Let's say I have a set:
S: Id X Counter
Id: \nat
Counter: \nat
Need help to define an operation filter which takes in two parameters, x:\nat and y:\nat which I can apply to S. This function will match on the first parameter with \exists a \in S # first(a) = x and then reduce second(a) with y. If the value is less than or equal to zero, a should be removed from the set S, otherwise (first(a), second(a) - y) should replace a.
If the above is indecipherable, just please give me any example definition that operates over a set and modifies it.
Thanks.
Related
For example, the analogous two dimensional question, x + y = 2n is easy to solve: one can simply consider pairs (i,2n-i) for i=1,2,...,n and thus index every solution, exactly once. We note that we have n such pairs solving x + y = 2n, for every fixed value of positive integer n, and so the cardinality of such a set is equal to n as expected.
However, trying to repeat the same problem for x + y + z = 2n, it is not clear to me how (or if it is possible) to write down a minimal set {(2n-i-j,i,j)} such that varying i and j over particular intervals precisely produces every such triplet, exactly once. It can be shown that the number of elements in such a minimal set would be equal to the nearest integer to n^2/3.
It is not hard to see how one can obtain such an indexing with repetitions, or how one can algorithmically remove repetitions, but what I would like to know is whether there is a clean, general construction, as for the x + y = 2n case. Is this possible, or will one always have to artificially restrict certain values of the parameters on the intervals for which they are defined?
I am working on a supply chain optimisation problem using Pyomo and I need to set up a constraint on specific variables in the model. The constraint is that the variable should be within the set (0,1) or (200, to infinity). However, when I try to set up that constraint I am getting a TypeError, here is my code:
def rail_rule(model):
for route in routes:
if "rail" in route[0].lower():
dest = route[0]
dg = route[1]
sg = route[2]
site = route[3]
return model.x[dest, dg, sg, site]>=200 or model.x[dest, dg, sg, site]<=1
model.railconst = Constraint(rule = rail_rule)
I get this error when I run it :
TypeError: Relational expression used in an unexpected Boolean context.
The inequality expression:
200.0 <= x[RAIL - KENSINGTON,8,8,BROCKLESBY]
contains non-constant terms (variables) that were evaluated in an
unexpected Boolean context at
File '<ipython-input-168-901363ebc86f>', line 8:
return model.x[dest, dg, sg, site]>=200 or model.x[dest, dg, sg, site]<=1
Evaluating Pyomo variables in a Boolean context, e.g.
if expression <= 5:
is generally invalid. If you want to obtain the Boolean value of the
expression based on the current variable values, explicitly evaluate the
expression using the value() function:
if value(expression) <= 5:
or
if value(expression <= 5):
So my understanding is that I cant give Pyomo a boolean expression as a constraint, but I am quite new to Pyomo and not too sure if that's what my issue is or if I am doing it correctly.
This constraint could also be implemented in the variable intialisation as boundaries, but I cant find a way to set up two boundaries on a single variable in Pyomo.
Thanks !
There are different ways to handle this:
(1) Use binary variables. Assume you have a good upper bound on x, i.e., x ∈ [0, U]. Then formulate the constraints
x ≤ 1 + (U-1) δ
x ≥ 200 δ
δ ∈ {0,1} (binary variable)
This is the easiest way.
(2) If you don't have a good upper bound on x, you can use a SOS1 set. (SOS1 means Special Ordered Set of type 1). Assume x,s1,s2 ≥ 0.
x ≤ 1 + s1
x ≥ 200 - s2
s1,s2 ∈ SOS1 (s1,s2 form a SOS1 set)
(3) Use disjunctive programming.
I need to rely on the fact that two Z3 variables
can not have the same name.
To be sure of that,
I've used tuple_example1() from test_capi.c in z3/examples/c and changed the original code from:
// some code before that ...
x = mk_real_var(ctx, "x");
y = mk_real_var(ctx, "y"); // originally y is called "y"
// some code after that ...
to:
// some code before that ...
x = mk_real_var(ctx, "x");
y = mk_real_var(ctx, "x"); // but now both x and y are called "x"
// some code after that ...
And (as expected) the output changed from:
tuple_example1
tuple_sort: (real, real)
prove: get_x(mk_pair(x, y)) = 1 implies x = 1
valid
disprove: get_x(mk_pair(x, y)) = 1 implies y = 1
invalid
counterexample:
y -> 0.0
x -> 1.0
to:
tuple_example1
tuple_sort: (real, real)
prove: get_x(mk_pair(x, y)) = 1 implies x = 1
valid
disprove: get_x(mk_pair(x, y)) = 1 implies y = 1
valid
BUG: unexpected result.
However, when I looked closer, I found out that Z3 did not really fail or anything, it is just a naive (driver) print out to console.
So I went ahead and wrote the exact same test with y being an int sort called "x".
To my surprise, Z3 could handle two variables with the same name when they have different sorts:
tuple_example1
tuple_sort: (real, real)
prove: get_x(mk_pair(x, y)) = 1 implies x = 1
valid
disprove: get_x(mk_pair(x, y)) = 1 implies y = 1
invalid
counterexample:
x -> 1.0
x -> 0
Is that really what's going on? or is it just a coincidence??
Any help is very much appreciated, thanks!
In general, SMT-Lib does allow repeated variable names, so long as they have different sorts. See page 27 of the standard. In particular, it says:
Concretely, a variable can be any symbol, while a function symbol
can be any identifier (i.e., a symbol or an indexed symbol). As a
consequence, contextual information is needed during parsing to know
whether an identifier is to be treated as a variable or a function
symbol. For variables, this information is provided by the three
binders which are the only mechanism to introduce variables. Function
symbols, in contrast, are predefined, as explained later. Recall that
every function symbol f is separately associated with one or more
ranks, each specifying the sorts of f’s arguments and result. To
simplify sort checking, a function symbol in a term can be annotated
with one of its result sorts σ. Such an annotated function symbol is a
qualified identifier of the form (as f σ).
Also on page 31 of the same document, it further clarifies "ambiguity" thusly:
Except for patterns in match expressions, every occurrence of an
ambiguous function symbol f in a term must occur as a qualified
identifier of the form (as f σ) where σ is the intended output sort of
that occurrence
So, in SMT-Lib lingo, you'd write like this:
(declare-fun x () Int)
(declare-fun x () Real)
(assert (= (as x Real) 2.5))
(assert (= (as x Int) 2))
(check-sat)
(get-model)
This produces:
sat
(model
(define-fun x () Int
2)
(define-fun x () Real
(/ 5.0 2.0))
)
What you are observing in the C-interface is essentially a rendering of the same. Of course, how much "checking" is enforced by the interface is totally solver specific as SMT-Lib says nothing about C API's or API's for other languages. That actually explains the BUG line you see in the output there. At this point, the behavior is entirely solver specific.
But long story short, SMT-Lib does indeed allow two variables have the same name used so long as they have different sorts.
Just started Haskell, it's said that everything in Haskell is "immutable" except IO package. So when I bind a name to something, it's always something immutable? Question, like below:
Prelude> let removeLower x=[c|c<-x, c `elem` ['A'..'Z']]
Prelude> removeLower "aseruiiUIUIdkf"
"UIUI"
So here:
1. “removeLower" is an immutable? Even it's a function object?
But I can still use "let" to assign something else to this name.
2. inside the function "c<-x" seems that "c" is a variable.
It is assigned by list x's values.
I'm using the word "variable" from C language, not sure how Haskell name all its names?
Thanks.
If you're familiar with C, think of the distinction between declaring a variable and assigning a value to it. For example, you can declare a variable on its own and later assign to it:
int i;
i = 7;
Or you can declare a variable and assign initial value at the same time:
int i = 7;
And in either case, you can mutate the value of a variable by assigning to it once more after the first initialization or assignment:
int i = 7; // Declaration and initial assignment
i = 5; // Mutation
Assignment in Haskell works exclusively like the second example—declaration with initialization:
You declare a variable;
Haskell doesn't allow uninitialized variables, so you are required to supply a value in the declaration;
There's no mutation, so the value given in the declaration will be the only value for that variable throughout its scope.
I bolded and hyperlinked "scope" because it's the second critical component here. This goes one of your questions:
“removeLower" is an immutable? Even it's a function object? But I can still use "let" to assign something else to this name.
After you bind removeLower to the function you define in your example, the name removeLower will always refer to that function within the scope of that definition. This is easy to demonstrate in the interpreter. First, let's define a function foo:
Prelude> let foo x = x + 2
Prelude> foo 4
6
Now we define an bar that uses foo:
Prelude> let bar x = foo (foo x)
Prelude> bar 4
8
And now we "redefine" foo to something different:
Prelude> let foo x = x + 3
Prelude> foo 4
7
Now what do you think happens to bar?
Prelude> bar 4
8
It remains the same! Because the "redefinition" of foo doesn't mutate anything—it just says that, in the new scope created by the "redefinition", the name foo stands for the function that adds three. The definition of bar was made in the earlier scope where foo x = x + 2, so that's the meaning that the name foo has in that definition of bar. The original value of foo was not destroyed or mutated by the "redefinition."
In a Haskell program as much as in a C program, the same name can still refer to different values in different scopes of the program. This is what makes "variables" variable. The difference is that in Haskell you can never mutate the value of a variable within one scope. You can shadow a definition, however—uses of a variable will refer to the "nearest" definition of that name in some sense. (In the case of the interpreter, the most recent let declaration for that variable.)
Now, with that out of the way, here are the syntaxes that exist in Haskell for variable binding ("assignment"). First, there's top-level declarations in a module:
module MyLibrary (addTwo) where
addTwo :: Int -> Int
addTwo x = x + 2
Here the name addTwo is declared with the given function as its value. A top level declaration can have private, auxiliary declarations in a where block:
addSquares :: Integer -> Integer
addSquares x y = squareOfX + squareOfY
where square z = z * z
squareOfX = square x
squareOfY = square y
Then there's the let ... in ... expression, that allows you to declare a local variable for any expression:
addSquares :: Integer -> Integer
addSquares x y =
let square z = z * z
squareOfX = square x
squareOfY = square y
in squareOfX + squareOfY
Then there's the do-notation that has its own syntax for declaring variables:
example :: IO ()
example = do
putStrLn "Enter your first name:"
firstName <- getLine
putStrLn "Enter your lasst name:"
lastName <- getLine
let fullName = firstName ++ " " ++ lastName
putStrLn ("Hello, " ++ fullName ++ "!")
The var <- action assigns a value that is produced by an action (e.g., reading a line from standard input), while let var = expr assigns a value that is produced by a function (e.g., concatenating some strings). Note that the let in a do block is not the same thing as the let ... in ... from above!
And finally, in list comprehension you get the same assignment syntax as in do-notation.
It's referring to the monadic bind operator >>=. You just don't need to explicitly write a lambda as right hand side parameter. The list comprension will be compiled down to the monadic actions defined. And by that it means exactly the same as in a monadic environment.
In fact you can replace the list comprension with a simple call to filter:
filter (`elem` ['A' .. 'Z']) x
To answer your question regarding the <- syntactic structure a bit clearer:
[c| c <- x]
is the same as
do c <- x
return c
is the same as
x >>= \c -> return c
is the same as
x >>= return
Consider the official documentation of Haskell for further reading: https://hackage.haskell.org/package/base-4.8.2.0/docs/Control-Monad.html#v:-62--62--61-
[c|c<-x, c `elem` ['A'..'Z']]
is a list comprehension, and c <- x is a generator where c is a pattern to be matched from the elements of the list x. c is a pattern which is successively bound to the elements of the input list x which are a, s, e, u, ... when you evaluate removeLower "aseruiiUIUIdkf".
c `elem` ['A'..'Z']
is a predicate which is applied to each successive binding of c inside the comprehension and an element of the input only appears in the output list if it passes this predicate.
In the below example code from a tutorial, I understand how the majority of it works. There are a few things I do not understand, though. Please help me to understand the purpose of some of these characters and the purpose of them. I will explain below what I don't understand.
Module paramByref
Sub swap(ByRef x As Integer, ByRef y As Integer)
Dim temp As Integer
temp = x ' save the value of x
x = y ' put y into x
y = temp 'put temp into y
End Sub
Sub Main()
' local variable definition
Dim a As Integer = 100
Dim b As Integer = 200
Console.WriteLine("Before swap, value of a : {0}", a)
Console.WriteLine("Before swap, value of b : {0}", b)
' calling a function to swap the values '
swap(a, b)
Console.WriteLine("After swap, value of a : {0}", a)
Console.WriteLine("After swap, value of b : {0}", b)
Console.ReadLine()
End Sub
End Module
What I don't understand is why the need for x, y, and temp. why not just keep declaration of a and b and swap the values?
Also, on the Console.WriteLine, I understand the {0} is an nth reference or index position, but the comma suggests to output something else. Is it saying to output the value of a in the 0 index? If so, there can only be one zero index, so why does the next line reference the value of the zero index? Or am I all wrong on this? Any explanation would be greatly appreciated...please dumb your answer down for this newbie.
The reason for x, y, and temp is to demonstrate how to do swapping. Your question asks
why not just keep declaration of a and b and swap the values
In this situation where you aren't doing any incrementing, sure you could just afterward say b = 100 and a = 200 but if you were doing this say in a loop and constantly swapping them, if you were to do
a = b
b = a
you would have a=100, b=100 because you set the value of b = a which is 100, then set a as the same value so it is still 100. That's why you need the temp value.
With regards to the console.writeLine, you are absolutely correct that it is indexing the 0th index, so in the first example, it is a, however it can be reused in the next line, because it is a totally separate line of code. Each of those console.writeLines can exist independently, so the values are indexed on a line by line basis
You need a temp to store the value from the first var so that you do not lose it when you overwrite it with the value from the second var. Once the second var's value has been transferred into the first var, the value stored in temp (from the original first var) can be transferred into the second var (overwriting the original value).
This is a common method when creating custom sorting routines that shuffle the values from various positions and ranks in an array around.
a and b are variables that are only visible in method Main(). They cannot be accessed from a different method. On the other hand, x, y, and temp are variables that are only visible in method swap(). More precisely, x and y are the method's parameters. The keyword ByRef means that the parameters are references to variables. I.e. when you call swap(a, b), x becomes a reference to variable a and y becomes a reference to variable b. You can work with references like with other variables. The difference is that when you change x in swap(), a in Main will change accordingly.
The Console.WriteLine() method takes a string parameter and an arbitrary number of additional parameters. The {0} is an index into the list of those additional parameters. E.g. Console.WriteLine("{0} {1} {2}", 100, 200, 300) would output 100 200 300. and Console.WriteLine("{2} {1} {0}", 100, 200, 300) would output 300 200 100.