Write a predicate weird_sum(List, Result) which takes a List of numbers, and computes the sum of the squares of the numbers in the list that are greater than or equal to 5, minus the sum of the absolute values of the numbers that are less than or equal to 2. For example:
?- weird_sum([3,6,2,-1], Result).
Result = 33
That’s 6×6 – 2 – 1.
Note:
(a) it is the value of the item, not its position in the list, that should be tested to see if it is greater than or equal to 5, or less than or equal to 2.
(b) you are not required to check whether the items in the list are numbers.
Think carefully about how the predicate should behave on the empty list — should it fail or is there a reasonable value that Result can be bound to?
Suppose that a set of family relationships have been loaded into Prolog using the same format as family.pl
parent(jim, brian).
parent(brian, jenny).
parent(pat, brian).
female(pat).
female(jenny).
male(jim).
male(brian).
NOTE: do not include these in your solution file.
We assume that each person will have the same family name as their father, but that married women retain their original birth name.
Write a predicate same_name(Person1,Person2) that succeeds if it can be deduced from the facts in the database that Person1 and Person2 will have the same family name. (It is ok if your code returns true multiple times). For example:
?- same_name(pat, brian).
false.
?- same_name(jenny, jim).
true
Note that your same_name predicate will be tested with different facts to those in family.pl
Write a predicate log_table(NumberList, ResultList) that binds ResultList to the list of pairs consisting of a number and its log, for each number in NumberList. For example:
?- log_table([1,3.7,5], Result).
Result = [[1, 0.0], [3.7, 1.308332819650179], [5, 1.6094379124341003]].
Note that the Prolog built-in function log computes the natural logarithm, and that it needs to be evaluated using is to actually compute the log:
?- X is log(3.7).
X = 1.308332819650179.
?- X = log(3.7).
X = log(3.7).
.
Any list of integers can (uniquely) be broken into “parity runs” where each run is a (maximal) sequence of consecutive even or odd numbers within the original list. For example, the list
List = [8,0,4,3,7,2,-1,9,9]
can be broken into [8, 0, 4], [3, 7], [2] and [-1, 9, 9]
Write a predicate paruns(List, RunList) that converts a list of numbers into the corresponding list of parity runs. For example:
?- paruns([8,0,4,3,7,2,-1,9,9], RunList).
RunList = [[8, 0, 4], [3, 7], [2], [-1, 9, 9]]
Note: you can find out how to test if a number is even or odd from the Prolog Dictionary
In this question we consider binary trees which are represented as either empty or tree(L, Num, R), where L and R are the left and right subtrees and Num is a number.
A binary tree of numbers is called a heap (or, it is said to satisfy the heap property) if, for every non-leaf node in the tree, the number stored at that node is less than or equal to the number stored at each of its children. For example, the following tree satisfies the heap property, because 3 ≤ 5, 5 ≤ 8 and 5 ≤ 7.
tree(empty,3,tree(tree(empty,8,empty),5,tree(empty,7,empty)))
On the other hand, the following tree does not satisfy the heap property, because 6 is not less than or equal to 5.
tree(tree(tree(empty,4,empty),
3,tree(empty,5,empty)),6,tree(tree(empty,9,empty),7,empty))
Write a predicate is_heap(Tree) which returns true if Tree satisfies the heap property, and false otherwise. For example:
?- is_heap(tree(tree(tree(empty,4,empty),
3,tree(empty,5,empty)),6,tree(tree(empty,9,empty),7,empty))).
false.
?- is_heap(tree(empty,3,tree(tree(empty,8,empty),5,tree(empty,7,empty)))).
true
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