L-99_Ninety-Nine_Lisp_Problems.html.txt

                       L-99: Ninety-Nine Lisp Problems

  Based on a Prolog problem list by werner.hett@hti.bfh.ch

Working with lists

   P01 (*) Find the last box of a list.
           Example:
           * (my-last '(a b c d))
           (D)

   P02 (*) Find the last but one box of a list.
           Example:
           * (my-but-last '(a b c d))
           (C D)

   P03 (*) Find the K'th element of a list.
           The first element in the list is number 1.
           Example:
           * (element-at '(a b c d e) 3)
           C

   P04 (*) Find the number of elements of a list.

   P05 (*) Reverse a list.

   P06 (*) Find out whether a list is a palindrome.
           A palindrome can be read forward or backward; e.g. (x a m a x).

   P07 (**) 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)))
           (A B C D E)

           Hint: Use the predefined functions list and append.

   P08 (**) 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))
           (A B C A D E)

   P09 (**) 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))
           ((A A A A) (B) (C C) (A A) (D) (E E E E))

   P10 (*) Run-length encoding of a list.
           Use the result of problem P09 to implement the so-called
           run-length encoding data compression method. Consecutive
           duplicates of elements are encoded as lists (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))
           ((4 A) (1 B) (2 C) (2 A) (1 D)(4 E))

   P11 (*) Modified run-length encoding.
           Modify the result of problem P10 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.

           Example:
           * (encode-modified '(a a a a b c c a a d e e e e))
           ((4 A) B (2 C) (2 A) D (4 E))

   P12 (**) Decode a run-length encoded list.
           Given a run-length code list generated as specified in problem
           P11. Construct its uncompressed version.

   P13 (**) Run-length encoding of a list (direct solution).
           Implement the so-called run-length encoding data compression
           method directly. I.e. don't explicitly create the sublists
           containing the duplicates, as in problem P09, but only count
           them. As in problem P11, simplify the result list by replacing
           the singleton lists (1 X) by X.

           Example:
           * (encode-direct '(a a a a b c c a a d e e e e))
           ((4 A) B (2 C) (2 A) D (4 E))

   P14 (*) Duplicate the elements of a list.
           Example:
           * (dupli '(a b c c d))
           (A A B B C C C C D D)

   P15 (**) Replicate the elements of a list a given number of times.
           Example:
           * (repli '(a b c) 3)
           (A A A B B B C C C)

   P16 (**) Drop every N'th element from a list.
           Example:
           * (drop '(a b c d e f g h i k) 3)
           (A B D E G H K)

   P17 (*) 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)
           ( (A B C) (D E F G H I K))

   P18 (**) 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)
           (C D E F G)

   P19 (**) Rotate a list N places to the left.
           Examples:
           * (rotate '(a b c d e f g h) 3)
           (D E F G H A B C)

           * (rotate '(a b c d e f g h) -2)
           (G H A B C D E F)

           Hint: Use the predefined functions length and append, as well as
           the result of problem P17.

   P20 (*) Remove the K'th element from a list.
           Example:
           * (remove-at '(a b c d) 2)
           (A C D)

   P21 (*) Insert an element at a given position into a list.
           Example:
           * (insert-at 'alfa '(a b c d) 2)
           (A ALFA B C D)

   P22 (*) Create a list containing all integers within a given range.
           If first argument is smaller than second, produce a list in
           decreasing order.
           Example:
           * (range 4 9)
           (4 5 6 7 8 9)

   P23 (**) Extract a given number of randomly selected elements from a
   list.
           The selected items shall be returned in a list.
           Example:
           * (rnd-select '(a b c d e f g h) 3)
           (E D A)

           Hint: Use the built-in random number generator and the result of
           problem P20.

   P24 (*) Lotto: Draw N different random numbers from the set 1..M.
           The selected numbers shall be returned in a list.
           Example:
           * (lotto-select 6 49)
           (23 1 17 33 21 37)

           Hint: Combine the solutions of problems P22 and P23.

   P25 (*) 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)

           Hint: Use the solution of problem P23.

   P26 (**) 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:
           * (combination 3 '(a b c d e f))
           ((A B C) (A B D) (A B E) ... )

   P27 (**) 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".

   P28 (**) 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))

           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))

           Note that in the above example, the first two lists in the
           result have length 4 and 1, both lengths appear just once. The
           third and forth list have length 3 which appears twice (there
           are two list of this length). And finally, the last three lists
           have length 2. This is the most frequent length.

Arithmetic

   P31 (**) Determine whether a given integer number is prime.
           Example:
           * (is-prime 7)
           T

   P32 (**) Determine the greatest common divisor of two positive integer
   numbers.
           Use Euclid's algorithm.
           Example:
           * (gcd 36 63)
           9

   P33 (*) Determine whether two positive integer numbers are coprime.
           Two numbers are coprime if their greatest common divisor equals
           1.
           Example:
           * (coprime 35 64)
           T

   P34 (**) Calculate Euler's totient function phi(m).
           Euler's so-called totient function phi(m) is defined as the
           number of positive integers r (1 <= r < m) that are coprime to
           m.

           Example: m = 10: r = 1,3,7,9; thus phi(m) = 4. Note the special
           case: phi(1) = 1.

           * (totient-phi 10)
           4

           Find out what the value of phi(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).

   P35 (**) Determine the prime factors of a given positive integer.
           Construct a flat list containing the prime factors in ascending
           order.
           Example:
           * (prime-factors 315)
           (3 3 5 7)

   P36 (**) Determine the prime factors of a given positive integer (2).
           Construct a list containing the prime factors and their
           multiplicity.
           Example:
           * (prime-factors-mult 315)
           ((3 2) (5 1) (7 1))

           Hint: The problem is similar to problem P13.

   P37 (**) Calculate Euler's totient function phi(m) (improved).
           See problem P34 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 problem P36 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 phi(m) can be calculated with the following
           formula:

           phi(m) = (p1 - 1) * p1 ** (m1 - 1) + (p2 - 1) * p2 ** (m2 - 1) +
           (p3 - 1) * p3 ** (m3 - 1) + ...

           Note that a ** b stands for the b'th power of a.

   P38 (*) Compare the two methods of calculating Euler's totient function.
           Use the solutions of problems P34 and P37 to compare the
           algorithms. Take the number of logical inferences as a measure
           for efficiency. Try to calculate phi(10090) as an example.

   P39 (*) 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.

   P40 (**) 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 (much larger
           than we can go with our Prolog system). Write a predicate to
           find the two prime numbers that sum up to a given even integer.

           Example:
           * (goldbach 28)
           (5 23)

   P41 (**) 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.

           Example:
           * (goldbach-list 9 20)
           10 = 3 + 7
           12 = 5 + 7
           14 = 3 + 11
           16 = 3 + 13
           18 = 5 + 13
           20 = 3 + 17

           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.

           Example (for a print limit of 50):
           * (goldbach-list 1 2000 50)
           992 = 73 + 919
           1382 = 61 + 1321
           1856 = 67 + 1789
           1928 = 61 + 1867

Logic and Codes

   P46 (**) Truth tables for logical expressions.
           Define predicates and/2, or/2, nand/2, nor/2, xor/2, impl/2 and
           equ/2 (for logical equivalence) which succeed or fail according
           to the result of their respective operations; e.g. and(A,B) will
           succeed, if and only if both A and B succeed. Note that A and B
           can be Prolog goals (not only the constants true and fail).

           A logical expression in two variables can then be written in
           prefix notation, as in the following example:
           and(or(A,B),nand(A,B)).

           Now, write a predicate table/3 which prints the truth table of a
           given logical expression in two variables.

           Example:
           * table(A,B,and(A,or(A,B))).
           true true true
           true fail true
           fail true fail
           fail fail fail

   P47 (*) Truth tables for logical expressions (2).
           Continue problem P46 by defining and/2, or/2, etc as being
           operators. This allows to write the logical expression in the
           more natural way, as in the example: A and (A or not B). Define
           operator precedence as usual; i.e. as in Java.

           Example:
           * table(A,B, A and (A or not B)).
           true true true
           true fail true
           fail true fail
           fail fail fail

   P48 (**) Truth tables for logical expressions (3).
           Generalize problem P47 in such a way that the logical expression
           may contain any number of logical variables. Define table/2 in a
           way that table(List,Expr) prints the truth table for the
           expression Expr, which contains the logical variables enumerated
           in List.

           Example:
           * table([A,B,C], A and (B or C) equ A and B or A and C).
           true true true true
           true true fail true
           true fail true true
           true fail fail true
           fail true true true
           fail true fail true
           fail fail true true
           fail fail fail true

   P49 (**) Gray code.
           An n-bit Gray code is a sequence of n-bit 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 predicate with the
           following specification:

           % gray(N,C) :- C is the N-bit Gray code

           Can you apply the method of "result caching" in order to make
           the predicate more efficient, when it is to be used repeatedly?

   P50 (***) Huffman code.
           First of all, consult a good book on discrete mathematics or
           algorithms for a detailed description of Huffman codes!

           We suppose a set of symbols with their frequencies, given as a
           list of fr(S,F) terms. Example:
           [fr(a,45),fr(b,13),fr(c,12),fr(d,16),fr(e,9),fr(f,5)]. Our
           objective is to construct a list hc(S,C) terms, where C is the
           Huffman code word for the symbol S. In our example, the result
           could be Hs = [hc(a,'0'), hc(b,'101'), hc(c,'100'), hc(d,'111'),
           hc(e,'1101'), hc(f,'1100')] [hc(a,'01'),...etc.]. The task shall
           be performed by the predicate huffman/2 defined as follows:

           % huffman(Fs,Hs) :- Hs is the Huffman code table for the
           frequency table Fs

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 Lisp we represent the empty tree by 'nil' and the non-empty tree by
   the list (X L R), where X denotes the root node and L and R denote the
   left and right subtree, respectively. The example tree depicted opposite
   is therefore represented by the following list:

   (a (b (d nil nil) (e nil nil)) (c nil (f (g nil nil) nil)))

   Other examples are a binary tree that consists of a root node only:

   (a nil nil) or an empty binary tree: nil.

   You can check your predicates using these example trees. They are given
   as test cases in p54.lisp.

   P54A (*) Check whether a given term represents a binary tree
           Write a predicate istree which returns true if and only if its
           argument is a list representing a binary tree.
           Example:
           * (istree (a (b nil nil) nil))
           T
           * (istree (a (b nil nil)))
           NIL

   P55 (**) Construct completely balanced binary trees
           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 predicate should
           generate all solutions via backtracking. Put the letter 'x' as
           information into all nodes of the tree.
           Example:
           * cbal-tree(4,T).
           T = t(x, t(x, nil, nil), t(x, nil, t(x, nil, nil))) ;
           T = t(x, t(x, nil, nil), t(x, t(x, nil, nil), nil)) ;
           etc......No

   P56 (**) 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 predicate symmetric/1
           to check whether a given binary tree is symmetric. Hint: Write a
           predicate mirror/2 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.

   P57 (**) Binary search trees (dictionaries)
           Use the predicate add/3, developed in chapter 4 of the course,
           to write a predicate to construct a binary search tree from a
           list of integer numbers.
           Example:
           * construct([3,2,5,7,1],T).
           T = t(3, t(2, t(1, nil, nil), nil), t(5, nil, t(7, nil, nil)))

           Then use this predicate to test the solution of the problem P56.
           Example:
           * test-symmetric([5,3,18,1,4,12,21]).
           Yes
           * test-symmetric([3,2,5,7,1]).
           No

   P58 (**) Generate-and-test paradigm
           Apply the generate-and-test paradigm to construct all symmetric,
           completely balanced binary trees with a given number of nodes.
           Example:
           * sym-cbal-trees(5,Ts).
           Ts = [t(x, t(x, nil, t(x, nil, nil)), t(x, t(x, nil, nil),
           nil)), t(x, t(x, t(x, nil, nil), nil), t(x, nil, t(x, nil,
           nil)))]

           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 predicate.

   P59 (**) Construct height-balanced binary trees
           In a height-balanced 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 predicate hbal-tree/2 to construct height-balanced
           binary trees for a given height. The predicate should generate
           all solutions via backtracking. Put the letter 'x' as
           information into all nodes of the tree.
           Example:
           * hbal-tree(3,T).
           T = t(x, t(x, t(x, nil, nil), t(x, nil, nil)), t(x, t(x, nil,
           nil), t(x, nil, nil))) ;
           T = t(x, t(x, t(x, nil, nil), t(x, nil, nil)), t(x, t(x, nil,
           nil), nil)) ;
           etc......No

   P60 (**) Construct height-balanced binary trees with a given number of
   nodes
           Consider a height-balanced binary tree of height H. What is the
           maximum number of nodes it can contain?
           Clearly, MaxN = 2**H - 1. However, what is the minimum number
           MinN? This question is more difficult. Try to find a recursive
           statement and turn it into a predicate minNodes/2 defined as
           follwos:

           % minNodes(H,N) :- N is the minimum number of nodes in a
           height-balanced binary tree of height H.
           (integer,integer), (+,?)

           On the other hand, we might ask: what is the maximum height H a
           height-balanced binary tree with N nodes can have?

           % maxHeight(N,H) :- H is the maximum height of a height-balanced
           binary tree with N nodes
           (integer,integer), (+,?)

           Now, we can attack the main problem: construct all the
           height-balanced binary trees with a given nuber of nodes.

           % hbal-tree-nodes(N,T) :- T is a height-balanced binary tree
           with N nodes.

           Find out how many height-balanced trees exist for N = 15.

   P61 (*) Count the leaves of a binary tree
           A leaf is a node with no successors. Write a predicate
           count-leaves/2 to count them.

           % count-leaves(T,N) :- the binary tree T has N leaves

   P61A (*) Collect the leaves of a binary tree in a list
           A leaf is a node with no successors. Write a predicate leaves/2
           to collect them in a list.

           % leaves(T,S) :- S is the list of all leaves of the binary tree
           T

   P62 (*) Collect the internal nodes of a binary tree in a list
           An internal node of a binary tree has either one or two
           non-empty successors. Write a predicate internals/2 to collect
           them in a list.

           % internals(T,S) :- S is the list of internal nodes of the
           binary tree T.

   P62B (*) 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 N-1. The root node is at level 1. Write a
           predicate atlevel/3 to collect all nodes at a given level in a
           list.

           % atlevel(T,L,S) :- S is the list of nodes of the binary tree T
           at level L

           Using atlevel/3 it is easy to construct a predicate levelorder/2
           which creates the level-order sequence of the nodes. However,
           there are more efficient ways to do that.

   P63 (**) Construct a complete binary tree
           A complete binary tree with height H is defined as follows: The
           levels 1,2,3,...,H-1 contain the maximum number of nodes (i.e
           2**(i-1) 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
           "left-adjusted". 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 predicate complete-binary-tree/2 with the following
           specification:

           % complete-binary-tree(N,T) :- T is a complete binary tree with
           N nodes. (+,?)

           Test your predicate in an appropriate way.

   P64 (**) Layout a binary tree (1)
           Given a binary tree as the usual Prolog term t(X,L,R) (or nil).
           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 below.

           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 extend the
           Prolog term representing a node (and its successors) as follows:

           % nil represents the empty tree (as usual)
           % t(W,X,Y,L,R) represents a (non-empty) binary tree with root W
           "positioned" at (X,Y), and subtrees L and R

           Write a predicate layout-binary-tree/2 with the following
           specification:

           % layout-binary-tree(T,PT) :- PT is the "positioned" binary tree
           obtained from the binary tree T. (+,?)

           Test your predicate in an appropriate way.

   P65 (**) Layout a binary tree (2)
           An alternative layout method is depicted in the illustration
           opposite. Find out the rules and write the corresponding Prolog
           predicate. Hint: On a given level, the horizontal distance
           between neighboring nodes is constant.

           Use the same conventions as in problem P64 and test your
           predicate in an appropriate way.

   P66 (***) Layout a binary tree (3)
           Yet another layout strategy is shown in the illustration
           opposite. The method yields a very compact layout while
           maintaining a certain symmetry in every node. Find out the rules
           and write the corresponding Prolog predicate. Hint: Consider the
           horizontal distance between a node and its successor nodes. How
           tight can you pack together two subtrees to construct the
           combined binary tree?

           Use the same conventions as in problem P64 and P65 and test your
           predicate in an appropriate way. Note: This is a difficult
           problem. Don't give up too early!

           Which layout do you like most?

   P67 (**) A string representation of binary trees
           Somebody represents binary trees as strings of the following
           type (see example opposite):

           a(b(d,e),c(,f(g,)))

           a) Write a Prolog predicate which generates this string
           representation, if the tree is given as usual (as nil or
           t(X,L,R) term). Then write a predicate which does this inverse;
           i.e. given the string representation, construct the tree in the
           usual form. Finally, combine the two predicates in a single
           predicate tree-string/2 which can be used in both directions.

           b) Write the same predicate tree-string/2 using difference lists
           and a single predicate tree-dlist/2 which does the conversion
           between a tree and a difference list in both directions.

           For simplicity, suppose the information in the nodes is a single
           letter and there are no spaces in the string.

   P68 (**) Preorder and inorder sequences of binary trees
           We consider binary trees with nodes that are identified by
           single lower-case letters, as in the example of problem P67.

           a) Write predicates preorder/2 and inorder/2 that construct the
           preorder and inorder sequence of a given binary tree,
           respectively. The results should be atoms, e.g. 'abdecfg' for
           the preorder sequence of the example in problem P67.

           b) Can you use preorder/2 from problem part a) in the reverse
           direction; i.e. given a preorder sequence, construct a
           corresponding tree? If not, make the necessary arrangements.

           c) If both the preorder sequence and the inorder sequence of the
           nodes of a binary tree are given, then the tree is determined
           unambiguously. Write a predicate pre-in-tree/3 that does the
           job.

           d) Solve problems a) to c) using difference lists. Cool! Use the
           predefined predicate time/1 to compare the solutions.

           What happens if the same character appears in more than one
           node. Try for instance pre-in-tree(aba,baa,T).

   P69 (**) Dotstring representation of binary trees
           We consider again binary trees with nodes that are identified by
           single lower-case letters, as in the example of problem P67.
           Such a tree can be represented by the preorder sequence of its
           nodes in which dots (.) are inserted where an empty subtree
           (nil) is encountered during the tree traversal. For example, the
           tree shown in problem P67 is represented as 'abd..e..c.fg...'.
           First, try to establish a syntax (BNF or syntax diagrams) and
           then write a predicate tree-dotstring/2 which does the
           conversion in both directions. Use difference lists.

Multiway Trees

   A multiway tree is composed of a root element and a (possibly empty) set
   of successors which are multiway trees themselves. A multiway tree is
   never empty. The set of successor trees is sometimes called a forest.

   In Prolog we represent a multiway tree by a term t(X,F), where X denotes
   the root node and F denotes the forest of successor trees (a Prolog
   list). The example tree depicted opposite is therefore represented by
   the following Prolog term:

 T = t(a,[t(f,[t(g,[])]),t(c,[]),t(b,[t(d,[]),t(e,[])])])

   P70B (*) Check whether a given term represents a multiway tree
           Write a predicate istree/1 which succeeds if and only if its
           argument is a Prolog term representing a multiway tree.
           Example:
           * istree(t(a,[t(f,[t(g,[])]),t(c,[]),t(b,[t(d,[]),t(e,[])])])).
           Yes

   P70C (*) Count the nodes of a multiway tree
           Write a predicate nnodes/1 which counts the nodes of a given
           multiway tree.
           Example:
           * nnodes(t(a,[t(f,[])]),N).
           N = 2

           Write another version of the predicate that allows for a flow
           pattern (o,i).

   P70 (**) Tree construction from a node string
           We suppose that the nodes of a multiway tree contain single
           characters. In the depth-first order sequence of its nodes, a
           special character ^ has been inserted whenever, during the tree
           traversal, the move is a backtrack to the previous level.

           By this rule, the tree in the figure opposite is represented as:
           afg^^c^bd^e^^^

           Define the syntax of the string and write a predicate
           tree(String,Tree) to construct the Tree when the String is
           given. Work with atoms (instead of strings). Make your predicate
           work in both directions.

   P71 (*) Determine the internal path length of a tree
           We define the internal path length of a multiway tree as the
           total sum of the path lengths from the root to all nodes of the
           tree. By this definition, the tree in the figure of problem P70
           has an internal path length of 9. Write a predicate
           ipl(Tree,IPL) for the flow pattern (+,-).

   P72 (*) Construct the bottom-up order sequence of the tree nodes
           Write a predicate bottom-up(Tree,Seq) which constructs the
           bottom-up sequence of the nodes of the multiway tree Tree. Seq
           should be a Prolog list. What happens if you run your predicate
           backwords?

   P73 (**) Lisp-like tree representation
           There is a particular notation for multiway trees in Lisp. Lisp
           is a prominent functional programming language, which is used
           primarily for artificial intelligence problems. As such it is
           one of the main competitors of Prolog. In Lisp almost everything
           is a list, just as in Prolog everything is a term.

           The following pictures show how multiway tree structures are
           represented in Lisp.
           Note that in the "lispy" notation a node with successors
           (children) in the tree is always the first element in a list,
           followed by its children. The "lispy" representation of a
           multiway tree is a sequence of atoms and parentheses '(' and
           ')', which we shall collectively call "tokens". We can represent
           this sequence of tokens as a Prolog list; e.g. the lispy
           expression (a (b c)) could be represented as the Prolog list
           ['(', a, '(', b, c, ')', ')']. Write a predicate tree-ltl(T,LTL)
           which constructs the "lispy token list" LTL if the tree is given
           as term T in the usual Prolog notation.

           Example:
           * tree-ltl(t(a,[t(b,[]),t(c,[])]),LTL).
           LTL = ['(', a, '(', b, c, ')', ')']

           As a second, even more interesting exercise try to rewrite
           tree-ltl/2 in a way that the inverse conversion is also
           possible: Given the list LTL, construct the Prolog tree T. Use
           difference lists.

Graphs

   A graph is defined as a set of nodes and a set of edges, where each edge
   is a pair of nodes.

   There are several ways to represent graphs in Prolog. One method is to
   represent each edge separately as one clause (fact). In this form, the
   graph depicted below is represented as the following predicate:

 edge(h,g).
 edge(k,f).
 edge(f,b).
 ...

   We call this edge-clause form. Obviously, isolated nodes cannot be
   represented. Another method is to represent the whole graph as one data
   object. According to the definition of the graph as a pair of two sets
   (nodes and edges), we may use the following Prolog term to represent the
   example graph:

 graph([b,c,d,f,g,h,k],[e(b,c),e(b,f),e(c,f),e(f,k),e(g,h)])

   We call this graph-term form. Note, that the lists are kept sorted, they
   are really sets, without duplicated elements. Each edge appears only
   once in the edge list; i.e. an edge from a node x to another node y is
   represented as e(x,y), the term e(y,x) is not present. The graph-term
   form is our default representation. In SWI-Prolog there are predefined
   predicates to work with sets.

   A third representation method is to associate with each node the set of
   nodes that are adjacent to that node. We call this the adjacency-list
   form. In our example:

 [n(b,[c,f]), n(c,[b,f]), n(d,[]), n(f,[b,c,k]), ...]

   The representations we introduced so far are Prolog terms and therefore
   well suited for automated processing, but their syntax is not very
   user-friendly. Typing the terms by hand is cumbersome and error-prone.
   We can define a more compact and "human-friendly" notation as follows: A
   graph is represented by a list of atoms and terms of the type X-Y (i.e.
   functor '-' and arity 2). The atoms stand for isolated nodes, the X-Y
   terms describe edges. If an X appears as an endpoint of an edge, it is
   automatically defined as a node. Our example could be written as:

 [b-c, f-c, g-h, d, f-b, k-f, h-g]

   We call this the human-friendly form. As the example shows, the list
   does not have to be sorted and may even contain the same edge multiple
   times. Notice the isolated node d. (Actually, isolated nodes do not even
   have to be atoms in the Prolog sense, they can be compound terms, as in
   d(3.75,blue) instead of d in the example).

   When the edges are directed we call them arcs. These are represented by
   ordered pairs. Such a graph is called directed graph. To represent a
   directed graph, the forms discussed above are slightly modified. The
   example graph opposite is represented as follows:

   Arc-clause form
           arc(s,u).
           arc(u,r).
           ...

   Graph-term form
           digraph([r,s,t,u,v],[a(s,r),a(s,u),a(u,r),a(u,s),a(v,u)])

   Adjacency-list form
           [n(r,[]),n(s,[r,u]),n(t,[]),n(u,[r]),n(v,[u])]
           Note that the adjacency-list does not have the information on
           whether it is a graph or a digraph.

   Human-friendly form
           [s > r, t, u > r, s > u, u > s, v > u]

   Finally, graphs and digraphs may have additional information attached to
   nodes and edges (arcs). For the nodes, this is no problem, as we can
   easily replace the single character identifiers with arbitrary compound
   terms, such as city('London',4711). On the other hand, for edges we have
   to extend our notation. Graphs with additional information attached to
   edges are called labelled graphs.

   Arc-clause form
           arc(m,q,7).
           arc(p,q,9).
           arc(p,m,5).

   Graph-term form
           digraph([k,m,p,q],[a(m,p,7),a(p,m,5),a(p,q,9)])

   Adjacency-list form
           [n(k,[]),n(m,[q/7]),n(p,[m/5,q/9]),n(q,[])]
           Notice how the edge information has been packed into a term with
           functor '/' and arity 2, together with the corresponding node.

   Human-friendly form
           [p>q/9, m>q/7, k, p>m/5]

   The notation for labelled graphs can also be used for so-called
   multi-graphs, where more than one edge (or arc) are allowed between two
   given nodes.

   P80 (***) Conversions
           Write predicates to convert between the different graph
           representations. With these predicates, all representations are
           equivalent; i.e. for the following problems you can always pick
           freely the most convenient form. The reason this problem is
           rated (***) is not because it's particularly difficult, but
           because it's a lot of work to deal with all the special cases.

   P81 (**) Path from one node to another one
           Write a predicate path(G,A,B,P) to find an acyclic path P from
           node A to node b in the graph G. The predicate should return all
           paths via backtracking.

   P82 (*) Cycle from a given node
           Write a predicate cycle(G,A,P) to find a closed path (cycle) P
           starting at a given node A in the graph G. The predicate should
           return all cycles via backtracking.

   P83 (**) Construct all spanning trees
           Write a predicate s-tree(Graph,Tree) to construct (by
           backtracking) all spanning trees of a given graph. With this
           predicate, find out how many spanning trees there are for the
           graph depicted to the left. The data of this example graph can
           be found in the file p83.dat. When you have a correct solution
           for the s-tree/2 predicate, use it to define two other useful
           predicates: is-tree(Graph) and is-connected(Graph). Both are
           five-minutes tasks!

   P84 (**) Construct the minimal spanning tree
           Write a predicate ms-tree(Graph,Tree,Sum) to construct the
           minimal spanning tree of a given labelled graph. Hint: Use the
           algorithm of Prim. A small modification of the solution of P83
           does the trick. The data of the example graph to the right can
           be found in the file p84.dat.

   P85 (**) Graph isomorphism
           Two graphs G1(N1,E1) and G2(N2,E2) are isomorphic if there is a
           bijection f: N1 -> N2 such that for any nodes X,Y of N1, X and Y
           are adjacent if and only if f(X) and f(Y) are adjacent.

           Write a predicate that determines whether two graphs are
           isomorphic. Hint: Use an open-ended list to represent the
           function f.

   P86 (**) Node degree and graph coloration
           a) Write a predicate degree(Graph,Node,Deg) that determines the
           degree of a given node.

           b) Write a predicate that generates a list of all nodes of a
           graph sorted according to decreasing degree.

           c) Use Welch-Powell's algorithm to paint the nodes of a graph in
           such a way that adjacent nodes have different colors.

   P87 (**) Depth-first order graph traversal (alternative solution)
           Write a predicate that generates a depth-first order graph
           traversal sequence. The starting point should be specified, and
           the output should be a list of nodes that are reachable from
           this starting point (in depth-first order).

   P88 (**) Connected components (alternative solution)
           Write a predicate that splits a graph into its connected
           components.

   P89 (**) Bipartite graphs
           Write a predicate that finds out whether a given graph is
           bipartite.

Miscellaneous Problems

   P90 (**) Eight queens problem
           This is a classical problem in computer science. The objective
           is to place eight queens on a chessboard so that no two queens
           are attacking each other; i.e., no two queens are in the same
           row, the same column, or on the same diagonal.

           Hint: Represent the positions of the queens as a list of numbers
           1..N. Example: [4,2,7,3,6,8,5,1] means that the queen in the
           first column is in row 4, the queen in the second column is in
           row 2, etc. Use the generate-and-test paradigm.

   P91 (**) Knight's tour
           Another famous problem is this one: How can a knight jump on an
           NxN chessboard in such a way that it visits every square exactly
           once?

           Hints: Represent the squares by pairs of their coordinates of
           the form X/Y, where both X and Y are integers between 1 and N.
           (Note that '/' is just a convenient functor, not division!)
           Define the relation jump(N,X/Y,U/V) to express the fact that a
           knight can jump from X/Y to U/V on a NxN chessboard. And
           finally, represent the solution of our problem as a list of N*N
           knight positions (the knight's tour).

   P92 (***) Von Koch's conjecture
           Several years ago I met a mathematician who was intrigued by a
           problem for which he didn't know a solution. His name was Von
           Koch, and I don't know whether the problem has been solved
           since.

           Anyway the puzzle goes like this: Given a tree with N nodes (and
           hence N-1 edges). Find a way to enumerate the nodes from 1 to N
           and, accordingly, the edges from 1 to N-1 in such a way, that
           for each edge K the difference of its node numbers equals to K.
           The conjecture is that this is always possible.

           For small trees the problem is easy to solve by hand. However,
           for larger trees, and 14 is already very large, it is extremely
           difficult to find a solution. And remember, we don't know for
           sure whether there is always a solution!

           Write a predicate that calculates a numbering scheme for a given
           tree. What is the solution for the larger tree pictured above?

   P93 (***) An arithmetic puzzle
           Given a list of integer numbers, find a correct way of inserting
           arithmetic signs (operators) such that the result is a correct
           equation. Example: With the list of numbers [2,3,5,7,11] we can
           form the equations 2-3+5+7 = 11 or 2 = (3*5+7)/11 (and ten
           others!).

   P94 (***) Generate K-regular simple graphs with N nodes
           In a K-regular graph all nodes have a degree of K; i.e. the
           number of edges incident in each node is K. How many
           (non-isomorphic!) 3-regular graphs with 6 nodes are there? See
           also a table of results and a Java applet that can represent
           graphs geometrically.

   P95 (**) English number words
           On financial documents, like cheques, numbers must sometimes be
           written in full words. Example: 175 must be written as
           one-seven-five. Write a predicate full-words/1 to print
           (non-negative) integer numbers in full words.

   P96 (**) Syntax checker (alternative solution with difference lists)
           In a certain programming language (Ada) identifiers are defined
           by the syntax diagram (railroad chart) opposite. Transform the
           syntax diagram into a system of syntax diagrams which do not
           contain loops; i.e. which are purely recursive. Using these
           modified diagrams, write a predicate identifier/1 that can check
           whether or not a given string is a legal identifier.

           % identifier(Str) :- Str is a legal identifier

   P97 (**) Sudoku
           Sudoku puzzles go like this:

    Problem statement                 Solution

     .  .  4 | 8  .  . | .  1  7      9  3  4 | 8  2  5 | 6  1  7            
             |         |                      |         |
     6  7  . | 9  .  . | .  .  .      6  7  2 | 9  1  4 | 8  5  3
             |         |                      |         |
     5  .  8 | .  3  . | .  .  4      5  1  8 | 6  3  7 | 9  2  4
     --------+---------+--------      --------+---------+--------
     3  .  . | 7  4  . | 1  .  .      3  2  5 | 7  4  8 | 1  6  9
             |         |                      |         |
     .  6  9 | .  .  . | 7  8  .      4  6  9 | 1  5  3 | 7  8  2
             |         |                      |         |
     .  .  1 | .  6  9 | .  .  5      7  8  1 | 2  6  9 | 4  3  5
     --------+---------+--------      --------+---------+--------
     1  .  . | .  8  . | 3  .  6      1  9  7 | 5  8  2 | 3  4  6
             |         |                      |         |
     .  .  . | .  .  6 | .  9  1      8  5  3 | 4  7  6 | 2  9  1
             |         |                      |         |
     2  4  . | .  .  1 | 5  .  .      2  4  6 | 3  9  1 | 5  7  8
   

           Every spot in the puzzle belongs to a (horizontal) row and a
           (vertical) column, as well as to one single 3x3 square (which we
           call "square" for short). At the beginning, some of the spots
           carry a single-digit number between 1 and 9. The problem is to
           fill the missing spots with digits in such a way that every
           number between 1 and 9 appears exactly once in each row, in each
           column, and in each square.

   P98 (***) Nonograms
           Around 1994, a certain kind of puzzles was very popular in
           England. The "Sunday Telegraph" newspaper wrote: "Nonograms are
           puzzles from Japan and are currently published each week only in
           The Sunday Telegraph. Simply use your logic and skill to
           complete the grid and reveal a picture or diagram." As a Prolog
           programmer, you are in a better situation: you can have your
           computer do the work! Just write a little program ;-).

           The puzzle goes like this: Essentially, each row and column of a
           rectangular bitmap is annotated with the respective lengths of
           its distinct strings of occupied cells. The person who solves
           the puzzle must complete the bitmap given only these lengths.

           Problem statement:          Solution:

           |_|_|_|_|_|_|_|_| 3         |_|X|X|X|_|_|_|_| 3          
           |_|_|_|_|_|_|_|_| 2 1       |X|X|_|X|_|_|_|_| 2 1        
           |_|_|_|_|_|_|_|_| 3 2       |_|X|X|X|_|_|X|X| 3 2        
           |_|_|_|_|_|_|_|_| 2 2       |_|_|X|X|_|_|X|X| 2 2        
           |_|_|_|_|_|_|_|_| 6         |_|_|X|X|X|X|X|X| 6          
           |_|_|_|_|_|_|_|_| 1 5       |X|_|X|X|X|X|X|_| 1 5        
           |_|_|_|_|_|_|_|_| 6         |X|X|X|X|X|X|_|_| 6          
           |_|_|_|_|_|_|_|_| 1         |_|_|_|_|X|_|_|_| 1          
           |_|_|_|_|_|_|_|_| 2         |_|_|_|X|X|_|_|_| 2          
            1 3 1 7 5 3 4 3             1 3 1 7 5 3 4 3             
            2 1 5 1                     2 1 5 1                     
   

           For the example above, the problem can be stated as the two
           lists [[3],[2,1],[3,2],[2,2],[6],[1,5],[6],[1],[2]] and
           [[1,2],[3,1],[1,5],[7,1],[5],[3],[4],[3]] which give the "solid"
           lengths of the rows and columns, top-to-bottom and
           left-to-right, respectively. Published puzzles are larger than
           this example, e.g. 25 x 20, and apparently always have unique
           solutions.

   P99 (***) Crossword puzzle
           Given an empty (or almost empty) framework of a crossword puzzle
           and a set of words. The problem is to place the words into the
           framework.

           The particular crossword puzzle is specified in a text file
           which first lists the words (one word per line) in an arbitrary
           order. Then, after an empty line, the crossword framework is
           defined. In this framework specification, an empty character
           location is represented by a dot (.). In order to make the
           solution easier, character locations can also contain predefined
           character values. The puzzle opposite is defined in the file
           p99a.dat, other examples are p99b.dat and p99d.dat. There is
           also an example of a puzzle (p99c.dat) which does not have a
           solution.

           Words are strings (character lists) of at least two characters.
           A horizontal or vertical sequence of character places in the
           crossword puzzle framework is called a site. Our problem is to
           find a compatible way of placing words onto sites.

           Hints: (1) The problem is not easy. You will need some time to
           thoroughly understand it. So, don't give up too early! And
           remember that the objective is a clean solution, not just a
           quick-and-dirty hack!
           (2) Reading the data file is a tricky problem for which a
           solution is provided in the file p99-readfile.lisp. Use the
           predicate read_lines/2.
           (3) For efficiency reasons it is important, at least for larger
           puzzles, to sort the words and the sites in a particular order.
           For this part of the problem, the solution of P28 may be very
           helpful.

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   Last modified: Mon Oct 16 21:23:19 BRT 2006