// [snippet: Ninety-Nine F# Problems - Problems 21 - 28 - Lists again] /// These are F# solutions of Ninety-Nine Haskell Problems /// (http://www.haskell.org/haskellwiki/H-99:_Ninety-Nine_Haskell_Problems), /// which are themselves translations of Ninety-Nine Lisp Problems /// (http://www.ic.unicamp.br/~meidanis/courses/mc336/2006s2/funcional/L-99_Ninety-Nine_Lisp_Problems.html) /// and Ninety-Nine Prolog Problems /// (https://sites.google.com/site/prologsite/prolog-problems). /// /// If you would like to contribute a solution or fix any bugs, send /// an email to paks at kitiara dot org with the subject "99 F# problems". /// I'll try to update the problem as soon as possible. /// /// The problems have different levels of difficulty. Those marked with a single asterisk (*) /// are easy. If you have successfully solved the preceeding problems you should be able to /// solve them within a few (say 15) minutes. Problems marked with two asterisks (**) are of /// intermediate difficulty. If you are a skilled F# programmer it shouldn't take you more than /// 30-90 minutes to solve them. Problems marked with three asterisks (***) are more difficult. /// You may need more time (i.e. a few hours or more) to find a good solution /// /// Though the problems number from 1 to 99, there are some gaps and some additions marked with /// letters. There are actually only 88 problems. // [/snippet] // [snippet: (*) Problem 21 : Insert an element at a given position into a list.] /// Example: /// * (insert-at 'alfa '(a b c d) 2) /// (A ALFA B C D) /// /// Example in F#: /// /// > insertAt 'X' (List.ofSeq "abcd") 2;; /// val it : char list = ['a'; 'X'; 'b'; 'c'; 'd'] (*[omit:(Solution 1)]*) let insertAt x xs n = let rec insAt acc xs n = match xs, n with | [], 1 -> List.rev (x::acc) | [], _ -> failwith "Empty list you fool!" | xs, 1 -> List.rev (x::acc) @ xs | x::xs, n -> insAt (x::acc) xs (n - 1) insAt [] xs n (*[/omit]*) (*[omit:(Solution 2)]*) let rec insertAt' x xs n = match xs, n with | [], 1 -> [x] | [], _ -> failwith "Empty list you fool!" | _, 1 -> x::xs | y::ys, n -> y::insertAt' x ys (n - 1) (*[/omit]*) // [/snippet] // [snippet: (*) Problem 22 : Create a list containing all integers within a given range.] /// Example: /// * (range 4 9) /// (4 5 6 7 8 9) /// /// Example in F#: /// /// > range 4 9;; /// val it : int list = [4; 5; 6; 7; 8; 9] (*[omit:(Solution)]*) let range a b = [a..b] (*[/omit]*) // [/snippet] // [snippet: (**) Problem 23 : Extract a given number of randomly selected elements from a list.] /// Example: /// * (rnd-select '(a b c d e f g h) 3) /// (E D A) /// /// Example in F#: /// /// > rnd_select "abcdefgh" 3;; /// val it : seq = seq ['e'; 'a'; 'h'] (*[omit:(Solution)]*) let rnd_select xs n = let rndSeq = let r = new System.Random() in seq { while true do yield r.Next() } xs |> Seq.zip rndSeq |> Seq.sortBy fst |> Seq.map snd |> Seq.take n (*[/omit]*) // [/snippet] // [snippet: (*) Problem 24 : Lotto: Draw N different random numbers from the set 1..M.] /// Example: /// * (rnd-select 6 49) /// (23 1 17 33 21 37) /// /// Example in F#: /// /// > diff_select 6 49;; /// val it : int list = [27; 20; 22; 9; 15; 29] // using problem 23 (*[omit:(Solution)]*) let diff_select n m = rnd_select (seq { 1 .. m }) n |> List.ofSeq (*[/omit]*) (*[omit:(Solution)]*) let diff_select' n m = let rndSeq = let r = new System.Random() in Seq.initInfinite(ignore >> r.Next ) seq { 1 .. m } |> Seq.zip rndSeq |> Seq.sortBy fst |> Seq.map snd |> Seq.take n |> List.ofSeq (*[/omit]*) // [/snippet] // [snippet: (*) Problem 25 : Generate a random permutation of the elements of a list.] /// Example: /// * (rnd-permu '(a b c d e f)) /// (B A D C E F) /// /// Example in F#: /// /// > rnd_permu <| List.ofSeq "abcdef";; /// val it : char list = ['b'; 'c'; 'd'; 'f'; 'e'; 'a'] (*[omit:(Solution 1)]*) // using problem 23 let rnd_permu xs = List.length xs |> rnd_select xs (*[/omit]*) (*[omit:(Solution 2)]*) let rnd_permu' xs = let rec rndSeq = let r = new System.Random() seq { while true do yield r.Next() } xs |> Seq.zip rndSeq |> Seq.sortBy fst |> Seq.map snd |> List.ofSeq (*[/omit]*) // [/snippet] // [snippet: (**) Problem 26 : Generate the combinations of K distinct objects chosen from the N elements of a list.] /// In how many ways can a committee of 3 be chosen from a group of 12 people? We all know that /// there are C(12,3) = 220 possibilities (C(N,K) denotes the well-known binomial coefficients). For /// pure mathematicians, this result may be great. But we want to really generate all the /// possibilities in a list. /// /// Example: /// * (combinations 3 '(a b c d e f)) /// ((A B C) (A B D) (A B E) ... ) /// /// Example in F#: /// /// > combinations 3 ['a' .. 'f'];; /// val it : char list list = /// [['a'; 'b'; 'c']; ['a'; 'b'; 'd']; ['a'; 'b'; 'e']; ['a'; 'b'; 'f']; /// ['a'; 'c'; 'd']; ['a'; 'c'; 'e']; ['a'; 'c'; 'f']; ['a'; 'd'; 'e']; /// ['a'; 'd'; 'f']; ['a'; 'e'; 'f']; ['b'; 'c'; 'd']; ['b'; 'c'; 'e']; /// ['b'; 'c'; 'f']; ['b'; 'd'; 'e']; ['b'; 'd'; 'f']; ['b'; 'e'; 'f']; /// ['c'; 'd'; 'e']; ['c'; 'd'; 'f']; ['c'; 'e'; 'f']; ['d'; 'e'; 'f']] (*[omit:(Solution 1)]*) // as a bonus you get the powerset of xs with combinations 0 xs let rec combinations n xs = match xs, n with | [],_ -> [[]] | xs, 1 -> [for x in xs do yield [x]] | x::xs, n -> [for ys in combinations (n-1) xs do yield x::ys if List.length xs > n then yield! combinations n xs else yield xs] (*[/omit]*) (*[omit:(Solution 2)]*) let rec combinations' n xs = let rec tails = function | [] -> [[]] | _::ys as xs -> xs::tails ys match xs, n with | _, 0 -> [[]] | xs, n -> [ for tail in tails xs do match tail with | [] -> () | y::xs' -> for ys in combinations' (n - 1) xs' do yield y::ys ] (*[/omit]*) // [/snippet] // [snippet: (**) Problem 27 : Group the elements of a set into disjoint subsets.] /// a) In how many ways can a group of 9 people work in 3 disjoint subgroups of 2, 3 /// and 4 persons? Write a function that generates all the possibilities and returns them /// in a list. /// /// Example: /// * (group3 '(aldo beat carla david evi flip gary hugo ida)) /// ( ( (ALDO BEAT) (CARLA DAVID EVI) (FLIP GARY HUGO IDA) ) /// ... ) /// /// b) Generalize the above predicate in a way that we can specify a list of group sizes /// and the predicate will return a list of groups. /// /// Example: /// * (group '(aldo beat carla david evi flip gary hugo ida) '(2 2 5)) /// ( ( (ALDO BEAT) (CARLA DAVID) (EVI FLIP GARY HUGO IDA) ) /// ... ) /// /// Note that we do not want permutations of the group members; i.e. ((ALDO BEAT) ...) /// is the same solution as ((BEAT ALDO) ...). However, we make a difference between /// ((ALDO BEAT) (CARLA DAVID) ...) and ((CARLA DAVID) (ALDO BEAT) ...). /// /// You may find more about this combinatorial problem in a good book on discrete /// mathematics under the term "multinomial coefficients". /// /// Example in F#: /// /// > group [2;3;4] ["aldo";"beat";"carla";"david";"evi";"flip";"gary";"hugo";"ida"];; /// val it : string list list list = /// [[["aldo"; "beat"]; ["carla"; "david"; "evi"]; /// ["flip"; "gary"; "hugo"; "ida"]];...] /// (altogether 1260 solutions) /// /// > group [2;2;5] ["aldo";"beat";"carla";"david";"evi";"flip";"gary";"hugo";"ida"];; /// val it : string list list list = /// [[["aldo"; "beat"]; ["carla"; "david"]; /// ["evi"; "flip"; "gary"; "hugo"; "ida"]];...] /// (altogether 756 solutions) (*[omit:(Solution)]*) let rec group ns xs = let rec combination n xs = match n,xs with | 0, xs -> [([], xs)] | _, [] -> [] | n, x::xs -> let ts = [ for ys, zs in combination (n-1) xs do yield (x::ys, zs)] let ds = [ for ys, zs in combination n xs do yield (ys, x::zs)] ts @ ds match ns,xs with | [], _ -> [[]] | n::ns, xs -> [ for g, rs in combination n xs do for gs in group ns rs do yield g::gs ] (*[/omit]*) // [/snippet] // [snippet: (**) Problem 28 : Sorting a list of lists according to length of sublists] /// a) We suppose that a list contains elements that are lists themselves. The objective /// is to sort the elements of this list according to their length. E.g. short lists first, /// longer lists later, or vice versa. /// /// Example: /// * (lsort '((a b c) (d e) (f g h) (d e) (i j k l) (m n) (o))) /// ((O) (D E) (D E) (M N) (A B C) (F G H) (I J K L)) /// /// Example in F#: /// /// > lsort ["abc";"de";"fgh";"de";"ijkl";"mn";"o"];; /// val it : string list = ["o"; "de"; "de"; "mn"; "abc"; "fgh"; "ijkl"] /// /// b) Again; we suppose that a list contains elements that are lists themselves. But this /// time the objective is to sort the elements of this list according to their length /// frequency; i.e.; in the default; where sorting is done ascendingly; lists with rare /// lengths are placed first; others with a more frequent length come later. /// /// Example: /// * (lfsort '((a b c) (d e) (f g h) (d e) (i j k l) (m n) (o))) /// ((i j k l) (o) (a b c) (f g h) (d e) (d e) (m n)) /// /// Example in F#: /// /// > lfsort ["abc"; "de"; "fgh"; "de"; "ijkl"; "mn"; "o"];; /// val it : string list = ["ijkl"; "o"; "abc"; "fgh"; "de"; "de"; "mn"] (*[omit:(Solution)]*) let lsort xss = xss |> List.sortBy Seq.length let lfsort xss = xss |> Seq.groupBy (Seq.length) |> Seq.sortBy (snd >> Seq.length) |> Seq.collect snd |> List.ofSeq (*[/omit]*) // [/snippet]