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

is equal to the position of the node in the 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.

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). ...

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

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]), ...]

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

d(3.75,blue)



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]

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