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Prolog Assignment

All skeleton files are provided in the assignments folder. Please fork this repo to get started. For what forking means or how to fork a repo, please visit this github help page for a detailed explanation.

Again, please feel free to post questions on github Issues page for the repo.

Q1: Flatten a Nested List Structure (33%)

Transform a list, possibly holding lists as elements into a 'flat' list by replacing each list with its elements (recursively).

Example:

?- flattenList(L, [a, [b, [c, d], e]]).
L = [a, b, c, d, e]

?- flattenList(L, [1, [1, 2], [3, 4], [5, [6, [7, 8]]]]).
L = [1,1,2,3,4,5,6,7,8]

?- flattenList(L, [[[[9, 10], 11], 12], [1, 2], [3, 4], [5, [6, [7, 8]]]]).
L = [9,10,11,12,1,2,3,4,5,6,7,8]

?- flattenList([1, 2, 3], [1, [2, 3]]).
true

Hint

  • Establish the base case for recursion first and build up the logic
  • Don't call your rule flatten/2, since it's a predefined predicate
  • Use [H|T] to destruct a list in to the head and tail
  • Use the predefined predicates is_list/1 and append/3

Q2: Duplicate the elements of a list (33%)

Produce a list that contains 2 copies of every element in the original list in order.

Example:

?- duplicate(L, [1, 2, a, b]).
L = [1,1,2,2,a,a,b,b]

?- duplicate([1,1,2,2,a,a,b,b], [1, 2, a, b]).
true

?- duplicate([1,1,2,2,a,a,b,b], L).
L = [1,2,a,b]

Hint

  • Establish the base case for recursion first and build up the logic
  • Use [H|T] to destruct a list in to the head and tail
  • Use the predefined predicate append/3

Q2.5: Duplicate the elements of a list N times [Optional]

If you are feeling ambitious, you can write a more generalized version of the function that allows you to specify how many copies of each element you want in the duplicate list.

Example:

?- duplicateN(L, [1, 2, a, b], 3).
L = [1,1,1,2,2,2,a,a,a,b,b,b]

?- duplicateN([1,1,1,2,2,2,a,a,a,b,b,b], [1, 2, a, b], 3).
true

?- duplicateN([1,1,1,2,2,2,a,a,a,b,b,b], L, 3).
L = [1,2,a,b]

Hint

  • Aside from the same hints above, you should define a rule that appends N copies on an item to a list. Something like prependN(NewList, N, List, Item)

Q3: Truth tables for logical expressions. (33%)

Define predicates and/2, or/2, nand/2, nor/2, xor/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 produces the truth table of a given logical expression in two variables.

Example:

?- table(A, B, and(A, B)).
A = true
B = true

?- table(A, B, and(or(A,B),nand(A,B))).

A = true
B = false

A = false
B = true

?- table(A, B, or(and(or(A,B),nand(A,B)), or(nor(A, B), xor(A, B)))).

A = true
B = false

A = true
B = false

A = false
B = true

A = false
B = true

A = false
B = false

%% Same Example as above, except we now restrict A to false
?- A = false, table(A, B, or(and(or(A,B),nand(A,B)), or(nor(A, B), xor(A, B)))).

A = false
B = true

A = false
B = true

A = false
B = false

Hint

  • the bind/1 rule defines the domain for A and B. In this case, we restrict both A and B to boolean values of true and false.

  • Remember from lecture that . in Prolog is equivalent to a logical or and , is equivalent to a logical and. So bind(A) specifies that A can be true or false

[Bonus] Degree of a Node in a Graph (50%)

Write a predicate degree(Deg, Node, Graph) that determines the degree, or the number of outgoing edges, of a given node.

Graph is a tuple of (Vertices, Edges), where Vertices is a list of atoms that represent vertices in the graph, and Edges is a list of tuples of 2 atoms that specify a directed edge.

Example:

?- degree(N, a, ([a, b, c, d], [(a, b), (b, c), (c, d), (d, a), (a, c), (b, d)])).
N = 2

The query asks what's the degree of atom a, in the graph that contains 4 nodes (a, b, c, d), and 6 directed edges (a -> b, b -> c, c -> d, d -> a, a -> c, b -> d)

In this case the a has 2 outgoing edges, a -> b and a -> c, so the answer should be 2.

Alteratively, you can also ask which nodes have a specific degree.

degree(2, N, ([a, b, c, d], [(a, b), (b, c), (c, d), (d, a), (a, c), (b, d)])).

N = a

N = b

degree(1, N, ([a, b, c, d], [(a, b), (b, c), (c, d), (d, a), (a, c), (b, d)])).

N = c

N = d

Hint

  • Establish the base case for recursion first and build up the logic
  • Use degree(N, Node, (Vertices, [(V1, V2)|Edges])) to deconstruct the arguments into variables you can use
  • Use member(N, List) predicate to check to see if Node is a valid vertex
  • Use is predicate to compute the degree, as we did in the lecture when computing the sums and averages of lists