I have a function that calculates f(n) in Haskell.
I have to write a loop so that it will start calculating values from f(0) to f(n), and will every time compare the value of f(i) with some fixed value.
I am an expert in OOP, hence I am finding it difficult to think in the functional way.
For example, I have to write something like
while (number < f(i))
i++
How would I write this in Haskell?
The standard approach here is
Create an infinite list containing all values of f(n).
Search this list until you find what you're after.
For example,
takeWhile (number <) $ map f [0..]
If you want to give up after you reach "n", you can easily add that as a separate step:
takeWhile (number <) $ take n $ map f [0..]
or, alternatively,
takeWhile (number <) $ map f [0 .. n]
You can do all sorts of other filtering, grouping and processing in this way. But it requires a mental shift. It's a bit like the difference between writing a for-loop to search a table, versus writing an SQL query. Think about Haskell as a bit like SQL, and you'll usually see how to structure your code.
You can generate the list of the is such that f i is larger than your number:
[ i | i<-[0..] , f i > number ]
Then, you can simply take the first one, if that's all you want:
head [ i | i<-[0..] , f i > number ]
Often, many idiomatic loops in imperative programming can be rephrased as list comprehensions, or expressed through map, filter, foldl, foldr. In the general case, when the loop is more complex, you can always exploit recursion instead.
Keep in mind that a "blind" translation from imperative to functional programming will often lead to non-idiomatic, hard-to-read code, as it would be the case when translating in the opposite direction. Still, I find it relieving that such translation is always possible.
If you are new to functional programming, I would advise against learning it by translating what you know about imperative programming. Rather, start from scratch following a good book (LYAH is a popular choice).
The first thing that's weird from a functional approach is that it's unclear what the result of your computation is. Do you care about the final result of f (i)? Perhaps you care about i itself. Without side effects everything neends to have a value.
Let's assume you want the final value of the function f (i) as soon as some comparison fails. You can simulate your own while loops using recursion and guards!
while :: Int -> Int -> (Int -> Int) -> Int
while start number f
| val >= number = val
| otherwise = while (start + 1) number f
where
val = f start
Instead of explicit recursion, you can use until e.g.
findGreaterThan :: (Int -> Int) -> Int -> Int -> (Int, Int)
findGreaterThan f init max = until (\(v, i) -> v >= max) (\(v, i) -> (f v, i + 1)) (init, 0)
this returns a pair containing the first value to fail the condition and the number of iterations of the given function.
Related
I have tried several things. Can I simply not match with less than?
The exception is redundant, to enable the matching loop. I can imagine all sorts of things are wrong and this bit of code. It would be nice if the vague less-than-you-less-than-me concept worked.
I admit I am frustrated. Why do I get an error at j < k every time, no matter where I improve or edit? Is that simply against the rules?
I was just looking at Array Length > Array Length as my while condition, that will have to change. Also, none of the Array Sorting methods seem to me to work in the smallest upwards routine as I understand the 0(n^2).
Can this code implement Array.for_all as a sorting algorithm?
UPDATE
let i = Array.length arr in
let sort (arr : array) (j : int) (k : int) =
Array.make n x = unsorted(arr : array) (n : int) (x : int)
while i > 0
match j < k with
| true -> a.(n) <- j
| false -> failwith (Printf.sprintf "Exception Found (%d)" i)
i - 1
;;
That was a lot of questions.
Thank you.
UPDATE:
I appreciate the help very much.
I know I need two arrays, not one. The finished code should read an array, and then give an array back. So, when k points to a smallest j the purpose of the algorithm is to kick j out of the first array, from wherever it it is, put it into the second array, and append every j after that one into the second array. The job is done when the first array is empty.
I see plenty of information about lists out there but not much about arrays. How can I remove the j_n-th element of an array?
This is a nice looking example of recursion used to search and delete an element from a list, can that work in an array?
Thank you again
The match in OCaml is restricted (in essence) to matching a value (an expression) against a set of structured constants that are known at compile time. If you want to test something like whether j < k you should just use an if expression, even if it's not as exotic and powerful seeming.
There are many other problems with your code, but this seems to answer your main question.
You can use an expression in a when clause to further constrain when a pattern matches. Simple example:
let f x =
match x with
| _ when x < 5 -> true
| _ -> false
In this trivial case, an if is of course better, but it can be useful in more complicated pattern matching with lots of options.
You'll have a lot of problems with your code but to answer your question, there are 3 ways I can see to do what you want to do:
match j < k with
| true -> a.(n) <- j
| false -> failwith (Printf.sprintf "Exception Found (%d)" i)
match j with
| _ when j < k -> a.(n) <- j
| _ -> failwith (Printf.sprintf "Exception Found (%d)" i)
if j < k
then a.(n) <- j
else failwith (Printf.sprintf "Exception Found (%d)" i)
The latter being the better. The only thing you need to remember is that pattern matching is done on patterns, it looks at what the values are, not their relations to other values. You can match an int with 1, 3 or 127 but not match this int with another one or compare it with another one.
On a side note, here are some problems in your code and some answers to your questions:
sort takes an array, a j, a k and a i but then
You use Array.make by checking if it's equal to unsorted (x = f is a boolean comparison unless you write let x = f in ...). It looks like you're trying to define an internal function but it's unclear how you're making it
if you want to define a randomly sorted array, you could just write Array.init 100 (fun _ -> Random.int 100)
you're clearly trying to sort your array between i and n (while a.(i) > a.(n) do ... done but then you're checking on j that is a parameter of your function and that never changes
Array.for_all checks that a predicate is true for every value in the array. If you want to apply a function with a different return type to your array you can either use Array.map:
map f a applies function f to all the elements of a, and builds an array with the results returned by f: [| f a.(0); f a.(1); ...; f a.(length a - 1) |].
or Array.fold_{left|right}:
fold_left f init a computes f (... (f (f init a.(0)) a.(1)) ...) a.(n-1), where n is the length of the array a.
fold_right f a init computes f a.(0) (f a.(1) ( ... (f a.(n-1) init) ...)), where n is the length of the array a.
As for the sorting functions, what makes you think they won't run in O(n^2)? You can look at the way they're implemented here
I'am learning OCaml and currently i'am trying to undertand how iteration works in OCaml and how to create a matrix. I want an array 5 x 5 filled with 0. I know there is an issue with shared references so i created a new array at each iteration however iam having issues in other places, specifically at line 6. Let me know of other issues like indentation practices.
open Array;;
let n = ref 5 and i = ref 0 in
let m = Array.make !n 0 in
while !i < !n do
m.(!i) <- Array.make !n 0;;
i := !i + 1;;
done
m;;
You are using ;; too much. Contrary to popular belief, ;; is not part of ordinary OCaml syntax (in my opinion anyway). It's just a special way to tell the toplevel (the REPL) that you want it to evaluate what you've typed so far.
Leave the ;; after open Array. But change all but the last ;; to ; instead.
(Since you reference the Array module by name in your code, which IMHO is good style, you can also just leave out the open Array;; altogether.)
You want the last ;; because you do want the toplevel to evaluate what you've typed so far.
Your syntax error is caused by the fact that your overall code is like this
let ... in
let ... in
while ... do
...
done
m
The while is one expression (in OCaml everything is an expression) and m is another expression. If you want to have two expressions in a row you need ; between them. So you need ; after done.
You also have a type error. When you create m you're creating an array of ints (your given initial value is 0). So you can't make it into a matrix (an array of arrays) later in the code.
Also (not trying to overload you with criticisms :-) this code reads like imperative code. It's not particularly idiomatic OCaml code. In most people's code, using ref is pretty rare. One immediate improvement I see would just be to say let n = 5. You're not changing the value of n anywhere that I see (though maybe this is part of a larger chunk of code). Another improvement would be to use for instead of while.
Finally, you can do this entire operation in one function call:
let n = 5 in
let m = Array.init n (fun i -> Array.make n 0) in
m
Using explicit loops is actually also quite rare in OCaml (at least in my code).
Or you could try this:
let n = 5 in
let m = Array.make_matrix n n 0 in
m
I am a SAS programmer learning R.
I have a matrix receivables read from a csv file.
I wish to read the value in the "transit" column, if the value of "id" column of a row is >= 1100000000.
I did this (loop 1):
x = vector()
for (i in 1:length(receivables[,"transit"])){
if(receivables[i,"id"] >= 1100000000){
append(x, receivables[i,"transit"]);
}
}
But, it does not work because after running the loop x is still empty.
>x
logical(0)
However, I was able to accomplish my task with (loop 2):
k=0
x=vector()
for (i in 1:length(receivables[,"transit"])){
if(receivables[i,"id"] >= 1100000000){
k=k+1
x[k] = receivables[i,"transit"]
}
}
Or, with (loop 3):
x = vector()
for (i in 1:length(receivables[,"transit"])){
if(receivables[i,"id"] >= 1100000000){
x <- append(x, receivables[i,"transit"]);
}
}
Why didn't the append function work in the loop as it would in command line?
Actually, to teach me how to fish, what is the attitude/beatitude one must bear in mind when operating functions in a loop as opposed to operating them in command line.
Which is more efficient? Loop 2 or loop 3?
Ok, a few things.
append didn't work in your first attempt because you did not assign the result to anything. In general, R does not work "in place". It is more of a functional language, which means that changes must always be assigned to something. (There are exceptions, but trying to bend this rule too early will get you in trouble.)
A bigger point is that "growing" objects in R is a big no-no. You will quickly start to wonder why anyone could possible use R, because growing objects like this will quickly become very, very slow.
Instead, learn to use vectorized operations, like:
receivables[receivables[,"id"] >= 1100000000,"transit"]
am using Data.Vector and am currently in need of computing the contents of a vector for use in computing a cryptographic hash(Sha1). I created the following code.
dynamic :: a -> Int -> (Int -> Vector a -> a) -> Vector a
dynamic e n f =
let
start = Data.Vector.replicate n e
in step start 0
where
step vector i = if i==n then vector
else step (vector // [(i,f i vector)]) (i+1)
I created this so that the function f filling out the vector has access to the partial
results along the way. Surely something like this must already exist in Data.Vector, no?
The problem statement is the following: You are to solve a dynamic programming problem where the finished result is an array. You know the size of the array size and you have a recursive function for filling it out.
You probably already saw the function generate, which takes a size n and a function f of type Int -> a and then produces a Vector a of size n. What you probably weren't aware of is that when using this function you actually do have access to the partial results.
What I mean to say is that inside the function you pass to generate you can refer to the vector you're defining and due to Haskell's laziness it will work fine (unless you make it so that the different items of the vector depend on each other in a circular fashion, of course).
Example:
import Data.Vector
tenFibs = generate 10 fib
where fib 0 = 0
fib 1 = 1
fib n = tenFibs ! (n-1) + tenFibs ! (n-2)
tenFibs is now a vector containing the first 10 Fibonacci numbers.
Maybe you could use one of Data.Vector's scan functions?
http://hackage.haskell.org/packages/archive/vector/0.6.0.2/doc/html/Data-Vector.html#32
OK, continuing with my solving of the problems on Project Euler, I am still beginning to learn Haskell and programming in general.
I need to find the lowest number divisible by the numbers 1:20
So I started with:
divides :: Int -> Int -> Bool
divides d n = rem n d == 0
divise n a | divides n a == 0 = n : divise n (a+1)
| otherwise = n : divise (n+1) a
What I want to happen is for it to keep moving up for values of n until one magically is evenly divisible by [1..20].
But this doesn't work and now I am stuck as from where to go from here. I assume I need to use:
[1..20]
for the value of a but I don't know how to implement this.
Well, having recently solved the Euler problem myself, I'm tempted to just post my answer for that, but for now I'll abstain. :)
Right now, the flow of your program is a bit chaotic, to sound like a feng-shui person. Basically, you're trying to do one thing: increment n until 1..20 divides n. But really, you should view it as two steps.
Currently, your code is saying: "if a doesn't divide n, increment n. If a does divide n, increment a". But that's not what you want it to say.
You want (I think) to say "increment n, and see if it divides [Edit: with ALL numbers 1..20]. If not, increment n again, and test again, etc." What you want to do, then, is have a sub-test: one that takes a number, and tests it against 1..20, and then returns a result.
Hope this helps! Have fun with the Euler problems!
Edit: I really, really should remember all the words.
Well, as an algorithm, this kinda sucks.
Sorry.
But you're getting misled by the list. I think what you're trying to do is iterate through all the available numbers, until you find one that everything in [1..20] divides. In your implementation above, if a doesn't divide n, you never go back and check b < a for n+1.
Any easy implementation of your algorithm would be:
lcmAll :: [Int] -> Maybe Int
lcmAll nums = find (\n -> all (divides n) nums) [1..]
(using Data.List.find and Data.List.all).
A better algorithm would be to find the lcm's pairwise, using foldl:
lcmAll :: [Int] -> Int
lcmAll = foldl lcmPair 1
lcmPair :: Int -> Int -> Int
lcmPair a b = lcmPair' a b
where lcmPair' a' b' | a' < b' = lcmPair' (a+a') b'
| a' > b' = lcmPair' a' (b + b')
| otherwise = a'
Of course, you could use the lcm function from the Prelude instead of lcmPair.
This works because the least common multiple of any set of numbers is the same as the least common multiple of [the least common multiple of two of those numbers] and [the rest of the numbers]
The function 'divise' never stops, it doesn't have a base case. Both branches calls divise, thus they are both recursive. Your also using the function divides as if it would return an int (like rem does), but it returns a Bool.
I see you have already started to divide the problem into parts, this is usually good for understanding and making it easier to read.
Another thing that can help is to write the types of the functions. If your function works but your not sure of its type, try :i myFunction in ghci. Here I've fixed the type error in divides (although other errors remains):
*Main> :i divise
divise :: Int -> Int -> [Int] -- Defined at divise.hs:4:0-5
Did you want it to return a list?
Leaving you to solve the problem, try to further divide the problem into parts. Here's a naive way to do it:
A function that checks if one number is evenly divisible by another. This is your divides function.
A function that checks if a number is dividable by all numbers [1..20].
A function that tries iterates all numbers and tries them on the function in #2.
Here's my quick, more Haskell-y approach, using your algorithm:
Prelude> let divisibleByUpTo i n = all (\x -> (i `rem` x) == 0) [1..n]
Prelude> take 1 $ filter (\x -> snd x == True) $ map (\x -> (x, divisibleByUpTo x 4)) [1..]
[(12,True)]
divisibleByUpTo returns a boolean if the number i is divisible by every integer up to and including n, similar to your divides function.
The next line probably looks pretty difficult to a Haskell newcomer, so I'll explain it bit-by-bit:
Starting from the right, we have map (\x -> (x, divisibleByUpTo x 4)) [1..] which says for every number x from 1 upwards, do divisibleByUpTo x 4 and return it in a tuple of (x, divisibleByUpTo x 4). I'm using a tuple so we know which number exactly divides.
Left of that, we have filter (\x -> snd x == True); meaning only return elements where the second item of the tuple is True.
And at the leftmost of the statement, we take 1 because otherwise we'd have an infinite list of results.
This will take quite a long time for a value of 20. Like others said, you need a better algorithm -- consider how for a value of 4, even though our "input" numbers were 1-2-3-4, ultimately the answer was only the product of 3*4. Think about why 1 and 2 were "dropped" from the equation.