Incomplete pattern match when two patterns share a `when` clause - f#

A common surprise for beginning F# programmers is the fact that the following is an incomplete match:
let x, y = 5, 10
match something with
| _ when x < y -> "Less than"
| _ when x = y -> "Equal"
| _ when x > y -> "Greater than"
But I just encountered a situation that surprised me. Here's a small bit of sample code to demonstrate it:
type Tree =
| Leaf of int
| Branch of Tree list
let sapling = Branch [Leaf 1] // Small tree with one leaf
let twoLeafTree = Branch [Leaf 1; Leaf 2]
let describe saplingsGetSpecialTreatment tree =
match tree with
| Leaf n
| Branch [Leaf n] when saplingsGetSpecialTreatment ->
sprintf "Either a leaf or a sapling containing %d" n
| Branch subTree ->
sprintf "Normal tree with sub-tree %A" subTree
describe true sapling // Result: "Either a leaf or a sapling containing 1"
describe false sapling // Result: "Normal tree with sub-tree [Leaf 1]"
describe true twoLeafTree // Result: "Normal tree with sub-tree [Leaf 1; Leaf 2]"
describe false twoLeafTree // Result: "Normal tree with sub-tree [Leaf 1; Leaf 2]"
This version of the describe function produced the "Incomplete pattern matches on this expression" warning, even though the pattern match is, in fact, complete. There are no possible trees that will not be matched by that pattern match, as can be seen by removing the specific branch of the match that had a when expression in it:
let describe tree =
match tree with
| Leaf n -> sprintf "Leaf containing %d" n
| Branch subTree ->
sprintf "Normal tree with sub-tree %A" subTree
This version of describe returns the "Normal tree" string for both the sapling and twoLeafTree trees.
In the case where the match expression contains nothing but when expressions (like the first example where x and y are being compared), it is reasonable that the F# compiler might not be able to tell whether the match will be complete. After all, x and y might be types with a "weird" implementation of comparison and equality where none of those three branches are true.*
But in cases like my describe function, why doesn't the F# compiler look at the pattern, say "If all the when expressions evaluated to false, there would still be a complete match" and skip the "incomplete pattern matches" warning? Is there some specific reason for this warning showing up here, or is it just a case of the F# compiler being just a little bit simplistic here and giving a false-positive warning because its code wasn't sophisticated enough?
* In fact, it is possible to set x and y to values such that x < y, x = y, and x > y are all false, without ever stepping outside the "normal" bounds of the standard .Net type system. As a special bonus question/puzzle, what are these values of x and y? No custom types needed to answer this puzzle; all you need is types provided in standard .Net.

In F# match syntax, the when guards apply to all cases enumerated just before it, not just to the last one.
In your specific scenario, the guard when saplingsGetSpecialTreatment applies to both Leaf n and Branch [Leaf n] cases. So this match will fail for the case when tree = Leaf 42 && saplingsGetSpecialTreatment = false
The following would be complete, since the Leaf case now has its own branch:
let describe saplingsGetSpecialTreatment tree =
match tree with
| Leaf n ->
sprintf "Either a leaf or a sapling containing %d" n
| Branch [Leaf n] when saplingsGetSpecialTreatment ->
sprintf "Either a leaf or a sapling containing %d" n
| Branch subTree ->
sprintf "Normal tree with sub-tree %A" subTree

Just clarifying Fyodor's post with an extra example. Think of it as a when y = 3 section, an otherwise section, and then, for everything else
let f y x =
match x with
| 0
| 1
| 2 when y = 3 -> "a"
| 0
| 1
| 2 -> "b"
| _ -> "c"
[0 .. 3] |> List.map (f 3)
[0 .. 3] |> List.map (f 2)
FSI
val f : y:int -> x:int -> string
> val it : string list = ["a"; "a"; "a"; "c"]
> val it : string list = ["b"; "b"; "b"; "c"]
So, is this a sensible default? I think so.
Here's a more explicit version:
let f2 y x =
match x,y with
| 0,3
| 0,3
| 0,3 -> "a"
| 0,_
| 1,_
| 2,_ -> "b"
| _ -> "c"
[0 .. 3] |> List.map (f2 3)
[0 .. 3] |> List.map (f2 2)
... and a more compact version:
let f3 y x = x |> function | 0 | 1 | 2 when y = 3 -> "a"
| 0 | 1 | 2 -> "b"
| _ -> "c"

Related

F# infinite stream of armstrong numbers

I'm trying to create an infinite Stream in F# that contains armstrong numbers. An armstrong number is one whose cubes of its digits add up to the number. For example, 153 is an armstrong number because 1^3 + 5^3 + 3^3 = 153. so far, I have created several functions to help me do so. They are:
type 'a stream = Cons of 'a * (unit -> 'a stream);;
let rec upfrom n = Cons (n, fun() -> upfrom (n+1));;
let rec toIntArray = function
| 0 -> []
| n -> n % 10 :: toIntArray (n / 10);;
 
let rec makearmstrong = function
| [] -> 0
| y::ys -> (y * y * y) + makearmstrong ys;;
let checkarmstrong n = n = makearmstrong(toIntArray n);;
let rec take n (Cons(x,xsf)) =
match n with
| 0 -> []
| _ -> x :: take (n-1)(xsf());;
let rec filter p (Cons (x, xsf)) =
if p x then Cons (x, fun() -> filter p (xsf()))
else filter p (xsf());;
And finally:
let armstrongs = filter (fun n -> checkarmstrong n)(upfrom 1);;
Now, when I do take 4 armstrongs;;, (or any number less than 4) this works perfectly and gives me [1;153;370;371] but if I do take 5 armstrongs;;nothing happens, it seems like the program freezes.
I believe the issue is that there are no numbers after 407 that are the sums of their cubes (see http://oeis.org/A046197), but when your code evaluates the equivalent of take 1 (Cons(407, filter checkarmstrong (upfrom 408))) it's going to force the evaluation of the tail and filter will recurse forever, never finding a matching next element. Also note that your definition of Armstrong numbers differs from, say, Wikipedia's, which states that the power the digits are raised to should be the number of digits in the number.

Printing tree list tuple elements in F#

I have a simple tree structure like this..
type Tree<'a,'b> =
| Node of list<'a * Tree<'a,'b>>
| Leaf of 'b
let phonebook = Node["MyPhonebook",
Node["Work",
Node["Company1", Node["Employee1", Leaf("phone#")];
"Company2", Leaf("phone#")];
"Private",
Node["Family", Node["Brother", Leaf("phone#");]; "Sister", Leaf("phone#")]
]
]
Im simply trying to print root folders (Work, Private) of this phone book but cant seem to get it right whatever I do...
let listelements tree =
match tree with
| Node[a, b] -> // OK - root node
printfn "%s" a // OK - root folder name
match b with
//| _ -> printf "%A" b // OK - prints whole node
| Node[a, b] -> printf "%A" a // Match failure!
listelements phonebook
After that I realized I've been matching tuple with list so I tried something a bit different but i get missmatches again.
let phonebook = [Node["Work",
Node["Company1", Node["Employee1", Leaf("phone#")];
"Company2", Leaf("phone#")];
"Private",
Node["Family", Node["Brother", Leaf("phone#");]; "Sister", Leaf("phone#")]
]
]
let listelements tree =
for i in tree do
match i with
| Node[a, b] -> printf "%s" a // Match fail
let listelements tree =
//tree |> List.iter (fun x -> printf "%A" x) // OK - prints nodes
tree |> List.iter (fun x ->
match x with
| Node[a, b] -> printf "%A" a) // Match fail
What the hell am I doing wrong here? There has to be a more elegant and simple way to do this. I come from C# and this is driving me insane :P
You are matching a list of a single element (a tuple) against a List, so in the first case the list contains only one element and it pass though the compiler warns you that the matching is incomplete.
But in the second match the list contains two tuples, so it fails.
Use instead a head::tail pattern, like this:
| Node((a, b)::xs)
or if you are not interested in the tail at all:
| Node((a, b)::_)
NOTE
Something that will help to clarify your code: tree is a list of tuples as from the definition. It's better if you write the pattern making the parenthesis of the tuple explicit, I mean if you write:
| Node[(a, b)] ->
instead of just
| Node[a, b] ->
Here the compiler understand that is a tuple, but a human may not, especially if it comes from a Haskell background.
So, writing the parenthesis will probably help you to figure out that the problem is that you are matching agains a list with only one tuple.
UPDATE
If you want to print all elements using List.iter:
let rec listelements tree =
let printNode (a, b) =
printfn "%s" a
listelements b
match tree with
| Leaf leaf -> printfn "Leaf %s" leaf
| Node(listOfNodes) -> List.iter printNode listOfNodes

i think i got infinite loop for this BST

this code i got is from Alexander Battisti about how to make a tree from a list of data:
let data = [4;3;8;7;10;1;9;6;5;0;2]
type Tree<'a> =
| Node of Tree<'a> * 'a * Tree<'a>
| Leaf
let rec insert tree element =
match element,tree with
| x,Leaf -> Node(Leaf,x,Leaf)
| x,Node(l,y,r) when x <= y -> Node((insert l x),y,r)
| x,Node(l,y,r) when x > y -> Node(l,y,(insert r x))
| _ -> Leaf
let makeTree = List.fold insert Leaf data
then i want to implement this code to my binary search tree code
let rec BinarySearch tree element =
match element,tree with
| x,Leaf -> BinarySearch (Node(Leaf,x,Leaf)) x
| x,Node(l,y,r) when x<=y ->
BinarySearch l y
| x,Node(l,y,r) when x>y ->
BinarySearch r y
| x,Node(l,y,r) when x=y ->
true
| _ -> false
then i use my search code like this:
> BinarySearch makeTree 5;;
and the result is none because it's like i got an infinite looping
can someone help me? if my code is wrong, please help me to correct it, thank you
The solution by Yin is how I would write it too.
Anyway, here is a solution that is closer to your version and (hopefully) explains what went wrong:
let rec BinarySearch tree element =
match element,tree with
| x, Leaf ->
// You originally called 'BinarySearch' here, but that's wrong - if we reach
// the leaf of the tree (on the path from root to leaf) then we know that the
// element is not in the tree so we return false
false
| x, Node(l,y,r) when x<y ->// This needs to be 'x<y', otherwise the clause would be
// matched when 'x=y' and we wouldn't find the element!
BinarySearch l element // Your recursive call was 'BinarySearch l y' but
// that's wrong - you want to search for 'element'
| x, Node(l,y,r) when x>y ->
BinarySearch r element
| x,Node(l,y,r) -> // You can simplify the code by omitting the 'when'
true // clause (because this will only be reached when
// x=y. Then you can omit the last (unreachable) case
let rec BinarySearch tree element =
match tree with
| Leaf -> false
| Node(l, v, r) ->
if v = element then
true
elif v < element then
BinarySearch r element
else
BinarySearch l element
BinarySearch makeTree 5

How to code Fizzbuzz in F#

I am currently learning F# and have tried (an extremely) simple example of FizzBuzz.
This is my initial attempt:
for x in 1..100 do
if x % 3 = 0 && x % 5 = 0 then printfn "FizzBuzz"
elif x % 3 = 0 then printfn "Fizz"
elif x % 5 = 0 then printfn "Buzz"
else printfn "%d" x
What solutions could be more elegant/simple/better (explaining why) using F# to solve this problem?
Note: The FizzBuzz problem is going through the numbers 1 to 100 and every multiple of 3 prints Fizz, every multiple of 5 prints Buzz, every multiple of both 3 AND 5 prints FizzBuzz. Otherwise, simple the number is displayed.
Thanks :)
I think you already have the "best" solution.
If you want to show off more functional/F#-isms, you could do e.g.
[1..100]
|> Seq.map (function
| x when x%5=0 && x%3=0 -> "FizzBuzz"
| x when x%3=0 -> "Fizz"
| x when x%5=0 -> "Buzz"
| x -> string x)
|> Seq.iter (printfn "%s")
and use lists, sequences, map, iter, patterns, and partial application.
[1..100] // I am the list of numbers 1-100.
// F# has immutable singly-linked lists.
// List literals use square brackets.
|> // I am the pipeline operator.
// "x |> f" is just another way to write "f x".
// It is a common idiom to "pipe" data through
// a bunch of transformative functions.
Seq.map // "Seq" means "sequence", in F# such sequences
// are just another name for IEnumerable<T>.
// "map" is a function in the "Seq" module that
// applies a function to every element of a
// sequence, returning a new sequence of results.
(function // The function keyword is one way to
// write a lambda, it means the same
// thing as "fun z -> match z with".
// "fun" starts a lambda.
// "match expr with" starts a pattern
// match, that then has |cases.
| x when x%5=0 && x%3=0
// I'm a pattern. The pattern is "x", which is
// just an identifier pattern that matches any
// value and binds the name (x) to that value.
// The "when" clause is a guard - the pattern
// will only match if the guard predicate is true.
-> "FizzBuzz"
// After each pattern is "-> expr" which is
// the thing evaluated if the pattern matches.
// If this pattern matches, we return that
// string literal "FizzBuzz".
| x when x%3=0 -> "Fizz"
// Patterns are evaluated in order, just like
// if...elif...elif...else, which is why we did
// the 'divisble-by-both' check first.
| x when x%5=0 -> "Buzz"
| x -> string x)
// "string" is a function that converts its argument
// to a string. F# is statically-typed, so all the
// patterns have to evaluate to the same type, so the
// return value of the map call can be e.g. an
// IEnumerable<string> (aka seq<string>).
|> // Another pipeline; pipe the prior sequence into...
Seq.iter // iter applies a function to every element of a
// sequence, but the function should return "unit"
// (like "void"), and iter itself returns unit.
// Whereas sequences are lazy, "iter" will "force"
// the sequence since it needs to apply the function
// to each element only for its effects.
(printfn "%s")
// F# has type-safe printing; printfn "%s" expr
// requires expr to have type string. Usual kind of
// %d for integers, etc. Here we have partially
// applied printfn, it's a function still expecting
// the string, so this is a one-argument function
// that is appropriate to hand to iter. Hurrah!
My example is just a minor improvement over the code posted by 'ssp'. It uses parameterized active patterns (which take the divisor as an argument). Here is a more in-depth explanation:
The following defines an active pattern that we can later use in the match
expression to test if a value i is divisible by a value divisor. When we write:
match 9 with
| DivisibleBy 3 -> ...
...it means that the value '9' will be passed to the following function as i and the value 3 will be passed as divisor. The name (|DivisibleBy|_|) is a special syntax, whith means that we're declaring an active pattern (and the name can appear in the
match on the left side of ->. The |_| bit means that the pattern can fail (our example fails when value is not divisible by divisor)
let (|DivisibleBy|_|) divisor i =
// If the value is divisible, then we return 'Some()' which
// represents that the active pattern succeeds - the '()' notation
// means that we don't return any value from the pattern (if we
// returned for example 'Some(i/divisor)' the use would be:
// match 6 with
// | DivisibleBy 3 res -> .. (res would be asigned value 2)
// None means that pattern failed and that the next clause should
// be tried (by the match expression)
if i % divisor = 0 then Some () else None
Now we can iterate over all the numbers and match them against patterns (our active pattern) using match (or using Seq.iter or some other technique as shown in other answers):
for i in 1..100 do
match i with
// & allows us to run more than one pattern on the argument 'i'
// so this calls 'DivisibleBy 3 i' and 'DivisibleBy 5 i' and it
// succeeds (and runs the body) only if both of them return 'Some()'
| DivisibleBy 3 & DivisibleBy 5 -> printfn "FizzBuzz"
| DivisibleBy 3 -> printfn "Fizz"
| DivisibleBy 5 -> printfn "Buzz"
| _ -> printfn "%d" i
For more information on F# active patterns, here is an MSDN documentation link. I think that if you remove all the comments, the code will be slightly more readable than the original version. It shows some quite useful tricks :-), but in your case, the task is relatively easy...
Yet one solution in F# style (i.e. with Active Patterns usage):
let (|P3|_|) i = if i % 3 = 0 then Some i else None
let (|P5|_|) i = if i % 5 = 0 then Some i else None
let f = function
| P3 _ & P5 _ -> printfn "FizzBuzz"
| P3 _ -> printfn "Fizz"
| P5 _ -> printfn "Buzz"
| x -> printfn "%d" x
Seq.iter f {1..100}
//or
for i in 1..100 do f i
To add one more possible answer - here is another approach without pattern matching. It uses the fact that Fizz + Buzz = FizzBuzz, so you don't actually need to test for all three cases, you only need to see if it is divisible by 3 (then print "Fizz") and also see if it is divisible by 5 (then print "Buzz") and finally, print a new line:
for i in 1..100 do
for divisor, str in [ (3, "Fizz"); (5, "Buzz") ] do
if i % divisor = 0 then printf "%s" str
printfn ""
The nested for loop assignes 3 and "Fizz" to divisor and str in the first iteration and then the second pair of values in the second iteration. The beneift is, you could easily add printing of "Jezz" when the value is divisible by 7 :-) ...in case that extensibility of the solution is a concern!
Here's one more:
let fizzy num =
match num%3, num%5 with
| 0,0 -> "fizzbuzz"
| 0,_ -> "fizz"
| _,0 -> "buzz"
| _,_ -> num.ToString()
[1..100]
|> List.map fizzy
|> List.iter (fun (s:string) -> printfn "%s" s)
I find this to be a bit more readable answer edited was inspired a bit by the others
let FizzBuzz n =
match n%3,n%5 with
| 0,0 -> "FizzBuzz"
| 0,_ -> "Fizz"
| _,0 -> "Buzz"
| _,_ -> string n
[1..100]
|> Seq.map (fun n -> FizzBuzz n)
|> Seq.iter (printfn "%s")
Here is my version:
//initialize array a with values from 1 to 100
let a = Array.init 100 (fun x -> x + 1)
//iterate over array and match *indexes* x
Array.iter (fun x ->
match x with
| _ when x % 15 = 0 -> printfn "FizzBuzz"
| _ when x % 5 = 0 -> printfn "Buzz"
| _ when x % 3 = 0 -> printfn "Fizz"
| _ -> printfn "%d" x
) a
This is my first program in F#.
It's not perfect, but I think someone who starts learning F# (like me :)) can figure out what happens here quite fast.
However I am wondering what is the difference between matching to any _ or to x itself in pattern matching above?
I couldn't find a working solution that didn't include testing for i % 15 = 0. I've always felt that not testing for that is part of this "stupid" assignment. Be aware that this is probably not idiomatic F# since it's my first program in the language.
for n in 1..100 do
let s = seq {
if n % 3 = 0 then yield "Fizz"
if n % 5 = 0 then yield "Buzz" }
if Seq.isEmpty s then printf "%d"n
printfn "%s"(s |> String.concat "")
Here's a version emphasizing a generic tuple list of carbonations:
let carbonations = [(3, "Spizz") ; (5, "Fuzz"); (15, "SpizzFuzz");
(30, "DIZZZZZZZZ"); (18, "WHIIIIII")]
let revCarbonated = carbonations |> List.sort |> List.rev
let carbonRoute someCarbonations findMe =
match(List.tryFind (fun (x,_) -> findMe % x = 0) someCarbonations) with
| Some x -> printfn "%d - %s" findMe (snd x)
| None -> printfn "%d" findMe
let composeCarbonRoute = carbonRoute revCarbonated
[1..100] |> List.iter composeCarbonRoute
I don't like all these repeated strings, here's mine:
open System
let ar = [| "Fizz"; "Buzz"; |]
[1..100] |> List.map (fun i ->
match i % 3 = 0, i % 5 = 0 with
| true, false -> ar.[0]
| false, true -> ar.[1]
| true, true -> ar |> String.Concat
| _ -> string i
|> printf "%s\n"
)
|> ignore
Here is an attempt that factors out the modulo checks
let DivisibleBy x y = y % x = 0
[ 1 .. 100 ]
|> List.map (function
| x when DivisibleBy (3 * 5) x -> "fizzbuzz"
| x when DivisibleBy 3 x -> "fizz"
| x when DivisibleBy 5 x -> "buzz"
| x -> string x)
|> List.iter (fun x -> printfn "%s" x)
Here is how I refined it

Avoiding code duplication in F#

I have two snippets of code that tries to convert a float list to a Vector3 or Vector2 list. The idea is to take 2/3 elements at a time from the list and combine them as a vector. The end result is a sequence of vectors.
let rec vec3Seq floatList =
seq {
match floatList with
| x::y::z::tail -> yield Vector3(x,y,z)
yield! vec3Seq tail
| [] -> ()
| _ -> failwith "float array not multiple of 3?"
}
let rec vec2Seq floatList =
seq {
match floatList with
| x::y::tail -> yield Vector2(x,y)
yield! vec2Seq tail
| [] -> ()
| _ -> failwith "float array not multiple of 2?"
}
The code looks very similiar and yet there seems to be no way to extract a common portion. Any ideas?
Here's one approach. I'm not sure how much simpler this really is, but it does abstract some of the repeated logic out.
let rec mkSeq (|P|_|) x =
seq {
match x with
| P(p,tail) ->
yield p
yield! mkSeq (|P|_|) tail
| [] -> ()
| _ -> failwith "List length mismatch" }
let vec3Seq =
mkSeq (function
| x::y::z::tail -> Some(Vector3(x,y,z), tail)
| _ -> None)
As Rex commented, if you want this only for two cases, then you probably won't have any problem if you leave the code as it is. However, if you want to extract a common pattern, then you can write a function that splits a list into sub-list of a specified length (2 or 3 or any other number). Once you do that, you'll only use map to turn each list of the specified length into Vector.
The function for splitting list isn't available in the F# library (as far as I can tell), so you'll have to implement it yourself. It can be done roughly like this:
let divideList n list =
// 'acc' - accumulates the resulting sub-lists (reversed order)
// 'tmp' - stores values of the current sub-list (reversed order)
// 'c' - the length of 'tmp' so far
// 'list' - the remaining elements to process
let rec divideListAux acc tmp c list =
match list with
| x::xs when c = n - 1 ->
// we're adding last element to 'tmp',
// so we reverse it and add it to accumulator
divideListAux ((List.rev (x::tmp))::acc) [] 0 xs
| x::xs ->
// add one more value to 'tmp'
divideListAux acc (x::tmp) (c+1) xs
| [] when c = 0 -> List.rev acc // no more elements and empty 'tmp'
| _ -> failwithf "not multiple of %d" n // non-empty 'tmp'
divideListAux [] [] 0 list
Now, you can use this function to implement your two conversions like this:
seq { for [x; y] in floatList |> divideList 2 -> Vector2(x,y) }
seq { for [x; y; z] in floatList |> divideList 3 -> Vector3(x,y,z) }
This will give a warning, because we're using an incomplete pattern that expects that the returned lists will be of length 2 or 3 respectively, but that's correct expectation, so the code will work fine. I'm also using a brief version of sequence expression the -> does the same thing as do yield, but it can be used only in simple cases like this one.
This is simular to kvb's solution but doesn't use a partial active pattern.
let rec listToSeq convert (list:list<_>) =
seq {
if not(List.isEmpty list) then
let list, vec = convert list
yield vec
yield! listToSeq convert list
}
let vec2Seq = listToSeq (function
| x::y::tail -> tail, Vector2(x,y)
| _ -> failwith "float array not multiple of 2?")
let vec3Seq = listToSeq (function
| x::y::z::tail -> tail, Vector3(x,y,z)
| _ -> failwith "float array not multiple of 3?")
Honestly, what you have is pretty much as good as it can get, although you might be able to make a little more compact using this:
// take 3 [1 .. 5] returns ([1; 2; 3], [4; 5])
let rec take count l =
match count, l with
| 0, xs -> [], xs
| n, x::xs -> let res, xs' = take (count - 1) xs in x::res, xs'
| n, [] -> failwith "Index out of range"
// split 3 [1 .. 6] returns [[1;2;3]; [4;5;6]]
let rec split count l =
seq { match take count l with
| xs, ys -> yield xs; if ys <> [] then yield! split count ys }
let vec3Seq l = split 3 l |> Seq.map (fun [x;y;z] -> Vector3(x, y, z))
let vec2Seq l = split 2 l |> Seq.map (fun [x;y] -> Vector2(x, y))
Now the process of breaking up your lists is moved into its own generic "take" and "split" functions, its much easier to map it to your desired type.

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