The python programming give wrong output. If any algorithm develop in python then, may be that will be affected by this.

There's no bug here.

"&" is the bitwise Boolean logical-and operator on integers. For example,

>>> 1 & 2
0
>>> 1 & 3
1

It binds more tightly than the "==" equality-testing operator. To get the result you want, you would need to add more parentheses to override the natural operator precedence:

>>> (a == 10) & (not(a!=b)) & (b == 10)
True

But, more to the point, what you probably really want is the "and" logical operator, not the bitwise integer "&" operator:

>>> a == 10 and (not(a!=b)) and b == 10
True

a = 10

b = 10


if (a >= b) & (a & b) & (a == b):

        print("Yes")

else:

        print("No")

Output: No

This is the bug because all then condition in if is true but it print No, This is the vulnerability of this code.

Please ask for help on StackOverflow or the general Python mailing list. Your understanding of the operations is incorrect.

"&" is NOT a logical operator ("and" is). It's a bitwise operator on integers.

>>> 10 & 10
10
>>> bin(10)
'0b1010'

The last bit is 0, so when you "&" the result with True (which is equivalent to integer 1) the result is 0:

>>> (10 & 10) & True
0

and the integer 0 is treated as False in a Boolean context.