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some new exercises and a pod version of the 99 problems document

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893 99-problems/99-problems.pod
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+=encoding utf8
+
+=head1 99 Perl 6 Problems
+
+Based on an original Prolog problem list by Werner Hett
+L<http://sites.google.com/site/prologsite/prolog-problems>.
+
+=head1 Working with lists
+
+
+=head2 P01 (*) Find the last box of a list.
+
+ Example:
+ > say my_last <a b c d>;
+ d
+
+=head2 P02 (*) Find the last but one box of a list.
+
+ Example:
+ > say ~my_but_last(<A B C D>);
+ C D
+
+=head2 P03 (*) Find the K'th element of a list.
+
+ Example:
+ > say element_at <a b c d e>,3;
+ c
+
+=head2 P04 (*) Find the number of elements of a list.
+
+=head2 P05 (*) Reverse a list.
+
+=head2 P06 (*) Find out whether a list is a palindrome.
+
+A palindrome can be read forward or backward; e.g. <x a m a x>.
+
+=head2 P07 (**) Flatten a nested list structure.
+
+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)
+
+=head2 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:
+ > say ~compress(<a a a a b c c a a d e e e e>)
+ a b c a d e
+
+=head2 P09 (**) Pack consecutive duplicates of list elements into sublists.
+
+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"]]
+
+=head2 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 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"]]
+
+=head2 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>).perl.say
+ ([4, "a"], "b", [2, "c"], [2, "a"], "d", [4, "e"])
+
+=head2 P12 (**) Decode a run-length encoded list.
+
+Given a run-length code list generated as specified in problem P11. Construct its uncompressed version.
+
+=head2 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 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"])
+
+=head2 P14 (*) Duplicate the elements of a list.
+
+ Example:
+ > say ~dupli(<a b c c d>);
+ a a b b c c c c d d
+
+=head2 P15 (**) Replicate the elements of a list a given number of times.
+
+ Example:
+ > say ~repli <a b c>,3;
+ a a a b b b c c c
+
+=head2 P16 (**) Drop every N'th element from a list.
+
+ Example:
+ > say ~drop(<a b c d e f g h i k>, 3);
+ a b d e g h k
+
+=head2 P17 (*) Split a list into two parts; the length of the first part is given.
+
+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"])
+
+=head2 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)
+
+=head2 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.
+
+=head2 P20 (*) Remove the K'th element from a list.
+
+ Example:
+ * (remove-at '(a b c d) 2)
+ (A C D)
+
+=head2 P21 (*) Insert an element at a given position into a list.
+
+You may choose to copy 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 (creating a copy):
+ > say ~insert_at_copy('alfa', <a b c d>, 2);
+ a alfa b c d
+
+=head2 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:
+ > say ~range(4, 9);
+ 4 5 6 7 8 9
+
+=head2 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.
+
+=head2 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.
+
+=head2 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.
+
+=head2 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) ... )
+
+=head2 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".
+
+=head2 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.
+
+=head1 Arithmetic
+
+=head2 P31 (**) Determine whether a given integer number is prime.
+
+ Example:
+ > say is_prime 7
+ 1
+
+=head2 P32 (**) Determine the greatest common divisor of two positive integer numbers.
+
+Use Euclid's algorithm.
+
+ Example:
+ > say gcd(36,63);
+ 9
+
+=head2 P33 (*) Determine whether two positive integer numbers are coprime.
+
+Two numbers are coprime if their greatest common divisor equals 1.
+
+ Example:
+ > say coprime(35,64)
+ 1
+
+=head2 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.
+ > 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).
+
+=head2 P35 (**) Determine the prime factors of a given positive integer.
+
+Construct a flat list containing the prime factors in ascending order.
+
+ Example:
+ > say ~prime_factors 315
+ 3 3 5 7
+
+=head2 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).perl.say
+ ([3,2],[5,1],[7,1])
+
+Hint: The problem is similar to problem P13.
+
+=head2 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) * ...
+
+=head2 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.
+
+=head2 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.
+
+=head2 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:
+ > say ~goldbach 28
+ 5 23
+
+=head2 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
+
+=head1 Logic and Codes
+
+=head2 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
+
+=head2 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
+
+=head2 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
+
+=head2 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?
+
+=head2 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
+
+=head1 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.
+
+=head2 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
+
+=head2 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
+
+=head2 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.
+
+=head2 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
+
+=head2 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.
+
+=head2 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
+
+=head2 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.
+
+=head2 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
+
+=head2 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
+
+=head2 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.
+
+=head2 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.
+
+=head2 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.
+
+=head2 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.
+
+=head2 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.
+
+=head2 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?
+
+=head2 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.
+
+=head2 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).
+
+=head2 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,[])])])
+
+
+=head2 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
+
+=head2 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).
+
+=head2 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.
+
+=head2 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 (+,-).
+
+=head2 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?
+
+=head2 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.
+
+=head1 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.
+
+=head2 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.
+
+=head2 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.
+
+=head2 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.
+
+=head2 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!
+
+=head2 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.
+
+
+=head2 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.
+
+=head2 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.
+
+=head2 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).
+
+=head2 P88 (**) Connected components (alternative solution)
+
+Write a predicate that splits a graph into its connected components.
+
+=head2 P89 (**) Bipartite graphs
+
+Write a predicate that finds out whether a given graph is bipartite.
+
+
+=head1 Miscellaneous Problems
+
+=head2 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.
+
+=head2 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).
+
+=head2 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?
+
+=head2 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!).
+
+=head2 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.
+
+=head2 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.
+
+=head2 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
+
+=head2 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.
+
+=head2 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.
+
+=head2 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.
+
View
115 99-problems/P36-ovid.pl
@@ -0,0 +1,115 @@
+use v6;
+
+# Specification:
+# 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).perl.say
+# (3 => 2, 5 => 1, 7 => 1)
+#
+# Hint: The problem is similar to problem P13.
+
+
+# This was originally a blog post:
+# http://blogs.perl.org/users/ovid/2010/08/prime-factors-in-perl-6.html
+
+# first some boring auxiliary stuff
+
+my @PRIMES = <2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53
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+167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269
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+5749 5779 5783 5791 5801 5807 5813 5821 5827 5839 5843 5849 5851 5857 5861
+5867 5869 5879 5881 5897 5903 5923 5927 5939 5953 5981 5987 6007 6011 6029
+6037 6043 6047 6053 6067 6073 6079 6089 6091 6101 6113 6121 6131 6133 6143
+6151 6163 6173 6197 6199 6203 6211 6217 6221 6229 6247 6257 6263 6269 6271
+6277 6287 6299 6301 6311 6317 6323 6329 6337 6343 6353 6359 6361 6367 6373
+6379 6389 6397 6421 6427 6449 6451 6469 6473 6481 6491 6521 6529 6547 6551
+6553 6563 6569 6571 6577 6581 6599 6607 6619 6637 6653 6659 6661 6673 6679
+6689 6691 6701 6703 6709 6719 6733 6737 6761 6763 6779 6781 6791 6793 6803
+6823 6827 6829 6833 6841 6857 6863 6869 6871 6883 6899 6907 6911 6917 6947
+6949 6959 6961 6967 6971 6977 6983 6991 6997 7001 7013 7019 7027 7039 7043
+7057 7069 7079 7103 7109 7121 7127 7129 7151 7159 7177 7187 7193 7207 7211
+7213 7219 7229 7237 7243 7247 7253 7283 7297 7307 7309 7321 7331 7333 7349
+7351 7369 7393 7411 7417 7433 7451 7457 7459 7477 7481 7487 7489 7499 7507
+7517 7523 7529 7537 7541 7547 7549 7559 7561 7573 7577 7583 7589 7591 7603
+7607 7621 7639 7643 7649 7669 7673 7681 7687 7691 7699 7703 7717 7723 7727
+7741 7753 7757 7759 7789 7793 7817 7823 7829 7841 7853 7867 7873 7877 7879
+7883 7901 7907 7919>;
+my %IS-PRIME = map {; $_ => 1 }, @PRIMES;
+
+sub is-prime(Int $number) {
+ return False if $number < 2; # special case
+ die "$number is too large" if $number > @PRIMES[*-1]**2;
+ return %IS-PRIME{$number};
+}
+
+
+# here is the main function
+sub prime-factors(Int $number is copy) {
+ # don't try to factor prime numbers
+ return { $number => 1 } if is-prime($number);
+
+ my %factors;
+ for @PRIMES -> $prime-number {
+ last if $prime-number ** 2 > $number;
+ while $number % $prime-number == 0 {
+ %factors{$prime-number} //= 0;
+ %factors{$prime-number}++;
+ $number /= $prime-number;
+ }
+ }
+ %factors{$number}++ if $number != 1; # we have a prime left over
+ return %factors;
+}
+
+for 17, 53, 90, 94, 200, 289, 62710561 -> $number {
+ say "Prime factors of $number are: ", prime-factors($number).perl;
+}
+
+
+# vim:ft=perl6
View
6 99-problems/P36-rhebus.pl
@@ -5,7 +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])
+# (3 => 2, 5 => 1, 7 => 1)
#
# Hint: The problem is similar to problem P13.
@@ -18,13 +18,13 @@ (Int $n)
$mult++;
$residue div= $k;
}
- take [$k, $mult] if $mult;
+ take $k => $mult if $mult;
last if $residue == 1;
# This if block is an optimisation which reduces number of iterations
# for numbers with large prime factors (such as large primes)
# It can be removed without affecting correctness.
if $k > sqrt $residue {
- take [$residue,1];
+ take $residue => 1;
last;
}
}
View
10 99-problems/P37-rhebus.pl
@@ -22,10 +22,10 @@ (Int $n)
$mult++;
$residue div= $k;
}
- take [$k, $mult] if $mult;
+ take $k => $mult if $mult;
last if $residue == 1;
if $k > sqrt $residue {
- take [$residue,1];
+ take $residue => 1;
last;
}
}
@@ -33,9 +33,9 @@ (Int $n)
# 1. One-liner version
-say "phi($_): ", [*] prime_factors_mult($_).map({ ($_[0]-1) * $_[0] ** ($_[1]-1) })
+say "phi($_): ", [*] prime_factors_mult($_).map({ (.key-1) * .key ** (.value-1) })
for 1..20;
-say [*] prime_factors_mult(315).map: { ($_[0]-1) * $_[0] ** ($_[1]-1) };
+say [*] prime_factors_mult(315).map: { (.key-1) * .key ** (.value-1) };
# 2. sub version
@@ -45,7 +45,7 @@ (Int $n)
sub totient (Int $n) {
my @factors = prime_factors_mult($n);
return [*] @factors.map: {
- ($_[0]-1) * $_[0] ** ($_[1]-1)
+ (.key-1) * .key ** (.value-1)
}
}
View
34 99-problems/P39-rhebus.pl
@@ -0,0 +1,34 @@
+use v6;
+
+# Specification:
+# 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.
+
+
+# Copied from P31-rhebus.pl
+sub is_prime (Int $n) {
+ for 2..sqrt $n -> $k {
+ return Bool::False if $n %% $k;
+ }
+ return Bool::True;
+}
+
+
+# *@range is a slurpy parameter - it will swallow all the arguments passed
+sub primes (*@range) {
+ gather for @range {
+ take $_ if is_prime $_;
+ }
+}
+
+# we can call it with a range, as in the specification...
+say ~primes(10..20);
+
+# or we can pass a list...
+say ~primes(3,5,17,257,65537);
+
+# or a series...
+say ~primes(1,2,*+*...100);
+
+# vim:ft=perl6
View
39 99-problems/P40-rhebus.pl
@@ -0,0 +1,39 @@
+use v6;
+
+# Specification:
+# 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:
+# > say ~goldbach 28
+# 5 23
+
+
+# From P31-rhebus.pl again
+sub is_prime (Int $n) {
+ for 2..sqrt $n -> $k {
+ return Bool::False if $n %% $k;
+ }
+ return Bool::True;
+}
+
+
+# require even arguments
+sub goldbach (Int $n where {$^a > 2 && $^a %% 2}) {
+ for 2..$n/2 -> $k {
+ if is_prime($k) && is_prime($n-$k) {
+ return ($k, $n-$k);
+ }
+ }
+ return; # fail
+}
+
+say ~goldbach $_ for 28, 36, 52, 110;
+
+# vim:ft=perl6
View
62 99-problems/P41-rhebus.pl
@@ -0,0 +1,62 @@
+use v6;
+
+# Specification:
+# 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
+
+
+# From P31-rhebus.pl again
+sub is_prime (Int $n) {
+ for 2..sqrt $n -> $k {
+ return Bool::False if $n %% $k;
+ }
+ return Bool::True;
+}
+
+
+# require even arguments
+sub goldbach (Int $n where {$^a > 2 && $^a %% 2}) {
+ for 2..$n/2 -> $k {
+ if is_prime($k) && is_prime($n-$k) {
+ return ($k, $n-$k);
+ }
+ }
+ # actually, it's more likely a logic error than a refutation :)
+ die "Goldbach's conjecture is false! $n cannot be separated into two primes!"
+}
+
+# Here we demonstrate an optional parameter with a default value
+sub goldbach-list (Int $low, Int $high, Int $limit = 1) {
+ for $low .. $high -> $n {
+ next if $n % 2; # skip invalid goldbach numbers
+ next if $n == 2;
+ my @pair = goldbach($n);
+ say "$n = ", @pair.join(' + ') if @pair[0] > $limit;
+ }
+}
+
+goldbach-list 9,20;
+goldbach-list 2,3000,10;
+
+# vim:ft=perl6
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