i am a beginner in ocaml and I am stuck in my project.
I would like to count the number of elements of a list contained in a list.
Then test if the list contains odd or even lists.
let listoflists = [[1;2] ; [3;4;5;6] ; [7;8;9]]
output
l1 = even
l2 = even
l3 = odd
The problem is that :
List.tl listoflists
Gives the length of the rest of the list
so 2
-> how can I calculate the length of the lists one by one ?
-> Or how could I get the lists and put them one by one in a variable ?
for the odd/even function, I have already done it !
Tell me if I'm not clear
and thank you for your help .
Unfortunately it's not really possible to help you very much because your question is unclear. Since this is obviously a homework problem I'll just make a few comments.
Since you talk about putting values in variables you seem to have some programming experience. But you should know that OCaml code tends to work with immutable variables and values, which means you have to look at things differently. You can have variables, but they will usually be represented as function parameters (which indeed take different values at different times).
If you have no experience at all with OCaml it is probably worth working through a tutorial. The OCaml.org website recommends the first 6 chapters of the OCaml manual here. In the long run this will probably get you up to speed faster than asking questions here.
You ask how to do a calculation on each list in a list of lists. But you don't say what the answer is supposed to look like. If you want separate answers, one for each sublist, the function to use is List.map. If instead you want one cumulative answer calculated from all the sublists, you want a fold function (like List.fold_left).
You say that List.tl calculates the length of a list, or at least that's what you seem to be saying. But of course that's not the case, List.tl returns all but the first element of a list. The length of a list is calculated by List.length.
If you give a clearer definition of your problem and particularly the desired output you will get better help here.
Use List.iter f xs to apply function f to each element of the list xs.
Use List.length to compute the length of each list.
Even numbers are integrally divisible by two, so if you divide an even number by two the remainder will be zero. Use the mod operator to get the remainder of the division. Alternatively, you can rely on the fact that in the binary representation the odd numbers always end with 1 so you can use land (logical and) to test the least significant bit.
If you need to refer to the position of the list element, use List.iteri f xs. The List.iteri function will apply f to two arguments, the first will be the position of the element (starting from 0) and the second will be the element itself.
Related
Say I have a unique list of length 9 of the values between 1 and 9 inclusive in a random order (think sudoku), and I want to extract a the sub-list of the items that occur between the values 1 and 9 (exclusive). IE: between1and9([1,3,5,4,2,9,7,8,6],[3,5,4,2]) should be true.
At the moment I'm trying to use flatten/2, but not having much luck. Here's my current tactic (assuming I enforce List ins 1..9, maplist(all_distinct, List), length(List, 9) elsewhere to keep it tidy here/seperation of concerns):
between1and9(List,Between) :-
flatten([_,[1],Between,[9],_], List);
flatten([_,[9],Between,[1],_], List).
This version fails though when 1 or 9 are at the first or last position in List, or if they're adjacent within List. between1and9([_,1,9,_,_,_,_,_,_],[]) is true, but between1and9([_,1,9,_,_,_,_,_,_],_) is false (and fails when I try to use it as a constraint to solve a bigger problem.)
It seems to be the same problem casuing both failures, flatten doesn't seem to like treating unknowns as empty lists unless they're made explicit somewhere.
I can see why that would potentially be, if flatten could "invent" empty lists in the first argument it would mean an infinite set of solutions for anything in the first argument. Although my full program has other constraints to prevent this, I can understand why flatten might not want to accomodate it.
I can account for the edge cases (pun intended) by matching every permutation with disjunctions (ie: flatten([_,1,B,9,_],L);flatten([_,9,B,1,_],L);flatten([_,1,B,9]);flatten..., And account for the Between as an empty list with: \*above permutations on flatten*\; ( Between = [], (\*permutations for either edge and 1/9*\) )
But that seems to be making an already longwinded solution (10 permutations of flatten in total) even worse (18) so I have two (strongly related) questions:
If I could do the following:
between1and9(L,B) :-
( ( X = 1, Y = 9 ); ( X = 9, Y = 1 ) ),
( ( Z1 = _; Z1 = [] ), ( Z2 = _ ; Z2 = [] ) ),
( B = _; B = [] ),
flatten([Z1,X,B,Y,Z2],L).
I wouldn't have to manually type out each permutation of match for flatten. Unfortunately this and a few variations on it all unilaterally fail. Am I missing somethign obvious here? (I suspect opperator precedence but I've tried a few different versions.)
Or am I doing this completely wrong? The flatten/2 documentation suggests that in most cases it's an anti-pattern, is there a more prolog-ish* way to go about solving this problem? Given all the pitfalls I'm realising as I go through this I'm almost certain there is.
(Sorry, I'm painfully aware that a lot of the terminology I'm using to describe things in this is probably very wrong, I'm only kind of familiar with predicate/formal logic and much more used-to describing control flow type programming. Even though I understand logic programming in practice reasonably well I'm struggling to find the language to talk about it robustly yet, I will amend this question with any corrections I get.)
Some background: I'm new to prolog and testing out my understanding by trying to extend one of the many sudoku solvers to solve a strange variety of sudoku I found in some puzzles I printed out years ago where you're shown the sum of all the numbers that appear between the 1 and the 9 in any given row or column as an extra hint, it's kind of like a mix of sudoku and picross. The solver as it stands now is on swish: SumSudoku(swish). Although it may be a mess when you get to it.
*Corollary quesiton: is there a prolog version of the word "pythonic?"
You could use good old append/3 for this. Is it possible that you wanted append/3 all along but somehow thought it is called flatten?
For the "1 comes before 9" case, you'd write:
between_1_and_9(List, Sublist) :-
append(_, [1|Rest], List),
append(Sublist, [9|_], Rest).
You need to swap 1 and 9 for the "9 comes before 1" case.
This also leaves a "spurious choice point" (Thank you #PauloMoura for the comment). Make sure to get rid of it somehow.
As for "Pythonic" (and this comes from a recovering Pythonista), I can only say, rest assured:
There is always more than one obvious way to do it in Prolog.
You don't even have to be Dutch.
I'd like to know what is the property testing aiming for, what is it's sweet point, where it should be used. Let' have an example function that I want to test:
f :: [Integer] -> [Integer]
This function, f, takes a list of numbers and will square the odd numbers and filter out the even numbers. I can state some properties about the function, like
Given a list of even numbers, return empty list.
Given a list of odd numbers, the result list will have the same size as input.
Given that I have a list of even numbers and a list of odd numbers, when I join them, shuffle and pass to the function, the length of the result will be the length of the list of odd numbers.
Given I provide a list of positive odd numbers, then each element in the result list at the same index will be greater than in the original list
Given I provide a list of odd numbers and even numbers, join and shuffle them, then I will get a list, where each number is odd
etc.
None of the properties test, that the function works for the simplest case, e.g. I can make a simple case, that will pass these properties if I implement the f incorrectly:
f = fmap (+2) . filter odd
So, If I want to cover some simple case, It looks like I either need to repeat a fundamental part of the algorithm in the property specification, or I need to use value based testing. The first option, that I have, to repeat the algorithm may be useful, If I plan to improve the algorithm if I plan to change it's implementation, for speed for example. In this way, I have a referential implementation, that I can use to test again.
If I want to check, that the algorithm doesn't fail for some trivial cases and I don't want to repeat the algorithm in the specification, it looks like I need some unit testing. I would write for example these checks:
f ([2,5]) == [25]
f (-8,-3,11,1) == [9,121,1]
Now I have a lot more confidence it the algorithm.
My question is, is the property based testing meant to replace the unit testing, or is it complementary? Is there some general idea, how to write the properties, so they are useful or it just totally depends on the understanding of the logic of the function? I mean, can one tell that writing the properties in some way is especially beneficial?
Also, should one strive to make the properties test every part of the algorithm? I could put the squaring out of the algorithm, and then test it elsewhere, let the properties test just the filtering part, which it looks like, that it covers it well.
f :: (Integer -> Integer) -> [Integer] -> [Integer]
f g = fmap g . filter odd
And then I can pass just Prelude.id and test the g elsewhere using unit testing.
How about the following properties:
For all odd numbers in the source list, its square is element of the result list.
For all numbers in the result list, there is a number in the source list whose square it is.
By the way, odd is easier to read than \x -> x % 2 == 1
Reference algorithm
It's very common to have a (possibly inefficient) reference implementation and test against that. In fact, that's one of the most common quickcheck strategies when implementing numeric algorithms. But not every part of the algorithm needs one. Sometimes there are some properties that characterize the algorithm completely.
Ingo's comment is spot on in that regard: These properties determine the results of your algorithm (up to order and duplicates). To recover order and duplicates you can modify the properties to include "in the resulting list truncated after the position of the source element" and vice versa in the other property.
Granularity of tests
Of course, given Haskell's composability it's nice to test each reasonable small part of an algorithm by itself. I trust e.g. \x -> x*x and filter odd as reference without looking twice.
Whether there should be properties for each part is not as clear as you might inline that part of the algorithm later and thus make the properties moot. Due to Haskell's laziness that's not a common thing to do, but it happens.
If I have a function
let rec function n =
if n<0 then []
else n-2 # function n-2 ;;
I get an error saying that the expression function n-2 is a list of int but it is expecting an int.
How do I concatenate the values to return all the n-2 values above zero as a list?
I cannot use the List module to fold.
Thanks
Your title asks how to concatenate lists, but your question seems rather different.
To concatenate lists, you can use the # operator. In many cases, code that depends on this operator is slower than it needs to be (something to keep in mind for later :-).
Here are some things I see wrong with the code you give:
a. You can't name a function function, because function is a keyword in OCaml.
b. If you use the # operator, you should have lists on both sides of it. As near as I can see, the thing on the left in your code is not a list.
c. Function calls have higher precedence than infix operators. So myfun n - 2 is parsed as (myfun n) - 2. You probably want something closer to myfun (n - 2).
Even with these changes, your code seems to generate a list of integers that are 2 apart, which isn't what you say you want. However, I can't understand what the function is actually supposed to return.
It seems like you are not concatenating lists, but concatenating ints instead. This is done by the :: operator. So your code would look like:
else (n-2)::(fun (n-2))
Although I could see this function possibly not producing the desired output if you put in negative numbers. For example if you pass through n = 1, n-2 will evaluate to -1 which is less than zero.
I'm a newbie to Haskell, I have a problem. I need to write a function that splits a list into a list of lists everywhere a 'separation' appears.
I will try to help you develop the understanding of how to develop functions that work on lists via recursion. It is helpful to learn how to do it first in a 'low-level' way so you can understand better what's happening in the 'high-level' ways that are more common in real code.
First, you must think about the nature of the type of data that you want to work with. The list is in some sense the canonical example of a recursively-defined type in Haskell: a list is either the empty list [] or it is some list element a combined with a list via a : list. Those are the only two possibilities. We call the empty list the base case because it is the one that does not refer to itself in its definition. If there were no base case, recursion would never "bottom out" and would continue indefinitely!
The fact that there are two cases in the definition of a list means that you must consider two cases in the definition of a function that works with lists. The canonical way to consider multiple cases in Haskell is pattern matching. Haskell syntax provides a number of ways to do pattern matching, but I'll just use the basic case expression for now:
case xs of
[] -> ...
x:xs' -> ...
Those are the two cases one must consider for a list. The first matches the literal empty list constructor; the second matches the element-adding constructor : and also binds two variables, x and xs', to the first element in the list and the sublist containing the rest of the elements.
If your function was passed a list that matches the first case, then you know that either the initial list was empty or that you have completed the recursion on the list all the way past its last element. Either way, there is no more list to process; you are either finished (if your calls were tail-recursive) or you need to pass the basic element of your answer construction back to the function that called this one (by returning it). In the case that your answer will be a list, the basic element will usually be the empty list again [].
If your function was passed a list that matches the second case, then you know that it was passed a non-empty list, and furthermore you have a couple of new variables bound to useful values. Based on these variables, you need to decide two things:
How do I do one step of my algorithm on that one element, assuming I have the correct answer from performing it on the rest of the list?
How do I combine the results of that one step with the results of performing it on the rest of the list?
Once you've figured the answers to those questions, you need to construct an expression that combines them; getting the answer for the rest of the list is just a matter of invoking the recursive call on the rest of the list, and then you need to perform the step for the first element and the combining.
Here's a simple example that finds the length of a list
listLength :: [a] -> Int
listLength as =
case as of
[] -> 0 -- The empty list has a length of 0
a:as' -> 1 + listlength as' -- If not empty, the length is one more than the
-- length of the rest of the list
Here's another example that removes matching elements from a list
listFilter :: Int -> [Int] -> Int
listFilter x ns =
case ns of
[] -> [] -- base element to build the answer on
n:ns' -> if n == x
then listFilter x ns' -- don't include n in the result list
else n : (listFilter x ns') -- include n in the result list
Now, the question you asked is a little bit more difficult, as it involves a secondary 'list matching' recursion to identify the separator within the basic recursion on the list. It is sometimes helpful to add extra parameters to your recursive function in order to hold extra information about where you are at in the problem. It's also possible to pattern match on two parameters at the same time by putting them in a tuple:
case (xs, ys) of
([] , [] ) -> ...
(x:xs', [] ) -> ...
([] , y:ys') -> ...
(x:xs', y:ys') -> ...
Hopefully these hints will help you to make some progress on your problem!
Let's see if the problem can be reduced in a obvious way.
Suppose splitList is called with xs to split and ys as the separator. If xs is empty, the problem is the smallest, so what's the answer to that problem? It is important to have the right answer here, because the inductive solution depends on this decision. But we can make this decision later.
Ok, so for problem to be reducable, the list xs is not empty. So, it has at least a head element h and the smaller problem t, the tail of the list: you can match xs#(h:t). How to obtain the solution to the smaller problem? Well, splitList can solve that problem by the definition of the function. So now the trick is to figure out how to build the solution for bigger problem (h:t), when we know the solution to the smaller problem zs=splitList t ys. Here we know that zs is the list of lists, [[a]], and because t may have been the smallest problem, zs may well be the solution to the smallest problem. So, whatever you do with zs, it must be valid even for the solution to the smallest problem.
splitList [] ys = ... -- some constant is the solution to the smallest problem
splitList xs#(h:t) ys = let zs = splitList t ys
in ... -- build a solution to (h:t) from solution to t
I don't know how to test it. Anybody tells me how to write a function to a .hs file and use winGHCi to run this function?
WinGHCi automatically associates with .hs files so just double-click on the file and ghci should start up. After making some changes to the file using your favourite editor you can write use the :r command in ghci to reload the file.
To test the program after fixing typos, type-errors, and ensuring correct indentation, try calling functions you have defined with different inputs (or use QuickCheck). Note Maybe is defined as Just x or Nothing. You can use fromMaybe to extract x (and provide default value for the Nothing case).
Also try to make sure that pattern matching is exhaustive.
So I'm reading http://learnyouahaskell.com/starting-out as it explains lists, and using ghci on Vista 64. It says that [2,4..20] steps by 2 from 4 to 20. This works. It says [20,19..1] goes from 20 to 1, but doesn't explain. I take it that the first number is NOT the step, the step is the difference between the 1st and 2nd number. This is confirmed by [4,4..20] which hangs (no error message, must kill console). This is unlike operators like !! and take which check the index's range and give an error message.
My question is: is this a bug on Vista port or is that the way it's supposed to be?
[x,y..z] does indeed step from x to z by step y-x. When y-x is 0 this leads to an infinite list. This is intended behavior.
Note that if you use the list in an expression like take 20 [2,2..20], ghci won't try to print the whole list (which is impossible with infinite lists of course) and it won't "hang".
Quoting this book:
[n,p..m] is the list of numbers from n to m in steps of p-n.
Your list [4,4..20] "hangs", because you have a step of 4-4=0, so it's an infinite list containing only the number 4 ([4, 4, 4, 4...]).
Haskell allows infinite lists and as the Haskell is the "lazy evaluation language", meaning it will only compute what is necessary to give you the result, so the infinite structures are allowed in Haskell.
In Haskell you could compute something like "head[1..]". This is because Haskell only calculates what is required for the result. So in the example above it would generate only the first element of the infinite list (number 1) and head would return you this element (number 1).
So, in that case program will terminate! However, if you calculate [1..] (infinite list) program won't terminate. Same applies to your example, you created an infinite list and there is no way of terminating it.
That syntax basically is derived from listing the whole list. [1,3,5,7,9,11,13,15,17,19] for example can be shortened by simply omitting the obvious parts. So you could say, if I specify the first two elements, it is clear how it would continue. So the above list equals to [1,3..19].
It's worth noting that the .. syntax in lists desugars to the enumFrom functions given by the Enum typeclass:
http://hackage.haskell.org/packages/archive/base/latest/doc/html/Prelude.html#t:Enum