I'm newbie with Mathematica, so probably my problem is very easy to be solved.
I want to solve a PDE:
The first problem is that the program is not substituting the values inside w[x,y] and its second derivatives, to use them as boundary conditions.
It tells me that Tag Equal in ((w1^(2,0))[x,y]==0)[x,y] is Protected
The other problem is that it gives me also another type of error:
"{NDsolve[{(w1^(0,4))[x,y]+(w1^(2,2))[x,y]+(w1^(4,0))[x,y]==0,{w1[0,y]==0,w1[5,y]==0,((w1^(2,0))[x,y]==0)[0,y]==0,((w1^(2,0))[x,y]==0)[5,y]==0},{w1[x,0]==3,w1[x,5]==3,((w1^(0,2))[x,y]==0)[x,0]==0,((w1^(0,2))[x,y]==0)[x,10]==0}},w1[x,y],{x,0,10},{y,0,5}]} is neither a list of replacement rules nor a valid dispatch table, and so cannot be used for replacing."
Thank you in advance.
I made some changes.
First of all I fixed the boundary conditions because they had a problem.
Moreover I changed also the equation.
Now it still gives me the error:
"{NDsolve[{(w^(0,4))[x,y]+(w^(2,2))[x,y]+(w^(4,0))[x,y]==2,{w[0,y]==0,w[5,y]==0,chix[0,y]==0,chix[5,y]==0},{w[x,0]==0,w[x,10]==0,chiy[x,0]==0,chiy[x,10]==0}},w,{x,0,5},{y,0,10}]} is neither a list of replacement rules nor a valid dispatch table, and so cannot be used for replacing."
If we succeed in fixing this problem, Mathematica should give us a non zero solution.
Thank you very much!
Rewriting it, it doesn't provide any solution. And it also gives no errors. It looks like it is not even substitutiing the boundary conditions. I decided to change them, solving in this way another kind of problem. It still doesn't work. I think that it is very strange, since if the boundary conditions were wrong it should have provided either an indeterminate condition or an impossible one. Any idea?
Thank you very much for your help!
(This question regards search/6.)
I was wondering if there is a way -rather than manual tracing- to pause the execution of search/6 every time a new solution for a single variable was found?
I would like to accomplish this to further investigate what is happening during search in constrained models.
For example, if you are trying to solve the classic sudoku problem, and you have written a set of constraints and a print method for your board, it can be useful to print the board after setting the constraints, but before searching, in order to evaluate the strongness of your constraints. However, once search is called to solve the sudoku, you don't really have an overview of the single results being built underneath unless you do a trace.
It would be very useful if something was possible in the likes of:
(this is just an abstract example)
% Let's imagine this is a (very poorly) constrained sudoku board
?- problem(Sudoku),constraint(Sudoku),print(Sudoku).
[[1,3,_,2,_,_,7,4,_],
[_,2,5,_,1,_,_,_,_],
[4,8,_,_,6,_,_,5,_],
[_,_,_,7,8,_,2,1,_],
[5,_,_,_,9,_,3,7,_],
[9,_,_,_,3,_,_,_,5],
[_,4,_,_,_,6,8,9,_],
[_,5,3,_,_,1,4,_,_],
[6,_,_,_,_,_,_,_,_]]
Now for the search:
?- problem(Sudoku),constraint(Sudoku),search_pause(Sudoku,BT),print(Sudoku,BT).
[[1,3,6,2,_,_,7,4,_],
[_,2,5,_,1,_,_,_,_],
[4,8,_,_,6,_,_,5,_],
[_,_,_,7,8,_,2,1,_],
[5,_,_,_,9,_,3,7,_],
[9,_,_,_,3,_,_,_,5],
[_,4,_,_,_,6,8,9,_],
[_,5,3,_,_,1,4,_,_],
[6,_,_,_,_,_,_,_,_]]
Board[1,3] = 6
Backtracks = 1
more ;
Using existing Visualization tools
Have a look at the Visualization Tools Manual. You can get the kind of matrix display you want by adding a viewable_create/2 annotation to your code, and launching a Visualisation Client from TkECLiPSe's Tools-menu.
Using your own instrumented search routine
You can replace the indomain_xxx choice methods in search/6 with a user-defined one where you can print information before and/or after propagation.
If that is not enough, you can replace the whole built-in search/6 with your own, which is not too difficult, see e.g. the ECLiPSe Tutorial chapter on tree search or my answer to this question.
Tracing using data-driven facilities
Using ECLiPSe's data-driven control facilities, you can quite easily display information when certain things happen to your variables. In the simplest case you do something on variable instantiation:
?- suspend(printf("X was instantiated to %w%n",[X]), 1, X->inst),
writeln(start), X=3, writeln(end).
start
X was instantiated to 3
end
Based on this idea, you can write code that allows you to follow labeling and propagation steps even when they happen inside a black-box search routine. See the link for details.
I was wondering what this code does:
:- set_prolog_flag(toplevel_print_options,
[quoted(true), portray(true), attributes(portray), max_depth(100)]).
I have seen it in some of the sample codes my prof has posted on his website but I have no clue what it does. Thanks for your help in advance.
I believe it might have something to do with making program output more formatted (and thus, more readable or accessible.)
See this article: "Help... Prolog writes [x, y, z|...], I want the whole answer".
Basically, in the case of your code's settings... it looks like the code is just setting some formatting for output. The max_depth setting means that anything that is nested more than (100, in your case,) levels will then be written as .... The rest of the options just enable normal output.
I edited the question after David Hanak's answer (thanks btw!). He helped with the syntax, but it appears that I wasn't using the right function to begin with.
Basically what I want is to let the compiler ignore multiple definitions of a certain label and just use the first. In order to do that, I thought I'd just do something like this:
\makeatletter
\newcommand{\mylabel}[1]{
\#ifundefined{#1}{\label{#1}}{X}
}
\makeatother
This does not work though, because the first option is always chosen (it doesn't matter if the label is defined or not). I think the \#ifundefined (and the suggested \ifundefined) only work for commands and not for labels, but I don't really know much about LaTeX. Any help with this would be great! Thanks!
Much later update:
I marked David Hanak's response as the correct answer to my question, but it isn't a complete solution, although it really helped me.
The problem is, I think but I'm no specialist, that even though David's code checks to see if a label is defined, it only works when the label was defined in a previous run (i.e. is in the .aux file). If two \mylabels with the same name are defined in the same run, the second will still be defined. Also, even if you manage to work around this, it will make LaTeX use the first label that you defined chronologically, and not necessarily the first in the text.
Anyway, below is my quick and dirty solution. It uses the fact that counters do seem to be defined right away.
\newcommand{\mylabel}[1]{%
\#ifundefined{c##1}{%
\newcounter{#1}%
\setcounter{#1}{0}%
}{}%
\ifthenelse{\value{#1} > 0}{}{%
\label{#1}%
\addtocounter{#1}{1}%
}%
}
I'm not sure if it is necessary to initialize the counter to 0, as it seems like a likely default, but I couldn't find if that was the case, so I'm just being safe.
Also, this uses the 'ifthen' package, which I'm not sure is necessary.
I am also not a LaTeX expert, however after one day of trying and searching the internet the following worked for me. I have used a dummy counter to solve the problem. Hopefully this helps, apparently not many people are looking for this.
\newcommand{\mylabel}[1]{
\ifcsname c##1\endcsname%
\else%
\newcounter{#1}\label{#1}%
\fi%
}
# is a special character in LaTeX. To make your declaration syntactically correct, you'll have to add two more lines:
\makeatletter
\newcommand{\mylabel}[1]{
\#ifundefined{#1}{\label{#1}}{X}
}
\makeatother
The first line turns # into a normal letter, the last line reverses its effect.
Update: You might also want to take a look at the "plain" \ifundefined LaTeX macro.
Update 2
Okay, I did some research to figure out the answer to the real problem. The thing is that defining a label does not create a macro by that name; it prepends a "r#" to it. So try the following:
\makeatletter
\newcommand{\mylabel}[1]{
\#ifundefined{r##1}{\label{#1}}{X}
}
\makeatother
For more technical details, refer to line 3863 of latex.ltx in your LaTeX distribution (where it says \def\newlabel{\#newl#bel r}).
By Victor Eijkhout, "TeX by Topic", p.143:
\def\ifUnDefinedCs#1{\expandafter\ifx\csname#1\endcsname\relax}
This can be used to check whether a label is defined; if not, the label is printed:
\newcommand{\myautoref}[1]{\ifUnDefinedCs{r##1}{\color{magenta}\IDontKnow\{#1\}}\else\autoref{#1}\fi}