----------------------------------- wtd Mon Jan 02, 2006 3:48 am Test your skills (2006) ----------------------------------- Io TYS. :) Io> foo(a, 42, a print) 42 ==> 42 Io> foo(i, "foo", i with("!") print) foo! ==> foo! Write a method which makes the above possible. Io> foo := method( /* your code here */ ) ----------------------------------- wtd Sun Jan 15, 2006 5:09 pm ----------------------------------- Write a function which appends lists. (defun list-append (...) ...) A few samples: (list-append '(1 2 3) '(4 5 6)) ; (1 2 3 4 5 6) (list-append '(1 2) '(3 4) '(5 6)) ; (1 2 3 4 5 6) You may use the following functions or macros. flet cons car cdr if defun (but only the one, in the code I already demonstrated) reduce You may very explicitly not use the loop macro. You may not have any code outside of the list-append function. ----------------------------------- wtd Fri Feb 03, 2006 2:27 am ----------------------------------- C++ gurus: a TYS just for you. Explain why this code will not compile. #include #include int main() { std::vector foo; return 0; } ----------------------------------- Martin Tue Feb 07, 2006 3:18 am ----------------------------------- C/C++ (or Java). Complete this function to determine if n is a power of 2. int isPowerOfTwo (int n) { return ; } bool isPowerOfTwo (int n) { return ; } boolean isPowerOfTwo (int n) { return ; } ----------------------------------- Andy Tue Feb 07, 2006 9:37 am ----------------------------------- bool isPowerOfTwo (int n) { return n % 2 ? n == 1 : isPowerOfTwo(n/2); } does that count as cheating? lol ----------------------------------- Martin Tue Feb 07, 2006 7:23 pm ----------------------------------- There's a much better way. Think bitwise. ----------------------------------- Andy Tue Feb 07, 2006 10:27 pm ----------------------------------- haha yea, i was going to do it bitwise, but then i got lazy =P, if i have time, i'll do it at work tomorow ----------------------------------- bugzpodder Wed Feb 08, 2006 12:53 pm ----------------------------------- bool isPowerOfTwo (int n) { return n> is an operator in C++. need a space in between. ----------------------------------- wtd Wed Feb 08, 2006 12:58 pm ----------------------------------- #include #include int main() { std::vector foo; return 0; } >> is an operator in C++. need a space in between. Yes, that is the solution. But why is it a problem? :) ----------------------------------- bugzpodder Wed Feb 08, 2006 1:07 pm ----------------------------------- ? its a problem because >> is a predefined operator in C++. But i thought I've already said that. ----------------------------------- wtd Wed Feb 08, 2006 1:08 pm ----------------------------------- But surely C++ compilers can tell the difference? ----------------------------------- bugzpodder Wed Feb 08, 2006 2:33 pm ----------------------------------- clearly not? lol ----------------------------------- Martin Wed Feb 08, 2006 6:26 pm ----------------------------------- bool isPowerOfTwo (int n) { return n> some_other_value Could clearly never be a type. So... why does C++ get confused? :) ----------------------------------- Martin Wed Feb 08, 2006 7:59 pm ----------------------------------- Would this be a problem relating to operator overloading? ----------------------------------- bugzpodder Wed Feb 08, 2006 8:02 pm ----------------------------------- bool isPowerOfTwo (int n) { return n x * 3) (fun x -> x + 1) And this is equivalent to: let my_func x = x * 3 + 1 Write a function or function object in C++ which replicates the functionality of the compose function demonstrated above. ----------------------------------- wtd Wed Feb 15, 2006 11:08 pm ----------------------------------- Give myList.ml as follows: module type LIST_ITEM_TYPE = sig type t end module MyList = functor (Elt : LIST_ITEM_TYPE) -> struct type element = Elt.t type t = | Empty | Node of element * t let empty = Empty let rec to_list = function | Empty -> [] | Node (v, rest) -> v :: to_list rest let rec from_list = function | [] -> Empty | x::xs -> Node (x, from_list xs) let rec fold_left f i = function | Empty -> i | Node (v, rest) -> fold_left f (f i v) rest let rec map f = function | Empty -> Empty | Node (v, rest) -> Node (f v, map f rest) let iter f = fold_left (fun _ b -> f b) () let length = fold_left (fun a _ -> a + 1) 0 let rec append lst1 lst2 = match lst1, lst2 with | Empty, lst2 -> lst2 | Node (v, rest), lst2 -> Node (v, append rest lst2) let reverse = fold_left (fun a b -> Node (b, a)) Empty end Complete the following: let (* your code *) in M.append (M.from_list [1; 2; 3]) (M.Node (4, M.Node (5, M.empty))) ----------------------------------- rizzix Thu Feb 16, 2006 7:20 pm ----------------------------------- Write the following expression in point-free form: f n = g (x n) (y n) ----------------------------------- wtd Wed Mar 01, 2006 9:53 pm ----------------------------------- Explain why: open Int64 let factorial n = let rec f n acc = if n = Int64.zero then acc else f (Int64.sub n Int64.one) (Int64.mul n acc) in f (Int64.of_int n) Int64.one Does not cause a stack overflow when called with a value of ten million. ----------------------------------- rizzix Sat Mar 18, 2006 1:11 am ----------------------------------- Write the following expression in point-free form: f n = g (x n) (y n) 500 bits to the first person who correctly answers this. Oh and you need to explain your solution. :) ----------------------------------- Martin Tue Mar 21, 2006 6:22 pm ----------------------------------- Rizzix - what is point free form? ----------------------------------- rizzix Wed Mar 22, 2006 1:58 am ----------------------------------- eta reduce it to the point where there is no "n" (function parameter) visible in that code. ----------------------------------- Martin Wed Mar 22, 2006 2:17 am ----------------------------------- Huh? Is there a website on this, I'm not following... ----------------------------------- rizzix Wed Mar 22, 2006 5:07 pm ----------------------------------- An example:f m n = g m (x n)f m n = ((g m) . x) nf m = (g m) . x Eta-reduced, yet it is still not point free. As you can see there is still one parameter visible: m. Another example: f m n = n + mf m n = (+m) n f m = (+) m f = (+) As you can see after completly eta-reducing it, there are no parameters visible. This final form is an example of point-free form. ----------------------------------- tez_h Mon Apr 03, 2006 9:13 pm ----------------------------------- Write the following expression in point-free form: f n = g (x n) (y n) First, some useful functions: (f . g) x = f (g x) flip f x y = f y x twice f x = f x x Now, we can derive a point-free expression of your function: f n = g (x n) (y n) {defn of (.)} = (g.x) n (y n) {defn of flip} = flip (g.x) (y n) n {defn of (.)} = (flip (g.x).y) n n {defn of twice} = twice (flip (g.x).y) n f = twice (flip (g.x).y) Is that the sort of answer you were looking for? -Tez ----------------------------------- rizzix Tue Apr 04, 2006 12:56 am ----------------------------------- Hey! Congrats and welcome to compsci! BTW: There is no need for you to redefine any of the Prelude functions (flip, composition). I was hoping you could do it without defining any of your own functions, like twice. The truth is, that is not the answer I was hoping for, but it is an answer I was expecting -- or one kind of answer anyway. You get 350 bits for that one :) Here's my solution that's synonymous to yours (i.e worth 350bits) Helper function: toPair n = (n, n)Eta-reduction: f n = g (x n) (y n) -- let a = g (x n) = (g (x n)) (y n) -- therefore, f n = (a . y) n = ((g (x n)) . y) n -- = ((. y) a) n = ((. y) (g (x n))) n -- therefore, f n = ((. y) (g (x n))) = ((. y) ((g . x) n)) n = ((. y) . g . x) n nTherefore,f = (uncurry $ (. y) . g . x) . toPair Note that if we used your function twice, it was just a matter of:f = twice $ (. y) . g . x Eitherway, this is still not what I'm looking for, for that extra 150bits. :) ----------------------------------- tez_h Tue Apr 04, 2006 5:31 am ----------------------------------- Hey! Congrats and welcome to compsci! BTW: There is no need for you to redefine any of the Prelude functions (flip, composition). I was hoping you could do it without defining any of your own functions, like twice. Hi, and thanks. Yeah, I thought I'd just write out the definitions for ease of reference. The truth is, that is not the answer I was hoping for, but it is an answer I was expecting -- or one kind of answer anyway. You get 350 bits for that one :) Eitherway, this is still not what I'm looking for, for that extra 150bits. :) Well, I haven't quite worked out if you're expecting some ultra-clever combination of (.), ($), and flip, say, but here's one that only uses standard prelude functions, starting from near the end of my derivation: f n = (flip (g.x).y) n n = foldr1 (flip (g.x).y) (replicate 2 n) = (foldr1 (flip (g.x).y) . replicate 2) n f = foldr1 (flip (g.x).y) . replicate 2 Not pretty, and certainly point-less :wink: . Similarly, starting from near the end of your derivation: f n = ((. y) . g . x) n n = foldr1 ((. y) . g . x) (replicate 2 n) = (foldr1 ((. y) . g . x) . replicate 2) n f = foldr1 ((. y) . g . x) . replicate 2 Is that more what you're looking for? -Tez ----------------------------------- rizzix Tue Apr 04, 2006 11:51 am ----------------------------------- No. Let me give you a hint: I'm looking for a simpler solution ;) Another hint: (->a) ----------------------------------- tez_h Wed Apr 05, 2006 4:27 am (-> a) monad ----------------------------------- No. Let me give you a hint: I'm looking for a simpler solution ;) Another hint: (->a) Yes, well I did see somewhere that there is a "(-> a)" monad that lambdabot (on #haskell on freenode) uses to transform point-wise expressions into point-free form. But since I haven't really done much proper haskell programming for a good few years, and googling for ' "(-> a)" monad ' turned up very little, I'm afraid I'm going to have to admit defeat here. I'd love it if you could provide a reference or something about the (-> a) monad, or even if it has a more googlable name. Or PM me a more obvious hint! Thanks for your time and the challenge. Now I need something to get me to learn about arrows! -Tez ----------------------------------- rizzix Thu Apr 06, 2006 1:04 am ----------------------------------- Just lookup the Reader Monad. (->a) is just an instance of Control.Monad.Reader. Where (->) is the function type constructor. http://www.nomaware.com/monads/html/readermonad.html ----------------------------------- wtd Thu Apr 20, 2006 2:34 pm ----------------------------------- C++ today. #include #include #include using namespace std; char upcase(const char original); string capitalize(const string& s); int main(int argc, char **argv) { string *args = new string[argc - 1]; copy(argv + 1, argv + argc, args); transform(args, args + argc - 1, args, capitalize); for (int i = 0; i < argc - 1; i++) { cout