for all $i$, so that $c_i=m_i$.  Then the original block can be decomposed into two subblocks by the hyperplane $\m_1=m_1$:
\begin{equation}\label{step1}
\bm \bb \\ \ba \ebm =
\bm m_1-1 & b_2&\cdots&b_n\\a_1&a_2&\cdots&a_n\ebm
\sqcup \bm b_1&b_2&b_3&\cdots&b_n\\c_1&a_2&a_3&\cdots&a_n\ebm.\notag
\end{equation}
(Note that $c_1$ occurs in the lower left corner of the second matrix where $m_1$ is expected, but these are equal.)  The first of these subblocks lies outside of $\langle \bx^{\bbm}\rangle$, and belongs to the difference we are computing; it equals the first block of (\ref{gnomonformula2}).  The second subblock is carried forward to the next step in a recursive process:  It is split by the hyperplane $\m_2=m_2$ into two subblocks, one having $a_2\leq \m_2<m_2$ and the other having $m_2\leq \m_2\leq b_2$.  Again, the first subblock is retained and the second carried forward to the third step, and so forth.

Finally, if $m_1\leq a_1$, then $m_1-1<a_1$ and the first subblock in (\ref{step1}) is empty.  Furthermore the lower left entry in the second matrix should be $a_1$ (rather than $m_1$, which would cause the block to include monomials that are not in the original block).  But in this case $c_1=a_1$, so (\ref{step1}) remains correct.  The same reasoning applies if the condition $m_i\leq a_i$ is encountered later in the recursion.  \epf

Now let $I=\langle \bbm^1,\dots,\bbm^r\rangle$ be a monomial ideal for which a block decomposition $B^1\sqcup\cdots\sqcup B^s$ is known, and let  $\langle \bbm \rangle$ be a principal monomial ideal.  Let $I'= I +
\langle \bbm\rangle$ be the sum of these ideals (in the usual sense), and note that $I'= \langle \bbm^1,\dots,\bbm^r,\bbm\rangle.$  We refer to this as {\bf adding a generator} to $I$.

\balg\label{addgenerator} A block decomposition for $S/I'$ is given by
\begin{equation}\label{addgenformula}
(B^1\smallsetminus \langle \bbm\rangle)\sqcup\cdots\sqcup(B^s\smallsetminus \langle \bbm\rangle).
\end{equation}
\ealg

\bpf  The standard monomials of $I'$ are the monomials that are standard for $I$ and, in addition, are not divisible by $\bx^{\bbm}$.  We remove the monomials divisible by $\bx^{\bbm}$ from each block $B^k$ of standard monomials for $I$ by finding the gnomons $B^k\smallsetminus \langle \bbm\rangle$.  Since these are disjoint, their union is a block decomposition.\epf

\brem\label{gnorder} For Section \ref{prime}, where the order of terms in a decomposition matters, we write the result of Algorithm \ref{addgenerator} in the form
\[
\bigsqcup_{k=1}^s \left( \bigsqcup_{i=1}^n B^{ki}\right)=
 (B^{11}\sqcup\dots\sqcup B^{1n})\sqcup\dots\sqcup (B^{s1}\sqcup\dots\sqcup B^{sn}).
\]
Here $B^{k1},\dots,B^{kn}$ are the terms of $B^k\smallsetminus \langle\bbm\rangle$ in the order shown in equation (\ref{gnomonformula2}), retaining any empty blocks, or, if (\ref{gnomonformula1}) applies, then $B^{k1}=B^k$ and $B^{ki}=\emptyset$ for $i=2,\dots,n$.  Retaining the empty sets allows a uniform notation in the proof of Theorem \ref{gnnice}.\erem

Let $I=\langle \bbm^1,\dots,\bbm^r \rangle$ be a monomial ideal given by its minimal generators.  The {\bf gnomon decomposition} of $S/I$, with respect to the standard order $x_1,\dots,x_n$ of the variables, is defined to be the result of applying Algorithm (\ref{gbd}) below.  The result may depend on the order in which the generators are indexed; this {\bf noncommutativity of the generators} is illustrated in the examples below.

\balg\label{gbd} Apply Algorithm \ref{addgenerator} repeatedly, beginning with $I=\emptyset$ and adding the generators $\bx^{\bbm^1},\dots,\bx^{\bbm^r}$ in that order.\ealg

Let $\pi$ be a permutation of the integers $(1,\dots,n)$, regarded as specifying an order $x_{\pi_1},\dots,x_{\pi_n}$ of the variables.  The $\pi$-{\bf gnomon decomposition} of $S/I$ is defined by changing variables temporarily to $y_i=x_{\pi_i}$, applying algorithm (\ref{gbd}) using the new variables, and then returning to the original variables.  (The generators are still taken in their indicated order.)

The blocks of the gnomon decomposition are unions of blocks of the elementary decomposition, because whenever a block is subdivided in Algorithm \ref{gbd}, the subdivision occurs along a portion of a hyperplane $\m_i=m^j_i$ for some $i$ and $j$.  The subdivisions for elementary blocks occur along these same hyperplanes (but along the entire hyperplane, not a portion of it.)

\bex\label{x1} For the example (\ref{firstexample0}), $I=\langle(3,9),(7,5)\rangle$.  According to Algorithm \ref{gbd}, we begin with the decomposition $\bm \infty&\infty\\0&0\ebm$ for $K[x,y]$ and add the generator $(3,9)$ using Algorithm \ref{addgenerator}, which in turn calls for using Lemma \ref{gnomonlem} to compute
\[
\bm \infty&\infty\\0&0\ebm \smallsetminus \langle (3,9)\rangle = \bm 2&\infty\\0&0\ebm \sqcup \bm \infty&8\\3&0\ebm.
\]
The first block contains the monomials that are not divisible by $x^3$, the second those that are divisible by $x^3$ but not by $y^9$.  Next we find
\[
\bm 2&\infty\\0&0\ebm \smallsetminus \langle (7,5)\rangle = \bm 2&\infty\\ 0&0\ebm
\]
since $2<7$ and $(7,5)$ is not in the block.  Finally
\[
\bm\infty&8\\3&0\ebm \smallsetminus \langle x^7y^5\rangle = \bm 6&8\\3&0\ebm\sqcup\bm \infty&4\\7&0\ebm.
\]
The result of these calculations is (\ref{firstexample2}).

The same problem works out as follows if we take $I=\langle (7,5),(3,9)\rangle$, that is, add the generators in the opposite order:
\begin{align*}
\bm \infty&\infty\\7&5\\0&0\ebm &= \bm 6&\infty\\0&0\ebm\sqcup\bm\infty&4\\7&0\ebm \\
\bm 6&\infty\\ 3&9\\ 0&0\ebm &= \bm 2&\infty\\0&0\ebm \sqcup \bm 6&8\\3&0\ebm\\
\bm \infty&4\\3&9\\7&0\ebm &= \bm\infty&4\\7&0\ebm.
\end{align*}
The last line is an instance of equation (\ref{gnomonformula1}).  The result is again (\ref{firstexample2}), so there is no noncommutativity in the generators in this example.  The $\pi$-gnomon decomposition with $\pi=(2,1)$ is (\ref{firstexample3}).\eex

\bex\label{x3} To show noncommutativity of generators, the gnomon decomposition of $K[x,y,z]/\langle (5,3,7),(10,6,2)\rangle$ is
\[
\bm 4&\infty&\infty\\0&0&0\ebm \sqcup \bm \infty&2&\infty\\5&0&0\ebm\sqcup \bm 9&\infty&6\\5&3&0\ebm \sqcup \bm\infty&5&6\\10&3&0\ebm \sqcup \bm\infty&\infty&1\\10&6&0\ebm,
\]
which is incompressible; the opposite order of generators gives instead
\begin{align*}
&\bm 4&\infty&\infty\\0&0&0\ebm \sqcup \bm 9&2&\infty\\5&0&0\ebm \sqcup \bm 9&\infty&6\\5&3&0\ebm\\
&\qquad \sqcup \bm \infty&2&\infty\\10&0&0\ebm \sqcup \bm \infty&5&6\\10&3&0\ebm \sqcup \bm\infty&\infty&1\\10&6&0\ebm.
\end{align*}
This is compressible (to the previous result), since
\[
\bm 9&2&\infty\\5&0&0\ebm \sqcup \bm \infty&2&\infty\\10&0&0\ebm = \bm \infty&2&\infty\\5&0&0\ebm.
\]
\eex

\bex\label{x4} The next example will be used in Remark \ref{x4cont} to illustrate Theorem \ref{gnnice}.  Let $S=K[x,y,z]$ and $I=\langle z^5,y^2z^3,x^3yz\rangle$.  Then
\begin{align*}
\bm \infty&\infty&\infty\\0&0&5\\0&0&0\ebm &= \bm \infty&\infty&4\\0&0&0\ebm\\
\bm \infty&\infty&4\\0&2&3\\0&0&0\ebm &= \bm \infty&1&4\\0&0&0\ebm\sqcup\bm\infty&\infty&2\\0&2&0\ebm\\
\bm \infty&1&4\\3&1&1\\0&0&0\ebm &= \bm 2&1&4\\0&0&0\ebm \sqcup \bm \infty&0&4\\3&0&0\ebm \sqcup \bm \infty&1&0\\3&1&0\ebm\\
\bm \infty&\infty&2\\3&1&1\\0&2&0\ebm &= \bm 2&\infty&2\\0&2&0\ebm \sqcup \quad\emptyset\quad \sqcup \bm \infty&\infty&0\\3&2&0\ebm.
\end{align*}
The gnomon decomposition for $S/I$ is the union of the last two lines.
\eex

We conclude this section with two lemmas that will be used in Section \ref{prime}.

\blem\label{intersect} With $c_i$ as in (\ref{defc}),
\begin{equation}\label{gnomonlem2}
\bm b_1&\cdots&b_n\\a_1&\cdots&a_n\ebm \cap \langle x_1^{m_1}\cdots x_n^{m_n}\rangle
  = \bm b_1&\cdots&b_n\\c_1&\cdots&c_n\ebm.
\end{equation}
\elem

\bpf If $m_i>b_i$ for any $i$, both sides are empty.  Otherwise, continue the reasoning in the proof of Lemma \ref{gnomonlem}.  Observe that the right-hand side of \ref{gnomonlem2} would be the next matrix added to (\ref{gnomonformula2}) if the pattern of (\ref{gnomonformula2}) is continued.\epf

\blem\label{gnlemma} Assume that $\bbm\preceq \bb$, so that (\ref{gnomonformula2}) is valid.  Let the blocks on the right-hand side of (\ref{gnomonformula2}) be labeled $B^1,\dots,B^n$.  Then for any $i=1,\dots,n$ we have
\[
B^i \sqcup\cdots\sqcup B^n \sqcup \bm \bb\\ \bc\ebm =
\bm  b_1 &\cdots &b_{i-1} &b_i &\cdots &b_n\\
     c_1 &\cdots &c_{i-1} &a_i &\cdots &a_n \ebm.
\]
\elem

\bpf Compute the unions successively starting from the right and working to the left, noticing that each union is between matrices that differ only in one column.  For instance, if $n=3$ the first step is
\[
\bm b_1&b_2&m_3-1\\c_1&c_2&a_3\ebm \sqcup \bm b_1&b_2&b_3\\c_1&c_2&c_3\ebm = \bm b_1&b_2&b_3\\c_1&c_2&a_3\ebm.
\]
If $c_3=m_3$ the result is clear; if $c_3=a_3$ then the left-hand block is empty and again the result is clear.  (The lemma and proof remain valid without assuming $\bbm\preceq\bb$, but in that case it does not apply to the gnomon decomposition, which is governed instead by (\ref{gnomonformula1}).)
\epf

\section{Subprime filtrations}\label{prime}

A {\bf filtration}  of $S$ by monomial ideals is a finite nested sequence
\begin{equation}\label{filt}
I=I_0\subset I_1\subset \cdots \subset I_s = S.
\end{equation}
This is often called a filtration of $S/I$, and it does induce an actual filtration of $S/I$ by subrings, namely $\{0\} \subset I_1/I\subset I_2/I\subset\cdots\subset S/I$.  If each set difference $\mon(I_k)\smallsetminus \mon(I_{k-1})$ is a block $B_k$, we call (\ref{filt}) a {\bf subprime filtration}.  If, further, each block $B_k$ is a Stanley block, we call (\ref{filt}) a {\bf prime filtration};  this is simpler than the usual definition in the literature, but is equivalent, as pointed out at the end of this section.  A block decomposition, {\it with the blocks indexed in a specific order}, will be called {\bf subprime} (or {\bf prime}) if it is associated in this way with a subprime (or prime) filtration.  In this section all block decompositions are taken as {\bf ordered block decompositions}.  This simplifies the discussion, in that we need not consider all possible orderings before deciding whether a given block decomposition is subprime.

The following equations hold trivially for subprime filtrations.  Quotients, direct sums, and isomorphisms are to be understood as quotients, direct sums, and isomorphisms of vector spaces, and $M$ is the set $\mon(S)\smallsetminus \mon(I)$ of standard monomials for $I$ in $S$.
\begin{align}
&I_k/I_{k-1} \cong \Span{B_k}\\
&S/I \cong \Span{B_1}\oplus\cdots\Span{B_s}\\
&M=B_1\sqcup\cdots\sqcup B_s\\
&\mon(I_{k-1})\sqcup B_k = \mon(I_k)\\
&I_{k-1}\oplus \Span(B_k)=I_k.
\end{align}
That is, with each subprime filtration of $S/I$ is associated a block decomposition in which the blocks are naturally ordered so that the $i$-th block $B_i=\mon(I_i)\smallsetminus \mon(I_{i-1})$, and when the block spaces are successively added (by direct sum) to the $I$ (considered as a vector space), the ideals of the filtration are reconstructed.  Thus the subprime filtration and the (ordered) block decomposition contain the same information.  Prime filtrations and (ordered) Stanley decompositions are related in the same way.

A convenient notation to exhibit the relationship between ideals and blocks in a subprime filtration is
\[
\begin{matrix} I=I_0 &\subset &I_1 &\subset &I_2 &\cdots &I_{s-1} &\subset &I_s=S.\\
{} &\vert &{} &\vert &{} &{} &{} &\vert \\
{} &B_1 &{} &B_2 &{} &{} &{} &B_s \end{matrix}
\]

For example, the block decomposition (\ref{bd1}) is associated with a subprime filtration as follows:
\[
\begin{matrix} \langle xy^3,x^3y\rangle &\subset &\langle xy \rangle &\subset &\langle y \rangle &\subset &\langle 1 \rangle,\\
{} &\vert &{} &\vert &{} &\vert &{}\\
{} &B_1 &{} &B_2 &{} &B_3 &{}
\end{matrix}
\]
with $B_1=B^3$, $B_2=B^2$, and $B_3=B^1$.

\brem\label{filtrationexample} Observe that the ideals in the filtration can be found by adding the inner corner of each block to the generators of the previous ideal.  For instance, adding the inner corner $(1,1)$ of $B_1=B^3$ to $\langle xy^3,x^3y\rangle$ gives $\langle xy^3,x^3y,xy\rangle=\langle xy\rangle$.  See Lemma \ref{obviouslemma} below.\erem

It is known that there exist Stanley decompositions that are not associated in this way with prime filtrations; thus there are also block decompositions that are not associated with subprime decompositions.  The standard example, introduced in \cite{McLagan-Smith}, is $S=K[x,y,z]$, $I=\langle xyz\rangle$,
\begin{equation}\label{MSexample}
S/I\cong K\oplus K[x,y]x\oplus K[y,z]y \oplus K[x,z]z.
\end{equation}
In this example, the Stanley blocks cannot be ordered in such a way that when added successively to $I$, they produce monomial ideals.  Here is our version of Soleyman-Jahan's theorem characterizing Stanley decompositions that come from prime filtrations.

\bthm\label{ASJ} An ordered block decomposition $B_1\sqcup\dots\sqcup B_s$ for $S/I$ is subprime if and only if $I\oplus \Span B_1 \oplus \cdots \oplus \Span(B_i)$ is a monomial ideal for each $i=1,\dots,s$.\ethm

\bpf The $\Rightarrow$ direction has already been proved.  For the $\Leftarrow$ direction, the hypothesis implies that $
\mon(I)\sqcup B_1\sqcup\cdots\sqcup B_k$ is a vector space basis for an ideal $I_k$ for each $k$, and that these ideals form a filtration.  It is equally clear that each $\mon(I_k)\smallsetminus \mon(I_{k-1})=B_k$, so the filtration is subprime.\epf

Note that when written as a block decomposition, the blocks of (\ref{MSexample}) are unions of blocks from the elementary decomposition of $S/I$, just as is the gnomon decomposition.  This shows that although the elementary decomposition itself is subprime (Theorem \ref{elemnice} below), the result of compressing an elementary decomposition is not in general subprime.  But the gnomon decomposition, which is a compression of the elementary decomposition, is subprime (Theorem \ref{gnnice}).

If $B$ is any block (\ref{defblock}), and $b_i<\infty$, we define the $i$th {\bf outer adjacent face} of $B$ to be the block
\[
F^i(B)=\bm b_1 &\dots &b_{i-1} &b_i +1 &b_{i+1} &\dots &b_n\\
           a_1 &\dots &a_{i-1} &b_i +1 &a_{i+1} &\dots &a_n\ebm.
\]
(This is not a subset of $B$, but is displaced by one step in the direction $i$ from the ordinary $i$th {\bf outer face} of $B$.)  If $b_i=\infty$ we set $F^i(B)=\emptyset$.

\blem\label{obviouslemma} Let $B$ be a block with inner corner $\ba$, and let $I$ be a monomial ideal disjoint from $B$.  Let $M=B\sqcup \mon(I)$.  Then $M$ spans a monomial ideal (or equivalently, $M= \mon(J)$ for some monomial ideal $J$) if and only if the outer adjacent faces of $B$ are contained in $I$.  In this case $B\sqcup \mon(I)=\mon(\langle \bx^{\ba}\rangle)\cup \mon(I)$ and $J=I+\langle \bx^{\ba}\rangle$.\elem

\bpf A set $M$ of monomials spans an ideal if and only if every monomial divisible by ($\succeq$) an element of $M$ belongs to $M$.  Let $M=B\sqcup \mon(I)$ and suppose that $M$ spans a monomial ideal.  Let $\bc\in F^i(B)$ for some $i$.  Then $\ba\prec\bc$, so $\bc\in M$.  But $\bc\notin B$, so $\bc\in I$.  Therefore $F^i(B)\subset I$ for each $i$.  Conversely, suppose that $F^i(B)\subset I$ for each $i$.  Since every element of $\langle\ba\rangle$ that is not in $B$ is divisible by some element of $F^i(B)$, and therefore belongs to $I$, we conclude that $\langle \ba\rangle \subset M$. Therefore $M= \mon(I) \cup \mon (\langle \ba\rangle)=\mon(I+\langle \bx^{\ba}\rangle)$, so $M$ spans an ideal.  \epf

We are now ready to prove that elementary and gnomon decompositions (with suitable orderings) are subprime.

\bthm\label{elemnice}  Let $I\subset S=K[x_1,\dots,x_n]$ be a monomial ideal and let $B^1\sqcup\dots\sqcup B^s$ be the elementary block decomposition of $S/I$, ordered according to Remark \ref{elemorder}.  Let $B_k=B^{s-k+1}$ for $k=1,\dots,s$; that is, reverse the ordering of the blocks.  Then $B_1\sqcup\dots\sqcup B_s$
is a subprime decomposition.\ethm

\bpf  We work in the superscript notation, restating the criterion of Theorem \ref{ASJ} as
\begin{equation}\label{subprimecriterion}
B^k \sqcup\dots\sqcup B^s\sqcup \mon(I)\quad \text{spans an ideal for each}\quad k=1,\dots,s.
\end{equation}
The proof begins with $k=s$ and works backward.  Let $A=\{\ba^1,\dots,\ba^s\}$ be the refined list of inner corners of elementary blocks for $S/I$, created in step \ref{refine} of Algorithm \ref{ebd}, in lexicographic order according to Remark \ref{elemorder}.  Since $\bbm \prec \bbm'$  implies $\bbm <_{lex} \bbm'$, $\ba^s$ is a maximal element of $A$ under $\prec$.

We claim that each outer adjacent face $F^i(B^s)$ is contained in $I$.  Let $\bc\in F^i(B^s)$.  Then $\ba^s\prec\bc$ and $\bc\notin B^s$.  But by Remark \ref{elemprop}, if $\bc\notin I$ then $\bc$ belongs to the block of $S/I$ whose inner corner is maximal among inner corners less than $\bc$.  But this inner corner must be $\ba^s$, so $\bc\in B^s$, a contradiction.  Therefore $\bc\in I$, proving the claim.

By Lemma \ref{obviouslemma} we conclude that
\[
B^s \sqcup I = I + \langle \ba^s \rangle = I'
\]
is an ideal.  Setting $A'=\{\ba^1,\dots,\ba^{s-1}\}$, we see that $A'$ is the list of inner corners of elementary blocks for $S/I'$, which has $B^1\sqcup\dots\sqcup B^{s-1}$ for its elementary decomposition.  Now the argument can be repeated, with $\ba^{s-1}$ as the maximal element.\epf

Lexicographic ordering is not the only ordering that gives a subprime decomposition.  It is only necessary to select a maximal element of $A$, remove it, and repeat.

\bthm\label{gnnice} The $\pi$-gnomon decomposition $B^1\sqcup\dots\sqcup B^s$ of $S/I$ is subprime, for any $\pi$, under the ordering $B_k=B^{s-k+1}$.\ethm

The proof of this theorem actually establishes more, namely, that if Algorithm \ref{addgenerator} is used to ``add a generator'' to a subprime block decomposition, the result will be subprime.

\bpf  Since any $\pi$-gnomon decomposition becomes a standard gnomon decomposition by renumbering the variables, it suffices to prove the standard case. The argument is by induction on the number $r$ of minimal generators of $I$.  If $r=1$, then
\[ \mon(I)=\bm \infty &\cdots&\infty\\ m_1&\cdots&m_n\ebm
\]
and the gnomon decomposition is
\[
\bm \infty&\cdots&\infty\\ m_1&\cdots&m_n\\0&\cdots&0\ebm.
\]
In this case Lemma \ref{gnlemma} shows that when the terms of the gnomon decomposition are added in reverse order to the ideal, an ideal is obtained at each stage.

Now make the induction hypothesis that the theorem is true for all monomial ideals $I$ having $r$ minimal generators.  Let $I'$ be a monomial ideal with $r+1$ generators, and write $I'=I + \langle \bbm\rangle$, where $I$ has $r$ generators.  Suppose that $B^1\sqcup\cdots\sqcup B^s$ is the gnomon decomposition for $S/I$.    The gnomon decomposition for $S/I'$ is
\[
(B^1\smallsetminus\langle\bbm\rangle)\sqcup\dots\sqcup(B^s\smallsetminus\langle\bbm\rangle)
 = \bigsqcup_{k=1}^s\left(\bigsqcup_{i=1}^n B^{ki}\right)
\]
in the notation of Remark \ref{gnorder}. We need to prove that for any $p\in\{1,\dots,s\}$ and $q\in\{1,\dots,n\}$,
\begin{equation}\label{toprove1}
(B^{pq}\sqcup\dots \sqcup B^{pn})\sqcup \bigsqcup_{k=p+1}^s\left(\bigsqcup_{i=1}^n B^{ki}\right) \sqcup \mon(I')
\end{equation}
spans an ideal.  From here on we assume $p$ and $q$ are fixed.  By the induction hypothesis,
\begin{equation}\label{indhyp}
B^k \sqcup\dots\sqcup B^s\sqcup \mon(I)
\end{equation}
spans an ideal for each $k=1,\dots,s$.  This will be used for $k=p+1$ and $k=p$.  For $k=p+1$, let $J$ be the ideal spanned by
\begin{equation}\label{defJ}
B^{p+1}\sqcup\dots\sqcup B^s\sqcup\mon(I).
\end{equation}
For $k=p$, (\ref{indhyp}) together with Lemma \ref{obviouslemma} implies that the outer adjacent faces of $B^p$ belong to $J$.

Let
\[
B^k=\bm \bb^k\\ \ba^k\ebm \qquad\text{and}\qquad c^k_i=\max\{a^k_i,m_i\}.
\]
By Lemma \ref{intersect} we have
\[
\bm \bb^k\\ \bc^k \ebm = B^k \cap \langle \bbm\rangle \subset I',
\]
so we can include these terms in (\ref{toprove1}) without changing the result, provided we change the last disjoint union to an ordinary union because at that point there is now an overlap.  Thus (\ref{toprove1}) equals
\begin{align}\label{toprove2}
&\left(B^{pq}\sqcup\dots \sqcup B^{pn}\sqcup \bm \bb^p\\ \bc^p\ebm \right)\\
&\qquad\qquad \sqcup \bigsqcup_{k=p+1}^s\left(\left(\bigsqcup_{i=1}^n B^{ki}\right)\sqcup \bm \bb^k\\ \bc^k\ebm \right) \cup \mon(I').\notag
\end{align}
By Lemmas \ref{gnomonlem} and \ref{intersect} we have
\[
\left(\bigsqcup_{i=1}^n B^{ki}\right)\sqcup \bm \bb^k\\ \bc^k\ebm = B^k.
\]
Since (\ref{defJ}) spans $J$ and $\mon(J)\cup\mon(I')=\mon(J+I')$,
\[
\left(\bigsqcup_{k=p+1}^s B^k\right) \cup \mon(I')
\]
spans $J'=J+I'$.  These calculations reduce (\ref{toprove2}) to
\begin{equation}\label{toprove3}
\left(B^{pq}\sqcup\dots \sqcup B^{pn}\sqcup \bm \bb^p\\ \bc^p\ebm \right) \cup \mon(J').
\end{equation}
The induction is completed by considering two cases.
\blist
\item If $\bbm^p\npreceq\bb^p$, then (\ref{gnomonformula1}) applies at the point when the generator $\bbm^p$ is added during the construction of the gnomon decomposition.  So by Remark \ref{gnorder}, $B^{p1}=B^p$ and $B^{pi}=\emptyset$ for $i=2,\dots,n$.  So (\ref{toprove3}) is trivial unless $p=1$.  But in that case, (\ref{toprove3}) reduces to $B^p\cup \mon(J')$.  The outer adjacent faces of $B^p$ belong to $J$, and therefore also to $J'$.  So (\ref{toprove3}) spans an ideal.

\item  If $\bbm^p\preceq\bb^p$, then Lemma \ref{gnlemma} implies
\begin{equation}\label{lastpiece}
B^{pq} \sqcup\cdots\sqcup B^{pn} \sqcup \bm \bb^p\\ \bc^p\ebm =
\bm  b^p_1 &\cdots &b^p_{q-1} &b^p_q &\cdots &b^p_n\\
     c^p_1 &\cdots &c^p_{q-1} &a^p_q &\cdots &a^p_n \ebm.
\end{equation}
The outer adjacent faces of this block are contained in those of
\[
B^p=\bm b^p_1&\cdots&b^p_n\\a^p_1&\cdots&a^p_n\ebm,
\]
but as noted above, these  belong to $J'$, so again, (\ref{toprove3}) spans an ideal.
\elist
\epf

\brem\label{x4cont} To find the subprime filtration associated with a gnomon decomposition, as in Remark \ref{filtrationexample}, one should delete any generators that are divisible by each new inner corner that is added.  The proof of Theorem \ref{gnnice} shows that when a sequence of blocks $B^{pn},\dots,B^{p1}$ is added  to $\mon(J')$, their inner corners decrease in $\prec$.  Therefore each new corner added will replace the previous one (and may also remove others).  For instance, in Example \ref{x4} the inner corners $(3,2,0),(0,2,0),(3,1,0),\newline (3,0,0),(0,0,0)$ are added to the starting ideal with generators $(0,0,5),\newline (0,2,3),(3,1,1)$.  The sequences $(3,2,0),(0,2,0)$ and $(3,1,0),(3,0,0),\newline (0,0,0)$ each arise from a single gnomon, so the inner corners decrease in $\prec$.   The subprime filtration is
\begin{align*}
&\langle (0,0,5),(0,2,3),(3,1,1)\rangle \subset \langle (0,0,5),(0,2,3),(3,1,1),(3,2,0)\rangle\\
&\qquad\subset\langle (0,0,5),(3,1,1),(0,2,0)\rangle
\subset \langle (0,0,5),(0,2,0), (3,1,0)\rangle\\
&\qquad \subset \langle (0,0,5),(0,2,0),(3,0,0)\rangle \subset \langle (0,0,0)\rangle.
\end{align*}
When $(0,2,0)$ is added, it eliminates both $(3,2,0)$, as expected, and also $(0,2,3)$, which is not predicted by this remark.
\erem

We conclude the section with some technical remarks concerning the usual definition of prime filtration.  Recall that a monomial ideal $P\subset S$ is prime if and only if it is generated by a subset $Y$ of the variables $x_1,\dots,x_n$.  Since $S$ is a $\Z^n$-graded algebra, and monomial ideals in $S$ inherit this multi-grading, quotients of monomial ideals are $\Z^n$-graded modules.  If $P$ is a monomial ideal and $\ba\in\N^n$, the symbol $S/P(-\ba)$ denotes $S/P$ with its $\Z^n$-grading shifted so that $(S/P)(-\ba)_{\bb} = (S/P)_{\bb+\ba}$.  If $P=\langle Y\rangle$ is a prime ideal, $S/P$ is isomorphic as a vector space to $K[X]$, where $X$ is the complementary set of variables to $Y$.   Then $S/P(-a)$ is isomorphic as a vector space to the Stanley space $K[X]\bx^{\ba}$.  Now (\ref{filt}) is defined to be a prime filtration if there exist monomial prime ideals $P_i$ and integer vectors $\ba_i\in\N^n$ such that $I_i/I_{i+1}$ is isomorphic as a $\Z^n$-graded $S$-module to $S/P(-\ba_i)$. It is therefore also isomorphic as a vector space to $K[X]\bx^{\ba}$, but this is not part of the definition.

Using these definitions, Soleyman-Jahan proved Theorem \ref{ASJ} above for the case of prime filtrations, with the blocks being Stanley blocks.  The difficulty of his proof seems to result from the fact that the association of $I_i/I_{i-1}$ with a particular $S/P_i(-\ba_i)$ is only specified by an isomorphism, and not by a concrete construction as we have done.  It seems to be the case that if $S/P(-\ba)$ is isomorphic as a $\Z^n$-graded $S$-module to $S/P'(-\ba')$, then $P=P'$ and $\ba=\ba'$.  In other words, $P_i$ and $\ba_i$ are uniquely determined by $I_i$ and $I_{i-1}$.  This observation (if correct) would bring his treatment closer to ours.  However, this uniqueness is only possible if the isomorphism $I_i/I_{i-1}\cong S/P_i(-\ba_i)$ is interpreted as an $S$-module homomorphism; neither algebra homomorphism nor vector space homomorphism is strong enough to achieve the desired uniqueness.  With our approach there is no need to bring in the $\Z^n$ grading or the $S$-module structure.

redpajama_set_name: RedPajamaArXiv


Pastor John MacArthur argued that evangelicalism's "newfound obsession" with social justice is a threat to the Gospel.
In a Monday blog post on his Grace to You ministry website, MacArthur, leader of Grace Community Church of Sun Valley, California, and president of The Master's College and Seminary, promised to tackle the notion of "social justice" in a series over the next two weeks.
The focus on social justice by evangelicals is shifting focus away from the Gospel, he claimed.
"Evangelicalism's newfound obsession with the notion of 'social justice' is a significant shift — and I'm convinced it's a shift that is moving many people (including some key evangelical leaders) off message, and onto a trajectory that many other movements and denominations have taken before, always with spiritually disastrous results," he wrote.
"Over the years, I've fought a number of polemical battles against ideas that threaten the gospel. This recent (and surprisingly sudden) detour in quest of 'social justice' is, I believe, the most subtle and dangerous threat so far," he said.
"I deplore racism and all the cruelty and strife it breeds. I am convinced the only long-term solution to every brand of ethnic animus is the gospel of Jesus Christ. In Christ alone are the barriers and dividing walls between people groups broken down, the enmity abolished, and differing cultures and ethnic groups bound together in one new people (Ephesians 2:14–15). The black leaders with whom I ministered during the civil rights movement shared that conviction," he said.
MacArthur was not immediately available for further comment when Grace to You was contacted Monday.
He argued without highlighting any specific names that the talking points of the most vocal evangelicals on social justice are too worldly.
Anthony Bradley, chairman of the program in religious and theological studies and associate professor of religious studies at King's College in New York City, called MacArthur's analysis "a pile of ideological conservative rubbish," in a Monday tweet.
"This. John MacArthur face plants in a pile of ideological conservative rubbish. If evangelicalism is going to have a future that reflects the Bible's teachings, nearly everything about this post must be in its past. Terrible," he wrote.
Are We on the Road to the Collapse of Civilization?


Please can you share the code you're using which generates those errors.
Hi @robertgauld, I've created a gist and updated the description.
Can you humor me and make the following chnage, it shouldn't make a difference but it just might.
+1 getting same error here. Also using separate date and time selects.
fyi: http://arjanvandergaag.nl/blog/rails_attributes_magic.html gives a good run down on whats going on, rails is just assigning these virtual attributes as strings.
Be sure to not have ignore_date as part of your options on your time select, or you will run into the same issue.


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Madappa N. Kundranda, M.D., Ph.D.
Copyright (©) 2017 Immunovative Therapies. All rights reserved.


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Suzuki KK, Yanagawa H (2019). Gliding patterns of Siberian flying squirrels in relation to forest structure. iForest 12: 114-117.
It is widely accepted that the evolution of gliding ability is correlated with forest environments, but differences in gliding locomotion in relation to forest structure remains poorly elucidated in mammals. Although the cost of gliding locomotion decreases with increasing glide distance per unit vertical drop (glide ratio), gliding mammals often use costly low-ratio glides and seldom exploit maximum-ratio glides. In this study, we evaluated our hypothesis that low-ratio glides are related to forest structure by measuring glide distance, vertical drops and landing tree heights in Siberian flying squirrels (Pteromys volans), and we also recorded their behaviour in landing trees. Glide ratio decreased with increasing landing tree height. Squirrels landed on taller trees using low-ratio glides and tended to depart from them quickly without spending much time there, but used high-ratio glides to land on shorter trees for foraging or nesting. Thus, flying squirrels use two different gliding behaviours depending on their immediate objective, where inefficient low-ratio glides are used to move to higher trees for continued gliding. This approach might be necessary for efficiency and safety in subsequent glides, because taller trees facilitate long-distance glides and significantly decrease energy costs and landing impact. Therefore, the location of tall trees in forests and/or average canopy height might alter glide path routes. This study provides important evidence that forest structure affects gliding patterns and provides insight on how forest management could influence the gliding locomotion of Siberian flying squirrels.
Gliding flight in the Japanese giant flying squirrel Petaurista leucogenys. Journal of Mammalogical Society of Japan 18: 19-32.
Glide performance and aerodynamics of non-equilibrium glides in northern flying squirrels (Glaucomys sabrinus). Journal of the Royal Society Interface 10 (80): 20120794-20120794.
Take-off and landing kinetics of a free-ranging gliding mammal, the Malayan colugo (Galeopterus variegatus). Proceedings of the Royal Society B 275: 1007-1013.
The physics of leaping animals and the evolution of preflight. American Naturalist 121: 455-467.
Gliding and the functional origins of flight: biomechanical novelty or necessity? Annual Review of Ecology, Evolution and Systematics 38: 179-201.
Tropical rain forest structure and the geographical distribution of gliding vertebrates. Biotropica 22: 432-434.
Locomotor performance in the squirrel glider, Petaurus norfolcensis, and the sugar glider, Petaurus breviceps. Australian Mammalogy 30: 25-35.
Gliding mammals of the world: diversity and ecological requirements. In: "Biology of Gliding Mammals" (Goldingay RL, Scheibe JS eds). Filander Press, Fürth, Germany, pp. 5-40.
Gliding performance in the yellow-bellied glider in low-canopy forest. Australian Mammalogy 36: 254-258.
Gliding performance and its relevance to gap crossing by the squirrel glider (Petaurus norfolcensis). Australian Journal of Zoology 57: 99-104.
A Jurassic gliding euharamiyidan mammal with an ear of five auditory bones. Nature 551: 451-456.
Evolution of gliding in Southeast Asian geckos and other vertebrates is temporally congruent with dipterocarp forest development. Biology Letters 8: 994-997.
Glide angle in the genus Petaurus and a review of gliding in mammals. Mammal Review 30: 9-30.
Gliding performance of the red giant gliding squirrel Petaurista petaurista in the tropical rainforest of Indian eastern Himalaya. Wildlife Biology 22: 7-12.
New gliding mammaliaforms from the Jurassic. Nature 548: 291-296.
Kinematics: gliding flight in the paradise tree snake. Nature 418: 603-604.
Gliding behavior of Japanese giant flying squirrels (Petaurista leucogenys). Journal of Mammalogy 83: 553-562.
Gliding locomotion of Siberian flying squirrels in low-canopy forests: the role of energy-inefficient short-distance glides. Acta Theriologica 57: 131-135.
Nest cavity selection by the Siberian flying squirrel Pteromys volans. Hystrix 24: 187-189.


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Home/News / USGLC Announces Cargill as Heartland Chair
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USGLC Announces Cargill as Heartland Chair
Two-year partnership harnesses the power of global development and diplomacy to advance America's leadership in the world economy
WASHINGTON, D.C (June 20, 2019) – The U.S. Global Leadership Coalition (USGLC) is pleased to announce the launch of an innovative partnership with Cargill to increase and amplify powerful conversations on global development and the critical role America's ag economy can play in advancing U.S. interests around the globe.
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Cargill's role as Heartland Chair kicked off at USGLC's State Leaders Summit on June 17, 2019 in Washington, D.C. Tour events, intended to strengthen rural leadership voices and build a network of military, business, NGO, faith and community leaders as advocates for the local economic and security benefits of global engagement, will resume in August 2019.
Jennie Bragg, USGLC, 202-730-4170 and April Nelson, Cargill, 952-913-5466
About USGLC
The U.S. Global Leadership Coalition (www.usglc.org) is a broad-based influential network of 500 businesses and NGOs; national security and foreign policy experts; and business, faith-based, academic, military, and community leaders in all 50 states who support strategic investments to elevate development and diplomacy alongside defense in order to build a better, safer world.
Cargill's 155,000 employees across 70 countries work relentlessly to achieve our purpose of nourishing the world in a safe, responsible and sustainable way. Every day, we connect farmers with markets, customers with ingredients, and people and animals with the food they need to thrive.
We combine 153 years of experience with new technologies and insights to serve as a trusted partner for food, agriculture, financial and industrial customers in more than 125 countries. Side-by-side, we are building a stronger, sustainable future for agriculture.

redpajama_set_name: RedPajamaCommonCrawl


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mz D i D i: SHOPPING IS GOOD!
Good for me but not so good for my savings!
I went to Bardot and as soon as I walked in I saw the accessories rack. And I came across this cute bow/ribbon bracelet! There was another one that had three rows of pearls with a bow attached to it but that looked too OTT so I went with the simpler one, easy to wear every day!
I'm still a lil confused about the actual bow part coz I duno if it's upside or not..
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text:
Cyber Industry Marketing Pharma Research Solutions
Artificial Intelligence, Real Concerns: Hype, Hope and the Hard Truth About AI
Artificial intelligence ( AI ) is generating both interest and investment from companies hoping to leverage the power of autonomous, self-learning solutions.
copyright by securityintelligence.com
Artificial intelligence ( AI ) is generating both interest and investment from companies hoping to leverage the power of autonomous, self-learning solutions. The Pentagon recently earmarked $2 billion in funding to help the Defense Advanced Research Projects Agency (DARPA) push AI forward, and artificially intelligent solutions are dominating industry subsets such as medical imaging , where AI companies raised a combined $130 million worth of investments from March 2017 through June 2018. Information security deployments are also on the rise as IT teams leverage AI to defeat evolving attack methods, and recent data suggests that AI implementation could both boost gross domestic product (GDP) and generate new jobs.
It's easy to see AI as a quick fix for everything from stagnating revenues to medical advancement to network protection. According to a recent survey from ESET , however, new increasing business expectations and misleading marketing terminology have generated significant hype around AI , to the point where 75 percent of IT decision-makers now see AI as the silver bullet for their security issues.
It's time for an artificial intelligence reality check. What's the hype, where's the hope and what does effective implementation really look like? What Are the Current Limitations of Artificial Intelligence?
AI already has a home in IT security. As noted in the Computer Weekly article, machine learning tools are "invaluable" for malware analysis since they're able to quickly learn the difference between clean and malicious data when fed correctly labeled samples. What's catching the attention of chief information security officers (CISOs) and chief information officers (CIOs) right now, however, is the prospect of AI tools that require minimal human interaction to improve network security .
This comes down to the difference between supervised and unsupervised machine learning — current tools and technologies empower the former, but the latter is still largely out of reach. Without humans to monitor the input and output of systems, it's possible for AI tools to capture and report basic system data, but beyond their scope to design intelligent threat response plans of the silver-bullet variety.
AI also has basic limitations that may be inviolate or may require a new research approach to solve. This is largely tied to experience: As noted by Pedro Domingos, professor of computer science at the University of Washington and author of "The Master Algorithm," machines don't learn from experience the same way humans do.
"A robot can learn to pick up a bottle, but if it has to pick up a cup, it starts from scratch," said Domingos, as reported by Wired.[…]
read more – copyright by securityintelligence.com
TagsAI • CMO • DL • ml • robot
$_kumar💫Saurabh💫
@SwissCognitive Really, I loved this.. 🤗🤗👍🎉🎊


Whether you are looking for a professional forester embossed seal, wet stamp or digital seal Acorn Sales has the right Forester Seal & Stamp for you. All of our professional stamps and seals are made to your state's specifications for your designation so they can vary from 1-5/8" to a 2" inch diameter impression area. We fully guarantee that our Professional Forester Seals and Professional Forester Stamps will meet the seal of approval for your state board.
Ordering online at Acorn Sales for your professional forester seal and stamp is both quick and easy! Have you heard about our extended long reach embosser? The Forester extended long reach desk seal embosser it is extremely heavy duty. It has a 4-1/2" reach, thus enabling you to emboss the center of an 8-1/2" wide piece of paper. This extended long reach embosser includes an adjustable paper guide that ensures your impression is right where you want it.
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Peter (Pete) D. Winiarski is a speaker, transformational leader, business consultant, results coach, and author of the # 1 International Best Seller Act Now! A Daily Action Log for Achieving Your Goals in 90 Days.
He is the founder and CEO of Win Enterprises, LLC and the creator of the Win Holistic Transformation Model, a framework to help business leaders achieve transformational results. In the business world, Pete works with management teams to improve the processes that shape their results and to shape the leadership and culture of the whole organization.
To help individuals maximize their results, Pete coaches them to clarify their goals, align their beliefs, and develop strategies to achieve them, including daily practices to create powerful success habits. Pete describes this methodology in his book, Act Now! a Daily Action Log For Achieving Your Goals in 90 Days.
The clients Pete has supported range from individual entrepreneurs and small business owners to international corporations with $80 Billion in revenue.
Pete is also the co-founder of the Business Consultant Institute.
Pete has degrees in Engineering and Business. He is also a graduate of Jack Canfield's elite Train the Trainer program, and is a practitioner of NLP (Neuro Lingustic Programming), RIM (Releasing Your Inner Magician), EFT (Emotional Freedom Technique), The Sedona Method, and other methods to help you and your team achieve your full potential.
Many of us set goals. We're driven by high aspirations of a better life with better results. We know that we need to convert our vision into discrete goals that will make our vision a reality. Despite the goals we set, we often find them elusive. No matter what books we've read or training we've attended, there's still something critical we need to do – Take Action! But… the type of action MATTERS!
By following the 5 C's of goal achievement and taking all three types of action we begin to move toward our goals rather than standing still and wondering why we have yet to hit them. Taking Action makes things happen. Action is Everything!
•	How to set goals and create plans to achieve them.
•	The most important Daily Habits to develop and incorporate into your daily routines.
•	How to recognize Intuitive Hits and know how to immediately take Inspired Actions when you do.
"Lean" has unfortunately received status as a buzzword, where everyone thinks they know what lean is yet few have truly been through a full transformation. Depending on research, 70%-90% of transformations fail. Engagement is a critical element to help drive your success rate up, yet most companies do a poor job here, too.
In this session you will learn to recognize the different mistakes companies make and also learn what systems to put in place to ensure your success.
•	How to integrate best practices to ensure that you obtain and sustain results that thrill you!
Win Conscious Leadership™ is leading from a higher state of personal and professional awareness, allowing you to find new and effective solutions to challenges. Conscious leaders have examined themselves sufficiently to know their strengths and weaknesses, and they have learned how to be more mindful and aware of themselves and those around them.
These leaders offer compelling visions that inspire their followers. Conscious Leaders focus on "we", rather than "me." Anyone knowing them would say they are authentic. They walk their talk.
Our lives as business leaders can be stressful as we strive for HUGE GOALS and manage the daily business dynamics with customers, employees, suppliers, and owners all demanding our time and energy as we attempt to deliver greater results. In our quest to achieve at the highest level and satisfy all the demands on our shoulders, we will either rise to the occasion or crumble under the pressure. Interestingly, appreciation and gratitude are potent forces you can put to work on your behalf to ensure you perform at your highest level and achieve the results you desire… without the negative impact of stress.
Countless studies have proven that expressing appreciation and gratitude helps you both improve your health and achieve your goals. You can learn to harness the power of appreciation and gratitude through simple, daily actions that will develop positive habits and will create shifts in your performance level and will ultimately shift the performance of your team and company.
Pete Winiarski is one of the most dedicated and knowledgeable professionals I've had the pleasure to work with in many years. His singular focus on client service and positive outcomes shines through everything he does. Give Pete a call - he'll work wonders for your business.
Pete is the go to authority on leadership and building a consulting practice. His experience and processes offer audiences rich and actionable information. Plus, he's a great guy to know with lots of resources and connections.


A kitchen is the "heart of the home" due to the versatility and the frequency the area is used. Having a functional and aesthetic kitchen design sets the appearance and style of the entire home. Kitchen spaces are multi-functional and are commonly used both by guests and residents for eating, cooking, and entertaining. People are usually drawn to the kitchen more than any other room in the home.
Having a kitchen space that is clean, visually pleasing, and highly functional can help promote a better overall atmosphere in the home. A custom kitchen makes cooking more efficient and enjoyable, as well as make entertaining easier and more natural. A kitchen renovation is one of the most valuable renovation projects a homeowner can undergo, greatly improving the resale value of a home.
There are so many reasons to renovate a kitchen space, to increase property value, improve the look and comfort of a home, improve the use and functionality. At Golden Acre Renovations, we can help with each step of your kitchen renovation in Calgary including the design, planning, construction, and cabinetry/appliance installation. Contact us today if you have any questions regarding your kitchen renovation project or if you'd like to book a consultation.
More Space – Older kitchen spaces can feel bulky and cluttered. A kitchen renovation can add valuable space and a natural flow to your current kitchen space.
Easier Home Cooking – With improved appliances and a new kitchen layout, your home kitchen can feel much more attractive and inviting for home cooking and eating. This can save you money on takeout and restaurants.
Resale Value – Improving your kitchen space increases the value of your home when it is time to sell, making it more appealing to prospective buyers.
Upgraded Appliances – New appliances can make cooking more fun and efficient, while making your home safer and "greener". New and matching appliances can also add a more consistent and upscale look and feel.
Backsplashes – Backsplashes can add a burst of colour, a unique texture or an artistic addition incorporated with your stove's hood vent.
Flooring – The flooring can be made from several materials and colours, and can be installed with custom finishings and transitions to create a unique design.
Cabinets – Cabinets are a key element in kitchen design and function. Cabinetry can often be the first thing individuals notice when they enter a kitchen and set the design and feel of the kitchen.
Countertops – Metallic, granite, marble, stained wood, and other materials can be used as kitchen countertops.
Paint Colour & Texture – The overarching colour and feel of the space can be defined by paint. Paint can compliment wood stains and can add a balance of colour in a mainly white space.
Lighting – There are numerous lighting options available for kitchen lighting. Natural light from large windows, LED lighting inside of cabinets, accent lighting, and custom lighting fixtures can all be installed to better illuminate a kitchen.
Contemporary Kitchens – The kitchen style of the 21st century. Contemporary kitchens blend elements of new and old to create a balanced kitchen design that is trendy and modern in look and feel. Contemporary kitchens use innovative appliances, a well-defined work triangle, open concept, and flat panel cabinets, while allowing for lots of room for seating and entertaining guests.
Modern Kitchens – Modern kitchen renovations create sleek and efficient kitchens that have flat and geometric designs. A modern kitchen often has flat-paneled cabinetry, horizontal lines and reflective or polished surfaces with metallic accents. These spaces forgo detailed designs and fabrics to maintain a highly modern and clean look.
Rustic Kitchens – Perfect for country living, a rustic kitchen design aims to resemble a western feel that is common in rural properties and farmhouses. These spaces have a range of wooden elements and old-world accents to create a relaxed and "outdoorsy" atmosphere.
French-Country Kitchens – These kitchens are glamorous and highly detailed. Timeworn surfaces, made from detailed and/or distressed wood and upscale metallic finishes are common aspects of country kitchens. French-country kitchens resemble a look and feel similar to a villa in the Southern areas of France.
Traditional Kitchens – Traditional kitchen renovations create classic kitchen designs that are both clean and comfortable. Wooden or glass-paneled cabinets, custom woodwork, an emphasis on white, and vintage lighting fixtures commonly build-up this design.
The terms "renovation" and "remodel" are often used interchangeably. Though, the meanings of each term are not the same thing. A kitchen renovation aims to revitalize a kitchen space through material and appliance upgrades. A kitchen remodel alters the size and/or layout of the existing space to offer a new flow and more personalised look and feel to the kitchen. If your current kitchen layout is the main reason you want to have work done, a kitchen remodeling service will better suit your home improvement needs.
Your kitchen cabinetry is a key part of the design and function of your kitchen. Cabinets can be made with a range of flat or raised panels, using a variety of styles and materials. Cabinets can be made frameless and overhang or can offer exposed hinges in a framed-design. Cabinets can be made with shelving or drawers while being crafted using a variety of materials, colours, stains, and closing mechanisms. Cabinets improve the function and storage of a kitchen, where these storage options can be used inside the cabinet and on the backside of cabinet doors to increase the capacity and organizational capabilities of the home's kitchen. Renovating your kitchen often involves the replacement or refinishing of cabinets; these are considered kitchen cabinet renovations and are a key component of any kitchen renovation project.
Your countertops can provide increased space for food preparation or for serving guests. Kitchen countertops can be made from several different materials. These are 5 popular custom countertop materials found in Calgary kitchens.
Granite Countertops – Granite can offer an elegant appeal and is quite durable. Granite is a classically upscale and natural rock that is perfect for luxurious and traditional kitchens.
Marble Countertops – Marble offers a base tone, coupled with distinct veining that works together to offer natural and unique aesthetics.
Quartz Countertops – Unlike granite and marble, quartz is a man-made stone that offers a similar look to natural stones, while providing a more durable finished countertop.
Laminate Countertops – These are often the most affordable option for kitchen countertops. Though, this material cannot offer the same durability from impact or heat that a stone or metal-based countertop can. If customization and affordability are your goals, then a laminate countertop is a great option.
Butcher Block Countertops – Butcher-block countertops help create a warm and inviting atmosphere. Though, they can require annual maintenance with sealing and oil application to avoid water damage and staining on the natural wood. These can be ideal as a section of a countertop that utilizes stone, metal or laminate as the main material.
For more information about us and our services, please visit our contact page.
Ask Us Your Renovation Questions!
Send us an email regarding your renovation ideas or questions, and we will get back to you as soon as possible!


Our 100% SSD global network helps speed up your site. Our network is equal part site accelerator and real-time control center. It was built so that both your customers and DevOps will get the maximum benefit of a next-gen CDN.
FetchCDN's servers are strategically located for today's warp-speed internet—not dial-up—and have direct reach into over 90 countries. We provide a reliable CDN service with high peering capacity to handle any given load around the globe.
FetchCDN boasts intelligent best-path routing and on-the-fly analysis to monitor packet loss and latency. We're constantly adjusting to sustain site speed and keep your data flowing smoothly.
Our custom FastStack is powered by highly optimized NGINX caching on 100% SSD bare metal servers. Combine instant provisioning, purging, and reporting with smart caching directives to facilitate easy offloading of more static and dynamic content to the Edge.
We have put our edge locations in premium data centers with best-in-class hardware for increased speed and reliability.
Our globally placed Origin Shield POPs protect your origin from request overload giving you the full benefit of our CDN.
Each FetchCDN POP delivers the newest assets to your site users. If your cache is reset or you have a constant stream of new, uncached content, your origin would receive requests from every FetchCDN POP for the newest assets. That can quickly overload your origin.
Once it's enabled, all requests will go through one of our globally placed Origin Shield POPs instead of every Edge Server. This significantly reduces the number of requests going to your origin. Our EdgeServers will make requests to the closest Origin Shield instead.
If you have a lot of fresh content that needs to be provisioned on a regular basis, Origin Shield will protect you around the clock. If you have large assets on your site and require the occasional cache purge, Origin Shield will be there for those large request spikes.
For optimal performance our Origin Shields are located across multiple regions and are connected via 10Gbps data centers. We will take the most direct path to your origin possible so Origin Shield requests won't negatively impact your users experience.
This merges multiple requests for the same URI into one request to your origin server for even greater optimization. This also results in notifying the origin of a cache re-validation request before the object even becomes stale.
Each of our Origin Shield locations is equipped with hardware load balancers that provide equal hardware load and fail-over ability. Additionally, all of our Edge servers are configured to use 1 Origin Shield location as the primary but still have the others as a backup. If there is network connectivity issues, our Edge servers can intelligently re-route to the next nearest shield.
We're obsessed with speed! So should YOU!
Think of peering partner locations as accelerator POPs. They reduce the number of hops between ISPs (internet service providers) which also provides a more direct route to your mobile users. It's all about SPEED & DELIVERY!
Our API honors your purge calls instantly regardless of whether assets are purged through Control Panel or via your own custom built app.
All of our SSL options can be enabled instantly. Secure up and move on.
Quickly grasp the behavior of your content and traffic you make, without delays. As soon as traffic has been made, we'll digest data from all edge locations and record it for you to monitor from your end.
EdgeSSL uses our global architecture to handle the CPU intensive security your site needs and can be enabled in real-time.
Simply click "Enable" in our Control Panel and your SSL will start working in real-time. No manual installations or waiting to deploy.
We use the power of our global architecture to secure your site right where your customers are. Not only is this faster, but it also frees up your origin server from this CPU intensive task.
FetchCDN's optimized SSL stack, advanced hardware and always-on provisioning make our SSLs fast and cost-effective.
All users can bring their own SSL certificates bought from any third-party and install them on any pull zones. It can be easily done using our Control Panel where we simply ask to paste the SSL certificate.
All pull zones can be enabled to use FetchCDN's wildcard SSL for the domain *.netdna-ssl.com. This way we offer a free and easy way for websites to get started with HTTPS enabled traffic.
We are using Nginx to serve all our traffic with a heavily tweaked an optimized configuration for HTTPS. Any SSL related change is instantly propagated to all of our servers in real time saving you time.
FetchCDN integrates easily with your system, no matter if it's a popular CMS or a full-blown custom setup.
FetchCDN integrates easily with many popular CMS, including WordPress, Joomla, Drupal and much more.
If you have a custom built system for your website you can easily deploy your site's static content through FetchCDN.
Need more assistance? Call us for a free 1-on-1 integration call.
Fetch CDN, a service of DMI Web Services, delivers upon more than a decade of hosting industry experience and customer care. Our team has done the research to ensure that the latest and most stable technology is made available to our clients and that you won't have to do any heavy lifting. We provide you direct access to the best-in-class edge network to ensure your content delivers on-time, every time! You'll have peace of mind knowing that your content is hosted on the same network as some of the fastest responding large blogs, ad networks, online gaming and widget hosting services on the Web. Contact us today to get started.
Did you know a 1 second delay in your website page load time can cause a 7% reduction in conversion†? This can have a huge impact on your bottom line which is why we work hard to maintain top page load times.
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text:
Hans Putz (November 17, 1920 in Vienna – January 31, 1990 in Hamburg) was an Austrian actor.  He had a stage career, including working at the Schauspielhaus Zürich and the Volkstheater in Vienna; he also appeared in a number of films.

Partial filmography

 Gottes Engel sind überall (1948) - Franz Wudra
 Arlberg Express (1948) - Flori Reutner
 Lambert fühlt sich bedroht (1949) - (uncredited)
 Rosen der Liebe (1949) - Praterbesucher Ferdl
 Der Schuß durchs Fenster (1950) - Chauffeur Strinzel
 The Fourth Commandment (1950) - Martin Schalanter - beider Sohn
 Großstadtnacht (1950)
 Four in a Jeep (1951) - Karl Idinger
 Gangsterpremiere (1951) - Willi
 Wienerinnen (1952) - Paul Rosenauer
 Verlorene Melodie (1952) - Toni
 Lavender (1953) - Kramer
 Anna Louise and Anton (1953) - Robert
 König der Manege (1954) - Jack Holl, Aktrobat, Tänzer
 Bruder Martin (1954) - Steighofer
 Hoheit lassen bitten (1954) - Lembke
 The Eternal Waltz (1954) - Alexander Girardi
 The Doctor's Secret (1955) - Charly
 His Daughter is Called Peter (1955)
 The Dairymaid of St. Kathrein (1955) - Franz
 Ein Herz schlägt für Erika (1956)
 Teenage Wolfpack (1956)
 Ein Mann muß nicht immer schön sein (1956) - Taschenkrebs
 For Love and Others (1959) - Dr. Morsky
 Im Namen einer Mutter (1960) - Wendland
 Hin und her (1963)
 In der Sache J. Robert Oppenheimer (1964) - Borris T. Pasch
 I Am Looking for a Man (1966) - Bauunternehmer Märtens
 The Murderer with the Silk Scarf (1966) - Toni Stein (voice, uncredited)
 The Smooth Career (1967) - Wolf Kamper
 Sir Roger Casement (1968, TV Series) - Arthur Conan Doyle
  (1968, TV Movie) - Oberkommissar Leo Klipp
 St. Pauli Report (1971) - Paul Besser
 Der letzte Werkelmann (1972) - Leo Wessely
 Die Konsequenz (1976, TV Movie) - Dusterer
 Moselbrück (1987–1993, TV Series) - Ludwig Zerfass (final appearance)

References

1920 births
1990 deaths
Austrian male stage actors
Male actors from Vienna
Austrian male film actors
20th-century Austrian male actors

redpajama_set_name: RedPajamaWikipedia


text:
Q: react native webview, open google maps URL with the google maps application In the WebView of my react-native application, I have an embedded google maps url like:
<WebView
   source={{html: `<iframe src="https://www.google.com/maps/d/embed?mid=1r7Yvqak34P-HrL8uZLOivB720i4dpXah" `}}
/>

this embedded maps view contains a full-screen/expand button, where the RN app on Android pressing that expand button opens the URL in the google map application, but on iOS it just turns to full screen within the iframe
extra info: the content of WebView is fetched from the server and created by users, so it is not always google maps URL

*

*is there any way to open the URL with google maps on iOS too?

*if I want to use Linking how should I intercept the link inside that iframe

redpajama_set_name: RedPajamaStackExchange