Solutions can be found here.
A list is either empty or it is composed of a first element
(head) and a tail, which is a list itself.
In Prolog we represent the empty list by the atom [] and a nonempty list
by a term [HT] where H denotes the head and T denotes the tail.
 1.01
(*) Find the last element of a list.
 Example:
? my_last(X,[a,b,c,d]).
X = d
 1.02
(*) Find the last but one element of a list.
 (de: zweitletztes Element, fr: avantdernier élément)
 1.03
(*) Find the K'th element of a list.
 The first element in the list is number 1.
Example:
? element_at(X,[a,b,c,d,e],3).
X = c
 1.04
(*) Find the number of elements of a list.
 1.05
(*) Reverse a list.
 1.06
(*) Find out whether a list is a palindrome.
 A palindrome can be read forward or backward; e.g. [x,a,m,a,x].
 1.07
(**) Flatten a nested list structure.
 Transform a list, possibly holding lists as elements into a 'flat'
list by replacing each list with its elements (recursively).
Example:
? my_flatten([a, [b, [c, d], e]], X).
X = [a, b, c, d, e]
Hint: Use the predefined predicates is_list/1 and append/3
 1.08
(**) Eliminate consecutive duplicates of list elements.
 If a list contains repeated elements they should be replaced
with a single copy of the element. The order of the elements should
not be changed.
Example:
? compress([a,a,a,a,b,c,c,a,a,d,e,e,e,e],X).
X = [a,b,c,a,d,e]
 1.09
(**) Pack consecutive duplicates of list elements into sublists.
 If a list contains repeated elements they should be placed
in separate sublists.
Example:
? pack([a,a,a,a,b,c,c,a,a,d,e,e,e,e],X).
X = [[a,a,a,a],[b],[c,c],[a,a],[d],[e,e,e,e]]
 1.10
(*) Runlength encoding of a list.
 Use the result of problem 1.09 to implement the socalled
runlength encoding data compression method. Consecutive duplicates
of elements are encoded as terms [N,E] where N is the number
of duplicates of the element E.
Example:
? encode([a,a,a,a,b,c,c,a,a,d,e,e,e,e],X).
X = [[4,a],[1,b],[2,c],[2,a],[1,d][4,e]]
 1.11
(*) Modified runlength encoding.
 Modify the result of problem 1.10 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] terms.
Example:
? encode_modified([a,a,a,a,b,c,c,a,a,d,e,e,e,e],X).
X = [[4,a],b,[2,c],[2,a],d,[4,e]]
 1.12
(**) Decode a runlength encoded list.
 Given a runlength code list generated as specified
in problem 1.11. Construct its uncompressed version.
 1.13
(**) 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 1.09, but only count them.
As in problem 1.11, simplify the result list by replacing the singleton
terms [1,X] by X.
Example:
? encode_direct([a,a,a,a,b,c,c,a,a,d,e,e,e,e],X).
X = [[4,a],b,[2,c],[2,a],d,[4,e]]
 1.14
(*) Duplicate the elements of a list.
 Example:
? dupli([a,b,c,c,d],X).
X = [a,a,b,b,c,c,c,c,d,d]
 1.15
(**) Duplicate the elements of a list a given number of times.
 Example:
? dupli([a,b,c],3,X).
X = [a,a,a,b,b,b,c,c,c]
What are the results of the goal:
? dupli(X,3,Y).
 1.16
(**) Drop every N'th element from a list.
 Example:
? drop([a,b,c,d,e,f,g,h,i,k],3,X).
X = [a,b,d,e,g,h,k]
 1.17
(*) Split a list into two parts; the length of the first
part is given.
 Do not use any predefined predicates.
Example:
? split([a,b,c,d,e,f,g,h,i,k],3,L1,L2).
L1 = [a,b,c]
L2 = [d,e,f,g,h,i,k]
 1.18
(**) 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 1.
Example:
? slice([a,b,c,d,e,f,g,h,i,k],3,7,L).
X = [c,d,e,f,g]
 1.19
(**) Rotate a list N places to the left.
 Examples:
? rotate([a,b,c,d,e,f,g,h],3,X).
X = [d,e,f,g,h,a,b,c]
? rotate([a,b,c,d,e,f,g,h],2,X).
X = [g,h,a,b,c,d,e,f]
Hint: Use the predefined predicates length/2 and append/3,
as well as the result of problem 1.17.
 1.20
(*) Remove the K'th element from a list.
 Example:
? remove_at(X,[a,b,c,d],2,R).
X = b
R = [a,c,d]
 1.21
(*) Insert an element at a given position into a list.
 Example:
? insert_at(alfa,[a,b,c,d],2,L).
L = [a,alfa,b,c,d]
 1.22
(*) Create a list containing all integers within a given range.
 Example:
? range(4,9,L).
L = [4,5,6,7,8,9]
 1.23
(**) Extract a given number of randomly selected elements from a
list.
 The selected items shall be put into a result list.
Example:
? rnd_select([a,b,c,d,e,f,g,h],3,L).
L = [e,d,a]
Hint: Use the builtin random number generator random/2 and the
result of problem 1.20.
 1.24
(*) Lotto: Draw N different random numbers from the set 1..M.
 The selected numbers shall be put into a result list.
Example:
? rnd_select(6,49,L).
L = [23,1,17,33,21,37]
Hint: Combine the solutions of problems 1.22 and 1.23.
 1.25
(*) Generate a random permutation of the elements of a list.
 Example:
? rnd_permu([a,b,c,d,e,f],L).
L = [b,a,d,c,e,f]
Hint: Use the solution of problem 1.23.
 1.26
(**) 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 wellknown binomial coefficients). For pure
mathematicians, this result may be great. But we want to
really generate all the possibilities (via backtracking).
Example:
? combination(3,[a,b,c,d,e,f],L).
L = [a,b,c] ;
L = [a,b,d] ;
L = [a,b,e] ;
...
 1.27
(**) 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 predicate that generates all the
possibilities via backtracking.
Example:
? group3([aldo,beat,carla,david,evi,flip,gary,hugo,ida],G1,G2,G3).
G1 = [aldo,beat], G2 = [carla,david,evi], G3 = [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],Gs).
Gs = [[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".
 1.28
(**) Sorting a list of lists according to length of sublists

a) We suppose that a list (InList) contains elements that are
lists themselves. The objective is to sort the elements of InList
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]],L).
L = [[o], [d, e], [d, e], [m, n], [a, b, c], [f, g, h], [i, j, k, l]]
b) Again, we suppose that a list (InList) contains elements that are
lists themselves. But this time the objective is to sort the elements
of InList 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]],L).
L = [[i, j, k, l], [o], [a, b, c], [f, g, h], [d, e], [d, e], [m, n]]
Note that in the above example, the first two lists in the result L
have length 4 and 1, both lengths appear just once. The third and
forth list have length 3; there are two list of this
length. And finally, the last three lists have length 2. This is
the most frequent length.
