I am creating an MSetList P with elements of type String, and I would like to obtain the Powerset of P. I am not being able to figure it out.
Code below.
Thanks for your help :-)
Require Import
Coq.MSets.MSetList
Coq.Strings.String
Coq.Structures.OrdersEx.
Module set := Make OrdersEx.String_as_OT.
Definition P := set.add "A"%string (set.add "B"%string (set.add "C"%string (set.add "D"%string set.empty))).
Compute P.
Answer was provided on the Coq Discourse Forum by Yves Bertot
The construction of set shows that you are using the Make function
with a module as argument. If you type Print Make., you see that
this is a functor taking a module of type OrderedType as argument,
and it produces a module with many fields, among which t, eq,
eq_equiv, lt, lt_strorder, lt_compat, compare,
compare_spec, and eq_dec. If you type Print OrderedType., you
see that these field are the ones required to make an OrderedType.
So Make constructs all the fields that would be required to call Make
again, thus producing the powerset.
You can just type :
Module pset := Make set.
and you will simply obtain a structure of sets, whose elements are in
set.t. The following example was tested with coq 8.15.
Require Import
Coq.MSets.MSetList
Coq.Strings.String
Coq.Structures.OrdersEx.
Module set := Make OrdersEx.String_as_OT.
Module pset := Make set.
Definition set1 := set.add "A"%string set.empty.
Definition set2 := set.add "B"%string set.empty.
Definition set3 := set.add "C"%string set1.
Definition pset1 := pset.add set1 (pset.add set2 pset.empty).
Compute pset.mem set3 pset1.
Compute pset.mem set2 pset1.
I'm trying to define an entity named isVector using the following syntax
Require Export Setoid.
Require Export Coq.Reals.Reals.
Require Export ArithRing.
Definition Point := Type.
Record MassPoint:Type:= cons{number : R ; point: Point}.
Variable add_MP : MassPoint -> MassPoint -> MassPoint .
Variable mult_MP : R -> MassPoint -> MassPoint .
Variable orthogonalProjection : Point -> Point -> Point -> Point.
Definition isVector (v:MassPoint):= exists A, B :Point , v= add_MP((−1)A)(1B).
And the Coq IDE keeps complaining that there's a syntax error for the definition. Currently, I haven't figured it out.
There are many problems here.
First, you'd write:
exists A B : Point, …
with no comma between the different variables.
But then, you also have syntax errors on the right-hand side. First, I'm not sure those 1 and -1 operations exist. Second, function calls would be written this way:
add_MP A B
You can write them the way you do:
add_MP(A)(B)
But in the long run you should probably become used to the syntax of function calls being just a whitespace! You might need to axiomatize this -1 operation the way you axiomatized other operations, unless they are a notation that you defined somewhere but did not post here.
Thanks for the help.
After experimenting a little bit. Below is the solution that works.
Definition Point:= Type.
Record massPoint: Type := cons{number: R; point: Point}.
Variable add_MP: massPoint -> massPoint -> massPoint.
Variable mult_MP: R -> massPoint -> massPoint.
Definition tp (p:Point) := cons (-1) p.
Definition isVector(v:massPoint):= exists A B : Point, v = add_MP(cons (-1) A)(cons 1 B).
I want to translate the Maude code I wrote before to Coq since Coq has more powerful expression than Maude.
I have no idea how to represent the following code:
As shown above, the GuardPostfix has three subsorts: GuardPostfix1, GuardPostfix2 and GuardPostfix3.
I use inductive type to denote the GCcomponent. However, when defining the type GComponent, GComponent1 has two constructors that use the constructor GuardPostfix1 and GuardPostfix2 respectively.
How can I define these types in Coq?
I think the general recipe is to translate all Maude sorts into Coq types, and to represent Maude subsort relationships as the supersort having constructors for all of its subsorts (eg, subsort A B < C becomes Definition C := A + B). Following that general rule here's how I translated your example. Note that I don't know Maude, so it's possibly I missed something about the example itself. Also, I would strongly consider using a new inductive rather than the generic sum type (+) for GuardPostfix and GComponent, but that's up to you.
Module GuardedComponent.
Parameter BoolExp:Type.
Parameter Assignment:Type.
Parameter Program:Type.
Parameter Index:Type.
Parameter EndPoint:Type.
Parameter Null:Type.
Parameter EventGuard:Type.
Parameter TimeControl:Type.
Inductive AssignmentGuard :=
| mkAssignmentGuard (_:BoolExp) (_:Assignment).
Inductive GuardPostfix1 :=
| mkGuardPostfix1 (_:Program) (_:Index).
Inductive GuardPostfix2 :=
| mkGuardPostfix2 (_:Program) (_:EndPoint).
Inductive GuardPostfix3 :=
| mkGuardPostfix3 (_:Program) (_:Null).
Definition GuardPostfix : Type :=
GuardPostfix1 + GuardPostfix2 + GuardPostfix3.
Inductive GComponent1 :=
| comp1_post1 (_:GuardPostfix1)
| comp1_post2 (_:GuardPostfix2).
Inductive GComponent2 :=
| mkGComponent2 (_:EventGuard) (_:GuardPostfix3).
Inductive GComponent3 :=
| mkGComponent3 (_:TimeControl) (_:GuardPostfix3).
Definition GComponent : Type :=
GComponent1 + GComponent2 + GComponent3.
End GuardedComponent.
I'd like my own scope, to play around with long distfixes.
Declare Scope my_scope.
Delimit Scope my_scope with my.
Open Scope my_scope.
Definition f (x y a b : nat) : nat := x+y+a+b.
Notation "x < y * a = b" := (f x y a b)
(at level 100, no associativity) : my_scope.
Check (1 < 2 * 3 = 4)%my.
How do you make a new scope?
EDIT: I chose "x < y * a = b" to override Coq's operators (each with a different precedence).
The command Declare Scope does not exist. The various commands about scopes are described in section 12.2 of the Coq manual.
Your choice of an example notation has inherent problems, because it clashes with pre-defined notations, which seem to be used before your notation.
When looking at the first components the parser sees _ < _ and thinks that you are actually talking about comparison of integers, then it sees the second part as being an instance of the notation _ * _, then it sees that all that is the left hand side of an equality. And all along the parser is happy, it constructs an expression of the form:
(1 < (2 * 3)) = 4
This is constructed by the parser, and the type system has not been solicited yet. The type checker sees a natural number as the first child of (_ < _) and is happy. It sees (_ * _) as the second child and it is happy, it now knows that the first child of that product should be a nat number and it is still happy; in the end it has an equality, and the first component of this equality is in type Prop, but the second component is in type nat.
If you type Locate "_ < _ * _ = _". the answer tells you that you did define a new notation. The problem is that this notation never gets used, because the parser always finds another notation it can use before. Understanding why a notation is preferred to another one requires more knowledge of parsing technology, as alluded to in Coq's manual, chapter 12, in the sentence (obscure to me):
Coq extensible parsing is performed by Camlp5 which is essentially a LL1 parser.
You have to choose the levels of the various variables, x, y, a, and b so that none of these variables will be able to match too much of the text. For instance, I tried defining a notation close to yours, but with a starting and an ending bracket (and I guess this simplifies the task greatly).
Notation "<< x < y * a = b >>" := (f x y a b)
(x at level 59, y at level 39, a at level 59) : my_scope.
The level of x is chosen to be lower than the level of =, the level of y is chosen to be lower than the level of *, the level of a is chosen to be lower than =. The levels were obtained by reading the answer of the command Print Grammar constr. It seems to work, as the following command is accepted.
Check << 1 < 2 * 3 = 4 >>.
But you may need to include a little more engineering to have a really good notation.
To answer the actual question in your title:
The new scope gets created when you declare a notation that uses it. That is, you don’t declare a new scope my_scope separately. You just write
Notation "x <<< y" := (f x y) (at level 100, no associativity) : my_scope.
and that declares a new scope my_scope.
The answers for this question only apply to older versions of Coq. I'm not sure when it started but in at least Coq 8.13.2, Coq prefers the user to first use Declare Scope create a new scope. What the OP has in their code is Coq's preferred way to declare scopes now.
See the current manual
So Mathematica is different from other dialects of lisp because it blurs the lines between functions and macros. In Mathematica if a user wanted to write a mathematical function they would likely use pattern matching like f[x_]:= x*x instead of f=Function[{x},x*x] though both would return the same result when called with f[x]. My understanding is that the first approach is something equivalent to a lisp macro and in my experience is favored because of the more concise syntax.
So I have two questions, is there a performance difference between executing functions versus the pattern matching/macro approach? Though part of me wouldn't be surprised if functions were actually transformed into some version of macros to allow features like Listable to be implemented.
The reason I care about this question is because of the recent set of questions (1) (2) about trying to catch Mathematica errors in large programs. If most of the computations were defined in terms of Functions, it seems to me that keeping track of the order of evaluation and where the error originated would be easier than trying to catch the error after the input has been rewritten by the successive application of macros/patterns.
The way I understand Mathematica is that it is one giant search replace engine. All functions, variables, and other assignments are essentially stored as rules and during evaluation Mathematica goes through this global rule base and applies them until the resulting expression stops changing.
It follows that the fewer times you have to go through the list of rules the faster the evaluation. Looking at what happens using Trace (using gdelfino's function g and h)
In[1]:= Trace#(#*#)&#x
Out[1]= {x x,x^2}
In[2]:= Trace#g#x
Out[2]= {g[x],x x,x^2}
In[3]:= Trace#h#x
Out[3]= {{h,Function[{x},x x]},Function[{x},x x][x],x x,x^2}
it becomes clear why anonymous functions are fastest and why using Function introduces additional overhead over a simple SetDelayed. I recommend looking at the introduction of Leonid Shifrin's excellent book, where these concepts are explained in some detail.
I have on occasion constructed a Dispatch table of all the functions I need and manually applied it to my starting expression. This provides a significant speed increase over normal evaluation as none of Mathematica's inbuilt functions need to be matched against my expression.
My understanding is that the first approach is something equivalent to a lisp macro and in my experience is favored because of the more concise syntax.
Not really. Mathematica is a term rewriter, as are Lisp macros.
So I have two questions, is there a performance difference between executing functions versus the pattern matching/macro approach?
Yes. Note that you are never really "executing functions" in Mathematica. You are just applying rewrite rules to change one expression into another.
Consider mapping the Sqrt function over a packed array of floating point numbers. The fastest solution in Mathematica is to apply the Sqrt function directly to the packed array because it happens to implement exactly what we want and is optimized for this special case:
In[1] := N#Range[100000];
In[2] := Sqrt[xs]; // AbsoluteTiming
Out[2] = {0.0060000, Null}
We might define a global rewrite rule that has terms of the form sqrt[x] rewritten to Sqrt[x] such that the square root will be calculated:
In[3] := Clear[sqrt];
sqrt[x_] := Sqrt[x];
Map[sqrt, xs]; // AbsoluteTiming
Out[3] = {0.4800007, Null}
Note that this is ~100× slower than the previous solution.
Alternatively, we might define a global rewrite rule that replaces the symbol sqrt with a lambda function that invokes Sqrt:
In[4] := Clear[sqrt];
sqrt = Function[{x}, Sqrt[x]];
Map[sqrt, xs]; // AbsoluteTiming
Out[4] = {0.0500000, Null}
Note that this is ~10× faster than the previous solution.
Why? Because the slow second solution is looking up the rewrite rule sqrt[x_] :> Sqrt[x] in the inner loop (for each element of the array) whereas the fast third solution looks up the value Function[...] of the symbol sqrt once and then applies that lambda function repeatedly. In contrast, the fastest first solution is a loop calling sqrt written in C. So searching the global rewrite rules is extremely expensive and term rewriting is expensive.
If so, why is Sqrt ever fast? You might expect a 2× slowdown instead of 10× because we've replaced one lookup for Sqrt with two lookups for sqrt and Sqrt in the inner loop but this is not so because Sqrt has the special status of being a built-in function that will be matched in the core of the Mathematica term rewriter itself rather than via the general-purpose global rewrite table.
Other people have described much smaller performance differences between similar functions. I believe the performance differences in those cases are just minor differences in the exact implementation of Mathematica's internals. The biggest issue with Mathematica is the global rewrite table. In particular, this is where Mathematica diverges from traditional term-level interpreters.
You can learn a lot about Mathematica's performance by writing mini Mathematica implementations. In this case, the above solutions might be compiled to (for example) F#. The array may be created like this:
> let xs = [|1.0..100000.0|];;
...
The built-in sqrt function can be converted into a closure and given to the map function like this:
> Array.map sqrt xs;;
Real: 00:00:00.006, CPU: 00:00:00.015, GC gen0: 0, gen1: 0, gen2: 0
...
This takes 6ms just like Sqrt[xs] in Mathematica. But that is to be expected because this code has been JIT compiled down to machine code by .NET for fast evaluation.
Looking up rewrite rules in Mathematica's global rewrite table is similar to looking up the closure in a dictionary keyed on its function name. Such a dictionary can be constructed like this in F#:
> open System.Collections.Generic;;
> let fns = Dictionary<string, (obj -> obj)>(dict["sqrt", unbox >> sqrt >> box]);;
This is similar to the DownValues data structure in Mathematica, except that we aren't searching multiple resulting rules for the first to match on the function arguments.
The program then becomes:
> Array.map (fun x -> fns.["sqrt"] (box x)) xs;;
Real: 00:00:00.044, CPU: 00:00:00.031, GC gen0: 0, gen1: 0, gen2: 0
...
Note that we get a similar 10× performance degradation due to the hash table lookup in the inner loop.
An alternative would be to store the DownValues associated with a symbol in the symbol itself in order to avoid the hash table lookup.
We can even write a complete term rewriter in just a few lines of code. Terms may be expressed as values of the following type:
> type expr =
| Float of float
| Symbol of string
| Packed of float []
| Apply of expr * expr [];;
Note that Packed implements Mathematica's packed lists, i.e. unboxed arrays.
The following init function constructs a List with n elements using the function f, returning a Packed if every return value was a Float or a more general Apply(Symbol "List", ...) otherwise:
> let init n f =
let rec packed ys i =
if i=n then Packed ys else
match f i with
| Float y ->
ys.[i] <- y
packed ys (i+1)
| y ->
Apply(Symbol "List", Array.init n (fun j ->
if j<i then Float ys.[i]
elif j=i then y
else f j))
packed (Array.zeroCreate n) 0;;
val init : int -> (int -> expr) -> expr
The following rule function uses pattern matching to identify expressions that it can understand and replaces them with other expressions:
> let rec rule = function
| Apply(Symbol "Sqrt", [|Float x|]) ->
Float(sqrt x)
| Apply(Symbol "Map", [|f; Packed xs|]) ->
init xs.Length (fun i -> rule(Apply(f, [|Float xs.[i]|])))
| f -> f;;
val rule : expr -> expr
Note that the type of this function expr -> expr is characteristic of term rewriting: rewriting replaces expressions with other expressions rather than reducing them to values.
Our program can now be defined and executed by our custom term rewriter:
> rule (Apply(Symbol "Map", [|Symbol "Sqrt"; Packed xs|]));;
Real: 00:00:00.049, CPU: 00:00:00.046, GC gen0: 24, gen1: 0, gen2: 0
We've recovered the performance of Map[Sqrt, xs] in Mathematica!
We can even recover the performance of Sqrt[xs] by adding an appropriate rule:
| Apply(Symbol "Sqrt", [|Packed xs|]) ->
Packed(Array.map sqrt xs)
I wrote an article on term rewriting in F#.
Some measurements
Based on #gdelfino answer and comments by #rcollyer I made this small program:
j = # # + # # &;
g[x_] := x x + x x ;
h = Function[{x}, x x + x x ];
anon = Table[Timing[Do[ # # + # # &[i], {i, k}]][[1]], {k, 10^5, 10^6, 10^5}];
jj = Table[Timing[Do[ j[i], {i, k}]][[1]], {k, 10^5, 10^6, 10^5}];
gg = Table[Timing[Do[ g[i], {i, k}]][[1]], {k, 10^5, 10^6, 10^5}];
hh = Table[Timing[Do[ h[i], {i, k}]][[1]], {k, 10^5, 10^6, 10^5}];
ListLinePlot[ {anon, jj, gg, hh},
PlotStyle -> {Black, Red, Green, Blue},
PlotRange -> All]
The results are, at least for me, very surprising:
Any explanations? Please feel free to edit this answer (comments are a mess for long text)
Edit
Tested with the identity function f[x] = x to isolate the parsing from the actual evaluation. Results (same colors):
Note: results are very similar to this Plot for constant functions (f[x]:=1);
Pattern matching seems faster:
In[1]:= g[x_] := x*x
In[2]:= h = Function[{x}, x*x];
In[3]:= Do[h[RandomInteger[100]], {1000000}] // Timing
Out[3]= {1.53927, Null}
In[4]:= Do[g[RandomInteger[100]], {1000000}] // Timing
Out[4]= {1.15919, Null}
Pattern matching is also more flexible as it allows you to overload a definition:
In[5]:= g[x_] := x * x
In[6]:= g[x_,y_] := x * y
For simple functions you can compile to get the best performance:
In[7]:= k[x_] = Compile[{x}, x*x]
In[8]:= Do[k[RandomInteger[100]], {100000}] // Timing
Out[8]= {0.083517, Null}
You can use function recordSteps in previous answer to see what Mathematica actually does with Functions. It treats it just like any other Head. IE, suppose you have the following
f = Function[{x}, x + 2];
f[2]
It first transforms f[2] into
Function[{x}, x + 2][2]
At the next step, x+2 is transformed into 2+2. Essentially, "Function" evaluation behaves like an application of pattern matching rules, so it shouldn't be surprising that it's not faster.
You can think of everything in Mathematica as an expression, where evaluation is the process of rewriting parts of the expression in a predefined sequence, this applies to Function like to any other head