Exercises
Tail of a list
Write a function last : 'a list > 'a option
that returns the last element of a list
# last ["a" ; "b" ; "c" ; "d"];;
 : string option = Some "d"
# last [];;
 : 'a option = None
# let rec last = function
 [] > None
 [ x ] > Some x
 _ :: t > last t;;
val last : 'a list > 'a option = <fun>
Last two elements of a list
Find the last but one (last and penultimate) elements of a list.
# last_two ["a"; "b"; "c"; "d"];;
 : (string * string) option = Some ("c", "d")
# last_two ["a"];;
 : (string * string) option = None
# let rec last_two = function
 []  [_] > None
 [x; y] > Some (x,y)
 _ :: t > last_two t;;
val last_two : 'a list > ('a * 'a) option = <fun>
N'th element of a list
Find the N'th element of a list.
REMARK: OCaml has
List.nth
which numbers elements from0
and raises an exception if the index is out of bounds.
# List.nth ["a"; "b"; "c"; "d"; "e"] 2;;
 : string = "c"
# List.nth ["a"] 2;;
Exception: Failure "nth".
# let rec at k = function
 [] > None
 h :: t > if k = 0 then Some h else at (k  1) t;;
val at : int > 'a list > 'a option = <fun>
Length of a list
Find the number of elements of a list.
OCaml standard library has List.length
but we ask that you reimplement
it. Bonus for a tail recursive
solution.
# length ["a"; "b"; "c"];;
 : int = 3
# length [];;
 : int = 0
This function is tailrecursive: it uses a constant amount of stack memory regardless of list size.
# let length list =
let rec aux n = function
 [] > n
 _ :: t > aux (n + 1) t
in
aux 0 list;;
val length : 'a list > int = <fun>
Reverse a list
Reverse a list.
OCaml standard library has List.rev
but we ask that you reimplement
it.
# rev ["a"; "b"; "c"];;
 : string list = ["c"; "b"; "a"]
# let rev list =
let rec aux acc = function
 [] > acc
 h :: t > aux (h :: acc) t
in
aux [] list;;
val rev : 'a list > 'a list = <fun>
Palindrome
Find out whether a list is a palindrome.
HINT: a palindrome is its own reverse.
# is_palindrome ["x"; "a"; "m"; "a"; "x"];;
 : bool = true
# not (is_palindrome ["a"; "b"]);;
 : bool = true
# let is_palindrome list =
(* One can use either the rev function from the previous problem, or the builtin List.rev *)
list = List.rev list;;
val is_palindrome : 'a list > bool = <fun>
Flatten a list
Flatten a nested list structure.
type 'a node =
 One of 'a
 Many of 'a node list
# flatten [One "a"; Many [One "b"; Many [One "c" ;One "d"]; One "e"]];;
 : string list = ["a"; "b"; "c"; "d"; "e"]
# type 'a node =
 One of 'a
 Many of 'a node list;;
type 'a node = One of 'a  Many of 'a node list
# (* This function traverses the list, prepending any encountered elements
to an accumulator, which flattens the list in inverse order. It can
then be reversed to obtain the actual flattened list. *);;
# let flatten list =
let rec aux acc = function
 [] > acc
 One x :: t > aux (x :: acc) t
 Many l :: t > aux (aux acc l) t
in
List.rev (aux [] list);;
val flatten : 'a node list > 'a list = <fun>
Eliminate duplicates
Eliminate consecutive duplicates of list elements.
# compress ["a"; "a"; "a"; "a"; "b"; "c"; "c"; "a"; "a"; "d"; "e"; "e"; "e"; "e"];;
 : string list = ["a"; "b"; "c"; "a"; "d"; "e"]
# let rec compress = function
 a :: (b :: _ as t) > if a = b then compress t else a :: compress t
 smaller > smaller;;
val compress : 'a list > 'a list = <fun>
Pack consecutive duplicates
Pack consecutive duplicates of list elements into sublists.
# pack ["a"; "a"; "a"; "a"; "b"; "c"; "c"; "a"; "a"; "d"; "d"; "e"; "e"; "e"; "e"];;
 : string list list =
[["a"; "a"; "a"; "a"]; ["b"]; ["c"; "c"]; ["a"; "a"]; ["d"; "d"];
["e"; "e"; "e"; "e"]]
# let pack list =
let rec aux current acc = function
 [] > [] (* Can only be reached if original list is empty *)
 [x] > (x :: current) :: acc
 a :: (b :: _ as t) >
if a = b then aux (a :: current) acc t
else aux [] ((a :: current) :: acc) t in
List.rev (aux [] [] list);;
val pack : 'a list > 'a list list = <fun>
Runlength encoding
If you need so, refresh your memory about runlength encoding.
Here is an example:
# encode ["a"; "a"; "a"; "a"; "b"; "c"; "c"; "a"; "a"; "d"; "e"; "e"; "e"; "e"];;
 : (int * string) list =
[(4, "a"); (1, "b"); (2, "c"); (2, "a"); (1, "d"); (4, "e")]
# let encode list =
let rec aux count acc = function
 [] > [] (* Can only be reached if original list is empty *)
 [x] > (count + 1, x) :: acc
 a :: (b :: _ as t) > if a = b then aux (count + 1) acc t
else aux 0 ((count + 1, a) :: acc) t in
List.rev (aux 0 [] list);;
val encode : 'a list > (int * 'a) list = <fun>
An alternative solution, which is shorter but requires more memory, is to use
the pack
function declared in problem 9:
# let pack list =
let rec aux current acc = function
 [] > [] (* Can only be reached if original list is empty *)
 [x] > (x :: current) :: acc
 a :: (b :: _ as t) >
if a = b then aux (a :: current) acc t
else aux [] ((a :: current) :: acc) t in
List.rev (aux [] [] list);;
val pack : 'a list > 'a list list = <fun>
# let encode list =
List.map (fun l > (List.length l, List.hd l)) (pack list);;
val encode : 'a list > (int * 'a) list = <fun>
Modified runlength encoding
Modify the result of the previous problem in such a way that if an element has no duplicates it is simply copied into the result list. Only elements with duplicates are transferred as (N E) lists.
Since OCaml lists are homogeneous, one needs to define a type to hold both single elements and sublists.
type 'a rle =
 One of 'a
 Many of int * 'a
# encode ["a"; "a"; "a"; "a"; "b"; "c"; "c"; "a"; "a"; "d"; "e"; "e"; "e"; "e"];;
 : string rle list =
[Many (4, "a"); One "b"; Many (2, "c"); Many (2, "a"); One "d";
Many (4, "e")]
# type 'a rle =
 One of 'a
 Many of int * 'a;;
type 'a rle = One of 'a  Many of int * 'a
# let encode l =
let create_tuple cnt elem =
if cnt = 1 then One elem
else Many (cnt, elem) in
let rec aux count acc = function
 [] > []
 [x] > (create_tuple (count + 1) x) :: acc
 hd :: (snd :: _ as tl) >
if hd = snd then aux (count + 1) acc tl
else aux 0 ((create_tuple (count + 1) hd) :: acc) tl in
List.rev (aux 0 [] l);;
val encode : 'a list > 'a rle list = <fun>
Decode a runlength encoded list
Given a runlength code list generated as specified in the previous problem, construct its uncompressed version.
# decode [Many (4, "a"); One "b"; Many (2, "c"); Many (2, "a"); One "d"; Many (4, "e")];;
 : string list =
["a"; "a"; "a"; "a"; "b"; "c"; "c"; "a"; "a"; "d"; "e"; "e"; "e"; "e"]
# let decode list =
let rec many acc n x =
if n = 0 then acc else many (x :: acc) (n  1) x
in
let rec aux acc = function
 [] > acc
 One x :: t > aux (x :: acc) t
 Many (n, x) :: t > aux (many acc n x) t
in
aux [] (List.rev list);;
val decode : 'a rle list > 'a list = <fun>
Runlength encoding of a list (direct solution)
Implement the socalled runlength encoding data compression method directly. I.e. don't explicitly create the sublists containing the duplicates, as in problem "Pack consecutive duplicates of list elements into sublists", but only count them. As in problem "Modified runlength encoding", simplify the result list by replacing the singleton lists (1 X) by X.
# encode ["a";"a";"a";"a";"b";"c";"c";"a";"a";"d";"e";"e";"e";"e"];;
 : string rle list =
[Many (4, "a"); One "b"; Many (2, "c"); Many (2, "a"); One "d";
Many (4, "e")]
# let encode list =
let rle count x = if count = 0 then One x else Many (count + 1, x) in
let rec aux count acc = function
 [] > [] (* Can only be reached if original list is empty *)
 [x] > rle count x :: acc
 a :: (b :: _ as t) > if a = b then aux (count + 1) acc t
else aux 0 (rle count a :: acc) t
in
List.rev (aux 0 [] list);;
val encode : 'a list > 'a rle list = <fun>
Duplicate the elements of a list
Duplicate the elements of a list.
# duplicate ["a"; "b"; "c"; "c"; "d"];;
 : string list = ["a"; "a"; "b"; "b"; "c"; "c"; "c"; "c"; "d"; "d"]
# let rec duplicate = function
 [] > []
 h :: t > h :: h :: duplicate t;;
val duplicate : 'a list > 'a list = <fun>
Remark: this function is not tail recursive. Can you modify it so it becomes so?
Replicate the elements of a list a given number of times
Replicate the elements of a list a given number of times.
# replicate ["a"; "b"; "c"] 3;;
 : string list = ["a"; "a"; "a"; "b"; "b"; "b"; "c"; "c"; "c"]
# let replicate list n =
let rec prepend n acc x =
if n = 0 then acc else prepend (n1) (x :: acc) x in
let rec aux acc = function
 [] > acc
 h :: t > aux (prepend n acc h) t in
(* This could also be written as:
List.fold_left (prepend n) [] (List.rev list) *)
aux [] (List.rev list);;
val replicate : 'a list > int > 'a list = <fun>
Note that
List.rev list
is needed only because we wantaux
to be tail recursive.
Drop every N'th element from a list
Drop every N'th element from a list.
# drop ["a"; "b"; "c"; "d"; "e"; "f"; "g"; "h"; "i"; "j"] 3;;
 : string list = ["a"; "b"; "d"; "e"; "g"; "h"; "j"]
# let drop list n =
let rec aux i = function
 [] > []
 h :: t > if i = n then aux 1 t else h :: aux (i + 1) t in
aux 1 list;;
val drop : 'a list > int > 'a list = <fun>
Split a list into two parts; the length of the first part is given
Split a list into two parts; the length of the first part is given.
If the length of the first part is longer than the entire list, then the first part is the list and the second part is empty.
# split ["a"; "b"; "c"; "d"; "e"; "f"; "g"; "h"; "i"; "j"] 3;;
 : string list * string list =
(["a"; "b"; "c"], ["d"; "e"; "f"; "g"; "h"; "i"; "j"])
# split ["a"; "b"; "c"; "d"] 5;;
 : string list * string list = (["a"; "b"; "c"; "d"], [])
# let split list n =
let rec aux i acc = function
 [] > List.rev acc, []
 h :: t as l > if i = 0 then List.rev acc, l
else aux (i  1) (h :: acc) t
in
aux n [] list;;
val split : 'a list > int > 'a list * 'a list = <fun>
Extract a slice from a list
Given two indices, i
and k
, the slice is the list containing the
elements between the i
'th and k
'th element of the original list
(both limits included). Start counting the elements with 0 (this is the
way the List
module numbers elements).
# slice ["a"; "b"; "c"; "d"; "e"; "f"; "g"; "h"; "i"; "j"] 2 6;;
 : string list = ["c"; "d"; "e"; "f"; "g"]
# let slice list i k =
let rec take n = function
 [] > []
 h :: t > if n = 0 then [] else h :: take (n  1) t
in
let rec drop n = function
 [] > []
 h :: t as l > if n = 0 then l else drop (n  1) t
in
take (k  i + 1) (drop i list);;
val slice : 'a list > int > int > 'a list = <fun>
This solution has a drawback, namely that the take
function is not
tail recursive so it may
exhaust the stack when given a very long list. You may also notice that
the structure of take
and drop
is similar and you may want to
abstract their common skeleton in a single function. Here is a solution.
# let rec fold_until f acc n = function
 [] > (acc, [])
 h :: t as l > if n = 0 then (acc, l)
else fold_until f (f acc h) (n  1) t
let slice list i k =
let _, list = fold_until (fun _ _ > []) [] i list in
let taken, _ = fold_until (fun acc h > h :: acc) [] (k  i + 1) list in
List.rev taken;;
val fold_until : ('a > 'b > 'a) > 'a > int > 'b list > 'a * 'b list =
<fun>
val slice : 'a list > int > int > 'a list = <fun>
Rotate a list N places to the left
Rotate a list N places to the left.
# rotate ["a"; "b"; "c"; "d"; "e"; "f"; "g"; "h"] 3;;
 : string list = ["d"; "e"; "f"; "g"; "h"; "a"; "b"; "c"]
# let split list n =
let rec aux i acc = function
 [] > List.rev acc, []
 h :: t as l > if i = 0 then List.rev acc, l
else aux (i  1) (h :: acc) t in
aux n [] list
let rotate list n =
let len = List.length list in
(* Compute a rotation value between 0 and len  1 *)
let n = if len = 0 then 0 else (n mod len + len) mod len in
if n = 0 then list
else let a, b = split list n in b @ a;;
val split : 'a list > int > 'a list * 'a list = <fun>
val rotate : 'a list > int > 'a list = <fun>
Remove the K'th element from a list
Remove the K'th element from a list.
The first element of the list is numbered 0, the second 1,...
# remove_at 1 ["a"; "b"; "c"; "d"];;
 : string list = ["a"; "c"; "d"]
# let rec remove_at n = function
 [] > []
 h :: t > if n = 0 then t else h :: remove_at (n  1) t;;
val remove_at : int > 'a list > 'a list = <fun>
Insert an element at a given position into a list
Start counting list elements with 0. If the position is larger or equal to the length of the list, insert the element at the end. (The behavior is unspecified if the position is negative.)
# insert_at "alfa" 1 ["a"; "b"; "c"; "d"];;
 : string list = ["a"; "alfa"; "b"; "c"; "d"]
# let rec insert_at x n = function
 [] > [x]
 h :: t as l > if n = 0 then x :: l else h :: insert_at x (n  1) t;;
val insert_at : 'a > int > 'a list > 'a list = <fun>
Create a list containing all integers within a given range
If first argument is greater than second, produce a list in decreasing order.
# range 4 9;;
 : int list = [4; 5; 6; 7; 8; 9]
# let range a b =
let rec aux a b =
if a > b then [] else a :: aux (a + 1) b
in
if a > b then List.rev (aux b a) else aux a b;;
val range : int > int > int list = <fun>
A tail recursive implementation:
# let range a b =
let rec aux acc high low =
if high >= low then
aux (high :: acc) (high  1) low
else acc
in
if a < b then aux [] b a else List.rev (aux [] a b);;
val range : int > int > int list = <fun>
Extract a given number of randomly selected elements from a list
The selected items shall be returned in a list. We use the Random
module but do not initialize it with Random.self_init
for
reproducibility.
# rand_select ["a"; "b"; "c"; "d"; "e"; "f"; "g"; "h"] 3;;
 : string list = ["g"; "d"; "a"]
# let rand_select list n =
let rec extract acc n = function
 [] > raise Not_found
 h :: t > if n = 0 then (h, acc @ t) else extract (h :: acc) (n  1) t
in
let extract_rand list len =
extract [] (Random.int len) list
in
let rec aux n acc list len =
if n = 0 then acc else
let picked, rest = extract_rand list len in
aux (n  1) (picked :: acc) rest (len  1)
in
let len = List.length list in
aux (min n len) [] list len;;
val rand_select : 'a list > int > 'a list = <fun>
Lotto: Draw N different random numbers from the set 1..M
Draw N different random numbers from the set 1..M
.
The selected numbers shall be returned in a list.
# lotto_select 6 49;;
 : int list = [20; 28; 45; 16; 24; 38]
# (* [range] and [rand_select] defined in problems above *)
let lotto_select n m = rand_select (range 1 m) n;;
val lotto_select : int > int > int list = <fun>
Generate a random permutation of the elements of a list
Generate a random permutation of the elements of a list.
# permutation ["a"; "b"; "c"; "d"; "e"; "f"];;
 : string list = ["c"; "d"; "f"; "e"; "b"; "a"]
# let rec permutation list =
let rec extract acc n = function
 [] > raise Not_found
 h :: t > if n = 0 then (h, acc @ t) else extract (h :: acc) (n  1) t
in
let extract_rand list len =
extract [] (Random.int len) list
in
let rec aux acc list len =
if len = 0 then acc else
let picked, rest = extract_rand list len in
aux (picked :: acc) rest (len  1)
in
aux [] list (List.length list);;
val permutation : 'a list > 'a list = <fun>
Generate the combinations of K distinct objects chosen from the N elements of a list
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 wellknown binomial coefficients). For pure mathematicians, this result may be great. But we want to really generate all the possibilities in a list.
# extract 2 ["a"; "b"; "c"; "d"];;
 : string list list =
[["a"; "b"]; ["a"; "c"]; ["a"; "d"]; ["b"; "c"]; ["b"; "d"]; ["c"; "d"]]
# let rec extract k list =
if k <= 0 then [[]]
else match list with
 [] > []
 h :: tl >
let with_h = List.map (fun l > h :: l) (extract (k  1) tl) in
let without_h = extract k tl in
with_h @ without_h;;
val extract : int > 'a list > 'a list list = <fun>
Group the elements of a set into disjoint subsets
Group the elements of a set into disjoint subsets
 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.
 Generalize the above function in a way that we can specify a list of group sizes and the function will return a list of groups.
# group ["a"; "b"; "c"; "d"] [2; 1];;
 : string list list list =
[[["a"; "b"]; ["c"]]; [["a"; "c"]; ["b"]]; [["b"; "c"]; ["a"]];
[["a"; "b"]; ["d"]]; [["a"; "c"]; ["d"]]; [["b"; "c"]; ["d"]];
[["a"; "d"]; ["b"]]; [["b"; "d"]; ["a"]]; [["a"; "d"]; ["c"]];
[["b"; "d"]; ["c"]]; [["c"; "d"]; ["a"]]; [["c"; "d"]; ["b"]]]
# (* This implementation is less streamlined than the oneextraction
version, because more work is done on the lists after each
transform to prepend the actual items. The end result is cleaner
in terms of code, though. *)
let group list sizes =
let initial = List.map (fun size > size, []) sizes in
(* The core of the function. Prepend accepts a list of groups,
each with the number of items that should be added, and
prepends the item to every group that can support it, thus
turning [1,a ; 2,b ; 0,c] into [ [0,x::a ; 2,b ; 0,c ];
[1,a ; 1,x::b ; 0,c]; [ 1,a ; 2,b ; 0,c ]]
Again, in the prolog language (for which these questions are
originally intended), this function is a whole lot simpler. *)
let prepend p list =
let emit l acc = l :: acc in
let rec aux emit acc = function
 [] > emit [] acc
 (n, l) as h :: t >
let acc = if n > 0 then emit ((n  1, p :: l) :: t) acc
else acc in
aux (fun l acc > emit (h :: l) acc) acc t
in
aux emit [] list
in
let rec aux = function
 [] > [initial]
 h :: t > List.concat (List.map (prepend h) (aux t))
in
let all = aux list in
(* Don't forget to eliminate all group sets that have nonfull
groups *)
let complete = List.filter (List.for_all (fun (x, _) > x = 0)) all in
List.map (List.map snd) complete;;
val group : 'a list > int list > 'a list list list = <fun>
Sorting a list of lists according to length of sublists
Sorting a list of lists according to length of sublists.

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.

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.
# length_sort [["a"; "b"; "c"]; ["d"; "e"]; ["f"; "g"; "h"]; ["d"; "e"];
["i"; "j"; "k"; "l"]; ["m"; "n"]; ["o"]];;
 : string list list =
[["o"]; ["d"; "e"]; ["d"; "e"]; ["m"; "n"]; ["a"; "b"; "c"]; ["f"; "g"; "h"];
["i"; "j"; "k"; "l"]]
# frequency_sort [["a"; "b"; "c"]; ["d"; "e"]; ["f"; "g"; "h"]; ["d"; "e"];
["i"; "j"; "k"; "l"]; ["m"; "n"]; ["o"]];;
 : string list list =
[["i"; "j"; "k"; "l"]; ["o"]; ["a"; "b"; "c"]; ["f"; "g"; "h"]; ["d"; "e"];
["d"; "e"]; ["m"; "n"]]
(* We might not be allowed to use builtin List.sort, so here's an
eightline implementation of insertion sort — O(n²) time
complexity. *)
let rec insert cmp e = function
 [] > [e]
 h :: t as l > if cmp e h <= 0 then e :: l else h :: insert cmp e t
let rec sort cmp = function
 [] > []
 h :: t > insert cmp h (sort cmp t)
(* Sorting according to length : prepend length, sort, remove length *)
let length_sort lists =
let lists = List.map (fun list > List.length list, list) lists in
let lists = sort (fun a b > compare (fst a) (fst b)) lists in
List.map snd lists
;;
(* Sorting according to length frequency : prepend frequency, sort,
remove frequency. Frequencies are extracted by sorting lengths
and applying RLE to count occurrences of each length (see problem
"Runlength encoding of a list.") *)
let rle list =
let rec aux count acc = function
 [] > [] (* Can only be reached if original list is empty *)
 [x] > (x, count + 1) :: acc
 a :: (b :: _ as t) >
if a = b then aux (count + 1) acc t
else aux 0 ((a, count + 1) :: acc) t in
aux 0 [] list
let frequency_sort lists =
let lengths = List.map List.length lists in
let freq = rle (sort compare lengths) in
let by_freq =
List.map (fun list > List.assoc (List.length list) freq , list) lists in
let sorted = sort (fun a b > compare (fst a) (fst b)) by_freq in
List.map snd sorted
Determine whether a given integer number is prime
Determine whether a given integer number is prime.
# not (is_prime 1);;
 : bool = true
# is_prime 7;;
 : bool = true
# not (is_prime 12);;
 : bool = true
Recall that d
divides n
iff n mod d = 0
. This is a naive
solution. See the Sieve of
Eratosthenes for a
more clever one.
# let is_prime n =
let n = abs n in
let rec is_not_divisor d =
d * d > n  (n mod d <> 0 && is_not_divisor (d + 1)) in
n <> 1 && is_not_divisor 2;;
val is_prime : int > bool = <fun>
Determine the greatest common divisor of two positive integer numbers
Determine the greatest common divisor of two positive integer numbers.
Use Euclid's algorithm.
# gcd 13 27;;
 : int = 1
# gcd 20536 7826;;
 : int = 2
# let rec gcd a b =
if b = 0 then a else gcd b (a mod b);;
val gcd : int > int > int = <fun>
Determine whether two positive integer numbers are coprime
Determine whether two positive integer numbers are coprime.
Two numbers are coprime if their greatest common divisor equals 1.
# coprime 13 27;;
 : bool = true
# not (coprime 20536 7826);;
 : bool = true
# (* [gcd] is defined in the previous question *)
let coprime a b = gcd a b = 1;;
val coprime : int > int > bool = <fun>
Calculate Euler's totient function φ(m)
Euler's socalled totient function φ(m) is defined as the number of positive integers r (1 ≤ r < m) that are coprime to m. We let φ(1) = 1.
Find out what the value of φ(m) is if m is a prime number. Euler's totient function plays an important role in one of the most widely used public key cryptography methods (RSA). In this exercise you should use the most primitive method to calculate this function (there are smarter ways that we shall discuss later).
# phi 10;;
 : int = 4
# (* [coprime] is defined in the previous question *)
let phi n =
let rec count_coprime acc d =
if d < n then
count_coprime (if coprime n d then acc + 1 else acc) (d + 1)
else acc
in
if n = 1 then 1 else count_coprime 0 1;;
val phi : int > int = <fun>
Determine the prime factors of a given positive integer
Construct a flat list containing the prime factors in ascending order.
# factors 315;;
 : int list = [3; 3; 5; 7]
# (* Recall that d divides n iff [n mod d = 0] *)
let factors n =
let rec aux d n =
if n = 1 then [] else
if n mod d = 0 then d :: aux d (n / d) else aux (d + 1) n
in
aux 2 n;;
val factors : int > int list = <fun>
Determine the prime factors of a given positive integer (2)
Construct a list containing the prime factors and their multiplicity. Hint: The problem is similar to problem Runlength encoding of a list (direct solution).
# factors 315;;
 : (int * int) list = [(3, 2); (5, 1); (7, 1)]
# let factors n =
let rec aux d n =
if n = 1 then [] else
if n mod d = 0 then
match aux d (n / d) with
 (h, n) :: t when h = d > (h, n + 1) :: t
 l > (d, 1) :: l
else aux (d + 1) n
in
aux 2 n;;
val factors : int > (int * int) list = <fun>
Calculate Euler's totient function φ(m) (improved)
See problem "Calculate Euler's totient function φ(m)" for
the definition of Euler's totient function. If the list of the prime
factors of a number m is known in the form of the previous problem then
the function phi(m) can be efficiently calculated as follows: Let
[(p1, m1); (p2, m2); (p3, m3); ...]
be the list of prime factors
(and their multiplicities) of a given number m. Then φ(m) can be
calculated with the following formula:
φ(m) = (p1  1) × p1^{m1  1} × (p2  1) × p2^{m2  1} × (p3  1) × p3^{m3  1} × ⋯
# phi_improved 10;;
 : int = 4
# phi_improved 13;;
 : int = 12
(* Naive power function. *)
let rec pow n p = if p < 1 then 1 else n * pow n (p  1)
(* [factors] is defined in the previous question. *)
let phi_improved n =
let rec aux acc = function
 [] > acc
 (p, m) :: t > aux ((p  1) * pow p (m  1) * acc) t
in
aux 1 (factors n)
Compare the two methods of calculating Euler's totient function
Use the solutions of problems "[Calculate Euler's totient function φ(m)][totient]" and "Calculate Euler's totient function φ(m) (improved)" to compare the algorithms. Take the number of logical inferences as a measure for efficiency. Try to calculate φ(10090) as an example.
timeit phi 10090
# (* Naive [timeit] function. It requires the [Unix] module to be loaded. *)
let timeit f a =
let t0 = Unix.gettimeofday() in
ignore (f a);
let t1 = Unix.gettimeofday() in
t1 . t0;;
val timeit : ('a > 'b) > 'a > float = <fun>
A list of prime numbers
Given a range of integers by its lower and upper limit, construct a list of all prime numbers in that range.
# List.length (all_primes 2 7920);;
 : int = 1000
# let is_prime n =
let n = max n (n) in
let rec is_not_divisor d =
d * d > n  (n mod d <> 0 && is_not_divisor (d + 1))
in
is_not_divisor 2
let rec all_primes a b =
if a > b then [] else
let rest = all_primes (a + 1) b in
if is_prime a then a :: rest else rest;;
val is_prime : int > bool = <fun>
val all_primes : int > int > int list = <fun>
Goldbach's conjecture
Goldbach's conjecture says that every positive even number greater than 2 is the sum of two prime numbers. Example: 28 = 5 + 23. It is one of the most famous facts in number theory that has not been proved to be correct in the general case. It has been numerically confirmed up to very large numbers. Write a function to find the two prime numbers that sum up to a given even integer.
# goldbach 28;;
 : int * int = (5, 23)
# (* [is_prime] is defined in the previous solution *)
let goldbach n =
let rec aux d =
if is_prime d && is_prime (n  d) then (d, n  d)
else aux (d + 1)
in
aux 2;;
val goldbach : int > int * int = <fun>
A list of Goldbach compositions
Given a range of integers by its lower and upper limit, print a list of all even numbers and their Goldbach composition.
In most cases, if an even number is written as the sum of two prime numbers, one of them is very small. Very rarely, the primes are both bigger than say 50. Try to find out how many such cases there are in the range 2..3000.
# goldbach_list 9 20;;
 : (int * (int * int)) list =
[(10, (3, 7)); (12, (5, 7)); (14, (3, 11)); (16, (3, 13)); (18, (5, 13));
(20, (3, 17))]
# (* [goldbach] is defined in the previous question. *)
let rec goldbach_list a b =
if a > b then [] else
if a mod 2 = 1 then goldbach_list (a + 1) b
else (a, goldbach a) :: goldbach_list (a + 2) b
let goldbach_limit a b lim =
List.filter (fun (_, (a, b)) > a > lim && b > lim) (goldbach_list a b);;
val goldbach_list : int > int > (int * (int * int)) list = <fun>
val goldbach_limit : int > int > int > (int * (int * int)) list = <fun>
Truth tables for logical expressions (2 variables)
Let us define a small "language" for boolean expressions containing variables:
# type bool_expr =
 Var of string
 Not of bool_expr
 And of bool_expr * bool_expr
 Or of bool_expr * bool_expr;;
type bool_expr =
Var of string
 Not of bool_expr
 And of bool_expr * bool_expr
 Or of bool_expr * bool_expr
A logical expression in two variables can then be written in prefix
notation. For example, (a ∨ b) ∧ (a ∧ b)
is written:
# And (Or (Var "a", Var "b"), And (Var "a", Var "b"));;
 : bool_expr = And (Or (Var "a", Var "b"), And (Var "a", Var "b"))
Define a function, table2
which returns the truth table of a given
logical expression in two variables (specified as arguments). The return
value must be a list of triples containing
(value_of_a, value_of_b, value_of_expr)
.
# table2 "a" "b" (And (Var "a", Or (Var "a", Var "b")));;
 : (bool * bool * bool) list =
[(true, true, true); (true, false, true); (false, true, false);
(false, false, false)]
# let rec eval2 a val_a b val_b = function
 Var x > if x = a then val_a
else if x = b then val_b
else failwith "The expression contains an invalid variable"
 Not e > not (eval2 a val_a b val_b e)
 And(e1, e2) > eval2 a val_a b val_b e1 && eval2 a val_a b val_b e2
 Or(e1, e2) > eval2 a val_a b val_b e1  eval2 a val_a b val_b e2
let table2 a b expr =
[(true, true, eval2 a true b true expr);
(true, false, eval2 a true b false expr);
(false, true, eval2 a false b true expr);
(false, false, eval2 a false b false expr)];;
val eval2 : string > bool > string > bool > bool_expr > bool = <fun>
val table2 : string > string > bool_expr > (bool * bool * bool) list =
<fun>
Truth tables for logical expressions
Generalize the previous problem in such a way that the logical
expression may contain any number of logical variables. Define table
in a way that table variables expr
returns the truth table for the
expression expr
, which contains the logical variables enumerated in
variables
.
# table ["a"; "b"] (And (Var "a", Or (Var "a", Var "b")));;
 : ((string * bool) list * bool) list =
[([("a", true); ("b", true)], true); ([("a", true); ("b", false)], true);
([("a", false); ("b", true)], false); ([("a", false); ("b", false)], false)]
# (* [val_vars] is an associative list containing the truth value of
each variable. For efficiency, a Map or a Hashtlb should be
preferred. *)
let rec eval val_vars = function
 Var x > List.assoc x val_vars
 Not e > not (eval val_vars e)
 And(e1, e2) > eval val_vars e1 && eval val_vars e2
 Or(e1, e2) > eval val_vars e1  eval val_vars e2
(* Again, this is an easy and short implementation rather than an
efficient one. *)
let rec table_make val_vars vars expr =
match vars with
 [] > [(List.rev val_vars, eval val_vars expr)]
 v :: tl >
table_make ((v, true) :: val_vars) tl expr
@ table_make ((v, false) :: val_vars) tl expr
let table vars expr = table_make [] vars expr;;
val eval : (string * bool) list > bool_expr > bool = <fun>
val table_make :
(string * bool) list >
string list > bool_expr > ((string * bool) list * bool) list = <fun>
val table : string list > bool_expr > ((string * bool) list * bool) list =
<fun>
Gray code
An nbit Gray code is a sequence of nbit strings constructed according to certain rules. For example,
n = 1: C(1) = ['0', '1'].
n = 2: C(2) = ['00', '01', '11', '10'].
n = 3: C(3) = ['000', '001', '011', '010', '110', '111', '101', '100'].
Find out the construction rules and write a function with the following
specification: gray n
returns the n
bit Gray code.
# gray 1;;
 : string list = ["0"; "1"]
# gray 2;;
 : string list = ["00"; "01"; "11"; "10"]
# gray 3;;
 : string list = ["000"; "001"; "011"; "010"; "110"; "111"; "101"; "100"]
# let gray n =
let rec gray_next_level k l =
if k < n then
(* This is the core part of the Gray code construction.
* first_half is reversed and has a "0" attached to every element.
* Second part is reversed (it must be reversed for correct gray code).
* Every element has "1" attached to the front.*)
let (first_half,second_half) =
List.fold_left (fun (acc1,acc2) x >
(("0" ^ x) :: acc1, ("1" ^ x) :: acc2)) ([], []) l
in
(* List.rev_append turns first_half around and attaches it to second_half.
* The result is the modified first_half in correct order attached to
* the second_half modified in reversed order.*)
gray_next_level (k + 1) (List.rev_append first_half second_half)
else l
in
gray_next_level 1 ["0"; "1"];;
val gray : int > string list = <fun>
Huffman code
First of all, consult a good book on discrete mathematics or algorithms for a detailed description of Huffman codes (you can start with the Wikipedia page)!
We consider a set of symbols with their frequencies.
For example, if the alphabet is "a"
,..., "f"
(represented as the positions 0,...5) and
respective frequencies are 45, 13, 12, 16, 9, 5:
# let fs = [("a", 45); ("b", 13); ("c", 12); ("d", 16);
("e", 9); ("f", 5)];;
val fs : (string * int) list =
[("a", 45); ("b", 13); ("c", 12); ("d", 16); ("e", 9); ("f", 5)]
Our objective is to construct the
Huffman code c
word for all symbols s
. In our example, the result could
be
hs = [("a", "0"); ("b", "101"); ("c", "100"); ("d", "111"); ("e", "1101"); ("f", "1100")]
(or hs = [("a", "1");...]
). The task shall be performed by the function
huffman
defined as follows: huffman(fs)
returns the Huffman code
table for the frequency table fs
# huffman fs;;
 : (string * string) list =
[("a", "0"); ("c", "100"); ("b", "101"); ("f", "1100"); ("e", "1101");
("d", "111")]
# (* Simple priority queue where the priorities are integers 0..100.
The node with the lowest probability comes first. *)
module Pq = struct
type 'a t = {data: 'a list array; mutable first: int}
let make() = {data = Array.make 101 []; first = 101}
let add q p x =
q.data.(p) < x :: q.data.(p); q.first < min p q.first
let get_min q =
if q.first = 101 then None else
match q.data.(q.first) with
 [] > assert false
 x :: tl >
let p = q.first in
q.data.(q.first) < tl;
while q.first < 101 && q.data.(q.first) = [] do
q.first < q.first + 1
done;
Some(p, x)
end
type tree =
 Leaf of string
 Node of tree * tree
let rec huffman_tree q =
match Pq.get_min q, Pq.get_min q with
 Some(p1, t1), Some(p2, t2) > Pq.add q (p1 + p2) (Node(t1, t2));
huffman_tree q
 Some(_, t), None  None, Some(_, t) > t
 None, None > assert false
(* Build the prefixfree binary code from the tree *)
let rec prefixes_of_tree prefix = function
 Leaf s > [(s, prefix)]
 Node(t0, t1) > prefixes_of_tree (prefix ^ "0") t0
@ prefixes_of_tree (prefix ^ "1") t1
let huffman fs =
if List.fold_left (fun s (_, p) > s + p) 0 fs <> 100 then
failwith "huffman: sum of weights must be 100";
let q = Pq.make () in
List.iter (fun (s, f) > Pq.add q f (Leaf s)) fs;
prefixes_of_tree "" (huffman_tree q);;
module Pq :
sig
type 'a t = { data : 'a list array; mutable first : int; }
val make : unit > 'a t
val add : 'a t > int > 'a > unit
val get_min : 'a t > (int * 'a) option
end
type tree = Leaf of string  Node of tree * tree
val huffman_tree : tree Pq.t > tree = <fun>
val prefixes_of_tree : string > tree > (string * string) list = <fun>
val huffman : (string * int) list > (string * string) list = <fun>
Construct completely balanced binary trees
A binary tree is either empty or it is composed of a root element and two successors, which are binary trees themselves.
In OCaml, one can define a new type binary_tree
that carries an
arbitrary value of type 'a
(thus is polymorphic) at each node.
# type 'a binary_tree =
 Empty
 Node of 'a * 'a binary_tree * 'a binary_tree;;
type 'a binary_tree = Empty  Node of 'a * 'a binary_tree * 'a binary_tree
An example of tree carrying char
data is:
# let example_tree =
Node ('a', Node ('b', Node ('d', Empty, Empty), Node ('e', Empty, Empty)),
Node ('c', Empty, Node ('f', Node ('g', Empty, Empty), Empty)));;
val example_tree : char binary_tree =
Node ('a', Node ('b', Node ('d', Empty, Empty), Node ('e', Empty, Empty)),
Node ('c', Empty, Node ('f', Node ('g', Empty, Empty), Empty)))
In OCaml, the strict type discipline guarantees that, if you get a
value of type binary_tree
, then it must have been created with the two
constructors Empty
and Node
.
In a completely balanced binary tree, the following property holds for every node: The number of nodes in its left subtree and the number of nodes in its right subtree are almost equal, which means their difference is not greater than one.
Write a function cbal_tree
to construct completely balanced binary
trees for a given number of nodes. The function should generate all
solutions via backtracking. Put the letter 'x'
as information into all
nodes of the tree.
# cbal_tree 4;;
 : char binary_tree/2 list =
[Node ('x', Node ('x', Empty, Empty),
Node ('x', Node ('x', Empty, Empty), Empty));
Node ('x', Node ('x', Empty, Empty),
Node ('x', Empty, Node ('x', Empty, Empty)));
Node ('x', Node ('x', Node ('x', Empty, Empty), Empty),
Node ('x', Empty, Empty));
Node ('x', Node ('x', Empty, Node ('x', Empty, Empty)),
Node ('x', Empty, Empty))]
# (* Build all trees with given [left] and [right] subtrees. *)
let add_trees_with left right all =
let add_right_tree all l =
List.fold_left (fun a r > Node ('x', l, r) :: a) all right in
List.fold_left add_right_tree all left
let rec cbal_tree n =
if n = 0 then [Empty]
else if n mod 2 = 1 then
let t = cbal_tree (n / 2) in
add_trees_with t t []
else (* n even: n1 nodes for the left & right subtrees altogether. *)
let t1 = cbal_tree (n / 2  1) in
let t2 = cbal_tree (n / 2) in
add_trees_with t1 t2 (add_trees_with t2 t1 []);;
val add_trees_with :
char binary_tree list >
char binary_tree list > char binary_tree list > char binary_tree list =
<fun>
val cbal_tree : int > char binary_tree list = <fun>
Symmetric binary trees
Let us call a binary tree symmetric if you can draw a vertical line
through the root node and then the right subtree is the mirror image of
the left subtree. Write a function is_symmetric
to check whether a
given binary tree is symmetric.
Hint: Write a function
is_mirror
first to check whether one tree is the mirror image of another. We are only interested in the structure, not in the contents of the nodes.
# let rec is_mirror t1 t2 =
match t1, t2 with
 Empty, Empty > true
 Node(_, l1, r1), Node(_, l2, r2) >
is_mirror l1 r2 && is_mirror r1 l2
 _ > false
let is_symmetric = function
 Empty > true
 Node(_, l, r) > is_mirror l r;;
val is_mirror : 'a binary_tree > 'b binary_tree > bool = <fun>
val is_symmetric : 'a binary_tree > bool = <fun>
Binary search trees (dictionaries)
Construct a binary search tree from a list of integer numbers.
# construct [3; 2; 5; 7; 1];;
 : int binary_tree =
Node (3, Node (2, Node (1, Empty, Empty), Empty),
Node (5, Empty, Node (7, Empty, Empty)))
Then use this function to test the solution of the previous problem.
# is_symmetric (construct [5; 3; 18; 1; 4; 12; 21]);;
 : bool = true
# not (is_symmetric (construct [3; 2; 5; 7; 4]));;
 : bool = true
# let rec insert tree x = match tree with
 Empty > Node (x, Empty, Empty)
 Node (y, l, r) >
if x = y then tree
else if x < y then Node (y, insert l x, r)
else Node (y, l, insert r x)
let construct l = List.fold_left insert Empty l;;
val insert : 'a binary_tree > 'a > 'a binary_tree = <fun>
val construct : 'a list > 'a binary_tree = <fun>
Generateandtest paradigm
Apply the generateandtest paradigm to construct all symmetric, completely balanced binary trees with a given number of nodes.
# sym_cbal_trees 5;;
 : char binary_tree list =
[Node ('x', Node ('x', Node ('x', Empty, Empty), Empty),
Node ('x', Empty, Node ('x', Empty, Empty)));
Node ('x', Node ('x', Empty, Node ('x', Empty, Empty)),
Node ('x', Node ('x', Empty, Empty), Empty))]
How many such trees are there with 57 nodes? Investigate about how many solutions there are for a given number of nodes? What if the number is even? Write an appropriate function.
# List.length (sym_cbal_trees 57);;
 : int = 256
# let sym_cbal_trees n =
List.filter is_symmetric (cbal_tree n);;
val sym_cbal_trees : int > char binary_tree list = <fun>
Construct heightbalanced binary trees
In a heightbalanced binary tree, the following property holds for every node: The height of its left subtree and the height of its right subtree are almost equal, which means their difference is not greater than one.
Write a function hbal_tree
to construct heightbalanced binary trees
for a given height. The function should generate all solutions via
backtracking. Put the letter 'x'
as information into all nodes of the
tree.
# let t = hbal_tree 3;;
val t : char binary_tree list =
[Node ('x', Node ('x', Empty, Node ('x', Empty, Empty)),
Node ('x', Empty, Node ('x', Empty, Empty)));
Node ('x', Node ('x', Empty, Node ('x', Empty, Empty)),
Node ('x', Node ('x', Empty, Empty), Empty));
Node ('x', Node ('x', Empty, Node ('x', Empty, Empty)),
Node ('x', Node ('x', Empty, Empty), Node ('x', Empty, Empty)));
Node ('x', Node ('x', Node ('x', Empty, Empty), Empty),
Node ('x', Empty, Node ('x', Empty, Empty)));
Node ('x', Node ('x', Node ('x', Empty, Empty), Empty),
Node ('x', Node ('x', Empty, Empty), Empty));
Node ('x', Node ('x', Node ('x', Empty, Empty), Empty),
Node ('x', Node ('x', Empty, Empty), Node ('x', Empty, Empty)));
Node ('x', Node ('x', Node ('x', Empty, Empty), Node ('x', Empty, Empty)),
Node ('x', Empty, Node ('x', Empty, Empty)));
Node ('x', Node ('x', Node ('x', Empty, Empty), Node ('x', Empty, Empty)),
Node ('x', Node ('x', Empty, Empty), Empty));
Node ('x', Node ('x', Node ('x', Empty, Empty), Node ('x', Empty, Empty)),
Node ('x', Node ('x', Empty, Empty), Node ('x', Empty, Empty)));
Node ('x', Node ('x', Empty, Node ('x', Empty, Empty)),
Node ('x', Empty, Empty));
Node ('x', Node ('x', Node ('x', Empty, Empty), Empty),
Node ('x', Empty, Empty));
Node ('x', Node ('x', Node ('x', Empty, Empty), Node ('x', Empty, Empty)),
Node ('x', Empty, Empty));
Node ('x', Node ('x', Empty, Empty),
Node ('x', Empty, Node ('x', Empty, Empty)));
Node ('x', Node ('x', Empty, Empty),
Node ('x', Node ('x', Empty, Empty), Empty));
Node ('x', Node ('x', Empty, Empty),
Node ('x', Node ('x', Empty, Empty), Node ('x', Empty, Empty)))]
The function add_trees_with
is defined in the solution of
Construct completely balanced binary trees.
# let rec hbal_tree n =
if n = 0 then [Empty]
else if n = 1 then [Node ('x', Empty, Empty)]
else
(* [add_trees_with left right trees] is defined in a question above. *)
let t1 = hbal_tree (n  1)
and t2 = hbal_tree (n  2) in
add_trees_with t1 t1 (add_trees_with t1 t2 (add_trees_with t2 t1 []));;
val hbal_tree : int > char binary_tree list = <fun>
Construct heightbalanced binary trees with a given number of nodes
Consider a heightbalanced binary tree of height h
. What is the
maximum number of nodes it can contain? Clearly,
max_nodes = 2^{h}  1.
# let max_nodes h = 1 lsl h  1;;
val max_nodes : int > int = <fun>
Minimum of nodes
However, what is the minimum number min_nodes? This question is more
difficult. Try to find a recursive statement and turn it into a function
min_nodes
defined as follows: min_nodes h
returns the minimum number
of nodes in a heightbalanced binary tree of height h
.
Minimum height
On the other hand, we might ask: what are the minimum (resp. maximum)
height H a
heightbalanced binary tree with N nodes can have?
min_height
(resp. max_height n
) returns
the minimum (resp. maximum) height of a heightbalanced binary tree
with n
nodes.
Constructing trees
Now, we can attack the main problem: construct all the heightbalanced
binary trees with a given number of nodes. hbal_tree_nodes n
returns a
list of all heightbalanced binary tree with n
nodes.
Find out how many heightbalanced trees exist for n = 15
.
# List.length (hbal_tree_nodes 15);;
 : int = 1553
Minimum of nodes
The following solution comes directly from translating the question.
# let rec min_nodes h =
if h <= 0 then 0
else if h = 1 then 1
else min_nodes (h  1) + min_nodes (h  2) + 1;;
val min_nodes : int > int = <fun>
It is not the more efficient one however. One should use the last two values as the state to avoid the double recursion.
# let rec min_nodes_loop m0 m1 h =
if h <= 1 then m1
else min_nodes_loop m1 (m1 + m0 + 1) (h  1)
let min_nodes h =
if h <= 0 then 0 else min_nodes_loop 0 1 h;;
val min_nodes_loop : int > int > int > int = <fun>
val min_nodes : int > int = <fun>
It is not difficult to show that min_nodes h
= F_{h+2}  1,
where (F_{n}) is the
Fibonacci sequence.
Minimum height
Inverting the formula max_nodes = 2^{h}  1, one directly find that Hₘᵢₙ(n) = ⌈log₂(n+1)⌉ which is readily implemented:
# let min_height n = int_of_float (ceil (log (float(n + 1)) /. log 2.));;
val min_height : int > int = <fun>
Let us give a proof that the formula for Hₘᵢₙ is valid. First, if h
= min_height
n, there exists a heightbalanced tree of height h
with n nodes. Thus 2ʰ  1 = max_nodes h
≥ n i.e., h ≥ log₂(n+1).
To establish equality for Hₘᵢₙ(n), one has to show that, for any n,
there exists a heightbalanced tree with height Hₘᵢₙ(n). This is
due to the relation Hₘᵢₙ(n) = 1 + Hₘᵢₙ(n/2) where n/2 is the integer
division. For n odd, this is readily proved — so one can build a
tree with a top node and two subtrees with n/2 nodes of height
Hₘᵢₙ(n)  1. For n even, the same proof works if one first remarks
that, in that case, ⌈log₂(n+2)⌉ = ⌈log₂(n+1)⌉ — use log₂(n+1) ≤ h ∈
ℕ ⇔ 2ʰ ≥ n + 1 and the fact that 2ʰ is even for that. This allows
to have a subtree with n/2 nodes. For the other subtree with
n/21 nodes, one has to establish that Hₘᵢₙ(n/21) ≥ Hₘᵢₙ(n)  2
which is easy because, if h = Hₘᵢₙ(n/21), then h+2 ≥ log₂(2n) ≥
log₂(n+1).
The above function is not the best one however. Indeed, not every 64 bits integer can be represented exactly as a floating point number. Here is one that only uses integer operations:
# let rec ceil_log2_loop log plus1 n =
if n = 1 then if plus1 then log + 1 else log
else ceil_log2_loop (log + 1) (plus1  n land 1 <> 0) (n / 2)
let ceil_log2 n = ceil_log2_loop 0 false n;;
val ceil_log2_loop : int > bool > int > int = <fun>
val ceil_log2 : int > int = <fun>
This algorithm is still not the fastest however. See for example the Hacker's Delight, section 53 (and 114).
Following the same idea as above, if h = max_height
n, then one
easily deduces that min_nodes
h ≤ n < min_nodes
(h+1). This
yields the following code:
# let rec max_height_search h n =
if min_nodes h <= n then max_height_search (h + 1) n else h  1
let max_height n = max_height_search 0 n;;
val max_height_search : int > int > int = <fun>
val max_height : int > int = <fun>
Of course, since min_nodes
is computed recursively, there is no
need to recompute everything to go from min_nodes h
to
min_nodes(h+1)
:
# let rec max_height_search h m_h m_h1 n =
if m_h <= n then max_height_search (h + 1) m_h1 (m_h1 + m_h + 1) n else h  1
let max_height n = max_height_search 0 0 1 n;;
val max_height_search : int > int > int > int > int = <fun>
val max_height : int > int = <fun>
Constructing trees
First, we define some convenience functions fold_range
that folds
a function f
on the range n0
...n1
i.e., it computes
f (... f (f (f init n0) (n0+1)) (n0+2) ...) n1
. You can think it
as performing the assignment init ← f init n
for n = n0,..., n1
except that there is no mutable variable in the code.
# let rec fold_range ~f ~init n0 n1 =
if n0 > n1 then init else fold_range ~f ~init:(f init n0) (n0 + 1) n1;;
val fold_range : f:('a > int > 'a) > init:'a > int > int > 'a = <fun>
When constructing trees, there is an obvious symmetry: if one swaps
the left and right subtrees of a balanced tree, we still have a
balanced tree. The following function returns all trees in trees
together with their permutation.
# let rec add_swap_left_right trees =
List.fold_left (fun a n > match n with
 Node (v, t1, t2) > Node (v, t2, t1) :: a
 Empty > a) trees trees;;
val add_swap_left_right : 'a binary_tree list > 'a binary_tree list = <fun>
Finally we generate all trees recursively, using a priori the bounds computed above. It could be further optimized but our aim is to straightforwardly express the idea.
# let rec hbal_tree_nodes_height h n =
assert(min_nodes h <= n && n <= max_nodes h);
if h = 0 then [Empty]
else
let acc = add_hbal_tree_node [] (h  1) (h  2) n in
let acc = add_swap_left_right acc in
add_hbal_tree_node acc (h  1) (h  1) n
and add_hbal_tree_node l h1 h2 n =
let min_n1 = max (min_nodes h1) (n  1  max_nodes h2) in
let max_n1 = min (max_nodes h1) (n  1  min_nodes h2) in
fold_range min_n1 max_n1 ~init:l ~f:(fun l n1 >
let t1 = hbal_tree_nodes_height h1 n1 in
let t2 = hbal_tree_nodes_height h2 (n  1  n1) in
List.fold_left (fun l t1 >
List.fold_left (fun l t2 > Node ('x', t1, t2) :: l) l t2) l t1
)
let hbal_tree_nodes n =
fold_range (min_height n) (max_height n) ~init:[] ~f:(fun l h >
List.rev_append (hbal_tree_nodes_height h n) l);;
val hbal_tree_nodes_height : int > int > char binary_tree list = <fun>
val add_hbal_tree_node :
char binary_tree list > int > int > int > char binary_tree list = <fun>
val hbal_tree_nodes : int > char binary_tree list = <fun>
Collect the leaves of a binary tree in a list
A leaf is a node with no successors. Write a function leaves
to
collect them in a list.
# leaves Empty;;
 : 'a list = []
# (* Having an accumulator acc prevents using inefficient List.append.
* Every Leaf will be pushed directly into accumulator.
* Not tailrecursive, but that is no problem since we have a binary tree and
* and stack depth is logarithmic. *)
let leaves t =
let rec leaves_aux t acc = match t with
 Empty > acc
 Node (x, Empty, Empty) > x :: acc
 Node (x, l, r) > leaves_aux l (leaves_aux r acc)
in
leaves_aux t [];;
val leaves : 'a binary_tree > 'a list = <fun>
Count the leaves of a binary tree
A leaf is a node with no successors. Write a function count_leaves
to
count them.
# count_leaves Empty;;
 : int = 0
# let rec count_leaves = function
 Empty > 0
 Node (_, Empty, Empty) > 1
 Node (_, l, r) > count_leaves l + count_leaves r;;
val count_leaves : 'a binary_tree > int = <fun>
Collect the nodes at a given level in a list
A node of a binary tree is at level N if the path from the root to the
node has length N1. The root node is at level 1. Write a function
at_level t l
to collect all nodes of the tree t
at level l
in a
list.
# let example_tree =
Node ('a', Node ('b', Node ('d', Empty, Empty), Node ('e', Empty, Empty)),
Node ('c', Empty, Node ('f', Node ('g', Empty, Empty), Empty)));;
val example_tree : char binary_tree =
Node ('a', Node ('b', Node ('d', Empty, Empty), Node ('e', Empty, Empty)),
Node ('c', Empty, Node ('f', Node ('g', Empty, Empty), Empty)))
# at_level example_tree 2;;
 : char list = ['b'; 'c']
Using at_level
it is easy to construct a function levelorder
which
creates the levelorder sequence of the nodes. However, there are more
efficient ways to do that.
# (* Having an accumulator acc prevents using inefficient List.append.
* Every node at level N will be pushed directly into accumulator.
* Not tailrecursive, but that is no problem since we have a binary tree and
* and stack depth is logarithmic. *)
let at_level t level =
let rec at_level_aux t acc counter = match t with
 Empty > acc
 Node (x, l, r) >
if counter=level then
x :: acc
else
at_level_aux l (at_level_aux r acc (counter + 1)) (counter + 1)
in
at_level_aux t [] 1;;
val at_level : 'a binary_tree > int > 'a list = <fun>
Collect the internal nodes of a binary tree in a list
An internal node of a binary tree has either one or two nonempty
successors. Write a function internals
to collect them in a list.
# internals (Node ('a', Empty, Empty));;
 : char list = []
# (* Having an accumulator acc prevents using inefficient List.append.
* Every internal node will be pushed directly into accumulator.
* Not tailrecursive, but that is no problem since we have a binary tree and
* and stack depth is logarithmic. *)
let internals t =
let rec internals_aux t acc = match t with
 Empty > acc
 Node (x, Empty, Empty) > acc
 Node (x, l, r) > internals_aux l (x :: internals_aux r acc)
in
internals_aux t [];;
val internals : 'a binary_tree > 'a list = <fun>
Construct a complete binary tree
A complete binary tree with height H is defined as follows: The levels 1,2,3,...,H1 contain the maximum number of nodes (i.e 2^{i1} at the level i, note that we start counting the levels from 1 at the root). In level H, which may contain less than the maximum possible number of nodes, all the nodes are "leftadjusted". This means that in a levelorder tree traversal all internal nodes come first, the leaves come second, and empty successors (the nil's which are not really nodes!) come last.
Particularly, complete binary trees are used as data structures (or addressing schemes) for heaps.
We can assign an address number to each node in a complete binary tree
by enumerating the nodes in levelorder, starting at the root with
number 1. In doing so, we realize that for every node X with address A
the following property holds: The address of X's left and right
successors are 2*A and 2*A+1, respectively, supposed the successors do
exist. This fact can be used to elegantly construct a complete binary
tree structure. Write a function is_complete_binary_tree
with the
following specification: is_complete_binary_tree n t
returns true
iff t
is a complete binary tree with n
nodes.
# complete_binary_tree [1; 2; 3; 4; 5; 6];;
 : int binary_tree =
Node (1, Node (2, Node (4, Empty, Empty), Node (5, Empty, Empty)),
Node (3, Node (6, Empty, Empty), Empty))
# let rec split_n lst acc n = match (n, lst) with
 (0, _) > (List.rev acc, lst)
 (_, []) > (List.rev acc, [])
 (_, h :: t) > split_n t (h :: acc) (n1)
let rec myflatten p c =
match (p, c) with
 (p, []) > List.map (fun x > Node (x, Empty, Empty)) p
 (x :: t, [y]) > Node (x, y, Empty) :: myflatten t []
 (ph :: pt, x :: y :: t) > (Node (ph, x, y)) :: myflatten pt t
 _ > invalid_arg "myflatten"
let complete_binary_tree = function
 [] > Empty
 lst >
let rec aux l = function
 [] > []
 lst > let p, c = split_n lst [] (1 lsl l) in
myflatten p (aux (l + 1) c)
in
List.hd (aux 0 lst);;
val split_n : 'a list > 'a list > int > 'a list * 'a list = <fun>
val myflatten : 'a list > 'a binary_tree list > 'a binary_tree list = <fun>
val complete_binary_tree : 'a list > 'a binary_tree = <fun>
Layout a binary tree (1)
As a preparation for drawing the tree, a layout algorithm is required to determine the position of each node in a rectangular grid. Several layout methods are conceivable, one of them is shown in the illustration.
In this layout strategy, the position of a node v is obtained by the following two rules:
 x(v) is equal to the position of the node v in the inorder sequence;
 y(v) is equal to the depth of the node v in the tree.
In order to store the position of the nodes, we will enrich the value
at each node with the position (x,y)
.
The tree pictured above is
# let example_layout_tree =
let leaf x = Node (x, Empty, Empty) in
Node ('n', Node ('k', Node ('c', leaf 'a',
Node ('h', Node ('g', leaf 'e', Empty), Empty)),
leaf 'm'),
Node ('u', Node ('p', Empty, Node ('s', leaf 'q', Empty)), Empty));;
val example_layout_tree : char binary_tree =
Node ('n',
Node ('k',
Node ('c', Node ('a', Empty, Empty),
Node ('h', Node ('g', Node ('e', Empty, Empty), Empty), Empty)),
Node ('m', Empty, Empty)),
Node ('u', Node ('p', Empty, Node ('s', Node ('q', Empty, Empty), Empty)),
Empty))
# layout_binary_tree_1 example_layout_tree;;
 : (char * int * int) binary_tree =
Node (('n', 8, 1),
Node (('k', 6, 2),
Node (('c', 2, 3), Node (('a', 1, 4), Empty, Empty),
Node (('h', 5, 4),
Node (('g', 4, 5), Node (('e', 3, 6), Empty, Empty), Empty), Empty)),
Node (('m', 7, 3), Empty, Empty)),
Node (('u', 12, 2),
Node (('p', 9, 3), Empty,
Node (('s', 11, 4), Node (('q', 10, 5), Empty, Empty), Empty)),
Empty))
# let layout_binary_tree_1 t =
let rec layout depth x_left = function
(* This function returns a pair: the laid out tree and the first
* free x location *)
 Empty > (Empty, x_left)
 Node (v,l,r) >
let (l', l_x_max) = layout (depth + 1) x_left l in
let (r', r_x_max) = layout (depth + 1) (l_x_max + 1) r in
(Node ((v, l_x_max, depth), l', r'), r_x_max)
in
fst (layout 1 1 t);;
val layout_binary_tree_1 : 'a binary_tree > ('a * int * int) binary_tree =
<fun>
Layout a binary tree (2)
An alternative layout method is depicted in this illustration. Find out the rules and write the corresponding OCaml function.
Hint: On a given level, the horizontal distance between neighbouring nodes is constant.
The tree shown is
# let example_layout_tree =
let leaf x = Node (x, Empty, Empty) in
Node ('n', Node ('k', Node ('c', leaf 'a',
Node ('e', leaf 'd', leaf 'g')),
leaf 'm'),
Node ('u'