Here is an example from the "Programming with Erlang" (2nd edition):
count_chars([], Result) ->
Result;
count_chars([C|Substring], #{C := N}=Result) ->
count_chars(Substring, Result#{C := N+1 });
count_chars([C|Substring], Result) ->
count_chars(Substring, Result#{C => 1}).
..which mercylessly yields the following error:
variable 'C' is unbound
So I am kind of stuck here; to my view, variable 'C' is bound, namely it must be a head of the string (just a linked list of chars, right?). Yet Erlang disagrees with me, breaking example from the (probably, outdated?) book I'am reading right now.
So what's wrong? What's the right way to pattern-match in this particular example?
P.S. A screenshot from the book. Pay attention at slightly different syntax, which also doesn't work for me:
P.P.S. I am using the latest version of Erlang I've managed to download from the official site.
C must be bound before the expression #{C := N}=Result is evaluated.
You consider that C is bound since the first parameter [C|Substring] was evaluated before: #{C := N}=Result. In fact it is not the case. There is no real assignment until a head evaluation succeed and the function enters the body.
Writing count_chars([C|Substring], #{C := N}=Result) -> is exactly the same as count_chars([C1|Substring], #{C2 := N}=Result) when C1 =:= C2 ->
During the head evaluation, each element is stored in a different element (a place in the heap) to check if all the parameters match the head definition. In your case the compiler want store the value C in an element, let's say x1 and the key C? in another element, let's say x2, and then verify that x1 and x2 are equals. the second operation is not possible without a deep modification of the compiler behavior.
I wrote a small example to show how it works, and compiled it with the option 'S' to see the result of the compilation:
test([K|_],K,M) -> % to see how the test of parameter equality is done
#{K := V} = M, % to verify that this pattern works when K is bound.
V.
the assembly result is :
{function, test, 3, 33}.
{label,32}.
{line,[{location,"mod.erl",64}]}.
{func_info,{atom,mod},{atom,test},3}.
{label,33}.
{test,is_nonempty_list,{f,32},[{x,0}]}. % the whole list is assigned to x0
{get_hd,{x,0},{x,3}}. % the list's head is assigned to x3
{test,is_eq_exact,{f,32},[{x,1},{x,3}]}. % the second parameter K is assigned to x1, verify if x1 =:= x3
{test,is_map,{f,34},[{x,2}]}. % the third parameter M is assigned to x2, check if it is a map if not go to label 34
{get_map_elements,{f,34},{x,2},{list,[{x,3},{x,0}]}}. % get the value associated to key x3 in the map x2 and store it into x0, if no suck key go to label 34
return.
{label,34}. % throw a badmatch error
{line,[{location,"mod.erl",65}]}.
{badmatch,{x,2}}.
Now, to code your function you can simply write:
count_chars([], Result) ->
Result;
count_chars([C|Substring], Result) ->
N = maps:get(C, Result, 0) +1,
count_chars(Substring, Result#{C => N }).
Related
Sometimes I need to use a declaration of a function inside another function. For example, I made the following in Mathematica:
Ie: There is a function f and when I compute f[Cos], it declares h as Cos[x]. Observe that I can't do the same by computing f[x] because it does not recognize $x$ as a function, although I can circumvent that by using Mathematica's notation for pure function: f[#&], now it works perfectly fine.
I noticed I can do the same with Maxima:
Although I don't know how to do the f[#&] in Maxima. Ie: Tell Maxima "x" is a function just as I did in Mathematica. Is there a way to do the same in Maxima? Also If I try to compile:
Without asking it to compute g(x), f won't work in the same way it worked before. It's not clear to me why this happens.
:= always defines a global function, even if it's within another function or block. As it stands, when you call f twice, the second definition of g clobbers the first one -- you can't have two different g functions.
I think what you want is an unnamed function, lambda([x], ...). The tricky part is that the body of lambda is not evaluated, so when you write lambda([x], y(x)), the value of y doesn't appear in the result. There are a few different ways to achieve that; I'll describe one way using subst.
subst('y = something, lambda([x], y(x))) constructs an unnamed function and then pastes something into it, replacing y. The result is lambda([x], something(x)) which I think is what you want.
To put this in the framework you outlined,
(%i3) f(x, y) := block ([g: subst ('y = y, lambda ([x], y(x)))], [g, g(x)]);
(%o3) f(x, y) := block([g : subst('y = y, lambda([x], y(x)))], [g, g(x)])
(%i4) f(1/2, cos);
1
(%o4) [lambda([x], cos(x)), cos(-)]
2
Given a function:
f(r, phi) := [r * cos(phi), r * sin(phi)];
I can find the derivative w.r.t. the first argument by providing arguments that turn it into an expression, like this:
f1(a, b) := diff(f(a, b), a);
But that's kind of wordy and round about. Is there a way to directly say "derivative of f w.r.t first argument?"
I'm trying to do this exercise:
I'm not sure how to use Type in F#, in F# interactive, I wrote type term = Term of float *int, Then I tried to create a value of type term by let x: term = (3.5,8);;But it gives an error.
Then I tried let x: term = Term (3.5,8);; and it worked. So Why is that?
For the first function, I tried:
let multiplyPolyByTerm (x:term, p:poly)=
match p with
|[]->[]
But that gives an error on the line |[]->[] saying that the expression is expecting a type poly, but poly is a in fact a list right? So why is it wrong here? I fixed it by |Poly[]->Poly[]. Then I tried to finish the function by giving the recursive definition of multiplying each term of the polynomial by the given term: |Poly a::af-> This gives an error so I'm stuck on trying to break down the Poly list.
If anyone has suggestion on good readings about Type in F#, please share it.
I got all the methods now, However,I find myself unable to throw an exception when the polynomial is an empty list as the base case of my recursive function is an empty list. Also, I don't know how to group common term together, Please help, Here are my codes:
type poly=Poly of (float*int) list
type term = Term of float *int
exception EmptyList
(*
let rec mergeCommonTerm(p:poly)=
let rec iterator ((a: float,b: int ), k: (float*int) list)=
match k with
|[]->(a,b)
|ki::kf-> if b= snd ki then (a+ fst ki,b)
match p with
|Poly [] -> Poly []
|Poly (a::af)-> match af with
|[]-> Poly [a]
|b::bf -> if snd a =snd b then Poly (fst a +fst b,snd a)::bf
else
*)
let rec multiplyPolyByTerm (x:term, p:poly)=
match x with
| Term (coe,deg) -> match p with
|Poly[] -> Poly []
|Poly (a::af) -> match multiplyPolyByTerm (x,Poly af) with
|Poly recusivep-> Poly ((fst a *coe,snd a + deg)::recusivep)
let rec addTermToPoly (x:term, p:poly)=
match x with
|Term (coe, deg)-> match p with
|Poly[] -> Poly [(coe,deg)]
|Poly (a::af)-> if snd a=deg then Poly ((fst a+coe,deg)::af)
else match addTermToPoly (x,Poly af) with
|Poly recusivep-> Poly (a::recusivep)
let rec addPolys (x:poly, y: poly)=
match x with
|Poly []->y
|Poly (xh::xt)-> addPolys(Poly xt,addTermToPoly(Term xh, y))
let rec multPolys (x:poly,y:poly)=
match x with
|Poly []-> Poly[]
|Poly (xh::xt)->addPolys (multiplyPolyByTerm(Term xh,y),multPolys(Poly xt,y))
let evalTerm (values:float) (termmm : term) :float=
match termmm with
|Term (coe,deg)->coe*(values**float(deg))
let rec evalPoly (polyn : poly, v: float) :float=
match polyn with
|Poly []->0.0
|Poly (ph::pt)-> (evalTerm v (Term ph)) + evalPoly (Poly pt,v)
let rec diffPoly (p:poly) :poly=
match p with
|Poly []->Poly []
|Poly (ah::at)-> match diffPoly (Poly at) with
|Poly [] -> if snd ah = 0 then Poly []
else Poly [(float(snd ah)*fst ah,snd ah - 1)]
|Poly (bh::bt)->Poly ((float(snd ah)*fst ah,snd ah - 1)::bh::bt)
As I mentioned in a comment, reading https://fsharpforfunandprofit.com/posts/discriminated-unions/ will be very helpful for you. But let me give you some quick help to get you unstuck and starting to solve your immediate problems. You're on the right track, you're just struggling a little with the syntax (and operator precedence, which is part of the syntax).
First, load the MSDN operator precedence documentation in another tab while you read the rest of this answer. You'll want to look at it later on, but first I'll explain a subtlety of how F# treats discriminated unions that you probably haven't understood yet.
When you define a discriminated union type like poly, the name Poly acts like a constructor for the type. In F#, constructors are functions. So when you write Poly (something), the F# parser interprets this as "take the value (something) and pass it to the function named Poly". Here, the function Poly isn't one you had to define explicitly; it was implicitly defined as part of your type definition. To really make this clear, consider this example:
type Example =
| Number of int
| Text of string
5 // This has type int
Number 5 // This has type Example
Number // This has type (int -> Example), i.e. a function
"foo" // This has type string
Text "foo" // This has type Example
Text // This has type (string -> Example), i.e. a function
Now look at the operator precedence list that you loaded in another tab. Lowest precedence is at the top of the table, and highest precedence is at the bottom; in other words, the lower something is on the table, the more "tightly" it binds. As you can see, function application (f x, calling f with parameter x) binds very tightly, more tightly than the :: operator. So when you write f a::b, that is not read as f (a::b), but rather as (f a)::b. In other words, f a::b reads as "Item b is a list of some type which we'll call T, and the function call f a produces an item of type T that should go in front of list b". If you instead meant "take the list formed by putting item a at the head of list b, and then call f with the resulting list", then that needs parentheses: you have to write f (a::b) to get that meaning.
So when you write Poly a::af, that's interpreted as (Poly a)::af, which means "Here is a list. The first item is a Poly a, which means that a is a (float * int) list. The rest of the list will be called af". And since the value your passing into it is not a list, but rather a poly type, that is a type mismatch. (Note that items of type poly contain lists, but they are not themselves lists). What you needed to write was Poly (a::af), which would have meant "Here is an item of type poly that contains a list. That list should be split into the head, a, and the rest, af."
I hope that helped rather than muddle the waters further. If you didn't understand any part of this, let me know and I'll try to make it clearer.
P.S. Another point of syntax you might want to know: F# gives you many ways to signal an error condition (like an empty list in this assignment), but your professor has asked you to use exception EmptyList when invalid input is given. That means he expects your code to "throw" or "raise" an exception when you encounter an error. In C# the term is "throw", but in F# the term is "raise", and the syntax looks like this:
if someErrorCondition then
raise EmptyList
// Or ...
match listThatShouldNotBeEmpty with
| [] -> raise EmptyList
| head::rest -> // Do something with head, etc.
That should take care of the next question you would have needed to ask. :-)
Update 2: You've edited your question to clarify another issue you're having, where your recursive function boils down to an empty list as the base case — yet your professor asked you to consider an empty list as an invalid input. There are two ways to solve this. I'll discuss the more complicated one first, then I'll discuss the easier one.
The more complicated way to solve this is to have two separate functions, an "outer" one and an "inner" one, for each of the functions you have been asked to define. In each case, the "outer" one checks whether the input is an empty list and throws an exception if that's the case. If the input is not an empty list, then it passes the input to the "inner" function, which does the recursive algorithm (and does NOT consider an empty list to be an error). So the "outer" function is basically only doing error-checking, and the "inner" function is doing all the work. This is a VERY common approach in professional programming, where all your error-checking is done at the "edges" of your code, while the "inner" code never has to deal with errors. It's therefore a good approach to know about — but in your particular case, I think it's more complicated than you need.
The easier solution is to rewrite your functions to consider a single-item list as the base case, so that your recursive functions never go all the way to an empty list. Then you can always consider an empty list to be an error. Since this is homework I won't give you an example based on your actual code, but rather an example based on a simple "take the sum of a list of integers" exercise where an empty list would be considered an error:
let rec sumNonEmptyList (input : int list) : int =
match input with
| [] -> raise EmptyList
| [x] -> x
| x::rest -> x + sumNonEmptyList rest
The syntax [x] in a match expression means "This matches a list with exactly one item in it, and assigns the name x to the value of that item". In your case, you'd probably be matching against Poly [] to raise an exception, Poly [a] as the base case, and Poly (a::af) as the "more than one item" case. (That's as much of a clue as I think I should give you; you'll learn better if you work out the rest yourself).
Consider the binary tree algebraic datatype
type btree = Empty | Node of btree * int * btree
and a new datatype defined as follows:
type finding = NotFound | Found of int
Heres my code so far:
let s = Node (Node(Empty, 5, Node(Empty, 2, Empty)), 3, Node (Empty, 6, Empty))
(*
(3)
/ \
(5) (6)
/ \ | \
() (2) () ()
/ \
() ()
*)
(* size: btree -> int *)
let rec size t =
match t with
Empty -> false
| Node (t1, m, t2) -> if (m != Empty) then sum+1 || (size t1) || (size t2)
let num = occurs s
printfn "There are %i nodes in the tree" num
This probably isn't close, I took a function that would find if an integer existed in a tree and tried changing the code for what I was trying to do.
I am very new to using F# and would appreciate any help. I am trying to count all non empty nodes in the tree. For example the tree I'm using should print the value 4.
I did not run the compiler on your code, but I believe this does even compile.
However your idea to use a pattern match in a recursive function is good.
As rmunn commented, you want to determine the number of nodes in each case:
An empty tree has no nodes, hence the result is zero.
A non-empty tree, has at least the root node plus the count of its left and right subtrees.
So something along the lines of the following should work
let rec size t =
match t with
| Empty -> 0
| Node (t1, _, t2) -> 1 + (size t1) + (size t2)
The most important detail here is, that you do not need a global variable sum to store any intermediate values. The whole idea of a recursive function is that those intermediate values are the results of recursive calls.
As a remark, your tree in the comment should look like this, I believe.
(*
(3)
/ \
(5) (6)
/ \ | \
() (2) () ()
/ \
() ()
*)
Edit: I misread the misaligned () as leaves of an empty tree, where in fact they are leaves of the subtree (2). So it was just an ASCII art issue :-)
Friedrich already posted a simple version of the size function that will work for most trees. However, the solution is not "tail-recursive", so it can cause a Stack Overflow for large trees. In functional programming languages like F#, recursion is often the preferred technique for things like counting and other aggregate functions. However, recursive functions generally consume a stack frame for each recursive call. This means that for large structures, the call stack can be exhausted before the function completes. In order to avoid this problem, compilers can optimize functions that are considered "tail-recursive" so that they use only one stack frame regardless of how many times they recurse. Unfortunately, this optimization cannot just be implemented for any recursive algorithm. It requires that the recursive call be the last thing that the function does, thereby ensuring that the compiler does not have to worry about jumping back into the function after the call, allowing it to overwrite the stack frame instead of adding another one.
In order to change the size function to be tail-recursive, we need some way to avoid having to call it twice in the case of a non-empty node, so that the call can be the last step of the function, instead of the addition between the two calls in Friedrich's solution. This can be accomplished using a couple different techniques, generally either using an accumulator or using Continuation Passing Style. The simpler solution is often to use an accumulator to keep track of the total size instead of having it be the return value, while Continuation Passing Style is a more general solution that can handle more complex recursive algorithms.
In order to make an accumulator pattern work for a tree where we have to sum both the left and right sub-trees, we need some way to make one tail-call at the end of the function, while still making sure that both sub-trees are evaluated. A simple way to do that is to also accumulate the right sub-trees in addition to the total count, so we can make subsequent tail-calls to evaluate those trees while evaluating the left sub-trees first. That solution might look something like this:
let size t =
let rec size acc ts = function
| Empty ->
match ts with
| [] -> acc
| head :: tail -> head |> size acc tail
| Node (t1, _, t2) ->
t1 |> size (acc + 1) (t2 :: ts)
t |> size 0 []
This adds the acc parameter and the ts parameter to represent the total count and remaining unevaluated sub-trees. When we hit a populated node, we evaluate the left sub-tree while adding the right sub-tree to our list of trees to evaluate later. When we hit the an empty node, we start evaluating any ts we've accumulated, until we have no further populated nodes or unevaluated sub-trees. This isn't the best possible solution for computing the tree-size, and most real solutions would use Continuation Passing Style to make it tail-recusive, but that should make a good exercise as you get more familiar with the language.
(define make (lambda (x) (lambda (y) (cons x (list y)))))
(let ((x 7)
(p (make 4)))
(cons x (p 0)))
I'm new to Scheme and functional program, so I am a bit clunky with walking through programs, but I get that if I used deep binding this program will return (7 4 0). Makes sense. What would this program do using shallow binding? I get this may sound dumb but is the p in the line with cons a redefinition? So in that case, we would return (7 0)?
Basically, I understand the concept of deep v. shallow binding, but I feel like I'm jumbling it up when looking at Scheme because I'm not crazy familiar with it.
Deep or shallow binding is an implementational technique and can not be observed from inside the program. The difference for the programmer is between lexical and dynamic scoping rules, but both can be implemented with any of the two techniques (i.e. one notion has got nothing to do with the other).
Deep or shallow refers to the choice of stack frame to hold a given outer scoped variable's binding. In deep binding there is a chain of frames to be accessed until the correct frame is entered holding the record for the variable; in shallow binding all bindings are present in one, shallow environment. See also "rerooting" (which only makes sense in the context of shallow binding implementation of lexical scoping).
To your specific question, under lexical scoping rules your code would return (7 4 0) and under dynamic - (7 7 0), because the call ((lambda(y) (list x y)) 0) is done inside the dynamic scope of x=7 binding (as a side note, (cons x (list y)) is the same as (list x y)):
x = 7
p = (lambda (y) (list x y)) ; x=4 is unused, in p=(make 4)
(cons 7 (p 0)) == (list 7 7 0) ; 'x' in this line and in lambda body for p
; both refer to same binding that is
; in effect, i.e. x=7
NB same terms (deep/shallow binding) are used in other language(s) now with completely different meaning (they do have something to do with the scoping rules there), which I don't care to fully understand. This answer is given in the context of Scheme.
Reference: Shallow Binding in LISP 1.5 by Baker, Henry G. Jr., 1977.
See this wikipedia article for a discussion on scoping (it mentions lexical/dynamic scoping and deep/shallow binding) bearing in mind that Scheme is lexically scoped. Will Ness' answer provides additional information.
For now, let's see step-by-step what's happening in this snippet of code:
; a variable called x is defined and assigned the value 7
(let ((x 7)
; make is called and returns a procedure p, inside its x variable has value 4
(p (make 4)))
; 7 is appended at the head of the result of calling p with y = 0
(cons x (p 0)))
=> '(7 4 0)
Notice that in the second line a closure is created in the lambda returned by make, and the variable x inside will be assigned the value 4. This x has nothing to do with the outer x, because Scheme is lexically scoped.
The last line is not a redefinition, as mentioned in the previous paragraph the x inside make is different from the x defined in the let expression.