99 Raku Problems
Working with lists
P01 (*) Find the last box of a list.
P02 (*) Find the last but one box of a list.
P03 (*) Find the K'th element of a list.
P04 (*) Find the number of elements of a list.
P05 (*) Reverse a list.
P06 (*) Find out whether a list is a palindrome.
P07 (**) Flatten a nested list structure.
P08 (**) Eliminate consecutive duplicates of list elements.
P09 (**) Pack consecutive duplicates of list elements into sublists.
P10 (*) Run-length encoding of a list.
P11 (*) Modified run-length encoding.
P12 (**) Decode a run-length encoded list.
P13 (**) Run-length encoding of a list (direct solution).
P14 (*) Duplicate the elements of a list.
P15 (**) Replicate the elements of a list a given number of times.
P16 (**) Drop every N'th element from a list.
P17 (*) Split a list into two parts; the length of the first part is given.
P19 (**) Rotate a list N places to the left.
P20 (*) Remove the K'th element from a list.
P21 (*) Insert an element at a given position into a list.
P22 (*) Create a list containing all integers within a given range.
P23 (**) Extract a given number of randomly selected elements from a list.
P24 (*) Lotto: Draw N different random numbers from the set 1..M.
P25 (*) Generate a random permutation of the elements of a list.
P26 (**) Generate the combinations of K distinct objects chosen from the N elements of a list
P27 (**) Group the elements of a set into disjoint subsets.
P28 (**) Sorting a list of lists according to length of sublists
Arithmetic
P31 (**) Determine whether a given integer number is prime.
P32 (**) Determine the greatest common divisor of two positive integer numbers.
P33 (*) Determine whether two positive integer numbers are coprime.
P34 (**) Calculate Euler's totient function phi(m).
P35 (**) Determine the prime factors of a given positive integer.
P36 (**) Determine the prime factors of a given positive integer (2).
P37 (**) Calculate Euler's totient function phi(m) (improved).
P38 (*) Compare the two methods of calculating Euler's totient function.
P39 (*) A list of prime numbers.
P40 (**) Goldbach's conjecture.
P41 (**) A list of Goldbach compositions.
Logic and Codes
P46 (**) Truth tables for logical expressions.
P47 (*) Truth tables for logical expressions (2).
P48 (**) Truth tables for logical expressions (3).
P49 (**) Gray code.
P50 (***) Huffman code.
Binary Trees
P54A (*) Check whether a given term represents a binary tree
P55 (**) Construct completely balanced binary trees
P56 (**) Symmetric binary trees
P57 (**) Binary search trees (dictionaries)
P58 (**) Generate-and-test paradigm
P59 (**) Construct height-balanced binary trees
P60 (**) Construct height-balanced binary trees with a given number of nodes
P61 (*) Count the leaves of a binary tree
P61A (*) Collect the leaves of a binary tree in a list
P62 (*) Collect the internal nodes of a binary tree in a list
P62B (*) Collect the nodes at a given level in a list
P63 (**) Construct a complete binary tree
P64 (**) Layout a binary tree (1)
P65 (**) Layout a binary tree (2)
P66 (***) Layout a binary tree (3)
P67 (**) A string representation of binary trees
P68 (**) Preorder and inorder sequences of binary trees
P69 (**) Dotstring representation of binary trees
P70B (*) Check whether a given term represents a multiway tree
P70C (*) Count the nodes of a multiway tree
P70 (**) Tree construction from a node string
P71 (*) Determine the internal path length of a tree
P72 (*) Construct the bottom-up order sequence of the tree nodes
P73 (**) Lisp-like tree representation
Graphs
P80 (***) Conversions
P81 (**) Path from one node to another one
P82 (*) Cycle from a given node
P83 (**) Construct all spanning trees
P84 (**) Construct the minimal spanning tree
P85 (**) Graph isomorphism
P86 (**) Node degree and graph coloration
P87 (**) Depth-first order graph traversal (alternative solution)
P88 (**) Connected components (alternative solution)
P89 (**) Bipartite graphs
Miscellaneous Problems
P90 (**) Eight queens problem
P91 (**) Knight's tour
P92 (***) Von Koch's conjecture
P93 (***) An arithmetic puzzle
P94 (***) Generate K-regular simple graphs with N nodes
P95 (**) English number words
P96 (**) Syntax checker (alternative solution with difference lists)
P97 (**) Sudoku
P98 (***) Nonograms
P99 (***) Crossword puzzle
=encoding utf8
Based on an original Prolog problem list by Werner Hett
L<http://sites.google.com/site/prologsite/prolog-problems>.
Example:
> say my-last <a b c d>;
d
Example:
> say ~my-but-last(<A B C D>);
C D
Example:
> say element-at <a b c d e>,3;
c
A palindrome can be read forward or backward; e.g. <x a m a x>.
Transform an array, possibly holding arrays as elements into a `flat'
list by replacing each array with its elements (recursively).
Example:
> splat([1,[2,[3,4],5]]).perl.say;
(1, 2, 3, 4, 5)
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:
> say ~compress(<a a a a b c c a a d e e e e>)
a b c a d e
If a list contains repeated elements they should be placed in separate sublists.
Example:
> pack-dup(<a a a a b c c a a d e e e e>).perl.say
[["a","a","a","a"],["b"],["c","c"],["a","a"],["d"],["e","e","e","e"]]
Use the result of problem P09 to implement the so-called run-length
encoding data compression method. Consecutive duplicates of elements
are encoded as arrays [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>).perl.say
[[4, "a"], [1, "b"], [2, "c"], [2, "a"], [1, "d"], [4, "e"]]
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>).perl.say
([4, "a"], "b", [2, "c"], [2, "a"], "d", [4, "e"])
Given a run-length code list generated as specified in problem P11.
Construct its uncompressed version.
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 singletons [1,X] by X.
Example:
> encode-direct(<a a a a b c c a a d e e e e>).perl.say
([4, "a"], "b", [2, "c"], [2, "a"], "d", [4, "e"])
Example:
> say ~dupli(<a b c c d>);
a a b b c c c c d d
Example:
> say ~repli <a b c>,3;
a a a b b b c c c
Example:
> say ~drop(<a b c d e f g h i k>, 3);
a b d e g h k
Do not use any predefined predicates.
Example:
> say bisect(<a b c d e f g h i k>,3).perl
(["a", "b", "c"], ["d", "e", "f", "g", "h", "i", "k"])
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)
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.
You may choose to mutate the array in-place or create a new sequence and
return it.
Example 1 (mutating in-place):
> my @l = <a b c d>;
> remove-at(@l,2);
> say ~@l;
a c d
Example 2 (returning a copy):
> say ~remove-at-copy(<a b c d>, 2);
a c d
You may choose to mutate the array in-place or to create a new
sequence and return it.
Example 1 (mutating in-place);
> my @l = <a b c d>;
> insert-at('alfa',@l,2);
> say ~@l;
a alfa b c d
Example 2 (returning a copy):
> say ~insert-at-copy('alfa', <a b c d>, 2);
a alfa b c d
If first argument is smaller than second, produce a list in decreasing order.
Example:
> say ~range(4, 9);
4 5 6 7 8 9
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.
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.
Example:
* (rnd-permu '(a b c d e f))
(B A D C E F)
Hint: Use the solution of problem P23.
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) ... )
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".
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.
Example:
> say is-prime 7
1
Use Euclid's algorithm.
Example:
> say gcd(36,63);
9
Two numbers are coprime if their greatest common divisor equals 1.
Example:
> say coprime(35,64)
1
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.
> say 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).
Construct a flat list containing the prime factors in ascending order.
Example:
> say ~prime-factors 315
3 3 5 7
Construct a list containing the prime factors and their multiplicity.
Example:
> prime-factors-mult(315).perl.say
([3,2],[5,1],[7,1])
Hint: The problem is similar to problem P13.
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) * ...
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.
Given a range of integers by its lower and upper limit, construct a list of
all prime numbers in that range.
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:
> say ~goldbach 28
5 23
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
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
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
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
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',Ā“110Ā“,Ā“111Ā“,Ā“101Ā“,Ā“100Ā“].
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?
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
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.
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
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
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.
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
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.
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
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.
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
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
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.
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.
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.
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.
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.
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?
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.
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).
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,[])])])
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
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).
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.
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 (+,-).
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?
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.
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.
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.
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.
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.
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!
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.
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.
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.
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).
Write a predicate that splits a graph into its connected components.
Write a predicate that finds out whether a given graph is bipartite.
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.
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).
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?
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!).
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.
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.
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
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.
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.
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.
The Camelia image is copyright 2009 by Larry Wall. "Raku" is trademark of the Yet Another Society.
All rights reserved.