Picat List Comprehensions in Picat 3

Question 1

What is the Picat list comprehension that flattens [[1,2],[3,4],[5]] into [1,2,3,4,5]?

Accepted Answer

[X : L in [[1,2],[3,4],[5]], X in L]

Answer

Using two generators — L iterates over the outer list of sublists, then X iterates over each sublist L — naturally flattens the nested structure into a single list.

Question 2

How do you generate the 7-times multiplication table (7×1 through 7×10) as a list using a Picat list comprehension?

Accepted Answer

[7*X : X in 1..10]

Answer

The comprehension [7*X : X in 1..10] iterates X over the range 1 to 10 and multiplies each by 7, producing [7,14,21,28,35,42,49,56,63,70].

Question 3

What does [X : X in [a,b,c,d,e], X \= b, X \= d] produce in Picat?

Accepted Answer

[a, c, e]

Answer

The two inequality conditions X \= b and X \= d both use structural non-unification, excluding the atoms b and d and leaving [a, c, e].

Question 4

Which expression correctly sums all elements in a list List using a list comprehension in Picat?

Accepted Answer

sum([X : X in List])

Answer

sum/1 accepts any list and returns its arithmetic total; [X : X in List] evaluates to a copy of the same list, so sum([X : X in List]) is equivalent to sum(List).

Question 5

Given a user-defined predicate is_prime/1, which Picat list comprehension generates all prime numbers up to 20?

Accepted Answer

[X : X in 2..20, is_prime(X)]

Answer

The comprehension [X : X in 2..20, is_prime(X)] generates candidates from 2 to 20 and keeps only those for which is_prime(X) succeeds.

Picat Practice Test

Picat Practice Test

Picat List Comprehensions in Picat 3