Question

Write a function in python that takes a set A and a relation R(x, y) on A (as a python function such that R(x, y) returns true if and only if the relation xRy holds), and returns True if and only if the relation R is reflexive. Here is the function signature you need to use.

def is reflexive(A, R): You can test your code as follows. s = {1,2,3} def y(x, y): return x == y def n(x, y): return x == y and x != 3 is_reflexive(s, y) is reflexive(s, n) # returns True # returns False

Answer 1

ANSWER:-

CODE:-

def y(x, y):
    return x == y
def n(x, y):
    return x == y and x != 3

def is_reflexive(A, R):
    setA = []
    for i in A:
        setA.append(i)
    s = []
    for i in setA:
        for j in setA:
            if R(i,j):
                s.append((i,j))
    for item in s:
        if item[0] == item[1]:
            setA.remove(item[0])
    if len(setA) == 0:
        return True
    return False

s = {1,2,3}
is_reflexive(s,y)
is_reflexive(s,n)

NOTE:- If you need any modifications in the code,please comment below.Please give positive rating.THUMBS UP.

         THANK YOU!!!

OUTPUT:-

>>> is_reflexive (s, y) True >>> is_reflexive (s, n) False >>> |

CODE SCREENSHOT:-

def y(x, y): return x == y def n(x, y): return x == y and x != 3 def is reflexive (A, R): seta = [] for i in A: setA.append(i) S = [] for i in seta: for j in setA: if R(i,j): s.append((i,j)) for item in s: if item [0] == item[1]: setA.remove (item [0]) if len (setA) == 0: return True return false s = {1,2,3} is reflexive (s,y) is reflexive (s,n)