I have 3 variables in functions section:
(x)
(y)
(z)
I have initialized all of them to 0 in my problem init.
An action in my domain changes the value of x and y.
Now when I try to assign the value of x/y to z , it says it is not a linear task.
(assign (z) (/ (x) (y) ) )
But when I write (assign (z) (/ (x) 2 ) ) it assigned correct value which is x/2, similarly (assign (z) (/ (y) 2 ) ) it is assigned correct value (y/2). This ensured that x and y are not zeroes.
Now why (assign (z) (/ (x) (y) ) ) is not working? Please help me out.
Metric-FF does not support non-linear expressions, as the one you are using to assign the value to variable z. Try making x and y constant (remove any action modifying them), and see that it will work, yet that may not be what you are looking for. For non-linear expressions in numeric planning problems have a look at ENHSP: https://sites.google.com/view/enhsp/
Related
I am working on a MonteCarlo simulation model and part of it is to calculate the following formula:
X = Sqr(1-p)Y + Sqr(p)Z,
Where:
Y and Z are randomly obtained values based (idiosyncratic and systematic factors, respectviely) on a standard normal (inv.) distribution, calculated as:
Application.WorksheetFunction.NormInv (Rnd(), mean, sd)
p represents a correlation factor.
My aim is to square root a recalled formula, however when I try the following (inserting the first Sqr), it does not work and gives an error:
Matrix (n, sims) = (R * Sqr(Application.WorksheetFunction.NormInv(Rnd(), mean, sd))) + (Sqr(1 - R) * RandomS(s, x))
where:
R: Correlation factor
RandomS(s,x): generated matrix with Z values.
I don't want to go into too much details about the background and other variables, as the only problem I am getting is with Square Rooting the equation.
Error message I recieve reads:
Run-time error '5':
Invalid procedure call or argument
When I click debug it takes me to the formula, therefore there must be something wrong with the syntax.
Can you help with directly squaring the formula?
Thank you!
Andrew
Square root is simply Sqr.
It works fine in Excel VBA, so for example:
MsgBox Sqr(144)
...returns 12.
Just don't confuse it with the syntax for a worksheet function with is SQRT.
If you're still having an issue with your formula, tit must be with something other than the Square Root function, and I'd suggest you check the values of your variable, and make sure they are properly declared (preferably with Option Explicit at the top of the module).
Also make sure that you're passing Sqr a positive number.
Documentation: Sqr Function
I'm not a math major, but with your formula:
X = Sqr(1-p)Y + Sqr(p)Z,
...you specified how Y and Z are calculated, so calculate them separately to keep it simple:
Dim X as Double, Y as Double, Z as Double
Y = Application.WorksheetFunction.NormInv (Rnd(), mean, sd)
Z = Application.WorksheetFunction.NormInv (Rnd(), mean, sd)
Assuming the comma is not supposed to be in the formula, and having no idea what p is, your final code to calculate X is:
X = Sqr(1-p) * Y + Sqr(p) * Z
I am reading text from spatstat textbook, and trying to learn model fit using ppm.
I created a model with carteisan coordinates as the covariates. And then I wanted to see the effect of only one covariate on the model,
model1 = ppm(chicago_ppp ~ x+y)
plot(effectfun(model1, covname = "x"))
but I get the error
Error in effectfun(model1, covname = "x") : A value for the covariate “y” must be provided (as an argument to effect fun)
The same happens if I use covname "y" it asks for "x"
Can someone please show me what is my mistake. Thank you.
UPDATE: When I use only one covariate, and I use effectfun with that one covariate, there is no error. When I use two covariates and I want to check effectfun of one covariate, I get this error in the question.
To be able to calculate the estimated intensity for different values of
x you need to fix a value for y like this:
library(spatstat)
model <- ppm(cells ~ x + y)
plot(effectfun(model, covname = "x", y = 0.1))
plot(effectfun(model, covname = "x", y = 0.9))
Is it safe to use bindings that depend on each other in let? For example:
let x = 1
y = x + 2
in y
Is it possible that they are evaluated in parallel? my ghci shows that it is evaluated correctly, but will this be always the case?
Haskell is lazy evaluated. This means that expressions are only evaluated by necessity. Let's start with your example.
let x = 1
y = x + 2
in y
The system looks at the y part (the expression), and says "Hey. I know what y equals. It equals x + 2" But it can only evaluate x + 2 if it has a definition for x, so it does the same thing and decides that y is 1 + 2 which is of course 3. Now this is only a small portion of the power of lazy evaluation. The next example shows it more fully.
let x = 0 : y
y = 1 : x
in take 50 x
This expression will evaluate correctly, yielding the first fifty elements on the list x, which will be 0 and 1 alternating. Both of the values x and y are infinite lists that depend on each other, and indeed in most languages this would overflow the stack. However, the evaluation rules allow the system to only see the parts it needs, which in that example is the first fifty elements.
So to answer your question about evaluation order, the system evaluates what it sees. If the function is supposed to return y, it evaluates y and then as necessary x. If you had put x in your first example, it would have evaluated x and left y alone. For example, this code will not err.
let x = 1
y = error "I'm an error message!"
in x
This is because the y form is never needed, so the piece of code that would crash the program is never even looked at.
In Haskell (regardless of whether you use a single let, multiple lets, case, where or function parameters) an expression is evaluated when the evaluation of another expression depends on its value.
So in your case, as soon as y's value is required (which of course depends on the surrounding program), y will be evaluated and x will be evaluated as part of y's evaluation.
Another way to think of it is this: whenever you use the value of a variable, it will be evaluated at that point at the latest. That is it might have been evaluated previously (if it was needed previously) or it might be evaluated now, but, as long as you do use the value, it will never not be evaluated. So, except for performance reasons, there's no need to worry about when a variable will be evaluated.
I'm making a predicate distance/3 that calculates the distance between 2 points on a 2d plane. For example :
?- distance((0,0), (3,4), X).
X = 5
Yes
My predicate only works if (0,0) is the list [0,0]. Is there a way to make this conversion?
You can do this with a simple rule that unifies its left and right sides:
convert((A,B), [A,B]).
Demo.
Although the others have answered, keep in mind that (a,b) in Prolog is actually not what you might think it is:
?- write_canonical((a,b)).
','(a,b)
true.
So this is the term ','/2. If you are working with pairs, you can do two things that are probably "prettier":
Keep them as a "pair", a-b:
?- write_canonical(a-b).
-(a,b)
true.
The advantage here is that pairs like this can be manipulated with a bunch of de-facto standard predicates, for example keysort, as well as library(pairs).
Or, if they are actually a data structure that is part of your program, you might as well make that explicit, as in coor(a, b) for example. A distance in two-dimensional space will then take two coor/2 terms:
distance(coor(X1, Y1), coor(X2, Y2), D) :-
D is sqrt((X1-X2)^2 + (Y1-Y2)^2).
If you don't know how many dimensions you have, you can then indeed keep the coordinates of each point in a list. The message here is that lists are meant for things that can have 0 or more elements in them, while pairs, or other terms with arity 2, or any term with a known arity, are more explicit about the number of elements they have.
If you just have a simple pair, you can use the univ operator and simply say something like:
X = (a,b) ,
X =.. [_|Y] .
which produces
X = (a,b) .
Y = [a,b] .
This doesn't work if X is something like (a,b,c), producing as it does
X = (a,b,c) .
Y = [a,(b,c)] .
[probably not what you want].
The more general case is pretty simple:
csv2list( X , [X] ) :- % We have a list of length 1
var(X) . % - if X is UNbound
csv2list( X , [X] ) :- % We have a list of length 1
nonvar(X) , % - if X is bound, and
X \= (_,_) . % - X is not a (_,_) term.
cs22list( Xs , [A|Ys] ) :- % otherwise (the general case) ,
nonvar(Xs) , % - if X is bound, and
Xs = (A,Bs) , % - X is a (_,) term,
csv2list(Bs,Ys % - recurse down added the first item to result list.
. % Easy!
My book presents a simple example which I'm a bit confused about:
It says, "consider the following program, and assume that the fine-grained atomic actions are reading and writing the variables:"
int y = 0, z = 0;
co x = y+z; // y=1; z=2; oc;
"If x = y + z is implemented by loading a register with y and then adding z to it, the final value of x can be 0,1,2, or 3. "
2? How does 2 work?
Note: co starts a concurrent process and // denote parallel-running statements
In your program there are two parallel sequences:
Sequence 1: x = y+z;
Sequence 2: y=1; z=2;
The operations of sequence 1 are:
y Copy the value of y into a register.
+ z Add the value of z to the value in the register.
x = Copy the value of the register into x.
The operations of sequence 2 are:
y=1; Set the value of y to 1.
z=2; Set the value of z to 2.
These two sequences are running at the same time, though the steps within a sequence must occur in order. Therefore, you can get an x value of '2' in the following sequence:
y=0
z=0
y Copy the value of y into a register. (register value is now '0')
y=1; Set the value of y to 1. (has no effect on the result, we've already copied y to the register)
z=2; Set the value of z to 2.
+ z Add the value of z to the value in the register. (register value is now '2')
x = Copy the value of the register into x. (the value of x is now '2')
Since they are assumed to run in parallel, I think an even simpler case could be y=0, z=2 when the assignment x = y + z occurs.