Injective, surjective, bijective

A function f: A → B on finite sets is given by the image of each element of A. Is it injective (one-to-one), surjective (onto) and bijective?

Input

Three lines of words separated by spaces: the elements of A, the elements of B, and the images f(a) of A's elements, in the same order as A. A and B have 1 to 50 elements each, and no element is listed twice.

Output

If some image is not an element of B, print not a function. Otherwise print three lines: injective: yes (or no), surjective: ... and bijective: ....

Examples

Example 1 · Input

1 2 3
x y z
y z x

Expected output

injective: yes
surjective: yes
bijective: yes

Example 2 · Input

1 2 3
x y
x y x

Expected output

injective: no
surjective: yes
bijective: no

Prompts inside input() are ignored when checking, so input("n: ") is fine.

Hints

Hint 1

Pair each domain element with its image: dict(zip(domain, images)).

Hint 2

If some image is not in the codomain, f is not a function from A to B.

Hint 3

Injective: no two images are equal (len(set(images)) == len(images)). Surjective: set(images) == set(codomain).

Solution

Show one correct solution (try it yourself first)

A = input().split()
B = input().split()
images = input().split()
if any(y not in B for y in images):
    print("not a function")
else:
    injective = len(set(images)) == len(images)
    surjective = set(images) == set(B)
    yes = lambda b: "yes" if b else "no"
    print("injective:", yes(injective))
    print("surjective:", yes(surjective))
    print("bijective:", yes(injective and surjective))

Lesson: Functions →