I am really new to F# and am struggling to adapt to the whole functional programming mindset. Basically I am trying to figure out how to iterate through a list of numbers and add them up while the sum is less than a certain number (200 for instance).
At the moment I have something along the lines of the following:
let nums = [20 .. 50]
let mutable total = 0
let addUp (num, (total : byref<int>)) =
if (num + total < 200) then
total <- total + num
for num in nums do
addUp (num, &total)
This does get the job done, but I feel like there must be a more appropriate way to do this while still remaining within the style of functional programming, and without out having to rely on a mutable. I still feel like I'm approaching things from an imperative mindset.
I was thinking I could somehow use List.fold to do this but after some experimentation I was unable to figure out how to utilize it in conjunction with the whole while < 200 thing. Any help would be hugely appreciated and I apologize in advance if this is a commonly asked and answered question (I couldn't find anything through my own search).
For educational purposes, the answer above is fine as it lets you see what exactly happens within the fold or reduce.
For a real project, it is better to use existing library functions. You can even separate summation and bound checking. Consider this:
let addUp threshold xs =
xs
|> Seq.scan (+) 0 // line 1
|> Seq.takeWhile (fun x -> x < threshold) // line 2
|> Seq.last // line 3
// Usage:
let nums = Seq.initInfinite ( (+) 20) // line 4
nums
|> addUp 200
|> printfn "%d" // prints "188"
A bit of explanation:
line 1 is a contraction of Seq.scan (fun state x -> state + x) 0 so it actually returns a sequence of intermediate sums (20, 41, 63, ...);
in line 2, we take only the elements that match or filtering predicate;
in line 3, we simply take the last element (that matched the filtering above);
in line 4, again, it's a contraction of Seq.initInfinite (fun x -> 20+x). I took my liberty to make your data an infinite sequence (20, 21, 22, ...), but it still works.
Note, the code looks like three calls, but the sequence is evaluated only once.
Note, in line 2, don't try contraction like above. The reason is that (<) threshold evaluates to fun x -> threshold < x which is wrong. If you. however, use (>) threshold, it reads counter-intuitive and confusing.
Note, there's no error checking in the function. Calling it with an empty source sequence will crash at Seq.last.
First, let's try to avoid mutable variable and one way to do it is to create a recursive function to iterate through the list with the latest total.
let rec addUp total nums =
match nums with
| [] -> total
| num::tl ->
let newTotal = num + total
if newTotal < 200 then
addUp newTotal tl
else
total
addUp 0 nums
We create a recursive function addUp which accepts total and the list of numbers. The function extracts the first number of the list and adds it to the total if the new total is less than 200. Since we might still have more numbers on the list, we call addUp again with the new total and the rest of the numbers. Otherwise, we stop and return the new total.
We can also use List.fold which makes the code cleaner, but the whole list will be iterated. Here is the code:
List.fold (fun total num ->
let newTotal = num + total
if newTotal < 200 then
newTotal
else
total
) 0 nums
Since the output is the same type as the member of the input, we can use List.reduce and get rid of initial state (0).
List.reduce (fun total num ->
let newTotal = num + total
if newTotal < 200 then
newTotal
else
total
) nums
Hope this helps.
Related
I disgress, I am stuck and I really can't wrap my head around what's wrong. The problem reads.
Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?
Now I store the digit as a string, I only fetch the numbers, I loop through and grab each 13-digit "substring" as an array, multiply them together and compare them. Now I have verified that I am only getting a 1000-digit char array, I have verified that I get exactly 75 char arrays of equal size. Yet I am not getting the right answer.
Here is the code
let problem8() =
let str = #"731671765313306..."
|> Seq.filter (Char.IsDigit)
|> Seq.toArray
(* We only need to go to 987 because 1000 isn't divisble by 13 and if we were to take the last 11 digits from 987
we would end up with 0 anyhow. *)
seq { for i in 0.. 13 ..987 -> str.[i..i + 12] }
|> Seq.map (Seq.fold (fun acc chr -> acc * int64 (Char.GetNumericValue(chr))) 1L)
|> Seq.max
problem8()
|> printfn "%d"
If you are using 0 .. 13 .. 987 in the sequence expression, then you are partitioning the array like this (for simplicity, using 3-sized blocks in 10 digits):
[012][345][678]9
I suppose that the question wants you to look for all possible sub-strings, i.e.
[012][345][678]9
0[123][456][789]
01[234][567]89
So, you probably need to try all indices using 0 .. 987.
By the way, I'd suspect that converting char to int64 will be faster using int64 c - 48L.
IE,
What am I doing wrong here? Does it have to to with lists, sequences and arrays and the way the limitations work?
So here is the setup: I'm trying to generate some primes. I see that there are a billion text files of a billion primes. The question isn't why...the question is how are the guys using python calculating all of the primes below 1,000,000 in milliseconds on this post...and what am I doing wrong with the following F# code?
let sieve_primes2 top_number =
let numbers = [ for i in 2 .. top_number do yield i ]
let sieve (n:int list) =
match n with
| [x] -> x,[]
| hd :: tl -> hd, List.choose(fun x -> if x%hd = 0 then None else Some(x)) tl
| _ -> failwith "Pernicious list error."
let rec sieve_prime (p:int list) (n:int list) =
match (sieve n) with
| i,[] -> i::p
| i,n' -> sieve_prime (i::p) n'
sieve_prime [1;0] numbers
With the timer on in FSI, I get 4.33 seconds worth of CPU for 100000... after that, it all just blows up.
Your sieve function is slow because you tried to filter out composite numbers up to top_number. With Sieve of Eratosthenes, you only need to do so until sqrt(top_number) and remaining numbers are inherently prime. Suppose we havetop_number = 1,000,000, your function does 78498 rounds of filtering (the number of primes until 1,000,000) while the original sieve only does so 168 times (the number of primes until 1,000).
You can avoid generating even numbers except 2 which cannot be prime from the beginning. Moreover, sieve and sieve_prime can be merged into a recursive function. And you could use lightweight List.filter instead of List.choose.
Incorporating above suggestions:
let sieve_primes top_number =
let numbers = [ yield 2
for i in 3..2..top_number -> i ]
let rec sieve ns =
match ns with
| [] -> []
| x::xs when x*x > top_number -> ns
| x::xs -> x::sieve (List.filter(fun y -> y%x <> 0) xs)
sieve numbers
In my machine, the updated version is very fast and it completes within 0.6s for top_number = 1,000,000.
Based on my code here: stackoverflow.com/a/8371684/124259
Gets the first 1 million primes in 22 milliseconds in fsi - a significant part is probably compiling the code at this point.
#time "on"
let limit = 1000000
//returns an array of all the primes up to limit
let table =
let table = Array.create limit true //use bools in the table to save on memory
let tlimit = int (sqrt (float limit)) //max test no for table, ints should be fine
let mutable curfactor = 1;
while curfactor < tlimit-2 do
curfactor <- curfactor+2
if table.[curfactor] then //simple optimisation
let mutable v = curfactor*2
while v < limit do
table.[v] <- false
v <- v + curfactor
let out = Array.create (100000) 0 //this needs to be greater than pi(limit)
let mutable idx = 1
out.[0]<-2
let mutable curx=1
while curx < limit-2 do
curx <- curx + 2
if table.[curx] then
out.[idx]<-curx
idx <- idx+1
out
There have been several good answers both as to general trial division algorithm using lists (#pad) and in choice of an array for a sieving data structure using the Sieve of Eratosthenes (SoE) (#John Palmer and #Jon Harrop). However, #pad's list algorithm isn't particularly fast and will "blow up" for larger sieving ranges and #John Palmer's array solution is somewhat more complex, uses more memory than necessary, and uses external mutable state so is not different than if the program were written in an imperative language such as C#.
EDIT_ADD: I've edited the below code (old code with line comments) modifying the sequence expression to avoid some function calls so as to reflect more of an "iterator style" and while it saved 20% of the speed it still doesn't come close to that of a true C# iterator which is about the same speed as the "roll your own enumerator" final F# code. I've modified the timing information below accordingly. END_EDIT
The following true SoE program only uses 64 KBytes of memory to sieve primes up to a million (due to only considering odd numbers and using the packed bit BitArray) and still is almost as fast as #John Palmer's program at about 40 milliseconds to sieve to one million on a i7 2700K (3.5 GHz), with only a few lines of code:
open System.Collections
let primesSoE top_number=
let BFLMT = int((top_number-3u)/2u) in let buf = BitArray(BFLMT+1,true)
let SQRTLMT = (int(sqrt (double top_number))-3)/2
let rec cullp i p = if i <= BFLMT then (buf.[i] <- false; cullp (i+p) p)
for i = 0 to SQRTLMT do if buf.[i] then let p = i+i+3 in cullp (p*(i+1)+i) p
seq { for i = -1 to BFLMT do if i<0 then yield 2u
elif buf.[i] then yield uint32(3+i+i) }
// seq { yield 2u; yield! seq { 0..BFLMT } |> Seq.filter (fun i->buf.[i])
// |> Seq.map (fun i->uint32 (i+i+3)) }
primesSOE 1000000u |> Seq.length;;
Almost all of the elapsed time is spent in the last two lines enumerating the found primes due to the inefficiency of the sequence run time library as well as the cost of enumerating itself at about 28 clock cycles per function call and return with about 16 function calls per iteration. This could be reduced to only a few function calls per iteration by rolling our own iterator, but the code is not as concise; note that in the following code there is no mutable state exposed other than the contents of the sieving array and the reference variable necessary for the iterator implementation using object expressions:
open System
open System.Collections
open System.Collections.Generic
let primesSoE top_number=
let BFLMT = int((top_number-3u)/2u) in let buf = BitArray(BFLMT+1,true)
let SQRTLMT = (int(sqrt (double top_number))-3)/2
let rec cullp i p = if i <= BFLMT then (buf.[i] <- false; cullp (i+p) p)
for i = 0 to SQRTLMT do if buf.[i] then let p = i+i+3 in cullp (p*(i+1)+i) p
let nmrtr() =
let i = ref -2
let rec nxti() = i:=!i+1;if !i<=BFLMT && not buf.[!i] then nxti() else !i<=BFLMT
let inline curr() = if !i<0 then (if !i= -1 then 2u else failwith "Enumeration not started!!!")
else let v = uint32 !i in v+v+3u
{ new IEnumerator<_> with
member this.Current = curr()
interface IEnumerator with
member this.Current = box (curr())
member this.MoveNext() = if !i< -1 then i:=!i+1;true else nxti()
member this.Reset() = failwith "IEnumerator.Reset() not implemented!!!"
interface IDisposable with
member this.Dispose() = () }
{ new IEnumerable<_> with
member this.GetEnumerator() = nmrtr()
interface IEnumerable with
member this.GetEnumerator() = nmrtr() :> IEnumerator }
primesSOE 1000000u |> Seq.length;;
The above code takes about 8.5 milliseconds to sieve the primes to a million on the same machine due to greatly reducing the number of function calls per iteration to about three from about 16. This is about the same speed as C# code written in the same style. It's too bad that F#'s iterator style as I used in the first example doesn't automatically generate the IEnumerable boiler plate code as C# iterators do, but I guess that is the intention of sequences - just that they are so damned inefficient as to speed performance due to being implemented as sequence computation expressions.
Now, less than half of the time is expended in enumerating the prime results for a much better use of CPU time.
What am I doing wrong here?
You've implemented a different algorithm that goes through each possible value and uses % to determine if it needs to be removed. What you're supposed to be doing is stepping through with a fixed increment removing multiples. That would be asymptotically.
You cannot step through lists efficiently because they don't support random access so use arrays.
I have this code written for a Project Euler problem in c++:
int sum = 0;
for(int i =0; i < 1000; i++)
{
//Check if multiple of 3 but not multiple of 5 to prevent duplicate
sum += i % 3 == 0 && i % 5 != 0 ? i: 0;
//check for all multiple of 5, including those of 3
sum += i % 5 == 0 ? i: 0;
}
cout << sum;
I'm trying to learn f# and rewriting this in f#. This is what I have so far:
open System
//function to calculate the multiples
let multiple3v5 num =
num
//function to calculate sum of list items
let rec SumList xs =
match xs with
| [] -> 0
| y::ys -> y + SumList ys
let sum = Array.map multiple3v5 [|1 .. 1000|]
What I have may be complete nonsense...so help please?
Your sumList function is a good start. It already iterates (recursively) over the entire list, so you don't need to wrap it in an additional Array.map. You just need to extend your sumList so that it adds the number only when it matches the specified condition.
Here is a solution to a simplified problem - add all numbers that are divisible by 3:
open System
let rec sumList xs =
match xs with
| [] -> 0 // If the list is empty, the sum is zero
| y::ys when y % 3 = 0 ->
// If the list starts with y that is divisible by 3, then we add 'y' to the
// sum that we get by recursively processing the rest of the list
y + sumList ys
| y::ys ->
// This will only execute when y is not divisible by 3, so we just
// recursively process the rest of the list and return
/// that (without adding current value)
sumList ys
// For testing, let's sum all numbers divisble by 3 between 1 and 10.
let sum = sumList [ 1 .. 10 ]
This is the basic way of writing the function using explicit recursion. In practice, the solution by jpalmer is how I'd solve it too, but it is useful to write a few recursive functions yourself if you're learning F#.
The accumulator parameter mentioned by sashang is a more advanced way to write this. You'll need to do that if you want to run the function on large inputs (which is likely the case in Euler problem). When using accumulator parameter, the function can be written using tail recursion, so it avoids stack overflow even when processing long lists.
The idea of a accumulator-based version is that the function takes additional parameter, which represents the sum calculated so far.
let rec sumList xs sumSoFar = ...
When you call it initially, you write sumList [ ... ] 0. The recursive calls will not call y + sumList xs, but will instead add y to the accumulator and then make the recursive call sumList xs (y + sumSoFar). This way, the F# compiler can do tail-call optimization and it will translate code to a loop (similar to the C++ version).
I'm not sure if translating from an imperative language solution is a good approach to developing a functional mindset as instrument (C++ in your case) had already defined an (imperative) approach to solution, so it's better sticking to original problem outlay.
Overall tasks from Project Euler are excellent for mastering many F# facilities. For example, you may use list comprehensions like in the snippet below
// multipleOf3Or5 function definition is left for your exercise
let sumOfMultiples n =
[ for x in 1 .. n do if multipleOf3Or5 x then yield x] |> List.sum
sumOfMultiples 999
or you can a bit generalize the solution suggested by #jpalmer by exploiting laziness:
Seq.initInfinite id
|> Seq.filter multipleOf3Or5
|> Seq.takeWhile ((>) 1000)
|> Seq.sum
or you may even use this opportunity to master active patterns:
let (|DivisibleBy|_) divisior num = if num % divisor = 0 the Some(num) else None
{1..999}
|> Seq.map (fun i ->
match i with | DivisibleBy 3 i -> i | DivisibleBy 5 i -> i | _ -> 0)
|> Seq.sum
All three variations above implement a common pattern of making a sequence of members with sought property and then folding it by calculating sum.
F# has many more functions than just map - this problem suggests using filter and sum, my approach would be something like
let valid n = Left as an exercise
let r =
[1..1000]
|> List.filter valid
|> List.sum
printfn "%i" r
I didn't want to do the whole problem, but filling in the missing function shouldn't be too hard
This is how you turn a loop with a counter into a recursive function. You do this by passing an accumulator parameter to the loop function that holds the current loop count.
For example:
let rec loop acc =
if acc = 10 then
printfn "endloop"
else
printfn "%d" acc
loop (acc + 1)
loop 0
This will stop when acc is 10.
I am trying to learn F# so I paid a visit to Project Euler and I am currently working on Problem 3.
The prime factors of 13195 are 5, 7,
13 and 29.
What is the largest prime
factor of the number 600851475143?
Some things to consider:
My first priority is to learn good functional habits.
My second priority is I would like it to be fast and efficient.
Within the following code I have marked the section this question is regarding.
let isPrime(n:int64) =
let rec check(i:int64) =
i > n / 2L or (n % i <> 0L && check(i + 1L))
check(2L)
let greatestPrimeFactor(n:int64) =
let nextPrime(prime:int64):int64 =
seq { for i = prime + 1L to System.Int64.MaxValue do if isPrime(i) then yield i }
|> Seq.skipWhile(fun v -> n % v <> 0L)
|> Seq.hd
let rec findNextPrimeFactor(number:int64, prime:int64):int64 =
if number = 1L then prime else
//************* No variable
(fun p -> findNextPrimeFactor(number / p, p))(nextPrime(prime))
//*************
//************* Variable
let p = nextPrime(prime)
findNextPrimeFactor(number / p, p)
//*************
findNextPrimeFactor(n, 2L)
Update
Based off some of the feedback I have refactored the code to be 10 times faster.
module Problem3
module private Internal =
let execute(number:int64):int64 =
let rec isPrime(value:int64, current:int64) =
current > value / 2L or (value % current <> 0L && isPrime(value, current + 1L))
let rec nextPrime(prime:int64):int64 =
if number % prime = 0L && isPrime(prime, 2L) then prime else nextPrime(prime + 1L)
let rec greatestPrimeFactor(current:int64, prime:int64):int64 =
if current = 1L then prime else nextPrime(prime + 1L) |> fun p -> greatestPrimeFactor(current / p, p)
greatestPrimeFactor(number, 2L)
let execute() = Internal.execute(600851475143L)
Update
I would like to thank everyone for there advice. This latest version is a compilation of all the advice I received.
module Problem3
module private Internal =
let largestPrimeFactor number =
let rec isPrime value current =
current > value / 2L || (value % current <> 0L && isPrime value (current + 1L))
let rec nextPrime value =
if number % value = 0L && isPrime value 2L then value else nextPrime (value + 1L)
let rec find current prime =
match current / prime with
| 1L -> prime
| current -> nextPrime (prime + 1L) |> find current
find number (nextPrime 2L)
let execute() = Internal.largestPrimeFactor 600851475143L
Functional programming becomes easier and more automatic with practice, so don't sweat it if you don't get it absolutely right on the first try.
With that in mind, let's take your sample code:
let rec findNextPrimeFactor(number:int64, prime:int64):int64 =
if number = 1L then prime else
//************* No variable
(fun p -> findNextPrimeFactor(number / p, p))(nextPrime(prime))
//*************
//************* Variable
let p = nextPrime(prime)
findNextPrimeFactor(number / p, p)
//*************
Your no variable version is just weird, don't use it. I like your version with the explicit let binding.
Another way to write it would be:
nextPrime(prime) |> fun p -> findNextPrimeFactor(number / p, p)
Its ok and occasionally useful to write it like this, but still comes across as a little weird. Most of the time, we use |> to curry values without needing to name our variables (in "pointfree" style). Try to anticipate how your function will be used, and if possible, re-write it so you can use it with the pipe operator without explicit declared variables. For example:
let rec findNextPrimeFactor number prime =
match number / prime with
| 1L -> prime
| number' -> nextPrime(prime) |> findNextPrimeFactor number'
No more named args :)
Ok, now that we have that out of the way, let's look at your isPrime function:
let isPrime(n:int64) =
let rec check(i:int64) =
i > n / 2L or (n % i <> 0L && check(i + 1L))
check(2L)
You've probably heard to use recursion instead of loops, and that much is right. But, wherever possible, you should abstract away recursion with folds, maps, or higher order functions. Two reasons for this:
its a little more readable, and
improperly written recursion will result in a stack overflow. For example, your function is not tail recursive, so it'll blow up on large values of n.
I'd rewrite isPrime like this:
let isPrime n = seq { 2L .. n / 2L } |> Seq.exists (fun i -> n % i = 0L) |> not
Most of the time, if you can abstract away your explicit looping, then you're just applying transformations to your input sequence until you get your results:
let maxFactor n =
seq { 2L .. n - 1L } // test inputs
|> Seq.filter isPrime // primes
|> Seq.filter (fun x -> n % x = 0L) // factors
|> Seq.max // result
We don't even have intermediate variables in this version. Coolness!
My second priority is I would like it
to be fast and efficient.
Most of the time, F# is going to be pretty comparable with C# in terms of speed, or it's going to be "fast enough". If you find your code takes a long time to execute, it probably means you're using the wrong data structure or a bad algorithm. For a concrete example, read the comments on this question.
So, the code I've written is "elegant" in the sense that its concise, gives the correct results, and doesn't rely on any trickery. Unfortunately, its not very fast. For start:
it uses trial division to create a sequence of primes, when the Sieve of Eratosthenes would be much faster. [Edit: I wrote a somewhat naive version of this sieve which didn't work for numbers larger than Int32.MaxValue, so I've removed the code.]
read Wikipedia's article on the prime counting function, it'll give you pointers on calculating the first n primes as well as estimating the upper and lower bounds for the nth prime.
[Edit: I included some code with a somewhat naive implementation of a sieve of eratosthenes. It only works for inputs less than int32.MaxValue, so it probably isn't suitable for project euler.]
Concerning "good functional habit" or rather good practice I see three minor things. Using the yield in your sequence is a little harder to read than just filter. Unnecessary type annotations in a type inferred language leads to difficult refactoring and makes the code harder to read. Don't go overboard and try to remove every type annotation though if you're finding it difficult. Lastly making a lambda function which only takes a value to use as a temp variable reduces readability.
As far as personal style goes I prefer more spaces and only using tupled arguments when the data makes sense being grouped together.
I'd write your original code like this.
let isPrime n =
let rec check i =
i > n / 2L || (n % i <> 0L && check (i + 1L))
check 2L
let greatestPrimeFactor n =
let nextPrime prime =
seq {prime + 1L .. System.Int64.MaxValue}
|> Seq.filter isPrime
|> Seq.skipWhile (fun v -> n % v <> 0L)
|> Seq.head
let rec findNextPrimeFactor number prime =
if number = 1L then
prime
else
let p = nextPrime(prime)
findNextPrimeFactor (number / p) p
findNextPrimeFactor n 2L
Your updated code is optimal for your approach. You would have to use a different algorithm like Yin Zhu answer to go faster. I wrote a test to check to see if F# makes the "check" function tail recursive and it does.
the variable p is actually a name binding, not a variable. Using name binding is not a bad style. And it is more readable. The lazy style of nextPrime is good, and it actually prime-test each number only once during the whole program.
My Solution
let problem3 =
let num = 600851475143L
let rec findMax (n:int64) (i:int64) =
if n=i || n<i then
n
elif n%i=0L then
findMax (n/i) i
else
findMax n (i+1L)
findMax num 2L
I basically divides num from 2, 3, 4.. and don't consider any prime numbers. Because if we divides all 2 from num, then we won't be able to divide it by 4,8, etc.
on this number, my solution is quicker:
> greatestPrimeFactor 600851475143L;;
Real: 00:00:01.110, CPU: 00:00:00.702, GC gen0: 1, gen1: 1, gen2: 0
val it : int64 = 6857L
>
Real: 00:00:00.001, CPU: 00:00:00.000, GC gen0: 0, gen1: 0, gen2: 0
val problem3 : int64 = 6857L
I think that the code with the temporary binding is significantly easier to read. It's pretty unusual to create an anonymous function and then immediately apply it to a value as you do in the other case. If you really want to avoid using a temporary value, I think that the most idiomatic way to do that in F# would be to use the (|>) operator to pipe the value into the anonymous function, but I still think that this isn't quite as readable.
I have this "learning code" I wrote for the morris seq in f# that suffers from stack overflow that I don't know how to avoid. "morris" returns an infinite sequence of "see and say" sequences (i.e., {{1}, {1,1}, {2,1}, {1,2,1,1}, {1,1,1,2,2,1}, {3,1,2,2,1,1},...}).
let printList l =
Seq.iter (fun n -> printf "%i" n) l
printfn ""
let rec morris s =
let next str = seq {
let cnt = ref 1 // Stack overflow is below when enumerating
for cur in [|0|] |> Seq.append str |> Seq.windowed 2 do
if cur.[0] <> cur.[1] then
yield!( [!cnt ; cur.[0]] )
cnt := 0
incr cnt
}
seq {
yield s
yield! morris (next s) // tail recursion, no stack overflow
}
// "main"
// Print the nth iteration
let _ = [1] |> morris |> Seq.nth 3125 |> printList
You can pick off the nth iteration using Seq.nth but you can only get so far before you hit a stack overflow. The one bit of recursion I have is tail recursion and it in essence builds a linked set of enumerators. That's not where the problem is. It's when "enum" is called on the say the 4000th sequence. Note that's with F# 1.9.6.16, the previous version topped out above 14000). It's because the way the linked sequences are resolved. The sequences are lazy and so the "recursion" is lazy. That is, seq n calls seq n-1 which calls seq n-2 and so forth to get the first item (the very first # is the worst case).
I understand that [|0|] |> Seq.append str |> Seq.windowed 2, is making my problem worse and I could triple the # I could generate if I eliminated that. Practically speaking the code works well enough. The 3125th iteration of morris would be over 10^359 characters in length.
The problem I'm really trying to solve is how to retain the lazy eval and have a no limit based on stack size for the iteration I can pick off. I'm looking for the proper F# idiom to make the limit based on memory size.
Update Oct '10
After learning F# a bit better, a tiny bit of Haskell, thinking & investigating this problem for over year, I finally can answer my own question. But as always with difficult problems, the problem starts with it being the wrong question. The problem isn't sequences of sequences - it's really because of a recursively defined sequence. My functional programming skills are a little better now and so it's easier to see what's going on with the version below, which still gets a stackoverflow
let next str =
Seq.append str [0]
|> Seq.pairwise
|> Seq.scan (fun (n,_) (c,v) ->
if (c = v) then (n+1,Seq.empty)
else (1,Seq.ofList [n;c]) ) (1,Seq.empty)
|> Seq.collect snd
let morris = Seq.unfold(fun sq -> Some(sq,next sq))
That basicially creates a really long chain of Seq processing function calls to generate the sequnces. The Seq module that comes with F# is what can't follow the chain without using the stack. There's an optimization it uses for append and recursively defined sequences, but that optimization only works if the recursion is implementing an append.
So this will work
let rec ints n = seq { yield n; yield! ints (n+1) }
printf "%A" (ints 0 |> Seq.nth 100000);;
And this one will get a stackoverflow.
let rec ints n = seq { yield n; yield! (ints (n+1)|> Seq.map id) }
printf "%A" (ints 0 |> Seq.nth 100000);;
To prove the F# libary was the issue, I wrote my own Seq module that implemented append, pairwise, scan and collect using continutions and now I can begin generating and printing out the 50,000 seq without a problem (it'll never finish since it's over 10^5697 digits long).
Some additional notes:
Continuations were the idiom I was looking for, but in this case, they had to go into the F# library, not my code. I learned about continuations in F# from Tomas Petricek's Real-World Functional Programming book.
The lazy list answer that I accepted held the other idiom; lazy evaluation. In my rewritten library, I also had to leverage the lazy type to avoid stackoverflow.
The lazy list version sorta of works by luck (maybe by design but that's beyond my current ability to determine) - the active-pattern matching it uses while it's constructing and iterating causes the lists to calculate values before the required recursion gets too deep, so it's lazy, but not so lazy it needs continuations to avoid stackoverflow. For example, by the time the 2nd sequence needs a digit from the 1st sequence, it's already been calculated. In other words, the LL version is not strictly JIT lazy for sequence generation, only list management.
You should definitely check out
http://research.microsoft.com/en-us/um/cambridge/projects/fsharp/manual/FSharp.PowerPack/Microsoft.FSharp.Collections.LazyList.html
but I will try to post a more comprehensive answer later.
UPDATE
Ok, a solution is below. It represents the Morris sequence as a LazyList of LazyLists of int, since I presume you want it to be lazy in 'both directions'.
The F# LazyList (in the FSharp.PowerPack.dll) has three useful properties:
it is lazy (evaluation of the nth element will not happen until it is first demanded)
it does not recompute (re-evaluation of the nth element on the same object instance will not recompute it - it caches each element after it's first computed)
you can 'forget' prefixes (as you 'tail' into the list, the no-longer-referenced prefix is available for garbage collection)
The first property is common with seq (IEnumerable), but the other two are unique to LazyList and very useful for computational problems such as the one posed in this question.
Without further ado, the code:
// print a lazy list up to some max depth
let rec PrintList n ll =
match n with
| 0 -> printfn ""
| _ -> match ll with
| LazyList.Nil -> printfn ""
| LazyList.Cons(x,xs) ->
printf "%d" x
PrintList (n-1) xs
// NextMorris : LazyList<int> -> LazyList<int>
let rec NextMorris (LazyList.Cons(cur,rest)) =
let count = ref 1
let ll = ref rest
while LazyList.nonempty !ll && (LazyList.hd !ll) = cur do
ll := LazyList.tl !ll
incr count
LazyList.cons !count
(LazyList.consf cur (fun() ->
if LazyList.nonempty !ll then
NextMorris !ll
else
LazyList.empty()))
// Morris : LazyList<int> -> LazyList<LazyList<int>>
let Morris s =
let rec MakeMorris ll =
LazyList.consf ll (fun () ->
let next = NextMorris ll
MakeMorris next
)
MakeMorris s
// "main"
// Print the nth iteration, up to a certain depth
[1] |> LazyList.of_list |> Morris |> Seq.nth 3125 |> PrintList 10
[1] |> LazyList.of_list |> Morris |> Seq.nth 3126 |> PrintList 10
[1] |> LazyList.of_list |> Morris |> Seq.nth 100000 |> PrintList 35
[1] |> LazyList.of_list |> Morris |> Seq.nth 100001 |> PrintList 35
UPDATE2
If you just want to count, that's fine too:
let LLLength ll =
let rec Loop ll acc =
match ll with
| LazyList.Cons(_,rest) -> Loop rest (acc+1N)
| _ -> acc
Loop ll 0N
let Main() =
// don't do line below, it leaks
//let hundredth = [1] |> LazyList.of_list |> Morris |> Seq.nth 100
// if we only want to count length, make sure we throw away the only
// copy as we traverse it to count
[1] |> LazyList.of_list |> Morris |> Seq.nth 100
|> LLLength |> printfn "%A"
Main()
The memory usage stays flat (under 16M on my box)... hasn't finished running yet, but I computed the 55th length fast, even on my slow box, so I think this should work just fine. Note also that I used 'bignum's for the length, since I think this will overflow an 'int'.
I believe there are two main problems here:
Laziness is very inefficient so you can expect a lazy functional implementation to run orders of magnitude slower. For example, the Haskell implementation described here is 2,400× slower than the F# I give below. If you want a workaround, your best bet is probably to amortize the computations by bunching them together into eager batches where the batches are produced on-demand.
The Seq.append function is actually calling into C# code from IEnumerable and, consequently, its tail call doesn't get eliminated and you leak a bit more stack space every time you go through it. This shows up when you come to enumerate over the sequence.
The following is over 80× faster than your implementation at computing the length of the 50th subsequence but perhaps it is not lazy enough for you:
let next (xs: ResizeArray<_>) =
let ys = ResizeArray()
let add n x =
if n > 0 then
ys.Add n
ys.Add x
let mutable n = 0
let mutable x = 0
for i=0 to xs.Count-1 do
let x' = xs.[i]
if x=x' then
n <- n + 1
else
add n x
n <- 1
x <- x'
add n x
ys
let morris =
Seq.unfold (fun xs -> Some(xs, next xs)) (ResizeArray [1])
The core of this function is a fold over a ResizeArray that could be factored out and used functionally without too much performance degradation if you used a struct as the accumulator.
Just save the previous element that you looked for.
let morris2 data = seq {
let cnt = ref 0
let prev = ref (data |> Seq.nth 0)
for cur in data do
if cur <> !prev then
yield! [!cnt; !prev]
cnt := 1
prev := cur
else
cnt := !cnt + 1
yield! [!cnt; !prev]
}
let rec morrisSeq2 cur = seq {
yield cur
yield! morrisSeq2 (morris2 cur)
}