In following examples z3 assumes that 2 symbols with same name are equal.
solve(Int('z')<Int('z'))
returns: "no solution"
from z3 import *
x1=Int('x')
x2=Int('x')
x3=Int('x')
solve(x1*x2*x3==27)
returns:"[x = 3]"
Can I assume that z3 always treats symbols with same name and type as equals? For example if I have an equation with variable Int type variable z - can replace "z" with "Int('z')" everywhere in the equation and be sure that output will be the same?
Can I assume that z3 always treats function-symbols with same name and same type of arguments as equals like in following example:
S1 = DeclareSort('S1')
S2 = DeclareSort('S2')
x1=Const("x",S1)
x2=Const("x",S2)
f1=Function('F', S1, S1,S2,IntSort())
f2=Function('F', S1, S1,S2,IntSort())
solve(f1(x1,x1,x2)<f2(x1,x1,x2))
returns:"no solution"
Can I assume that z3 never treats function-symbols with same name, but different type or number of arguments as equals like in following example:
S1 = DeclareSort('S1')
S2 = DeclareSort('S2')
x1=Const("x",S1)
x2=Const("x",S2)
f1=Function('F', S1, S1,S2,IntSort())
f2=Function('F', S1, S2,S2,IntSort())
solve(f1(x1,x1,x2)<f2(x1,x2,x2))
returns:"
[x = S2!val!0,
x = S1!val!0,
F = [else -> 1],
F = [else -> 0]]
"
In Z3 variables are identified by name and sort. So, if two variables have the same name and sort then they are the same. Correspondingly, if they either have different names OR sorts, then they are different. The sort is the type and for function names the type is a function type. (i.e. there’s no special treatment for function names. You can think of ordinary variables as functions that take 0 arguments.)
In particular, two different variables can have the same name so long as they have different sorts. This can be rather confusing, however, since when z3 prints the corresponding SMTLib or models, you won’t have an obvious indication showing which one is which. So I’d advise against using the same name even if the sorts differ.
Related
The trace processing of the functor 'flatten2' with input list: "[4, [3, [2, [1,[ ] ] ] ] ]" (Screenshot of the tracing process)
The above is the screenshot of me calling functor flatten2 with input list "[4, [3, [2, [1,[ ] ] ] ] ]", and a variable 'X'.
The below is the function that I was tracing, stolen from this question: Flatten a list in Prolog.
So my essential question is that during the recursion, what happens to the variable X and its value? Why is the 'X' showed as '_295' in the "call 1", how does prolog grammar calculating this value?
flatten2([], []) :- !.
flatten2([L|Ls], FlatL) :-
!,
flatten2(L, NewL),
flatten2(Ls, NewLs),
append(NewL, NewLs, FlatL).
flatten2(L, [L]).
X is just a local name for "globally visible stuff" that is held in the "term store".
X designates (or names or denotes) either:
Concrete content: A term, which a generally tree. The variable is called "bound" or "instantiated". The leaves of the tree are either concrete content (atoms, numbers etc.) or cells with no content (see below). The inner nodes are called compound terms. If X designates a term, then the query nonvar(X) succeeds. When X is printed, it "disappears": The content is printed instead.
A cell with no content (I like to call this a "hole"), which is meant to be take up a term eventually. In that case the variable is called "unbound" or "uninstantiated". If X is such an unbound variable, i.e. if it designates a hole, then the query var(X) succeeds. When X is printed, its name is printed.
Confusingly (and I should add, rather sloppily), a "variable name" is commonly also called a "term". That's a Prolog tripmine, these two concepts should be held apart but they are not.
If you write your Prolog clause, you will use nice variable names.
flatten2(L, [L]).
Remember these variables names have no particular significance and they are local to the clause.
When Prolog runs, it has to pull new variable names that are distinct from any other names "out of the hat". These fresh variable names look like _295. Different implementations have different conventions here.
An example where new variable names have to be created to describe a list that contains a member foo somewhere (on at least one place). List templates of increasing length are generated. At each place of the list except the place holding foo (a concrete term), there is a "cell without content"/"hole". To express this, a random new variable name distinct from any other variable name is generated and printed. The variable name is probably directly derived from the hole address.
?- member(foo,L).
L = [foo|_23302] ;
L = [_23300, foo|_24038] ;
L = [_23300, _24036, foo|_24774] ;
L = [_23300, _24036, _24772, foo|_25510] ;
L = [_23300, _24036, _24772, _25508, foo|_26246] ;
L = [_23300, _24036, _24772, _25508, _26244, foo|_26982]
enum sup;
sup=['a','b','c'];
enum sup2;
sup2=['d','e','f'];
enum sup3;
sup3=sup++sup2;
I want to get an new enumerated type sup3 with all a,b,c,d,e,f.Is there any way in minizinc we can do this.
The short answer is no, this is currently not supported. The main issue with the concatenation of enumerated types comes from the fact we are not just concatenating two lists of things, but we are combining types. Take your example:
enum sup = {A, B, C};
enum sup2 = {D, E, F};
enum sup3 = sup ++ sup2;
When I now write E somewhere in an expression, I no longer know if it has type sup2 or sup3. As you might imagine, there is no guarantee that E would have the same value (for the solver) in the two enumerated types, so this can be a big problem.
To shine a glimmer of hope, the MiniZinc team has been working on a similar approach to make this possible (but not yet formally announced). Instead of your syntax, one would write:
enum X = {A, B, C};
enum Y = {D, E, F} ++ F(X);
The idea behind this is that F(X) now gives a constructor for the usage of X in Y. This means that if we see just A, we know it's of type X, but if we see F(A), then it's of type Y. Again, this is not yet possible, but will hopefully end up in the language soon.
More of a comment but here is my example of my need. When doing code coverage and FSM transition analysis I am forced to use exclusion to not analyze some transitions for the return_to_state, in the code below. If instead I could use concatenated types as shown, I would have more control over the tools reporting missing transitions.
type Read_states is (ST1);
type Write_states is (ST2, ST3, ST4);
type SPI_states is (SPI_write);
type All_States is Read_states & Write_states & SPI_states;
I could make return_to_state of type Write_states and FRAM_state of type All_states and then not have to put in exclusions in my FSM analysis.
I want to make a list with its elements representing the logic map given by
x_{n+1} = a*x_n(1-x_n)
I tried the following code (which adds stuff manually instead of a For loop):
x0 = Input["Enter x0"]
a = Input["a"]
M = {x0}
L[n_] := If[n < 1, x0, a*M[[n]]*(1 - M[[n]])]
Print[L[1]]
Append[M, L[1]]
Print[M]
Append[M, L[2]]
Print[M]
The output is as follows:
0.3
2
{0.3}
0.42
{0.3,0.42}
{0.3}
Part::partw: Part 2 of {0.3`} does not exist. >>
Part::partw: Part 2 of {0.3`} does not exist. >>
{0.3, 2 (1 - {0.3}[[2]]) {0.3}[[2]]}
{0.3}
It seems that, when the function definition is being called in Append[M,L[2]], L[2] is calling M[[2]] in the older definition of M, which clearly does not exist.
How can I make L use the newer, bigger version of M?
After doing this I could use a For loop to generate the entire list up to a certain index.
P.S. I apologise for the poor formatting but I could find out how to make Latex code work here.
Other minor question: What are the allowed names for functions and lists? Are underscores allowed in names?
It looks to me as if you are trying to compute the result of
FixedPointList[a*#*(1-#)&, x0]
Note:
Building lists element-by-element, whether you use a loop or some other construct, is almost always a bad idea in Mathematica. To use the system productively you need to learn some of the basic functional constructs, of which FixedPointList is one.
I'm not providing any explanation of the function I've used, nor of the interpretation of symbols such as # and &. This is all covered in the documentation which explains matters better than I can and with which you ought to become familiar.
Mathematica allows alphanumeric (only) names and they must start with a letter. Of course, Mathematic recognises many Unicode characters other than the 26 letters in the English alphabet as alphabetic. By convention (only) intrinsic names start with an upper-case letter and your own with a lower-case.
The underscore is most definitely not allowed in Mathematica names, it has a specific and widely-used interpretation as a short form of the Blank symbol.
Oh, LaTeX formatting doesn't work hereabouts, but Mathematica code is plenty readable enough.
It seems that, when the function definition is being called in
Append[M,L2], L2 is calling M[2] in the older definition of M,
which clearly does not exist.
How can I make L use the newer, bigger version of M?
M is never getting updated here. Append does not modify the parameters you pass to it; it returns the concatenated value of the arrays.
So, the following code:
A={1,2,3}
B=Append[A,5]
Will end up with B={1,2,3,5} and A={1,2,3}. A is not modfied.
To analyse your output,
0.3 // Output of x0 = Input["Enter x0"]. Note that the assignment operator returns the the assignment value.
2 // Output of a= Input["a"]
{0.3} // Output of M = {x0}
0.42 // Output of Print[L[1]]
{0.3,0.42} // Output of Append[M, L[1]]. This is the *return value*, not the new value of M
{0.3} // Output of Print[M]
Part::partw: Part 2 of {0.3`} does not exist. >> // M has only one element, so M[[2]] doesn't make sense
Part::partw: Part 2 of {0.3`} does not exist. >> // ditto
{0.3, 2 (1 - {0.3}[[2]]) {0.3}[[2]]} (* Output of Append[M, L[2]]. Again, *not* the new value of M *)
{0.3} // Output of Print[M]
The simple fix here is to use M=Append[M, L[1]].
To do it in a single for loop:
xn=x0;
For[i = 0, i < n, i++,
M = Append[M, xn];
xn = A*xn (1 - xn)
];
A faster method would be to use NestList[a*#*(1-#)&, x0,n] as a variation of the method mentioned by Mark above.
Here, the expression a*#*(1-#)& is basically an anonymous function (# is its parameter, the & is a shorthand for enclosing it in Function[]). The NestList method takes a function as one argument and recursively applies it starting with x0, for n iterations.
Other minor question: What are the allowed names for functions and lists? Are underscores allowed in names?
No underscores, they're used for pattern matching. Otherwise a variable can contain alphabets and special characters (like theta and all), but no characters that have a meaning in mathematica (parentheses/braces/brackets, the at symbol, the hash symbol, an ampersand, a period, arithmetic symbols, underscores, etc). They may contain a dollar sign but preferably not start with one (these are usually reserved for system variables and all, though you can define a variable starting with a dollar sign without breaking anything).
Where can I find a list of Scala's "magic" functions, such as apply, unapply, update, +=, etc.?
By magic-functions I mean functions which are used by some syntactic sugar of the compiler, for example
o.update(x,y) <=> o(x) = y
I googled for some combination of scala magic and synonyms of functions, but I didn't find anything.
I'm not interested with the usage of magic functions in the standard library, but in which magic functions exists.
As far as I know:
Getters/setters related:
apply
update
identifier_=
Pattern matching:
unapply
unapplySeq
For-comprehensions:
map
flatMap
filter
withFilter
foreach
Prefixed operators:
unary_+
unary_-
unary_!
unary_~
Beyond that, any implicit from A to B. Scala will also convert A <op>= B into A = A <op> B, if the former operator isn't defined, "op" is not alphanumeric, and <op>= isn't !=, ==, <= or >=.
And I don't believe there's any single place where all of Scala's syntactic sugars are listed.
In addition to update and apply, there are also a number of unary operators which (I believe) qualify as magical:
unary_+
unary_-
unary_!
unary_~
Add to that the regular infix/suffix operators (which can be almost anything) and you've got yourself the complete package.
You really should take a look at the Scala Language Specification. It is the only authoritative source on this stuff. It's not that hard to read (as long as you're comfortable with context-free grammars), and very easily searchable. The only thing it doesn't specify well is the XML support.
Sorry if it's not exactly answering your question, but my favorite WTF moment so far is # as assignment operator inside pattern match. Thanks to soft copy of "Programming in Scala" I found out what it was pretty quickly.
Using # we can bind any part of a pattern to a variable, and if the pattern match succeeds, the variable will capture the value of the sub-pattern. Here's the example from Programming in Scala (Section 15.2 - Variable Binding):
expr match {
case UnOp("abs", e # UnOp("abs", _)) => e
case _ =>
}
If the entire pattern match succeeds,
then the portion that matched the
UnOp("abs", _) part is made available
as variable e.
And here's what Programming Scala says about it.
That link no longer works. Here is one that does.
I'll also add _* for pattern matching on an arbitrary number of parameters like
case x: A(_*)
And operator associativity rule, from Odersky-Spoon-Venners book:
The associativity of an operator in Scala is determined by its last
character. As mentioned on <...>, any method that ends
in a ‘:’ character is invoked on its right operand, passing in the
left operand. Methods that end in any other character are the other
way around. They are invoked on their left operand, passing in the
right operand. So a * b yields a.*(b), but a ::: b yields b.:::(a).
Maybe we should also mention syntactic desugaring of for expressions which can be found here
And (of course!), alternative syntax for pairs
a -> b //converted to (a, b), where a and b are instances
(as correctly pointed out, this one is just an implicit conversion done through a library, so it's probably not eligible, but I find it's a common puzzler for newcomers)
I'd like to add that there is also a "magic" trait - scala.Dynamic:
A marker trait that enables dynamic invocations. Instances x of this trait allow method invocations x.meth(args) for arbitrary method names meth and argument lists args as well as field accesses x.field for arbitrary field names field.
If a call is not natively supported by x (i.e. if type checking fails), it is rewritten according to the following rules:
foo.method("blah") ~~> foo.applyDynamic("method")("blah")
foo.method(x = "blah") ~~> foo.applyDynamicNamed("method")(("x", "blah"))
foo.method(x = 1, 2) ~~> foo.applyDynamicNamed("method")(("x", 1), ("", 2))
foo.field ~~> foo.selectDynamic("field")
foo.varia = 10 ~~> foo.updateDynamic("varia")(10)
foo.arr(10) = 13 ~~> foo.selectDynamic("arr").update(10, 13)
foo.arr(10) ~~> foo.applyDynamic("arr")(10)
As of Scala 2.10, defining direct or indirect subclasses of this trait is only possible if the language feature dynamics is enabled.
So you can do stuff like
import scala.language.dynamics
object Dyn extends Dynamic {
def applyDynamic(name: String)(a1: Int, a2: String) {
println("Invoked " + name + " on (" + a1 + "," + a2 + ")");
}
}
Dyn.foo(3, "x");
Dyn.bar(3, "y");
They are defined in the Scala Language Specification.
As far as I know, there are just three "magic" functions as you mentioned.
Scalas Getter and Setter may also relate to your "magic":
scala> class Magic {
| private var x :Int = _
| override def toString = "Magic(%d)".format(x)
| def member = x
| def member_=(m :Int){ x = m }
| }
defined class Magic
scala> val m = new Magic
m: Magic = Magic(0)
scala> m.member
res14: Int = 0
scala> m.member = 100
scala> m
res15: Magic = Magic(100)
scala> m.member += 99
scala> m
res17: Magic = Magic(199)
I would like to pass the parameter values in meters or kilometers (both possible) and get the result in meters/second.
I've tried to do this in the following example:
u = 3.986*10^14 Meter^3/Second^2;
v[r_, a_] := Sqrt[u (2/r - 1/a)];
Convert[r, Meter];
Convert[a, Meter];
If I try to use the defined function and conversion:
a = 24503 Kilo Meter;
s = 10198.5 Meter/Second;
r = 6620 Kilo Meter;
Solve[v[r, x] == s, x]
The function returns the following:
{x -> (3310. Kilo Meter^3)/(Meter^2 - 0.000863701 Kilo Meter^2)}
which is not the user-friendly format.
Anyway I would like to define a and r in meters or kilometers and get the result s in meters/second (Meter/Second).
I would be very thankful if anyone of you could correct the given function definition and other statements in order to get the wanted result.
Here's one way of doing it, where you use the fact that Solve returns a list of rules to substitute a value for x into v[r, x], and then use Convert, which will do the necessary simplification of the resulting algebraic expression as well:
With[{rule = First#Solve[v[r,x]==s,x]
(* Solve always returns a list of rules, because algebraic
equations may have multiple solutions. *)},
Convert[v[r,x] /. rule, Meter/Second]]
This will return (10198.5 Meter)/Second as your answer.
You just need to tell Mathematica to simplify the expression assuming that the units are "possitive", which is the reason why it doesn't do the simplifications itself. So, something like
SimplifyWithUnits[blabla_, unit_List]:= Simplify[blalba, (#>0)&/#unit];
So if you get that ugly thing, you then just type %~SimplifyWithUnits~{Meter} or whatever.