Mini-Trabalho

Um mini-trabalho consiste em resolver pelo menos dois problemas dos apresentados nas listas, e apresentar a solução até uma data pré-definida. Há 3 fases:

Material de Apoio

Problemas sobre listas

?- my_last(X,[a,b,c,d]).
X = d
?- element_at(X,[a,b,c,d,e],3).
X = c
?- compress([a,a,a,a,b,c,c,a,a,d,e,e,e,e],X).
X = [a,b,c,a,d,e]
?- pack([a,a,a,a,b,c,c,a,a,d,e,e,e,e],X).
X = [[a,a,a,a],[b],[c,c],[a,a],[d],[e,e,e,e]]
?- encode([a,a,a,a,b,c,c,a,a,d,e,e,e,e],X).
X = [[4,a],[1,b],[2,c],[2,a],[1,d][4,e]]
?-  encode_modified([a,a,a,a,b,c,c,a,a,d,e,e,e,e],X).
X = [[4,a],b,[2,c],[2,a],d,[4,e]]

Mini-Trabalho I: Estruturas de Dados

 L= [[],  [jim(z),  [2,  [b,  c]]],  [[]],  Z]].

F should unify with

 [jim(z), 2, b, c, Z].
 L= [1, 2, 3, 4, 5, 6].

unify A with

[[1, 2], [3, 4], [5, 6]].
X = [[4,a],b,[2,c],[2,a],d,[4,e]]

Example:

?- dupli([a,b,c,c,d],X).
X = [a,a,b,b,c,c,c,c,d,d]

Example:

?- dupli([a,b,c],3,X).
X = [a,a,a,b,b,b,c,c,c]

What are the results of the goal:

?- dupli(X,3,Y).
?- drop([a,b,c,d,e,f,g,h,i,k],3,X).
X = [a,b,d,e,g,h,k]

Example:

?- split([a,b,c,d,e,f,g,h,i,k],3,L1,L2).
L1 = [a,b,c]
L2 = [d,e,f,g,h,i,k]

Example:

?- slice([a,b,c,d,e,f,g,h,i,k],3,7,L).
X = [c,d,e,f,g]
?- rotate([a,b,c,d,e,f,g,h],3,X).
X = [d,e,f,g,h,a,b,c]

?- rotate([a,b,c,d,e,f,g,h],-2,X).
X = [g,h,a,b,c,d,e,f]

Hint: Use the predefined predicates length/2 and append/3.

?- remove_at(X,[a,b,c,d],2,R).
X = b
R = [a,c,d]
?- insert_at(alfa,[a,b,c,d],2,L).
L = [a,alfa,b,c,d]
?- range(4,9,L).
L = [4,5,6,7,8,9]

Example:

?- rnd_select([a,b,c,d,e,f,g,h],3,L).
L = [e,d,a]

Hint: Use the built-in random number generator random/2.

?- rnd_select(6,49,L).
L = [23,1,17,33,21,37]
?- rnd_permu([a,b,c,d,e,f],L).
L = [b,a,d,c,e,f]

Hint: Use the solution of a previous problems.

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 (via backtracking).

Example:

?- combination(3,[a,b,c,d,e,f],L).
L = [a,b,c] ;
L = [a,b,d] ;
L = [a,b,e] ;
...

Example:

?- group3([aldo,beat,carla,david,evi,flip,gary,hugo,ida],G1,G2,G3).
G1 = [aldo,beat], G2 = [carla,david,evi], G3 = [flip,gary,hugo,ida]
...

b) Generalize the above predicate in a way that we can specify a list of group sizes and the predicate will return a list of groups.

Example:

?- group([aldo,beat,carla,david,evi,flip,gary,hugo,ida],[2,2,5],Gs).
Gs = [[aldo,beat],[carla,david],[evi,flip,gary,hugo,ida]]
...

Note that we do not want permutations of the group members; i.e. [[aldo,beat],...] is the same solution as [[beat,aldo],...]. However, we make a difference between [[aldo,beat],[carla,david],...] and [[carla,david],[aldo,beat],...].

You may find more about this combinatorial problem in a good book on discrete mathematics under the term "multinomial coefficients".

Example:

?- lsort([[a,b,c],[d,e],[f,g,h],[d,e],[i,j,k,l],[m,n],[o]],L).
L = [[o], [d, e], [d, e], [m, n], [a, b, c], [f, g, h], [i, j, k, l]]

b) Again, we suppose that a list (InList) contains elements that are lists themselves. But this time the objective is to sort the elements of InList according to their length frequency; i.e. in the default, where sorting is done ascendingly, lists with rare lengths are placed first, others with a more frequent length come later.

Example:

?- lfsort([[a,b,c],[d,e],[f,g,h],[d,e],[i,j,k,l],[m,n],[o]],L).
L = [[i, j, k, l], [o], [a, b, c], [f, g, h], [d, e], [d, e], [m, n]]

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

?- is_prime(7).
Yes
?- gcd(36, 63, G).
G = 9

Define gcd as an arithmetic function; so you can use it like this:

?- G is gcd(36,63).
G = 9
?- coprime(35, 64).
Yes
m = 10: r = 1,3,7,9; thus phi(m) = 4. Note the special case: phi(1) = 1.
?- Phi is totient_phi(10).
Phi = 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).

Example:

?- prime_factors(315, L).
L = [3,3,5,7]
?- prime_factors_mult(315, L).
L = [[3,2],[5,1],[7,1]]

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

28 = 5 + 23.

It is one of the most famous facts in number theory that has not been proved to be correct in the general case. It has been numerically confirmed up to very large numbers (much larger than we can go with our Prolog system). Write a predicate to find the two prime numbers that sum up to a given even integer. Example:

?- goldbach(28, L).
L = [5,23]
?- goldbach_list(9,20).
10 = 3 + 7
12 = 5 + 7
14 = 3 + 11
16 = 3 + 13
18 = 5 + 13
20 = 3 + 17

In most cases, if an even number is written as the sum of two prime numbers, one of them is very small. Very rarely, the primes are both bigger than say 50. Try to find out how many such cases there are in the range 2..3000.

Example (for a print limit of 50): ?- goldbach_list(1,2000,50). 992 = 73 + 919 1382 = 61 + 1321 1856 = 67 + 1789 1928 = 61 + 1867

Logic and Codes

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
?- table(A,B, A and (A or not B)).
true true true
true fail true
fail true fail
fail fail fail
?- 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

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?

We suppose a set of symbols with their frequencies, given as a list of fr(S,F) terms. Example:

[fr(a,45),fr(b,13),fr(c,12),fr(d,16),fr(e,9),fr(f,5)].

Our objective is to construct a list hc(S,C) terms, where C is the Huffman code word for the symbol S. In our example, the result could be Hs = [hc(a,'0'), hc(b,'101'), hc(c,'100'), hc(d,'111'), hc(e,'1101'), hc(f,'1100')] [hc(a,'01'),...etc.]. The task shall be performed by the predicate huffman/2 defined as follows:

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

Binary Trees

A binary tree is either empty or it is composed of a root element and two successors, which are binary trees themselves.
In Prolog we represent the empty tree by the atom 'nil' and the non-empty tree by the term t(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 Prolog term:

T1 = t(a,t(b,t(d,nil,nil),t(e,nil,nil)),t(c,nil,t(f,t(g,nil,nil),nil)))

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

T2 = t(a,nil,nil) or an empty binary tree: T3 = nil

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

?- istree(t(a,t(b,nil,nil),nil)).
Yes
?- istree(t(a,t(b,nil,nil))).
No

Write a predicate cbal_tree/2 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
?- 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([3,2,5,7,4]).
No
?- 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.

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

% 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.

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

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

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

% 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.

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 2A and 2A+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.

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.

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

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.

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).

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,[])])])

?- istree(t(a,[t(f,[t(g,[])]),t(c,[]),t(b,[t(d,[]),t(e,[])])])).
Yes
?- nnodes(t(a,[t(f,[])]),N).
N = 2

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

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.

The following pictures show how multiway tree structures are represented in Lisp.

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

Example:

?- tree_ltl(t(a,[t(b,[]),t(c,[])]),LTL).
LTL = ['(', a, '(', b, c, ')', ')']
As a second, even more interesting exercise try to rewrite tree_ltl/2 in a way that the inverse conversion is also possible: Given the list LTL, construct the Prolog tree T. Use difference lists.

Graphs

A graph is defined as a set of nodes and a set of edges, where each edge is a pair of nodes. There are several ways to represent graphs in Prolog. One method is to represent each edge separately as one clause (fact). In this form, the graph depicted below is represented as the following predicate:

edge(h,g). edge(k,f). edge(f,b). ... We call this edge-clause form. Obviously, isolated nodes cannot be represented. Another method is to represent the whole graph as one data object. According to the definition of the graph as a pair of two sets (nodes and edges), we may use the following Prolog term to represent the example graph: graph([b,c,d,f,g,h,k],[e(b,c),e(b,f),e(c,f),e(f,k),e(g,h)]) We call this graph-term form. Note, that the lists are kept sorted, they are really sets, without duplicated elements. Each edge appears only once in the edge list; i.e. an edge from a node x to another node y is represented as e(x,y), the term e(y,x) is not present. The graph-term form is our default representation. In SWI-Prolog there are predefined predicates to work with sets. A third representation method is to associate with each node the set of nodes that are adjacent to that node. We call this the adjacency-list form. In our example:

[n(b,[c,f]), n(c,[b,f]), n(d,[]), n(f,[b,c,k]), ...] The representations we introduced so far are Prolog terms and therefore well suited for automated processing, but their syntax is not very user-friendly. Typing the terms by hand is cumbersome and error-prone. We can define a more compact and "human-friendly" notation as follows: A graph is represented by a list of atoms and terms of the type X-Y (i.e. functor '-' and arity 2). The atoms stand for isolated nodes, the X-Y terms describe edges. If an X appears as an endpoint of an edge, it is automatically defined as a node. Our example could be written as: [b-c, f-c, g-h, d, f-b, k-f, h-g] We call this the human-friendly form. As the example shows, the list does not have to be sorted and may even contain the same edge multiple times. Notice the isolated node d. (Actually, isolated nodes do not even have to be atoms in the Prolog sense, they can be compound terms, as in d(3.75,blue) instead of d in the example).

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

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

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

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

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

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

Miscellaneous Problems

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.

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).

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?

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!).

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

   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.

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.

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.prolog. 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.