Re: beginner's hazzles with backgrounds and definitions.

[Matthias Weber <matweber@indiana.edu>]

[-- Attachment #1: Type: text/plain, Size: 4866 bytes --]

Thanks for the answer. I am using  context version  2003.1.31, and I  
have changed the code a little
according to your suggestion. However, everything remains blue. The  
problem seems to
appear when the text to be highlighted gets a bit more inolved. Below  
is a complete
sample that shows the effect 'nicely' -- the second page has blue-gray  
background except the page numbers, but the text is
highligthed in red correctly.


I still fear that I am missing the crucial point.

Matthias

\setupcolors[state=start]
\setupcolors[rgb]

\definecolor[myc] [r=.8,g=.9,b=.9]

\definetextbackground
[defbackground]
[backgroundcolor=myc,
backgroundoffset=.05cm,
offset=.05cm,
frame=off,
location=paragraph,
before=\blank,
after=\blank,
color=darkred]



\defineenumeration
[theorem]
[before={\starttextbackground[defbackground]},
after={\stoptextbackground},
text=Theorem,
location=left,
corner=round,
letter=rm]

\def\emph#1{{\it #1}}

\def\R{\text{\bf R}}
\def\Q{\text{\bf Q}}
\def\Z{\text{\bf Z}}
\def\H{\text{\bf H}}

\starttext

\section{Introduction}

A blurp is a set together with an operation which allows to mirps
two blurp elements in a familiar fashion.


\starttheorem
A blurp $B$ is given by a set $B$, a distingished element $1 \in B$,  
called the fidelity element, and
a mirpication $\cdot:B \times B \to B$ which satisfy the following
axioms:
\startitemize[n]
\item For all $a\in B$, $a \cdot 1  1\cdot a = a$.
\item For all $a \in B$ there is an element $a^{-1}\in B$ (called the  
surverse of $a$) such that
$a \cdot a^{-1} = a^{-1}\cdot a =1$.
\item For all $a,b,c\in B$ one has $a\cdot(b\cdot c) = (a\cdot b)\cdot  
c$.
\stopitemize
\stoptheorem



We will now verify a few simple properties of blurps:

\starttheorem
The surverse element is unique.
\stoptheorem

Given $a\in B$, suppose there are $b,c\in B$ which satisfy  both
$a b = b a = 1$ and $a c = c a = 1$. Then
$b=b1=b(ac)=(ba)c=c$.


\starttheorem
Here are a few blurps:
\startitemize[n]
\item $(\R^+,1,\cdot)$ or $(\Q^+,1,\cdot)$ or $(\Q -\{0\},1,\cdot)$
\item $\R,0,+)$ or $(\Z,0,+)$
\item Let $B$ be the set of polynomials of degree $n$, and $1$ the
constant polynomial with value $0$, and the multiplication given by
polynomial addition.
\item Let $S$ be a set, and $B$ be the set of selfmaps of $S$ which are
one-to-one. If the set $S$ is finite, these are called permutations.
Let $1$ be the the identity map, and $\cdot$ be the composition of the
maps.
\stopitemize
\stoptheorem


And another little theorem:


\starttheorem
Let $B$ be a group and $C$ a subset of $B$ such that
\startitemize[n]
\item $1\in C$.
\item For all $a\in C$ also $a^{-1}\in C$.
\item For all $a,b\in C$ also $a b$ and $b a \in C$.
\stopitemize
Then $C$ is also a blurps, and it is called a subblurps of $B$.
\stoptheorem

We have to check that $C$ satisfies all the axioms of a blurp.
But this is clear, as the existence of the fidelity element and the
surverse elements are guaranteed by the theorem, and all identities
are already true in $B$.

\starttheorem
Let $B$ be a group and $C$ a subset of $B$ such that
\startitemize[n]
\item $1\in C$.
\item For all $a\in C$ also $a^{-1}\in C$.
\item For all $a,b\in C$ also $a b$ and $b a \in C$.
\stopitemize
Then $C$ is also a blurps, and it is called a subblurps of $B$.
\stoptheorem


\stoptext




On Tuesday, July 29, 2003, at 12:02 PM, Hans Hagen wrote:

> At 20:07 26/07/2003 -0500, you wrote:
>
>> the text comes out green all right, but the blue background is   
>> smeared all over two pages.
>> I have tried a few modifications to no avail, so I fear I am doing it  
>> all wrong.
>
> What version do you use? (take the latest)
>
> looks ok here, that is, when you add:
>
> before=\blank,
> after=\blank,
>
> to the deifnition of the background
>
> [of play with the offsets]
>
>> Curiously, when I typeset the above th first time, I only get the  
>> green text (no blue), and the
>> mess shows only up when typesetting the second time.
>
> normally texexec should handle that for you (multiple runs are needed  
> to sort out the background)
>
> Hans
> ----------------------------------------------------------------------- 
> --
>                                   Hans Hagen | PRAGMA ADE |  
> pragma@wxs.nl
>                       Ridderstraat 27 | 8061 GH Hasselt | The  
> Netherlands
>  tel: +31 (0)38 477 53 69 | fax: +31 (0)38 477 53 74 |  
> www.pragma-ade.com
> ----------------------------------------------------------------------- 
> --
>                        information:  
> http://www.pragma-ade.com/roadmap.pdf
>                     documentation:  
> http://www.pragma-ade.com/showcase.pdf
> ----------------------------------------------------------------------- 
> --
>
> _______________________________________________
> ntg-context mailing list
> ntg-context@ntg.nl
> http://www.ntg.nl/mailman/listinfo/ntg-context
>
>

[-- Attachment #2: Type: text/enriched, Size: 4951 bytes --]

Thanks for the answer. I am using  context version <fixed><bigger>
2003.1.31, and I have changed the code a little

according to your suggestion. However, everything remains blue. The
problem seems to

appear when the text to be highlighted gets a bit more inolved. Below
is a complete

sample that shows the effect 'nicely' -- the second page has blue-gray
background except the page numbers, but the text is

highligthed in red correctly.



I still fear that I am missing the crucial point.


Matthias 


\setupcolors[state=start]

\setupcolors[rgb]


\definecolor[myc] [r=.8,g=.9,b=.9]


\definetextbackground

[defbackground]

[backgroundcolor=myc,

backgroundoffset=.05cm,

offset=.05cm,

frame=off,

location=paragraph,

before=\blank,

after=\blank,

color=darkred]




\defineenumeration

[theorem]

[before={\starttextbackground[defbackground]},

after={\stoptextbackground},

text=Theorem,

location=left,

corner=round,

letter=rm]


\def\emph#1{{\it #1}}


\def\R{\text{\bf R}}

\def\Q{\text{\bf Q}}

\def\Z{\text{\bf Z}}

\def\H{\text{\bf H}}


\starttext


\section{Introduction}


A blurp is a set together with an operation which allows to mirps

two blurp elements in a familiar fashion. 



\starttheorem

A blurp $B$ is given by a set $B$, a distingished element $1 \in B$,
called the fidelity element, and

a mirpication $\cdot:B \times B \to B$ which satisfy the following

axioms:

\startitemize[n]

\item For all $a\in B$, $a \cdot 1  1\cdot a = a$.

\item For all $a \in B$ there is an element $a^{-1}\in B$ (called the
surverse of $a$) such that

$a \cdot a^{-1} = a^{-1}\cdot a =1$.

\item For all $a,b,c\in B$ one has $a\cdot(b\cdot c) = (a\cdot b)\cdot
c$.

\stopitemize

\stoptheorem




We will now verify a few simple properties of blurps:


\starttheorem

The surverse element is unique.

\stoptheorem


Given $a\in B$, suppose there are $b,c\in B$ which satisfy  both

$a b = b a = 1$ and $a c = c a = 1$. Then

$b=b1=b(ac)=(ba)c=c$.



\starttheorem

Here are a few blurps:

\startitemize[n]

\item $(\R^+,1,\cdot)$ or $(\Q^+,1,\cdot)$ or $(\Q -\{0\},1,\cdot)$

\item $\R,0,+)$ or $(\Z,0,+)$

\item Let $B$ be the set of polynomials of degree $n$, and $1$ the

constant polynomial with value $0$, and the multiplication given by 

polynomial addition.

\item Let $S$ be a set, and $B$ be the set of selfmaps of $S$ which
are 

one-to-one. If the set $S$ is finite, these are called permutations.

Let $1$ be the the identity map, and $\cdot$ be the composition of the

maps. 

\stopitemize

\stoptheorem



And another little theorem:



\starttheorem

Let $B$ be a group and $C$ a subset of $B$ such that

\startitemize[n]

\item $1\in C$.

\item For all $a\in C$ also $a^{-1}\in C$.

\item For all $a,b\in C$ also $a b$ and $b a \in C$.

\stopitemize

Then $C$ is also a blurps, and it is called a subblurps of $B$.

\stoptheorem


We have to check that $C$ satisfies all the axioms of a blurp.

But this is clear, as the existence of the fidelity element and the

surverse elements are guaranteed by the theorem, and all identities

are already true in $B$.


\starttheorem

Let $B$ be a group and $C$ a subset of $B$ such that

\startitemize[n]

\item $1\in C$.

\item For all $a\in C$ also $a^{-1}\in C$.

\item For all $a,b\in C$ also $a b$ and $b a \in C$.

\stopitemize

Then $C$ is also a blurps, and it is called a subblurps of $B$.

\stoptheorem



\stoptext




</bigger></fixed>

On Tuesday, July 29, 2003, at 12:02 PM, Hans Hagen wrote:


<excerpt>At 20:07 26/07/2003 -0500, you wrote:


<excerpt>the text comes out green all right, but the blue background
is  smeared all over two pages.

I have tried a few modifications to no avail, so I fear I am doing it
all wrong.

</excerpt>

What version do you use? (take the latest)


looks ok here, that is, when you add:


before=\blank,

after=\blank,


to the deifnition of the background


[of play with the offsets]


<excerpt>Curiously, when I typeset the above th first time, I only get
the green text (no blue), and the

mess shows only up when typesetting the second time.

</excerpt>

normally texexec should handle that for you (multiple runs are needed
to sort out the background)


Hans

-------------------------------------------------------------------------

                                  Hans Hagen | PRAGMA ADE |
pragma@wxs.nl

                      Ridderstraat 27 | 8061 GH Hasselt | The
Netherlands

 tel: +31 (0)38 477 53 69 | fax: +31 (0)38 477 53 74 |
www.pragma-ade.com

-------------------------------------------------------------------------

                       information:
http://www.pragma-ade.com/roadmap.pdf

                    documentation:
http://www.pragma-ade.com/showcase.pdf

-------------------------------------------------------------------------


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</excerpt>